Cesiege of Architecture Library
Cornell University

 

 

Goruell University Library
Sthaca, New York

 

BOUGHT WITH THE INCOME OF THE

SAGE ENDOWMENT FUND

THE GIFT OF

HENRY W. SAGE

1891
tig
STRUCTURAL ENGINEERING

By
J. E. KIRKHAM

Professor of Structural Engineering, Iowa State College. Consulting
Bridge Engineer, lowa Highway Commission. Formerly
Designing Engineer with American Bridge Co.

4

First Epirion

SEconD IMPRESSION

McGRAW-HILL BOOK COMPANY, Inc.
NEW YORK: 239 WEST 39TH STREET
LONDON: 6 & 8 BOUVERIE ST., E. C. 4
1914
ASogq\2

CoPpYRIGHT 1914
BY
THE MYRON C. CLARK PUBLISHING CO.
CHIcaco
TABLE OF CONTENTS

CHAPTER I.
Preliminary.
PAGES
Definitions, structural material, eiennek method of Pe in designing, and
fabrhea thon acencessuctarsnuriea ie Broseracth orice athiangaied a Die cate aga ee aa BiG fh lac ansns ete say ses 1-8
CHAPTER II.
Structural Draughting.
Equipment, free-hand lettering, drawings, and hints Remnisinanid shop drawings
ANd OILS) «<6 esse orsrsvsaione es Je Beis Rew Ea IES a wetwans aiayenime hs Sa gree se RtaNS 9-13
CHAPTER III.
Fundamental Elements of Structural Mechanics.
Definitions, forces, elasticity, distortion, equilibrium polygon, center of gravity,
moment of inertia, radius of gyration ............-..008. OU ERs eee 14-50
CHAPTER IV.
Theoretical Treatment of Beams.
Shearing stresses, bending stresses, reactions and deflections of cantilever, simple,
over-hanging, fixed at one end and supported at other, fixed and continuous
DOAMS: canscs svi sees dicey FR see SFE ETE PRS eee See eee ee 51-105
CHAPTER V.
Theoretical Treatment of Columns.
Rankine’s formula, straight line formula, eccentrically loaded columns, and ex-
QWGH .sacacees seen eet evieee es creer eRe revty tr ce ee 106-114
CHAPTER VI.
Rivets, Pins, Rollers and Shafting.
Bearing, shearing and bending stresses on rivets and pins, allowable pressure
on rollers and stresses on shafting ...... doe eee ereeerionciie Be ee Sa S ae nent 115-123

CHAPTER VII.

Maximum Reactions, Shears, and Bending Moments on Beams and Trusses and
Stresses in Trusses.

Maximum reactions, shears and bending moments on simple beams and trusses
due to both uniform and concentrated loads, and stresses in simple trusses
due to panel loads ..........- Shediac Wh Ow Rite hardy sr aN 6 ASS aes sich beads 124-143
iv CONTENTS

CHAPTER VIII.

Graphic Statics.
PAGES
Definition and limitation, graphical determination of reactions and bending
moments on simple beams, graphical determination of stresses in trusses
and drawing of an equilibrium polygon through two ana three points....144-161

CHAPTER IX.
Influence Lines,

Definition, influence lines for determining reactions shears and moments on
simple beams and trusses and stresses in truSS€S ...++..sseeeeeeeenee 162-171

CHAPTER X.
Design of I-Beams and Plate Girders.

Data and. method of procedure, section modulus, economic depth of girders,
designing of flanges and’ webs, length of cover platcs, flange increments,
rivet spacing in flanges, web splice, stiffening angles, and examples......172-194

CHAPTER XI.
Design of Simple Railroad Bridges.

Types, loadings, design of beam, plate girder and truss bridges and viaducts and
deflection and camber of trusses ..... ‘

CHAPTER XII.
Design of Simple Highway Bridges.

Types, loadings, specifications and design of beam, pony truss and high truss
DEVAS ES Sirsa, Hencrerocdunsedorenscanseetucealnnlasesnvantromenanonannlia aadeo nee asker ere anenetenstsldrevenend Bis eyes 533-569

CHAPTER XIII.

Skew Bridges, Bridges on Curve, Economic Height and Length of russes and
Stresses in Portals.

Types, determination of stresses in skew bridges, determination of stresses due
to centrifugal force, economic depth and length of ordinary trusses and
determination of stresses in various types of portals ..............+.4-. 570-582

CHAPTER XIV.
Design of Buildings.

Description, loadings, and designing of various types of mill buildings and the
analysis and design of high buildings ............. as # Sudiiieta te xesuoueace,’) eS DOOS O02
PREFACE

This book is intended as a textbook for college students and as a
self-explanatory manual of structural engineering for practical men.

During the author’s nineteen years of engineering experience (the
greater part of which was spent in actual practice) it fell to his lot to
“break in” students from most all of our engineering schools, and he has
no apology to offer for the elementary mechanics given in this volume, as
he is convinced that no matter how thorough the course in mathematics
and theoretical mechanics may be it is quite desirable that students have
a short review of the materialistic phase of mechanics at the beginning
of the subject of structures to whet their appetites for the work.

As regards the designing given, the author has endeavored to present
the usual methods, and this in such a fashion that the average engineering
student as well as the practical man can read, understandingly, without
having a dictionary, glossary, or a compendium on theoretical mechanics
at his elbow.

The author’s aim has been to arrange for college use the drawing
room exercises throughout the text, so that the work in the classroom
and drawing room will go hand in hand. The appreciation of this
feature will depend a great deal upon the allotment of hours for the two
classes of work. The ratio of two hours recitation to six in drawing is
recommended. “Drawing Room Exercise No. 1,” at the end of Chapter
II, can be started at the very first drawing room period without wasting
any time. Chapters I, II, VII, X, and Chapter XI as far as “Deck
Plate Girder Bridges,’ should be read by the time “Drawing Room
Exercise No. 1” is finished, so that “Drawing Room Exercise No. 2” can
be taken up. During the time spent on “Drawing Room Exercises No. 2,
No. 3, and No. 4” ample time will be found for the necessary advance
reading—up to “Through Plate Girder Bridges’—and the reading of
Chapters III to VI and also VIII and IX. Beyond this the author
refrains from making suggestions as to assignments as he has confidence
in the ability of instructors to assign correctly the work to suit conditions.

The practical men of limited theoretical training and others desiring
a review of the subject treated, will find the book well suited to their case
if the chapters be taken up in consecutive order.

The designs given in this book are entirely the work of the author
and are designed especially for this work, yet he has endeavored to make
them as general as possible.

This book, which treats only of simple structures, is the author’s
first volume on Structural Engineering. A second volume, which will
be known as Higher Structures, is in preparation; this will treat of
Movable Bridges, Cantilever, Arch, and Suspension Bridges, Secondary
Stresses, ete.

The author desires to acknowledge his appreciation of the work of
his assistant, Mr. B. S. Myers, in reading and checking proof and pre-
paring and checking drawings, and to thank Prof. F. O. Dufour for
valuable suggestions and criticisms. J. E. KIRKHAM.

Ames, Towa, Sept. 5, 1914.
CHAPTER I

PRELIMINARY

1. Structural Engineering.—The part of Civil Engineering per-
taining to the designing of steel structures, such as bridges, buildings,
towers, etc., is known as Structural Engineering. The work involved
consists, principally, in determining the stresses, selecting the material
to be used, known as sections, and the contriving and drawing of the
details. But, in addition to this work, the structural engineer has, as a
rule, the designing of a great deal of incidental construction, such as
foundations, concrete floors, roofs, etc.

In order to design structures properly, one must be perfectly
familiar with the material used in their construction and have a clear
understanding of the manner in which their manufacture and erection
are accomplished as well as have a thorough knowledge of the mechanical
principles involved throughout. A properly designed structure is one
wherein no mechanical principles are seriously violated and which at the
same time is economic in material and easily manufactured and erected.

2. Structural Material.—Steel, concrete, stone, and wood are the
principal materials used by the structural engineer in the construction of
modern structures. However, cast iron, wrought iron, and a few other
materials are used in a few cases, as will be designated when these cases
present themselves in the text. Concrete and stone will not be treated in
this book further than to designate certain allowable pressures.

3. Manufacture of Steel—tIn describing in a general way the
usual process of making steel, we can say that steel is made from iron ore
which is mined and carried to a blast furnace through which it is run
together with coke and limestone and thus converted into cast iron. The
cast iron is then taken to either an open-hearth furnace or to a Bessemer
furnace, usually spoken of as a Bessemer converter, and there converted
into steel. When the conversion is complete, the molten steel is run into
ingot molds, which are rectangular cast-iron molds, and molded into
ingots. These ingots are allowed to cool to some extent. Then the molds
are pulled off and the ingots are taken to a rolling mill, which is known
as a slab or blooming mill, where they are heated in what is known as a
soaking pit until they have the proper temperature for rolling. Then
they are rolled to a convenient rectangular cross-section and cut up into
convenient-sized pieces known as slabs or billets. These billets or slabs,
as the case may be, are then taken to another rolling mill and reheated
very much the same as the ingots just described, and then rolled into
structural shapes, plates, rails, etc., ready for the market.

Sometimes the cast iron from the blast furnace is cast into rough
bars, known as pig iron, which may be stored and converted into steel
or molded into castings at any desired time, or it may be shipped to some
distant point to be utilized in the same manner. At the most modern
plants the molten metal is taken directly from the blast furnace and
converted into steel without letting it cool to any great extent.

1
2, STRUCTURAL ENGINEERING

The steel produced in an open-hearth furnace is known as “‘open-
hearth steel,” while the steel produced in a Bessemer converter is known
as “Bessemer steel.” Open-hearth steel is considered more reliable in
every respect than Bessemer, and, consequently, the Bessemer steel is
being fast replaced by the open-hearth product. In fact, most engineers
at present exclude the use of Bessemer steel in structural work altogether.

4, Grades of Common Structural Steel— There are three recog-
nized grades of common structural steel: Medium Steel; Soft Steel;
Rivet Steel.

Medium Steel is a medium-hard steel which has practically replaced
all other grades of steel in structural work. It has an ultimate strength
of 60,000 to 70,000 pounds per square inch, and an elastic limit of 30,000
to 35,000 pounds per square inch.

Soft Steel is softer than medium steel. It was the grade of steel
first used in structural work, but has been practically replaced by medium
steel. It has an ultimate strength of 52,000 to 62,000 pounds per square
inch and an elastic limit of 26,000 to 31,000 pounds per square inch.

Rivet Steel is a very soft steel used almost exclusively for making
rivets and bolts. It has-about the same chemical composition as wrought
iron, but has a higher ultimate strength and elastic limit. It has an
ultimate strength of 48,000 to 58,000 pounds per square inch and an
elastic limit of 24,000 to 29,000 pounds per square inch.

5. Nickel Steel is a steel containing from 2 per cent to 34 per cent.
of nickel. It has an ultimate strength of 80,000 to 112,000 pounds per
square inch and has a very high elastic limit—something like 60,000
pounds per square inch. This metal has been used recently in some of
the large bridges in this country.

6. High Carbon Steel is a hard steel which contains more com-
bined carbon than the ordinary medium steel. It has practically as high
ultimate strength and elastic limit as nickel steel. It is used almost exclu-
sively in the form of rods to reinforce concrete.

7. Wood, or timber, as it is usually spoken of, is used in structural
work principally for floor and roof covering. The timber mostly used is
white oak and yellow pine, both of which have an average ultimate crush-
ing strength of about 7,000 pounds per square inch, and an ultimate
tensile strength of about twice that amount.

8. A Casting is made by running molten metal into an impression
made in sand by means of a wooden pattern which is shaped to the size of
the metal piece desired. The castings used in structural work are made
of either cast iron or steel. The castings made of steel are much stronger
than those made of cast iron but cost more. The steel used for making
castings has a somewhat different chemical composition than that of rolled
steel, referred to above. It is produced in the same way, however, the
chemical composition being controlled mostly in the mixing of the charge.

9. Structural Steel Shapes.—The I-beam, channel, angle, Z-bar,
and T-shape, shown in Fig. 1, are the principal steel shapes used in
structural ‘work. The T-shape is not used very extensively except for
very special work. Plates, while they are not classified as structural
shapes, are really used in a more general way than the shapes.
PRELIMINARY 3

The tables in the back’ of this book give the properties, gauges, etc.,
of the shapes and plates in general use. The use of these tables will be
explained as the occasion for their
use occurs. Either a Carnegie or
Cambria handbook is a convenient
book to have—in fact, practically
indispensable in structural work
—but the student must bear in
mind that he is not at liberty to
use just any section therein that
he happens to pick out, as a great
many specials which are not in
general use are listed, which will
be furnished only when the order
for such becomes large enough to
warrant the rolling, and, conse-
quently, in the case of ordinary

orders containing these special

sections, an unusual delay in <f

obtaining the material will likely en Fillet7|_

be experienced. a
10. Rivets and Riveting.— ti :

Rivets are used in structural work

to connect the structural shapes and plates together. They are simple bits
of metal in themselves, but, indirectly, they are the source of a great deal

 

patente Lora __

button Head Rivet — Countersunk Head River
Fig. 2

 

 

of trouble. Practically, there are but two kinds: the buttonhead rivet
which has a hemispherical head, and the countersunk rivet which has a
countersunk head. Each kind is shown in Fig. 2, including the names of
their parts. The buttonhead rivet is the kind most used; in fact, the
countersunk rivet is never used except where it is absolutely necessary
to do so.

As a rule, each structural company manufactures its own rivets.
They are made by feeding red-hot rods into a machine which cuts the rods
into pieces of proper length and at the same time forms a head on each
piece, thus completing the rivets as they are shown in Fig. 2. The way
in which this is done can be seen by referring to Fig. 3, where the parts
of the machine that directly form the rivets are shown. Let us first
consider the plan diagram at (a), where A represents the “header,”
B a fixed die, C a moving die, D a forked bar, and § a bar known as the
gauge. While the parts are in the position shown at (a) the heated rod
is pushed into the machine, passing through the slot in the bar D until it
comes in contact with the gauge S, as indicated by the dotted outline
e-n-o-k. Then the die C moves toward B and cuts the rod off at o by
4 STRUCTURAL ENGINEERING

shearing with the bar D, and at the same time the piece of the rod thus
cut off is caught in the cylindrical grooves in the dies (see section m-m)
and held firmly while the gauge S moves up out of the way of the header
A which then moves until it

 

 

 

 

 

 

 

 

 

 

 

 

ra
comes in contact with the dies iki (P
B and C, upsetting the part e-n dD): J
of the piece of the rod into the = B C
hemispherical cup in the header Bi C : i
A, thus forming the head of the m ‘ m wh
rivet. The parts of the ma- LETHE f | A | (db)

 

 

chine are then in the position wees > (a 1)
shown at (b). The-header A

and then the die C and the A

gauge S move back to the posi-

tion shown at (a), while the D

rivet just formed drops into a
cooling basin. Then to obtain aa C
the next rivet, the rod is fed E

into the machine as before. “
Thus rivets are made at the Section-mm. is
rate of about one per second.

The rivets formed by the machine just described are the buttonhead
rivets. However, countersunk rivets can be made on the same machine by
replacing the header shown by a header having a plane end (without any
cup), and the dies shown by dies which have the cylindrical groove
chamfered out on the end next to the header so as to form the counter-
sunk head. In that case the heads are formed in the dies instead of in
the header as in the case of the buttonhead rivet.

After the rivet holes have been either punched or drilled into the
plates and shapes to be riveted together, and the same have been
assembled, each in its proper place, rivets are heated, placed in the holes,
and “driven.” The driving of a rivet consists principally of forming a
head on the plane end, usually the same as the one on the other end
formed by the rivet machine just described.

Rivets are driven by machines known as “riveters”’ except in a few
cases where it is necessary to drive them “by hand.” The riveter, or
rivet-driving machine, has two headers known as tools, one being fixed
and the other movable, each of which is very similar to the header used
in the rivet-manufacturing machine described above.

In order to show the working of

a viveter, let m (Fig. 4) represent an

angle which is to be riveted to a plate n.

q The rivet holes are either punched or
drilled in the two so as to match. Then

h 3

‘s Q) oe rs Lb ») the plate and the angle are placed to

gether as shown, and the driving of the
rivets proceeds as follows: A rivet is
heated and placed in one of the holes, as
shown at (a). Then the fixed tool B of the riveter is placed against the
head of the rivet and the power is applied (which may be air, steam or
water) which moves the tool A toward B whereby the projecting part of

Fig. 4
PRELIMINARY 5

the rivet is upset into the cup in the end of A, thus forming the head on
that side. When the tool 4 moves until it practically touches the plate n,
the rivet is fully driven, as shown at (b). The tool 4 is then brought
back to its former position, and the riveter is placed on the next rivet by
either moving the pieces being riveted together, or by moving the riveter,
and the operation of driving is repeated, and so on. .
In case a countersunk rivet is to be driven, the rivet hole is either
punched or drilled the same as though a buttonhead rivet were to be used.
Then the hole is countersunk, which means that it is reamed out cone-
shaped on the side where the countersunk head is.to come so that the
head will just fit in. The countersinking of the hole is done, usually,
with a flat, diamond point drill, as shown at (a), Fig. 5. The driving of
a countersunk rivet with a riveter is very much the same as driving a
_buttonhead rivet. For example,
suppose a rivet having a counter- m LL m
sunk head is to be used to connect
the angle m to the plate n as shown
at (b), Fig. 5, where the counter-
ae ang on the same side as i a aa 1
the.angle. The rivet hole will be |(Z lO |B
countersunk in the angle as shown.
The countersunk rivet is heated
and placed in the hole from the
same side as the angle and
driven: just the same as was explained above in the case of the button-
head rivet, but the tool held against the countersunk head would be a
plane tool without a cup. If the other tool has a cup in it, the head on
that side would be an ordinary buttonhead, as shown at (c), Fig 5. If it
were necessary to have a countersunk head on each side, the hole would
- be countersunk in the plate as well as in the angle, as shown at (d),
Fig. 5. The rivet before driving would have a countersunk head, the
same as before, and the driving would take place in the same manner, but
each of the tools would have a plane end.

In cases where it is either impossible or impracticable to use the
riveter, the rivets are driven by hand. This, in the case of a buttonhead
rivet, consists of heating the rivet and placing it in the hole, and while
the-end of a round bar which contains a-cup (known as a dolly bar) is
held firmly (by hand) against the head, the projecting end of the rivet is
‘battered down with a hammer and then formed into a finished head by a
heading hammer (known as a snap) which is held against the battered
end of the rivet and struck by a sledge. A heading hammer, or snap, is
really a two-faced hammer with a cup in one end used to form the head
of the rivet. In the case of a countersunk rivet, the driving by hand is
practically the same as in the case of the buttonhead rivet, but the tools
used on the countersunk heads will be plane, that is, without cups.

There are but very few cases where it is necessary to drive rivets by
hand in the shop, but practically all of the rivets connecting the individual
members in a structure to one another are driven by hand as the structure
is being erected. Such rivets are known in structural engineering as
“field rivets.” Holes left open in the shop for such rivets are known as
open holes or field holes.

Fig. 5
6 STRUCTURAL ENGINEERING

The thickness of metal connected by a rivet is known as the grip of
the rivet. The diameter of the shank of a rivet is known as the diameter
of the rivet. It is also known as the size of the rivet.

The part of the rivet projecting beyond the material to be riveted
together is just a little longer than that necessary to form a head, for, in
driving, some of this is taken up in upsetting the shank of the rivet into
the hole which is always about 7g in. larger in diameter than the shank.

The size of the rivets used depends upon the material to be connected.
A common rule is that no rivet shall be less in diameter than the thickness
of any single piece connected. Rivets vary in size from 4 in. in diameter
to 1} ins. The rivets most often used are the g- and j-in. diameter.

11. Method of Procedure in the Designing of Steel Structures.
—As a rule, parties desiring structures built specify the location and pur-
pose, and limit to some extent the amount they shall cost. The engineering
work really begins with the survey of the site, which, however, does not
necessarily require the services of a structural engineer, yet a structural
engineer should know how to do such work as it is sometimes required of
him; and also such knowledge is often essential in working out the design
of a structure. The structural engineering work really begins after the
drawings showing the site are. made from which the structural engineer
works out the preliminary plans of the structure. These preliminary
plans show, in a general way, what is desired. They usually consist of
what are known as “Stress Sheets” and “General Drawings.” A Stress
Sheet is usually a single-line drawing of the structure wherein each
member is represented by a single line upon which the corresponding
stresses and sections are written, and the general dimensions of the struc-.
ture are given in reference to centers of bearing and to centers of gravity.
A General Drawing is usually intended to show the structure as a whole.
In many cases it is a general picture of the structure wherein the details
are shown in a general way but not fully dimensioned. In the case of
very important structures, the general drawings usually include a general
picture of the proposed structure showing no specific details at all. Such
drawings may be prepared by an engineer or by an architect. The
purpose is to present the general appearance of the structure. That such
drawings are included does not lessen in the least the necessity of the
other general drawings.

After the general drawings have been completed to the satisfaction
of all parties concerned, they are used as a guide in making the “Shop
Drawings” (sometimes called “Working Drawings”) which show the
spacing of rivets and all holes, cuts, etc., for each individual member of
the structure. After these shop drawings are made and thoroughly
checked and are approved by all parties concerned, they are sent to the
oe shops where each member of the structure is fabricated accord-
ingly.

All of the drawings referred to above are, as a rule, made on tracing
cloth and blueprints are made from these which are furnished to all
parties concerned instead of actual drawings. These prints, as a rule,
are what are referred to as drawings.

12. A Bridge Company is an organization which designs, manu-
factures and erects structural work. However, the name is appropriated
PRELIMINARY q

by various concerns which in many cases are capable of doing only part
of this work. A full organization really consists of an operating depart-
ment and an engineering department. The operating department has the
general management of the company, especially the commercial end,
while the engineering department attends to everything pertaining to the
engineering. It suffices here for us to consider only the engineering
department. This department consists of a designing and estimating
department, a draughting department, a shop organization, and an erect-
ing department.

The work of the designing and estimating department consists mainly
in determining the stresses in structures, the selecting of the required
sections, drawing up the stress sheets showing the same, making prelimi-
nary estimates of the weight of the material to be used, and computing
the cost of the work proposed to be manufactured by the company.

The work of the draughting department consists of the working out
of the complete details of structures, using the stresses and sections as
specified on the stress sheets (which are furnished by either the designing
and estimating department of the company or by outside parties) and
making the necessary general and shop drawings. In addition, the
draughting department makes out bills of the material required from
which the material is ordered from the rolling mills. These bills when
completed are known as “Shop Bills” and are sent to the shops along
with the shop drawings.

The work of the shop organization consists in fabricating the
individual members of the structures in the shops according to the shop
drawings furnished by the draughting department.

The work of the erecting department consists in erecting the
structures in their final position after the individual members are fully
fabricated in the shops.

The shops are divided into departments which are referred to as
separate shops. These different shops, or departments, are as follows:
Templet Shop, Pattern Shop, Laying-off Shop, Punch Shop, Rivet Shop,
Finishing Shop, Forge Shop, Machine Shop, Paint Shop, Foundry, etc.,
the name of each clearly indicating the nature of the work done therein.
Some departments occupy separate buildings, while in some cases several
departments are under one roof.

13. Fabrication of Steel Structures—The draughting depart-
ment of a bridge company usually orders the material for any structure
that the company is to fabricate just as soon as the preliminary work on
the shop drawings is far enough along. The steel mills roll this material
and ship it to the shops where it is unloaded and placed in the ‘“‘receiving
yards,” where it remains until the shops need it. After the shop drawings
are completed, blueprints are made of them which are sent to the shops,
each department receiving the prints showing the part of the work
required of it. The fabrication of riveted work, as a rule, begins in the
templet shop where full-sized wooden templets are made for most of the
shapes and plates in the structure. These templets are made so that each
hole and cut shown on the shop drawings can be located on the actual
pieces of metal used. These templets, as a rule, are made of one-inch
white pine boards, well seasoned. The templets are either made of one
8 STRUCTURAL ENGINEERING

board or of two or more boards connected together, quite often forming
a frame.. Pasteboard is being used of late, to some extent, in making
templets.

The templets are taken to the laying-off shop where they are clamped
to the material:for which they were made to lay off. Then the workmen
mark where each hole is to be in the metal by placing a center punch in
each hole in the templet and striking the punch with a hammer, and
indicate the cuts required by marking the outline of the templet on the
metal. Then the templets are unclamped and thrown to one side and the
material thus laid off is taken to the shearing and punch shop where all
cuts are made in accordance with the marks, and the rivet holes are
punched or drilled as indicated. After this work is completed, the
material is taken to the assembling shop where the pieces are assembled
so as to form individual members of the structure and bolted together
temporarily, just a few bolts being used in each member. Then these
members are taken to the rivet shop where the rivet holes are reamed, if
such is called for, and all the rivets indicated on the shop drawings are
driven. Then the members are taken to the finishing shop where pin holes
are bored and all ‘surfaces and joints are finished as called for on the shop
drawings. When this work is completed the members are taken to the
paint shop where they are cleaned and painted. Then they are taken to
the loading yards where they are loaded on cars for shipment to the site.
Thus the fabrication is completed.

Usually when plates and shapes are duplicated a great many times,
templets are not used. The duplicated pieces, in that case, are run
through a multiple punch which is so constructed that the operator can
punch the rivet holes by referring directly to the shop drawings.

In addition to the riveted work referred to above, there are usually
castings, forged’ work and machine-shop work included in each structure
which are gotten out by the foundry, forge shop and machine shop
independently of the other shops, except that the patterns used in
molding the castings are made in the pattern shop, which is usually very
closely associated with the templet shop. This work, when completed, is
loaded on cars and shipped to the site, the same as the riveted work, often
going on the same cars.
CHAPTER II
STRUCTURAL DRAUGHTING

14. Preliminary.—It is very essential that all young engineers
should be good draughtsmen, as they are usually called upon to do
considerable draughting, and good draughting is highly appreciated
throughout the engineering profession. Good draughting is an accom-
plishment; however, one sometimes hears of a ‘“‘natural born’ draughts-
man. A newspaper reporter while talking to one of our greatest inventors
referred to him as a genius. The inventor retorted that genius and
perspiration were synonymous terms. The inventor may have been
slightly mistaken in that particular case, but in this case there is no
question: anyone can acquire the art of making good drawings, that is,
good, neat, plain, practical drawings, nothing of an artistic nature being
considered, by simply going about it with a will, being careful, each time
trying to do a little better than before.

Structural drawings consist of lines, letters, and figures. The lines
should be true and distinct. The letters and figures should be plain, neat,
free-hand letters and figures. Each can be practiced independently of
the others, but after a fair amount of such practice, good draughtsmanship
can be most readily acquired by making exact copies of some good draw-
ings, providing the letters and figures be of the same type as those being
practiced by the student, as it is not advisable to auth from one type to
another until one type is fairly mastered.

15. The Necessary Equipment for Structural Draughting
consists of a drawing board, T-square, two triangles, one decimal scale,
one duodecimal scale, one first class, medium size, right line drawing pen,
one first class, small ink compass, one large combined ink and pencil com-
pass, one pair of medium size dividers, a slide rule, either a Carnegie or
Cambria handbook, one small bottle of black waterproof drawing ink, a
writing pen and holder, a scratch pad, drawing pencils, thumb tacks,
erasers, drawing paper, and tracing cloth.

There are several very useful things which could be added to the
above list, such as logarithmic tables, book of squares, beam compass, etc.
There are no objections to a full set of drawing instruments instead of the
few instruments mentioned above, but in either case, the instruments
should be first class. It is better to have a few good instruments than a
full set of poor ones.

16. Free-Hand Letters and Figures——Some distinct character-
istics will always be seen in the free-hand lettering of each individual the
same as in the case of ordinary writing, but the type of letters and figures
should always conform to some recognized standard. Poor lettering
should not be recognized as a characteristic, but as an indication of lack
of practice.

9
10 STRUCTURAL ENGINEERING

Most of the free-hand lettering on drawings is done on tracing cloth,
so it is best to practice upon the same. The small remnants of cloth
which would otherwise be wasted can often be utilized for that purpose.
The student can have the necessary material, which includes a small

Systemor Lettering
Smal! Le#fers
ols bi te AEA CD FPS Gk hl i 0)
Si) EDO MED Nb 03 pG¢9 GEP r¢5
8(9) LILI OED VU) WUAAXA Yiff Z(2) %(85)
Figures
1h 2Q3R) WBS ED CC) TAB C5) (1069
Capita! Letfers
AMI BUSCCDES LU FUY GCAO IOI
KORAGIMED NODO PCS YORE) HS)
TOUED VED WUADX OP YORZAZ)

AU mokeriol mecliunn stze/ unless othernrise
noted. Hl rivets sot? steel and “la,
Allrive? holes punched ¥¢'dia.

GENERAL DRAWING
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Fig. 6

bottle of drawing ink, writing pen, and tracing cloth, at his private room
and practice during spare moments as a diversion. Opinions will differ
somewhat as to the kind of pen to use in making free-hand letters and

figures. The author prefers Gillott’s No. 303.
The letters and figures shown in Fig. 6 are of about the type used on
structural drawings. Just how each letter and figure is made is fully
indicated, the arrows indicating the direction of

WH MY Mt MU the strokes, and the small figures above the

letters and figures indicating the order in which
90000 the strokes are made. ‘

The student can best practice making the
letters and figures shown in Fig. 6 by making as
nearly as possible exact copies of the work shown there. In addition, at
intervals he should practice drawing parallel lines (free hand) having the
same slope as the letters, and curves which form the letter O, all of which
is outlined in Fig. 7. If there is any letter or figure which gives the

Fig. 7
STRUCTURAL DRAUGHTING 11

student particular trouble, he should keep practicing upon it at intervals
until he has mastered it.

17. Size of Drawings.— The usual practice in structural work is
to make the drawings 23 x 35 inches inside the border line with a one-half
inch margin on all sides. This, however, is not an absolutely fixed size.
In some cases it is necessary to make larger drawings, and in other cases
it is convenient to make smaller ones. Drawings 18 x 24 inches is a very
convenient size in some cases. The border line should be an ordinary
plain, single line.

18. Tracings.—It is the usual practice to pencil out all drawings
upon ordinary drawing paper, and then make tracings of these pencil
drawings upon tracing cloth. The drawing on the tracing cloth should be
upon the dull side. Before starting the tracing, the cloth should be
rubbed over with taleum powder or with chalk, which should then be
brushed off with a cloth or brush. A knife should never be used to make
erasures on the cloth. Repeated erasures can be made by using a
comparatively soft rubber, as the Ruby Eberhard Faber No. 112. No
other than this quality of eraser should be used.

19. General Hints Regarding Shop Drawings.—The shop draw-
ings for a structure are usually the last drawings made, but in order to
‘work up the preliminary drawings satisfactorily, knowledge of shop draw-
ings is indispensable, which makes it necessary that the student take up the
making of some simple shop drawings as preliminary work. As stated
above, the shop drawings show the complete details of the individual
members of structures, that is, all rivets, holes, and cuts are located, and
sizes of all shapes and plates are given, and in addition, the kinds and

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sizes of rivets and holes and also the nature of the machine work desired
are indicated.

In order to indicate the kind of rivets desired, certain conventional
signs are used, most of which are shown in Fig. 8. Shop rivets are those
driven by riveters while the members are in the shops, and the field rivets
are those driven at the site as the structures are being erected. Counter-
sunk rivets are used on the account of clearance, but sometimes sufficient
12 STRUCTURAL ENGINEERING

clearance will be obtained by just mashing down the heads of the rivets.
In such cases the heads are said to be flattened.

Rivet holes are usually 7g inch larger than the rivets to’be driven
into them. The sizes of the heads of rivets are given in Fig. 9. The
sizes of the heads shown on drawings should correspond with the sizes
given here.

In locating rivets, the following rules should be practically followed:
Never space rivets closer together, that is, center to center, than three

 

 

 

 

 

 

 

 

 

 

 

 

 

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times the diameter of the rivets, nor farther apart than six inches. Never
space rivets closer to the edge or end of a shape or plate than two times
their diameter. These rules are not iron-clad, but the deviation from
them should be slight. For example: in practice it is customary to space
Z-inch rivets 14 inches from the edge or end of a shape or plate,and $-inch
rivets 14 inches. It is also customary to limit the minimum distance
between #-inch rivets to 3 inches, and }-inch rivets to 24 inches.

The spacing of rivets or holes in reference to one another along one
direction is known as the pitch. Rivets or holes passing through the
flange of any shape are located in reference to the shape by what is known
as the gauge. For example, the distance between any two consecutive
rivets or holes along the angles shown in Fig. 10 is the pitch at that
point, while the distance g from the back of the angles to the line o-o,
passing through the rivets, is the gauge. The distances marked e are

 

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known as the end distances. In case the flange of a shape has double
gauge lines, the pitch is the distance between the rivets measured along
the piece and not the distance between ‘the rivets on one gauge line.

The standard gauges for shapes are given in the tables in the back
of this book. The same are to be found in the various “standards” and
handbooks gotten out by structural companies.

No rivet or hole should be located in reference to the edge of a
flange, as would be done by giving the distance 2 in Fig. 10.

All lines upon which rivets are located are known, in general, as
rivet lines, and the lines used in giving the spacing of the rivets, as well
as all other distances, are known as dimension lines. All rivet lines
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STRUCTURAL DRAUGHTING 13

with the lines used to outline the members. All should be in black ink.

The duo-decimal scale is used in laying out structural drawings.
The usual scale for shop drawings is three-quarters inch equals one foot,
but in some cases a scale of one-half inch equals one foot is used. In
light work it is sometimes advisable to use a larger scale, in which case
a scale of one inch equals one foot or one and one-half inches equals one
foot is used. It is advisable for the student to use a large scale at the
beginning.

In making structural drawings, it is customary to represent feet by
putting one dot above the figures specifying the number of feet, and two
dots in the case of inches. Thus 4’ is read as four feet, and 4” is read as
four inches. 2’-2’ is read as two feet and two inches; 6’-74” is read
as six feet and seven and one-half inches, and so on. Square feet and
square inches are represented by placing a square in front of the dots.
Thus 40’ is read as four square feet, and 40” is read as four square
inches.

In specifying the material on structural drawings it is customary to
represent the shapes by symbols instead of writing the name. Thus, the
I-beam, channel, Z-bar, and angle are represented by I, [, Z, and L,
respectively, or in case of plural, as Is, [s, Zs, and Ls, respectively.

All dimensions and sizes given on structural drawings are in feet
and inches and fractional parts of inches. The fractional parts are
expressed in halves, fourths, eighths, sixteenths, thirty-seconds, and
sometimes in sixty-fourths of an inch.

Depths, widths, and thicknesses are always expressed in inches,
while length is always expressed in feet and inches.

In the case of I-beams and channels, the number required, depth,
weight and length are given thus:

1—I 12” x 30# x 207-4”; 1—[ 15” x 33# x 6’-935;

2—Ts 15” x 42# x 21’-84"; 6—[s 12” x 35# x 12’-64”; etc.; ete.

In the case of angles, the number required, width of flanges (known
as legs), thickness of metal, and length are given, thus:

1—_ 5” x 34” x 4” x 6’-104”; I—L 5* x 6” x a” x 97-74" ;

4—1 5 6% x4" x 3" x 12-84"; T—Ls 8” x 3 x 3% x 4’-0” ete.

In case of plates, the number required, width, thickness, and length
are given, thus:

1—Plate 36’x4"x14’-21”; 6—Plates 14” x54" x4’-1”.

In detailing a member, it is the usual practice to show an elevation,
top view, and a section through the member, looking downward. How-
ever, there should always be as many sections and views taken as is
necessary to clearly show the member in detail.

DRAWING ROOM EXERCISE NO. 1

Reproduce Plates 1, 2, and 3 each to a scale of 14’°=1'-0”. Each
drawing should first be reproduced in pencil upon an 18 x 24-inch sheet
of ordinary drawing paper, then traced upon a sheet of tracing cloth of
the same size. In making these drawings, the student should study each
detail as he goes along. All necessary information concerning the
material shown on these drawings will be found in the tables in the back
of this book or in a Carnegie or Cambria handbook.
CHAPTER III
FUNDAMENTAL ELEMENTS OF STRUCTURAL MECHANICS

20. Structural Mechanics.— Mechanics, in general, is the science
which treats of the effects produced upon bodies by force, while Structural
Mechanics, as the name would imply, is simply that part of mechanics
that applies directly to structures.

21. Body.—In common usage, the word “body” refers to what is
known as an ordinary solid body. But in reality we are unable to think
of material (matter) at all other than as bodies. This is most readily
realized by attempting to conceive of material in a rarefied state, as in the
composition of a gas. We can conceive of it only as being made up of
very small bodies, known as particles, molecules, etc. Neither can we
conceive of the composition of solid bodies or liquids otherwise if thought
of in detail. Such being the case, it is evident that any part of a recog-
nized body, to us, is really a body in itself; and as we are not limited as to
subdivision, it is evident that any part of any body may, at will, be
treated as an independent body.

22. Force.—In mechanics, the word “force” refers to that which
we recognize as the push or pull that bodies exert upon each other, which
invariably produces motion or a tendency of motion of the bodies con-
cerned.

Forces exerted by bodies at rest produce only a tendency to motion,
and are known as static forces, and the part of mechanics treating of such
forces is known as ‘Statics. Forces exerted by moving bodies are known
as dynamic forces, and the part of mechanics treating of such forces is
known as Dynamics. Most of the mechanics known as Structural
Mechanics can be classed under the head of Statics.

Force is known by various names, as pressure, load, stress, reaction,
etc. These names are used to indicate certain existing conditions; how-
ever, force is the same wherever we find it.

23. Measure of Forces.— Measuring forces is the same as measur-
ing any other thing; it is simply a matter of comparison, that is, we com-
pare forces with some one force which is taken as the basis. We are all
familiar with the practice of measuring forces in terms of the common
gravity unit, in which case the pull of gravity upon a certain body selected
as a standard is taken as the basis of comparison and designated as a
pound and known as the unit of weight. It is readily seen that this unit
is an arbitrary one, as the pull of gravity upon most any body could have
been selected as the unit, although in most cases convenience would have
been sacrificed.

In speaking of a force, we say a force of so many pounds, or the
intensity of a force is so many pounds. But what we really mean is that
the push or pull (as the case may be) exerted upon a certain body by

14
FUNDAMENTAL ELEMENTS 15

some other body is so many times greater or less than the pull of gravity
upon the standard body, which pull, as stated above, is known as a pound.
It is evident that any static force is directly measurable as compara-
tive weight, but dynamic forces are more readily determined by consider-
ing the change of motion produced by them. It is known by experiment
that the force of gravity will cause the velocity of a body falling freely
to increase about 32.2 feet per second for each second of time. This
increment of velocity, 32.2 feet, is known as the acceleration of gravity
and is usually represented by the letter “g.” Then using the force of
gravity expressed in the pound unit as our basis of comparison (as in
static forces) we can say that the intensity of any dynamic force is to the
force of gravity as the acceleration produced by the dynamic force is to
the acceleration produced by gravity, regardless of whether the body
considered moves horizontally, vertically, or in any other direction. For
example, suppose a dynamic force of an unknown intensity F by acting
continuously upon a body weighing W pounds increases its velocity a feet
per second for each second of time. Then by direct proportion we have
F:W::a: 49.
Expressing this in words we say: The unknown force (F) in pounds is
to the force of gravity (W) in pounds as the acceleration (a) in feet per
second caused by the unknown force (fF) is to the acceleration (g) caused
by gravity. From which we have

which is one of the fundamental equations of mechanics. The equivalent
formula is often written: F=ma, where m=W/g, and is known as the
mass of the body considered. But Formula (A) is the one mostly used
in practical work. The mass of a body is the quantity of material it
contains; but in a practical sense, the mass of a body is equivalent to what
its weight would be if the acceleration of gravity were one foot (consider-
ing a foot as unity) per second instead of 32.2 feet,—which it really is.
Then by direct proportion, using the force of gravity as our basis as
before, we have

Ww
m:W ::1+:g, or m= —

which is the expression for mass given above.

The above discussion refers to constant forces, as the forces encoun-
tered in structural engineering are mostly constant forces. In case of
variable forces it is necessary to know the rate and the limit of variation
first, then the intensity can be determined.

Problem 1. A body weighing 200 lbs. is acted upon by a constant
force which increases its velocity from 0 to 64.4 ft. in 1 second of time.
What is the intensity of the force?

Solution:. Substituting in Formula (A), we have

200
==. > 45 lbs.
F 39.9 * Ot4 400 lbs

Problem 2. What would be the intensity of the force in Problem 1
if the velocity of the body were increased in 5 seconds from 50 ft. per
second to 150 ft. per second?
16 STRUCTURAL ENGINEERING

Solution: Here the increment of velocity or acceleration per second

is (150-50) +5=20 ft. Then substituting in Formula (A), we have
200
Ese < 20=124 Ibs. (about).

Problem 8. If the body in Problem 1 had a velocity of 150 ft. per
second, what would be the intensity of a constant force required to stop
the body in 3 seconds?

24. Direction of Action of a Force.—Whenever a force is applied
to a body it either moves the body or has a tendency to move it. The
direction in which the body moves or has a tendency to move is said to be
the direction of action of the force causing the motion or tendency.

25. Line of Action of a Force.—Whenever a body receives a force
by coming into contact with another body, which is the most common case,
there is always a surface upon which the force is exerted. We cannot
conceive of this force being transmitted to the surface other than that
each infinitesimal area of the surface receives a small amount of the total
force transmitted. So it is evident that what we usually consider a force
is really made up of an infinite number of forces. But it would be
impossible to deal with forces in mechanics when considered in this way,
since there would be an infinite number of forces to consider in every case.
We avoid this difficulty by assuming the total force to be applied along a
line which passes through the surface at the center of mean intensity of
the force and in the direction of its action. This line is known as the line
of action of the force. The line of action of a force is unlimited in length,
and the force may be considered to be applied at any point along the line
of action as far as motion or tendency of motion of the body upon which
it acts is concerned.

When a force is distributed over quite a large surface it is usually
necessary to consider the surface divided up into small portions, say,
portions one foot square, or one inch square, as the case may require, and
to treat the part of the total force exerted upon each portion as a separate
force. This means that the force exerted upon any portion will have a
line of action of its own passing through that portion at the center of the
mean intensity of the force exerted upon it. In any case we can obtain
only an approximation, but the approximation should come within the
limits of practicability. ,

In case of such forces as gravity and magnetism, where each particle
of material is assumed to be equally affected, the line of action passes
through what is known as the center of gravity of the body. However, if
the bodies be large, as is the usual case of structures, it often becomes
necessary to consider the body to be divided up into smaller ones and to
consider that the force acting upon each of the smaller bodies has a-line
of action of its own, similar to the case of surfaces.

The plane in which the line of action of a force lies is known as the
plane of the force, and the direction of the line of action is known as the
direction of the force, and the direction of the action along the line of
action is known as the direction of the action of the force. Any number’
of forces could act upon a body and each force could be in a different
plane, but in most of the problems in structural engineering the conditions
are such that we can consider them as acting in the same plane.
FUNDAMENTAL ELEMENTS 17

26. Graphical Indication of Forces.—When considering the effects
produced upon a body by forces, for convenience we graphically represent
the forces applied without reference to the bodies applying them. This is
done by drawing an arrow in the line of action of each force. The arrow
point in each case indicates the direction
of action, and the point upon the body
where the arrow touches indicates the
point of application. Thus in Fig. 11, the
arrows P1, P2, P3 and P4 indicate forces
applied upon the body AB at the points
a, b, c, and d, respectively. The arrows
would be spoken of as forces P1, P2, etc.
And we would say that the forces P1, P2, and P4 act toward the body,
while the force P3 acts away from it, this being indicated in each case
by the arrow points.

27. Applied Forces and Reactions.—An applied force is any force
which we conceive of as being applied to a body from without by some
other body, and at the same time as being the initiative of the outward
activity. A reaction is the same as an applied force, except We conceive
of it as resisting the activity instead of being the cause of it. Both are
spoken of as external forces. For example, a body placed at C upon the
bar AB (Fig. 12) by virtue of its own weight will exert a force P (pres-
sure) upon the bar at that point. This

 

Fig. 11

F force P will cause the bar to have a ten-

A = 8 dency to move downward, which is the ac-
RI R2 tivity produced. Actual motion is prevented
Fig. 12 by the supports at A and B exerting the

forces R1 and R2 acting upward on the
bar. We would call P an applied force, as we conceive of it as being
the initiative of the activity of the bar, while we would call R1 and R2
reactions, as we conceive of them as resisting the activity, which, in this
case, as stated above, is the tendency of motion caused by the force P.
But regardless of name, all three of these forces are exactly the same
in character. This can be seen very readily
by imagining the bar to be turned upside RI R2
down and supported only at C, while all
three of the forces continue to act the same
in reference to the bar. Then the case .
would be as shown in Fig. 13. This could Fig. 18
be actually accomplished by placing a body
weighing the same in pounds as the reaction R1 where R1 is indicated
to act and another body weighing the same in pounds as the reaction R2
where R2 is indicated to act. Then R1 and R2 would become applied
forces, while P would become a reaction.

28. Stress.—Stress is the push or pull which the constituent parts
of a body exert upon each other due to the application of external forces.
Stresses are known as internal forces, while applied forces and reactions
are known as external forces.

As a simple case, suppose a steel rod having a uniform cross-section
of A square inches is suspended from one end and a weight of P pounds
is hung on the other. It is known from experience that the pull P exerted
18 STRUCTURAL ENGINEERING

at the lower end of the rod by the weight is in turn exerted at the upper
end. So this pull is, as we say, transmitted through the rod, and we
cannot conceive of this transmission taking place other than that the.
material particles exert a pull upon each other in the direction of the
length of the rod. A practical idea of the stress in the rod can be obtained
by first imagining the weight P to be supported by a very small wire, so
small in cross-section that the wire would be really a row of individual
molecules; then we can conceive of the pull P being transmitted along the
wire from molecule to molecule, thus producing a stress of P pounds upon
each molecule. Next imagine another wire of the same size suspended
along the side of the wire just considered, and suppose this second wire
now supports half of the weight P: then the stress on the molecules in
either wire would be P/2; and by adding another wire of the same size
the stress on the molecules in each wire would be P/3; and by adding
another wire it would be P/4; and so on. So if there be m number of
wires, the stress on each would be P/n. Now imagine the rod made up
of n number of such wires and let da be the area of the cross-section of
each wire in square inches, and let s be the stress per square inch on each
wire; also let p be the stress per square inch on the rod at any cross-
section. It is obvious that the stress per square inch on the rod at any
cross-section will be p=P/A. Then, likewise, the stress on each wire
being P/n, the stress per square inch in each wire will be

i) id

 

da nda
But nda=A;
therefore, we have
= A = p-

That is, the stress on the rod per square inch is the same whether we
consider the entire cross-section or a portion or a mere molecule in the
cross-section. When we say that the stress on a body is so much per
square inch, we do not necessarily mean that there is a square inch of
material in the body that actually has that stress. We may simply mean
that there is some material in the body that has that stress; it may be a
square foot or a millionth part of a square inch. The stress per square
inch on a body is known in structural mechanics as Unit Stress.

If the above rod were stood up vertically—say, upon a floor—and
the weight placed upon the top end, the direct stress produced by the
weight would be the same in intensity as if suspended as above, but the
stress would be produced by a push instead of a pull. When the stress in
a body is produced by a direct pull, it is known as Tensile Stress; but
when it is produced by a direct push, it is known as a Compressive Stress.

Whenever a body is subjected to one kind of stress uniformly dis-
tributed over its cross-section throughout its length, as was considered in
the case of the above rod, the stress is known as Simple Stress. The unit
stress in that case at any cross-section is always equal to the total stress
on the cross-section divided by the area of the cross-section, or, in general,
we have p=P/4, as given above.
FUNDAMENTAL ELEMENTS 19

From the above discussion of stress one may be led to think of all
bodies as being fibrous in structure. It is true thet wood, wrought iron,
and some other materials manifest this structure to some extent, but not
wholly, while some other materials, as steel and concrete for example,
seem to be devoid of it. But in any case we cannot conceive of the
arrangement of the ultimate parts of a body other than in some consecu-
tive order, and consequently our demonstration is not faulty, as we only
conform to that conception.

Stress, known as shear, cross-bending, and torsion, will be treated
later under these specific heads as the occasion for their treatment occurs.

29. Elasticity.—It is an observed fact that whenever a force is
applied to a body, the dimensions of tlie body are changed, and that when
the force causing such a change is released, the body has a tendency to
regain its original dimensions, and will regain them unless the force be so
great as to break or injure the body. This property of regaining their
original dimensions which bodies manifest is known as their elasticity.
We can conceive of this regaining of dimensions as being due to a stress-
resisting force inherent in the material and which we can designate as the
force of elasticity, yet the nature of the force is unknown. As an example,
suppose a steel rod to be in tension. We can conceive of the material
particles exerting a pull upon each other: This we call stress. And we
can conceive of this pull being resisted by a force inherent in each particle.
This inherent force is the force of elasticity, the intensity of which, in
accordance with Newton’s Law, is necessarily equal to the stress which it
resists.

30. Distortion—%In this book the total change of form of a body
due to external forces will be known as distortion, while the amount of
change per unit of dimension of a body will be known as unit distortion.
What really takes place in every case is cubic distortion, but, in deter-
mining the distortion of structures, it is usually necessary to consider only
the distortion of the length of their parts, and hereafter the word dis-
tortion in this book will refer only to the distortion of length unless other-
wise stated. Suppose a steel rod, ten feet long, is suspended from one
end and a weight is hung on the other end. If the weight causes the rod
to stretch one-tenth of'an inch in length, the distortion of its length or
simple distortion, in that case, is one-tenth of an inch, while the unit
distortion, considering an inch as the unit of length, is 1/120 of the
distortion, the rod being ten feet long or 120 inches.

31. Elastic Limit. Whenever a body is distorted the maximum
amount that it will sustain and yet return to its original form when
permitted to do so, it is said to be distorted to the elastic limit. If a body
is distorted beyond the elastic limit, it will remain shorter or longer than
its original length. Then we say that the body has taken set. If a body
be in tension, its length would remain longer, and if in compression it
would remain shorter.

32. Relation of Stress and Distortion.—Hooke (in 1678) was the
first to state that the distortion of a body is directly proportional to
the stress. This is known as Hooke’s Law, and has been proven by
experience to be practically true, provided the distortion does not exceed
the elastic limit. Whenever a body is distorted beyond the elastic limit,
the distortion increases more rapidly than the stress.
20 STRUCTURAL ENGINEERING

Suppose a steel rod having a length L and a uniform cross-section
A is distorted an amount D when subjected te a simple stress P; then,
according to Hooke’s Law, if the stress were 2P, the corresponding dis-
tortion of its length would be 2D; if 3P, it would be 83D; and so on.
While this direct proportion holds good in all cases as long as the stresses
are within the elastic limit, yet the actual value of D in any case will
depend upon the value of L and A, being directly proportional to L, and
inversely proportional to 4. This is readily seen, for it is obvious that
if the rod were two feet long it would be distorted twice as much when
subjected to the same stress as a rod of the same section only one foot
long, since each foot of length would be distorted the same in amount in
either case.

In regard to the cross-section, it is obvious that if the area is two
square inches, the distortion of the rod will be only one-half as much as
when its area is one square inch, as the actual stress on the material in the
rod in the first case is only one-half as much as it is in the second case.

33. Modulus of Elasticity and Determination of Simple Dis-
tortion.— According to Hooke’s Law, if we know the distortion of any one
piece of: material subjected to a known stress, we can determine the
distortion of any other piece of the same kind of material subjected to any
known stress simply by direct proportion.

Suppose it is found by experiment that a rod of some unknown
material, having a length of L inches and a uniform cross-section of 4
square inches, is distorted D inches in length when subjected to a simple
stress of P pounds. What would be the distortion of a rod of the same
kind of material having a length of L’ inches and a uniform cross-section
of A’ square inches, if subjected to a simple stress of P’ pounds? Let
D’ be the distortion required. Reducing everything concerned to its
lowest terms, which is done for convenience, we have

The unit stress in the first rod = p = P/A,
while the unit distortion in inches = d = D/L.
The unit stress in the second rod = p’ = P’/A’,
while the unit distortion in inches = d’ = D’/L’.
By direct proportion we have
d '
ye soul ccukewaae sh a ae cia) ala Alot ce ainda (1).

Expressing this in words, we say: The stress (p) per square inch

in the first rod is to the stress (p’) per square inch in the second rod as

the distortion (d) of the first rod per inch of length is to the distortion

(d’) of the second rod per inch of length; from which we have d’=dp’/p,

which is the distortion of each inch of the second rod; then, of course, the
total distortion of the rod would be L’ times this, so we have

D’=a'L’ <9 Ie ht lel eek fal) cee ce ae (2).

Now d/p is the ratio of the unit distortion to the unit stress, given
for the first rod, but according to Hooke’s Law, this ratio is the same for
any piece of this same kind of material. So if this ratio is determined
for any one piece of this material, the elastic property of the material is
known, and the distortion of any piece of the material can then be deter-
mined if its length, area of cross-section, and stress are known.
FUNDAMENTAL ELEMENTS 21

The ratio of the unit distortion to the unit stress, as expressed above,
would be a very small fraction in any case, so, for convenience, we can
invert the expression so as to have a whole number. Then we have
p/d=p’/d’=E, which is a constant for any piece of the same kind of
material composing the two rods. Now substituting 1/E for d/p in
equation (2), we have D’=p’L’/E. E would be known as the Modulus
of Elasticity of the material composing the two rods. For the modulus of
elasticity of any material in general we have

_p__ unit stress

d unit distortion’
which is usually designated as Young’s Modulus. It varies with the
different kinds of material, but is practically constant for all pieces of
the same kind. For the simple distortion of any piece of any kind of
material having a uniform cross-section and being subjected to a simple
stress, we have the general equation
PL pL
D TR Bn eee (B),

where D =distortion of the piece in inches (or feet if L be taken in fect) ;

P=total stress on the cross-section of the piece in pounds;

A=area of cross-section in square inches;

L=total length of piece in inches or feet;

p=unit stress on the cross-section of the piece;

E=modulus of elasticity of the material composing the piece.

The modulus of elasticity of a material is determined by experiment.
The average values of the moduli of elasticity of the principal materials
used in structural engineering have been found to be as follows:

29,000,000 for steel;
26,000,000 for wrought iron;
30,000,000 for cast steel;
18,000,000 for cast iron;
720,000 for long-leaf yellow pine;
555,000, for white oak.

In determining the modulus of elasticity, it is laboratory practice to
measure the distortion in inches, in which case the length of the test piece
must always be reduced to inches; but in using Formula (B), L may be
taken either in feet or inches. If taken in feet, the distortion will be
given in feet, and if taken in inches, it will be given in inches.

Problem 4. Suppose a steel rod having a length of 10 feet and a
uniform cross-section of 2 square inches is suspended from one end and
supports a weight of 30,000 pounds hung on the lower end:

(a) What will be the distortion of the rod due to the 30,000 pounds?

Referring to Formula (B), given above, we have in this case
P=30,000; p=30,000+2=15,000; E=29,000,000. Substituting these
values in the formula, we-have for the distortion
D= 15,000 x 10

29,000,000

(b) What will be the distortion of the rod due to its own weight?

In that case the stress varies from zero at the bottom of the rod to a
maximum at the top. For convenience, instead of using the numerals

= 0.00518 ft.= 4; inch (about).
92 STRUCTURAL ENGINEERING

given above, let L be the length of the rod and 4 the cross-section ; and
let w be the weight of the rod per foot of length. Then the stress on the
cross-section at a point b (Fig. 14), « feet from the
lower end of the rod, will be P = wa, which is really the
weight of the rod below b.

For the distortion of the length of the material in
the rod between the cross-section at b and the cross-
section at c,-which is an infinitesimal distance dx above
b, we have

 

ia wads
Fig. 14 AE’

which corresponds to Formula (B), given above.

Now as a is a variable, varying from 0 to L, the last expression will
give the distortion of any infinitesimal part of the length of the rod any-
where between the ends. Then the total distortion of the rod will be equal
to the summation of the distortion of these parts; so, for the total distor-
tion of the rod, we have

L 2
b= [ Be ooete inners tens (c).
oO

 

 

AE 2 AE

Now let wL, which is the total weight of the rod, be represented by P.
Then substituting in (c) we have

a ah ih NN ii ca eagle ta (a).

This shows that the distortion of the rod due to its own weight is one-half
of what it would be for an equal weight suspended from the lower end.
The weight of the rod = 6.8 x 10 = 68 Ibs, Substituting in (d) we have
for the distortion
1 68 x 10
D=5 Rs 29,000,000 = 0.000006 ft. (about).

Problem 5. What will be the distortion of length of a solid, circular,
cast-iron column having a diameter of 6 ins. and a length of 18 ft., when
supporting a load of 282,000 Ibs.?

Problem 6. What will be the distortion of an eye-bar 8” x 14” x 39’-0”
(c.c. end holes) when subjected to a stress of 16,000#0”?

34, Motion.—Strictly speaking, a body can have but two distinct
kinds of motion, one known as translation and the other as rotation.
When a body moves as a whole along an imaginary line, known as its
path, its motion is translation, particularly in reference to any relatively
fixed position in the immediate vicinity of the body; but if it either moves
around or turns about an imaginary line, known as its axis, its motion is
rotation in reference to that line.

There is one fact regarding motion which should never be overlooked,
that is, motion is absolutely relative, and that the kind of motion is always
dependent upon the position of reference. For instance, the motion of a
particle in a body rotating about an axis is rotation in reference to that
axis; yet at the same time its motion (the particles in the axis excepted)
is translation when referred to any relatively fixed position in the imme-
diate vicinity of the particle.
FUNDAMENTAL ELEMENTS 23

In reference to time, motion is either uniform or variable; in refer-
ence to the character of path described it is either rectilinear or curvi-
linear, being rectilinear when the path is a straight line and curvilinear
when a curve.

35. Moment of a Force.—The moment of a force about any point
is a measure of its tendency to produce rotation of the body upon which
it acts about the point chosen and is equal to the perpendicular distance
from the point to the line of action of the force multiplied by the intensity
of the force. The point is known as the center of moment, while the
perpendicular distance is known as the lever arm of the force, or simply
“arm.”

In the case shown in Fig. 15, the moments of the forces P1, P2, and
P3 about the point O would be expressed as aP1, bP2, and cP3, respect-
ively. As is readily seen, the forces Pl
and P2 would have a tendency to rotate
the body A clock-wise about O, while P3
would have a tendency to rotate it counter
clock-wise. Now, taking one direction as
minus and the other direction as plus, say,
counter clock-wise minus, and letting M
be the algebraic sum of the moments of
the three forces about O, we have M =
aP1+bP2-—cP3, which is the equation

ay of moments of the three forces
ig. 15
about O.

In case of forces not in the same plane, the moments are usually
taken about a line, in which case the center of moment of each force is the
point on the line nearest the force. The tendency of rotation about the
line is what is really measured in that case.

It is obvious that the moment of a force about any point in its line
of action is zero, for in that case its arm is zero. If the arm is taken in
feet and the force in pounds, the moment will be expressed in what is
known as foot pounds, but if the arm is taken in inches and the force in
pounds, the moment will be expressed in what is known as inch pounds.

36. A Couple and Moment of Same:—Two equal and opposite
parallel forces acting upon a body (other than along the same line of
action) form what is known as a couple. Thus in Fig. 16, the force P
applied at ¢ upon the body AB and the equal and opposite force P applied
at b form a couple.

The moment of a couple is equal to the distance between the forces

 

multiplied by the intensity of one of the s 3
forces. For example, the moment of the ©
couple shown in Fig. 16 is M = Pd. eee ee ey

!

1
The moment of a couple is not changed
by changing the center of moments, that {
is, the moment about one point is the same AL_ib c |B
as about any other point in the plane of
the forces. This is readily seen, for tak-
ing moments about e or g, we have M = h Q
(es + sg)P = Pd; taking moments about s,

 

 

 

 
94 STRUCTURAL ENGINEERING

we have M = P(se) + P(sg) = P(se + sg) = Pd; and, again, taking
moments about 0, we have M = P(ho) — P(ko) = P(ho — ko) = P(hk) =
Pd. Now, evidently, if the moment of a couple is the same for all points
in the plane of the forces, a couple can be considered to be applied
anywhere in the plane of its forces, which means that we can consider
any couple as being moved bodily to any desired position in the plane of
the forces. Of course, the forces must act upon the body concerned.

Any two couples are equivalent when their moments are equal. But
the forces and arms in the two cases are not necessarily equal. Rotation
or tendency of rotation is always due to the action of a couple. No single
force can producc rotation.

37. Concurrent and Non-Concurrent Forces.— When the lines of
action of two or more forces acting upon a body meet at a common point,
the forces are said to be concurrent, while if their lines of action do not
so meet, the forces are said to be non-concurrent.

38. Conspiring Forces.— When the lines of action of two or more
forces acting upon a body coincide, the forces are known as conspiring
forces.

39. Resultant and Component Forces and Graphical Repre-
sentation of Same.—A body acted upon by two or more forces may,
thereby, have a tendency to move in more than one direction at the same
instant; but as a body cannot occupy two positions at the same time, it is
evident that motion can take place in but one direction, which is known as
the direction of the resultant of the two or more forces. Now, it is evident
that a single force acting in this direction with a certain required intensity
could produce the same effect as that produced by the two or more forces
combined. Such a force would be known as the resultant of the two or
more forces, while each of the two or more forces would be known as a
component of the resultant force.

For example, if two concurrent forces Pl and P2 (Fig. 17) be

applied simultaneously to a small

/6 Cc body resting upon a smooth, horizon-

/ wo tal plane at A, we can readily see that

/ yar the force P1 would have a tendency

ae /\$ to move the body along its line of

Ph oN “ _\_¢ action Ac, while at the same time the
Ra 8 force P2 would have a tendency to
g move the body along its own line of
Fig. 17 action Ab. Now we know, from

reasons given above, that the body
cannot move along both of the lines at the same time, and we can readily
see that it would not move along cither of the lines, as the two tendencies
of motion are not in the same direction; so, undoubtedly, if the two forces
moved the body, the motion would take place along some line different
from either of the two.

Suppose the force P1, acting alone and continuously for one second,
moved the body from 4 to B, so that at the end of the second the body
would be at B. Then suppose the force P1 ceased to act at that instant,
and suppose the body stopped instantly at that point and the force P2
were then applied and in the same manner moved the body from B to C in
FUNDAMENTAL ELEMENTS 25

one second, so that at the end of the next second the body would be at C.
Then, the two forces P1 and P2, each acting separately for one second,
would have moved the body from A to C in two seconds, but if they had
acted simultaneously upon the body they would undoubtedly have accom-
plished the same thing in one second, for the simple reason that each
force would have been acting for one second, as in the case when acting
separately, and with the same intensity, for otherwise they would lose
their identity. By this it is not meant that the two forces acting simul-
taneously would have moved the body along the same paths as when
acting separately, but that they would have moved it from A to C along
some line in one second. We cannot conceive of these forces doing any-
thing more than they would have to do in moving the body from A to C;
then, undoubtedly, they would move it along the straight line AC joining
the two positions, as that would be the least required of them in the
transaction. Then the line AC would be the direction of the resultant of
the forces P1 and P2.

Now it is evident that the two forces P1 and P2 acting simulta-
neously upon the body at A would have the same effect as a single force
R (Fig. 17) would have if acting along the line AC with such an intensity
as to move the body from 4 to C in one second. Therefore, the two forces
P1 and P2 could be replaced by this single force R, which is their
resultant. Each of the forces P1 and P2 are, in that case, components
of the resultant force R.

Assuming the resistance to be the same everywhere on the plane—
which was really the assumption at the.start—the distance AB will be
directly proportional to the intensity of force P1, which moved the body
over that distance, and, likewise, the distance BC will be directly propor-
tional to the force P2. Then as the sides AB and BC of the triangle
ABC (Fig. 17) are directly proportional to the forces P1 and P2,
respectively, it is evident that we can construct a similar triangle such
that the sides will represent to convenient scale the intensities of these
forces. Such a triangle is shown in Fig. 18, where the side P1, parallel
to the side AB in Fig. 17, is drawn to represent the intensity of the force
P1 to scale (so many pounds per inch) and the side P2 parallel to the
side BC in Fig. 17 is drawn to represent the
intensity of the force P2 to the same scale as P1.
Then the side R, which will be parallel to the
side AC in Fig. 17, will represent the intensity
of the resultant force R to the same scale as Pl
and P2. Now it is evident that any two concur- Fig. 18
rent forces with their resultant can be repre-
sented in magnitude and direction by the sides of a triangle which is
known as a Force Triangle. It is important to note that the forces act
in one direction around the triangle, while their resultant acts in the
opposite direction.

Suppose, instead of the two forces Pl and P2 acting upon the small
body shown in Fig. 17, that there be four concurrent forces P1, P2, P3,
and P4, as shown in Fig. 19. The resultant R of the two forces P1 and
P2 is determined by constructing the force triangle ABC in the same
manner as was shown above. Then as RB is the resultant of the two forces

 
26 STRUCTURAL ENGINEERING

P1 and P2, it will replace them, and the remaining forces to be considered
are P3, P4, and R. Now constructing the force triangle ACD (Fig. 19)
for the force R and -P3, we have their re-
sultant R1. Next, constructing the force
triangle ADE for the two remaining ‘forces,
R1 and P4, we have their resultant, R2,
which is the final resultant, or, in other
words, is the resultant force of all the
forces P1, P2, P3, and P4, and as such can
replace them. We now have a polygon
ABCDEA wherein the forces are seen to
act in the same direction around the poly-
gon, while their final resultant R2 acts in
the opposite direction. Such a polygon is
known as a Force Polygon. The force
polygon ABCDEA in this case was con-
Fig. 19 structed by combining force triangles; but
it is evident that the same could have been
constructed by laying off the forces P1, P2, P3, and P4 in consecutive
order, thus obtaining the part ABCDE of the polygon, and then joining
EA for completion as shown in Fig. 20. ‘
While it is advisable to begin with some force as P1 and lay the
forces off in consecutive order around the body as we have done here, yet
it is not absolutely neces-
sary to do so, for it will

E
E be seen upon inspection
4 that the forces may be
g taken in any order (ex-
D y| cept that they must be
wi °3 . laid off so that they act
/ NS in the same direction
Cc yx around the polygon) as
shown in Fig. 21, where
the line AE, which is the
final resultant, is the
A ‘2B same in length and direc-

Zs tion as the final resultant
Fig. 20 Fig. 21 AE shown in Fig. 19 and
Fig. 20.

Now it is evident from the above that any number of concurrent
forces with their resultant can be represented in magnitude and direction
by the sides of a closed polygon known as a force polygon, wherein the
forces act in one direction around the polygon while their resultant acts
in the opposite direction, the direction of action in each case being indi-
cated by the arrow points. The force triangle is really a force polygon
wherein the component forces are only two in number.

As any number of concurrent forces with their resultant can be
represented in magnitude and direction by a closed polygon, it follows
that any force may be resolved into any number of components, the only
requirements being that the force and its components form a closed
polygon. Thus we can construct any number of components as Aa, ab,

 

 

    
FUNDAMENTAL ELEMENTS 27

be, cd, and dB for the force AB, as shown in Fig. 22, by simply drawing
the components practically at will, the only restrictions being that they
form a closed polygon with the force 4B. Here the force AB is consid-
ered as a resultant, which is obviously permissible in any case. We would
say that the force AB was resolved into its component forces Aa, ab, bc,
ed, and dB. Resolving forces into components is known in mechanics as
the Resolution of Forces. The most convenient components into which a
force can be resolved are two components whose directions are at right

b

  

A ; 3 A
Fig. 22 Fig. 23

angles to each other, and are known as rectangular components. Thus
AD and DB or AE and EB (Fig. 23) are rectangular components of the
force AB. It is obvious that a force can be resolved into any number of
pairs of rectangular components—or any number of any other kind.

It will now be shown that the above is as true for non-concurrent
forces as it is for concurrent forces. Let P1, P2, and P3 (Fig. 24)
represent three non-con-

current forces acting IS 6, x
upon the body AB. As _ /

Pp
far as motion or ten- ine x IE
N /
A

 

 

dency of motion are con- et
cerned, any of these San 7
forces may be considered \e

as applied at any point 4

along their line of action; me c &
however, in making such \
assumptions in any case, : oN |

we are compelled to con- : >
ceive that there always if

exists the necessary ma- x/

terialistie connection be- F
tween such points of ap- \y
plication and the body
concerned. Then, if we
prolong the lines of action of the two forces P1 and P2 until they inter-

sect at O, we can consider them as being two concurrent forces applied
at O, and by constructing the force triangle OCD (where OC = P2 and
CD = P1 to scale) we have their resultant given in amount and direction
by the line OD. Now this resultant, which we will designate as Rl, can be
considered as being applied at any point along yy, its line of action (the:
same as any other force). Then prolonging the line of action yy of this
resultant R1 until it intersects the line of action of the third force P3 at
O’, and constructing the force triangle O’EF, we have the resultant R2,
which is the final resultant, or, in other words, the resultant of all three

Fig. 24
28 STRUCTURAL ENGINEERING

forces, P1, P2, and P3. This final resultant R2 can be considered as
being applied at any point upon its line of action xa.

From the above it is evident that the resultant of any number of
intersecting non-concurrent forces can be represented graphically by first
selecting any two forces and prolonging their lines of action until they
intersect; then constructing a force triangle and obtaining the resultant
of these two forces; then prolonging the line of action of this resultant
until it intersects with the line of action of a third force; and then again
constructing a force triangle and obtaining the resultant of the first
resultant and the third force; and again prolonging the line of action of
this second resultant until it intersects the line of action of a fourth force;
and so on until all of the forces are combined. The last resultant will be
the resultant of all of the forces, or the final resultant.

As an additional example, let P1, P2, P3, P4, and P5 represent five
non-concurrent forces acting upon the body AB, as shown in Fig. 25 (a).
Let us first take the two forces P1 and P2 and prolong their lines of action
until they intersect at O. Then by constructing the force triangle ODC
we obtain their resultant R1. Then by prolonging the line of action of
this resultant until it intersects the line of action of a third force P3 at O’,
and constructing the force triangle O’EF, we obtain R2, the resultant of
R1 and P3. Then prolonging the line of action of R2 until it intersects
the line of action of a fourth force P4 at O”, and constructing the force

a is
7 t Ec
ay Ee
Ne @ x dle Sk
SKF a PC x : oer (b)
or an 7
(ee ee ORC

Fig. 25

 

triangle O’ GH, we obtain R3, the resultant of R2 and P4. Then pro-
longing the line of action of R3 until it intersects the line of action of a
fifth force P5 at O”’, and constructing the force triangle O’’ KL, we
obtain the resultant R4, which is the final resultant, or, in other words,
the resultant of all the forces Pl... P5.

So far our treatment of non-concurrent forces does not seem to have
much in common with our treatment of concurrent forces, but let us go a
little further and see what we can observe. The force triangles ODC,
O’EF, O”’GH, and O”’ KL are the same in kind as we found in our
treatment of concurrent forces. Now suppose the force triangle ODC in
Fig. 25 (a) be moved bodily and parallel to itself to the position ODC in
Fig. 25 (b). This would not change the triangle in the least. Next
suppose the force triangle O’EF be moved in the same manner to Fig. 25
(b), taking the position OCF, the vertex O’ coinciding with O in Fig.
25 (b) and E with C. Next, suppose the force triangles O”GH and
O” ’ KL (in Fig. 25 (a)) to be moved in the same manner to the position
in Fig. 25 (b) so that O” and O”’ would fall at O. Then these two
triangles would take the positions OF K and OKL, respectively. By thus
grouping the force triangles we have constructed the force polygon
FUNDAMENTAL ELEMENTS 29

ODCFKLO, which is the same kind of a force polygon as was constructed
for concurrent forces in Fig. 19, and it can be drawn in the same manner.
This shows that any number of non-concurrent forces with their resultant,
the same as concurrent forces, forms a closed polygon, known as a force
polygon, and hence the resultant of any number of any kind of external
forces acting upon any body whatever can be determined simply by con-
structing a force polygon representing each in intensity and direction, at
the same time placing the forces so that they act in the same direction
around the polygon.

As a general case, let Fig. 26 represent a body acted upon by four
non-concurrent forces Pl... P4. We can determine the resultant of
these four forces by simply taking any one of them, say, P4, and drawing
a line as AB equal and parallel to it. Then from B draw BC equal and
parallel to P38. Then from C draw CD equal and parallel to P2.. Then
from D draw DE equal and parallel to Pl.

Then joining E and A we have EA, which 4

represents the resultant of the four forces in @/ [p3\ge x
intensity and in direction. This same me‘hod i B
of procedure holds for both concurrent and E P3
non-concurrent and also for conspiring forces. es 6, JC

The conspiring forces would simply be laid
off in one straight line. Their resultant would
be represented by the distance from the last rare tyix, SE
force laid off to the starting point. mae

40. Analytical Representation of Resultant and Component.
Forces.—From the force triangle shown in Fig. 18 (Art, 39) we have

R?=P1?4 P2242 cos PIX P2.cccccevcecusccuevessess (1),

where ¢ is the angle which the lines of action of the forces P1 and P2
make with each other. Any one of the forces P1, P2, R, and the angle
can be determined from the above equation, provided the other three are
given. When ¢=90°, the above equation becomes

R?= P1?+ P2?

which is the “rectangular equation” that results whenever a force is
resolved into rectangular components. As a rule, it is much more con-
venient to express these rectangular components in terms of the force
and angles which they make with the force. Thus, from Tig. 27, we have

Reos¢=P1; = Pl = .
Reos 6=P2; i cosh PL Mate Pi
Rsing=P2; P2

ER sin 6=P1; R= og 7 Pexsecd.

The most useful thing to be observed from the above is that the
rectangular component of a force along any line is equal to the force
multiplied by the cosine of the angle which the force makes with the line.

In the case of conspiring forces, it is evident that the resultant of
two or more conspiring forces is equal to their algebraic sum, for we can
consider the forces acting in one direction as plus and those acting in the
opposite direction as minus. Then by adding all of the plus forces
together and all of the minus forces together and subtracting the lesser
30 STRUCTURAL ENGINEERING

from the greater sum, we shall thus obtain the intensity and direction of
action of their resultant.

The resultant of three or more forces
can be determined analytically most readily
by resolving the forces into their horizontal
and vertical components, thus obtaining a
rectangular equation where the square of
the algebraic sum of the horizontal com-
ponents plus the square of the algebraic
sum of the vertical components is equal to
the square of the resultant. Thus from Fig. 19 (Art. 39) we have

R2? =aE? + Aa’.
But aE =ab+be+cE = P2cos6, + P3cosd, + P4cos6,,

which is the algebraic sum of the vertical components of the forces
Pl, P2, P3, and P4;

and Aa=AB+Bd-Cf-Dc=P1+ P2sin6, — P3sind, — PAsiné,,

which is the algebraic sum of the horizontal components of the same
forces, as is readily seen from Fig. 19. 6,, 6,, etc., are the angles which
the forces make with the vertical line aE. P1 has no vertical component,
as the cosines of the angles it makes with the vertical is O—the angles
being 90°—and, the sine of 90° being 1, it is evident that its horizontal
component is equal to Pl.
. 41. Proposition.—The moment of the resultant of two or more
forces about any point in the plane of the
forces is equal to the algebraic sum of the
moments of the two or more forces them-
selves, about the same point.

As the resultant of two or more forces
is a force which will produce the same.effect
(as far as motion or tendency of motion are .

°

 

concerned) as the two or more forces com-
bined, it is practically self-evident that the
above proposition is true. However, it is
seen to be true from the following demon- Fig. 28

stration:

Let R (Fig. 28) represent the resultant of two concurrent forces P1
and P2, acting upon a body at A, and let the intensity of these forces and
their resultant be represented by the sides of the force triangle ABC.
Select any point O as the center of moments. Then we are to prove that
the moment of R about O is equal to the algebraic sum of the moments of
the component forces P1 and P2 about the same point. Now, it makes no
difference where O is taken, we can always draw a line through O parallel
to the line of action of the resultant R. Then in this case draw the line
OO’ parallel to & and from 4 draw h’ perpendicular to the line OO’;
then the moment of the resultant R about O’ is the same as it is about O,
since h’ =h.

For the equation of moments about O’, if the above proposition be
true, we have Rh’=a’P1+d’P2; but a’=h’cos¢ and d’=h’cos 0, and
substituting we have Rh’ =(Plcosd + P2cos0)h’, or R= P1cosd + P2cos6h.

 

 

 

 
FUNDAMENTAL ELEMENTS 31

But from the force triangle ABC we have R=Plcos¢+P2cos 6; then
substituting this value for R in the last equation, we have (Pleos¢ +
P2cos6) = (Plcos$+P2cos#) an identity, which proves the proposition
when 0’ is the center of moments. Now, as the moment of R about O is
the same as it is about O’, it remains for us to prove that the change in
the algebraic sum of the moments of P1 and P2, due to the changing of
the center of moments from O’ to O is zero. That is, we are to prove that
(a-a’)P1-(d’-d)P2=0, or (a—a’)P1=(d’ —d)P2. Now from Fig. 28
we have a—a’=c sin # and d’-d=c sin 6. Then substituting these
values, and cancelling c, we have

Pisin ¢=P2sin 6,

which is readily seen to be true from the force triangle ABC, where we
have
Pisin ¢ = p= P2sin 6.

So we have thus proven our proposition for the case of two forces. Now
as it was shown in Art. 39 that the force polygon representing any
number of forces was really made up by combining force triangles, it is
obvious that we could show that the moment of the final resultant of any
number of forces about any. point is equal to the algebraic sum of the
moments of the forces about the same point by simply considering the
resultants and components in each triangle in consecutive order.

42. Condition of Equilibrium.—We say a body is in equilibrium
whenever it is either at rest or in uniform motion. The only state of
equilibrium here considered shall be that of rest, as uniform motion is of |
little interest to us in structural mechanics. As there are but two kinds
of motion, translation and rotation, it is evident that a body will be at rest
if neither of these motions takes place, and that the problem regarding
equilibrium thus reduces to the investigation of the conditions conducive
to these two motions.

Our only conception of equilibrium of forces is that of two equal and
opposite conspiring forces balancing each other through the body upon
which they act. Then, in order that a body be in equilibrium, it is obvious
that the forces acting upon it must reduce to that condition. The forces
themselves, or resultants, or a combination of the forces and. resultants,
may form a pair or several pairs of equal and opposite conspiring forces,
so that all of the forces acting upon a body will thus be balanced, or, as
we say, be in equilibrium, and such being the case, the body upon which
they act will be in equilibrium. It makes no difference what combination
we choose to consider the forces forming, for so long as we can show that
the whole system of forces acting upon a body reduces to equal and
opposite conspiring forces, we are assured that the body upon which the
system acts is in equilibrium, and of course the converse will be true.

Referring to Fig. 17 (Art. 39), it is obvious that if a force equal in
intensity to the resultant of the forces P1 and P2 be applied to the body
at A so that it would act along the line AC in the direction from C to 4,
it would balance the forces P1 and P2 and the body would thus be in
equilibrium under the action of these three forces. Here it is seen that
the resultant of the two forces P1 and P2 and the balancing force would
be two equal and opposite conspiring forces to which the system reduces
when the body is in equilibrium.
82 STRUCTURAL ENGINEERING

And, referring to Fig. 19 (Art. 39), it is obvious that if a force equal
in intensity to the final resultant R2 be applied to the body at A so that
it would act along the line EA in the direction from E to A, it would
balance all of the forces P1, P2, P3, and P4, and thus the body would be
in equilibrium under the action of the five forces P1, P2, P3, P4, and the
balancing force. Here it is seen that the resultant of the forces Pl, P2,
P3, and P4 and the balancing force are two equal and opposite conspiring
forces to which the system reduces when the body is in equilibrium.

Further, referring to Fig. 25 (a) (Art. 39), it is obvious that if a
force equal and opposite to R4 be applied along the line $’O” ’ it would
balance all of the forces P1 ...P5, acting upon the body AB, and thus
the body would be in equilibrium. Here again it is seen that the system
reduces to equal and opposite conspiring forces when equilibrium exists.
So it is in all cases.

It will be observed from the above that any force acting upon a body
in equilibrium will always form an equal and opposite conspiring force
with the resultant of all the other forces acting upon the body.

Beyond a question, a body will move in translation if the forces
acting upon it have a resultant. Then, evidently, a body will be in
equilibrium, as far as translation is concerned, whenever the forces acting
upon it have no resultant. Then as the forces acting upon a body in
equilibrium must have no resultant—in order that no translation may
take place—it is obvious that the force polygon representing them, instead
of having a final resultant, as R2 in Fig. 19, will simply extend on around
and close on the starting point without a resultant. This would give us a
force polygon wherein all the forces are indicated to act in the same
direction around the polygon. From this the following practical state-
ment can be made: A body will be in equilibrium as far as translation is
concerned whenever the forces acting upon it (when represented graph-
ically to scale) will form a closed polygon wherein all of the forces act
in the same direction around the polygon, or, in other words, the force
polygon will close without a resultant, and the converse is just as true,
that is, if a body be in equilibrium the polygon will close. This is true
regardless of whether the forces be non-concurrent, conspiring, or con-
current forces, for it is evident that a body will have motion of translation
if a resultant exists, as the resultant is equivalent to a single force, in
which case motion is inevitable; and if no resultant exists, motion of
translation will not take place, as there would be no force to produce it.
All this is manifestly true regardless of condition. ,

As the resultant of concurrent or conspiring forces always passes
through a common point of action, and the resultant of all the forces
except one always forms an opposite and equal conspiring force with the
remaining one, in case of equilibrium, and the moments about any point
of these two equal and opposite conspiring forces are equal and have
opposite signs, it is evident that rotation will not take place if the
equilibrium polygon closes. So in the case of concurrent or conspiring
forces, if the force polygon closes, we need go no further in our investi-
gations regarding equilibrium, for if no resultant exists, no motion of
rotation, as well as no motion of translation, will take place.

But in regard to non-concurrent forces the case is different, for here
the forces may reduce to two equal and opposite forces, which indicates
FUNDAMENTAL ELEMENTS 33

that no resultant exists, and thus the force polygon would close, indicating
no motion of translation; but at the same time the two forces might not be
conspiring forces, in which case, while motion of translation would not
take place, motion of rotation would, as the two forces would form a
couple which would produce motion of rotation. For example, the forces
acting upon the body AB shown at (a) (Fig. 29) could form a closed
polygon as shown at (b), which indicates

that no resultant exists, and such being pil fr2 °? 8

the case, no translation would take place; A B PA
but by mere inspection of the figure at 5 2
(a) we can see that the forces would - (a) 7 |p4 i
form a couple which would rotate the (b)
body AB counter clock-wise and hence Fig. 29

equilibrium would not exist. So to state

the condition of equilibrium fully, we say: A body is in equilibrium when
both the resultant and the algebraic sum of the moments about every point
of the external forces acting upon it, equal zero.

In addition to the above statement, we may add that a body is in
equilibrium, as far as motion of translation is concerned, when the
algebraic sum of both the horizontal and vertical components of the forces
acting upon it is equal to zero, for it was shown in Art. 40 that the sum
of the squares of the horizontal and vertical components of any number
of forces is equal to the square of their resultant, and, of course, if the
components are equal to zero, the resultant will be equal to zero. Further,
we can state that the sum of the components of the forces acting upon a
body in equilibrium is zero along any line whatsoever, for it is evident
that if a component exists along any line, a resultant would exist in some
direction, and hence motion of translation would surely take place in that
direction.

43. Point of Application of a Force in Relation to Stress and
Equilibrium.—tIn order to determine the nature of the stress produced in
any case it is absolutely necessary to know the actual point of application
of the force producing the stress. Each force in that case is indicated to
act at some particular point as shown in Art. 26, while in reference to
equilibrium (as stated in Art. 39), a force may be considered as being
applied at any point along its line of action, in which case, of course, we
are compelled to conceive of there always being the necessary materialistic
connection between the point of application and the body concerned.

In the case of simple stress, whenever a force acts toward a body,
the stress produced in the body by the force will always be compression,
but if a force acts away from a body, the stress will be tension.

As a simple case, let AB (Fig. 30) represent a steel rod acted upon
by two equal and opposite conspiring forces, Pl and P2. If Pl be
applied at a and P?2 at b, as indicated, it is obvious that the rod would be
in tension. Now let us suppose the forces interchanged, P1 being applied
at b and P2 at a, then it is obvious that the rod would be in compression.
Thus, we sce that by changing the point of application of the forces, the
stress in the rod would be completely reversed, while the state of
equilibrium would not be disturbed.

Then again, suppose P1 be applied at a and P2 at c, then it is
obvious that the stress in the rod between a and c would be the same as
34 STRUCTURAL ENGINEERING

when P2 is applied at b and P1 at a, while the stress in the rod between
ce and b would be zero. So it is readily seen that any change in the point
of application of either of the forces will produce a corresponding change

  

FI

  

  

A

Fig. 30

upon the rod in reference to the stress, but it is just as readily seen that
any such change will not disturb the state of equilibrium, for it is evident
that the forces P1 and P2 will balance each other so long as their points
of application remain in the body anywhere on their common line of action
between the points a and b. As for that, either of the forces can even be
considered as being applied at any point, as 0, off of the body- altogether,
in which case, of course, we would have to consider the rod as extending
to o or being rigidly connected to it in some way.

As another case, let three non-concurrent forces, P1, P2, and P3
(Fig. 31) be applied to the body ZZ at the points a, b, and c, respectively.
Now, as far as equilibrium is concerned, we can
consider any of these forces as being applied at
any point along their line of action, so prolong
the lines of action of the two forces P1 and P2
(see Art. 39) until they meet at d. Then we can
consider these two forces as two concurrent forces
applied at A, and constructing the force triangle
ABC, we have the intensity and direction of their
resultant #1, which can be considered as a force
being applied at any point along its line of action
Cy. Then prolong the line of actjon of the force
P3 until it meets the line Cy at D. Then we can

Fig. 31 consider that R1 and P3 are two concurrent

forces applied at D. Then constructing the force

triangle DEF, we have their resultant #2, which is the final resultant of
the three forces P1, P2, and P3.

Now it is evident that if a force equal to R1 be applied to the body
along the line Cy it would produce the same motion as P1 and P2 if it
acted in the same direction as indicated by R1, but if it acted in the
opposite direction it would balance them. Then again, it is evident that if
a force equal to the resultant R2 be applied to the body along the line Dz
it would produce the same motion of translation as the forces P1, P2, and
P3 combined, providing it acted along the line Dz, in the same direction
as indicated by R2; but if it acted in the opposite direction, it would
balance the three forces P1, P2, and P3.

Now in regard to stresses, it is obvious that the stress produced upon
the body ZZ by the action of the three forces P1, P2 and P3 will be of a
very complicated nature. We can readily see that a force equal to the
resultant I1, if applied at o or at any other point in or on the body along
the line Cy, would not produce the same stress as the two forces them-
selves produce. And it is more readily seen that a force equal to the
resultant R2 applied along the line Dx would not produce the same stress
in the body as the forces P1, P2, and P3 actually produce. So it is

 
FUNDAMENTAL ELEMENTS 35

evident that a force representing the resultant of two or more forces,
while it will produce the same effect as the forces themselves as far as
motion or tendency of motion is concerned, yet will not necessarily produce
the same effect in reference to stress. However, in the case of concurrent
forces, we assume the stress produced by a force representing a resultant
to be the same as that produced by the components, especially when the
forces meet at the surface of the body considered. Some cases are very
complicated and require very careful consideration, but we can always
obtain a reasonable approximation which will come within the limits of
practicability, and with this we must be contented.

44, Equilibrium of Couples.—In regard to equilibrium of couples,
there is one important fact that should always be borne in mind, and that
is, it always requires another couple to balance a couple, and the couples
must be equivalent to each other and act in opposite direction to each
other. As to the proof that it requires another couple to balance a couple,
we have the following:

We know from the definition that a couple can have no single
resultant, as the resultant always reduces to zero, and therefore would
not balance a single force, and hence any single force or any system of
forces reducing to a single resultant could not balance a couple. Then,
evidently, the forces balancing a couple must have a resultant equal to
zero and an equal and opposite moment, which would undoubtedly con-
stitute another couple.

45. Graphical Determination of the Resultant of Parallel Forces
or Those Nearly Parallel—First Method: In the case of intersecting
forces, the final resultant can be fully determined in its true position by
prolonging in consecutive order the lines of action of both the forces and
their successive resultants to interséction, and constructing a force triangle
at each intersection, as shown in Art. 39, but it is obvious that in the case
of parallel forces or forces nearly parallel the same method will not apply
as such forces either do not intersect at all or their intersection is beyond
practical limits. However, by resorting to the resolution of forces, the
resultant of parallel forces and those nearly parallel can be determined
as readily in its true position as the resultant of intersecting forces. As
an example, let AB at (a) in Fig. 32 represent a body acted upon by
three non-concurrent forces P1, P2, and P3, which are nearly parallel.
Select some convenient point on the line of action of one of the forces,
say, point O on the line of action of Pl. Then at this point O resolve P1
into any two component forces as ¢ and cl by constructing (at will) a
force triangle as abO. This is accomplished by assuming the intensity of
one of the components, say c, and laying off aO from O in any convenient
direction to represent this intensity, and then from a laying off ab equal
and parallel to P1, and drawing bO to complete the triangle, we have cl,
the other component, fully given by this last line bO. Then from O pro-
long the line of action of the component cl until it intersects the line of
action of the force P2 at O’. Then at this point resolve P2 into two
components such that one of them will be equal and opposite to the com-
ponent cl at O and act along the same line OO’. Then there will be a
component force cl at O and an equal and opposite component force cl
at O’, which will balance each other. The component cl at O’ being
designated, we obtain the other component of P2 which we will call 2,
36 STRUCTURAL ENGINEERING

by constructing the force triangle deO’, which is constructed by laying off
dO’ equal and parallel to cl and de equal and parallel to P2, and drawing
eO’ for completion. Then c2 is given by this last line e0’. Now from
O’ prolong the line of action of the component force c2 until it intersects
the line of action of the force P3 at O”. Then at this point (O’) resolve
P3 into two components such that one of them will be equal and opposite

 

 

 

to the component c2 at O’ and act along the same line O’O”. Then there
will be a component force c? at O’ and an equal and opposite component
force c2 at O” which will balance each other. The one component force
c2 at O” being designated, the other component of P3, which we will call
c3, we obtain by constructing another force triangle ghO”, where c3 is
given by the line hO”.

Now it will be observed that the three forces P1, P2, and P3 at (a)
FUNDAMENTAL ELEMENTS 37

have each been resolved into two component forces which replace the
original force in each case, and that these component forces are all mutu-
ally balanced, except c (at O) and c3 (at O”), so that the three original
forces, P1, P2, and P3, are really replaced by these two component forces,
ec and c3. So, evidently, if we determine the resultant of these two com-
ponent forces in its true position, we will have the resultant of the three
original forces, P1, P2, and P3, fully determined in its true position, as
evidently the resultant in the two cases must be identical.

Now as tlie two component forces ¢ and c3 are intersecting forces,
their resultant is readily determined in its true position by prolonging
their lines of action until they intersect at I and constructing the force
triangle Ikm, where their resultant is fully given in its true position by
the line Im. This resultant is the same in every respect as the resultant
of the original forces P1, P2, and P3, and it may be assumed to be applied
at any point along its line of action yy.

Another way of determining the resultant is as follows:

Instead of resolving the forces P1, P2, and P3 into components; as
was done above, we can select any point on the line of action of one of the
forces, as point S (really the same as we did before) on the line of action
of P1 (at (a)) and apply two forces f and f1 such that the force P1 will
be balanced by them. This is accomplished by assuming the intensity
and direction of one of the forces and then constructing a force triangle
to determine the intensity and direction of the other. Thus, beginning at
S, lay off Sn (at will) to represent f in intensity and direction. Then
from n lay off no equal and parallel to P1 and draw oS for completion of
the triangle and we have f1 represented in intensity and direction by this
last line oS. As the three forces at S are in equilibrium, they will act in
the same direction around the triangle Sno as indicated, which is in
accordance with Art. 42. Next, from S prolong the line of action of the
force f1 until it intersects the line of action of the force P2 at S’. Then
at this point apply two forces such that they will balance P2 and one of
them be equal and opposite to the force f1 at S and act along the same
line SS’. Then there will be a force {1 at S and an equal and opposite
force f1 at S’ which will balance each other. The force f1 at S’ being
known, the other balancing force at S’, which we will designate as f2, we
determine by constructing the force triangle S’pr in the same manner as
was explained in the case of triangle snO at S. Then from S’ prolong
the line of action of the force f2 until it intersects the line of action of
P3 at S’’. Then at this point (S”) apply two forces such that they will
balance P3 and one of them be equal and opposite to f2 at S’ and act
along the same line §’8”. Then there will be a force f2 at S’ and an
equal and opposite force f2 at S” which will balance each other. The
force f2 at S” being known, the other balancing force, which we will
designate as f3, at S”, we determine by constructing another force
triangle S’’st, where f3 is given by the line St.

Now it will be observed that each of the three forces P1, P2, and P3
is balanced, that is, held in equilibrium by two forces applied in each
case for that purpose, and such being the case, the system will be in
equilibrium. But it will be observed that all of the balancing forces are
mutually balanced except the force f at S and f3 at S”. Then, evidently,
these two forces, f and f3, hold the three forces Pl, P2, and P3 in
38 STRUCTURAL ENGINEERING

equilibrium, in which case the resultant of the forces f and f3 will be
an equal and opposite conspiring force to the. resultant of the forces
P1, P2, and P3 (see Art. 42). So if we determine the resultant of the
two forces f and f3 in its true position, we will have a force the same in
position and in intensity as the resultant of P1, P2, and P3, but which
acts in the opposite direction, so that the resultant of the forces P1, P2,
and P3 will thus be relatively determined in its true position. To deter-
mine the resultant of the forces f and f3, prolong their lines of action
until they intersect at J’ and construct the force triangle’ I’k’m’ and we
have their resultant fully represented by the line I’m’.

The only difference in the two methods shown above is due to the com-
ponents being reversed in direction in the second method so that they are
really balancing forces instead of actual components. Otherwise the two
methods are identical. Either of the broken lines rOO’O’’z or vSS’S’u
would be known as an Equilibrium Polygon, and the lines 20, OO’,
0’0”, and O”z, or vS, SS’, S’S”, and S’’u would be known as their
respective segments.

It is evident that the resultant of any number of such forces as P1,
P2 and P3 can be determined in its true position in the same manner as
shown above for these three forces, but in practice the same thing is
accomplished with less work by a somewhat different method of pro-
cedure, as will now be shown.

Imagine the triangle abO, at (a) Fig. 32, moved bodily and parallel
to itself to the position abP at (b). Next imagine the triangle deQ’ at
(a) moved in the same manner to the position beP, d coinciding with b
and O’ with P, and likewise imagine the triangle ghO”” moved in the same
manner to the position ehP, g coinciding with e and O” with P. Now we
have the three force triangles abO, deO’, and ghO” collected into a single
diagram PabehP (at (b)) which we will call a Ray Diagram, where the
line abeh is the load line, and the lines aP, bP, eP, and hP are the rays,
and the point P is the pole. It will be observed that these rays, aP, bP,
eP, and hP are parallel, respectively, to the segments x0, OO’, 0’0”,
and O’’z of the equilibrium polygon 200’0”sz, at (a). So, evidently, if
this ray diagram, PabehP, had been drawn beforehand, the equilibrium
polygon 200’0”z could have been constructed by beginning at O and
drawing xO parallel to the ray aP, then from O drawing OO’ parallel to
the ray bP, and likewise from O’ drawing O’O” parallel to the ray eP,
and then from O” drawing O”z parallel to the ray hP. Then after
having drawn the equilibrium polygon 200’O”z in this manner, the
segments rO and 20” could be prolonged until they intersect at I, and
thus the point I would be located, which is one point upon the line of
action of the desired resultant. But the things lacking would be the
intensity, direction, and direction of action of the resultant. But it will
be observed at (b) that by drawing the line ah we will have a force
polygon abeha which represents the forces P1, P2, and P3, and their
resultant, each in intensity, direction, and direction of action. Then the
intensity, direction, and direction of action of the desired resultant is
given by the line ah. So by drawing through J a line parallel to this line
ah we would have the line of action of the desired resultant, and hence
the resultant would be fully determined, as its intensity and direction of
FUNDAMENTAL ELEMENTS 39

action, as stated above, as well as its direction, are given by the line ah
in the force polygon abeha at (b).

Now, according to Art. 39, in any case the force polygon represent-
ing any two or more forces and their resultant can always be drawn. So
suppose in the above case the force polygon abeha at (b) be the very
first thing drawn. Then we would have the intensity, direction, and
direction of action of the resultant of the three forces P1, P2, and P3
given by the line ah, and the only thing lacking would then be the loca-
tion of the line of action of the resultant, which can be obtained as above
by constructing an equilibrium polygon. But we would have so far only
the force polygon abeha constructed. Now if the point P at (b) and the
point O at (a) were given, we could draw the rays aP, bP, etc., and
beginning at O we could draw the equilibrium polygon x00’O”:, as
explained above, and thus obtain the desired line of action yy. However,
it is not necessary to have the location of either of these particular points
given, as the location of O was arbitrarily assumed in the first place, and
the position taken by the equilibrium polygon «00’O”’z was governed by
the arbitrarily constructed triangle abO. So, evidently, we might just as
well assume the pole P and draw the rays of a ray diagram and construct
an equilibrium polygon accordingly by starting at any convenient point
on the line of action of one of the forces.

As an example, let AB, Fig. 33, represent a body acted upon by four
forces, P1, P2, P3, and P4. First construct the force polygon abcdea, at
(b), by drawing ab equal and parallel
to P1, and be equal and parallel to Ne \Pe fe - if 3 ‘/[P#
P2, and so on as explained in Art. 39, ,;-t Yip 1B
and we have the resultant of the four y pe ts i
forces P1 ... P4 given in intensity, ‘ ek ; (a
direction, and direction of action by mt6ertoress. | !)
the line ae. Then to obtain its line yo ' 0 ~Ao"!
of action, take any convenient point £ a“ J PI .
P as a pole (at (b)) and draw the b “he
rays aP, bP, etc., thereby obtaining p2 z
the ray diagram PabedeP. Then Pp ic
construct a corresponding equilibrium P3 ( b)
polygon at (a) as #200’0”0'"s,
which is accomplished in the follow- id
ing manner: First select any con- P4.
venient point on the line of action of e
one of the forces, say, point O on the Fig. 33
line of action of P1. Then, from this
point draw the segments zO and OO” parallel to the rays aP and bP,
respectively. Then from O’ draw the segment O’O” parallel to the ray
cP. Then from O” draw the segment O”0O’” parallel to the ray dP.
Then from O’” draw the segment 20’” parallel to the ray eP. Then
prolong the segments xO and zO’” until they intersect at I, and through
this point I draw the line yy parallel to the line ae in the force polygon
abcdea, and it will be the desired line of action of the resultant of the
four forces, and hence the resultant is fully determined.

If the forces be absolutely parallel, the load line will be a straight

‘
40 STRUCTURAL ENGINEERING

line, and the line representing their resultant will coincide with this line,
but, however, the work of determining the resultant is the same as shown
above. For example, let AB, at (a) Fig. 34, represent a body acted upon
by four parallel forces Pl... P4. We construct the load line abcde at
(b) by laying off the forces in consecutive order as shown. Then assume
the pole P and construct the ray diagram PabcdeP and draw the
equilibrium polygon x00’0”0’” g at (a) as before and locate the point
I, through which draw the line of action yy of the resultant parallel to the
load line ae.

 

 

 

 

 

 

y a
|p P2 4% |P3 PA PI
AC a a 1B b
A RON P2
' el 4 Sy
ee ee ee :
1 et seals. | P3
Ke uta ee (bore
4 “Nz P4
é
Fig. 34

The method of procedure just shown is quite satisfactory in practice,
but the student should not acquire the habit of constructing the force
polygon and the ray diagram and then the corresponding equilibrium
polygon without fully recognizing the exact significance of each step.

In Fig. 82 the force triangles were constructed on the segments of
the equilibrium polygons, while in Figs. 33 and 34 the force triangles
were constructed on the load lines of the force polygons, but the principle
involved is just the same in either case. In the case shown in Figs. 33
and 34 we really resolved each of the forces P1, P2, P3, and P4 into two
components by constructing, respectively, the triangles abP, bcP, cdP,
and deP. However, it is easy to overlook this fact when the triangles
are constructed by merely drawing the rays aP, bP, etc. If we consider
the forces resolved into components, as is the usual custom, the com-
ponent in each case will act around the force triangle in the opposite
direction to that of the force, while if we reverse the direction of the
components, they bécome balancing forces, in which case they will act in
the same direction around the triangle as the force (see Arts. 39 and 42).
Take, for example, the case shown in Fig. 33; the two sides aP and bP of
the force triangle abP, at (b), give directly either the intensity and
direction of two components of P1 or of two balancing forces. If we
assume components, the one represented by the side bP will act from P
to b, but if we assume balancing forces, the one represented by the side
aP will act from P to a, while the one represented by the side bP will act
from b to P. The same is true of all the other force triangles beP, cdP,
and deP shown at (b). In constructing an equilibrium polygon it really
makes no difference whether we consider the forces resolved into com-
ponents or balanced by forces applied for that purpose, as the final result
will be just the same.
FUNDAMENTAL ELEMENTS 41

In constructing an equilibrium polygon after the ray diagram is
completed, we virtually transfer the two components or the two balancing
forces, as the case may be, of each force to a point upon its actual line of
action. In order to show what is really involved in the construction, take
for an example the case shown in Tig. 33. As the forces here were
assumed to be resolved into components, the components will in each case
act around the force triangle in the opposite direction to that of the force.
Now beginning with P1, we can assume its own components, represented
by the sides aP and bP of the force triangle abP at (b), as applied at any
point O upon the line of action of P1. Then as the component represented
by the side aP acts from a to P, a line drawn from O in this direction and
parallel to aP will represent that component as being applied at O, and
we thus obtain the segment 20 of the equilibrium polygon. The other
component of P1, which is represented by the side bP, acts from P to b.
Then a line drawn from O in this direction and parallel to bP will repre-
sent that component as being applied at O, and we thus obtain the seg-
ment OO’, which is really the line of action of this other component of P1
drawn only to O’, where it intersects the force P2. We can assume the
two components of P2, represented by the sides bP and cP of the force
triangle bcP at (b), applied at any point upon the line of action of P2 the
same as in the case of the two components of P1. Then, evidently, we
can assume them applied at O’ as well as at any other point. Now the
component of P2 represented by the side bP of the force triangle bcP
at (b) acts from 6 to P. Then a line drawn from O’ in this direction and
parallel to bP will represent that component of P2 as being applied at
O’, and we thus obtain the segment OO’ again. The other component of
P2 represented by the side cP acts from P toc. Then a line drawn in
this direction from O’ and parallel to cP will represent that component
applied at O’, and we thus obtain the segment 0’0”, which is really the
line of action of this other component of P2 drawn only to O” where it
intersects the force P3. The two components of P3, represented by the
sides cP and dP of the force triangle cdP at (b), can be assumed to be
applied at O”, and then the two components of P4, represented by the
sides dP and eP of the force triangle deP, can be assumed to be applied
at O’’, and the discussion of the previous cases will apply to each of
these cases.

The direction of action of the components of each force can be
indicated on the rays of the ray polygon, and on the segments of the
equilibrium polygon as in Fig. 33, but usually we omit such indications,
because such are usually not necessary to any great extent—however, that
is a minor item.

After an equilibrium polygon is drawn for a system of forces as
shown in Fig. 33, the resultant of any number of these forces, if taken in
consecutive order, can be determined as well as the resultant of all of
them. For example, take the three consecutive forces P1, P2, and P3
(Fig. 33). By drawing the line ad (at (b)), we have the force polygon
abcda, wherein the resultant of the three forces is given in intensity,
direction, and direction of action by this line ad. By prolonging the
segment 0’0’” until it intersects the segment +O at I’, we have one
point on the line of action of the resultant of these three forces, P1, P2,
and P3. Then by drawing a line through I’ parallel to the line ad, we
42 STRUCTURAL ENGINEERING

obtain the desired line of action of their resultant, and thus we have the
resultant of the three forces determined. As another case, the resultant
of the two forces P2 and P38 can be determined by drawing a line bd
(at (b)) and prolonging the segments OO’ ” and OO’ until they intersect
at I”.
Second Method: The following graphical method, which may be
designated as the Proportional Triangle Method, can be used to advantage,
quite often, in determining the resultant of parallel forces:

Let P1 and P2 (Fig. 35) represent any two parallel forces, and let

 

 

Fig. 35

R represent their resultant. Let a line ab be drawn perpendicular to the
two forces and their resultant, intersecting the resultant at 0. As there is
no question as to the intensity of the resultant being equal to the sum of
the forces, that is, R=P1+P2, and the direction being the same as the
forces, it remains only to determine the point o through which the line of
action of the resultant passes. This can be obtained by the following
construction: From b draw a line bz, making any convenient angle with
the line ab. Then from b lay off, to any convenient scale, be-=P1; and
from c lay off, to the same scale, cd=P2. Then draw ad and through c
draw a line parallel to ad, and where it intersects the line ab will be the
required point o through which the resultant passes. The same can be
accomplished by drawing the line ay from a and laying off the loads as
indicated and then drawing the lines fb and eo.

The above is readily seen to be true, for taking moments about b,
according to Art. 41, we have

Pixab
P1+P2

which is the distance of the point o from b; but from similar triangles,
bco and bda, we have

obxR=P1xab or ob=

ob ch SPL
ab cb+cd P1+P2’
or ober Xe ,
P14+P2

and thus we have an identity.
FUNDAMENTAL ELEMENTS 43

In case there are more than two forces, the resultant of any two is
determined as above and this resultant is then considered with one of the
remaining forces, and so on. For example, let P1, P2, and P3 (Fig. 36)

|r |re [rs
k t ° -

a

 
 

Fig. 36

represent three parallel forces. First draw a line as abc perpendicular to
the forces. Then from one of the points a, b, or c, say a, draw a line az,
making any convenient angle with the line abe. Then lay off ae=P2 and
eg=P1. Then draw gb and through e draw‘a line parallel to gb, and the
point & where this line intersects the line abe is a point on the line of
action of the resultant of the two forces Pl and P2. Then from this
point draw a line ky, making any convenient angle with the line abc.
Then lay off kh=P3 and hm=P1+P2. Then draw cm and through h
draw a line parallel to cm, and the point o where this line intersects the
line abe is a point on the line of action of the resultant of the three
parallel forces.

It is important to note that the force from which the diagonal line
is drawn is laid off last on that line. Thus in Fig. 36, eg = P1 and
hm = P1 + P2.

‘In cases where the forces are in groups, as locomotive wheel loads,
the method is a very convenient one, as the resultant of each group is
known from observation and thus the forces really considered in the con-
struction are quickly reduced to only a few.

46, Analytical Determination of the Resultant of Parallel
Forces.—It is evident that the resultant of two or more parallel forces
must act in the same diréction as the forces, and with an intensity equal
to their algebraic sum; otherwise it would not produce the same effect as
the forces themselves. So the intensity, direction, and direction of action
of the resultant of two or more parallel forces are really known directly
from the forces themselves, and such being the case, the problem involved
in determining their resultant really reduces to the locating of the line of
action of the resultant.

Let Pl... P5 represent five parallel forces acting upon the body
AB (Fig. 37), and let R represent their resultant. Then we have
R=P1+P24+P3+P4+4+P5. According to Art. 41, the moment of this
resultant about any point is equal to the algebraic sum of the moments of
the five forces about the same points. Then taking moments about some
point O, we have ‘ i
aP1+6P24+cP34+dP4+eP5=Rz,
from which we get
aP1+bP2+cP3+dP4+eP5_ aP1+bP2+cP3+dP4+eP5

R ~  P1+P2+P3+P44+P5

 

r=
44 STRUCTURAL ENGINEERING

So then we have the line of action of R located in reference to O.
In practice it is usual to take moments about one of the forces,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4
d
. :
4S
a
ch NR ae
ea; gt gf oy le
AL ; j8--i—lo
Fig. 37

 

thereby shortening the work by eliminating the moment of that force.
Thus taking moments about Pl (Fig. 38), we have

6P24+cP3+dP4+eP5 :
P1+ P24+P3+P4+P5

In case the moments about one point are known, the moments about any
other point can be determined by either increasing or diminishing the
known moments by the sum of the forces multiplied by the common dif-
ference of lever arms. For example, the moments about the point z
(Fig. 38) are equal to the moments about P1 plus the sum of the forces

@=

 

b '
Q M = Ne
— 18

Z

 

 

 

 

Fig. 38 Fig. 39

Pl... P5 multiplied by k. That is, the moments about z are equal to
bP2+cP3+dP4+eP5+ (P1+P2+P3+P4+P5)k. This is readily seen
to be true, for let M be the moment of a force P about A (Fig. 39) and let
M’ be its moment about B; then we have M=aP and M’=P(a+b)=
aP+bP=M+bP. It is readily seen that this is as true for a number
of forces as it is for one. The product of the sum of the forces and
common difference of lever arms would be subtracted from the moments
in case the lever arms became shortened by the transferring of the center
of moments. ‘

47. Center of Gravity—The most common case of parallel forces
is that of gravity, which acts with equal intensity upon each ultimate
part of material contained in a body, regardless of the kind of material.
In case of an individual body, the line of action of the resultant of the
gravity forces passes through what is known as the center of gravity of
FUNDAMENTAL ELEMENTS 45

the body, but when more than one bedy is considered, the center of gravity
of the botlies considered as a group is the same as the resultant of so many
parallel forces, the weight of each body being considered as a force acting
through the center of gravity of the respective body.

We can consider symmetrical bodics as being made up of parallel
layers of homogeneous material wherein the centers of gravity of the
layers coincide each with each, and such being the case, we need treat
only one layer, which we may consider as a plane without thickness, from
which results the common practice of treating area instead of volume,
weight, or mass.

In case the material composing a body be symmetrical about a plane,
there is no question but that the center of gravity of the body will be in
that plane; and hence, if the material be symmetrical about three or more
planes intersecting in a point, their point of intersection will undoubtedly
be the center of gravity of the body. So it is evident that the centers of
gravity of many bodies and plane figures are known from mere inspection
of their state of symmetry. Thus, the center of gravity of a sphere is at
its center, a line at its middle point, and we know the center of gravity
of a circle, cylinder, cube, an ellipse, etc., from the same deduction. In
fact, whatever the method employed in determining the centers of gravity
of bodies and plane figures, we assume the center of gravity of certain
elements as being predetermined by mere symmetry.

The determination of the center of gravity of loads in groups and the
center of gravity of the cross-section of individual members of structures
are the principal center of gravity problems occurring in structural
engineering.

We can determine the center of gravity of any group of loads in the
same manner as the resultants of parallel forces were determined in Arts.

 

  

 

 

 

 

f |
wy wr
y qi ¢g
Ef.
Uz sy ly
4 8 t
Fig. 40 Fig. 41

45 and 46. The analytical method outlined in Art. 46 is most used as it is
usually more convenient than any other method. As an example, let
Pl... P4 (Fig. 40) represent four loads. Taking moments about one
of the end loads, either P1 or P4, say P4, we have

cP14+bP2+aP3
P1+P2+P3+P4?

where 2 is the distance the center of gravity of the four loads is from P4.

In case of the cross-section of a member, we deal with the area
instead of the weight. For example, let the sketch in Fig. 41 represent
the cross-section of a plate and an angle which is riveted to the plate.
Let A be the area of the plate, and let A’ be the area of the angle, and let
a and b be the distance from the bottom edge of the plate to the center
of gravity of the plate and angle, respectively. Then taking moments
about the bottom edge of the plate, we have

r=
46 STRUCTURAL ENGINEERING

A’b+Aa
A’+A

where @ is the distance from the bottom edge of the plate to the center of
gravity of the total cross-section of the plate and angle combined. The
work could be shortened by taking moments about the center of gravity
of the plate or angle, thus following out the same scheme as was used in
the case of loads (Fig. 40).

The center of gravity of plain areas can be determined quite readily
by the aid of calculus, wherein the general formula is

yda
A

For example, let Fig. 42 represent a rectangle having a height h and
width b. Taking moments about the lower edge of the rectangle, we have

we [yda_ [Pbydy bh? dhe _h

A Jo A 24 2h 2°
That is, the center of gravity of the rectangle from its lower edge is equal
to one-half of its height, which we really knew from mere inspection.

=@7,

A’b+ Aa=2(A’ +A) or

=2.

 

 

 

 

 

 

 

 

 

 

 

Al =z
¥
&
tg
we cinewe= dy ___ | aa N
pa n-n- === eer Es3
Pe re cee
IX
a : 8B A
Pig: 42 Fig. 43

As another example, let Fig. 43 represent a triangle. Taking
moments about a line AB parallel to its base, we have

ae [ate = [eae ee
4 Jo hA  3bh? 3°”

which is the distance of the center of gravity of the triangle from the
axis AB.

Problem 7. Determine the center of gravity in reference to the
horizontal axis of the section shown in Fig. 44, which is composed of
38—14” x }” plates and 2—6” x 6” x 4” angles.

Solution: Taking moments of the areas about the center of gravity
of the top plate, we have

32.5 7=0.5x74+1x7 + 2.93 11.5,

oe = 44.19
82.5

Hence the center of gravity of the section is 1.36’ below the center of the
top plate, or 0.11” below the back of the angles. The numerical quanti-

 

= 1.36”.

 
FUNDAMENTAL ELEMENTS AT

ties in this and the preceding problem can be verified by referring to th
tables in the back of this book. power

: Problem 8. Determine the center of gravity in reference to the
horizontal axis of the section shown in Fig. 45, which is composed of
*—15” x 33* channels and one cover plate 22” x 4”.

"

%

cS Cov, 22%. £"
—_——=5t a =F
= iy

— Seat y
we gh Srvity Axis lg MS

ir Tr

—
42-[5 15°\3 3%

 

 

x
Int

 

2.93]

 

 

 

Que

 

 

 

 

 

Vv
Fig. 44 Fig, 45

Solution: Taking moments of the areas about the center of gravity
of the cover plate, we have

30.8% = 7.75 x 19.8,

= PGR TOR os,
or wo = 4.98",

That is, the center of gravity of the entire section is 4.98” below the
center of the cover plate, or 4.73” from the back of the top flanges of the

_channels. The center of gravity here lies on the horizontal line marked
gg» which is known as the gravity line or gravity axis.

It is readily seen that the center of gravity of the above section in
reference to the vertical axis would lie on the vertical line vv passing
through the center of the cover plate; however, the center of gravity of
any section in reference to a vertical axis can be obtained by taking
moments about some vertical line in the same manner as shown above for
the horizontal axis.

48. Inertia.—It is an observed fact that it always requires a force
to move a body when at rest, to stop it when in motion, or to change its
motion in any respect. This would be true even if friction and all other
external resistances were removed. Hence, it is evident that a body in
itself offers resistance to every change of motion; otherwise, it would
require no force to change the motion of the body if all the external
resistances were absent. This property which bodies have in themselves
of offering resistance to change of motion is known as Inertia. As an
example, a body lying upon an absolutely smooth horizontal plane would
offer resistance to any change of motion along the plane simply by virtue
of its inertia. The actual force exerted would be directly proportional
to the mass of the body concerned, and to the change of motion, or in
other words to the acceleration produced. Let W be the weight of the
body concerned, and let a be the acceleration produced in feet per second.
48 STRUCTURAL ENGINEERING

Then for the intensity of the force (in pounds) exerted, we have the
proportion
F:W::4a:4,

from which we obtain

F=

Ww
Pei.
g
which is Formula (A) given in Art. 23.
49, Moment of Inertia and Radius of Gyration.—Let AB (Fig.
46) represent a very thin rectangular plate of homogeneous material.
Suppose the plate starts to rotating about
y a vertical axis YY. The resistance offered
E by all such strips as cd (which will be
x due to their inertia) will be directly pro-
portional to their distance out from the
Lola B axis YY, as their relative increment of
A Fall? velocity or acceleration will be directly
d proportional to that distance. Then the
Q intensity of the resistance of any strip
will be directly proportional to the dis-
Fig. 46 tance of the strip from the axis and to the
mass of the strip. Suppose the plate made
up of infinitesimal strips as cd, and let m be the mass of each strip. Then
the force resisting the motion of the strip out unit distance from the axis
will be (m) 1, while the force out x distance will be mz, and the moment of
it about the axis would be (mz)(«)=mz?. Then evidently the moments
about the axis YY of all the resisting forces of these strips will be 3mz?
where m is the mass of each strip, which was assumed to be the same for
each and every one, and z the distance out to each strip, no two 2’s being
the same. The expression 2ma? is known as the Moment of Inertia, which
is usually indicated by I, and we have in general

T=3m2?.

In the case of the above plate it is evident that all of its mass could
be concentrated at some point out from the axis YY so that its moment
of inertia about the axis would be the same as that of the plate. This
distance out would be known as the Radius of Gyration of the plate in
reference to the axis YY. Then we have the general equation

I=Mr’*,

 

 

 

 

 

 

 

 

 

 

Y

where M is the total mass of the plate and r the radius of gyration.

The mass of any one of the infinitesimal strips of the above plate is
directly proportional to da, the area of the strip; then if we imagine the
plate to diminish in thickness until it becomes a plane without thickness,
we have for its moment of inertia

I= (2da)r*= Ar’,

aa
"NA

from which we obtain
FUNDAMENTAL ELEMENTS 49

Let a be the length of the plate and b its width; then considering the
plate as a plane, its moment of inertia about the axis YY is

sg 3
= [ pie
oO 3

Knowing its moment of inertia, we can then determine its radius of
gyration from the above formula, J=Ar*. Thus substituting, we have

ba’ _y 2 = i
3 = oar” or r=a 3

The determination of the moment of inertia and radius of gyration
of the cross-section of individual members of structures are the most
common problems under this head met with in structural engineering. It
is really the case of determining the moment of inertia and the radius of
' gyration of a plane, or as we may put it, the material cut by a plane.
There are a few cases, however, where it is necessary to consider the
mass, but in such cases the student will have no trouble if the general
formula, I= Mr’, given above, is applied.

 

 

 

 

 

 

 

 

 

 

 

 

» Bas
rv
etre Ne >
ope eRe ty Si s
G eee eae oe ae ‘ ———y _
e Sn y
b L » |
Fig. 47 Fig. 48
PROBLEMS

1) Moment of inertia of a rectangle in reference to an axis GG
(Fig. 47) through its center of gravity.
Solution: Here we have
+h
2 [2 7.2, _bh*® bh* _ bhi
I= 3day’ = on by dy=o4 $4 "3
(2) Moment of inertia of a triangle about its base. (Fig. 48.)
Here we have

("thw a bh?
1=sday’= [ o( Gt ay =F (hy? ~ y*)dy = 3+
oO oO

In case the axis is perpendicular to the plane, we have what is known
as the Polar Moment of Inertia. For example, for the polar moment of
inertia of a circle about its center we have from Fig. 49

R R R*
T=3dax?= [ dast=2n [ a?da= 2 :

A more general case of the polar moment of inertia would be that of
a rectangle. Here let O (Fig. 50) be the axis and draw XX and YY
through O at right angles to each other. Then we have

. [=Xdap? = 3da(a? + y?) =32daa? + Seday?,

 

 
50 STRUCTURAL ENGINEERING

which is the moment of inertia of the rectangle about the axis XX plus
the moment of inertia about the axis YY. This gives the general formula
which will apply to any plane figure, and further discussion of the polar
moment of inertia is unnecessary. Its principal use is in the designing of
shafting. The polar moment of inertia is usually designated by the letter
J instead of I, which is known as the Rectangular Moment of Inertia.

50. Proposition.— The rectangular moment of inertia of any plane
figure about any axis in the same plane as the figure is equal to its moment
of inertia about a parallel axis through its center of gravity, plus its area
multiplied by the square of the distance between the two axes. As proof
of this, let ABCD (Fig. 51) represent a rectangle whose area is 4 =bh

I

     

“~g e

dad
eA ,
y
x Ov=x x
J .

Fig. 50 Fig. 51

 

 

 

 

 

 

 

 

and whose moment of inertia about its gravity axis gg is I’. Then,
according to the above, the moment of inertia of the rectangle about any
axis as YY is I= I’ + d?A. This is shown to be true in the following
manner: The moment of inertia of the rectangle about the axis YY is

I= ["batde=5 (ae) Nuss ie teagan eee Rid Atte k (1).

Now from Fig. 51 we have a=d+(h/2) and e=d-(h/2). Substituting
these values of a and e in (1) we have
b h\* b h\?_ bAS ‘

1=$ (445) aC *) =p + bha?.
But bh?/12 equals the moment of inertia of the rectangle about its gravity
axis gg. Then substituting I’ for bh?/12 and A for bh, we have

I=I’'+ Ad’,

which proves the above proposition in the case of a rectangle.

The above proposition is true of any plane figure. This can be shown
by following out the same scheme as above.: In case of bodies, the only
thing different is that we would consider mass instead of area and write
the formula

T=1’+Md?
instead of
I=I’+Ad’,
CHAPTER IV
THEORETICAL TREATMENT OF BEAMS

51. Classification Beams are, as a rule, classified according io
the number and kind of supports they have. Whenever a beam simply
rests upon a support, the support is known as a simple support, but when
the beam is held rigidly by a support, the support is known as a fixed
support.

Figures 52 to 55 show the most common types of beams. The canti-
lever beam shown in Fig. 52 has one fixed support, while the fixed beam

 

 

 

alae
Fig. 52

A 7
Fig. 54 Fig. 55

shown in Fig. 53 has two fixed supports. The simple beam shown in
Fig. 54, which is the most common type, has two simple supports. The.
continuous beam shown in Fig. 55 has four simple supports; however, any
beam with more than two supports of any kind is a continuous beam.

There are various other types of beams, such as overhanging, inclined,
etc., but they are all really modifications of the above types and will be
readily recognized as such without any preliminary description.

Beams are usually supported in a horizontal position and support
vertical loads which produce vertical reactions. This will be understood
to be the case unless otherwise stated.

52. Shearing Stress on Beams.—Let AB (Fig. 56) represent an
ordinary rectangular wooden cantilever beam supported in a horizontal
position by having the end B built into a wall. Let P represent a load at
A which the beam supports in addition to its own weight. Let ab and cd
represent the traces of two imaginary planes passing perpendicularly to
the longitudinal axis of the beam, mentally separating the very short
rectangular block abcd from the other parts of the beam, so that we can
think of this block as being a separate body.

By imagining the beam to be cut off instantly along the section ab we
readily realize that the part abA of the beam has a tendency to move
down vertically and the load P with it. As the material in the block abcd
prevents the motion, evidently the part abd of the beam exerts a down-
ward force along the section ab upon the block equal to the combined
weight of the part abA of the beam and the load P. Let S be this force.

“51
52 STRUCTURAL ENGINEERING

Now it is evident that the part cdB of the beam to the right of the block
must exert an equal force S (neglecting the weight of the block abcd)
upward along the section cd upon the block in order to prevent the block
from being pulled downward by the force S acting downward upon it
along the section ab. It is readily seen that the downward force S acting
along the section ab upon the block and the equal and opposite force S
acting upward along the section cd have a tendency to break the material
in the block off vertically. This action is most readily
comprehended by imagining the sections ab and cd to
approach each other until the block abcd is really a
single layer of molecules. Then there would be a
force S acting downward along ab upon the left side
of the molecules and an equal and opposite force S
acting upward along cd upon the right side of the
molecules, thus tending to break the molecules off crosswise instead of
tending to crush or pull them apart as in the case of simple stress, and
the stress thus produced would be known as the shearing stress on the
beam at that point. Whenever forces act in this crosswise manner, in any
case, the stress directly produced is known as shearing stress.

In speaking of the shear on a beam, or on any other body, we refer
to it as being x pounds along a certain section of # pounds cut by an
imaginary plane; but what we really mean is that a stress of x pounds is
produced upon a strip of material infinitesimal in thickness adjoining
that section.

It is evident that in the above case the shearing stress on the section
ab (really the shear on the block abcd) of the beam is equal to the load P
and to the weight of the part abA of the beam. The part of the shear due
to the load P is the same for all vertical sections of the beam between A
and B, while the part due to the weight of the beam will evidently be
directly proportional to the length of the part abd. Then, if w be the
weight of the beam per foot of length, the shear at any vertical section «
feet from the end 4 willbe S=P+ we. Here it is seen that the shear on
any vertical section of the above beam is equal to the algebraic sum of the
forces between the section and the end. If additional forces were applied,
they would likewise be included in the summation, provided they were to
the left of the section considered. As the forces upon the part of the
beam to the right of the block abcd must exert an'upward force upon the
block along the section cd equal to the downward force along ab exerted
by the forces to the left, it is evident that the shear on the block is equal to
the algebraic summation of the forces on either side of it. So is the case
of all beams, where the forces are applied perpendicularly to the longi-
tudinal axis, and hence we have in general: The shear on any section of
a beam perpendicular to its longitudinal axis is equal to the algebraic
summation of the forces on either side of the section, provided all of the
forces act perpendicularly to such axis.

This is readily seen to be true. For, let AB (Fig. 57) represent a
beam in equilibrium acted upon by six forces Pl ... P6, no reference
being made as to which are reactions; let abed be an imaginary block
cut through the beam, the same as the block abed shown in Fig. 56: It is
obvious that the forces P1 and P2 would cause the part abd of the beam

 

 
THEORETICAL TREATMENT OF BEAMS 53

to exert a force S upon the block along ab as shown, and it is readily seen
that this force S, which is the shear on the block (neglecting the weight of
the beam) would be equal to P1 + P2, that is, their sum. As the block is
in equilibrium, undoubtedly there must be ap equal and opposite force S
exerted upon the block along cd, as shown, by the part cdB of the beam
due to the forces acting upon that part. Then we have P1+P2=+P6-
P5+P4-—-P3=S; that is, the shear on the block
abed is equal to the algebraic summation of the
forces on either side of it. Likewise, the shear on
8 any section between P6 and P35 is equal to either
peek, «= ira eo, PG or P1+P2—P3+P4-—P5; between P5 and
Fig. 67 P4 it is equal to either P6—P5 or P1+P2—P3+
P4; and between P4 and P3 it is equal to either
P6—-P5+P4 or P1+P2-P3. This means that the shear at any cross-
section of any beam (as stated above) is equal to the algebraic summa-
tion of the forces on either side of the section, provided the forces are
perpendicular to the beam.

It is common practice to assume this shearing stress to be uniformly
distributed over the cross-section of the beam. Then according to this
assumption, the shearing stress per square unit on any cross-section of
the beam is equal to the shear at the section divided by the area of the
cross-section. So if » = the shearing stress in pounds per square inch of
cross-section, S = the total shear on the cross-section in pounds, and A =
the area of the cross-section in square inches, we have for the shearing
stress per square inch the general formula

af
eZ

|p3 |rs

iQ kn

 

 

 

 

w----|

 

The assumption that the shearing stress is uniformly distributed over the
cross-section is not absolutely true, as will be shown later; however, it
meets the requirements of practical designing in most cases.

53. Bending Stress on Beams.— Imagine for the time being that
the block abcd (Fig. 56) be removed and suppose instead the parts abd
and dcB of the beam to be connected by a hinge placed in the center of
the cross-section of the beam as shown in Fig. 58. Now it is evident that
this hinge could be so constructed that it would prevent the part abd of
the beam from moving down vertically. This means

   

that the hinge would resist the shearing force or ad
“shear” which was resisted by the block abed, but a BoB BUY
it is evident that the part abA would now fall by ih

rotating downward to the left about the hinge. So it Fig. 58
is seen that the part abd of the beam has a tendency
to rotate about the block abed, and hence will be subjected to stresses
therefrom. All stresses produced in this manner in beams, or any other
bodies, are known as bending stresses, or as stresses due to cross bending.
To determine the bending stress in the block abcd of the above beam
(Fig. 56), let us assume all of the block removed except a thin horizontal
strip ad at the top of the beam and-an equal strip be at the bottom, as
shown in Fig. 59. For the sake of simplicity let us first consider the
stress on these strips due to the load P only. The load P would cause the
54 STRUCTURAL ENGINEERING

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

part abA and the load P form a couple whose moment is 2sx or Pz, as
couple+o balance a couple (see Art. 44) the stress produced upon the two
the beam. Let F be the stress produced on each
on each strip. So far our problem is very simple.
it is evident that the additional strips will help to resist the moment of
Here it is seen that we have one equation and two
will occur, so our problem is not as simple as it
in tension while those at the bottom are in compression, consequently the
top strips are in tension and the bottom ones in compression, it is evident
location of this strip which is known
it is obvious that g is the center of
Fig. 61
distance out from og. Now as the stress on any strip, according to

part abA of the beam to exert a downward vertical force s upon each strip
(see Fig. 59) which must be resisted by an upward force s exerted upon
the part abA in turn by each strip. These two upward forces s upon the
2s = P. This couple tends to rotate the part abd of the beam counter
clock-wise, but actual rotation is prevented by the strips ad and be, and
consequently each will be subjected to a stress. Now as it requires a
strips must form a couple and the stress in strip
P t ad must act to the right, while the stress in the
Z strip bc must act to the left upon the part abd of
x :
A
= Ia 3 strip and let A be the vertical distance between
ae their centers of cross-section. Then we have Fh=
Fig. 59 Px, from which we obtain F =(Pzx)/h as the stress
But suppose we add two other strips, mm and mn, the same in section as
the two just considered and which are also symmetrically arranged in
reference to the longitudinal axis of the beam as shown in Fig. 60. Now
the couple Pz, so that the stress F on the two strips just considered will
be reduced. Let f and f’ now be the stresses on the strips, and h and h’
their lever arms, as indicated in Fig. 60. Then we have Pa=fh+f'h’.
unknown quantities, f and f’, and it is readily seen
that for each additional pair of such strips com-
posing the block added, another unknown force
first appeared. However, we get out of our diffi-
culty by resorting to Hooke’s Law (see Art. 32). Fig. 60
It is seen that the strips at the top of the beam are
strips at the top will be lengthened while those at the bottom will be
shortened, and the block abcd, which is rectangular when not subjected to
bending stress, will really be wedge-shaped as shown in Fig. 61. As the
that there must be a horizontal strip of fibers somewhere between the top
and bottom of the beam which has neither tension nor compression; that
is, no stress at all. Let og be the
as the neutral plane or neutral axis.
As all the strips above og are in ten-
sion and all below are in compression,
tendency of rotation of the part abA
; ; of the beam. Then, evidently, the
» distortion of the strips both above and below og will vary directly as their
Hooke’s Law, will be directly proportional to its distortion, the stress per
square inch on any strip will be directly proportional to its distance out
from og, and hence the stress on the strips may be represented by the
THEORETICAL TREATMENT OF BEAMS 55

arrows enclosed in the triangles nkr and mpr as shown in Fig. 61, where
the lines kr and mr have the same slope with the vertical.

Let f be the stress per square inch on any horizontal strip of fibers
y distance out from og, and let f’ be the stress per square inch on a strip
out unit distance from og. Then we have the proportion

Lig fs f,
from which we get

head
Py

Now as f/y is the stress on a strip out unit distance from og, the stress on
a strip out any distance zg from og would be 2 times f/y or (f/y)z, and
the moment of it about g would be 2 times this or (f/y)s?. Now f was
taken as so many pounds per square inch, but as the stress varies con-
tinuously from og outwardly, there will not be a square inch of material
anywhere having the same stress, so that each strip must really be con-
sidered as being a horizontal element; that is, a mere plane of fibers
having an infinitesimal area of cross-section da. For the sake of concep-
tion we may say that da is 1/1,000,000 of a square inch. Then the actual
stress or force out z distance from og would be 1/1,000,000 of (f/y)z or
(f/y)2da (da = dzb) (see section Fig. 61), and the moment of it about g
would be 2 times that, or (f/y)2°da. Now, undoubtedly the summation of
these moments (f/y)2?da about g for all of the strips in the block abcd,
which we can express as (f/y)32?da, must be equal to the moment of the
vertical couple Pz. So we have

Pa=(f/y)32°da.
But S2*da=I

(see Art. 49), the moment of inertia of the cross-section of the beam.
Then we have the equation

Px et I.
y

It will be observed very readily that Px is simply the moment of the
load P about the section ab (ab being considered vertical, and practically
it is); that’ is, Px is the measure of the tendency of rotation of the part
abA of the beam about the section ab due to the load P. Then evidently
the moment of any other load to the left of the section ab about the section
would be the measure of the tendency of rotation of the part abd of the
beam about the section due to any such loads. Then if =Pw be the alge-
braic summation of the moments of any number of loads on the left of the
section about the section, we have the formula

from which the stress f on any horizontal element of the block abcd due to
the loads can be computed by substituting for y the distance of the element
out from og, and f would be known as the bending stress on the element,
the block abcd being considered infinitesimal in length.

Now it is obvious that the above formula will apply to any section
of the above beam if SPz be taken as the algebraic summation of the
56 STRUCTURAL ENGINEERING

moments of the loads to the left of that section about the section. But it
is readily seen that the }Px on one side of any section must be equal and
opposite to the %Pz on the other side in order that the beam may not
rotate about the section. This is undoubtedly true in the case of any
beam whatever. That means that the above formula applies to any
section of any beam if 3Pzx be taken as the algebraic summation of the
moments of the forces (both applied forces and reactions) on either side
of the section about the section. This summation is known as the bend-
ing moment at the section and is usually represented by the letter M.
Then we can write the general equation as

I
M ee erin ea cidsa ee eee Wea eeIs yes (C),
y
and by transposing we have
M
f= a betes oleh fos Meena Sci Diet Nahe hate oe (Dy,

which is the stress on any horizontal element y distance from the neutral
axis either above or below it. It will be observed that the neutral axis
extends all the way along the beam.

It is seen that y varies from zero to either +e or -¢ (Fig. 61) and
that f will be a maximum when y is a maximum, as M and J are constants
for each specific section. So if y be taken as the distance to the farthest
element out from g the maximum value of f will be obtained. This is what
is usually desired, and for that reason y is usually spoken of as the dis-
tance to the extreme fiber and f as the stress on the extreme fiber.

The summation of the bending stresses above the neutral axis is
always equal to the summation of the bending stresses below it, but the
sums of their moments about the neutral axis are not equal except in the
case of symmetrical section. But each force on one side is always paired
with an opposite and equal force on the other side, thus forming couples
throughout the cross-section.

54. Reaction on Simple Beams.—In order to determine the shear-
ing and bending stress in a beam, it is first necessary to know all of the
external forces acting upon it. The applied forces, or loads as they are
known, are usually given, but the reactions are
usually unknown and hence have to be deter-
mined before the stresses in the beam can be
determined. We usually resort to the equations
of moments to determine the reactions on beams.
However, they can be graphically determined,
as will be shown later.

Fig. 62 Let AB (Fig. 62) represent a beam which

is supported at A and B and which in turn sup-

ports the loads P1, P2, P3, and P4. The reaction R at A due to the

above loads can be obtained by taking moments about B, and the reaction

R1 at B can be obtained by taking moments about 4. Thus, taking
moments about B, we have

+RL-aP1-bP2-cP3-dP4=0,

 

 
THEORETICAL TREATMENT OF BEAMS 57

from which we obtain
aP14bP24+¢P3+dPt
L

and by taking moments about A we can determine R1 in the same manner.
It will be observed that by taking moments about the line of action of one
of the reactions we eliminate that reaction from the equation of moments,
thus reducing the number of unknowns in the equation to one. The sum
of the moments being equal to zero is, in accordance with the laws of
equilibrium, that the moments of the external forces acting upon a body in
equilibrium about every point must be equal to zero (see Art. 42). The
above method is quite general, provided the loads are fully known and the
reactions are limited to two in number, which is always the case for
simple beams.

55. Graphical Representation of Shear on Simple Beams.—
The shear on a beam at any section is equal to the algebraic sum of the
loads and reaction (as stated in Art. 52), summed up to the section, begin-
ning at either end. For example, the shear at every section between the
loads P3 and P2 (Fig. 62) is equal to either R- P4—-P3 or R1-P1-P2;
between A and P4 it is equal to either R or R1-P1-P2-P3-P4;
between B and P1 it is equal to either Rl or R-P4-P3-P2-P1.

This shear on the above beam can be graphically represented in the
following manner: /

Draw a horizontal line' ab (Fig. 63) equal to L, the length of the
beam (between centers of end bearings), to some convenient scale. Then
draw the lines of action of the forces in their relative positions as shown
at 1, 2, 8, and 4. Then beginning at one end,
¢ d say, at a, draw ac=R. Then through c¢ and
é oh |. parallel to ab draw cd and from d lay off de=

Pie ; P14. Then from e draw ef and from f lay off
a AR re 5 fg =P3, and so on, thus obtaining the zigzag line
k i Ri cde ...mn. Any ordinate as rr from the line
m—n ab to the zigzag line represents the shear on
Fig. 68 the beam at that point.

The value of the shear on any vertical sec-
tion of a beam will depend upon the weight of the load or loads it sup-
ports and upon the position they occupy in reference to the section.
Loads are known -as dead load and live load.

The dead load upon a beam is the weight it supports that is fixed in
position, while the live load is the weight it supports that is not fixed in
position but may move to any position upon it. The dead load of struc-
tures usually consists of the weight of the structure itself, while the live
load consists of loads which it supports but which at the same time move
over it, as a locomotive, train of cars, wagons, etc. The dead load is
usually considered as a uniformly disttibuted load. The live load is
sometimes considered as a uniformly distributed load, but more often as
concentrated loads.

If the load be a uniformly distributed dead load, each reaction will
be R=wL/2, where L is the length of the span and w the weight of the
dead load per foot of span, and the shear out any distance z from either
end will be S=R-2zw=wL/2-2w. It is seen that when z=L/2 the

R=

 

 

 

 

 

 

 

 

 
58 STRUCTURAL ENGINEERING

.

shear is equal to zero, and when 2=0 the shear is equal to wL/2. This
shows that the shear due to such a load is a maximum at the ends and
zero at the center. :

The above equation for shear, S=wL/2-2w, is an equation to a
straight line, so evidently the shear can be represented graphically by the
ordinates to a straight line, as acb (Fig. 64), where 4B represents the
length of span and aA and bB are equal to
wL/2. The shear, beginning at A, decreases
until c, the center of the span, is reached,
where it is zero; then beyond that point, still
summing up to the right, the shear will be
minus, increasing toward B, being equal to
~wL/2 at B. So it is seen that the ordinates

Fig. 64 to the line acb truly represent shear for dead

load uniformly distributed. For example,

the ordinate fe represents the shear at f, which is positive, while kh rep-

resents the negative shear at k. The sign of the shear has no practical
significance other than algebraic.

In case of uniform live load moving over a beam, the shear at any
section will be a maximum when the load just extends from one of the
ends up to the section. For example, suppose a live load of p pounds per
foot extends from B up to some section O, a point « feet from B. Then
the shear at O is equal to the reaction at A, which expressed in terms of
the load is (pa)a/2+L=p27/2L. It is evident that the shear produced
at section O, as the load moves from B to O, keeps increasing as the load
approaches O, as it is equal to the reaction at A, which keeps increasing
as the load moves from B to O. Now if the load extends past O to any
section m, the reaction at A will be increased a certain amount by the
additional load py, but in obtaining the shear at O we subtract all of the
load py from the reaction at A, and as all of the additional load py is not,
as we say, transmitted to the end 4, it is evident that the load extending
beyond O will diminish the shear at that section. Therefore the above
statement is shown to be true, and the equation S= px?/2L is the general
equation for the maximum shear produced on a beam by a uniform live
load moving over it. The equation is that of a parabola. Then, evidently,
if we construct a parabola as auB so that Aa is equal to pL /2, the shear
at any section will be represented by the ordinate to the parabola at that
section. For example, the shear at ¢ is represented by the ordinate ut.

The shear curve auB (Fig. 64), as stated above, is an absolute
parabola for a uniform live load, but for such loads as the wheels of a
locomotive it would be a curve made up of a series of straight lines,
however, if the loads are nearly equal in weight and quite uniform in
spacing, the curve will approach a parabola, and in most cases it is near
enough to be considered one for practical’ purposes.

56. Graphical Representation of Bending Moments on Simple
Beams.— :

Case'I. When the load is uniformly distributed over the entire
length.

Let AB (Fig. 65) represent a beam supported at 4 and B which
in turn supports a uniform load of w pounds per lineal foot of length.
The bending moment at any section ss, # feet from A due to this load is

 
THEORETICAL TREATMENT OF BEAMS 59

a wa,
M=Re- (wz) a =Re—s-
woL
But R Zs
so we have
whe wx
tS ge

for the bending moment at any section of the beam. When #=L/2, we
have

wl? wl?
f=— _——_
a 4 8?
from which we obtain
wL?
M=—; “Sats tesla as Ga laste eatroiac ona hee abeeaa ore snatenole eigen mero ees (E),

which is a formula of great practical value as it expresses the moment at
the center of any simple beam uniformly loaded, which is the maximum
moment that can occur under such loading, and usually this is what we
desire to know.

The above formula, M = wla2/2 - wa’/2, is an equation to a
parabola. Then, evidently, the bending moments at the different sections
along the beam vary as the ordinates to a par-
abola. It is seen from the equation that M=0
when #=0, and also when «=L; and when 2
=L/2,M=wL?/8. Then if we draw CO equal to
wL?/8 to scale, and perpendicular to the beam
at its center and pass a parabola through 4, O,
and B as shown in Fig. 65, any ordinate as rr
will represent the bending moment on the beam

 

at that point.

Case II, When the loads are concentrated at different points along
the beam.

Let 4B (Fig. 66) represent a beam supported at A and B which
in turn supports the loads Pl... P5, as shown. The bending moment
anywhere between A and P5, due to these loads, is M=Rz, which is an
equation to a straight line. When «=0, M=0, and when x=a, M=Ra.

Then, if we draw cd equal to Ra, to scale, and

 

 

 

 

 

 

 

 

 

join A and d, any ordinate as ss to the line Ad L

will represent the bending moment on the beam = a mo Moe

at that point. The bending moment anywhere Pie | hg
between P5 and P4 is M=R(a+a’)—-P5x’, ° me i
which is also an equation to a straight line. When R a pe

 

 

«z’=0, M=Ra, so, evidently, one point in the Fig 66
line represented by the equation M=Rf (a+2’)

—P52’ isd, as cd=Ra. When x’ =b, M=R(a+b)—-P5b. So if wedraw
ef equal to R(a+b)—P5b to scale, using the same scale as in the case
of cd, we obtain the line df, the ordinates to which evidently represent
the bending moments between P5 and P4. Now it is evident that the
lines fg, gh, hk, and kB can be similarly constructed, the ordinates to
60 STRUCTURAL ENGINEERING

which represent the bending moments between P4 and P3, P3 and P2,
P2 and Pi, P1 and B, respectively. So it is seen that the bending
moments on a simple beam due to any number of concentrated loads
can be graphically represented by the ordinates to a series of straight
lines forming a closed polygon with the ends of the beam. The nearer
the loads approach a uniform load, the nearer this polygon approaches
a parabola.

It is important to note that the maximum moment always occurs
under a load. This is readily seen, for the maximum ordinate to any
segment of the above polygon is under a load and hence the maximum
of them all will be under a load.

57. Graphical Representation of the Bending Moments and
Shears on Cantilever Beams.—

Case I. When the load is uniformly distributed
over the entire length.

Let AB (Fig. 67) represent a cantilever beam
supporting a uniform load of w pounds per foot of
length. The bending moment at any section ss, « feet
from A, is M=(wa)x/2=wx?/2, which is an equation
to a parabola. Making #=L, we have M=wL?/2,
which is the bending moment at B. Then if we draw
the horizontal line ab=Z, and the vertical line cb=

Fig. 67 wL?/2, to scale, and construct the parabola ac, the

ordinates between the line ab and the curve ac will
graphically represent the bending moments on the beam due to the above
load. For example, the moment at the section ss is represented by the
ordinate rr.

The shear at any section ss is S=wze, which is an equation to a
straight line. When r=0, S=0, and when rx=L, S=wL, which is the
shear at B. Then, if we draw bo equal to
wL, to scale, and join a and 0, the shear at
any section in the beam is represented by
the ordinate between the line ab and ao
immediately under the section. For exam-
ple, the shear at the section ss is repre-
sented by the ordinate er.

Case II. When the loads are concen-
trated at different points along the beam.

Let AB (Fig. 68) represent a canti-
lever beam supporting the concentrated
loads P1 . .. P4 as shown. Asa general 9
example, the bending moment due to the
above loads at a section as ss, expressed (P2)
analytically, is M=nP1+bP2+tP3. If we f &
proceed in the same manner as in the case (P3)
of a simple beam, making the horizontal Ff
line ab=L, and the verticals wa, w’a’,
etc., under each load, equal to the bending Fig. 68
moment about each, respectively, making
bv equal to the moment ‘at B, and drawing the broken line aww’w’’s,
we have the bending moments on the beam graphically represented by

 

 

 

 

 

 

 

 

xe act a

Pl)

 

 

vn

 

 

 

(P4)
C Mm

 

 

 
THEORETICAL TREATMENT OF BEAMS 61

the ordinates between the line ab and this broken line. For example,
the moment at the section ss is represented by the ordinate ro. The
shear at the section ss is S=P1+P2+P3, expressed analytically. Con-
structing the zigzag line acd...m, the same as in the case of simple
beams (Art. 55), we have the shear at any section represented by the
ordinate between the line ab and this zigzag line.

58. Graphical Construction of a Parabola.— The following method
of constructing a parabola will be found to be quite convenient. It really
consists in passing a parabola through two points, when one of the points
is taken at the vertex. Let A and B (Fig. 69) be any two given points
and let A be at the vertex of the parabola desired. Draw the line XX
through A as the X-axis of the parabola. Then draw AC perpendicular

 

 

 

 

 

B . 4
oe AAS
a 4
& 3
4 :

t A =
C i: go © dg ae C¢

Fig. 69

to XX and BC perpendicular to AC. Divide AC and BC into an equal
number of equal parts, say, six, as shown. Then from 4 draw the radial
lines 1-A, 2-4, etc., and where the radial line 1-A intersects the vertical
aa is one point on the curve of the required parabola, and where the next
radial line 2-A intersects the next vertical bb is another point on the
curve, and soon. The other side AB’ of the parabola can be constructed
-by laying off AC’=AC and dividing it into the same number of equal
parts as AC and drawing verticals as shown and then projecting the
corresponding points over from the curve AB just constructed, or by
dividing B’C’ into the same number of equal parts as AC’ and proceeding
as explained for the other side of the curve.

59. Relation of Shear to Bending Moment.—Let AB (Fig. 70)
represent a simple rectangular wooden beam supported at A and B and
let P be a load which the beam supports at mid-span, and let R and #1
be the reactions (which will be equal) at A and B, respectively, due
to this load P. Suppose the reactions applied exactly at the ends of
the beams as indicated, which is not at all an imaginary case, and imagine
the beam divided up into short rectangular blocks as shown, by imaginary
-planes passing perpendicularly to the longitudinal axis of the beam. Let
Ax be the length of each block. The reaction R at A applied along the
end of the first block, beginning at A, tends to move that block upward.
If the block does not move undoubtedly the second block prevents it by
exerting an equal and downward force R where the two blocks join as
indicated. As the first block is held entirely by the second block it is
obvious that the first block will exert an upward force upon the second
block where the two blocks join, and if the second block does not move
upward undoubtedly the third block prevents it by exerting a downward
62. STRUCTURAL ENGINEERING

force R where the second and third blocks join, and hence it is seen that
the blocks to the left of the load resist each other in transmitting the
shear, each thereby receiving an upward force R upon one end and an
equal downward force upon the other end.
as is indicated. The same is true of the
cel « } __g#d, blocks to the right of the load P except
oa SHAE 4 Hs the forces are reversed in order. These

 

 

 

 

 

 

 

 

 

 

 

 

hha el vertical forces not only tend to shear the
Al blocks off vertically but tend to rotate them
Fig. 70 as well, as the two forces on each block

form a vertical couple.

Now beginning at A and considering the blocks to the left of P only,
the bending moment at the right end of the first block is RAz; at the right
end of the second block it is R(2Ax); and at the right end of the third
block it is R(3Ax); and so on, being R(nAz) at the right end of the nth
block from A, which is seen at a glance to be nothing more than the
summation of the moments of the vertical couples on the blocks beginning
at A. This means that the bending moment beginning at the end 4
increases toward the load P by one RAz at each block. For example, the
bending moment at kk is one RAw greater than it is at hh. So, evidently,
RAz is the increment of the bending moment to the left of the load P in
the case of the above beam. By the same reasoning we have R1Az as the
_ increment of the bending moment to the right of the load P.

It will be observed that all of the blocks to the left of the load P
tend to rotate clock-wise while all of the blocks to the right of P tend to
rotate in the opposite direction or counter clock-wise, so that the sign of
the increments of the bending moment on one side will be different from
those on the other side of the load. The bending moment at any point
to the left of the load P is equal to the sum of the increments RAz
between A and the point and the bending moment at any point to the
right of the load is equal to the sum of the increments R1lAz between B
and the point, although the bending moment at any point is equal to the
sum of the increments summed up algebraically from either end of the
beam. For example, the bending moment at gm is equal to the sum of
the increments to the right of gm or to the sum of the increments to the
left of P minus the sum of the increments between P and gm, as the signs
are different.

In the case shown in Fig. 70 the increments of the bending moment
are all equal, but suppose we consider the case shown in Fig. 71 where
R and R1 are the reactions at A and B, respectively, due to the three loads
Pi, P2, and P3. Here it is readily seen
that the increment of the bending moment i |Pe ie
between 4 and P1 due to Pl, P2, and P3 yCEREFEF EE EEE b
is RAx; between P1 and P2 it is (R-P1) +H HEHE B

Ri

 

 

 

 

 

 

Aa; between P2 and P3 it is (R-P1-P2)
Az; and between P3 and B it is (R-P1
—P2-—P3)Ax. That is, the increment of
the bending moment at any point is equal to the shear at that point multi-
plied by Aw. It will be observed that this was true in the case shown in
Fig. 70, for R is the shear to the left of P and R1 the shear to the right
of P, which are equal in that case. Now, the vertical couples on any such

 

 

w

Fig. 71
THEORETICAL TREATMENT OF BEAMS 63

blocks or strips, as here considered, in any beam, could not be anything
other than the vertical shear couples on them, so we have, in general,

AM =SAz

for the increment of the bending moment, where § is the shear on the
block in question and Az its length. Now, Az could have any practical
value but the ultimate increment will be when Az is an infinitesimal.
Then assigning the smallest possible value to Az making it an infinitesimal
dz, we obtain

dM =Sdzx

as the ultimate increment of the bending moment, as it is the smallest
increment possible. That is, the differential of the bending moment at
any section of any beam is equal to the vertical shear couple on an
imaginary vertical block of infinitesimal length. The preceding formula
is usually written

dM _

dz
Expressing this in words, we say that the first derivative of the bending
moment in reference to 2 is equal to the shear, where 2 is the distance
from the end of the beam to the section considered.

The above equation (F) is sometimes quite useful, as the determining
of the shear from the bending moment is sometimes desirable. The
equation, or formula, is very readily applied, as is seen from the follow-
ing, although its true worth is more readily recognized in more compli-
cated cases.

Considering both the load P and the weight of the beam acting upon
the beam shown in Fig. 70, the bending moment at any section x distance
from the end 4 is

2
M=Rx-—~

a

where w represents the weight of the beam per foot. Differentiating
both sides of the equation and dividing through by dx we have

dM

Ve =R- Wr,
which, as is readily seen, is the shear at any cross-section 2 distance from
the end A and to the left of the load.

The bending moment at any section x distance from the end of a
simple beam having a length L and supporting a uniform load’ of w pounds
per lineal foot is

wLe wx?
oe a
(see Art. 56, Case I). Differentiating toth sides of this equation, are
dividing through by dz, we have

dz 2

we,
64 STRUCTURAL ENGINEERING

which, as is readily seen, is the general expression for the shear at any
section of the beam. If 2=0, the shear is equal to wl. /2, and if «=L, it
becomes —(wL/2), each of which is recognized as an end shear or reac-
tion. If e=L/2 the shear is equal to 0, all of which goes to show how
readily Formula (F) can be applied, even for ordinary cases.

60. Increment of the Bending Stress.—Let ab (Fig. 70) represent
a very thin horizontal strip through the beam y distance above the neutral
axis 00. The end block at A, owing to its tendency to rotate clock-wise,
due to the vertical shear couple RAz, will exert a force to the right upon
all horizontal elements in the second block above the neutral axis and a
force to the left upon all such elements in the second block below the
neutral axis. Then undoubtedly the first block will exert a force to the
right upon the strip ab. Let Af be this force. The end block at B has
the same tendency to rotate as the end block at A but in the opposite
direction, and hence the end block at B will exert a force Af to the left
upon the strip ab. Then the strip ab will have a compressive stress of
Af in it from c to d due to the end blocks. The second block from the
end 4 having the same tendency to rotate as the end block will likewise
exert a force Af to the right upon the strip ab and the second block from
the end B will exert a force Af to the left upon the strip ab, and hence the
strip ab will have a compressive stress of 2Af from e to f due to the first
and second blocks from the ends. The third block from the end 4 will
exert a force Af to the right upon the strip ab and the third block from B
will exert a force Af in the opposite direction upon the strip ab and hence
the strip ab will have a compressive stress of 3Af from 1 to g due to the
first, second, and third blocks from the ends. Thus it is seen that the
stress in the strip ab is increased at each block by one Af as we pass from
either end toward the load P where the stress is a maximum. So,
evidently, Af is the increment of the bending stress in the strip ab. As
the strip ab could be any horizontal element out any distance y from the
neutral axis, either above or below, it is evident that the bending stress on
any horizontal element in the beam will vary by such increments as Af
the same as in the case of the strip ab; of course, their real values will be
different owing to their distance out from the neutral axis being different.
the value of Af at any block to the left of the load P (Fig. 70) we.

ave

ag =EAe)Y ,

and for its value to the right we have

oe
where I is the moment of inertia of the cross-section of the beam in
reference to the horizontal gravity axis of the beam; y the distance out
from the neutral axis to the strip or element considered; and RAz and
F1Azx the vertical couple in each respective case. But as the vertical
couple on any imaginary vertical block or strip in any beam whatever is
simply the shear on the block multiplied by its length, we can write the
general expression for the increment of the bending stress at any point in
THEORETICAL TREATMENT OF BEAMS 65

any beam as

Then beginning at either end of a beam we can obtain the stress on any
horizontal element « distance from the end by summing up the stress
increments as expressed by (a). This would give us

SAf=3SAx 7

But SAf =f and 3SAz=M, the bending moment (see Art. 59) at a point «
distance from the end, when Az becomes an infinitesimal. Then substi-
tuting f for SAf and M for’ SSAz in the last equation we have

M
fost,

which is Formula (D), Art. 53.

Owing to the shear being constant the increments of the bending
stress in the beam shown in Fig. 70 are constant throughout fer each
horizontal element and are equal to

But in such cases, as shown in Fig. 71, they will be different owing to the
variation of the shear. The increment between 4 and the load Pl, in
that case, due to the loads P1, P2, and P3 is expressed as

—_(RAz)y |
Af= T >
between P1 and P2 as
hes (Rat ey :

between P2 and P3 as
R-P1-—P2)Axr
apa : Ary

and between P3 and B as
(R -P1-P2-P3)Ary (R1Azr)y
T or = T

It is seen from the above that the increment of the bending stress
varies directly as the shear and hence in most cases the greatest increment
will be at the ends of a beam, especially when the weight of the beam is
considered.

61. Horizontal Shearing Stress in Beams.—Let AB (Fig. 72)
represent a portion of the same beam shown in Fig. 56 (Art. 52), which
is here drawn so as to show the imaginary block abcd to a large scale.
Conceive of the original imaginary block abcd sub-divided into smaller
imaginary blocks as egkh, gmnk, etc. We can now conceive of the shear-
ing force exerted upon the block abcd along ab and cd as being distributed
to each of the smaller blocks, whereby each is seen to have a vertical
couple which we can conceive of as being an increment couple of the

 

Af=
66 STRUCTURAL ENGINEERING

vertical couple acting upon the block abcd as a whole. Now, evidently,
each of the smaller blocks has a tendency to rotate owing to this vertical
couple, and if rotation does not take place, evidently it is prevented by
the material joining the blocks horizontally. Then the forces resisting the
vertical couple on each of the smaller blocks must act through this mate-
rial, forming a horizontal couple on each block as shown, which must
necessarily be equivalent and opposite to the vertical couple that it resists.
Now, it is readily seen that these horizontal couples tend to shear the
smaller blocks off horizontally the same as the vertical couples tend to
shear the same blocks off vertically. If a block be square, it is readily
seen that the vertical and horizontal couples on it will not only be
equivalent, -but will be equal and opposite, which means that each of the
forces in the horizontal couple is equal to each of the forces in the vertical
couple; that is, the four forces acting upon the block are equal. This
means that the horizontal shear on the smaller blocks is equal to the
vertical shear on the same. But if the smaller blocks are considered

ad

 

Crs}
okt
me

3I>>

 

 

 

 

 

 

 

bc

Fig. 72

 

shorter or longer in one direction than in the other, the unit shear is not
changed. The larger sides having the greater area will simply have a
greater force acting, but the force per square unit will be just the same,
as our mentally changing the block will not alter the unit shear. As far
as the two couples on any block remaining equivalent is concerned, it is
evident they will, for as the force on one side is increased, the lever arm
of the other is increased, so that the moments of the couples are con-
tinually equivalent. So it really does not matter what shape our fancy
molds the imaginary parts, for it remains evident that the horizontal and
vertical shear on any particle in any beam are equal. Then, if we know
the horizontal shear on any particle, we know the vertical, as the two are
equal, and vice versa. The determining of the total vertical shear at any
vertical section of a beam is an easy matter, but just what the intensity
of it is on any particular particle is another thing. What we really do is
to find the horizontal shear and consider the vertical shear equal to the
same. In order to do this, let abed, Fig. 73, represent the imaginary block
abcd of Fig. 56, as an independent body where the other parts of the beam
are replaced by the forces which those parts exert upon the block. Let
go be the neutral plane. The part of the beam to the left of the block
will exert upon it the bending stresses represented by the arrows in
triangles agm and bgm’, and also the vertical shearing force S along ab,
while the part of the beam to the right will exert upon it the forces
THEORETICAL TREATMENT OF BEAMS 67

represented by the arrows in the triangles dok and cok’, which resist the
bending stresses represented by the arrows in the triangles agm and bgm’,
and hence are equal and opposite to them, and also the shearing force §
along de, which is equal and opposite to the force § acting along ab, and
also the bending stress increments represented by the arrows in the
triangles don and con’.

The bending stresses represented by the arrows in triangles agm and
bgm’ are transmitted directly through the block and are balanced directly
by the forces represented by the arrows in the triangles dok and cok’, so,
evidently, none of these forces causes horizontal shear on the block and
can be disregarded. The force § acting along ab andthe equal and
opposite force § acting along dc and the bending stress increments repre-
sented by the arrows in the triangles don and con’ are the only forces
remaining. But as each of the forces S acts vertically, it is evident that
the horizontal shear in the block is due entirely to the forces represented
by the arrows in the triangles don and con’; that is, to the increments of
the bending stresses on the block, and hence it remains for us to consider
these forces only.

‘Now, evidently, the horizontal shear on any element as ee of the
block abcd is equal to the algebraic summation of these increment forces
on either side of ee, the same as in the case of any body. Therefore, the
horizontal shear at ee is equal to the sum of these forces between d and e,
or between e and ec.

The moment of these increment forces is equal to SAz, which they
resist. Then it is readily seen that the stress at d, represented by nd, is

Af= (Baey :
and at e it is r
af’=4 Aey :

where I is the moment of inertia of the cross-section of the beam at the
block abcd. Then the average increment stress between d and e is

Multiplying this by bxde, where b=width of beam, we obtain the total
horizontal shear along ee, which is

(Fe) Pat)(s-9)

Now this shear is distributed over the horizontal strip through the block
at ee. If b be the width of the beam, the area of this horizontal strip will
be bAr. Then the shear per square inch upon it will be

_ [ Sdx\ (h+y
a= ( ) (4) (+-y) b,
from which we obtain the formula

h? — 2
5-3 ("5") reeee sai akaale Sissies 9-00 Wie ere: o0 a ate chee wieie 8G)

 
68 STRUCTURAL ENGINEERING

From Formula (G) the horizontal shear (and incidentally the vertical)
can be obtained at any point in any rectangular beam either above or
below the neutral axis.

It is readily seen that Formula (G) is an equation to a parabola,
which means that the horizontal shear on any rectangular beam, and
likewise the vertical shear, varies on any vertical section as the ordinates
to a parabola, as represented by the horizontal ordinates to the curve
axb shown in Fig. 73. If y=h, the shear’ S, is equal to 0. If y=0,
S,=38/A, which is the maximum, where A=area of cross’ section
of beam. So it is seen that both the vertical and horizontal shear is 0 at
the top and likewise at the bottom of any beam, and a maximum at the
neutral axis. The same is true of all beams.

In case the width of a beam varies the general equation

8.=(F)m.. eaaeeat’ Fascucineepindatik Game

can be used for determining the horizontal shear, where m=the moment
about the neutral axis of the part of the cross section above ee or below
in case ee is below the neutral axis and b=width of beam at ee.

62. Maximum Stress.—It is seen from the preceding article that
every particle in a loaded beam, except the particles at the top and bottom
edges, and those in the neutral axis, is subjected to horizontal and vertical

shear and to a direct horizontal stress which may be
“ig either tension or compression, depending upon the
‘position of the particle in the beam. In the case of
~ ° a cantilever beam the particles at the top edge are
B subjected to direct tension only, while all particles
A =@F between the top edge and neutral axis are subjected
Ee to horizontal and vertical shear and also to direct
Fig. 74 tension, and the particles in the neutral axis are sub-
jected only to horizontal and vertical shear. At the
bottom edge of a cantilever beam the particles are
subjected to a direct. compression only, while the
particles between the bottom edge and neutral axis
are subjected to horizontal and vertical shear and
also to direct compression. In the case of a simple
beam we have exactly the reverse of a cantilever.
beam. As an illustration, let AB (Fig. 74) represent
a short portion of a simple beam where 00 represents
the neutral axis. The particles at the top edge are subjected to direct
compression only, as indicated at a, while the particles between the top
edge and neutral axis are subjected to horizontal and vertical shear and
also to direct compression as indicated at b, the particles in the neutral
axis are subjected only to horizontal vertical shear as indicated at c. At
the bottom edge the particles are subjected to direct tension only, as in-
dicated at e, while the particles between the bottom edge and the neutral
axis are subjected to horizontal and vertical shear and also to direct
tension, as indicated at d. ;

It is seen from Fig. 74 that the intensity of the maximum stress on
any particle at the top or bottom edge of the beam is simply the direct
stress on it, but the intensity of the maximum stress on any other particle

 

 

 

 

 

 

 

 

 

 

 
THEORETICAL TREATMENT OF BEAMS 69

is not so evident, for herc the maximum stress is due to the combined
action of the shearing forces and direct stress. Let Fig. 75 represent a
small imaginary block taken from a simple beam, corresponding to the
block at d, Fig. 74. For convenience consider the block to be a cube
having sides of unit length, and let h represent the horizontal shear and
v the vertical shear. It is readily seen that the kh along fe and the v along
ek acting against the v along fg and the hk along kg produce tension upon
all such strips through the block as fk, while the h along fe and the v
along fg acting against the v along ek and the h along kg will produce
compression upon all such strips through the block as eg. Now it is
evident that the maximum tension and compression as the case may be,
due to the shearing forces, will be upon strips perpendicular to the
resultant of the shearing forces. As the horizontal and vertical shear are
equal it is obvious that their resultant will always be at 45° with the
horizontal and vertical and hence the maximum tension or compression
stresses due to these forces will be upon strips perpendicular to this
direction, that is, 45° from the direction of the shearing forces them-
selves. It is readily seen that the direct tensile stress indicated as p on
the block will increase the tension due to the shearing forces on the strip
fk and decrease the compression on the strip eg; while if p were a com-
pressive stress just the reverse would be true. While the tensile stress p
would increase the tension on the strip fk, yet this would not be the
maximum tension on the block, for evidently the maximum would be on a
strip through the block perpendicular to the resultant of the stress p and
the shearing forces v and h. This resultant would have some position as
Rz, if p were tension, and Re, if p were compression, and the maximum
tensile stress then would be upon strips through the block perpendicular
to Rt; and if p were compressive the maximum compressive stress would
be upon strips through the block perpendicular to Re.

If the resultant of the shearing forces on the particles of any beam
were graphically combined throughout the beam with the direct stresses
on the particles, the balanced resultants obtained would form curves
known as the lines of maximum

 

 

stress. As an example, let 4B (Fig. lp |rz [rs

76) represent a simple beam. By yo =
combining the forces at e, d, c, ete., | ie pe ae a ee
as indicated at e’, d’, and c’, theajae “—_< < 7 — At B
curve ede would be obtained and by gM C4

 

combining the forces on the particles
throughout the entire beam we would |R
obtain some such lines as 7, #1, #2,
etc., representing the direction of the
action of the maximum tensile
stresses in the beam. Now, evi- Fig. 76
dently, if any particle in the beam
failed in tension the failure would take place perpendicular to these lines
of maximum tension. “

If the maximum compressive stresses were platted in the same
manner they would take some such position as indicated by the dotted
lines on the beam, and if any particle failed in compression the failure

 

 
70 STRUCTURAL ENGINEERING

would evidently take place perpendicular to these lines of maximum
compression.

It is readily seen from Fig. 75 that the shear due to the shearing
forces v and h along a strip as eg, 45° with the horizontal or vertical,
will be zero, for the two shearing forces h and v on each side will just
balance each other when-resolved along that direction. Then the only
shear along that direction will be due to the direct stress p. If we
imagine the strip rotated about o it is readily seen that the shearing
forces will produce shear along the strip when it slopes other than 45°
with the horizontal and the maximum shear on it due to the shearing
forces and direct stress combined will occur when the angle of slope has
some particular value. But this angle of slope depends upon the relative
intensity of the shearing forces and the direct stress. The direction of
maximum tension and compression stresses also depends upon the relative
intensity of the shearing forces and direct stress. So we will now deter-
mine these relations.

Let Fig. 77 represent a material particle assumed rectangular,
subjected to horizontal and vertical shear of s pounds per square inch
and to a tensile stress of f pounds per square inch. Suppose the particle
to be one unit in thickness, and let a be its width
and b its length. Then sa will be the total vertical
shear and sb the total horizontal shear, and fa the
total tensile stress, as indicated in the figure. Let
dd be an imaginary diagonal strip through the par-
ticle. If the shearing and tensile forces be resolved
perpendicularly and parallel to this imaginary strip
dd, the components perpendicular to the strip will
produce tension on it, while the components along the strip will produce
shear. Let t be the tensile stress per square inch perpendicular to the
strip dd and let v be the shearing stress per square inch along the strip,
and let @ be the angle that the strip dd makes with the horizontal axis
: the particle; then, considering the forces on one side of dd only, we

ave

 

(1) tdd=fa sin6+ sb sin6+sa cos for the tension,
and
(2) vdd=fa cosé + sb cos6— sa sin6 for the shear.

Dividing (1) and (2) by dd and substituting sin for a/dd and cos@ for
b/dd, we have

(3) t=f sin?0+ 2s cosOsiné = f (1 - cos26) +s sin26,
and

(4) v=f sinécos6+s (cos?6— sin?6) = f sin26@ +s cos26.

It is evident that t will be a maximum when @ has a certain value and v a
maximum when @ has a certain other value. Differentiating (3) we have

dt=f sin26d6 + 2s cos26dé.
THEORETICAL TREATMENT OF BEAMS 71

Now, ¢ will be a maximum when
=f sin26 + 2s cos26=0,
from which we obtain
(5) tan be= 2!
as the value of 6 when ¢ is a maximum. :
Treating (4) in the same manner we obtain

(6) tan ee
as the value of 6 when v is a maximum. Expressing the value of the
tangent in (5) in terms of the sine, we have

sin26 s&s

V1-sin? 200
Squaring and reducing, we have

(a) cao — =

VPP ide

Then expressing the value of the tangent in (5) in terms of the cosine,

we have
V 1-cos? 26 2s

= —_?>

cos 20. f
from which we obtain

f
b) cos 26= + ———_..
(®) Vf? +48?
Referring to equation (5), as tan 26=~—sin 26/cos 26, a minus quantity, it
is evident that sin 26 and cos 26 will have opposite signs, that is, when
one is plus the other will be minus. Then substituting the value of sin 26
and cos 26 given in (a) and (b), respectively, in (3), taking one plus and
the other minus, we obtain

al x \ Be a kd Diol betes oak eee Seo H).
t= 5" sey 7 (
By treating (6) in the same manner as (5) was treated above, we obtain
f
ch 5a
sin 26= V fi 4st
and

cos 26= + eae
Vf? + 48?

Substituting these values, which will have like sign, in (4), we obtain

v=tq feel pa cate esta irs ol DC Ee (1).

The maximum or minimum tensile or compressive stresses on any
material particle subjected to shear and direct stress can be computed
from (H). If f be tension, ¢ will be tension when the radical is taken as
7? STRUCTURAL ENGINEERING

plus and compression when taken as minus. Thus both tension and
compression stresses are obtained. If f be compression, by using the plus
sign before the radical we obtain the compressive stress and by using the
minus sign we obtain the tension. As the plus sign in (H) gives the
maximum stress in all cases it is rarely necessary to use the minus sign.

The maximum shear on any material particle subjected to shear and
direct stress can be obtained from (I). It is immaterial whether the
plus or minus sign before the radical be used as the two values will be
the same.

The direction of the maximum shearing stress can be obtained from
(6) and a direction perpendicular to the maximum and minimum tensile
and compressive stresses can be obtained from (5).

There are only a few cases where the above formulas need to be
applied. They need not be applied in the case of ordinary beams, for
here both the maximum tension and compression stresses occur on the
outer elements where the shearing stress is zero, and the maximum shear
occurs at the neutral axis where the direct stress is zero.

63. Deflection of Beams. —Beams
supporting loads deflect or bend ow-
ing to the distortion of their parts which
elated _L_Le results from the loading. Let Fig. 78 rep-

ay Tk resent an unloaded rectangular wooden

cantilever beam which we will assume to
Zn straight and absolutely horizontal.
Imagine this beam divided into very short
rectangular blocks by imaginary planes

1 passing perpendicularly to its longitudinal
| axis. Now if loads were applied to this
beam, these rectangular blocks would be-
come wedge-shaped as shown in Fig. 79,
and the imaginary planes dividing the beam
into blocks instead of being parallel would
become tangent to some curve as ss’s’’, and
the neutral axis 00 instead of being straight
would become a curve which would be
known as the “elastic curve” of the beam.
The distance from the intersection of any
two adjacent imaginary planes to the neu-
tral axis 00 would be known as the radius
of curvature of the elastic curve at the in-
tercepted block. The vertical distance that
any point on the neutral axis would move
down from its original position would be
known as the deflection of the point and of
the beam at that point, and the deflection
of any one point on the neutral axis in
reference to another would be the differ-
ence of their deflections. For example, the
distance y (Fig. 79) from the horizontal
line Ho to the point g would be the deflec-
Fig. 79 tion of point g, and likewise the deflection

 

 

 

 

 

 

 

 

 

 

 

Fig. 78

 
THEORETICAL TREATMENT OF BEAMS 73

of the beam at that point, while the distance z would be the deflection of
point g in reference to point k.

It is readily seen that the curving of the neutral axis results from
the distortion of the blocks. Then it is evident that if the slope of the
blocks in reference to one another be determined, an equation of the
elastic curve expressing this relation can be derived, from which the
deflection of the beam at any point in reference to any other point can be
computed. So we shall now proceed to derive such an equation of the
elastic curve.

Let Fig. 80 represent any four consecutive distorted blocks of the
beam shown in Fig. 79. By drawing the center line of each block we
have the broken line curve BabcA which has for its limit the elastic
curve SS as the lengths of the blocks decrease; that is, become infini-
tesimal. It is readily seen that the radius of curvature at any block is
perpendicular to its center line
at midpoint. So the slope of the
center line of any block with the
horizontal or X-axis is the same
as that of the tangent of the
elastic curve at that point. In
fact, the two coincide.

Let HB be a horizontal
line. Then ¢ is the angle that
the tangent to the elastic curve
at the center of the block vurw
makes with the horizontal or X-
axis. Now, as the length of the
block vurw decreases, the center
line aB comes nearer and nearer
to coinciding with the arc of the
elastic curve passing through the
block. So if the length of the
block be infinitesimal the points
a and B will be on the elastic
curve and the vertical distance
ad will be the deflection of point Fig. 80
a in reference to B, and ad
would be a first differential of the vertical or y ordinate to the elastic
curve at that point. Likewise, the distance eb, ge, and kA will each be
a first differential of a y ordinate to the élastic curve when the length of
each of the other blocks becomes infinitesimal. Letting A represent the
deflection of point A in reference to B, we have

A=ad+eb+gcet+kA,

 

 

which is simply the summation of the first differentials summed up from
B to A. Let dB=ea=gb=kc=dz, an infinitesimal, which would really
be the case when the lengths of the blocks are infinitesimal. Then by
prolonging aB, the center line of block vurw, to f we have ef=ad. Now
as ad and eb are each a first differential of a y ordinate to the elastic
curve, fb is a second differential of the same, being the difference between
two consecutive first differentials; and similarly hc and mA are also
14 STRUCTURAL ENGINEERING

second differentials of y ordinates to the elastic curve. It is readily seen
that gc=ad+fb+he and that kA=ad+fb+hce+mA, etc.; that is, each
first differential is equal to the summation of the second differentials plus
the first differential ad summed up from B to the point considered. Now
if the summation be from A to B instead of B to A, as above, we have the
case shown in Fig. 81, where ce, bf, ak, and Bn are first differentials, and
gb, ha, and mB are second differentials. Taking A as the origin, we have
ak=ce-—gb-—ha, and Bn=ce-gb-ha-mB; that is, any first differential
of a y ordinate to the elastic curve is equal to the summation of the second
differentials summed up from the origin to the point considered, plus the
first differential ce. (Fig. 81.) The second differentials are minus in the
last case, as they are measured down to the curve; while the first dif-
ferentials are measured up to the curve. For the deflection of point 4 in
reference to B we have A=ce+bf
+ak+Bn, which is simply the sum-
mation of the first differentials
summed up from 4 to B. Now, as
stated above, any first differential
at any point is equal to the summa-
tion of the second differentials
summed up from 4 to the point,
plus the first differential ce, while
in the case shown in Fig. 80 the
first differential at any point is
equal to the summation of the sec-
ond differentials summed up from B
‘ Fig. 81 to the point, plus the first differen-
tial ad. It is readily seen that as
far as the deflection of point A in reference to point B is concerned,
both of the first differentials ad (Fig. 80) and ce (Fig. 81) are
constants, and being such they will appear as constants of integration in
the summation of the second differentials. It is first necessary to derive
an expression for the second differential of any y ordinate to the elastic
curve in terms of known quantities. Referring to Fig. 80, if the length
of the two blocks stwo and vurw be infinitesimal, the angle subtended by
their radii of curvature will be an infinitesimal angle d6, and as one radius
is perpendicular to aB, the center line of block vurw, and the other one
to ab, the center line of block stuv, the angle fab will be equal to that
subtended by the radii, or d#. Then, as the angle d@ (Fig. 80) is
infinitesimal, we have

foSnbdl om diy = abdR. 6255 oak pee ade aie kenwaes (1).

The permissible deflection of a beam in any case is very small compared

to its length (1/1,000), so that no appreciable error will be made if we

“oti the slope distance ab=ea=dz. Then by substituting in (1), we
ave

 

 

DYSON aie Wi Pree eo since ee pa eso kw Kes oeEl eeKGunwle ke

Now d6 is the slope that the block stwv makes with the block OUT,
which results entirely from the distortion of the block stwv. Then,
evidently, the. value of d@ will be directly proportional to the bending
THEORETICAL TREATMENT OF BEAMS 75

moment at the block stuv and inversely proportional to the modulus of
elasticity of the material composing the block and also inversely propor-
tional to the moment of inertia of its cross-section. Then d@ can be
expressed in terms of these quantities.

Through a (Fig. 80) draw the line s’t’ parallel to st. Then the
angle s‘av=d6. Let a = the longitudinal distortion of an element of the
block (stuv) distance z above ab. Then we have

a=2d0.
But, according to Art. 33,
_Mz_ab_ Made

POE eB
(Mz/I=stress on element) (see Formulas B and D, Arts. 33 and 53),
where M = the bending moment of the block stuv and I = moment of
inertia of the cross-section of the block and E the modulus of elasticity
of the material composing it. Now, combining these two equations, we
have

Mdzr
db=— .
Substituting this value of d# in (2) we have
»  Mdz?
» ae

which is the expression for the second differential of the y ordinate to the
elastic curve at block stuv in terms of known quantities, but which at the
same time is really the expression for the vertical drop of the elastic
curve from a to 6 due wholly to the distortion of block stuv. Now, this
same expression will hold in the case of any of the blocks, as no special
case was taken. Then, evidently, by substituting the proper value of M
in the expression and integrating twice, the deflection of point A in
reference to B will be obtained. It is evident that the above discussion
will apply to any number of blocks as well as to four, so our expression
for the second differential of the y ordinates to the elastic curve will apply
to the entire beam or any part of it, as A and B can be any two points,
and as the bending in the case of any beam can be nothing different from
the bending of a cantilever beam here considered, the expression is
evidently applicable in the case of any beam whatever, loaded in any
manner, and is hence a general differential equation of the elastic curve
of any beam or any body whatever, subjected to bending stresses. For
convenience, the equation is usually written

d*4
EI pa Site aie Pale natant ante tates (K).

Referring to Fig. 79, it will be seen that the curve ss’s” is an
evolute, while the elastic curve oo is an involute. In case of a simple
beam there would be two branches to the evolute.

64. Deflection of Cantilever Beams.—

Case I. When the beam supports a single load at its free end.
76 STRUCTURAL ENGINEERING \

Let AB (Fig. 82) represent the beam of length L, P the load, and
let 2 and y be the co-ordinates to the elastic curve at any point when the
origin is taken at B, and z, and y, the co-ordi-
NS nates when the origin is taken at A.
RSSséF rst, taking B as the origin, we have M+P
| (L-«) for the bending moment at any point «#
A distance from B. Now substituting this value of
aes M in Formula (K) (Art. 63), we have

 
  

L
ae SS

 

   

2
Er} = PUL - 2) dese nehaoaaneeeatnclnoeee cat

Integrating once, we have

dy _ a
E152 =PLa-P = +0,

Now, dy,/dz, is the expression for the tangent of the angle that the elastic
curve makes with the horizontal or 2 axis at any point « distance from B.
It is readily seen that dy/dx = 0 at B, the origin, as the elastic curve is
horizontal at that point. But « = 0 also at that point. Then substituting
0 for dy/dx and for « in the above equation, we have C,=0. Now as
C, = 0, the preceding equation becomes

dy x?
a HP er gs See Re RV aw we Rw Es i petere easels 2
EI 5 =PL2-P> (2);
which is the general equation for the slope of the elastic curve at any
point between B and A with B as the origin. From this equation, the
slope of the beam (= slope of the elastic curve) at any point can be
computed by substituting for x its numerical value. For example, the
slope of the beam at a point b distance from B is
bPL-P
dy 2.
dx EI
which is the tangent of the angle that the tangent to the elastic curve at
that point makes with the a-axis.
Next integrating (2) we have

xz xz
EIy=PL— -P GE tee
Now at B both y and «=0. Then substituting 0 for 2 and also for y, we
have C,=0. Hence, we have

Plow 2
i> mle : +):
which is the algebraic equation of the elastic curve when the origin is at
B. From this equation the deflection of the beam at any point can be
computed by substituting for «x its numerical value. It is readily seen

that the deflection will be a maximum when « = L. Let A = the maximum
deflection; then we have
THEORETICAL TREATMENT OF BEAMS 12

yea B(E 1) _ PL

 

EI\2 6) 3ET
Now taking 4 as the origin, we have for the bending moment at any

point 2, distance from 4, M=Pz,. Then substituting this value of M in
Formula (K) (63), we have
d?y,

EI-—=Pzu,.
dz,”
Integrating once, we have
dG, a Be ay
EI i P ao Cc’.

Now, dy,/dz,=0 when #=L. Then substituting these values for dy,/dz,
and 2, respectively, in the preceding equation, we have C’=-PL?/2.
Then substituting this value of C’ in the equation, we have

 

 

for the general equation for the slope of the elastic curve, from which the
slope of the curve at any point 2, distance from A can be computed by
‘substituting for «, its numerical value.

Next, integrating (4), we have

Pe? PL?2,
6 2
Now, when z, = 0, y,= 0. Substituting 0 for x, and also for y, we have

c’=0. Then we have

P /a} Le,
va % 5 ) Dehra see Se crete ener tees (5)

for the equation of the elastic curve when the origin is at A, from which
the vertical ordinate y, at any point 2, distance from A can be computed
by substituting for z, its numerical value. Then, by subtracting this y,
ordinate from the maximum deflection A the deflection of the point in
question is obtained. It is readily seen from Fig. 82 that the y, ordinate
is equal to A when x,=L. So substituting in (5) we have

x P/L? ZL PL’

ye -a(§ 3)” BEF’
which is the same as found above except for the sign, which is due to the
ordinates being measured in the opposite direction.

Case II. When the beam supports two loads.

Let AB (Fig. 83) represent’ the beam, P
and P’ the loads, and let a be the distance that
load P is from B, b the distance between the
loads and d the distance that the load P’ is from
A, the end of the beam. Take B as the origin,
and let « and y represent the co-ordinates to
the elastic curve. When «x is less than a, we 4
have for the bending moment at a point 2 dis- Fig. 88

 

Ely,= +C”,

 

 

 

SX

   

 
78 STRUCTURAL ENGINEERING

tance from B
M=(a-«)P+(b+a-2)P’,

2
therefore, BITS =(a-2)P+ (b+a-2)P’.
Integrating, we have
dy _ x ‘ ‘ _ Pa?
EI = Pax-P> +P’b2+ P’ax 5
But dy/dz = 0 when « = 0; therefore, C’ = 0, and we have

+C’,

 

P’ 2x?

ag 2" + P'b2+ Plas 6
EI 7) =Par-P > + P’ba+ P’ax Be eta eede Keema (6).

Now integrating (6), we have
za? P2? P’ba? P’ax? P’x®
— — —-—— —— eee GC".
Se age ee 67
But y = 0 when « = 0; therefore, C” = 0, and we have

oe Pz? P’b2? Pon Pe)

 

 

 

t—F7\ a & ° 2 * 2 6

Now when z is greater than a and less than a + b, we have
M=P’(at+b-2)

for the bending moment at any point between P and P’, x distance from
B. Therefore,
d’y - - a
EI —,=P’a+P’b-P’e.
da?
Integrating, we have

P’ x?

2
Now it is readily seen that equations (6) and (8) are equal when

z=a in each. So, substituting a for z in each, equating and cancelling,
we have

 

ert! = P'an + P’bex -
dz

 

_Pas
. rs 2 ,
Substituting this value of C,in (8) we have
dy 4, ; P’x? Pa?
EIT =P ax+P’be- Qt rt tte eee e eee eee (9).

Integrating, we have

Plax? P’bx? P’a? Pa’e
ge oe
Now it is readily seen that equations (7) and (10) are likewise

equal when « =a in each. So substituting u for x in each, equating and
cancelling, we have

 

Ely=

Pa®
iC; =- e ’
THEORETICAL TREATMENT OF BEAMS 73

Substituting this value of C,, in (10), we have
_1(Ptaz® P’ba® P’x* Patz Pa’
I~ ET =

 

 

 

ot rg eRe oe

Now the slope of the elastic curve at any point between B and the
load P can be computed from equation (6) and at any point between the
loads from (9), while the deflection at any ‘point between B and the load
P can be computed from equation (7) and at any point between the loads
from (11) by substituting for x its numerical value in each case.

The maximum deflection will be at A. This can be determined in
the following way:

First compute the deflection of the beam at the load P’ from equa-
tion (11) (substituting (a+b) for «). Then compute the slope, that is,
the tangent of the slope angle at that point from (9) and multiply this
slope by the distance d, which will give the deflection of the point A in
. reference to the point at P’. Then add this deflection to the deflection of
the beam at P’, and we will have the total deflection at A.

Case III. When the beam supports a uniform
load.

Let AB (Fig. 84) represent the beam support-
ing a uniform load of w pounds per foot of length.
Take B as the origin, and let 2 and y represent the

 

 

co-ordinates to the elastic curve. Then the bending Fig. 84
moment at any point z distance from B is
M=w(L-2) 5° ) =a(L -2Lx +2"),

therefore,

Q

2
EI G4 => (L?-2Le+2*).
&

Integrating once, we have

dy _ w

When z = 0, dy/dz = 0; therefore, C’ = 0, and we have

3
(x0 ~ La? + =) C".

dy _© (roy Tar 4™ 12
erl=S(z a2—-Le ws cB: aie fen ietien tere dar ueriecah leap Nenerte hs a cae ae eUeE ee ( )
Integrating again, we have

wf{L?2? La xt
ep yet
Hy =( 2 3 +z)+

Now, when +=0, y=0; therefore, C’=0, and we have

w (L?x? La? xt
Se a lh mayo mawatiog Uaeeuea ensue 13).
y=ser( 2. 3 +B) ~

From equation (12) the slope at any point of the beam can be
computed, and from (13) the deflection at any point can be computed by
substituting for x its numerical value in each case. It is evident that the
maximum deflection will be at A, the free end. For that point z=L.
80 STRUCTURAL ENGINEERING

Then substituting L for x in equation (13), we have

for the maximum deflection.
6), Deflection of Simple Beams.—
Case 1. hen the beam supports a single load at mid-span.
‘ Let AB (Fig. 85) represent a simple beam of
L length L supporting a single load P at mid-span.
Let R and R’ represent the reactions at A and B,
4S Sn eB 3 : i :
» ', respectively, which are equal in this case. Take A
Fig. 85 as the origin, and let x and y represent the co-ordi-
nates to the elastie curve. For the bending moment
at any point to the left of the load, x distance from 4, we have

M=Re=* 2.

therefore, we have
2y P
EI —= 3 a,

x

QQ

Integrating once, we have
dy Px? ,,
EI dz = a +0.

Now dy/dx = 0 when « = L/2. Substituting this value for z, we have
eet
7 16
and substituting this value of C’ in the above equation, we have
dy Pa? PL?
ee Di pease ae ips cata lath Sasa as ach eal ovate Cio.
Integrating this equation, we have
Pst PLits
12 16
Now, y = 0 when 2 = 0; therefore, C’” = 0, and we have

P/# Lx
y alt = =) Sy Gres hues eal’ wie ae ASL esas aha ae ae Side Se ele eee (2);

which is the equation of the elastic curve to the left of the load. As the
elastic curve is symmetrical about the load, there is no need for deriving
the equation for the curve to the right of the load. However, this can be
readily accomplished by substituting in Formula (K) (Art. 63), the
expression for the bending moment to the right of the load, which is

P L
M=F2-P(« 5):

It is readily seen that the maximum deflection will occur under the —

EI

Ely = per,

load. So letting A represent the maximum deflection, and substituting - - -

£72 for x in equation (2), we have
THEORETICAL TREATMENT OF BEAMS 81

PL’

48ET

The slope of the elastic curve at any point between A and the load can be
computed from equation (1).

Case II. When the beam supports a single load at any point.
Let AB (Fig. 86) represent, a simple beam

 

 

 

 

supporting a load P at any point z distance x =
from A. Let R and R’ represent the reactions e iP

at A and B, respectively, due to this load P. ee ee
Take A as the origin, and let 2 and y represent S Hic. 86 fx

the co-ordinates to the elastic curve. For the
bending moment at any point to the left of the load, we have

M= Ra=* (Le- ax),

therefore, we have

d*y P
EIT) = (Lez - zx).
Integrating, we have
2
tones aoe aan (3)

dx an.
Integrating again, we have

_Pa® Pax _, o
Ely= oe BE +C’a+C”,
Now, y = 0 when a = 0; therefore, C” = 0, and we have
Px? Pzx®
oh ihs RMA SAGE HOS DERE aS 4).
ae ee “

      

Now for the bending moment at any point to the right of the load, that is,
when 2>s. we have

M’ =Re-P(2-2)=-"2* +Pz,

“ therefore, we have ,
d*y Psx

EI de? =Pz- “EE
Integrating, we have
dy Pzx*
—2 = SH Oy a gaa ele RaSs MEU R AS Bes 5).
EI Pzx art © (5)
Integrating again, we have
2 8
Hips ee © oe = 102) ©, suaneninimneeee (6).

2

It is readily seen that equations (3) and (5) are equal when 2=2
in each. Then substituting = for 2 in each equation, equating and
cancelling, we have
82 STRUCTURAL ENGINEERING

 

oe Ce esis ie area nS ct hes teal Lda 7)
Now substituting this value of C, in (6), we have
Pez? Pee® ., Para
Ely =—>- 2 “ay eae 2 +C, Awe DAN ae Be Se we ee (8)

It is readily seen that equations (4) and (8) are equal when «=z
in each case. Then substituting z for 2 in each, equating and cancelling,
we have

Now it is readily seen that y in equation (8) equals zero when
zw = L. Then substituting L for x and Pz*/6 for C, in that equation,
equating and reducing, we have

Substituting this value of C’ in equations (3) and (4), we have,
respectively,

ou Ps? Pex? Pe? Pel Pz®

 

EIT. e = ae ts oe er, Oa ia Site fs das sav ates oot (11),
Pe? Pze® Ps?x PsaL Pzia
Ely=— _ 6L . a ey 6 Baap aay eee eee tec (12).

AANVAAAAAN

Now, from equation (11) the slope of the elastic curve at any
point to the left of the load can be computed, while the deflection at any
point to the left of the load can be computed from (12).

From equations (7) and (10) we have

 

 

 

Cc a Pek Fe
ae BE
and from (9) we have
P3?
eg
Substituting these values in equations (5) and (6), we have, respectively,
dy Pzx? PzL Ps?
EIT = Pzxr - OL ~ 37 7 BE CY (13),

Now, from equation (13) the slope of the elastic curve at any point to
the right of the load can be computed, while the deflection at any point to
the right of the load can be computed from (14).

Equations (11) and (12) apply only to the part of the elastic curve
to the left of the load, while equations (13) and (14) apply only to the
‘YHEORETICAL TREATMENT OF BEAMS 83

part to the right of the load, and hence we can consider the elastic curve
to be composed of two separate curves which are tangent at the load.
Now, if the beam supported two loads some distance apart, the elastic
curve would be composed of three separate curves; and if it supported
three loads, the elastic curve would be composed of four separate curves;
and so on. That is to say, if there be n loads, there will be (n+1)
separate curves composing the elastic curve. The equations for each of
these curves could be derived as readily as the above equations, and then
from these equations the slopes and deflections of the elastic curve could
be computed. The main labor is the determining of the constants of
integration. It will be observed that, at the point of maximum deflection,
dy/dx = 0, as the tangent to the curve will be horizontal at that point.

Case III. When the beam supports a uni-
form load.

Let AB (Fig. 87) represent a simple beam,
length L, supporting a uniform load of w pounds 2
per foot of length. Let R and R1 represent the Fig. 87
reactions at A and B, respectively, and let 2 and
y represent the co-ordinates of the elastic curve, A being taken as the
origin. Then the bending moment at any point « distance from 4 is

 

 

Ta ae
therefore,
: d*y wlhrx we"
nag eg

Integrating once, we have

Now, dy/dz = 0 when x = L/2, as the tangent to the elastic curve is
horizontal at that point. Now substituting this value of 2, we have
wL'

Dies A
C’= 24”

and substituting this value of C’ in the above equation, we have
dy wLhx wa? wl?

Integrating this equation, we have
wLe® wat whiz

12 24 24
But y = 0 when 2 = 0; therefore, C’’ = 0, and we have

 

 

‘EIy= +C%,
84 STRUCTURAL ENGINEERING

YEI\ 12 244
from which the deflection at any point can be computed.

It is evident that the deflection will be a maximum at midspan; that
is, when # = L/2. So letting A represent the maximum deflection, and
substituting L/2 for # in (2), we have

5wL*
~ 384ET

66. Proposition.—The bending moment at any point in a beam is
equal to the bending moment at any other point plus the shear at the
other point multiplied by the distance between the points, plus the
algebraic sum of the moments of the forces between the points about the
point in question. ;

Let AB (Fig. 88) represent a simple beam supporting the loads
P, P1, and P2 as shown. Now, according to the above proposition, the

ps 4 3
: (a5 ewe =) aie dtee siete eee eee (2),

y=A=

 

 

 

 

 

1 € Fg
ay i le c
Z ;
é iP R
Fig. 88 Fig. 89

bending moment at D is equal to the bending moment at E, plus the shear
at E multiplied by c, plus the sum of the moments of the forces between
the points about D. This is readily seen to be true, for, taking moments
about E, and letting R represent the reaction at A due to the three loads,
we have

M,=R(a+b)-Pb

for the bending moment at that point, and taking moments about D, we
have

M,=R(a+b+c)—-P(c+b) —-dP1-hP2
=R(a+b) -Pb+0e(R-P) ~dP1- hP2,

which is seen to agree with the above proposition, as the part R(a+b) -
Pb is the bending moment at E, R-P is the shear at E, which is multi-
plied by c, the distance between E and D, and -dP1—hP2 is the sum of
the moments of the forces between E and D, about D. The minus signs
in the last are simply the signs of the moments.

The above proposition is readily proven in the case of any beam
whatever, but regardless of its simplicity, it is quite useful.

67. Reactions, Shears, and Bending Moments on Overhanging
Beams.—Let ABC (Fig. 89) represent an overhanging beam supported
at B and C and which in turn supports the loads P and P’, as shown.
The part AB is known as the cantilever arm, while the part BC is known
as the anchor arm. Let & represent the reaction at B due to the two
loads, P and P’, and let R’ represent the reaction at C due to the same.
Now, taking moments about C, we have

RL-aP -bP’=0.
THEORETICAL TREATMENT OF BEAMS 85

from which we obtain

Then taking moments about B, we have
R’L+eP-dP’=0,

from which we obtain
+ R’=

If eP be greater than dP’ it is evident that R’ will act downward
upon the beam, and hence the beam would pull upward upon the support
at C, which would require that the beam be anchored in some manner to
the support. In all such cases the reaction is spoken of as being negative.
If dP’ be greater than eP, of course R’ will be positive, the same as for
simple beams.

In case of a greater number of loads, the reactions are determined
in the same manner as shown above for the two loads. We simply include
the moment of each load in the equation of moments. In the case of a
uniform load the reactions are determined practically in the same manner,
except that we use average lever arms for the uniform load. For example,
suppose the beam ABC (Fig. 89) supports a uniform load of w pounds
per foot of length, extending from 4 to C. Let R, represent the reaction
at B due to this uniform load, and let R,, represent the reaction at C due
to the same. Then taking moments about C, we have

_w(L,+L)*
ie 2L
and taking moments about B, we have

l/oL? wl?
2 a et Se ES hw
R, iS s|

2

In case the beam overhangs two supports the reactions are deter-
mined as above by taking moments about the supports and simply
including the moment of each force acting upon
the beam in the equation of moments. As an ob d _@
example, let ABCD (Fig. 90) represent a beam I If

 

 

 

 

 

Li p
overhanging two supports. Let P, P’, and P” Bis jP' ic D
represent three loads supported by the beam as | «4, fr > 1 Rei |
shown, and let R represent the reaction at B Fig. 90

due to these three loads, and let R’ represent
the reaction at C due to the same. Now, taking moments about B, we have

aP—bP’ + R’L-(e+L)P”=0,
from which we obtain

_ bP’ + (e+ L)P”—-aP

+ R’ 7
86 STRUCTURAL ENGINEERING

Then taking moments about C, we have
—(L+a)P + RL-dP’ +eP”=0,
from which we obtain
i poor.

Now suppose the beam represented in Fig. 90 supports a uniform
load of w pounds per foot of length extending from 4 to D, and let R,
represent the reaction at B due to this uniform load and let R,, represent
the reaction at C due to the same. Then taking moments about B, we
have

_w(L+L,,)? wh?

Rus QL ~ 2b?

and taking moments about C, we have
w(L+L,)? wh,”
2L 2h

The determining of the shear at any section of an overhanging
beam is just the same as for any other beam; that is, the shear at any
section is equal to the algebraic summation of the forces on either side of
the section summed up from the end of the beam in each case. For
example, the shear at section ss of the beam ABCD (Fig. 90) due to the
three loads P, P’, and P”, is

S=P+RorP”+R’'+P’.

 

R,=

The determining of the bending moment at any section of an over-
hanging beam is just the same as for any other beam; that is, the bending
moment at any section is equal to the algebraic summation of the moments
of the forces on either side of the section about the section. For example,
the bending moment about the section ss of the beam ABCD (Fig. 90)
due to the three loads P, P’, and P” is

M=(a+g)P + Rg or (L-gt+e)P” + (L-g)R’ +hP’.
All of the above holds as well for cantilever arms as for anchor arms,

but it is more convenient to treat the cantilever arms as independent
cantilever beams.

68. Reactions, Shears, and Bending Moments on Beams Fixed
at One End and Supported at the Other.—Let AB (Fig. 91) repre-
sent such a beam which has the simple support at A and the fixed support
at B. As the beam is fixed at B it is evident that any
load on the beam will produce tension in the elements
above the neutral axis and compression below the
neutral axis at that point. But it is equally evident
that the reverse is true just to the right of the simple
Fig. 91 support at A ; that is, the top elements of the beam just

to the right of A will be in compression and the bot-
tom ones in tension. Then, evidently, there will be some section be-
tween A and B where the bending reverses; that is, a section where
there is no bending, and hence the bending moment at that point of the

    

 

A= Yy
y
THEORETICAL TREATMENT OF BEAMS 87

beam will be equal to zero. Let S be such a point. Then the part SB of
the beam, to the right of S, would be simply a cantilever beam, while the
part SA of the beam, to the left of S, would be a simple beam, whence the
bending or curving of the beam due to loads would be as indicated in
Fig. 91. .The point S, where the bending moment equals zero, would be
known as the point of “‘contra-flexure,” also as the “point of inflection.”

As a case of concentratedeloads, let R (Fig. 91) represent the
reaction of A due to a single load P at any distance kL from A, k being
any fraction less than unity. Owing ‘to the unknown forces acting upon
the beam to the right of B, this reaction R, at A, cannot be determined in
the usual way by taking moments about B. Take 4 as the origin, and let
xz and y represent the co-ordinates to the elastic curve. Then, for the

bending moment at any point do the left of the load, that is, when z is
less than kL, we have

M=Rz,
therefore,
d?y _
EI Fe Re
Integrating once, we have
dy Rz*
zis = — BAO aioe a lab aS cree tah aaa (1).

Integrating again, we have
Rz* _, i
Ely= “er C’x+C”,

But, y = 0 when « = 0; therefore, C’”’ = 0, and we have

Now for the bending moment at any point to the right of the load,
that is, where 2 is greater than kL, we have

M,=R2z-P2r+PkL,

 

therefore,
d’y
EI 5 =Ra-Pa2+PkL.
dz
Integrating once, we have
dy Rz? Px
Ee et ay

But, dy/dz = 0 when x = L. Then substituting L for 2, and equating,
we have

 

RL? PL? .
- —— = =).
ye PkL -
Substituting this value of C, in the last equation, we have
dy Rx? Px? RL? PL?

 

of — ——. — ——- aya ah 2
BIGh ==> + Pha - "> 4 - PREP, (3).
88 STRUCTURAL ENGINEERING

Then integrating (3), we have

Re Pa? PkLa? RL _ Pla
[ee ee = ae 2
But here y = 0 when z = L. Then substituting L for 2, equating and
cancelling, we have

-PkL?2+C,.

 

 

Ely=

ef
oe ae

Substituting this value of C, in the last equation, we have

 

 

 

 

3 Px? PkLe? RL22 PL?
Ely="2 Fe ee kts
RL? PL? PkL*
eg gg Seen RRS BTR ESS eR ERM RE RHE Ys (4).

It is readily seen that equations (1) and (3) are equal when 2=kL
in each. Then substituting kL for x in each equation, equating and
reducing, we have

 

LPL? RL* | PL?
<2 ee
Then substituting this value of C’ in (2) we have
Re? PkL?e RL?x PL ec
= 4+ = PREG occas esas :
67 2 or) 2 : ()
It is readily seen, also, that equations (4) and (5) are equal when

«= kL in each. Then substituting kL for x in each of these equations,
equating and reducing, we have

Cc’ —PkL?’.

 

Ely

= (H-3k+2) annals Santee caeaipaiaele (6),

from which the reaction at A due to a load at any point on the beam can
be computed. For example, suppose the load P is at mid-span. Then
k=4. Substituting this $ in equation (6) we have

P/1 3 5
R=F(5-$+2) = 79 P:

Again, suppose the load is at the quarter point nearest A. Then
k =4. Substituting this } for & in (6) we have

P/1 38 81
R-F(G-E+?) = [28 P.

If there be more than one load on the beam, the reaction at A due to
each load would be computed separately and all added together and we
would thus obtain the reaction at A due to all the loads. Of course the k
for each would be different from the others; that is, no two k’s would be
the same.

In case the beam supports a uniform load of w pounds per foot of
length extending over the full length of the beam, that is, from 4 to B,
THEORETICAL TREATMENT OF BEAMS 89

we have
wx"
M=Rz-——
orp

for the moment at any point 2 distance from A, where R represents the
reaction at A due to this uniform load. Therefore, we have

d?y a
EI dx? = Rex - 3
Integrating once, we have
dy Ra? wz? _,
fn 3 a

But dy/dz = 0 when « = L. So substituting L for x in the preceding
equation, we have

RL? wl
c—-_ ———
Se ae

dy Raz? we? RL? | wl!

 

 

Integrating this equation, we have
Re? wet RL?x wlir
aS Rh eg. te
But here y = 0 when 2 = 0; therefore, C’ = 0, and we have

Re? wet RL?x whiz
alate yeltdue oe Grants avsnetuessiec tate (8).

ee Gag ge .
Now y = 0 in (8) when 2 = L. Then substituting L for 2 (8), we have
RL? wl* RL‘ wl

t cv

 

 

from which we obtain
R =3 ol,

That is, when a beam supported at one end and fixed at the other
supports a uniform load, the reaction at the supported end is three-
eighths of the total load.

After the reaction at the supported end due to the loads supported
is determined, the shear and bending moment at any point in the beam
are obtained just the same as in the case of a simple beam, except we deal
only with the supported end. The shear at any point is equal to the
reaction at the supported end minus all intervening loads, while the
bending moment at any point is equal to the reaction at the supported
end multiplied by its lever arm, minus the moment of all intervening
loads about the point. For example, the shear at any point x distance
from the supported end due to a uniform load of w pounds per foot of
span is

S =< wl —w2,
90 STRUCTURAL ENGINEERING

while the bending moment is
2

 

3 Wr
M=—wLe - 3
If « = L, we have
M=- pw, a

which is the bending moment at the support B or the fixed end.

Deflections can be computed from equations (4), (5), and (8).

69. Shears and Bending Moments on Fixed Beams.—Let AB
(Fig. 92) represent a fixed beam. The supports being fixed, any load on
the beam, as is seen, will produce tension in the
top elements of the beam and compression in the
bottom elements of the beam at the supports,

YZ while the reverse will be the case out some dis-
BY tance from the supports. Then, evidently, there
Fig. 92 will be two points of contra-flexure and the
bending of the beam due to loads supported will
be as indicated in Fig. 92, where S’ and S are the points of contra-flexure.
It is evident that the parts 4S’ and SB are cantilevers, while the part
S’S is a simple beam. In the case of fixed beams, the point of applica-
tion of the reactions cannot be definitely fixed, so we deal with the end
shears instead of the reactions.

As a case of concentrated loads \et P represent a single load upon
the beam at any distance kL from A. Now, when we proceed to deter-
mine the shears and bending moments on the beam due to this load P, we
quickly realize that we have but little to start with, and it is only through
the application of the general differential equation of the elastic curve
(K) that we are able to obtain either. Let /’’ and V represent the end
shears at A and B, respectively, due to the load P, and let M’ and M”
represent the bending moments at A and B, respectively. Take A as the
origin and let « and y represent the co-ordinates to the elastic curve.
Then, according to Art. 66, the bending moment at any point to the left
of the load any distance « from A is

 

 

 

M=M’+V's.
Therefore, we have
Ered ou 4y"
dx?” Ks

and integrating once, we have

+C’,

 

dy_y,,. Vx?

But dy/dz = 0 when x = 0; therefore, C’ = 0, and we have

 

 

dy ,. Vix?
Ely =M’s a Day ear terre tay nto ae save paren eee cea Ske Ose (1).
Integrating this equation, we have
Pm Ind
Hijet ES sox

 

2 6
THEORETICAL TREATMENT OF BEAMS 91

But here y = 0 when a = 0; therefore, C’” = 0, and we have

Mx? 123

Ely=— ee (2).

 

 

+

Now the bending moment at any point to the right of the load »
distance from 4 is

M,=M’'+V’a—- Px+ PRL.

 

Therefore,
d*y
EI 3 = M’+V'a— Part PkL,
and integrating this equation, we have a
pret ae, CF ppiecc
Bee ag ete

Now, dy/dx = 0 when « = L. So substituting L for x and equating, we
have

 

 

 

 

 

 

77 2 2
sc! Po ae: :
2 2
Then substituting this value of C, in the last equation, we have
dy V'x? Px? VL? PL?
— = 4 Set PkL2z - M’L- _ Bs .
oF M’a + 2 2 x os PkL?. .(3)
Integrating (3), we have
Id Td P 3 PkL 2 'T? PL?
Ely=" +o A SS Whe -" fa: <* _ PkL?a+Cy.

But y = 0 when « = L. So, substituting L for 2 in the last equation,
equating and reducing, we have
M’L* i V'L> PL* i. PKL?
2 3 3 2
Then substituting this value of C, in the last equation, we have

 

 

 

=C,.

 

 

 

 

 

 

Pd TP 8 3 2 17 2 2
Ely= “3 oe -75 ee -M'Le- VS? PEs
a M’I? V’Ls PL? PkL*
PRES aE =e GSR ee © aoe (4).

Now it is evident that equations (1) and (3) are equal when c=kL
in each. Then by substituting kL for « in each, equating and reducing,
we have

PRL V’'L PL
= Se PRE isis ee atk nee Se ieee teed ‘

a a (9)

Then by substituting this value of M’ in each of the equations (2)
and (4) and then substituting kL for z in each, we have the two equal.
Then by equating these equations each to each and reducing, we have

PRP =i ene er geared ae lst (6).

M’

 
92 STRUCTURAL ENGINEERING

By substituting this value of V’’ in (5), and reducing, we have
ME SPGQ eo TA) asic asad atesaacn ag Rag waesiareew eae RS (7).

Then, having V’ and M’ determined from (6) and (7) for the point
A, the bending moment at any other point in the beam, due to the load P,
can be determined according to Art. 66, and the shear at any point can
be determined, as it is equal to the summation of the forces summed up
from A beginning with V’.

If there be more than one load on the beam, the shear and bending
moment at A due to each can be determined separately from (6) and (7).
Then adding these shears together and these bending moments together,
we obtain the shear and bending moment at A due to all of the loads.
Then the bending moment and shear at any point of the beam, due to all
of the loads, can be determined. That is, if the shear and bending
moment at one support of a fixed beam are known, the shear and bending
moment at any other point can be determined, and hence. the equations
(6) and (7%) are all that are absolutely necessary. But from these
equations we can readily derive equations for the shear and bending.
moment at the other support, if such be desired. For example, the shear
at B, Fig. 92, is equal to the load P, producing the shear, minus the
shear at A. Then subtracting the value of V’, given in (6), from P; we
have

V=P-P(1+2k' - 3k’),

and reducing, we have

PSP Cah? BRP ick sas waiatocs vy anlar wateuy veya ana wes Means (8).

The bending moment at B (Fig. 92) is
M”=M’+P’L-P(L-kL).

Then by substituting the value of J’ and M’ given in (6) and (7),
respectively, we have

M” = PL (2k? —k°—k) + PL(1+2k* - 3k?) -P(L—kL),
and reducing, we have

NESTE SI leases peup renamed ened baeiee sas baen (9).

As an example of application, suppose the load P (Fig. 92) to be at
mid-span. Then k = 1/2. Now substituting this 1/2 for k in (6) and
(7), we have : ,

; i 8\ Pp
pr=P (14h -2) a2,
pope AN PE

and m’=PL(}-4-3)=-7%,

respectively, which is the shear and bending moment at 4, For the
bending moment under the load we have
PL PB ) _PL

M=M’ ’kEL=— =
+V/hL 8 a) 5 8

 
THEORETICAL TREATMENT OF BEAMS 93

/

In case a uniform load extends over the entire length of a fixed beam,
the end shears are each equal to one-half of the total load on the beam.
Suppose the beam shown in Fig. 92 supports a uniform load of w pounds
per foot of length extending over the entire span. Let V’ and V’” repre-
sent the end shears at A and B, respectively, due to this uniform load,
and let M’ and M” represent the bending moments at A and B, respect-
ively, due to the same.

Taking A as the origin as before, we have for the bending moment
at any point « distance from 4
whe wat

a
M=M'4V'2—-o = M+

o

 

2 2 2
therefore,
d*y , whe wa"
Ble eg eS

Integrating this equation, we have

 

 

dy ‘5 whe wa*
Ag ee fe ge
But dy/dz = 0 when a = 0; therefore, C, = 0, and we have
dy ,. wLhr? wa
EIT =M @& + 4 = Go PEE TEE DE Teas (10).

Integrating again, we have

M’2? wz? wat
ay ae ae

But y = 0 when 2 = 0; therefore, C, = 0, and we have
M’e? whe wet

3 ego gp thts heehee Biase eww eens (11).

+C,.

Ely=

 

 

But y = 0 also when « = L. Then substituting L for x in (11), and
reducing, we have

wL?

a aed See: ale nseedoi sue aa e a seeanet ee wee eae ord aes

3 (12),
which is the bending moment ‘at 4. Then knowing the bending moment
and shear at A, the bending moment and shear at any other point in the
beam can be determined. For example, the bending moment at the
center is
PL wl? wel wh? woh 1

- 8° i kB ee

To determine the points of contra-flexure, we simply write the
equation for the bending moment at any point and equate it to zero. For
example, the bending moment at any point « distance from A, when the
beam supports a uniform load of w pounds per foot (Fig. 92), is

 

Mc=M’+

wa? wl? wher wa? *
RE Ta es gg er ag

 

 
94 STRUCTURAL ENGINEERING

Then by equating this to zero, we have

wh? whe wa? _

 

ie ge
and solving for 2, we have ‘
e=< xD reid or 0.79 L (about).

That is, one point of contra-flexure is 0.21 of L from A—considered
in practice as being } of L—and the other 0.79 of L from A—considered
in practice as being 3 of L. The points of contra-flexure can be deter-
mined in a similar manner in case of any other kind of loading.

Deflections, if desired, can be computed from equations (2), (4),
and (11). ;

70. Bending Moments, Shears, and Reactions on Continuous
Beams.—Let ABC (Fig. 93) represent two consecutive spans of a con-
tinuous beam of n spans. The bending or curving of the beam in the two

nu Zr

 

 

 

 

 

 

 

Fig. 93

spans is shown to be similar to that of two fixed beams, each span being
considered a beam. If the spans were equal in length and symmetrically
loaded, this would practically be the case, but otherwise it would not be,
for it is readily seen that the loads in one span would tend to produce
reverse bending in the adjacent span, which would curve it up instead
of down. For example, the loads in span AB could be such that the span
BC would be curved upward instead of downward, as it is shown. So it is
evident that the tangent to the elastic curve at any support is horizontal
only when the spans are equal in length and rigidity, and symmetrically
loaded. Hence the slope of the elastic curve at the supports cannot, in
general, be utilized in preliminary investigations, as in the case of fixed
beams.

In the following treatment it is assumed that the beams have a
uniform cross-section and are homogeneous throughout, and are supported
upon simple supports of equal elevation. This is an assumption usually
made in practice. The equation employed in analyzing continuous beams
is known as the “Three-Moment Equation,” which we shall now proceed
to derive.

Case I. When the beam supports concentrated loads.

Referring to Fig. 93 let L be the length of the span 4B and L, the
length of the span BC. Let P represent a load at any point kL distance
from A in span AB and let P’ represent a load at any point k,L, distance
from B in span BC. Let M’, M”, and M’” represent the bending
THEORETICAL TREATMENT OF BEAMS 95

moments at the supports 4, B, and C, respectively, due to the loads P
and P’, and let V and V’ represent the shears just to the right of the
supports A and B, respectively, due to these same loads.

Considering the span AB, and taking A as the origin, the bending
moment (according to Art. 66) at any point to the left of the load P, z
distance from 4, is

M=M’+Vz,
therefore,

d?
EIS 4 =M’+Vz.

Integrating once, we have

 

dy _,,,. Vax? .
Eig =M’e+s + Cem ce eer ere ces sees ee nesanerenees (1).
Integrating again, we have
Pat 3
Ely= le 22 y0tes oF,
2 6
But y = 0 when « = 0; therefore, C” = 0, and we have
Pine 3
Ely="* re C's caucasian aaa eeaIS (2);

Now, the bending moment at any point to the right of the load P,
a distance from A, is

M=M’+Va-P(a-kL),

therefore, °
d*y
EI 73 =M’+Va-P2+PkL.
Integrating once, we have
GY = spre a = EE oe
EI = M'n + Be t+ PRLG AC ccc ccccceevenvnes (3).

Integrating again, we have
M’s? Ve® Pa? PhL2?

a

ge a: Me On cea sacs w(4).

It is readily seen that equations (1) and (3) are equal when «=kL
in each. Then substituting kL for « in each and equating, we have

 

 

 

Ely=

 

 

 

272 272 kL?
me + 2 = $C =MRL +E a ~ + PkL?+C,,
from which we obtain
272 i
Ces a ORR Reto HS OES RATE RES (5)

Now substituting this value of C’ in (2), we have
M's? V2? ‘ PRL? 2 se
ool 2

 

 
96 STRUCTURAL ENGINEERING

Now in equation (4) y=0 when x=L. Then substituting LZ for ez
in that equation, we have
M’L? VL PL? PkL®
= 4 4S I
la 5 te 6% 3 +C,L+C,,,

from which we obtain

C= M’L? VL? PL? PLE

Pr Cae wie ips ae
Substituting this value of C,, in equation (4), we have
M's? Va? P2® PkLz?

a ge Ge
M’L? VIF PL? PRL

Ee ee
Equation (6) applies to the part of the elastic curve to the left of the
load P, while (8) applies to the part of the curve to the right. Then these

two equations are equal when x = kL in each. So substituting kL for 2 in
each, equating and reducing, we have

PkeL? M’L VL? PL? PkL?

 

Secon Meteo: (7).

Ely= +C x

 

 

 

 

 

 

 

 

 

 

 

oe ce ae 6 + 6 gg Tees eee e eee. (9).
Then substituting this value of C, in (5), we have
, PRL? PRL? M’L VL? PL? PkL
C’= 9 2G Sg ge ee See (10),
and substituting the value of C, given by (9) in (7) and reducing, we have
PRL
C,,= GET T Tne eee eee es (11).

So we have thus determined all of the constants of integration so far
involved.
Now substituting in (1) the value of C’ given in (10), we have

 

 

 

 

 

 

 

dy_,,, Va? Pk?L? Pk?L? M’L
ag a a a
VL? PL? PkL?
GT GT TTT ttt eee (12),
and substituting the same value of C’ in (2), we have
Ely= M’x? i va’ Pk°L?2 PkL?x M’Le
COS gee Wagner, — = age oe
VIPx PL?2 PkLiz
GT tet e eee en ees (18).

Next substituting in (3) and (4) the value of C, and C,,, respectively,
given in (9) and (11), we have
THEORETICAL TREATMENT OF BEAMS 97

 

 

 

 

 

 

 

 

2 2 37,2
pr! yey PE py Ee
dz 2 2 6
M’L VI? PL? Pk?
gy gee TTT eet e ee eee eee eeeeaes (14)
and
pra Mis? Va Pst PkEot  PkeLte
Ca ee ag ee
= M’Le o VL?2 PL?x PkL?x PL? 15
2 Zot ge Rg (15).

Equations (12) and (13) apply to the part of the elastic curve to the
left of the load P, while equations (14) and (15) apply to the part of the
elastic curve to the right of the load P. Now it is evident that by pro-
ceeding in-the same manner with span BC, four equations corresponding
to (12), (18), (14), and (15) could be derived for the elastic curve in
that span. But it is readily seen that the equations would differ from the
above equations only in the marking of the M’s, V’s, P’s, k’s, and L’s. So
for span BC, we can write

 

 

 

 

 

 

 

 

 

 

G0 og PR PEGS PEL?
EI = Mn+ a + 3 eee gem
M’L, V’'L? P’L,? P’k,L?
Sage te og ARNE (16).
EL M's? P's? P'k?AL ia PRADA 2
of) eee ¢
M’L,« V’L?2 P’LZ2 P’k,L, 2
pe eg Eg eeeetbterean (17)
for the part of the elastic curve to the left of the load P’, and
FD yl 4 2 r 8 2
pp ewig! per ye Pele
da 2 6
M’L, V’L, P’L,? P’'k,L?
ye og Qt testes eeeress (18),
EI _M"2? ie Via? Plo? Ph Lo? Pek sb ML
Ce : ¢” 8 6 2
V'L{2 P'LZ2 Ph Lex P’k SL?
- : i _ ee ee ee 19
6 6 z° 8 (19);

for the part of the elastic curve to the right of the load P’.
Now taking moments about B, according to Art. 66, we have

M”’=M’+VL-P(L-KkL),

from which we obtain
98 STRUCTURAL ENGINEERING

Then substituting this value of V in (14) and reducing, we have

 

 

 

dy», Mx? M'x? Pha? PEL?
EIS! = M’a+— 7 - py -- y+ Pha --G
M’L M’L M’'L PkL?
ee ar eared ed am 21).
po. Ge ee 1)

Likewise, taking moments about C, we have
M’"=M"+V’L,-P’(L,-4,L,),
from which we obtain
M’”_—M”
= =
Then substituting this value of V’ in (16), we have

v’ ORG dela Sta a enn (22),

 

 

 

dy «5, 0" 2" Mee Pat | P’k,2? = P’k?L?
AD 8? a, oe, 2. 1 2 .
P’k3L,2 P’k,L M"L, M’’L, ML, P’k,L? * 93
= fu ey = a8 = a ee ( ).
6 6 2 6 6 2

Equation (21) applies to the part of the elastic curve to the right of
the load P in span AB, and equation (23) applies to the part of the
elastic curve to the left of the load P’ in span BC. Now it is readily seen
that these two equations are equal when z = L in (21) and 0 in (28), as
the two curves have a common tangent at B. Then substituting L for 2
in (21), and 0 for w in (23), equating and reducing, we have

M’L M’L PRL? PRL? M’L, M’'"L,

 

 

 

ca a mel 6
Pk2L? P’k3L2 Pk,L,?
—a Oe ee

from which we obtain
M’L+2M"(L4+L,)4+M’'” L,=-PL*(k—k*) —P’L? (2k,- 3k? +k) .(L),

which is the three-moment equation when the loads are concentrated loads.
The more general equation, as usually written, is

M’L+2M” (L+L,)+M’L,= —3PL?(k-k®) —3 P’L,(2k,-3k2 +k).

Case II. When the beam supports uniform loads.

Let w be the uniform load per foot on span AB (Fig. 93) and let w’
be the uniform load per foot in span BC. Let M’, M”, and M’” now
represent the bending moments at the supports 4, B, and C, respectively, .
due to the above uniform loads, and let V and V’ now represerm th-
shears just, to the right of supports A and B, respectively, due to these
same loads. Considering span AB, and taking A as the origin, the bend-
ing moment at any point « distance from 4 is

2

M=M’+Vo-"—.
THEORETICAL TREATMENT OF BEAMS 99

 

 

therefore,

d*y ‘ wa

EIT a=M’+Va-=—.
Integrating once, we have
dy _,,, Va? wa?
i eS A lar aS esac mlietenet we (olele set ceecceeeee ee (4),

Integrating again, we have

fd 3 4

EI a ee ae bi Car O,

But y = 0 when x = 0; therefore, C’ = 0, and we have
M’2? Va? wet

a ge ee
But also y = 0 when # = L. Then substituting L for « in the last
equation, we have

Ely= + Ce.

 

ML? VL? wl
: ° € 8a

 

 

1 LC=0,

from which we obtain
ou WL _VEE wl?
2 6 2+

Substituting this value of C in equation (24), we have

d

EIS 2 6 2 6 * 94

Similarly, if the origin be taken at the support B, for span BC, we have
for the bending moment at any point x distance from B

 

 

M=M"+V'e- =,
therefore,
EI 4 =M"+V'2- “2
Integrating once, we have
sep png Oa eisinmomenneeebens (26).

Integrating again, we have
Ely =—>— + ——- 37 + C #+C,,.
But 2 = 0 when y = 0; therefore, C,, = 0, and we have

PF 42 Vy’ 8 ? mt
Ely="5* 4 7 Sr +Cn.

 

But y = 0 when « = L,. Then substituting L, for 2 in the last
equation, we have
100 STRUCTURAL ENGINEERING

M’L2 V’L! wL
> ' § B42

 

—— +C,L,=0,

from which we obtain
M’"L, V’'L? 2®’L}
2” @ * og
Now substituting this value of C, in equation (26), we have
Vix? w’t? M’L, V'L? w’L?

 

 

 

C,=-

dy _ ayn
EIT =M Be a é aac
It is readily seen that equations (25) and (27) would be equal
when x« = L in (25) and zx = 0 in (27), the two elastic curves having a
common tangent at B. Then substituting LZ for x in (25) and 0 for x in
(27), equating and reducing, we have

12M’L + 8VL? - 3wL’ = -12M"L, - 4V’L7? + ’L,’..... (28).

Now (referring to span 4B) taking moments about the support B,
we have, according to Art. 66,

 

 

 

M” 2M’ .7L- 2.

Transposing and dividing by L, we have
M”—-M’ wL

 

(a ee eS
V= L Po eee eee e es (29).
Next (referring to span BC) taking moments about C, we have
w’L,?
M’”=M"+V’L,- 5 “

Transposing and dividing by L,, we have

M’"—™M” w’L,

Ee 4 9 ‘ eee rene
Now substituting the values of /’ and V’ as given by (29) and (30),

respectively, in equation (28), and reducing, we have

3 , 3
M’L + 2M"(L+L,)+M’"L, = _" a oe ete (M),

y=

 

 

 

which is the three-moment equation when the loads are uniformly
distributed.

The three-moment equation, as is readily seen, is an expression for
the bending moments at any three consecutive supports in terms of the

Mm” ? FP mt
a b c a 2 A

Se a ee fi
Fig. 94 Fie. 95

lengths of the two intervening spans and the loads supported in these
spans. The lengths and loads, of course, are known quantities. Now
for any continuous beam of n spans, there are always (x + 1) supports,

  

 

 
THEORETICAL TREATMENT OF BEAMS 101

and for cach three consecutive supports a threc-moment equation can be
written. Then, (x — 1) three-moment equations can be written for any
continuous beam. For cxample, let abcdef (Fig. 94) represent a con-
tinuous beam of five spans. Here we can write four (which is (n — 1))
three-moment equations: one for supports a, b, and c; one for supports
b, c, and d; one for supports c, d, and e; and one for supports d, e, and f.
Now the bending moment at each of the supports would be included in
these four three-moment equations, being (n + 1) moments in all. But,
as the bending moments at both a and f are equal to zero, the four
unknown moments at b, c, d, and e can be determined, as we have four
equations and four unknown moments. Then, after these unknown
moments at the supports are determined, the reactions at each support
can be determined and then the shear and the bending moment at any
point of the beam can be determined.

As an example, let ABC (Fig. 95) represent a continuous beam
having three supports which would be known as a beam continuous over
three supports. Let M’, M”’, and M’” represent the bending moments
at the supports 4, B, and C, respectively, due to any loads that we choose
to consider, and let R1, R2, and R3 represent the reactions at the sup-
ports A, B, and C, respectively, due also to any loads that we choose to
consider. The intensity of these moments and reactions would, of course,
vary with the loads.

Let us first consider two concentrated loads P and P’ on the beam as
shown in Fig. 95. Now, as there are but three supports, only one three-
moment equation can be written. So writing this one, which is really
Formula (L), given on page 98, we have

M’L + 2M"(L+2L,) + M’" L,=-PL?(k - k®) - P’L,?(2k, - 3k? +k).
But in this case M’ = 0 and M’” = 0. So the equation becomes
2M" (L + L,) =—-PL?(k — k°) — P’L? (2k, - 3k + 1,3),

from which we obtain

 

M’ = MLtLy (31).

Now considering points A and B, and taking moments about B,
according to Art. 66, we have

M” =R1L-P(L-kL).

Transposing and reducing, we have

”

M
R1l= LZ

 

+P(1-k).

Now, by substituting the value of M”, as given in equation (31), in
this last equation, the value of the reaction R1 at A is obtained. Then,
after R1 is known, the bending moment at any point in the span AB is
obtained as readily as in the case of a simple beam, as it is equal to the
algebraic sum of the moments about the point considered of the forces to
the left of the point, which are now all known.
102 STRUCTURAL ENGINEERING

Now, similarly, considering points B and C, and taking moments
about B, we have

M” =R3L,—P’k,L,.
Transposing and reducing, we have
M”
L

Then by substituting the value of M”, as given in equation (31), in this
last equation, the value of the reaction R3 at C is obtained. Then after
R3 is known, the bending moment at any point in the span BC is readily
obtained, as it is equal to the algebraic sum of the moments about the
point considered of the forces to the right of it, which are now all known.
The shear at any point in either span is obtained by simply adding up
(algebraically in each case) the forces between the end and the point
considered. R2, as is readily seen, is equal to the sum of the loads minus
the two reactions R1 and R3. It is also equal to the shear just to the
right of B plus the shear just to the left of B. By taking moments about
both 4A and B, equations can be derived from which these shears can be
obtained, and consequently R2.

Now it is evident that the bending moments, shears, and reactions on
the above beam due to any number of such loads could be determined by
considering the loads in pairs, as in the above case. But it is more con-
venient to derive equations expressing the values of these for one single
load at any point. So let P (Fig. 95) be the only load on the beam.
Then P’ and k, will not occur in the three-moment equation, that is, they
will be equal to zero, and hence the three-moment equation becomes

2M” (L+L,) =-PL*(k-k*),

 

R3= +P’k,.

¥

from which we obtain
PL? (k—k?®)

SSS SEE) cd (33).

Now taking moments about B, considering points A and B, we have
M”=LR1-P(L-KkL).

Then substituting this value of M’”, as given in (33), in this last equation,
and reducing, we have

PL(k-k*)

R1=P(1-k) -——__. ..............
(1-k) (LEE) {PRP Petre ee iaaene ens (34).
Then taking moments about B, considering points B and C, we have
M” =L,R3.

Substituting this value of M”, as given in (33), in this last equation, and
dividing by L,, we have
_ PL? (k -k*)

BL (LAL Jee seen swe saga reside ewe (35).

Now, R1 and R3 can be determined for any load in the span AB
from equations (34) and (35), respectively, and likewise for any load in

R3=
THEORETICAL TREATMENT OF BEAMS 103

span BC by interchanging the R’s, that is, R1 would in that case be at C
and R3 at d—or just turn the beam end for end. If there are several
loads in the two spans, the reactions #1 and #3 due to each can be deter-
mined separately and added algebraically (as R3 is always minus), and
thus the total reaction at both 4 and C would be determined. Then the
bending moment and shear at any point in the beam is readily determined,
as explained above.

Now if the two spans are equal, we have L = L, in both equations

(34) and (35). When (34) reduces to

Me CERES WY ciccteteatdslnasinl ee ee (36),
and (35) reduces to
RB = (k-) renders! Sih lla Se Gua sapien ik oes Seg a adey Secaters (37)

Then adding (36) and (37) and subtracting the result from the load P
and reducing, we have the formula

Re =; et leiden case een tetees (38).

Now the reactions due to a load at any point on any beam continuous
over three supports and having equal spans, can be determined from
equations (36), (37), and (38). Then the bending moments and shears
are readily determined as explained above.

As a practical example, let L and L, in Fig. 95, each be equal to 12
feet, and let P be 4 feet from A and let P’ be 6 feet from C. Then
k = 4/12 =1/8, and k, = 6/12 = 1/2.

First considering P alone and substituting 1/3 for k in (36) and
reducing, we have

for the value of the reaction at A due to the load P. Next substituting
1/3 for k in (37). and reducing, we have
13
R3= Ei P

for the reaction at C, which is minus; that is, it pulls down upon the beam
instead of pushing up. Now after these reactions are determined, the
bending moment and shear at any point in the beam due to the load P
alone can be determined as explained above. The bending moments in
span BC would be negative, as R3 is negative; that is, the top elements in
the beam would be in tension while the bottom ones: would be in com-
pression.

Now considering P’ alone, and substituting 1/2 for k (=k,) in (36)
and reducing, we have

pies:
R1i=35P’,
104 STRUCTURAL ENGINEERING

which is the reaction at C due to load P’ in span BC. Next, substituting
1/2 for k in (37) and reducing, we have

- 32 p
R3=- 5 P’,

which is the reaction at A due to the load P’ in span BC. Then the
reaction at A due to the two loads is

1 BS
a7 3a?
and the reaction at C due to the same is
3, 1
ag? ba"

After these reactions are thus determined, the bending moment and shear
at any point in the beam are readily determined, as explained above.

Now instead of the concentrated loads just considered, suppose the
beam shown in Fig. 95 supported a uniform load of w pounds per foot in
span AB and w’ pounds per foot in span BC, extending over the entire
span in each case. In this case, as the loads are uniformly distributed,
we would use the three-moment equation (M), given above, which, as M’
and M” are equal to zero, reduces to

wl? w’L,5

 

A a —_ —
2M" (L+L,)= Z DTT TEE e ees (39).
Then transposing and dividing through by 2(Z + L,), we have
aa wl? +0’'L,?
M”= "REG, rte eehenet watiesamewesnenvesict (40)

Now taking moments about B, and considering points A and B, according
to Art. 66, we have
43 wl?
M”=LR1- “=
Then substituting the value of M”, as given in (40), in this last equation,
transposing, and reducing, we have

aL wh +w’L}

 

R1= a = SL(LiZ,) CC (41).
Now taking moments about B, and considering points B and C, we have
, 2
M"=LB3 = ,

and substituting the value of M’ , » as given in (40), in this last equation,
transposing and reducing, we have
R3 wh, wl? +0’L}
=" 7 GO alla
After these reactions are obtained, the bending moment and shear at any

point in the beam can be determined very readily as explained above.
Suppose the spans to be equal: in length and suppose the loads are
THEORETICAL TREATMENT OF BEAMS 105

equal. Then L = ZL, and w = w’, and substituting L for L, and w for w’
in both (41) and (42), and reducing, we have

3 3
Rl = 3 wl and R3 = g wh.

that is, the reactions at A and (' are equal to 3} of the load on one span.
Then, evidently, the shear just to the right and also just to the left of B
is equal to 8(wL) or $ of the load on one span, and hence for the reaction
at B we have

 

10
R2 er wl.
Substituting L for L, and w for w’ in (39) and reducing, we have
iS wL? :
eee

which is the bending moment at B, the central support. This being minus
shows that the top elements of the beam are in tension while the bottom
ones are in compression. This same moment can be obtained by taking
moments about B and considering either span, say, span AB. Then we
have

‘ 2
wrens 22.

But R1 = 3 (wL), as shown above. So substituting this value for R1 in
this last equation, and reducing, we have

 

pie wl? d
ale 8
Now taking moments about the center of span AB, we have
L wLhL\L L wL?
uage-(S aes BF
Substituting 3 (wL) for 21, we have for the bending moment at mid-span
_ 3 , #L? 1
M=; wl? — 3 =7g 2h’,

which is one-half of what it is at B, the central support.

Beams having four or more supports can be analyzed in a similar
manner. Of course, there would be more three-moment equations involved
—being (n — 1) in each case. However, the same general method as
outlined above would hold for all cases.
CHAPTER V
THEORETICAL TREATMENT OF COLUMNS

71. Preliminary.—Let CB (Fig. 96) represent a round steel rod
of uniform cross-section A and length L standing vertically upon a
smooth, rigid base at B and supporting a load P at C which is sym-
metrically applied in reference to the longitudinal axis of
lp the rod; that is, the load is applied in the center of gravity
of the cross-section of the rod. It is evident that the load
P will produce a direct simple compressive stress P at
every cross-section of the rod throughout its length, and
hence the direct compressive unit stress at every section
“| will be equal to P/A. All bodies having a length much
greater than their width, as the above rod, and loaded in
the same manner, thereby being in compression, are known
in general as columns. .
Now suppose the above rod to be 2 inches in diameter
and, say, 4 inches long. Then the unit compressive stress
PIES would be equal to

 

 

 

 

 

o
ed ape ee eee Ree
a -
2

P P

-(y
2

which corresponds to P/A given above. It is evident that this short rod
would support a very heavy load without failure, in fact, the compressive
stress P/a could be as great as if it were a tensile stress.
But suppose the rod to be 20 feet long instead of 4 inches.
We know that the rod, if 20 feet long, would not support as
heavy a load as it would if only 4 inches long, as the longer
rod would be too limber, that is, it would fail by bending
transversely; of course, the direct compressive stress would \
be acting just the same in either case.

If a column were absolutely straight and the load 2.
applied absolutely in the longitudinal axis, the bending re-
ferred to above would not occur. But such conditions do not Hn
exist, as we know from experience. So in the designing of
columns the stress due to this bending must be taken into
account in addition to the direct compressive stress.

72. Rankine’s Formula.—Let ED (Fig. 97) repre-
sent a column which bends as indicated under a load P. It
is evident that the elements on the concave side of the column

. 5; ‘ : ‘ D
are in compression, owing to the bending, while the elements ATE
on the other side are in tension, exactly as in the case of a
loaded beam. Now it is readily seen that the maximum stress Fig. 97

106

 

 

 

 

 

 
THEORETICAL TREATMENT OF COLUMNS 107

on the column will occur on the compression or concave side, for here the
direct and bending stresses combine, while the tensile stress on the other
side reduces the direct stress. This maximum compressive stress will, as
is readily observed, occur at the middle of the column, where the deflection
is greatest, which is represented as A. The bending moment on the
column at any point 2 distance from E due to the load is

M=Pz,

where g is the deflection. This moment will be a maximum when z = A.
Then we have
M=PA
for the maximum bending moment on the column. But from (D), Art.
_ 58, we have f = My/I, and substituting PA for M, we have

_ PAy

ot
(where y = the distance from neutral axis of column to the extreme

elements) for the maximum compressive stress on the column due to
bending. Now if we add this to the direct stress, P/A, we have

which is the maximum compression unit stress on the column.
Now—as is proven by experiment—A will vary directly as L? and
inversely as y; that is,
L?
y
Then if L?/y be multiplied by some constant c, we have

2
A=c—-

Aw

 

Substituting this value of A in (1), we have

P PcL*

Gana

But I = Ar? (Art. 49). Then substituting this value of I in the last
equation and reducing, we have

 

P L?
p=3(1+6%5) BGs Shoei hogar andes Meaty inte Hehe eee s% (N).
‘This can be written in the form
P p
eu :
tee jowtse ie ge Maree ee a en ada eu tee ty +++(O),
r

which is known as Rankine’s Formula, where

p=maximum compressive unit stress on the column;
P=)load on the column;
L=length of column in inches-
108 STRUCTURAL ENGINEERING

A =area of cross-section of the column in square inches;

r=least radius of gyration of the cross-section in inches in
reference to the neutral axis of the column;

ce=a constant which is determined by experiment, and depends
upon the material composing the column and upon end
conditions.

The maximum unit stress on any known column due to a given load
can be determined from Formula (N) and the allowable direct or average
compressive unit stress from (O), providing the proper value of c is
known, which is determined from experiment.

The following practical values of ¢ are recommended:

For timber columns c = 4/3,000;

For cast-iron columns ¢ = 4/5,000;

For (medium) steel columns ¢ = 1/11,000.
(American Bridge Co.)

If p be taken as the elastic limit of the material in the column, P/A
will be the direct compressive unit stress that the column would be sub-
jected to when the elastic limit of the column was reached, but if p bé
taken as the ultimate strength of the material, P/A would be the direct
compressive unit stress that the column would be subjected to at failure.
If p be given either of these values, a factor of safety must be used in
determining the “working stress” for any column. If 32,000 lbs. be
taken as the value of p at the elastic limit of steel, a factor of safety of
2 should be used, and if 64,000 lbs. be taken as the value of p when the
column fails, a factor of 4 should be used. For practical designing, it is
more convenient to take p as the allowable or working stress of the
material in tensian. This for medium steel may be taken as 16,000 lbs.,
in accordance with the specifications of A. R. E. Ass’n. Now, substituting
this value of p and the above value of ¢ in Rankine’s Formula (QO), we
have

16,000

i Ts Pe cae ncbusleaerehees abana (P).
1+ 7,000 (=)

This Formula (P) may be used for designing any steel column, but
a Straight Line Formula is preferable, as it is more readily applied and
the results obtained are practically as accurate.

73. Straight Line Formula.—The ordinates to the curve abcd
(Fig. '98) measured from the horizontal line OH give the value of P/
corresponding to the different values of L/r, as shown. This curve is
obtained by substituting the different values of (L/r)® in Formula (P).
For example, the ordinate eb is equal to

16,000
1 2
Ti,000 (2°)

It is readily seen from Formula (P) that the ordinate Oa is equal to
16,000 lbs., as L/r =0. Now, if some line as ak be drawn such that the
ordinates to it will be practically the same as to the curve, it is obvious

Ee
>

 

fe = 14,800 1
aS = 14,800 lbs.

 

1+
THEORETICAL TREATMENT OF COLUMNS 109

Volues of £ (10 Scale)

 

Valves of £ (20 Scale)
Fig. 98

that the equation to this line can be used as a column formula instead of
Formula (P), and the equation to the line would be known as a “Straight
Line Formula.”

The ordinate Hk to this line is taken here as 7,600 Ibs., and the
ordinate Oa is 16,000 lbs. Then, as OH is known, the slope of the line ak
can.be determined, and consequently its equation in reference to 0 can be
derived.

Let 6 be the angle that the line makes with the horizontal. Then we
have any ordinate y, distance « from 0, given by the formula

y = 16,000 — wtand. 2... cee eee ee eee (1).

Now tan 0= (Oa ~ Hk) + OH=0.14, as Oa and Hk are equal to 16,000 and
7,600 lbs., respectively, and OH, to the same scale, is equal to 60,000 lbs.
Now for z, in terms of L/r, we have

 

10,000 /L L
BD (7) ree
Then substituting these values of tan and 2 in (1), we have
y=" = 16,000 - 707 Ab sAnd ae he ae aherraanien Geese i Mise wbaded (Q),

which is the straight line formula as given in the A. R. E. Ass’n Speci-
fications.

As is obvious, the author just deliberately took the ordinate Hk so
that the above formula would result. If the line ak be drawn to conform
to mere observation, a slightly different formula would, very likely, be
obtained. -

It is readily seen that a straight line formula for any other material:
can be determined in the same manner by simply using the proper c and
working stress in each case.

74, Examples in the Application of Column Formulas.—

Lzample 1.. What will be the maximum compressive unit stress
produced in a wooden column 10x10 ins. in cross-section and 12 ft.
long by a load of 20,000 lbs.?

Here A.= 100, L = 144, r = 2.88, and c = 4/3,000. The value of p is
desired. Then substituting the above values in Formula (N), we have

_ 20,000 4 f 144\?7_
P= 700 [:+5h0 (a) |=80s re

In this example, the direct unit stress is 20,000 + 100 = 200 Ibs. So
the stress due to cross bending is 868 — 200 = 668 lbs. The working unit

 
110 STRUCTURAL ENGINEERING

stress on timber should be taken at about 1,200 lbs. So the above column
will safely support more than 20,000 Ibs.

Example 2. What direct unit stress would the above column safely
sustain if the working unit stress be taken as 1,200 lbs.? Here P/A is
desired. Then using Formula (O), we have

PB 1,200
A” | 1 4 /144 y|
* 3,000 (5
Then the safe load, or P, would be 277 x 100 = 27,700 lbs.

What is usually given is the load and the length of the column. In
that case the problem is to design a column outright that will safely carry
this known load. In all such cases it is practically a matter of first
guessing a column which we think will satisfy the conditions and then
modifying our guess until the conditions are satisfied, as will be shown in
the following problems.

Ezample 3. Design a timber column 10 ft. long to carry a load of

20,000 lbs. Taking 1,200 lbs. as the working unit stress, then p will be
1,200. Using Formula (N), we have

20,000 4 /120\?
1200-2 Ta a(t) |:

Here it is seen that A and r are unknown. First assume that a column
9x9 ins. willdo. Then A = 81 and r= 2.6. Substituting these values in
the last equation, we have

20,000 4 120\?
200 == eet ee =
1,200 81 E +3900 (3) | 938 lbs.,

which is too small, or, in other words, the column is too large. So next
try an 8x 8-in. section. Then we have

20,000 4 /120\2] _
1,200= 20 | one (Fs) | = 1,430 lbs.,

which is too large, that is, this column is too small. So the correct size is
between the two. However, a 9x 9-in. section would be used, as more
than likely nothing between the two could be obtained.

In case of steel columns, the straight line Formula (Q) will be used,
although the Formula (P) could be used.

Example 4. Design a steel column 25 ft. long to support a load of
160,000 Ibs. Using Formula (Q), we have

p=16,000 - 70 ne

= 277 Ibs. per sq. in.

 

 

 

The direct unit stress in accordance with economy should not be less than
9,000 lbs., and it rarely ever exceeds 14,000 lbs. So, guessing in this
case, as the column is short, that it will be 13,000 lbs., and dividing the
load by that, we have

160,000

13,000
From Table 3 it is seen that 2—[s 12” x 20.5% have about this

=12.3 sq. ins.
THEORETICAL TREATMENT OF COLUMNS 111

section. The r for these two channels in reference to the gravity axis
perpendicular to their webs will be the same for the two as it is for one
channel, which is given in the same table as 4.61. The radius in ref-
erence to the gravity axis parallel to the webs can always be made equal
to the one given in the table by moving the channels far enough apart.
Now substituting the above value of r in the above formula, we have

300
p=16,000- 70 a6i* 11,450 Ibs.

Then dividing this into the load, we have
160,000
11,450

which is more than the area of the channels assumed. So try
2—[s 12” x 25% = 14.70”, Then r, from Table 3, is 4.43, and our
formula becomes

=14.0 sq. in. (about),

300

p=16,000- 70 7 = 11,260 lbs.
Then dividing this into the load, we have
160,000 _ ,
“71,260 = 14.2 Sq. ins.,

which agrees very closely with the area of the 2—[s 12” x 25#, which
would therefore be used. The channels composing a column as just
designed would be latticed together.

The method of applying the column formula in the last example is
general, but the form of column is only one of many. It is beyond the
limits of practicability to have the radii of gyration tabulated for each
form. However, the approximate value of the radius for any form in
general use can be obtained from Table 10, where it is given in terms of
the height and width of the section, as shown. After the approximate
area of the cross-section is obtained by dividing the load by an assumed
intensity, as above, the section can be selected and then the corresponding
approximate radius of gyration can be obtained from Table 10. In this
way the assumed section can be tested, and, if found to be about correct,
the exact radius can be computed and substituted in the column formula,
and the section modified slightly, if necessary, to satisfy this last require-
ment. The form of steel columns depends upon the kind of structure and
the position they occupy in the structure as well as to the loads they
support.

The value of p for the different values of L/r could be taken from
such a diagram as shown in Fig. 98 if it be drawn to a sufficiently large
scale, or a table giving the same can be computed. The latter is often
used in designing offices.

In case a column is fixed at the ends, one-half of the length of the
column is taken as L, and if fixed at one end only, two-thirds of the
length is taken as Z in the column formulas. The reason for so doing is
readily seen, as the actual length of the column as far as bending is
concerned in the first case is the distance between the points of contra-
flexure, which is about one-half of the length of the column; and in the
112 STRUCTURAL ENGINEERING

stcond case it is the distance from the free end to the point of contra-
flexure, which is about three-fourths of the length of the column, the same
asin the case of beams. (See Arts. 68 and 69.)

75. Columns Eccentrically Loaded.—Whenever

 

 

 

P possible, the load on any column should be applied in the
A__ center of gravity of its cross-section. However, in some
Le| 4 cases this is not possible. In such cases the bending
y stresses due to the eccentric loading must be considered.

2 Let AB (Fig. 99) represent a column, which bends
y . as indicated, due to the eccentrically applied load P. Let
e represent the eccentric distance, A the maximum deflec-

A tion, and z and y the co-ordinates to the elastic curve re-

N] ferred to A as origin. Now the bending moment at any
point « distance from 4 is equal to

 

 

 

ay _
Elva =-P(e+y).
ef IB For convenience let k = \/P/EI. Then we have
Lee d?

Fig. 99 50 =-h*(e+y).

Now multiplying both sides of the equation by 2dy, we have the form
- 2dyd(dy)_ ,,
ae =—k?(e+y)2dy.
Integrating (dx constant), we have

He Gaye , 1)
ae e+y DMG Read ane en ee Renan (1).

Now, as is readily seen, dy/dz=0 when y=A. Therefore, C,=k?(e+A)?.

Then substituting this value of C, in (1), we have
d 2
Gor =~ W(e+y)?+h*(e +A)?

Now extracting the square root and transposing, we have
dz= gp ee ree,
k/(e+A)?-(e+y)?

Integrating this equation, we have

_1lf. ety
e=7 (sin 8).0, Cua a Get iele Berson as Taha s vata lskctt eee iets (2).

Now y = A when 2 = L/2; therefore, C,, = (L/2 — x/2k). Then substi-
tuting this value in (2), we have

olf at Fa L ie
o=z(sin +2). (5-3) Sy esa wins Carwin wee (3).

N.w y = 0 when 2 = 0. So, substituting 0 in (2) for both y and «, we

bare

 

 
THEORETICAL TREATMENT OF COLUMNS 113

ae Dee
¢,,= 7 (sin x).

Then by equating the two values of C,,, we have

set gee fe
k etd) \2° 2k)?

from which we obtain

 

 

é
A=— -e.
kL
cos ~>-
A

Then substituting this value of A in (3), we have

_1/.. _ (e+y) coskL/2 Loa
o=7(sin forgone?) + (F-x) oe Oe de ee Sree (4).

Now transposing and reducing, we have

sin | (He es 5\-¢ cos “E °

But, according to trigonometry,

kL
sin| (te -"2) |= sin (ie - eos + cos (ie 2%) sin—
2 2
=fty kL
cos

But, cos 7/2 = 0, and sin 7/2 = 1. Then the last equation reduces tc
( =) ety kL
cos 7 :

 

ka —-—

2

But again, according to trigonometry,

 

   

 

kL kL, . kL
cos ¢ , ) =coskxcos ae + sinkzsin ‘got a = Das “.
Dividing through by coskL/2, we have
coska + sinketan *E aetd.

But, tan kL/2 = 1 - coskL/sinkL. Then substituting this value of
tan &L/2 in the last equation, and reducing, we have
e[sink(L—a)+sinkz]
sinkL
Now from this last equation, y can be computed for any point along

the column, and then the lever arm (e + y) is known and hence the stress
due to bending at any point along the column can be determined, as we

have
I
P(e+y) =H
14 STRUCTURAL ENGINEERING

according to Art. 53. (d is here the distance from the neutral axis to the
outermost element in compression. )

Then by adding the unit compressive stress (f) thus obtained to the
direct unit stress (P/A), we have the maximum unit stress at the point
considered. This, of course, will be a maximum when y = A. Now
y = A when x = L/2. Then, substituting this value of « in (5) and

reducing, we have
e(see LS Acer es seein s Rds Milne Wa a goals es (6).

This last equation (6) can be used for determining A, the maximum
deflection. After this is determined, the maximum stress can readily be
determined as stated above. For the maximum compressive stress we
then have the formula

Pd(e+A) P
ea

3

where p is the maximum compressive unit stress in pounds, I the moment
of inertia of the cross-section of the column, 4 the area of the cross-
section in square inches (which is assumed to be constant), and P, e, and
A are as stated above.
CHAPTER VI
RIVETS, PINS, ROLLERS, AND SHAFTING

RIVETS

76. Kinds of Stress.—Rivets may fail by shearing off transversely,
by bearing, that is, by crushing against the metal through which they
pass, ov by bending, as a beam. So, therefore, we have to consider shear-
ing, oearing, and bending stresses on them.

77. Shearing Stress.—The shear on a rivet at any cross-section is
equal to the algebraic sum of the forces on either side of the section—
just the same case as a beam. Let A and B (Fig. 100) represent two
bars held together by one rivet, as shown. Let P be the tensile stress on
each bar. Then the bar B would exert forces along
ab upon the rivet, the sum of which (resolved along the B
bar) would be equal to P, while the bar A would exert
forces upon the rivet along cd, the sum of which (re-
solved along the bar) would be equal to P- The forces
along ab, acting against the forces along cd, tend to

   

 

shear the rivet off transversely. This shear, as is (P)
readily seen, varies from O at each end of the shank to Thu

a maximum at the cross-section bc. If it were possible
to determine the intensity of these forces at all points, \\
the shear at any cross-section along the rivet could be
obtained by beginning at either end of the shank and

simply summing up the forces to the section consid- 8
ered. However, it is not practical to do so, neither is
it necessary, for it is readily seen that the maximum

shear occurs just between the two bars at cross-section |
be. This shear, as is evident, is equal to P, and as it P
is the maximum, it is the only shear we need consider.
The shear between the pieces connected is what is generally referred to as
the shear on rivets, and it is what we shall consider as the shear.

The unit shearing stress, that is, the stress per square inch on a rivet,
is obtained by dividing the shear (in pounds) by the area (in square
inches) of the cross-section of the rivet. Thus if S be the shear in pounds
at a given section of a rivet having a cross-section of A square inches, and
V the unit shearing stress, we have

S

V= 7
The allowable shear on a given rivet is what we usually desire. This is
obtained by multiplying the allowable intensity per square inch by the
area of the cross-section of the rivet. We shall take this allowable
intensity as 12,000 pounds per square inch for shop rivets and 10,000
pounds for field rivets, as specified by the A. R. E. Ass’n in their speci-

115

 

 

Fig. 100
116 STRUCTURAL ENGINEERING

fications for railroad bridges. Then if s be the allowable shearing stress,
we have s = 12,000 x A for shop rivets, and s = 10,000 x A for field rivets.
Substituting in these formulas we have for the allowable shear on a {”
shop rivet, s=12,000 x 0.6013=7,216 lbs., and for a %” field rivet, we
have s,=10,000 x 0.6013=6,013 Ibs. These values for the different size
rivets are given in column 3 of Table 11.

A rivet is in single shear when there is (so to speak) one tendency
to shear it off, double shear when there are two tendencies, triple shear
when there are three, and so on. If two pieces be riveted together, as
indicated in the sketch at (a) (Fig. 101), the rivets will be in single
shear; and if three pieces be riveted together as indicated in the sketch
at (b), the rivets will be in double shear; and if there were four pieces,
the rivets would be in triple shear. If the pieces riveted together are
properly designed, rivets in double shear will have twice the allowable

PF aa pee 2%
et Se

oT
(Q) (b)

Fig. 101 Fig. 102

 

 

shearing strength of rivets in single shear, and three times in the case of
triple shear, and so on. For example, the shear on the rivet shown at
(a) (Fig. 101) would be equal to P, while in the case shown at (b) it
would be equal to P/2, providing e and d have equal cross-section.

78. Bearing Stress on Rivets.—Let Fig. 102 represent a bar
pulling with a force of P pounds against a rivet which passes through it.
The force transmitted to the rivet thereby will be a normal force
uniformly distributed from b around to a, as indicated. Let ¢ be the
thickness of the bar, D the diameter of the rivet, and let p be the uniform
bearing force per square inch. Now, the bearing force on any infini-
tesimal strip through the plate, as tds, where ¢ is the thickness of the
plate, would be ptds. Now this force, resolved along the bar, is equal to
ptds cos0, where 6 is the angle the normal force makes with that direc-
tion, as indicated. Then if the normal forces on all such strips, from
b around to a, be resolved along the bar and summed up, we would have
pt3dscosé = P. But Xdscosé = ab = D, the diameter of the rivet. So we
have

To obtain the value of P, such that the rivet would not be too highly
stressed, we would substitute the working value of p in the above formula,
in which case p would be known as the allowable unit-bearing stress on
the rivet and P as the allowable bearing of the rivet on the plate of
thickness t. For the unit value we will take 24,000 Ibs. per square inch
for shop rivets and 20,000 Ibs. for field rivets, as specified by the A. R. E.
Ass’n in their specifications for railroad bridges.

Now suppose the rivet shown in Fig. 102 to be a {” shop rivet, and
RIVETS, PINS, ROLLERS, AND SHAFTING 117

suppose the plate to be 4” thick. Then substituting in (1), we have

24,000 x i x £ =10,500 Ibs. = P,
that is, the allowable bearing of a $” rivet on a 3” plate is 10,500 lbs.
If the thickness of the plate be }” instead of 4”, we would have

1_7
24,000 x7 Xe 5,250=P,

In this way the allowable bearing for rivets of different diameters on
different thicknesses of plates can be computed. These values are given
in Table 11.

79. Bending Stress on Rivets.—If rivets are properly driven,
bending stresses can be ignored, except in the case of loose fillers as is
illustrated in Fig. 103. Here the bars c, d, and e exert double shear on
the rivet just the same as the case shown at Pp
(b), Fig. 101, and the bearing stresses are se 7 Se
just the same, but the idle fillers, f and g, * %
hold the bars apart so that the rivet is really Fig. 103
a beam loaded in the center and supported at
the ends. Let & be the distance from the center of the bar c to the center
of the bar d, and h the distance from the center of the bar c to the center
of the bar e. Then the maximum bending moment on the rivet is (P/2)k
or (P/2)h. Now, according to (D), Art. 53, the maximum bending
stress on the rivet per square inch is

 

(2x)2

_My_\2"/2 _16Pk

cay nD? 7D?’
“64,

 

where D is the diameter of the rivet.

The bending stress on a rivet should always be combined with the
shearing stress according to Art. 62 so as to obtain the maximum stress.

Details should be so contrived as to avoid bending on rivets. In
fact, there is practically no excuse for placing rivets so that they are
subjected to bending of any consequence.

80. Examples in Determining Number of Rivets. —Suppose
that each of the bars a and b, shown at (a), Fig. 101, be }” thick, and
suppose the stress P to be 50,000 lbs., how many 2” shop rivets would be
required to transmit this stress?

The allowable single shearing stress on one 3”. shop rivet, from Art.
77, is 12,000 x (2)? + 4 = 12,000 x 0.4418 = 5,301 lbs. Then the number
of rivets required for shear is 50,000 + 5,301 = 9.4 rivets (use 10).

The allowable bearing stress on one 3” shop rivet, from (1), Art. 78,
is 24,000 x 4.x 2 = 4,500 lbs. Then the number required for bearing is
50,000 + 4,500 = 11 rivets (about). So 11 rivets is the number required.

Now suppose the bars to be 3” thick instead of 3”. The allowable
shearing stress on each rivet would not change. So 10 rivets would be
required for shear the same as before. But the allowable bearing stress
is now 24,000 x 3 x } = 6,750 lbs, Then the number of rivets required
118 STRUCTURAL ENGINEERING

for bearing is 50,000 + 6,750 = 7.4 (use 8). So in that case 10 rivets, as
required for shear, would be used, as it is the greater number.

As another example, suppose the bar c, shown at (b), Fig. 101, to be
4” thick, and each of the bars d and e to be 3” thick, and suppose the
stress P to be 60,000 Ibs.; how many 8” shop rivets would be required to
transmit this stress? The allowable double-shearing stress on one 3”
shop rivet is 2 x 12,000 x 0.3068 = 7,364 lbs. Then the number of rivets
required for shear is 60,000 + 7,364 = 8.15 rivets (use 9). The bearing
here, in one direction, is on a $” bar, while in the other direction it is on
two 2” bars, which is equivalent to one }” bar. So the bearing stress on
the 4” bar (c) will be the greater. The allowable bearing stress on one
rivet is 24,000 x 4 x £ = 7,500 lbs. So the number of rivets required for
bearing is 60,000 + 7,500 = 8 rivets. Here the number of rivets used
would be 9, the number required for shear.

The allowable bearing and shearing intensities for rivets are given
in Table 11. However, the student should know how to compute these
intensities.

PINS

81. Kinds of Stress.—The same kind of stresses occur on pins as
on rivets. However, in the case of pins, the bending stresses are, as a
rule, the most serious.

82. Shearing and Bearing Stresses.—The shearing and bearing
stresses on pins are determined exactly as on rivets. The allowable unit
intensities for shear and bearing are 12,000 and 24,000 lbs., respectively
—the same as for shop rivets. The shearing stress very rarely affects
the design of a pin, while the same is practically true of the bearing stress.
However, the bearing affects the details of the other parts of a structure
as each member connected must have the proper thickness of bearing on
the pin. As an example, what would be the required thickness of bearing
to transmit a stress of 250,000 lbs. to a 6-in. pin? The allowable bearing
of a plate 1 in. thick on this pin, according to Art. 78, is 24,000 x 1 x 6=
144,000 lbs. Then the thickness of the bearing required is 250,000
+ 144,000 = 1.76 ins.

83. Bending Stresses on Pins——The forces applied to pins are
assumed to be applied at the center of bearings of the pieces connected.

The bending moments on pins can be computed most

z m5 [o-1—~& readily by utilizing the proposition in Art. 66. For ex-
b<— pb "bi ample, let AB (Fig. 104) represent a pin acted upon by
Mey le ¢ forces Fa, Fp, etc.,-applied at sections a, b, c, etc., as
4 ab ,- shown, and let My, Mp, etc., and S,, Sp, etc., represent
& tlnes the bending moments and shears, respectively, at the sec-
* tions a, b, c, ete. It is obvious that the bending moments

 

 

 

 

 

 

 

 

",S,\2t— § at sections a and g are equal to zero. Then, by starting

B at either of these sections we can determine the bending

Fig. 104 moment at each of the other sections, according to
Art. 66.

So, starting from a, the bending moment at b is

My=M,(=0) + (ab) S,= (ab) Fi;
RIVETS, PINS, ROLLERS, AND SHAFTING 119

at c it is
M,=My+ (be) 8)=My+be(Fa-Fy1);
at d it is f
Mai=M,+ (cd)S,=M,t+ed(Fy-Fy+F);
at e¢ it is
M,.=Ma+ (de)Sa=Mat+de(Fa-Fot+Fo-Fa);
at f it is

M;=M,+ (ef)Sc=Met ef(Fa-Fot Fo-Fat+Fc);
at g it is
Myg=M;+ (gf)S;=My+fo(Fa-Fo+Fe-Fat Fe-F,) =0.

In determining the bending moments in this manner, the maximum
is readily observed. Care should be taken in regard to algebraic signs.
The shear in some cases will be minus, as is readily seen, and consequently
the moments in some cases may be minus.

As a numerical example, let the lever arms be as follows:

ab=2 in., de=1 in,
be=14 in, ef =1} in.,
ed=1 in., fg=2 in.;
and let
F,=150,000 Ibs., F,=125,000 Ibs.,
F,=-200,000 Ibs., * F,=-200,000 lbs.,
F,=125,000 Ibs., Fg=150,000 Ibs. ;
Fg=-150,000 Ibs.,
then
Sq= 150,000 lbs., Sa= -75,000 Ibs.,
S,=-50,000 lbs., S, = 50,000 lbs.,
S,= 175,000 lbs., S;=-150,000 Ibs.
Now, by substituting these values in the above equations, we have
M,=2 x 150,000 = 300,000 in. Ibs.,

M,=300,000— %5,000 = 225,000 in. lbs.,
My=225,000+ 5,000 = 300,000 in. Ibs.,
M,=300,000— 5,000 = 225,000 in. lbs.,
M;,=225,000+ 5,000 =300,000 in. Ibs.,
M, = 300,000 — 300,000 =0.

Here it is seen that the maximum moment on the pin would be
300,000 inch pounds, which happens to occur at three points—b, d, and f.

After. the maximum bending moment is thus obtained, the bending
stress is computed according to (D), Art. 53, just the same as if the pin
were a beam. Thus, in the last example, suppose the pin to be 6” in
diameter. Then we would have

f = 300,000 x3+ nes = 14,150 Ibs. (aboat).

The allowable bending stress intensity for pins, as specified by the
A. R. E. Ass’n, is 25,000 lbs. per square inch.

Let D be the diameter of any pin, M, the bending moment, and we
have
120 STRUCTURAL ENGINEERING
D _7D* 32M

Transposing, we have

D=,/2M need ied talent hae, tO Ote ead).
fr
Now by substituting 25,000 lbs. for f in this last equation, the diameter
of pin required for any given bending moment can be determined. The
diameters required for given moments are given in Table 12. Here the
moments are given in inch pounds, and f taken as 25,000 pounds.
Often the members bearing on a pin act upon it from different
directions, as shown in Fig. 105. In such cases the stresses in the
members are resolved horizontally and vertically and the bending moments
determined in each plane, and the maximum or resultant moment is then
obtained by extracting the square root of the sum

 

. zt of the squares of these maximum horizontal and

‘ vertical moments. Thus, in the case shown in Fig.

Fl Ee) FZ 105, F3 would be resolved horizontally and ver-
Fig. 105 tically. Then the horizontal component would be

included with the forces F1 and F2 in the computa-

tions for the horizontal moments, and the vertical component would be
included with the force F4 in the computations for the vertical moments.

Let M;, represent the maximum horizontal moment, and let Mv

represent the maximum vertical moment; then the maximum or resultant

moment would be
M=\/Mi Mi,
for which the pin would be designed.

ROLLERS

84, Allowable Pressure.— In modern practice, the allowable pres-
sure on rollers is obtained by the use of an empirical formula, which has
the form

pa DD sme cai Seciee ah vine desea Raiaie hcl oe (1),

where p = the allowable pressure per lineal inch of roller and D = the
diameter of the roller in inches, and c = a constant.

Prof. A. Marston* found that for the elastic limit of soft steel rollers
c was about 880. (See paper Trans. A. S. C. E., Vol. 32.) For the
allowable pressure, c should be taken as about one-half of this. Then for
the allowable pressure per lineal inch on soft steel rollers, we have

ree aamiausstose aatras eae alte aah wakes & (2).

The formula, for medium steel rollers as specified by the A. R. E.
Ass’n, is
P= 000 Ds cad niceties Wa be eeheedadawws aes (8).

This is the formula now in general use, as rollers are, as a rule, made
of medium steel.

*Dean, Engineering Division, lowa State College.

 
RIVETS, PINS, ROLLERS, AND SHAFTING 121

The formula is very readily applied. As a practical example, sup-
pose a load of 600,000 pounds be supported upon six 6” rollers; what will
be the required length of each?

Substituting in Formula (3), we have p = 600 x 6 = 3,600 Ibs. Then
the total linear inches required is 600,000 + 3,600 = 166.6 ins.; and as
there are six rollers, the length of each will be 166.6 + 6 = 27.7 ins.

Rollers should always be as large in diameter as is consistent with
accompanying details.

SHAFTING

85. Stress Due to Torsion.—Shafting used to transmit power for
operating movable bridges, as well as in all other machinery, is subjected
to a twisting about the longitudinal axis known as torsion. The stresses
resulting therefrom are known as torsion stresses. This stress is the same
as the shearing stress in a beam, except the action on each particle is
perpendicular to a radius through the center of rotation, and the stress
varies directly as the distance out from this center of rotation. The
difference results owing to the stress being caused by a tendency of
rotation instead of a tendency of translation, as in the case of a beam.

Let AB (Fig. 106) represent a round steel shaft subjected to torsion.
Suppose this shaft be cut off at ab and spliced by means of inserted steel
pins which fit tightly into drilled holes.
Now it is readily perceived that the tor- ef}
sion (twisting of the shaft) tends merely
to shear each of these pins off transversely
and perpendicularly to a radius through
the center of rotation, and as far as the Fig. 106
torsion is concerned, the pins have only a
direct shearing stress on them. Now, evidently, the material particles in
every cross-section of the shaft are subjected to exactly the same kind of
stress from torsion as the pins, as the action causing the stress is the same.

This shearing stress varies directly as the distance out from the
center of the shaft. This is readily seen by considering again the pins at
section ab. Each pin will be distorted a certain amount, and as the
tendency of rotation is about the center of the shaft, it is evident that the
distortion of the pins will vary directly as their distance out from the
center of the shaft, and hence the stress on them will so vary. Then,
evidently, if the stress on these pins varies directly as their distance out
from the center of the shaft, the stress on the material particles at every
cross-section of the shaft will vary directly as their distance out from the
center of the shaft.

So we have thus far shown that the stress on the shaft due to torsion is
a transverse shearing stress acting perpendicularly to the radii of rotation,
and that it varies directly as the distance out from the center of the shaft.
Now it remains yet to determine the intensity of the stress. The stress,
of course, will depend upon the torsion, which is directly proportional to
the moment about the center of the shaft of the force producing the
torsion. Let P represent the force producing the torsion of the shaft AB
(Fig. 106) and let d be its lever arm. Then the moment of this force,
which produces the torsion, is Pd. This would be known as the “torque.”
Now it is evident that this moment Pd, or torque, must be balanced at

 

 
 

 

 

LP (Sect ab)
122 STRUCTURAL ENGINEERING

each cross-section of the shaft by the moments of the torsion stresses, at
each section, about the center of the shaft.

Let Fig. 107 represent an enlarged cross-section of the shaft AB
(Fig. 106). As the torsion stresses vary directly as the distance out from
the center. of the shaft, the maximum stress will evidently occur at the

outermost elements. Let S be this stress per square inch,

and let 7 be the radius of the shaft. Then the stress per

> square inch out unit distance will be S/r, and the stress. per

(4 square inch out any distance z will be (S/r)z. But as the

A stress varies continuously outward from the center of the

shaft it is evident that there will not be a square inch of

Fig. 107 + material out 2 distance having a stress of (S/r)z, but only

an infinitesimal area da of material having that stress. So

the actual stress or force out any distance z is (S/r)zda, and, of course,

the moment of this about the center of the shaft is (S/r)z°da. Now it is

evident that the stress is the same on all material-equal distances out from

the center of the shaft, so da can be taken as a circular strip of width dz

and length 27z. Then we have da=2rzdz. Then the moment of the tor-
sion stress out g distance from the center of the shaft is

Ss

Ss
—2°da=2n— 2°dz,
r r

and summing up this for the entire cross-section, we have

 

S fr
Pa=2e {sas maar a Gada tilafa sk 3 ab dats, Soak nae: gaan tenes AINA ve oteoa Aaya (1).
Oo
Integrating, we have Pd=Szr*/2, and transposing,
2Pd
S= pre (2),

from which the maximum torsion stress per square inch on any round
solid shaft, as the above, can be computed. S = stress per square inch;
Pd = torque; and r = radius of the shaft.

In the case of a hollow shaft, the integration of (1) would be between
the limits of the internal and external radii. Thus, let r be the external
radius and r, the internal radius of a hollow shaft. Then we would have

Pd=27 oF dz,

vi

Integrating, we have
RIVETS, PINS, ROLLERS, AND SHAFTING 123

from which the maximum torsion stress per square inch on any hollow
shaft can be computed. S = stress per square inch; r = external radius;
r, = internal radius; and Pd the torque.

In case of a shaft having a square cross-section, da

 

would be taken as any infinitesimal rectangular area. da
For instance, let Fig. 108 represent the cross-section of a Wily
square shaft, the center of which is at O. Then, for the 0

moment of the torsion stress on an infinitesimal area da
out any distance z from O, we have (S/e)2’da. Here S
is the stress on the outermost element, which is distance
e from O. Let Pd be the torque, and we have

 

 

 

Fig. 108

Pd= Sata,
e
Expressing z in terms of the rectangular co-ordinates 2 and y, we have
pa=* (3a°da + Sy?da).

But (32?da+3y?da) is the polar moment of inertia of the cross-section,
which is designated as J in Art. 49. Then we have

Pd=—-
e
Transposing, we have

gaFde Final ot Medea Raaeeacuesemeaaaeeas (4),

from which the maximum torsion stress, not only on square shafting but on
any shaft whatever, can be computed. S = stress per square.inch; Pd =
torque; J = polar moment of inertia; and ¢ = the distance to the outermost
element. Formula (4) is a general formula, as it applies to any form of
cross-section.

 
CHAPTER VII

MAXIMUM REACTIONS, SHEARS, AND BENDING MOMENTS
ON SIMPLE BEAMS AND TRUSSES, AND
STRESSES IN TRUSSES

86. Maximum Reactions on Simple Beams.—The reactions on
simple beams in all cases can be determined by taking moments about the
supports as explained in Art. 54. In the case of uniform dead load, the
two reactions are equal and each is known directly as being equal to half
of the dead load supported. In case the dead load is not uniformly
distributed, the reactions are obtained by taking moments about the
supports, as stated above. In all cases of dead load, the reactions are
fixed in intensity, as the loads are fixed in position, but for live load the
case is different as the reactions vary with the position of the loads, and
the maximum reaction occurs when the loads are in one certain position.
If the live load be a uniform load, the reactions at the two supports will
be a maximum at the same time and that will be when the load extends
over the full length of the beam, in which case each reaction is known
directly as being equal to half of the load supported. The live load often
consists of wheel loads, as in the case of locomotives, trains, wagons,
traction engines, etc., where the loads or wheels are a fixed distance apart.
The maximum reaction due to such loading will occur, as is readily seen,
when the beam is fully loaded and the heaviest loads are as near as
possible to ‘the support considered.

87, Maximum Shear on Simple Beams.—The reactions on simple
beams, in all cases, can be determined by taking moments about the

L
ae 2

 

 

 

 

 

RI

 

 

 

 

Fig. 109

supports, as stated in the preceding article, and after the reactions are
known the shear at any vertical section of a beam is determined by simply
adding up the forces on either side of the section, beginning at the end of
the beam, as explained in Art. 52. This much applies in all cases whether

124
MAXIMUM REACTIONS, ETC. 125

the shear be a maximum or not. The maximum shear on any vertical
section of a beam, due to dead load, is readily obtained as such loads are
fixed in position and hence the shear obtained by adding up the forces
on either side of any section is the maximum for that section, but in the
case of live load the maximum shear is not so readily obtained for the
shear on every vertical section of a beam changes continually as such
loads move over the beam, and hence the maximum shear will occur when
the load is in one certain position. As an example of live load, let AB
(Fig. 109) represent a simple beam supporting a single moving load
P, and.let R and R1 represent the reactions at A and B, respectively, due
to this load when at any point on the beam.

The shear on any short strip through the beam, as abcd, due to P,
can be expressed as

Px

S=> =k,

when the load is at any point to the right of the strip. The shearing
force exerted upon the strip will then act as indicated. It is readily seen
that this shear on the strip varies directly as « and hence will be a
maximum when «=m. Then if we lay off the vertical line gh equal (by
scale) to Pm/L and draw the line hk, any ordinate to this line hk graph-
ically represents the shear on the strip abcd when the load is directly over
the ordinate. For example, the ordinate y represents the shear on the
strip when the load is in the position shown, and any other ordinate as 2
represents the shear on the strip when the load is over that ordinate.

When the load is at any point to the left of the strip, the shear on
the strip can be expressed as

oo.
S’=3" -P=(R-P),

but the shearing forces exerted upon the strip will then act in the oppo-
site direction to those shown. This change in the direction of action of
the shearing forces takes place the instant the load passes the strip. We
would say that the shear changes signs as the load passes the strip and
that the shear on the strip is positive when the load is on one side, and
negative when on the other.

It is seen, from the last equation, that S’ will be a maximum when
R has its least value. This, as is seen, occurs when the load is just to
the left of the strip, that is, when e=m+dz, but, as dx can be taken as
an infinitesimal, we would say when c= m. Then if we lay off the vertical
line fg equal to (Pm/L) -P=(R-P) and draw the line fe we have the
shear on the strip graphically represented for the load P at any point to
the left of the strip, as any ordinate 1, to the line fe, represents the shear
on the strip when the load is over that ordinate. Then we have the variation
of the shear on the strip abcd due to the load P, as it moves over the beam
from B to A, fully represented by the ordinates to the broken line efhk.
It is seen from this that the loads on the two sides of any vertical section
of a beam really neutralize each other as regards the shear they produce
on the section, and hence it appears that the maximum shear would occur
when the loads are on one side only and extending on that side from the
126 STRUCTURAL ENGINEERING

end of the beam up to the section. This is absolutely true in the case of a
uniform live load, as shown in Art. 55, and is practically true for prac-
tically all classes of loading, yet it is not absolutely true in some cases of
concentrated live loads, for sometimes the maximum shear on a vertical
section of a beam, due to such loads, occurs when the loading extends
from one end of the beam to a little beyond the section considered. The
most common instance of this is that of the. typical system of locomotive
wheel loads. In such cases the exact position required to produce maxi-
mum shear must be determined by trial.

y2" As a general case of uniform load,
: let it be required to determine the maxi-
mum positive and negative shear on a
beam 12 ft. long at a point 4 ft. from
the end due to a uniform dead load of

 

 

 

 

A a B 100 lbs. per foot and a uniform live load
7B 7 of 1,500 lbs. per foot. Let AB (Fig.
Rg gs’ Rl 110) represent the beam, and let ba

 

represent the section at which the shear
is required.

 

 

 

 

 

 

 

 

 

| i As the dead load is symmetrically
located in reference to the center of the

beam, the two reactions due to dead

( (b ) load are equal to each other, and each

eon is equal to 6x 100=600 lbs. Then for

the dead-load shear at the section ba, adding up from 4 and taking R
(=600) as positive, we have

600 -(4 x 100)=+ 200 Ibs.

When the uniform live load extends from the end B up to the section, as
shown, one of the maximum live-load shears will occur. Taking moments
about B when the load is in this position, we have

_ 1,500x 8x4 _
-——-*-

for the live-load reaction at A. Then for the shear due to this load at the
section ba, adding up from A, we have +4,000 Ibs. as there is no inter-
vening live load. This is the maximum positive shear at the section ba
due to the live load. It will be readily seen that the dead load and the
live load, where the live load extends from B to the section ba, both exert
shearing forces at the section which act in the direction shown at (a),
hence the two will add, and we have 200+4,000=+ 4,200 Ibs. for the
maximum positive shear at the section ba, due to the dead and live load
combined. Now if the live load extends from A to the section ba instead
of from B, the reaction at A, due to the live load in that position, would be

1,500 x 4x10
12
Then for the shear at the section ba due to this load, adding up from 4,

we have 5,000 — 6,000 =— 1,000 lbs. which is the maximum negative shear
at the section due to the live load.

R +4,000 Ibs.

R= = 5,000 Ibs.
MAXIMUM REACTIONS, ETC, 12,

It will be readily seen that the live load when extending from A te:
the section ba will exert shearing forces at the section which will act as
shown at (b), and as this direction is opposite to the direction of action
of the shearing forces exerted by the dead load, the shear at the section,
due to dead and live load combined, will undoubtedly be equal to the
difference of the two shears. So we have 200—1,000 =— 800 lbs. for the
maximum negative shear at the section ba due to dead and live load
combined.

Now it is seen that in the above example the maximum positive and
negative shear at the section ba, due to dead and live load combined, is
simply the algebraic summation of the two simultaneous shears in each
case. This will hold in all cases provided the adding up of the forces be
in reference to the same end of the beam in each case. As an illustration,
let us add up the forces in the above example from the end B instead of
from A, as we did above. The reaction at B, due to the dead load, is
600 lbs. Then for the dead-load shear at the section ab, adding up from
B, we have, 600-8 x 100 =— 200 lbs.

For the live-load reaction at B, when the live load extends from B to
the section ba, we have

Rl= 1,500x8x8

12

Then for the shear at the section, adding up from B, we have 8,000-
12,000=-4,000 lbs. Adding this to the dead-load shear, we have
— 200 — 4,000 =- 4,200 lbs., which is the same as we obtained by adding
up from 4, except the sign is negative instead of positive. If the live
load extends from 4 to the section ba instead of from B to the section,
we have

= 8,000 Ibs.

_ 1,500x4x2
onl ae

for the reaction at B due to the live load. Then for the shear at the
section ba, adding up from B, we have +1,000 lbs. Now adding this to
the dead load, as just determined by adding up the forces from B, we
have — 200+1,000=+800 lbs., which is the same as we obtained avove
by adding up from 4A, except that the sign is positive instead of negative.

Thus we see that by adding up the forces from one end we obtain
positive shear and by adding up the forces from the other end we obtain
negative shear. There is no mystery about this at all, as is readily
perceived by referring to Fig. 111 where AB represents a simple beam
supporting a number of loads, and & and R1 represent the reaction at
A and B, respectively, due to the loads, and abcd represents a very short
strip through the beam. If we take the reactions as positive, then any
force acting upward will be positive. By adding up the forces from A to
the strip abcd we really obtain the shearing force acting upon the strip
along ab and by adding up the forces from B to the strip we really obtair
the shearing force acting upon the strip along dc. As these two shearing
forces on any vertical section of a beam are always equal and opposite, it
is evident that the sum obtained by adding up the forces from one end
will be positive, while the equal sum obtained by adding up the forces
from the other end will be negative.

Ri = 1,000 lbs.
128 STRUCTURAL ENGINEERING

There will be no confusion as regards the sign of the shear on any
sertion of a beam in any case, providing the adding up of the forces, that
is. the algebraic summation, be started from the same support.

| bed | |
A N |B

R 2 RI

Fig. 111

 

 

 

 

 

 

88. Maximum Bending Moments on Simple Beams.—The bend-
ing moment at any cross-section of a beam, in all cases, is equal to the
algebraic sum of the moments about the section, of the forces on either
side of the section. In case the load be uniformly distributed over the
entire length of the beam, the maximum moment will occur at the center
of the span, and will be equal to wL*/8 as shown in Art. 56. This is true
for all uniformly distributed loads, either live or dead. In the case of
fixed concentrated loads, the maximum moment will occur under one of
the loads which can be ascertained readily by trial. However, it can be
determined otherwise as shown later. /

If the live load be wheel loads, it is always necessary to first deter-
mine the position of the loads on the beam for maximum moment. As
wheel loads roll over a beam it is evident that the moment at every section
is continually changing, ‘and it is the absolute maximum of these various
moments produced that we desire, although it may last only for an instant
during the passage of the load.

5 a x Let AB (Fig. 112) represent

i : ‘
Fk eo a simple beam supporting a system
b rr of wheel loads. According to Art.

56, the maximum moment will occur

© O © e © © under a wheel. Let this wheel be

 

 

 

 

 

 

 

 

AC . WS the one marked C. Let W be the
IR P dw RI weight of all the wheels on the
L L_' beam, and x the distance from the

Fig. 112 right support’ to their center of

gravity. Further, let P be the
weight of the wheels to the left of wheel C, and b the distance of their
center of gravity from wheel C, and let g be the distance from the left
support to wheel C. Then the bending moment under wheel C can be.
expressed as

Max FONE py (Re— PRY cee cc cece ccc vevveasseens (1).

_ Now from Fig. 112 it is scen that e=L-—a-z. Substituting this
value of x in (1), we have

Now any slight movement of the loads either to the right or Jeft will
cause a small change in z and M. Let dz and dM represent this incre-
MAXIMUM REACTIONS, ETC. 129

ment of z and M, respectively, due to a slight movement of the loads.
Adding the increments in equation (2), we have

M+dM= + [(#+dz)L—(z+dz)a-(z+dz)?|~Pb,
and reducing we have
Ww Ww
M+dM= r [2L - az—-27]+ Lr [dzL - adz —2zdz]—Pb.
Now subtracting (2) from this last equation and reducing we obtain

dM = Was? adzg — 7 C802 via ve se iW Os eT RE RETO eas (3)
which is undoubtedly the expression for the increment of the bending
moment under wheel C due to any slight movement of the loads. It is
readily seen that if wheel C is to the right of the point of maximum
moment any slight movement of the loads to the left will increase the
moment M by the amount dM—in other words, dM will be positive—and
that all such movements of the loads to the left will increase M-the amount
dM until wheel C arrives at the point of maximum moment and from there
on any slight movement of the loads to the left will decrease M by the
amount dM; in other words, dM is negative in that case. Evidently the
increment dM is zero just as wheel C arrives at the point of maximum
moment as it is positive when the wheel is just to the right of the point
of maximum moment and negative when to the left and hence changes
signs at the point. So we have from (3), when M is a maximum,

Ww Ww
dM = Wdz-> adz— Ll 22dz=0.
Reducing, we obtain
a

2

which is the value of s when the maximum bending moment occurs under
wheel C. This last equation shows that the center of the beam bisects
the distance a. From this we see, that the point of mazimum moment and
the center of gravity of all the wheels are equidistant from the center of
the beam and on opposite sides of the center. The maximum moment will
occur under one of the two wheels adjacent to the center of gravity,
practically always under the wheel nearer the center of gravity. So, as a
rule, all we have to do to locate a system of wheels on a beam for
maximum moment is to find their center of gravity and then place the
center of the beam half way between this center of gravity and the wheel
nearest to it. For example, take the case shown in Fig. 112. By taking
moments about either of the end wheels, either wheel D or E, we can
determine the center of gravity of all the wheels which comes nearest to
wheel C and hence the center of the beam will come half way between
the center of gravity and this wheel. In any case where the center of
gravity comes near half way between two wheels, the moment under each
of the wheels should be determined in order to ascertain which is the
maximum.

aL
aa)
180 STRUCTURAL ENGINEERING

Usually the only bending moment used in designing beams is the
absolute maximum referred to above, yet there are a few cases where the
maximum at other points, other than the point where the absolute maxi-
mum occurs, is needed, and which will occur when the wheels are in one
certain position in reference to the point considered.

Let AB (Fig. 113) represent a simple

z beam supporting a system of wheel loads

e l-é as shown, and let D be a point distance e
from A where the maximum bending mo-
2 x ment, due to these wheels as they move
' over the beam, is desired. Let W repre-
ac o dd +O O © 1B sent the weight of all of the wheels on the
D beam and 2 the distance of their center of

If Pow Rl gravity from B. Further, let P represent
Fig. 113 the weight of the wheels to the left of
point D and z the distance of their center

 

 

 

 

 

 

 

 

 

 

of gravity from D.

According to Art. 56, the maximum moment at D will occur when
a wheel is at that point. So let a wheel be at the point as shown. Now
taking moments about D and considering the forces to the left, we have

M="- ex ~Ps=(Re-Ps) sd aCe iw aloes aie 4igihns aw etal (1)
for the bending moment at that point. Now, if the wheels are in the
position for maximum moment at D, the increment of the moment that
would be caused by the wheels moving to the right or left an infinitesimal
distance would be equal to zero. Then assuming we have such a move-
ment, M, x, and z will have corresponding increments which we will
designate, respectively, as dM, dx, and dz. Now adding these in equation
(1), we have

Ww WwW
M+dM= Terr L edz —Pz—Pdz,

and subtracting equation (1), we have

W
dM= zed —Pdz.

But, as is readily seen, dzs=dx, so we have
W
dM= Tr edx—Pdx

for the increment of the bending moment, which will be equal to zero
when the loads are in the position for maximum moment at D. Then we
have

am =" edz —Pdzx=0.

Reducing, we have
MAXIMUM REACTIONS, ETC. 131

In the last equation we have the total load on the beam divided by
the length of the beam equal to the load on the left of the point D
divided by the distance from the point to the left support. That is, we
have the average unit load on the beam equal to the average unit load to
the left of the point. Then, to obtain the maximum moment at any point
in a simple beam, due to wheel loads, the loads must be so placed that
there will be a load at the point considered and at the same time the
average unit load on the entire beam will be equal to the average unit
load to the left of the point.

It is shown, in Art. 59, that the first derivative of the bending
moment, in terms of x, at any section of a beam, is equal to the shear.
But at the point of maximum moment the first derivative of the bending
moment will be equal to zero. So, it follows that the shear at the point
of maximum bending moment will be zero. Then, as the point of maxi-
mum bending moment is at the point of zero shear, or where the shear
changes signs, which is the same thing, the point of maximum moment
can be obtained by adding up the forces from one end of the beam until
the point where the shear is zero, or changes signs, is reached, and this
point will be the point of maximum bending moment. This method of
determining the point of maximum moment is quite convenient in the case
of fixed loads, especially where the loads are mixed, uniform, and
concentrated.

As an example, let it be required to determine the point of maximum
bending moment on the beam shown in Fig. 114, due to the concentrated
and uniform load indicated. After determining the reactions shown at
A and RB, by taking moments about the supports, we can begin at one: end
of the beam and ascertain the point of zero shear by adding up the forces

 

 
  

3’ 2° =a eS
%| y *) *
9
9 9 9 #
8 y % eoo ber tt

 

 

 

 

 

 

 

Aft {8
x
g
w
Q

7d.

 

Fig. 114

and noting the sign of the shear as we progress. Thus, beginning at A,
we have 10,766 — 1,800 — 3,000 =+5,966 lbs. for the shear just to the right
of the 3,000-pound load, and 10,766 - 3,000 —7,000=+766 Ibs. for the
shear just to the right of the 4,000-pound load. This last shear shows us
that the point of zero shear is just a little to the right of the 4,000-pound
load. Let 2 be the distance. Then we have

600 « = 766,
from which we obtain

_ 166

#= Bo =1.27 ft.
132 STRUCTURAL ENGINEERING

So the point of maximum moment is 1.27 ft. to the right of the 4,000-
pound load. If the 4,000-pound load were something over 766 pounds
heavier, it is evident that the point of maximum moment would be under
that load. For example, suppose that load weighed 6,000 pounds instead
of 4,000; then, all the other loads remaining the same, the reaction at A
would be 11,932 lbs. instead of 10,766. Adding up from A we have
11,932 — 3,000 — 3,000 =+5,932 Ibs. for the shear just to the left of the
second load from A, which is now 6,000 Ibs., and 11,932 — 3,000 — 3,000 —
6,000 =—68 lbs. for the shear just to the right of the load. As the shear
is positive on one side of the load and negative on the other, the shear
undoubtedly passes through zero at the load, and hence the point of
maximum moment is at the load.

89. Maximum Reaction on Simple Trusses.— The loads on trusses
are transmitted to the joints, either directly or indirectly; directly when
applied at the joints, and indirectly when applied between the joints,
that is, in the panels. As an example, let the diagram in Fig. 115

 

 

 

 

 

 

 

 

 

 

 

 

; L
a Zz
Pa era
| ih] kt.
l |
Pl ‘ f P2
Aj > C f fF B
ik ri ‘ras "73 TRI
Fig. 115

represent a truss of length L supporting two loads, P1 and P2, as shown.
These loads, as is readily seen, would be transmitted to the joints c, d,
and e. Let rl, r2, and r3 represent the amount transmitted to each as
indicated. These would be known as concentrations or panel loads. The
intensity of rl and r3 can be determined by taking moments about d and
treating cd and de as simple beams. Thus we have

rl= LL 3 pone’

cd de
By taking moments about e we can determine the concentration at d due
to P2, and by taking moments about ¢ we can determine the concentration

at d due to Pl. Then, by adding these two concentrations together, we
obtain the concentration r2._ Thus we have

cd-h de—k
r2=( cd )pr+( ae ) pa,

In this manner the concentrations on the joints of any truss, loaded in
any manner, can be determined. As the concentrations are really. com-
ponents of the loads, either may be used in determining reactions on
trusses. Thus, for the reaction at A (Fig. 115), due to the two loads
P1 and P2, we have

_2P1+yP2 ps (Be)r3 + (Bd)r2+(Be)r1,

L, L

 

 

 

R
MAXIMUM REACTIONS, ETC. 133

As a rule, the loads are used instead of the concentrations (owing to
convenience), in which case any concentration at the end point due to
loads in the end panel must always be subtracted from the reaction found.
For instance, if there were loads in the panel 4b, we would take moments
about b and determine the concentration at A due to the loads in that
panel, and subtract this from the reaction found at A by taking moments
of all the loads about B. Other than this one step, the determination of
reactions on trusses is exactly the same as for beams.

Dead load (which is usually taken as a uniform load) and uniform
live load are always considered as concentrated at the joints. In such
cases we deal only with joint loads, or panel loads, which is the same
thing. If the panels be of equal length, the panel loads will all be equal,
and in such cases the reactions can be determined practically by inspec-
tion. For example, let the diagram in Fig. 116 represent a truss of 6

 

 

 

 

 

 

 

B
b e d é Ff,
‘ \P P P Pp tip
R 4 3 2 4 Tk

5
Fig. 116

equal panels supporting a uniform. load of w pounds per foot of truss.
Let L be the length of the truss and d the length of each panel. If the
load extends over the full length of the span there would be five panel
loads, represented as P, each equal to wd. In that case we can readily
see that each reaction would be equal to 24 P. This, of course, is mere
inspection. In case the panel loads are unequal or some of the joints
not loaded, we can obtain the reactions by simply taking moments about
the supports just the same as in the case of beams, except as stated above,
but if the panels are of equal length it is more convenient to deal with
panel lengths instead of feet. For example, to obtain the reaction at A
due to a load at f, we would take moments about B and the load at f
would be multiplied by d and divided by 6d, and hence the reaction at
A would be 4 of the load. Similarly a load at e would be multiplied by
2d and divided by 6d, and hence the reaction at A, due to this load,
would be 2 of the load at e, and likewise the reaction at A, due to a load
at d, would be # of the load at d, and ¢ and % for a load at ¢ and },
respectively. From this it is seen that the reactions on trusses of equal
panels can be obtained directly by numbering the panels 1, 2, 3, ete., as
shown (just below the diagram) in Fig. 116, and multiplying the load on
each joint by the number under it, and then adding all the results together
and dividing by the number of panels in the truss. If the reaction at the
other support be desired, the numbers would simply be reversed in order.
If the panel loads are all equal, the work will be shortened, of course, as
the numbers under all the joints loaded can be added together and multi-
plied by one panel load, divided by the number of panels. As an example,
suppose the joints f, e, d, and c each support a load P. The reaction at
A due to these four loads would be

R=" 424344)=% 10.
134 STRUCTURAL ENGINEERING

If the joint b were loaded also, we would have

R= (1 $2484445)= 2 15,
If joints e, c, and b alone were loaded, we would have

ee

R=G (e+ 445)=¢ 11.

If joint ¢ alone were loaded, we would have
P P
Rao (= 54

and so on. In general, the maximum reaction on a truss, the same as a
beam, will occur when the span is fully loaded and the heaviest loads
are near the support considered. Specific cases will be taken up later.

90. Maximum Shear on Simple Trusses.— The determination of
the shear on trusses is the same as that of beams, except the loads are con-
sidered to be applied to trusses only at the panel points, or joints. Let
the diagram in Fig. 117 represent a simple truss supporting a load P at
each bottom joint, as indicated, and let R and #1 represent the reactions
due to these loads. Then the shear in the first panel from the left end is
R; in the second it is R-—P; in the third it is R-—2P; in the fourth
it is R-38P; and so on. This case is that of dead load or a uniform live

 

 

 

 

 

 

 

qR P P P P yP TR

Fig. 117

load extending all the way across the span. In case of dead load, this
shear would be a maximum. It is customary to assume that a uniform
live load moving over a truss increases by panel loads, in which case the
maximum shear in any panel will occur when the joints from one end of
the truss up to the panel considered are loaded. By loading from one
end we obtain, as we may say, the maximum positive shear, and loading
from the other end we obtain the maximum negative shear. For example,
one maximum shear would occur in the panel be of the truss shown in
Fig. 118, when the joints e, d, and ¢ alone are loaded, and the other
maximum would occur when the joints a and b alone are loaded. Instead
of considering the load to move on from the two directions, we usually
consider it to move on from only one direction, and by determining the
shear in each panel as the load reaches it we obtain all the shears we
need, as the positive shears are determined in the panels on one side of
the center of the truss while the negative shears are determined in the
corresponding panels on the other side of the center.

As the maximum shear in any panel, due to a uniform live load,
occurs when the load just reaches that panel, it is readily seen that this
shear, in all cases, will be equal to the reaction at the unloaded end of the
truss. Thus, the maximum shear in panel ed (Fig. 118) due to a uniform
live load moving on from the right, will be equal to the reaction at A due
MAXIMUM REACTIONS, ETC. ~ 135

to the load at e; and, similarly, the maximum shear in the panel de will
be equal to the reaction at A due to the loads at e and d; and the maxi-
mum shear in panel cb will be equal to the reaction at A due to the loads
at e, d, and c; and so on for the other panels. Now if the panels be of
equal length, these reactions can be determined by numbering the joints
1, 2,3, ete., and adding these together for the loaded joints and multiply-
ing the sum by one panel load divided by the total number of panels in
the truss, as explained in the last article. As an example, let the diagram
in Fig. 118 represent a truss of six equal panels, each of length d (the

 

 

 

 

 

 

 

d Zz
Afg. a6 ¢ é y
| Ss 4 3 2 /
® © @ |
L. L= 6d al
Fig. 118

length of the span then will be L=6d), and let it be required to deter-
mine the maximum shear in each panel due to a uniform live load of w
pounds per foot of truss moving over it from right to left. Let W (=wd)
represent the panel load. Then numbering the joints 1, 2, 3, ete., from
right to left, as shown, we have

Ww
GM
for the maximum shear in the panel ed, when joint e alone is loaded,
W
5 (1+2)
for panel de when joints e and d are loaded,
# (1eoe3)
for panel cb when joints e, d, and c are loaded,
* (14+24+344)
for panel ba when joints e, d, c, and b are loaded, and
iF
=a +24+34+4+5)
for panel aA when joints e, d, c, b, and a are loaded.
From this it is seen that the maximum shear on any truss of equal

panels, due to a uniform live load ‘moving over the truss, can be expressed
by the general formula,

where n is the number of panels in the truss considered.
It will be found quite convenient to write the summation of the
136 STRUCTURAL ENGINEERING

numbers in the parentheses under the corresponding panel points as
shown in Fig. 118 where the incircled numbers are the respective summa-
tions. These summations, as is readily seen, can be made in any case by
simply referring directly to the diagram of the truss considered. Then
the shear in the panels can be read off of a slide rule directly as,

W Ww

Ww W
ae =, 15);

¥ (6),

n

.

and so on.

In the case of uniform live load, just considered, we usually assume,
as stated above, that the load is applied to the joints of a truss in maxi-
mum panel loads. This is not possible, however, as such loads are con-
tinuous and must necessarily extend beyond the last loaded joint in order
to fully load it. For example, to fully load the joints e and d (Fig. 118)
the uniform live load would really have to extend from B to the joint c,
as is readily seen. However, as will be shown later, the maximum shear
would occur in panel de when the load extended a short distance z beyond
joint d. In that case there would be less than a full panel load at d and
a small concentration at c due to the load in the panel dc, and the shear
in the panel de would really be equal to the reaction at A minus the
concentration atc. The error resulting from the assumption stated above
is small in the case of a uniform live load, as will be seen later, and hence
the assumption in that case is permissible, but in the case of wheel loads
no such assumption can be made as any slight shifting of the loads may
materially change the shear in a panel and consequently it is necessary
for us to determine the exact position of such loads for maximum shear.

As a general case, let the diagram in Fig. 119 represent a truss

x
d t
Zz |
Af 5 Q POO od OODOONS
R d é F
T [3 Tri

 

 

 

 

 

 

 

 

 

W
Le L= ed ra
Fig. 119

supporting a system of wheel loads. Now let it be required to determine
the position of these wheels for maximum shear in panel bc. Let W be
the weight of all the wheels and x the distance from B to their center of
gravity, and let P be the weight of the loads in the panel be, and g the
distance from c to their center of gravity. Now if & be the reaction at A
due to W, and r the concentration at b due to P, the shear in the panel
be can be expressed as
MAXIMUM REACTIONS, ETC. 137

Ihe a toda te aco Jase Ge dee avasgy etans ae eke pia eve le Sisawe Siete Ja
We Pz
But R= >- and De
Then substituting these values, we have
We Ps
S= To ees (2)

With P and W constant, any shifting of the wheels to the right or left
will affect R and r, not the same, but similarly. The weight of P could
be such that any movement to the left would increase the value of R
more than r. Then, evidently, any such movements to the left would
continue to increase the shear § until another load rolled past c into the
panel bc, when the rate of increase of r would be greater than just before
the load passed joint c; but R would continue to increase all the while,
and, consequently, the shear S would continue to increase, unless the load
P were increased so much by the additional load that r would be increased
at a greater rate than R for each movement to the left. So it is seen that
the shear S will be a maximum when the total load in the panel be is such
that the increment of R is equal to the increment of r. That is, any
shifting of the loads with W and P constant (for that instant) would not
affect the shear S in panel bc; or, in other words, the increment of the
shear would be zero. Now, suppose the loads move an infinitesimal
distance to the left; then equation (2) would become

We Wdxr Pz Pdz

 

S+dS=——+ aa Tg (dee),
and subtracting (2) from this, we have
Wdz Pdz
dS = L = “"_ Cy (3),

which is the increment of the shear S in panel bc due to the movement.
But this increment will be zero when S is a maximum, and then we
would have
Wdx Pdx
= = EN a Naina a eS eae oes 4),
dS =—, 7a (4)

It is readily seen that this occurs when W/L = P/d, that is, the shear in
panel be is a maximum when the average unit load on the truss is equal to
the average unit load in the panel. This will hold in the case of any truss
of the type shown as the case being considered is general. In the case of
a truss of » equal panels, each of length d, we have L=nd. Substituting
this value of L in the last equation, and reducing, we have

 

That is, the shear in any panel of a truss of n equal panels will be a
maximum when the load in the panel is equal to the total load on the
truss divided by the number of panels in the truss. This will hold for
any class of live load either concentrated or uniform.

It is readily seen that the increment of the shear in panel bc,
expressed by equation (3), will pass through zero only as a load passes
138 STRUCTURAL ENGINEERING

joint c, so there will be a wheel at joint ¢ when the maximum shear in
panel be occurs. Thus it is in all cases—a wheel will be at the joint on
the loaded side of the panel considered when the maximum shear in the
panel occurs.

91. Maximum Bending Moments on Simple Trusses.—In the
case of bending moments on trusses, the moments are usually taken about
the joints only, otherwise the case is the same as that of beams. As a gen-
eral case of dead load or uniform live load, let the diagram in Fig. 120
represent a truss of six panels supporting a load at each bottom joint as

 

 

 

 

 

B
4, ie try

 

 

 

 

Fig. 120

indicated. Let R and R1 represent the reaction at A and B, respectively,
due to these loads. Then the bending moment at joint b or g is M’=Rz;
at joint c or h it is M’=R(e+y)—Ply; at joint & or d it is M”’=
R(e+y+2)—-P1(y+2)-P2(z); and so on.

In the case of uniform live load, the maximum moment at any joint
will oceur when the load extends the full length of the truss, and hence
the moment at all joints, the same as in the case of dead load, is a
maximum. In the case of wheel loads it is necessary to determine the
exact position of the wheels for the maximum bending moment at each
joint. Let the diagram in Fig. 121 represent a truss of six panels
supporting a system of whcel loads, as indicated. Now let it be required
to determine the position of these wheels for maximum bending moment
at joint d.

Let W be the weight of all the wheels on the truss, and 2 the
distance from B to their center of gravity, and let P be the weight of the

 

 

 

 

 

 

 

 

 

 

 

 

z
oO lod Ovo O1O
A R 8 «| d é F 8
P =e Rl
LA) Pw
L
Fig. 121

wheels to the left of joint d and z the distance from d to their center of
gravity, and also let k be the distance from A to d and R the reaction
at A due to all the wheels when in any position. Then the bending
moment at d can be expressed as
MAXIMUM REACTIONS, ETC. 139

We
But =
u R L
so, substituting this value of R, we have
We
M=k-~>——- Ps abate arias Sele er anecasondeate wells (PRES Sgeeane Ss (2).

Now as the loads move to the left, R will be increased, and likewise
the moment M, until P becomes so great that the rate of increase of Pz
is greater than the rate of increase of (Wx /L)k; then, evidently, the
moment M at d will be a maximum when its increment is zero, which, as is
readily seen, will occur when the increment of (Wx /L)k is equal to the
increment of Pz. Now suppose the loads move an infinitesimal distance to
the left, then equation (2) becomes

 

We Wde
M+dM=k a +k a, 7 Ps - Pda
Then subtracting (2) we have
BN hse sess east ate ett (3),

which is the increment of the bending moment at d due to the movement.
But this increment will be equal to zero when M is a maximum, and hence
we would have

 

am=k™* _pdo=0 husmeaemeudtiee Raintsebetyoas (4).
It is readily seen that this occurs when

Ww oP

Prk ttt pee atents iisbuerecGamaw es (5),

that is, the bending moment at joint d is a maximum when the average
unit load on the truss is equal to the average unit load on the left of the
joint d. Now, as the above is a general case, we have: The mazimum
bending at any joint of a truss of the type shown will occur when the
average unit load on the truss is equal to the average unit load to the left
of the joint considered. It will be seen that this is the same as found for
beams in Art. 88.

92. Stresses in Trusses.—As the loads supported by trusses are
applied at the joints, the stress produced in each member is simple
stress, and consequently the unit stress in any member is equal to the
total stress divided by the area of its cross section. The stresses in the
individual members can readily be obtained by resorting to the resolution
of forces and to the equation of moments.

As an example, let the diagram at (a) in Fig. 122 represent a truss
of six equal panels, supporting a load at each lower joint as indicatec.
Let d be the length of each panel, & the height of the truss center to
center of chords, L the total length of span center to center of end
bearings, and let R and R1 represent the reactions as indicated. To
140 STRUCTURAL ENGINEERING

obtain the stress in the end post LOU1 and in the bottom chord LOLI,
let S and S1 be the stress in each, respectively, and suppose the two
members cut off along the section mm, and imagine the part of the truss
to the left of this section moved bodily to (b) without any stresses or
forces being affected in the least. Then we simply have an independent

 

 

 

 

 

 

 

 

 

 

 

+ me
mn a
wr | { 2 | us =e
f p
|
<0 “ at ty LS a “Tr
al ja Bm fi ir ee
! ‘yy mu
=z 26
en S44 hoy)
Z fut |
1 1 s2
! Sx tC)
LO R The
!
sue.
S7
Lo . Z iz
PI
'
‘ ul U2 s/0
, lh (F)
$84
oe oT #33
P i)
Fig. 122

structure at (b) held in equilibrium by the three external forces R, S,
and S1—the stresses § and S1 being unknown. Now, as these three
forces are in equilibrium, the algebraic sum of their components along
any direction is equal to zero. So resolving the forces vertically, we have

= SC080-= 0g rea solani as eee wee alee alee ee (1),
from which we obtain

S=R/cosd=Rsecb ....... Mie aeumoteudeus aries (2);
and resolving the forces horizontally, we have

SL Ssind = 0) ea wield ates div weve seater nuadie ead Re dale eevee (3),

from which we obtain

NSS el eae ate ace eg ter aes ane (4).
MAXIMUM REACTIONS, ETC. 141

But, according to (2), S=R/cosé. Then, substituting in (4), we have

sind
SI=R — Vg = Rtand Se aioe es eae waves aahateioteue al omrimand (5).

Equation (2) shows that the stress § in LOU1 is equal to the shear
(=) in the end panel multiplied by the secant of the angle 6 which the
member makes with the vertical; and equation (5) shows that the stress
S1 in LOL1 is equal to the shear in the end panel multiplied by the
tangent of angle 6. It should be observed that R has no horizontal com-
ponent while $1 has no vertical—the cosine of 90° being zero.

Again referring to the figure at (b), by taking moments about any
point O on the line of action of S, we have

Rea -2zS1=0,
from which we obtain

S1=R2.-
2

But it is obvious that it is more convenient to take moments about the
panel point U1, for in that case the lever arms are directly known, and
we have

Rd-hS1=0,

from which we obtain
si=R4 =Rtand disap Sp teletdal ey Gece: ice Bae: Se SaoBie AGRA aI Tree eae (6).

The stress S could be computed by taking moments about a point in the
line of action of the stress S1, but it is obvious that S can be computed
more readily from equation (2).

It is also obvious that the stress in the hanger U1L1 is equal to the
load P which it supports directly at L1.

To obtain the stress in the chords U1U2, L1L2, and diagonal U1L2,
let S2, 83, and S4 be the stress in each, respectively, and suppose the
three members cut off along the section m’m’, and imagine the part of
the truss to the left of this section moved bodily to (c) without any
stresses or forces being affected in the least. Then we have an inde-
pendent structure at (c) held in equilibrium by the external forces KR, P,
S2, 83, and S4—the stresses $2, $3, and $4 being unknown.

As S2 and S4 intersect at Ul, S3 can be determined by taking
moments about that point. So taking U1 as the center of moments
we have

Rd-hS3=0,
from which we obtain
d
S3=R RTE neta (7),

which is the same as we obtained for the stress in LOL1 in equation (6).
As S4 and S3 intersect at L2, S2 can be determined by taking moments
about L2. Therefore, taking L2 as the center of moments, we have

2dR —dP -hS2=0,
142 STRUCTURAL ENGINEERING

from which we obtain

d d

= 2 ee oP pssdalvignd wai eee wd Ge ktie es sane ae ees 8).
S2=2 k R k P (8)
$4, the stress in the diagonal U1L2, can be determined most readily
from the algebraic sum of the vertical components of the five forces, for
which we have '
R-P-S4cos6=0,

from which we obtain
(R-P) _

cosé

This shows that the stress S4 in the diagonal U1L2 is equal to the shear
in the panel L1L2 multiplied by the secant of angle 6.

To obtain the stress in the vertical post U2L2, let S6 be the stress,
and suppose the three. members U1U2, U2L2, and L2L3 be cut off along
the section mm”, and imagine the part of the truss to the left of this
section moved bodily to (e¢) without any forces or stresses being affected
in the least. Then we have an independent structure at (e) held in
equilibrium by six external forces—R, $7, S6, S5, P, and Pl. The
stresses S7%, S6, and S5 may all be unknown. As S7 and S5 have no
vertical components (being horizontal members), S6 can be determined
directly from the algebraic sum of the vertical components of all the
forces, for which we have

S4= (ea Py 868 race eee eis es (9).

R-2P-S6=0,
from which we obtain
SO S00 = 2D eeu ey wus a Wave ese we ORAS VA soe (10).

This shows that $6 is equal to the shear in the panel L2L8.

To obtain the stress in the chords U2U3, L2L3, and the diagonal
U2L3, let $10, $8, and $9 be the stress in each, respectively, and sup-
pose the three members cut off along the section m”’m’’’, and imagine
the part of the truss to the left of this section moved bodily to (f) with-
out any forces or stresses being affected in the least. Then we have an
independent structure at (f) held in equilibrium by six forces, R, S10,
S9, 88, P, and P1, the stresses S10, 89, and S8 being unknown. Taking
moments about U2, we have

2dk -dP -hS8=0,
from which we obtain

2dR - dP
Se ara eaaTaRteaaes (11).

Then taking moments about L3, we have

3d — 2dP -dP1-hS10=0,

from which we obtain

d
S105 ak -EP— Pl) sraeiwiaimaieanveweevavenigess (12).
MAXIMUM REACTIONS, ETC. 143

Resolving the forces vertically, we have
R-P-P1-S9cosé=0,
from which we obtain

_R-P-P1
~  cos6

89 =secO(R-P-P1) ........0005. ilies (13).

The above loads do not produce any stress in the post U3L3. This
member could be omitted as far as loads applied at the lower panel points
are concerned, but it will support directly any load applied at U3.

Stress in the members to the right of U3L3 can be determined in
the same manner as shown above for the members to the left.

The above shows how readily the stresses in a truss can be deter-
mined by considering portions of the truss as independent structures, the
portion being selected each time to suit the case considered; the portion
may be a single joint or several panels of a structure. This manner of
treatment is known as the “Method of Section.” It will apply in the case
of any truss and may be carried to any desired extent, but after one
becomes familiar with a certain type of truss, the analysis is often made
without any thought of the method of section, as will be seen later.
However, the method is really implied.

Problem 1. Determine the stresses in the members of the truss
shown at (a), Fig. 120, due to a load of 20,000 lbs., on each lower panel
point, assuming the length of each panel to be 25’-0”, and the height of
the truss 30’-0”, center to center of chords.
CHAPTER VIII
GRAPHIC STATICS

93. Definition and Limitation of Graphic Statics.— Graphic
Statics is that part of Mechanics which treats of the graphical determina-
tion of static forces. It results from the fact that forces acting upon a
body in equilibrium will form a closed polygon (see Art. 42) wherein all
the forces act in the same direction around the polygon. This knowledge
applied to geometrical construction constitutes the method.

A force is fully known when its point of application, its intensity,
the direction and location of its line of action, and its direction of action
(along the line of action) are known.

The determination of the point of application does not come within
the scope of Graphic Statics, for it is one of the many things that can be
ascertained only from knowledge of actual conditions, and consequently
is a presupposed knowledge in graphic problems. The intensity, direc-
tion, and location of the line of action and the direction of action can be
graphically determined. However, the method is practically limited to
the two following cases:

Case I. The intensity and direction of action of two forces can be
graphically determined provided all of the other forces acting upon the
same body are fully known and the lines of action of the two are known.
This will hold in the case of one unknown force as well as for two.

Case II. The intensity, direction of action, and the direction and
the location of the line of action of one force can be graphically deter-
mined provided all of the other forces acting upon the same body are
fully known.

Concisely expressed, the intensity and direction of action of two
forces can be found when the lines of action of all the forces are known
and the intensity and direction of action of all except those two are
known; and the intensity, direction of action, and the direction and
location of the line of action of one force can, be found when all the other
forces are known. As a rule, the intensity and direction of action are all
that are desired.

94. Preliminary Application —Let AB, at (a), Fig. 123, represent
a body in equilibrium under the action of the five forces PI—P5. Sup-
pose all of the forces are known except P4 and P5, which are unknown
in intensity and direction of action only, which means their lines of action
are known, so that the problem comes under Case I.

To determine their intensity and direction of action, select some
convenient scale and lay off AB at (b), equal and parallel to P1; and
from B lay off BC equal and parallel to P2; and from C lay off CD equal
and parallel to P3. Then, to complete the polygon, P4 and P5 must
close on D and A. So from D draw Dy parallel to P4 and from 4 draw
Az parallel to P5, intersecting Dy at E. Then the length of the lines
DE and AE represents the intensity of P4 and P5, respectively. Now, as

144
GRAPHIC STATICS 45

all of the forces will act in the same direction around the polygon. P4
will act in the direction from D to E and P5 from E to A, as the arrow
points indicate. So we have their direction of action, which is toward
the body in each case. Then the forces P4 and P5 are fully known. The
two forces P4 and P5 could close on A and D, as AO and DO, just as
well as DE and AE. KEither is correct.

It is usually convenient to go
around a body laying off the forces
in consecutive order as was done at
(b), but in some cases it is not pos-
sible to do so, and in fact it is not
at all necessary. For example, in
the above case we could lay off the
forces as they are shown at (c) by
laying off P1 first, then P3, then P2,
and close on K and H with P4 and
P5; or we could lay off either P2 or
P3 first, just so the forces are joined
so that they act in the same direc-
tion around the polygon, which is the
important thing to watch. However,
any mistake in that respect is readily
detected. For instance, suppose we
were to lay off Pl as AB (at d) and
then lay off P3 from 4; we could
readily see that the forces would not Fig. 128
be acting in the same direction
around the polygon being constructed, and consequently the polygon
would not be correct.

Now suppose P2 and P4 at (a) are the two forces unknown in
intensity and direction of action instead of P4 and P5 as considered
above. In that case there would be a known force in between the two
unknown forces, and consequently it would be impossible to lay off the
forces in consecutive ‘order as we pass around the body. However, the
case presents no trouble at all, as we can simply lay off the known forces
in any order, so long as they are placed so that they all act in the same
direction around the polygon. For example, to determine the intensity
and direction of action of P2 and P4, we could lay off P5 as AB at (e),
then from B lay off P1 as BC, and from C lay off P3 as CD, then closing
on 4 and D with P4 and P2 by drawing Dz parallel to P4 and Ay
parallel to P2, intersecting Dx at N. Then DN and NA represent the
intensity of P4 and P2, respectively, and following around the polygon
from A to B, C, etc., it is readily seen that P4 acts from D to N, and
P2 from N to A.

It is evident that if only one of the five forces at (a2) were unknown
in intensity and direction of action, our problem would be quite simple.
For example, suppose P5 were the only unknown force. We would simply
construct a polygon as ABCDE, as shown at (6), and join E to A and
P5 would thus be determined in intensity and direction of action.

As an example, under Case II, suppose all of the forces shown at
(a), Fig. 123, are completely known except P5, which we will assume is

 
146 STRUCTURAL ENGINEERING

completely unknown. By constructing the force polygon ABCDEA -at
(f) we have P5 given in intensity, direction, and direction of action by
the line EA, but the location of its line of action is unknown. To locate
its line of action select any point O, at (f), as a pole, and draw the ray
diagram as shown, and construct at (a) the equilibrium polygon fabcdf’
(as explained in Art. 45). Then prolonging the segments df’ and fa
until they intersect at J and through J drawing Im parallel to EA, we
have the line of action of the resultant of the four forces P1, P2, P3,
and P4, Now, as all five of the forces are in equilibrium, the resultant
just found must balance P5, which requires that P5 must act along the
same line as the resultant. (See Art. 42.) Therefore, the line Im is
the line of action of P5, and thus P5 is fully determined. P5 may be
applied upon either side of the body as far as equilibrium is concerned.

Referring to the force polygon at (b), Fig. 123, it is evident that
if P4 and P5 were parallel, a straight line joining A and D would
represent the direction and intensity of both, but it would be impossible

c

 

 

 

Fig. 124

to determine the intensity of either directly from the polygon, as there
would be no break in the line indicating where the two join. However,
their intensities are readily determined by the aid of the equilibrium
polygon. For example, let AB, at (a), Fig. 124, represent a body held
in equilibrium by the five forces P1...P5 as indicated. Let P1 and P5
be the two unknown parallel forces. By constructing the force polygon
ABCDA, at (b), we have the combined intensities of the two given by the
line DA. Next select any point as O, at (b), as a pole and draw the ray
diagram ABCDOA. Then, beginning at any point on the line of action
of one of the known forces, say at b on the line of action of P2, construct
the equilibrium polygon fbcdf’, as explained in Art. 45. Then the three
known forces P2, P3, and P4 are replaced by components, all’of which
are balanced except f at b and f’ at d. As the next step, prolong the
GRAPHIC STATICS 147

segment bf from 6 until it intersects the line of action of the unknown
force P1 at a, and likewise prolong the segment df’ from d until it
intersects the line of action of the other unknown force P5 at e. In order
to extend the equilibrium polygon farther, it is necessary to resolve P1
into two components such that one of them will be equal and opposite to
the known component f at b, and P5 into two components such that one
of them will be equal and opposite to the known component f’ at d. This
could be readily accomplished if the forces P1 and P5 were known, but
as they are unknown we have only one component in each case to start
with; that is, f at a and f’ at e.

But as the five forces are in equilibrium, their components will be in
equilibrium. Therefore, the unknown component f” of P1 must balance
the unknown component of P5, and consequently they will act along the
same line. So, by drawing the line ea, we have the line of action of the
two unknown components, as this is the only line that both could act
along. Now at e we have one known force f’ and the direction of the
two unknown forces P5 and f”, each of which can be graphically deter-
mined by drawing the force triangle DEO as shown at (c). P1 and f”
at a can be graphically determined in the same way. But it is not neces-
sary to construct the triangle at (c) as f’ is given in the diagram at (b)
by the line OD, and by drawing OE parallel to the line ae we would have
the same forces given by the equal triangle ODE at (b), and the forces
at a by the triangle OFA, as indicated; and thus P1 and P5 are fully
determined as EA and ED, respectively, in the diagram at (6b).

Referring to Fig. 124, it will be observed that the equilibrium
polygon closes, which always occurs in the case of any system of forces
in equilibrium. The line ea is known as the closing line.

95. Graphical Determination
of Reactions and Bending Mo-
ments on Simple Beams.—Let AB,
at (a), Fig. 125, represent a simple
beam supporting the four vertical
loads P1...P4 as indicated, and sup-
pose the reactions R and #1, due to
these loads, are unknown in intensity.
First lay off the load line AB as
shown at (b). Then select any point
O as a pole and draw the ray dia-
gram as shown and construct the
equilibrium polygon A’abcdB’ at (c).
Then from O draw the line OE par-
allel to the closing line A’B’, and,
according to the last article, we have
R given by the line AE and R1 by
the line BE, and thus the reactions
are determined.

Conceive of the closing line A’B’ Fig. 125
at (c) as being a beam and the
other part of the equilibrium polygon as a rope suspended from the ends
of this beam, and imagine the loads supported upon the rope as indicated.
Then the equilibrium polygon really becomes a structure upon which the

 

 

 
148 STRUCTURAL ENGINEERING

bending moment at any vertical section will be the same as for the beam,
and consequently it may be treated instead of the beam for the purpose
of finding these moments.

Then to obtain the bending moment at the section mm of the beam,
extend this section on down to the equilibrium polygon, cutting it off
along the section ee’. Then imagine the part of the equilibrium polygon
to the left of this section moved bodily to (d) without any loads or forces
being affected in the least. Then taking moments about e, we have

aR -bP1-cP2=M.

Now, this moment is balanced by the moment of f2 about e, then we have

M=d(f2),
and taking moments about ec’, we have
M=k(f5).

Both f2 and f5 (f2 being the stress in the rope from b to ¢, and f5 the
stress in the beam throughout) are given in the ray diagram, and d and k
can be ascertained by scale. A better way to obtain the moment is to
resolve f2 and f5 into horizontal and vertical components as shown, and
then taking moments at either e or e’, we have

M=Hh.

HI is the same for all segments, as is seen from the ray diagram at (b),
and is known as the “pole distance.’ Then, evidently, the bending
moment at any vertical section of the beam is equal to this pole distance
(H) multiplied by the vertical ordinate between the closing line and the
broken line of the equilibrium polygon at the same vertical section. For
example, the bending moment at the section m’m’ is equal to Hy, at
mm’, Hz, and so on. The ordinates are measured in feet or inches to
the same scale as the beam, while H is in pounds, so many thousand
pounds per inch. In laying out the ray diagram, H can always be taken
as some convenient number, as 100,000 Ibs.

96. Graphical Analysis of Trusses consists in graphically deter-
mining the reactions and stresses. The reactions are determined by
means of the equilibrium polygon, while the stresses are determined as
concurrent forces at the joints by means of force polygons, one polygon
for each joint, the joints being considered in the order that the applica-
tion of the method requires, being usually the consecutive order.

Example 1. Let the diagram at (a), Fig. 126, represent a truss
supporting the three loads P1, P2, and P3, as indicated, and let R and
F1 represent the reaction at A and B, respectively, due to the three loads.

Let it be required to determine the reactions on the above truss and
the stresses in the members due to the three loads P1, P2, and P3.

To determine the reactions, first construct the ray diagram at (b),
which is accomplished by laying off the line 1-2 to any convenient scale,
equal and parallel to P1, and line 2-3 equal and parallel to P2, and line
3-4 equal and parallel to P3, thus obtaining the load line 1-2-3-4, and
then taking any point O as a pole and drawing the rays 1-0, 2-0, ete.
Then construct the equilibrium polygon CghkD, which is accomplished
GRAPHIC STATICS 149

by taking any point on the line of action of one of the forces, say point
g, on the line of action of P1, and resolving P1 into two components by
drawing Cg parallel to the ray 1-O and gh parallel to 2-O and virtually
resolving each of the other two loads into two components by drawing hk
parallel to 3-O and kD parallel to 4-O, and drawing the closing line CD

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 126

for completion. Then by drawing OF parallel to the closing line CD,
we have the reaction R given by the line 1-E and R1 by the line 4-E.
These lines can be scaled and thus the reactions will be fully known as
we know their direction of action will be upward, in order to balance the
loads.

After the reactions are determined all of the external forces are
known and we can proceed to determine the stresses in the truss members.

Any joint of any truss can be considered as being a small body
acted upon by the truss members, meeting at the joint, and any load
there applied. The details should be such that these will really be con-
current forces in every case and are so considered in graphic analysis.

Then at joint 4 we have three concurrent forces: the reaction R, the
force exerted by the member Ac, and the one exerted by the member Ab.
At joint c we have four concurrent forces, the load P1 and one for each
of the three members meeting at that joint; at joint b we have four con-
current forces, one for each of the members meeting at that joint; at joint
d we have five concurrent forces, the load P2 and one for each of the
150 STRUCTURAL ENGINEERING

members meeting at that joint; and likewise at joints e, B, and f we
-have, respectively, four, three, and four concurrent forces, as is readily
seen.

The intensity of the force exerted by any member on a joint will be
equal to the stress in the member exerting the force, and if the force acts
toward the joint the member exerting it will evidently be in compression
and hence the stress in the member will be compression, while if the force
acts away from the joint the member exerting it will be in tension and
hence the stress in the member will be tension. So if the force exerted
at a joint by a member be known, the stress in the member will be known.

Now it is evident that the graphical determination of the stresses
in the truss members is simply a matter of determining the concurrent
forces at each joint, which we can do by simply drawing a force polygon
for each joint. However, we cannot begin with just any joint, as the
application of the method is limited, as stated in Art. 86. It will be seen
upon inspection of the diagram at (a) that the only joints that can be
analyzed at the beginning are joints 4 and B. Here we have, in each
case, all forces known in direction and only two unknown in intensity and
direction of action. So we have simply Case I, Art. 86. Then let us take
joint A to begin with:

Laying off 1-2 at (c), to any convenient scale, equal and parallel to
R, and drawing 2-3 parallel to the member Ac and 1-3 parallel to the
member Ab, we have the force diagram 1-2-3-1 representing the concur-
rent forces at joint A, where 2-3 represents the force exerted by the
member Ac, and 1-3 represents the force exerted by the member Ab.
By scaling 2-3 and 1-3 we obtain the intensity of these forces, respect-
ively. We know that R acts upward and hence will act from 1 to 2 in
the force polygon. Then as all of the forces must act in the same direc~
tion around the polygon, we have the force exerted by Ac acting from
2 to 3 (in the polygon) and the one exerted by 4b acting from 3 to 1.
Now transferring these directions to joint 4 we have the force exerted
by the member 4c acting away from the joint and the force exerted by
the member 4b acting toward it, as indicated. Then the stress in Ac will
be tension and the stress in 4b compression, and as the intensity of the
first is given by the line 2-3, and the intensity of the second by. the line
1-3, in the force polygon, we have the stresses in the two members Ac
and Ab fully determined.

Knowing the force exerted by the member Ac at joint 4, we know
the force it exerts at joint c, as the two will, undoubtedly, be equal and
opposite, and hence the one at ¢ will act away from the joint, as indicated,
and likewise, knowing the force exerted at joint 4 by the member Ab,
we know the force it exerts at joint b, which acts toward b.

Joint b cannot be analyzed as yet for there are still three unknown
forces acting at that joint, but joint c can be, as there are only two
unknown forces there. Then laying off 2-3 at (d) equal and parallel to
the line 2-3 at (c) and 2-4 equal and parallel to Pl and drawing 4-5
parallel to the member cd and 3-5 parallel to cb, we have the force
polygon 2-3-5-4-2 representing the forces at joint c, where the force
exerted by the member cd is represented by the line 4-5, and the force
exerted by the member cb is represented by the line 3-5, which is equiva-
lent to saying that the intensity of the stress in cd is given by the line 4-5,
GRAPHIC STATICS 151

and the intensity of the stress in cb is given by the line 3-5, and by
scaling these lines we have the intensity of the stress in each of the two
members. As the force exerted at joint c by the member Ac acts to the
left from the joint, it will act from 3 to 2 in the polygon at (d), and as
the other forces must act in the same direction around the polygon, it is
readily seen that the forces exerted by the members cd and cb both act
away from the joint, and, hence, the stress in each of the two members
will be tension and thus we have the stresses in the members cd and cb
fully determined.

Now as the force exerted at c by the member cb is fully determined,
the force exerted at b by the same member is known and acts away from
the joint, as indicated. Then we have only two unknown forces at joint b
and hence the joint can now be analyzed. Then laying off 1-3 at (e),
equal and parallel to the line 1-3 at (c), and 3-5 equal and parallel to
the line 3-5 at (d), and drawing 5-6 parallel to the member bd, and 1-6
parallel to the member be, we have the force polygon 1-3-5-6-1 repre-
senting the forces at joint b, where 5-6 and 1-6 represent the intensity of
the force exerted by the member bd and be, respectively. Then by
scaling these lines we obtain the intensity of the stress in the members
bd and be.

Now the force exerted upon the joint b by the member Ab, as
previously stated, acts toward the joint and the force exerted by cb acts
away from the joint. Then transferring these directions to the polygon
at (e), it is readily seen that the forces will act around the polygon, as
indicated, and that the force exerted by the member bd acts in the polygon
from 5 to 6, while the force exerted by the member be acts from 6 to 1,
and transferring these directions to joint b, we have the first acting away
from the joint and the second toward it, hence the stress in bd is tension
while the stress in be is compression, and thus we have the stresses in the
two members bd and be fully determined.

The force exerted at b by the member bd being determined, the force
exerted by the same member at d is fully known and the same is true of
the force exerted at d by the member ed. Then there are but two unknown
forces at d and hence the joint can be analyzed by drawing the force
polygon shown at (f) and thus the stresses in the members de and df can
be determined. Then next the joints f and e can be analyzed in the same
way and thus the stresses in all the members in the truss can be deter-
mined, but one continuous diagram as shown at (h), wherein the forces at
each joint and likewise the stress in each member are represented, can be
constructed more readily than the separate diagrams, providing we pass
around all the joints in the same direction. Either direction, clock-wise
or counter clock-wise, can be taken for the first joint analyzed, but when
the direction is once taken we should pass around each of the other joints
in the same direction, for otherwise there is apt to be confusion.

For convenience, let S, S1, S2, and so on, represent the stresses in
the various members of the truss, as indicated at (a).

To construct the continuous diagram at (h), let us begin at joint 4,
as before, and pass around the joint counter clock-wise. Beginning with
the known force R we lay off 1-2 at (hk) to any convenient scale, equal
and parallel to R. The next force we come to is the one exerted by the
member 4c, So from 2 draw a line 2-x parallel to the member 4c. Then
152 STRUCTURAL ENGINEERING

the next force we come to is the one exerted by the member Ab. So from
1 draw a line parallel to this member, intersecting the line 2-« at 3, and
we have the force polygon 1-2-3-1 representing the forces at joint A,
where the line 2-3 represents the intensity of the stress S in the member
Ac and the line 1-3 represents the intensity of the stress S1 in the
member Ab. As R acts upward the force exerted by 4c acts from 2 to 3
(in the polygon), and the force exerted by the member Ab acts from
3 to 1,—all in the same direction around the polygon. Transferring these
directions to joint A, we see that S is tension and $1 compression, as
explained above, and thus we have S and S§1 fully determined. Passing
on to joint c, we have the force exerted by the member Ac represented by
the line 2-3 and acting from 3 to 2 as S is tension. The next force we
come. to, passing around the joint counter clock-wise, is P1, which we lay
off downward as 2-4. Then from 4 draw a line 4-z parallel to the
member cd, and from 3 draw a line parallel to the member cb, and we
have the force polygon 3-2-4-5-3, representing the forces at joint c, where
the line 4-5 represents the intensity of the stress S2 in the member cd,
and the line 5-3 represents the intensity of the stress S3 in the member
cb. As the force exerted by the member 4c acts (in the diagram) from
3 to 2, and Pl from 2 to 4, the force exerted by the member cd will act
from 4 to 5 and the force exerted by the member cb will act from 5 to 3.
Transferring these directions to joint c, we have the force exerted by the
member cd acting away from the joint and the force exerted by the
member cb acting away from it also. Then $2 and S3 are both tension
and thus we have both fully determined. At joint b the force exerted by
the member 4b is known, as its intensity was determined at joint A, and
its direction of action at b will be in the opposite direction to that of the
equal force exerted by the same member at A, and hence toward the joint
b, as indicated, and the force exerted at b by the member cd is known as
its intensity was determined at joint c, and its direction of action will be
in the opposite direction to that of the equal force exerted by the same
member at joint c, and hence away from joint b. Then only the forces
exerted by the members bd and be remain to be determined at joint b.
Beginning with the force exerted by the member 4b we have its intensity
represented by the line 1-3 and it acts from 1 to 3 in reference to joint b.
Passing on around the joint counter clock-wise, we have next the force
exerted by the member cb, the intensity of which is represented by the
line 5-3, and it acts from 3 to 5. The next force we come to is the one
exerted by the member bd. So from 5 draw a line 5-r parallel to the
member bd and close the polygon by drawing a line from 1 parallel to
the member be, intersecting the line 5-r at 6, and we have the polygon
1-3-5-6-1 representing the forces at joint b. As the force exerted by the
member Ab acts (in the polygon) from 1 to 3 and the one exerted by cb
acts from 3 to 5, the force exerted by the member bd will act from 5 to 6,
and the one exerted by the member be from 6 to 1. Transferring these
directions to joint b we have the force exerted by bd acting away from
the joint and the force exerted by be acting toward it. Hence the stress
S84 in bd is tension and the stress $5 in be is compression, and as the
intensity of S4 is given by the line 5-6 and that of S5 by the line 6-1, we
have the stress in the members bd and be fully determined.

Passing on now to joint d we have the forces exerted by the members
GRAPHIC STATICS 153

bd and cd and also the load P1 known, to determine the forces exerted by
the members de and df. Beginning with the force exerted by the member
bd, and passing around the joint counter clock-wise, we have that force
represented by the line 6-5 and it acts from 6 to 5, and the force exerted
by the member cd represented by the line 5-4 and it acts from 5 to 4.
Now, as P2 acts downward, from 4 lay off 4-7 downward and equal and
parallel to P2 and from 7 draw a line 7-n parallel to the member df and
close the polygon by drawing from 6 a line parallel to the member de
intersecting the line ?-n at 8, and we have the polygon 6-5-4-7-8-6 repre-
senting the forces at joint d. Following around this polygon in the
direction of the action of the known forces, that is, from 6 to 5, from
5 to 4, on around to 6, we see that the forces exerted by the member df
and de each act away from joint d and hence the stress in each of the
members df and de will be tension, and, as the intensity of the stress S6
in the member df is given by the line 7-8 and the intensity of the stress
S7 in the member de. by the line 8-6, we have the stress in each of the
members df and de fully determined.

Passing on to joint f we have the forces exerted by the member df
and the load P3 known, to determine the forces exerted by the members
fB and fe. The force exerted by the member df is represented by the
line 8-7 and acts from 8 to 7. Then from 7 lay off 7-9 downward and
equal and parallel to P3 and drawing 9-10 from 9 parallel to the member.
‘{B and 8-10 from 8 parallel to the member fe, we have the force polygon
8-7-9-10-8 representing the forces at joint f where the line 9-10 repre-
sents the force exerted by the member fB and 10-8 represents the force
exerted by the member fe. Following around the polygon in the direction
of the action of the known forces, that is, from 8 to 7, 7 to 9, on around
to 8, we have the force exerted by the member fB and by the member fe
both acting away from the joint f, and hence the stress $8 in fB and the
stress S9 in fe are both tension and as the line 9-10 gives the intensity of
§8, and 10-8 the intensity of §9, we have the stresses in the two members
{B and fe fully determined.

Now, passing on to joint e, we have here all of the forces known
except the one exerted by the member eB. Starting with the force
exerted by the member be, and passing around the joint counter clock-
wise, we have the force exerted by the member (be) given by the line 1-6
and it acts from 1 to 6; the force exerted by the member de given by the
line 6-8 and it acts from 6 to 8; and the force exerted by the member fe
given by the line 8-10 and it acts from 8 to 10. Then a line drawn from
10 parallel to the member eB should close the polygon 1-6-8-10-1 repre-
senting the forces at joint e. If this polygon should not close it shows
that the continuous diagram has not been accurately drawn and hence
must be wholly redrawn to insure that the intensity of the stresses thus
obtained is correct. The intensity of the stress $10, in the member eB,
is given by the line 10-1, and as the force exerted by that member at joint
e acts from 10 to 1 in the polygon 1-6-8-10-1, S10 is seen to be compres-
sion and hence is fully determined, and thus the stresses in all of the truss
members are found.

For joint B we have the force polygon 10-9-1-10, where the line 9-1
represents the reaction R1, and 1-10 and 10-9 the stress in eB and fB,
respectively. The continuous diagram at (h) would be known as a
154 STRUCTURAL ENGINEERING

“stress diagram,” in fact all such diagrams representing stresses are
known as stress diagrams.

It will be observed that all of the external forces, which consist of
the two reactions and the three loads, are laid off on the same line 2-9,
which is known as the load line; yet these forces, as they hold the truss
in equilibrium, really form a closed polygon, so to speak, where the
reaction R is laid off from 1 to 2 and the load P1 from 2 down to 4, P2
from 4 to 7, P3 from 7 to 9, and R1 from 9 back to 1—the starting point
—all acting, as we may say, in the same direction around the polygon
1-2-4-7-9-1, which would really be a polygon if the forces were not
parallel. In constructing any stress diagram the force polygon, known
as the load line, formed by the external forces can be laid off first and the
remainder of the diagram added to this polygon, as shown in the follow-
ing example:

Example 2. Let it be required to determine the stresses in the
members of the truss shown at (a), Fig. 127, due to the three loads P1,
P2, and P3. Let R and R1 represent the reaction at A and B, respect-
ively, due to the three loads.

    

 

 

yy”

Fig. 127

Beginning with R and passing around the truss as a whole counter
clock-wise, and taking the forces in consecutive order, we obtain at (b)
the force polygon 1-2-3-4-5-1 by laying off 1-2 equal and parallel to FR,
2-3 equal and parallel to P1, 3-4 equal and parallel to P2, 4-5 equal and
parallel to P3, and 5-1 equal and parallel to #1, thus closing the polygon.
As is readily seen, the forces act in the order 1-2-3-4-5-1.

To obtain the stresses, we can begin at either 4 or B, say, B, and
passing around that joint counter clock-wise we obtain the polygon
1-11-5-1, representing the forces at that joint, by drawing 1-11 parallel
to gB and 5-11 parallel to fB. The line 1-11 gives the stress in gB and
5-11 the stress in fB. As R1 acts upward from 5 to 1, the force exerted
by gB will act from 1 to 11, and hence toward the joint, and the force
exerted by fB will act from 11 to 5, and hence away from the joint.
Thus it is seen that the stress $12, in the member gB, is compression,
while the stress $10, in fB, is tension.

Passing on to joint /, we obtain the polygon 4-5-11-10-4 by drawing
11-10 parallel to gf and 10-4 parallel to fd. We really begin with the
load P3, which acts from 4 to 5. Then the force exerted by fB comes
next, which acts from 5 to 11, and next is the force exerted by the
GRAPHIC STATICS 155

member gf which we can lay off from 11 only as a line parallel to gf,
and its length is then obtained by drawing a line from 4 parallel to fd,
thus closing the polygon. In a similar manner the polygon 10-11-1-9-10
for joint g is obtained, and likewise the polygon 9-1-8-9 for joint e,
3-4-10-9-8-7-3 for joint d, 1-8-7-6-1 for joint b, and 3-7-6-2-3 for joint c.
By scaling in each case the line representing the intensity of stress and
observing at the same time the direction of action of the force exerted at
the joints, the stress in each member is fully determined.

If the direction around the joints in examples 1 and 2 be taken
clock-wise instead of counter clock-wise, as above, the stress diagram
would simply fall on the left side of the load line instead of the right as
shown at (h), Fig. 126, and (b), Fig. 127.

Stress diagrams should be drawn without the aid of any letters or
marking of any kind other than the forces on the load line, and the
stresses in the members may be indicated in the way shown above; how-
ever, after a little practice, even this marking will be found to be
superfluous. The student should not acquire the habit of laying off the
load line and drawing the stress diagram without fully analyzing each
joint of the truss, and if this is done, no marking of the diagram is needed.

Example 3. So far we have considered loads only at the bottom
of the trusses, which is often the case for live load, but for dead load,
especially, we have loads applied at both the top and bottom of the
trusses as shown at (a), Fig. 128. Let it be required to determine the
reactions on the truss shown there, and also the stresses in the members,
due to the six loads indicated. Let R and R1 represent the reaction at
A and B, respectively, due to these loads.

The most practical way of determining the reactions is to add the
top load at each panel point to the bottom load and treat the sum as a
single load in each case. Then the ray diagram is constructed as shown
at (b) and an equilibrium polygon as shown at (c) can be drawn. How-
ever, the loads can be laid off on a load line in any order and a ray
diagram constructed and a corresponding equilibrium polygon drawn,
For example, suppose the loads be laid off on the load line in consecutive
order, passing around the truss, as a whole, counter clock-wise, as shown
at (d). Then taking G as a pole and drawing the ray diagram at (d) and
beginning at any convenient point on the line of action of one of the
forces, say, point 1, on the line of action of Pl, resolve P1 into two
components f1 and f2, which is accomplished by drawing f1 (at point 1)
parallel to the ray 1-G and f2 parallel to ray 2-G. As fl and f2 are
components of P1, the first will act in the ray diagram from 1 to G and
the second from G to 2, and transferring these directions'to point 1 we
have them acting as shown. By prolonging the line of action of f2 from
point 1 till it intersects the line of action of P2 at point 2, and resolving
P2 into two components f2 and f3 by drawing f2 parallel to the ray 2-G
and f3 parallel to the ray 3-G, we have the component at point 2 balanc-
ing the component f2 at point 1, and drawing the line 2-3 parallel to the
ray 3-G, 3-2 parallel to the ray 4-G, 3-4 parallel to the ray 5-G, and so
on, we obtain the polygon y-1-2-3-«-3-4-5-z, wherein the forces, or com-
ponents, are all balanced except f1 and f%. Then prolonging the line of
action of f1 until it intersects the line of action of R1 at K and prolonging
the line of action of f7 until it intersects the line of action of R at H, and
156 STRUCTURAL ENGINEERING

drawing the closing line HK, we have the equilibrium polygon K-y-1-2-3-
«-3-4-5-H-K closed, and by drawing E’-G, at (d), parallel to the closing
line HK we have the reaction R given by the line E’-1 and the reaction
R1 by the line E’-7. The polygon could be closed just as well by draw-
ing 3-K’ and 1-H’, and then H’-K’, which is the closing line in that case,
and which is parallel to H-K.

As another case, suppose the loads are laid off on the load line, as
we may, just in any order as shown at (e). Then constructing the ray
diagram as shown we can draw the equilibrium polygon V-6-7-8-9-10-
t-10-U-V as shown at (f), where UV is the closing line. Then drawing

  

 

 

4
4
f,
2
4
! eo 76_ss elf s7
d inl s@ 3
9 yg es K
A= S25 _s8 lf sio Nig
fe HE rz ges Ng

 

 

 

Fig. 128

E”-P parallel to U-V we have the reaction 2 given by the line E”-F and
Rl by the line E”-T7.

Complicated diagrams and polygons should be avoided as much as
possible. We can accomplish this result only by judicious selection of
order and position, for which no fixed rule can be given, and one must be
guided by experience and common judgment. Complicated diagrams and
polygons are more objectionable on account of inaccuracy, as a rule, than
on account of difficult construction. For example, it is obvious that there
is more chance of error in constructing the polygon at (f) than there is
in the case of the one at (c).

The most practical way of obtaining the stresses in the truss shown
at (a) is to lay off the load line as shown at (g), which is accomplished
GRAPHIC STATICS 157

by beginning with R and passing around the truss counter clock-wise,
taking the forces in consecutive order. Thus we lay off at (g) the line
1-2 upward, equal and parallel to R. Then from 2 lay off 2-3 downward,
equal and parallel to P1, 3-4 equal and parallel to P2, 4-5 equal and
parallel to P3, and from 5 lay off 5-6 upward, equal and parallel to R1,
and from 6 lay off 6-7 downward equal and parallel to P4, 7-8 equal and
parallel to P5, and the line 8-1 closing the polygon 1-2-3-4-5-6-7-8-1
should be equal and parallel to P6. After the load line is laid off, we
can start either from joint A or B, and, passing around each joint counter
clock-wise, the stress diagram shown at (g) can be readily constructed.

97. Oblique Reactions.— As a rule, the dead and live load upon
structures act vertically, producing vertical reactions, but the pressure
due to the wind, known as wind load, produces reactions which are not
vertical. However, the method of determining such reactions is prac-

 

 

Fig. 129

tically the same as for vertical reactions. As an example, let the diagram
at (a), Fig. 129, represent a truss supporting five wind loads, P1—P3,
applied normally to the sloping side of the truss as indicated. In case
the truss be equally fixed at its supports A and B, the reactions FR and R1
will be parallel to the loads as indicated, or in case the loads are not all
parallel, the reactions will be parallel to their resultant.

To determine the reactions R and R1, first lay off the load line CD
at (b) and draw the ray diagram as shown and construct the equilibrium
polygon abcdega at (a) the same as for any other forces. Then by
drawing OF in the ray diagram parallel to the closing line ag of the
158 STRUCTURAL ENGINEERING

equilibrium polygon, we have the reaction R given by the line EC and R1
by the line ED.

There is nothing about the equilibrium polygon at (a) out of the
ordinary, except at point a. Ate the force P5 is resolved into two com-
ponents, and the line of action of the one to the right (f5) is prolonged
until it intersects the line of action of R1 at g. Now, at a exactly the
same method of procedure is followed, but as P1 and # have the same
line of action, the segment parallel to the ray CO does not appear as its
length is zero.

After the reactions are known, the stress in the individual members
of the truss can. be graphically determined, as explained above, by begin-
ning at either joint A or B.

In case one end of a truss be supported upon rollers, the reaction at
that end will be vertical, as the rollers will not resist any horizontal force.
In such cases we always know the direction of the one reaction, but, as a
rule, the intensity of that one and the intensity and direction of the other
remain to be determined.

Let the diagram at (a), Fig. 130, represent the same truss and loads
as represented at (a), Fig. 129. First, suppose the truss supported upon

 

 

Fig. 130

rollers at B and upon a fixed bearing at 4. Then, the end B being upon
rollers, the reaction there, indicated by R, will act vertically and, conse-
quently, we know its line of action, while the reaction at A, indicated as
R1, will evidently act obliquely, as it resists the horizontal component of
the wind; but, further than this, R1 is unknown.

Lay off the load line CD at (b), Fig. 130, and draw the ray diagram
the same as in any other case, and construct the equilibrium polygon
AabcdeA at (a) by beginning at dA, the point of application of the
oblique reaction. Then treating this equilibrium polygon as a truss, we
can obtain R by analyzing joint e, as the stress in the member ed is
given by ray 6 in the rav diagram at (b). Then by drawing from D, at
(6), the line DE parallel to R and closing on O with line OE drawn from
O parallel to the member eA (ab e), we have the intensity of R given
by the line DE. Then, as the forces P1—P5 and the two reactions R
and R1 are a system in equilibrium, they will form a closed polygon,
GRAPHIC STATICS 159

DECD, wherein the forces act in the same direction around the polygon.
So, evidently, the line EC, at (b), represents the other reaction R1 in
intensity and direction. However, the analysis of joint A will show this,
for beginning with P1, we have, at (b), the force polygon CmOEC.

In case the truss were supported upon rollers at A and upon a fixed
bearing at B, we would begin at B, as that would be the point of applica-
tion of the oblique reaction, and construct an equilibrium polygon as
Ba’b’c’d’e’H’ and then analyze the points H’ and B to obtain the reac-
tions R2 and R3, whence their intensity and direction would be given, at
(b), by the lines CF and FD, respectively.

98. Method of Drawing an Equilibrium Polygon through Two
and Three Given Points.— As a case of two points, let P1, P2, and P3,
Fig. 131, represent three given forces and let A and B be the two given
points. Imagine the line AB, joining the two points, as being a beam
supporting the three forces. Then the reactions on this beam, due to the
three forces, will be parallel to the resultant of the forces. By con-
structing the force polygon CabDC, at (b), we have the resultant of the
forces given by the line CD. Then the reaction at 4d and B, represented,
respectively, as R and R1, can
be drawn, as shown, parallel to
the line CD. Taking any point
O, at (b), as a pole and con-
structing the ray diagram as
shown, we can draw the equi-
librium polygon 4A-1-2-3-B’-A,
as shown at (a). Then drawing
OE parallel to the closing line
AB’ we have R given by the
line EC and R1 by the line ED.
These reactions will, of course,
be constant regardless of the
slope of the closing line of the
equilibrium polygon used to de-
termine them. So, regardless of
the location of the pole in the
ray diagram, the Jine drawn
through it and parallel to the
closing line of the correspond-
ing equilibrium polygon will :
pass through E. Then evidently the pole of any ray diagram, where the
corresponding equilibrium polygon passes through both 4 and B will be
on a line through E parallel to AB as Ez, and as any equilibrium polygon
drawn from A, having a closing line parallel to AB, will pass through
both A and B, it follows that any point on the line Ex can be taken as a
pole and the corresponding equilibrium polygon will pass through the two
points A and B. Thus, taking O1 on the line Ez, at (b), as a pole, and
constructing the ray diagram, as shown, the equilibrium polygon 4-4-5-6-B
can be drawn, and taking O2 as a pole and constructing the ray diagram
to the left of the load line, as shown, the equilibrium polygon 4-7-8-9-B
can be drawn. _

 

 
160 STRUCTURAL ENGINEERING

As a case of three points, let P1—P5, shown at (a), Fig. 132, repre-
sent five given forces and let A, B, and C be three given points through
which an equilibrium polygon is to be drawn.

First construct the ray diagram at (b) by laying off the forces in
consecutive order, to any convenient scale, thus obtaining the load line
DabGcF, and taking any point O as a pole and drawing the rays.

Imagine the line AB, joining the two points A and B, as being a
beam supporting the three forces P1, P2, and P3 (as loads). Then the
reactions on beam AB, at A and B, due to the three forces, will be
parallel to the resultant of the three forces, which is represented by the
line DG in. the ray diagram. Then by drawing a line, as yy, through
each of the points A and B, parallel to DG, we have the lines of action
of the two reactions on the beam AB which are indicated as & and Rl,

We can consider the part ODG of the ray diagram as pertaining to
beam AB. By drawing the equilibrium polygon A-1-2-3-m-A, as shown

 

Fig. 132

at (a), and then drawing OE parallel to the closing line 4m, and from
E drawing a line parallel to 4B, we have the line Ez, any point of which
could be taken as a pole for a ray diagram, and the corresponding
equilibrium polygon would pass through the points 4 and B as shown
above for the case of two points.

Next, imagine the line BC as being a beam supporting the two forces
P4 and P5. Then the reactions on this beam at B and C would be
parallel to GF (shown at (b)) and by drawing the lines tt through B
and C parallel to GF we have the lines of action of the reactions on the
beam BC which are represented as R2 and R3. The part OGF of the ray
diagram can be considered as pertaining to the beam BC. Then drawing
the equilibrium polygon n-4-5-h-n and next the line OF1 parallel to the
closing line nh and from E1 drawing a line parallel to BC we have the
GRAPHIC STATICS 161

line E1-z, any point of which could be taken as a pole for a ray diagram,
and the corresponding equilibrium polygon would pass through the two
points B and C, provided, of course, that the polygon be started from
one of the points. But, as any point on the line E- can be taken as a
pole for a ray diagram, and the corresponding equilibrium polygon would
pass through points 4 and B, provided it be started from one of the
points, undoubtedly if the point 7, the intersection of E-x and E1-z, be
taken as a pole of a ray diagram, the corresponding equilibrium polygon
will pass through all three points 4, B, and C, if started at one of the
points.

Problem 1. Determine graphically the bending moment on the beam
AB, as shown in Fig. 125, at load P3 due to the loads P1, P2, P3, and P4,
assuming that each of the loads weighs 8,000 lbs. and that all are spaced 4
feet apart along the beam and load P1 4 feet from R, and load P4 5 feet
from Rl.

Problem 2. Determine graphically the stress in the members of the
truss shown in Fig. 126 due to the loads indicated. Assuming:

P1 = 8,000 lbs., P2 = 10,000 lbs., and P3 = 12,000 Ibs.
Length of span = 4 panels @ 25’ = 100’.
Height of truss = 28’.
Problem 8. Determine graphically the stress in the members of the
truss shown in Fig. 127 due to the loads indicated. Assuming:
P1 = 20,000 Ibs., P2 = 30,000 lbs., and P3 = 20,000 lbs.
Length of span = 4 panels @ 26’ = 104’.
Height at c and f = 28’.
Height at d = 34’.
Problem 4. Determine graphically the stress in the members of the
truss shown in Fig. 128 due to the loads indicated. Assuming:

P1 = 22,000 Ibs., P2 = 24,000 Ibs., P3 = 25,000 lbs.
P4 = 8,000 lbs., P5 = 7,000 lbs., P6 = 9,000 lbs.
Length of span = 4 panels @ 26’ = 104’.

Height of truss = 31’.
CHAPTER IX
INFLUENCE LINES

99. Definition.—An influence line is a line showing the intensity
and variation of reactions, shears, moments, and stresses, produced on
beams or trusses by a single moving load. Owing to convenience the single
moving load is taken as a unit load.

100. Influence Line for Reactions and Shears on a Simple
Beam.—Let AB, Fig. 133, represent a simple beam supporting a single
moving load P and let R and R1 represent the reaction at A and B,
respectively, due to this load P when at any point on the beam.

Suppose that the load P moves over the
beam from B to 4; the shear on all vertical
sections of the beam between A and the load P,
due to that load, at all times will be equal to R,
which varies, of course, with the position of
the load. Then for the shear at all points be-
tween A and the load, including the point at
the load, we have R= Px/L, which is an equa-
tion to a straight line.

Fig. 133 From the equation we have R=P when
v=L, and R=0 when «=0. Then if we draw
the horizontal line ab (=L) and erect the perpendicular ac equal to P
(by scale) and draw the line cb, any ordinate as y to the line cb will be
equal to the reaction at A and also to the shear on the vertical section of
the beam just over the ordinate when the load P is at that point, and from
this it is seen that the reaction # and the shear in the beam vary from
0 to P as the load moves from B to A. So then the line cb is the influence
line for the reaction at A and also for the shear in the beam when the
single moving load moves from B to A. In case the load moved from A
to B, the line ad would be the influence line for the reaction at B and for
the shear in the beam if bd be laid off equal to P.

As an example of application, suppose we wish to determine the
reaction at A and B and the shear on the beam due to a load of 20,000
Ibs. at C and a load of 30,000 Ibs. at M. For the sake of illustration,
suppose P, the single moving load referred to above (which is really a
mythical load and is not on the beam at all and is used only to construct
the influence line), is equal to 1,000 lbs. Then the ordinate z represents
(by scale) what the reaction R would be if the 1,000-lb. load were at C.
Then, evidently, the reaction at A, due to the 20,000-Ib. load at C, would
be 20 times as much or 202, and similarly the reaction at A due to the
30,000-lb. load at M will be 30s, and hence the reaction at A due to both
the 20,000+ and 30,000-Ib. loads is equal to 20z + 30s, which is also equal
to the shear anywhere between A and C, while the reaction at B and also
the shear anywhere between B and M is equal to 200 + 30¢ and the shear

162

 

 

 

 

 

 

 
INFLUENCE LINES 163

anywhere between M and C is equal to 20z + 30s — 20,000 or 20u + 30¢-
30,000.

” ‘The value of any ordinate, as 2, s, etc., will always be given to the
same scale'as that used in laying off the single moving load P. It is
always convenient to take this single moving load P as unity, for in that
case it does not appear in the computations at all.

Example 1. Determine the maximum reaction and shear on a simple
beam 30 ft. long due to the wheel loads as per diagram.

The placing of the wheels in
order to obtain the maximum re-
action or shear on a simple beam is
really a matter of trial, yet in most
cases we can determine the position
by mere inspection. As in this case,
we can readily see that the maximum reaction will occur when wheel 2
is at the end of the beam and wheel 1 off of the beam altogether. Then,
to obtain the maximum reaction, draw the line ab (Fig. 134) to represent
the length of the beam to some convenient scale, say, 7;’" = 1’-0’, and
space the loads to the same scale, placing wheel 2 at a. Then wheels 2, 3,
4, 5, 6, and 7 will be on the beam as shown in Fig. 134. Next draw the
vertical ordinate ac to represent 1 Ib. to a convenient scale, say, 1 in. =
1 lb., and then draw the influence line cb. Next draw the vertical lines
through the loads as shown and we are then ready to find the reaction at a,
which we do in the following manner:

Take a pair of dividers, put one of the points at c and bring the other
point to a, then we have the length ac on our dividers; then set one point
at e and rest the other point at o (eo = ca); hold the point firmly at o and

open the dividers, bringing

o” the point e to f; then we will
of have the length of ca and ef
on our dividers. Then put

one point of the dividers at g

and rest the other point at 0’

(o’g =ca+ef=of); holding
the point firmly at o’ bring

the point at g down to A.
6 Then put one point of the
sto"| sto"| sto”| 910 | sfo“la” dividers at k and rest the
other at 0” (o”’k=ac+ef+
gh) ; holding the point firmly

 

 

 

 

ae
Gy

Ry
a

 

oO
a

 

 

 

 

 

 

 

 

 

s “USO S x at o” bring the point at k

X down to m, and the distance
2 3 4 5 6 Gy o’’m on the dividers equals
ber the sum of the ordinates ac,
Ir RI ef, gh, and km, each of which

Fig. 134 is an ordinate under a 20,000-

lb. load. Then lay the di-

viders on a scale divided to 1/10 inches and suppose we find that we have
3.03 inches (=0’m). Then the reaction at a due to the four 20,000-lb.
loads is equal to 20,000 x 3.03=60,600 lbs. Then with our dividers we
get, in the same manner, the distance nt and rp, the sum of which suppose
164 STRUCTURAL ENGINEERING

we find is 0.25 of aninch. Then the reaction at a due to the two 13,000-
Ib. loads is equal to 13,000 x 0.25 =3,250 lbs. Then the total reaction at a
is 60,600 + 3,250=63,850 lbs. In the case of wheel loads, as a rule, time
can be saved by using the ordinates at the centers of gravity‘ of the dif-
ferent groups of wheels instead of the ordinates directly under the wheels.
For example, the maximum reaction in the above case is equal to

80,000 x g’ + 26,000 x 9”.

The maximum reaction at b due to the above wheel loads would be
the same as at a, and would be determined in the same manner, but of
course the loads and the influence line would be reversed, end for end,
from the position shown in Fig. 134.

To determine the maximum shear at any intermediate point in the
beam, the first thing to do after constructing the influence line is to place
some wheel which we think will be at the point in question when the
maximum shear occurs. Then determine the end reaction in the same
manner as was shown above and subtract the intervening loads, if any,
from the reaction, and the difference will be the shear at the point.

Thus, let ab (Fig. 135) represent a beam to a convenient scale, and
let it be required to find the maximum shear at point d due to a system of
wheel loads as shown. First construct the influence line cb, making ae

 

 

 

 

 

 

Q * B
x
Ic Ph B
Au fs a}
R I! IRI
( to i
A : B
Y & y
c @ x
” (b)
n
a 5
Fig. 135 Fig. 136

equal unity. Say we place wheel 2 at d, as shown, and find the reaction at
a in the same manner as was shown above. Then subtract wheel 1 from
this reaction and the difference will be the shear at d. If there be any
question as to this being the maximum shear at d, wheels 1 and 3, and
possibly 4, should each in turn be placed at d and the shear computed.
In this way we can readily ascertain the wheel to place at d for maximum
shear at that point.

101. Influence Line for Bending Moments on Simple Beams.—
As a general case, let AB (Fig. 136) represent a simple beam of length
L and let cc be any section a distance from A and b distance from B.
Imagine a single load P moving over the beam from B to A. The reaction
at A due to this load P, at any time, will be R = Px/L. When P is to the
right of cc, the moment at that section is

M,=Ra =P> a,

which is an equation to a straight line wherein M, = 0 when x = 0, and
M, = P(b/L)a when « = b. Then drawing 4’B’ (=L) as a reference
INFLUENCE LINES 165

line, and laying off om = P(b/L)a, we can draw the line mB’, which
evidently is the influence line for the bending moment at the section cc
for loads on the right of the section, as any ordinate y represents the
bending moment at the section ce due to P when P is directly over that
ordinate, and the moment for any other load can be obtained by direct
proportion. When P is to the left of cc, the moment at that section is

M, =Ra-P(2-b)=P > a-P(2-b),

which is also an equation to a straight line wherein M, = 0 when 2 = L and
M,=P(b/L)a when « =b. Then, evidently, the line mA’ is the influence
line for the bending moment at the section cc for loads to the left of the
section, as any ordinate y’ represents the bending moment at the section ce
due to P when P is directly over that ordinate, and the moment for any
other load can be obtained by direct proportion.

The two lines B’m and mA’ combined would be known as the
influence line, which would be referred to as influence line A’mB’. By
prolonging the line mB’ until it intersects the line of action of FR at n, we
have two similar triangles, 4’B’n and oB’m. From these similar triangles
we have

A’n _A’B’

om oB’

or

 

from which we obtain A’n = Pa. But if P = unity, we would have A’n =a,
and the influence line for unit load could then be constructed by making
A’n = a, drawing B’n, dropping the perpendicular om, and drawing A’m.
Or the same thing could be accomplished by laying off B’r = b, drawing
A’r, dropping the perpendicular om, and drawing mB’.

Example 1. Let it be required to determine the maximum bending
moment on a 40-ft. girder due to the loading given in Example 1 of the
. preceding article. Wheels 1 to 6 are the heaviest that can be placed on

the girder, and consequently will very likely produce the maximum
moment—this much being determined by mere inspection. It is first
necessary to determine the center of gravity of these wheels in order to
place them on the girder so as to satisfy the position for maximum
moment. (See Art. 88.) In such problems it is quite convenient to
determine the center of gravity by means of proportional triangles. (See
Art. 45.) This is accomplished in the following manner:

First lay out the diagram of the loads to a convenient scale (say,
}/’ =1’-0”) as shown on line AB in Fig. 137. We know from observation
that the center of gravity of wheels 2 to 5 inclusive is at d. Then the
center of gravity of wheels 2 to 5 inclusive being at d, the problem reduces
to finding the center of gravity of wheels 1, 6, and the 80,000 pounds
acting through d. From wheel 1 draw the line az, making a convenient
angle with line 4B and lay off ab = 80,000 lbs. and be = 10,000 Ibs.
(weight of wheel 1) to any convenient scale. Then join c and d and
166 STRUCTURAL ENGINEERING

through b draw a line parallel to cd and the point e where it intersects
the line AB will be the center of gravity of all of the wheels from 1 to 5
inclusive. Next, from e draw the line ez’ at random, making any con-
venient angle with AB, and lay off ef = 13,000 lbs. (weight of wheel 6)

 

 

 

 

 

 

 

 

 

 

 

 

 

4 80" | sto” sto” sto” gto” *
* . re %
gy 8 Sais” 8 8 §
8 s a 3
2 OAS ® oo,
a A?
¥
ee
aS
2d | Pe
t
a4
tu tae x
20-0 2010
ony : : Z E
, 5
4
m
F
Fig. 137

and fg=90,000 lbs. (weight of wheels 1 to 5). Then join g and h, and
through f draw a line parallel to gh and the point k where it intersects the
line AB is the center of gravity of all of the wheels from 1 to 6 inclusive,
which was desired.

Now, as this point & is nearest wheel 4, the maximum moment will
(very likely) occur under that wheel, and the line SS bisecting the dis-
tance from k to wheel 4 will be the center of the span. Then, by laying

 

 

 

Z off 20 ft. on each side of this line

a b SS to the same scale as the load
x dx diagram, we have the girder in
position represented as DE. By

 

 

   

dropping the perpendicular from
a g wheel 4 we have the point o on the
Ri beam where the maximum moment
occurs. The influence line DmE
for the moment at that point is then
constructed by laying off EF =oE
Fig. 138 and drawing FD, om, and mE, as
explained above. Then by drop-

ping perpendiculars down from the loads, we have the ordinates 1, 2, 3,
. 6, and by multiplying each by the load above it, and adding these
products, we will have the bending moment at 0, which is the maximum
bending moment on the beam. The sum of the ordinates 2, 3, 4, and 5
INFLUENCE LINES 167

ean be obtained by the use of dividers as-explained in the preceding
article. ;

Example 2. As a case of uniform load, let 4B (Fig. 138) represent
a simple beam supporting a uniform load of p pounds per foot, and let o
be any section a distance from A and b distance from B. The influence
line AmB for the bending moment at o is constructed in the same manner
as explained above by laying off An = Ao =a, then drawing nB, om, and
Am in consecutive order. From proportional triangles we obtain

ese a
=7 4
We can consider the uniform load as made up of infinitesimal loads
each weighing pdx pounds. Then for the bending moment at o due to any
such infinitesimal load at « distance to the right of o we have

 

EME dy ai as ish iatadit Lean 2 Gia dented jak ge apo anslagehare ae or apaae a (1).
But from proportional triangles we have
y _b-#
a, F
zr) 4
from which we obtain
a
y=z (6-2).

Now substituting this value of y in (1), we have
dM = a(b-a)de ea iuhe cmrashivaikexsmeersaksines (2).

Then for the bending moment at o due to all such loads to the right, we

have
M= a b-« dx Pp x a b

But $(b/L)axb = area of the triangle omB, since the ordinate om =
(b/L)a. So we have the bending moment at o due to the uniform load to
the right equal to the area of the triangle omB multiplied by the uniform
load per linear foot of span. In a similar manner it can be shown that
the bending moment at o due to the uniform load to the left of o is equal
to the area of the triangle omA multiplied by the uniform load per linear
foot. Then letting M represent the total bending moment at o due to the
total uniform load on the girder, we have

b b b a b bra
MiP (Fex3)+0(zpax$)-0( a Cr)

= area of triangle AmB x p...... (3),

that is, the total bending moment at o is equal to the area of the triangle
AmB (formed by the influence line and the reference line) multiplied by
the uniform load per linear foot of span.
168 STRUCTURAL ENGINEERING

If the section be taken at the center of the span, we have b=a=L/2.
Substituting in equation (3), we have
L L\ pL?
M= Pp ( 4 x ) a 2
which is readily recognized as the formula for the bending moment at the
center of a beam uniformly loaded.

102. Influence Lines for Shears and Bending Moments on
Trusses.—Let the diagram at (a), Fig. 139, represent a simple truss and
let it be required to construct an influence line for the shear in any panel
as CD. It is evident that any load as W when at D or to the right of D
will produce tension in the diagonal UD, while any load as W1 when at
C or to the left of C will produce compression in the same diagonal.

 

P= ™*

    
    

   

  

| Yr
Y3 (9)
m

Fig. 139

Then evidently there must be some point between C and D where, if a
load were placed, the stress in the diagonal due to the load would be zero.
Then evidently the influence line for the shear in the panel will be of the
form EdeF shown at (b), which is readily constructed for a unit load by
laying off En and Fn’ each equal to unity and dropping the perpen-
diculars ef and hd under the panel points and drawing ed. The shear
in the panel producing tension in the diagonal UD, due to any load as W
when at any point « distance from B, is equal to Wy,, y, being the
ordinate under the load as shown, while the shear in the panel producing
compression in the diagonal due to any load as W1 is equal to W1y,.

It is evident then that to obtain the maximum shear in panel CD
producing tension in the diagonal UD, which we will call positive shear,
there should be no loads to the left of O, and to obtain the maximum
shear producing compression in the same diagonal, which we will call
negative shear, there should be no loads to the right of O. That is, one
kind of shear will be a maximum when the loads extend from B to O, and
the other kind will be a maximum when the loads extend from 4 to O.
INFLUENCE LINES 169

In case of a uniform live load, the maximum positive shear would be equal
to the area of the triangle Fke multiplied by the load per foot, while the
maximum negative shear would be equal to the area of the triangle Edk
multiplied by the load per foot. In case of uniform dead load, the shear
is equal to the difference of the areas of the two triangles Edk and Fhe
multiplied by the dead load per foot. In the case of concentrated live
loads, as wheel loads, to obtain the maximum shear in the panel the loads
would be so placed that a load would be at the panel point and the front
load as near the point O as possible (not beyond). For example, to obtain
the maximum positive shear in panel CD a load would be placed at D, the
one that would bring the front load closest to the point O, and to obtain
the maximum negative shear the load would be placed at C that would
bring the front load as near O as possible. The same result is obtained
in this manner (graphically) as would be obtained by satisfying the
criterion of Art. 90.

The infiuence line for the shear in each of the other panels can be
drawn on the same preliminary construction at (b) as used for panel CD.
For examiple, the influence lines for panel MC, LN, NB are Ed’e’F,
Ed’e”F, and Ed” ’ F, respectively. The construction of each is accom-
plished in the same manner as explained above in the case of panel CD.

In the case of trusses, the bending moments are desired only at the
panel points. The influence lines for the moments at these points are
constructed the same as though the points were on a simple beam, which
was treated in the preceding article. For example, the influence line for
bending moment at the panel point C of the truss represented at (a)
(same for point U) is constructed as shown at (c) by laying off Gs =a,
the distance of the panel point from A, and drawing sH, mm’, and mG
in consecutive order, as explained in the last article for simple beams.
The influence line for the bending moment at any other panel point would
be constructed in the same manner. The moment at any panel point
would be obtained in the same way as explained in the preceding article
for the case of any point on a simple beam. The exact placing of concen-
trated loads for maximum moments at the different panel points would be
according to Art. 91.

103. Influence Lines for Stresses in Truss Members.—Instead
of constructing influence lines for the shears and moments as in the last
article and then determining the stresses in the truss members from these,
it is more convenient to construct influence lines for the stresses in the
members, thereby obtaining the maximum stresses directly from the
influence lines without dealing with either shears or moments.

Let the diagram at (a), Fig. 140, represent a truss and let it be
required to construct an influence line for the stress in the diagonal ED.
First draw the line 4’B’ at (b) (=Z) and lay off A’n and B’n’ each equal
to unity and draw nB’ and A’n’. Then by drawing the ordinates de and
fb and the line db we have the influence line A’bdB’ for the shear in the
panel CD the same as in the preceding article. The ordinate ed is equal
to the positive shear that would be produced in the panel by a unit load
at D, and the ordinate fb the negative shear in the panel that would be
produced by a unit load at C. Now, as the stress in the diagonal ED
varies as the shear in the panel, it is evident that if the ordinates de and
bf were laid off to represent the stress in the diagonal that would be
170 STRUCTURAL ENGINEERING

produced by a unit load at these same points and an influence line be
constructed accordingly, we would have an influence line for the stress in
the diagonal. ;

For example, if e’d’, at (c), were equal to the stress produced in the
diagonal ED by a unit load at D, and f’b’ were equal to the stress pro-
duced in the same by a unit load at C, the line A”b’d’B” would be the
influence line for the stress in the diagonal. And likewise, if f’t’’, at (c),
and r’’e’ were equal, respectively, to the stress produced in the post EC
by a unit load at C and D, the line A’'t’'r’B” would be the influence line
for the stress in the post.

Laying off B’k at (b), equal to

 

 

 

 

 

 

 

 

 

 

 

 

2 rT - ==? BD and drawing kA’, ea, and aB’ in
2p 7 consecutive order, we have the influ-

\ g ence line for the bending moment at
o gle panel point D, as explained in the
! last article. Now, as the stress in

Sn id the top chord EF varies directly as
& t (b) this moment, it is evident that if ev,
A wri 8’ at (b), were equal to the stress pro-
$ 3 =| duced in the chord by a unit load at
= «| —_-D, the line A’vB’ would be the influ-
re a’ ence line for the stress in the chord

EF, Fyrom the above it is evident
; that an influence line for the stress in
|
1
4;

 

any truss member can be drawn ‘as

readily as the influence lines fer the

moments and shears on the truss by

computing the maximum stress in the

member considered, due to a unit

Fig. 140 load and using this value as the ordi-

nate in establishing the influence line.

The stresses due to the unit load can be determined most readily by
graphics.

As an example, let us first construct the influence line for the stress
in the top chord EF. The first thing to do is to determine the stress in
the member due to a unit load at D. Imagine OO’ (drawn through the
top chord) to be a beam, and cach of the lines OD and DO’ to be a rod
or arope. Then we would have a structure OO’D such that the stress in
the member OO’, produced by a unit load at D, is the same as the stress
produced in the top chord EF by a unit load at the same point. Now this
structure OO’D is very readily analyzed graphically. The reaction at O
due to a unit load at D is given by the ordinate de, at (b). Then con-
sidering the point O and drawing from d a line parallel to OO’ and from
e a line parallel to OD intersecting the first line at r, we have the stress
in OO’ given by the line dr which is also the stress in the top chord EF.
Then from e lay off ev equal to dr, just found, and the influence line
B’vA’ for the stress in the top chord EF is drawn.

Next, let us construct the influence line for the stress in the diagonal
ED. The stress in the diagonal corresponding to positive shear due to a
unit load at D is given by the line rs, at (b), which is drawn from r
parallel to the diagonal, and as bf is equal to the negative shear in the

 

 
INFLUENCE LINES 171

panel that would be produced by a unit load at C, by drawing the lines
OC and CO’ and br’ and r’f parallel, respectively, to OO’ and CO’, and
r’s’ parallel to the diagonal, we will have the stress in the diagonal that
would be produced by a unit load at C given by this last line r’s’.. Then
by laying off e’d’ = rs, at (c), and f’b’ = r’s’, the influence line A’’b’d’B”
is readily constructed. In like manner the influence line A’’t’’r’’B” for
the stress in the post EC is readily constructed by laying off the ordinates
e’r’ =rt (given at (b)) and t’’f’ = r’t’, as it will be readily seen that rt
is equal to the stress in the post due to a unit load at D, and r’t’ is equal
to the stress in the same due to a unit load at C. The influence line for
the stress in any member of any truss can be readily constructed in the
same manner as shown for the above cases.

There are other methods employed to determine the stresses in the
truss members due to the unit load, and these will be presented later as
the practical application of influence lines is taken up.
CHAPTER X
DESIGN OF I-BEAMS AND PLATE GIRDERS
104, General Data.—

Specifications, A. R. E. Ass’n.* (For steel railroad bridges.)
Allowable bending stress on beam and girder flanges. 16,000 lbs. per sq. in.

Allowable shearing stress on shop rivets.........-. 12,000 Ibs. per sq. in.
Allowable shearing stress on field rivets and webs... . 10,000 Ibs. per sq. in.
Allowable bearing stress on shop rivets............ 24,000 lbs. per sq. in.
Allowable bearing stress on field rivets............. 20,000 Ibs. per sq. in.
Allowable bearing on masonry.........--+2+seeees 600 Ibs. per sq. in.

DESIGN OF I-BEAMS
105. Outline of Usual Method of Procedure.—In designing

I-beams the first thing to do is to determine the maximum bending mo-
ment in inch pounds. Then, using Formula C, Art. 53, we have the
bending moment

sath
¥y
dividing through by f, we have
a
fy

The quantity I/y is known as the “Section Modulus,” which is given in
Tables 1 and 2 in the back of this book, for practically all I-beams, and
the same will be found in structural handbooks, such as Carnegie,
Cambria, etc. So the size of an I-beam required to resist a known
bending moment can be determined by simply dividing the bending
moment (in inch pounds) by the allowable stress, which is specified above
as 16,000 lbs., thus obtaining the required section modulus, and from the
tables we find the lightest I-beam having this or approximately this
modulus and that will be the beam to use.

For example, suppose the bending moment for a given span is
433,000 inch pounds. Then the section modulus of the beam requited
for the span is

433,000

16,000

Glancing over column 10 of Table 1 we find that a 10-inch by 30-pound
beam, which has a section modulus of 26.8, is the nearest to the required
beam, and hence would be used.

After an I-beam is designed to resist the maximum bending moment,
it can be tested for shear, by dividing the maximum shear by the area of

 

=27.1.

* These specifications can be obtained from the Secretary, American Railway En-
gineering Association, 900 South Michigan Avenue, Chicago, Il]. Twenty-five cents per
copy, or less if several copies are ordered at one time.

172
DESIGN OF I-BEAMS AND PLATE GIRDERS 173

the cross-section of the beam. If this should exceed 10,000 Ibs. per
square inch, the permissible intensity, a heavier beam should be used or
the web of the beam should be reinforced. However, there are but very
few cases where the shear affects the design of an I-beam, owing to the
webs of I-beams being comparatively thick.

An I-beam can be designed by using Formula D,

My
f=;

and finding values for y and I that will give f=16,000 (approximately).
But, as is readily seen, this is not a very convenient method of procedure.
106. Example 1.—Design an I-beam of 15-ft. span to support a
total uniform load of 800 lbs. per foot.
The maximum bending moment (=pL?/8) = 4x800x15?x12 =
270,000 inch Ibs.

Then, 270,000

16,000

The I-beam having a section modulus nearest this value is the 8-in.

by 25.5-lb., but the 9-in. by 21-lb. would be used as it is lighter. It is
seen that the 8-in. beams are all too small except the 25.5-lb. beam.

107. Example 2.—Determine the size of I-beam required in the
case of the simple beam shown in Fig. 141, where AB represents the

 

= 16.8=section modulus.

 

 

 

 

 

 

oa
| ye

. MM OM ick PELL Le
iz

13/- Oo”

Fig. 141

 

=

beam supporting the loads indicated. By taking moments about B we
find R=19,630 lbs. Adding up the forces, beginning at A, we find that
the shear passes through zero at the 2,000-lb. load, hence the maximum
bending moment occurs at that load. (See Art. 88.) Then for the
maximum bending moment we have

M =[19,630 x 6 -(1,600 x 6 x 3)—(9,000 x 3) ]12= 743,760 inch Ibs.

Then, 743,760
16,000

So a 15-in. by 42-lb. I-beam would be used. The 12-in. by 40-lb. is too
light and 12-in. by 45-Ib. is heavier than the 15-in. by 42-lb. beam.

108. Example 3.—Determine the size of I-beam required in the
case of the simple beam shown in Fig. 142, where AB represents the
beam supporting a load which varies from 0 at A to 2,000 Ibs. at B, as
indicated.

For convenience, let p (=2,000 Ibs.) represent the intensity of the

= 46.4 =Section Modulus.
174 STRUCTURAL ENGINEERING

load at B. Then for the total load on the beam (neglecting the weight
of the beam) we have

 

 

 

 

 

 

 

 

 

a
x i
x x
| A
ly Mf
lL, Ley /)} Wy
Al JB
R ! SL ’
L=/2!' 4
>
Fig. 142
L
w=Pe
2
and for the intensity of the load at any point x distance from A we have
i
L
Now, taking moments about B we have
L
Ww
R= 3. W_ pL
a Se en)

for the reaction at A.
Now for the bending moment at any point z distance from A, we have

en
M= GT BE tenn ee ete ee (1).
When this moment is a maximum
dM pL px =
dz 6. nt OL = 0 ee ewe meee wee we wee wee ee me eee oa (2)

(which is obtained by differentiating (1)). This, as is readily seen, will
occur when L?=3.27, from which we obtain

e=Ly|} = 0.58Z (approximately).

So the point of maximum bending moment is 0.582 from A. Then
substituting 0.58Z for « in equation (1) we have for the maximum
bending moment

M= (2) o.ssz* S (2) 0.192? =2 (0.39L*),
and substituting the numerical value of L and p, and multiplying by 12,
to reduce the moment to inch pounds, we have

2,000

M= ra (0.389 x 144 x 12)= 224,600 inch lbs.
DESIGN OF I-BEAMS AND PLATE GIRDERS 175

for the maximum bending moment.

224,600
16,000

So an 8’ x 18# I-beam would be used.

In case the weight of the beam were considered, we would simply
include the moment of this weight in an equation for the bending moment,
corresponding to (1). Thus, if the weight of the beam be w pounds per
foot, the equation for the bending moment at any point x distance from A
would then be

pL pe wh we?

6 az 6L Eg ge eee eee ee ee ee ee (3).

Then by placing dM/dxr=0, the point of maximum moment can be
determined and then the maximum bending moment at that point due to
the combined loads can be obtained and the beam designed accordingly.

109. Example 4.—Determine the size of I-beam required in the
case of the overhanging beam shown in Fig. 143, where ABC represents
the beam supporting the loads indicated.

=14=section modulus.

Then,

 

 

 

 

 

3f é/__. 4!
%. t e te
is} 3
3 i 8
600*ber tr
A 1c
*
s° 10

Fig. 143

In this case there are two bending moments to consider; one at B
and the other at the point of zero shear in the span BC.
For the bending moment at B we have

M = (4,000x3 + 600x5x2.5)12 = 234,000 inch Ibs.

Then taking moments about C we have
R= a (600x15x7.5 + 8,000x4 + 4,000x13)= 15,150 lbs.

for the reaction at B. Then for the reaction at C we have the total load
on the beam minus R, that is,

R1=21,000 — 15,150 =+5,850 Ibs.

Now beginning at C and adding up the forces toward the left we
find that the shear changes signs, that is, passes through zero, at the
8,000-Ib. load. So the maximum bending moment in span BC occurs at
that load. Then taking moments about that load and considering the
forces to the right of it, we have

M’=(5,850x4 — 600x4x2) 12 = 223,200 inch lbs.

for the maximum bending moment in span BC. The bending moment at
B is the greater and hence the I-beam must be designed for that moment.
176 STRUCTURAL ENGINEERING

‘So, using the moment at B, we have
234,000
16,000
and hence an 8-in. by 18-lb. I-beam would be used.
110. Example 5.—Determine the size of I-beam required in the
case of the continuous beam shown in Fig. 144, where ABC represents the

beam supporting 4 ft. of uniform load in span AB and a single load of
18,000 lbs. in span BC.

= 14,.6=section modulus,

y

AZ

 

Fig. 144

Applying the three-moment equation, (L),
M’L+2M" (L+L,)+M”’ L,=-PL?(k-k*)- P’L? (2k,-3k2+k8),
given in Art. 70, we have both M’ and M”’=0, and the quantity

> a2 b°L bt a’®L at
-PL*(k-W)=— J, pae(5 -F5)=~P("o aL 2 +a)

and the quantity

 

3
-P’L? (2k, — 3k? +k?) =-P’ (zez,- 3445)
1

Then substituting these values in the general three-moment equation,
we have
b°L bt aL at 8
” ae Sis = |} = (Pp? = 2 a
2M” (L+L,)= »( $ 46° 8 +n) P (2er, 3¢ +=). .-(1).

Now everything in this equation is known except M”, the bending moment
at support B. Then by substituting the numerical values of the known
quantities, as given in Fig. 144, and reducing, we have

M’’=-25,363 ft. Ibs. =-304,356 inch lbs.

for the bending moment at B.
Then taking moments about B, we have

-25,363 =10R1 -— 4,000 x 4,
from which we obtain
R1=-936 lbs.

for the reaction at A, and taking moments about B and considering span

BC, we have
—25,363=12 R3- 18,000 x6,
DESIGN OF I-BEAMS AND PLATE GIRDERS 17’

from which we obtain
R3=+6,884 lbs.

for the reaction at C.
Now, as the three reactions must be equal to the total load on the
beam, we have

22,000 =-936 + R2+ 6,886,
from which we obtain
R2=+16,050 lbs.

for the reaction at B.

Now, as all of the reactions are determined, the moments and stresses
on the above beam can be determined as readily as for any beam. There
are three bending moments to consider in the above case; the maximum in
each of the spans and the one at support B.

The reaction 1 being minus, it is readily seen that the maximum
bending moment in span AB will occur at support B, and as span BC
supports only the one load it is evident that the maximum bending
moment in that span will occur under the load. Then we really have only
the moment under the 18,000-lb. load yet to determine as the moment at
B is determined above.

Then, taking moments about the 18,000-Ib. load, we have

M, = 6xR3 = 6x6,884= 41,316 foot lbs.=41,316x12=495,792 inch Ibs.

for the bending moment at that load, which is greater than the moment at
B and hence is the maximum bending moment on the beam, and conse-
quently the beam must be designed to resist this moment. Thus we have

495,792

16,000
So a 12-in. by 31.5-Ib. I-beam would be used.

= 31.0 =section modulus.

 

DESIGN OF PLATE GIRDERS

111. Description.—A plate girder is, for the most part, a built
I-beam composed of a plate web and angle flanges which are riveted to
the edges of the plate as shown at (a), Fig. 145. Yet in addition to these
parts there are usually vertical angles, known as stiffeners, riveted to the
web, as shown at (b), to stiffen the web against buckling. The stiffeners
are either bent around the flange angles as shown at (c) or filler plates
are placed between them and the web, as shown at (d). When they are
bent around the flange angles we say they are crimped. It is practice to
limit the thickness of metal to about 3 of an inch and when the area
required in a flange is greater than that of two 6” x 6” x 3” Zs, plates are
riveted to the backs of the outstanding legs of the flange angles as shown
at (e), to provide for the additional area required. These plates are
known as cover plates, or flange plates. The area of the cover plates in
any flange should never be much greater than that of the two flange
angles, and when they exceed the area of the angles very much, plates,
known as side plates, are placed between the flange angles and the web
178 STRUCTURAL ENGINEERING

as shown at (f) so that the area of the cover plates can be reduced. In
this case the side plates are considered as part of the flange. There are
other types of flanges used occasionally which will be shown later.

Web-~

 

 

 

Fig. 145

 

112. Stress and Area in Flange.—In case of a beam composed of
one piece, as an I-beam, the stress due to cross bending is obtained

through the application of For-
mula D, Art. 53, but in the case
of a plate girder, while the same
method would apply, quite a dif-
ferent one is used, wherein the web
and flanges are treated separately.
The cross bending is resisted by
the flanges and web combined, but
the resistance of the web is ignored
by some engineers, while others
consider it. So we really have two
cases to consider.

In case the resistance of the
web be ignored, the stresses in the
two flanges form a couple which
will be equal and opposite to the
algebraic sum of the external cou-

Cc

 

c

Fig. 146
ples on either side of any cross-section—which is the same thing as the

bending moment.
DESIGN OF I-BEAMS AND PLATE GIRDERS 179

Let Fig. 146 represent a portion of a plate girder:
Let M = bending moment at section cc;
F = total stress in each flange at that section;
d = distance between the centers of gravity of the flanges,
which is known as the effective depth.
Then, in accordance with the condition of equilibrium, we have

M

M=Fd or Barc ene e tte e eee nen (1).
If f be the allowable unit stress, the area required for each flange will be
BA a lainitstieadenseeaxieais iisleseira Renae (2).

This is assuming that the unit stress is the same over the entire cross-
section of each flange—an assumption invariably made, although not
absolutely true in any case, but quite accurate enough for practical
designing.

In case the resistance of the web be considered, the bending moment
will be resisted by the flanges, the same as shown above, aided by the
web, which is really a rectangular beam. Then for the bending moment
we have

M= rat

where f’ represents the unit stress on the extreme elements of the web,
and hf and I represent, respectively, the height and moment of inertia of
the web, while the other letters signify the same as they do in equation
(2), except F, of course, is less.

Now, let ¢ be the thickness of the web, A’ its area of cross-section
and 4 the area of each flange; then we have

r= th, T= 7-13
A’=th, T= 75th

and as the unit stress on the extreme elements of the web is practically
the same as that of the flanges, we have f’=f and also F=f’A=fA. Then
by substituting these values in (3), we have
M=fAd+ (an.
But h = d, practically, so we have
M=fad+l4* = fa (4 +4),

from which we obtain

(4 + *) =F sitawa wiek ene oie tal oa eee (4),

which shows that one-sixth of the area of the web appears as flange area,
but owing to the moment of inertia of the web being reduced in most cases

 
180 STRUCTURAL ENGINEERING

by rivet holes, one-eighth is assumed in practice instead of one-sixth, in
which case we have

A’ M
eaters iiss isha aris cave atonacareet Weveslacateine OCD)
Ata jd seas olay (4),
from which we obtain
M A’
ee ee er ere eee re ee ot fake eeees 6

for the area required in each flange.

118. Economic Depth.—lIt is seen from the preceding article that
the area, and consequently the weight, of the flanges of a plate girder
varies inversely as the depth of the girder and it is evident that the
weight of the web, stiffeners, and fillers varies directly as the depth. Then
evidently a plate girder will have a theoretical economic depth when the
weight of the two flanges’equals the combined weight of the web, stif-
feners, and fillers. There are really two cases to consider, which are as
follows:

Case I. When the web is not considered to resist cross bending.

Let M =maximum bending moment in inch pounds;
L=length of girder in feet;
f =allowable unit stress on flanges;
'¢ =thickness of web in inches;
W =total weight of girder in pounds;
zw =depth of girder in inches back to back of, flange angles.
Girders without cover plates. For the area of the cross-section of

the two flanges we have
2M

fe
assuming depth and effective depth as being equal, and for the weight of
the two flanges we have

2 us x 3.40,
fe

As a bar of steel one square inch in cross-section and one foot long
weighs 3.4 pounds, the total weight of the web alone is equal to 3.4Ltz
and the total weight of the stiffeners, fillers, splices, etc., usually runs
about -60 per cent of the weight of the web, so the total weight of the web,
stiffeners, fillers, etc., can be taken as 1.6(3.4Liz).

Now adding the weight of the web, stiffeners, fillers, etc., to that of
the flanges, we have

M
Hae, eee ESE) aieptito\escer Gueldhe Malaya ne.sedat ateler acer btmseea te (1)
for the total weight of the girder.
This will be a minimum when

om gue
ns =) x 3.40 + 1.6(3.4Lt)=0,
which is obtained be pac wae (1).
DESIGN OF I-BEAMS AND PLATE GIRDERS 181

From this we obtain for the economic depth

e=Lite Iie ape Sco yram eee a gore a cheesa (2).

Girders with cover plates. In this case the area of the flanges varies
from the center to the ends of the span. If the cover plates have theo-
retical lengths, the average cross-section of each flange will be about 0.75
of the maximum area. So for the total weight of the two flanges we have

M
1.555 (3.4L).

Now substituting this instead of 2(M/fx) x (3.4L) in (1) and differ-
entiating and reducing we obtain for the economic depth

ae ote /gniSenhneieeew wa eeRlenn PAAM deareee eae (3).

Case II. When the web is considered to resist bending moment.

By assuming the web to resist bending moment each flange can be
reduced in area to the amount of one-eighth of the area of the cross-
section of the web, and hence the total weight of the girder would be
reduced an amount equal to 2(4tz)3.4L.

Then subtracting this from (1) we have

W=2 (=) 3.4L +1.6(3.4Lt2)- tx) BA lahore. thea (4)

fe
for the total weight of the girder. Differentiating this equation and
placing the first derivative = 0, and reducing, we obtain

2=1224(% ee wm eae eee ee ee eo eee em me oO eo he eo (5)

for the economic depth of girders without cover plates, and by substituting
1.5(M/fz) x (8.4L) for 2(M/fax) x (3.4L) in (4) and differentiating and
placing the first derivative = 0, and reducing, we obtain

for the economic depth of girders with cover plates.

The theoretical economic depth of plate girders can be computed
from the above formulas. An inch or two either way from the theoretical
depth will affect the design but little.

114, Length of Cover Plates.—A flange of a plate girder varies in
area along the girder practically as the ordinates of a parabola, the same
as the bending moment. (See Art. 56.) Then, if AB (Fig. 147) repre-
sents the length of a plate girder and OC the total area of the cross-
section of one flange at the center of the span, any ordinate x to the
parabola ABC will represent (approximately) the area of the flange at
that point.

Suppose the flange of this girder to be composed of two angles and
three cover plates.
182 STRUCTURAL ENGINEERING

Let al=azea of top plate;
a2=area of second plate from the top;
a3=area of third plate from the top;
a4=area of the two angles; and
A=total area of the flange.

Theoretically, net areas should be used throughout and one-eighth of
the area of the web should be included with the area of the angles, but to
provide against discrepancies that may result by assuming that the bend-
ing moment varies along the girder as the ordinates to a parabola (which
is not absolutely true in the case of concentrated loads), we will use gross
areas throughout and neglect the one-eighth of the web in determining
the length of cover plates.

 

‘LZ
Fig. 147

Now, if these areas be laid off to scale, on the line OC (Fig. 147),
the theoretical diagram of the cover plates can be drawn as shown and
then the theoretical length of each plate can be determined by scale. In
this manner the length of the cover plates on any plate girder can be
graphically determined, but in practice the lengths are usually computed
from the formulas given below, as that is the more convenient way.

In accordance with the properties of the parabola we have, referring
to Fig. 147,

from which we obtain

for the half length of the top cover plate, and multiplying this by 2, we
have
DESIGN OF I-BEAMS AND PLATE GIRDERS 183

for the full theoretical length. Likewise we have
yh al+ a2

(Ey 4
2
; © jal + a2
Yu 9 A

for the half length of the second cover plate from the top, and multiplying
this by 2, we have

al a2
sei Ee a ak aces ween meee as ees 2
Qy,,=L (2)

for the full theoretical length of that plate.
In the same manner we obtain

al+a2+a3
ee ial eee eee (3)

for the full length of the third cover plate from the top, and further, we

have
L fal + a2 +43 + a4

for the full length of the angles.

Now from the above it is readily seen that the general formula for
the theoretical length of cover plates can be written as

l= [a a Sa ee adaiag: s senawtbsesstie (a).

 

from which we obtain

 

 

Then for the length of the first or outside plate, in either the top or bottom
flange, we have

Pad, eb
BY seesaw aheniasanis cutebasabinen tite (b),

for the second plate from the top or bottom we have

wv al+a2

W=L eo ok ew Oe Ie 0h ie 88 @ Wie eee Wee ei ew ees wee 0 (c),
and for the third we have

‘, al+a2+a3

Vv aL[itaiea! sods lie eine tonne cam uo eaaals (a),

and so on.

The theoretical lengths of cover plates can be determined very
readily by the use of the ordinary slide rule. When a slide rule is used,
Formula (a) should be ace thereby obtaining

Pat (al +a. . tan).
184 STRUCTURAL ENGINEERING

Then, first the quantity L’/A can be set off on the upper scale, once for all.
By multiplying this by a1, (a1+.a2), (al+a2+a3), and so on, we obtain,
respectively, the square of the lengths of the first, second, third, etc.,
cover plate, and at the same time the square root of this in each case
(which is the length desired) is read off on the bottom scale. Thus the
length of each cover plate on a girder can be determined by one setting of
the rule.

115. Increment of the Flange Stress.— Imagine the web of the
girder shown at (a), Fig. 145, to be made up of vertical strips each of

   

fr ae.

  

APA HA i | ar

   

R Ri

  

Fig. 148

which is connected to each flange’ by one rivet, as shown in Fig. 148.

Suppose the girder supports a number of loads and let R and R1 be
the reactions due to these loads and for the sake of simplicity suppose
both of the reactions and all of the loads to be applied directly to the web.

The loads tend to move the girder downward as a whole; this motion
is prevented by the reactions, and through the balancing of this activity
results first a vertical shearing couple on each of the imaginary strips
which tends to rotate each strip, but rotation of each is prevented by the
rivets connecting it to the flange angles, whereby the flange stress results.
To show this, let us first consider the end strip at A. The shearing couple
on that strip tends to rotate it clock-wise which causes the strip to exert
(through the connecting rivet) a force to the right upon the top flange
angles and an equal force to the left upon the bottom flange angles, and
the same is true of the second strip from the end A, and of the third, and
so on, while, as is readily seen, the shearing couple on each of the strips
near end B tends to rotate each of those strips in the opposite direction, or
counter clock-wise, and hence the force exerted on the top flange angles
by each will be to the left, while the equal force in each case will be
exerted to the right upon the bottom flange angles.

The end rivet in the top flange angles at A and the corresponding end
rivet at B, acting toward each other, produce a simple compressive stress
in the flange angles throughout the distance between these rivets, and,
similarly, the second rivets from the ends, acting toward each other, will
produce a like stress in the angles throughout the distance between those
rivets, and the third rivets from the ends will produce a like compressive
stress in the flange angles from one rivet to the other, and so on. The
same is true of the bottom flange except the direct stress produced there is
tension.

Thus it is seen that the stress in the flanges of a simple plate girder
DESIGN OF I-BEAMS AND PLATE GIRDERS 185

is increased by cach flange rivet as we pass from either end up to the
point of maximum moment. (See Art. 60.) This increase at each rivet
is the increment of the flange stress, which is often referred to as the
“flange increment.”” The summation of these increments between any two
points would be known as the increment of the flange stress between the
two points.

116. Spacing of Rivets in the Vertical Legs of Flange Angles.
—The rivets in a plate girder are practically always the same size
throughout the girder, in which case the rivets in the vertical legs of the
flange angles should be so spaced that the pressure against the flange
angles would be the same for each rivet throughout the girder, and hence
the stress on each rivet would be the same throughout. So the problem
involved is to determine the pitch of the rivets so that the increment of the
flange stress is just equal to the allowable stress on the flange rivet at all
points along the flange.

Let S be the shear on any strip abcd (Fig. 148) of longitudinal
length p, and let r represent the increment of the flange stress at that
strip. Now as the couple rh is equal to the equal and opposite couple
reacting on the strip and balancing the shearing couple Sp, we have

Sp=rh,
from which we obtain
rh
Py pene ser eeee eer scars . . ores eee tee eee (1)

From this the length of any strip can be computed for any desired value
of r. But the length of any strip can, in all cases, be taken as the pitch
of the rivets at the same point and hence if r be taken as the allowable
stress on one rivet, the required pitch of the rivets in the vertical legs of
the flange angles at any point can be computed from this formula.

It is readily seen (from Fig. 148) that these rivets would fail either
in double shear or in bearing on the web, and r, in any case, should be
taken equal to whichever is the least. The bearing on the web is usually
the least.

There are really four cases to consider:

Case I. When the loads are applied directly to the web and the
resistance of the web to bending is neglected. In this case the pitch of
the rivets at any point of the flange is determined from the above equation,

where p= pitch;

r=allowable stress on one rivet in either double shear or bearing

on the web, whichever is the least;
h=vertical distance between the rivets in the two flanges;
S=shear on the girder at the point considered.
Case II. When the loads are applied directly to the web and the

resistance of the web to bending is considered. In this case a certain part
of the shear couple Sp is resisted by the web.

Let S =shear on the girder at any point; __
A =area of the cross-section of one flange at any point;
186 STRUCTURAL ENGINEERING

A’ =area of the cross-section of the web;
f=stress per square inch transmitted to each flange by each
rivet; :
r =pressure exerted upon the flange by each rivet;
h =vertical distance between the rivets in the two flanges;
p=pitch of flange rivets.

It is shown in Art. 112 that the resistance of the web to cross bending
can be accounted for by considering one-eighth of its area as being con-
centrated in each flange. So the portion of the shearing couple Sp resisted
by the web can be approximately expressed as f4’h/8 and the remaining
portion of the couple which is resisted by the flanges can be approximately
expressed as fdh. Then adding these two expressions, we have

,
pans tA? sp, :

r

But, [a5 ;

then substituting this value of f in the last equation and reducing, we have

rh(, A’
p=F (1+) palguia eae aeatd: Oi Gstaad Nie ce eea nas (3)

for the pitch of the rivets.

 

 

 

 

 

 

 

 

 

 

 

 

 

Case III. When the loads are applied di-

rectly to the flange angles and the resistance of | | |
the web to bending is neglected. In this case the a
loads will exert a vertical force upon the rivets in Or ee Ve
addition to the horizontal force considered above. ! XR
Let v (Fig. 149) represent the vertical force per y VP
linear inch of flange and r the horizontal force on SQ te |
the rivet (the same as above). Then for the i '
resultant force we have t

Rie opie ot-+o-+6o 3

oe Fig. 149

But from (2) we have
_SP
“hh

and substituting this in the last equation, we have
S 2
R?= (op)? + (=) :

Then transposing and extracting the square root, we have

Tr

for the pitch of the rivets, where
p=pitch;
S=shear on the girder at point considered;
v=load on the flange angles per linear inch;
DESIGN OF I-BEAMS AND PLATE GIRDERS 187

Re=stress on rivet, which should be taken as the allowable stress on
one rivet; ‘

h=vertical distance between the rivets in the two flanges.

Case IV. When the loads are applied directly to the flange angles
and the resistance of the web to bending is considered. We have here,
the same as in Case ITI,

R? = (op)? +7’,

r= Sp

except 7 fav
= ory

which is obtained from (3).
Then substituting this value of r, we have

R?= (vp) (Tz) ;
“omrlsGoa))

from which we obtain

R

Pe AT rrtrttttteeeteteee eee (5)
“3

for the pitch, where the letters signify the same as specified above.
In case the flange angles have double rows of rivets, h should be
taken as the mean vertical distance between the rivets in the two flanges.

117. Web Splice.— It is always desirable that the web of any plate
girder be one continuous piece, but the length of such plates is limited by
the steel mills and whenever the length of a web exceeds these limits it
becomes necessary to splice it.

# ‘

   

c

(Q) (b)

Fig. 150

In practice there are two standard ways of splicing a web, both of
which are shown in Fig. 150. The splice shown at (a) is made up of six
188 . STRUCTURAL ENGINEERING

splice plates, three on each side of the web, while the splice shown at (6)
is made up of only two plates, one on each side of the web. The splice
plates should have sufficient material to transmit the shear at the point of
splice, and also the cross bending on the web. -The maximum shearing
and maximum bending stress at any point in a simple girder do not occur
at the same time, as is readily seen, yet in designing web splices it is
practice to consider the two occurring simultaneously at the splice—which
is an assumption entirely on the side of safety.

In the case of the splice shown at (a) the plates marked p1 are
usually assumed to resist only the cross bending on the web, while the
plates marked p2 are assumed to resist only the shear on the web. That
is, the plates p1 and their connection to the web are designed as if no
shear occurred at the splice and plates p2 and their connection to the web
are designed as if no cross bending occurred. To illustrate this method of
designing the.splice:

Let f=allowable stress per square inch in each flange;

A’=area of the cross-section of the web;

d=vertical distance between the centers of gravity of the flanges;
d’=vertical distance center to center of plates p1;

F =stress in each pair of plates p1;

S= maximum shear at the splice.

Then the resistance of the web to bending is represented by the
couple (f4’/8)d which must be equal to Fd’ and hence we have

_(fA\d

r-(5)e
for the direct stress in each pair of plates pl, due to the cross bending on
the web. Now, evidently, these plates should be designed to take this
stress F and the number of rivets on each side of the splice connecting
each pair of these plates to the web should be sufficient to transmit the
stress F in either double shear or bearing on the web, whichever requires
the greater number of rivets. However, as the bending stress varies
directly as the distance out from the center of the web, the intensities used
in designing the plates pl and also the stress allowed on the rivets con-
necting them to the web must be proportional to the intensities allowed in
the flange. For example, if the allowable stress in the flange is f per
square inch, the intensity to allow on the plates pl would be (f) d’/d,
and if r is the stress allowed on a rivet, say, jn bearing on the web at the
flange, the same size rivet through the plates would have an allowable

bearing on the web of (r) d’/d. ;

Then for the net area of cross-section of each pair of plates pl we

have
=(F\4.
VF at

And for the number of rivets required on each side of the splice, we have
F
(r)d””
ds

n=

 
DESIGN OF I-BEAMS AND PLATE GIRDERS 189

In designing the plates p2, all we have to do is to select two plates
that have sufficient net area along the vertical section to transmit the
shear and wide enough to admit the necessary rivets on each side of the
splice. However, as a matter of fact, there is, as a rule, more metal in
the two splice plates than is necessary to carry the shear, as each plate is
usually as thick as the web, in which case the two plates would have twice
the net cross-section of the web. The number of rivets connecting the
plates p2 to each side of the splice should be sufficient to transmit the
shear. If r be the allowable stress on each rivet—and we will assume
that each resists the same amount of stress—-we have

n= .
A
for the number of rivets on each side of the splice in plates p2. The
value of r in this case is the full allowable intensity on a rivet in double
shear or bearing on the web, using, of course, whichever is the least.

The assumption that plates pl resist only the cross bending on the
web and plates p2 only the shear is quite a reasonable assumption as
plates p2 are too near the center of the web to resist so very much of the
cross bending and plates pl are really so narrow and too far from the
center of the web to transmit so very much of the shear. As a matter of
fact, however, the plates p2 do resist some of the cross bending and plates
pl resist some of the shear.

In the case of the splice shown at (b) the two plates should have
sufficient net area of cross-section along the line cc or c’c’ to resist the
cross bending on the web and also the shear.

Let f’ =stress per square inch on the outer edges of the splice plates;
f =the allowable stress per square inch in the flanges;
A’=area of cross-section of the web;
h=height of each splice plate;
I=net moment of inertia of the two splice plates along the
section cc or c’c’ (deducting the moment of inertia of the
rivet holes). Then we have

(ajo ee
i ( 8 )a * QI
for the stress per square inch on the outer edges of the splice plates. This
stress should not exceed fh/d. As a matter of fact it never does, as the
two splice plates, as a rule, have at least twice as much area of cross-
section as the web. However, the stress f’ should be considered in doubt-
ful cases. The net area of the cross-section of the two splice plates along
section cc or c’c’ should be sufficient to transmit the shear. That is, the
shear divided by the net area should not exceed the allowable unit shear-
ing stress on the web, say 10,000 lbs. per square inch, as specified in the
A. R. E. Ass’n Specifications.

In case the web be spliced as shown at (b) we can not well consider
other than that the maximum stress on each rivet is due to the cross
bending on the web and shear combined. The cross bending produces a
horizontal force on each rivet while the shear produces a vertical force
and, as is evident, the maximum stress on each rivet will be the resultant
of these two forces. The vertical shearing stress will be practically the
190 STRUCTURAL ENGINEERING

same for each rivet, but the stress due to cross bending, which (as stated
above) is transmitted to each horizontally, will vary directly as their
distance out from the center of the web.

Let s be the horizontal stress due to cross bending on each of the
rivets farthest out, as indicated at (b), and let e be the distance of these
rivets from the center of the web. Then if there were a rivet out unit
distance from the center of the web the stress on it would be equal to s/e
and evidently the stress on any rivet out z distance from the center of the
web would be (s/e)z and the moment of this force or stress about the
center of the web would be

(e)=@)*

Now it is evident that if this moment be determined for each and
every rivet on one side of the splice, and these be added together, their
sum would be equal to the couple (f4’/8)d. So we have

C)e= (2)

From this equation s can be determined and if v be the vertical shear on
each rivet, which is readily computed, we have

R=V 048?
for the maximum resultant stress on each outer rivet which is the absolute
maximum.

As an example, let a, b, c, d, and e represent, respectively, the dis-
tance of the first, second, third, fourth, and fifth horizontal row of rivets
out from the center of the web, shown at (b), Fig. 150, then we have

2 b? 2 d? A?
o( 2s ae + Be + 2se) i (ya.

e e 8

 

 

 

From this we obtain

Then combining this with the vertical shear, as shown above, the maximum
stress on the rivets can be obtained. However, s should not exceed the
allowable at the flange multiplied by 2e/d.

The maximum stress on the rivets in the type of splice shown at (a)
can be determined in the same manner. In fact, the rivets in all web
splices should be tested in this manner.

118. The Stiffening Angles on a plate girder really have two func-
tions: one is to stiffen the web against buckling, and the other is to
transfer loads from the flanges directly to the web. It is obvious that the
closer the stiffeners are spaced along a web the stronger the web is
against buckling, and consequently the higher the web can be stressed.
So the design of the web is influenced by the spacing of the stiffening
angles.

There is no theoretical way of determining the spacing of stiffeners.
A practical rule is to space them so that their distance apart along the
DESIGN OF I-BEAMS AND PLATE GIRDERS 191

girder is equal to the depth of the girder, but in no case farther apart
than five feet or a little over. However, to satisfy modern requirements,
it is usually necessary to space them closer together near the ends of the
girders than this practical rule calls for in order to obtain an economic
web. The allowable spacing of the stiffeners for a given stress on the
web is specified as

t
dS 10 00= a) aw hieitataaveeieales wexcenexie wees (1)

in the Specifications prepared by the A. R. E. Ass’n,
where d=distance between the stiffeners (as shown at (b), Fig. 145);
t=thickness of web;
s=shearing stress per square inch in the web, which is obtained
by dividing the shear at the section considered by the gross
area of the cross-section of the web.

Equation (1) is really an empirical formula.

It is not economic, as a rule, to space stiffeners closer together than
half the depth of the girder. When a spacing less than that at the ends.
of a girder is given by the above formula, a thicker web should be used.

There. should always be stiffeners at any point where excessive
concentrated loads are applied. These stiffeners should be designed as a
column to support the load, the effective length being taken as half the
depth of the girder.

119. Example 1.—Let it be required to design a girder 30 ft. long
(c.c. end bearings), to support a uniform dead load of 400 lbs. per linear
foot of girder, and a live load of 4,000 lbs. per foot.

For the maximum bending moment we have from dead load

M =5 x 400 x 30° x 12 = 540,000” Ibs.

and from live load
M’= r x 4,000 x30 x 12=5,400,000” Ibs.,

making a total of 5,940,000” Ibs. bending momert.

Let us assume the web to be #” thick, and that the loads are applied
directly to the top flange, and that the web resists bending. Having made
these assumptions we can proceed with the determination of the economic
depth. Formula (5), Art. 113, will be used, as girders of such short
lengths usually have no cover plates. So, substituting in this formula we

have
5,940,000 _ao,.
@= 1.224) S000 = 38.4 ins.

for the economic depth of the girder. So we will assume a 38” x 3” web.
The next thing, we will test this web to see if it is satisfactory. For the
maximum end shear we have from dead load

S=400 x 15=6,000 lbs.
192 STRUCTURAL ENGINEERING

and from live load
S’=4,000 x 15 = 60,000 Ibs.,

making a total of 66,000 Ibs. end shear.
Now dividing this by the area of cross-section of the web
(=88” x 2”), we have
66,000
14.25
for the actual average unit shearing stress on the web.
Next, substituting this in Formula (1), Art. 117, we have

= 4,630 lbs.

dix x (12,000 — 4,630)=69.2 ins.

for the maximum theoretical distance between stiffeners at the ends of the
girder; and as this spacing is not less than half the depth of the girder
the web itself is satisfactory, but common practice is not to permit the
distance between the stiffeners to be greater than the depth of the girder.
So we will space them so as not to violate this practice, that is, the clear
horizontal distance between the stiffeners will not be made more than 38”
in any case. ;
Then substituting 38 for d in Formula (1), Art. 117, we have

38 = i (12,000 ~s),
from which we obtain

s=7,946 Ibs.

as the allowable unit shearing stress on the web.
Now dividing this into the shear we have

66,000
7,946

for the required area (of cross-section) of the web, which is 5.950”
(=14.25-8.3) less than the area of the assumed web. So theoretically
the web can be thinner than 3”, but we will use the assumed web as the
specifications limit the thickness to ?’’.. However, in the cases of build-
ings and highway bridges, the web would likely be reduced in thickness,
but in no case should it be thinner than }’”.

As an example in such cases, let us assume a 38” x 1” web which has
an area of cross-section of 9.50” (=38’x4”). Then for the actual unit
shearing stress on the web we have

66,000
9.5

Substituting this in Formula (1), Art. 117, we have

=8.3 sq. ins.

= 6,952 Ibs.

d= a (12,000 - 6,952)=31.5 ins.

for the required spacing of the stiffeners at the end of the girders, and as
DESIGN OF I-BEAMS AND PLATE GIRDERS 193

this is not less than half the depth of the girder the web is satisfactory
and hence would be used.

We will next take up the designing of the flanges. Let us first try
2—Ls 6”x 6” x4” for each flange. From Table 6 we find that the dis-
tance from the back of these angles to their center of gravity is 1.68’.
So if the girder be made 38} deep back to back of flange angles, that is,
}” greater than the depth of the web, which is usual practice, we have
38.25 - 1.68 x 2=34.89” for the effective depth. Now, by dividing this
into the maximum bending moment we have

5,940,000
34.89

for the stress in each flange, and by dividing this by 16,000, the allowable
unit stress, we have

=170,000 Ibs.

170,000
16,000

for the required net flange area.
We have

=10.6 sq. ins.

2—Ls 6x6" x4/=11.50-1 =10.500” net
4 of web=(14.25x4)= 1.780” net

12.280” net

This is too large, so let us try
2—Ls 6” x 6” x 74” =10.12 — 87 =9.250” net
4 of web= 1.789” net

11.030” net

This flange section is about right, being only about 0.49” more than
required, but the specifications require that the thickness of these flange
angles be at least one-twelfth of the width of the outstanding legs which
are 6”, So these 7%” angles are too thin and consequently if 6” x 6”
angles be used at all the 6” x 6” x 4” angles would have to be used, whict.
gives an excess of metal.

Let us try 2—Ls 6”%x4”x }” (the 4” legs outstanding) for each
flange.

The distance from the back of the 4” leg to the center of gravity of
each of these angles is (given in Table 4) 1.99”, say, 2’. Then for
effective depth we have 38.25 — 2x 2=384.25”.

Now dividing this into the maximum bending moment we have

 

5,940,000 _
34.95 = 173,000 Ibs.
for the stress in each flange, and dividing this by 16,000 we have
173,000 _ ‘
“16,000 =10.8 Sq. Ins.
for the required net flange area.
We have
2—Ls 67x4” x4” = 9.50 —- 1 = 8.500” net
4 of web = 1.780” net

 

10.285” net,
194 STRUCTURAL ENGINEERING

This flange section is 0.520” less than the required, so let us try the
following:

2—Ls 6% x4” x 3%” = 10.62 -1.12=9.500” net

4 of web= 1.780” net

11.280” net.

This section is 0.480” more than required. So either of the last two
sections may be used. We will use the latter (6” x4” x 7%” angles) as
that is on the side of safety.

The size of the intermediate stiffeners is governed practically by the
specifications which require that they be no less than 3” thick and the
width of their outstanding legs be no less than #5 of the depth of the
girder plus 2 inches. So in this case we will use 34’ x 34” x 3” angles
for stiffeners throughout as this is the minimum size allowed. Thickness
of the end stiffeners would very likely be 7%” or $”, depending upon
conditions. If the girder be supported upon masonry, these stiffeners
would be designed as columns to take the end shear or reaction, one-half
of the depth of the girder being taken as the length of the column.

If the girder be riveted at the ends to other girders or to columns
which wholly support the girder, the end stiffenérs in that case would be
at least 7'5’”’ thick to provide for the facing of the ends of the girder
which is usually required in modern practice.

To obtain the spacing of the flange rivets at the ends of the girder,
we have

 

R='1,880 lbs. allowable bearing of a }” rivet on the 2” web;
v= 333 Ibs. =4,000 = 12;

S = 66,000 lbs.;

h= 81.5 = (388.25 — 6.75).

Then substituting these values in Formula (4), Art. 116, (the formula
required by the specifications), we obtain

7,880
PE i, = 3.7 ins.

BOP aes
-or the pitch of the flange rivets at the ends of the girder. This pitch
should be used for a distance out from the ends of the girder equal to the
depth of the girder and varied from there on toward the center of the
span to suit the shear at the different points, no pitch, however, being
greater than 6” in accordance with the specifications.

From the above calculations the necessary information for making
the detail drawing of the girder is obtained, and when this drawing is

finished the design of the girder is complete. All plate girders are
designed in this manner.

 
CHAPTER XI
DESIGN OF SIMPLE RAILROAD BRIDGES

120. Types.—Simple railroad bridges can be divided into four
general types: beam bridges, plate girders, viaducts, and truss bridges.
Plate girders and truss bridges can be further divided into deck and
through bridges; deck when the track is supported on the top of the
structure; and through when the track is between the main girders or
trusses. Beam bridges and viaducts are always deck bridges.

121. The Specifications prepared by the American Railway Engi-
neering Association, referred to in the preceding chapter, will be used
throughout this work.

122. The Live Load usually specified for railroad bridges consists
of two typical consolidated locomotives and tenders coupled in tandem
followed by a uniform train load, and a special load concentrated on two
axles which is used in case of very short spans. The most common load-
ing of this type is that known as “Cooper’s Loading,” devised by Theodore
Cooper, M. A. Soc. C. E. Cooper’s loadings vary in weight and are desig-
nated, beginning with the lightest, as £30, E35, E40, £45, E50, E55, and
E60. These loadings vary by a certain ratio so that if any stress, shear,
bending moment, etc., due to any one of the loadings be known the intensi-
ties of the same due to any one other of the loadings can be obtained by
direct proportion, and hence any tables or diagrams giving the shears,
moments, etc., for any one of the loadings can be used for any of the other
loadings. The figures following the letter E are-the indices of the ratio
referred to above. Thus, for example, if any stress, shear, etc., due to
E40 be known, the intensity of the same for £50 will be 50/40 of that
due to E40; 60/40 for E60; 35/40 for E35; and so on. Most of the
railroads in this country have adopted some one of Cooper’s loadings,
either exactly or slightly modified. The trend has been toward the using

LAP Fe
60 FEE SCS B B'S SSO SESS

offs. 3 "aT eS ¥S

S} Sas 3 3
u TO S000 pe

40 OOOO 000040 OOOO 00 00 Zz ror

wpeciol/load

    

32501

 

Fig. 151

of the heavier loadings. At present E50, represented in Fig. 151, is used
quite extensively. The spacing of the wheels is the same for all of the
loadings.

The E40 loading is the most convenient to use owing to the loads
being in even thousands of pounds.

Table A* gives for this loading moments about any wheel of the
wheels to the right or to the left of it, as will be seen upon inspection.

 

*The student shonld verify the results given in this table.

195
196 STRUCTURAL ENGINEERING

This table is quite convenient in determining the centers of gravity,
shears, and moments, as will be shown later.

123. ‘‘An Equivalent Uniform Live Load’’ is sometimes used
instead of the wheel loads described above. In the case of beams and deck
girders this will give exactly the same results as the wheel loads, while in
the case of trusses and through girders the results obtained are only
approximately the same as for the wheels.

To obtain an equivalent uniform live load for the bending moment
on a beam or deck girder of length L, the maximum bending moment M
due to the wheel loads is computed, and then we have

er.
° M= 8
from which we obtain
8M
Do ae

for the equivalent uniform load per foot which will produce the same
maximum moment on the beam or girder as the wheels.

To obtain an equivalent uniform live load for the end shear or
reaction on a beam or deck girder of length L, the maximum reaction R
due to the wheel loads is computed and then we have
p’L

2

 

R=

2

from which we obtain
,

_2R
UE

for the equivalent uniform live load which will produce the same maxi-
mum end shear or reaction as the wheels.

In practice the equivalent uniform live load for trusses is usually
taken either as the uniform load that will produce as great a bending
moment at the quarter point of the span as the wheel loads or the uniform
load that will produce as great a shear in the end panel as the wheel
loads.

The first is obtained by computing the maximum bending moment
M’ at the quarter point due to the wheel loads, the span being considered
as a simple deck beam, and then we have :

 

,_ 3pL*
M = 32 -
from which we obtain
_ 32M’
p= 372

for the equivalent uniform live load per foot where L is the length of
span in feet.

The second equivalent load is obtained by computing the maximum
shear S in the end panel due to the wheel loads and then we have

,{(L-d
sp’ (434).

 
DESIGN OF SIMPLE RAILROAD BRIDGES 197

from which we obtain

for the equivalent uniform live load where L is the length of span and d
the panel length in feet.

The first loading gives stresses too low for web members and chords
near the end and too high for the chords near the center of the span,
while the second loading gives stresses about correct in all web members,
but considerably too high in the chord members, except the chords in the
end panels. To correct for this discrepancy the author has proposed the
reduction of the stresses in the chords (except the chords in the end
panels) by the per cent given by the following empirical formula :*

(Jo + 2.5) per cent
where L is the length of span in feet.

For example, the chord stresses in a 300-ft. span would be reduced
by 5.5 per cent. This does not apply to chords in end panels.

124. Dead Load consists of the weight of the metal in the structure
(except the metal at the supports), and the weight of the track, known
as the deck. The approximate weight of metal per foot of single-track
bridges designed for Cooper’s E50 loading can be obtained from the
following formulas wherein

p=weight of metal per foot of span,
L=length of span in feet, c. c. end bearing:
For beam spans without lateral bracing, and stringers,

PRL +100: ieee s ienie tases eda seucae arcs s yew eee (1),
For deck girder spans and beam spans with lateral bracing,

Dea TOO assis ic ie coarse east eee we iduh reece ccerietets (2),
For through plate girder spans,

P= TSE 600) sess esl ay a ace sede sates ANS OA Sle ee wes (3),
For truss spans,

P= TE C60: ca wienawae wa wuisies pa M ews Minis eae as eat (4).

The weight obtained from the above formulas is for the metal alone,
and to this must be added the weight of deck which in the case of ordinary
wooden decks can be taken at 400 Ibs. per foot of track. The weight
obtained from the above formulas is usually correct enough for computing
stresses, as 10 per cent variation is permissible, but it should not be used
in making estimates of cost.

The approximate dead weight of metal in bridges designed to carry
Cooper’s E60 loading is about one-eighth more than given by the above
formulas, and the weight of those designed to carry the E40 loading is
about one-eighth less than that given by the above formulas.

The weight of metal in double-track bridges depends upon their
construction. Their weight as for metal is about 70 per cent heavier than
that of single-track bridges if three or two main girders or trusses are
used, but about 100 per cent if four trusses or girders are used.

* See Engineering News, September 13, 1906.
198 STRUCTURAL ENGINEERING

In case of concrete or metal floors the floor should be designed and
then the weight of it computed, and then the weight of the main girders
or trusses can be assumed and added to this weight and in this way the
approximate dead load can be determined, which should be within 10 per
cent of the actual weight; if not, the dead load stresses should be redeter-
mined, using a revised dead load.

125. Impact.—Rapidly moving trains will produce greater stresses
in bridges than the same load when simply standing on a structure, as we
consider it when computing the live-load stress, and to provide for this
additional stress a certain amount of the live-load stress in each member
is taken as the impact stress which is added to the corresponding live-load
wtress. The impact stress is obtained by simply multiplying the maximum
live-load stress by a coefficient obtained from an empirical formula. The
following formula is the one most used:

we ..
~L+300

Then for the impact stress in any member we have the formula

300
P=8 (“mn)

as specified by the A. R. E. Ass’n in their Specifications referred to above,
where
I =impact stress,
S=maximum computed live-load stress,
L=length of track (in feet) loaded when the maximum live-load
stress occurs.

126. Wind Loads.— The horizontal pressure exerted on bridges by
the wind is known as the “Wind Load.” As a rule this load can be safely
taken at 30 lbs. per square foot of the horizontal projection of both the
structure and the live load carried. But, in addition to this, some pro-
vision should be made for lateral vibration due to the live load. To pro-
vide for this and the wind load a greater pressure than 30 lbs. probably
should be used in some cases.

The lateral force, which includes the wind pressure, specified in the
A. R. E. Ass’n Specifications, will be used in this book.

Cc

BEAM BRIDGES

127. Preliminary. Beam bridges are used only for short spans,
rarely ever exceeding 20 ft. in length. There are usually at least two
I-beams under each rail connected to each other by diaphragms, similar to
that shown in Figs. 158 and 159, thus forming a compound girder. These
compound girders should be braced to each other by diagonal and trans-
verse bracing, as shown in Fig. 159, whenever the span exceeds 12 ft.
When the span length exceeds 15 ft., four panels of bracing should be
used, in which case a horizontal diaphragm is needed at each lateral
connection. In case there be more than two beams in each compound
girder the beams can be placed somewhat closer together than in the case
of two beams.
DESIGN OF SIMPLE RAILROAD BRIDGES 199

Complete Design of a 10-ft. Span
128. Data.—

Length=10’ c.c. end bearings.
Dead load = 220 + 400 = 620 lbs. per foot of span.
(From (1), Art. 123.)
Live load, Cooper’s £50 (shown in Fig. 151).
Specifications A. R. E. Ass’n.
129. Calculations.—For the maximum bending moment due to dead
load we have

M= ; x ae x 10° x 12 =46,500 inch Ibs.

From mere inspection of the live-load diagram, in Fig. 151, it is seen
that the maximum moment will occur either when one 62,500-lb. axle load
(of the special load) is on the span or when two of the 50,000-Ib. axle
loads are on. So we have two cases to try.

The maximum moment due to the 62,500-lb. axle, one-half of which
goes to each girder, will occur (according to Art. 88) when the load is at
the center of the spen, as shown in Fig. 152. Each of the reactions will

 

 

  

Fig. 153

be equal to one-half of the load on each girder, or 15,625 lbs. Then we
have

15,625 x 5 x 12 =937,500 inch lbs.
for the maximum bending moment due to this load.

The maximum due to the two 50,000-lb. axles will occur when the
loads are in the position shown in Fig. 153 (according to Art. 88) and it
will occur under the wheel marked 3.

Taking moments about A (Fig. 153) we have

R1x10- 25,000 x 1.25 — 25,000 x 6.25=0,
from which we obtain the reaction

R1=18,750 Ibs.

Then for the bending moment at wheel 3 (which is the maximum)
we have
M’ =18,750 x 3.75 x 12 = 848,700 inch lbs.,

which is less than that produced by the one wheel of the special load, as
200 STRUCTURAL ENGINEERING

is seen above, and hence the moment due to the special load will be used.
Then for the maximum impact moment we have

300 :
l= 937,500 (35-00) = 907,000 inch lbs.
Now adding the maximum dead- and live-load moments and impact

together we have
46,500 + 937,500 + 907,000 = 1,891,000 inch Ibs.,

which is the total moment which the girder under each rail must be
designed to resist.

Dividing this by 16,000 (see Art. 105) we have
1,891,000
16,000

From Table i we find that this calls for 2—Is 15” x 42# under
each rail.

We can now make a preliminary estimate of the dead load. The
weight of the four Is is 168 lbs. per foot of span, and the details can be
neglected as the channel diaphragm (see Fig. 158) at the center of the
span is all the detail there is to consider. Then adding this 168 lbs. to
the weight of the deck we have 568 lbs., which is 52 lbs. less than the‘
assumed dead load, but as this is within 10 per cent of the assumed load
no recalculations are necessary.

For the end shear or reaction due to dead load we have

ee 2 =1,550 Ibs.

=118.2 for the section modulus.

The maximum end shear or reaction due to the live load will occur
when the two 62,500-lb. axles are on the span and in the position indi-
cated in Fig. 154.

  

Taking moments about the support B we have Ky *
R, x10-31,250x10-31,250x3=0, & W
from which we obtain
RF, = 40,600 lbs., Ae B
which is the maximum live-load end shear or reaction. © 10!
For the impact we have Fig. 154
300

Now adding the above reactions and impact together we have
1,550 + 40,600 + 39,300 = 81,450 Ibs.
for the total end shear or reaction. Dividing this by 600 we have
81,450
600
for the required area of bearing on the masonry for each of the four

supports. This completes the necessary calculations, and next the general
drawing, as shown in Fig. 158, or a shop drawing can be made for. the

 

=1836 sq. ins.
DESIGN OF SIMPLE RAILROAD BRIDGES 901

span. The details shown in Fig. 158 should be studied by the student
until thoroughly understood.

Complete Design of a 15-Ft. Span
130. Data.—
Length = 15’ c.c. end bearings.
Dead load = 330 + 400 = 730 lbs. per ft. of span.
(From (2), Art. 123.)
Live load, Cooper’s £50.
Specifications, A. R. E. Ass’n.
131. Calculations.—For the maximum bending moment due to dead
load we have
M= x me x 15* x 12 = 128,000 inch Ibs.

It is readily seen, from the diagram
in Fig. 151, that the heaviest loading pos-
sible for this span is three 50,000-Ib. axle
loads, and, according to Art. 88, these will
produce the maximum moment when
placed in the position shown in Fig. 155,
where the loads are indicated for one
girder only.

Then taking moments about B (Fig. 155) we have
Rx 15 — 25,000 x 2.5 — 25,000 x 7.5 - 25,000 x 12.50=0,

from which we obtain

 

Fig. 155

R=37,500 lbs.
for the reaction at 4. Then taking moments about the center load\we have
M’ = (87,500 x 7.5 — 25,000 x 5) 12 =1,875,000 inch Ibs.,
and for the impact we have

300 :
I=1,875,000 (23m) = 1,785,000 inch lbs.
Now adding the dead and live moments and impact together we have
3,783,000 inch lbs. for the maximum bending moment. Then dividing
by 16,000 we have

38,783,000
16,000

for the section modulus which calls for 2—Is 20” x 70# under each rail.

This beam appears a little heavy, from the section modulus, but the
material cut out of the web along the vertical row of rivets at the center
of the span must be taken into account. This is done by subtracting the
moment of inertia of the material cut out of the web from the moment of

= 236.4
202 STRUCTURAL ENGINEERING

inertia of the beam. Then substituting this net moment of inertia in the
formula

ily

~

which is Formula D, Art. 53, we obtain the maximum stress on the beam,
which should not.exceed 16,000 lbs.

The area of cross-section cut out of the web at each rivet hole,
assuming that 3” rivets are used, is }# x 7%=0.46 square inches. Then
taking moments about the neutral axis, which we will assume is at the
third rivet from the bottom of the beam, which is only approximately true
(see drawing, Fig. 160), we have

2 a8
0.46x6.75 +0.46x2.5 =23.8

for the moment of inertia of the material cut out above the neutral axis,
and

fs 2,
0.46x2.5 +0.46 x 6?=19.4

for the moment of inertia of the material cut out below the neutral axis.
Then adding these together we have 43.2 for the total moment of inertia
of the material cut out. Subtracting this from the moment of inertia of
one beam we have

1219.9 —43.2=1176.7

for the net moment of inertia of one beam.
Now substituting in the above formula we have

fol 3,783,000 x 10
“9% 4176.7

which is very nearly the allowable stress. So the beam has pzactically
the correct section. .

The rivet holes in the beams shown in Fig. 158 are so near the
neutral. axis that the moment of inertia is affected but little, and hence
the material cut out was not considered in designing those beams.

We can now make a preliminary estimate of the dead weight.

The four beams will weigh 280 lbs. per foot of span and the laterals
and details, including lateral connections and diaphragms, will be about
30 lbs. per foot of span, so the total effective weight of metal per foot is
about 280+30=310 lbs. Adding this to the weight of the deck we have
710 lbs., which is 20 lbs. less than the assumed dead weight, but this is
much less than the allowable 10 per cent deviation, so recalculation is
unnecessary.

The weight of details can be correctly assumed only by experienced
designers. It is necessary that the student draw out the details to some
extent in order to get the weight of metal to any reasonable degree of
accuracy.

For the maximum end shear or reaction due to dead load we have

R= = x 7.5=2,740 lbs.

The maximum live-load end shear or reaction wit] occur at A when

the wheels are in the position shown in Fig. 156.

=16,100 Ibs.,
DESIGN OF SIMPLE RAILROAD BRIDGES 203

Then taking moments about B we have
R’ x 15 — (25,000 x 15 + 25,000 x 10 + 25,000 x 5) =0,

from which we obtain
R’ =50,000 Ibs.

for the maximum live-load end shear or reaction. For the impact we have

300
I=50,000 (he

Adding these reactions and impact together we have
2,740 + 50,000 + 47,600 = 100,340 Ibs.

for the maximum end shear or reaction.
Dividing this by 600 we have
100,840
600
for the required bearing on the masonry, for each of the four supports.

According to the A. R. E. Ass’n Specifications the lateral force on
the laterals will be

200 + 5,000 x 0.10= 700 Ibs. per foot of span.
This force is applied to the laterals only at the central connection. Let

Fig. 157 represent the plan of the laterals. There will be a load of
100 x 7.5= 5,250 lbs. applied at the central connection which must be

**

) = 47,600.

= 167.2 sq. ins.

S250

 

 

 

 

 

 

2ee5®
.

 

2625

Fig. 156 Fig. 157

transmitted to the ends by the laterals, one-half going each way. So,
evidently, the maximum shear in each panel will be 2,625 Ibs., which is
one-half of 5,250 lbs. The stress in each lateral will then be equal to
2,625 x secO. SecO is equal to about 2.6. Then we have 2,625 x 2.6 =6,825
lbs.-for the stress in each lateral. If this stress be tension, as it would be
in the case shown in Fig. 157, the net area required in each lateral would
be equal to
6,825

16,000

but when the lateral force is exerted in the opposite direction the laterals
would be subjected to stress of the same intensity, but it would be com-
pression, in which case the area required would be equal to

6,825
(16,000 - 0 “)

= 0.43 sq. ins.
204 STRUCTURAL ENGINEERING

So the laterals must be designed as compression members as well as
tension members. Now as the stress is so very small the ratio of length
to radius of gyration will really govern. This should not be over 120,
As regards construction, a single angle for each lateral is the best section.
Then if we use an angle the first question is—which radius of gyration
shall we use in designing it? The laterals are fixed at their ends in the
horizontal plane so that if the radius about the vertical axis be used, only
one-half of the length of each lateral should be taken as the length. In
the vertical plane the laterals are only partially fixed at their ends, as the
connection plates resist bending upward and downward only to a limited
extent so that if the radius about the horizontal axis be taken, the full
length of the lateral should be taken, which, to be sure, is on the side of
safety. As the laterals are fixed in the horizontal plane and partially
fixed in the vertical plane they cannot fail readily about the 45° axis, so
it would not be correct to use the least radius of gyration of an angle.

From this it is seen that the radius about the horizontal axis is the
one to use. The length of each of the laterals is about 8 ft. or 96 ins.
Then the radius of gyration about the horizontal axis must not be less than

96

120 ae,
which permits the use of a 3x3” angle, but the specifications limit use
to 34” x 34” x 2” angles, so this size will be used.

In the designing of the transverse struts, sense of fitness must govern
to some extent as rigidity is the thing sought. However, the stress is
readily determined. The one at the center of the span has the greatest
load which is 5,250 lbs., which produces a compression stress in the strut
of that amount, as is readily seen from Fig. 157. A 9” [ is the maximum
size that will give us a convenient three-rivet connection. This size also
looks about correct as far as rigidity is concerned.

The length of these struts is about 42 ins. and the least radius of
gyration of the 9” [ is 0.64. Then we have

42 L
ieee ee

which is quite low. So the 9” [ appears to be satisfactory, and we will
use 1—[ 9” x 20# for each transverse strut.

This completes the necessary calculations and next a general draw-
ing as shown in Fig. 159 can be made. This drawing shows the design only
in a general way and the bridge could not be fabricated from it. Such
drawings are usually made in railroad offices and in the offices of con-
sulting engineers. A bridge company would make a shop drawing as
shown in Fig. 160, and, in addition, the necessary shop bills from which
the bridge would be fabricated.

The following shop bills are about what a bridge company would
require for the work. However, bridge companies use printed forms.

Page No. 1 of the shop bills, which would accompany the shop
drawing shown in Fig. 160, is principally for the templet and bridge shop.

Pages No. 2 and No. 5 are principally for the forge shop.

Page No. 3 is principally for the pattern shop and foundry.

Pages No. 4 and No. 5 would be used by the shipping department.
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GENERAL BRIDGE GOMPANY
SUMMARY OF FIELD RIVETS AND BOLTS

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213
214 STRUCTURAL ENGINEERING

The superintendent of the shops, the inspectors, the erectors and
others would receive the shop drawing and bills, depending upon the way
the work is handled. The different companies do not have exactly the
same kind of bills nor do they handle the work exactly in the same way.
The above bills are intended only as a fair example of shop bills.

DRAWING ROOM EXERCISE NO. 2

Design a 12-ft. I-beam span and make a general drawing to a 1”
scale and a tracing of the same. The details to be similar to those shown
in Fig. 159.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Data:
Length =12’ 0” c.c. end bearings.
Width=5’ 0” c.c. of girders.
Dead load to be assumed.
Live load, Cooper’s £50.
Specifications, A. R. E. Ass’n.
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Fig. 161

A layout to a 13” or 3” scale similar to the one shown in Fig. 161
should be made before beginning the final drawing for the span in order
to determine the correct dimensions of details.
DESIGN OF SIMPLE RAILROAD BRIDGiS 215

DECK PLATE GIRDER BRIDGES

132. Preliminary.— Deck plate girder bridges are used for spans
ranging from 25 ft. to 110 ft. There have been a few spans longer than
110 ft. built, but such lengths are not economic, as a rule.

_ A single-track deck plate girder bridge is composed of two main
girders connected to each other by cross frames and laterals. An isometric
view of a typical single-track span is shown in Fig. 162 where the names
of the different parts of the structure are given.

Double-track deck plate girder spans are, as a rule, composed of two
single-track spans placed side by side.

   

Fig. 162

Complete Design of 50-ft. Single-Track Deck Plate Girder Span
133. Data.—

Length=50’-0” c.c. end bearings.
Width = 6’-6” c.c. girders.

Dead load = (750 +400) =1,150 lbs. per ft. of span (Art. 124).
Live load, Cooper’s £50.
Specifications, A. R. E. Ass’n.
216 STRUCTURAL ENGINEERING

134, Calculations for Main Girders.—For the maximum bending
moment due to dead load we have

M =i x 55 x 50 x 12=2,156,000 inch Ibs.

From inspection of the diagram of the loading in Table A, we can see
that the maximum moment due to live load will, very likely, occur under
one of the heavy wheels, either wheel 12 or 13. Let us assume wheel 13,
and let us consider wheels 9 to 16 on the span. Taking moments about
wheel 16 (using Table A) we have

— 2,740
7 155-26

 

= 21.24 ft.

for the distance that the center of gravity of these wheels (9 to 16
inclusive) is to the left of wheel 16. This shows that the center of gravity
comes 2.24 ft. to the left of wheel 13. Then the maximum bending
moment under wheel 13 will occur when the loads are in the position
shown in Fig. 163. (See Art. 88.)

 

 

2.24
“2 yehbue 488
(ie oe DiQ1 9:0 eee
ii 25’ a 2s! i
q *|
Fig. 163

Taking moment about B (Fig. 163), and using Table A, we have
Rx 50-2,740 - (155 - 26) 4.88 =0,

from which we obtain
R= 67,390 lbs.

for the reaction at 4.
Then taking moments about wheel 13, considering the forces to the
left, we have

M’ = 67,390 x 26.12 — 818,000 = 942,200 foot lbs.

or 11,306,000 inch Ibs. for the bending moment at that wheel when wheels
9 to 16 are on the span. Next, suppose wheels 10 to 16 on the span.
Taking moments of these wheels about 16 (using Table A) we have

Ea (2388

ae) = 18.57 ft.
DESIGN OF SIMPLE RAILROAD BRIDGES O17

for the distance that the center of gravity of the wheels (10 to 16
inclusive) is to the left of wheel 16. Therefore, the center of gravity is
0.43 ft. to the right of wheel 13. But when these wheels are placed on
the span for maximum moment, wheel 17 comes on from the right. So
we will consider wheels 10 to 17 on the span. Taking moments about
wheel 17 we find that the center of gravity of wheels 10 to 17 is 2.9 ft.
to the right of wheel 13. So this group of wheels will be in the position
shown in Fig. 164 when the maximum under wheel 13 occurs.

 

 

 

 

 

 

2.9
23.55! 23.55! ;
IAS PS 145 ahs
a -@ 8" QD: @:@i@ 9! @@6'da
R es5' Hk es’ RI
Fig. 164

Then taking moments about B (Fig. 164) (using Table A) we have
Rx 50-2,851- (142-13) 1.45=0, .
from which we obtain
R=60,760 lbs. for the reaction at A.
Then taking moments about wheel 13 we have
M” = 60,760 x 23.55 — 480,000 = 950,900 foot Ibs.

or 11,411,000 inch pounds for the maximum bending moment under wheel
13 for this group of wheels. This is the absolute maximum as will be
found by further trials. It is seen that the center of gravity is a little
nearer to wheel 14 than it is to 13, yet the maximum moment occurs under
wheel 13. This is one of the few cases where the maximum does not
occur under the wheel nearer the center of gravity. (See Art. 88.)
Generally, the position of the wheels for maximum moment can be ascer-
tained the first trial. The above span is about as troublesome as any
found.

For the maximum impact we have

300 aA
1=11,411,000 x (aon) = 9,780,000 inch lbs.

Now multiplying the above maximum moment and impact each by
50/40, as the loading specified is E50, and adding these results to the
dead-load moment, we have

2,156,000 + 14,263,000 + 12,220,000 = 28,639,000 inch lbs.
for the total maximum moment on the span.

The next thing is to determine the economic depth. Assuming the
thickness of the web to be 3” and substituting in equation (6), Art. 113,

we have
[28,639,000 = .
2=1.055 16,000x = = 72.7 ins.,

so we will use a 72” x 2” web.
218 STRUCTURAL ENGINEERING

We can now determine the flange area required. It is readily seen
that the effective depth will likely be a little less than the depth of the
web, so let us assume 71” as the effective depth. Then we have

28,639,000 + 71 = 403,000 Ibs.
for the flange stress.

Then 403,000 + 16,000 = 25.2 sq. ins.,

the net area required for each flange.
Using the following:
2—Ls 67x 6” x zh” = 12.86 — 2.25=10.610"” net
1—cover plate 14” x 4” = 7.00 -1= 6.004” net
1—cover plate 14/x7%""= 6.12-0.87= 5.250” net
+ of the web = 3.370” net
25.230” net
for each flange we have practically the required area.

We can now determine the actual effective depth. Let Fig. 165
represent the cross-section of the above flange.

 

 

 

 

218656 x2"

 

 

 

 

UU
aa
Fig. 165

3"
8

Then taking moments about the center of the top cover plate (see
Art. 47), we have

—_ 12.86 x 2.42 +7.00 x 0.46
i 25.98
for the distance from the center of the top cover plate to the center of

gravity of the flange.
Then we have

=1.32 ins.

1.32-0.71=0.61 ins.

as the distance from the back of the angles to the center of gravity of the
flange.

The web is assumed 72” deep and according to practice the vertical
distance from the back of top flange angles to the back of the bottom
flange angles will be 0.25” more or 72.25”. Then for the actual effective
depth we have

d= 72.25 — (0.61 x 2) =71.03 ins.,

which is almost what we assumed, so recalculation of the flange is
unnecessary. As a rule, the assumed effective depth does not come so
close to the actual; however, 4” either way will not materially affect final
results. If a greater difference than 4’ is obtained the stress and section
of the flange should be recalculated, using the computed effective depth.
DESIGN OF SIMPLE RAILROAD BRIDGES 919

The thickest cover plates should always be placed next to the flange
angles. So in this case the 4” plate will be next to the angles, as shown
in Fig. 165.

For the length of the 75” cover plate, or outside plate, we have

6.12
50 25.98 =24,3 ft.

(see Art. 114) and for the length of the $” cover plate we have

. 13.12
50 oo = 85.5 ft.

It is customary to make cover plates from two to three feet longer
than the theoretical length, so we will make the above plates 27 and 37
feet long, instead of 24.3 and 35.5.

The next thing to determine is the stiffening angles and the web. To
do this the first thing is to calculate the maximum end shear.

For maximum reaction or end shear due to dead load we have

R = nelbO BO = 14,400 Ibs
ae ge oe

The maximum live-load reaction or end shear will occur when the
wheels are in the position shown in Fig. 166. (See Art. 86.)

2!

COO) 2’ QO. @2@ 2 Ol,
R 50’

Fig. 166

Taking moments about B, Fig. 166, and using Table A, we have
R’ x 50- 4,072 —142 x 2=0,
from which we obtain

R’ =87,100 lbs.

for the maximum reaction or end shear at A due to E40 loading. Then
for E50 we have

50
87,100 x a 108,900 Ibs.

For impact we have

300 _
T= 108,900 x 35, = 93,400 Ibs.

Now adding the above dead- and live-load reactions and impact
together we have
14,400 + 108,900 + 93,400 = 216,700 Ibs.

for the maximum reaction or end shear on each girder.
According to the specifications the outstanding legs of intermediate
stiffeners must not be less than one-thirtieth of the depth of the girder

plus two inches. So we have
72

30 +2=4.4 ins.
220 STRUCTURAL ENGINEERING

for the required width of their outstanding legs. The standard angle
coming nearest to this requirement is a 5” x33”,

y which will be used throughout.
The end stiffeners must be designed to transmit

et let st the total maximum reaction, each pair being consid-
| ered as a column having a length equal to one-half
Noy the depth of the girder. (See specifications.), Then
| s assuming Ls— 5’ x 34”x4” used, we have for each
eat pair a column having a cross-section as shown in
y

Fig. 167. For the radius of gyration of this column
Fig. 167 in reference to axis y-y, we have

_ 2472 G
r= Skeets =2.88.

Then substituting this radius and 36” for the length in Formula Q,
Art. 73, we have
36

16,000 — 70 ——— 3.88 = 15,100 Ibs.

for the allowable compressive unit stress on the end stiffeners.
Now dividing this stress into the maximum reaction we have
216,700
15,100

for the required area of cross-section of the end stiffeners. So we will
use two pairs, or

= 14.35 sq. ins.

4—ts 5” x33” x4”=16.0 sq. ins.,

the 5x 34 x 7g angles being a little too small.

There is no theoretical way of computing the area of the inter-
mediate stiffeners. It is practice to make them as thin as the specifica-
tions will permit. So we will use 5”x33”x 3” angles throughout for
intermediate stiffeners.

Using the assumed web, 72” x 3”, we have

216,700 _
s= oy = 8,020 lbs.
for the maximum average unit shearing stress in the web.
Then substituting in Formula (1), Art. 118, we have

= é (12,000 — 8,020) = 37 ins. (about)

for the required distance between the stiffeners near the ends of the
girders. Now as this spacing is not less than half the depth of the web
the assumed web is economic and hence will be used.

For the bearing on the masonry we have

216,700
600

Each support must be so designed that there will be at least this much
area of bearing on the masonry.

= 361 sq. ins.
DESIGN OF SIMPLE RAILROAD BRIDGES 921

This completes the necessary calculations for the main girders, as
far as the general design is concerned, and the next thing is the designing
of the lateral bracing, that is, the laterals and frames.

135. Calculations for Lateral Bracing.—The lateral bracing
should always be symmetrical about the center of the span. The laterals
should have a slope as near 45° as is practicable. The distance between
cross-frames should never be over 15 ft. In accordance with this there
must be three intermediate cross-frames in a 50-ft. span, as a less number
would place them more than 15 ft. apart. So the lateral bracing will be
as shown in Fig. 168.

 

 

 

 

 

 

 

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12:6'(obout)

Z Of WO) @ iP
w e.g u oe rl Wl 4 8
é 7 ye fF 7 if «

7 6 5&5 4 383 2 /
Bea” @ © © ©
Fig. 168

According to the specifications the laterals must resist a uniform
lateral force of 200 4-0.10 (5,000) = 700 Ibs. per ft. of span, considered
as a moving load.

Suppose this force or load acts from the direction indicated by the
arrows, and suppose it moves on to the span from the right, as a uniform
live load. The panel load at any point h, g, etc., will be

P=00x6.2=4,340 Ibs. (about).

Then for the maximum shear in the different panels, according to
Art. 90, we have
4340

Shear in panel hg=—.- x 1= 540 lbs.;
4340

“ “ “ of Soa x B= 1,630 Ibs.;
“ “ “ fe = x 6= 3,260 Ibs. 3
“é « “e eda. 10= 5,420 Ibs. ;

“ “c se de= SX, 15= 8,140 Ibs. ;

_ 4340

“ “ “ cb

x 21=11,400 Ibs.;

“ “ “ ba= ex 28 =15,200 Ibs.

Now if each of these shears be multiplied by the secant of the angle
marked 6 we shall obtain the stress in the corresponding diagonals (see

Art. 92).
999, STRUCTURAL ENGINEERING

The tangent of angle 6, for the purpose of determining stresses,
can be taken as
6.25 |
6.50 —
and the corresponding secant can then be found in almost any table of
natural trigonometrical functions (see Carnegie or Cambria handbook),
or the secant can be determined directly by arithmetic.
For the above assumed figures we have

Sec 9 = 1.39.

Then for the maximum stress in the diagonals we have
5,420 x 1.89=— 7,500 lbs. for diagonal nd,
8,140 x 1.89 =+11,300 lbs. for diagonal dm,
11,400 x 1.39 =-15,800 lbs. for diagonal mb,
15,200 x 1.39 =+21,000 lbs. for diagonal bl.

These are all of the stresses that we need determine with the load
applied as indicated as the corresponding diagonals in the right half of
the span will have the same stress when the force moves on to the span
from left to right.

If the lateral force comes from the other direction than that indi-
cated by the arrows, the panel loads will be twice as great as given above,
as there are only half as many panels considered.

Then the stress in each of the diagonals nd and dm is

0.962 (about),

Pee x3x 1.39 =+ 9,050 Ibs.
and pases x6x1.39=+ 18,100 Ibs.

in each of the diagonals mb and bl.

We now have the maximum stresses in the diagonals determined
which are indicated in the diagram in Fig. 168.

The plus and minus signs above signify compression and tension,
respectively.

It is seen from the above that the diagonals have to resist both
tension and compression. Compression will likely govern.

Let us try a single angle, say, 1—L 34” x 34” x2”, as this is the
smallest that the specifications will permit. From Table 6, or from a
Carnegie or Cambria handbook, we have 1.07 for the radius of gyration
of this angle about the horizontal axis (see Art. 130) and the length of
each lateral will be about 8 ft. or 96 ins., which can be determined accu-
rately enough by scale. Then substituting in Formula Q, Art. 73, we
have

 

16,000 — 70 “

oF 722700 Ibs.

iS
7
for the allowable unit stress.
Dividing the greatest compressive stress, which occurs in the Jateral
bl, by this intensity we have
21,000

9,700 = 2.16 sq. ins.
DESIGN OF SIMPLE RAILROAD BRIDGES 223

for the required cross-section of the lateral, and the assumed angle has
2.480”, Then, as the ratio of L/r is only about 90, the assumed angle is
safe in the case of the greatest compressive stress.

As is seen above the greatest tension is 18,100 lbs., which occurs also
in lateral bl. Then we have

18,100
16,000

for the required net cross-section, and as the angle assumed has
[2.48 -— (3 x1”) ] 2.114” it is quite safe for the greatest tension. So
we will use 1—L 34” x34” x 3” for each end lateral. But as this angle
is the smallest permitted in this work the laterals throughout will each
be made of 1—3}4” x 34” x 2” angle.

The intermediate cross-frames are not subject to theoretical analyses
and hence their design is governed principally by experience and sense of
fitness.

The end cross-frames can be fairly well analyzed. The stress in the
top angle of the frame can be taken equal to one-half of the lateral force
per foot of span multiplied by one-half the length of the span, and this
force multiplied by the secant of the slope of the diagonals in the frame
is equal to the stress in each of these diagonals. The bottom angle of the
frame has no stress (theoretically).

Then according to the above we have

~ x 25=8,750 Ibs.
for the stress in the top angle of the end frame, and as the diagonals of
the frame slope about 45° with the horizontal we have

8,750 x 1.4=12,250 Ibs.

for the stress in the diagonals of the end frame. From this it is seen that
very small angles are required theoretically for the end frames which
are subjected to greater stresses (apparently) than the intermediate
frames, but as 34” x33” x2” angles are practically the minimum size
used in railroad bridges we will use this size in all of the frames.

This completes the necessary preliminary calculations except for the
preliminary estimate of the dead weight. The stress sheet for the span,
as shown in Fig. 169, can be drawn from the information given in the
above calculations. The estimate of the dead weight in this case will be
deferred until after the detail shop drawing of the span is considered so
as to familiarize the student with details before taking up preliminary
estimates of dead weight.

136. Making of the Shop Drawing.—After the stress sheet
(Fig. 169) is completed the shop drawing for the span, as shown in Fig.
171, can be made. This work is known as detailing and includes not only
the drawing but the calculations for the details as well.

To evolve this drawing (Fig. 171) the details are first drawn in
pencil to a 3” scale upon a 24” x 36” sheet of ordinary drawing paper.
After selecting the relative positions for the different views, so that no
part of the sheet will be crowded, we start the drawing, as shown in Fig.
170, by drawing the center line CC of the span. Then scaling off 25 ft.,

=1.13 sq. ins.
 

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DESIGN OF SIMPLE RAILROAD BRIDGES 295

the line BB through the center of bearing is drawn. Next, the line EE,
through the end of the span, must be located and drawn. The end
stiffeners should be placed 6” back to back as shown, this being about the
minimum distance permissible on account of driving rivets in the out-
standing legs of the stiffeners connecting to the end cross-frames. Then
allowing, say, 2’ from the end pair of stiffeners to the end of the girders,
we have 3” +337+2”=84” for the distance from the line BB to line

€ 8 ie

 

 

 

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20;
Re

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 170

EE, and the line EE locating the end of the span can then be drawn.
Then the top plan of each girder is drawn as shown in Fig. 170. The
next thing after that is to locate the cross-frames ds shown, so that the
lateral bracing will be symmetrical about the center line CC, and the
laterals will all have the same length (thus saving templets). Now there
will be a pair of stiffeners on each girder at each cross-frame, one of
which, in each case, will be connected to a frame, and when the position
of the cross-frames is fixed these stiffeners to which the frames connect
are also fixed in position. Then the elevation of one of the girders,
always the far girder, can be drawn and the drawing completed as far as
is shown in Fig. 170.

The next thing is to draw the laterals and the lateral plates connect-
ing the laterals to the girders (as shown in Fig. 171). As the laterals
are all of the same section and the strength of each is greater than is
required by the stress, each end connection must contain enough rivets
to develop the strength of the lateral in compression (see specifications),
Now, as shown above, each lateral will safely resist 9,700 lbs. per square
inch of cross-section. Then, as 2.480” is the area of cross-section of
each Jateral, we have 2.48 x 9,700 =24,000 lbs. for the total allowable
compressive stress in each lateral. Then using {” field rivets for the
 

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226
GENERAL BRIDGE COMPANY
SuHop BILL.

Name of StRucTURE 20% S7 Deck Plate Girder Bridge, tor ANEYRR.
r ' Gale’d Order

. ‘ Ite
Kind {Section Remarks] we} Se .

of

717,07 Bot

Drawing No,--4 ef f.__~____..----------Contract No 382_.
Made by--¢:0.m.%2 /9/0. Checked by. 464 %4/2Fage No.-2_...

 

227
GENERAL BRipGe GoMPANY
SHOP BILL.

M a : Galed rder

Kind ction ; Remarks Item

of ,

Drawing No,--L -Contract No, 382
Made by _¢:0.. V2 /910.Checked by @k4"%Page No._2

 

228
 

GENERAL BripeGe ComPany
MISCELLANEOUS

NAMEOF STRUCTURE $0/t Deck Plate Girder Bridge tor AN. AY RR. _

 

 

 

 

 

 

 

 

 

 

 

 

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229

 
GENERAL BRIDGE COMPANY
SKETCH SHEET.

 

Name of STRUCTURE 50/25. 7-Deck Plate Girder Bridge for AMBYRR.__.

 

    

3”
62" fe

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2oxe”

     
       
  

  

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Contract No. 382
Made by ¢.2.m. Z21910___Checked by 628. %410__ Page No. 4

 

 

 

230
GENERAL BrRinGE Company
EREC TORS LIST OF FIELD RIVETS AND BOLTS

No. n Bo 3
of |6rip |i Under H Pieces Connected

Drawing No../0f/__..___
Made by Checked by GAR4” Page No..F___

 

231
GENERAL BRIDGE COMPANY
SUMMARY OF FIELD RIVETS AND BOLTS

; ; Ie'd
oF | Dia, Lof ivet Head ee Weight

Drawing No, - Contract No, 282._
Made by ¢.2.7%. Checked by @2.2G" Page No. _.©

 

232
GENERAL BRIDGE COMPANY
SHIPPING BILL

NAME OF STRUCTURE SOHES. 7. Deck Plate Girder Bria

moer: led
a Name Mark |Sheet Is ve of Remarks |Weight

6 §/-ex.Nofon

Contract No.-38E.
Made by -:0:t..%19/0__ Checked by €8L 4”. Page No.-7__...

 

233
234 STRUCTURAL ENGINEERING

lateral connections, the value of which in single shear will govern, 6,000
‘bs. being the allowable value of each, we have

24,000 _ 4
6,000 —

for the number of rivets required in each end of each lateral.

Now, the number of rivets connecting the lateral plates to the girders
must be sufficient to take the component along the girder of the lateral or
laterals. Twenty-four thousand pounds is the strength of each lateral,
and using {” field rivets in single shear, the number of rivets for inter-
mediate points will be

24,000 x sind 24,000 x .695\ _
2( 6,000 )- 2( 6,000 ) ae

The number used would be not less than 6, or 3 for each lateral.

The laterals and lateral plates can now be drawn in the top plan
(shown in Fig. 171), giving the spacing of the rivets at each connection,
and determining clearances and the sizes of plates by scale as the work
progresses, using separate sketches for each point drawn, say, to 14” or
3” scale. After this work is completed the stiffeners between the cross-
frames can be drawn on the elevation of the girder, care being taken to
space the stiffeners so as to miss the lateral plates as much as possible.
Then the longitudinal section showing the bottom flange can be drawn.

After this the next thing is to space the rivets in the vertical legs of
the flange angles. As the live load is applied directly to the top flanges
and the specifications ignore the one-eighth of the web in determining the
spacing of flange rivets, ‘the problem of determining the spacing comes
under Case III, Art. 116.

We assume that each wheel of the live load is distributed over three
ties and assuming ties 8’ wide and spaced 6” apart (face to face) we
have 42” for the length over which each wheel is distributed. Then for
the greatest vertical load per linear inch of flange due to the live load
(Cooper’s £50), when the maximum shear occurs, we have

25,000
42
per linear inch, and adding a 100 per cent for impact will make a total
of 1,200 lbs. per inch.

Now referring to Case III, Art. 116, we have

v= 1,200 lbs.

R= 17,880 Ibs., which is the allowable bearing of a 3’ shop
rivet on the 2” web.

S= 216,700 lbs., which is the maximum shear at either end of

each girder.
h= 66 ins.

Then substituting these values in Formula (4), Art. 116, we have
7,880

a ‘ (216,700 \2
1,200? + 66

 

= 595 Ibs., say 600 Ibs.,

=2.247 ins. (about)

 
DESIGN OF SIMPLE RAILROAD BRILGES 235

for the theoretical spacing of the rivets near the ends of the girders, and
the theoretical spacing at other points along the girders can be deter-
mined by computing the maximum shear at each of the points and apply-
ing Formula (4), Art. 116, in each case. However, the following graph-
ical method of determining the spacing is preferable as it saves time and
is accurate enough for practical work:

The shear due to dead load is quite small compared with that due to
live load and impact, so that the maximum shear can be assumed to vary
across the span as the ordinates to a parabola (see Art. 55).

Now the above spacing (2.27) gives about 5.3 rivets per foot of
girder. Then if we lay off a vertical line AB (Fig. 172) equal to 5.3”,
and taking a pair of dividers divide the line AB into any convenient

  
 

 

 

 

 

 

6:38 x y
ee e a
‘ 3
A. | a
\
%,
ps3 S
y '
\
2
<
pf
y
XN
he
a
| Spon Length
| I
Fig. 172

number of equal parts and from A step off the same number of equal
parts on the line AC (the equal parts on the line AC can be taken any
length, just so A to C is a convenient distance), and construct the
parabola CDB (see Art. 58) and draw the horizontal lines to the curve
through the points 1, 2, 3, ete., we have a diagram from which the
required spacing of the rivets at any point between the end and the
center of the span can be obtained, as the distance AC corresponds to
the length of the span and the line AB gives the shear (correspondingly)
in number of rivets per foot of girder. From the diagram (Fig. 172) it
is seen that about one and one-third rivets per foot are required at the
center of the span, and practically three rivets per foot at the quarter
point. The required spacing at all other points from 4 to the center of
the span is seen at a glance.

It is not practical, if not impossible, to space the rivets throughout
a girder according to the theoretical requirements, as no two spaces
would be the same (which is impractical) and the spaces next to stif-
feners can not be less than about 34’, which in some cases is greater than
the required; so the best we can do is to fit the rivets in between the
stiffeners to about the theoretical spacing.
236 STRUCTURAL ENGINEERING

The theoretical spacing of the flange rivets as determined in the
manner shown above will satisfy the requirements of the specifications,
which are as follows:

“The flanges of plate girders shall be connected to the web with
sufficient number of rivets to transfer the total shear at any point in a
distance equal to the effective depth of the girder at that point combined
with any load that is applied directly to the flange.”

This is simply a practical manner of obtaining approximately the
theoretical spacing. To show the justification of the above requirement
given in the specifications let AB, Fig. 173, represent a plate girder.
Let M and S be the bending moment and shear, respectively, at section
C. Then for the bending moment at section D we have

M’=M+Szr-3% Pz. (See Art. 66.)

Now, Sx-% Pz is the difference between the moments at the two sections.
If the intervening loads be neglected, which is an error on the side of
safety, the difference between the two
moments is Sx, and if this be divided by
the effective depth the result would be
the difference between the flange stresses
at the two sections, which would be
equal to the shear when x is equal to the
Fig. 178 effective depth, as is readily seen. This
difference of flange stress is simply the
increment of the flange stress between the two sections and of course
there should be a sufficient number of rivets to transfer this from the
web to the flange and at the same time support whatever vertical load
there is applied to the flange. This is entirely in accord with Arts. 115
and 116.

Now, having the stiffeners spaced as shown in Fig. 171 we can begin
spacing the rivets in the vertical legs of the flange angles. Beginning at
the end of the girder, the theoretical spacing as given above is 2.27’, but
we will use 24” in order to have practical spaces. So we obtain the
spacing from the end of the girder to the first intermediate stiffener as
shown in Fig. 171. Between the first and second stiffeners we can
increase the spacing a little as shown by the diagram in Fig. 172, and
between the second and third stiffeners we can increase the spacing a
little more and so on towards the center of the span until the limit of 6”
spacing is reached, which according to the specifications must not be
exceeded. It is practice to limit the fractions in the spacing to no less
than }”. If necessary the stiffeners can be shifted slightly to suit the
spacing. :

After the rivets are spaced in the vertical legs of the flange angles
the rivets in the horizontal legs, connecting the cover plates to the angles,
can be spaced. The theoretical requirement in this case is that the rivets
be spaced so that the number of rivets per foot will be sufficient to
transmit the part of the flange increment taken by the cover plates. The
part of the flange increment taken at any point by the cover plates will
be to the total flange increment at that point as the area of the cross-
section of the cover plates is to the total area of the cross-section of the
flange at that point. As an example, let M be the bending moment found

=x

 
DESIGN OF SIMPLE RAILROAD BRIDGES 937

at section C (Fig. 173) and M’ the bending moment at section D, which
is a short distance to the right of C (say, a foot). Then for the total
flange increment between the two sections we have

AL —M’
h

where h is the effective depth of the girder. . Now it is obvious that there
must be a sufficient number of rivets in the vertical legs of the flange
angles between the two sections to take all of this increment as the total
increment is transmitted through these rivets, but as the cover plates
take only their proportional part of the increment, the number of rivets
connecting them to the flange angles need be only sufficient to take their
part. So if r be the allowable stress on each rivet connecting the cover
plates to the flange angles, and A the total area of cross-section of the
flange between section C and D, and A”, the area of the cover plates, we
have

=F,

FA”
ne RT

for the number of rivets required in the cover plates between the two
sections and the number of rivets required to connect the cover plates to
the flange angles at any other point along the girder can be determined
in the same manner, but the same result can be obtained in any case with
much less work by using such a diagram as shown in Fig. 172. So we
will use the above diagram (Fig. 172) in this case. Beginning at the
center of the span, the number of rivets per foot required to take the total
flange increment is 14. Then as the net area of cross-section of the
flange at that point, not including the one-eighth of the web, is 21.860”,
and the net area of the cover plates is 11.254’’, we have

1s 11.25
21.86

for the number of rivets per foot in the cover plates, which gives a
spacing of 17.6’. This would be the spacing if the allowable stress on
the rivets in the horizontal legs were the same as in the vertical. But as
the rivets in the horizontal legs must be considered in single shear and
the rivets in the vertical legs in bearing on the 3” web, the above number
(0.68) must be multiplied by 7,880/7,200. So we have
11.25 — 7,880
nali*o7 oe © F200

for the number of rivets per foot in the cover plates at the center of the
span, which gives (12/0.75) 16” spacing, but as there are two rows of
rivets in the cover plates the spacing of the rivets in each flange angle
would really be twice this, or 32”.

Again, from the diagram (Fig. 172) the required number of the
rivets in the vertical legs of the flange angles at the quarter point is 3 per
foot. Then for the number of rivets required at that point in the cover
plate (as the 7%” plate only extends to about that point), we have

6.00 _ 7,880

3X 7661 * 7,200

1 =0.68 (about)

 

=1.18 per foot,

.
’
238 STRUCTURAL ENGINEERING

which gives a spacing of 10.1” or 20.2” along each angle. Now, if
other points along the girder be considered it will likewise be found that
the required spacing of the rivets in the cover plate or plates will exceed
the 6” maximum allowed. So that if we were to space the rivets in the
cover plates, using the 6” maximum allowable throughout, there would yet
be an excess of rivets. Now the spacing of the rivets in the lateral plates
is determined when the laterals are detailed, as stated above, and as
these rivets must be spaced to suit those details it only remains to fit the
rivets in between the lateral connections using as near the maximum 6”
spacing as will fit in nicely without interfering with the stiffeners. So
we obtain the spacing shown in the plan of the top flanges (Fig. 171) and
the same spacing can be used in the bottom flange, as shown in the longi-
tudinal section below the girder. As is evident, the spacing selected at
the different points depends entirely upon individual judgment and hence
no two persons are likely to make the same selection. However, this is
not a serious matter at all as any reasonable variation is permissible.

We shall next space the rivets in the stiffeners. Now according to
shop practice the rivets in the stiffeners should all line up horizontally
all the way across the girders.

There is no definite rule for determining the number of rivets
required in the intermediate stiffeners other than that there should be a
sufficient number to clamp the stiffeners firmly to the web, and the
maximum 6” spacing usually suffices for this, but the end stiffeners should
be connected with a sufficient number to transmit the total end shear
from the web to the masonry, as these stiffeners are really columns bear-
ing against the bottom flange of the girder and really transmit this force.
It is customary to space the rivets in the end stiffeners so that about
every other rivet of this spacing can be omitted for the spacing in the
intermediate stiffeners whereby an economic spacing that lines up hori-
zontally throughout the full length of the girder is obtained. It is
economic to crimp all stiffeners on all girders over three feet deep, but it
is usual shop practice to put fillers under end stiffeners and under the
stiffeners at all points where cross-frames connect. So we shall put
fillers under the end stiffeners and under the stiffeners at cross-frames.
The rivets in the stiffeners are in double shear and bearing on the 2” web.
Then, using {’ shop rivets, we have

12,000 x 0.6 x 2=14,400 Ibs.
for the allowable shear on each rivet and
24,000 x $x 2=7,880 lbs.

for the allowable bearing on each rivet. So the bearing governs the
number of rivets.

Now, taking the case of the end stiffeners, the maximum end shear
being 216,700 lbs., we have

216,700
7,880

for the number of rivets required in the two pairs of end stiffeners at each
support. But, as the end stiffeners are on fillers there must be an excess
of 50 per cent, according to the specification, so 42 rivets will be used as

= 27.5, say, 28,
DESIGN OF SIMPLE RAILROAD BRIDGES 239

shown (Fig. 171), and by omitting every other rivet in the spacing shown
in the end stiffeners we obtain the spacing shown in the intermediate
stiffeners. The rivets in the stiffeners can be made to line up horizontally
without any trouble if 3” spacing be used in the end stiffeners. In case
this spacing gives more rivets in the end stiffeners than are required,
some of the rivets can be omitted, thus making some 6” spaces.

The distance from the toe of any flange angle to the first rivet out
(in the stiffener) from the flange, according to shop practice, should be
equal to not less than twice the thickness of the flange angle plus 14”.
So in this case we have 14 +147 +14” =41” for the minimum allowable
distance from the outer gauge line in the flange angles to the first rivet
out from the flange. These distances are given in the spacing at the end
of the girder (Fig. 171) as 48”, which exceeds the minimum allowed
by 3”. These spaces should not be less in any case nor much more than
4” greater than the minimum allowed in conformity with the practice
mentioned above.

After the rivets are spaced in the stiffeners the web splices can be
calculated and drawn to conform to this spacing. The number of web
splices is always limited in all cases to as few as is possible, which
depends altogether upon the maximum length of the web plates obtainable
from the steel mills. Each splice should be at a stiffener. The maximum
length of 72” x” plates is given in Table 9 as 30 ft. So one splice is
necessary in the above 50-ft. span. ;

Let us use the type of splice shown at (a), Fig. 150, Art. 117. The
first thing to do is to determine the size of the longitudinal splice plates
next to the flanges. These plates should have at least three horizontal
rows or rivets in them in order to get a good hold on the web. This much
is simply a matter of judgment. So let us assume that there are three
horizontal rows of rivets in each pair, then the distance from the center
of gravity of the rivets in the top pair of plates to the center of gravity
of the rivets in the bottom pair is equal to the distance from the second
rivet, from the top flange, to the second rivet from the bottom flange,
which, as seen from the spacing on the girder (Fig. 171) is 4 ft. or 48”.
Now, according to Art. 117, using the figures given above, we have

Gass x =t) 11=48xF,

from which we obtain

F=79,600 lbs.

for the direct longitudinal stress on each pair of plates.

Then we have 79,600 + (16,000) 48/71 =7.350” for the required net
area of each pair of plates, or about 3.675” for each plate. Allowing 3”
clearance between these plates and the flange angles and the same clear-
ance between them and the vertical splice plates, we obtain 103” for
their width. Now, let us assume that they are 4” thick, and that their
cross-section is reduced by three rivet holes (for {’’ rivets), then we have

(104 x4) -1x4$x3= 3.630”

for the net area of cross-section of each plate, which is about equal to the
required area. But if the plates be $” thick a 7%” filler would be
240 STRUCTURAL ENGINEERING

required under each of the stiffeners at the splice and to avoid these thin
fillers we will make the plates as thick as the flange angles, which is 7%”.
This, of course, gives a little excess area, but the result obtained justifies
its use.

The next thing is to determine the number of rivets in these longi-
tudinal plates. As the number of rivets depends upon their allowable
bearing upon the 2” web, which is

(7,880) = = 5,330 Ibs.,

we have
79,600
5,330

for the number of rivets required on each side of the splice in each pair of
plates; and spacing them in the plates, so as to conform to the spacing in
the stiffeners and flange angles previously established, we obtain the
spacing shown in Fig. 171.

The next thing is to determine the size of the vertical splice plates
and the number of rivets required in them to transmit the maximum shear
at the splice as explained in Art. 117.

: Oe gag 4,

IR 2s’
S0!/ x]
Fig. 174

=15 (about)

 

 

 

 

Placing the wheel loads on the span so that wheel 2 will be at the
splice, as shown in Fig. 174, and taking moments about the support B
(using Table A), we have

1,640 + 103 x1
R= eee
(ea

for the reaction at A, then, we have

) 1,000 = 34,900 Ibs.

(84,900 — 10,000) a = 81,200 Ibs.

for the maximum live-load shear at the splice.
For the impact we have

300
ae (; 5 +300

Now, as the dead-load shear at the splice is zero, we have
31,200 + 28,800 = 60,000 Ibs.

for the total maximum shear at the splice.
Then we have

) = 28,800 lbs.

60,606
10,000

for the required net area of the vertical section of the vertical splice

 

=6.0 sq. ins.
DESIGN OF SIMPLE RAILROAD BRIDGES 941

plates, but as the splice plates are made 7,” thick in order to avoid the
qs” fillers under the stiffeners, referred ta vabiies there will be consider-
able more metal in them than is required for shear, so they will be amply
strong as far as shear is concerned. This.is practically always the case,
for the splice plates cannot, according to the specifications, be less than
#” in thickness and the two of them will always be thicker than the web
they splice. The number of rivets required to connect these plates to the
web will depend upon their allowable bearing on the 3” web as shown
above. So we have

60,000

7,880

for the required number of rivets on each side of the splice, but a greater
number will be used as there must be two rows of rivets on each side of
the splice in order to securely clamp the plates to the web and the spacing
must conform to that already established in the stiffeners. So following
out these requirements without spacing the rivets farther apart than 6”,
we obtain the spacing shown in Fig. 171.

Now to test the splice as a whole, let S =the horizontal stress on each
of the rivets in the horizontal rows farthest out from the center of the
web, that is, in the rows next to the flanges, and let V = the vertical shear
on each rivet, which we will assume to be the same for each rivet in the
splice, that is,

= 7.6, say, 8,

60,000
V =F — =1,764 lbs.

Now taking moments about the center of the web, as explained in
Art. 117, we have

2 = [324 624 92+(12)2+ 2(15)24+ 2(18)%+ 5(21)2+ 5(24)2+ 5(27)?]=
16,000 x 3.37 71,

from which we obtain

§=5,118 Ibs.

Then for the maximum resultant stress which occurs on the outer

rivets, we have

R=5118?+1,7642=5,413 (about),
which is about 567 Ibs. less than the allowable, which is 5,980 = (7,880)
27x 2/71. So the splice is all right and the drawing of it can be com-
pleted as shown.

It is seen from the above equation that the rivets near the center of
the web are of little value, in resisting bending.

We will next take up the detailing of the cross-frames. The first
thing to do is to draw the vertical lines representing the centers of the
girders as shown in the detail to the right of the girder. Next the top
and bottom horizontal angles should be located and the position of the
diagonal determined. Then the next thing is to determine the size of
each of the connection plates marked ak. The size of each of these plates
will depend upon the number of rivets in the ends of the diagonals. As
the diagonals take both tension and compression we consider them as
942 STRUCTURAL ENGINEERING

compression numbers, and assuming them independent of each other, each
has a length of 97” (about). Then we have
97

16,000 - 70 Loy = 9,650 lbs.

for the allowable unit compressive stress on each diagonal,
and 9,650 x 2.48 = 23,900 Ibs.
for the allowable stress on each. Then using §” shop rivets, we have

23,900
7,220
for the number of rivets required in each end of the diagonals.

Now a large scale drawing of one of the four plates ak, showing the
entire detail at that point (girder and all) can be made from which the
location of the rivets and all necessary clearances can be determined.
Then the cross-frame and the cross-section of the girder can be drawn as
shown and the pencil drawing is then complete and the tracing of the
same can be made and by adding the required list and writing on the
title and general notes we have the shop drawing for the span completed
except for the anchor bolts and cast pedestals which are detailed on
sketch sheets included in the above shop bills which are made after
the shop drawing is completed. After the shop drawing and bills are
checked and approved blueprints are made of them which are used in the
fabricating, shipping, and erecting of the span.

137. Estimate of Dead Weight and Cost of Span.—From the
shop drawing (Fig. 171) we have the following weight of metal:

= 3.3, say, 4,

Estimate of the weight of main girders.

1—web 72 x $x 91.8 lbs. x 51.4’... 0... eee 4,718 Ibs.
418 6x 6x gx 21.9 Ibs. x 514’. cece eanageetaane 4,503 Ibs.
2—cov. pls. 14x 7% x 20.82 Ibs. x 27.0’ (top and bot.)....... 1,124 Ibs.
1—cov. pl. 14x 4 x 23.80 Ibs. x 51.4’ (top).............0.. 1,223 Ibs.
1—cov. pl. 14 x4 x 23.80 lbs. x 37.0’ (bot.)............... 881 Ibs.
8—Ls 5x 34x4x 18.6 lbs. x 6.0’ (end stiff.).............. 653 Ibs.
26—Ls 5x 34x23x 10.4 lbs. x 6.0’ (int. stiff.)............... 1,622 Ibs.
12—fills. 34.x 7% x 6.7 Ibs. x 5.07.0... ee cece eee tenes 402 Ibs.
2—sp. pls. 16 x 7% x 30.6 Ibs. x 3.2’... ce 196 Ibs.
4—sp. pls. 10x 7x 19.14 Ibs. x 2.1’.... 0... eee 161 Ibs.
2—sole pls. 14x $x 35.71 Ibs. x 1.8’... ce eee 93 Ibs.
15,576 Ibs.

1,015* rivets @ .44 lbs. per pair of heads (}” rivets)...... 446 lbs.
Total weight of one main girder................ 16,022 Ibs.

Total weight of two main girders............... 82,044 Ibs.

Total weight of main girders per foot of span equals

32,044
50

*¥or weight of rivet heads see Carnegie Handbook.

= 641 Ibs.

 

 
DESIGN OF SIMPLE RAILROAD BRIDGES 943

It -vill be seen from the above that the weight of the rivet heads is
about 3 per cent ot the weight of the other material in the girders.

Estimate of the weight of laterals and lateral plates.

 

8—Ls 34 x 3 x 3 x 8.5 Ibs. x 8.2’. eee 558 lbs.
W—pls. 12x 3x 15.3 lbs. x 2.67... cece eee 278 lbs.
2—pls. 12x$x15.3bs.x 2.0’... eee ee 61 lbs.
2—pls. 12x $x15.3lbs.x 1.38’... cece 40 Ibs.
§—pls. 12x 2 x15.8 lbs. x 1.0'. occ eie csenta dread eres cues 137 Ibs.
4—pls. 12x3x15.3lbs.x 0.87.00... eee eee 49 Ibs.
Total weight of laterals and lateral plates........... 1,123 Ibs.
Estimate of the weight of cross-frames.
2—Ls 84x 34x%2x8.5 Ibs. x 6.2’... cece ee eee 105 Ibs.
2—Ls 84x 3$x9x8.5 Ibs. x 7.87. ee ees 124 Ibs.
4—ple: 148 x.17.86 bss ell cag d alain da aos wea eeale 78 lbs.
1—pl. 84 x $x 10.84 Ibs. x 0.87.0... cece 9 Ibs.
316 lbs
33 Tivets @ 0.44 Wb. iassgier evr e cen ie ew seen sas ns ee -.. 14 Ibs.
Total weight of one frame.......... 0.0 e eee eee 330 Ibs.
Total weight of five frames..................05. 1,650 lbs.
Total weight of lateral system = 1,123 + 1,650 = 2,773 Ibs.
Total weight of lateral system per foot = a ce = 55 lbs. per ft. of span.
Estimate of the total effective dead load per ft. of span.
PD) PAPC OES 5 ecoud a aia sNasaea Waka aaaeie Mess acdn chs Glanayeles Hac brome ess 641 Ibs.
Tateral Sy Stemicccuetecisiacare-tiuala ately a each ease ean ness 55 lbs.
Decl: (track ):y.ceciwsoos es (etek a tease ceases sk 400 Ibs.
"Potal: igdexuxca'ant aod pacer abu emeatslare saan we wees 1,096 lbs.

Difference between the assumed dead load and estimated = 1,150 -
1,096 = 54 lbs., which is less than 10 per cent, so no recalculation is
necessary on account of error in assumed dead load.

Summary of total weight of metal in span.

OU PIT CYS scien eA TEESE OS ete Gs HEME RE eS 82,044 Ibs.
Lateral ‘system... 20 a. sidered eee Rea he eae ace 2,773 Ibs.
4 pedestals @ 360 Ibs...... eee eee eee ee eens 1,440 Ibs.
Anchor bolts and nuts........-. 02: e cece erect eee eee 61 Ibs.

Total ivinhsxes cases ee see eee we ee eek ne Ha Se 36,318 Ibs

Approximate Cost of Span Erected: 36,318 lbs. @ 34¢ = $1,180.
Three and one-quarter cents is only a fair average price for this class of
work erected. The price will vary from 24¢ to 44¢. It depends upon the
market price of steel and the freight.
ea

244 STRUCTURAL ENGINEERING

It will be seen from the above that the weight of rivet heads (=1,000

Ibs., approximately) is equal to about 3 per cent of the weight of the
other metal (=32,700 lbs.) in the span, not considering the pedestals. So
in making preliminary estimates of such structures the weight of the rivet
heads can be taken at about 3 per cent of the weight of the other metal.

138. Hints Regarding Shop Bills.— The above shop bills are for
the most part self-explanatory. Pages Nos. 1 and 2 are for the structural
material proper. The finished material is given on the material side of
the bills exactly as it appears on the shop drawing. The length of
material as obtained from the rolling mills is likely to vary slightly from
the length ordered, so in case of long pieces where an under run would be
objectionable a 4” or so is added in the length given on the mill order side
as shown. In case of short pieces having a length of 10 ft. and under,
and having the same section, they are ordered in multiple, that is, they
are combined and ordered as one or more pieces, as is seen.

All pieces planed (marked fin.) on one side are ordered 7g” thicker
than the finished thickness and }” thicker if planed on two opposite sides,
All pieces planed on the ends are ordered }” to 4” longer than the
finished length. This is in addition to the amount added for under runs.

The dimensions appearing on the mill order side are only those
differing from the finished dimensions. In cases where the material is to
be ordered exactly as the finished material, it is not repeated on the mill
order side, as a rule.

The cast-steel pedestals are detailed on page No. 4. The required
area of bearing on the masonry of each pedestal is given in Art. 134 as
3610”. The actual area is

204 x 18 = 364.5 sq. ins.,

which is about the correct area. The thickness of metal in such pedestals
should not be less than 1)” in order to insure the molds to properly fill.
The top of each pedestal (where the girder rests) should be as small as is
consistent with good details.

Everything shown on such bills as the above is, as a rule, written on
by the draughtsman making the shop drawing. The item numbers are
given by the order clerk as the final mill order blanks are made out in the
order department, which, as a rule, is a separate department from the
drawing room. The bills not mentioned are considered as being entirely
self-explanatory.

Partial Design of a 75-Ft. Single-Track Deck Plate Girder Span

139. Data.—

Length = 75’ 0” c.c. end bearings.

Dead Load = 12x 75 +550 = 1,450 lbs. per ft. of span.
Live Load, Cooper’s £50.

Specifications, A. R. E. Ass’n.

140. Calculations.—In a manner similar to that shown in Art. 134
for the 50-ft. span, we obtain the following for the 75-ft. span:
DESIGN OF SIMPLE RAILROAD BRIDGES

Maximum End Shear:

D= 27,200 Ibs.
L=147,200 Ibs.
I=117,700 Ibs.

292,100 Ibs.

Maximum Moment:

D= 6,117,000 in. Ibs.
L=28,875,000 in. Ibs.
I=23,100,000 in. Ibs.

58,092,000 in. lbs.

Assuming the web to be 7% in. thick we have :

2= 1.055. | 58,092,000
ze x 16,000

for the economic depth. So we will try a 96x74 web. Using the
ordinary flange the effective depth is about 95” (merely an assumption).
Then we have 58,092,000 + 95 = 611,500 Ibs. for the flange stress, and
611,500 + 16,000 = 38.20” for the net area of the flange. As stated
before, about half of this area (minus 4 of the area of the web) should
be in each pair of flange angles. This would require angles which would
be entirely too thick if 6” x 6” angles be used. So either 6” x 6” angles
with side plates and cover plates, or 8’ x 8% angles with cover plates
should be used. The last-mentioned section is really the ordinary flange
wherein the angles are 8” x 8”.

Let us first try the flange made up of 6” x 6” angles, side plates
and cover plates. As the effective depth in this case is less than in the
case of ordinary flanges, the area of the flange will be a little larger than
indicated above. ‘ So let us assume the following section:

Gross Area Net Area
2—Ls 6” x 6” x 3” = 14.220” 2.5 0”=11.720”
2—Side plates 18% x 3” =13.000”—3.0 0”=10.000”

= 96.1 ins. (about)

1—Cover plate 16% x 4” = 8.000”-1.0 0”= 7,000”
1—Cover plate 16” x 74” = 7.000”-0.8707= 6.130”
4 of web = 5.250”

42.220” 40.100”

The flange is assembled as shown in Fig. 175.
By taking moments about the center of the top cover plate we obtain

 
     
 
    
 

 

eu 8x$+414.22 x 2.48413 x7} waaay.
42.22
for the distance from the center of the top . §
cover plate to the center of gravity of the Swe%,, ¢ —+
flange. Then we have © © ig x
Ve See Xs
3.240.175 =2.49 ins. 520 8 &
OO x 9|
for the distance from the back of the flange _ Sa ie :
angles to the center of gravity of the 2 al ¥
flange, and a oI »
sD i
96.25 —4.98 = 91.27 ins. N |
&

for the effective depth. Fig. 175
246 STRUCTURAL ENGINEERING

Then using this effective depth we have

58,092,000 _
91.27 = 637,000 lbs. (about)
for flange stress, and
637,000 _ 2
“76,000 = 389.8 Sq. Ins.

for the required net area of each flange. This shows that the above
assumed flange is about correct.
Next let us try a flange made up of 8” x 8” angles and cover plates.
First assume the following section:

Gross Area Net Area
2—Ls 8” x8” x 4h” = 21.060” — 2.750” = 18.310”
2—cov. pls. 18” x4” = 18.000” — 2.008” = 16.000”
4 of web = 5,250”
39.060” 39.560”
Taking moments about the center of the top cover plate we obtain
Bet 28S 8 ae:

39

for the distance from the center of the top cover plate to the center of
gravity of the ‘lange, and hence we have

1.73 - 0.75 =0.98 ins., say 1in.,

for the distance from the back of the angles to the center of gravity of
the flange. This gives 96.25 — 2 = 94.25” for the effective depth. Then
we have
58,092,000
94.25 x 16,000

for the required net area of each flange. This shows that the section
assumed above is a little too large. By making one of the cover plates
qq” thick (instead of 4’’) the section will be about correct.

It is seen from the above that the flange made up of 8” x 8” angles
and cover plates is more economic than the flange made up of 6” x 6”
angles, side plates, and cover plates, and hence apparently should be.
used. However, this will depend mostly upon whether the 8” x 8” angles
can be readily obtained from the rolling mills at the same price as the
other sections. In case several spans are required it would very likely
pay to use 8” x 8” angles, but in case of only one span it would likely
be best to use the 6” x 6” angles. side plates, and cover plates.

The calculations for the remainder of this span would be quite
similar to that shown above for the 50-ft. span, and when completed the
corresponding stress sheet, detail shop drawing, and shop bills could be
made. Fig. 176 shows a partial detail of the girder where the flanges
are made up of 6” x 6” angles, side plates, and cover plates.

The student will have no trouble in designing and detailing this type
of girder if the outline given above for the 50-ft. span be followed.

= 38.55 sq. ins. (about)
            

 

 

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248
DESIGN OF SIMPLE RAILROAD BRIDGES QAO

Partial Design of a 90-Ft. Single-Track Deck Plate Girder Span
141. Data.—

Length = 90’-0” ¢.c. end bearings.

Dead Load = 12 x 90 + 550 = 1,630 lbs. per ft. of span.
Live Load, Cooper’s E50.

Specifications, A. R. E. Ass’n.

142. Calculations——For the main girders of this span we have:

Maximum End Shear: Maximum Moment:
D= 36,700 Ibs. D= 9,902,000 in. lbs.
£L=171,500 lbs. L£=40,057,000 in. Ibs.
T=132,000 Ibs. I= 30,813,000 in. Ibs.

340,200 Ibs. 80,772,000 in. lbs.

Assuming the web to be 7%” thick, we have

80,772,000
= 1.055, /—- _ ——_ = ins.
x 7x 16,000 113 ins. (about)

for the economic depth. But, as a few inches either way from the theo-
retical depth in case of such deep girders will not materially affect the
design, it will be better to take 108” as the depth, as this will give us a
108” web, which is a very common plate, much more so than the 112”
or 113’. So we will assume a 108” x 7%” web.

There are three types of flanges suitable for this girder: One made
up of 6” x 6” angles, side plates, and cover plates, as shown in Fig. 176;
one made up of 8” x 8” angles and cover plates, as ordinary flanges; or
one made up of 4—6” x 6” angles and side plates for the top flange and
8” x 8” angles and cover plates for the bottom flange, as shown in
Fig. 177. ‘

The flanges made up of 8” x 8” angles and cover plates the author
thinks preferable. However, the design shown in Fig. 177 has some
desirable features. For instance, there are no rivet heads on the top
flange to interfere with the ties, as there are no cover plates and the top
laterals are connected to the bottom angles of the top flange.

The stress sheet and detail shop drawing and bills for this span
can be worked up in the same manner as shown above for the 50-ft. span.

All spans over 75 ft. long should be supported at one end upon
rollers. Fig. 178 shows the general details for such a bearing, which is
for the above 90-ft. span. Fig. 179 shows the general details for the
corresponding fixed support for the same span. Figures 180 and 181
show the shop details for the girder shoe, roller shoe, roller nest, and
pedestal.

In designing and drawing up such a bearing as that shown in Fig.
178, the first thing is to determine the space taken up by the rollers. For
the above 90-ft. span we have 340,200# for the maximum reaction. The
minimum size of rollers is limited by the specifications to 6” diameter,
and as plate girder bridges are comparatively small bridges we will use
the minimum size roller. The allowable pressure per linear inch on
S/F

    

HOR 10yrugy v

Fig. 178

 

wOr2 XP XZEXGT

    

Fig. 179

 

sMIy

72 XOE XO ST -$

250
DESIGN OF SIMPLE RAILROAD BRIDGES 251

these rollers is 600 x 6 = 3,600#. (See specificati oe
We then have ( P ations, also Art. 84.)

340,200

~~ 3,600 = 94.5

for the linear inches of roller required. If we use five rollers we have

94.5
Oe 19 ins. (about)

for the required net length of each roller, and adding 44” to each to
provide for the slots in the ped-
estal we have 234” for the total
length of each roller, as shown in
Fig. 178. All rollers having a
diameter of 6’ and over are
usually flattened on the sides like
those shown in Figs. 178 and 181,
and are known as_ segmental
rollers. These rollers should be
placed quite close to each other
so that their sides will come to-
gether whenever they are rolled
beyond the amount desired. Thus (Cast Steel)
the rollers are prevented from fall- :
ing over. The distance between
the rollers (between the sides) is
usually about 4”, as shown in Fig.
181. After determining the size
of the roller nest, a preliminary
drawing of the shoes and pedestal
can be made similar to that shown
in Fig. 178, wherein the general
dimensions are made to conform
to those of the roller nest, while
the correct size of the pin passing
through the shoes and the thick-
ness of metal throughout are de-
termined as the work progresses.
We begin by deciding upon the Fig. 180

general form of the shoes. Then

we assume the size of the pin passing through them. Next we calculate
the required thickness of the bearings of the shoes for the pin assumed.
If this seems satisfactory we proceed with the work of calculating the
stress on the pin to determine whether the pin assumed is of correct size,
and if the size is found to be correct we proceed farther with the design,
but if found to be incorrect we start over and make some new assumption,
and so on, until the design is worked out satisfactorily.

  

   

433 Fin Holes

 
  

  
    

2- Roller Shoes- RS.
(Cas? Stee/)
252 STRUCTURAL ENGINEERING

  
 
  

 

SB io |B

bolts as shown
Side bors >

fe Tap
he.
~ 23s, 2

RE
oF" ie
56 ‘Seg. Follers.
- f !
fire
a

 

 

 

 

 

 

 

° # tnilled bolesin bors,

2- Roller Nest -RN,

   

Cored holes

 
 

7

& Pedestals -RP
(Cast Stee/)

Fig. 181

     
  
  

Taking the case shown in Fig.
178, let us assume the pin to be 44”
in diameter. The total thickness of
the required bearing on each shoe is
then

840,200

24,000 x 44

The central bearing of each shoe
will likely be subjected to a little
more pressure than either of the side
bearings. So we will make them a
little thicker than the side bearings.
By making the central bearing of
each shoe 14” thick and each side
bearing 14” we have 33” for the
total thickness of bearing on each
shoe, which is about equal to that
required, and as the side bearings
are of minimum thickness for such
castings these thicknesses are quite
satisfactory and hence will be used.

The maximum shear and cross
bending on the pin occur at the side
bearings. For the maximum shear
we have

340,200 x = 109,300 Ibs.

= 34 ins.

and for the maximum shearing unit stress on the 44” pin we have

» = 109,300 _ 109,300
= ( )" “14,18
wit

2

= 1,700 Ibs.

For the maximum cross bending on the pin we have

109,300 x 12” = 150,000 inch lbs. (about)

and for the maximum fiber stress due to cross bending we have

fe 150,000 x 24
64

= 20,000 Ibs. (about).

Now, as the allowable shear on pins is 12,000 Ibs. per square inch and
the allowable fiber stress is 24,000 lbs. per square inch (see specifica-
tions), it is seen from the above that the 44” pin assumed is amply strong;
in fact it could be reduced in size, but as the saving in cost would be
insignificant we will use the size assumed. It also appears to be about

the correct size.
DESIGN OF SIMPLE RAILROAD BRIDGES 953

In addition to having sufficient bearing on the pin the shoes should
be deep enough and contain enough metal to resist the cross bending to
which they are subjected.

Considering the girder shoe (GS) we have a case of cross bending
as indicated in Fig. 182. For the maximum moment we have

_ 340,200
15

which occurs at the vertical transverse section c-c through the pin hole.
The moment of inertia of this cross-section of the shoe (not subtracting

the pin hole) is about 390. Then for the maximum compressive unit
stress due to cross bending we have

_ 638,000x6§ _,

M,. x 74 x 32 = 638,000 inch lbs. (about),

9
fe 390 2,200 lbs.
and for the maximum tensile unit stress we have
_ 638,000 x 33 |
f.= so 5,500 lbs.

*

This shows that the girder shoe is amply strong, as far as cross bending
is concerned, as 16,000 lbs. unit stress is permissible. The bending on
the roller shoe (RS) can be determined in the same manner. Further
designing of the shoes is very much a matter of common judgment.

Regarding the above calculations
it may first appear that the material cut
out of the vertical section c-c by the pin
hole should be deducted from the sec-
tion in determining the moment of
inertia. But this is far from being cor-
rect, as the top half of the pin is
assumed to transmit 24,000 lbs. per
square inch against the shoe—this we
can rely upon—and owing to the lateral
deformation of the pin there will be
some horizontal pressure (perhaps equal
to Poisson’s ratio); so, taking all
in all, the above assumption is fairly
correct.

The designing of the pedestal (RP) is very much a matter of com-
mon judgment. The pedestal should be at least 4 ins. or 5 ins. high in
order to have the rollers above snow, slush, and dirt, and should be broad
and long enough to support the load coming on them without producing
a greater pressure upon the masonry than 600 lbs. per square inch.
There should be open slots extending down from the top as shown in
Fig. 178, so that dirt will not accumulate between the rollers. The
longitudinal circular holes should be cored in the casting and the slots
cut from the solid when the pedestals are being finished in the machine
shop. This will guard against the pedestal warping when cooling. The
details of the fixed bearing shown in Fig. 179 are considered to be self-
explanatory.

3.

HAlus
oO

 

17000"

 

 

 

 

 

 

Fig. 182
254 STRUCTURAL ENGINEERING

143. Flange Splices.—The flanges of all plate girders over 90 ft.
long must be spliced, as that is about the maximum length of angles and
cover plates obtainable. These splices should preferably be symmetrically
arranged in reference to the center of the span. The flange angle splices
should, as a rule, be nearer to the ends of the girder than to the center
of the span, and only one flange angle should be cut off at a splice. That
is, one flange angle should be spliced on one side of the girder on one
side of the center of the span and the other flange angle should be
spliced on the other side of the girder at a corresponding point on the
other side of the center of the span, as shown in Fig. 183. Here the
near flange angle is spliced at D, one part extending from D to B, and
the other part from D to A, while the far flange angle is spliced at E,
one part extending from E to A and the other part from E to B.

 

 

ot

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 183

The splice of a flange angle should be made by means of an angle
having the same net area of cross-section, as the flange angle. The splice
angle must be ground to fit the fillet of the flange angle and its legs cut
so as not to project beyond the legs of the flange angle, as indicated at
section S-S. (Fig. 183.) It is general practice to use two splice angles
at each splice, one on each side of the girder. The splice angle on the
side of the flange angle spliced should be of sufficient cross-section in
every case to splice that angle. The splice angle on the other side of
the girder is used simply to balance the flange section at that point.

The cover plates are usually spliced by extending the adjacent cover
plates a short distance beyond their theoretical length, as shown in Fig.
183. As a rule, the cover plates next to the flange angles are the only
ones that need to be spliced. Cover plates, of course, can be spliced by
simply using a short plate having the same area of cross-section as the
plate spliced, but usually this presents an unsightly appearance.

As an example in figuring splices, let the two flange angles shown in
Fig. 183 be 2—Ls 6” x 6” x 3”, as indicated. For the net area of either
of these flange angles we have (using #” rivets) 7.11—1.25=5.860”. Then
for the allowable stress in each flange angle we have 5.86x16,000=
93,700#. Then for the number of }” shop rivets required on each side
of the splice (in the splice angle) we have 93,700+17,200=13 rivets (in
single shear). As is seen, 14 are used—7 in the vertical leg and 7 in the
horizontal leg. The splice angle at each splice should have 5.864” net.
Using a 6” x6” x44” angle, and deducting 1.370” for rivet holes and
0.864” for the cutting of the legs and for the grinding to fit fillet, we
have 7.78-1.87-0.86=5.5509” net, which is about correct.
DESIGN OF SIMPLE RAILROAD BRIDGES 255

The calculation for the splicing of a cover plate is simply a matter
of determining the number of rivets required in single shear to develop
the strength of the plate spliced. The student should have no trouble
in doing that.

144, Graphical Determination of Live-Load Shear and Bending
Moment on Deck Plate Girder Bridges.— The analytical method out-
lined in Art. 133 is generally used, as it is simple in application and the
results obtained are absolutely numerically accurate, yet the graphical
method is fully as simple in application and the results obtained are
quite accurate enough. The graphical method at least affords an excel-
lent means of checking results obtained analytically, and if for no other
reason than this the engineer should be familiar with the method. In
fact, there are two methods: that of influence lines and the one wherein
the equilibrium polygon is used.

50'- 0"

 

 

 

 

 

 

 

Fig. 184

The application of influence lines is fully given in Arts. 100 and
101. As an application of the equilibrium polygon let us take the case
of the 50-ft. span treated analytically in Art. 133, where Cooper's E50
loading is used. Let AB (Fig. 184) represent the span drawn, say, to
4’ ascale. From inspection of the diagram in Table A (in the back of
this book) we can see that the maximum end shear will occur when
wheel 2 is at the end of the girder and wheels 2 to 10 (inclusive) are
on the span, as shown in Fig. 184. After the loads are thus placed, the
next thing to do is to lay off the load line wv (say, to a py-in. scale) at
(b), and construct the ray diagram, as explained in Art. 95, and draw
the corresponding equilibrium polygon ab...na at (a). Then by draw-
ing from O (at b) the line OF parallel to the closing line an, we have
the maximum end shear given by the part of the load line extending from
E to u which can be scaled, using the same unit of scale as was used in
laying off the load line.

To determine the shear at any point K of the girder, place the point
K under wheel 2 by moving the girder, so to speak, to the left, so that
256 STRUCTUKAL ENGINEERING

the end A will be at A’ and the other end will be at B’ and wheels 1 to 7
will be on the span. Then add wheel 1 to the load line and draw the
ray wO and extend the equilibrium polygon on to s and we have the
equilibrium polygon sta...fs’s. Then draw OE’ in the ray diagram
parallel to the closing line ss’ and we have the maximum shear at point
K given by the part of the load line extending from £’ to u. In this
manner the maximum shear at any point on the span can be determined.

To determine the maximum bending moment place the loads on the
span, as shown in Fig. 185, so as to satisfy the criterion for maximum
bending moment as per Art. 88. Then construct the ray diagram as

H= 100000"

 

Afr

0

They Fin

 

 

 

    

 

WV:

Fig. 185

shown at (b) and draw the corresponding equilibrium polygon ab... ma,
and we obtain the maximum moment by multiplying the ordinate e’e
under wheel 13 (e’e must be scaled off in feet or inches to the same
scale as was used in laying off the length of*span at (a) and spacing
the loads) by H, the pole distance which is laid off in pounds to the
same scale as the load line. If the ordinate e’e is taken in feet the
bending moment will be in foot pounds, and if taken in inches the bending
moment will be in inch pounds.

Such a diagram as shown in Fig. 186 is very convenient if the work
to be done is extensive enough to warrant the drawing of it. In offices
where one loading is used for all bridges it is advisable to draw up such
a diagram of the loading on a sheet of thick cardboard, inking in all
lines shown in Fig. 186 except the closing lines. The closing lines can
be drawn lightly in pencil as needed and then erased. In this way | the
diagram can be used in determining the shears and moments for ‘any
number of bridges. The diagram shown in Fig. 186 is for Cooper’s E40
loading. The loads are spaced off to j5-in. scale on the horizontal line
at the top of the sheet. Then the load line LL is laid off to a #y-in. scale
and the ray diagram is constructed, as shown, and the corresponding
equilibrium polygon ABC is drawn, and then the scale lines are drawn
below this, as shown.

To show the manner of using the diagram in Fig. 186, let ug take
the case of a 50-ft. span. The maximum moment due to Cooper’s loading
in the case of most deck plate girder bridges will occur under one of the
four wheels 11 to 14 (this we obtain from experience). So, beginning with
zero at wheel 13 we lay off the scale line HH into 5-ft. units. To determine
the maximum bending moment in this 50-ft. span let us start by assuming
DESIGN OF SIMPLE RAILROAD BRIDGES Y57

that the span extended out 25 ft. cach side of wheel 13. Then drawing
the closing line 1-1 we have the equilibrium polygon 1-B-1, and by
scaling off the maximum ordinate between the closing line 1-1 and the
curve 1-B-1 and multiplying it by the pole distance given in the ray
diagram we obtain the maximum bending moment for the span in that
position. By moving the span 5 ft. to the right and drawing the closing
line 2-2 we have the equilibrium polygon 2-B-2 and by scaling off the
maximum ordinate and multiplying it by the pole distance we obtain
the maximum bending moment for the span in that position, and moving
the span to the left the equilibrium polygon 3-B-3 is drawn and the
maximum bending moment for the span in that position can be deter-
mined. Now, by simply drawing a few of these sub-equilibrium polygons
the absolute maximum ordinate can be ascertained from the intersection
of the closing lines with the verticals through the wheels. The maximum

Ol 58 FG IN IS

m7 ; 1 a] x 7
\ | ee a
DOE © AG

   
  
  

 

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IS
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AA"
8
8
Pe
x
m7

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ARS AY

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Ss
ITS 1G" 15" 20" 25" 30 35" 40" 45" 50" 55° 60" 65

Fig. 186

ordinate will always be under a wheel, and after a sufficient number
of closing lines are drawn the maximum ordinate can be quickly deter-
mined by the aid ‘of a pair of dividers and then scaled off and multiplied
by the pole distance. In this way the maximum bending moments are
determined without bothering with the criterion for maximum moment,
as is necessary when other methods are used.

To determine the maximum end shear on the 50-ft. span, place
wheel 2 at the end of the span. Now using the scale line SS, the span
will extend 50 ft. to the right of wheel 2 and we obtain an equilibrium
polygon which has the closing line nn’. Then drawing Pr in the ray
diagram parallel to this closing line we have the maximum end shear
or reaction given by the part of the load line extending from r to e. By
 
DESIGN OF SIMPLE RAILROAD BRIDGES 959

the following construction the end shears can be obtained more quickly
than if scaled from the ray diagram: From n (on the equilibrium
polygon) lay off the horizontal line nm equal to the pole distance EP
and draw the vertical line kk’ through m and prolong the segment gn,
of the equilibrium polygon, until it intersects this vertical line at k’.
Then we have the triangle nmk’ equal to the triangle Pre (in the ray
diagram), and hence ck’ is equal (by scale) to the maximum end shear
on the 50-ft. span. Likewise, x’ k’ and 2k’ are equal, respectively, to
the maximum end shear on a 60-ft. and 75-ft. span.

The equilibrium polygon ABC can be used in general to determine
the shear at any point in a span in the manner explained in the case
shown in Fig. 184.

Problem 1. Construct a diagram for Cooper’s E50 loading similar
to the diagram shown in Fig. 186 and determine from it the maximum
end shears and bending moments on a 40-ft., 50-ft., 60-ft., 70-ft., 80-ft.,
and 90-ft. span.

DRAWING ROOM EXERCISE NO. 3

Design a 60-ft. single-track deck plate girder railroad bridge and
make a stress sheet for the same. The finished drawing, similar to that
shown in Fig. 169, is to be upon an 18” x 24” sheet of tracing cloth.

Data:

Length=60’-0” c.c. end bearings.

Width = 6’-6” c.c. girders.

Height, to be determined by student.

Dead Load, to be determined by the student.
Live Load, Cooper’s £50.

Specifications, A. R. E. Ass’n.

DRAWING ROOM EXERCISE NO. 4

Make a shop drawing and shop bills for the 60-ft. bridge specified
in Drawing Room Exercise No. 3. The finished drawing, similar to that
shown in Fig. 171, is to be upon a 24” x 36” sheet of tracing cloth.

THROUGH PLATE GIRDER BRIDGES

145. Preliminary. Through plate girder bridges are used, as a
rule, only when the under clearance will not permit of the use of a deck
span—owing to the cost of the through bridge being considerably more
than the deck bridge. The ordinary through plate girder bridge is com-
posed of two main girders and a floor system composed of longitudinal
beams (or girders), which support the ties and are known as stringers,
and cross beams, which support the stringers and are known as floor
beams.

An isometric view of an ordinarv single-track through plate girder
span is shown in Fig. 187, where the names of the different parts of
the structure are given.

Double-track through plate girder bridges are usually the same in
construction as the single-track bridges, except the main girders in the
double-track structures are ferther apart and the floor system provides
for the two tracks, which requires twice as many stringers.
260 STRUCTURAL ENGINEERING

Complete Design of a 60-Ft. Single-Track Through Plate Girder Span
146. Data— y

Length=4 panels @ 15’-0” =60’-0” c.c. end bearings.
Width=15’-6” c.c. main girders.
Assumed Dead Load:
For main girders, (13 x 60+ 600+400) = 1,780 Ibs. per ft.
of span.
For stringers, (12 x 15 +100 +400) =680 lbs., say 700 lbs.,
per ft. of span.
Live Load, Cooper’s £50 loading.
Specifications, A. R. E. Ass’n.

147. Design of 15-Ft. Stringers.—It is usually necessary to make
the stringers in through plate girders just as shallow as good details will
permit, owing to the under clearance being limited. This can be accom-
plished best by the use of I-beams, two or more under each rail. So we

will use I-beams.
For dead-load moment, using the load assumed in Art. 146, we have

ue 5x gex IB x 12118,000 ake Ta

For live-load moment we have
M’ =1,875,000 inch lbs. (same as Art. 130).
For impact we have
I=1,785,000 inch Ibs. (same as Art. 130).
Then we have for the total bending moment
118,000 + 1,875,000 + 1,785,000 = 3,778,000 inch lbs.
Then for the section modulus we have

3,778,000

16,000 =236. (See Art. 105.)

This calls for 2—Is, 20’ x 70#, under each rail. That is, each stringer
is to be composed of 2—20” x 70# I-beams.
For dead-load end shear on each stringer we have

R - x ~ = 2,625 lbs., say 2,600 lbs.

For live-load end shear we have
R’ =50,000 Ibs. (same as Art. 130)

and for impact we have
I=47,600 lbs.

Then, for the tctal end shear we have
2,600 + 50,000 + 47,600 = 100,200 Ibs.
DESIGN OF SIMPLE RAILROAD BRIDGES 261

Preliminary estimate of weight of stringers.

4—Is 20”x70#x15’-0” =4,200 lbs.
1—[ 9” x 207 x 3’-9” = 5 lbs.
2—end con.Ls on 9” [s = 21 Ibs.
2—[s 12”%x25#x1’-3” = 62 Ibs.

Total weight of one panel = 4,358 lbs.

Weight of stringers per ft. of span=4,358 + 15 = 289 lbs. metal.
Weight of deck = 400 lbs. metal.

Total dead load for stringers per ft. of span = 689 lbs. metal.

This shows that the 700 lbs. dead load assumed above is about
correct. :

148. Design of Intermediate Floor Beams.—The dead load on
these beams consists of the dead weight applied to them by the stringers
and also the weight of the floor beams themselves.

The dead load from the stringers is concentrated on the floor beams
at the points where the stringers are connected to them. This concen-
tration at each point is equal to twice the dead-load end shear on one
stringer which is given above as 2,600 lbs. So each concentration is
2,600 x2=5,200 lbs. The weight of the floor beam is a uniform load.
Such floor beams usually weigh about 3,600 Ibs. each. So we will assume
that weight. Then the dead load on any intermediate floor beam will be

as shown in Fig. 187A, where A-B represents the beam.
It is readily seen from Fig. 187A that

zero shear occurs at the center of the beam
and hence that is the point of maximum
moment. As the bending moment on the
floor beam due to the two concentrated loads
= = is constant between the stringers it is cus-
fee We eel tomary to multiply one concentration by its
Fig. 187A distance from the end of the beam to obtain
the bending moment due to these loads.
To show the correctness of this method let 4B, Fig. 188, represent
a floor beam supporting the two equal concentrated loads P, as shown.
Taking moments about the center of the beam we have, since R=P,
b b b b
m=R(a+5)-P3 =Pat+Ps P 3 =Pa.
That the moment between the loads is constant can also be shown very
readily by graphics: Lay off the load line VWS

 

 

 

 

 

 

(Fig. 188) and construct the ray diagram VOS apy PP 1
and draw the corresponding equilibrium polygon 4 cA iB
efghe. Now, as the reactions are equal for the *F Pa
two loads, the closing line eh must be parallel to det

the ray OW, and as the segment fg is parallel vi

to this ray also, it is seen that the ordinates of : 6

the equilibrium polygon are constant between a

the two loads, and hence the moment between 5

them must be constant. Now going back to the Fig. 188
262 STRUCTURAL ENGINEERING

problem in hand, we have the moment
M =5,200 x 63 = 327,600 inch lbs.
from the concentrated load and

,_1_ 3,600 — 7 :
M =9*% p85 x 15.5 x 12=83,700 inch lbs.
from the weight of the floor beam, making a total maximum of 411,300
inch-Ibs. dead-load bending moment.

It is obvious that the maximum live-load bending moment on the
floor beam will occur when the maximum live-load concentrations from
the stringers on to the floor beam occur. So it is first necessary to deter-
mine the position of the wheels when the maximum live-load concentra-
tions occur.

Let Fig. 189 represent two adjacent panels of stringers with a floor
beam at each of the ends, A, B,
4 x z and C. Let I be the length of
| r panel AB and l’ the length of

OO SL OO O q_ panel BC, as indicated. Let W
a , e ae 1° be the total weight of the wheels
in panel AB, the center of gravity
of which we will assume is # dis-
tance from A, and let W’ be the total weight of the wheels in panel BC,
the center of gravity of which we will assume is z distance from C.
Let r be the concentration on the floor beam at B due to the wheels in
the panel AB, and let r’ be the concentration due to the wheels in the
panel BC, and let R be the concentration on the floor beam at B from
the wheels in both panels. Then we have

 

 

 

 

 

 

 

 

 

Fig. 189

 

RSP br ica essen sieve ieee ee aes wasreleaeareenees™ (1)
Taking moments about A we have
pos
od
and taking moments about C we have
,_ W's
=F ‘
Then substituting these values in (1) we have the general expression
We W
R= 7 4+ eh eRe e BY Wa ork © a aie SWOT a deals ale Beate e ee whey oes (2)

for the concentration on the floor beam at B. Now, if the wheels roll
to the left, r will decrease and r’ increase provided no wheels pass B,
and just the reverse is true if the wheels roll to the right. Now if the
weight of the wheels in the two panels is such that the increment of r
is just equal to the increment of r’ for a very slight movement, it is evi-
dent that the increment of the concentration R will be zero and hence R
will then be a maximum, for otherwise it would be either increasing or
decreasing. If increasing, it evidently has not reached the maximum;
and if decreasing, it has passed the maximum. So, differentiating equa-
tion (2) we have
DESIGN OF SIMPLE RAILROAD BRIDGES 963

 

_Wde W'dz 7

 

dR a oe =0.
But dz=dz. Then we have
an WW’, | W_W
i a ee Oe

That is, the unit load in one panel will be equal to the unit load in the
other panel when the maximum concentration on the floor beam occurs.
If the panels are of equal length, as they are here, we have

W=W’.

That is, the maximum concentration on the floor beam will occur when
the load in one panel is equal to the load in the other. Now the increment
of R can change sign only when a wheel passes the floor beam at B.
So there will be a load at that point when the maximum concentration
occurs. It is evident that this criterion for maximum concentration on the
floor beam, so far established, can be satisfied by most any group of
wheels, but it is also evident that the absolute maximum concentration will
occur when the heaviest group of wheels is near the floor beam. So the
whole criterion for maximum concentration on an intermediate floor beam
can be stated as follows:

The maximum concentration will occur when the heaviest group of
wheels is near the floor beam and one wheel at it, and when the load in
one panel is equal to the load in the other.

The next thing in our case is to satisfy this criterion. Referring to
Table A we can see that wheels 1 to 5 are as heavy a group of wheels as
can be placed on two adjacent panels.

Let us place them as shown in Fig. 190. Then the load in panel
AB=30,000 Ibs., and that in
BC= 40,000 lbs. This does not & % ‘ 5

 

* § 8s § 8
seem to be very close to the re- 8 g gy Q Q
quirement—that the loads in the g 4 . , % : \
two panels be equal. But by trial 2' 3" 5(3) s'(4)5 5"
it will be found to be as close as A a Re a c

 

 

 

‘possible and still satisfy the first
part of the criterion at the same Fig. 190
time. It is not often that this
criterion can be satisfied absolutely as to the loads in the panels being
equal. We should, however, place the loads so that the criterion is as
nearly satisfied as possible. In testing for the criterion the load at the
floor beam can be considered as being equally divided between the two
panels; that is, one-half of its weight being considered in each panel,
or it can be ignored as was done above. It will make no difference which
way it is considered. bon as

So, taking the position shown in Fig. 190 as satisfying the criterion and
taking moments about C, we have

5+10
15

 

20,000 ( ) = 20,000 Ibs.,
264 STRUCTURAL ENGINEERING

and taking moments about A we have
10,000 x 2 + 20,000 x 10
15

Then adding these two concentrations to the 20,000-Ib. load at the floor
beam, we have

= 14,660 lbs.

R= 20,000 + 14,660 + 20,000 = 54,600 Ibs.

for the maximum floor beam concentration due to Cooper’s £40 loading,
and multiplying this by 50/40 to reduce it to E50, we have

* x 54,600 = 68,200 Ibs. (about)

for the maximum concentration.

8 8 Then the maximum live load on the floor
® , beam will be as shown in Fig. 191.

  

 

A B For the maximum live-load bending mo-
*S $ ment, we have
8 § M = 68,200 x 63 = 4,297,000 inch Ibs.,
tas and for the impact we have
300 :
Dapp % 1:297,000 = 8,906,000 inch Ibs.

Now adding the above maximum dead- and live-load bending moments
and impact together we have

411,000 + 4,297,000 +3,906,000 = 8,614,000 inch lbs.

for the total maximum bending moment on the floor beam for which the

beam must be designed to resist.
For the maximum end shear on the floor beam due to dead load we

have
3,600

2

From live load we have 68,200 lbs., as shown above (Fig. 191), and from
impact we have

5,200 +

 

= 7,000 Ibs. (See Fig. 187A.)

800

68,200 x 30 +300

= 62,000 Ibs.

Then adding the above dead- and live-load end
shears and impact together, we have Ze

7,000 + 68,200 + 62,000 = 137,200 Ibs.

for the total maximum end shear.

The next thing is to design the floor beam.
The flanges of such floor beams are usually com- 5
posed of 6”x6” angles, and the stringers should Fig. 192
fit in between the flanges as shown in Fig. 192 so
as to avoid fillers. So giving a }” clearance at both the top and bottom
of the stringers, as shown, we obtain a floor beam 724” deep. The

 

a

 

 

 

 

 

poets?
pe
=
bz| 207 be
322

 
DESIGN OF SIMPLE RAILROAD BkIDGES 265

distance from the back of the 6x6” flange angles tu their center of
gravity could be obtained from a handbook or from Table 6 provided
the exact weight of them were known. However, we can obtain the
average from this source. So let us take 1.78” as the average. Then
for the approximate effective depth we have

32.5 ~1.78 x 2= 28.94 ins.
Now dividing this into the maximum bending moment given above we have

8,614,000

saga 298,000 Ibs. (about)

for the fiange stress. Then we have

298,000
16,000

for the net flange area required minus one-eighth of the area of the cross-
section. of the web. Let us assume the web to be 32’x4. Then one-
eighth of the area of web = 20”,

Then for the make-up of each flange we have

=18.6 sq. ins.

Gross Net
2—Ls 6” x 6’ x 42” =18.189” —1.62=16.560”
4 area of web = 2.000”

18.560”

Now for the actual effective depth (since we know the weight of the
flange angles) we have

82.5 —1.8 x 2=28.9 ins.,

which is practically the same as assumed above.

As the floor beam has no regular stiffeners (the stringers, however,
stiffen it) the vertical distance between the flange angles can be taken as
din Formula (1), Art. 118. Then we have

ae
20.5 = 75 (12,000-s),

from which we obtain
s = 10,360 Ibs., say, 10,000 Ibs.

Now dividing this into the maximum end shear given above we have

137,200
10,000

for the required area of the cross-section of the web. The web assumed
above has 164”, which is a little more than required, but if a thinner web
be used the flange rivets would be quite close together, perhaps too close.
This trouble often occurs in such shallow floor beams. So we will use the

assumed web 32” x 3”.

= 13.72 sq. in.
266 STRUCTURAL ENGINEERING

Preliminary estimate of the weight of an intermediate floor beam.*

1—web 32” x4” x 15.5’ x BL4F eee ee = 843 Ibs,
dhs 6 6" x 28g ISD © BLP ci careteiane ye eaeeees = 1,922 lbs.
Gusset plates (1 plate 60” x $” x 4’-0” for the two) @ 102#
WOME bese Aeer lobes a enain es Darah eee. dy hav eter uals een reaae a = 408 lbs.
4—1s 34” x 34% x 8” x 4/-0” x 8.5% (on gusset plates or
brackets): ¢vvoseeuweuak Mie pi eee < gaara eee lage = 1386 lbs.
2—Ls 34” x 34” x 8” x 2’-6” x 8.5% (end con. angles)...... = 42\bs.
4—special plates 16” x #/” x 1/-8” x 27.20%. ...... eee eee = 181 Ibs.
3,532 Ibs.
PLD o LOR TIVES ec oa ego's & Siete Gow We Desa wl ne alee Sade Hees rw ace 88 lbs.
Totaly ¢tasaechaae nae ahie Sa real GeEG ek Sa Oe aL 3,620 lbs.

This shows that the 3,600 Ibs. assumed weight of the floor beam is
very close to the actual weight and hence no recalculations are necessary
as 10 per cent variation either way is permissible.

149. Design of End Floor Beam.—The designing of end floor
beams is very much the same as that of intermediate floor beams. The
concentrations, however, are different and the weight of the beams them-
selves is a little less.

Each concentration from dead load is equal
to the dead-load end shear on a stringer and
each live-load concentration is equal to the maxi-
mum live-load end shear on the same. Then
assuming the weight of an end floor beam to be
3,000 Ibs. we have the loading on the beam as
shown in Fig. 193, using the end shears given
in Art. 146. Then we have for the dead-load

 

 

 

 

15-6"
Fig. 1938

 

moment.
M =2,600 x 63 = 163,800 inch lbs.
from the concentrated dead load and
M’ =4x 3,000 x 15.5 x 12=69,750 inch lbs.

from the weight of the floor beam, making a total of 233,550 inch Ibs. for
the total maximum dead-load bending moment.
For the live-load bending moment we have
M’”’= 50,000 x 63 = 3,150,000 inch lbs.,
and for the impact we have
= 300 \_ :
I= 3,150,000 (on) = 3,000,000 inch lbs.
Now adding these moments and impact together, we have
233,550 + 3,150,000 + 3,000,000 = 6,383,550 inch lbs.
for the total maximum bending moment.
Assuming an effective depth of 29” we have
6,383,550

59 = 220,000 Ibs. (about)

* These preliminary estimates are made before the detail drawings are made and
are intended to be only approximately correct.
DESIGN OF SIMPLE RAILROAD BRIDGES 267

for the flange stress. Then we have

220,000

“16,000 =13.75 Sq. ins.

for the net area required for each flange minus one-eighth of the area of
the cross-section of the web.

Assuming the web to be 32” x # = 120”, one-eighth of the area of
the web is 1.55”. Then for each flange we have

21s 6” x 6” x we” = 12.860" — 1.12 =11.740” net
+ area of web = 1.500” net

13.240” net

As is seen, the net area of this flange is about 0.510” less than required,
but the 6” x 6” x §” angles give a flange about 0.724” too large, so the
above will be used. /

For the maximum end shear, as is readily seen from Fig. 193, we have

-

D = 1,500 + 2,600 = 4,100 Ibs.

be 50,000 Ibs.

1 = 50,000 (20°) = 47,600 »
es: ang
Total 101,700 Ibs.

For the allowable stress on the web we have
20.5 + (12,000 -s),

from which we obtain
s=9,815 lbs., say, 10,000 Ibs.

Then dividing this into the above shear we have

101,700

[0,000 =10.17 sq. ins.

for the required area of the web. So the assumed web 32” x 3’ =120”
is satisfactory, as the specifications permit of no thinner web.

The flange angles and web as specified above for the end floor beam
weigh about 2,000 lbs., which is about 765 Ibs. less than the flange angles
and web in the intermediate floor beam. Then as the other parts are about
the same as for an intermediate floor beam we have 3,621 — 765 = 2,856
lbs. for the weight of the end floor beam. So the assumed 3,000 lbs. for
the weight of an end floor beam is within the 10 per cent limit.

150. Design of the Main Girders.—In the case of through plate
girder bridges the live load is applied to the main girders only at the
panel points, that is, where the floor beams connect to the main girders,
just the same as in the case of through truss bridges. The dead load is
applied to the main girders in the same way except for the weight of the
girders themselves which is uniformly distributed along the girders. But
268 STRUCTURAL ENGINEERING

it is usual practice to consider the dead load as wholly applied at the
panel points as about the same result is obtained as if the more exact
conditions were considered.

Taking the assumed dead load given in Art. 146 we have

aie x 15=13,350 lbs.

for the panel load of dead load per girder. Then the dead load per girder
will be as shown in Fig. 194, where AB represents the girder. Now it is
obvious that the maximum bending moment due to this dead load will
occur under the load at C. So taking moments at C we have

M = (20,025 x 30 — 13,350 x 15) 12=4,806,000 inch Ibs.

for the maximum bending moment due to dead load.
The live load will be ap-

 

 

 

 

 

 

 

3 : % ‘

plied at the panel points, but «> *s 8 8 ">

of course the panel loads will % . q i

not be equal to each other as, a 5 =8

in the case of dead load. The *u) yszo” | ystov £ sstov | usta” 18
u 9

maximum bending moment & G0t0" 8

will occur at a panel point as“ Fig. 104

in the case of a truss. It is
readily seen that the maximum moment will occur at the panel point C,
at the center of the span. Then the first thing to do is to place the
wheel loads so as to obtain the maximum moment at that point which is
simply the satisfying of the criterion for maximum moment as given in
Art. 91. That is, the maximum live-load bending moment on the girder
(which will oceur at C) will occur when the average load to the left of C
is equal to the average load on the span.

Now referring to Table
A, Jet us try wheel 13 at C.
Then the loads will be in the
position shown in Fig. 195,

 

 

 

60% 0% 4
Fig. 195 and for the average unit load

on the left, considering half

of wheel 13 as being on the left, we have
73,000

30
and for the average unit load on the whole span, we have

142,000
60

 

= 2,430 Ibs. (about),

= 2,370 Ibs.

Now it is quickly seen that this position will give the maximum moment
at C. For if wheel 14 be placed at C there would be 80,000 Ibs. to the
left which would give 2,660 lbs. average unit load to the left and the
average unit load on the span would be the same as given above, and if
wheel 12 be placed at C the average unit load on the left will be 53,000 +
30 = 1,760 lbs. and the average unit load on the span will remain the same
as before.. So the wheels when in the position shown in Fig. 195 will
‘produce the maximum bending moment on the girder.
DESIGN OF SIMPLE RAILROAD BRIDGES 269

Then, the next thing is to determine the bending moment about wheel
18 as that is the moment desired. To do this we must first determine the
reaction & at A which we do by taking moments about B. As wheel 18
just happens to come exactly at B we can obtain the moments of all the
wheels on the span directly from Table A. Passing down the line through
wheel 18 (in the table) until we come to the zigzag line, then passing to
the left until we come to the vertical line through wheel 9, we find the
moment 4,224 which is the moment in thousands of foot pounds of all the
wheels, 9 to 17 inclusive, about wheel 18, which happens to be at the right

 

support.
Then for the reaction R we have
(*35*) =70.4 (thousands).

Then taking moments about wheel 13 we have (using Table A)
M = (70.4 x 30 — 818) 1,000 x 12 = 15,528,000 inch lbs.
Then multiplying this by 50/40 to reduce it to Cooper’s £50, we have

50
40

for the maximum live-load bending moment on the girder.
Then for the impact we have

300
60 + 300

Now adding the above dead- and live-load moments and impact together
we have

4,806,000 + 19,410,000 + 16,175,000 = 40,391,000 inch Ibs.

for the total maximum bending moment on the girder. Now, assuming
the web will be 2” thick, we have

[40,391,000 on,
2=1.055 16,000x = 86.5 ins. (about)

for the economic depth. So we will use a web 84” deep (as this is a
common plate) and assume it to be 3” thick.

The total depth of the girder back to back of angles will then be
84.25”. Then subtracting about an inch we have 83.2” for an assumed
effective depth. Dividing this into the maximum moment given above we
have

15,528,000 x 2~ = 19,410,000 inch Ibs.

19,410,000 x ( ) = 16,175,000 inch Ibs.

40,391,000
eee . (about
33.2 486,000 lbs. (about)
for the flange stress, and then we have
486,000

76,000 = 30.4 sq. ins. (about)

for the required net area of the cross-section of each flange.
270 STRUCTURAL ENGINEERING

The area of the cross-section of the 84” x 2” web = 31.50”, and
4 =3.90”. Then for the make-up of each flange we have

Gross (rivets) Net
2—Ls 6” x 6” x 447 =15.560” — 2.750” = 12.800”
1—cov. pl. 14” x 8%= 8.750”-1.2507= 7.500”
1—cov. pl. 14” x #”= 7.000”-1.000”=. 6.000”
$ area of web = 3.900”

31.310” 30.200”

Now taking moments about the center of the top cover plate (see
Fig. 196) we have

ne 2.62 Xx ea x 0.56 SHRP tae

 

for the distance from the center of the top cover
plate to the center of gravity of the whole flange.
Then we have

1.46 —-0.87=0.59 ins.

for the distance from the back of the flange
angles to the center of gravity of the flange. Then we have

84.25 — (0.59 x 2) =83.07 ins.

 

Fig. 196

for the actual effective depth of the girder, which is quite close to that
assumed above, so no recalculation on account of the assumed effective
depth is necessary.

For the length of the $” cover plates, which will be the outside cover
plates on the girder, we have

y =60 as = 28’-6” + 1’-6” =30’,

For the 2” cover plate we have

Ee ai 15.7 = Few FREE ,
y 60/25 42 67 + 17-67 =44’.

The maximum end shear due to dead load is equal to the reaction
shown in Fig. 194 plus the weight of the girder for half of the length of
the end panel. The four flange angles of the girder, as given above,
weight 106# per ft. of girder, the web weighs 108# per ft., and the weight
of stiffeners and details at this point on the girder will be about 90 per
cent of the weight of the web per ft., or 97#, and the two 3” cover plates
weigh about 60#, making in all about 370#, say, 400# per ft. of girder.
Then we have

S=20,025 + 400 x 7.5 = 23,000 lbs. (about)

for the maximum end shear on the girder.

The maximum end shear on the girder due to live load is equal to
the maximum shear in the end panel due to that load. Now according to
Art. 90 the maximum shear in the end panel (the same as in a truss
DESIGN OF SIMPLE RAILROAD BRIDGES 271

bridge) will occur when the load in the panel is equal to the total load on
the bridge divided by the number of panels. So the first thing is to
satisfy this criterion for maxi-
mum shear in the end panel.
Placing wheel 3 at the first
panel point out from the end of
the girder as shown in Fig. 197
we have (using Table A) Fig. 197

10,000 + 20,000 + 10,000 = 40,000 Ibs.

for the load in the end panel AD, considering one-half of wheel 3 in the
panel. With the wheels in this position the total load on the span is
152,000#. Then we have

152,000
4

which to satisfy the criterion should be 40,000#. But this is as close as
the criterion can be satisfied, as can be verified by trial.

The next thing is to determine the shear in the end panel AD with
the wheels in the position shown in Fig. 197. First of all the shear in the
panel is equal to the reaction R at A, due to all of the loads on the span,
minus the concentration r from the stringers, due to the loads in the panel
AD, In fewer words, the shear in the end panel AD is

S’=R-r.
Taking moments about the right support B (using Table A) we have
_ 4,632,000 + 152,000 x 2

2°
i HO g als
“Ks

 

 

 

 

 

= 38,000 lbs.,

R 60 = 82,300 Ibs.,
and taking moments about the panel point D we have
. 230,000
[> ae. = 15,300 lbs.

Then for the maximum shear in the end panel AD we have
S’ = 82,300 — 15,300 = 67,000 lbs.,
and multiplying this by 50/40 we have

67,000 x a = 84,000 Ibs. (about)

for the maximum live-load shear on the girder.

Then for the impact we have

300
Now adding together the dead-load and live-load end shears, given above,
and the impact, we have

23,000 + 84,000 + 73,000 = 180,000 Ibs.
972, STRUCTURAL ENGINEERING

for the total maximum end shear on the girder, which is practically the
shear throughout the end panel.
Now for the unit-shearing stress on the assumed 84” x 2” web we

have 180,000

eo 5,710 lbs.

Then substituting this for s in Formula (1), Art. 117, we have
d = (12,000 5,710) =59 ins.

for the maximum spacing of the stiffeners at the ends of the girders.
Now, as this is not less than the half depth of the girder, the assumed
web is satisfactory. The specification will not permit of the web being
made thinner.

The outstanding flanges or legs of the stiffeners, to satisfy the speci-
ficatious, should not be less than 1/30 x 84+2=4.6”. This requires the
use of 5” x 34” angles, the 5” leg outstanding. All intermediate stiffeners
can be taken as 5” x 34” x 3” angles without hesitating—this being the
minimum thickness allowed—but the end stiffeners must have sufficient
area of cross-section to take the maximum reaction on the girder when
considered as columns, as per specifications. The cross-section of a pair
of these angles, considered as a column, is shown in Fig. 198.

 

 

 

 

v 2"
(- CIOs 2' O50 ¢@£@ @s eH 2
sane | aa Rage | 15! Ne gen i. wee
Golo”
ly
Fig. 198 Fig. ‘199

Let us first assume them to be 5” x 3)” x 3” angles. Then for the
radius of gyration about the axis y-y we have .

2 x.98 +948 x 61
r= NI

64 = 2.96 ins.

Then taking one-half the depth of the girder as the length, as per speci-
fication, we have

p=16,000-70 = = 15,000 Ibs. (about)

for the allowable unit stress on such a column.
Now from Fig. 194 it is readily seen that the total maximum dead-
load reaction is

20,025 + 6,675 = 26,750 Ibs., say, 27,000 Ibs.

The maximum live-load reaction, as is readily seen, will occur when
the wheels are in the position shown in Fig. 199, where AB represents the
span.
DESIGN OF SIMPLE RAILROAD BRIDGES 273

Taking moments about B (using Table A) we have

ei (2208 +162 x4

a ) 1,000 = 97,600 lbs.

for the maximum live-load reaction at A due to Cooper’s E40, and multi-
plying this by 50/40, to reduce it to Cooper’s E50 loading, we have

97,600 x ae = 122,000 lbs.

for the maximum live-load reaction desired.
Then for the impact we have

300
T=122,000 —_ 604300 = 101,000 Ibs.

Now adding together the above maximum dead- and live-load reactions
and the impact we have

27,000 + 122,000 + 101,000 = 250,000 Ibs.

for the total maximum reaction. Now dividing this by the allowable unit
stress as found above for the end stiffeners we have

250,000
15,000

for the required area of the end stiffeners.

It is necessary to have two pairs of end stiffeners at each end of the
girders in order to properly distribute the pressure over the pedestal.
Two pairs of the stiffeners assumed above have 12.25” cross-section,
which is about 44” less than required. So thicker angles than those
assumed will have to be used. The allowable unit stress will not be
materially affected by using thicker angles as the radius of gyration varies
but little with the thickness. So we can use any two pairs of 5” x 34”
angles that have the proper area. It is seen from a handbook, or from
Table 4, that 4—Ls 5” x 34” x 4” have 164” cross-section, and that
4t—tLs 5” x 34” x 3%” have 17.880’. As is seen, the 4” angles have an
area of cross-section that is 0.64’” smaller than required, and the 7%”
angles have-about 1.280” more than required. So let us use 4—Ls 5” x
34” x 4” = 16.00” at each support for end stiffeners.

For area of bearing on the masonry we have required

250,000
600

The pedestal should be made to suit.

= 16.6 sq. ins.

=417 sq. ins.

Preliminary estimate of the weight of one main girder. |

4—Ls 6” x 6” x 41” x 69-0” x 26.5% (combined
with end Ls) LEA eave hese weenie edeweR 7,314 Ibs.
1—web 84” x 8” x 62’-0” x 107.12%............. 6,641 lbs.
1—-cov. pl. 14” x 4” x 60’-0” x 23.80% (2 combined) 1, ‘428 lbs.
1—cov. pl. 14” x 3” x 118’-0” x 29.75%.......... 3,510 Ibs. 18,893 lbs.
274 STRUCTURAL ENGINEERING

8—end stiff. Ls 5” x 834’ x4 x 13.64% x 7’-0”...... 762 Ibs.
26—int. stiff. Ls 5” x 34” x 8” x 10.44% x 7’-0”..... 1,893 lbs.
7—fillers 8” x 42” x 6’-0” x 18.74 (at floor beams). 785 lbs.

8—-splice pls. 10” x 44” x 2’-3” x 23.387.......... 421 Ibs.
4—-splice pls. 14” x 44” x 4’-4” x 32.72#......... 568 Ibs.
38—fills. 34’ x 44” x 6’-0” x 8.187............. _.. 147lIbs. 4,576 Ibs.
23,469 lbs.
3% for rivet heads....... 00sec cece ee cece eee eee eeee 700 Ibs.
Total weight of one main girder................. 24,169 lbs.
Total weight of two main girders................ 48,338 lbs.

Total weight of main girders per ft. of span =48,338/60=805 lbs.

To make this estimate the student would have to sketch the details of
the girders to some extent.

151. Design of Lateral System.—The laterals with the floor
beams and main girders form a horizontal double truss as shown in Fig.
200, where AB represents one main girder
and CD the other. The laterals, according
to the specifications, must be designed to
resist a lateral force of 200+5,000 x 0.10
=00# per ft. of span considered as a live
load. One system can be considered as
taking the force when applied from one

Fig. 200 direction and the other system when it is
applied from the other direction. So the
laterals may be considered as taking tension only.

For a panel load we have

P=700 x 15=10,500 Ibs.

Then assuming the lateral force to be applied from the direction indicated
by the arrows and suppose it moves onto the structure from right to left,
loading panel points G and F, we have

 

Ss == x 10,500 = 1,800 Ibs.

for the maximum shear in panel FE (see Art. 90), and for the maximum
shear in panel EA, we have

s’ =t x 10,500 = 15,700 Ibs.

By multiplying each of these shears by the secant of the angle 6 we will
obtain the stress in the corresponding laterals. As angle 6 is about 45
degrees we can take the secant as 1.4. Then for the stress in the diagonal
marked (a) we have

T =15,700 x 1.4 = 22,000 Ibs.,
and for the stress in the diagonal marked (b) we have

T’ =7,800 x 1.4=10,900 Ibs.

If the lateral force were applied in the opposite direction to that
indicated by the arrows, the lateral marked (c) would be subjected to a
maximum stress of 22,000# and the lateral marked (d) would be subjected
DESIGN OF SIMPLE RAILROAD BRIDGES 275

to a maximum stress of 10,900# and Jaterals (a) and (b) in that case
would have no stress. The same maximum stresses would be found in the
laterals in the right half of the span if the lateral force were considered
to move onto the span from left to right.

The stresses in the oor beams due to the lateral force (=700#) can
be neglected, as the stress produced is simple compression in the bottom
flanges, which are in tension from dead and live load. It is readily seen
that the stresses calculated above are all that we need for designing the
laterals.

For the laterals marked (a) and (c) we have

22,000

16,000 =1.37 sq. ins.

for the required area of cross-section.

Use 1—L 34” x 3” x 3” = 2.30 — 0.37 = 1.930” net. This is the
smallest angle allowed by the specifications and therefore the same size
will be used for each of the other laterals.

Preliminary estimate of weight of lateral system.

Bis 317 x a4 x 8" x 7.98 & 20 evs vecew ices 1,264 Ibs.
lat. pls, 19" x As 16H Bf. ea ceases 339 Ibs.
dat, pla 17? x BY @ D16F 224... een casens 173 Ibs.
4—sp. pls. 9” x 8” x 11.5% x 3’... ee. 138 lbs.
16—Ls 6” x4" x4" x 16.2% x1... eee 259 Ibs.
rivets: ssacesaceeds 0% lah Ae sla sala Wada ais ew Mee a 70 Ibs.
2,293 Ibs.

Total weight of laterals per ft. of span = 2,293/60 = 38% per ft.
152. Preliminary Estimate of Weight and Cost.—
Estimates of effective dead weight per ft. of span.

Weight of stringers per ft. of span (Art.147).. 289 lbs.
Weight of intermediate floor beam per ft. of

span (3,621/15) (Art. 148)............. 242 lbs.

: Weight of main girders per ft. of span (Art.150) 805 lbs.
Weight of laterals per ft. of span (Art. 151).... 38 lbs.
Weight of deck per ft. of span........-...++4. 400 lbs.

Total weight of dead load per ft. of span... 1,774 Ibs.

As 1,780# was assumed no recalculations are necessary.

Estimate of total weight of metal in span.

2 main girders at 24,169%...............-4-- 48,338 lbs.
3 intermediate floor beams at 3,621#.......... 10,863 Ibs.
2 end floor beams at 2,8564...........0.0005 5,712 Ibs.
4 panels of stringers and details at 4,3587..... 17,432 Ibs.
8 laterals and details........... 0c cece eeeee 2,293 lbs.
4 pedestals and sole plates..........+eeeeeees 2,200 lbs.

86,838 lbs.
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280 STRUCTURAL ENGINEERING

For the cast at 34¢ we have 86,838 x 0.035 = $3,039.33. 34¢ per Ib.
is a common price for this class of work erected. However, the pound
price will vary from 24¢ to 4$¢. It depends upon the market price of
metal and freight.

This completes the necessary preliminary calculations for the span.
Next a stress sheet as shown in Fig. 201 can be drawn. Then the shop
drawings (Figs. 202 and 203) for the span can be made.

/ Ww

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Fig. 205

The work up to the completion of the stress sheet, as a rule, is done
in the designing office of a bridge company, or in the office of a railroad
company, or in the office of a consulting engineer. Consulting engineers
as a rule (and sometimes railroad companies) make general drawings,
similar to the one shown in Fig. 204, instead of stress sheets. In that
case the general drawing for the span is submitted to the different bridge
companies for use in preparing their bids for fabricating and (usually)
erecting the structure. After the contract is awarded, the general draw-
ing, instead of a stress sheet, is used as a guide in making the shop draw-
DESIGN OF SIMPLE RAILROAD BRIDGES 281

ings by the bridge company obtaining the contract. As a rule bridge
companies do not make general drawings of such structures—only the
stress sheets and shop drawings.

1538. Making of the Detail Drawings.—In beginning the work,
the first thing to do is to draw a half cross-section of the span, as shown
in Fig. 205, to a large scale, say 14” scale. In making this sketch (Fig.
205) first draw the cross-section of the main girder and the stiffeners.
Then the next thing to do is to draw the floor beam. Counting from the
back of the bottom flange angles of the main girder, we have 44” for the
thickness of the flange angle, and 2” for the thickness of the lateral plate,
making in all 44 + 3 = 175” for the distance from the back of the flange
angles of the main girder to the back of the bottom flange angles of the
floor beam. This distance being determined, the bottom flange of the floor
beam is located and can be drawn. The distance from the back of the
bottom flange angles to the back of the top flange angles of such shallow
floor beams is usually made 4” more than the depth of the web. So in this
case, as the web is 32” deep, we have 324” for the distance from the back
of the bottom flange angles to the back of the top flange angles, and thus
the top flange is located and can be drawn. Next the cross-sections of the
two I-beams composing the stringer can be drawn at their proper location.
Then by drawing a horizontal line 93’ above the top of these beams
(using 10” ties dapped 4” on the stringer) we have the base of rail
located, and having the base of rail located the clearance line can be drawn
as shown. This is about as far as we can proceed with the drawing (Fig.
205) until the required spacing of the rivets in the floor beam is deter-
mined.

For the shear on the part of the floor beam between the end and
where the stringer connects, we have 137,200# (see stress sheet). The
vertical distance from the center of rivets in the top flange to center of
rivets in the bottom flange is about 25.5”, and the allowable bearing of a
4” shop rivet on the 4” web is 10,500#. So, substituting 137,200 for S,
25.5 for hk, and 10,500 for r in Formula (2) (Art. 116), we have

_ 10,500 x 25.5
P= —~ 337,200

say, 2”, for the theoretical spacing of the flange rivets in the part of the
floor beam between the end and the stringer.

The shear on the part of the floor beam between the stringers is very
slight. In fact there is no shear at all except that due to the weight of the
intervening part of the floor beam itself. This being the case, the flange
rivets in that part of the floor beam can be spaced 6” apart, which is the
maximum spacing allowed.

Now beginning at the end of the floor beam (Fig. 205) we space the
rivets 2” apart (as required) in the flange angles out to the splice (where
the gusset and web meet) and the only break in the 2” spacing there is
that necessary to fit the details of the splice. The splice is located so
as to bring the bracket just inside the clearance line as shown.

The rivets in the splice should line up with those in the stringer
connection and at the same time with those in the end details of the floor

beam.
There must be enough rivets in each stringer connection to transmit,

= 1.95 ins.,
282 STRUCTURAL ENGINEERING

in bearing on the web, the maximum floor beam concentration. For the
concentration we have 5,200# from the dead load, 68,200# from the live
load as given in Art. 148, and 62,0004 [=300 + (30 +300) x 68,200] from
the impact, making a total of 135,400#. The allowable bearing of a {”’
field rivet on the 4” web is % x 20,000 x 4 = 8,750#. Then for the number
of rivets required in each stringer connection we have

135,400
8,750

It is seen from this that a single line of rivets in each of the angles
connecting the stringers to the floor beam is sufficient. So we have this
much data for later use.

Now the next thing to do is to determine the vertical spacing of the
rivets in the floor beam. It appears that there is no question but that the
rivets should be spaced as closely as is permissible in the vertical direction
in the case of such shallow floor beams. So we will simply try putting in
as many rivets in the vertical direction as will fit in. Beginning at the
bottom of the floor beam, we will make the first gauge in the flange angle
24”, as the angle is quite thick (42), and the second gauge 24”, which
leaves 13” edge distance on the angle. For the next space we have 13”
(edge distance on the angle) plus 3” (clearance) plus 14” (edge distance
for fillers and splice plates), making in all 34’. Now these same spaces
will be used at the top of the floor beam in order to have symmetrical
spacing. So we have

824 - 2(2$4 24434) =17 ins.

for the remaining distance in which rivets are to be spaced. It is readily
seen that 3” spacing will not fit in this distance and that five spaces is the
maximum number possible. So we will space the rivets as shown.

We will now see how this spacing fits the detail at the end of the floor
beam, at the splice, and at the stringer connection. There should be
enough rivets in the end of the floor beam proper to transmit the greater
part of the maximum end shear on the beam to the main girder. In the
detail shown we have 8 field rivets in double shear or bearing on 175”
metal. The double shear being the least we have 8 x 0.6 x (10,000 x 2) =
96,000 lbs. for the amount that these rivets will be considered to transmit.
This leaves 41,200 lbs. (=137,200- 96,000) to be taken by the rivets in
the bracket above the floor beam proper. These rivets are in single shear
and hence the allowable stress on each is 6,000 lbs. Then we have

41,200
6,000

required in the bracket above the beam proper. So the detail as shown for
the end of the floor beam is quite satisfactory, as the extra rivets at the top
of the bracket are really needed to hold the bracket which stiffens the top
flange of the main girder. Now by prolonging the lines giving the vertical
spacing on to the stringer connection, it is seen that this spacing fits nicely
at that point, as we obtain the detail shown. The connecting angles
between the I-beams are shop riveted to the floor beam to avoid difficult
field riveting, as it would be difficult work to drive the rivets connecting
these angles to the floor beams in the field.

=15.4, say 16, rivets.

 

=6.8 rivets, say 7,
DESIGN OF SIMPLE RAILROAD BRIDGES 283

Let us now turn our attention to the splice. The most satisfactory
way of designing such splices is to draw in what looks to be about correct,
and then determine the stress on the rivets assumed, and make whatever
modifications are necessary. So let us assume we need three rows of rivets
on each side of the splice as shown. First suppose that the 3” plates
extending over the flange angles are omitted and that there is just a single
3” splice plate on each side of the web.

; Now, for the vertical force on each rivet due to the vertical shear we
have

137,200
ee = 7,620 Ibs.,
and let s be the maximum horizontal force on each of the rivets farthest
out from the center of the web due to cross bending. Then, according to
Art. 117, we have the equation

2x3 (LYS +5.25 +85 ) xs+8.5=1/8 area of web

multiplied by 16,000 x h (=2 x 16,000 x 29) =928,000 in. Ibs., from which
we obtain
_ 928,000
~ 72.6
This horizontal force should not exceed ($ x 24,000 x $)17/25.5 = 7,000
lbs., so the splice as just considered would not be sufficient.

As one more line of rivets on each side of the splice will not be
sufficient, it is obvious that it will be best to use the 2” plates extending
over the flange angles as shown. So, considering the splice as it is shown
in Fig. 205, we have

= 12,800 Ibs.

s’

ee a, er |
2 [sca7s +5.25 +85 )4+2 (11.6 +13.75 Nee = 928,000 in. Ibs.,

from which we obtain
, _ 928,000
~ 139.0

for the horizontal force on the outer rivets (which are those in the outer
gauge lines in the flange angles). These rivets are so near the top and
bottom edge of the beam that the vertical shear on them can be neglected
(see Art. 61).

The above 6,700# force can be considered as applied to the rivets by
the web and transmitted from there to the angles and on to the 2” plates
producing a shear of 3,330# on each rivet at each 2” plate. The flange
increment can be considered as being transmitted by the 2” plates and
web combined. Taking the flange increment as 10,500# (allowable bear-
ing of a }” rivet on the 4” web) and assuming that one-third is trans-
mitted to the flange angles by the web and one-third by each of the 3”

plates, we have

= 6,700 Ibs. (about)

s

10,500 5 330 =6,830 Ibs.

3
284 STRUCTURAL ENGINEERING

for the shear on each rivet at each 3” plate and

10,500
3

 

+6,700 = 10,200 Ibs.

for the bearing stress on the web. Now, as %,200# is allowed in the
former case and 10,500% in the latter, the top and bottom part of the
splice is about correct.
The rivets connecting the splice plates to the web directly, not passing
through the flange angles, are in double shear and bearing on the web.
The allowable bearing on the web, which is 10,500 lbs., is less than double
shear, and hence the maximum stress on these rivets results from the
bearing on the web. Now, as a test for these rivets, let us consider the
ones farthest out from the center of the web, which are 84” out. Let
s’”’ be the horizontal force on each. Then we have
a a oe ain [gl zs
2 [s (1.75 +5.25 48.5 )42 (11.6 + 13.75 15 = 928,000 in. lbs.,

from which we obtain
» . 928,000
3! =
224
(this could be as great as 7,000 Ibs.).
Then for the resultant force on each of these rivets we have

R’ =/ 4,140? + 7,6202 = 8,670 lbs.,

which is less than the allowable bearing of each on the $” web, and hence
the splice as shown is amply strong.

The splice, as shown in Fig. 205, is designed to develop the strength
of the beam (as should be done), while as a matter of fact the flange at
the point of splice is practically twice as strong as need be, as the bending
moment at that point is only about one-half the maximum on the beam
and if the splice were designed to carry only the shear no actual weakness
would be detected in it.

This completes the necessary calculation in connection with the
drawing of the general cross-section shown in Fig. 205. Next a sketch
to 14” scale for each of the lateral connections, and also a sketch (to a
large scale) of the curved end of the main girder, should be made. After
these preliminary sketches are made, we can proceed with the making of
either the general drawing (Fig. 204) or the shop drawings (Figs. 202
and 203). We simply transfer the dimensions on these sketches to the
final drawings.

Most of the calculations for the details shown on these drawings are
quite similar to those given in Art. 135 for the 50-ft. deck plate girder.
The main difference is in the determination of the flange rivets in the
main girders. In the case of the through girder, here considered, the
loads are applied directly to the web of the main girders (by the floor
beams) instead of to the flanges as in the case of the deck girder treated
in Art. 135.

So Formula (2) of Art. 116 is used to determine the pitch of the
flange rivets instead of Formula (4). The shear in each panel is prac-
tically constant throughout the panel, and consequently the rivet spacing

= 4,140 Ibs. (about)
DESIGN OF SIMPLE RAILROAD BRIDGES 285

should be constant throughout each respective panel. Taking the case of
the end panel, we have

S=180,000% (maximum end shear) ;
r=7,880% (= allowable bearing of a #” rivet on the 3” web);

hat?”,
Now substituting these values in Formula (2) (Art. 116), we have
7,880 x 77
— 6880 x17 4 a6 ; es
180,000 3.36 ins., say, 34 ins.,

for the theoretical pitch of the flange rivets in the end panels. The shear
on the girder is reduced some by the weight of the girder itself, as we
pass from the end toward the center, and, consequently, the pitch can be
increased slightly as the first intermediate floor beam is approached, as is
shown in Fig. 202.

The dead-load shear in the second panel from the end, as seen from
Fig. 194, Art. 150, is

20,025 — 13,350 =6,670 Ibs.

The maximum live-load shear occurs in the second panel when the
wheel loads are in the position shown in Fig. 206, as the criterion of Art.

 

 

 

90 is satisfied. l
Taking moments about 4 TOs @s@i@s@ 2' OO),
the end B of the span (Fig. Pee los 16 eases
206) we have (using Ta- foo Eman
ble A) * Fig. 206
Re 2,155,000 oa x1 =37,900 Ibs.

for the reaction at A. Then taking moments about the floor beam at C
we have

_ 80.000

ee 15

for the stringer concentration on the floor beam at D. Then we have
R -— r = 37,900 — 5,330 = 32,570 lbs. for the live-load shear in the second
panel from the end of the span due to Cooper’s E40; and for Cooper's
E50 we have

 

= 5,330 Ibs.

"82,570 x = =40,700 lbs.,
and for impact we have
309

I=40,700 (55-300) = 37,000 lbs.

Now adding the above dead- and live-load shears and the impact
together, we have
6,670 + 40,700 + 37,000 = 84,370 Ibs.

for the total maximum shear in the second panel from the end of the span.
Then using Formula (2) (Art. 116) we have
_ 7,880 x 74

=6.9 ins. (ab
81,370 6.9 ins. (about)
286 STRUCTURAL ENGINEERING

for the theoretical spacing of the flange rivets in the main girders through-
out the second panel from the end of the span. So 6”, the maximum
spacing allowed, will be used.

Further analyses of the details of the above span are considered
unnecessary, for if the details of the deck plate girder spans previously
treated are thoroughly understood, the student should have little trouble in
making the detail drawings for through plate girder spans, and especially
if the drawings shown in Figs. 202, 203, and 204 be observed closely as
the work progresses. The shop bills for this work are practically the
same as for deck spans.

154. General Remarks.—The general discussion given above,
regarding the design of flanges and end bearings for deck plate girder
bridges, holds in the case of through plate girder bridges.

The floor beams for through plate girder bridges are sometimes made
as shown in Fig. 207. This type of detail eliminates the web splice in the

 

 

Le 5*

5

’ RTT
xg 1} | ey

Wwe the
TLRS
aS | |S
q 4
x x

° oe 0-0 ©

 

 

 

 

 

 

 

 

 

 

Fig. 207

floor beam. The brackets are usually shipped loose and riveted to the
floor beams and main girders in the field, especially when the main
girders are quite deep. While this type simplifies the details of the floor
beam, it complicates the details of the main girders and there are con-
siderably more field rivets than there are in the type used above for the
60-ft. span. The details of the two types vary considerably in practice,
and the details given here are intended only as a fair example.

The stringers for through plate girder bridges are sometimes plate
girders (where the under clearance will permit), in which case two
stringers per track are used.

155. Graphical Determination of Live-Load Shears and Bend-
ing Moments on Main Girders of Through Plate Girder Bridges,—
The “influence line” method, outlined in Arts. 100 and 101, is the most
convenient graphical method in the case of through plate girder bridges.
However, the equilibrium polygon can be used if one so desires.

As an illustration, let it be required to determine the maximum
reaction, end shear, and bending moment on a 75-ft. single-track through
plate girder span of five 15-ft. panels, due to Cooper’s E50 loading. (See
diagram, Fig. 151.)
DESIGN OF SIMPLE RAILROAD BRIDGES 287

To determine the maximum reaction, lay off the span AB (Fig. 208)
to, say, }” scale, and place the loads as shown. © ,

Then draw the base line ab and lay off ac = 1” = 1# and draw the
influence line cb, and draw the ordinates 1, 2, ... 13 under the loads, as
shown, and all is ready for determining the reaction. Take a pair of
dividers and step off the ordinates 1 to 8 as explained in Example 1, Art.
100, and multiply their sum, in inches, by 25,000#. Next, in the same
way, step off the ordinates 9 to 12 and multiply their sum by 16,250#,

“9 %, % hy
8 { OP ars 8 .
YJ sl sl oO S} Ss J] gl ©
a 8 8S uy 8 S] s|
qf g FF 3 qv o} 3 8 v
e '

APELEDSNS) 9° @5sQe'@s@ 8’ © 8'OsQs@s@

R ’ 1 1

| ; i 8
15 D9 i." | 5% 1st
T

| (For one Girder))

 

   

9 0 lw Ye

Fig. 208

Then multiply the length of ordinate 13 by 12,500 and add all three of
the results together, and the desired reaction is thus obtained.

To determine the maximum shear in the end panel AD (Fig. 209),
place the wheels as shown, thus satisfying the criterion for maximum
shear (Art. 90). Then draw the reaction influence line cb for the span
and cd for the panel dD. Then by multiplying the sum of the ordinates
(obtained as explained above) ef, 1, 2, 3, 4, and 5 by 25,000, 6 to 9 by
16,250#, and h and 10 by 12,500*, and adding these products together,
the maximum shear in the end panel AD is obtained.

To determine the maximum shear in any of the intermediate panels

 

 

Fig. 209

as DE (Fig. 210), first draw the reaction influence lines eb and af and
then the influence line adcb for shear in the panel, as explained for trusses
in Art. 102, and ‘place the wheels as shown for maximum shear in the
panel, as per Art. 90. Then by multiplying the sum of the ordinates
2 to 5 by 25,000#, 6 to 9 by 16,250#, and ordinate 1 by 12,500#, and
adding the products together, the maximum shear in the panel DE is
obtained.

It should be noted in the last case that the position of the wheels for
maximum shear in the panel can be determined directly from the influence
line. For, according to Art. 102, no load should pass the point O (Fig.
210), and there must be a load at E, so by placing wheel 1 as near O as
possible, with a wheel at E, we have the position of the wheels for
9&8 STRUCTURAL ENGINEERING

maximum shear in the panel. The position of the wheels for maximum
shear in any of the intermediate panels can be determined in this manner.

The maximum bending moment will undoubtedly occur at the floor
beam nearest the center of the span, and, as there are two floor beams
live load (in reference to C) as shown, thus satisfying the criterion for

@ 8 OOOO » ‘Ose @s® 8

ft nee

Fig. 210

equally near the center, the location of either beam can be taken as the
point of maximum moment. So, if dB (Fig. 211) represents the span,
either C or D can be taken as the point of maximum bending moment.
Let us take point C. Then to determine the maximum moment, place the
maximum moment given for trusses in Art. 91. Next construct the
influence line aOb for the bending moment at C as explained in Art. 101,
and also in Art. 102. Then by multiplying the sum of the ordinates 1 to 4
by 25,000#, 5 to 9 by 16,250#, 10 by 12,500#, and one-half of 11 by

 

% 3’ © 2 OBO 3! @:@ Qs@ s W7zaBe
LE2LEE B
ot

 

 

 

 

 

 

 

AfS
; |
15! 1s L 1st L ws’ | yst
7Lo" 7
4 Fe 6
7 Wa {n
: &
Fig. 211

(2,500# x 10’) and adding these products together, we obtain the maxi-
mum bending moment on the main girders.

To determine the maximum shear in any panel, as CD (Fig. 212),
by means of an equilibrium polygon, first place the load for maximum
shear in the panel, as shown, and draw the ray diagram MNO, and the
corresponding equilibrium polygon a-c-b ...a. Then by drawing OE,
in the ray diagram, parallel to the closing line ac we have the reaction
(RB) at A due to the loads given by the line EM. The shear in panel CD
is equal to R minus the stringer reaction (r) at C due to the load in panel
CD. By drawing the closing line ek we have the equilibrium polygon
ekhe for the panel CD. Then by drawing E’O (in the ray diagram)
parallel to the closing line ek we have the stringer reaction r, at C, due to
wheel 1 (as that is the only load in the panel), given by the line E’M.
Then, evidently, ne maximum shear in panel CD is given by the line EE’,
DESIGN OF SIMPLE RAILROAD BRIDGES 289

The maximum shear in the other panels can be determined in the same
manner.

To determine the maximum bending moment on the main girders by
means of an equilibrium polygon, first place the loads on the span for

ir ">
A De VG: s G:@£@s6

15°

 

 

|

 

 

yr"

 

(2

 

 

 

 

Fig. 212

maximum moment at any panel point as C (Fig. 213), as explained above,
and draw the ray diagram STP and the corresponding equilibrium polygon
acda. Then the maximum moment at C is equal to the ordinate y multi-
plied by the pole distance H. H is measured in pounds to the same scale
as used in drawing the ray diagram and y is measured in feet or inches to

 

Fig. 213

the same scale as used in drawing the span AB and in spacing the loads.
If y is taken in feet the moment will be in foot pounds, and if taken in
inches the moment will be in inch pounds.

In case an equilibrium polygon similar to the one shown in Fig. 186
(Art. 144) be drawn the shears and bending moments for any number of
through plate girder spans can be determined from it.
290 STRUCTURAL ENGINEERING

The shear would be determined as shown in Fig. 212, the wheels
being placed for maximum shear as per criterion, Art. 90. The maximum
moment at any panel point C can, however, be determined without refer-
ence to the criterion for maximum bending moment, very much the same
as the case of deck spans treated in Art. 144, the main difference being
that the moment here is for a certain point on the girder.

As an illustration let UVW (Fig. 214) represent an equilibrium
polygon, similar to the one shown in Fig. 186. We start, say, by first
placing the point C, the panel point where the moment is desired, under

a4 W
Lo
- *

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig, 214

. wheel 11 and drawing the closing line 1-1, we obtain the ordinate y. Next
placing the point C under wheel 12 and drawing the closing line 2-2 we
obtain the ordinate y’, and placing C under wheel 13 and drawing the
closing line 3-3 we obtain the ordinate y”, and so on. In this manner the
maximum ordinate for point C is obtained by trial, and by multiplying
this maximum ordinate by the pole distance the maximum bending moment
is obtained.

The maximum bending moment at any panel point can be obtained in
this manner.

156. Plate Girder Bridges with Solid Floors.— Plate girder
bridges, especially through spans, sometimes have solid floors, particularly
those over streets in cities where an ordinary open floor is undesirable
owing to the dripping of water from the floor upon the sidewalk and
street below due to rain and melting snow.

These structures, in reference to floor, taking the most common con-
struction, can be divided into two general types: those having solid metal
floors covered with concrete and ballast, and those having reinforced con-
crete floors covered with ballast. The most common of the solid metal
floors is that known as the trough floor shown in Fig. 215 where the floor
proper is made up of vertical and horizontal plates connected by angles
so as to form successive troughs. In designing such bridges the first thing
to do is to draw a longitudinal section of the floor, enough to include three
ties 18”’ on centers, as shown to a large scale in Fig. 215. At the start the
whole thing is an assumption, that is, we draw in what looks to be correct.
Siz ‘Sta
Uolt2aG soulpnybuo7z

 

|

        

|

 

cal
een ie
WBE

x
«7
‘7%
‘Sh

 

 

 

"COWMIICSSOs1Y

0 Ex ex -7.

7

(7,928 Wolf yn)
“ o a

isi {2090 syay201g
Bf

 

 

 

291

 
292 STRUCTURAL ENGINEERING

From this sketch we determine the dead weight of the floor, including
metal, concrete, ballast and track. Then we compute the dead- and live-
load stresses on this assumed floor and if necessary modify the design until
it agrees with the final computation.

As an example let it be required to design the through plate girder
span indicated in Fig. 215, using Cooper’s £50 loading. Estimating from
the longitudinal section of the floor (Fig. 215), taking concrete to weigh
150# and ballast 125# per cu. ft., we have for a strip through the floor one
foot wide and two feet long, along the track, the following weights:

14 cn. ft. of concrete at 150#...... 225 Ibs.
1 cu. ft. of ballast at 1257........ 125 Ibs.
Tie (6 ft. B. M.) at 6x4d#...... 27 Ibs.
34 ft. of 12” x 2” plates at 15.34.. 53 Ibs.
4 ft. 34x 34x23 Ls at 8.54%........ 34 Ibs.
Rails (90#) and spikes........ .. 12 Ibs.
Tar, stone dust, and burlap....... 20 Ibs.

Total ayeds Caen e aes +496 Ibs

for two square feet of floor, or 248% for each square
foot (horizontal projection). The floor can be con-
sidered as being made up of independent Z-shape sec-
tions, each composed of one vertical plate, two angles
and two halves of horizontal plates as indicated in
Fig. 216. This section can be treated as an inde-
pendent transverse beam. For the moment of inertia
of this beam with reference to axis O-O, we have

   

Fig. 216

2x G43 x6" x eS iendik patesn Glen = 186 (hor. pls.)
soWeeIe wiguealeresosas = 54 (vert. pl.)
2 (See xh EBB i ccseuee = 281 (angles)
“B21
—(1.50” x (25) (rivets staggered).. —60
TL Otalis. sc wisn ogapanninh egal Sek “461

If the main girders are 15’-6” apart
the dead load on our Z-beam will really be
as indicated in Fig. 217, as the 248-lb. load
computed above holds only for the part of | |III, LEEELEIT ITI HT
the floor directly under the track. But as ehige
the final result will not be affected mate- Fig. 217
rially if we assume the 248 lbs. to extend
the full length of the beam, we shall so consider it.

Then for the maximum bending moment due to dead load we have

M=4x248x 15.5 x12=89,370 inch Ibs.

The next thing is to determine the maximum bending moment due to
live load. In accordance with the usual practice we will assume the

é #
“9@160" 12'@ 248%verln. ff. ——'a'e1co"

 

 

 

5 Lig

 

 

 

 
DESIGN OF SIMPLE RAILROAD BRIDGES 993

maximum live load in this case to be the heaviest axle, distributed over
three ties. So we have (special load, see diagram, Fig. 151)

62,500
3

 

= 20,833 lbs.

for the total load on each tie. Now from the longitudinal section of the
floor (Fig. 215) we can see that our Z-beams carry from half of this load
to practically all of it. So to be sure we will assume that each carries the
20,833 lbs. uniformly distributed along the beam by the tie.

Then for the maximum bending moment due to live load we have

M’ =4x 20,833 x 15.5 x 12 = 484,300 inch lbs.

Now adding together the above dead- and live-load moments and allowing
100 per cent for impact we have

M” = 89,370 +2(484,300)= 1,057,970 inch Ibs.

for the total maximum bending moment on our Z-beam.
Then substituting the above values in Formula (D), (Art. 53), we
have
pa cLie x 6.6
461
for the maximum bending stress on our Z-beam which is the maximum
bending stress on the floor. This shows that the floor is but little heavier
than necessary for cross bending, 16,000 lbs. being the allowable bending

stress.
For the end shear on our Z-beam we have
15.5

248 x2 = 1,920 Ibs.,

=15,100 Ibs.

from dead load and

20,833
2

from live load and the same from impact, making 22,700 Ibs. in all.
Then to resist this shear we should have

22,700
10,000

of cross-section in the vertical plate of our Z-beam. This shows that the
shear on the floor is amply provided for, as each vertical plate contains
4.5 sq. ins.

. The above calculations show that the floor shown in Fig. 215 is about
as correct as we can design it, as 3” metal is the minimum thickness
allowed. :

Having the floor designed, the dead weight coming onto the main
girders from the floor can be computed and by adding this per foot of
girder to the assumed weight of girder itself (per ft.) we obtain the
dead load for determining the dead-load stress in the main girders. Then,
using the formula pL?/8, the maximum dead-load bending moment on the
main girders can be determined.

 

= 10,416 Ibs.

= 2.27 sq. ins.
294 STRUCTURAL ENGINEERING

To obtain the maximum ‘live-load shear and bending moment on the
main girders the live load is placed just the same as in the case of deck
spans.

. The trough floor shown in Fig. 215, while it is the most common type
and a good design, is only one of several types of solid floors in use. For
example, there is the buckle plate floor composed of buckle plates (see
manufacturers’ handbooks) laid upon I-beam or plate girder stringers,
flat plates laid upon I-beams used either as stringers or transverse beams,
the rolled trough section, etc. In addition to these types there are the
I-beam and concrete floors shown in Fig. 218, in which case the I-beams
are designed to carry all of the load. All of these types are used mostly

Ballast

  
   
 

 

Cross Sechon, 1 ‘Concrete Slab.

 

 

 

eC Ballast.
Biel teats Concrete
=y

pee sa)

 

 

 

 

  

le"

Fig. 218 Fig. 219

for through plate girder bridges. The trough floor is sometimes used on
deck spans, in which case the troughs are placed transversely resting
directly upon the main girders.

The out and out reinforced concrete floors are used on deck plate
girders. These floors are simple concrete slabs resting directly upon the
top of the main girders and turned up at each end so as to hold the
ballast, as shown in Fig. 219.

In designing all plate girder bridges having solid floors, the first
thing to do is to design the floor and determine its weight. Then the
weight of the main girders can be assumed and added to the weight of the
floor and we have the dead weight for determining the dead-load stresses
in the main girders. The other work involved is practically the same as
given above for ordinary bridges.

157. Double-Track Spans.—Double-track deck plate girder bridges
are practically always composed of two simple single-track spans placed
side by side, as previously stated. Double-track through spans are usually
composed of two main girders placed far enough apart to admit the two
parallel tracks 13-ft. centers. In the case of ordinary open floors there
are usually four lines of stringers. The load in that case on each stringer
is just the same as in the case of single-track spans, while the floor beams
have four equal concentrations, each concentration being the same as for a
single-track span. The dead load on the main girders, as stated in Art.
124, is about 70 per cent more than for single-track spans, while the live
load is just twice as much, two trains being considered as moving abreast
over the structure. In the case of double-track through plate girder
bridges composed of three main girders, there are two independent single-
track floor systems. The loads in that case carried by each floor and by
each outside main girder are the same as for a single-track span, while the
central girder carries practically twice as much as each of the outside
DESIGN OF SIMPLE RAILROAD BRIDGES 295

girders. The objection to the three-girder type is that it is usually
necessary to “spread” the tracks at the location of these bridges in order
to obtain the necessary side clearance.

DRAWING ROOM EXERCISE NO. 5

_ Design a 64-ft. single-track through plate girder bridge and make a
stress sheet for same upon a 24” x 18” sheet and tracing of same.
Length of span = 4 panels at 16’-0” = 64’-0” c.c. end bearings.
Width = 15’-8” e.c. girders.
Live load, Cooper’s E50 loading.
Dead load, to be assumed by student.
Specifications, A. R. E. Ass’n.
The required work consists of making the calculations and the

drawing of a stress sheet for the span, similar to the one shown in Fig.
201 for the 60-ft. span.

VIADUCTS

158. Preliminary.—The ordinary railroad viaduct is a bridge com-
posed of a series of deck plate girder spans supported upon towers. How-
ever, as a rule any bridge supported upon towers is classed as a viaduct.
Viaducts are used where the crossings are so deep that ordinary concrete
or masonry piers are impracticable as to cost. The most common type of
viaduct is shown in Fig. 220(a). The type shown in Fig. 220(b) is
built to some extent. The principal difference in the two types is in the

ra

 

 

 

 

 

 

|
(Q)

Cross Section

 

  

 

(b)

Cross Sechon

Fig. 220

tower bracing. Two columns connected together transversely by bracing
constitutes what is known as a bent (see Fig. 221). Two bents connected
together by longitudinal bracing constitutes what is known as a tower. It
is usual practice to make the spans between the towers twice as long as
the tower spans, except where the height of the viaduct is 40 ft. and
under, in which case the spans are usually made equal in length.

From actual calculations the economic lengths of spans are found for
ordinary cases to be as follows: 30’ and 30’ for viaducts 30 to 40 ft.
high; 30’ and 60’ for a height of 40 to 80 ft.; and 40’ and 80’ for a
height of 80 ft. and over.

The cost of erection governs the lengths of spans to quite an extent.

The approximate weight of metal in towers of ordinary single-track
viaducts is given on the diagram in Fig. 222. This diagram is self-
explanatory.
296 STRUCTURAL ENGINEERING

Complete Design of an Ordinary Single-Track Viaduct
159. Data.—Specifications, A. R. E. Ass’n.
Live Load, Cooper’s £50.

160. General Layout of Structure.—The first thing the designer
needs is a good profile of the crossing, which is usually furnished by the
railroad company. Let the profile shown in Fig. 223 be such a profile of
the crossing for which we are to design a viaduct.

 
  

for details see fig.

    
    
    

Note:

x

   

Longitudinal View of Tower Transverse View of Bent.
Fig. 221

Usually (the first thing) we would redraw this profile carefully to a
convenient scale (as a rule to a sy scale) so that it can be traced on the
general stress sheet for the structure. Then upon this profile the positions
of the towers are chosen, avoiding the stream and obtaining as many
towers and columns of equal length as possible, using the length of spans
that is economic for the height of the viaduct. Working in this manner
we obtain the general layout of our structure as shown at the top of the
general stress sheet, Fig. 224. After this preliminary work we start the
detail design by taking up the spans first. :

161. Designing of the Spans.—This work is just the same prac-
tically as previously outlined for deck plate girder bridges, except the
50-ft. spans in this case are made the same depth as the 60-ft. spans in
order to simplify construction, the 60-ft. spans being of economic depth.
The complete stress diagram, or stress sheet, as we might say, for the
three different lengths of spans is given on the general stress sheet
(Fig. 224).

162. Designing of the Columns and Tower Bracing.—In design-
ing a column the first thing to do is to determine the maximum concentra-
DESIGN OF SIMPLE RAILROAD BRIDGES 297

tions on the top of it due to dead and live load. The total dead load
applied to the top of each column is equal to the sum of the dead-load
reactions on the two girders supported and the greatest live load applied
Weightat Single Track Viaduct lowers.
Coopers Esoloading

a A.RE ASSN. Specs.

 
  

ya 1000 lb.

He1ght of Towers{Frombaseof rail, masonry)

40°% 80"

17 1000

  

150°

TY)

| 90" 100". 1 d
Height of Jowers (from base of rail tamason
Fig. 222

to the top of each is the maximum live-load concentration coming from
the live load in the two adjacent spans, just the same as in the case of an
intermediate floor beam. (Art. 148.) These loads are the same for all
columns in bents 1 to 9 (Fig. 224), as each supports an end of a 30- and

   

42/37

Fig. 223

60-ft. span. The load on the columns in bent 10 is that coming from a
30- and 50-ft. span and from the two 50-ft. spans in the case of bent 11.

Using the dead-load reactions given for the girders on the stress sheet,
Fig. 224, we have

19,000 + 6,800 = 25,800 Ibs.
for the dead-load concentration on all columns supporting 30-ft. and
 

 

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298

Fig. 224
SIMPLE RAILROAD BRIDGES 299

60-ft. spans, which includes the columns in bents 1 to 9, and
6,800 + 14,400 = 21,200 lbs.

for columns in bent 10 and

14,400 x 2 =28,800 lbs.
for those in bent 11.
To obtain the maximum live-load concentration on a column we place
the loading so that the heaviest loads (wheels) are near the column with
one load at the column and the unit load in one adjacent span equal to

 

 

 

Fig. 225

the unit load in the other, which is in accordance with Art. 148; the spans
in this case being of unequal length.

Taking first the columns supporting 30-ft. and 60-ft. spans, let C
(Fig. 225) represent the column considered. By placing the loads as
shown in Fig. 225, we have (see Table A) (considering one-half of wheel
12 in each span)

152,000

60 = 2,533 Ibs.
for the average unit load in the 60-ft. span, and
76,000 |
30 = 2,533 Ibs.

for the average unit load in the 30-ft. span.
This shows that the criterion is exactly satisfied, and we need go no
farther with the work of satisfying the criterion. Taking moments about
«A (using Table A) we have
(8,154 + 142 x 4)1,000
60°
for the reaction on the column at C due to loads 3 to 11 (inclusive), and
taking moments about B we have
1,121 x 1,000
380
for the reaction on the column at U due to wheels 13 to 16 (inclusive).
Now adding these two reactions and the weight of wheel 12 together we
have

= 62,000 Ibs.

= 37,300 Ibs.

62,000 + 37,300 + 20,000 = 119,300 Ibs.

for the maximum live-load concentration on the column for Cooper’s £40
loading and multiplying this by 50/40 we have

119,300 x 2 = 149,000 Ibs.
300 STRUCTURAL ENGINEERING

for the maximum live-load concentration due to the E50 loading which
is the concentration desired.
Then for the impact we have

300
149,000 x (so300) = 115,000 Ibs.

Now owing to the columns sloping transversely, known as the batter,
the actual stress in the columns due to the above concentrations is equal to
the concentrations multiplied by the secant of the slope angle. The slope,
or batter, will be taken as 2 in 12, which is the usual batter for such
columns. So we have 1.014 for the secant of the slope angle. Then we
have

25,800 x 1.014=26,000 Ibs.
in bents 1 to 9 (inclusive) for the dead-load stress in the top portion of

the columns, and
149,000 x 1.014= 151,000 Ibs.

for the live-load stress in these columns throughout their whole length and
115,000 x 1.014=116,000 lbs.
for the impact.
Now adding these together we have
26,000 + 151,000 + 116,000 = 293,000 Ibs.

for the total maximum stress in the top sections of all columns in bents
1 to 9 (inclusive).

We will next determine the dead- and live-load stresses in the
columns in bent 10. The dead-load concentration on each of these

 

 

 

Fig. 226

columns is given above as 21,200 lbs., and we will proceed to determine
the live-load concentration on them. Placing the loads as shown in Fig.
226 we have (considering one-half of wheel 13 in each span)

 

112,000 |
=50 — = 2,240 Ibs.
for the average unit load in the 50-ft. span, and
69,000 _
30 =2,300 Ibs.

for the average unit load in the 30-ft. span.
This shows that the load in the 30-ft. span is a little too great. So
let us place wheel 14 at column C. Then we have

132,000

50 = 2,640 lbs.
DESIGN OF SIMPLE RAILROAD BRIDGES 301

for the average unit load in the 50-ft. span, and

62,000
30

 

=2,066 lbs.

for the average unit load in the 30-ft. span. So it is seen that the
position of the wheels shown in Fig. 226 will give the maximum concen-
tration on the column. So taking moments about A (using Table A)
we have

916 x 1,000

30 = 30,500 lbs.
for the reaction on the column at C due to loads 17 to 14 (inclusive), and
taking moments about B we have

(2,036 + 102 x 8)1,000

50 = 57,000 Ibs.

for the reaction due to wheels 6 to 12 (inclusive).
Now adding these reactions and the weight of wheel 13 together we
have

30,500 + 57,000 + 20,000 = 107,500 Ibs.

for the maximum live-load concentration on each column in bent 10 due
to the £40 loading. Then for the £50 loading we have

a x 107,500 = 134,000 Ibs.

and for the impact we have

300
134,000 x (as) = 105,000 Ibs.

Now multiplying the dead- and live-load concentrations and impact
by the secant of the slope angle of the column, as was done above, and
adding all three of the results together we obtain 264,000 lbs. for the

8

15

 

 

Ben#il
Fig. 227

total maximum stress in each of the columns in bent 10. Next placing
the wheel loads as shown in Fig. 227 and proceeding in the same manner
as above, in the case of the other columns, we obtain

50
128,600 x = = 161,000 Ibs.
302 STRUCTURAL ENGINEERING

for the live-load concentration on each column in bent 11, and

300
161,000 x (eran) = 121,000 Ibs.

for the impact concentration.

Then, taking the dead-load concentration found above for these
columns, we have

28,800 x 1.014=29,200 Ibs.
for the dead-load stress, and.

161,000 x 1.014= 163,000 lbs.

t

e
ez000"
Lene

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

y
4 A r 8
3 2 YG) * 2 —"
3 a.
(f), yf i (h)
é ys 5 gy x} 6 y 0
8 % &
: a
9
3 Wk:
i, yf * 3
wy,
wy
Fig. 228

for the live-load stress, and
121,000 x 1.014 = 122,000 Ibs.

for the impact, making in all 314,000 lbs. stress in each of the columns
in bent 11.

Having the stresses in the columns this far determined we can now
write the dead, live and impact stress on them at the cross-sections, as
shown in Fig. 224, increasing the dead-load stress in the lower sections
of the columns where the weight of the part of the tower above increases
the stress an appreciable amount. The dead weight per ft. of height of
towers 1-2 and 7-8 as seen from the diagram in Fig. 222 is about 1,000
DESIGN OF SIMPLE RAILROAD BRIDGES 303

Ibs. So the weight per ft. of column would be 250 lbs. Then the dead-
load stress in the lower portion (about one-half of the height) of the
columns in these towers will be increased (in the worst case) about
250 x 28 = 7,000 Ibs., about 11,000 lbs. in the case of towers 3-4 and 5-6,
In tower 9-10 and bent 11 this increased dead weight is ignored as it is
inappreciable.

We shall next determine the stresses in the columns and transverse
bracing due to wind load. According to the specifications: ‘Viaduct
towers shall be designed for a force of 50 lbs. per sq. ft. on one and
one-half times the vertical projection of the structure unloaded; or 30 lbs.
per sq. ft. on the same surface plus 400 lbs. per linear ft. of structure
applied 7 ft. above the rail for assumed wind load on the train when the
structure is fully loaded with empty cars assumed to weigh 1,200 lbs. per
linear ft. of track.”

Let us first consider the towers 3-4 and 5-6 (see Fig. 224), and let
the diagram at (a), Fig. 228, represent the elevation of one of these
towers drawn to, say, a yy scale. The wind load coming onto a single
bent as AB would be that applied between the vertical sections SS and
S8’S’. This load will be considered (in accordance with practice) to be
applied at the joints of the bents and on the girders and train as shown
in the cross-section at (b), Fig. 228. To obtain the intensities of these
forces it is first necessary to estimate the areas of the vertical projection
of the structure at the different points. The area of the vertical projec-
tion of the train need not be considered as the load on it is given in the
specification as 400 lbs. per ft. of track. However, the train is usually
considered to be 10 ft. high.

For the area contributing to force Fl we have 1.5 x 45 = 670’ in
round numbers from ties and guard rail; 3.5 x 30 = 1050’ from the 60-ft.
girder, and 2.25 x 15 = 334” from the 30-ft. tower girder, making in all
205’ (= 67H’ + 1050’ + 330’). (See Figs. 221 and 232.)

In the case of F2 we have the same area from the girders (= 105 +
33 = 1384’) and 9 ft. of tower. From the tower, we have 0.8 x 15 = 120”
from the top longitudinal strut (see detail of tower, Fig. 232, also see
Fig. 221), 0.5 x 24=120’ from one-half of the longitudinal diagonal, and
1.5 x 9 = 140’ from the column, making in all 1760” (=1380" + 120” +
120” + 140’),

For F3 we have 1.5 x 18 = 27” from the column (only).

For F4 we have 1.5 x 27.7 = 42 from the column; 0.8 x 15 = 120”
from the longitudinal strut and 0.5 x 48 = 240” from the two longitudinal
diagonals (one-half of each), making in all 785” (= 420" + 120” + 2407”),

Now by increasing each of the above areas by one-half (as per
specification) we obtain the areas (to the nearest square feet) indicated at
(b). The wind load on the‘lower part of the bent is transmitted directly
to the masonry and hence is not considered.

Taking first the case of the 50-lb. load (when the train is not on the
structure) we obtain the forces indicated at (c) by multiplying the given
areas by 50. Then laying off these forces on the load line BC, at (f),
we obtain the diagram of the stresses in the bent, as shown, by beginning
at a and passing around each joint counter clock-wise. The intensities
of the stresses thus obtained for the different members are given on the
bent at (c). Next for the case of the 30-Ib. load on the structure and the
304 STRUCTUARL ENGINEERING

400-lb. per linear ft. on train, we have the forces shown at (d). Those
applied to the structure here are obtained by multiplying the correspond-
ing areas given at (b) by 30 and the load on the train by multiplying
400 by 45’, the length of the train contributing pressure to bent AB.
By drawing on and np we have the forces acting upon a continuous frame
which can be graphically analyzed and the addition of the members on
and np will not affect the stresses in the members of the bent in the least.
Then by laying off the forces on the load line BD and beginning at joint
n and passing around each joint counter clock-wise we obtain the diagram
of the stresses shown at (g). The intensities of the stresses thus obtained
for the different members are given on the bent at (d).

This work completes the determination of the wind stresses in the
towers 3-4 and 5-6, and we shall next take up the determination of the
stresses in these same towers due to the longitudinal force from the train.
This force is known as traction. It is the force exerted along the rails
when the brakes are applied to the wheels of a moving train causing the
wheels to slide on the rails. The intensity of this force, as is evident, is

2Q5@s@s@©_2' ©5@6'@s@ 2' @ 8' OIS@2@s@ 9' @5@
30/ 60"

Fig. 229

 

 

 

equal to the coefficient of friction (of the wheels on the rails) times the
vertical load on the wheels. This coefficient is designated in the specifica-
tions as 0.20. We will consider the traction on each of the towers, 3-4
and 5-6, as coming from the loads on the girders rigidly connected to the
tower in each case, or, in other words, the loads between the expansion
points of the girders. This, as is seen from Fig. 224 (the points of
expansion being marked Ex), will be the loads on a 30-ft. and 60-ft. span,
making in all 90 ft. of continuous load. The load will be a maximum
when the wheels are in the position shown in Fig. 229. This gives a
load of 248,000 Ibs. (see Table A) for Cooper’s E40 and 310,000 Ibs.
(= 248,000 x 50/40) for E50. Then for the traction force on each tower
we have |

T = 310,000 x 0.2 = 62,000 Ibs.,

which is applied at the top of the rail. By drawing mk and kn at the top
of the tower as shown at (e) we have the load applied to a continuous
frame which can be graphically analyzed. The diagram of stresses in
the tower shown at (h), due to this 62,000-Ib. force is obtained by laying
off AB (representing the force) as a load line and then beginning at k
and passing around each joint counter clock-wise. The intensities of the
stresses thus obtained are given on the elevation of the tower at (e). The
exact values of the stresses, however, are obtained by multiplying each of
the given stresses by the secant of the slope angle of the columns, but as
each stress is increased but a small amount we will ignore this correction
(as is usual practice) and use the stresses shown.
DESIGN OF SIMPLE RAILROAD BRIDGES 305

We now have all of the stresses determined in the towers 3-4 and 5-6
except the dead, live, impact, and traction stresses in the top transverse
struts (see Fig. 221) of the bents. These stresses are due to the columns
being inclined. Owing to this inclination the columns exert a horizontal
thrust toward each other upon this strut causing a compressive stress in
it equal to the vertical load at the top of the column multiplied by the
tangent of the slope angle of the columns. The slope of the columns is
2 in 12, so the tangent of the slope angle is 0.1666. Then for the top
transverse strut we have the following stresses, in round numbers:

D= 25,800x0.1666= 4,000 Ibs.
L=149,000 x 0.1666=25,000 Ibs. } +48,000 Ibs.
I=115,000 x 0.1666 =19,000 Ibs.
T= 88,000 x 0.1666=15,000 Ibs.
W= 26,000 Ibs.

+ 89,000 Ibs.

In order to design the column bases and anchorage it is necessary to
know the positive and negative reactions on each column. The maximum
positive reaction on each is equal to the maximum combined stress in the
bottom section of the column divided by the secant of the slope angle of
the column plus one-fourth of the weight of the lower half of the tower,
which is about 11,000 lbs. So we have (see Fig. 224)

564,000

+h = 11,000+ ay

=+567,000 lbs.

for the maximum positive reaction.

The maximum negative reaction, if there be such, will occur when
the structure’ is loaded with a train of empty box cars which is assumed
to weigh 1,200 lbs. per ft. of track, as per specification.

We have —98,000 lbs. for the negative reaction due to wind load as
given at (g), Fig. 228. For the traction force along each rail due to the
1,200 Ibs. live load we have

1,200

3% 0.2 x 90 = 10,800 Ibs.

 

Now this 10,800-lb. force can be applied to the top of the tower as
was done with the 62,000-lb. force (at (e), Fig. 228) and the reaction
due to it determined graphically. But as the reaction due to the 62,000-lb.
force is already determined (at (h)), the one due to the 10,800-lb. force
can be determined very readily by proportion as the reactions are directly
proportional to the forces. Hence we have

_ 10,800
~ 62,000

,

x 160,000 =-28,000 Ibs.

 

for the negative reaction due to traction.
306 STRUCTURAL ENGINEERING

For the positive reaction due to the 1,200-lb. live load we have

+R” = ie x 45 =+27,000 Ibs.

Now adding (algebraically) this and the dead-load reaction, which is

37,000
11,000 + —-- L014 = 48,000 Ibs.
(see cross-section of bents, Fig. 224), to the above negative reactions we

have
27,000 + 48,000 — 98,000 — 28,000 =-51,000 Ibs.

for the maximum negative reaction that can occur on each column and for
which anchorage must be provided.
This method of determining these
ix | negative reactions is not absolutely correct
' as the effect of the wind and traction forces
i being in different planes is not taken into
"#4 account. The tendency of rotation of each
: tower, as is obvious, will really be about
x axes perpendicular to the resultant of the
ests oal.ar” two kinds of forces. However, owing to
a" the uncertainty of the intensities of the
ee forces in the first place and to the result-
7 ing error being small, the more exact anal-
ysis is not really justifiable and hence will

 

 

 

 

 

 

 

Fig. 230

not be given.

As we now have all of the stresses determined for towers 3-4 and 5-6
we can write the same on the stress sheet (Fig. 224) as shown, the stresses
in the longitudinal bracing on the elevation of the towers and those in the
transverse bracing and columns on the cross-section of the bents, and we
can then proceed with the designing of the sections of the different mem-
bers in these towers.

Taking the columns first (bents 3, 4, 5, and 6) let us assume the
following section for the top portion of each column:

1—cov. pl. 18” x 3” = 11.250”
2—[s 15” x 40# =23.520”
34.770”
Taking moments about the center of the cover plate (see Fig. 230) we have
— 23.52 x 7.81 F
t= 3477 = 5.29 ins.

for the distance to the center of gravity of the assumed section from the
center of the cover plate (axis x-zx).
Now for the moment of inertia about axis 2-2 we have
ay ee]
T=347.5x 2+ (2.52 x 23.52) + (5.29 x 11.25) =1,159,
and hence for the radius of gyration about axis x-2 we have
1,159
34.77

 

= 5.7%.

 
DESIGN OF SIMPLE RAILROAD BRIDGES 307

For the moment of inertia about axis y-y we have
’=9.89x2+ (6.03 x 23.52) + 303.75 = 1,178,
and for the radius of gyration about the same axis we have

1,178
34.77

f—

= 5.82.

 

The distance along the column between the points of connection of the
longitudinal bracing is the greatest unsupported length (as is seen from
Fig. 224) and hence will be taken as the length of column. This length
(as obtained by scale) is about 36’ or 432”. Then substituting this
length and the radius r’ (as the column would fail about axis y-y) in the
column formula, we have

p=16,000-70 5° = 10,800 Ibs.

for the allowable unit stress in the case of the combined dead- and live-
load and impact stresses. As the maximum stresses due to live load, wind
and traction are not likely to occur simultaneously the unit stress can be
increased 25 per cent for the case of combined dead- and live-load, impact,
wind, and traction stresses, as per specifications.
So we have
293,000

10,800

for the area required for the combined dead- and live-load and impact
stresses, and for the combined dead, live, impact, wind, and traction
stresses we have

= 27.1 sq. ins.

468,000 _ 468,000
10,800 (1+0.25) 13,500

which is the actual required area, being the greater. The section assumed
is satisfactory as it is about as near t,‘e required area as we can get it.

Taking next the bottom portion of the column, let us assume the
following section:

= 34.66 sq. ins.,

1—eov. pl. 18x 4) =12.370”
2—[s 15” x 50# =29.420”

41.790”
Taking moments about the center of the cover plate (see Fig. 231) we have
— 29.42 x 7.84 ‘
TS aa = 5.52 ins.

for the distance to the center of gravity of the section from the center of

the cover plate.
For the moment of inertia about axis 2-2 we have

7 = 402.7 x2 + (2.32 x 29.42) + (5.52 x 12.37) =1,341,
308 STRUCTURAL ENGINEERING

and for the radius of gyration about the same axis we have

[1,342 a
r= Z.79 = 5.66.

For the moment of inertia about axis y-y we have
I’ = (6.05 "x 29.42) + (11.22 x 2) +. 334.13 = 1,483,
and for the radius of gyration we have

,_ {1,483 _
r = 4/489 = 5.86.

The length of the bottom portion of the columns is about 36 ft., or
432 ins., and as this length is the same 2s regards the two axes, r the
smaller radius will be used in the column formula. So we have
432
p=16,000 -70 5.667 10,680
for the allowable unit stress. Dividing this into the combined dead, live,
and impact stress we have

304,000
10,680
for the required area of the column, and by increasing this unit stress

25 per cent and dividing it into the combined dead, live, impact, wind, and
traction stress we have

 

 

= 28.46 sq. ins.

564,000
13,360
for the required area which is the greater

oc of the two. The assumed section is about
as near to the required section as is pos-

= 42.2 sq. ins.

 

|

| sible to obtain, so it will be used.

dS “yy Rs 4 We will next take up the designing
of the transverse bracing. Beginning
2c with the top diagonals (see cross-section of
232 518-34"  bents 3, 4, 5, and 6, Fig. 224) we have
zaalt| 5 a stress of 38,000 lbs., which we will
Fig. 231 assume to be carried in tension by one
system, so we have

38,000
16,000

for the required net area of each of the top diagonals. It is necessary to
use two angles for each diagonal in order to obtain good details. It is
seen that comparatively small angles could be used as far as the area of
cross-section is concerned, but as 34” x 34’ x 3” angles are about as small
as is consistent with good practice in the design of railroad bridges we
will use 2—Ls 84” x 34” x 3” =4.98 —- 0.87 =4.61 sq. ins. net for each diag- |
onal.

‘wok ”
Ss.

 

 

 

6.05

IS
|

 

 

 

 

 

 

 

=2.87 sq. ins.
DESIGN OF SIMPLE RAILROAD BRIDGES 309

Now as these angles are the minimum size used and the stresses being
Jess in the other transverse diagonal than in the ones just considered,
2—Ls 34x 3$”x8”" will be used for each of the other transverse
diagonals.

In designing the transverse struts L/r should not be greater than
120, and 100 would be better. The stresses, as is seen, are so small in
these struts that the designing of the sections really consists in selecting
sections that will be sufficiently rigid and provide good details.

The bottom strut is about 31’, or 372”, long. So we have

372
100

for the required radius of gyration to give L/r = 100. We will use
2—[s 10” x 20% which have a radius of 3.66 and considerable more
section than required, but it is about as satisfactory a section as is obtain-
able and hence will be used for the bottom transverse strut.

The second transverse strut from the bottom is about 18’-6”, or 222”,
long, so we have

= 3.72

222
100

for the required radius. Here we will use 2—[s 8” x 13.75# which have
a radius of 2.98. We will use the same. section for the next strut above
also, in order to obtain uniform details.

Let us next take the case of the top strut. This strut has a length of
about 78”. Let us assume 2—Ls 5” x 34” x 3” = 6.100”. The least -
radius of gyration of these two angles taken as a strut is 1.60. Then
substituting in the column formula we have

z 18
p= 16,000-70 5 = 12,600 Ibs.

2.2

for the allowable unit stress for the combined dead, live, and impact
stresses. So we have
48,000
12,600

for the required area of cross-section, and increasing the above unit stress
25 per cent and dividing it into the combined dead, live, impact, wind, and
traction stresses we have

89,000
15,700

for the required area which is the greater. So our assumed section is
about correct and hence will be used.

We will next take up the designing of the longitudinal bracing in
towers 3-4 and 5-6. Here each diagonal will be designed to carry the
96,000-lb. stress in tension, assuming only one system to act at a time.
So we have

= 3.8 sq. ins.

= 5.6 sq. ins.

96,000

16,000 ~° sq. ins. (net)
   

  
 

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Tig 232 .
310
DESIGN OF SIMPLE RAILROAD BRIDGES 311

for the required net area of cross-section of each diagonal. Then for each
diagonal we can use (counting out of each angle one rivet hole) 2—Ls
6” x4” x 8” = 7.22 — 0.75 = 6.47 sq. ins. (net).

We will next consider the longitudinal struts, which are really
columns 30’ or 360” long. The maximum stress of 62,000# occurs on

the intermediate one. Let us assume a section composed of 2—[s 10x
20# = 11.760”, Then we have

EL 360

= 2 a = 985,
and we also have
360
p=16,000-70 3.667 9,110 lbs.
for the allowable unit.stress. Dividing this into the stress we have
62,000 _
“9,110 = 6.79 sq. ins.

for the required area which is much less than the area of the assumed
section, yet we will use the assumed section for each of the longitudinal
struts as the L/r is about correct. This completes the design for the
towers 3-4 and 5-6, and the sections can be written on the stress sheet as
shown (Fig. 224).

The other towers can be designed in the same manner as shown above
for towers 3-4 and 5-6 and then the stress sheet can be finished as shown
in Fig. 224.

163, Detail Drawings——After the stress sheet (Fig. 224) is
completed a general drawing of one (at least) of the towers should be
made in order to show the kind of details desired unless standard draw-
ings for such details, previously made, are available. The tower selected
for detailing should be one of medium height so as to obtain average
details. In this case we will select tower 1-2 (see stress sheet, Fig. 224).
The general details for the tower will be sufficiently shown by drawing
the details of one column and the principal details of the bracing connect-
ing to the column. The first thing to do is to make large scale detail
sketches of the column cap, base, and of the principal intermediate points.
From these sketches we determine the exact slopes and clearances for the
bracing, in fact all essential details are worked out on these sketches and
simply transferred to the finished general drawing. Working in this
manner we obtain the general drawing for the tower 1-2 shown in
Fig. 232.

The calculations for these details are quite simple. The cap is made
as small as is consistent with good details. The I-beam diaphragm at
the top of the column is for the purpose of distributing the loads from the
girders equally to the two channels. There should be a sufficient number
of rivets in each side of this diaphragm to transmit one-half of the
maximum end shear on the longer girder minus the dead-load end shear
on the shorter girder. So in this case we have (see stress sheet, Fig. 224)

243,000

9 — 6,800 = 114,700 lbs.
312 STRUCTURAL ENGINEERING

for the shear that these rivets are to transmit. Dividing this by 7,200
(the value of a 4” shop rivet in single shear) we obtain practically 16
rivets. Sixteen is the number used.

The bolts connecting the girders to the columns should be sufficiently
large to transmit the shear due to traction, neglecting the friction on the
expansion end. Taking the case of the 60-ft. span, the heaviest load will
occur when wheels 2 to 11 (inclusive) are on the span. This (for Cooper’s
E50, see Fig. 151) gives a load of 205,000 lbs. per girder.

Then, we have 0.2 x 205,000 = 41,000 Ibs. for the horizontal thrust
on each girder which must be taken by the bolts at the fixed end of the
girder. We have four bolts taking this, so each must take one-fourth of
it or 10,250 lbs. This calls for 14” bolts, as the area of cross-section of
each is 0.9944”, and stressing them 10,000 lbs. per square inch gives
9,940 lbs., which is about the required value. The same size bolts are
used for the tower spans in order to have uniform sizes, although the
shear due to traction in that case does not require them to be so large.

The column base should have sufficient area so as not to stress the
masonry more than 600 Ibs. per square inch (see specifications). The
maximum reaction on bents 1 and 2 (see stress sheet, Fig. 224) is 512,000
Ibs. Dividing this by 600 we obtain 853 sq. ins. for the required area of
bearing on the masonry. The masonry plate or base plate used has
(30” x 29’) 870 sq. ins., which is about the correct area. The masonry
plate should be symmetrically placed in reference to the center of gravity
of the column and the details should be such that the pressure from the
column is quite uniformly distributed over the plate.

The anchor bolts should be large enough to take the maximum nega-
tive reaction (uplift) on the column at 16,000 lbs. per square inch. The
splice in the column is considered to be a butt joint, that is, the top section
of the column is considered to bear firmly against the bottom section so
that the stress is transmitted from one to the other without the aid of the
rivets in the splice. In that case there need be only enough rivets in the
splice to hold the column in line. Just how many to use in such cases
must be determined by mere judgment.

The calculation for the other details is mostly a matter of develop-
ing the sections in the bracing which the student should have no trouble
in doing. Take, for example, the longitudinal diagonals. Each of these
diagonals is composed of 2—Ls 6” x 4” x 3” = 7.22 — 0.75 = 6.47 sq. ins. (net).
Multiplying this net area (of the two angles) by 16,000 lbs. we have

6.47 x 16,000 = 103,500 Ibs.

for the tensile strength of each diagonal. Dividing this by 6,000 lbs. we
obtain about 17—{” field rivets to develop the section. Eight on a side
are used (in the end connections), which is about correct.

Take next the longitudinal struts which are composed of 2—[s
10” x 20#=11.760". The strength of these struts as given in the last
Article is 9,110 Ibs. per square inch. Then for the total strength of each
we have

11.76 x 9,110 = 107,000 Ibs.
DESIGN OF SIMPLE RAILROAD BRIDGES 313

Dividing this by 6,000 lbs. we obtain about 18—”’ field rivets. Nine on
a side are used (in the end connections), which is correct. The riveting
in the end connections and splices of the transverse bracing is obtained in
the same manner.

The stress sheet (Fig. 224) and the general drawing of the tower 1-2
(Fig. 232) are quite sufficient for general drawings in this case as the
work is quite similar throughout. However, there should always be a
sufficient number of general drawings to fully show the character of the
work. All special towers should really be detailed.

From these general drawings the bridge company, obtaining the
contract for fabricating the structure, works up the shop drawings for
the work.

In working up the shop drawings usually the first thing, after large
scale sketches of the principal details of the towers are made, is to make
a drawing showing the location of the anchor bolts. This is known as the
masonry plan. This drawing is used by the party building the sub-
structure and is usually the first drawing called for. The material as a
rule is next ordered and line drawings of the towers are made upon which
the lengths of all members are given, which are usually computed by the
aid of logarithmic tables. After this the making of the shop drawings
proper (and bills) proceeds. The shop drawings for the girders are prac-
tically the same as previously shown for deck-plate girder bridges. The
shop drawings for the towers are quite easy to make; the columns should
be on separate sheets from the bracing. All should be drawn as much in
their relative position as possible to aid in checking and also in identifying
the members.

We will next make a preliminary estimate of the weight and cost of
metal in the structure.

164. Preliminary Estimate of Weight of Metal and Cost.—
Taking up the weight of the girders first we have the following:

Weight of metal in one 60-ft. span.
Weight of one girder.

 

1—web 84” x 2” x 107.1% x G07... 2. eee 6,426 lbs.
Abe 68 eB? BE 6 P42 BD races Soncansanosoures 5,808 Ibs.
1—cov. pl. 14” x 8” x 29.767 x 60)... ccc eee es 1,786 lbs.
1—eov. pl. 14” x 8” x 29.76% x 44/0 eee 1,309 Ibs.
2—cov. pls. 14” x 4” x 23.8% x29. eee eee 1,380 Ibs.
26—stiff. Ls 5” x 347’ x 8% x 10.44 x. eee 1,893 Ibs.
8—end stiff. Ls 5’ x 947 x wy? KIO RT cece c cece ees 672 Ibs.
4—fillers 7” x 8” x 14.88% x 67.2... 357 Ibs.
8—sp. pls. 10” x 8% x 21.257 x 24. 408 Ibs.
4—sp, pls. 14” x £" x 29,767 24.3". esanreranevreaeree 512 Ibs.
2—sole pls. 12” x 3” x 80.6% x 1.2/7.0... cece ee eee ee %4 Ibs.
6—fillers 347 x 87 x 7.444 x 6... eee ees 268 Ibs.
20,893 Ibs.

pivekay HOG, (20,890 #00) se ceorasaese tr atures 627 Ibs.
Total weight of 1 girder..,..........: see e ee eens cas lbs.

 

Total weight of 2 girders.............seeeeeeees 43,040 Ibs.
314 STRUCTURAL ENGINEERING

Weight of one frame.

 

 

 

2—Ls 84” x 34’ xB" x BSF XG eee eee *.. 104 Ibs.
2—Ls 34" x 84" x BY x 8.57 x 8.2" caste ease ee nena 140 Ibs.
4—pls. 124” x 8% x 15.944 x1. eee 70 Ibs.
1— pl. 97! 8° x ALAR se 0:8) oie dion Sele cen vise 9 Ibs.
323 lbs
TIVCLS ocd Geese ewe Gs AGE Wowe ea see ee eee 2S 22 lbs
+ Total weight of 1 frame............. 0.02 e ee eee 345 lbs.
5
Total weight of 5 frames.............0. cece ues 1,725 Ibs.
Weight of laterals and lateral plates.
SLs BE! 8k" x BY K-80 cece cade bee se yp eseen'e 612 lbs.
Y=ple:.12” x8" x 158% 2.8" vera ncere sede seee eras 300 lbs.
15—pls. 127 % 8% x 15:87 x 1.08 oo cee eee ee teen as 230 Ibs.
2— pls. 127 8" x 15.37 & 2.0! oo. cee a eee wa ee Ree 61 lbs.
TLV EUS. sd cisicks Gran sngta da nara tight a Sag tgee can Sin ne anne aig eee 27 Ibs.
Total weight of laterals and lateral plates........ 1,230 Ibs.
Summary of weight.
2—-BITdElS: casi epore sere si wea g eres ae reeae wees eee 43,040 Ibs.
= — ERAN ES cis feta sad Seta calle ti easase ee boa emiceord Twist nbn ees Goce at noes SUSUR 1,720 Ibs.
laterals and plates........ 6... c cece eee cee tence ene 1,230 Ibs.
Total weight of metal in 1—G60-ft. span........... 45,990 lbs.

Weight of metal in one 30-ft. span.
Weight of one girder.

 

 

 

1—web 54” x 8” x 68.85% x 25’. cece eee 1,721 Ibs.
4—1s 6” x 6” x BY x 24.2% x BO. eee eee 2,904 Ibs.
Q—end pls. 307 x 8’ x 38.25% x67. cece cee eee eee 459 Ibs.
4—sp. pls. 12” x 8” x 25.5% x BB’... eee 357 lbs.
4—1 5 6” x 6” x ¥” x 19.6% x 4’ (bent)........ 2 eee. 314 Ibs.
Beasl giGl 30 OM x AY ae 19 OF x Ye ekiains vas oh PREM EOS 549 Ibs.
d= fillers6” x8" x 120 6 wrave sy eran eee Aidwie 305 lbs.
141s 57 x BAY x AY x 104E XA ccc eee nee 655 Ibs.
Qa psy 12" xB! QB ee a oo. os esi casecneg ve swe ales ads 102 Ibs
2—sole pls. 12” x 3” x 30.6% x 1.27... eee cee eee 74 lbs
7,440 Ibs.
rivets; 22 Vo seies ae Ghee eae se Peau tee wes 150 Ibs.
Total weight of 1 girder..............-.. 0.0 eeu 7,590 Ibs.

2
Total weight of 2 girders..................00005 15,180 Ibs.

Weight of frames.

2—end frames = 345 x 2....... ....ee 690 Ibs. (from 60-ft. span)

1—interior frame..............6 rae 320 lbs. (computed)

Total weight of 3 frames......... 1,010 Ibs.
DESIGN OF SIMPLE RAILROAD BRIDGES 315

 

 

 

 

Weight of laterals and plates nt - zl x 30 = 615% (from 60-ft. span)
Summary of weight.
A PIGELS. wccy ny au awa ow Se AEE Sele AIS sass Mac actin a 15,180 Ibs.
OS ERAMOS aoa yea sia a ER e cael vk od nba Wromowase eon ouan 1,010 lbs.
laterals and plates......... 0c. cece eee ence ee ennees 615 lbs
Total weight of metal in 1—30’ tower span........ 16,805 Ibs.
Weight of metal in one 50-ft. span. .
1—web 84” x 8” x 10717 x 50’... eee ees 5,355 lbs.
4—1s 6” x 6” x f” x 83.17 x 50. ccc ees 6,620 Ibs.
22—Ls bY % Bh" xe Bw LOAF OY eee cee ea crete cays 1,602 lbs.
4—1 8 5” x Bb" x ge” x ROR KM eee eee 336 Ibs.
4—fillers 7” x 2” x 20.88% x 6’... oceans 500 Ibs.
4—fillers 34” xP” x 10.42F x Occ eee eee 250 Ibs.
splice plates (see 60-ft. span)............... 2c eee 650 Ibs.
sole plates (see 60-ft. span).......... 0. cece ence eee 74 Ibs.
15,387 lbs.
ML VEES) ots Howe ayioia is hea atencene treed aaah cele ahadiaen anit wbeveancnane Baaens 450 lbs.
Total weight of 1 girder............. 0.00.02 eee 15,837 Ibs.
: 2
Total weight of 2 girders................0.0-08- 31,664 Ibs.

Lateral system = (about same as 50-ft. span, Art. 137).. 2,776 lbs.
Total weight of metal in 1—50-ft. span........... 34,440 Ibs.

Weight of metal in tower 1-2.
(Detailed, Fig. 232.)
One Column.

 

 

 

2— [6 15" «OB & OBIE sim caein nein ie ceannnw enn ans 1,568 lbs.
1—cov. pl. 18” x 8” x 22.95% x 23.1... eee eee 544 Ibs.
Ba [6 15% me DOF Se QOL. ois. k soi wesacein solace dieanwih ned ante bees 2,660 lbs.
1—coy;, pl, 18 x ag” E DOT BE BOO. we ce niwnwis vans 712 Ibs.
Total weight of main section................0005 5,484 Ibs.

Cap.
1—cap plate 204” x 2” x 52.28% xB... eee 162 Ibs.
VB 10 SEBO Fe Oo isso os dastnstcndar tna oainbecada ht Mev anglesatandigiad Ae 69 Ibs.
Q— fills.” ah! ABP x28 oe ca use a seen wen 20 lbs.
Bag 6 eA ae BP OOF x De oa ee wendievegadwereeeuets 68 Ibs.
2— ls 6" eA" xe 8" x 208 XD occa ein ies seems aaees 40 lbs.
*1—gus. pl. 29” x 8” x 86.977 x 24/0. Loc 89 Ibs.
dps, pls BO" 8" eb BOP BO sci ianie din simn niece ny 120 lbs.
2—Ls B4” x BA x BY x BB LM eee ees 24 Ibs.
2—pls. 16” x 3” x 20.4% x 1.75. cee 71 Ibs.
2—L 5 Be 85" x 8" we 104 & LU ccc aresonvsnacensaesas 21 Ibs.
684 lbs.

* Average width and length of plates are given.
316 STRUCTURAL ENGINEERING

Intermediate details.

 

3—tie pls. 18” x 3” x 22.95% x 1.5’..........4. sated hearers 103 Ibs.
2—lat. pls, 14” = §? e 17.85" x 2"... ue evectewesneasees 107 Ibs.
4—sp. pls. 12” x 87 x 15.3% x 1.85’... . cc cece 113 Ibs.
2—ap. ple: 18? xe” x 28908 = 20 scare gernnvscnamnn 92 Ibs.
2—lats pla: 10” #4" x 204 BO cvcewsceedsarencsemess 122 Ibs.
2—lat. pls. 20% x 8” x 25.57 2 BA .cevesavessnesnsmewnss 173 Ibs.
128 ft. of 24” x 2” latt. bars @ 3.19% per ft........... 408 lbs.
1,118 lbs

Base.
1—base plate 30” x #/” x 89.25% x 2.57.0... cece eee ees 223 Ibs.
2—18 6% x 6" x 8" x QADR ELD ccc eee eee eee 53 Ibs.
2—Ls 6% x44 xB" x EXTOL cece cece eee 64 lbs.
2—Ls 6” x 6" x BY x 24.2% KOA. eee eee ee 19 Ibs.
2—Ls 6” x 6" x 8" x 24.2% 0.8’... eee eee eee 39 Ibs.
dg A op BE AP ce 10AE x 1D swe any caveeanes aceon 42 Ibs,
2—s 5” x BE" x 10.4% & 2.0 eesarrenesoureeereneeris 42 lbs.
IL 6” x 6" x 8" x 24.27 wd cece teens 58 Ibs.
2—fills. 84” x 8’ x VASP RLS eee eee 22 Ibs.
Be Oy Be Bo cco ceveinbemcemaesan 17 Ibs.
Dom 9 BR ox BA” eB ce: BDF DD ccccmntirwans mower emus 26 Ibs.
1—gus. pl. 30” x 8” x 38.25% x 3.3’... eee es 126 Ibs.
1—gus. pl. 287 x #7 x SB.7% x 1.8... c cess ec ee sce eeee 64 Ibs.
2—gus. pls. 30” x 8” x 88.254 x 1.75’... eee eee 134 Ibs.
929 lbs

Summary of weight of details.

 

Cap sececsiule cs mares eM ee RRR nE aaa ewe nae 684 Ibs.
Int; detail sinc saaviuisa cs dieciine ce eiwestwsc oi eae waecaeare 1,118 Ibs
BB aSOs ec v. iene! ace: sheets iaviastniactaana seats ad on ca ueene aie usb tnen deere ces 929 lbs
RIVES iasecogsccueiaceiaes SravaneaaSeeatdi ic quansemarecauateserden ake eammiotetiansiend 200 Ibs.

2,931 lbs.

2,931
5,484

 

Percentage of details = = 53% of main section.

Summary of weight of one column (bents 1 and 2).

Details: cnc ayeeee ts 2,931 Ibs.
Main section......... 5,484 lbs.

u

8,415 lbs. = total weight of one column.
4

 

33,660 Ibs.

total weight of four columns.
DESIGN OF SIMPLE RAILROAD BRIDGES 317

Longitudinal bracing.
Longitudinal strut.

 

 

 

 

 

e810" DOF eS AT Gea nerguoae Suapg nee eekweseaews 1,136 lbs.
4—tie pls. 12” x 3” x 15.8% x 1.25’... cece v7 Ibs.
164 ft. latt. bars of 24” x 2” @ 2.87# per ft........... 470 lbs.
TV EtS sia seus us ee ORs ee Ma ea Ae RO Cre 53 Ibs.
(Details = 447%.) 1,736 Ibs.
6
Weight of metal in six struts................ 10,416 Ibs.
Diagonal.
R16 6? x A ae 8? we 128 £ Ob cee es ae easecausesexes 873 Ibs.
$—tie pla, 12% x 2" = 15.37 1.26" pv iwaisiocaveesawewes 38 Ibs.
48 ft. latt. bars 24” x 2” @ 2.87% per ft.............. 138 Ibs.
FIVElS. Wave ewe vie darrian Sole wineh Aaa ee Nays als 20 lbs.
(Details = 22%.) 1,069 lbs
4
Weight of metal in four diagonals............ 4,276 Ibs.
Diagonal (spliced).
4—Ls 6% x4" x BY x 12BF LTS ccc ccc eee 861 Ibs.
4—tie pls. 12” x 8” x 15.3% x 1.25’.......... Ciiaratecwsana tease 77 Ibs.
2—sp. pls. 13” x 8” x 16.57F x 3.5’... ee eee eee 116 Ibs.
44 ft, latt. bars 24” x 8” @ 2.87% per ft............... 126 Ibs.
PIVEUS: 646 shin an Ret wen Re eee wad jRRieeemaste eG 24 lbs.
(Details = 40%.) 1,204 Ibs
4
Weight of metal in four diagonals............ 4,816 Ibs.
Total weight of longitudinal bracing......... 19,508 Ibs.
Transverse bracing (bents 1 and 2).
Strut (beginning at top of bent).
2—Ls 5” x 84" x 8" x 10.4% x 4.0’... ee, 83 Ibs.
7 ft. 24” x 2” lat. bars @ 2.87%............. 20 Ibs.
TiVElS sc aa awe a a a MWS 2 lbs. 105 Ibs.
(Details, 27%.)
Diagonal.
2—Ls 34” x 34” x 9” x 8.5% x 12.75'........... 217 Ibs.
2—tie pls. 9” x 8” x 11.48% x1.0’...........0.. 23 Ibs.
16 ft. lat. bars 24” x 8” @ 2.874%............ 46 lbs.
FivetS ...cceecereees divi Gandia, ewisig ave tenn aa eueae 11 lbs. 297 Ibs.

(Details, 36%.)
318 STRUCTURAL ENGINEERING

Diagonal.

Q—Ls 34” x BA" x 2% x 8.54 2 B.S.
2—Ls 34” x 34” x io OF @ BAY ee seb wey
4—tie pls. 9” x 8’ x 11.48# x LO’. ..... eee,
2—sp. pls. 12” x §" x 10.8" = 8.0" as oe ten em
14 ft. latt. bars 24” x 2” @ 2. td sein Qtadeus raves
PIV ELS. B30 420 5 ate eke, es tere aoe ter arerneereta hare enero es

2—[s 8” x 13. OPTERON ae aaa en ona at os
4—tie pls. 12” x 3” x 15.8% x 0.8’... 2.2... cee
20 ft. lat. bare 24” x2” @ EDU Picctierar ecu ninin a 9
PIV OES yy bse se eek os da a terete SH HR hw eae Oe

Diagonal.
2—Ls 34” x 84” x BY x B.5F x 17.2’..000.. pigkake
2—tie pls. 9” x 8” x 11.48% x 1.0’...........0..
34 ft. lat. bars 247 x 8” @ 2.878% ............
TIVOUS £5 cs ai OMe cain ees Was eaue sis oes

Diagonal. *
2—Ls 84” x BY” x BY x 8.5% x 6.6"..........0..
2—Ls 384” x 384” x 8’ x 8.5% x 10.0’............
4—tie pls. 9” x 8” x 1148# x 1.0’..............
30 ft. lat. bars 24” x 2” @ 2.87%............
2—sp. pls. 12” x a” x 15.38% 3.0 2s fetuses
TIVEUS ch genre sada te eka eae dee eee ka

2 [9:8 x ISSF £18 Saas eeras casey aes
4—ie pls. 12” x 8” x 15.8% x 0.8’....... .....
40 ft. Jat. bars 247 x 8” @ 2.877 ...... 02...
FIVEUS sas acne s Kesiewaed Salaried anti agen oe

Diagonal.
2—Ls 34” x 84 x BY’ x B.BF x 27.2’..0.........
2—tie pls. 9” x 8” x 11.48% x 1.0’.............
48 ft. lat. bars 24” x 8” @ 2.877% ............

EV EUS? soscetitcer eres olidreaie angel anced ad wh deo ede Gey

Diagonal.
2—Ls 34” x 34” x 2" x 8.5% x 10.6’............
2—Ls 34” x 3Y’ x 8’ x 8.5% x 16.0’............
4—tie pls. 9” x 2” x 11.484 x 1.0’.............
44 ft. lat. bats 247 x 8” @ Q8TF. Le.
2—sp. pls. 12” x 8” x 15.38% x 3.0’.............
TIVEUS: Ware s SENG wee AS oie Saar eaes

16 Ibs. 401 Ibe:
(Details, 93%.)

240 lbs.
49 lbs.
57 Ibs.
12 lbs. 358 lbs.

(Details, 49%.)

292 lbs.
23 Ibs.
97 Ibs.
16 lbs. 428 lbs.

(Details, 46%.)

112 lbs.
170 Ibs.
46 lbs.
86 lbs.
92 lbs.
20 Ibs. 526 lbs.

(Details, 87%.)

366 lbs.
49 lbs.
115 Ibs.
40 lbs. 570 Ibs.

(Details, 56%.)

462 lbs.
23 lbs.
137 lbs.
26 lbs. 648 Ibs.

(Details, 40%.)

180 Ibs.

272 Ibs.

46 lbs.
126 lbs.

92 lbs.

30 Ibs. 746 Ibs.

(Details, 65%.)
DESIGN OF SIMPLE RAILROAD BRIDGES 319

Strut. *
2—[s 10” x 20F x QOD ccc cee cee ees 820 Ibs.
4—tie pls. 12” x 8” x 15.3% x 0.8’..........000, 49 lbs.
72 ft. lat. bars 24” x 2” @ 2.87#............ 206 lbs.
TIVEUS 4. serena Seaweed aes awe ee eee 50 Ibs. 1,125 Ibs.

(Details, 37%.)

Weight of transverse bracing in one bent = 5,204 lbs.
Weight of transverse bracing in two bents or tower = 10,408 lbs.

Summary of weight of metal in tower 1-2.

A COlNMNS sé sie ck cea eee RCNA a Ae Re aCh A Ea ees 33,660 Ibs.
Teoria: Bra CAM pias vce 8s lolaud tac dydenco atte as ba ase desu beatae lore mine 19,508 lbs.
“Drang; Bracing cso bine g ail ches flee Dae ey Osea eS 10,408 Ibs.

otal! es mise eleh acd ees Bae BONES Ha EN eS 63,676 lbs.

Proceeding in the same manner the following weights are found:

Weight of metal in each of the towers 3-4 and 5-6.

4 columns (386% details). ... 0.0... ccc cece eee 52,200 lbs.
Long: DRAG ge aos cns asin oe ewes dea it eaw aca 21,820 Ibs.
Evang. DISC 2eccsyopaeeeiaes cece eign weesev ase alge Ube,
Motal ie ctawiyedk nine Sede gens Ga Seta eew aS 88,460 lbs.
Weight of metal in tower 7-8...... 06.06 cece eee eens 64,900 lbs.
Weight of metal in tower 9-10............ 0.0. eee 42,380 Ibs.
Weight of metal in bent 11............ 0. cece eee 8,420 Ibs.
‘ Summary of weight of metal in structure.
5—60-ft. spans @ 45,990 Ibs............. 229,950 int
5—30-ft. spans @ 16,805 lbs............. 84,025 Ibs. ¢ 382,855 Ibs.
2—50-ft. spans @ 84,440 lbs............. 68,880 lbs. J
I—tOWeF (1-2 sido teas ara cs awleeeee BAe 63,676 lbs.
2—towers (8-4 and 5-6) @ 88,460 lbs..... 176,920 Ibs.
l—tower (7-8) oscess eee viiceeee yeetens 64,900 Ibs. ( 356,296 lbs.
I—tower (9-10). ccs iaise case tees enes 42,380 lbs.
Te pene Chesed toes tiered ess 8,420 Ibs. J
Total weight of metal............ 739,151 Ibs.
Cost of structure (superstructure).
Girders, 382,855 Ibs. @ 3¢.........-. cece ence eee eee $11,486
Towers, 356,296 Ibs. @©@ 3h¢....... 0. cece eee eee 12,470
Total cost of steel work (except anchor bolts)........ $23,956

This price is only a fair average pound price.

165. Double-Track Viaducts.—The ordinary double-track viaduct
has a double line of spans, that is, four lines of track girders, which, as
a rule, are supported upon cross-girders as shown in Fig. 233 instead’ of
320 STRUCTURAL ENGINEERING

resting directly upon the top of columns. The towers, as for general
design, are practically the same as for single-track viaducts. The concen-
trations at the top of the towers due to dead and live load and impact are
just twice as much as for single-track viaducts and consequently the
stresses in the columns due to the same will be twice as much, provided
of course, the columns in the two cases have the same batter. We assume
that two trains move abreast over the double-track structure. The trac-
tion is just twice as much for a double-track as it is for a single-track
viaduct, while the wind pressure is practically the same for the two struc-
tures. (See specifications for wind load.)

The method of procedure in designing double-track viaducts is the
same as for single-track viaducts except the transverse bracing is sub-

 

 

 

 

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u
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y S 8 %
oe 2 C 2 \g
| S S 9 S
: ae oll —
l.
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Fig. 233

jected to live-load stress and impact when only one track is loaded. This
is due to the unequal thrusts exerted by the two columns upon the top
strut, or cross-girder. As an illustration, suppose the right-hand track of
a double-track viaduct (see Fig. 233) to be loaded with the maximum live
load and no live load on the left-hand track. The right-hand column will
receive most of the load, as is evident. Let V = the concentration on the
right-hand column and V’ the concentration on the left-hand column and
let @ represent the slope angle of the columns. The horizontal thrust on
the cross-girder exerted by the right-hand column is equal to Vtand, and
that exerted by the left-hand column is equal to V’tand. Now if these
two thrusts were equal they would just balance each other and there
would be no live-load stress in the transverse bracing. But as they are
unequal the transverse bracing must transmit the difference (which pro-
duces a horizontal shear) down to the masonry. This force (= Vtand —
V’tand) can be assumed as applied at the bottom flange (considered as a
strut) of the cross-girder (just exactly as a wind load) and the stresses
in’ the transverse bracing due to it graphically determined.
DESIGN OF SIMPLE RAILROAD BRIDGES 321

The maximum stress in the columns occurs when the two tracks are
fully loaded, at which time no live-load stress occurs in the transverse
bracing.

166. Analytical Method of Determining Stresses in Tower
Bracing.—Let Fig. 234 represent a transverse view of a bent of an
ordinary viaduct, acted upon by forces as indi-
cated. Let us first suppose that the horizontal
wind forces F, F1, F2, and F3 alone are acting
and that the stress in the diagonals AD, CN, and
in the strut CD, due to these forces, is to be deter-
mined,

To determine the stress in the diagonal AD,
first assume the part of the bent (forces and all)
below the section mm removed. Then the part
above this section mm is held in equilibrium by
the forces F, F1, F2, S, 83, and S4, where S, S3,
and $4 represent the stress in members AD, AC,
and BD, respectively. The stress S is what we
desire. By prolonging the lines of action of
the forces (stresses) S3 and $4 and taking mo-
ments about their point of intersection, O, we
eliminate these forces from the equation of moments Fig. 234
and we have

 

 

 

 

cF +hF1+kF2+b6S8=0,
from which we obtain (in known quantities)
ck +hF1+kF2 |
b
Similarly, to determine the stress S2 in diagonal CN, assume the

part of the bent below the section nn removed and taking moments about
O we have

S=

cP +hF1+khF24+7F3 + d82=0,
from which we obtain the required stress
ch + hF1+kF2 + rF3

Bo eect teeter etre teens (18).

To determine the stress S1 in the strut CD assume the part of the
bent below the section 00 removed and then taking moments about O we

have
cf +hF14+kF2+rF3+7S81=0,

from which we obtain the required stress
CF +hF1 + kF2 + rF3

r

S1

Instead of wind loads, as just considered, suppose there be an eccen-
trically applied load W acting upon the bent. Then taking moments
about O, as before, we have

eW+bS=0,
    

             

   

 

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DESIGN OF SIMPLE RAILROAD BRIDGES 33s

from which we obtain

and similarly we obtain
We
a.

Eccentrically applied loads, as just considered, occur mostly on
viaducts built on curve and on double-track viaducts when only one track
is loaded with live load.

If desired, the stresses in the columns can be determined by taking
m=:ments about panel points. For example, the stress S4 in BD can be

obtained by passing the section mm and taking
moments abeut Ad. By passing section oo and
taking moments about D the stress $3 in AC can
Web be determined, and, by taking moments at the same
eS time about C the stress S6 in DN can be determined.
Similarly, by passing the section nn and taking
moments about N the stress S5 in CM can be
Fig. 236 determined.

The only objection to the above method is that
the length of some of the lever arms is troublesome to determine unless
it be determined by scale, which, as a rule, is not a very satisfactory
manner unless special care be taken.

167. General Remarks. —Viaduct columns are sometimes built
without cover plates as shown in Fig. 235. There is no question but that
a column with a cover plate is better than one without. Whenever a
greater area is required than can be obtained by using two channels and a
cover plate a section like the one shown in Fig. 236 should be used.
However, in case there be just a few column sections in a structure
requiring more area than is contained in two
channels and a cover plate, the area may be 6 Anchor bolts
increased so that that type of section can be
used by riveting a plate to the back ‘of each
channel.

The batter of the columns in single-
track viaducts varies from 1} in 12 to 3 in 12,
for very low bents. Two in 12 is a very
common batter. The batter of the columns in
double-track viaducts, as a rule, is less than
in the case of single-track structures. The
batter should be sufficient to limit the negative
reactions at the bases of the columns to a
reasonable intensity, at least. In fact it Fig. 287
would be better to have no negative reaction
at all, but as a rule it is not practical to have that condition in the case of
single-track viaducts.

There should be sufficient material in each pedestal (which is usually
made of concrete) to properly distribute the positive reaction of the
column supported and at the same time have sufficient weight to resist

si-7* and §2=

  

 

 

 

 

 

 
324 STRUCTURAL ENGINEERING

the negative reaction of the column. The anchor bolts should extend well
down into the pedestals as indicated in Fig. 237.

_ In the case of viaduct towers having no horizontal struts, as shown
in Fig. 221, the two systems of bracing are usually considered to act
simultaneously, which requires that each member of the bracing be
designed for compression as well as for tension, but at the same time,
however, the stress in each is only half as much as obtains in the type
of bracing shown in Fig. 220, assuming only one system to act at a time.

DRAWING ROOM EXERCISE NO. 6

Design and make stress sheet of a single-track viaduct for the
following crossing:
Elevation of

Station Elevation of Ground Base of Rail
400 640 648
400+81 588 648
401 579 648
401+ 70 568 648
402413 564 648
402 +20 554 648
402 +40 554 } Creek 648
402 +49 566 648
403 567 x 648
403 +44 564 648
404+ 5 608 648
404441 613 648
404+ 80 626 648
405+ 2 646 648
Data:

Live load, Cooper’s E50 loading.
Specifications, A. R. E. Ass’n.
Dead load, to be determined by student.

TRUSS BRIDGES

168. Preliminary—Truss bridges are used for spans 100 ft. in
length and over. With regard to the location of the track there are two
types of truss bridges, the through and the deck bridge, the same as in

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Pratt Truss curved Chord Prottirvss.
NI c= -
Pettit Truss (a)
(Q)
S :
Pelt Trust) (6)

Fig. 238 Fig. 239
DESIGN OF SIMPLE RAILROAD BRIDGES 325

the case of plate girders. The tnrough type is used wherever the under
clearance will not permit of the use of the deck type.

The four trusses shown in Fig. 238 are the most common types used
for through bridges. Pratt trusses are used for spans up to 175 ft. in
length and as a rule are riveted trusses. Curved chord Pratt trusses are
used for spans from 200 ft. to 325 ft. long and as a rule are pin-connected

 

Fig. 240

trusses. Pettit trusses, both (a) and (b) types, are used for long spans,
350 ft. in length and over.

There are other types of bridge trusses which are used to some
extent. These will be considered later.

In the case of deck bridges, the Pratt truss is practically always
326 STRUCTURAL ENGINEERING

used. The end panels, however, are modified cither as shown at (a) or
(b), Fig. 239.

The diagram, Fig. 240, gives the names of the different parts of an
ordinary Pratt truss bridge, but the same names hold for bridge trusses
in general.

Complete Design of a 150-Ft. Single-Track Through Riveted
Pratt Truss Span
169. Data.—

Length=6 panels @ 25’-0” =150’-0” c.c. end pins.
Width = 16’-0” c.c. of trusses.
Height = 30’-0” c.c. chords.

Stringers spaced, 6’—6” c.c.

Live Load, Cooper’s £50 loading.

Specifications, A. R. E. Ass’n.

170. Design of 25-Ft. Stringers.—(For detail of stringers, see
Fig. 290.) For dead load on stringers we have, from (1) Art. 124,
w=12x25+100+400=800lbs. per ft. of span or 400 Ibs. per ft. of
stringer. Then, using this load, we have

M’=4x 400 x25" x 18=375,000 inch Ibs.

for the maximum bending moment due to dead load.

It is seen from Table A that the maximum live-load moment will
likely occur when wheels 2 to 5 are on the stringer, and, according to Art.
88, they will be in the position shown in Fig. 241.

Taking moments about A (Fig. 241), to
find the reaction R (using Table A), we find
first the moment of wheels 8, 4 and 5 about 2

| 8 by passing down the vertical line through

2.5’ ‘Yt 42.5' * wheel 2 to the zig-zag line, then to the right

22 to the vertical line through wheel 5 and the

al figure 600, just to the right of this line, is the
moment of wheels 3, 4 and 5 about 2, in thousands of foot pounds.

Then multiplying the total weight, in thousands of pounds, of the
wheels, 2 to 5, by 8.75 (see Art. 46) and adding this product to the 600,
we have the moment of the wheels, 2 to 5, about 4. So we can write
the equation of moment about 4 as

600 + (80 x 3.75) — (R x 25) =0,
from which we obtain
R= 600+ (80x 3.75)
25

28!
3.15 AGS tg 6.25"
@ 5 @dsl@ *@

IR!

 

>

 

 

 

 

 

x 1,000 = 36,000 Ibs.,

for the reaction at B.

Then taking moments about 4, as the maximum moment occurs under
that wheel (see Art. 88), we have the moment of R about wheel 4 minus
the moment of wheel 5 about wheel +: for the maximum bending moment;
that is, we have

M” = 36,000 x 11.25 — 5 x 20,000 = 305,000 ft. lbs.,

for the maximum live-load bending moment due to Cooper’s F40 loading,
DESIGN OF SIMPLE RAILROAD BRIDGES 327

and multiplying this by 50/40 and by 12 we have 4,575,000’# for the
desired maximum live-load moment in inch pounds, provided the correct
group of wheels were selected, which can be ascertained by trial, in case
of doubt. .

Then for the impact moment we have

300 :
T=4,575,000 x 305 = 4,223,000 inch lbs.
Now adding the above moments together we have
M =3875,000 + 4,575,000 + 4,223,000 = 9,173,000 inch lbs.

for the total maximum bending moment on the stringer.
Next let us assume the web as 3” thick. Then substituting in (5)
of Art. 113 (stringers, as a rule, never have cover plates), we have

9,173,000 4. ».
r= 1.02/27 N00 =47.7 ins.

«for the economic depth. So we will use a web 48”
@Qi@Qs@@ 2’ @|, deer.
Atr 25! For the dead-load end shear on the stringer,

Fig. 242 we have

400 x 12.5 =5,000 lbs.,

and placing the wheels as shown in Fig. 242 and taking moments about

the support B (using Table A) we have

_ 1,820+ (93x 1)
25

for the maximum end shear due to Cooper’s £40 and

x x 56,500 = 70,650 Ibs., say 71,000 Ibs.,

due to Cooper’s £50 loading, and for impact we have
300

71,000 x == = 66,000 Ibs. (about)

R x 1,000 = 56,520 Ibs.

and adding we have
5,000 + 71,000 + 66,000 = 142,000 lbs.

for the total maximum end shear on the stringer.
Now for the unit shear on the assumed web we have

142,000
3x48

and substituting this in (1) of Art. 118, we have

= 1,888 lbs.,

a= 3 (12,000 — 7,888) = 39 ins. (about)
for the maximum distance allowed between the stiffeners near the ends
of the stringer, and, as this is more than half the depth of the girder
and as the minimum thickness of web allowed by the specifications is

8 in., we will use a 48” x 3” web.
328 STRUCTURAL ENGINEERING

To determine the area of flanges the first thing to do is to approxi-
mate the effective depth of the stringer. Assuming that 6” x6” angles
will be required we find from Table 6, or from some handbook, that the
average distance from the back to the ‘center of gravity of these angles
is about 1.75 ins., and taking the depth of the girder as 3” more than
the web, we have

48.25 -1.75 x 2 = 44.75, say 45 ins.,

for the approximate effective depth. Dividing this into the maximum
total bending moment, we have
9,173,000
45

for the stress in each flange (see Art. 112), and dividing this by 16,000
we have

= 204,000 Ibs. (about),

204,000
16,000
for the net area of each flange. Let us try

2-156" x 6" x 4 = (5.75 x 2) —1= 10.500” net
4 area of web = 2.2507 net

12.750” net

= 12.75 sq. ins.,

This gives exactly the area indicated. Now checking back, we have
48.25 — 1.68 x 2= 44.89 ins.

for the actual effective depth, which differs so little from the assumed
that the area of the flanges would be changed but very little, and hence
recalculation of flanges is unnecessary.

The stiffeners on stringers are usually made of 34x34” angles.
There is no rational way of determining them. Owing to the planing off
of the ends of the stringers to obtain exact length, the end stiffeners are,
as a rule, made 7,” to }” thicker than the intermediate stiffeners, which
are of minimum (3”) thickness.

The laterals—bracing for stringers—are designed the same as for
any deck plate girder. (See Art. 135.)

171. Design of Intermediate Floor Beams.—(For details of
floor beams, see Fig. 291.) The dead load on an intermediate floor beam
consists of the dead-load concentrations from the stringers and the weight
of the floor beam itself.

Each concentration from the stringers is equal to twice the end shear
on one stringer and such floor beams weigh, depending on details, from
3,000# to 3,800# each. So let us, in this case, assume 3,600# as the
weight of one intermediate floor beam complete, or 225# per ft. of floor
beam. Then, for dead load we have the case shown in Fig. 243, and
for the maximum bending moment due to dead load, we have

10,000 x 4.75 x 12 =570,000 in. Ibs.
$x 225 x 162x12 = 86,000 in. lbs.

656,000 in. Ibs.
DESIGN OF SIMPLE RAILROAD BRIDGES 399

and for the maximum end shear, due to the same load, we have

— + 10,000 = 11,800 lbs.

To determine the maximum live-load bending moment and end shear
on the floor beam we must first satisfy the criterion of Art. 148 for
maximum live-load concentration on the floor beam. That is, the loads in
the two adjacent panels must be as nearly equal as possible and at the
same time the heaviest loads must be near the floor beam, with one load
exactly at it.

Ky 9
q

  
 
 

ss
B [ ® & ME@SOsS@ 2' @5@6O
A 1¢

5 sorBeam __,
25

   

Ch ccc CCE

R g-9"| 66"
25’

Fig. 243 Fig. 244

      
 

 

=I

By trial (using Table A) it will be found that the maximum con-
centration on the floor beams will occur when the wheels are in the
position shown in Fig. 244, as the criterion is most nearly satisfied with
them in that position, that is, the heaviest wheels are near the floor beam,
wheel 13 at it, and the load in panel BC, consisting of wheels 14, 15, and
16 (wheel 17 is exactly over the floor beam at C and hence is not in the
panel BC), weighs 46,000 lbs., while the load in panel AB, consisting of
wheels 10, 11, and 12, weighs 50,000 lbs. This is as close as the crite-
rion can be satisfied, as will be found by further investigation. So,
taking moments (using Table A) about 4 to find the concentration at B
due to wheels 10, 11, and 12, we have

1,000
25
and taking moments about C to find the concentration at B due to wheels

14, 15, and 16, we have

~ 25
Now adding these concentrations and the weight of wheel 13 together,
we have

R=r+r’ +20,000 = 30,800 + 24,850 + 20,000 = 75,650 Ibs.,

for the maximum live-load concentration due to Cooper’s E40 loading,
and multiplying this by 50/40 we have 94,562#, say 94,500#, for the
maximum live-load concentration due.to Cooper’s E50 loading.

Then for the live load on each intermediate floor beam we have the
case shown in Fig. 245, and for the maximum live-load moment, taking
moment about either C or D, we have

4.15 x 94,500 x 12 = 5,386,000 inch Ibs.,
and for the impact we have
300

5,386,000 x 3507 4,616,800, say, 4,620,000 inch lbs.

ry

r= (420+50x7) = 30,800 lbs.,

x 1,000 = 24,850 Ibs.
330 STRUCTURAL ENGINEERING

(The load extends over two panel lengths, or 50 ft.)
Now adding the above dead, live, and impact moments, we have

M = 656,000 + 5,386,000 + 4,620,000 = 10,662,000 inch Ibs.,

for the total maximum bending moment on the floor beam, and, as is
seen from Fig. 245, the maximum live-load shear is 94,500# and the
impact = 94,500 x 300/350=81,000#. So by adding these to the above
dead-load shear we have 11,8004 94,500 + 81,000=187,300# for the total
maximum end shear on the floor beam.

The depth of the floor beam is governed practically by the depth
of the stringers. The bottom laterals and the bottom flanges of the floor
beams should be in the same plane, as the bottom flanges of the beams
act as struts in the bottom lateral system, and there should be sufficient
distance from the bottom of the stringers down to the bottom of the floor
beam to permit the laterals to pass beneath the stringers, which usually
requires from 4” to 6”, depending on the size of the laterals, and the dis-
tance from the top of the floor beams down to the top of the stringers
is usually about 3’. In this case we will assume 3” as the distance
from the top of the floor beams down to the top of the stringers and
5” from the bottom of the stringers down to the bottom of the floor
beams, which gives 564” for the depth of the floor beam as shown in
Fig. 246. It will be found upon investigation that floor beams, as a rule,
are deeper than the economic depth.

 

 

 

 

 

 

 

 

 

 

Floor Beam.
* % Lo a
8 Q <tr a
° ¥ ‘Ww z
$ 9% 8 of Mi
eT]
9p — te ct? s
Sf 4-9" | ote” 49" KR WS %
< 1620" 8 s r
S x %
Fig. 245 Fig. 246

After having decided upon the depth of the floor beam, the next
thing is to approximate the effective depth. Assuming that each flange
of the floor beam will be composed of 2—6”x 6” angles, we can take
about the average distance from the back of the angles to their centers
of gravity, which is given in Table 6 as, say, 1.75’. Then, for the
approximate effective depth, we have

56.25 — (1.75 x 2) = 52.75, say 53 ins.

Dividing this into the above maximum bending moment, we have
10,662,000 + 53 =201,000 Ibs. (about)
for the flange stress, and dividing this by 16,000 we have
201,000 + 16,000 = 12.56 sq. in.
for the required net area of each flange.

Assuming a 56” x 75” web which has a section of 24.54” (=56x 47%),
let us try

2—Ls 6” x 6” x $”=(5.75 x2) -1=10.500” net
4 area of cross-section of web = 3.060” net

13.560” net
DESIGN OF SIMPLE RAILROAD BRIDGES 331

This gives 15” more than is required. 2—Ls 6” x6” x 7%” have a
section more nearly equal to the required area, but the specifications
require the angles to be 3” thick if the outstanding leg is 6’, so, to com-
ply with the specifications, the above section will be used.

It will be seen that the true effective depth here is slightly less than
the assumed, as the distance from the back of the 6” x 6”x 4” angles is
1.68”. So, for the actual effective depth, we have

56.25 — (1.68 x 2) =52.89 ins,

This, however, is so near the assumed effective depth that recalculations
are unnecessary.
For the maximum unit shear on the floor beam we have

187,300 + 24.5 = 7,644 lbs. per sq. in.
Now substituting this value for s in (1), Art. 118, we have

7
d= a (12,000 a 7,644) =47.6 ins.,

for the allowable distance between stiffeners. The shear is practically
zero between the stringers, so that no stiffeners are required there, and
the clear distance from stringer to truss is 4’-9” minus about 7” for
half width of truss and 6” (at least) for connections, leaving 3’— 8”, or
44”, for the actual value of d, while 47.6” is allowed, so the assumed
web is quite thick enough. In fact it could be thinner, but the rivets,
spaced along the flanges between the stringers and trusses, would be
closer than desired, if a thinner web be used, and hence the assumed
56” x 7%’ web will be used, even though it is a little thicker than
required.

The end stiffeners on the floor beams are usually 6” x 6” x 4” angles,
placed upon fillers as shown in Fig. 291.

172. Design of End Floor Beams.—The dead load on an end
floor beam consists of the two dead-load concentrations from the stringers,

 

 

 

 

 

 

 

 

 

 

¥
« % %

y 8 § 8

8 & 8 y
91 sch por fF ty ss %
A 7 ? Cen B ¥y Po s Zz ig
4-9" | ete” | ato” S479" | 646” 4to"\8
4620" x 60" x

Fig. 247 Fig. 248

each of which is equal to the dead-load end shear on one stringer, and the
weight of the floor beam itself. The dead-load end shear on a stringer, as
given in Art. 170, is 5,000%, and such end floor beams weigh about 2,400#,
or, say, 150# per foot. Then the dead weight on each end floor beam is
as shown in Fig. 247.

For the maximum bending moment, due to this dead load, we have

5,000 x 4.75 x12 =285,000 inch lbs.

4x 150x 16 x12 = 57,600 inch Ibs.
342,600, say 343,000 inch lbs.,
332 STRUCTURAL ENGINEERING

and for the maximum end shear, due to the same dead load, we have
5,000 + 150 x 8 =6,200 lbs.

The maximum live-load concentrations from the stringers on an end
floor beam are each equal to the maximum live-load end shear on one
stringer, which is given in Art. 170 as 71,000*. So the live load on an
end floor beam is as shown in Fig. 248.

For maximum bending moment, due to this live load, we have

71,000 x 4.75 x 12=4,047,000 inch lbs.,

and for the impact we have

4,047,000 x ee = 3,735,000 inch Ibs.,

and for maximum end shear, due to live load, we have 71,000, and for
end shear, due to impact, we have

300
825
Now adding the above moments, we have

343,000 + 4,047,000 + 3,735,000 = 8,125,000 inch lbs.,

for the total maximum bending moment, and adding together the above
end shears we have ;

6,200 + 71,000 + 66,000 = 143,200 lbs.,

for the total maximum end shear on an end floor beam.
Now assuming 53” for the effective depth, the same as for the inter-
mediate floor beam, and dividing it into the above moment, we have

8,125,000

71,000 x 2— = 66,000 Ibs. (about).

53 = 153,000 lbs.
for the flange stress, and dividing this by 16,0060, we have
153,000 _ :
16,000 = 9.56 sq. in.,

for the required net area of each flange. Now it is readily seen, from
Table 6, that 6’ x 6” angles are too large for these flanges and hence we
will use 6’ x4” angles with the 6” leg along the web to provide for the
flange rivets.
Let us try, assuming a 56” x 3” web (same depth as the intermediate
floor beams),
2—Ls 6” x4" x qe” =4.18 x 2-0.87= 7.490” net
4 area of web =(56x8)+8 = 2.620” net

10.110” net
This flange section is about correct. It is 0.555” greater than called for

above, but the assumed effective depth is greater than the actual, which is
really 56.25 - (21.96) =52.33”.
DESIGN OF SIMPLE RAILROAD BRIDGES 333

Dividing the moment by this, we have
8,125,000 + 52.33 = 155,000,

and dividing this by 16,000#, we have

155,000
16,000

for the actual flange area required, which is 0.430” less than the above
assumed section, but the above section will be used, as 6” x 4” x 3” angles
are too small, giving a net area of 0.590” less than required.

Dividing the end shear by the area of cross-section of the web,
we have

 

= 9.68 sq. ins.

143,200
21

for the actual unit shear on the web. Substituting this value for ¢ in (1),
Art. 118, we have 3

= 6,819 lbs.,

d= & (12,000 - 6,819) =48.5ins.,

for the allowable distance between stiffeners, but as the unsupported dis-
tance between stringers and truss is about 3’-—8”, or 44”, according to
the preceding article, no stiffeners are needed and the assumed 56” x 3”
web will be used. The actual allowable stress on it is determined by sub-
stituting 44” for d in (1), Art. 118, and solving for s. Thus we have

3
44= 5 (12,000-s),

from which we obtain
s = 17,306 lbs.,

for the actual allowable unit stress. Now dividing this into the maximum
end shear, we have

143,200
7,306

for the required area of cross-section of the web. The area of the
56” x 8” web, which we propose to use, is 210”, which, as is seen, is
more than is required, but the area of a 56” x 7;”’ web, which is the next
in size, contains only 17.50”. Therefore, a thinner web could not be
used even if the specifications would permit it. From this it is seen that
the assumed 56” x3” web is as near the correct section as is possible
to obtain and hence will be used.

This completes the design of the floor system, and next we will take
up the design of the trusses.

173. Determination of Dead-Load Stresses in Trusses.—The
dead load of an ordinary bridge is considered as uniformly distributed
along the span but applied to the trusses only at the panel points or
joints, one-third at the top joints and two-thirds at the bottom joints, in
the case of through bridges, and just the reverse in the case of deck spans.

 

= 19.57 sq. ins.,
334 STRUCTURAL ENGINEERING

This load consists of the weight of metal in the span and the weight
of the deck.
From (4), Art. 124, we have

p=7x150 + 660=1,710 Ibs.,

for the approximate weight of the metal per foot of span, and adding
400# for the weight of the deck, we have

1,710 +400 = 2,110 Ibs.,

for the total assumed dead load per foot of span, or 2,110/2 =1,055# per
foot of truss.
Multiplying this by 25, the panel length in feet, we have

W =1,055 x 25 =26,375 lbs.,

for the panel load on each truss. One-third of this is considered as
applied at each top joint and two-thirds at each bottom joint. However,
this assumption affects only the verticals, and, as far as the stresses in
the other members are concerned, the full panel load could be considered
at the bottom joints, or top either, as for that.

After having thus computed the panel load, the next thing to do is
to draw a sketch of the truss, as shown in Fig. 249 (as a guide), and then
compute the values of tan 8 and sec 6.

Tan p= 2B 0.8333 and sec = 1.299, say 1.3.

In determining the dead-load stresses in the web members (diagonals
and verticals), we can begin either at the center of the span or at the
end. The dead-load stresses in these members can really be determined
by mere inspection.

”

 

 

 

 

 

 

 

 

d= 25- : 1, he ay is
ald 200" cla +99000 dhg 42W’ tanee| aWione s\% =
tA A % @ 2 v
67 ere, S| Ne 8 8 8
9 8 S of y 2 2.
d ON ©, May h, ol
xf S On et z \ ty} 3
‘ - 55000* |-55000*\| _ sgo00*\\/awiane “|//2zW 1006 | 23Wrane
2 ot Bl2 Cle Oe Ele Fle G
Ir aw Fl 2w 2w Bw aw 2w Ir
L 6 Panels @25-0"2 150-0” 4
Fig. 249

Beginning at the center of the span and considering the post dD, it
is obvious that this post can do nothing more than carry the load applied
to the top of it, which is one-third of the panel load; so, evidently, the
stress in it is 26,375+3=8,791#, say 9,000#, compression.

The one-third of a panel load transmitted by the post dD down to
joint D combines with the two-thirds at D and undoubtedly one-half of
this combination is transmitted to the two ends of the truss. Therefore,
the shear in panel CD is one-half of a panel load and the stress in the
diagonal cD is

W 26,375

g see d= gn 1.3=17,143, say 17,000 Ibs. tension.
DESIGN OF SIMPLE RAILROAD BRIDGES 335

The stress in the post cC is equal to the half panel load coming from panel
point D, or the vertical pull of diagonal cD, plus the one-third panel
load at c. So, for the stress in post cC, we have

WW o5W 5
ot3=@ 76* 26,375 = 21,980, say 22,000 Ibs.

Likewise, it is evident that the diagonal bC must transmit, so to
speak, the 4 W coming from the central panel point D, the 4 W from ec,
and the $ W from C, making 14 W in all, which is the shear in panel BC
and must, undoubtedly, be equal to the vertical component of the stress in
the diagonal 6C, and hence the stress in the diagonal is equal to 14x
26,375 x sec 0= 51,430#, say 51,000#, tension.

The member 6B, known as a hanger or hip vertical, as is readily
seen, can carry nothing more than the load hung on (so to speak) at the
joint B, and hence the stress in it is 3x 26,375=17,582#, say 18,000#,
tension.

It is evident that the end post bA must transmit 4 W from the cen-
tral panel point D, the 4 W from c, the 3 W from C, the 3 W from B,
and the 4 W from b, making in all 24 W, which is the shear in panel AB
and equal to the vertical component of the stress in this member bA. So
the stress in it is equal to 24 x 26,375 x sec 0= 85,7184, say 85,000#, com-
pression.

Now, if we begin at the end of the span instead of at the center, as
we did above, we have first the reaction R=24 W at A. (This is obtained
from mere observation.) As this reaction acts vertically the end post
bA must resist it, as the end post is the only member at point A having a
vertical component. So we have 24 W sec 6@ for the stress in the end post,
the same as found above for this member. The stress in the hanger bB
is W, as found before. The shear in panel BC is equal to the reaction
R (=24W) minus the 3 W at b and 3W at B, so that we have 24 W-
W=14 W for the shear in panel BC. This, evidently, must be equal to
the vertical component of the stress in diagonal bC, as that member is the
only member in that panel having a vertical component force, and hence
the stress in diagonal bC is equal to 14 W sec 0, the same as found before.
Likewise, the stress in diagonal cD is equal to the shear in panel CD
times sec 6. The shear is equal to R-2 W=4 W, and hence the stress is
equal to 4 W sec 6, the same as found before. The stress in the post eC
is equal to the vertical component of the diagonal cD, which is 4 W (the
shear in panel CD) plus the 4W at c, making ¢/V, the same as found
before, or it is equal to the vertical component of diagonal bC, which is
14 W (the shear in panel BC), minus the 32 W at C.

The expression for the stresses in the web members can be written
on the members, as shown on the right half of the truss (Fig. 249), and
the numerical results of the same may be found very quickly by the use
of the slide rule.

For the chord stresses we can assume the full panel loads as applied
at the bottom panel points, in order to simplify the work.

Let us first write out the formulas for the chord stresses. For the
stress S in bottom chords AB and BC, we have, taking moments about
point b (see Fig. 249),

Rd- Sh=24 Wd-Sh=0.
436 STRUCTURAL ENGINEERING

Transposing, we have
S=24 w8 = (24) W tail coos encibeece aaaeaeae =O:

For the stress S,, in chord bc, we have, taking moments about C,
R (2d) -Wd-S,xh=2x24 Wd-Wd-(S,xh)=0,
from which we obtain

8,=5 WS WE =5 Wtan0-W tan d=4 W tan 6 swhauee aaaiets (2)

for the stress in chord be. We obtain the same thing for bottom chord
CD by taking moments about c.
For the stress S,, in top chord cd, we have, taking moments about D,

Rx3d-(Wx2d) -Wd-(S,xh) =3x24Wd-2Wd—-Wd-(S,xh) =0,
from which we obtain

8,=1474 -29s- w 4 =14W tan 6-30 tan 6=41 W tan 6 (3).
We thus have an expression for the stress in each chord member.
Substituting the numerical values in (1), we have

S =24 x 26,375 x 0.8333 = 54,945, say 55,000 lbs.

tension for the stress in chords 4B and BC, and by substituting, like-
wise, the numerical values in (2) and (3) we obtain 87,672#, say
88,000%, for the stress in the chord members be and CD, and 98,631%,
say 99,000#, for the stress in the chord member cd. This completes the
calculations for the dead-load stresses in the trusses, as the structure is
symmetrical about the center of the span. '

 

 

 

 

 

 

 

 

bt(y ¢ (9) d (10) @ (10) £ $ A. ee
x
Fi 2 2) \u0IN ait 2 +4
a c a Z| w
Re aw iw wo lw w w |r
Fig. 250

It will be seen from Formulas (1), (2), and (3) that Wtan6 is a
constant, and after the coefficients are written out the stresses in the
chords could be quickly determined by the use of the slide rule.

These coefficients can be written down on the chord members by
mere inspection. For example, let Fig. 250 represent a truss of nine
equal panels.

Taking moments about b for the stress in the bottom chords AB and
BC, we have 4Wd/h, so the coefficient for chords AB and BC is 4.

Taking moments about C for stress in top chord bc, we have
(4W)2d/h-—Wd/h, so the coefficient for chord bc is 7. It is the same for
the bottom chord CD, as the center of moments in that case is at c, which
is in the same vertical line as point C. Taking moments about either d
or D, for the stress in chords cd and DE, we have the 4 at A multiplied
by 3d minus the W at B multiplied by 2d and minus the W at C multi-
DESIGN OF SIMPLE RAILROAD BRIDGES : 337

a by 1d, so the coefficients for chords cd and DE are each (4x3) -
2-1=9.

Similarly, taking moments about e or E, for the stress in chords de
and EF, also ef, we have (4x4) -3~2-—1=10 for the coefficient to be
used in determining the stress in these chords.

After a little practice the student can write these coefficients down
without hesitating.

If the slide rule is used, about all that need be written down, out-
side of the stresses (as they are determined), are the numerical values
of the chord coefficients and that of W, tan 6, and sec@, and if found
convenient or necessary, W tan@ and W sec 6.

174,—Determination of Live-Load Stresses in Trusses— First,
draw a sketch of the truss, as shown in Fig. 251, for reference.

Member bA. Suppose we start by determining the maximumi stress
in the end post bA. As is readily seen, the stress in this member will be

 

 

 

    

 

 

 

b +260000¢ +294000d e £
*
oA ey <
of 8 wy xt
wy nN \ 9
* ay %
c »« O & S G
6 Panels @ 25-0"= 15070”

 

 

Fig. 251

a maximum when the shear in the panel AB is a maximum, and this will
occur when the wheels are in the position that most nearly satisfies the
criterion of Art. 90, which is: The load in the panel AB must be as
nearly equal as possible to the total load on the bridge divided by the
number of panels, and, at the same time, a wheel must be at B. Now
placing wheel 4 at B, as shown in Fig. 252, we have (using Table A)
125 — 91 =34 ft. of uniform load on the bridge, making a total of 352,000#
= (2844 34x2) 1,000 per truss, and dividing by the number of panels
we have

352,000

6
The load in the panel including one-half of wheel 4 is 60,000¥. So this

position comes as near to satisfying the criterion for shear as is possible,
as will be found by trying the other positions.

© @@@ f G
a B c D E * oa!

Fig, 252

= 58,666 lbs.

 

Taking moments about G (Fig. 252), we have (using Table A)

_2z
R =( 16,364. 284 3844+2%x 5 ara 181,172 lbs.

for the total reaction at A. Then taking moments about B, we have

r= (35) 1,000 = 19,200 Ibs.
338 STRUCTURAL ENGINEERING

for the floor beam concentration at A. So we then have

R-r=181,172 —19,200=161,972, say 162,000 lbs.
for the maximum shear in the end panel AB, due to Cooper’s E40 loading
and for £50 we have

50
Jo 7 202,800 Ibs.

Then for the maximum live-load stress in the end post bA we have
202,500 x 1.3 = 263,250, say 263,000 lbs. (compression)

Member bB. The maximum live-load stress on the hanger bB, as is
obvious, is equal to the maximum floor beam concentration found in Art.
171 to be 94,5004, and hence we will not recalculate it.

ro @@@
B é D E F ans

Fig. 253

162,000 x

 

Afr

Member bC. The maximum stress in the diagonal bC will occur
when the shear in panel BC is a maximum, as this member carries the
shear in that panel. Placing wheel 3 at panel point C, as shown in Fig.
253, we have 100 ~96=4 ft. of uniform load on the bridge, making in all
292,000# = (284+2x4)1,000 on one truss.

Dividing this by the number of panels, we have 292,000 + 6 =48,666#,
and the load in panel BC is 40,000*, including one-half of the wheel 3,
at C (see Fig. 253). This does not seem to satisfy the criterion very
closely, so we will try wheel 4 at C, in which case we have 9 ft. of uni-
form load on the bridge, making a total load of 302,000# per truss, and
dividing by the number of panels we have 302,000 +6 =50,333#, and the
load in the panel is 60,000. This position does not satisfy the criterion
as closely as the first, and hence the first position, with wheel 3 at C,
will be taken. It is evident that no other positions need be tried, as the
load in the panel in the first case was too light and in the second it was
too heavy.

Then taking moments about G (Fig. 253), we have (using Table A)

 

 

R= (16,3644 2sixd4) a =116,773
for the reaction at A, and taking moments about C we have
1,000 _
r=280 x- 35 = 9,200 lbs.

for the concentration at point B.
Then, for the maximum shear in the panel BC, due to the E40
loading, we have
R-r=116,773 — 9,200 = 107,573,
and for the E50 loading we have 134,466, say 134,500#. Then multi-
plying this by sec 6, we have
134,500 x 1.8 =174,850, say 175,000 lbs.

for the maximum tensile stress in the diagonal bC.
DESIGN OF SIMPLE RAILROAD BRIDGES 339

Member cD. The maximum live-load stress in the diagonal cD will
occur when the live-load shear in the panel CD is a maximum. To
satisfy the criterion for maximum shear in that panel, let us try wheel 3
at point D (see Fig. 254). This position of the wheels brings wheel 15
exactly at G, the right support.

 

/ © @@@®@
* B ¢ ° = F 46
Fig. 254

Then the load on the bridge (not including wheel 15) is 232,000%.
Dividing this by the number of panels, we have 232,000 +6=38, 666%,
and the load in the panel is 40,0004 (including one-half of wheel 3).
This comes as near satisfying the criterion as is possible, as will be seen
by trying other positions.

Now, by taking moments about G (using Table A) we obtain

1, a

R=10,816 x 150

 

= 12,107 Ibs.

for the reaction at A, and taking moments about D we obtain

1,000

=23
r Ox 35

 

= 9,200 Ibs.

for the concentration at C.
Then, for the shear in panel CD, due to the E40 loading, we have

R-r=172,107 — 9,200 = 62,907 lbs.

and for the E50 loading we have 78,633#. Multiplying this by sec,
we have
78,633 x 1.3 = 102,223, say 102,000 Ibs.

for the maximum live-load tensile stress in the diagonal cD.

Member cC. As the live load is applied wholly on the bottom chord
joints, the post cC will have a maximum compression stress when the
diagonal cD has maximum tension, ‘as the post takes only the vertical
component of the stress in that diagonal. The maximum vertical com-
ponent of the stress in the diagonal cD, as seen above, is equal to the
maximum shear in the panel CD and given there as 78,633. So the
maximum live-load compressive stress in the post cC is, say, 78,000#,
using round numbers.

Member dD. There is no live-load stress in the post dD at all, as
there is no member (as a diagonal) connecting to the top of it that will
resist vertical action. For the sake of illustration, let us assume a live-
load stress, S, in this post. Then the components of this stress must be
taken by the top chords, and we would have

S cos 90° =C=0, as cos of 90° =0.

Members eE and eD. If the span is loaded from the right end so
that maximum shear occurs in panel DE, the post e£ will be subjected to
maximum live-load tensile stress and the eae eD will be subjected
to maximum compressive stress.
340 STRUCTURAL ENGINEERING

If wheel 2 be placed at E, wheel 10 will be just 2 ft. from the right
end of the span, as shown in Fig. 255, and the total load on the bridge
will be 152,000%.

Dividing this by the number of panels we have 152,000 + 6 = 25,333#,
and the load in the panel DE is equal to 20,000#, including one-half of
wheel 2. Now, if wheel 3 be placed at E, the total load on the bridge

2” ’

{ @ @@0
a B S D = = G

Fig. 255

 

will be 152,000#. Dividing this by the number of panels, we have
152,000 + 6=25,333#%, and the load in the panel DE is then 40,000%.
So it is seen that the position of the wheels, shown in Fig. 255, comes
nearest to satisfying the criterion for shear in panel DE.

Taking moments about G (Fig. 255), we obtain

R=( 4,632 +152x 2) = 32,906 lbs.

for the reaction at 4, and taking moments about E we obtain

1,000
r=80x “oe 3,200 Ibs.
for the concentration at D.

Then, for the shear in panel DE, due to the E40 loading, we have
R-r= 32,906 — 3,200 =29,706 lbs.,

and multiplying by 50/40 we have 37,100# for the shear due to the E50
loading, which is the maximum live-load tensile stress in post eZ, and
multiplying this by sec 6 we obtain 48,230#, say 48,000#, for the maxi-
mum live-load compression stress in diagonal eD.

5

|" p @@O
Ata B é D E F A,
Fig. 256

 

Member fE. Now placing wheel 2 at F, wheel 6 will be just 1 ft.
to the left of the right end of the span as shown in Fig. 256. Then the
load on the span is 103,000#. Dividing this by 6 we have 17,166#, and
the load in panel EF is 20,000#. This position comes the nearest to
satisfying the criterion, as will be found by trial.

Taking moments about G (Fig. 256), we obtain

 

1,000 _
R= (1,010 + 103) aoe = 11,620 Ibs.
for the reaction at A, and taking moments about F we obtain
1,000 _
r=80x oa 3,200 Ibs,

for the concentration at E.
DESIGN OF SIMPLE RAILROAD BRIDGES 341

Then, for the shear in panel EF, due to the E50 loading, we have

50
(x- rage (11, 620-3 200) 40 = 10,525 Ibs.

Multiplying this by sec@ we have 13,682#, say 14,000#, for the
maximum live-load compressive stress in diagonal fE.

It is seen, from Fig. 251, that diagonal bC has 175,000# tension in
it, while the corresponding diagonal fE, on the right half of the span,
has 14,000# compression, that the post cC has 78,000# compression,
while the corresponding post e£ has 37,000# tension and diagonal cD has
102,000# tension while the corresponding diagonal on the right half of
the span has 48,000# compression. Now it is evident that, if the span
is loaded from the left end instead of the right end, as it was above, and
the maximum stresses in the web members computed, just the opposite
results will be obtained, that is, for example, diagonal bC would be sub-
jected to 14,000# compression, while diagonal fE would be subjected
to 175,000# tension.

So it is seen that we have now determined both maximum tensile
and compressive live-load stresses in the web members throughout the
structure, and we will next determine the chord stresses.

Members AB and BC. It is obvious that, if either of the members
AB or BC (Fig. 251) be cut, rotation would take place instantly about

b

3g!
Joana a
AR c 2 é F

Fig. 257

as

 

joint b. Therefore these members undoubtedly prevent rotation about
that joint, and hence the greater the tendency of rotation the greater the
stress in these two members; so undoubtedly the maximum live-load stress
in them will occur when the live-load bending moment about b is a
maximum.

According to Art. 91, the moment about b will be a maximum when
the average unit load on the left of the panel point is equal to the average
unit load on the bridge.

Placing wheel 4 at B (Fig. 257), we have 34 ft. of uniform load on
the bridge (see Table A). Then for the average unit load on the span
we have

1,000
150

and for the average unit load on the left of B (or b) we have, including
one-half of wheel 4, 60,000 + 25 = 2,400.

This position of the wheels satisfies the criterion as nearly as pos-
sible, as will be found by trying other positions of the wheels.

Now, taking moments about G, we have

R=181,173 lbs.

 

(284+34x2) = 2,346 lbs.
3492 STRUCTURAL ENGINEERING

for the reaction at A. Then taking moments about b we have, using
Table A,

181,173 x 25 — 480 x 1,000 = 4,049,300 ft. Ibs.

for the maximum bending moment about b, and multiplying this by 50/40
and dividing by the height of span (see Art. 92), we obtain 168,000#
tension for the maximum live-load stress in member AB, and also in BC.

Members bec and CD. By imagining the top chord be (Fig. 258) to
be cut, it is readily observed that this member resists rotation about joint
C. So the live-load stress in the top chord be will, evidently, be a maxi-
mum when the load is in the position for maximum moment about panel
point C, and a maximum in the bottom chord CD when the load is in the
position for maximum moment about panel point c. But as joints C and c
are in the same vertical line the position of the wheels will be the same
in the two cases, and it will make no difference which joint be taken as
the center of moments.

 

20’

 
 

 

 

© ®@ @ WMA

 
 

 

G

 

Fig. 258
By placing wheel 7 at panel point C, there will be 28 ft. of uniform
load on the bridge, as shown in Fig. 258, and we have
(284,000 + 56,000) +150 =2,266 lbs.
for the average unit load on the bridge, and
13,000

2

 

(103,000 + )s 50 = 2,190 lbs.

for the average unit load to the left of joint C. This position of the
wheels comes as near to satisfying the criterion for maximum moment

about joint C or ¢ as possible. Then, taking moments about G, we have
(using Table A)

 

R= (16,3614 284 x 28+ 38") — = 167,333 Ibs.

oO
for the reaction at A (Fig. 258), and then taking moments about C
we have

167,333 x 50 — (2,155 x 1,000) =6,211,650 ft. Ibs.

for the bending moment about joint C. Now, it is a question as to whether
the actual maximum moments occur at C or at the corresponding joint E
on the right half of the structure, as the load passes to the left over the
bridge. Therefore, to make sure, we will test for joint E.

By placing wheel 14 at E, we have 20 ft. of uniform load on the
bridge, as shown in Fig, 259, and for the average unit load on the span
DESIGN OF SIMPLE RAILROAD BRIDGES 343

 

 

 

 

 

 

 

we have
1,000
(284440) 150 = 2,160,
and for the average unit load to the left of E we have
222,000 _
—T090  = 22220.
_ x
2r’ 9 ’
cy % 20
@ 6
A R 8 S Do 7 a G
) Fig. 259

This position of the wheels comes nearest to satisfying the criterion for
maximum moment about E. Taking moments about G, with the wheels
in this position, we obtain

R= (16,304 +284 x20 +20") 200? 149,600
for the reaction at A, and taking moments about E of the forces to the
left we obtain

149,600 x 100 — 8,728 x 1,000 = 6,232,000 ft. Ibs.

for the maximum moment about EF, which is greater than found above for
point C, and consequently will be taken as the maximum bending moment
for determining the stress in chords be and CD or for the corresponding
chords ef and DE on the right half of the bridge—-which amounts to the
same. Multiplying this moment by 50/40 and dividing by 30, the depth
of the truss (in feet), we have 259,666, say 260,000#, for the stress in
top chord be and also in bottom chord CD, being compression in be and
tension in CD.

It is obvious from Fig. 259 that the bending moment about E would
be increased if a uniform load be placed to the left of wheel 1, in which
case a uniform load would be preceding the engines as well as following
them. This loading is used in some cases, but, when such is used, the
impact added should be materially less than when the engines are not
preceded by a uniform load, for the reason that the speed of the train
is not likely to be very high when the engines are pushing as well as
pulling a load. In other words, a train so loaded is not likely to reach
the “critical speed.”

' As an illustration, let us assume wheel 13-at E and 18 ft. of uniform
load (2,000# per ft.) ahead of the engines, as indicated in Fig. 260.
Then, for the average unit load on the bridge, we have

1,000 _

+30) =~ = 2,333 Ibs.
(56 +284+30) Tap = 25838 Ibs.,
344 STRUCTURAL ENGINEERING

and for the average unit load to the left of E we have

1,000
= =2 3
(304202) 100 3880 Ibs

This position of the loading comes nearest to satisfying the criterion
for the maximum moment about E. Taking moments about G, we obtain

 

 

 

 

 

 

 

“—2\ 1,000
R= (30 x 1414 16,364 + 284x154 is’) 150 = 172,833 lbs.
for the reaction at A.
F
26’ : is
it a, my}
a R a c D & &
Fig. 260

Now taking moments about E, of the forces to the left, we obtain
172,833 x 100 — 7,668 x 1,000 — 36 x 91 x 1,000 = 6,339,300 ft. Ibs.,

for the maximum bending moment due to Cooper’s E40 loading. Multi-
plying by 50/40 and dividing by 30 we obtain 264,137#, say 264,000#,
for the stress in chords bc and CD, which, as is seen from the above, is
about 4,000# greater than for the usual load, but if the impact is-reduced,
say, 50 per cent below the usual amount, the final result would be much
less than that obtained from using the usual loading.

Member cd. It is evident that the maximum live-load stress will
occur in the top chord cd when the live-load moment about joint D is a
maximum. By placing wheel 12 at D we have 35 ft. of uniform load, as
shown in Fig. 261, and for the average unit load on the bridge we have

1,000 _
(284 + 70) ue = 2,360 lbs.,

and for the average unit load on the left of D we have (including one-
half of wheel 12)

‘

 

 

 

 

 

182,000 _
ee 2,427 Ibs.
Bb c —f
6 i.
Qq 3. 5?
Ale B Cc } E F G
Fig. 261

This position of the loading comes the nearest to satisfying the crite-
rion for maximum moment about joint D. Taking moments about G, we
obtain

1,000

R= (16,364+ 284 x 35 +55") LO 183,526 Ibs.
DESIGN OF SIMPLE RAILROAD BRIDGES 345

for the reaction at 4, and taking moments about joint D we obtain
183,526 x 75 — 6,708 x 1,000 = 7,056,450 ft. Ibs.

for the maximum live-load moment about that point, due to the E40 load-
ing. Then multiplying this by 50/40 and dividing by 30 we have 294,018,
say 294,000* compression stress in top chord cd.

As the bridge is symmetrical about the center of span we have all
of the live-load stresses now determined and we will next determine the
impact and make'a summary of all the stresses in the trusses.

175. Summarizing of Stresses and Determination of Impact in
the Trusses.— The dead- and live-load stresses can be written on the
members, as shown in Fig. 262, directly from Figs. 249 and 251, and we
have only the impact to determine.

 

 

      
 

 

 

 

 

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+ 26000084 + 29400082
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6 Panels @ 25°0 2150’

 

 

Fig. 262

The live load extends practically over the entire length of the bridge,
as seen above, when the chords and end posts receive their maximum live-
load stress. So, in determining the impact.stress in these members, L
in the impact formula (Art. 125) will be taken as the total length of the
span, or 150 ft.

Then we have

800

~ 750+ 300

for the coefficient of impact for these members, and multiplying the live-
load stress in each by this coefficient we obtain the impact given on them
in Fig. 262. As an example, the impact in the end post bd is 263,000 x
0.6666 = 175,315, say 175,000#; in chord be it is 260,000 x 0.6666 =173,-
316, say 173,000#; and so on for the other chord members.

The maximum live-load stress in hanger bB results from loads in
the two adjacent panels, AB and BC, alone. So the L in the impact
formula would be taken as 50 ft., the sum of the lengths of the two panels,
in determining the impact stress in that member. Then for the impact
stress in hanger bB we have

Cc = 0.6666

300
—~~__} =81,000 lbs.
94,500 x (soxso0) 81,000 Ibs

When the maximum live-load tensile stress occurs in the diagonal
bC the live load extends from the right support to a little beyond panel
346 STRUCTURAL ENGINEERING

point C (see Fig. 253). In the case of a uniform live load the distance
from the right support C would be taken as L in the impact formula, as
in that case the loads are considered to be only at panel points, and as the
result is affected but little we will use the same distance in the case of
the concentrated loads. So for the impact stress in diagonal bC we have

300

TES OUOS (asi +300

The maximum live-load compression occurs in post cC when the load
extends from the right support to a little beyond panel point D, and the
maximum tension occurs in diagonal cD (see Fig. 254) at the same time.
So the distance from the right support to panel point D (75 ft.) will be
taken as L in determining the impact in these members. Then for the
impact (compression) in post cC we have

300
718,000 x (acum) = 62,400, say 62,000 Ibs.,

and for the impact (tension) in diagonal cD we have

300

102,000 x (sam

The maximum live-load tension in post cC and maximum compression
in diagonal cD occur when the load extends from the left support to a
little beyond panel point C. (See case of corresponding members (eF
and eD) on right half of truss, Fig. 255.) So the distance from the left
support to panel point C will be taken as L in determining the impact
in this case, and we have

) = 131,250, say 131,000 lbs. tension.

) = 81,600, say 82,000 lbs.

300
=) =! 2 ;
37,000 x (sors) 31,685, say 32,000 Ibs
for the impact (tension) in post cC and
300
48,000 x (so =41,146, say 41,000 Ibs.

for the impact (compression) in diagonal cD.

The impact stress found above for each respective member can now
be written on each member, and adding them in each case to the dead-
and live-load stresses the total combined stresses, as shown in Fig. 262,
are obtained.

In the case of members cC and cD we have what is known as
reversal of stress.” The dead-load stress, which acts at all times, is com-
pression in post cC and tension in diagonal cD. When the live load
moves onto the structure from the left a tensile stress of 37,000# occurs
in post cC and a compression stress of 48,000# in diagonal cD, as shown
for the corresponding members (eE and cD) on the right half of the
bridge (Fig. 251). Now it is evident that, if the dead-load compression in
post cC were just equal to the 37,000# live-load tension, the two stresses
would just balance each other and consequently the actual stress in the
post would be zero when the maximum live-load tension occurred in the
post; but, if the dead-load stress were less than the live, the actual stress
would be the difference between the two. In any case, before the dead
DESIGN OF SIMPLE RAILROAD BRIDGES 347

load is subtracted the impact should be added to the live-load stress. To
be sure that reversal is well provided for, as a rule not all of the dead-
load stress is subtracted from the sum of the live-load and impact stresses.
The A. R. E. Ass’n specification, the one we are using, calls for 4%.
Therefore, for the tensile stress in-post cC, we have

-37,000 - 32,000 + ox 22,000 =—54,334, say 54,000 Ibs.,

and for the compressive stress in diagonal cD we have

+48,000 +41,000 -£ x 17,000 =+77,667, say 78,000 Ibs.

In case % of the dead-load stress in any member is greater than
the sum of the reverse live-load stress and impact, the member is not
reversed and, consequently, the reverse stress is ignored, as in the case
of diagonal bC. In bC we have 14,000# (see Fig. 251) live-load com-
pression and 14,000 x 300 + (25 +300) =13,000# (about) impact, making
in all a stress of 14,000+13,000=27,000# compression, but 3 of the
dead-load tension is 51,000 x 2/3=33,666# and hence no reversal takes
place, that is, the diagonal is always in tension, due to dead load, and,
consequently, the reversal stress can be ignored. The same is true in all
such cases.

176. Designing of Members in Trusses After all of the maxi-
mum stresses are computed in the members of the trusses and combined
as shown in Fig. 262, the next thing, logically, to do is to design the
sections of these members. The first thing to do in such work is to
glance, so to speak, over the structure and associate the details of the
different members and, as far as possible, ascertain the governing
members.

In this case, as in all such bridges, the intermediate posts are the
governing members, as their width governs the width of practically all
of the other members of the trusses. So we will first design the inter-
mediate post cC (Fig. 262).

Member cC. (See Fig. 262.) Let us assume

2X
each of these posts to be composed of 2—15” [s “|
placed, in reference to each other, as shown in Fig. |

 

263. The distance d between the toes of the flanges | y NI
is really the governing distance. This distance :

should not be less than 54” in any case, and a little I
more is preferable. If this distance is less than x

54” it is impossible to get the jaw of an ordinary Fig. 263

riveter inside of the member and consequently the

rivets would have to be hand driven, which is expensive work. From
table 3 it is seen that the average width of flange of 15” [s is about 33”,
making 7” for the two flanges, and if we make d=6}”, we have

w=644+7=13f ins.
for the width of the posts. The least average radius of gyration, which

is in reference to axis 2-x, as seen from Table 3, is about 5.5”. Then
substituting this value of r in the column formula of Art. 73, and taking
348 STRUCTURAL ENGINEERING

L as 30 ft., or 360 ins., we have

16,000 — 70 oe = 11,420 Ibs.

for the allowable unit compressive stress on the post.

For the stress in the post we have 162,000# compression and 54,000#
tension, as given in Fig. 262. Now, according to the specifications, each
of these stresses must be increased by 0.5 of the lesser. So we have
162,000 + 0.5 x54,000 =189,000# compression and 54,000 +0.5 x 54,000=
81,000# tension which the post must be designed to carry.

Taking, first, the case of compression, we have

189,000
11,420

for the area required for compression and

81,000
16,000

for the net area required for tension. The area, 16.64”, required for
compression governs. The nearest to this (considering 15” channels) is
2—[s 15’ x 33#, which have an area of 19.85”. These channels have
3.20” more area than required, so let us try 2—[s 12”x30#,. Then
we have

=16.6 sq. ins. (about)

= 5.06 sq. ins.

360
16,000 — 70 4987 10,100 lbs.

for the allowable unit stress on the post, and dividing this into the stress
we have

189,000 + 10,100 = 18.7 sq. ins.

for the required area. This shows that the 12” x 30# channels are too
light and that the 12’ x 35% channels would have to be used, which are
heavier than the 15” x 33# channels; and, besides, better details are ob-
tained by using the 15” channels, so we will use the 2—[s 15” x 33%.
The radius of gyration of these channels is 5.62, within 0.12 of the
assumed radius, and hence recalculations are unnecessary.

Member dD. The post dD carries nothing but the 9,000# of dead-
load compression and theoretically could -be made of very light section,
but to keep the floor beams the same throughout the span and other details
constant, as well as for general appearance, we will make this post of
2—[s 15” x 33# (the same as post cC), that is, each of these posts will
be made of 2—[s 15” x 33# (the lightest 15” channels) regardless of
the excess of metal.

Member bB. The hangers bB, or hip verticals as they are often
called, are not posts but wholly tension members, and it is not necessary
that they be of the same type of section as the posts; however, they are
often made so, but more often they are made of four angles built into an
I-section. (See Fig. 298.) The total stress, as given in Fig. 262, divided
by the allowable unit intensity gives 193,500+16,000=12.092” (net)
for the required area of cross-section. We will-use 4—Ls 67x 4” x §”=
DESIGN OF SIMPLE RAILROAD BRIDGES 349

14.440” —1.50”=12.940” (net), deducting a one-inch hole (for 4-in.
rivet) out of each angle. This type of member is cheaper than the two
channel sections, as used in the posts, as there are fewer details and, con-
sequently, less shop work, but it is not an economic column section, as the
radius of gyration is small in comparison with the area of cross-section
contained, as in comparison with two channels of an equal cross-section.

Member bC. The diagonal is subjected to tension only, and, using
the maximum stress given in Fig. 262, we have

357,000 + 16,000 = 22.31 sq. ins. (net)

for the required area of cross-section. We could use either four angles,
as in the case of hanger bB, or two channels, as the member is subjected
to tension only, but the two-channel section is better in this case as the
member is subjected to some cross bending due to its own weight, and
as the channel section is the more capable of resisting this it will be used.
Let us try 2—[s 15’x40#. We can consider either two rivet holes cut
out of each web or one hole cut through each flange, whichever is the
greater. The web of the 15”x40# [ is 49 thick (see Table 3) and
the grip thickness (g) on the flange is 34”, hence the holes through
the flanges govern. Then we have 2—[s 15” x 40# =23.52-4 x 21/32=
20.900”, As is seen, these two channels are too light. So let us try
.2—[s 15”x45#. In this case the thickness of the web and the grip
through the flanges are the same, and hence either can be taken in deter-
mining the net section of member. So we have 2—[s 15”x45¥#=
26.48 -4x%=23.980” (net), which is 1.670” more than required, but
this is the best we can do if we use channels, so we will use this section
for diagonal bC.

Member cD. This diagonal is subjected to 201,000# tension and
78,000# compression, as given in Fig. 262. Now increasing each by 0.5
of the lesser, as we did above in the case of post cC (according to the
specifications), we have

201,000 + 0.5 x 78,000 = 240,000 lbs. tension

and

78,000 + 0.5 x 78,000 = 117,000 lbs. compression.

Then dividing the tensile stress by 16,000# (the allowable unit stress)
we have

240,000 + 16,000 = 15.0 sq. ins.
for the net area of cross-section required to carry the tension. Now
assuming the member to be made of two 15” channels having a radius of
gyration of 5.6 (a mere guess) and taking the length of the diagonal as
39 ft., or 468 ins., we have

16,000-—70 = = 10,150 lbs.

for the allowable compressive unit stress. Dividing this into the com-
pressive stress, we have

"117,000 + 10,150 =11.5 sq. ins.

for the required area of cross-section to carry the compression. Deduct-
ing for four rivet holes, out of the flanges of the two lightest 15” channels,
350 ; STRUCTURAL ENGINEERING

we have 2—[s 15” x 334% =19.8-4 x 21/32=17.20” (about) for the net
area to carry tension, which is 2.24” more than required. The total,
19.80” is available for compression, so the section is larger than is neces-
sary for compression as well as for tension, but this section will be used
as it is desirable to have all of the diagonals of the same width. We now
have all of the web members designed and we will next take up the
designing of the chord members.

Members AB and BC. The bottom chords are purely tension mem-
bers; so, dividing the maximum stress in chord AB or BC by 16,000,
we have

335,000 + 16,000 = 20.93 sq. ins.

for the net area required. By glancing over the table of channels (see
Table 3) we see that two 15” channels can be used as section for these
members. Deducting four rivet holes from the flanges we have 2—[s
15” x 40# = 23.52 —2.6= 20.90”, which is practically the section required,
and hence these channels will be used.

Member CD. For the net area of cross-section required for chord
CD, we have 521,000 +16,000=32.56 sq.ins. Now, from the table of
channels (see Table 3 or any structural handbook, as Cambria or Car-
negie) it is seen that no two 15” channels will have the required net sec-
tion. So, if channels are used, they will have to be reinforced by riveting
plates to their webs. Let us try the following sections:

2—[s15”x45# =26.48-4x5/8 =23.980” net
2—pls. 12” x 7/16 =10.50-4x7/16= 8.750” net
82.730” net

 

The net area of this is practically equal to the area of the required
section and hence will be used.
Top chords. The top chords and end posts of such bridges are usu-
ally made up of either two channels and a cover plate as shown at (a),
Fig. 264, or of two built chan-
nels and a cover plate as shown
at (b). The rolled channels are
used whenever that section is
sufficient. It is not desirable to
use the heaviest channels, which
are of 42” metal, as the shop
work is rather expensive on ac-
count of the thick metal, and
whenever the end post or heav-
Fig. 264 iest chord members exceed the
area of the cover plate and, say,
two 15” x45# channels, a section similar to the one shown at (6), Fig.
264, is used throughout the structure, as it would be unsightly construction
to build part of the chord members of a bridge of rolled channels and
part of plates and angles. So, in this case, we will first ascertain as to
whether we can or cannot use the rolled channel section as shown at (a),
Fig. 264, for the top chord. The member cd, as is seen from Fig. 262, will
be the heaviest member, that is, the one requiring the greatest area of

 

 

 

 

(a)
DESIGN OF SIMPLE RAILROAD BRIDGES 351

cross-section. In preliminary calculations the least radius of gyration of
this type of section can be taken as 0.40 of its depth. Then, assuming 15”
channels, we have 15x 0.40=6.0 for the approximate radius of gyration
about axis z-x, which is usually the least radius. Now substituting this
value of r and the length of the member, which is 25 ft., or 300 ins., into
the column formula, we have

16,000 - 0 = 12,500 Ibs.

for the allowable unit stress on the member. Then dividing the maximum
stress, as given in Fig. 262, by this, we have

589,000
12,500

for the required area of cross-section of the member. The width of the
intermediate posts, as found above, is 134’, and allowing for two 3”
gusset plates and 7” (=34%x2) for the flanges of the channels, we have

134 +14+7=213, say 22 ins.

for the width of the cover plate. The cover plate should be as thin as is
consistent with good practice. The specifications limit this to zy of
the distance (shown as d at (a), Fig. 264) between the lines of rivets
connecting the plate to the other parts of the member. That distance
here, using 2” gauge in channel flanges, is 1834+144+4=183”, say 19”.
One-fortieth of this is practically 4’, so the cover plate would be 22” x 4”.
Now, using the heaviest channels that it is practical to use, we have the
following sections:

=47,1 sq. ins.

2Q—[s15”x454#  =26.4890”"
1—cov. pl. 22” x #”=11.000”
37.480”

This is about 104” less than the required area found above, and hence
the rolled channel section has not sufficient area; and, therefore, the type
of section shown at (b), Fig. 264, will be used.

In designing the top chords it is best to begin with the lightest sec-
tion, which is the member having the least stress, and make it of minimum
thickness of metal and then increase the area of cross-section of the
other top chord members by increasing the thickness of webs, leaving the
other parts about constant. For by so doing the centers of gravity of
the chords will practically be in the same plane throughout. Therefore,
we will take up the designing of chord bc first—it being the lightest.

Member bc. The maximum stress in chord bc, as given in Fig. 262,
is 521,000#. Then assuming the allowable unit stress to be 12,500” (see
Example 4, Art. 74), we have

521,000
12,500
tor the approximate required area. Next draw a sketch of the cross-
section of the chord as shown in Fig. 265. The cover plate and top
angles should be made as light as the specifications will permit in order

that the gravity axis 2-x be as near the center of the web as possible. As
seen above, the cover plate must be about 4” thick and the smallest angles

= 41.68 sq. ins.
352 STRUCTURAL ENGINEERING

that can be used at the top, considering details and all, are 34” x 34” x3”.
The bottom angles should be as heavy as is practical to use for the same
reason that the top ones are made light. So for the bottom angles we
will use 6’ x4” x 2”, as that is about the maximum size and thickness
used for such work. The cover plate, top angles, and the bottom angles
are usually about constant throughout the chord.
Allowing for 2—%” gusset plates and for 2—}” webs (a mere guess),
we have
184 4+1$4147=22f ins.

for the width of cover plate. To insure a neat finish, the cover plate
should project 4” or }’ beyond the angles. So we will make the cover
plate 23” wide. However, the details

 

 

 

 

 

 

 

 

 

 

 

 

 

y : should always be such that the width of

8 la Jo cover plate-would be in even inches.
My iLeze", % The depth of webs should be such
vs) xl. a mila %. §,| that the radius of gyration about the
wn oe | &| horizontal axis (a-«, Fig. 265) is about
& wae) s equal to the radius about the vertical
g | Sm axis (y-y). For preliminary calcula-
ie ity tions the radius of gyration in reference
I ase! 7 a.70" to axis 2-x of such chords can be taken
L275" as 0.4 of the depth of the web, and the
Fig. 265 radius in reference to axis y-y can be

taken as the horizontal distance out
from the axis to the outer face of the web. So, in this case, we have
(assuming 4” webs)

1344+144+1

3 =7.87

for the approximate radius about axis y-y. Then for equal radii about
the two axes, we have 0.4h=7.87, from which we obtain for the depth
of the web

p= 8" rb ins,

As the work here is only fairly approximate this figure shows only about
what the depth of the web should be, and hence any web about this depth
can be used. A 19” web is an odd width and a 20” web appears a little
deep for this section, so we will use an 18” web. Now the radius of
gyration of the section about the horizental gravity axis, which is usu-
ally the least radius, is about 0.4 of the depth of the web, as stated above,
so we have

0.418 = 7.2 ins,

for the approximate value of the least radius of gyration of the section.
Then substituting this value of r in the column formula, we have

16,000 —70 = 13,084, say 13,000 Ibs.
DESIGN OF SIMPLE RAILROAD BRIDGES 353

for the approximate allowable unit stress. Then dividing this into the
stress we have

521,000 + 13,000 = 40.07 sq. ins.

for the approximate required area. Taking this as a guide, let us assume
the following section for chord be:

1—cov. pl. 23” x4” =11.500”
2—web pls. 18” x 3” =13.500”
Q—1s 34" x34" x9" = 4,960”
2-18 6x4" x 3%” =10.620”

40.580”

The next thing in order is to determine the center of gravity, moment
of inertia and radius of gyration of this section and check back to see if
the section is actually the correct one to use.

Taking moments about the center of the cover plate (Fig. 265),
we have

13.50 x 9.37 + 4.96 x 1.26 + 10.62 x 16.49

r= 40.58 = 7,59 ins.

 

for the distance from the center of the cover plate to the horizontal grav-
ity axis a-r, and for the moment of inertia about this axis we have the
following:

cover plate 11.50 x 759 +0 = 662.49
webs 13.50x 1.78 +182.25x2 = 407.27
top angles 4.96x6.33 +2.87x2 = 204.48
bot. angles 10.62x8.87 +19.26x2 = 874.07

2,148.31

Then for the radius of gyration about this same axis z-x we have

2,148.31
ai (8B
e \ 1058

which is within 0.08 of 0.4 of the depth of the web.

As the section is symmetrical about the vertical plane through the
center of the cover plate, the vertical gravity axis y-y will pass through
the center of the cover plate and for the moment of inertia of the section
about this axis y-y we have the following:

cover plate 0+ 506.96 506.96
webs 13.50x 7.56 +0 YY15Y
top angles 4.96x3.76 +2.87x2 = 386.36

bot. angles 10.62x8.78 +6.91x2= 832.50
2,497.39
354 STRUCTURAL ENGINEERING

Then for the radius of gyration about this axis y-y we have

rhe [Peete =1.84ins.,

which is a little larger than the radius about the 2-2 axis.
Now substituting the value of r (the least radius) in the column
formula, we have

300
16,000 - 70 7.38 = 13,116 lbs.
for the actual allowable unit stress on the chord. Dividing this into the
stress, we have

521,000
13,116

which is 0.865” less than the section assumed above, but as this is about
as close as we can obtain we will use the assumed section for chord be.

Member cd. There will be such a small difference between the allow-
able unit stress for the other top chord members and that found above
for chord be (as can be verified by actual calculations) that the allowable
unit stress found for that member will be used in designing the others.
So, dividing the maximum stress in chord cd, as given in Fig. 262, by
that unit stress, we have

= 39.72 sq. ins.,

589,000
13,116

for the required area of cross-section of the member cd.
Then using the same size angles and cover plate as for chord be
and increasing the thickness of the web to 4”, we have :

1—cover pl. 237’ x4/” =11.500”
2—web pls. 18”%x4” =18.000”
2—Ls 347x337" x3” = 4.960”
2—Ls 67x 4"%xi%"” =10.620”

45.080”

This section is quite close to the required section, being only 0.180”
larger, and hence will be used for chord cd.

This completes the design of the main truss members, as the bridge
is symmetrical about the center of the span, except the end posts, which
will be designed later, after the bending moment on them due to the
wind pressure is determined.

177. Designing of the Bottom Lateral System.—The bottom
lateral system is really a horizontal truss in the plane of the bottom chord.
It is for the purpose of resisting the wind pressure on the lower portion
of the structure, and, in the case of through bridges (such as the one we
are now designing), the wind pressure on the train as well as the vibra-
tion due to the swaying of the train. A double system of diagonals, or
laterals, is practically always used, which is as indicated in Fig. 266
where the bottom lateral system is shown in plain view, one system of
laterals being dotted to avoid confusion.

= 44.9 sq. ins.

 
DESIGN OF SIMPLE RAILROAD BRIDGES 355

The bottom chords of the main trusses act as the chords in this lat-
eral system and the floor beams as posts. The stress in the chords due
to the wind and vibration is, as a rule, ignored unless it reverses the dead-
load stress, taking the wind pressure in that case on the unloaded struc-
ture as 350# per foot of span or where the maximum exceeds 25 per cent
of the maximum combined dead, live, and impact stresses (see specifica-
tions), in which case it is combined with the dead, live, and impact

stresses as in the case of viaduct columns (Art. 162). This combination,
however, is very rarely necessary in ordinary bridges. The stress in the
floor beams (acting as columns), due to wind and vibration, is ignored
entirely, as the laterals are connected to the bottom flange of the beams,

le "li "le "l@ wl
7 ¢ d

 

 

 

 

 

 

 

 

 

 

. e Bot Chord. f g
me al # ie ,
% tL | exe “oO wee ae Strnger 7 2. f
A a 7 oe 7 7 7
OL 22a} & ire “ao: y Zz <
mL L-7 ae a le S17 ob. a
2
Ri? a c a é& Gi,
tne 107
ae ---  -.--6 Panels @ 2810"=150*0”

 

 

 

Fig, 266

which are in tension due to dead and live load, and hence the compres-
sion in these same flanges, due to wind and vibration, only tends to
reverse the dead- and live-load stress in them.

There are two ways of considering the bottom lateral system: One
is to consider the double system, in which case the two diagonals in any
panel resist equally the shear in the panel, that is, one diagonal is con-
sidered as having the same intensity of stress as the other, one being in
compression and the other in tension; the other way is to consider only
a single system, one system being in action when the pressure comes
from one direction and the other system acting when the pressure comes
from the opposite direction. We will consider the single system, in which
case the diagonals carry only tension, and then investigate for rigidity
after the sections are determined for the single system.

According to the specifications, the pressure or horizontal load on
the bottom lateral system, here considered, is 200 +0.10x 5,000 =700#
per foot of span. So, for a panel load we have

W =00 x 25 =17,500 Ibs.

This is, according to the specifications, to be considered a moving load
(live load). Let us assume that this load acts in the direction indicated
by the arrows (Fig. 266) and that it moves over the structure from right
to left, that is, from G to A. Now, from Art. 90, we have

s-7[1s2....(m-0]

for the maximum shear in the different panels, and hence for the maximum
stress in the corresponding diagonal (considering single system) we have

r=" | 142. 5 (0-1) | see.
356 STRUCTURAL ENGINEERING

Tan w= 25/16 = 1.56, and see w=1.85, which is most readily obtained
from a table of natural functions after the tangent is computed.
Then for the maximum stress in diagonal cD we have

S= 7 (14248) seew= a x 6x 1.85 = 32,375, °

say 32,000 Ibs. (tension),

 

and for the maximum stress in bC we have

17,500
6

              

S’= ” (1424344) seew=
say 54,000 lbs. (tension),

and for the maximum stress in aB we have

s” =” (142484445) seow= ox 15 x 1.85 = 80,937,

say 81,000 lbs. (tension).

This completes the necessary determination of stress in the diagonals or
laterals (as they are called) due to wind and vibration, as the structure
is symmetrical about the center of the span, and we will next design the
sections for these members.

Taking the first lateral aB, we have

81,000
16,000

for the net area of cross-section required. Using 1—L 5” x34” x3}’=
5.81-0.75 =5.060”, we have exactly the correct net area for that diag-
onal. For lateral bC, we have

54,000
16,000

for the net area of cross-section required and by using 1—L 5” x 34” x4”
=4-0.5=3.50”, we have about the correct area, at least about as near
the correct area as is possible to obtain. For lateral cD, we have

32,000
16,000

for the net area of cross-section required, and by using 1—_ 5” x 34” x 3”
=3.05 —- 0.387=2.684” we have 0.685” more net area than required, but
we will use this angle so as to have the bottom laterals made of 5” x 34”
angles throughout.

As the structure is symmetrical about the center of the span, we
now have all of the laterals designed, considering the single system, and
we will now investigate the double system. The laterals are connected
to the bottom of the stringers at the points of intersection and hence
the longest unsupported length of lateral is between the stringer and
truss. This maximum unsupported length is about 8’-0”, or 96”. Now
taking the end lateral aB, which is composed of 1—L 5” x34” x2”, we
have L/r=96/0.96 =100, which is quite satisfying as the maximum value

 

= 5.06 sq. ins.

 

= 3.37 sq. ins.

 

=2 sq. ins.
DESIGN OF SIMPLE RAILROAD BRIDGES 357

for L/r is 120, and substituting the value of L and r in the column
formula we have

96
0.96

for the allowable compressive unit stress, and dividing this into one-half
of the stress found above (the stress that it would be considered to resist
if acting in the double system) we have
40,500
9,000

for the required area of cross-section, which is considerably less than the
cross-section of the angle. So this lateral is sufficient, considering either
a single or double system, and the same is true for the other laterals, as
will be found upon investigation. So the bottom laterals as designed are
capable of resisting the forces coming on them from wind and vibration
when considered either as a single or double system.

The bottom laterals are subjected to stress from traction in addition
to the stress resulting from wind and vibration. As an illustration, sup-
pose a rapidly moving train to come onto the structure and the brakes to

16,000 - 70 ~~ =9,000 Ibs.

= 4.5 sq. ins.

 

 

 

 

 

 

 

 

 

 

A loor Bear Bottom Chora ; Stringer)
{ L I
fea
Z \
Fig. 267

be suddenly applied—the wheels would skid or tend to skid along the
rails, whereby a longitudinal thrust would be exerted along the stringers
which would tend to bend the floor beams transversely as indicated in
Fig. 267 (where the train is assumed to be moving from right to left).
Now this bending of the floor beams, which would cause serious stresses
in them, is prevented by attaching the laterals to the bottom of the string-
ers so that the longitudinal thrust exerted by the stringers is transmitted
directly to the laterals, and from there to the bottom chords of the trusses
and on out to the end supports. As the stringers are connected to one
another rigidly, end to end, throughout the length of the structure, it is
evident that the traction from any loading would be transferred more or
less from stringer to stringer, and hence would be distributed to some
extent to all of the laterals; so in computing the stress in the laterals
due to traction we will consider that the laterals in any panel resist
the traction from an average panel load of fully loaded bridge. Then
placing wheel 1 (see Table A) at one end of the bridge, we have

2 (284441 x2) 1,000 5 = 915,000 lbs.

for the weight of the full live load (Cooper’s E50 loading) on the bridge.
Dividing this by the number of panels, we have

see = 152,500 Ibs.
358 STRUCTURAL ENGINEERING

for the average panel load. Two-tenths (0.2) of this vertical load is
considered as traction and is exerted along the track, that is, the coeffi-
cient of friction of the wheels on the rails is taken as 0.2. So we have

152,500 x 0.2 = 30,500 Ibs.

as the traction force exerted

on the two stringers in each

panel, or 15,250# to each

# # stringer. This force on each

Stringer ae ge adnees is transmitted to

: ¢ . the laterals at two points,

where the laterals and

#* stringers connect, so we can

do. Lt iagso* ohgre ne the case as indi-

cated in Fig. 268. One

component of the traction

force, 15,250# +2 applied at

B D each of the points ¢ will, in

Fig. 268 each case, act along the lat-

eral, causing compression in

the case of the part Be and tension in the part De, and the other com-
ponent at each point ¢ will act along member cc.

Without the members cc, there would be transverse bending on the
stringers from traction, and hence cc is an essential member in the
system.*

Then for the stress in a lateral, due to traction, we have

15,250 ie 15,250

3 xsee $= —a— x 1.18 =8,997, say 9,000 lbs.,

 

 

 

Floor Beam

 

 

 

 

 

 

 

which is compression in the part cB and tension in part cD, and just the
reverse if the train were moving in the opposite direction. For the stress
in member cc, which can be called a tie or strut, we have

15,250 _ 15,250
9 xtang= 3

 

 

x 0.64 =4,880 Ibs.,

which is compression in one and tension in the other.

These stresses are not very great in the first place, as is seen, and
they are not very likely to occur at the same time that the maximum
wind stress occurs and hence, as a rule, are ignored in the designing of
the sections of the laterals. However, the traction should be provided
for in so far as making the laterals capable of taking compression
(L/r<120) and the inserting of the ties cc. The tie cc does not neces-
sarily have to be a very heavy section. They are usually made of 1—L
34” x 8” x 3”, in which case we have

78
16,000 — 105-99 = 9934 Ibs.

* As far as the author knows, Dr. J. A. L. Waddell, M.A.Soc.C.E., was the first to
point out the necessity of these struts,
DESIGN OF SIMPLE RAILROAD BRIDGES 359

for the allowable unit stress for compression. Then we have
4,880
9,934
for the required area in the case of compression, and less is required for
tension, as is obvious. So it is seen that the theoretical required area of
these struts (cc) is not’ the governing feature, but that it is a matter of
obtaining sufficient rigidity. That means that L/r should not exceed 120
in any case. The L/r for a 34” x3” x3” angle is about 0.87, which is
reasonably low. This angle is used as it is a very common angle in this
class of work, and it is about the minimum size permitted and it fits the
case wherein common judgment decides.
For the wind stress in the bottom chord ab or BC (Fig. 266) we have

24 W tan w = 24 x 17,500 x 1.56 = 68,250 Ibs.,

which is compression in ab and tension in BC, in this case. Now, as
shown in Fig. 262, the maximum combined stress, dead, live and impact,
is 335,000# tension. Twenty-five per cent of this is 83,750#, so the maxi-
mum wind stress in these chords can be ignored. (See specifications.)
In case of combined dead, live, impact, and wind stresses, the allowable
unit stress is raised 25 per cent, as the probability of all of these stresses
being a maximum at the same time is very remote. The same is true
of the other bottom chords, as will be found upon investigation. Now,
so far, the wind stresses in the bottom chords can be ignored; but further,
suppose the bridge to be unloaded, that is, no live load on it, the wind
pressure of course will be less. Let us then assume a pressure of 350#
per foot of span, which is about correct, instead of 700#. Then the
stress in the chord ab or BC is :

 

= 0.49 sq. ins.

350
68,250 x 7007 84,125 Ibs.,

which here, in the case of ab, is compression. But, as this does not
reverse the dead-load tension in the member, which is 55,000#, the wind-
load stress need not be considered. The same will be found to be true
for the other bottom chords, so the stresses in the bottom chords of this
bridge, due to wind—or lateral pressure, as the specifications call it—can
be ignored. The same is true for most all ordinary railroad bridges.

178. Designing of the Top Lateral System.—The top lateral
* * system is really a horizontal truss in the plane of the top chords. In the
case of through bridges (such as we are designing) it is for the purpose
of holding the top chords transversely and resisting the wind pressure
coming on the top portion of the structure, but in the case of deck bridges,
where the track is on the top of the structure, the wind pressure on the
train and vibration due to the train must be provided for also, in which
case the designing of the top lateral system would be the same as shown
in the last article for the bottom lateral system. The top lateral system,
as a rule, has a double system of diagonals, the same as the bottom lat-
erals, except in the case of through bridges a portal is placed in each
end panel in the plane of the inclined end posts which takes the place
of diagonals in those panels. So, following the usual practice, the top
lateral system in this case will be made as shown in Fig. 269, where one
system is dotted to avoid confusion.
366 STRUCTURAL ENGINEERING

The lateral pressure or load which this system must be designed
to carry, according to the specifications, is 200# per foot of span. This
horizontal load should be considered as a moving load, that is, as a uni-
form live load. Then for a panel load we have

P=25 x 200=5,000 lbs.,
one-half of which is to be considered as applied to each truss, that is,
wherever it is necessary to do so.

2540%= 15040"

wie io 4.

 

 

 

 

 

 

 

7 - a“ #£
27 OO SPSS eee KE
NOON
+07 ne CY
oO ae Git
AU 2829 2
259" Lom

Fig. 269

Now, assuming this load as moving onto the bridge from the right
end, and acting in the direction indicated by the arrows (Fig. 269), we
obtain

6x7 see w= 6x

for the maximum tensile stress in diagonal cD, and

10 «22 x 1.85=15,416, say 15,400 Ibs.

 

a 1.85 =9,250, say 9,200 Ibs.

for the maximum tensile stress in diagonal bC.

Assuming a full panel load at each of the points F and E and a half
panel load at d (none at D), we have
o 5,000 7% 5,000

6 2
for the maximum compressive stress in strut dD, and assuming a full

panel load at each of the points F, E, and D, and a half panel load at ¢
(none at C), we have

6x

3

 

= 5,000 Ibs.

5,000 5,000 _
WE + y= 7,550 Ibs.

for the maximum compression in strut cC.

This really completes the necessary calculation for the stresses in
the top lateral system, as will be seen upon investigation. For example,
when panel point F alone is loaded, and the forces acting as indicated,
the stress in diagonal eF will be tension, as is obvious, but this diagonal
has maximum tension when the panel points are loaded from the left
end up to and including point E with forces acting in the opposite
direction, in which case the stress in the diagonal would be 15,400#, the
same as found for diagonal bC and loading points F and E, and the
forces acting as indicated, diagonal dE would be in tension; but the
maximum tension will occur in this diagonal when the span is loaded
DESIGN OF SIMPLE RAILROAD BRIDGES 361

from the left end up to and including point D and forces acting in the
opposite direction, in which case the tension in diagonal dE would be
9,200#, the same as found for diagonal cD. So it is seen that one sys-
tem can be considered in action when the span is loaded from one direc-
tion and the other system when loaded from the opposite direction, and
as the lateral system is symmetrical about the center of span it is obvious
that the stresses determined above are sufficient for designing all the
members in ‘the system.

The members (diagonals and struts) should be as deep as the top
chords, in order to hold the top chords rigidly. The diagonals, which
are usually designed as tension members, are as a rule made of two
angles latticed together, one angle being in the plane of the top of the
chord and the other being in the plane of the bottom of the chord. The
struts (members cC, dD, and eE) are compression members and are, as
a rule, made of four angles, two at the top of the chord and two at the
bottom. These are latticed together in the vertical plane so as to form
an I-section.

Now, beginning with diagonal bC (Fig. 269), we have

15,400
16,000

 

=0.9 sq. ins.

for the required net area of cross-section of the member, considered as a
tension member. This, as is seen, (theoretically) calls for two very

- small angles, but (practically) the angles should be such that L/r be not
less than 120, to insure against vibration. The length of L can be taken
as one-fourth of the total length of the member.

(The diagonals are connected to each other at their intersection;
this alone reduces the length one-half, and each half can be considered
as a fixed column, thus reducing the length to one-fourth of the total
length.) The total length of the diagonal center to center of end con-
nections is about 30 ft., or 360 ins. Then using one-fourth of this,
we have

90

=720 =0.75

r

for the required radius of gyration.

According to this and to the required area of cross-section, found
above, 24x24” or 3x3” angles could be used, but as much better
details can be obtained by using 34” x 34” angles—and the weight of the
top lateral system is a small item—we will use 2—Ls 34” x34” x 3” for
each of the diagonals throughout, as cB, eF, and fE require the same
section as bC, and the others, theoretically, require less.

The designing of the struts cC, dD, and eE is very similar to the
designing of the diagonals; it is mostly a matter of obtaining rigidity
and the selecting of sections that are satisfactory as to details. But in
all such cases the theoretical requirement should always be determined
before the selection of section is made.

These struts are more or less fixed at their ends, but as they are
wholly compression members we will take the distance center to center
of trusses, as their length, which is 16 ft., or 192 ins. In order to insure
362 STRUCTURAL ENGINEERING

rigidity, L/r should not exceed 120. Then, taking L as 192, we have

192
=o
for the allowable maximum’ radius of gyration. The most satisfactory
section for these struts is four unequal leg angles, arranged in reference
to one another as shown in Fig. 270. The long legs of the angles are
turned out so as to make the radius of gyration about axis y-y as large
as is possible, for, as is obvious, the least radius is about that axis. The
lattice bars hold the angles about 4” apart, and, as the radius of gyration
about axis y-y is the same for one pair of angles as it is for the two
pairs, we can select the angles from Table 5 by using the angles having
r3 equal to 1.6—the required radius. As is seen in this table, the 3” x 24”
angles are too small, as r3 for them is less than 1.6; the 34’ x 24” are
large enough, but to use a 24” leg would necessitate the using of 3”
rivets, while {” would be used in all the other members; so we will use
the 34” x 3” x 2” angles, in which case r3 is a little more than 1.71, as
seen in Table 5. Now substituting this value for r in the column formula,
we have
192
16,000 — rma = 8,140 lbs. (about)

for the allowable unit stress on a strut made of these angles. Dividing
this into the greatest stress found in these struts, which is 17,2504, we
obtain less than a square inch of metal for the required section. So it
is seen that the stress does not really influence the design and that each

 

 

 

 

 

 

 

 

“ay
&
v ’ e
a : ”
Se 3 y x
S| 5
4 5 <
TON 2
fH pO S 2 .
4
_ Ks
If ~ SKE y)
' <5 0
t gy a SY a Ky e
* v 2
t-Lattice LY VI
| ks vy
oul i
\
ly :
Fig. 270 Fig. 271

strut will be made of 4—1s 34” x3” x3”=9.20", as this fits the case
most satisfactorily as regards rigidity and details.
179. Design of Portals.—There are several types of portals in
use. The type shown in Figs. 269 and 271 will be used in this case, as
it is a type commonly used; it is quite simple, very rigid, and in fact
quite satisfactory for ordinary bridges.
In determining the stresses in portals, the first thing to do, the same
as in the case of all frames, is to locate all of the applied forces affecting * °°
it. As is obvious, the maximum stresses in the portals will occur when
DESIGN OF SIMPLE RAILROAD BRIDGES 363

all of the panel points B to F, inclusive (Fig. 269), are loaded. As a
specific case, let us consider the portal at the left end of the bridge
(at bB) and let us assume that all of the panel points are loaded and
that the loads act as indicated in Fig. 269. Then assuming that the
diagonals carry only tension, the diagonal bC (the diagonal connecting
to the portal at 6) will be in tension and there will be no stress in
diagonal cB. One component of the stress in bC will be taken by the
portal and the other by the top chord be. The component taken by the
portal is applied (as is obvious) at b and is equal to the shear in panel
BC, the panel next to the portal. This force we will designate as R.
In addition, there will be one-half of a panel load at b and also one-half
of a panel load at B. Designating the panel load as P, we then have the
loading on the portal as shown in Fig. 271: (R+P/2)at point b and
P/2 at point B. If the pressure, or load, acted in the opposite direction,
the loading of the portal would be just the reverse, that is, the force
(R+P/2) would be applied at B and P/2 at b. These forces applied
to the portal are held horizontally by the two equal horizontal reactions,
one at the bottom of each end post. Let each of these reactions be repre-
sented by H and we have

H= [(#+F)+F]3-* shbph aud o) ocacaaays Gai aha garage anctale (1).

These two horizontal reactions at the bottom of the end posts and the
horizontal forces, (R+P/2) and P/2, at the top of the portal form a
couple which must be balanced by two equal and opposite reactions, V
applied along and at the bottom of the end posts. Now, as the forces
(R+P/2), P/2, the two H’s, and the two V’s balance, evidently, the
frame formed by the end posts and portal can be treated as an inde-
pendent structure in equilibrium under the action of these forces. So,
taking moments about u (Fig. 271), and considering the end posts and
portal combined as a frame, we have

+V= [(# +o)eel =(R+P)%,

 

or taking moments about w, we have
L
V= (R+P) 5 solo aia: secbie ta wa Us Ske oe eve eta rade ark (2).

In the designing of portals for bridges, there are two cases: one
when the end posts are considered as hinged at the bottom ends, and
the other is when the end posts are considered as fixed at the bottom ends.
The top ends of the end posts are practically always considered to be fixed
by the portal. In the case of light bridges, and especially when end
floor beams are not used, the end posts are not very rigidly fixed at the
bottom ends and the actual condition is most closely met by considering
the bottom ends of the end posts hinged, while in the case of heavy
bridges, with end floor beams and with wide shoes, there is no question
but that the bottom ends of the end posts are fixed.

In the case of the bridge we are designing, there is no question but
that the end posts will be quite rigidly fixed at the bottom ends, but for
the sake of illustration we will first determine the stresses in the portals.
364 STRUCTURAL ENGINEERING

assuming the bottom ends of the end posts to be hinged, and then deter-
mine them, assuming the bottom ends of the end posts to be fixed.

By imagining the part to the left of section n-n (Fig. 271), including
the end post bu and a portion of the portal, moved bodily (forces and all)
to a new position we obtain the structure shown in Fig. 272 which is
acted upon by the known forces (R+P/2), H, —-V, and by the unknown
forces § and S1, S being the stress in member bo and §1 the stress in po.
The dotted members shown in the portal are assumed to have no. stress
in them as they are redundant members intended only to brace the main
members of the portal, which are shown in full lines, and hence the dotted
members are ignored in the calculations. By taking moments about p
(Fig. 272) we eliminate $1 and -V from the equation of moments, and
as the moments of H and (R+P/2) (about p) have the same sign, both
tending to produce clock-wise rotation about p, it is evident that the sum
of the moments of the two forces about p must be equal but of opposite
sign to the moment of S about p. So we have

Sk= (e+$) b+ Hm,

from which we obtain

P Hm
Sa hits Bett Ra a a db aoa ta at te Uh aca bere aig (3)

for the stress in portal member bo. Now it is readily seen from Fig. 272
that this stress Sis compression, as H and (R+P/2) both tend to produce
clock-wise rotation about p, and as S alone prevents this rotation it must

oO
Ke

  

Fig. 272 Fig. 273

evidently act to the left toward b, and hence is compression. In other
words, the member bo pushes to the left against the end post and un-
doubtedly is in compression.

To determine the stress, $1, in the portal member po, let us con-
sider it resolved at point p into two components, one (f) along the end
post and the other (4) perpendicular to the end post. Then taking
moments about b (Fig. 272), we eliminate -V, f, S, (R+P/2), and P/2
DESIGN OF SIMPLE RAILROAD BRIDGES 365

from the equation of moments and we have
(Hx L) -(hxk)=0,
from which we obtain

 

 

Fa Axl
k
for the component of $1 perpendicular to the end post. Then we have
Sl=hsec6= EXE seo lake ce eye eteass enagh (4)'

for the stress in the portal member (knee brace) po. Stress $1 is ten-
sion, as is seen by considering point b as the center of rotation; the force
H acting to the left would necessitate h acting in the opposite direction
or away from the post at point p, which indicates the component h to be
tension, and hence the stress $1 is tension.

The stress S1 can also be determined from the summation of the
vertical forces. In which case we have

V+f=0
(-V and f being the only vertical forces). From this we have
f=V,
and multiplying by sec we obtain

S1=fseep=V seep=(R+P)% sees sist Saye Pater als ore ace biena alee (5).

f equals V, and as these two forces must balance each other it is seen
that f acts as indicated in Fig. 272, which shows S1 to be tension.

By imagining the part to the right of section n’-n’ (Fig. 271), in-
cluding end post wB and a portion of the portal, moved bodily (forces
and all) to a new position we obtain the independent structure shown in
Fig. 273 acted upon by the known forces H, V, and P/2 and by the
unknown forées (stresses) S2 and 3. Taking t as the center of moments,
it is seen that P/2 and H both tend to produce clock-wise rotation, and as
this rotation is prevented by $2 alone, we have

Soxk= 7 k+Hm,

from which we obtain

for the stress in portal member Bo. This stress is tension, as is readily
seen, as it acts away from the end post at point B, that is, the member
Bo pulls on the end post at point B and hence is in tension.

Resolving (at point ¢) the stress $3 into two components h’ and f”’,
and taking moments about B, we have

HxL=h’xk,
366 STRUCTURAL ENGINEERING

from which we obtain

 

 

,_HxL
h’= — %
and multiplying by sec @ we have
S3=2' seod=2** seo Hees ee Aah ag tie ted acs (7)

for the stress in portal member ot.

By considering point B (Fig. 273) as the center of moments, the
stress S3 is seen to be compressive, as H acting to the left requires the
member ot to act to the right against the end post, and hence the member
ot is in compression.

As is‘seen by comparison, (4) and (7) are exactly alike. That
means that the stress in po is equal to the stress in ot, but, as stated
above, one is tension and the other is compression.

Now, having derived a formula for the stress in each member of the
portal, we will proceed with the determination of the stress in the mem-
bers. To do this it is first necessary to determine the value of P, R, H,
L, m, k, and sec@. P is given in the last article as 5,000#. R is equal
to the shear in the panel BC (Fig. 269) when all the panels are loaded.
Beginning at the center of the span, we have one-half of a panel load at
D and a full panel load at C, making in all one and one-half panel loads
as the shear in the panel BC, hence we have

 

R= - + 5,000 = 7,500 Ibs.
for this shear. Then from (1) we have
B= WOT EO00 = 6,250 Ibs

L is about 39 ft., as the height of the span is 30 ft. and the panel length
is 25 ft. It will be close enough, for designing, to take @ as 45 degrees.
Then, k=8 ft., m=31, and sec#=1.4. Now, using the above values,
from (3) we obtain

ee + 6,250 x a = 34,218, say 34,000 Ibs.

(compression) for the stress in portal member bo. From (4) or (7)
we obtain

S=7,500+

 

6,250 x 39
8

for the tensile stress in portal member po and compressive stress in ot.
From (6) we obtain
5,000 6,250 x 31
sa= + aa eae = 26,718, say 27,000 Ibs.
for the tensile stress in the portal member Bo. By comparing (8) and
(6) it is seen that the stress in member Bo differs from the stress in
member bo by the value of R.

S1=S3= x 1.4 = 42,655, say 43,000 lbs.
DESIGN OF SIMPLE RAILROAD BRIDGES 367

This completes the calculations for the stresses in the portals under
the assumption that the bottom ends of the end posts are hinged, and
we will now determine the stress in the portal, assuming the bottom ends
of the end posts fixed. To illustrate conditions, let e and g at (a) (Fig.
274) represent two thin wooden strips connected rigidly by another piece h.
If this frame be supported as shown at (6b) and a force F be applied,
the pieces e and g would bend as indicated at (b), and we thus have
an illustration of the case just considered .
where the end posts were assumed hinged at xf ft he
the bottom ends. Now, suppose the bottom
ends of the pieces to be buried in concrete as y é
indicated at (c) and the concrete be allowed 8
to set and the force F' be then applied. The
pieces e and g would then bend as indicated at
(c), and we thus have an illustration of the
case to be considered where the bottom ends
of the end posts are assumed to be fixed. The =
points o’ are the points of contra-flexure, where :
the bending moment on the pieces e and g is
zero, as is readily seen from the sketch. The ( b ))
points o’ are at mid-distance from the con-
crete to the piece h, or m/2 from h, neglecting
the portion buried in the concrete as being too
small to materially change the length of the
part marked m.

Now, as there is no bending at the points
o’ (points of contra-flexure), the frame can be
assumed as cut off at those points and we have
the independent structure shown at (d). From
this it is seen that the problem of determin-
ing the stresses in portals, when the bottom
ends of the end posts are considered fixed, is
the same as in the case of hinged ends, except
the distance m (Figs. 271 and 273) is one-
half as great, and hence the above formulas for
determining the stresses in the portals are ap- Fig. 274
plicable if m/2 be substituted for m and
(L—m/2) or (k+m/2) for L. In an imaginary sense, we simply move
the bottom supports of the end posts half way up their unsupported
length, and proceed with the determination of the stress in the portals in
the same manner as if the bottom ends of the end posts were hinged.
Thus substituting in (3), we have

3
S=7,500+ oe 6,250 x = = 22,109, say 22,000 lbs. (compression)

 

 

 

 

 

 

for the stress in portal member bo (Fig. 271), assuming the bottom ends
of the end posts fixed; and substituting in (4), we have

Si= 6.250% GO) x 1.4= 25,703, say 26,000 Ibs.

for the tensile stress in portal member po and also for the compressive
368 STRUCTURAL ENGINEERING

stress in portal member ot, and likewise substituting in (6) we have

BL
s2= a+ Hm/2 — 5,000 2 6,250 x 33

 

 

ko ~ 2 8
for the tensile stress in portal member Bo. Thus we have all the stresses
in the portals determined when assuming the bottom ends of the end posts
to be fixed. These stresses will be used in designing the sections of the
portal members as the bridge considered is designed for a heavy loading;
and we have provided end floor beams, consequently the bottom ends of
the end posts will be fixed quite securely.

Each member of the portals will be as deep as the end posts and
composed of two angles, latticed together vertically, similar to the top
laterals.

As seen above, the members are subjected to one kind of stress
when the wind load comes from one direction and just the reverse when
it comes from the opposite direction, that is, each portal member is sub-
jected to reversal of stress, and hence the members must be designed to
carry both compression and tension.

The members po and ot (known as knee braces) are each subjected
to 26,000# stress, considered to be either tension or compression. Taking
the case of tension first, we have

26,000
16,000

for the required net area of cross-section of each, which shows that two
very small angles could be used (theoretically) as far as tension is con-
cerned. As for compression, the length of each member is about 11.3 ft.
(9=45°), but as the redundant members (those dotted, in Fig. 271) sup-
port each at mid-point, the length L can be taken as 11.3/2=5.6 ft., say
66 ins. As the members are subjected to reversal of stress, L/r should
be reasonably low, say 80. Then we have
_ 66
7 20
for the required value of the radius of gyration. .

Now, glancing over the table of angles (Tables 5 and 6) it is seen
that this calls for very small angles. So the case here is the same as
that of the top lateral struts—very much a matter of judgment. 34”x
34” x 3” angles are as small as are used in this class of work (as it is
desirable to use {” rivets throughout), so we will try that section. Sub-

stituting the radius, found in Table 6, for this angle in the column
formula, we have

= 14,609, say 15,000 Ibs.

 

= 1.62 sq. ins.

= 0.82

66
1.07
for the allowable unit compressive stress in each of the members po and
ot. Dividing this into the stress, we have

26,000 + 11,682 = 2.22 sq. ins.

for the required area, while 2—Ls 34” x 34” x 2” have 4.960”, more than
twice the amount required for compression, and, as found above, only
1.624” is required for tension. So it is seen that 34x 33” x 2” angles

16,000- 70 ——. =11,682 Ibs.
DESIGN OF SIMPLE RAILROAD BRIDGES 369

are much heavier than required, theoretically, yet smaller angles would
give the portals a flimsy appearance even if satisfactory details were
obtained by their use. So we are now to the point where the selection
must be made mostly upon appearance. As the stress in the members
bo and Bo is less than in members po and ot, and their lengths are also
less, it is not necessary to compute the required sections for those mem-
bers, but use the same as used for members po and ot. As the 34” x34”
angles are our minimum size we could use two of these for each portal
member, but for the sake of details and appearance we will use 2—Ls
4” x4” x2” for each of the main members and 2—Ls 34x34” x 3” for
each of the redundant members, thus giving the main members a dis-
tinctive outline.

In the case of large bridges these members can be designed theoretic-
ally correct, as the required area of cross-section in those cases is such
that sections can be found which are both theoretically and practically
correct, but in the case of small bridges it is very much a matter of
judgment, as is seen from the above calculations.

180. Design of End Posts—The designing of the section of the
end posts is here taken as a special problem for the reason that, unlike
the other main truss members, it is subjected to bending stress in addi-
tion to direct stress. To correspond with the top chord, it will be made
of 18” webs, 23” cover plate and four angles. The length of the end

 

 

 

 

 

8
\
Ys 8
0
A mM
Fig. 275

post, considering the 2-« axis (see Fig. 265) is the total length L (shown
in Fig. 275 to equal AB) unless a collision strut OM be inserted, in which
case the length is L/2 and the length, considering axis y-y, is equal to m.
The collision strut should be used as a safeguard in case a derailed train
should strike the end post, and therefore we will use one in this case.
Then the length of the post in reference to the z-x axis is 19.5 ft. or 234
ins., and 31 ft. (=m) or 372 ins. in reference to axis y-y.

Using the radii found for top chord be (Art. 176), which will be
about the same in this case, we obtain

234
16,000 — 70 7287 13,750 Ibs.

for the allowable unit direct compressive stress in the end post, consider-

ing axis 2-2, and
372
16,000 — 70 7.84= 12,678 lbs,
370 STRUCTURAL ENGINEERING

for the allowable unit direct compressive stress considering axis y-y.
Now dividing the latter, as it is the smaller, into the total direct stress
in the end post, as given in Fig. 262, we obtain

523,000 ‘

12,678 41.2 sq. ins. (about)
for the required area of cross-section of each end post, providing there
were no cross bending, due to the wind load, on the posts. But as the
bending stress on the compression side of the posts adds to the direct
stress, the two unit stresses must be combined. However, in case these
stresses be combined, the allowable unit stress can be increased 25 per
cent, so the above area may be about correct. As preliminary let us try
the section used for top chord bc, which contains 40.580”.

In determining the bending moment on the end posts, which is due
to wind—and consequently a transverse moment—we can consider each
post as composed of two cantilevers as indicated at (a) in Fig. 275.
Then the moment on each post at any point either above or below the
point of contra-flexure (0’) is Hz. Now it is seen that this moment is a
maximum at the points p and ¢ and at the bottom of each post, and that
the maximum in each case is equal to Hm/2. As the points p and ¢ are
the farthest out from the ends of the member, where the column formula
really applies, we will consider the bending at those points only. From
the last article, we have H=6,250 lbs. and m=31 ft. or 372 ins. Then
for the moment at p or t (Fig. 275 (a)) we have

M =6,250 e = 1,162,500 inch Ibs.

Now using the moment of inertia found in Art. 176 for chord be in
reference to axis y-y and taking the horizontal distance to the extreme
fiber as 7.75+4=11.75” (see Fig. 265), we obtain

re 1,162,500 x 11.75 é

~ 2,497

for the unit stress due to cross bending, assuming that the same section
as used for chord be is used for the end post.

Now dividing the area of the section into the total direct stress,
we have

=5,470 lbs.

523,000
40.58

for the actual direct unit stress, and adding this to the unit stress due
to cross bending we have

5,470 + 12,880 = 18,350 Ibs.

for the combined unit stress. The allowable unit stress as given above is
12,678. But this can be increased 25 per cent in the case of combined
stress, as per specifications. So we have

(12,678) 1.25 =15,847 Ibs.
for the actual allowable combined unit stress, and as it is less than the

above (18,350*) a larger section will have to be used for each end post
than was just considered.

= 12,880 Ibs. (about)
DESIGN OF SIMPLE RAILROAD BRIDGES 371

Let us assume the following sections:

1—coy. pl. 23’ x4” =11.500”
2—web pls. 18” x 4” =18.900”
2—Ls 34” x34"x3”= 4.960”
2—1s 67% x4"’x 8" =11.720”

46.180”

As the post is stronger in reference to the z-# axis than in reference to
the y-y axis, owing to the length being greater in the last case (using
the collision strut), we need consider the above section only in reference
to the y-y axis.

For the moment of inertia about the y-y axis (see Fig. 276), we have

cov. pl. 507 +0 = 507 (about)
web pls. 0+7.62 x9x2 = 1,045
top angles 2.87x2+8.88 x2.48x2= 397
bot. angles 7.52x2+8.90 x5.86x2= 943

2,892

Then for the radius of gyration about axis y-y we have

va, P82 7.92
46.18

Now substituting this radius in the column formula, we have
p=16,000—70 2" ~ 12,710 (about)
7.92

for the allowable unit.stress for direct compression,
and increasing this 25 per cent we obtain 15,890#
for the allowable unit stress in the case of direct
compression and bending stress combined. Then,
using the last assumed section, we have

523,000
46.18

for the actual direct unit compressive stress, and

p= 162,500 x 11.75
= 2,892

for the unit stress due to bending. Now, adding these two stresses
together, we have 11,320+4,730=16,050# for the combined unit stress,
which is only 160# (=16,050-—15,890) greater than the 15,890# allowed,
and as this is about as close as we can obtain we will use the above
section for each end post.

181. Designing of Collision Struts—These members should at
least have sufficient section to resist the thrust that an ordinary loaded
car moving at an ordinary rate would exert upon them in case the car
were derailed. The problem involved is of such a nature that the entire

 

 

   

= 11,320 Ibs. (about)

403
Fig. 276

= 4,730 (about)
372 STRUCTURAL ENGINEERING

premises must rest upon practical assumptions. We will assume that the
car and its load weigh 100,000# and that it has a velocity of 30 ft. per
second, which is a little over 20 miles per hour, and, further, we will
assume the car brought to rest in one second. Then substituting in
Formula A (Art. 23), we have

_ 100,000 x 30
=

for the constant force required to stop the car in one second. But at
the beginning of contact the force would be zero and a maximum when the
car was brought to rest, so the maximum force exerted by the car would
really be twice the constant force required to stop it, or 94,000x2=
188,000#. This force would be exerted on the end post at a point about
9.5 ft. above the bottom chord (see Fig. 277), so that the horizontal force
exerted against the collision strut would be about 9.5/15 of the 188,000,
or 119,000#.

Then multiplying one-half of this by sec 8, which we will take as
1.56, B being the complement of 6, we have

119,000
2

for the stress in the collision strut. It may appear that the collision strut

should connect to the end post at the point where the 188,000# is applied,

but this is not the case, for if the blow

2 of 188,000# were struck at the point

where the strut connects to the end

«Ae , | post there would be but little resilience

vest > 208 0} and, consequently, the structure would
E 1880 6 ty

F = 93,750, say 94,000 Ibs.,

S= x 1.56 = 92,820, say 93,000 Ibs.

be subjected to severe shock.

% It is seen from the amount of siress
oy

15707

that the section of the collision struts
25’ need not be large as far as direct stress
Fig. 277 is concerned, but the value of L/r should
not exceed 80 in order to obtain rigidity.
The best section for these struts is two channels latticed together, and
as 8” channels are about the smallest used in this class of work we will
try 2—[s 8”x 16.25%. The length of each strut, center to center of
end connections, is about 19.5 ft., or 234 ins., and, as seen from Table
3, the radius of gyration of the channels is 2.89, so we have
L234 _
r 2.89
Then for the allowable unit compressive stress on each strut we have
p=16,000 -70 x 81= 10,330 lbs.
Now, dividing this into the stress, we have

93,000
10,330

for the section required. The area of the section of the two channels

 

 

 

 

81.

 

= 9.0 sq. ins.
DESIGN OF SIMPLE RAILROAD BRIDGES 873

assumed, as seen from Table 3, is 9.565”, which is 0.560” more than the
required area, but we will use this section as it is about as close to the
required area as we can come and at the same time obtain good details.

The above analysis is only an attempt to provide for reasonable
accidents, as we might term it. The actual intensity of the blow struck
by a derailed car is distinctively problematic for the reason that there
could be so many different conditions assumed. As an illustration, sup-
pose a derailed car in the middle of a train should strike the end post
of a bridge. How much would the cars in the rear of the derailed car
increase the blow? And how much would the engine pulling on the
train increase it? And how much would the resistance of the wheels of
the derailed car bumping over the ties diminish it? And, further, sup-
pose a train traveling 50 miles per hour should jump the track just as
it gets to the bridge and the engine should strike the end post. There is
little doubt but that the bridge would be destroyed, yet it might glance
off and do but little damage. It is obvious that it would be absolutely
impracticable to provide for resisting the blow in the last case; and prac-
tically the same is true in several cases that could be assumed, yet it
would not be good engineering to make no provision for réasonable acci-
dents, as past experience has taught us, and that is what we have
endeavored to do above.

182. Maximum Reaction on Shoe.—For the dead-load reaction on
each shoe we have one-half of a panel load from D (Fig. 249), a full
panel luad from B and C each, and one-half (so considered) of a panel
load from A, making in all three panel loads on each shoe. Then taking
the dead-load panel load as found in Art. 173, we have

R=3x 26,375 =79,125, say 80,000 Ibs.

for the dead-load reaction on each shoe.
As is evident, the maximum live-load reaction on a shoe will occur
when the span is fully loaded and the heaviest wheels are near the shoe

& ec o C

 

49! '

 

 

 

 

 

 

 

f |
3 TTL
A lr B Cc D E F G

Fig. 278

and one directly over it—that is, over the end floor beam. So, placing
the live load as shown in Fig. 278 with wheel (2) at A and taking
moments about G, using Table A, we have

—2
ae (eS )1,000 <2 SS = 259,175, say 259,000 Ibs.

for the maximum live-load reaction on the shoe at A, which is the same
as for the others, and, for the impact, we have
300

259,000 x $35 = 172,666, say 173,000 Ibs.
613 “Sh

 

 

  

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874

 
DESIGN O! SIMPLE RAILROAD BRIDGES 875

Now adding the above dead, live, and impact reactions together, we have
80,000 + 259,000 + 173,000 = 512,000 Ibs.
for the maximum reaction on each shoe.

183. Stress Sheet.—After having completed the calculations as
above for the span the “stress sheet,” Fig. 279, can be made. This draw-
ing contains all the information resulting from the calculations; in fact,
the drawing is really a summary of the calculations. After the stress
sheet (Fig. 279) is completed the detail drawings can be made.

184. Calculations of Details—In starting the detail drawings, a
large scale pencil sketch (14, 2” or 3” scale) of each joint should be
made first. Such a drawing for joint LO is shown in Fig. 282. The
calculations of the details are made as the sketches are being drawn.

Joint LO. Taking first the case of joint LO (Fig. 282), the first
thing to do is to locate the center line of the pin in the end post. The
section of the end post is given on the stress sheet (Fig. 279), also in
Art. 180.

Taking moments about the center of the cover plate of the section

(see Fig. 280) we have

1.26 x 4.96 +18xX9.37411.72x 16.47 367.9
46.18 ~ 46.18

for the distance from the center of the cover plate down to the gravity axis
z-2 of the end post. Now, as far as the direct stress in the end post is
concerned, the pin could be properly placed on this gravity axis 2-2; but
the member tends to bend downward as shown in Fig. 281, owing to its
own weight, and it is necessary to place the pin a short distance below
the gravity axis so as to counterbalance this bending. The pin being
placed below the gravity axis 2-x, the direct stress tends to bend the
member upward as indicated by the dotted line, and, as is evident, if the
pin be placed just the correct distance below the gravity axis the end post
will be straight when the maximum stress in it occurs. For the weight of

the end post per foot of length we have
Y le ‘

 

z= = 7.96 ins.

 

 
 

i
2796
citi

a

£8.25
[ .
ad LE-F

 

 

 

 

 

 

 

Fig. 280 Fig. 281

W = 46.180” x 3.4# = 157 Ibs.,

and increasing this one-third to provide for details we have 210# for the
total weight of the member per foot of length. This weight acts vertically
and, as the member is inclined, only the component perpendicular to the
member causes transverse bending. So, for the bending moment, we have
 

 

   

 

         

2E4%45 BF

   

»

 

 

Fig. 282

376

 
DESIGN OF SIMPLE RAILROAD BRIDGES 317

M =4x 210 x sind x 19.5° x 12 = 76,658 inch lbs. (Sind = 0.64.)

Now by placing the pin a distance 2 below the gravity axis 2-2, we
have 2x 523,000 for the moment caused by the direct stress which tends
to bend the end post upward, and hence the end post will be straight
when g X 523,000 = 76,658, from which we obtain

76,658 :
s= 523,000 ~ 0.1466 ins.,

which is the distance that the pin should be placed below the gravity axis
z-a. Now adding this to x (given above) we have 7.96 + 0.1466 = 8.1066”
for the distance from the center of the cover plate down to the center of
the pin, and substracting one-half of the thickness of the cover plate from
this we have 8.1066 — 0.25 = 7.8566” for the distance from the under side
of the cover plate to the pin. The distance from the under side of the
cover plate to the working line in the top chord, which would be the cen-
ter line of the pins if the top chord were pin-connected, should be the
same as for the end post in order to obtain uniform construction of these
members, and hence at this juncture it is necessary to examine the top
chord sections and select a distance suitable for all top chord members
and end posts.

Considering the case of the top chord section U2-U3 (Fig. 279), and
taking moments about the cover plate as was shown above in the case of
the end post we obtain

1.26 x 4.96 + 18 x 9.37 + 10.62 x 16.49

r= 45.08 =17.76 ins.

for the distance from the center of the cover plate down to the horizontal
gravity axis. For the weight of the section we have

45.08 x 3.4 = 153 lbs.

per foot and adding one-third for details we have 204# for the weight of
the member per foot of length. Then we have

M =4x204x 25 x 12=191,000 lbs. (about)

for the bending on the chord due to its own weight, and for the distance

z’ below the horizontal gravity axis, where pins would have to be placed
to balance the above moment, we have

, . 191,000

2! =

589,000

Adding this to the above 7.76” and subtracting one-half of the thickness

of the cover plate we have
7.76 + 0.325 — 0.25 = 7.84 ins.

for the distance down from the under side of the cover plate to the work-

ing line. In the same manner we find that the corresponding distance in

the case of chord U1-U2 is 7.75”. So it is seen that 7%” is about the cor-

rect value to use for the distance from the under side of the cover plate to

the pin center or working line, as the case may be, and hence this distance

will be taken in the case of all top chord sections and end posts throughout

the structure. Thus we have the pin LO located and the outline of the

portion of the end post shown (Fig. 282),can be drawn and the rivets

 

= 0.3825 ins.
378 STRUCTURAL ENGINEERING

spaced transversely in it as shown and then the drawing of the other
details can proceed. The end post should extend far enough beyond the
pin so that the bottom chord and end floor beam connections balance fairly
well about the pin. This part is at first determined merely by appearance.
After this the shoe can be sketched so as to clear the end post, the number
of rollers required is determined by sketching roughly their length, and at
the same time the length of the pedestal can be determined. After this is
done the required pin bearing on the end post can be figured. This bear-
ing must be sufficient to carry the maximum vertical reaction, which is
512,000# (see stress sheet, Fig. 279). Let us assume a 54” pin. Then
we have

t = 512,000 + (54 x 24,000) = 3.87 ins.

for the required thickness of bearing on the two sides of the end post or
1.93” (=14%) for each side. As shown, we have a 2” gusset plate, 3”
web, §” filler, and a 7%” plate, making in all, 3+ 44+ 3+ gg = 23%”, which
is 4” thicker than required. But this is about as close as we can come,
as the gusset plate should be 3” thick (as will be seen later) and the 3”
filler can not be reduced, as the bottom angles on the end post have that
thickness and the 7%” plate is as thin as should be used, owing to the coun-
tersunk rivets (%” rivets should not be countersunk in metal less than
qg’’ thick). So we will use the metal shown. There should be enough
rivets above the pin to hold these plates. First, there should be enough
rivets connecting the 2” filler and the 7%” plate to the end post and gus-
set plate to transmit the pin pressure coming on the two. For this pres-
sure we have (3+ 7) 54 x 24,000 = 140,000# (about). The rivets are {”
shop rivets in single shear, each of which is good for 12,000 x 0.6 = 7,200#.
So we have 140,000+7,200=19.5, say 20 rivets. We have just about
20 rivets above the pin, so the detail is satisfactory so far. The rivets in
the 4” filler are not included, as they simply hold that narrow filler.
There should be enough rivets in the 7%” plate to take the pin pressure
upon it, which is 7% x 54 x 24,000 =57,750%. Dividing this by 7,200 we
have 57,750 + 7,200 =8 rivets—whereas we have over twice that number,
and that part is amply strong. To be on the safe side there should
be enough rivets connecting the end post to the gusset plate to trans-
mit, in single shear, one-half of the stress in the end post, which is
523,000 + 2 = 261,500 (see Fig. 279). Dividing this by 7200 we have
261,500 + 7,200 =36.3, say 37 rivets. We have about 38 above the pin,
so that part of the detail is satisfactory. We will next consider the bot-
tom chord connection. The net area of the bottom chord is 20.90” (see
Fig. 279), and for its strength we have 20.9 x 16,000 = 334,400#. Then
as the rivets are {’ field rivets, we have 334,400 + 6,000 = 55.7, say, 56
rivets, or 28 on each side. We have 30, so that detail is correct.

The end floor beam is cut, as shown (Fig. 282), to clear the shoe, pin,
and the details on the end post. Dividing the end shear on the end floor
beam (see Fig. 279) by 6,000 we have 143,200+6,000= 23.8, say 24
rivets for the number of {” field rivets required in each end connection.
As is seen, 24 rivets are used. The thickness of pin bearing on the shoe
need be (theoretically) but 148” on a side, as found above for the end
post, but there should be a point of bearing on each side of each web of
the end post (as shown) so as to distribute the pressure well over the
DESIGN OF SIMPLE RAILROAD BRIDGES 379

rollers and at the same time lighten the stress on the pin, and as every
part of such castings should be at least 14” in thickness, we will make
each of these bearings 14” thick, although we obtain an excess of bearing.
However, as the shoe is subjected to both bending and shear from the
direct load (see Art. 141) and wind, the metal is not so overly excessive
as at first appears.

Assuming one-fourth. of the maximum reaction on the shoe, which is
equal to 512,000 + 4=128,000#, as coming on each bearing of the shoe,
and taking moments about the center of bearing of the end post, we have

128,000 x 23 = 304,000 inch Ibs.

for the maximum bending moment on the 5$” pin. The lever arm 23”
can be either computed or scaled from the drawing. For the bending
stress on the pin, due to this moment, we have
fe 304,000 x 24 _
© (7/64) (53)*~
which shows the pin to be amply strong, as 25,000# per square inch is
allowed. It requires a moment of 408,350’’# to stress the 54” pin 25,000#
per square inch. The maximum bending moment allowed on the different
sizes of pins is to be found in tables of practically all structural hand-
books, and, in designing pins, usually these tables are consulted instead
of computing the fiber stress as is done above.
For the maximum shearing stress per square inch on the pin we have

128,000 + ar? = 128,000 + 23.75 =5,390 lbs. per sq. in.,

which is quite low, as 12,000# is allowed. As seen from above, the pin is
really larger than required and, theoretically, should be made smaller;
but if reduced in size the bearing would increase, little would be saved,
and, judging from general appearance, a 54” pin is as small as should
be used in this class of work; hence the assumed 54” pin will be used.

Six-inch rollers are the minimum size permitted by the specifications,
and as that size appears to be about the correct size for this case, they
will be used. Then, for the linear inches of roller required, we have

512,000 _ a2 :
G00 x6 > 142.2 ins.

Now, making each roller 30” long and deducting for the 7—3” slots in
pedestals, we have (30-7 x 2) x 6=148.5”, which is a few inches more
than is required, but is about as close as we can come and at the same time
obtain good details all around. So 6—6” rollers, each 30” long, as shown,
“will be used. The rollers are notched }” into the cast shoe and pedestal,
so as to prevent their sliding transversely (due to wind), and the side
bars on their two ends control their relative longitudinal motion. It is
necessary that the pedestal and the base of the shoe project to clear the
side bars and bolt heads. In some cases it is necessary for the pedestal
and the base to project out farther to obtain sufficient bearing on the
masonry. In this case we have

512,000
600

18,600 lbs. per sq. in. (about)

 

A= = 853 sq. ins.
380 STRUCTURAL ENGINEERING

for the required area of bearing and we have, as shown, (3’ — 44”) x
(2’ — 44’) =1,1549”, which is about 3004” more than required, but the
roller nest requires about the same size of pedestal as shown and there
would be but little saving, if any, in reducing it to exactly the theoretical
size. So the size shown will be used.

In drawing the lateral connection at LO, we first determine the num-
ber of %” field rivets necessary to develop the lateral in tension. The
lateral is a 5” x 34” x 3” angle, which has a net area of 5.060” (allowing
for 1—1” rivet hole). Multiplying this area by 16,000 and dividing by
6,000 we obtain 13.5, say 14 rivets. As is seen, 14 are used. The bolts
connecting the 7%” lateral plate to the shoe should be sufficient to carry
the component (along the end floor beam) of the stress in the lateral.
This component is equal to 81,000 + seew = 81,000 + 1.85 = 43,800# (about)
(see Art. 177). Dividing this by 6,000 we obtain 7.3 for the number of
¥’ turned bolts required. As is seen, 7 are used. Part of the component
(43,800#) may be transferred along the floor beam to the other shoe, in
which case less than 7 bolts would be required; but by using the 7 we are
sure that the detail is sufficiently strong.

For the component of the stress in the lateral along the bottom chord
we have

81,000 x sinw = 81,000 x 0.8418 = 68,200 Ibs. (about).

Dividing this by 6,000 (the allowable shear on a ¥” field rivet) we obtain
11.3, say 12 rivets. We have, as shown, 14. However, some of these
are also needed to transmit the direct stress of the bottom chord into the
gusset plate, but as the maximum stress in the bottom chord and lateral
are not likely to occur at the same time, the detail as shown will be
considered sufficient. The shop rivets connecting the 2—4” x 4” angles
to the 37” lateral plate should be sufficient to transmit the 68,200# com-
ponent in bearing on the 7%” plate.

One {” rivet at 24,000# bearing on a 7%” plate will transmit
2x qe x 24,000 = 9,188%. Then we have 68,200 + 9,188 = 7.4 for the num-
ber of rivets required, and, as is seen, 8 are used. The number of rivets
required to connect the 7” lateral plate to the bottom of the end floor
beam should be sufficient to take at least one-half of the component along
the floor beam of the stress in the lateral. This requires about 4 rivets,
but to obtain a good, rigid connection, more are needed. So 8 are used,
wherein we are governed very much by appearance and sense of fitness.

The 34” longitudinal holes in the cast-steel pedestal should be cored
in casting and the }” slots cut at the machine shop. If the slots were to
be made in casting, the pedestal would be likely to warp in cooling. These
slots and holes in the pedestal are for two purposes: one is to save metal;
and the other one is to permit dust and dirt that would accumulate between
the rollers to fall down into the holes. In other words, this arrangement
permits the dirt that is blown into the roller nest to be blown out, and thus
the roller nest is not likely to become clogged with dirt and rust. The
side bars at each end of the rollers should be at least £” thick to be of
much service. ;

Joint L1. Taking the next lower chord joint L1 (Fig. 283), the
cross-section of the floor beam and end connection angles of the same can
be drawn as shown, and the outline of the bottom chord collision strut and
88% “31a

 

 

 

 

 

 

381
382. : STRUCTURAL ENGINEERING

the hanger can be sketched—care being taken to provide about 34” clear-
ance between these members. Then the rivet lines can be drawn and the
rivets spaced as shown. The stress in the collision strut as given in Art.
181 is 93,000#. Dividing this by 6,000# we obtain 15.5, say, 16—3”
field rivets or bolts, as bolts should be used in this particular connection.
As is seen, 16 are used. Dividing the maximum end shear on the floor
beam, which is given on the stress sheet (Fig. 279) as 187,300, by 6,000
we obtain about 31 rivets for the number of {” field rivets required to con-
nect the floor beam to the truss. All of these should be above the bottom
chord, as there is no diaphragm in the bottom chord to transmit the
pressure across to the other side of the hanger and, consequently, the
rivets shown connecting the floor beam to the bottom chord must not be
considered to carry any of the end shear on the floor beam. As is seen,
there are 32 rivets connecting the floor beam directly to the hanger,
which is really one more than required theoretically.

After spacing the rivets in the collision strut and floor beam, the
gusset plate can be drawn to suit this spacing as shown. The rivets
connecting the gusset plate to the bottom chord should at least be suffi-
cient to transmit the component of the stress in the collision strut which
is cqual to 98,000 x sind =59,500#. Dividing this by 7,200 we obtain a
little over 8—{” shop rivets, or 4 on a side, whereas 12 (not including
the field rivets passing through the floor beam) are used, but the detail
is satisfactory, as this number is necessary to obtain a well-balanced
joint. Next, the bottom lateral connections can be drawn. The number of
rivets required in the lateral to the left of Z1 is 14—¥{” field rivets, as
determined at LO, and the detail can be drawn as shown. The lateral on
the right-hand side of Z1 (Fig. 283) is made of 1—5” x 34’ x}” angle,
which will have a net area of 3.54”. Then for the strength of this lat-
eral we have 3.5 x 16,000 =56,000#. Dividing this by 6,000 we obtain
about 9—¥” field rivets. This, as is seen, is the number used. The num-
ber of rivets connecting the lateral plate to the floor beam should be suffi-
cient to transmit the component of the lateral to the left of L1 along the
floor beam. This component is equal to 5.06 x 16,000 x cosw = 80,960 x
0.5397 =43,700#. Dividing this by 6,000 we obtain about 7—3” field
rivets, whereas 10 are used. After the rivets are spaced in the laterals
the lateral plate can be drawn to suit; that is, so the edges at no point
will be less than 13” from the rivets. Then the rivets connecting the lat-
eral plate to the floor beam can be spaced as shown.

Joint L2. We will next consider lower chord joint L2 (Fig. 283).
The floor beam connection at this joint must be exactly the same as al
L1, as the floor beams should be the same throughout. However, if the
spacing established at L1 does not suit at L2, it must be changed so it
will be satisfactory for the two joints and, likewise, for the other joints.
In each case there must be 32 rivets above the chord as was determined
at L1. After drawing the cross-section of the floor beam and the end
connection, as shown (Fig. 283), the outlines of the bottom chord ana
diagonal can be drawn. The net area of the main diagonal, as given on
the stress sheet (Fig. 279), is 23.99”. Multiplying this by 16,000 and
dividing by 6,000 we obtain about 64— 7” field rivets for the end con-
nection, or 32 on a side, which, as is seen, is the number used. There
should be a sufficient number of rivets connecting the gusset plates to
DESIGN OF SIMPLE RAILROAD BRIDGES 383

the post to transmit the vertical component of the stress in the diagonal,
which is equal to 23.9 x 16,000 + sec6=294,000#, Dividing this by 6,000
we obtain 49—j” field rivets or, say, 24 on a side. As is seen, 24 on a
side are used.

There should be a sufficient number of rivets to the left of the floor
beam connecting the gusset plate to the chord (at L2) to transmit the
horizontal component of the stress in the diagonal. For this component
we have 23.9 x 16,000 x sind = 244,700, and dividing by 7,200 we obtain
34—” rivets (shop), or 17 rivets on each side of the chord, whereas 16
are used, but the rivets passing through the floor beam and those to the
right of the floor beam can be counted on for more than making up for the
one rivet missing.

The splice in the bottom chord just to the left of L2 should be suffi-
cient to develop the strength of the bottom chord member L1-L2. The
chord is spliced, as is seen, by the 12” x 7%” inside plates, by the 12” x 4”
outside splice plates and the 2’ tie plates. The rivets through the web
of the channels to the left of the splice, which are field rivets, are in double
shear and bearing on the web of the channels. The strength of the chan-
nels is equal to 20.9 x 16,000 = 334,000#, or 167,000# for each channel.
Then considering one channel, the 16 field rivets in the web are good for
192,000# (=2 x 6,000 x16) in double shear, and 145,600# (=0.52x $x
16 x 20,000) in bearing on the web of the channel; and hence the latter
governs. The 6 rivets in the flanges of the channel are good for
6 x 6,000 =36,000#. Adding this to the 145,600# we have 181,600#,
which is 14,600# more than is necessary to provide for; or, in other words,
there is at least one rivet too many; but the rivets in the tie plates are none
too numerous to hold those plates and to omit one rivet in the web would
leave the spacing unbalanced; that is, unsymmetrical, and to omit two
would be too much of a reduction, so the rivets as shown will be used.
There should be enough rivets in the 12” x 4” splice plate on each side
of the splice to develop the strength of that plate. The net section of the
plate is equal to 12x4-1=50”. Multiplying this by 16,000 we have
80,000# for the strength of the plate. Dividing this by 7,200 we obtain
11 for the number of required shop rivets, and dividing it by 6,000 we
obtain 13.3, say 14, for the number of required field rivets. As is seen,
we have 12 shop rivets on one side of the splice and 16 field rivets on
the other side, so that the riveting as far as the outside splice plates are
concerned is quite satisfactory. .

The net area of the 12” x7,” plate is 5.25-0.87=4.389”, and the
net area of the 12” x4” plate is 55”, making a total of 4.38+5=9.380”
in the splice plates, while the net area of the 15”x40* channel is
10.450”; but the area of the tie plates can be counted to the extent of 6
field rivets. This is equal to 6,000 x6+16,000=2.250”. Adding this to
the area of the splice plates we obtain 11.634” for the total net area of
the plates splicing each channel—which is quite sufficient.

The number of rivets in the laterals, shown at L2 (Fig. 283), is
obtained by developing the laterals. For example, the lateral to the left
of the floor beam is a 5”x34”x4” angle which has a net section of
4-—0.5=3.50". Multiplying this by 16,000 we obtain 56,000# for the
strength of the lateral, and dividing this by 6,000 we obtain 9—{” field
¥8o “StL

 

 

 

 

 

384
DESIGN OF SIMPLE RAILROAD BRIDGES 885

rivets, whereas 9 are used. The numbcr of rivets in the other laterals is
obtained in the same manner.

Joint L3. Next we will make a sketch of the details at the joint
3, as shown in Fig. 284. In drawing this sketch, we can draw the
outline of the bottem chord, diagonals and post. Then we can draw the
end connections of the floor beam onto the sketch and draw the rivet
lines in all the members, and then we are. ready to determine the number
of rivets and spacing of same for each member connecting at that joint.

The rivets in the connection of the two diagonals will be the same
for each. These diagonals are subjected to both tension and compres-
sion, and, according to the specifications, the sum of the two stresses
should be used in determining the required number of rivets in the end
connections. So we have (201,000+ 78,000) =279,000# for the stress to
be considered (see Fig. 279). Dividing this by 6,000 we obtain 46.5—{”
field rivets or, say, 24 on each side of each diagonal, and, as is seen, 24
are used. The rivets on each side of the floor beam connecting the
gusset plate to the bottom chord should be sufficient to transmit the com-
ponent of the stress in the diagonal along the bottom chord. This com-
ponent is equal to 279,000# x sin 6=279,000 x 0.64=178,000# (about).
Dividing this by 7,200# (the allowable single shear on a $” shop rivet)
we obtain 24.8 rivets or, say, 13 on a side and, as is seen, 14 are used.
The rivets in the diagonals should be spaced first and the edge of the
gusset plate located and drawn down to the bottom chord, and the rivets
arranged in the bottom chord to suit the gusset plate. Then the next
thing to do is to determine the rivets connecting the floor beam to the
truss, as was explained above in the case of panel points L1 and L2.

After having the rivets spaced in the floor beam connection, the
gusset plate can be drawn completely; however, the spacing in the floor
beam connection selected for this joint must be the same as for the other
intermediate lower chord joints. The bottom lateral connections can
next be drawn. The rivets in each lateral are determined by developing
the strength of each lateral as was shown above for the other joints.
After the rivets are spaced in the laterals the lateral plate can be drawn
to suit the spacing; however, the rivets in the bottom of the floor beam
should be the same as required at L1, in order to have all of the inter-
mediate floor beams alike.

Referring to the sketch of the intermediate floor beam, drawn just
to the right of L3 (Fig. 284), the rivets connecting the end connection
angles to the floor beam must be sufficient to transmit the maximum end
shear on the floor beam in double shear or bearing, whichever requires
the greater number. These end angles are placed upon 4” fillers (the
fillers being just as thick as the flange angles). These fillers are held
by rivets placed outside of the connection angles, and hence the thickness
of bearing on the rivets through the angles can be taken as the total
thickness of the 2—}” fillers and web, making 13%”, or the combined
thickness of the two angles, which is the least thickness—as it is 1”. So
for the allowable bearing on each #’ shop rivet in this connection we
have 7x 1x 24,000 =21,000#, and for the allowable double shear on the
same we have 2x0.6 x 12,000 =14,400#, which is less than the bearing
and hence governs. Then dividing the maximum end shear by the last
386 STRUCTURAL ENGINEERING

figure we have 187,300 + 14,100=13 rivets, and, as is seen, 13 rivets are
used, not counting those passing through the flange angles.

There should be enough rivets connecting the stringer to the floor
beam to transmit the maximum end shear on a stringer. So we have
142,000 + 6,000 = 23.6 rivets in single shear, and, as is seen, 24 are used.
The bearing of these rivets is upon the 2—}” fillers and the 7%” web, so
that the shear governs. There should be enough rivets through these
fillers, outside of ‘the stringer connection, to carry the total end shear on
a stringer and also the concentration on the floor beam. In the first
case, the rivets are in single shear and for the number required we have
142,000 + 7,200=about 20 shop rivets, whereas we have 20. In the sec-
ond case, these rivets are in bearing on the 7%’’ web and for the number
required we have 187,300 +9,180=about 20 rivets, so the correct number
is used. As a matter of fact, the field rivets in the stringer connection
(theoretically) bear on the web also and hence we have an excess of 24
field rivets in bearing on the web in the connection shown, but it is a
question in the case of such long rivets, and especially when field driven,
just how much bearing they will exert on the web, and there is some
question as to the bending, and to be on the safe side it is advisable to
compute the number on this important connection as is here shown.

It will be seen that in the case of the end connection of the floor
beam there is an excess of 5 rivets in bearing on the web (not considering
the ones passing through the flange—they resist the flange increment)
and consequently 5 rivets could (theoretically) be omitted from the fillers,
but this would make the rivets look rather sparing, so the number shown
will be used.

The rivets through the fillers of the stringer connection are counter-
sunk on one side of the connection to permit the swinging of the stringers
in position during the erecting of the structure. The small angles under
each stringer (connected to the floor beam) are for erection purposcs
and are not assumed to carry any load from the stringers.

Joint U1. We will next consider the drawing of the hip joint Ul
shown in Fig. 285. The first thing to do in making this sketch is to
select the center point of the joint and draw the center lines of the erd
post and top chord. Then bisect the angle between these lines, and this
bisector will be the joint line—where the two members meet. Now, using
the same transverse spacing as was used in the end post at LO (Fig.
282), the outlines of the end post and top chord can be drawn and thic
rivets spaced. There should be enough rivets connecting the end post
to the gusset plates to transmit the total stress in the member. C£o
we have

523,000 +6,000=87, say 88,—2” field rivets,

or 44 on a side. As is seen, there are 38 in single shear, which provides
for 6,000 x 38=228,000* of the stress, and 4 passing throngh the small
outside splice plate which can be considered for bearing on the $” web
and hence provide for 8,750 x4=35,000# of the stress in the end post.
Adding these two stresses we have 228,000 + 35,000 =263,000%, and mul-
tiplying this by 2 we have 526,000, which is 3,000# more than the stress
in the end post, and hence the riveting as shown in the end post is suffi-
cient. There should be enotgh rivets connecting the top chord to the
S8z “St

 

 

 

 

 

387
388 STRUCTURAL ENGINEERING

gusset plates to transmit the total stress in that member. As is shown,
there are 33 shop rivets (}” diameter) which provide for 33 x 7,200=
237,600# of the stress in the top chord, and the 4 field rivets in bearing
on the 3” web which provide for 6,563 x 4=26,250# stress. Adding these
two quantities together and multiplying by 2 we have 263,850x2=
527,700#, which shows this riveting to be satisfactory, as the stress in
the top chord U1-L2 is 521,000# (see Fig. 279). Next, the hanger
U1-L1 and the diagonal U1-L2 can be dfawn and the rivets connecting
them to the gusset plates spaced as shown. The rivets connecting the
hanger to the gusset plates should be sufficient to develop the strength
of the hanger, which is 12.94x 16,000=207,000# (see Fig. 279). Then
we have

207,000 + 6,000 =34.5, say 36,—}” field rivets,

or 18 on a side. As is seen, 18 are used. The number of rivets con-
necting the diagonal U1-L2 to the gusset plates should be sufficient to
develop the strength of the diagonal, which is 23.9 x 16,000 = 382,000#,
Dividing this by 6,000# we obtain 64—{” field rivets, or 32 on a side;
whereas 32 are used, as is seen.

There should be enough metal in the gusset plates to properly
transmit all of the forces acting in the plane of the truss at the joint.

 

 

 

Fig. 286

These forces are as indicated in Fig. 286. As is readily seen, these
forces will not all be a maximum at the same time. About the most
severe stress will occur on the plates when the end post has a maximum
stress, as the top chord U1-U2 has about the maximum stress at that time
and the hanger U1-L1 has quite a large stress, and the diagonal U1-L2
has a low stress. Combining S and $3 (Fig. 286) we obtain the resultant
R2, and combining $1 and S2 we obtain the resultant R1, which is equal
and opposite to R2. So, evidently, the maximum tension on the plates
will be on same section as cd, and combining S and S2 we obtain the
resultant R3; and, likewise, combining S1 and $3 we obtain the resultant
R4, which will ke equal and opposite to R3. So, evidently, the maximum
DESIGN OF SIMPLE RAILROAD BRIDGES 389

compression on the gusset plates will be on some section as ab perpen-
dicular to these resultants (R3 and R4), due mostly to S and S3.' A
very satisfactory approximation can be obtained by making the plates
thick enough so that two-thirds of the metal along the section ab is suffi-
cient to transmit the maximum stress in the end post. By making the
plate 8” thick, two-thirds of the section along section ab is about 48 x 3x
-%x2=400”. Dividing this into the maximum stress in the end post, we
have 523,000+400”=13,100# (about) for the maximum compressive
unit stress in the plates, which is about the correct value, as it is about
the same as allowed in the chord U1-U2. So the gusset plates will be
made 3” thick. The same thickness will be used at all of the other
joints, whether required or not, in order to have the sides of the truss
members in the same plane throughout without the use of fillers.

After the connections of the main members at joint U1 are drawn
as shown (Fig. 285), the portal can be drawn. In drawing the portal,
the first thing to do is to draw a top plan of the end post and locate
the center of the portal. Then locate the clearance line as shown and
draw the cross-section of the portal on the elevation of the end. post (as
shown), and then draw the plan of the portal, keeping inside the clear-
ance line at every point.

The required number of rivets in the end connections of the portal
members is obtained by developing the strength of each of the members.
After the portal is drawn the top lateral and the bent lateral plates can
be drawn as shown, and thus the sketch of joint U1 is completed.

Joint U2. The details of this joint are shown in Fig. 287. The
outlines of the top chord as seen in elevation should be drawn first, and
then the outlincs of the diagonal (U2-L3) and the post (U2-L2) can be
drawn as shown. The next thing after this is to locate the rivets in the
end connection of the diagonal. As this diagonal is subjected to both
tension and compression, the number of rivets in each end connection
must, according to the specifications, be sufficient to transmit the sum of
the two stresses. So we have

(78,000 + 201,000) + 6,000 =46.5—2” rivets

for the required number, or say 24 on a side, and, as is seen, 24 are
used, the same as in the end connection of this same diagonal at L3 (see
Fig. 284).

Next, the cross-section of the transverse strut can be drawn on to
the chord and then the transverse view, shown to the right, can be
drawn. In drawing this view it is best to draw first the cross-section
of the top chord, as shown, and then the elevation of the transverse strut.
Next the clearance line should be located and dotted in, as shown, and
then the knee brace and sub-strut can be drawn, at which time the con-
nections of the knee brace and sub-strut to the post are determined and
can be projected over to the other view of the post as shown. The next
thing to do is to determine the number of rivets required to connect the
post (U2-L2) to the gusset plates.

This member is subjected to 162,000# compression and 54,0004
tension. So, according to the specifications, there should be a sufficient
number of rivets connecting the post to the gusset plates at U2 to trans-
mit the sum of these two stresses. So we have (162,000 +54,000) +
 

 

 

Fig. 287

390

 

LV eee
DESIGN OF SIMPLE RAILROAD BRIDGES 391

6,000 = 36—”" field rivets, or 18 on each side of the post, whereas 20
are used.

There should be enough rivets (at least) connecting the gusset
plates to the top chord to transmit the horizontal component of diagonal
U2-L3, This component, considering the sum of the stresses in the diag-
onal, is equal to (201,000+ 78,000) x sin 6=279,000 x 0.64=178,560#.
Dividing this by 7,200 we obtain about 25 shop rivets, say 13 on a side,
whereas there are almost three times this number shown (Fig. 287).
But the number cannot well be reduced, as the spacing of the rivets in
the chord angles is about what the specifications require, and the rivets
between the chord angles cannot be more sparingly spaced. So the rivet-
ng of the gusset plates to the chord will be considered satisfactory as
shown.

The next thing, the plan of the top chord, strut and laterals can be
drawn. There should be enough rivets in each lateral to develop the
strength of the lateral in tension. Each lateral is composed of 2—Ls
34” x34” x3”. So for the strength of each we have (4.96 —0.75) x
16,000 = 67,360%, Dividing this by 6,000 we obtain about 11 field rivets,
say 5 in each angle, whereas 5 are used—5 in the angle shown and 5 in
the angle at the bottom of the chord.

There should be enough rivets in each transverse strut to develop
the strength of the strut in compression. Each of these struts is com-
posed of 4—1Ls 34” x3” x2” arranged so as to form an I-section. So,
considering the 34” legs turned horizontally and the vertical legs }”,
back to back, we have

a 192 _
p=16,000-70 > =8,140 Ibs.

for the allowable unit stress. Then multiplying this by the area of the
4+ angles in each strut we obtain 8,140 x 9.2=74,888. Dividing this by
6,000 we obtain about 13 field rivets, say 7 in the top angles and 7 in
the bottom angles, whereas 8 are used so as to have symmetrical spacing.

The chord splice just to the left of joint U2 (Fig. 287) is what is
known as a butt joint. The chord sections are planed so they fit per-
fectly against each other and consequently the stress can be considered
as being transmitied from one chord segment to the other without the aid
of the splice plates, and hence the splice plates are considered as merely
holding the chord segments in position; and, as is evident, the size of these
plates and the number of rivets connecting them to the chord is mostly
a matter of judgment.

Joint U3. The details at this joint will be practically the same as
at U2, except the connection of the diagonal and chord splice are
omitted, and consequently no larger scale sketch is necessary.

This completes the necessary large scale sketches and next the gen-
eral drawing, Fig. 288, and the shop drawings, Figs. 289 to 303, can be
made, wherein the details are, as we may say, simply transferred from
the above sketches to these drawings.

185. Camber.—To prevent bridge trusses from sagging they are
“cambered,” that is, they are built so that they curve upward. In the
case of an ordinary truss bridge the cambering of the trusses is accom-
 
     
   
  

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408
DESIGN OF SIMPLE RAILROAD BRIDGES 409

plished by simply making the top chord longer than the bottom chord
and increasing the length of the diagonals to correspond.

The top chords are usually increased 3’ for each 10 feet of length
(horizontal). The length of each diagonal is computed by taking the
mean of the top and bottom chord lengths (in the corresponding panel)
as the base of a right-angled triangle, and the height of the adjoining
post as the altitude, and the diagonal as the hypotenuse. This method
of cambering is satisfactory for trusses up to 300 feet in length, and
hence is used in the case of the above bridge. The panel length here is
25’-0”, so by increasing the top chord 4” for each 10 feet of length we
obtain 25’-03%;” for the cambered length of the top chord in each panel
as is shown for the chord sections in Figs. 292 and 294. The end posts
are not increased in length. For the mean lengths of the top and bottom
chords in each panel we have

| (25’-07)+(287-04%”) | $=25' —09%,”
Then for the cambered length of each diagonal we have
V/ (80)? + (257 — 035;”)? = 39’ — 028”,

This length is used for the diagonals, as is seen in Fig. 297.

The camber affects the lengths of only the top chords, diagonals,
and top laterals.

186. Hints Regarding Shop Drawings.—Drawing E1 (Fig. 289)
is the erecting diagram. This drawing is intended principally for the
use of the party erecting the structure. The mark (as is seen) of each
piece, or member, and the general dimensions of the structure are shown
and in addition a list of the shop drawings is given.

Drawing 1 (Fig. 290) shows the complete details of the stringers
and stringer bracing. The work of detailing in this case is practically
the same as for deck plate girders (Art. 136). There should be enough
rivets in the flanges to transmit the flange increment and at the same
time support the vertical load transmitted to them from the ties. To
obtain the spacing of the flange rivets at the ends of the stringers we have

R=1,880#, which is the allowable bearing of a {” rivet on the
8” web;
25,000
= # ==
v=1,200 2
S=142,000%;

h=41.5.

 

+ 100% impact (see Art. 136) ;

Then substituting these values in Formula (4), Art. 116 (the formula
required by the specifications), we obtain

7,880

2
p =\/12008 + () = 2:18 (about), say 24, ins,
410 STRUCTURAL ENGINEERING

for the spacing of the flange rivets near the ends of the stringers, and,
as is seen, this spacing is used. The spacing at intermediate points is
obtained as explained in Art. 136.

The rivets connecting each pair of end stiffeners to the stringer
should be sufficient to transmit the maximum end shear on the stringer.
The two angles have a combined thickness of 1” in bearing on the rivets,
and the two fillers and web combined have a thickness of 13” in bearing.
Then, using {” rivets, we have

24,000 x 1x $= 21,000 Ibs.

for the allowable bearing on each rivet. The rivets are in double shear,
so for the allowable shear on each rivet we have 12,000 x 0.6 x 2=14,400.
As is seen, double shear governs. Then for the number of rivets required
in the angles we have 142,000 + 14,400=9.8 (about), whereas 11 are used,
but the rivets in the flanges should not be counted for very much, as they
take the flange increment. So the riveting, as shown, is shout correct.
There should be a sufficient number of rivets connecting the 4” fillers to
the web to transmit the maximum end shear in bearing on the 3” web.
Hence, for the number required we have 142,000+7,880=18 (about),
and, as is seen, 18 are used.

The intermediate stiffeners are spaced in accordance with Art. 118
(see Art. 170). The spacing of the rivets in the intermediate stiffeners,
as previously stated, is mostly a matter of judgment (see Art. 136).

There should be a sufficient number of rivets in the end of each
lateral to develop the strength of the lateral in compression. The length
of each lateral is about 98”. Then for the allowable unit stress on each
lateral we have

98
90

Then for the strength of each we have 8,400 x 2.48=20,800#. Dividing
this by 6,000 we obtain a little over 3—{” field rivets, and, as is seen,
3 are used. There should be enough rivets connecting each lateral plate
to the stringer to transmit the sum of the components of the two laterals
along the stringers, as the two act in the same direction, one being in
compression and the other in tension. Each field rivet passing through
the flange will take the component exerted on it directly and hence it is
a matter of taking care of the component of the rivets in the lateral
plate outside of the flange angle. Two of these are good for 12,000#.
Resolving this along the stringer we have 12,000x 0.69=8,280#%. This
requires about 1 shop rivet, and, as is seen, 1 shop rivet is used.

The connection of the bottom laterals and struts to the bottom
flanges of the stringers is mostly a matter of judgment, as the stresses
due to traction are quite low (see Art. 177). However, there should be
enough rivets in these connections to insure rigidity.

Drawing 2 (Fig. 291) shows the complete details of the floor beams.
The end connections and the stringer connections, in the case of the
intermediate floor beams, were previously designed (see Art. 184) and
will not be considered here. The number of flange rivets in the inter-
mediate floor beams should be sufficient to transmit the flange increment.
To obtain the spacing of these rivets between the truss and the stringer

p=16,000- 70 — =8,400 lbs. (about).
DESIGN OF SIMPLE RAILROAD BRIDGES All

we have
r=9,190#, which is the allowable bearing of a }” rivet on the
tx”’ web;
S=187,300% ;
h=49.5.

Substituting these values in Formula (2) (Art. 116) we obtain

_ 9,190x49.5 5 ,.
=—"T87,300 = 2.4 ins. (about)

for the required spacing, whereas 24” spacing is used.

Apparently this spacing could be a little more than 2}, but as the
flange stress is zero at the end of the beam and practically a maximum
at the stringer connection, it is obvious that there should be a sufficient
number of rivets in each flange, between the end of the beam and the
stringer connection, to transmit the flange stress. Then dividing the
flange stress (see stress sheet, Fig. 279) by 9,190 we have 201,000 +
9,190=22 (about) for the number of rivets required, whereas 20 are
used. From this it is seen that the number shown is about the correct
number. In all such cases, where the distance is short, the rivet spacing
should be verified in this manner. In most girders, however, the spacing
required by the flange increment governs.

The spacing of the flange rivets between the stringer connections
can be 6”, the maximum spacing allowed, as the shear between the string-
ers is (theoretically) zero except for the small amount due to the weight
of the part of the floor beam intervening.

To obtain the spacing of the flange rivets in the end floor beam to
transmit the flange increment, between the end of the beam and stringer
connection, we have ‘ :

r=1%,880%, which is the allowable bearing of a {” rivet on the
3” web;

S= 143,200# (see stress sheet, Fig. 279);

h=49.5.

Then substituting these values in Formula (2) (Art. 116) we obtain
_ 7,880 x 49.5
P~~ "143,200

for the required spacing. For the number required to transmit the flange
stress we have 153,000 + 7,880 = 20 (about), and, as is seen, 20 rivets are
used in the top flange and hence the spacing shown is correct, although
it is less than required by the flange increment.

In the bottom flange there are 14 rivets between the end of the
beam and stringer connection, all of which can be considered as bearing
on the 3” web, and those passing through the plates extending over the
flange angles can be considered as being in double shear in addition to
their bearing: on the 3” web. So we have

(14x 7,880) + (4x 14,400) = 167,900 Ibs.
for the value of the rivets transmitting the flange stress, whereas this
stress is 153,000#, and hence the rivets shown in the bottom flange,
between the end of the beam and stringer connection, are quite satis-
factory.

=2.7ins. (about)
412 STRUCTURAL ENGINEERING

The spacing of the flange rivets between the stringer connections
can be 6”, as the shear between the stringers is practically zero.

There should be sufficient metal through the reduced portion, near
the end of the end floor beam, to resist the cross bending at that point.
Taking a section c-c (Fig. 282) we have the metal shown in Fig. 304.

ae moments about the back of the top flange angles we have

_ (83. 62 x 19.75) + (12.37 x 16.62) + (8.36 x 1.96)

44.35 = 15.52 ins.

 

for the distance from the back of the angles to the center of gravity of
the section.

 

 

 

 

 

 

 

 

 

Fig. 304

Then for the moment of inertia about the gravity axis «-2 we have
the following:

8.36x 13.56 +30.9 =1,568 for the angles;

12.3Yx 11 +1,122 =1,137 for the 2” web;

23.62x 423° +1,435 = ae sa for the zy” plates;

; .  — 295 ‘
Total moment of inertia = £368 for rivet holes.

Then for the maximum stress on the outer element we have

_ (148,200 17) 17.7 _
f= 7,268 = 10,090 lbs.
This shows that the metal at the reduced portion of the end floor beams
is quite sufficient, as the allowable stress is 16,000#.

Drawings 8 to 14 (Figs. 292 to 303) are self-explanatory. The
making of these drawings is mostly a matter of copying the details from
the large scale sketches (igs. 282 to 285 and 287).
DESIGN OF SIMPLE RAILROAD BRIDGES 413

187. Summary of Weight of Metal and Cost of the Above
150-Ft. Span.—

12—Stringers ................. @ 4,300#.. 51,600 lbs.
Stringer bracing ..... Creer ee re ere 2,120 lbs.
2—End floor beams............ @ 3,010#.. 6,020 Ibs.
5—Intermediate floor beams..... @ 3,470#.. 17,350 Ibs.
4—End posts U1-LO.......... @ 8,850#.. 35,400 Ibs.
4—Top chords U1-U2......... @ 4,8307.. 19,320 lbs.
2—Top chords U2-U2.......... @ 10,890#.. 21,780 Ibs.
4—Bottom chords LO-L2....... @ 4,730#.. 18,920 Ibs.
2—Bottom chords L2-L2....... @ 10,740#.. 21,480 Ibs.
4—Hangers UI-L1 ........... @ 1,780#.. 17,120 lbs.
6—Int. posts U2-L2 and U3-L3 @ 2,840#.. 17,040 Ibs.
4—Diagonals U1-L2 .......... @ 3,760#.. 15,040 Ibs.
4—Diagonals U2-L3 .......... @ 2,880#.. 11,520 Ibs.
4—Collision struts M1-L1...... @ 950%... 3,800 lbs.
2—Portal struts .............. @ 2,270#.. 4,540 Ibs.
Lop: laterals: diwi0 es scwe ee uaedabenegsss 5,630 lbs.
Transverse struls ..........ecceceeeeee 2,770 lbs.
Bottom laterals ............ 00.0000 ee ee 6,460 lbs.
4 Shoes gvavesie ee aesee anes @ 940#.. 3,760 lbs.
2—Roller nests ........2.0000. @ 1,180#.. 2,360 lbs.
2—Cast pedestals ............ @ 1,130#.. 2,260 ]bs.
2—Pedestals ...........00000: @ 1,090#.. 2,180 Ibs.
PADS: xia wwe gates ete aed eae won . 640 Ibs.
Anchor bolts ....... 0... cece cece ences . 130 Ibs.
Total weight of bridge............. 279,240 Ibs.

Cost of span @ 34¢ per pound is 279,240 x 0.0325 =$9,075.30. This
price (84¢) per pound is only a fair average price for this class of work.
The price will vary from 23¢ to 4¢ per pound.

The effective dead load from the metal is equal to the total weight
of metal in the span minus the weight of the end floor beams, shoes.
roller nests, pedestals, pins, and anchor bolts. So we have ?79,240-
17,350=261,890# for the effective dead weight of metal in the span.
Dividing this by 150, the length of the span, we have 261,890+150=
1,745# per foot of span, which is only 35% more than assumed above (see
stress sheet, Fig. 279). So the dead load assumed is quite satisfactory,
as 10 per cent variation would not materially affect the design.

DRAWING ROOM EXERCISE NO. 7

Design a single-track through riveted railroad bridge and make a
stress sheet for same—the stress sheet to be upon tracing cloth.
Data :—
Length of span = 6 panels at 267-0” = 156’-0”.
Height of trusses = 31’-0” e.c. of chords.
Live load, Cooper’s E50.
Dead load, to be assumed.
Specifications, A. R. E. Ass’n.
 

 

 

J

© ae (End Post) +

 

 

 

 

Bose of Roil |

; 206565 F516"

TS LIBESS a 137

   

- Sola PLIG% Bx 1-6"
) Mas. PLIE% Ix 1-62

 

   
  

Bot Chord LoL! % “le,
2B 15793" 45-8
2F°.-g" Lott Bors,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  
 

 

 
 
  

Srrptok (a | ee

 

 

 

 

ois:

 

 

21e Pls, 134-6"

L OQ istatsgirn :
Cast Stee! Show.

 

Zhe Z “tat Bars.

 

 

Boller Dest,
SPollers 4 2-4'lg. 9

Side Bars Fxg

 

 

 

 

‘$Pins. !

 

 

 

 

 

 

 

 

% Torned Bolts: \i
aca |

 

 

Leif’ Hole~ |

 

 

 

Fin)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

: {
ee etew yg

 

Fig. 305

414

 
DESIGN OF SIMPLE RAILROAD BRIDGES A15

DRAWING ROOM EXERCISE NO. 8

Make a general drawing of the above 156-ft. bridge—the finished
drawing to be upon tracing cloth and similar to the one shown in

Fig. 288.

188. Remarks.—End floor beams are sometimes omitted in through
truss bridges in which case the end stringer rests directly upon the piers
as shown in Fig. 805. The ends of the stringers are usually constructed
as shown at (a). The stringers, as is seen, are braced to each other by
a cross-frame and they are connected to the shoes and trusses by a trans-
verse strut which forms the lower part of the cross-frame.

Circular rollers are sometimes used. The roller nests in that case
are usually constructed as shown in Fig. 305. The A. R. E. Association
Specifications practically prohibit the use of circular rollers as the mini-
mum diameter is specified as 6” and this size of rollers would place them
so far apart that the shoes would be unduly large. It is not considered
good practice in any case to use rollers less than 4” in diameter.

There is no question but that it is better practice to use end floor
beams, as the bottom ends of the end posts are held better than when they
are omitted and the thermal expansion and contraction are better pro-
vided for, as the entire structure is supported upon the shoes and will
move as a whole. Im fact, there is no excuse for omitting the end floor
beams other than the slight saving in cost.

189. Dead-Load Stresses in Curved Chord Pratt Trusses —
As previously stated, the dead load is considered as uniformly distributed
over the span. The approximate dead weight per foot of span can be
obtained from (4), Art. 124, and adding 400 lbs. to this to provide for
the weight of the deck (track) the total dead weight per foot of span
for single-track bridges is obtained, and by dividing this by 2 and mul-
tiplying by the panel length the panel load per truss is obtained. One-
third of this is considered applied at the top chord joints and two-thirds
at the bottom chord joints.

Let it be required to determine the dead-load stresses in the truss
shown at (a), Fig. 806, where P represents the panel load per truss.

Members bA and AB. Considering the part of the truss to the left
of section 1-1, as shown at (b), and resolving the forces vertically we

have
S1cos#-R=0,

from which we obtain
S1=R secé

for the stress in the end post bd, and resolving the forces horizontally

we have
S1 sind —S2=R secd x sind -S2=0,

from which we obtain
S2=R tan6é=38P tand

for the stress in the bottom chord AB.
Members BC, bce and bC. Considering the part of the truss to the
left of section 2-2, as shown at (c), and taking moments about joint b,
416 STRUCTURAL ENGINEERING

we obtain

d d
=R-=9P—
S5=K i 3 i
for the stress in bottom chord BC. Next, resolving the stress $3 (in top
chord bc) into vertical and horizontal components at c and taking moments
about C we have

Rx2d—2 Pd} Pd=H3xi,
from which we obtain
8Px2d Pd d
Sa ae ok
for the horizontal component of the stress S3 in the top chord be, and
multiplying this by secf we obtain

83 = H3 x secd

for the stress in top chord be.

H3=

 

 

 

 

 

 

fe.
A2.

oy Lsw

 

 

AL fl, 4. 2
R “lap “ep lee
s! h daz adai? (fp

Fig. 306

 

 

 

Resolving the forces vertically and horizontally and summing up the
vertical we have 5

R-P-V3-V4=0,
from which we obtain ;
V4=(R-P)-V3
for the vertical component of the stress S4 in diagonal bC and multiply-
ing this by sec we have
S4=[(R-P) -V3] secd
for the stress in that member. As (R-P) is the shear in panel BC, it
DESIGN OF SIMPLE RAILROAD BRIDGES 417

is seen that the vertical component of the stress in diagonal bC is equal
to the shear in panel BC minus the vertical component of the stress in
the top chord be. Another way to obtain the stress in diagonal bC is
to prolong the top chord be until it intersects the bottom chord at O [see
the diagram at (c)] and take moments about O. Thus, taking moments
about O we have

Rs-P (s+d)=S84xt,

s sid
S4=R oor ( ; )

from which we obtain

 

for the stress in the diagonal bC.

In case the last method be used, the distances s and ¢ can be deter-
mined in the following manner: Triangles cOC and cbm being similar
we have

842d _ Al
d “hi-h

2h—-h1
s= (=) d.
Similarly, since triangles OnC and bBC are similar we have

t s42d

hk bC’

s+2d
1 (284),

Member cC. Considering the part of the truss to the left of section
3-8, as shown at (d), and resolving the forces vertically we obtain

S6-R-P-2P_v3=(R-15P)-V3

 

>

from which we obtain

 

from which we obtain

 

for the stress in the post cC. Also, taking moments about O [at (d) |
we have

S6 (s+2d)=Rs—P (s 1-d) -=P (s+2d),
from which we obtain
Rs—P(s+d)- =P (s+2d)

nbn s+2d
for the stress in post eC.
Members CD, cd and cD. Considering the part of the truss to the

left of section 4-4, as shown at (e), and taking moments about c we have

SYxhl1=Rx2d-Pd,
from which we obtain

-Rx2d Pd d
S7= hA — oP = 8

 
418 STRUCTURAL ENGINEERING

(as seen above) for the stress in the bottom chord CD. Next, resolving
the stress in the top chord cd into horizontal and vertical components at
d and taking moments about D, we obtain

= Rx3d—-Px2d-Pxd 6p;

ca h2 h2

for the horizontal component and multiplying this by sec@l we obtain

6Pd
S8= h2 secol

for the stress in the top chord cd.
Resolving the forees horizontally and vertically and summing up
the vertical we have

R-2P-V8-V9I=0,
from which we obtain
V9=R-2P-V8

for the vertical] component of the stress in diagonal cD and hence mul-
tiplying this by sec91 we obtain

S9=[(#-2P)-P8] sech1
for the stress in that member. Also, taking moments about O’ we obtain
_Rs’—P (s’+d)—P (s’+2d)
ji —. = =
, Members dE and eD. Adding up from either end of the truss we
ave

S89

R-3P=0

for the shear in panel DE and as the top chord de is parallel to the
bottom DE it is evident that the diagonals dE and eD have no stress
from dead load and hence in determining the dead-load stresses in the
truss we can ignore these members altogether.

Members dD, DE and de. Considering the part of the truss to the
left of section 5-5, as shown at (f), (ignoring diagonal eD) and summing
up the vertical components and forces we have

R-22P_78 910 =0,

from which we obtain

ai S10=R-22P-78

for the stress in post dD.

If R-2(2/3 P) -V8 is minus $10 will be tension, as it would have to
act in the same direction as R in that case in order that the part of the
truss to the left of section 5-5 be in equilibrium, while if R —2(2/3 P) —- V8
is plus $10 will be compression, as it would be acting in the opposite
direction to that of PR.
DESIGN OF SIMPLE RAILROAD BRIDGES 419
The stress in dD can also be obtained by taking moments about O”
(see diagram at (f) ). Thus, taking moments about O’ we obtain

_ Bs’ ~P(s’ +d) +P(s’ + 2d) +4 P(s’ + 3d)
(s’ + 3d) ,

Taking moments about d we obtain

S10

d
for the stress in the bottom chord DE, which, as seen above, is equal
to H8, the horizontal component of the stress in the top chord cd.
Considering the diagonals eD and dE omitted from panel DE, as
there is no dead-load stress in them, it is obvious that the stress in top
chord de must be equal and opposite to the stress in the bottom chord
DE as these are the only horizontal forces in the panel and hence must
balance each other. So we have

§12=S11=88

for the stress in the top chord de.

Member bB. The only load on the bridge that could affect the
hanger bB is the one applied at B. This load, as is obvious, is supported
directly by the hanger and hence the dead-load stress in it is equal to 39.

As the truss is symmetrical about the center of the span and sym-
metrically loaded we need not consider further the dead-load stresses, as
we have now fully considered one-half of the span.

It is usual practice to determine graphically the dead-load stresses
in curved chord Pratt trusses.

190. Live-Load Stresses in Curved Chord Pratt Trusses.—
Chords. The criterion for the placing of wheel loads for maximum
stress in the chords is exactly the same as given in Art. 91 for simple
parallel chord trusses. That is, the maximum moment about any panel
point will occur when the average oe
unit load to the left of the point is 6
equal to the average unit load on the Kh
bridge. For example, to determine e
the maximum live-load stress in chord “f° & © 9 & Ff @  f
CD (Fig. 307) we would place a Fig. 307
wheel at C such that the average unit
load to the left of joint C would be equal (as nearly as possible) to the
average unit load on the bridge. Then by taking moments about ¢ of
the oe to the left and dividing this moment by h we would obtain
the nlaximum live-load stress in the bottom chord CD and by multiplying
this stress by secé we would obtain the maximum live-load stress in the
top chord be. The live-load stresses in the other chord members are
obtained in a similar manner.

Web Members. Let it be required to place the wheel loads so that
maximum stress will occur in diagonal cD (Fig. 308). If the top chord
cd and bottom chord CD were parallel maximum stress would occur in
diagonal cD when the shear in panel CD was a maximum, as the diagonal
in that case would carry all of the shear in the panel and hence the

s11=3(R-P) “=P

 

3

 

 

 
420 STRUCTURAL ENGINEERING

criterion for maximum shear given in Art. 90 would apply; but as the top
chord cd is inclined, and consequently carries some of the shear in panel
CD, the shear criterion will not apply exactly and hence we shall pro-
ceed to determine a criterion for the placing of the wheels for maximum
stress in the diagonal cD. Suppose the span is loaded from H (Fig.
308) up to panel CD as shown, which is about the same position as for
maximum shear in panel CD. Let P be the total load in panel CD, the
center of gravity of which is z distance from D, and let W be the total
load on the bridge, the center of gravity of which is x distance from H.

Considering the part of the truss to the left of section 1-1 and taking
moments about O we obtain

R
S1 == Z 7 (s +a)
for the stress in diagonal cD.
We
But R= ae
Pz
and Dieta e

Substituting these values of R and r we obtain

Ws Ps Pa

Sl=y 8-7 e~Fd = ew wm ee ww we rman reer ears ene (1).

 

Now suppose that there be a slight movement of the loads to the right
or to the left, say, to the left (W and P remaining constant), then $1,
x, and z will each receive an increment, AS1, Ax, and Az, respectively.
Now adding these increments in (1) we have

i 7. P P
sisasi=(Ties Tia.) - (FF + Fas) - (Fr e+ Peas).
Subtracting (1) we have

Ws Ps Pa

AS1=>— Ar ze Feed = td

for the increment of the stress in the diagonal cD due to the slight move-

ment of the loads. But Az=Az, as is obvious. So substituting Az for Az
in (2) we have

Ws Ps Pa

ASL=77 Ag a ~ Fg Ae Ce me meee cee rm nee tves (3).

Now S1, the stress in the diagonal cD, will be a maximum when

AS1=0. So placing (3) equals 0 and reducing and transposing we

obtain
Ww P a
nea (1 + <\ eho any ais wig OY SR nd BREA Wa a HE arvana elas Sia oe (4).

Expressing this equation in words we have: The average unit load
on the bridge is equal to the average unit load in panel CD multiplied by

a
DESIGN OF SIMPLE RAILROAD BRIDGES 421

(1+a/s) when the maximum stress in diagonal cD occurs. This can be
taken as the criterion for placing the loads.

The increment of the stress in diagonal cD (W remaining constant)
can change signs only when a wheel passes the panel point D, so it is seen
that a load will be at that point when the maximum stress in the diagonal
occurs. One-half of this load should be considered in panel CD in com-
puting the value of P.

Next, let it be required to place the wheel loads so that maximum
stress will occur in post cC (Fig. 308). Considering the part of the

 

 

 

 

 

 

 

 

Fig. 308

truss to the left of section 2-2 with the wheel loads in about the position
shown and taking moments about O’ we have
§2(s’ +a) =Rs’-r(s’+a),
from which we obtain
Rs’
= rr
(s’ +a)
for the stress in the post cC. Substituting (W/L)a, and (P/d)z for
R and r, respectively, we have

page
“L(s’+a) d

Now, $2, 2, and z will each receive a small increment if the loads
move slightly to the right or left. Let AS2, Az, and Az be the increment,
respectively, that each receives due to a slight movement to the left.
Then substituting these increments in (5) we have

Ws’ Ws’ A Pe -P. ‘Re
—_—_—_—_— n= _-
L(s’ ta)" L(s’ +a) dd

S82

82

§24+AS82=

and subtracting (5) from this we have

Ws’ P
ee hee Ay
L(s’+a) od

and substituting Az for Az, since Ar=Az, we obtain

AS2

Ws’ P
= ENGIN aia tee Gin wee wa ea eee a Sales 6\

AS2=7 7g) SP gee (6)

for the increment of the stress S2 in the post cC due to any slight move-

ment of the loads. Now, S2 will be a maximum when the increment
429 STRUCTURAL ENGINEERING

AS2=0 (see discussion of Art. 90). So placing (6) equal to zero and
reducing we obtain

P= 7 (ite) Monsees fl On ae (7)

which is exactly the same as (4) except s’ appears instead of s. Again,
suppose it be required to place the wheel loads so that maximum stress
will occur in diagonals bC. By loading panel BC, very much the same
as we did CD (above), and taking moments about O’ and adding incre-
ments, in the equation of moments we would obtain

WP la ®
GL 2c er

which expresses the criterion for the placing of the wheel loads for
maximum stress in the diagonal bC. This last equation is of exactly
the same form as (4); the only difference is the symbols have not the
same value as they have in (4). From this it is seen that equation (4)
expresses the general criterion for the placing of the wheel loads for
maximum stress in diagonals and intermediate posts.

In case the chords are parallel, a/s=0 as s in that case is equal to
infinity. The equation (4) then becomes

W_P
Ee

Reducing and substituting i for L we obtain

q (1+0).

ep
n
which is equation (5) of Art. 90.
It is evident that a live load moving onto the bridge from the right
would cause compression in diagonal fE (Fig. 308) and that this com-

 

 

 

A B c

 

 

 

 

 

oa

 

 

Fig. 809

pression would continue to increase until the load extended from H to a
short distance beyond F. Now if this compression were greater than
the dead-load tension in the diagonal we would have what is known as a
DESIGN OF SIMPLE RAILROAD BRIDGES 423

f

reversal of stress and the diagonal fE in that case would have to be
designed to carry both tension and compression or the member eF’, which
would be known as a counter, would have to be inserted to carry the
reverse stress. In case the diagonal fE be composed of eye-bars, which
are often used in the case of long span bridges, the counter eF would be
used, as the bars would not carry compression; but if the diagonal fE be
a rigid member—that is, capable of carrying compression—the counter
would be omitted and the diagonal would be made sufficient to carry
both the tensile and compressive stresses in it, previously explained in
Art. 176 for diagonal cD of the 150-ft. span.

To determine the maximum live-load tension in the counter eF let
the bridge be loaded as shown at (a), Fig. 309. Let P represent the
load in the panel EF and W the total load on the bridge, and 2 the dis-
tance from F to the center of gravity of P and z the distance from H to
the center of gravity of W. Now, considering the part of the truss to
the left of section 6-6 as shown at (b) and taking moments about O we
have

R(L+s)-r(at+s) -St’-Slt=0,

from which we obtain

 

for the stress in the counter eF. Substituting (W/L)z and (P/d)z for R
and r, respectively, we obtain

W (/Lis\ P bias t
8-7 (A) Fs (4) -s15 Ae ee a a ee eco eee eee QO).

S1, as is readily seen, is equal and opposite to the dead-load stress in
diagonal fE and hence is a known quantity.

Now, adding increments to S, 2, and z in (9) and proceeding in the
same manner as shown above in the case of equations (1) and (5) we

obtain

as-7 (45*) ae 5 (Se) ae stake gideindueears ts (10)
L t

for the increment of the stress in the counter eF. The stress in the coun-

ter will be a maximum when this increment is equal to zero, as previously
explained. So placing (10) equal to zero and transposing we obtain

W Pfa+s
—=F PCRS eRe OS RAE EA Bw OS 11).
L 7 (44) GY)

Expressing this in words we have: The average unit load on the
bridge is equal to the average unit load in the panel EF multiplied by
(a+s)+(L+s) when the maximum stress in the counter occurs. This
can be taken as the criterion for placing the loads.

To determine the maximum stress in the counter we would first
determine the value of s, also of ¢ and ¢’, which can be determined suf-
ficiently accurately by scale—in case of counters. Then we would place
the loading so as to satisfy equation (11) (as nearly as possible) and

 

 

 
424 STRUCTURAL ENGINEERING

determine R and r by taking moments about H and F. Then the stress S
in the counter is readily found by substituting in equation (8). It can
also be determined by summing up the vertical forces and components
to the left of section 6-6. To determine it in this way, we would first
take moments about F (Fig. 309) and determine the horizontal com-
ponent of the stress in the top chord ef. Then multiplying this by tan¢
we would have the vertical component of the stress in chord ef, which we
will designate as 1. Then by determining the vertical component of
the dead-load stress in diagonal fE, which we will designate as V2, we
would have all of the vertical forces to the left of section 6-6 determined
except the vertical component of the stress in the counter eF. Let V3
represent this vertical component. Then summing up the vertical forces
and components to the left of section 6-6 we have

R-r+V14+02-V3=0,
from which we obtain
V3=R-r+V1+VP2

for the vertical component of the stress in the counter eF and multiply-
ing this by secO we would obtain the desired stress S in the counter. The
stress in other counters is determined in a like manner.

The maximum live-load stress in hanger bB (Fig. 310) is equal to
the maximum floor beam concentration at B. This, as explained in Arts.
148 and 171, is obtained by placing a wheel at B, such that the load in
panel AB will be equal (as nearly as possible) to the load in panel BC.
After having the loads thus placed the concentration at B, which is equal
to the stress in hanger bB, is obtained by taking moments about both
A and C. The maximum live-load stress in hanger gG is obtained in
the same manner as for bB.

Owing to the top chord segments having different slopes at the joints,
the intermediate posts of curved chord bridges are subjected to relatively
greater tensile stress from live load than the intermediate posts of parallel
chord bridges. For example, let us consider the case of post cC (Fig.

 

 

 

 

& e g & ¢
oe Ss. ‘ Z
o. oO. m oN
Af, @ ™C\WOVEO FG #

1 2
lw Iw live Load

el dg is lee dead load

 

 

 

Fig. 310

310). If panel points B and C alone were loaded with live load (ignor-
ing counter dC) it is readily seen that post cC would be in tension—very
much the same as in the case of parallel chord bridges—and it is obvious
that this tensile stress would be greater than it would be if the chords
be and cd had the same slope at c, as the chords, owing to their slopes
being different, pull upward (so to speak) on the post cC and also on
diagonal cD. Now, it is evident that the tensile stress in post cC will
increase as the stress in the top chords be and cd increases, and that it
DESIGN OF SIMPLE RAILROAD BRIDGES 425

will be a maximum when the stress in diagonal cD is zero, for in that
case the post alone would resist the upward pull from the chords be
and cd. So the problem in placing the live load for maximum tension
in post cC, is to place it so as to obtain zero stress in diagonal cD ana
at the same time as great a stress as possible in the top chords be and ed.
It is customary in practice to use an equivalent uniform live load (see
Art. 123) in determining the tension in intermediate posts of curved
chord bridges, as the work is very tedious if wheel loads be used. So
we will consider a uniform live load in this case. Suppose that this live
load moves onto the bridge from the left and loads it from A to m, just
so the dead-load tension in diagonal cD is reversed. Then the stress in
diagonal cD and also in counter dC would be zero. Now, as the load
continues to move on to the right past m, the counter dC will be in tension
and this tension will increase steadily, and likewise the stress in the top
chords be and cd (the stress in diagonal cD remaining zero) until some
point n is reached when the counter dC will have maximum tension and
the stress in diagonal cD will still be zero. Then, as the load continues
to move on to the right (past 7), while the stress in the top chords be and
cd will steadily increase, the tension in counter dC will steadily decrease
until some point 0 is reached when the stress in the counter dC is zero.
This position is the one for maximum tension in the post cC as any
further movement of the load to the right would produce tension in diago-
nal cD (which has zero stress when the load extends from A to 0) and
compression in post cC and hence the tension in the post would be rapidly
reduced.

The determination, as to location, of point o (the head of the uniform
load) in any case is very much a matter of trial. We could first load
from A to D, for instance. Then compute the stress in the counter dC
(or diagonal cD in case no counter is used) and if a stress occurs with
the load in that position we could move the load to the right or left, as
the case may require, until the stress in the counter is found to be zero
and this position of the load is the one required.

After the position of the load for maximum tension in the post is
found, as Ao, we can take moments about H and obtain R, the live-load
reaction at 4. Then we can readily obtain the maximum live-load tension
in the post cC by taking moments about O and considering all of the forces
to the left of section 2-2. Thus, taking moments about O, we have

Rs—W(st+d)—-W(s+2d) +8(s+2d)=0,
from which we obtain
W(s+d)+W(s+2d)—-Rs
sare
(s+2d)
for the maximum live-load tension in post cC, where W=panel of live
load.

To this tension should be added the dead-load tension that occurs in
the post at the same time. The dead-load tension in the post is readily
obtained in the same manner as shown above for the live-load tension.
Thus, taking moments about O we have

R's — P(s +d) —3P(s+2d) +8’(s+2d) =0,
426 STRUCTURAL ENGINEERING

from which we obtain
gr oP +4) — §P (st 2d) -R’s
(s+2d)
for the dead-load tension that occurs in the post at the same time the
maximum live-load tension occurs in it. P=panel of dead load and R’=
total dead-load reaction at d—the same as occurs at H.

The maximum live-load tension in any other intermediate post is
obtained in a manner similar to that shown above for post cC. For
example, to obtain the maximum live-load tension in post dD, we would
load the bridge from A on to the right, until we obtained zero stress in
eD and then we would determine the maximum tension in the post by
taking moments about O’. In the case of post dD the maximum tension
would occur in the member when the span was fully loaded, for in that
case both diagonals, eD and dE, would have zero stress.

Complete Design of a 225-Ft. Single-Track Through Pin-Connected
Curved Chord Pratt Truss Span
191. Data.—
Lerfgth = 9 panels @ 25’-0” = 2257-0” c.c. end pins.
Width=17’-0” c.c. trusses.
Height =45’-0” at center and 31’-0” at hip.
Stringers spaced 6’6” c.c.
Live load, Cooper’s E50 loading.
opecifications, A. R. E. Association.

192. Design of 25-Ft. Stringers.—Proceeding in the manner out-
lined in Art. 170 we obtain the following for the stringers:

Maximum End Shear: Maximum Moment:

Dz= _ 5,000 lbs. D= 375,000 in. lbs.

L= %1,000 Ibs. L=4,575,000 in. lbs.

I= 66,000 lbs. L=4,223,000 in. lbs.

142,000 lbs. 9,173,000 in. lbs.
142,000 + 7,888 = 180” 9,173,000 + 45 =204,000#
1—web 48” x 3’ =180” 204,000+16,000= 12.750”
End stiff—Ls 34” x 34x 4” 2—Ls 6” x6" x3”= 10.500”
Int. stiff—Ls 34” x 84x 3” sareaofweb= 2.250”
12.750”

193. Design of the Intermediate Floor Beams.— Proceeding in
the manner outlined in Art. 171 we obtain the following for the inter-
mediate floor beams:

Maximum End Shear Maximum Moment
D= 12,000 lbs. D= 721,000 in. Ibs.
L= 94,500 Ibs. L= 5,953,000 in. lbs.
f= 81,000 lbs. T= 5,104,000in. Ibs. ,

187,500 lbs. 11,778,000 in. lbs.
DESIGN OF SIMPLE RAILROAD BRIDGES 427

187,500 + 7,644 = 24,50” 11,778,000 + 53 =222,0004
1—web 56” x 74” = 24.50” 222,000+16,000= 13.870”
2—Ls 6”x 6" x4”’= 10.500”

gZareaofweb= 3.060”

13.560)”

194. Design of End Floor Beams.— Proceeding in the manner
outlined in Art. 172 and- assuming the length of each beam to be 14’-6”,
as they will rest upon the shoes in this case, we obtain the following for
the end floor beams:

Maximum End Shear: Maximum Moments:
D=_ 6,500 lbs. D= 305,000 in. lbs.
L= 1,000 lbs. L= 3,408,000 in. lbs.
I= 66,000 lbs. I =3,168,000 in. Ibs.
143,500 lbs. 6,881,000 in. lbs.
143,500 + 8,266 =17.370” 6,881,000 + 52.37 =131,000#
1—web 56” x 2” =21.000” 1381,000+16,000= 8.198”

ae 2—Ls 6" x4" x 27 =7.22-0.75= 6.470”

B85 t/e = ‘ ‘ . a
4g 3 ae’) gareaofweb= 2.620”
9.090”

195. Determination of Dead-Load Stresses in Trusses.—From
(4), Art. 124, we have

pH=tx 225 + 660 =2,235 lbs.

for the approximate weight of metal per ft. of span and adding 400 lbs.
for the weight of the deck, we have

2,235 + 400 = 2,635 lbs.

for the total assumed dead load per ft. of span or 2,635 + 2 =1,318 Ibs. per
ft. of truss.
Multiplying this by 25, the panel length in feet, we have

W =1,318 x 25 = 32,950, say, 33,000 Ibs.

for the panel load on each truss. One-third of this will be considered at
the top joints and two-thirds at the bottom joints. The dead-load stresses
can be determined most readily by graphics.

Before we can proceed farther it is necessary that we determine the
exact outline of the truss. The panel points of the top chord will be on
the arc of a parabola. The height of the truss at the hip is 31 ft. and
45 ft. at the center of the span, making a drop of 14 ft. from the center of
the span to the hip. Laying off the panel lengths along the bottom chord
LO-D (Fig. 311) and taking C as the vertex of the parabola, CD as the
«z—axis and considering half panel lengths we have

ze 1?
i>
from which we obtain
14

=—= x 1= 0.2857 ft.
a= Fy x 1= 0.2857 ft
428 STRUCTURAL ENGINEERING

for the vertical distance from the horizontal line through C down to
panel point U4.
Similarly, we have

ae
14° «7
from which we obtain
a= is x9=2.5718 ft.

for the vertical distance down to panel point U3, and similarly we have

a’ 5?

i2” 7
from which we obtain

= 7.1428 ft.

for the vertical distance down to panel point U2, and thus we have all of
the panel points or joints of the top chord located and the outline of the

 

 
    
   

  

 

 

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truss can be drawn as shown (Fig. 311). After the outline of the truss is
carefully drawn (say 7%; scale), the dead-load stress in each member
can be graphically determined as shown in Fig. 311; one-third of a panel
load of dead load, which is 33,000x4=11,000#, being considered as
applied at each top joint, and two-thirds, which is 33,000 x =22,000#,
at each bottom joint.

The reaction =4 x 33,000 =132,000#, which is applied at LO. We
DESIGN OF SIMPLE RAILROAD BRIDGES 429

obtain the diagram of the stresses shown (Fig. 311) by first laying off 1-2
(to, say, a gy scale) equal to the reaction and passing around joint LO
counter clock-wise we obtain the polygon 1-2-3-1 for joint LO. Then
starting with LO-L1, at L1, and passing around joint £1 counter clock-
wise we obtain the polygon 3-2-4-5-3. Considering U1 and passing around
counter clock-wise, beginning with the 11,000# load, we obtain the dia-
gram 6-1-3-5-7-6, and so on for the other joints as fully explained in Ex-
ample 2, Art. 96.

196. Determination of Live-Load Stresses and Impact in the
Trusses.—In beginning the work of determining the live-load stresses
in the truss, a diagram showing the entire truss should be drawn carefully
to scale (say, toa fy scale) and upon this diagram the important lengths
and the calculated values of the different angles should be given as shown
in Fig. 312.

End Post bA. The maximum stress will occur in this member when
the loading is placed for maximum shear in panel AB. By placing wheel

 

 

 

 

 

 

 

MAVEN VEL
Sm Es SVSq SV3s &
Bao Tyee eae S
in iy FRA ISR B
09 oS Ro Re SESS g
oe 99.5
ALL _¢ AiG
as v4 \ s
~<a ay NY
ex ae t \
") Ze \
y
3 A 2 G 2 E G W =z J
S = 87
$'2 156" 9 Panels @ 25° 0"s 225*%0"
“= 378"
Fig. 312

4 at B (using Table A) we obtain 2,231 Ibs. for the average unit load on
the bridge and, considering one-half of the load at B as being in panel
AB, we obtain 2,400 Ibs. for the average unit load in the panel 4B. This
position comes nearest to satisfying the criterion of Art. 90. So taking
moments about J (Fig. 312) we obtain (using Table A)

—,. 1,000
R= [16,364 + (284 x 109) + 109?] _ 263,100 lbs.

for the reaction at A, and taking moments about B we obtain

1,000
25

for the stringer reaction at A. Then for the maximum shear in panel
AB we have

r= (480)

 

= 19,200 Ibs.

S§ = 263,100 — 19,200 = 243,900 Ibs.
for Cooper’s E40 loading and

243,900 x 5 = 804,875, say, 805,000 Ibs.
430 STRUCTURAL ENGINEERING

for Cooper’s £50. Now multiplying this by sec@ we have
805,000 x 1.28 = 390,400, say, 390,000 lbs.

for the maximum live-load stress in end post bd.

The load extends over practically the entire span when this maximum
stress occurs, and hence the L in the impact formula (Art. 125) will be
taken as 225. So we have

300
I= (soo ras) 390,000 = 223,000 lbs.

for the maximum impact stress in bd.
Now adding the above live-load stress and impact and the dead-load
stress, given in Fig. 311, together we have

390,000 + 223,000 + 169,000 = 782,000 Ibs.

for the total maximum stress in the end post bd.

4 = 225"
Fig. 313

 

Diagonal bC. By placing wheel 3 at C (Fig. 313) and applying
Formula 4 of Art. 190 we obtain

442 40 /, 25
sae7 L97 ands (1 +3 )- 2.06.

By placing wheel 2 at C we obtain

432 20 25
995 ee an nds z (1 +) 1.03.

From this it is seen that the criterion for placing the wheels for maxi-
mum stress iri diagonal bC is nearest satisfied when wheel 3 is at C. Then
placing wheel 3 at C as shown in Fig. 313 and taking moments about J we
obtain 1,000

R= (16,364 + 22 436 +79?) = 200,180 lbs.

for the reaction at 4. Then taking moments about C of the forces to left
we obtain

H1=[ (200,180 x 50) — 230,000] + 37.86 = 258,290 Ibs.
for the horizontal component of the stress in the top chord be and multi-
plying this by tan¢ we have
V1=258,290 x 0.2744 = 70,875 lbs.

for the vertical component of the stress in the top chord be.
Taking moments about C of wheels 1 and 2 we obtain

r=9,200 Ibs.
DESIGN OF SIMPLE RAILROAD BRIDGES 431

Now adding up all of the vertical forces and components to the left
of sections 2-2 we have
R-r-V-V1=0,
from which we obtain
V=R-r-V1
and substituting in the numerical values given above we have

V = 200,180 - 9,200 — 70,875 = 120,105 lbs.

for the vertical component of the stress in diagonal bC and multiplying
this by sec@ we obtain

$= 120,105 x 1.28 = 153,734 Ibs.

for the stress in the diagonal due to Cooper’s £40 and multiplying this
by 50/40 we obtain 192,167, say 192,000 lbs. for the maximum live-load
stress in diagonal bC.

When this maximum stress occurs the span is loaded practically from
C to J and hence the L in the impact formula will be taken as 175 ft. So
we have

300
Ps (3c) 192,000 = 121,000 Ibs.

for the impact stress in diagonal bC.
Now adding together the above live-load stress and impact and the
dead-load stress, given in Fig. 311, we have

192,000 + 121,000 + 73,000 = 386,000 lbs.

for the total maximum tension in diagonal bC.
Intermediate Post cC. Placing wheel 3 at D and applying Formula
4 of Art. 190 we obtain
392 40

50
295 1.74 and 25 € an a) 2.52

and placing wheel 2 at D we obtain

382
225

50

20
=1.7 and 25 (1 + at) = 1.26.

 

 

 

 

: : °
Ale E F G H z i"
Ig’
[ L= 225"
. ~!
Fig. 314

From this it is seen that the criterion for placing the wheel loads for
maximum stress in post cC is nearest satisfied when wheel 2 is at D.
Then placing wheel 2 at D, as shown in Fig. 314, and taking moments

about J we obtain
‘R=145,200 lbs.
432 STRUCTURAL ENGINEERING

for the reaction at 4 and taking moments about D of wheel 1 we obtain
r= (10,000 x 8) +25 =3,200 lbs.

for the floor beam concentration at C. Then taking moments abouc C
of all the forces and components to the left of section 3-3 we obtain

H=(Rx50) +h,
and substituting in the numerical values given above we have
H = (145,200 x 50) + 37.86 = 191,700 Ibs.

for the horizontal component of the stress in the top chord bc. Then
multiplying this component by tang we obtain

V =191,700 x 0.2744 = 52,600 Ibs.,

for the vertical comporfent of the stress in the top chord be. Now adding
up (algebraically) all of the vertical forces and components to the left of
section 3-3 we have

R-r-V-S=0,

and substituting in the numerical values given above and transposing
we obtain

§ = 145,200 — 3,200 — 52,600 = 89,400 lbs.

for the stress in post cC due to Cooper’s. E40, and multiplying this by
50/40 we obtain 111,700, say, 112,000 lbs. for the maximum live-load
compression in post eC.

When this maximum stress occurs the span is loaded practically from
D to J and hence the L in the impact formula will be taken as 150. So
we have

300
3 (ss5s0s) 112,000 = 75,000 Ibs. (about)
for the impact stress in post cC. Now adding together the above live-load
stress and impact and the dead-load stress, given in Fig. 311, we have

112,000 +'75,000 + 34,000 = 221,000 Ibs.

for the total maximum compression in post cC.
Diagonal cD. Placing wheel 3 at D (Fig. 312) and applying equa-
tion 4 of Art. 190 we obtain

392 40 50
395 =1.%4a nd 55 (14355 )= 2.11.

This position, as can be verified by trial, comes nearest to satisfying
the criterion for maximum stress in diagonal cD. So placing wheel 3 at
D, as shown in Fig. 315, and taking moments about J we obtain

R= (16,364 + 15,336 + Ba) = 153,700 lbs.
for the reaction at A, Taking moments about D of the wheels to the left
we obtain
DESIGN OF SIMPLE RAILROAD BRIDGES 433

1,000 _
595 = 92200 Ibs.

for the floor beam concentration at C. Next, taking moments about D of
the forces and components to the left of section 4-4 we obtain

171 = (153,700 x 50) ~ (9,200 x 25)

 

r=(230)

 

 

 

 

 

ree = 266,200 lbs.
4
L L= 225"
Fig. 315

for the horizontal component of the stress in the top chord cd. Multiply-
ing this by tané1 we obtain

V1=H1 xtan¢l1 = 266,200 x 0.1828 = 48.660 lbs.
for the vertical component of the stress in top chord cd. Now summing up
the vertical forces and components to the left of section 4-4 we have
R-r-V-V1=0.
Substituting the numerical values given above and transposing we obtain
V =153,700 — 9,200 — 48,660 = 95,840 lbs.

for the vertical component of the stress in diagonal cD and multiplying
this by sec#1 we have

S=V sec61 = 95,840 x 1.199 = 114,912 Ibs.

for the stress in diagonal cD due to Cooper’s £40 loading. Then by mul-
tiplying this by 50/40 we obtain 143,635, say, 144,000 lbs. for the maxi-
mum tension in diagonal cD due to the F'50 loading.

When this maximum stress occurs the span is practically loaded from
D to J and hence the L in the impact formula will be taken as 150 ft. So
we have 7

300
1=(55 = a0) 144,000 = 96,000 Ibs.

for the impact stress in diagonal cD.

Now adding together the above live-load stress and impact and the
dead-load stress, given in Fig. 311, we obtain

144,000 + 96,000 + 40,000 = 280,000 Ibs.

for the total maximum tension in diagonal cD.
Intermediate Post dD. Placing wheel 2 at E (Fig. 312) and apply-
ing equation 4 of Art. 190 we obtain

332 20/,  %5\_
sag zl? and 55 (1+ 355)=148
434 STRUCTURAL ENGINEERING

This position of the wheels practically satisfies the criterion for maxi-
mum stress in post dD. So placing wheel 2 at E, as shown in Fig. 316,
and taking moments about J we obtain (using Table A)

R=105,500 lbs.

 

 

for the reaction at A, and taking moments about E of wheel 1 we obtain
r= 8,200 lbs.

for the floor beam concentration at D. Then taking moments about D
of the forces and components to the left of section 5-5 we obtain

H1=186,500 lbs.

for the horizontal component of the stress in top chord cd and multiply-
ing this by tang1 we obtain

V1=186,500 x 0.1828 = 34,092 Ibs.

for the vertical components of the stress in top chord cd.
Now, by summing up (algebraically) the vertical forces to the left
of section 5-5 we obtain

S = 105,500 — 3,200 — 34,092 = 68,208 lbs.

for the stress in post dD due to Cooper’s £40 and multiplying this by
50/40 we obtain 85,260, say, 85,000 lbs. for the maximum compression in
post dD due to the £50 loading.

When this maximum stress occurs the span is practically loaded from
E to J and hence the L in the impact formula will be taken as 125 ft. So
we have

300
I- (ar) 85,000 = 60,000 Ibs.

for the impact stress in post dD.
Now, adding together the above live-load stress and impact, and the
dead-load stress, given in Fig. 311, we obtain

85,000 + 60,000 + 12,000 = 157,000 Ibs.

for the total maximum compression in post dD.
Diagonal dE. Placing wheel 3 at E (Fig. 312) we obtain

ate = 1.52 and 53 (1+ oe )=109.

 

 

225 25 378
DESIGN OF SIMPLE RAILROAD BRIDGES 435

This position of the load comes nearest to satisfying the criterion for
maximum stress in diagonal dE. So placing wheel 3 at E, as shown in
Fig. 317, and taking moments about J we obtain

R =113,200 lbs.

for the reaction at A, and taking moments about E of the wheels to the
left we obtain
r= 9,200 Ibs.

for the floor beam concentration at D. Then taking moments about E
of the forces and components to the left we obtain

H1= 248,100 lbs.

for the horizontal component of the stress in the top chord de and mul-
tiplying this by tan¢2 we obtain

V1=248,100 x 0.0912 = 22,626 lbs.
for the vertical component of the stress in top chord de.

Now, summing up the vertical forces and components to the left of
section 6-6 we obtain

V =113,200 — 9,200 - 22,626 = 81,374 lbs.

for the vertical component of the stress in diagonal dE, and multiplying
this by sec62 we obtain 94,393 lbs. for the stress in diagonal dE due to

 

 

Fig. 317

E40 loading; and multiplying this by 50/40 we obtain 117,991, say, 118,-
000 Ibs. for the maximum live-load stress in the member. When this maxi-
mum stress occurs, the span is loaded practically from E to J and hence
the L in the impact formula will be taken as 125 ft. So we have

300
I= (a3oa00) 118,000 = 83,200, say, $3,000 Ibs.

for the impact stress in diagonal dE.
Now, adding together the above live-load stress and impact and the
dead-load stress given in Fig. 311, we obtain

118,000 + 83,000 + 21,000 = 222,000 Ibs.

for the total maximum stress in diagonal dE.
436 STRUCTURAL ENGINEERING

Intermediate Post cE. sa wheel 2 at F (Fig. 312) we obtain

284 100\_

and placing wheel 3 at F we pie

292 40/, 100
On = = 2.16.
bop and oe (1+ 3)

From this it is seen that the criterion (Art. 190) for maximum stress
in post eE is the nearest satisfied with wheel 2 at F. So placing wheel 2
at F, as shown in Fig. 318, and taking moments about J we obtain
(using Table A)
R=71,466 lbs.

 

 

 

7, vr
S/ Gy, HI
c |
b eS
S 1
fe !
wa
so
@ J
A a Cc o Ee F G 4 Z 4
B ae
L=225' J
Fig. 318

for the reaction at A and taking moments about F of the wheel to the
left we obtain .
: r= 3,200 lbs.

for the floor beam concentration at E. Then taking moments about E
of the forces and components to the left of section 7-7 we obtain

H1=159,843 lbs.

for the horizontal component of the stress in the top chord de and multi-
plying this by tan¢2? we obtain

V1=14,577 lbs.

for the vertical component of the stress in top chord de. Then adding
algebraically the vertical forces and components to the left of section
7-7 we obtain me

S =71,466 — 3,200 - 14,577 = 53,689 Ibs.

for the maximum compression in post eE due to the E40 loading and mul-
tiplying this by 50/40 we obtain 67,111, say, 67,000 lbs. for the maximum
live-load compression in post eZ.

When this maximum stress occurs the span is loaded practically
from F to J and hence the L in the impact formula will be taken as 100.
So we have

300
I= (sorcsur) 67,000 = 50,250, say, 50,000 Ibs.

for the impact stress in post eF.
DESIGN OF SIMPLE RAILROAD BRIDGES 437

Now, adding the above live-load stress and impact, and the dead-
load stress given in Fig. 311 we obtain

67,000 + 50,000 — 5,000 = 112,000 Ibs.

for the total maximum compression in post eE.

Diagonal eF. As the top chord ef (Fig. 312) is parallel to bottom
chord EF the diagonal eF will carry all of the maximum shear in panel
EF when the load moves onto the bridge from the right and hence the
ordinary criterion of Art. 90 for the maximum shear in panel EF applies.
According to this criterion, the maximum shear will occur in panel EF
when the average unit load on the bridge is equal to the average unit load
in the panel.

This is the nearest satisfied when wheel 3 is at F. So placing wheel
3 at F as shown in Fig. 319 and taking moments about J we obtain

R="7,851 lbs.

 

G H Z E

 

Fig. 319

for the reaction at A and taking moments about F of the wheels to the
left we obtain
r=9,200 lbs.

for the floor beam concentration at E. Then for the maximum shear in
panel EF we have

(R -r) = (77,851 — 9,200) =68,651 lbs.
This is assumed to be carried altogether by the diagonal eF,, so, evi-
dently the vertical component of the stress in that member must be equal
to this shear; that is,
V = 68,651 lbs.

Then multiplying this by sec03 we obtain
S=68,652 x 1.145 = 78,605 lbs.

for the stress in the diagonal eF due-to £40 loading and multiplying this
by 50/40 we obtain 98,256, say 98,000 Ibs. for the maximum live-load
stress in diagonal eF’.
When this maximum stress occurs the span is loaded practically from
F to J and hence the L in the impact formula will be taken as 100.
Then we have

‘/ 300 és
1=(sqps400) 98,000 = 73,500, say 73,000 lbs.

for the impact in diagonal eI’.
438 STRUCTURAL ENGINEERING

Now, adding the above live-load stress and impact, and the dead-load
stress given in Fig. 311 we obtain

98,000 + 73,000 + 00,000 = 171,000 lbs.

for the total maximum stress in diagonal eF.

 

 

 

 

 

 

 

 

 

 

 

 

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3] Pt :
" ; ¥ a
Spake vi |
vy. $3! 3
D E E q H I lk
Tr
S=378' a= soo’ (6)
45225"
Fig. 320

Counter eD. Placing wheel 3 at D as shown at (a), Fig. 320, and
applying Formula 11 of Art. 190 we have (not considering wheel 15 on
the bridge)

232 40 (1004+ 378
—— L — ( —_—__ J= 1.26.
pag Pando, (soecaie)

This position of the wheels comes nearest to satisfying the criterion
for maximum stress in counter eD. So placing wheel 3 at D and taking
moments about 4 we obtain

1,000
R= (10,816) “gap 7 48,071 lbs.
for the reaction at J and taking moments about D of the wheels to the

right we obtain
r=9,200 lbs.

for the floor beam concentration at E.

§1 is equal and opposite to the dead-load stress in diagonal dE which
is given in Fig. 311 as 21,000 lbs. Then, resolving S1 into horizontal
and vertical components at E, we obtain

V1=81 + sec$2 = 21,000 + 1.16 =18,103 Ibs.

for the vertical component of the stress in diagonal dE.
Taking moments about D of the forces and components to the right

we obtain
(Fx 150) - (V1 x 25) — (rx 25) — (2 x 42.43) =0.
DESIGN OF SIMPLE RAILROAD BRIDGES 439

Substituting in the numerical values given above and transposing and
reducing we obtain

H2=153,853 lbs.

for the horizontal component of the stress in top chord de and multiply -
ing this by tan¢@2 we obtain

V2=153,853 x 0.0912 =14,031 lbs.

for the vertical component of the stress in top chord de.
Now, summing up all the vertical forces and components to the
right of section 9-9 we obtain

V=R-r-V1+V2
and substituting in the numerical values given above we have
V =48,071 — 9,200 - 18,103 + 14,031 = 34,799 lbs.

for the vertical component of the stress in counter eD and multiplying
this by sec03 we have

S= 34,799 x 1.145 = 39,844 lbs.

for the maximum stress in counter eD due to the £40 loading and mul-
tiplying this by 50/40 we obtain 49,805, say, 50,000 lbs. for the maximum
live-load stress in counter eD due to E50 loading.

When this maximum stress occurs the span is loaded practically from
D to A and hence the L in the impact formula will be taken as 75.

Then we have

800
for the impact stress in counter eD.
Now, adding the above live-load stress and impact together we have
(counters carry no dead load)

50,000 + 40,000 =90,000 Ibs.

for the total maximum stress in counter eD.

No counter is needed in panel CD as the live-load shear is not suf-
ficient to reverse the dead-load shear; or, in other words, the dead-load
tension in diagonal cD is greater than the live-load compression in it.

Tension in Post dD. The maximum shear in end panel AB was
found above (in determining the stress in end post bA) to be 305,000 Ibs.

Substituting this shear in the formula given in Art. 123 we obtain

305,000
P| 225-25
2
for the equivalent uniform live load per foot of truss for determining the

stress in the web members and hence we can use this load for determin-
ing the tension in intermediate posts. Then for the panel load we have

W =3,050 x 25 = 76,250, say 76,000 Ibs.

The first part of our problem now is to place this uniform load so
as to obtain zero stress in counter eD and also in diagonal dE and at the

= 3,050 lbs.
440 STRUCTURAL ENGINEERING

same time as great a stress in chords cd and de as possible (see Art. 190).

The equivalent uniform live load will produce practically the same
stress in the counter eD (Fig. 320) as the wheel loads. So, if panel
points B, C, and D are loaded with the above uniform load the counter
eD will have a maximum tensile stress of 50,000 Ibs., the same as pre-
viously found for wheel loads. Dividing this by sec63 we obtain

50,000 + 1.145 = 44,000 Ibs. (about)

for the vertical component of the maximum tensile stress in counter eD.
Now, any load at E, F, or any panel point to the right of panel DE will
tend to reduce the tension in counter eD. What we desire is to place the
load just so that the 50,000 lbs. tension in the counter is exactly counter-
acted, for then the counter eD and diagonal dE will have zero stress and
hence the post dE will have maximum live-load tension.

By loading panel point E alone we obtain a reaction at A of %6,-
000 x 5/9 =42,000 lbs. Now, as the vertical component of the 50,000 lbs.

 

 

 

 

= [s}
Ss 8
8 39
©
# _sneah—— I 2
420000. t
(d)
(b)
dy) 1 9 :
Ce A
B : a3 x
*
A B ie D iE F G H x Z,
lr 20' Tr
8 7 6 $ 4 3 2 ,
‘ 2 3 4 5 6 7 8
Fig. 321

tensile stress in the counter eD alone is about 44,000 lbs., it is obvious
that a panel load at E will not reverse the 50,000 lbs. tension in the
counter. So next, let us place a panel load at both E and F. Then for
the reaction at A, which is equal to the negative shear in panel DE due
to the loads at E and F, we obtain

=
9
Taking moments about D (panel points E and F alone being loaded) we
obtain

76,000 xo4 76,000 x — = 76,000 Ibs. (about).

H1 = (76,000 x 75) + 42.43 = 135,000 Ibs. (about)

for the horizontal component of the stress in top chord de. Then mul-
tiplying this by tan¢2 we obtain

V1=135,000 x 0.0912 = 12,300 Ibs. (about)

for the vertical component of the stress in ton chord de due to the panel
loads at E and F.
DESIGN OF SIMPLE RAILROAD BRIDGES 441

Then subtracting this from the reaction we obtain
76,000 — 12,300 = 63,700 lbs.

for the vertical component in counter eD. This is too great, as the coun-
ter would be more than reversed and hence the diagonal dE would then
be in tension.

Next, suppose the load extends just up to panel point F’ so that the
load at F will be only a half panel load or 38,000 lbs. Then for the
reaction at A we have

76,000 x + 38,000 x <= 59,000 lbs. (about).

Now, taking moments about D we obtain
59,000 x 75
te ( 42.43

for the vertical component of the stress in the top chord de.
Then subtracting this from the reaction at A we have

59,000 — 9,500 = 49,500 Ibs.

for the vertical component of the compressive stress in counter eD. As is
seen, this is too large. So let us extend the load just 20 ft. beyond panel
point E as shown. Then the load at E will be 75,000 lbs. and at F it
will be 24,000 lbs. The reaction at A due to these two loads will be

5

( 75,000 x 3) + (24,000 <t) = 52,300 Ibs. (about).

Then taking moments about D we obtain
9
Vi= coe x75

) 0.0912 = 9,500 Ibs. (about)

243 ) 0.0912 = 8,400 lbs. (about)
for the vertical component of the stress in the top chord de. Now, sub-
tracting this from the reaction at A we obtain

52,300 — 8,400 = 43,900 Ibs.

for the vertical component of the compressive stress in the counter eD
due to the loads at E and F which just reverses the maximum tension
in the counter. So this is the position of the load being sought for. That
is, by placing the load so that it extends from 4 to 20 ft. beyond E the
stress in both the counter eD and diagonal dE will be zero and the chords
cd and de will have the greatest stress possible with that condition and
hence the tension in post dD will be a maximum.

So placing the load in that position and taking moments about 4
we obtain

R’ = (3,050 x 120)22.- 97,644 lbs.
225

for the reaction at J.

Then taking moments about D we obtain

_ 97,644 x 150 — 3,050 x 45 x 45/2 = 418,000 Ibs. (about)

Hil 42.43
442 STRUCTURAL ENGINEERING

for the horizontal component of the stress in top chord de and multiplying
this by secp2 we obtain

418,000 x 1.004=420,000 Ibs. (about)

for the stress in top chord de. Now as the stress in diagonal dE is zero,
the tensile stress in the post dD can be obtained very quickly by drawing
the diagram shown at (b). From this diagram the tension in post dD
is found to be about 38,000 lbs.

As is seen, when this maximum tension occurs in the post dD there
is 120 ft. of load on the bridge, so we have

800
T= (s5caa) 38,000 = 26,000 Ibs. (about)
for the impact stress.

Now, assuming diagonal dE as having zero stress we can obtain
the dead-load tension in post dD by drawing the diagram shown at (d).

The stress in top chord'cd is given in Fig. 311 as 177,000 lbs. Then
drawing 1-3 equal (by scale) to 177,000 and parallel to cd, and 1-2 and
3-2 parallel to de and dD, respectively, we have the dead-load tension
in the post given by the line 3-2. This tension is found in this manner
to be 16,000 Ibs.

Now adding together the above live- and dead-load stresses and
impact we obtain

38,000 + 26,000 + 16,000 = 80,000 Ibs.

for the total maximum tension in post dD.

Tension in Post eE. As is obvious, the maximum tension will occur
in post e£ when the span is fully loaded, for in that case both of the
diagonals eD and fE have zero stress. The counter eD also has zero
stress at the same time.

Now, as we are using a uniform live load we can readily obtain the
live-load tension in post eE by direct proportion as it will be propor-
tional to the dead-load tension given in Fig. 311.

So, letting S represent the live-load tension, we have

Ss 3,050
5,000 1,318”
from which we obtain
S=11,570, say, 12,000 Ibs.

for the maximum live-load tension in post eE where 3,050 is the uniform
live load and 1,318 the dead load per ft. of truss.
For the impact we have

225 +300

Now, adding together the above live-load stress and impact, and the
dead-load stress given in Fig. 311, we obtain

12,000 +7,000 + 5,000 = 24,000 Ibs.

for the total maximum tension in post eE.

I= (3.22) 12,000 = 6,857, say, 7,000 lbs.
DESIGN OF SIMPLE RAILROAD BRIDGES 443

Bottom Chords AB and BC. The maximum stress can be obtained
in these members by placing a load at B such that the average unit load
to the left is equal to the average unit load on the bridge (see Art. 90)
and taking moments about b; but the same can be found by simply mul-
tiplying the maximum shear in panel 4B by tané. The maximum shear
was found above (in determining the stress in the end post b4) to be
305,000 lbs. So we have

305,000 x 0.8066 = 246,013, say, 246,000 Ibs.

for the live-load stress in cach of the bottom chords AB and BC.
When this maximum stress occurs, the load extends practically over
the entire span, so for the impact stress in each of these chords we have

800
f= (aera) 246,000 = 140,571, say, 141,000 lbs.

Now, adding together the above live-load stress and impact, and
the dead-load stress, given in Fig. 311, we obtain

246,000 + 141,000 + 106,000 = 493,000 Ibs.

for the total maximum stress in each of the bottom chords AB and BC.
Bottom Chord CD. The stress in this member (see Fig. 312) is
found by taking moments about panel point c. Then, evidently, the
stress will be a maximum when the load is placed so that this moment
is a maximum. According to Art. 90, the moment will be a maximum
when a load is at C such that the average unit load to the left is equal

   

QB) WAZA,

 

Fig, 322

to the average unit load on the bridge. Placing wheel 7 at C, as shown

in Fig. 322, we have

1,000 _
Kp” = 2s190 Ibs.

for the average unit load to the left of C (considering one-half of wheel
7 as being to the left of C) and

[2844 (103 x 2)]

 

(103 + 6.5)

1,000
225
for the average unit load on the bridge. This position of the load comes
nearest to satisfying the criterion for maximum moment about ¢ (or C).
Then by taking moments about J (using Table A) we obtain
1,000
225

for the reaction at A. Next, taking moments about c of the forces to the
left and dividing by the height of the truss at that point we obtain

= 2,177 Ibs.

*

R= [16,364 + (284x103) +103] = 249,880 Ibs. (about)

 
444 STRUCTURAL ENGINEERING

S = [ (249,880 x 50) - (2,155 x 1,000) ] =.-5, = 273,000 lbs. (about)

37. "ai
for the maximum stress in the bottom chord CD due to the £40 loading
and multiplying this by. 50/40 we obtain 341,250, say 341,000 lbs. due to
the E50 loading which is the maximum live-load stress desired,
When this maximum stress occurs the live load extends over practi-
cally the entire span and hence the L in the impact formula will be taken
as 225. So for the impact stress we have

300
= (sesa00) 341,000 = 194,000 Ibs. (about).

Now, adding together the above live-load stress and impact, and the
dead-load stress given in Fig. 311, we obtain

341,000 + 194,000 + 152,000 = 687,000 Ibs.

for the total maximum stress in bottom chord CD.

Top Chord be. Considering the forces to the left of section 1-1 (Fig.
322) it is readily seen that the horizontal component of the maximum
stress in top chord bc is equal to the maximum stress in bottom chord CD.
So by multiplying the stress in CD (found above) by sec (see Art. 190)
we obtain

S81 = 341,000 x 1.037 = 353,617, say, 854,000 lbs.
for the maximum live-load stress in top chord be. For the impact we
have

225 + 3800

Now, adding together the above live-load stress and impact, and
the dead-load stress given in Fig. 311, we obtain

354,000 + 202,000 + 157,000 = 713,000 Ibs.

for the total maximum stress in top chord be.

Bottom Chord DE. The stress in this member is found by taking
moments about panel point d and hence the stress in the member will be
a maximum when the moment about d is a maximum. Placing wheel 11
at D, as shown in Fig. 323, we have

I= Gotha) 354,000 = 202,000 lbs. (about).

(152+10) we = 2,160
for the average unit load to the left of D (or d) and
1,000
(284 + 210)=—— 905 = 2,195

for the average unit load on the bridge. This position of the load comes
the nearest to satisfying the criterion for maximum moment about d
(or D).

Then by taking moments about J we obtain

ae 254,260 Ibs. (about)

                            
DESIGN OF SIMPLE RAILROAD BRIDGES 445

for the reaction at 4d. Next, taking moments about d of the forces to the
left and dividing by the height of the truss at that point we obtain

S=[ (254,260 x 75) — (5,848 x 1,000) Inn = 311,600 lbs. (about)

for the maximum stress in chord DE due to the E40 loading and multiply-
ing this by 50/40 we obtain 389,500, say 390,000 lbs. for the maximum
stress due to the £50 loading.

For the impact we have

300
I= (seca) 390,000 = 222,856, say, 223,000 lbs.

Now, adding together the above live-load stress and impact, and the
dead-load stress given in Fig. 311, we obtain

390,000 + 223,000 + 174,000 = 787,000 lbs.
for the total maximum stress in bottom chord DE.

Lop Chord cd. Considering the forces to the left of section 2-2 (Fig.
823) it is readily seen that the horizontal component of the maximum

 

 

Fig. 323

stress in top chord cd is equal to the maximum stress in bottom chord DE.
So by multiplying the maximum stress in DE by secdl we obtain

390,000 x 1.016 = 396,240, say, 396,000 Ibs.

for the maximum live-load stress in top chord cd.
For the impact we have

300
I= (mn) 396,000 = 226,000 Ibs. (about).

Now, adding together the above live-load stress and impact, and the
dead-load stress given in Fig. 311, we obtain

396,000 + 226,000 + 177,000 = 799,000 lbs.

for the total maximum stress in top chord cd.

Bottom Chord EF. The stress is found in this member by taking
moments about panel point e and hence the stress in the member will
be a maximum when the moment about e is a maximum.

Placing wheel 13 at E, as shown in Fig. 324, we have 2,020 lbs. for
average unit load to the left of E (or e) and 2,062 lbs. for the average
unit load on the bridge. This position of the load comes the nearest to
satisfying the criterion for maximum moment about e (or E).
446 STRUCTURAL ENGINEERING

Then taking moments about J we obtain
1,000
225

for the reaction at A. Next, taking moments about e of the forces to the
left and dividing by the height of the truss at that point we obtain

 

R=[16,364-+ (284 x 90) x 90°] = 222,320 Ibs. (about)

S = [ (222,320 x 100) - (7,668 x 1,000) laa = 325,743 Ibs.

for the maximum stress in bottom chord EF due to the E40 loading and

325,743 x x = 407,178, say, 407,000 Ibs.

for the maximum stress due to the E50 loading.
For the impact we have

I= (ee sa) 407,000 = 232,571, say 233,000 Ibs.
Now adding together the above live-load stress and impact, and the
dead-load stress given in Fig. 311 we obtain
407,000 + 233,000 + 184,000 = 824,000
for the total maximum stress in bottom chord EF.

Top Chord de. Considering the forces to the left of section 3-3 (Fig.
824) it is readily seen that the horizontal component of the maximum

 

 

 

Fig. 324

stress in top chord de is equal to the maximum stress in bottom chord EF.
So by multiplying the maximum stress in the bottom chord EF, which,
as found above, is 407,000 lbs., by sec#2 we obtain

407,000 x 1.004 = 408,628, say, 409,000 lbs.

for the maximum live-load stress in top chord de.
For the impact we have

I= (095/00) 409,000 = 234,000 Ibs. (about).
Now, adding together the above live-load stress and impact, and the
dead-load stress given in Fig. 311, we obtain
409,000 + 234,000 + 185,000 = 828,000 lbs.
for the total maximum stress in top chord de.

Top Chord ef. In accordance with usual practice we will consider
the stress in top chord ef to be equal and opposite to the stress in bottom
DESIGN OF SIMPLE RAILROAD BRIDGES 447

chord EF. This would be absolutely true if the shear in panel EF were
zero. If a uniform load be used the shear would be zero and in case
wheel loads are used the shear is slight and hence the usual practice is
not far wrong in any case.

We have now determined all of the stresses in one-half of the truss
and as the structure is symmetrical about the center of span these are all
the stresses necessary for designing the trusses. We can next write the
above stresses on the stress sheet, Fig. 328.

197. Designing of Members in Trusses.—As the intermediate
posts are thé governing members (see Art. 176) we will consider them
first. We will first ascertain as to whether channels can be used. The
longest post is U4-L4 (Fig. 328), which has a length of about 536 ins.
The average radius of gyration of 15” channels is about 5.4 (see Table
3). Then for the maximum L/r we have

536 +5.4=99.2.

The maximum allowed for L/r for main members by the specifications
is 100. So, as far as L/r is concerned, 15” channels can be used. We
will next ascertain as to whether the area of the channels is sufficient.

Post U2-L2 has the greatest stress and hence will require the great-
est area. This post is about 454 ins. long. Now substituting in the
column formula, using the average radius, we have

p=16,000-705— = 10,115 Ibs.

for the allowable unit stress. Dividing this into the stress in the column

we obtain
221,000 + 10,115 = 21.74 sq. ins.

for the required area. From this it is seen that channels can be used.
Post U2-L2. The area just found for this post is 21.740”. Use
2—[s 15” x 40% =23.520". (The 15”x35% channels have 1.164’ less
area than required.)
Post U8-L8. This member has a maximum compressive stress of
157,000 lbs. and a maximum tensile stress of 80,000 lbs. Combining,
according to the specifications, we have

157,000 SOO 197,000 Ibs.

for the maximum compression and

80,000

80,000 +—5

= 120,000 Ibs.

for the maximum tension to be considered in designing the member.
The length of the member is about 509 ins. So substituting in the
column formula, using the average radius (as preliminary), we have

p= 16,000 —- 0 = 9,400 lbs. (about)
448 _ STRUCTURAL ENGINEERING

for the allowable compressive unit stress. Then we have
197,000 + 9,400 = 20.96 sq. ins.

for the area of cross-section required for compression, and for the area
required for tension we have

120,000 +16,000 = 7.5 sq. ins.

As is seen, the compression governs. Referring to Table 3, it is seen
that 2—[s 15” x 35# =20 580” have the nearest to the required area for
compression. So substituting the radius of these channels in the column
formula we have

509

16,000 —- 70 ——~ 558 = 9,615 Ibs.

for the required unit stress. Dividing this into the maximum compres-
sion we obtain
197,000 + 9,615 = 20.49 sq. ins.

for the required area. This is very close to the area of the
2—[s 15” x 35# and hence these channels will be used.

Post U4-L4. This member has a maximum compressive stress of
112,000 lbs. and a maximum tensile stress of 24,000 lbs. Combining,
we have 124,000 lbs. compression and 36,000 Ibs. tension to be consid-
ered in designing the member. The length of the member is about 536 ins.
Then, using the radius of the lightest 15” channel (as the stresses are
low), we have

16,000 — 70 ee = 9,324 Ibs.

for the allowable unit compressive stress on the member. Dividing this
into the above combined stress we have

124,000 + 9,324=13.3 sq. ins. (about)

for the required area for compression. For the area required for tension
we have

36,000 = 16,000 = 2.25 sq. ins.

As is seen, the compression governs. 2—[s 15” x 33#=19.80” (which
are the lightest 15” channels) will be used for this member. As is
seen, there is an excess of 6.54” of metal, yet we cannot do better if
channels are used. The L/r would be too great if 12” channels were
used instead of the 15” channels.

Hanger U1-L1. This is entirely a tension member. For the re-
quired net area of cross-section we have

198,000 + 16,000 = 13.37 sq. ins.
We will use 2—[s 15” x 33# =19.8 — (21/32 x 4) =17.180” net.
DESIGN OF SIMPLE RAILROAD BRIDGES 449

(In this case the metal cut-out of the flanges is considered.) This is more
section than needed but 15” channels are used for the intermediate posts
and for the sake of appearance they will be used for this member.

Diagonal U1-L2. For the required net area we have 386,000+
16,000 = 24.120” net. Use 2—bars 8” x 14” = 24.00”,

Diagonal U2-L3. For the required net area we have 280,000+
16,000 =17.50” net. Use 2—bars 7” x 14/=17.50”.

Diagonal U3-L4. For the required net area we have 222,000 +
16,000 =13.870” net. Use 2—bars 7” x1=140” net.

Diagonal U4-L4. For the required net area we have 171,000+
16,000 = 10.684” net. Use 4—1s 5” x 34 x 3=12.20—-1.50=10.70” net.

Counter U4-L3. For the required area of cross-section we have
90,000 + 16,000=5.620”. Use 1—bar 23” x 28” = 5.640” (standard bar).

Bottom Chords LO-L1 and L1-L2. For the required net area we
have 493,000 + 16,000 = 30.85”. Use the following section:

2—pls. 20” x §”=25.0 — 3.75 = 21.250” net
4—|_5 34" x 84” x fe" =11.48-1.75= 9.730” net

30.980” net

 

Bottom Chord L2-L3. For the required area of cross-section we
have 687,000 + 16,000 =42.920”. Use the following section:

2—bars 8” x 13” =22.000”
2—bars 8” x 1455/7 = 21.000”

' 43.000”

Bottom Chord L8-L4. For the required area of cross-section we
have 787,000 +16,000=49.185”. Use the following section:

2—bars 8” x1)” =24.000”
2—bars 8” x 13%” = 25.000”

49.000”

Bottom Chord L4-L4. For the required area of cross-section we
have 824,000+16,000=51.55”. Use 4—bars 8” x 1$=52.0”.

In designing the top chords, it is best to design the lightest section
first, using minimum thicknesses of web (if sufficient) so that the area of
cross-section can be increased for the others by merely increasing the
thickness of the webs. So in this case we will first consider top chord
U1-U2, as it has the least stress and consequently will have the lightest
section.

Top Chord U1-U2. The first thing to do is to determine the gen-
eral dimensions of the section. The width of the posts will be 124”
(see Art. 176) and the maximum thickness of eye-bars is 3” (2—8” x 14”
bars at U1) and allowing, say, 2” for pin plates and clearance we have
1244+3+4+2=174” for the required distance between the webs. Then if
the webs be 2” thick (each) and the top angles be 4” x4” we obtain
17$ +144 8=26%, say, 27” for the width of the cover plate. The radius
of gyration in reference to the y-y axis (see Art. 176) is approximately
450 STRUCTURAL ENGINEERING

equal to the distance from the center of the cover plate out to the outer
face of the web, which in this case is (17$+1})+2=9.87. Now, in
order that the section be of economic depth, that is, so that the radius
of gyration about the 2-a axis -is the same as about the y-y axis, the webs
should have a depth 0.4 of which should equal-the above radius. So for
the economic depth of the web we have
h= oat = 23.4 ins.

So we will make the webs 24” deep. According to the specifications,
the thickness of the cover plate should not be less than ~y of the dis-
tance between the rivet lines. The distance between the rivet lines in
this case will be about 17$+14+4$=234”". dy of this distance is
about 3% of an inch. So we will make the cover plate 7%” thick. The
thickness of the webs, according to the specifications, should not be less
than 1, of the distance between the flanges. If the top angles be
4” x 4” and the bottom angles 6” x4” (the 6” leg along the web), we
have 24-(4+6)=14” for the distance between the flanges. Then for
the minimum thickness of the webs we have 14x1/30=0.46, which is
about 7% of an inch.

The top angles should be of minimum thickness and the bottom
angles should be of maximum thickness in order that the center of gravity
of the section be as near the center of the web as possible.

The length of the member (U1-U2) is about 312 ins. Now substi-
tuting this length and the approximate radius given above in the column
formula we obtain

16,000 — 70 —— aE 13,670 Ib
ee Cay ee
for the approximate allowable unit stress. Dividing this into the maxi-
mum stress in the member we obtain

713,000 + 13,670 = 52.15 sq. ins.
for the approximate required area of cross-section.
In accordance with the above we obtain the following section:

1—cover pl. 27” x 7%/” = 15.180”
2—web pls. 24’ x 3’ = 21.000”

2—Ls 4” x47 x 3” = 5.720”
2—Ls 6” x 4” x 3%” = 10.620”
52.520”

This section is really the minimum and is very near the approxi-
mate section found above.

Now taking moments about the cover plate, in the manner shown
in Art. 176 we obtain 9.38” for the distance from the cover plate down
to the horizontal gravity axis of the section.

The next thing is to see if the eye-bars will fit into the top chord.
The largest pin, which will be at U1, will be about 7” in diameter. From
the table of eye-bars, to be found in practically all structural hand-
DESIGN OF SIMPLE RAILROAD BRIDGES 451

books, it is seen that the 8” bars require 174” head for a 7” diameter
pin. Half of this is 82”. So it is seen that the eye-bars fit into the
chord satisfactorily, as it is 9.38” from the gravity axis to the cover
plate and hence the bars will not interfere with the plate.

In the same manner as shown in Art. 176 the radius of gyration
about the horizontal gravity axis, or x-2 axis, is found to be about 9.5 and
9.3 about the y-y axis. Now using the least radius we obtain

312

16,000 — 70 = 13,652 Ibs.

for the actual allowable unit stress.
Then dividing this into the stress we obtain

713,000 + 13,652 =52.22

for the required area of cross-section which is practically equal to the
above assumed section and hence that section will be used.

Top Chord U2-U3. Dividing the maximum stress in this member
by the allowable unit stress found for U1-U2 (which will be practically
the same as for this member) we obtain

799,000 + 13,652 = 58.53 sq. ins.

for the approximate area of cross-section required. The following section
has about this area:

1—cov. pl. 27” x 3%” =15.180”
2—web pls. 24” x 4%" =27.000”
2184" x4"x8" = §,7Q0”
2Q186"x4" x53" =10.620”

58.520”

By taking moments about the cover plate we obtain 9.66” for the
distance from the cover plate down to the horizontal gravity, or 2-x axis.
The radius of gyrations (found as previously explained) in reference
to the 2-x axis is 9.3 and the radius in reference to the y-y axis is 9.2.

The length of the member is about 305 ins. Now, using the least
radius we obtain

16,000-70 x = 13,680 Ibs.

for the allowable unit stress. Dividing this into the maximum stress
in the member we obtain

799,000 + 13,680 = 58.40 sq. ins.

for the required area of cross-section. This is practically the same
as the approximate scction found above and hence that section will be
used.

Top Chord U3-U4. Dividing the maximum stress in the member by
the allowable unit stress found for U2-U3 we obtain

828,000 + 13,680 = 60.52 sq, ins.
452 STRUCTURAL ENGINEERING

for the approximate area of cross-section required. The following section
has about this area:

1—cov. pl. 27” x 7%” =15.180”

2—web pls. 24” x8” =30.000”

Q—1s4"x4"%x8" = = 5.720”
2-186" x4" x" =10.620”
61.520”

The allowable unit stress for this member is practically the same
as for U2-U3 and the above section is as near the required section as we
can obtain without changing the thickness of the bottom angles which
would be an objectionable thing to do, as the center of gravity would be
shifted, and hence the above section will be used.

Top Chord U4-U4. As the allowable unit stress for this member
is practically the same as for U3-U4 and the stress is practically the same
the same section will be used as for U3-U4.

198. Designing of the Bottom Lateral System—The load
specified by the specifications for the bottom laterals is 700 lbs. per foot
of span. This load, according to the specifications, is to be considered
as a live load.

For the panel load we have 700 x 25=17,500 Ibs., and for determin-
ing the stresses we have the following:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

17,500+9 = 1,944.44,
Seew = 1.78.
1,944.44x1.78= 3,461.1.
| |
A B 3K 2 a Fi : : -
Pee : Pees] ae eo ee ee
fe a S S $ + 3 2 ‘
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Fig. 325

Lateral Ab. For the maximum tensile stress in this member (see
Fig. 325, also Art. 177) we have

36 x 3,461.1= 124,600, say 125,000 Ibs.
For the area of cross-section required we have
125,000 + 16,000 =7.81 sq. ins.

Use 2—Ls 6” x 84” x 4” =8.00” net.
Lateral Bc. For the maximum tensile stress in this member we have

28 x 3,461.1 = 96,910, say 97,000 Ibs.
For the area of cross-section required we have
97,000 + 16,000 = 6.06 sq. ins.
Use 2—Ls 6” x 34” x 2” = 6.09” net.
DESIGN OF SIMPLE RAILROAD BRIDGES 453

Lateral Cd. For the maximum tensile stress in this member we
have
21x 3,461.1 = 72,683, say 73,000 lbs.

For the area of cross-section required we have
73,000 + 16,000 = 4.56 sq. ins.

Use 2—Ls 34” x 34” x 7,” =4.870” net.
Lateral De. Yor the maximum tensile stress in this member we have

15 x 3,461.1 = 51,916, say 52,000 Ibs.
For the area of cross-section required we have
52,000 + 16,000 = 3.25 sq. ins.

Use 2—Ls 34” x 34” x 3” = 4.210” net.
Lateral Ef. For the maximum tensile stress in this member we have

10 x 3,461.1 = 34,611, say 35,000 Ibs.
For the area of cross-section required we have
35,000 + 16,000 = 2.18 sq. ins.

Use 2—Ls 34” x 34 x $7” =4.210” net.

This completes the designing of the bottom laterals as the structure
is symmetrical about the center of span.

The stringer bracing is designed as previously explained for deck
plate girder bridges and the transverse struts connecting to the stringers
at the points of intersection of the laterals are designed as explained in
Art. 177 for the 150-ft. riveted bridge.

199. Designing of Top Lateral System.—The load specified by
the specifications for the top laterals is 200 lbs. per foot of span. This
load, according to the specifications, is to be considered as a live load,

For the panel load we have P=25 x 200=5,000 lbs., and for deter-
mining the stresses we have the following:

5,000 +9 =555.55.

 

 

 

 

 

 

 

 

 

 

Seewl = 1.82.
Seew2 =1.79.
Seew3 = 1.78.
Secw =1.78.
fa. 8 Cc D E | F A
= Me a7 wa Rag | . a 2 a i 2
Beg NE NL Le 4
a d .
é 7 6 5 4 3 2
@ @ © ®
Fig. 326

Lateral Be. For maximum tensile stress in this member (see Fig.
326) we have (see Art. 178)

‘ secol x 28 = 555.55 x 1.82 x 28 = 28,310, say, 28,000 lbs.
454 STRUCTURAL ENGINEERING

For the area of cross-section required we have
28,000 + 16,000 = 1.75 sq. ins.

Use 2—Ls 34” x 34” x 38 =4.220” net. These are about the smallest an-
gles used in railroad work and hence are used in this case.
Lateral Cd. For the maximum tensile stress in this member we have

B

9 secw2 X 21= 555.55 x 1.79 x 21 = 20,883, say 21,000 lbs.

For the area of cross-section required we have
21,000 + 16,000 = 1.31 sq. ins.
Use 2—Ls 34” x 34” x 8” =4.220” net—same as for Be.
Lateral De. For the maximum tensile stress in this member we have

- secw3 x 15 =555.55 x 1.78 x 15 = 14,883, say 15,000 Ibs.

Use the same section as used for Bc—2—Ls 34” x 34” x 2”.
Lateral Ef. For the maximum tensile stress in this member we have

- secw X 10 = 555.55 x 1.78 x 10 = 9,888, say 10,000 lbs.

Use the same section as used for Bc—2—Ls 34” x 34” x }”.
Strut Cc. For the maximum compression in this member we have

(see Art. 178)

21x 555.55 + CS

2

 

) = 14,166, say, 14,000 Ibs.

The length of member (c.c. of chords) is 17 ft., or 204 ins. If the strut
be composed of 4—Ls 34” x 3” x 2” the least radius of gyration will be
about 1.7. Then L/r=204+1.7=120, which is just the maximum limit
permitted by the specifications and hence the above angles will be used.
It is seen that the stress in this strut is so low that it does not really
influence the design, and, as the stress in the other struts—Dd, Ee, ete.—
are less yet, there is really no need of computing it. We will simply
make each of the other struts of 4—Ls 34” x 3” x 2” and let it go at that.

200. Design of Portal—Assuming the ends of the end posts fixed
and using the same symbols as used in Art. 179 (see Fig. 271) we have
in this case

P=5,000 lbs.,

5,000
9

R= x 28= 15,555 lbs.,

 

H=10,277 lbs.,
AV = (20,555 x 24.16) +17 =29,212, say 29,000 lbs.
sec = 1.42,
DESIGN OF SIMPLE RAILROAD BRIDGES 455
Taking moments about p (Fig. 327) we have

P m
Sk= (e+ Pera?

and substituting the numerical values given above and reducing we obtain
$=18,055 + (10,277 x 15,665) + 8.5 = 36,994, say 37,000 Ibs.

for the compressive stress in the part Bo of the portal.
In a similar manner taking moments about ¢ we obtain

S1=2,500 + 18,939 = 21,439, say 22,000 lbs.

for the tensile stress in the part ob of the portal.
For the stress in the parts po and ot of the portal we have

S2=V sec = 29,000 x 1.42 = 41,180, say 41,000 lbs.

This is compression in one and tension in the other. With the applied
forces acting as indicated in Fig. 3827, op would be in tension and of in
compression.

The designing of the sections of the members of the portal in this
case is mostly a matter of obtaining rigidity as the stresses are relatively
low. For each of the members
Bb, op and ot we will use
2—Ls 4” x 4” x 3” and for eac
of the secondary members, the
ones shown dotted in Fig. 327,
we will use 2—Ls 34x3$/'x?””.
These sections are obtained in
the manner shown in Art. 179.

201. Design of End Post.
—As stated in Art. 180, colli-
sion struts are desirable, but
they are sometimes omitted and
for the sake of variety we will
omit them in this bridge. The
cross-section of the end posts
will be the same in form as
that of the top chords. The
length of the member as re-
gards the 2-2 axis (see Fig. 265) will be taken as the full length,
which is 478 ins., and as regards the y-y axis the length will be taken as
the distance from the lower end of the end post to the portal (distance
pu, Fig. 327). This distance is about 376 ins.

Using the radius in reference to the 2-2 axis found for the chord
section U2-U3 as preliminary we obtain

p=16,000-70 = = 12,403

 

for the approximate allowable unit stress, provided there were no cross
bending on the member. Dividing this unit stress into the maximum
456 STRUCTURAL ENGINEERING :

stress in the member we obtain 782,000 + 12,403 =63.040”. Now as the
member is subjected to cross bending, due to wind, the section should
very likely be larger than this. Let us assume the following section:

1—cov. pl. 27” x 3%” =15.190”
2—web pls. 24” x 34” =33.000”
Q184"x4"x4" = 7,500”
Q136"x4"x 8" = -=11,7201”

67.400”

Taking moments about the cover plate of this section it is found
that the distance from the-cover plate to the gravity axis z-a is 10.14 ins.

The radius of gyration in reference to the 2-2 axis is 9.22. The
moment of inertia in reference to the y-y axis is 5,742 and the radius in
reference to the same axis is 9.23.

Then for the allowable unit stress, provided there were no cross bend-
ing we would have

478
16,000 - 70 9.237 12,371 lbs.

But in case of direct and bending stresses combined this can be increased
25 per cent (see specifications). So we have

12,371 x 1.25 =15,464 lbs.

for the allowable unit stress.

The maximum bending moment on the end post due to wind, consid-
ering the posts fixed at the ends, will occur at the bottom of the portal;
that is, at points p and t (Fig. 327). For this moment we have

M=10,277 x 15.665 x 12 = 1,931,870 inch-Ibs.
(see Fig. 327).

Then for the maximum bending stress we have
_ 1,931,870

f= Faia *18.5=4,542 lbs. per sq. in,

For the actual direct unit stress we have
782,000 + 67.4 = 11,602 lbs.

Now adding the bending stress to the direct stress we have
4,542 + 11,602 = 16,144 lbs.

for the actual unit stress on the member due to direct stress and cross
bending combined. As this stress is very near the allowed, being only
680 Ibs. more (16,144-15,464=680), the above assumed section will
be used.

202. Maximum Reaction on Shoe.—The dead-load reaction on the
shoe is the same as for the truss except for the half panel load at the end.
DESIGN OF SIMPLE RAILROAD BRIDGES 457

The panel load of dead load is 33,000 Ibs. (see Art. 195). Then for
the dead-load reaction on the shoe we have

44 x 33,000 = 148,500 lbs.

Placing wheel 2 at one end of the span, as shown in Fig. 278 (Art.
182) for the 150-ft. span, and taking moments about the other end we
obtain 365,000 Ibs. for the maximum live-load reaction on the shoe. Multi-
plying this by 300+ (225+300) we obtain 208,500 Ibs. for the impact.
Then adding together the above dead- and live-load reactions and impact
we obtain

148,500 + 365,000 + 208,500 = 722,000 Ibs.

for the total maximum reaction on the shoe.
The stress sheet, Fig. 328, can now be completed and then the work
of designing the details can proceed.

203. Details. The general details of the entire span are shown in
Figs. 329, 330 and 331. These details are drawn to a 3” scale on 3”
layout; that is, the outline of the truss is drawn to a 2’ scale and the
details are drawn on this layout to a 3” scale.

Joint LO. The detail of this joint is shown in Fig. 329. In
drawing this joint, first the joint line should be located, which is done
by bisecting the angle between the end post and shoe. Then the distance
from the cover plate of the end post to the center line of pin should be
determined as explained in Art. 184. Then the outline of the lower
end of the end post can be drawn and next the outline of the shoe can
be sketched. The shoe should be deep enough to resist the cross bend-
ing on it, as explained in Art. 142, and it should be long enough to cover
the rollers. The general dimensions of the shoe and number of rollers
are first assumed and then modified, if necessary, to agree with the
calculated requirements.

By drawing the end view and part cross-section of the joint, as
shown, to the left of LO, about the desired length of the rollers is obtained.
The size of the rails in the pedestal is selected, the rails are spaced and
then a suitable diameter of roller can be determined. In this case, as is
seen, 70-lb. (per yd.) rails are used. Allowing for one-eighth of an inch
to be planed off the top of these rails, the width of bearing of each
against a roller is 24’ (see Carnegie and the Illinois Steel Company
catalogue).

Resolving the stress in the end post vertically (which can be done
very rapidly by graphics) we obtain 612,000 Ibs. (about). One-half of
this, which is 306,000 lbs., should be supported by the rails to the left
of the center line of the end post. Assuming 7” rollers (as shown), the
allowable bearing per inch of length of roller is 7 x 600=4,200 lbs. Then
as there are five rails (not considering the 90-lb. guide rail at the center
line) to the left of the center line of the end post and there being seven
rollers, we have

(5 x 24) 7x 4,200 = 330,700 Ibs.

for the allowable pressure on the part of the rollers to the left of the
center line of the end post, which, as is seen, is about 24,000 lbs. more
than necessary. The first five rails to the right of the center line of the
458 STRUCTURAL ENGINEERING

end post will support the other half of the vertical component of the end
post and the other two rails, directly under the end floor beam, should
be about sufficient to support the maximum reaction of the beam. This
reaction (see Fig. 328) is 143,500 Ibs. For the allowable pressure on
the two rails we have

(2x 24) x 4,200 = 132,300 Ibs.,

which is 11,200 Ibs. less than required, but as the maximum stress in
the end post and the maximum reaction on the end floor beam do not occur
at the same time (and the other rails are understressed), the design of
the roller pedestal, as shown, is satisfactory.

The shoe proper must be made to conform with the details of the
end post. Assuming a 7” pin, we obtain

782,000 + (24,000 x 7) =4.65 ins.

for the required thickness of pin bearing on the end post, or 2.32” on

each side. We have on each side of the post, as shown, a 7%”, $” and 3”

plate and an 44” web, making in all 23” (2.375) of bearing, which is the

thickness required. ;
For the required thickness of bearing on the shoe we have

612;000 + (24,000 x 7) =3.64 ins.,

or 1.82” at each side. We have at each side, as shown, 3—2” plates,
making 1{” (1.875) bearing which is about the required bearing. It
will be seen that in addition to this bearing there is a 3” filler and a
qe jaw plate, but as these extend only a short distance below the pin,
they will not be considered for bearing.

The bottom edges of the vertical bearing plates of the shoe are
planed so that they bear firmly against the sole plate and hence the rivets
passing through the plates simply hold them together. However, there
should be as many rivets in the vertical legs of the angles connecting
these plates to the sole plate as is possible to put in so that the pressure
will be well distributed over the sole plate by the outstanding legs of
the angles.

The pin plates on the lower end of the end post should be arranged
so that they firmly grip the post. For the allowable bearing on the 3%,”
inside plate against the 7” pin we have

fs X 7 x 24,000 = 94,500 lbs.

Then the number of {” shop rivets required in single shear to hold this
plate is
94,500 + 7,200 =13.
As is seen, there are 21. For the allowable bearing on the }” outside
plate, we have
4x7 x 24,000 = 84,000 Ibs.
Then for the number of ” shop rivets required in single shear to hold

this plate. we have
84,000 + 7,200 = 12.
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462
DESIGN OF SIMPLE RAILROAD BRIDGES 463

This plate should have a sufficient number of rivets connecting it directly
to the top and bottom angles of the end post to transmit the total pin
pressure on it to these angles. There are 14 passing through it and the
angles, so the riveting is satisfactory, although counting the other rivets
passing through the plate there is an excess of 15 rivets. For the allow-
able bearing on the 8” filler we have

8x 7x 24,000 = 105,000 lbs.

Then for the number of 4” shop rivets required in single shear to hold
the filler we have

105,000 +.7,200 = 15.

As is seen, there are 23 passing through the filler, which is an excess of
8 rivets. The 4” outside plate and the 2” filler acting together tend to
shear the rivets off against the web and angles of the end post. The
combined allowable bearing of these two plates against the 7” pin is

84,000 + 105,000 = 189,000 Ibs.

Then for the number of §’” shop rivets required to hold these two plates
when considered as acting together we have

189,000 + 7,200 =27.

As is seen, there are 3% — an excess of 10 rivets.

Although from the above there appears to be an excess of rivets in
these pin plates, nevertheless, the rivets are spaced about as sparingly
as possible, while yet obtaining good details, so the riveting is satisfactory.

According to the specifications, the net area of cross-section through
the pin hole of the bottom chord (at LO) should be 25 per cent more
than the net section of the member. The net section of the member (see
Fig. 328) is 30.985”. Then the area of section through the pin hole
should be 38.720”. i

As is seen, the 7%’ plates have a net area through the pin hole of
5.250”, the 4” plates have 13.000”, and the web has 16.254”, making a
total of 34.510” for the plates. In addition, each of the angles can be
considered to the extent of four rivets in shear. Each angle then is
equivalent to (4x 7,200) +16,000=1.89”, or 7.20” for the four angles.
Adding this to the net area of the plates we obtain 34.51 + 7.2=41.710”,
which is 2.990” more than required.

According to the specifications (A. R. E. Ass’n), the net area along
the center line between the pin and the end of the member must be equal
to the net area of cross-section of the member. As is seen, we have

(4+ ge +8) (144-34) — (1x 34) = 31.25 sq. ins.

for the net area along the center line between the pin hole and the end
of the member. As this is practically equal to the net area of cross-sec-
tion of the member, which is 30.984”, the detail of the end of the bottom
chord, as shown, is satisfactory as far as sections are concerned.
Considering the net area of cross-section through the pin hole, the
strength of the 7%” outside plate in tension is 6x 7% x 16,000 =42,000
464. STRUCTURAL ENGINEERING

lbs. Then for the number of }” shop rivets required to the right of the
pin hole to transmit this stress in single shear we have ,

42,000 +7,200=5.8, say 6.

As is seen, 6 rivets are used.

The strength of the 4” inside plate in tension is 13x4x 16,000
=104,000 lbs. Then for the number of $” shop rivets required to the
right of the pin hole to transmit this stress in single shear, we have

104,000 + 7,200 =15.

As is seen, 16 rivets are used. From the above, it is seen that the detail
of the end of the bottom chord at LO is correct both as to section and
riveting and hence the detail is satisfactory.

Resolving the forces in the
end post horizontally and _ verti-
cally, we have the loading on the
pin at LO shown at (a), Fig. 332.
Considering first the bending mo-

*

)pa1Z000"

 

 

 

 

 

 

 

 

ws a“
423000 ( 493000 ment on the pin due to the hori-
4at(@) zontal forces, shown at (b), the

8 . :

S bending moment at d is zero and
\ at e it is equal to 0+ 246,500x2=
493,000’#, which is the maximum
2165004246500", horizontal moment, the forces
FA Yu being symmetrical in reference to

= s the center line ec.
rere ae) The maximum bending mo-
246500"\ 2 en oe nee on fan pin due to the vhost
(Hor) orces shown at (c), as is readily

Fig. 382 seen, occurs at both g and h and is
equal to 306,000 x #=229,500’#.
Then for the resultant maximum bending moment we have

M= 93,000" + 229,500" = 543,000 inch lbs.

Applying Formula (1), Art. 83, taking f as 25,000 Ibs., we find
that a 64” pin is required for bending. For the maximum shear on this
pin we have

 

‘ 2
306,000 +7 (2) = 10,385 Ibs. per sq. in.

So the 64” is large enough as far as shear is concerned, as 12,000 lbs.
per square inch is allowed. The bearing as computed above is for the 7”
pin and to use a 64” pin would necessitate increasing the bearing by
about one-fourth, which would increase the weight of metal twice as much
as would be saved on the pin by reducing it to 64 diameter. So there
is really no economy in using the smaller pin and therefore the 7” pin
will be used.

The other details at LO are considered to be self-explanatory, espe-
cially after Arts. 184 and 186 are carefully read.
DESIGN OF SIMPLE RAILROAD BRIDGES 465

Joint U1. In drawing this joint the first thing to do is to locate
the joint line by bisecting the angle between the end post and top chord
and then the outlines of the members can be drawn and the portal located
as explained in the case given in Art. 184; and next the required bearing
on the pin can be calculated.

We will assume a 7” pin—the size used at LO; then for the required
thickness of pin bearing on the end post we have

782,000 + (24,000 x 7) = 4.65 ins.,

which is about 443”. As is seen, we have on each side of the post a
qs” and a 8” plate and a 2” filler. This with the 44” web makes 23”
bearing on a side or 43” for the member, which is practically the required
bearing.

For the required thickness of pin bearing on the top chord we have

713,000 + 168,000 = 4.24 ins.

As is seen, we have on each side of the chord 1—}” and 2— 7” plates
and 1— 3%” filler. This, with the 7%’ web, makes 22” bearing on a side
or 43” for the member. This is about 4” more than required, but it is
necessary to use this thickness in order to pack the joint.

At least one pin plate in each case should extend at least 6 ins.
beyond the end of the tie plate.

The number of rivets required to connect the pin plates to the end
post is determined in the same manner as shown above for joint LO.
For the allowable pin pressure on the 7%” plate we have

x qq x 24,000 = 73,500 lbs.

Then for the number of §” shop rivets required to transmit this in single
shear, we have
73,500 + 7,200 =11 (about).

It is obvious that most of the pin pressure on this 7%” outside plate will

be first transmitted to the 3” plate, then to the §” filler, and on to the
web and angles of the end post instead of being transmitted directly to
the web and angles by the rivets, as the rivets are quite long and conse-
quently will bend to some extent. Let us assume that the total pin
pressure on the 7%” plate is transmitted to the 3” plate. Then for the
total pressure on the 3” plate we have

73,500 +7 x 8 x 24,000 = 178,500 Ibs.

For the number of }” shop rivets required to transmit this force, we
have .
178,500 + 7,200 =25 (about).

There should be about enough rivets connecting this 8” plate to the angles
of the end post to transmit this 178,500 Ibs. directly to the angles in
order to distribute the pin pressure well over the cross-section of the
member. As is seen, there are 22 rivets, which is about the correct num-
ber. For the allowable pin pressure (bearing) on the 3” filler we have

7x 8 x 24,000 = 105,000 Ibs.
466 STRUCTURAL ENGINEERING

Then for the number of #” shop rivets required to transmit this pressure
in single shear, we have

105,000 + 7,200=14.5, say 15.

As is seen, there are 16 rivets between the lower end of the filler and
the 7%”’ plate, which we can consider as holding the filler.

As is seen from the above, if the rivets just below the pin be consid-
ered only to take the pressure on the 7,”’ outside plate, the riveting of
the pin plates to the end post at U1 is about correct.

The allowable pin pressure on the 7” inside pin plate of the top
chord is 7x 7q_x24,000=73,500 lbs. It requires about 10—{%” shop
rivets to transmit this in singular shear. As is evident, practically all
of the 73,500 Ibs. will be transmitted first to the 4” inside pin plate
and from there on to the web, and hence there should be a sufficient
number of rivets connecting the 4” plate to the web to transmit the
combined pin pressure exerted on both the 7%” and 4” plate.

For the allowable combined pin pressure of the two, we have
73,500 + (7x 4x 24,000) =157,500 lbs.

It requires about 22—-2” shop rivets to transmit this in single shear. As
is seen, there are 38 passing through the 3” inside plate. But 10 of
these should be considered as holding the 7%” inside plate, thus really
leaving 28, which is 6 more than required.

The allowable pin pressure on the 7%’ outside pin plate is 73,500
Ibs. To transmit this in single shear requires about 10— %” shop rivets.
As this plate connects directly to the angles of the chord, there should
be enough rivets connecting it to the angles to transmit the entire 73,500

lbs. As is seen, there are 14 rivets connecting the plate to the angles. --—

So the riveting is satisfactory as far as the 7%’’ outside plate is con-
cerned. The allowable pin pressure on the 7%,” filler is

7 x 4% x 24,000 = 94,500 Ibs.

To transmit this in single shear requires about 13 — }’”” shop rivets.
As is seen there are 26—twice as many as needed. As the web of the
top chord is only 7%” thick, it is a question as to whether the rivets
connecting the pin plates to the member are not over-stressed in bearing
on the web.

The 3” inside plate has a total pressure, as given above, of 157,500
Ibs. There are 38-10 (the 10 being considered to hold only the 7”
inside plate) to transmit this, which stresses each rivet 157,500+28
= 5,625 lbs. The pin pressure on the 3%” outside filler is 94,500, as
given above, and this stresses each rivet 94,500 +26 =3,634 lbs. Adding
these values we have 5,625+3,634=9,259 lbs. for the maximum rivet

bearing on the 7%’’ web’ For the maximum allowable bearing of a {”

shop rivet on 7%” metal, we have

ais X $x 24,000 = 9,180 Ibs.,

which is about the bearing exerted. So, taking all in all, the riveting of
the pin plates of the top chord at U1 is about correct.
The calculations for the details at the end of the hanger at U1 are
DESIGN OF SIMPLE RAILROAD BRIDGES 467

made in the manner as shown above for the end of the bottom chord
at LO.

The net section of the member is 17.185’. Then the net section
through the pin hole should be 17.18 (1+ 0.25) =21.570”,

The net area of the two channels through the pin hole is
19.9-(%x0.4x2)=14.30". The net area of the 2—4” plates through
the pin hole is (12.5x4x2)-(7x4x2)=5.50” and of the 3” plates
it is 6.870”, making in all 14.34+5.54+6.87=26.670”, which is about
54” more than necessary.

The net section along the center line between the pin hole and the
end of the hanger should be equal to the net section of the member, or
17.180”, according to the specifications. We have (3.05x 9.25).
— (38.05 x 8.5) =17.540”, which is about the section required, and as
the distance shown as 94” can not be increased, on account of clearance,
the detail of the end of the hanger at U1 is satisfactory as far as the

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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section is concerned. The strength of the 4” plate in tension is
16,000 x 5.5 =88,000 lbs. To transmit this in single shear requires about
12— ¥” shop rivets, or 6 on a side, which should be below the pin. As
seen, there are 8. The strength of the §” plate is 16,000 x 6.87 =109,900
lbs. This requires about 15— {” shop rivets, or 7.5 on a side. As
seen, there are 8 passing through this plate outside of the 4” plate.
So, taking all in all, the detail of the end of the hanger, as shown at U1,
is satisfactory.

In determining the bending moment on the pin at LO, there is no
question concerning the live-load stress in the members as the maximum
in the end post and bottom chord occurs simultaneously, but at U1 the
case is different, for at that joint the maximum stress in the members
does not occur simultaneously, and, consequently, it is necessary to ascer-
tain the position of the live load for maximum bending moment on the
pin at that joint. This can be done only by trial, as the bending on the
468 STRUCTURAL ENGINEERING

pin due to any member not only depends upon the stress in the member,
but upon its lever arm as well.

It is seen from Fig. 329 that the hanger and the diagona: (at U1)
have the longest lever arms and hence most likely the pin will be sub-
jected to the maximum moment when these members have about the
maximum simultaneous stress. As this can be ascertained only by trial,
let us place the loads as shown at (a), Fig. 333, in which case wheel 4
is at B. Taking moments about J (using Table A) we obtain the reaction
263,100 lbs. at A, and taking moments about B we obtain the concentra-
tion 19,200 Ibs. at A. Further, taking moments about C and A we
obtain the concentration 75,600 Ibs. at B—all due to Cooper’s E40 load-
ing. Then beginning at joint 4 and drawing the stress diagram shown
at (b) we obtain the stresses in all the members at joint b (U1) due to
Cooper’s £40 loading and multiplying these by #§ we obtain the live-
load stresses shown at (a), which are due to the £50 loading. Next,
multiplying each of these stresses given at (a) (except the stress in the
hanger) by 300+(225+300) we obtain the impact stress in each
member. Then adding together these live-load stresses-and the impact
and the dead-load stresses (given in Fig. 328), we obtain the stresses
shown at (c).

There is some question as to how much impact should be added
to the hanger. Should ZL in the impact formula be taken as 50 or 225?
As is seen, the impact really added is just sufficient to balance up the
vertical components on the pin. This is a fair average for the impact
and seems to be a rational value.

The stresses given at (c) are resolved graphically into horizontal
and vertical components as shown. The horizontal components for half
of the pin are shown at (d) and the vertical components for half of
the pin are shown at (e).

Proceeding as outlined in Art. 83, for the bending moment on the
pin due to the horizontal components, we have

M,=0
M,=04 245,000 x 44 =+168,400 inch lbs.
M,=+168,400 + (-98,500 x 225) =—-34,700 inch lbs.

Similarly, for the bending moment, due to the vertical components,
we have
M.=0
M,=0+ 305,000 x 44 =+209,700 inch lbs.
M, = 209,700 + (212,000 x 2445) =+646,900 inch lbs.
M,, = 646,900 + (90,500 x 18) = +793,900 inch lbs.

As the members are symmetrically arranged in reference to the cen-
ter line of the chord, the moment at g, due to the horizontal forces, is
constant between the two diagonals and hence for the resultant maxi-
mum moment at h, we have

Mes [34,700" + 793,900 °='797,200 inch lbs.

This, as is readily seen, is the maximum resultant bending moment
on the pin when the live load is in the position shown at (a), Fig. 333.
DESIGN OF SIMPLE RAILROAD BRIDGES 469

In fact, it is about the maximum moment that can occur on the pin.
Applying (1), Art. 83, taking f as 25,000 lbs., we find that the above
moment (797,200”#) requires a 63” pin and hence the 7” pin assumed
is about the correct size as far as the bending moment is concerned.

In case of doubt, in any case, as to the position of the live load, the
loading can be placed in different positions and the moment on the pin cal-
culated for each position in the same manner as shown above for that one
position. This method of procedure, as is evident, is tedious. As a rule,
the position of the live load can be ascertained near enough (as the impact
is questionable) by mere inspection.

The maximum shear on the pin at each side of the chord is about
330,000 lbs., which is the resultant of the stresses in the hanger and
diagonal. Then for the maximum shearing stress on the 7” pin we have
330,000 + 38.4=8,590 lbs. per sq. in.

From this it is seen that the 7” pin is amply large as far as shear
is concerned, 12,000 lbs. being permissible, and as it is the correct size
for bending it will be used. .

The details of the portal and laterals at Ul are considered to be
self-explanatory. The number of rivets required in each connection is
obtained by developing the section of the member connected, as explained
in Art. 184.

Joint Li. All of the details at this joint are practically self-ex-
planatory. The pin supports only the weight of the bottom chord. The
pin plates on the bottom chord are intended solely to replace the metal
cut from the chord by the pin hole. The rivets connecting the bottom
of the hanger to the lateral plate should be sufficient to transmit the
component of the lateral in panel LO-L1 along the bottom chord. This
component is about 105,000 lbs. This requires

105,000 + 6,000 =17—-¥” field rivets.
16 are used.

The component of the stress in the same lateral along the floor
beam is about 70,000 lbs. This requires about 12—” field rivets. 12
are used.

The pin plates on the hanger are for the purpose of reinforcing the
hanger for bending. The bending on the hanger about the pin (at Z1),
which is due to the longitudinal component of the stress in the bottom
lateral, is 105,000 x 14.25=1,496,250’#. The moment of inertia of the
two channels is 625.2, and the moment of inertia of the 2— 7%” plates
is 246.1, making a total moment of inertia of 871.3. Then for the
maximum fiber stress in the hanger at L1, due to bending, we have

f= (1,496,250 x 7) +871.3 = 12,020 Ibs. per sq. in.

From this it is seen that the hanger is amply strong to resist the bending
at L1, .

The rivets connecting the lug angles to the bottom of the hanger
should be sufficient to transmit the longitudinal component of the stress
in the lateral. These rivets are in double shear and bearing on about
42” of metal. For the number of §” shop rivets required in double
shear, we have

105,000 + (7,200 x2) =17.3, say 8.
470 STRUCTURAL ENGINEERING

For the number required in bearing on the 4%” metal, we have
105,000 + 17,060=6.

As is seen, 8 are used—the number required for shear.

Joint L2. The details of this joint are shown in Fig. 330. The
calculations for the end of the bottom chord are just the same as given
above for the other end of that member (at LO).

The pin bearing on the post should be sufficient to transmit the ver-
tical component of the maximum stress in the diagonal U1-L2. This
component is equal to about 300,000#. Then for the required bearing
on the post, assuming a 7” pin, we have

300,000 + (7 x 24,000) =1.78 ins.

or 0.89” on a side. The thickness of the web of the 40# channel is
0.52” (about $3”). This with the 7%” pin plate makes $4” or 0.96”,
which is about 3%; more than needed, but as counter-sinking (as a rule)
is not permitted in metal less
than qe” thick the ae” plates

shown will be used.
About the maximum bend-
ing moment on the pin will occur
zy {0 when the bottom chord L2-L3
has maximum stress. (This can
a ¥¢ be determined in the manner

= SNeSSS -

   

~486000*

eee L

 

 

 

 

 

* shown above for joint U1.) The

eg 2 wheels are in the position shown
a) t in Fig. 322 when this stress

gs (b) occurs. Taking moments about
100500 6 te yegs00* J (Fig. 322) with wheel 7 at C
243000" > _ = we can obtain the reaction R at
“Qf £75000 A and taking moments about B

Fig. 334 we can obtain the concentration

at A. Then the stress in the
bottom chord AB can be obtained quickly by analyzing graphically the
joint A. Then the stress in chord BC is known, as it is equal to the stress
in chord AB. Next the stress in the diagonal bC (U1-L2) is readily de-
termined, as the horizontal component of its stress is equal to the difference
between the stress in chord BC and CD and as the vertical component of
the stress in the diagonal is equal to the stress in the post, we can readily
determine the stress in the post and thus we would have all the live-load
stresses in the members at L2 determined.

After determining these live-load stresses, the impact is determined
in each member by multiplying the live-load stress in it by
300+ (225+300). Then by adding together the. live-load stress and
impact and the dead-load stress, we obtain the total stress in each
member as given at (a), Fig. 334. Two-thirds of a panel load of dead
load is added to the post to balance up the vertical component of the
diagonal. However, this should really be added to the post as the pin
is below the floor beam connection and hence most of the weight at the
joint is transmitted to the lower end of the post.
DESIGN OF SIMPLE RAILROAD BRIDGES 471

The horizontal components of the stresses on one-half of the pin
are shown at (b) (Fig. 334), and the vertical components on one-half
of the pin are shown at (c).

For the bending moment on the pin, due to the horizontal com-
ponents, we have ;

M.=0
M,=0+ (175,000 x 13) = 306,250 inch lbs.

M, =+306,250 + (—68,000 x 22) = +144,750 inch Ibs.
My, =+144,750 + (100,500 x 14) =+295,500 inch Ibs.

For the bending moment on the pin, due to the vertical component,
we have

M,= 0
M,.=0 + (124,000 x 1445) = 178,200 inch lbs.

Then for the maximum resultant moment on the pin, we have

M=- /295,500° + 178,200” = 345,000 ineh lbs.

Then by applying (1), Art. 83, taking f as 25,000, we find that the
above moment (345,000’#) calls for a 54” pin (about).

The maximum shear on the pin is about 175,000 lbs., which requires
that the pin be only about 47%,” diameter.

From the above it is seen that the assumed 7” pin is larger than
necessary, but the 7” will be used, for little would be saved by reducing
the size, as the thickness of pin bearing on the bottom chord (L2-L0O)
and also on the post U2-L2 would have to be increased if a smaller pin
were used.

The other details at £2 are readily understood. The calculation
of the details of the bottom laterals, shown at this joint, is mostly a
matter of developing the members. The rivets connecting the bottom
of the post to the lateral plate should be sufficient to transmit the longi-
tudinal component of the stress in the lateral in panel L1-L2.

Joint U2. The top chord has a butt joint at this point; that is,
the ends of the members are planed so that they bear tightly against
each other and hence the stress is transmitted from one chord member
to the other chord member without passing through the pin. The splice
plates on the chord are intended only to hold the chords in line. How-
ever, there must be sufficient pin bearing on the chord to take the com-
ponent (along the chord U2-U3) of the stress in diagonal U2-L3. The
maximum stress in the diagonal (see Fig. 328) is 280,000 lbs. Resolv-
ing this along the chord U2-U3 (which can be done graphically) we
obtain a component of 159,000 Ibs. To transmit this component, assum-
ing a 6” pin, requires 159,000 + (24,000 x 6) =1.1” thickness of bearing,
or about 4” on each side of the chord. As is seen, there are 2— 3”
plates and a 3%” web, making in all 1,3,” bearing on each side of the
chord, which is quite excessive.

The pin bearing on the post should be sufficient to transmit the
maximum stress in the post, which is 221,000 lbs. (See Fig. 328.)
This requires a bearing (assuming a 6” pin) of 221,000 + (24,000 x6)
472 STRUCTURAL ENGINEERING

=1.54” or 0.72” on a side. The web of the 40# channel is 43” thick.
This with the 7” plate makes $4” thickness of bearing on each side of
the post, which is excessive.

The maximum bending moment on the pin will occur when the
diagonal (U2-L3) has maximum stress, as the post (U2-L2) has about
the maximum stress at the same time. By placing the live load for
maximum stress in the diagonal (as shown in Fig. 315) and determin-
ing the reaction at A and the concentration at C (these are given on
pages 432 and 433) the live-load stress in each of the members at U2
can be readily determined by graphics. Then determining the impact
by multiplying each stress by 300+ (150+300) and adding together
this impact and the live- and dead-load stresses and resolving the
resulting stresses horizontally and vertically, the maximum bending
moment on the pin can be determined in the same manner as shown above
for the other joints. This maximum moment is about 190,000 inch lbs,
This requires about a 44” pin. From the above, it is seen that the 6”
pin is larger than theoretically required. But it is usual practice to
limit the minimum diameter of a pin to 35 of the width of the smallest
eye-bar connected. The eye-bars at U2 (as seen) are 7” wide. So,
according to this requirement, the pin should be 5.6’ in diameter and
hence the assumed 6” pin is about the correct size, and will be used.
The other details at U2 are readily understood.

The details at U3 are similar to those at U2 and the details at L3
are very similar to those at L2 and hence the calculations of the details
at these joints are quite similar. to those given above and consequently
are omitted. The same can be said regarding joints U4 and L4 (Fig.
331). The details at these joints are readily understood.

The calculations of the details of the intermediate floor beams,
shown in Fig. 331, are almost exactly as given in Art. 186 (pages 410
to 412) for the floor beams in the 150-ft. riveted span.

After the general drawings (Figs. 329, 330 and 331) are completed,
the shop drawings and shop bills for the structure can be made by
proceeding in the same manner as outlined in Art. 186 for the 150-
ft. span.

204, Camber.—The camber of the trusses in this case is obtained
by lengthening the top chord 4” for each 10 feet of horizontal length,
as explained in Art. 185. The corresponding lengths of the diagonals
are calculated by using the mean lengths of the top and bottom chords
in each case. For example, the horizontal uncambered length of top
chord U2-U3 (Fig. 330) is 25 ft. So the cambered horizontal length
would be 25’— 03%”. Taking this as the base of a right angle triangle,
and the difference between the lengths of posts U3-L3 and U2-L2 as the
altitude, the cambered length of the top chord U2-U3 (given as 25’ — 54”)
is computed as the hypotenuse of the triangle. Then taking 25’ - 0.5,”
as the base of a right angle triangle and the length of post U2-L2 as the
altitude, the length of diagonal U2-L3 (given as 45’—44/’) is calculated
as the hypotenuse of that triangle. The length of the end post is not
increased.

The determination of actual amount of camber of bridge trusses and
also the determination of the theoretical camber of the same will be
given later.
DESIGN OF SIMPLE RAILROAD BRIDGES 473

205. Graphical Determination of Live-Load Stresses in Curved
Chord Pratt Trusses.—The influence line method, outlined in Art. 103,
is the most convenient graphical method to use in this case, as the stresses
in the truss members are thus determined directly without considering
either shears or moments.

Chord Members. Let it be required to determine the live-load stress
in the top chord cd of the truss shown in Fig. 335, due to Cooper’s
E40 loading.

It is obvious that a single load moving from J to the left over the span
will cause a stress in chord cd which will be zero when the load is at J and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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increase constantly until the load reaches point D and that this stress will
then decrease constantly as the load moves on to the left from D, becoming
zero when the load reaches A. Then evidently the influence line for
the stress in chord cd is of the form c’o’d’, shown at (c). Now, know-
ing the form of the influence line, the next thing is to construct it. This,
as is readily seen, is a simple matter after the ordinate 0’g’ is known.
This ordinate 0’g’, as we know, is the stress in the top chord cd, due
to a unit load at D. Then, of course, the first thing to do in construct-
ing the desired influence line is to determine the stress in the top chord
ed, due to a unit load at D. This stress can be determined very readily
ATA STRUCTURAL ENGINEERING

either analytically or graphically. The graphical determination is the
more convenient and hence that method will be used. By drawing, at
(a), the lines OO’, OD and O’D we obtain, as we may say, the truss
ODO’, which will have end reactions exactly equal to those of the bridge
truss, due to a unit load at D, and as the member OO’ coincides with
top chord cd, the stress, due to this unit load at D, will be the same
in the two. This is readily seen, for, taking moments about D, we
obtain
a

S=Rr

for the stress in each, where R represents the reaction at either A or
O due to a unit load at D.

By drawing mp (at (b)) and laying off nm (equal to one inch)
equal to one pound and drawing np, we have the influence line for the
reactions at A. Then gh is equal to the reaction at A and also at O, due
to a unit load at D.

Then by drawing from h a line parallel to OD and from g a line
parallel to OO’, we obtain the line og, which is equal (in inches) to the
stress in chord cd, due to the unit load at D. In other words, by con-
structing a diagram of the forces at O, we obtain the stress in OO’,
which is equal to the stress in the chord cd, due to a unit load at D.

Then by drawing c’d’ (at (c) ) and laying off 0’g’=og and draw-
ing c’o’ and o’d’, we have the desired influence line for the stress in top
chord cd.

Then placing the loading for the maximum moment about D, accord-
ing to Art. 91 (see Fig. 323), and drawing the ordinates 1, 2,.... 7, as
shown, and multiplying each by the load or group of loads at each, and
adding together the results, the desired maximum stress in top chord cd
will be obtained.

Ordinates 1 and 2 are drawn at the center of gravity of equal
groups. Then by adding together the ordinates 1 and 2 and multiply-
ing by the weight of one of the groups, the stress in top chord cd, due
to the two groups, will be obtained. Similarly, by adding together the
ordinates 3 and 4 and multiplying by the weight of one of the groups
(at either of the ordinates), the stress in the top chord cd, due to those
two groups, will be obtained. Next, adding together ordinates 5 and 6
and multiplying by the weight of one of the wheels, the stress in top
chord ed, due to the two wheels (1 and 10), will be obtained. Multiply-
ing the ordinate 7 (which is equal to one-half of ordinate 8) by the total
uniform load the stress in chord cd, due to the uniform load, is obtained.
Then adding together all of these results, the total maximum live-load
stress in top chord ed is obtained.

The combined length of ordinates 1 and 2 (which can be quickly
combined by the use of a pair of dividers, as shown in Example 1 of
Art. 100) scales 1.54 ins. This means that if (as nm=1”=1#) one
unit load (1 Ib.) be placed on the truss exactly over ordinate 1 and
another exactly over ordinate 2, these two unit loads would produce a
stress of 1.54 Ibs. in top chord cd. Then evidently the two groups of
wheels at these ordinates, each weighing 80,000 lbs., will produce a stress
DESIGN OF SIMPLE RAILROAD BRIDGES 475

in the top chord ed of
1.54 x 80,000 = 123,000 Ibs.

The combined length of ordinates 3 and 4 scales 1.73 ins. Then
for the stress in top chord cd due to the two groups of wheels at these
ordinates, we have

1.74 x 52,000 = 90,480, say 90,500 lbs.

Similarly for the stress due to wheels 1 and 10, the combined length
of ordinates 5 and 6 being 1.23 ins., we have

1.23 x 10,000 = 12,300 lbs.

Ordinate 7 scales 0.41 ins. Then for the stress in top chord cd,
due to the uniform load, we have

0.41 x (2,000 x 105) = 86,100 lbs.
Now, adding together the above, we obtain
123,000 + 90,500 + 12,300 + 86,100 = 311,900, say, 312,000 Ibs.

for the maximum stress in top chord cd due to Cooper’s E40 loading,
and multiplying this by £2 we obtain 390,000 lbs. for the stress due to
the E50 loading. As is seen from Fig. 328, this is 6,000 lbs. less than
the actual stress, but as this difference is less than 2 per cent, the result
is accurate enough. That is, the change of section of the top chord
resulting from the error would be negligible.

As the horizontal component of the stress in top chord cd is equal
to the stress in bottom chord DE, as previously shown, and the position
of the loading for maximum stress in the two members being the same,
the influence line for the stress in bottom chord DE can be quickly
constructed after the one for the top chord cd is completed. Thus: By
drawing or we obtain the stress in bottom chord DE, due to a unit
load at D. Then drawing D’E’ (at (d)) and laying off r’o’’=or and
drawing D’r’ and 7’E’, we obtain the influence line for the stress in
chord DE. The stresses can then be determined in this member in the
same manner as shown above for top chord cd. However, the stress
in bottom chord DE can be obtained more readily as the horizontal
component.of the stress in top chord cd. The method, of course, can
be reversed; that is, the influence line for the stress in bottom chord
DE could be constructed and the stress in DE determined and then the.
stress in top chord cd could be obtained by multiplying the stress found
in DE by the secant of the slope angle of chord cd.

The construction of the influence lines and determination of the
stresses in the other chord members are practically the same as shown
above for chords cd and DE.

For example, by drawing 0”0’’, O”’E and EO”’ and constructing
a diagram at ordinate z of the forces at O”, the stress in top chord de,
due to a unit load at EF, is obtained. Then the influence line for the
stress in top chord de, as well as for bottom chord EF, is readily
constructed.

As another example, by drawing tv (at (b) ) parallel to the end
476 STRUCTURAL ENGINEERING

post, we obtain vu, which is equal to the stress in bottom chord AB (also
BC), due to a unit load at B. Then drawing A’B’ (at (e) ) and laying
off v’u’ =u (which can be done by using a pair of dividers) and drawing
A’v’ and v’B’, we obtain the influence line for the stress in bottom chords
AB and BC, shown at (e).

Web Members. As an example, let it be required to determine the
stress in diagonal cD. The influence line for the stress in this member,
as explained in Art. 103, will be of the form A’s’o’B’, shown at (c) in
Fig. 336. To construct this influence line, first draw the influence lines
for reactions, as shown at (b). Then gh represents the reaction at A,
due to a unit load at D. Next, draw OO’, OD and O’D. Then, drawing
og and oh (at (b)) parallel, respectively, to OO’ and OD, and ko
parallel to diagonal cD, we have the stress in the diagonal cD, due to a
unit load at D, represented by this line ko. Then, by drawing A’B’
(at (c) ) and laying off k’o’ equal to ko, the part o’B’ of the influence
line can be drawn.

Next, draw OI and IDO’, as shown at (a). Then, drawing vs and
su parallel, respectively, to JO’ and OO’ and ts parallel to the diagonal
cD, we have the stress in the diagonal cD, due to a unit load at C, repre-
sented by this line ts. Then, by laying off t’s’ (at (c) ) equal to ¢s,
the parts s’A’ and s’o’ of the influence line can be drawn, which will
complete the construction of the influence line A’s’n’B’.

Any load at any point to the right of N will produce tension in
diagonal cD, and any load to the left will produce compression in the
same. Then, evidently, to obtain the maximum tension in the diagonal,
all the loads should be placed to the right of N. According to Art.
190, a wheel will be at D when the maximum tension in cD occurs. Then,
evidently, the loading will be in the position for maximum stress in
diagonal cD if the wheel be placed at k’ that brings wheel 1 the closest
to N, the limit being that wheel 1 should not be to the left of N. From
this it is seen that the position of the loading for maximum stress in
the diagonal can be determined directly from the influence line instead
of applying the criterion of Art. 190.

The position of the loading for maximum tension in diagonal cD is
shown at (c). By multiplying the ordinates by the loads, as explained
above, the maximum tension in diagonal cD will be obtained.

In any case when a group of wheels comes at a point where the
influence line changes direction, as wheels 2. . . 5 do in this case, the ordi-
nate at the center of gravity of the group can not be used. In such
eases the weight of each wheel can be multiplied by the ordinate at it,
or the ordinates at the center of gravity of the wheels to either side of
the change of slope of the influence line can be used, or the sum of all
of the ordinates can be obtained by the use of dividers (see Example
1, Art. 100), and this sum multiplied by the weight of one wheel. Thus,
in the case shown at (c) the tensile stress in diagonal cD, due to wheels
2... 5, can be obtained by multiplying ordinate 1 by the weight of wheel
2, and ordinate 3 by the weight of wheels 3, 4 and 5, or by multiplying
the sum of all the ordinates 1, 2, 3 and 4 by the weight of one wheel.

The maximum compression in diagonal cD can be obtained by revers-
ing the loading; that is, bringing it on from the left, and placing the
wheel at ¢’ that brings wheel 1 closest to N without passing that point.
DESIGN OF SIMPLE RAILROAD BRIDGES 477

The actual stress is then obtained by multiplying the loads by the ordi-
nates, as previously explained.

The influence line for the maximum compression in post dD is
shown at (d). To obtain this influence line OO’, OD, O’D, ODI’ and
O’I’ should be drawn. Then, by drawing fy parallel to OO’, yl parallel
to OL’, and yz parallel to post dD, we have the stress in post dD, due
to a unit load at E, represented by yz. Then, drawing C’D’ at (d),
and laying off y’z’ equal to yz, we can draw the part y’D’ of the influ-
ence line. Next, drawing gw parallel to OO’, wh parallel to O’D, and
ew parallel to post dD, we have the stress in post dD, due to a unit load
at D, represented by line ew. Then, laying off e’w’, at (d), equal to ew,
we can complete the influence line C’w’y’D’ by drawing C’w’ and w’y’.

 

 

 

 

 

 

 

 

lo"
-
pe
we
as
oe a
Cc / Ms “
’ \ “
/ \ Ce
On J < Bee
v7
/ t
| \ (a) L ;
SS T : ; oF \ i
A BS OP oe
wee 1 aa
oa NY = ' $ ie
\ {
& 8
a \ i iF 1
t ye\k! h , 2
m Z 7 ' m
TT (B)
: :
! oy
1
'
-

 

Fig. 336

The maximum compression in post dD can then be determined by
placing the loading on PD’ with the wheel at 2’ that brings wheel 1
the closest to P, and proceeding as explained above in the case of the
diagonal dD. Influence lines for maximum tension in the posts will
be considered later.

The influence lines for maximum stress in the other diagonals and
posts and determination of the stresses are similar to the above.

The influence lines can be constructed more readily in the following
manner, taking first the one shown at (c) in Fig. 336:

The ordinate o’k’ can be obtained in the manner explained above,
478 STRUCTURAL ENGINEERING

and point N can be located by drawing OCz and O’Dz, as x is directly
over N. Having the ordinate o’k’ determined and point N located, the
influence line A’s’o’B’ can be drawn. The ordinate y’z’, shown at (d),
can be obtained in the manner explained above, and point P can be
located by drawing ODa’ and O’E2a’, as 2’ is directly over P. Having
the ordinate y’z’ determined and the point P located, the influence line’
C’w’y’D’ can be drawn.

A better way to locate points N and P is to draw the line OJ.
Then point F, where OJ intersects the diagonal cD, is directly over
point N; and point G, where the line OJ intersects diagonal dE, is
directly over point P.

It can be readily shown that this construction is correct. Referring
to Fig. 337, the structure OCDO’cO, supporting the loads P1 and P2,
is a rigid structure. It is evident that the loads P1 and P2 could have

 

Fig. 337

such relative weights that the member cD (also cC) would have zero
stress. In that case, OCDO’ would be an equilibrium polygon. Then
by prolonging the segments O’D and OC to intersection, the point x
is obtained, which would be at the resultant, or center of gravity, of
the loads Pl and P2 (see Art. 45). Then both of the loads could
be placed at 2 and the stress in member cD (also cC) would be
zero. Now, it is evident that as long as Pl and P2 have this same
relative value, OCDO’ would remain an equilibrium polygon, regardless
of the actual weights of the loads P1 and P2; so, evidently, the stress
in cD due to any load at x will be zero, and hence 2 is the point of
zero stress for diagonal cD.
From similar triangles Dmx and DJO’, we have

zo
J

ll

oe

from which we obtain
DESIGN OF SIMPLE RAILROAD BRIDGES 475
From similar triangles OAC and «mC, we have
a

>

g
a

| 6

Yrom which we obtain

Now equating (1) and (2) equal, we have

a oe
b ae i:
from which we obtain
9. ae
OD TTT TEETER nena (3)
From similar triangles cFT and DFU, we have
Ew
FT TTT enn tee e ees (4)
But from similar triangles O’JO and cOT, we'have
7 2!
Ah LD’
from which we obtain
a
y= L h.
And from similar triangles JAO and JDU, we have
o¥
a
from which we obtain
= : ke
Now substituting these values of y and z in (4), we obtain
sah
E = bk eee ee eee eee eee Oe ee ht : oe (5).
Equating (3) and (5) equal, we obtain
Dg
DP TTT EEE (6)
But c=d-gandt=d-s.

So, substituting these values in (6) and reducing, we obtain

9-8,
which proves that points F and « are on the same vertical line.

Any of the other cases can be proved in a similar manner. Refer-
ring to Fig. 336, point 1 is the point of zero stress for diagonal bC and
480 STRUCTURAL ENGINEERING

point 2 is the point of zero stress for post cC. Point 3 is the point of
zero stress for diagonal dE and point 4 is the point of zero stress for
post eE. The point of zero stress for diagonal eF is at 5.

Tension in Posts. Let it be required to determine the maximum live-
load tensile stress in post dD (Fig. 338). First construct the influence
line, shown at (c), for stress in diagonal dE. As explained on pages
424 and 425 (Art. 190), the maximum live-load tension will occur in
post dD when the load is so placed that the stress in diagonal dE is
zero. The stress in the diagonal dE due to dead load is tension. A
live load moving onto the bridge from the left would begin reversing
this dead-load tension from the start and continue reversing it until a

 

oo

 

 

>
iw
Q

 

/21*
x,
Al
it
>
N

 

 

 

 

 

 

 

 

 

 

Fig. 338

point was reached where the stress in the diagonal would be zero and
from there on, as the load moved on to the right, the stress in the diagonal
would be compression and this compression would continue to increase
until the live-load extended from M to N (see the influence line at (c) )
at which time the live-load compression would be a‘maximum. Then as
the load moved on past N the compression in the diagonal would decrease
until a point was reached, as the load moved on to the right, where the
stress in the diagonal dE would be zero for the second time. This last
position is the one desired, as the maximum tension in post dD will
occur when the load is in that position. The first thing is to find that
position.

Let S represent the dead-load tension in diagonal dE and S’ the
live-load compression in the same when the load extends from M to N.
Then the load to the right of N must be just sufficient to produce a ten-
DESIGN OF SIMPLE RAILROAD BRIDGES 481

sion of S’-S pounds in diagonal dE. An equivalent uniform live load
of w pounds per foot of truss will be used. Let y be the distance from
H to the head of the uniform live load and let A represent the required
shaded area. Then the required tension in the diagonal is Aw, which
must be equal to S’—S, so we have

 

Aw=S'-S,
from which we obtain
_s’-S
ar)

for the required shaded area in square feet. The next thing is to find
the value of y when the shaded area is equal to (S’ —S)/w.
Referring to the influence line at (c), we have

(SH )e-a=1¢98) Sessa etic ae Gleave wees (”).

 

But g=*

So, substituting this value of z, and (S’—S) /w for A in (7), and reduc-
ing, we obtain

y=a(e-t— 2 (9”-8) ie- = (8"- tne Miniee eats tae ee (8).

From this equation (8) the head of the equivalent uniform live
load can be located. Then, having the load located, the influence line
for the maximum live-load tension in the post dD can be constructed
as shown at (d), and by multiplying the shaded area by w the desired
live-load tension in the post will be obtained.

To construct the influence line shown at (d): First assume zero
stress in diagonal dE. Placing a unit load at D and drawing kh parallel
to O’D and kg parallel to O’d, we have the stress in top chord cd given
by the line kg; and as the horizontal components of top chords cd
and de must be equal (since the stress in dE is zero) by drawing from
g a line parallel to top chord de intersecting the vertical mk at m, we
have the stress in chord de represented by the line gm and the stress in
the post dD represented by the line mk. Then laying off m’k’ at (d)
equal to mk the influence line as shown can be constructed.

In case the uniform live load does not extend to ordinate e, as shown
at (e), it is better to locate the head of the load with reference to P.
Let «z represent the distance from P to the head of the uniform live load.
Then we have

A:S watiot,
from which we obtain-

CL aI eee are Tee eee (9).

@= —
é

206. Determination of Dead-Load Stresses in Pettit Trusses.—
Let it be required to determine the dead-load stresses in the truss shown
482 STRUCTURAL ENGINEERING

at (a), Fig. 339. Let WV1, W2, and so on (all of which are equal)
represent the loads at the lower panel points; P1, P2, and so on (all
of which are equal) represent the loads at the upper panel points; and
let R represent the reaction due to these loads.

Members U1-LO, LO-L1, L1-L2 and U1-L1. As previously shown
(Art. 189), the dead-load stress in U1-LO is equal to RsecO, Rtané in
both LO-L1 and L1-L2, and W1 in U1-L1.

Member U1-U2. The stress in this member is readily obtained by
taking moments about L2. Thus, we obtain

[Rx2d-(Wi+P1)d]-=H

for the horizontal component of the stress in U1-U2. The stress in the
member is then obtained by multiplying H by secd.

P:
MBN (b) e ) (9)
L 23 4 4 wz
(P3+W3)
H W3 of 6 P g w3 é aie

&
és P3 7 U7 2 PA e8
23 < ak i eC
PI ; raf sie : na an! i ee Ph
a . t
i : : me (Q)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

UA [6 \ X° e
AN ANN \ ee 5 | He Vy OR” me m5 Ma.
yi ne NT
his iNT Kt NAT: a
7 i= rs i ee ee ; [67 1G i |e fad
R wi \ Ve Vie \ wa\ ws: y Ave \ | w? Wwe Me we w3 ive wi R
\s 19
d é
'

 

 

 

 

Fig. 339

Member U1-L2. Waving determined H, the horizontal component
of the stress in top chord U1-U2, the vertical component of the same
is equal to

Htand=V.

Now, considering the part of the structure to the left of section 1-1 and
summing up the vertical forces, we obtain

Vi=R-(W1+P1)-V

for the vertical component of the stress in U1-L2. Then we have V’1secé
for the stress in U1-L?2.

The stress in this diagonal can also be determined by taking moments
about the intersection of the chords as explained in Art. 189,
DESIGN OF SIMPLE RAILROAD BRIDGES 483

Members M8-L2 and M8-L8. In determining the stresses in these
two members the triangle L2-M3-L4 can be considered as a separate
truss, as shown at (b). The stress in M3-L2 is equal to $(P3 + W3)sec’,
and the stress in the hanger M3-L3 is simply equal to W3.

Members U2-L2. Considering the part of the structure to the left
of section 2-2 shown at (c) and resolving the stress in L2-M3 both
vertically and horizontally at L2 and resolving the stress in U1-U2 and
adding up the vertical forces, we have

R-[W1+W2+4(W3+ P3)+V+P1]-S81=0,
from which we obtain
S1=R-(W1+W2+4(W3+P3)+V+P1)

for the dead-load stress in post U2-L2. This stress can also be deter-
mined by taking moments about O. Thus, taking moments about O, we
have

Rs— (P1+W1) (s+d) -(W2+4W3+4P3) (s+ 2d) -S1(s+2d) =0,
from which we obtain
S1=[Rs—-(P1+W1) (s+d) -(W2+4W34+4P3) (s+2d)]

for the stress in U2-L2.

Member M2-M38. This member is not subjected to any direct stress.
It is for the purpose of stiffening post U2-L2.

Member U2-U8-U4. Considering the part of the structure to the
left of section 4-4, as shown at (d), and by taking moments about L4,
we obtain

1
s+2d

 

H1=4[(R x 4d) - (W1+P1)3d—- (W2+ P2)2d-(W3+P3)d]

for the horizontal component of the stress in top chord U2-U3-U4. Then
by multiplying H1 by secd’ the stress in the member is obtained.

Member M3-L4. Having H1, the horizontal component of the stress
in the top chord U2-U3-U4 determined, the vertical component V2 is
equal to H1 tand’. Then resolving the stress in M3-L4 vertically and
horizontally and summing up the vertical forces shown at (d) we obtain

V3=R—-(W1+W24-W3) -(P1+ P2+P3)-P2

for the vertical component of the stress in diagonal M3-L4. Then for
the stress in the member we have

S3=V3 x sec6’.

Member L2-L3-L4. Considering the part of the structure shown
at (d), and considering $2 and S3 instead of their components (see Art.
41), and taking moments about U2 of the forces shown at d, we obtain

S4= rR x 2d) — (W1+P1)d+ (WVW3+P3)d]

for the stress in bottom-chord L2-L3-L4.
484 STRUCTURAL ENGINEERING

Member U2-M3. The vertical component of M3-L2 is equal to
4(P3+W3); the vertical component of the top chord U2-U3-U4, as
found above, is equal to V2. Then considering the part of the structure
to the left of section 3-3 and summing up the vertical forces and
components on same, we obtain

V4=R-(W14+W2+P1+P2)-3$(P3+W3) - V2

for the vertical component of the stress in diagonal U2-M3. Then the
stress is equal to V4 x sec6’.

Member U3-M8. This member simply supports the load P3 and
hence the stress in it is equal to P3:

Members M5-L5 and M5-L4. The stress in hanger M5-L5 is equal
to W5. The vertical component of the stress in M5-L4 is equal to
$(P5+W5). Then the stress in M5-L4 is equal to $(P5+ W5)sec6”.

Member U4-L4. Considering the part of the structure to the left
of section 5-5 and resolving the stress in U4-U3-U2 and M5-L4 vertically
and horizontally and summing up the forces and components, we obtain

V5=R-(W1+W2...W4) -(P1+P24P3) -V2-4(P5+W5)

for the stress in post U4-L4.

Member U4-U5-U6. Considering the part of the structure to the
left of section 7-7 and taking moments about L6 we can obtain H2, the
horizontal component of the stress in top chord U4-U5-U6. Then the
stress in the member is equal to H2 x seco”.

Member L4-L5-L6. Considering again the part of the structure
to the left of section 7-7 and taking moments about U4, we obtain

ss=5 [ (Rx 4d) — (W1+ P1)3d — (W2+ P2)2d-
(W3+P3)d+(W5+P5)d]

for the stress in bottom chord L4-L5-L6.

Member M5-L6. Considering again the part of the structure to
the left of section 7-7 and letting V6 represent the vertical component of
the stress in top chord U4-U5-U6, we obtain

Vi=R-(W1+W2...W5)-(P1+P2...P5)-V6

for the vertical component of the stress in diagonal M5-L6. Then mul-
tiplying V7 by sec6” the stress in the member is obtained.

Members M4-M5 and M5-M6. These members have no direct stress
in them. They are for the purpose of stiffening the posts U4-L4 and
U6-L6.

‘Member U4-M5. Considering the part of the structure to the left
of section 6-6 and letting $(P5+W5) and V6 represent the vertical com-
ponent of the stress in M5-L4 and U4-U5-U6, respectively, and summing
up the vertical forces and components to the left of section 6-6, we
obtain

V8=R-(W1+W2...W4) -(P1+P2...P4) -4(P5+W5) - V6

for the vertical component of the stress in diagonal U4-M5. Then the
stress in the member is equal to V8 x sec”.
DESIGN OF SIMPLE RAILROAD BRIDGES 485

Member U5-M5. This member supports the load P5 and hence
thr stress in it is equal to P5.

Members U7-M7 and M7-L7. As is evident, the stress in U7-M? is
equal to PY and the stress in M7-L is equal to W7.

Members U6-M7 and M7-L6. The dead-load stress in these mem-
bers is due to the loads W7 and P?. We will assume that each is equally
stressed. Then the stress in each will be equal to }(W7+P7)sec6””’.

Member U6-L6. Considering the part of the structure to the left of
section 8-8 and letting 76 and V9 represent the vertical component of the
stress in members U4-U5-U6 and M%-L6, respectively, and summing up
ali the forces and components to the left of section 8-8, we obtain

SG=R-(W1+W2...W6)-(P1+P2...P5)-V6-V9

for the stress in U6-L6.
Members UG-U7 and L6-L7. Considering the part of the structure
to the left of section 9-9 and taking moments about L7, we obtain

Si =;5 [(Rx'd) - (W1+ P1)6d — (W2+ P2)5d— (W3+ P3)4d-
(W4.+ P4)3d-(W5 + P5)2d- (W6+P6)d]

for the stress in top chord U6-U7. It will be seen readily that the
moments of the stresses in U6-M7 and M7-L6 about L7 are equal and of
opposite signs and annul each other, and consequently are not given
in the equation of moments.

The stress in L6-L? is equal to the stress in top chord U6-U?%. How-
ever, the stress in L6-L7 can be obtained by taking moments about U7.

As the truss is symmetrical in reference to the center of span and
symmetrically loaded, we have now sufficiently considered the analytical
determination of the dead-load stresses in the truss.

As a matter of fact, the dead-load stresses in such trusses are
graphically determined. The diagram of the dead-load stresses for the
right half of the truss is shown at (e). This diagram is obtained by
beginning at N and passing around the joints clock-wise. Thus, begin-
ning at point N, we draw 1-2 equal to # and then 1-3 and 2-3 parallel,
respectively, to bN and BN. Then passing on to joint B we have the
diagram 3-2-4-5-3 for that joint. Then passing on to joint b we have the
diegram 6-1-3-5-7-6 for that joint. Now, at C there are three unknown
forces, the same is true of joint c and, consequently, before we can go
farther we must determine one of the unknown forces at one of these
jeints. By considering E-M3-C as a separate truss, as shown at (f), and
constructing the diagram at (g), we have the stress in M3-C given by
the line vu. Then considering joint C (at (a)) and passing around the
joint clock-wise, we have the part (starting at 7) 7-5-4-8 of the stress
diagram for that joint. We know that the stress in post cC will be rep-
resented by a vertical line from 7. So we can draw 7-y. We can also
draw 8-2. Then laying off 0-10 equal to 4(P3+V3) and drawing 9-10
parallel to M3-C, we have the stress in M3-C represented by this line
9-10 (see diagram at (g)) and thus we have the complete stress diagram
7-5-4-8-9-10-7 for joint C. For joint ¢ we obtain the stress diagram
486 STRUCTURAL ENGINEERING

(beginning at 11) 11-G6-7-10-12-11. For joint d we obtain the stress
diagram (beginning at 13) 13-11-12-14-13. For joint D we obtain the
diagram 9-8-15-16-9. For joint M3 we obtain the stress diagram 14-12-
10-9-16-14. The line 16-14, which represents the stress in member
M3-E, must be parallel to 12-10, which represents the stress in member
c-M3. This is the first check on the work.

By considering G-M5-E as a separate truss, the stress in M5-E can
be determined. Now, considering joint E and passing around clock-wise,
we obtain the part 14-16-15-17 of the stress diagram for joint E.

Then drawing 14-r and laying off r-19 equal to $(P5+W5) and
drawing 19-18 parallel to the member M5-E, we have the complete

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

8 7 te at
2. Ty 1
af bet i tol oo
; Ve { :
“|p ai ‘faz ee a am
ye PLN Ans
4 ey iL z "
a, !} et |
a t .
| | '
| 3
ry iJ L e
b Eo “Ip! \
JON i
I Y Naw it
| . VN
x DES
A B c] yt ‘ey
pea
I |
I 1
j
'
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Olack
de a adxtar
a tp soa
O° nce, y of
Fig. 340

stress diagram 14-16-15-17-18-19-14 for joint E. For joint e we
obtain the stress diagram 20-13-14-19-21-20. For joint f we obtain
the stress diagram 22-20-21-23-22. For joint F we obtain the diagram
18-17-24-25-18. For joint M5 we obtain the diagram 23-21-19-18-25-23.
The line 23-25 must be parallel to line 21-19. This is the second check
on the work. :

The dead-load stress is assumed to be the same in the four members,
M%-U6, M?-L6, M'-g and M’?-G. This stress is due to the loads W?
and P7. Then, by considering L6-M7-G as a separate truss, having a
reaction at each end of (W%+P7), the stress in M?-G can be graphically
determined.

Starting with 23, we have the part 23-25-24-26 of the stress diagram
for joint G. Then drawing a vertical line from 23 and a horizontal
line from 26 and laying off t-28 equal to }(W7+ PY) and drawing 28-27
parallel to M%-G, we obtain the complete stress diagram 23-25-24-26-27-
DESIGN OF SIMPLE RAILROAD BRIDGES 487

28-23 for joint G. For joint g we have the stress diagram 29-22-
28-28-30-29,

This completes the graphical determination of the dead-load stresses,
as the truss is symmetrical in reference to the center of span and the
load is symmetrical in reference to same. The distance 26-29 should
equal 4(P7+W7). This is the third and final check on the work.

207. Determination of Live-Load Stresses in Pettit Trusses.—
Let it be required to determine the live-load stresses in the truss shown
at (a), Fig. 340, due to Cooper’s £50 loading.

Members bA, AB and BC. The maximum live-load stress will occur
in these members when the wheel loads are placed for maximum shear in
panel AB. The placing of the loading will be in accordance with (5),
Art. 90. After the loads are thus placed, the next thing to do is to deter-
mine the reaction at A by taking moments about N. Let R represent this
reaction. The next thing to do is to determine the concentration at A
due to the loads in panel AB, which can be done by taking moments about
B. Let r represent this concentration. Then we have R-r for the shear
in panel AB. Then we obtain (R-r)secO for the maximum live-load stress
in end post bd and (R-r)tan@ for the maximum live-load stress in each
of the members AB and BC (see Art. 174).

Member bC. The maximum live-load stress will occur in diagonal
bC when the loads are placed according to the requirements of (4) of
Art. 190. After the loads are thus placed, the next thing to do is to
determine the reaction at A, which can be done by taking moments about
N. Let & represent this reaction. Next, by taking moments about C,
the horizontal component of the stress in top chord be is readily obtained.
Let H represent this component. Then the vertical component of the
stress in be is equal to Htan¢. Taking moments about C of the loads
in panel BC, the concentration at B is readily obtained. Let r repre-
sent this concentration. Now, for the vertical component of the stress
in diagonal bC, we have V=R-r-— Htan¢, and multiplying this by sec,
we obtain (R-—r—Htand)secé for the stress in the member bC.

Member bB. The maximum live-load stress in this hanger, as is
obvious, is equal to the maximum live-load floor beam concentration, which
is determined as explained in Art. 148.

Member be. The maximum live-load stress will occur in this member
when the loads are placed for maximum moment about C. The placing
of the loads will be in accordance with (5), Art. 91. That is, a wheel
must be at C and the average unit load to the left of C must equal
(approximately) the unitload onthespan. After the loads are thus placed
the next thing to do is to determine the reaction at 4, which can be done
by taking moments about N. Then, by taking moments about C of the
forces to the left, the horizontal component of the stress in be is readily
obtained, and multiplying this component by seed, the maximum live-load
stress in bc is obtained.

Members cC and c-M1. It is obvious if C-M1-E be considered a
separate truss, shown as C-o-E at (b), that the stress in cC and c-M1 will
not be affected in the least. The truss C-o-E really acts as a stringer
extending from C to E. Then, as is readily seen, the maximum live-load
stress in cC can be obtained by loading the span from the right, placing
488 STRUCTURAL. ENGINEERING

a wheel at E and loading panel CE (no loads to the left of C), all in
accordance with (4) of Art. 190.

In applying equation (4) s should be taken equal to the distance
IA, shown at (b), and a should be taken equal to distance AC.

After the loading is placed, the next thing to do is to determine the
reaction at A, which can be done by taking moments about N. Let R
represent this reaction. The next thing to do is to determine the con-
centration at C due to the loads in panel CE, which can be done by
taking moments about E, ignoring panel point D altogether; that is, CE
would be considered as a stringer. Let r represent this concentration
at C,

Next, by taking moments about C, the horizontal component of the
stress in top chord be is readily obtained, as previously explained. Let
H represent this component. Then the vertical component of the stress
in the top chord be is equal to Htand. Now, considering the part of the
structure to the left of section 1-1, and adding up the vertical components
and forces, we obtain R-r—Htand for the maximum live-load stress in
member cC. The member oC is entirely ignored, as the vertical com-
ponent of its stress is included in the concentration r.

To obtain the maximum live-load stress in c-M1, the loads would be
placed very much the same as for cC. In applying equation (4) of Art.
190 s would be taken equal to the distance I’A and a equal to AC. After
the loads are properly placed, the next thing to do is to determine the
reaction at A, which can be done by taking moments about N. Then
the next thing to do is to determine the concentration at C due to the
loads in panel CE, ignoring panel D. By taking moments about E, the
horizontal component of the stress in top chord ce is readily obtained, and
multiplying this component by tang’, the vertical component of the stress
in top chord ce is obtained. Then considering the part of the structure to
the left of section 2-2 and summing up the vertical components and forces
to the left which consist of the reaction at A, concentration at C, the
vertical component of the stress in ce, and the vertical component of the
stress in cM1, the vertical component of the stress in cM1 can be
obtained; and multiplying this component by sec6’, the stress in cM1
is obtained.

Members D-M1 and C-M1. As is obvious, the maximum live-load
stress in hanger D-M1 is equal to the maximum live-load floor beam concen-
tration, which is determined as explained in Art. 148. The maximum live-
load stress in sub-diagonal C-M1, as is readily seen, is equal to one-half
of the maximum floor beam concentration multiplied by sec6’.

Member ce. The maximum live-load stress in this member occurs
when a load is at E and the average unit load to the left of E is equal
to the average unit load on the span. This is in accordance with Art.
91. After the loads are properly placed the next thing to do is to
determine the reaction at A, which can be done by taking moments about
N. Then considering the part of the structure to the left of section 3-3
and taking moments about E, the horizontal component of the stress in
top chord ce is readily obtained, and then multiplying this component
by sec¢’, the maximum live-load stress in ce is obtained.

Member E-M1. It is readily seen that if oF (shown at (b)) were
combined with E-M1, any load at D would cause compression in £-M1
DESIGN OF SIMPLE RAILROAD BRIDGES 489

and hence the member E-M1, as far as maximum stress is concerned,
acts just the same as an ordinary diagonal where DE is the panel
length. Then the maximum stress in E-M1 is obtained by loading the
span from the right with a wheel at E, and loading panel DE (no loads
to the left of D) according to (4) of Art. 190. In applying equation
(4) s should be taken equal to the distance I’A and a equal to the dis-
tance AD. After the loads are properly placed, the next thing to do is
to determine the reaction at A, which can be done by taking moments
about N. Then the next thing to do is to determine the concentration
at D, due to the loads in panel DE. Then, considering the forces to the
left of section 3-3 (oF and E-M1 being considered as combined, that is,
one member) and taking moments about E, the horizontal component of
the stress in top chord ce is readily obtained; and multiplying this
horizontal component by tan¢’ the vertical component of the stress in the
top chord ce is obtained. Then summing up the vertical components and
forces to the left of section 3-3 the vertical component of the stress in
diagonal E-M1 is readily determined; and multiplying this component by
sec’ the maximum live-load stress E-M1 is obtained.

Member CE. It is readily seen that any load in panel CD or DE
will cause tension in CE, owing to the member CE acting as the bottom
chord of truss C-M1-E. This stress, as we may say, is in addition to
the stress produced in the member acting as a main bottom chord sec-
tion of the structure. Then it appears that the maximum live-load stress
in bottom chord CE will occur when the loads in panels CD and DE
have some certain value as compared to the other loads on the structure.
This can be investigated most satisfactorily by the use of influence lines.

The stress in CE is finally obtained by considering the part of the
structure to the left of section 3-3 and taking moments about c. Then
evidently the stress in CE will vary as the moments about ¢ and hence
the stress in the member can be investigated by constructing an influence
line for moments about c.

Let P represent a load at any point « distance from N. Then when
P is at any point to the right of panel point E we have M=(Px/L)a
for the bending moment about c. If «=b, the last equation becomes
M=(Pb/L)a, and if P=1 we have M=(b/L)a. So if we draw OO’
at (c) and lay off wo equal to (b/L)a we can draw uO’, which is the
influence line for moments about ¢ for loads to the right of E.

If the load P moves to any point in panel DE, the moment
about ¢ is

&

M’= (Pz) a+P(«—b).

If w=b we have the same as given above for ordinate uv. If
x=b+d and P=1 the equation reduces to

,_fo+d _atd _
m=(*7°) at+d= L (L-a).

 

Then if the ordinate nm be laid off equal to this last value of M’ the
line nw can be drawn, which is the influence line for moments about
ec for loads in panel DE.
490 STRUCTURAL ENGINEERING

If the load P moves to any point in panel CD, the moment about ¢
will be

M” = (P z) a+(b+2d—2)P.

L
This is the equation to the line tn, which is the influence line for moments
about c for loads in panel CD.

If the load P moves to any point to the left of C, the moment about
ce will be

M’”’= (77) a-(«-b-2d)P= (72) a+(b+2d—2)P,

which is the equation to line tO. But this value of M’’’ is exactly the
same as found above for M’”. Therefore, the lines tn and ¢O are in the
same straight line On and hence On is the influence line for moments
about c for loads to the left of D. Then evidently OnuO’ is the complete
influence line for moments about c. As is seen, this influence line has
three different slopes and consequently there will be three different
moment increments to consider.

Let W1 represent the resultant of the loads to the left of D, W2
the resultant of the loads in panel DE, W3 the resultant of the loads to
the right of E and let W represent the total load on the span; that is,
W=W14+W2+1V3. Then for the moment about c, we have

SyW ty We ty WS: vcseienscswes ar ere sabe dees ve (1).
Now if all loads move to the left the distance dx, we have
AM =~ WidatanB1+ W2datanB2+W2drtanB3 ........... (2).

for the increment of the moment about c. If the loads are in the posi-
tion for maximum moment about c this increment will be equal to 0 and,
hence, (2) would become

 

 

 

0=-WitanB1+ W2tanB2+W3tanB3 2.0.2.2... cess ee eee (3).
But, tanRl= Ee ~~
x b 1 Lia
tanga =[45" (L-a)-7 a |F= L 2
tan83 = -

Substituting these values of the tangents in (3), we obtain

L- Lt+a a
0=-171( L *) wi( eee E aE") +W3 5,

from which we obtain
(W2-W1)L+a(W1+W2+W3)=0,

from which we obtain

 
DESIGN OF SIMPLE RAILROAD BRIDGES 491

Expressing equation (4) in words, the average unit load on the span
is equal to the load to the left of D, minus the load in panel DE, divided
by a. This can be taken as the criterion for placing the load for maxi-
mum stress in bottom chord CE.

After the loads are placed in accordance with equation (4), the next
thing to do is to determine the reaction at 4, which can be done by taking
moments about N. Then the next thing is to determine the concentra-
tion at D, which can be done by taking moments about C and E.

Let R=the reaction at 4, r=the concentration at D (due to the
loads in panels CD and DE), m=moment of load to the left of C about
c, and h=height of post cC. Then considering the part of the structure
to the left of section 3-3 and taking moments about c, we obtain

S=(Ra-m+rd)+

for the maximum live-load stress in bottom chord CE.

The moment about c of course could be obtained by the use of influ-
ence line OnuO’.

If W remains constant, the increment of the moment would change
signs only as a load passed joint D, so there will be a load at that joint
when the maximum moment occurs at joint c. Then in placing the load-
ing for maximum moment about c, that is, satisfying (4), a wheel must
be at joint D.

Member eg. The maximum live-load stress in top chord eg is
obtained by placing a load at G such that the unit load to the left of G
is equal to the unit load on the span. Then by taking moments about G
the horizontal component of the stress in eg is readily determined and
multiplying this component by sec” the stress in the member is obtained.

Member eE. The maximum live-load stress in this member is
determined in the same manner as shown above for cC. The sub-diag-
onal E-M3 and hanger F-M3 are assumed to be omitted. The span is
loaded from the right, a wheel at G and panel EG loaded in accordance
with (4) of Art. 190.

The value of s (in equation (4)) is obtained by prolonging top chord
ce and a is taken equal to the distance AE. After the loads are prop-
erly placed, the next thing to do is to determine the reaction at A, which
can be done by taking moments about N. Next, the concentration at E
can be determined by taking moments about G. Then, the next thing
is to determine the vertical component in top chord ce, which can be done
by first taking moments about E, thereby obtaining the horizontal com-
ponent of the stress in the member and multiplying this component by
tang’, we obtain the vertical component of the stress in ce.

Let R represent the reaction at A, r the concentration at E, and V
the vertical component of the stress in top chord ce. Then considering
the part of the structure to the left of section 4-4 and summing up the
vertical components and forces, we have

S=R-r-P,
for the maximum live-load stress in post eZ.

Member e-M8. The maximum live-load stress in this member is
determined in the same manner as shown above for member c-M1. The
492 STRUCTURAL ENGINEERING

members E-M3 and F-M3 are assumed to be omitted. The span is loaded
from the right with a wheel at G and panel EG loaded in accordance
with (4) of Art. 190. Top chord eg in that case would be prolonged to
determine the value of s, and a would be the distance AE.

After the loads are properly placed for maximum stress in e-M3, the
next thing to do is to determine the reaction at 4, which can be done by
taking moments about N. The next thing to do is to determine the con-
centration at E, which can be done by taking moments about G. Then
the next thing in order is to determine the vertical component of the
stress in top chord eg, which can be done by first taking moments about
G, thus obtaining the horizontal component of the stress in top chord eg
and multiplying this component by tand’’, we obtain the vertical com-
ponent of top chord eg. Then considering the part of the structure to
the left of section 5-5, and summing up the vertical components and
forces, we obtain

V1=R-r-V

for the vertical component of the stress in diagonal e-M3, and multiply-
ing this component by sec6’’, we obtain the maximum live-load stress in
member e-M3.

Members F-M8 and E-M3. As is evident, the maximum live-load
stress in hanger F-M3 is equal to the,maximum floor beam concentration,
which is determined as explained in Art. 148. The maximum live-load
stress in sub-diagonal E-M3 is equal to one-half of this concentration
multiplied by sec6’’.

Member G-M8. The maximum live-load stress in this member is
determined in the same manner as explained above for diagonal E-M1.
The span is loaded from the right, a wheel is placed at G and panel
GF is loaded in accordance with (4) of Art. 190. In this case the value s
is obtained by prolonging top chord eg (see Art. 190) and a is equal to
the distance AF’. The reaction at A is obtained by taking moments about
N. The concentration at F is obtained by taking moments about G, and
the horizontal component of the stress in top chord eg is obtained by
taking moments about G, and the vertical component. of same is then
obtained by multiplying the horizontal component by tang”. After these
are determined the vertical component of the stress in diagonal G-M3 is
obtained by considering the part of the structure to the left of section 6-6
and summing up the vertical component and forces, and then the stress
in G-M3 is obtained by multiplying this component by sec6’”’.

Member EG. The maximum live-load stress in this member is deter-
mined in the same manner as explained above for bottom chord CE. In
applying equation (4) (given above) a would be taken equal to distance
AE. W1 would be the load between A and F, V2 the load in panel FG,
W3 the load between G and N, and W the total load on the span.

After the loads are properly placed, that is, in accordance with
equation (4), the stress in EG is readily obtained by considering the
part of the structure to the left’ of section 6-6 and taking moments
about e.

Member gG. The maximum live-load stress in this member is ob-
tained by considering members Gk and Hh as being omitted. The
span is then loaded from the right, a wheel at K and panel KG loaded
DESIGN OF SIMPLE RAILROAD BRIDGES 493

in accordance with (4) of Art. 190. The value of s in equation (4) of
Art. 190 is obtained by prolonging top chord eg and a is equal to the dis-
tance AG. After the loads are properly placed the reaction at A is
obtained by taking moments about N. The concentration at G is obtained
by taking moments about K. The horizontal component of the stress in
top chord eg is obtained by taking moments about G and the vertical
component of same is obtained by multiplying the horizontal component
by tang”. Having the above determined, the stress in gG is obtained
by summing up the vertical forces to the left of section 7-7 (ignoring
member G-M5).

Member GK. The maximum live-load stress in this member is de-
termined in the same manner as explained above for bottom chord CE.
In applying equation (4) (given above) a would be taken equal to dis-
tance AG, W1 would be the load between A and H, W2 the load in panel
HK, W3 the load between K and N. After the loads are properly placed,
that is, in accordance with equation (4), the stress in GK is readily ob-
tained by considering the part of the structure to the left of section 8-8
(ignoring member k-M5) and taking moments about g.

Member g-M5. To obtain the maximum live-load stress in this
member, we assume the members Gk and Hh to be omitted, and load
the span according to (5) of Art. 90. That is, the span would be loaded
from the right with a wheel at .K and the load in panel GK equal to the
load on the bridge divided by the number of panels. The panels con-
sidered in this case would be double panels—that is, the number would
be 7 instead of 14.

After having the loads thus placed, the next thing to do is to deter-
mine the reaction at A and the concentration at G. These can be ob-
tained by taking moments about N and K, respectively. Let & repre-
sent the reaction at A and 7 the concentration at G. Then we have

S=(R-r)sec6’’’

for the maximum live-load stress in diagonal g-M5.

Member K-M5. To obtain the maximum live-load stress in this
member we assume member &-M5 to be omitted and load the span accord-
ing to (5) of Art. 90. That is, the span would be loaded from the right
with a wheel at K, and the load in panel HK equal to the total load on
the span divided by the number of panels. In this case the number of
panels would be 14. After the loads are thus placed, the next thing
to do is to determine the reaction at A and the concentration at H. Let
R’ represent the reaction at 4, and 7’ the concentration at H. Then we
have

S’ = (R’ —-1’)sec6’””

for the maximum live-load stress in diagonal K-M5.

Members g-M5 and k-M5 would be assumed to have equal tensile
stress and likewise members K-M5 and G-M5. The latter members would
be subjected to compression from the floor beam concentration at H.
Assuming the concentration at H to be transmitted to panel points G and
K by sub-truss K-M5-G the maximum live-load compression in each of
the members G-M5 and K-M5 would be equal to one-half of the maximum
floor beam concentration at H multiplied by sec6’”’.
494 STRUCTURAL ENGINEERING

Member gk. The maximum live-load stress in this member is ob-
tained by placing the load for maximum moments about K. Then taking
moments about K (ignoring member k-M5) and dividing this moment by
the height of post kK, we would obtain the greater part of the maximum
stress in gk. Next the concentration at H, due to the loads in panels
GH and HK, can be determined by taking moments about G and K.
Then, multiplying one-fourth of this concentration at H by tan0’’’, and
adding the result to the stress found in gk, by taking moments about K,
the total maximum live-load stress in the member gk is obtained.

Mazimum Tension in Post gG. The maximum live-load tension will
occur in this post when the live load extends from A to some point beyond
G, so that the stress in diagonal g-M5 is zero. The position of the load

e)
&
g
SI

Z

 

Fig. 341

can be determined by trial as explained in Art. 205 for curved chord
Pratt trusses. After the position of the loading is found, the tension
in the member is readily determined by considering the part of the struc-
ture to the left of section 7-7. The use of an equivalent uniform live-
load will simplify the work very much and the result thus obtained
will be sufficiently accurate.

Stress in Counters. In case a member g-M3 were inserted, it would
be known as a counter. This member would not be stressed at all
unless the dead-load tension in either or both of the members e-M3 and
G-M3 were reversed. In case either, or both, of these members be sub-
jected to reversal, the counter g-M3 would be used or members e-M3 and
G-M3 would be designed to carry both tension and compression. In
DESIGN OF SIMPLE RAILROAD BRIDGES 495

case the counter g-M3 is required, the maximum live-load stress (tension)
in it can be determined by assuming members eG and F-M3 to be omitted
and loading the span from the left with a load at E and loading panel
LG in accordance with (4) of Art. 190. After the loading is thus placed,
the next thing to do is to determine the reaction at N and the concentra-
tion at G, which can be done by taking moments about A and E, re-
spectively. Then by considering the part of the structure to the right
of section 6-6, the stress is readily determined. If counters are found
necessary at other points the stress in them can be determined in a
similar manner.

The determination of the stresses in trusses having the sub-paneling
at the top of the truss, as shown at (a) in Fig. 341, is very similar to
that given above for the truss shown in Fig. 339 wherein the sub-paneling
is at the bottom of the truss.

Dead-Load Stresses. The graphical determination of the dead-load
stresses in the type of truss shown in Fig. 341 is simpler than in the
case of the type shown in Fig. 339, as the sub-paneling does not bother
at all in the case of the truss shown in Fig. 341.

The graphical diagram of the dead-load stresses for the right half
of the truss is shown at (b). This diagram is readily followed throughout.

The analytical determination of the dead-load stresses is just the
same in some cases and similar in other cases to that given above for the
truss shown in Fig. 339. To save space, just the formulas for the dead-
load stresses in part of the members in the right half of the truss shown
in Fig. 341 will be given:

Stress in sT equals Rsec9.

Stress in TS and OS equals Rtané.

Stress in sS equals W1.

Stress in os equals [ (Rx 2d) —(W1+P1)d]|secp/h1 (d=panel lgth.).

Stress in Os equals [R-(W1+P1)—V|secé, where V represents the
vertical component of the stress in top chord os.

The stress in member 00 equals R-(W1+W2+P1)-V. In this
case the part of the structure to the right of section 1-1 is considered.

The stress in member MO equals [Rx2d-(W1+P1)d]1/h1. This
is obtained by taking moments about o.

The stress in member mo equals [(Rx4d)—(8W1+3P1+2W2
+2P2)d|secfi/h2. This is obtained by considering the part of the
structure to the right of section 2-2 and taking moments about M.

The stress in member o-M3 equals [R-(W1+W2+P1+P2)
—V1]sec61, where V1 represents the vertical component of the stress in
top chord mo. In this last case the part of the structure to the right of
section 2-2 was considered.

The stress in sub-post n-M3 is equal to P3 and in N-M3 it is W3.

The stress in sub-diagonal m-M3 is determined by considering
n-M3-o as an independent truss supporting the load (W3+P3) at M3, as
shown at (c) in Fig. 341.

The stress in the member m-M3 equals (JV3+P3) + (cos62
+sin62cos61). :

The stress in member m-M3 can be determined most readily by
graphics (see the diagram at (d)).

The stress in M-M3 equals [R-(W1+W2+W3+P1+P2+P3)-
496 STRUCTURAL ENGINEERING

V'1+V2]sec61, where V1 and V2 represent, respectively, the vertical com-
ponent of the stress in mo and m-M3. In this case the part of the structure
to the right of section 3-3 is considered.

The stress in mM equals R-(W1+W2+W3+W4+P1+P2+P3)
-V1+P2. In this case the part of the structure to the right of section
4-4 is considered.

Following this method the dead-load stresses in the other members
of the truss are readily determined.

Live-Load Stresses. The determination of the live-load stresses in
the type of truss shown in Fig. 341 is very much the same as given above
for the truss shown in Fig. 340. The method of analysis will be suffi-
ciently shown by considering the members in panel EG.

Member eE. The maximum live-load stress will occur in this mem-
ber when the span is loaded from the right, a wheel at G and panel EG
loaded in accordance with (4) of Art. 190. In this case the top chord ce
would be prolonged to determine the value of s to be used in equation
(4) of Art. 190 and a would be taken equal to distance AE. After the
loads are properly placed for maximum stress in eZ, the next thing to
do is to determine the reaction at A and the concentration at E, which
can be done by taking moments about T' and G, respectively. The next
thing to do is to determine the vertical component of the stress in top
chord ce. This can be done by taking moments about E, thus obtaining
the horizontal component of the stress in ce (ignoring member e-M3, as
it is not subjected to any live-load stress at this time, the live load not
extending beyond E), and then multiplying this horizontal component
by tan¢1, the vertical component of the stress in ce is obtained. Let R
represent the reaction at A, r the concentration at E, and V the vertical
component of the stress in top chord ce. Then considering the part of
the structure to the left of section 5-5 and summing up the vertical com-
ponents and forces, we have

S=R-r-V

for the maximum live-load stress in post eE.

Member EG. The stress in this member is obtained, as is obvious,
by taking moments about joint e. Then by placing the loads so that
there is a wheel at E with the unit load to the left of E equal to the unit
load on the span, which is in accordance with Art. 91, and taking
moments about e of the forces to the left and dividing this moment by
the height of post ek, the maximum live-load stress in EG is readily
determined.

Member F-M5. The maximum live-load stress in this hanger is
equal to the maximum live-load floor beam concentration at F, which
is determined as explained in Art. 148.

Members M4-M5, M5-M6 and f-M5. These members are not sub-
jected to live-load stress at all. f-M5 is subjected to dead-load stress
from the load at f only. The other two members are not subjected to
any direct stress whatever. They are for the purpose of stiffening the
posts eE£ and gG. :

Member e-M5. The maximum live-load stress in this member will
occur when the span is loaded from the right, a wheel at F and panel
EF loaded in accordance with (4) of Art. 190.
DESIGN OF SIMPLE RAILROAD BRIDGES 497

The top chord eg would be prolonged to determine the value of s to
use in (4) of Art. 190, and a would be taken equal to distance AE.
After the loads are properly placed for maximum live-load stress in diag-
onal e-M5, the next thing to do is to determine the reaction at A and the
concentration at E, which can be done by taking moments about T and F,
respectively. Let R represent the reaction at A and r the concentration
at E. Then considering the part of the structure to the left of section
6-6 and summing up the vertical components and forces, we have

S=(R —-r-V)sec63,

for the maximum live-load stress in diagonal e-M5. V here represents
the vertical component of the stress in top chord eg, which can be deter-
mined by considering the part of the structure to the left of section 6-6,
and taking moments about G and dividing this moment by the height of
post gG, thus obtaining the horizontal component of the stress in top
chord eg, then multiplying this component by tan@2, the vertical com-
ponent V is obtained.

Member G-M5. In determining the maximum live-load stress in
this member, the members F-M5, g-M5 and f-M5 are assumed to be
omitted. The maximum stress will occur when the span is loaded from
the right, a wheel at G, and panel GE loaded according to (4) of Art.
190. After the loads are properly placed for maximum stress in diagonal
G-M5, the stress is obtained by considering the part of the structure
to the left of section 7-7 (ignoring member g-M5) and summing up the
vertical components and forces. Thus let R represent the reaction at 4,
r the concentration at E and V1 the vertical component of the stress in
top chord eg, then we have

S1=(R-r-V1)sec63

for the maximum live-load stress in diagonal G-M5.

Member g-M5. The maximum live-load stress will occur in this
member when the live-load concentration at F is a maximum. Let r
represent this concentration. Then, considering e-M5-g as an independent
truss, we obtain

S=r+ (cos§4+ sind4cos63)

for the maximum live-load stress in sub-diagonal g-M5. This stress in
g-M5 can be determined most readily by graphics, considering e-M5-g
as an independent truss.

Member eg. The live-load stress in this member can be investigated
most satisfactorily by the use of the influence line.

The stress in eg is finally obtained by considering the part. of the
structure to the left of section 6-6 and taking moments about G. Then
evidently the stress in eg will vary directly as the moments about G and,
hence, the stress in the member can be investigated by constructing an
influence line for moments about G. ’

Let P represent a load at any point « distance from T. Then
when P is at any point to the right of panel point G, we have

m-(P 5 )a
498 STRUCTURAL ENGINEERING

for the bending moment about G. If #=b, the above equation becomes

‘Pb
M= (Pe
and if P=1,
ba
as.

So if we draw OO’ at (e) and lay off ts=ba/L, we can draw tO’

which is the influence line for moments about G for loads to the right of G.
If the load P moves to any point in panel FG, the moment about G,

considering the part of the structure to the left of section 6-6, is

,_ {Px
men) ®

which is the same as found above for M, so the influence line vt for
loads in panel FG is simply a continuation of the line 10’. If #=b+d
and P=1, we obtain

,_({b+d
u’=(*T" a

which is equal to the ordinate vw.
If the load moves to any point in panel FE, the moment about G is

 

M’ = (=) a—P(x-b-d)2.

If c=b+2dand P=1, we obtain
oF. a _b
M = (b+2d) 7 —2d=5 (a-2d),

which is equal to ordinate uo, so the influence line vu for loads in panel
EF can be drawn.

If the load P moves to any point to the left of E, the moment about
G is M’’=(Pa2/L)a-P(«—b). If «=L, we obtain M’’=(PL/L)a—
P(L-b)=Pa-—Pa=0. If w=b+2d and P=1, we obtain M’”’=(b+
2d)a/L—2d=b/L(a-—2d), which is the same as found above for ordi-
nate uo. So the line Ou can be drawn, which is the influence line for
loads to the left of E. Then we have the complete influence line OuvtO’
for moments about G. As is seen, this influence line has three different
slopes and consequently there will be three different moment increments
to consider. Let W1 represent the resultant of the loads to the left of
E, W2 the resultant of the loads in panel EF, W3 the resultant of the
loads to the right of F, and let W represent the total load on the span;
that is, W=W14+W2+W3. Then for the moment about G, we have

Ma=yW1+y1W24+y2W8 .. 0 ccc cee eens (1).
Now if all loads move to the right a distance dx, we have
AM = WidatanB1+ W2tanB2-—W3tanB3 ......... eee eee (2),

for the increment of the bending moment about G. If the loads are in the
position for maximum moment about G, this increment will be equal to
zero and, hence, (2) would become
DESIGN OF SIMPLE RAILROAD BRIDGES 499

O=WltanBl+W2tanB2—W3tanB3 ............e. cee eee (3).
But, tanB1=b/L(a-—2d) + (a—2d) =b/L,

tang? = (4

tan83 =a/L.
Substituting these values of the tangents in (3), we have

L+b
0=Wie+ went? “7-3 F = bW1+LW2+b6W2-aW3.

Substituting L~—a for we have
0=LW1-aW1+ LW2+LW2-aW2-aW3=L(W1+2W2)-

 

 

 

at+2b L+b
Lo DL

’

)a—b/L(a—2a) 1/d=

a(W1+W2+W3),
from which we obtain
W1+2W2 wp
— FP etn ete teen eens (4)’.

That is, the load to the left of E plus twice the load in the panel EF
divided by the distance a must equal the total load on the span divided
by the length'of the span when the maximum stress in top chord eg
occurs.

W remaining constant, the increment of the bending moment about
G can change signs only as a load passes joint F’, hence there will be a
load at F when the maximum live-load stress in eg occurs. Then by
placing the live load according to the above equation (4)’ with a load at
F, and taking moments about G (considering the part of the structure
to the left of section 6-6) the maximum live-load stress in eg can be
readily determined.

The reaction at A would be determined first, then the concentration
at E, due to the loads in panel EF, which can be done by taking moments
about F. Let R=reaction at A, r=the concentration at E, due to the
loads in panel EF, and m=the moment of all loads to the left of E about
G. Then taking moments about G, we have

M=R(6d) —-m-r(2d).
Dividing this moment by h, the height of post gG, we obtain the hori-
zontal component of the stress in top chord eg and multiplying this com-
ponent by sec#2, we obtain the maximum live-load stress in top chord eg.

The stress in top chord ce is determined in a similar manner. The
above equation (4)’ would be applied to determine the position of the
loading. The value of a would be taken in that case equal to the dis-
tance AE.

208. Graphical Determination of Live-Load Stresses in Pettit
Trusses.—The influence line method outlined in Art. 103 is the most con-
venient graphical method to use. The work as a whole is practically
the same as shown in Art. 205 for curved chord Pratt trusses.

As an illustration, let it be required to determine the live-load stress
in top chord eg of the truss shown at (a) in Fig. 342. The first thing
to do is to draw the influence line for reaction, as shown at (b). The
500 STRUCTURAL ENGINEERING

stress in eg can be determined by taking moments about G and consider-
ing the part of the structure to the left of section 2-2. Then evidentiy
the influence line for the stress in eg will be of the form shown at (ec).
By placing a unit load at G and drawing ec (at (b)) and ed parallel,
- respectively, to SS’ and SG, we have the stress in top chord eg, due to
the unit load at G, represented by the line ec. Then drawing AB, at

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+ Prig
Ss & \ He
si Ip i a t
(i |
A oa {ari :
' |
: Shs | | a
. vy
a oe | ag || | g |
ae
| |
2
for sfress ine
tik : :
1 Ie
4 \ Ty |
I | |
eT
ee i
~=44 Ne
| | | (A) forstressi1-£EG
C 1 foo 4
ie
| I |
jl tet (e) for stress in-GH3
| ee
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! | |
“
| le Ff) For stress 11-CM3
Se : 4
i
t
fei
La le! (G)  forstressin-e£
x

 

 

 

Fig. 342

(c), and laying off e’c’=ec and drawing e’A and e’B, we obtain the
influence line Ae’B for the stress in top chord eg. Then placing the live
Soad on the span so that there is a load at G and the unit load to the left
of G equal to the unit load on the span, and multiplying the loads by
their respective ordinates, as previously explained, the maximum ive-
load stress in eg is obtained. In case an equivalent uniform live ioad be
nsed, the maximum stress in eg will be obtained by multiplying the area
DESIGN OF SIMPLE RAILROAD BRIDGES 501

of triangle A4e’B by the uniform load per foot, as previously explained.

The stress in bottom chord EG can be determined by taking moments
about e. Then evidently the influence line for the stress in EG will be
partly of the form Ck’D, shown at (d). But we found in Art. 207 (see
Fig. 340) that for moments the line Ck’n was a straight line and undoubt-
edly Ck’n will be a straight line if the influence line be constructed for

   
       
     

 

  
 
    

   

 
  
 
   
  

 

 

 

        

 

  
  

    
   

     
  
   

  
   
     
  
 

    

 

   
 
 

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= ae. “iy oP tle 2

oe pt Elta ce

FFIN Hath edssye!s” wPn.
Fig. 343

the stress in EG. So if k’m’ were known, k’D could be drawn, thus
locating point 0, and next Cn could be drawn and then the line on could
be drawn and thus we would have the complete influence line CnoD for
the stress in bottom chord EG. As is seen, the only thing needed to con-
struct this influence line is the ordinate k’m’. By placing a unit load at
E and taking moments about e (considering the part of the structure to
the left of section 2-2) and dividing by the height of post eZ the stress
502 STRUCTURAL ENGINEERING

in bottom chord EG, due to the unit load at E, is obtained, which is the
desired value of k’m’. By drawing Ae and eN we have the truss deNEA.
Now it is readily seen that a unit load at E will produce the same stress
in AN as would be found in EG by taking moments about e. By draw-
ing hm at (b), parallel to Ae, we have the stress in AN represented by
the line km. Then by laying off k’m’=km, the influence line CnoD can
be drawn as explained above. Then placing the live load on the span
in accordance with (4) of the last article, and multiplying the loads by
the proper ordinates to the influence line at (d), the maximum live-load
stress in bottom chord EG will be obtained. The other influence lines
shown in Fig. 342 are considered self-explanatory, provided Art. 205 is
thoroughly understood.

209. Remarks Concerning Pettit Trusses.—Pettit trusses, as pre-
viously stated, are used for long span bridges. These trusses should
have economic heights, in which case, if single paneling were used, the
floor system would be very heavy, or the diagonals would have an un-
economic and awkward slope. These undesirable features are eliminated
by the sub-paneling.

All diagonals and sub-diagonals can be made out-and-out tension
members (eye-bars) if the sub-paneling be at the top chord. This will
result in a lighter truss than if the paneling be at the bottom chord. But
experience seems to indicate that trusses with sub-paneling at the bot-
tom chord are more rigid than trusses with sub-paneling at the top chord.
It is a question whether the lack of rigidity, however, is not due more
to poor details than to the form of the truss.

The stress sheets and the detail drawings for Pettit trusses are
gotten out in the same manner as previously explained for parallel and
curved chord bridges and the calculations of the details are similar te
those previously shown for those bridges.

A fair idea of the details of a Pettit truss can be obtained from
Fig. 343, where the details of a panel of the bridge indicated in Fig. 339
are shown. These details are of the Bismarck bridge designed by Ralph
Modjeski and built by the American Bridge Company. It is a Northern
Pacific Railway bridge over the Missouri River at Bismarck, North
Dakota.

The designing of the floor and lateral systems and end bearings of
Pettit truss bridges is exactly the same as for the curved chord Pratt
truss bridges, previously given.

MISCELLANEOUS TRUSSES

210. Warren Trusses.—The truss shown in Fig. 344 is known as
the Warren truss. In case of a through bridge, there would be a floor
beam at each of the joints B, C, and D.

Dead-Load Stresses. Let P represent the dead load per panel at
each of the upper joints and W the dead load per panel at each of the
lower joints and let R represent the end reaction due to these loads.
The dead-load stresses in the web members are indicated at (a) in Fig.
344. These expressions can be written from mere inspection. As is
readily seen, :

R=(14W+2P).
DESIGN OF SIMPLE RAILROAD BRIDGES 503

Taking moments about b of the forces to the left of that joint and
dividing this moment by h, we obtain

S= (1 +2py$ = (14W +2P)tand

for the stress in bottom chord AB. Taking moments about B of the
forces to the left of that joint, we obtain

S1=(14W+ ep)24 -P£=3(W +P)tand

for the stress in top chord be.

 

P P Pp P
9 tat fore le +4lW+ P) fone ly [i ss
AK Y N&
2e Ne WY (Q)

—

1
i “(1g W+ BPN 3 We A am

 

 

 

 

 

 

R La’ Ee P
Ww
“ 4s 8d :
: 4 +3Wtone __¢_ +4Wtane of - '
0 *< wx
Sc! et ce. 2
a A ae |
0 -sWhote \*7 - gf Whave ‘
A a Cc oOo. fF
A 3 2 ‘
@
Fig. 344

In a similar manner, taking moments about c, we obtain
S2= (34W+4P)tand

for the stress in bottom chord BC and by taking moments about C, we
obtain
S3=4(W+P)tand

for the stress in top chord cd.
From this it is seen that the dead-load stress in any Warren truss

is easily determined.

Live-Load Stresses. In case the live load be a uniform load, the
stresses are very easily determined. Suppose the live load to be w
pounds per foot of truss. Then we have

QWdxw= W

for the live-load panel load.
Loading joint D (alone), we have

“sect

for the maximum live-load compression in diagonal dC (as indicated at
(b)). Loading joints D and C, we have
W

3 seed

for the maximum live-load stress in each of the diagonals Cc and cB,
tension in cC and compression in cB.
504 STRUCTURAL ENGINEERING

Loading joints D, C, and B, we have
6 W sect

for the maximum live-load stress in diagonal bB and end post bA, tension
in bB and compression in bd.

The chord stresses will be a maximum when the span is fully loaded;
that is, when joints D, C, and B are loaded. The reaction at A is then
equal to 14W. Then taking moments about b we have

1x4 =14tand

for the maximum live-load stress in bottom chord 4B. Taking moments
about B, we have

(1417) 2 ‘ = 3Wtand

for the maximum live-load stress in top chord be. Taking moments about
c, we have
d d
(14W)3 ko W , 7 34Wtand
for the maximum live-load stress in bottom chord BC. Taking moments
about joint C, we have

(ya -W2 ¢ _4Wtand

for the maximum live-load stress in top chord cd.

4

b c d e

e Zz oe a

20 OW6 ola ds% 00 8
ra

 

 

 

 

 

 

/2/®
i

 

147%

 

 

 

Fig. 345

From this it is scen that the stresses in any Warren truss due to a
uniform live load are easily determined.

The stresses in Warren trusses, due to wheel loads, are very readily
determined. As an example, let it be required to determine the maximum
live-load tensile stress in diagonal cC. The span would be loaded as
shown in Fig. 345. Let P represent the load in panel BC, the center of
DESIGN OF SIMPLE RAILROAD BRIDGES 505

gravity of which is 2 distance from C; and let W represent the total load
on the bridge, the center of gravity of which is x distance from E. Then
by taking moments about £, we obtain

We
eo
for the reaction at A and taking moments about C, we obtain
Pe
“2d

for the concentration at B.
Then for the live-load tensile stress in eC, we have

Wz P
S= (R-r)secd= 7 - 3) SECO caneu me ia solemn cenes (1).

Now suppose the loads move a very short distance to the right or
left, say to the left, we have

We Waa Pz Pag
$+ as= ( T. + L ) sec6 — (73 oe) sec0

 

 

for the stress in cC.
Subtracting (1) from this last equation, we obtain
Wax PAz
AS= ( L ) sec6 — (=) secd

for the increment of the stress in cC. Now this increment would be zero
if the stress in cC were a maximum and hence the last equation would
become

 

Waar PAs.
L ~ 2d
But Az= Az, as is readily seen, so we have

WwW P W_P
% o2° - 3d i (2),

0=

0=

when the stress in diagonal cC is a maximum. Expressing this in words,
the unit load on the bridge is equal to the unit load in panel BC when
the maximum live-load tensile stress in diagonal cC occurs. A load will
be at C. The maximum compression in diagonal cB, which is equal to
the tension in cC, occurs at the same time. To obtain maximum tension
in bB and maximum compression in bd, the span would be loaded from
the right with a wheel at B and panel AB loaded according to the above
equation (2).

Let it be required to determine the maximum live-load stress in bot-
tom chord BC. .The span would be loaded as shown in Fig. 346. Let
W represent the total load on the bridge, the center of gravity of which
is z distance from E. Let P1 represent the load to the left of B, the cen-
ter of gravity of which is z distance from B, and let P2 represent the
506 STRUCTURAL ENGINEERING

load in panel BC, the center of gravity of which is y distance from C.
Then we have
We

L

Taking moments about c, we obtain

s-| (7) o-Pice+a) = pe) 4]

for the stress in BC.

R= and r= (P2) 2.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A iE
R 2 PI 8 \yp2 ¢ x
ibe ctw
t |
i !
Ce
Qa ' 5 N
t Loess i sil
es er am
Me
a \
0 ‘0!
Fig, 346

Now, suppose the loads all move a short distance Az, then we would
have

 

as-| (7 )a-Pias-3P2)ay] 7

for the increment of the stress in chord BC. But if the stress in BC
were a maximum, this increment would be zero and we would have

o-| (" )a—Piaz—3(P2)ay | re

But Av =Az=Ay, so this last equation reduces to
Wa

 

0= ee Pi- $P2,
from which we obtain
W 4P2+P1
TEER nena (3).

This last equation shows clearly how the loads should be placed
for maximum stress in bottom chord BC. The case of the other bottom
chords is similar. The stress in the top chords is obtained by taking
moments about the bottom chord joints. The placing of the loads in that
case is simply a matter of applying (5) of Art. 91.
DESIGN OF SIMPLE RAILROAD BRIDGES 507

The influence line for stress in diagonal cC is shown in Fig. 345 and
the influence line for stress in bottom chord BC is shown in Fig. 346.
These are readily understood; in fact, the drawing of the influence line

b 4 f 2 x 7 ip

 

 

 

 

 

 

 

 

‘

A B C D & F G a

L Fig. 847
for any of the members in a Warren truss is a simple problem, provided
the work previously given on influence lines is understood.

There are very few out-and-out Warren trusses (such as shown in
Fig. 344) built. They usually have vertical posts and hangers, as shown
in Fig. 347. In that case the loading for maximum stress and the determi-
nation of the stresses are the same as for an ordinary Pratt truss.

Referring to Fig. 347, the stress in hangers dD and fF is the same as
in hangers bB and hH.

The posts cC, eE and gG in the case of through bridges have only
dead-load stress, which is due to the load at the top chord joints. To obtain
the maximum live-load stress in bottom Ghord CE, due to wheel loads, a
wheel would be placed at D, such that the average unit load to the left of
D would equal the average unit load on the bridge. The stress would then
be obtained by taking moments about d. For maximum stress in top chord
df the span would be loaded in reference to joint E.

That is, a wheel would be placed at E, such that the average unit
load to the left of E would be equal to the average unit load on the
bridge. The stress in df would then be determined by taking moments
about £.

212. Double-System Warren Truss.—The truss shown at (a),
Fig. 348, is known as a “double-system’ Warren truss. The dead-load
stresses and stresses due to a uniform live load can be determined very
readily by considering the truss composed of two independent trusses,
one of which is shown at (b) and the other one at (c).

The stresses are then determiried in each of these trusses and com-
bined; thus the stress in the structure as a whole is obtained.

Dead-Load Stresses. Let P represent the dead load per panel at
each top joint and let W represent the dead load per panel at each
bottom joint. The end reaction on the truss shown at (b) is

 

R1=(2W+14P).

Then the dead-load stress in the web members is as indicated.
Taking moments about b, we have

S=(2W+14P) = (2W +14P)tand
for the stress in bottom chord AC. Taking moments about C, we have

S1=(2W+ 1gP)2 4-307 + P) t (347 + 24P)tand
508 STRUCTURAL ENGINEERING

for the stress in top chord bd. Taking moments about d, we have

82=(2W+ 14P)3-4 3+ PF 2 we = (4W +34P)tan6

/
for the stress in bottom chord CE. Taking moments about E, we have
S3= (44W + 34P)tand

for the stress in top chord de. The dead-load stresses indicated on the
truss shown at (c) are

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

obtained in a similar b e a e ys 2 a
manner. By combining 4\)
the stresses given on (Q) h
the top and bottom A Z e D ¥ v7 oS 7) Ke
chords and end posts e La L= 8d
at (b) and (c) the ‘ e %
dead-load stresses in ; é b (3twr2bpyione. | (45weaspyione | f Zh
the truss as a whole w% RS 3 S (b)
are obtained. omit » ree (a mnsb Mane K
Live-Load Stresses. a y Slw ri aC y
—Let w represent a [rideme) “(ijhe2P)tone w 2
uniform live load per ee Pla (adweaPltane Pig ly
foot of truss. Then a : VY XS f, ol (c)
we have ‘ Wood robmesbeio Seo Y obs abPiton . ,
wd=W Re=(1h ¥ “ " a. we ¥
=(15W+2P)
for the live load per ;
panel. Referring to |
the truss shown at (d) | ee eee ae -
and considering $W at! OLE > id) a
H and W at G, we ob- | < Ne ty 2 \
tain A /antan soit 3 4wfone v “é LF K
W 3 6 4 2 2
2 (F) sec6 ow
a 8 © ) WSCCB C_ sations. eo g 7
for the maximum live- A ‘ i &., / vee .
load compressive stress “land” ekwiane NY” szwhane ,
in diagonal Ef, and by 4 3s z 2 %
considering 4W at H  jré=uw
and W at each of the Fig. 348

joints G and E, we ob-

tain 64(/8)sec@ for the maximum live stress in each of the diagonals dE
and dC, tension in dE, and compression in dC. Considering 4$W at H
and W at each of the joints G, E, and C, we obtain 124(W/8)sec0 for
the maximum live-load tensile stress in diagonal bC. Considering 4/V
at H and W at each of the joints G, E, and C and 4W at B, we obtain
2Wsecé for the live-load stress in end post bd. When the truss at (d) is
fully loaded, that is, W at each of the joints C, E, and G, and 4W at
each of the joints H and B, we obtain, by taking moments about b,
DESIGN OF SIMPLE RAILROAD BRIDGES 509

(QW) cs 2Wtand
for the stress in bottom chords ABC. Taking moments about C, we obtain
(2W)2 4 - (4) 4 34Wtand
for the stress in top chord bd. Taking moments about d, we obtain
(2W)3 é —(4W)2 Q -(W) d =4Wtand

for the live-load stress in bottom chord CE and taking moments about E,
we obtain 44Wtané for the live-load stress in top chord df. The stresses
shown at (e) are obtained in a similar manner. Then combining the
stresses given on the top and bottom chords and end post at (d) and (e)
the live-load stresses in the truss as a whole are obtained.

In case wheel load be used, the stresses can be determined most
readily by the use of influence lines. As an illustration, let it be required
to determine the stress in chord de (Fig. 349) due to wheel loads. First
construct the influence lines for shear as shown at (b). £ is the center
of moments when the system drawn in full is considered. Then draw-
ing ms parallel to OF, the distance ns is obtained, which gives the stress
in de, due to a unit load at E. Then, laying off at (c) the ordinate b=ns,
the influence line C-4-B is obtained. D is the center of moments when the
dotted system is considered. Then drawing ut parallel to OD the dis-
tance vt is obtained, which gives the stress in de due to a unit load at D.
Then laying off the ordinate a=vt, the influence line C-3-B is obtained.
Let us consider a single load passing over the span starting from K.
When it reaches joint H half of it is transmitted to the two systems.
Then, evidently, the point half way between the two influence lines at
that point will be on the influence line for stress in de. When the load
reaches joint G, the load will be supported by the system shown in full
and, hence, the point 6 will be on the influence line for stress in de. When
the load reaches joint F, the load will be supported by the dotted system
and, hence, the point 5 will be on the influence line for stress in de.
Tracing out in this manner, we obtain the influence line B-7-6-5-4-3-2-1-C
for stress in de. Then placing the wheel loads (mostly by trial) for
maximum stress in de and multiplying each load by its respective ordi-
nate to this zigzag influence line the maximum stress in de, due to wheel
loads is obtained. The maximum stress in the other top chord members
is obtained in the same manner.

The influence line for the stress in bottom chord DE is shown at (d).
Considering the system drawn in full, the center of moments in deter-
mining the stress in DE is at d. Drawing mz parallel to Ad, the dis-
tance nx is obtained, which gives the stress in DE, due to a unit load at
E and drawing or parallel to SA, the distance ko is obtained, which gives
the stress in DE, due to a unit load at C. Then laying off at (d) the
ordinates d and c equal, respectively, to nx and ko, the influence line
E-9-11-F is obtained. Considering the dotted system, e would be the
center of moments in determining the stress in DE. Drawing wy paral-
510 STRUCTURAL ENGINEERING

lel to Ae, the distance ey is obtained, which gives the stress in DE
due to a unit load at F, and drawing zc parallel to Ke, the distance vz is
‘obtained, which gives the stress in DE, due to a unit load at D. Now,
laying off at (d) the ordinates f and e equal, respectively, to ey and vz,
the influence line E-10-12-F is obtained. Then by considering a single
load to move over the span as in the above case, the influence line
F-14-13-12-11-10-9-8-E for the stress in bottom chord DE is obtained.
By placing the wheel loads (by trial) for maximum stress in DE and

 

 

 

 

 

 

   

 

 

 

 

 

 

 

 

a." | ; | | |
{ | | | | |
J 1 i | Fok stressin TEGIC
~ |
ee A ee
Ws Ib es a’ ee
Ge -snck * H
iis 27) (2)
Fig. 349

multiplying each load by its respective ordinate to this influence line, the
maximum live-load stress in bottom chord DE will be obtained. The
stress in any of the other bottom chord members can be determined in the
same manner.

Let it be required to determine the stress in diagonal dE. Draw-
ing am parallel to diagonal dE, we have the stress in the diagonal due
to a unit load at EF, and drawing bh parallel to dC, we have the stress
in diagonal dC, also diagonal dE, due to a unit load at C. Then laying
off at (e) the ordinates a’m’ and b’h’ equal, respectively, to am and
DESIGN OF SIMPLE RAILROAD BRIDGES "511

bh, the influence line H-m’-h’-G is obtained. Then by considering a
single load to move over the span, the influence line H-20-19-18-m’-16-
h’-15-G for the stress in diagonal dE is readily drawn. Then by placing
the wheel loads (by trial) on the part of this influence line to the right
of 16 and multiplying each load by its respective ordinate the maximum

 

c od :e ft g 4b k

bxe 6 XY (Q)

 

 

 

Af” 6 C D E& F G H I F kK LM WO P
Ld jv jaw lew lew lew yaw

 

 

 

 

fp Pei? PoP P e
@ -
aX (4)
Ale ¢)_ oOo 4) FF a Hw a Ky aa o 9:
ew 2w ew. 2W 2w 2w ew

 

 

 

 

forstressin en
«At

“ow

      

ax
‘

 

a a ao mes

 

“ “ a Gl |

4

5 aa ae Se a)
a
1
'

Sees
Y_!
1
t
I
!
=

 

 

14/8

 

(F)- ‘

 

Fig. 350

tensile stress in diagonal dE is obtained. At the same time the maximum
compression in diagonal dC is obtained, as the stress in dE and dC are
equal and opposite. To obtain the maximum compression in dE (also
maximum tension in dC) the wheel loads would be placed to the left of
16 and each load multiplied by its respective ordinate. The stress in
any of the other web members due to wheel loads can be determined in
the same manner.
512° STRUCTURAL ENGINEERING

In order to obtain short panels, double-system Warren trusses are
sometimes sub-paneled as shown at (a), Fig. 350. The stresses are
determined in very much the same manner as shown above for the trusses
without sub-paneling.

Considering the case of dead load, let P represent the panel load at
each upper panel point and W the load at each lower panel point. We
can consider the sub-paneling as being independent sub-trusses or trussed
stringers as indicated at (b) and that the load from these sub-trusses
is transmitted directly to the main lower panel points, in which case
there would be 2/V on each.

Then by considering the truss to be composed of two separate trusses
as shown in Fig. 348 the stress in all the main members can be readily
determined and then considering the additional stress in each bottom
chord and in the lower half of each diagonal due to the sub-trusses the
dead-load stresses throughout the truss are obtained.

Let it be required to determine the dead-load stress in diagonal eo.
The truss being symmetrical about J one-half of the load at that point
can be considered as transmitted by eo and also the load P at e. Then
we have (7+P)sec@ for the dead-load stress in eo. This same stress
will be in oF and, in addition, the load W at F will produce a compressive
stress of $WsecO in the member which should be added to the above. So
we have (W+P)sec6+4Wsecé for the total dead-load stress in diag-
onal oF.

For the dead-load stress in diagonal od, we have }PsecO from panel
point f and 2Wsecé from panel point G. So for the total dead-load
stress in od, we have 2Wsec6+4PsccO. This same stress would be in 0G
(so to speak) but the load W at F would produce a compressive stress of
4Vsec6, which should be subtracted from the above and hence we obtain
2Wsec8 + 4Psec6—4Wsec for the total dead-load stress in diagonal oG.

The dead-load stress in each of the hangers bB, mD, oF, etc., is equal
to W. The stress in the bottom chord of each sub-truss, as CmE, is equal
to 4#tan@. This must be added to the stress found in each bottom chord
when considering the truss as composed of two independent trusses.
That is, the stress in each bottom chord is determined by considering the
truss to be composed of two independent trusses (2 at each lower joint),
just the same as shown above in the case of trusses without sub-paneling,
and then the stress (4/tan@) in the bottom chord of the sub-truss is
added to this.

The stress in the top chords and upper half of the diagonals is just
the same as if there were no sub-paneling.

The stress in be (the upper half of the end post) is equal to the
shear in panel AC multiplied by secO. So we have ({W+34P) sec6 for
the dead-load stress in be. This same stress occurs in Ab and an addi-
tional amount of 4sec6, due to the load at B. So for the stress in Ab,
we have (7W+3}P)sec0+4WsecO. For the dead-load stress in bottom
chord AC, we have (7W+34P)tan6+4Wtané. The part, 4Vtan6, is
due to the load W at B. ‘4

In determining the stress in the hanger cC the load at c, E, e, and
I can be considered as not affecting that member. Then we have 4P
from f, 2W from G, P from d and 2W from C, making in all (4W +14P),
which can be considered to be the stress in cC.
DESIGN OF SIMPLE RAILROAD BRIDGES 513

In determining the live-load stresses the work is greatly simplified
by using an equivalent uniform load.

Let w represent the live load per foot of span. Then we have wd=W
for live load per panel. As an example, let it be required to deter-
mine the maximum live-load tension in diagonal od (shown at (a), Fig.
350). We would load all panels from P to F (inclusive). This would be
equivalent to placing 2W at each of the joints O, M, K, I, and G and 4{W
at E. The load at O, K, and G are the only ones that produce stress in
od. So we have [(W)1/8+ (2W)3/8+(2W)5/8]|secO for the maximum
live-load tension in diagonal od.

In determining the maximum live-load tensile stress in diagonal oG
the load at F would be omitted as it would cause compression in that
member. That is, the panels from P to G (inclusive) would be loaded
which would be equivalent to placing 2W at each of the joints O, M, K,
and I and 14W at G. Then ignoring the load at I and M, we have
[(V)1/8 + (2W)3/8+ (14W)5/8]secO for the maximum live-load tensile
stress in diagonal 0G.

To determine the maximum live-load compression in od, panel points
B, C, and D would be loaded, in which case the stress in od would
be 1/8(2W)sec6, which is due to the 2W at C.

In the case of diagonal oG, the maximum compression in it would
occur when panels B to F (inclusive) were loaded.

The load at E would produce no stress in oG. One-half of the load
at F would be transmitted to G. This would produce a compressive
stress of 4Wsecé in oG but five-eighths of this would be transmitted back,
so we would have (4Wsec@)? for the compressive stress in oG due to the
load W at F. The load 2W at C is the only load, except the one at F,
that affects diagonal 0G. So we have [1/8W+(4W)3/8]sec@ for the
maximum live-load compressive stress in diagonal oG.

The maximum live-load tensile and compressive stresses in any of
the diagonals due to a uniform live load can be determined in the same
manner as shown above for diagonals 0G, od, eo, and o£. The maximum
stress in the chord members due to a uniform live load occurs when all the
joints from B to P (inclusive) are loaded. The sub-paneling is first
ignored and a load of 2W is assumed to be at each of the joints C, E, G,
I, K, M, and O. Then the stresses in the chords are determined by con-
sidering the truss to be composed of two independent trusses as previously
explained. But to the stress in each bottom chord thus obtained the
stress }WtanO must be added. 4/WVtané is the stress in the bottom chord
of each sub-truss as CmE, EoG, etc.

In case wheel loads be used, the stresses can be determined most
readily by the use of influence lines. The influence lines are easily drawn,
some of which are shown in Fig. 350. The influence lines for the top
chords are the same as if the truss were not sub-paneled. The influence
line E-8-9-10-15-11-12-13-14-F, shown at (d), is for bottom chord GI.
The influence line G-15-h-16-m-18-19-20-H, shown at (e), is for diagonal
en, and G-15-h-16-r-m-18-19-20-H is for diagonal nI. The influence
line K-3-4-5-7-8-9-10-M, shown at (f), is for diagonal nf and K-3-4-5-6-
%-8-9-10-M is for diagonal Gn. The construction of these lines will be
readily understood by referring to the diagrams at (c) and (g).
514 STRUCTURAL ENGINEERING

213. Whipple Trusses.—The truss shown at (a), Fig. 351, is
known as a Whipple truss. Dead-load stresses and stresses due to uni-
form live load in this type of truss are readily determined by considering
the truss as composed of two independent trusses. In determining the
dead-load stresses in the truss shown at (a) the truss can be considered
as composed of two independent trusses, one of which is shown at (b)
and the other one at (c). Let W represent the dead load per panel, one-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

—a 1
a b Cc J e f 3 A k
“Se wid
() 8 go aes.
okay hy
“ “ * _ kK
a) B C 0 E F G 7 = T
wl diwsecd % +6swtond | ¥ ¥
(b)
- 62Whon i
ew 2w ew w 7
a ‘o 3 a
3 ewhond _\$ $ w
W/ lars % (Cc)
Maio %
& si Ne, & Sey &y
2whong| _2Wtang swrang
2
RaW ” a a” ¥ |
+4hPlan$ +64 Ptond +6$Ptang
& e- 440)
v ro} Ne. (a)
S oe uN =, ug S
Gy SN Noe YN Ss.
ve : Re Nts
tang VkPhang - 4$Ptang -6$Prlang
43 8 6 + z $ {
9 sSPlang 6Ptang
f/| Se wl & (2) | Ne
SY Uy Se sy Se, Se,
& i Ye 9 Ss Ww °
Ptangd| 2Ptang 5Piang NY
\ 48 2 a 3 5 t
Fig. 351

third applied at each top joint and two-thirds at each bottom joint. Tak-
ing moments about b and considering the truss at (b), we obtain

15 d 3
ast 9 ane
(GI) ® (5

=“
for the stress in CE. Likewise taking moments about a and considering
the truss at (c), we obtain

d 1
w)i=45 Wtand

(27) es 2W tang
DESIGN OF SIMPLE RAILROAD BRIDGES 515
for the stress in BD. Then the stress in bottom chord CD is equal to
(a" ) tand + 2Wtand = 65 Wtand.

Again taking moments about E and considering the truss at (b) we obtain
15 d 3 d d 1

for the stress in bd and taking moments about D and considering the
truss at (c), we obtain
1

@w)s4 ze G W)2¢ =51tang

for the stress in ac. Then we have ;
(55 W ) tand + 5Wtand = 115 Wtand

for the dead-load stress in top chord be. The dead-load stress in any
chord member can be readily determined in this manner.

 

 

 

 

 

 

 

 

 

 

 

 

o ab C f e f- g Ab k
oe ‘ age
{ Wo bes
: oS 3 i. (Q) a S
1 ‘ AON
1 > > K
A - 7 2 A
a ¢ 0 é G44 7
! t
i 1 1
*
x : 1 (b) '
ell I
N ! <

f=/*

TT
a

\
Y
\
\
\\
\\
Ser Sbieteee see ho
\
\
7
A
f
iw
‘
A
f
t

 

 

 

 

 

 

 

 

i
q
:
1
‘
1
i
t
1
4

Fig. 352

Considering the trusses at (b) and (c) the dead-load stresses in
the web members, as ‘indicated, are readily determined.

Let P represent the uniform live load per panel. The live-load
stresses in the chord members will then be as indicated on the trusses
BiG STRUCTURAL ENGINEERING

at (d) and (e). These are determined in exactly the same manner as
explained above for the dead-load stresses in the chords. The live-load
stresses in the web members are indicated at (d) and (e). These are
determined as previously explained. For example, considering the truss
at (d) and loading the panels from J to E (inclusive), we obtain

P fi 1

i0 (+2 +44 6) secd = 125 Psec#
for the maximum live-load tensile stress in diagonal bE and also
124x(P/10) for the maximum live-load stress in post bC. Likewise,

 

 

 

 

 

b c J e F g A A
4 \2
(a) h
A B Cc D E F G H Z K
8 7 6 my 4 3 2 /
/ 2 3 4 Ss 6 7 8

 

 

 

|
|

(Cc)

 

 

a 8 F
Fig. 353

loading the panels from J to F (inclusive) and referring to the truss at
(e), we obtain 7

P fi eee!

10 (F+3+5) sec) = 8 5 Psecd
for the maximum live-load stress in diagonal cF and also 84x (P/10) for
the maximum live-load stress in post cD.

If wheel loads are used the stresses can be determined most readily

by use of influence lines. Influence lines for stresses in Whipple trusses
are constructed practically in the same manner as shown above for the
DESIGN OF SIMPLE RAILROAD BRIDGES 517
double-system Warren trusses. Some of these influence lines are shown
in Fig. 352. The one at (c) is for top chord cd and the one at (d) is
for diagonal cF. The construction and use of influence lines in the
case of the Whipple truss is quite simple provided the work on influence
lines previously given is understood.

   

e

    

    
 

~,

    

     
     

3

2his%4oe
2h 435" Lott.

   

betoilof Cost [ren Bose for Fixed End.

 

 

  

 

 

 

 
  

 

 

  

 

 

 

    
 

 

 

 

 

 

 

 

 

 

   

(iftetel)
\
4a" , ye
Boseotka/} LAG SM ‘ r3g73°
~ ee A Oy TAGE
ie NI cf , SLID
te 's |} / “, Digh{ Ee ies
ep WL —
awe Olid : 2615.33
‘y s I f/ ot 2 GUS. FI 2h% Lo
% " % qo i EA ath al
& 1 Wim l lt gm
JA. be ial ey 2920 EL
aN.
44." STs $$ "Ah. B
Mfohot ad A Cc
q 2 6'Seg, Rollers.

isila@2p-rS stl
tees

j- 60 “Mov/s

Fig. 354

214. Lattice Truss.—The truss shown at (a), Fig. 353, is known

as a lattice truss.

It can be

considered composed of four independent

trusses, one of which is shown at (b) and another at (c) and by turning
these two end for end, we obtain the other two trusses, By determining
the dead and maximum live-load stress in each member of the trusses
shown at (b) and (c) and combining the stresses, the stress in the
518 STRUCTURAL ENGINEERING

truss shown at (a) is obtained. The determination of the dead-load
stress is a simple problem and the determination of the live-load stresses
is practically as simple provided a uniform live load be used. Let P
represent the uniform live load per panel. Considering the truss at (b)
and loading joint C (only) we obtain? Psec@ for the live-load tension
in eC and compression in eG. Loading joint G (only) we obtain
$Psec for the live-load tension in eG and compression in eC. Loading
both joint C and G we obtain4,° Psecd for the maximum live-load ten-
sion in bC and §Psecé for the maximum live-load tension in kG, also
10 Psecp and &Psecd for the live-load compression in bA and kK, re-
spectively. Loading joints C and G and taking moments about B, we

obtain
10 d 10
for the live-load stress in AC. Taking moments about C, we obtain

10 d 10
(FP) 2 Roe (FP) tand
for the live-load stress in be. Taking moments about k (considering
reaction at K), we obtain :

8 d 8
(GP )e- (EP )tang

for the live-load stress in GK. Taking moments about e (considering
the forces to the left), we obtain

10 d d 32
(3?) 45 ~Pxe Rag Ptane

for the live-load stress in CG. The stresses due to a uniform live load
in the trusses shown at (b) and (c) are readily determined in this man-
ner. Then by combining these carefully for the chords and end posts,
the stresses throughout the truss are obtained.

In case wheel loads are used, the stress therefrom can be most
readily determined by the use of influence lines. These are constructed
very much the same as previously shown for double-system Warren and
Whipple trusses.

Several railroad companies use the lattice truss shown at (a). The
details of part of one of these trusses are shown in Fig. 354. These
details are taken from the standard plans of the Chicago & North West-
ern Railway Company (W. C. Armstrong, engineer of bridges).

DEFLECTION AND CAMBER OF TRUSSES

215. Analytical Determination of Deflection —Let it be required
to determine the deflection of joint C of the truss shown in Fig. 355,
due to any loads. In the case of pin-connected trusses the deflection
is due to the distortion of the members and to a small clearance between
the pins and pin holes, while in the case of riveted trusses the deflection
is due wholly to the distortion of the members. We will first consider
the deflection of joint C, due entirely to the distortion of the members.
DESIGN OF SIMPLE RAILROAD BRIDGES 519

Let u represent the stress in any member as aB, due to a unit load
placed at C, and let § represent the stress in the same member due to
any loads placed at any panel points.

a b c

"| ° “le dl

Fig. 355

Let Z represent the length of member aB and A its gross area of
cross-section. Then for the work done on the member aB by the stress u,
we have ux3(uL/AE) which must undoubtedly be equal to $(Ad)1#,
where Ad represents the deflection of joint C, due to the stress u. So

we have
saz) u=5 (Ad)1Ib.,

uL
Ad= (33) u

for the deflection of joint C, due to stress u. Now, evidently, the deflec-
tion of joint C, due to any stress S in member aB, will be directly pro-
portional to the deflection Ad, due to stress u. Then let Ay represent
the deflection of joint C, due to stress S and we have

Ay _ a
Ad ~
and substituting (uL/ALE)u for Ad and efiies we obtain

SuL
49= 4E

from which we obtain

for the deflection of joint C, due to any stress S in member aB.

Now, evidently, if u’, u’’, etc., represent the stress in the other mem-
bers ofthe truss, due to a unit load at C, and S$’, 8”, etc., the stress
in these members, due to any loads placed at any joints, and Ay’, Ay’,
etc., the deflection of joint C, due to the stresses 8’, 8S”, etc., we can
write the formula

 

Sul S'u'L! | 8a" L” Seu"
= , u” RO am Freee oer eee By eet i ge
tha ay Sag he) = (a "EA pat Bae )

for the total deflection of joint C, due to any loads producing the simul-
taneous stresses S, S’, S’’, etc.
By, letting S=stress in any member due to any loading,
u=stress in any member due to a unit load at any point
considered,
L=length of any member in inches,
A =gross area of cross-section of any member in square
inches,
520 STRUCTURAL ENGINEERING

the general formula for deflection can be written as

y= (Fp) nitee lutea pe a yee ae Te Acne (1).

Example 1. Let it be required to determine the deflection of joint
L3 of the 150-ft. truss shown in Fig. 279, due to the given dead and live
loads.

The live-load stresses given for the truss members in Fig. 279 should
not be used in determining the deflection as they do not occur simul-
taneously. The maximum deflection will occur when the span is fully
loaded and a sufficiently accurate result will be obtained by using an
equivalent uniform live load, in which case the live-load stresses will be
directly proportional to the dead-load stresses, and hence they can
readily be determined directly from the dead-load stresses by the use of
a slide rule.

The maximum live-load stress in the end post (see Fig. 279) is
263,000 Ibs. Dividing this by the secant of the slope of the post, we
obtain

263,000 + 1.3 = 203,000 lbs.

for the maximum live-load shear in the end panel LO-L1 (see Art. 174).
Then substituting this value for S in the formula of Art. 123, we obtain

P’ = (203,000) +3(L-d) =203,000 + 62.5 = 3,250 lbs.

for the equivalent uniform live load per foot of truss for determining
the live-load stress in the web members, but as the chords contribute
most to the deflection we will use

8,250 — 3,250 (182 +2.5) % =3,120 Ibs.

per foot of truss for the equivalent uniform live load, which is really
the correct equivalent uniform live load for determining the maximum
chord stresses. (See Art. 123.) ‘

 

 

 

 

 

 

 

 

D- 38000" D- _ 399000*
L- 2600, < £- 222000%
- © =
ur 774800 2 39200 3
2 2
* wo x
0+ ake, a, i.
2; : SLOSL Aas
Se / a0 Se, Ot, NSS
PY mo O10 OSC8, ools Som 16
Ys 89g oPa* 100) eae %
Q/ n> ode aS Oo = ‘1
Vv; Sas cata},
Sok Volto Ss
+ 1
g af Qa

L0 “/ 2 Lg

2+ s5000% D+ eg000F

4+/63000* i+ 260000*%
+218000# + 348000"
Fig, 356

As seen from Fig. 279, the dead load per foot of truss is 2,110+2=
1,055 lbs. Then multiplying each of the dead-load stresses given for the
truss members in Fig. 279 by #422 the live-load stresses for determin-
ing the deflection of joint L3 are obtained. Then adding these to the
dead-load stresses, the total stress in each truss member, as shown in
Fig. 356, is obtained.
DESIGN OF SIMPLE RAILROAD BRIDGES 521

Then from Figs. 279 and 356 the following table can be computed:

L S A SL u SuL
Member Ins. Lbs. — Sq. Ins. EA Lbs, “EA
U1-LO ... 468 -337,000 46.18 -0.1138 —-0.6500 0.0739
L0-L2 ... 600 +218,000 23.52 +0.1853 +0.4166 0.0772
[2-L3 ... 300 +348,000 36.98  +0.0941 +0.8332 0.0784
U1-U2 ... 300 -348,000 40.58 -0.0857  -0.8332 0.0714
U2-U3 ... 300 -392,000 45.08  -0.0869  -1.2498 0.1086
U1-L2 ... 468 +203,000 26.48  +0.1196 40.6500 0.0777
U2-L3 ... 468 +67,000 19.80 40.0527  +0.6500 0.0343

U1-L1 ... 360  +71,000 14.44 40.0590 0 0
U2-L2 ... 360 87,000 19.80 0.0527 -0.5000 0.0263
U3-L3 ... 360  -9,000 19.80 0.0054 0 0
(E = 30,000,000)
wae y (S24) ~ 0.5498
AE 5

 

Total deflection of joint L3 =1.0996 ins.

As L3 is at the center of the span and the bridge symmetrically
loaded, the deflection of L3 can be obtained, as shown by the above table,
by determining the, deflection, due to just half of the truss members, and
then multiplying this deflection by two. In case of the determination of
the deflection of any other joint (except U3) all of the truss members
would have to be considered. For example, let it be required to deter-
mine the deflection of joint L2 (see Fig. 279). The table in that case
would be just the same as the above, except that the last two columns
on the right would have two values in each case as the u for the members
on one side of the center of span would be different from the u for the
corresponding members on the other side of the center of the span. For
example, placing a unit load at L2, we have four-sixths of a pound for
the reaction at one end of the truss, and two-sixths for the reaction at
the other end. Then for the u in one end post we would have é x 1.3=
0.8333, and ? x 1.3=0.4166 for the w in the other end post; and likewise
¢ X 0.8333 = 0.5555 for the u in one bottom chord LO-L2, and % x 0.8333 =
0.2777 for the uw in the other bottom chord LO-L2. Thus it is throughout
the span. The last two right-hand columns in the table would simply
have double numbers and the summation of the SuL/AE would not be
multiplied by two. Otherwise, the table for each of the other joints
would be just the same as shown above for joint L3.

In case the loading causing the stresses S were placed unsym-
metrically, each member would be listed.separately in the tables. In that
case it is advisable not to have two or more members with the same mark,
as in the case of the truss shown in Fig. 279.

It is customary to consider the distortion of compression members as
negative, and that of the tension members as positive. In that case the
tensile stresses, both u and S, should be indicated as plus, and the com-
pressive stress as negative, as shown in the above table.

In case the longitudinal displacement of any joint be desired, it
can be determined in the same manner as shown above for deflection,
except that the unit load applied at the joint in question would be con-
522 STRUCTURAL ENGINEERING

sidered as acting horizontally instead of vertically. In that case it is
necessary to pay strict attention to the signs of the u’s and S’s, as they
may not have like signs, which they ordinarily have in the case of
deflection.

As an example, let it be required to determine the horizontal dis-
placement of joint L3 of the truss shown in Fig. 279, due to the stresses.
The unit load would be applied at L3, as shown in Fig. 357. The end
L6 being fixed and the end LO being on rollers, the only members that
would really be stressed by the unit load would be L3-L4 and L4-L6,
and hence we need consider only these members in determining the hori-
zontal displacement of L3. The stress in each of these members, due to
the unit force, as is obvious, is 1 lb. Then making the following table, .
the horizontal displacement of L3 is obtained:

L KY A SL u Sub’
Member Ins. Lbs. Sq. Ins. EA Lbs. EA

L3-L4 ... 300  -348,000 36.98  -0.0941 -1.0000 0.0941
L4-L6 ... 600 -218,000 23.52 -0.1853  -1.0000 0.1853

Total horizontal displacement of joint L3 = 0.2794 ins.

This shows that joint £3 will move about }” to the left. It will be
seen that this table is really compiled from the table given above in
determining the deflection of L3.

The horizontal displacement of any of the other lower chord joints
can be determined in a similar manner, and just as readily after the
tables for the deflection of those joints are made.

The determination of the horizontal displacement of the top chord
joints, while similar to that shown for the bottom chord joints, is not so
simple, owing to the fact that more members are involved. As an illus-

 

 

 

 

 

 

 

 

 

ip U2 3 ff U4 US
S
fe e L6
Lo ja
3 L Ss
rr Zi £2 4 Z lr
'
L ‘ L J
Fig. 357

tration, let it be required to determine the horizontal displacement of
joint U4 (Fig. 357). The unit load would be applied at U4 as indicated.
The horizontal reaction of this load would be at L6, as indicated (the
end L6 being fixed), but in addition,-the unit load at U4 would produce
a positive reaction of r at LO and an equal negative reaction r at L6.
These vertical reactions and the 1 lb. horizontal reaction at L6 would
have to be considered in determining the u’s for the members throughout
the truss. For each of the vertical reactions we have r=h/L. After the
u’s are determined for the members throughout the truss, the horizontal
displacement of the top chord joints is determined by making a table for
each as explained above.
DESIGN OF SIMPLE RAILROAD BRIDGES 523

In determining the deflection of pin-connected bridges, the distor-
tion (SL/AE) of each member should be increased by wy in. to allow
for the clearance between the pins and pin holes. Otherwise the work
of determining the deflection of pin-connected bridges is exactly the same
as shown above for the riveted bridge.

216. Graphical Determination of Deflection.—Let ACB at (a),
Fig. 358, represent a cantilever frame and suppose a vertical load P be
applied at C. This load will cause tension in member AC and compres-
sion in member BC and hence the distortion in AC will increase its
length while the distortion in BC will decrease the length of that member.
Let Ca represent the distortion of AC and Ce the distortion of BC and
let us assume that the joints 4, C, and B are pin connected so that the
members AC and BC are free to turn about their ends. Now, undoubt-
edly, the distortion of the members AC and BC will cause the frame to
deflect to some position as
AC’B. As the joints 4
and B are fixed in posi-
tion the deflection of the
frame really consists of
joint C moving to C’ and,
hence, the determination
of the deflection is simply
a matter of locating C’ in
reference to joints A and
B, which remain fixed. By
taking A as a center and
Aa as a radius, the arc aC’
could be described, and
taking B as a center and
Be as a radius the are
eC’ could be described
and thus the point C’
would be located.

As another example, let D-F-G-E at (d) represent a cantilever frame
supporting a vertical load P at G. Let Fd, Ff, G’k, and Gm represent
the distortions of the members DF, FE, FG, and GE, respectively. Now,
owing to these distortions, the frame will deflect to some position as
D-F’-G”-E. As is obvious, the problem of determining the deflection of
the frame consists of locating F’ and G” in reference to the fixed joints
D and E. First, taking D as a center and Dd as a radius, the arc dF’
could be described; and taking E as a center and Ef as a radius the arc
fF’ could be described; and thus the position of F’ would be determined,
which is the deflected position of joint F. Now, having joint F located,
the deflected position of joint G can be determined in reference to E and
F. Suppose, for the time being, that joint G be disconnected and that
member FG moves to the parallel position F’G’, while EG remains in
position: Then, taking F’ as a center and F’k as a radius, the arc kG”
could be described; and taking FE as a center and Em as a radius, the are
mG” could be described; and thus the position of G’’ would be deter-
mined, which is the deflected position of joint G; and hence the deflected
position of the entire frame would be known.

 

 

 

 

 

  

 

 

u ae
[i

Fig. 358
524 STRUCTURAL ENGINEERING

Now it is evident that, theoretically, the deflection of any cantilever
frame could be determined in the same manner as shown for the above
two cases, and as any truss can be considered as being composed of one
or more cantilevers, the same is true of any truss. But, in practice, the
application of the method is practically impossible, as the distortions
of the members are so small compared with their lengths that if the
members be laid off to a reasonable scale it would be practically impos-
sible to measure the distortions and deflections. This very fact, how-
ever, makes possible the graphical method in general use, in which it is
assumed that the distortions are so small in comparison with the lengths
of the members that all such ares as aC’, eC’, dF’, mG”, etc., without
appreciable error can be considered straight lines; that is, the are and
its tangent, in each case, are assumed to coincide. One can satisfy him-
self as to the accuracy of this assumption by trial. As a suggestion, sup-
pose the distance AB (Fig. 358(a)) be laid off equal to 10 ft.; AC and
BC equal to 15 ft. and 18.5 ft., respectively; and Ca and Ce each equal
to 4 of an in. and the arcs AC’ and eC’ described, as explained above,
and the difference between these arcs and their tangents noted: By
assuming that all such arcs as AC’ and eC’, etc., coincide with their
tangents, the deflection of any truss can be determined by simply laying
off the distortions and drawing perpendiculars to these. As an example,
let us consider the case shown at (a), Fig. 358. The joints A and B
are considered fixed. Let us take any point A’ at (b), as origin, to
determine the deflection of the joints 4, C, and B. Now, as both A
and B are fixed, they will, as regards deflection, be at A’ and hence we -
have 4 and B located. Joint C in reference to A moves to the right the
distance Ca. Then at (b) lay off Ca from A’ (the location of joint 4)
to the right as shown. Joint C in reference to B moves toward B the
distance Ce. Then at (b) lay off Ce from A’ (the location of joint B),
as shown. Then, by drawing from e and a the perpendiculars aC’ and
eC’ we have C’ located, which is:the deflected position of joint C. That
is, as A’ is the origin (point of zero deflection), V represents the deflec-
tion of joint C, and h represents the horizontal movement of that joint.

Next let us consider the case shown at (d), Fig. 358. Here joints
D and E are considered fixed. Let D’ at (e) be the origin to determine
the deflection of joints D, E, F, and G. As D and E are fixed, they will
both be at D’, the origin. In reference to joint D joint F moves to the
right the distance Fd. So from D’ (the location of joint D) lay off Fd
to the right as shown. In reference to E joint F moves toward E the
distance Ff. So from D’ (the location of E) lay off Ff as shown. Then
by drawing the perpendiculars dF’ and fF’ the point F’ is located, which
is the deflected position of joint F. Now having F located, joint G can
be located in reference to joints F and E. In reference to joint F joint G
moves away from F the distance G’k, as shown. In reference to joint E,
joint G moves toward E, or to the left, the distance mG. So from D’
(the location of joint E) lay off mG to the left as shown. Then by draw-
ing the perpendiculars kG” and mG”, the point G” is located, which is
the deflected position of joint G. We then have the deflection of all
the joints determined. The distances V’ and V” represent the deflection
of joints F and G, respectively, and h’ and h” represent respectively the
horizontal movement of these joints.
DESIGN OF SIMPLE RAILROAD BRIDGES 525

As another example, let it be required to determine the deflection
of the joints of the cantilever frame shown at (a), Fig. 359, where
ATA? bene ees A10 represent the distortions of the members. The plus
sign signifies that the length of the member is increased by the distortion,
and the minus sign signifies just the opposite. First, select any point A’
at (b) asthe origin. The joints 4 and B are fixed so they will both be at
A’. In reference to A, joint C moves to the right the distance Al. So
from A’ (at (b)) draw Al to the right. In reference to B, joint C moves
toward B the distance A4. So from B’ (at (b)) draw A4 as shown, and
then drawing perpendiculars to these distortions the point C’ is located,
which shows the deflected position of joint C. That is, joint C has moved
from A’ to C’. Now, having C located, we can proceed to locate joint D
in reference to joints B and C. In reference to C, joint D moves down
the distance A5. So from C’ lay off A5 downward as shown. In refer-
ence to B, joint D will move toward B the distance A9. So from B’
(at (b)) lay off A9 to the left. Then, drawing perpendiculars to these
distortions (A5 and A9), we obtain D’ which shows the deflected position
of joint D. That is, g
joint D has moved from ‘vet
4’ to D’. Now, having eG
both C and D located, i Fa
we can next locate the 4a +a taz2__£ 443 ~G oon
joint E. In reference to 7
C, joint E moves to yi .
the right the distance
A2. So from C’ draw =49 aie
A2 to the right. In
reference to D, joint E (Q) “ (b)
moves toward D_ the 3
distance A6. So from
D’, draw A6 as sHown.
Then by drawing per-
Sscuillcl acs to aie ere
distortions (A2 and AG), we obtain E’, which shows the deflected posi-
tion of E. That is, joint E has moved from 4’ to E’. Now having
joints E and D located, we can next locate joint F. In reference to E,
joint F moves downward the distance A7. So from E’ draw AY down-
ward. In reference to joint D, joint F moves toward D the distance A10.
So from D’ draw A10 to the left. Then by drawing perpendiculars to
these distortions (AY and A10), we obtain point F’, which shows the
deflected position of joint F. That is, joint F has moved from 4’ to F’.
Now, having joints E and F located, we can locate joint G. In reference
to E, joint G moves to the right the distance A3. So from E’ draw A3
to the right. In reference to F, joint G moves toward F the distance A8.
So from F’ draw A8 as shown. Then by drawing perpendiculars to these
distortions (A8 and A3), we obtain the point G’, which shows the de-
flected position of joint G. That is, joint G has moved from J’ to G’.

Referring to the diagram at (b), we can consider all of the joints
as being at A’ before deflection, and that they deflect to the posi-
tions indicated. That is, C moves from A’ to C’, D from A’ to D’,
E from A’ to E’, F from A’ to F’, and G from 4’ to G’. By measuring

QO

 

+47

vo

 

 

 

a
an
526 STRUCTURAL ENGINEERING

the vertical distance down from A’ in each case, the deflections are ob-
tained and the horizontal movements are obtained by measuring the hori-
zontal distance in each case from a vertical line through 4’.

As all trusses can be considered as being composed of one or more
cantilever frames, it is obvious that the deflection of any truss can be
determined by constructing diagrams similar to those shown at (b) and
(e), Fig. 358, and at (b), Fig. 359.

These diagrams are known as “Williot diagrams,” being named after
the French engineer Williot, who proposed them.

Example 1. Let it be required to determine the deflection of the
truss shown in Fig. 279, due to the stresses given in Fig. 356. The

distortions of the mem-

 

 

 

 

 

 

“ Oi 28 /e.cite U3 bers, due to these stresses

Sy 4) See, a are given in Example 1 of
wil 4 <a Xe (a) Art. 316 (see table).

i First, draw the out-

Lola" i7 ag le we ast line of the truss as shown

at (a), Fig. 360. The dis-
tortions shown there for
ai : | (5 ) one-half of the members
(they are the same for the
Lo other half of the truss)
Ss are expressed in fractional
ay form for convenience.
As the truss is sym-
*. metrical about U3-L3 and
w? symmetrically loaded the
a deflection will be fully
7) known if it be determined
i for one-half of the truss.
Ag By considering joint L3
as fixed in position and
See member U3-L3 as fixed in
Fig. 360 direction either half of the
truss can be considered as
a cantilever and the deflection determined as explained above. Let us take
L3 at (c) as a starting point. Then the position of joint U3 is obtained
by laying off the distortion of U3-L3 (which is 4,’) downward, using a
seale of 7” = 7”. Then having joint U3 located, the position of U2
can be determined in reference to U3 and L3. Joint U2 moves towards
U3 and away from £3. Then laying off the distortion of member
U2-U3 to the right from U3 (at (c)) as shown, and the distortion of
member U2-L3 upward and parallel to U2-L3 from L3 and drawing
perpendiculars to these distortions, we obtain point U2, which shows the
position of joint U2. That is, joint U2 has moved from £3 to U2. Joint
L2 moves toward U2 and away from L3. Then laying off the distortion
of member U2-L2 upward from U2 and the distortion of L2-L3 to
the left from L3, and drawing perpendiculars to these distortions, we
obtain point L2, which shows the position of joint L2. Joint U1 moves
toward U2 and away from L2. Then laying off the distortion of member
U1-U2 to the right from U2, as shown, and the distortion of member U1-L2

 

 

 

 

 

 

oe
“64

 

 

 

 

 
DESIGN OF SIMPLE RAILROAD BRIDGES 527

upward and parallel to U1-L2 from L2, as shown, and drawing perpen-
diculars to these distortions, we obtain point U1, which shows the loca-
tion of joint U1. Joint L1 moves downward from U1 and to the left from
2. So laying off the distortion in U1-L1 downward from U1 (at (c) )
and the distortion of L1-L2 to the left from L2 as shown and drawing per-
pendiculars to these distortions we obtain point L1, which shows the loca-
tion of joint L1. Joint LO moves toward U1 and away from L1. Then
laying off the distortion of member U1-LO upward and parallel to U1-LO
from U1 and the distortion of member LO-L1 to the left from L1, and
drawing perpendiculars to these distortions, we obtain the point LO,
which shows the location of joint LO. We now have the position of the
joints of the left half of the truss determined at (c) in reference to L3.
As is obvious, in reference to that point, the left half of the truss would
be bent upward, as indicated by the dotted outline, and of course the
right half would be bent up the same amount. Now, it is evident that
the deflection of the bottom chord joints is the distance they are below a”
horizontal line through LO’. These distances are obtained for each joint
by measuring down from LO (at (c)). We thus obtain 7%”, 7”, and
1,4” for the deflection of joints L1, L2, and L3, respectively. Then a
diagram of these deflections can be drawn as shown at (b). The deflec-

U2_-s4 U3-23 b U4

 

& x
y ON

Az
az

 

 

 

 

 

AG SAI |
£2 43 oa L4

 

 

“ i

Vado vt 3

L3 , $43,441
U

Fig. 361

tion of any ordinary bridge truss can be determined in this manner. In
case of odd panels, it is best to begin at the middle of the center panel
in constructing the Williot diagram. For example, point a (Fig. 361(a))
would be considered fixed in position and line ab fixed in direction. The
diagonal U3-L4 and U4-L3 would be ignored, as they would have no
stress in them—assuming that a uniform live- and dead-load extended
over the entire span. The point b would move downward the distance
A2, which is the distortion of each of the posts U3-L3 and U4-L4. Then
taking a as the starting point (at (b)), the point b is located by laying
off A2 downward from a, as shown. Then joint L3 is located by laying
off Al, which is one-half of the distortion of L3-L4, to the left from a
and joint U3 is located by laying off A3, which is one-half of the distor-
tion of U8-U4, to the right from b. Then, having joints 23 and U3
located, the work of determining the deflection proceeds as previously
explained.
528 STRUCTURAL ENGINEERING

The above manner of determining deflection of trusses is quite
accurate as far as vertical movement is concerned, which is usually all
that is desired, but if the horizontal movement be desired, the work must
be carried out in a somewhat different manner. It will be seen that in
the above case we assumed the center of the truss to remain fixed and that
each end of the truss was free to move. This assumption is not true, as
we know, as one end of the truss is fixed and the other end is free to move.

gta...
gas “o

Ul by U2 73 Be’ 4 es

oA | News SNe Ne
Poh NT iy tA
{/ £ x (@ = >

‘428

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0,3" “+” L2 +e 13 +e L4 +2 75 ee Le
, a
40 @ LO Zi Le 23 L4 LS LE
Mesh gt 1 ge ea
Secale ig gr) ja" glk
Eire be LA | eee 3 :
e Fe alert ae #
O on a
2
Gi i +40”
:
= “yD
FE x)
hay
SI
U
= inches eg ae
La" ha
ea uf
/
q SS A | a "of u3"
* L2 i? A
x
N
us?
%
%
A
a us”
a :
ry
Fig. 362

Now, evidently the only way to locate correctly the deflected posi-
tion of the truss would be to begin at the fixed end, .

So let the diagram at (a), Fig. 362, represent the same truss as
considered above, and suppose the end L6 fixed and the end LO free (sup-
ported upon rollers). Then considering L6 fixed in position and member
L6-L5 as fixed in direction and taking L6 (at (b)) as the starting point
DESIGN OF SIMPLE RAILROAD BRIDGES 529

and constructing the Williot diagram as shown, we obtain the relative
position of the joints, but the truss will be bent upward (so to speak) as
indicated by the dotted outline. So, to obtain the correct position of the
joints, the truss must be rotated downward about LG, until the joint LO is
again in the same horizontal line as L6. This means that joint LO will be
rotated through the arc ed. This are is so nearly equal to the vertical
distance A that it can be considered equal to it. The distance A is equal
to the vertical distance LO-LO’, shown at (b). Now, evidently, all of the
other joints of the truss will be revolved through the same angle as LO
and the are they describe in each case will be directly proportional to
their radius. These arcs are so small that the distance from L6 to the
undeflected position of any joint can be taken as its radius. Then the
radius for any joint of the bottom chord is the horizontal distance from
L6 out to the joint. For example, the radius for joint L1 is L1-L6; for
£2 it is L2-L6; and so on. The radius for joint U1 is U1-L6; for U2
it is U2-L6; and so on. The arcs can be considered as straight lines
perpendicular to these radii.

Then the arcs through which the joints LO, L1, L2, L3, L4, and L5
are rotated are represented, respectively, by the lines A, 1, 2, 3, 4, and
5 and the ares through which joints U5, U4, U3, U2, and U1 are rotated
are represented, respectively, by the lines 6, 7, 8,9, and 10. Now, having
the lengths of the arcs through which the joints are rotated, it is an
easy matter to determine the true deflected position of any joint. As an
example, let us consider joint L4. By considering the member L5-L6
fixed in direction the joints would all move upward to the positions shown
on the Williot diagram at (b). That is, each would move from L6 to
the position indicated. Then laying off from L4 downward the arc 4,
we obtain the point L4’, which is the true deflected position of joint L4.
That is, joint L4 really deflects from L6 to L4’ and, hence, its vertical
movement is represented by the distance O-L4’ and its horizontal move-
ment by the distance L6-O. As another case, let us consider joint U5.
The line 6, which is perpendicular to the end post U5-L6, represents the
are through which this joint is rotated. So, by laying off this arc from
U5 the point U5’ is obtained, which is the true deflected position of
joint U5. That is, joint U5 really moves from L6 to U5’. The true
deflected position of each of the other joints can be determined in this
manner, but time can be saved by laying off all of the ares upward from
L6. In applying this method let us first consider joint L4. Laying off
are 4 upward from L6, we obtain point L4”. Now, the deflected posi-
tion of joint L4 is shown by the line L4’’-L4. This is readily seen to be
true as L6-L4” is the arc through which the joint is rotated, and as the
joint was already deflected up to L4, the difference between the two posi-
tions would undoubtedly be the actual deflected distance. In the same
manner, laying off arc 7, from L6, we obtain point U4’. Then the actual
movement of joint U4 is from U4” to U4. By laying off all the ares
described by the joints from L6 we will obtain the points LO”, L1”, U1”,
U2”, ete. If these be connected by lines, we obtain the truss shown, which
is perpendicular to the truss shown at (a). We then have the deflected dis-
tance of each joint given. For example, the joint LO moves from LO” to
LO, joint L1 from L1” to L1, joint U1 from U1” to U1, joint U2 from
U2” to U2, and so on. By measuring the vertical and horizontal com-
530 STRUCTURAL ENGINEERING

ponents of these distances the true deflected position of each joint can be
plotted as shown at (c), which in that case is for the bottom chord joints.

Now, it will be seen that all we need do to determine the deflection
of the truss is to construct the Williot diagram at (b) and draw the truss
LO”-U1” ..... L6. This truss can be drawn by dividing the line L6-LO”
into as many equal divisions as there are panels in the bridge and then
locating one of the points U1’’, U2”, etc., which is easily done. For in-
stance, U1” is located by drawing from LO” a line perpendicular to end
post LO-U1 and drawing from L1” a line perpendicular to L6-L0”. The
entire truss LO”-U1” ...L6 of course can be constructed in the manner
indicated at (b). That is, by projecting points L1”, £2”, etc., over from
the line C-L6 and then locating the points U1”, U2”, ete., by laying off
from LG the arcs 10, 9, ete.

That the extremities of the arcs laid off from L6 will form the truss
shown at (b) is readily seen. In the first place, there is no question as
to the location of the points LO”, L1”, L2” ...L6 and it is seen from
the construction that the two points L4” and U4”, L3” and U3’, and
so on, are in each case on the same horizontal line. So that in reality,
the only thing in question is as to whether the points U1", U2”, U3”, etc.,
are in the same vertical line. This, however, is readily shown to be true
by comparing the construction at (b) with the diagram at (a). The are
L6-U4” is perpendicular to the radius L6-U4, and, as is seen from the
construction £4’ and U4”, is on the same horizontal line and, hence, the
triangles L6-U4”-L4” and L6-U4-L4 are similar, the corresponding sides
being perpendicular each to each. In the same manner it can be shown
that triangle L6-U3”-L3” is similar to L6-U3-L3, and triangle L6-
U2”-L2” is similar to triangle L6-U2-L2. Then, evidently, as U4-L4,
U3-L3, U2-L2, ete., are of equal length at (a), the lines U4’-L4”,
U3"-L3” will be of equal length at (b) and, hence, the points U5”, UL",
etc., will be in the same vertical line.

The determination of deflection or displacement by the use of the
Williot diagram and the rotation diagram combined, as shown at (b),
was first proposed by
Professor Mohr,* and the
diagram LO”’-U1” ...L6
is known as Mohr’s rota-
tion diagram.”

The construction of
the Mohr rotation dia-
gram is based upon the
fact that the lines joining
the extremities of the arcs
described by the joints of

Fig. 363 a truss when rotated

through a_ small angle

about any point will, when the arcs are laid off from a point, form a truss
perpendicular to the first truss. For example, suppose the truss shown at
(a), Fig. 363, be revolved about any point O through the small angle 6.
Let Al, A2....A5 represent the arcs described by the joints. Then, by

   

 

*See Molitor’s Kinetic Theory of Engineering Structures,
DESIGN OF SIMPLE RAILROAD BRIDGES 531

laying off these arcs from any point P, as shown at (b), and connecting
their extremities, we obtain the truss shown dotted.- This truss is per-
pendicular to the truss at (a). All of this is readily shown to be true by
simple geometrical analysis.

217. Determination of Camber.—It is obvious that, if each com-
pression member in a truss were lengthened an amount equal to its dis-
tortion and each tension member were shortened an amount equal to its
distortion, that the truss when supporting no load would be cambered
(curved upward) to an extent equal to the deflection and that the truss
would be straight when supporting the maximum full load. Camber ob-
tained in this manner is known as “‘exact camber.”

In the case of trusses up to 300 ft. in length sufficient camber seems
to be obtained by merely increasing the length of the top chord about 4
of an inch for each 10 feet of length, as explained in Art. 185, but longer
spans should have exact camber.

The position of the joints due to camber, in the case of exact camber,
is determined exactly in the same manner as the deflection; in fact, the
movement of the joints is exactly equal but opposite in direction. In case
the camber is obtained by merely lengthening the top chord, the posi-
tion of the joints is most readily obtained by the graphical method, and
the work is practically the same as for the deflection. As an example,
let it be required to locate the cambered position of the joints of the truss
shown in Fig. 279, due to the cambered length specified for that struc-
ture. (See Art. 185.)

First draw the diagram of the truss shown at (a), Fig. 364, and
write on each of the members the amount its length is changed to obtain
the camber. As is seen,

 

only the lengths of the )_ +8" ue +8" us tie” ut +8" us
top chords and diagonals e “yi ‘
are changed, and as the = 7 | By
lengths are increased in (Q)
16

 

 

 

 

 

 

every case, the plus sign 16 ti 12 13 74 5
is used throughout. |
\

We draw the dia-

'
grams at (c) by taking ‘ :
£3 as the starting point | oA Se | (i)
and assuming joint L3 Bu

as fixed in position, and

 

member U38-L3 fixed in As
direction. Joint U3 does ate
not move in reference to tit fey
ow eeennne $l
3 (as the length of U3- ues ‘
L3 is not changed), so Saeed
U3 will be at the same Nive
point as L3. Joint U2 in Fig. 364

reference to L3 moves to

the left #; in. So from U3 (marked L3 also) lay off 7 in. to the left,
as shown. Joint U2 in reference to L3 moves away from L3 the dis-
tance 3%; in. So from U3 lay off 38 in. upward and to the left as shown.
Then drawing perpendiculars to these, we obtain point U2, which shows
the position of joint U2. Joint L2 does not move in reference to either
532 STRUCTURAL ENGINEERING

joint U2 or L3. So drawing from point U2 a perpendicular to member
U2-L2 and from L3 a perpendicular to member L3-L2, we obtain point
£2, which shows the location of joint L2. To aid in understanding this
step, we can imagine the change in length of each of the members
U2-L2 and L2-L3 as being infinitesimal and the lines U2-L2 and L2-L3
as being perpendicular, respectively, to these changes.

Joint U1 moves 3; in. away from U2 and 38; in. from L2. Then
from U2 (at (c)) lay off 75 in. to the left and from L2 lay off 4%; in.
upward and to the left, and then drawing perpendiculars we obtain point
U1, which shows the position of joint U1. Joint L1 does not move in
reference to either joint Ul or L2. Then by drawing from U1 a line
perpendicular to member U1-L1 and from L2 a line perpendicular to
member L1-L2, we obtain the point L1, which shows the location of joint
£1. Likewise, joint LO does not move in reference to either joint U1
or L1. Then by drawing from U1 a line perpendicular to member U1-LO
and from L1 a line perpendicular to member LO-L1, we obtain point
LO, which shows the location of joint LO. Now, we have the position of
the joints shown at (c) for the left half of the truss, and as the truss
is symmetrical about the center of the span the diagram at (c) really
shows the position of all the joints in the truss. The diagram at (b)
shows the positions of the joints of. the lower chord which are obtained
from the diagram at (c) by measuring upward from LO.

It is interesting to note how near the camber shown at (b) comes
to being equal to the deflection found for this same truss in Art. 215,

The cambered position of the joints, especially the lower chord
joints, of long spans is required mostly for locating the tops of the camber
blocks upon which the truss is supported while the structure is being
erected. These supports or blocks, placed under each panel point, should
have about the correct height, for otherwise the truss members will not
fit into place.
CHAPTER XII
DESIGN OF SIMPLE HIGHWAY BRIDGES

218. Types.—Steel highway bridges can be divided into the fol-
lowing types: beam bridges, plate girders, viaducts, pony truss, and high
truss bridges. Beam bridges and viaducts are usually deck structures,
while pony truss bridges are always through structures. Plate girders
and high truss bridges may be either through or deck structures.

219. Live Load.—The maximum live load depends, as a rule, upon
the locality of the structure. It varies from street cars, heavy traction
engines, trucks, and dense crowds of people in and near cities and towns,
to light traction engines and droves of live stock in the country districts.
So, in respect to the live load, highway bridges can be classed, except for
isolated cases, as city bridges and country bridges. As each city bridge,
as a rule, is a special problem, we shall consider only country bridges,
although the general principles involved in the design of the two classes
are identical. as

The live load for country bridges is specified as a uniform load of
so many pounds per square foot of floor—the longer the span, the less
the load—or a load of so many tons concentrated on two axles from 10
to 12 ft. apart, the transverse distance between the centers of the wheels
being from 5 to 7 ft. The concentrated load is used where it produces
a greater stress than the uniform load. The intensities of the loads used
vary for different localities depending, of course, in each case, upon the
actual loads found to exist and upon the probable future loads. As a rule,
the influence of the latter would, as is obvious, depend upon the financial
status of the locality paying for the structure. The live loads that the
author considers satisfactory for most country bridges are specified farther
on in the “Specifications for Steel Highway Bridges.”

220. Impact.—From actual tests made on highway bridges over the
State of Iowa, under the joint auspices of the State Highway Commis-
sion and the Iowa State College Engineering Experiment Station, it was
found that impact due to maximum loads is negligible. In some cases the
impact due to some light loads—for instance, a trotting horse—was
found to be quite high, especially for bridges having wood floors, but in
all such cases the stresses were so low that the impact in no way could
influence the design of such structures. “In the case of bridges having
concrete floors no impact of any consequence was found in any case.

These tests were conducted by the author, assisted by Mr. C. S.
Nichols, Assistant Director of the Experiment Station, and others. The
results were verified by tests made in the State of Illinois by Prof.
F. O. Dufour and State Engineer A. N. Johnson.

Judging from the above, no impact should be considered in designing
highway bridges.

533
594 STRUCTURAL ENGINEERING

221. Dead Load.—The dead load, as in the case of railroad
bridges, consists of the weight of metal in the structure and the weight
of the floor.

Owing to the wide variation of the weight of metal in highway
bridges due to the design, details, width of roadway, loading, and so on,
it is practically impossible to give formulas for the weight of metal in
general.

In the case of beam and plate girder bridges, the weights of metal
in the different parts are readily assumed and verified as the calculations
for the designs are made. The approximate weight of metal in the
structures designated can be obtained from the following formulas where
p=weight of metal in pounds per foot of bridge, and L=length of span
in feet, c.c. end bearings:

For pony trusses with concrete floors (without joists)

D=42L F220 sagcieiwinwe deed eee wens wih MYER Oe eR (1).
For through trusses with concrete floors (with joists)
Pa=RBEP 280 Sea SAR Ne ee Ae Sak he a ola eee ed oe » (2).

The above formulas are for 16-ft. roadways. To obtain the weight
of metal for wider roadways add 15 lbs. for each additional foot of width.

The weights obtained from the above formulas will, as a rule, be
within 10 per cent of the actual weight. The weight of reinforcing steel
in the concrete floors is not included. The weight of the floor in each
case must be computed and added to the weight of metal.

222. Specifications—Quite a number of very good specifications
for steel highway bridges have been written and published of late years,
but unfortunately these do not agree on some important points and for
that reason none of them will be designated here as a standard, but instead
the following specifications by the author, mostly a compilation, will be
used in this work:

SPECIFICATIONS FOR STEEL HIGHWAY BRIDGES
TypPrs oF BRIDGES

1. The following types are recommended:

Spans up to 30 feet—Rolled I-beams,
Spans 30 to 50 feet—Plate girders,

Spans 40 to 90 feet—Pony trusses,

Spans 90 to 150 feet—Riveted high trusses,
Spans over 150 feet—Pin-connected trusses.

LoaDING

2. Dead Load shall consist of the entire weight of structure, including the
metal, concrete floor or wood floor (when allowed), and the fill on concrete floors.

In estimating the dead load concrete shall be considered as weighing 150
Ibs. and fill 120 Ibs. per cu. ft., and timber as 4.25 lbs. per ft. B. M.

The dead load used in figuring the dead-load stresses must be within 10 per
cent of the actual dead weight.
DESIGN OF SIMPLE HIGHWAY BRIDGES 535

3. The Live Load shall be as follows:
The live load on the roadway shall be taken as a uniform load of

100 Ibs. per sq. ft. for all spans up to 50 ft.,
90 Ibs. per sq. ft. for spans 50 to 100 ft.,
80 Ibs. per sq. ft. for spans 100 to 150 ft.,
70 Ibs. per sq. ft. for spans 150 to 200 ft.,
60 Ibs. per sq. ft. for spans over 200 ft.,

or a 15-ton engine, as per Fig. 365, using the engine whenever greater stresses are
obtained than by using the uniform load.
+ The live load on sidewalks shall be taken as a uniform load of 50 lbs. per
sq. ft.
4. Wind Load for high truss bridges shall be 300 Ibs. per lineal ft. of span
on the loaded chord and 100 Ibs. per lineal ft. on the unloaded chord. The wind

a i ses irae Tas eso mT
f

. 8 ; .

5000 uf 19000%4 =
a al
Ss 16 to BO FE g
0

8 7ep_of| Floor
5 IP. or ese phce)
ie af [1a000%}

Fig. 365 Fig. 366

 

 

 

 

1540"

 

 

 

 

 

 

 

 

 

load on all other structures shall be taken as 30 Ibs. per sq. ft. on one and one-
half times the vertical projection of the structure, all of these loads to be consid-
ered as moving loads.

CLEARANCE

5. The Clearance shall not be less than shown on the diagram in Fig. 366.

MATERIAL

6. All material shall be of medium steel except rivets, bolts, rockers, shoes,
and pedestals. Rivets and bolts shall be of rivet steel. Rockers, shoes, and
pedestals shall be of cast iron or cast steel.

7. Medium Steel shall have an ultimate strength of 60,000 to 70,000 Ibs.
per sq. in.; an elastic limit of not less than one-half the ultimate strength; elonga-
tion twenty-two per cent (22%); bending test through 180 degrees to a diameter
equal to the thickness of the piece tested without fracture on outside of bent por-
tion. It shall not contain more than 0.05 per cent of sulphur and if basic shall
not contain more than 0.04 per cent of phosphorus or 0.06 per cent if acid.

8. Rivet Steel shall have an ultimate strength of 48,000 to 58,000 Ibs. per
sq. in.; elastic limit of not less than one-half of the ultimate strength; elongation
twenty-six per cent (26%) ; bending test 180 degrees flat on itself without fracture
on outside of bent portion. It shall not contain more than 0.04 per cent of either
sulphur or phosphorus.

9. Cast Steel shall have an ultimate strength of not less than 65,000 Ibs. per
sq. in.; elongation of fifteen per cent (15%); bending test through 90 degrees to
a diameter of three times the thickness of the piece tested without fracture on
outside of bent portion. It shall not contain more than 0.08 per cent phosphorus
nor more than 0.05 per cent of sulphur.

10. All of the above steel shall be made by the open hearth process.

11. Cast Iron shall be what is known as grey iron. A test piece one inch
square in cross-section, loaded in the middle between supports 12 ins. apart, shall
support at least 2,500 Ibs, and deflect at least 0.15 in, before rupture,
536 STRUCTURAL ENGINEERING

ALLOWABLE UNIT STRESSES

12. The Unit Stress in any part of any structure due to dead and live load
combined shall not exceed the following:

For Metal
Axial tension on net section...........eee ee eeeees ove tau 16,000 Ibs. per sq. in.
Axial compression on gross section where ZL is length of
member in ins. and r the least radius of gyration

IN; ANG: 52s chea yea wee eeaeles 6 es 34 ee) 16,000—70 (Z/r) Ibs. per sq. in.
Shear on shop rivets and pinS..........e cece esac ee reese 12,000 Ibs. per sq. in.
Shear on field rivets, webs of girders and rolled beams.... 10,000 Tbs. per sq. in.
Shear on bolts...... ii skein sucha -W aeay onedoro dig aliade Mea seth eu ereaene OAT 9,000 Ibs. per sq. in.
Bearing on shop rivets and pins.........+....ee0see eee . 24,000 Ibs. per sq. in.
Bearing on field rivets............64- ssa ste sixeSat ave rates STS 20,000 Ibs. per sq. in.
Bearing on bolts.........0ecccnerecceeen eee seetenerens 18,000 Ibs. per sq. in.
Bending on pins, rivets and bolts...........- esses eeeees 25,000 Ibs. per sq. in.
Bending on beams and girders...........00.ceeeeeeees 16,000 Ibs. per sq. in.
Bearing on rockers or rollers where d is the diameter of

rocker or roller in ins..............e eee enone (600 X d) Ibs. per lin. in.
Pin-bearing on rockersS....... cee ee cence een e eee renee 12,000 Ibs. per sq. in.

For Timber

/ (Douglas fir, white oak, and long leaf yellow pine)

Bending 2. nivs csing a5 taser Wem ew ae aera eutewlns eaneu le 8 1,500 Ibs. per sq. in.

Tension With grain... .. 2... cece cee ee eee eee en eee 1,500 Ibs. per sq. in.

Compression with grain... ...... ccc ccc cee cee tence ees 1,500 Ibs. per sq. in.

Shear across grain. .... 2... cece eee eee eee tent eees 800 Ibs. per sq. in.
Concrete

Compression due to cross bending.............. 00.0 cece 700 Ibs. per sq. in.

Direct compression and bearing of shoes and pedestals on

MASONCY 6 cae sess gees ois arin sone ba aeoe eves 600 Ibs. per sq. in.
Tension ......ccecseccceree SS OOH Eee ea dues agonal 000 Ibs. per sq. in.

PROPORTION OF Parts

13. The lengths of compression members shall not exceed 120 times their
least radius of gyration in case of pin-connected members nor more than 125 times
in case of members having riveted end connections.

14, Members subjected to alternate stresses of tension and compression shall
be proportioned for each kind of stress.

15. Whenever the live- and dead-load stresses are of opposite signs only two-
thirds of the dead-load stress shall be considered as counteracting the live-load
stress.

16. Members subjected to both axial and bending stresses shall be propor-
tioned so that the combined stresses will not exceed the allowed axial stress,
although if the bending is due to wind load the axial stress can be increased 25
per cent over the allowable specified above.

17. In proportioning tension members, the net area of cross-section shall
be considered as the effective section and in deducting for rivet holes the diameter
of each hole shall be considered as being 4.in. larger than the nominal diameter of
the rivet.

18. In determining the number of rivets the nominal diameter shall be
used,

19. Pin-connected riveted tension members shall have a net section through
the pin hole at least 25 per cent in excess of the net section of the body of the
member, and the net section back of the pin hole, parallel with the axis of the
member, shall be not less than the net section of the body of the member.

20. Plate girders shall be proportioned either by the moment of inertia of
their net section, or by assuming that the areas of the flanges are concentrated
at their centers of gravity, in which case one-eighth of the gross section of the
web, if properly spliced, may be used as flange section.

‘21, The gross section of compression flanges of plate girders shall be not
less than the gross section of the tension flange and the unsupported distance of
any compression flange shall be not greater than 15 times the width of the flange.
DESIGN OF SIMPLE HIGHWAY BRIDGES 537

22, The flanges of plate girders shall be connected to the web with a suf-
ficient number of rivets to transmit the flange increment combined ‘with any load
applied directly on the flange and, in case cover plates are used, the cross-section
of these plates must not exceed the cross-section of the flange angles.

DeEtTAILs or DESIGN

23. The strength of end connections shall be sufficient to develop the full
strength of the member, even though the computed stress is less, the kind of stress
to which the member is subjected being considered.

24, The minimum thickness of metal shall be 4 in. except for fillers and
webs of channels. ;

The minimum distance between centers of rivet holes shall be three diameters
of the rivet; but preferably the distance shall be not less than 3 ins. for $-in.
rivets, 24 ins. for {-in. rivets, and 2 ins. for §-in. rivets, The maximum pitch shall
be 6 ins. for f-in. rivets, 5 ins. for 3-in. rivets, and 44 ins, for §-in. and 4-in. rivets.
Where two or more plates are used in contact, rivets not more than 12 ins. apart
in either direction shall be considered sufficient to hold the plates well together.

25. The minimum distance from the center of any rivet hole to a sheared
edge shall be 1} ins. for {-in. rivets, 14 ins. for #-in. rivets, and 14 ins, for §-in.
and 4-in. rivets. The maximum distance from any edge shall be eight times the
thickness of the plate, but shall not exceed 6 ins. in any case.

26. The pitch of rivets at the ends of built compression members shall not
exceed four diameters for a distance equal to one and one-half times the maximum
width of the member. In compression members the metal shall be concentrated
as much as possible in webs and flanges, The thickness of each web shall not be
less than one-fortieth of the distance between its connections to the flanges. Cover
plates shall have a thickness not less than one-fiftieth of the distance between
rivet lines.

27. Flanges of girders without cover plates shall have a minimum thickness
of one-twelfth of the width of the outstanding legs.

28. The open sides of compression members shall be provided with lattice
and shall have tie plates as near each end as practicable. Tie plates shall be
provided at intermediate points where the lattice is interrupted. In main members
the end tie plates shall have a length not less than their width and intermediate
ones not less than one-half their width. Tie plates shall have a thickness of not
less than one-fiftieth of the distance between rivet lines.

29. The minimum width of lattice bars shall not be less than 24 ins. for
f-in. rivets, 23 ins. for }-in. rivets, 2 ins. for §-in. rivets, and 12 ins. for 4-in.
rivets. The thickness of lattice bars shall not be less than one-fiftieth of the dis-
tance between end rivets for single lattice and one-sixtieth for double lattice.

30. The inclination of lattice bars with the axis of the member shall be
not less than 45 degrees, and when the distance between rivet lines in the flanges is
more than 15 ins., double lattice shall be used which shall be riveted at inter-
sections.

31. Abutting joints in top chords of high trusses when faced for bearing
shall be spliced on four sides sufficiently to hold the connecting members accurately
in place. All other joints in riveted work, whether in tension or compression, shall
‘be fully spliced.

82. Pin-holes shall be reinforced by plates where necessary, and at least one
plate shall be as wide as the flanges will allow and be on the same side as the
flange angles. They shall contain sufficient rivets to distribute their portion of the
pin pressure to the full cross-section of the member.

33. In case special permission be given, the field connections of small bridges
may be bolted. This refers to the connections of laterals, transverse bracing, joists,
floor beams, hand rails, and main splices in trusses. The holes in the connections
of the floor beams and joists (when the joists connect directly to the web of
the floor beams) must, if bolted, be sub-punched and reamed to iron templets and
the open holes in the main splices of trusses must be sub-punched and reamed to
size at the shop while the trusses are assembled. The bolts used must have hexa-
gonal heads and nuts and standard washers and must fit tightly into the holes.
If a desired fit can not be obtained without, turned bolts must be used. In no
case will the bolting of the connections of trusses shipped ‘‘knocked down’’ be
permitted. Rivets are preferable to bolts.
538 STRUCTURAL ENGINEERING

34. Rivets carrying stress and passing through fillers shall be increased 50
per cent in number; and the excess rivets, where possible, shall be outside of the
connected member.

35. Provision for expansion and contraction to the extent of 4 in. for each
10 ft. of length shall be made for all bridges.

Spans under 70 ft. in length shall be fixed at one end and allowed to expand
and contract upon planed surfaces at the other end. Both ends shall be anchored,
Slotted holes for the anchor bolts shall be provided at the sliding end.

The masonry plates or pedestals shall be cast iron in all cases.

36. The ends of 65-ft. spans and over shall be pin-bearing and have cast
rockers at one end and fixed cast shoes at the other. The rockers shall be of the
style shown in Fig. 367. i

The fixed shoes shall be securely anchored to the masonry and the rockers
shall be anchored as indicated in Fig. 367.

 

  

Fig. 367

37. All laterals and transverse bracing in all structures shall be composed
of rigid members.

38. The webs of plate girders shall be stiffened by stiffening angles gen-
erally in pairs over bearings at points of concentrated loads and at other points
where the thickness of the web is less than one-sixtieth of the unsupported distance
between flange angles. The distance between stiffeners shall not exceed that given
by the following formula, with a maximum limit of 6 ft., but in no case exceeding
the depth of the girder:

i
d=5, (12,000 ~ s)

where d = clear distance between stiffeners or flange angles,
= thickness of web,
s§ = shear per square inch.

39. The stiffeners at ends and at points of concentrated loads shall be pro-
portioned by the formula of Paragraph 12—(16,000—70 (L/r) )—the effective
length being taken as one-half the depth of the girder. End stiffeners and those
under concentrated loads shall be on fillers and have their outstanding legs as wide
as the flange angles will allow. Intermediate stiffeners may be on fillers or be
crimped and their outstanding legs shall be not less than one-thirtieth of the depth
of the girder plus 2 ins. All stiffeners shall fit tightly against the flange angles.

40. Through plate girders shall have their top flanges stayed at each end
of every floor beam, or in the case of solid floors at a distance not to exceed 12 ft.,
by brackets or gusset plates.

41. Pony trusses shall be riveted structures, with double-webbed chords and
shall have web members latticed or otherwise effectively stiffened.

42. The top chords of pony trusses under 70 ft. in length shall be securely
held in position at the panel points by gusset plates, knee braces, or solid webbed
vertical posts connected rigidly to the floor beams. The top chords of pony trusses
70 ft. in length and over shall be stiffened by knee braces in addition to having
solid webbed posts rigidly connected to the floor beams,
DESIGN OF SIMPLE HIGHWAY BRIDGES 539

43, Trusses supporting wooden floors shall be given a camber by increasing
the length of the top chord 4 in. for each 10 ft. of length and trusses supporting
concrete floors shall be given a camber by increasing the length of the top chord
ys in. for each 10 ft. of length.

Hip verticals and similar members and the two end panels of the bottom
chords of pin-connected bridges shall be rigid members.

44, All web splices in plate girders shall be sufficient to transmit the shear
and bending on the web at the point of splice.

45. The eye-bars composing a member shall be so arranged that adjacent
bars shall not have their surfaces in contact; they shall be as nearly parallel to
the axis of the truss as possible. The maximum inclination of any bar is limited
to ; in, in 16 ft,, and the bars composing a member must all have about the same
inclination.

WorKMANSHIP

46. All parts forming a structure shall be built in accordance with approved
selieg The workmanship shall be equal to the best practice in modern bridge
works.

47, Material shall be thoroughly straightened in the shop, by methods that
will not injure it before being laid off or worked in any way.

48. The size of rivets called for on the plans shall be understood to mean
the actual size of the cold rivet.

49. The diameter of the punch shall not be more than yy in. greater than
the diameter of the rivet; nor the diameter of the die more than 3 in. greater than
the diameter of the punch.

50. Punching shall be accurately done. Drifting to enlarge unfair holes
will not be allowed. If the holes must be enlarged to admit the rivet they shall be
reamed. Poor matching of holes will be cause for rejection.

51. Where sub-punching and reaming are required, the punch used shall
have a diameter of 3 in. smaller than the nominal diameter of the rivet. The
holes shall then be reamed to a diameter 7; in. larger than the nominal diameter
of the rivet used.

If necessary to take the pieces apart for shipping and handling, the re-
spective pieces reamed together shall be match-marked so that they may be re-
assembled in the same position in the final setting up. That is, no interchanging
of reamed parts will be permitted.

52. Riveted members shall have all parts well assembled and firmly drawn
together with bolts before riveting is commenced.

53. Web plates of girders, which have no cover plates, shall be flush with
the backs of the flange angles.

54. Rivets shall look neat and finished with heads of approved shape, full
and of equal size. They shall be central on shank and grip the assembled pieces
firmly. Recupping and ecalking will not be allowed. Loose, burned, or otherwise
defective rivets shall be cut out and replaced. All rivets shall be driven by
pressure tools wherever possible. Pneumatic hammers shall be used in preference
to hand-driving.

55. Eye-bars shall be straight and true to size and shall be free from twists,
folds in the neck or head, or any other defects. The heads shall be made by up-
setting, rolling, or forging. Welding will not be allowed. The heads shall be so
proportioned that the bars will break in the body when tested to rupture.

56. Before boring the pin-holes each eye-bar shall be properly annealed and
carefully straightened. The pin-holes shall be in the center line of the bars and
in the center of the heads. Bars of the same length shall be bored so accurately
that, when placed together, pins yy in. smaller in diameter than the pin-holes can
Ls passed through the holes at both ends of the bars at the same time without

orcing.

57. The distance center to center of pin-holes shall be correct within gz
in., and the diameter of the holes not more than yg in. larger than that of the pin,
for pins up to 5 ins. in diameter, and gy in. for larger pins.

58. Pins shall be accurately turned and shall be straight and smooth and
entirely free from flaws.
540 STRUCTURAL ENGINEERING

PAINTING

59. All metal work before leaving the shops shall be thoroughly cleaned from
all loose scale and rust and be given one coat of pure, boiled linseed oil, well
worked into all joints and open spaces, except pin-holes and finished surfaces,
which shall be given one coat of white lead and tallow.

All surfaces of riveted work coming into contact shall be painted before
being riveted together with one good coat of pure, boiled linseed oil and red lead.
After the structure is erected the metal work shall be thoroughly and evenly
painted (cleaning parts when necessary) with two coats of either red lead and
linseed oil or sublimed blue lead and linseed oil, tinted as directed.

BEAM BRIDGES

Complete Design of an 18-Ft. Beam Span with Wood Floor

223. Data.—
Length of span=18’0” c.c. of end bearings,
Roadway =16’0” clear,
Live load, 15-ton traction engine as per Fig. 365, Art. 222.
Dead load as hereafter determined.

224. Design of Floor.—The beams will
be spaced about 27” center to center. The
structure must be designed so that planks 16’
long can be used, as odd lengths are usually ex-
pensive and difficult to obtain. Let us assume
that 3”x12” yellow pine plank will be used.
The large wheels on a traction engine are
about 20” wide. Owing to the plank deflect-
ing, the wheel of a traction engine will bear
mostly at its edges and consequently the pres-
sure on the plank will be somewhat as indi-
cated at (a), Fig. 368. From this sketch it is
seen that we can justly assume the load on
the wheel to be applied at two points, a and b,
one-half of the total load at each point. As-
suming the plank as forming a series of fixed
beams, we have at (a) the case of a fixed beam
supporting two equal loads of 5,000# each.
In applying the formulas of Art. 69, L equals
: 27”, and with reference to support A, kL
138 | 13% equals 7” for the load at a and 20” for the
Z _ load at b. Then we have k equals 3% in the

() first case and 39 in the second case.
Fig. 368 Now applying (7) of Art. 69, we have

nj Ne 7X % 20\2
M’ =5,000x27| 2(4) ~(4) _~£ aN
x | () (3) | +5,00027[ 2 ()

20\? 20
(=) — Fy] =-25020

for the moment at A. Substituting in a similar manner in (6), Art. 69,
we obtain 5,000 for the shear at A. Now, having the bending moment
and shear at A determined, the bending moment on the plank at any

 

 

 

 

 

 

 

 

 

(Q)

   

 
DESIGN OF SIMPLE HIGHWAY BRIDGES BAL

point between the beams is readily determined. The maximum will be
at the loads, being the same at each load. Then we have

M =-25,920 +5,000 x 7=9,080 inch Ibs.

for the moment at a. In the case of heavy wagons the loading on the
plank would be as shown at (b), Fig. 368. Considering the plank to be a
fixed beam, as before, and assuming the load on the wheel to be 2,000%,
we have

(2,000 x 27) +8=6,750 inch lbs.

for the maximum bending moment at A and also at the center of the
span, it being a negative moment at A and a positive moment at the cen-
ter of the span.

As seen from above, the moment —25,920’# due to the traction
engine ,is the greatest bending moment on the plank. Assuming each
plank to be 3” x 12”, we have

=(15;2)=
r= (56m) =27

for the moment of inertia. Then for the maximum fiber stress we have
fe (25,020 x 15) +27=1,440 Ibs.

(see Art. 53). This stress is for live load alone. The dead load—the
weight of the plank—will raise this slightly. The 1,440# stress given
above is about the allowable intensity, but owing to the fact that lumber
under-runs in thickness and that the thickness is reduced also from
wear, we will use 4” x12” plank? The weight of the plank can be taken
as 4.25% per ft. B. M. Then for the weight per ft. of a 4”%x 12” plank
we have 4.25x4=17#. Considering a fixed beam, as before, we obtain

—2
(17x 24 )+12=7.17 foot Ibs. or 86 inch lbs.

for the dead-load bending moment at the supports (see Art. 69).
Then adding this to the above maximum live-load moment we ob-
tain
25,920 + 86 =26,006 inch Ibs.
for the total maximum bending moment on the floor plank. Now, apply-
ing Formula D of Art. 53, we have

f’ = (26,006 x 2) + 64=812 Ibs.

for the maximum bending stress on the 4’x12” floor plank. This is
rather a low stress, but when we consider the wear from traffic and that
some plank may be only 8” wide and the thickness in some cases may
really be only 34” it is evident that 4” is none too thick and hence we
will use that thickness.

It is true that, by planking for heavy loads, planks 3” thick could
be used. However, there is no question but that it would have to be
replaced sooner than if 4” plank were used and consequently the saving
is not as great as it appears.

225. Design of Beams.—The beams will be about 27’ =2}' apart.
The 4” floor will weigh 4x44=17# per sq. ft. Then we have
17x 2}=38.25% for the dead load on a beam from the floor planks and
542 STRUCTURAL ENGINEERING

assuming the beam to weigh 25* per ft. we have 38.25 + 25 = 63.25, say
64% for the total dead load per ft. of beam. Then we have

M =5 x 64x18" x12=31,104, say 31,000 inch Ibs.

for the maximum dead-load bending moment.

Prof. F. 0. Dufour and the author, working independently of each
other, found that the wheel concentra-
tion on joists (spaced about 2’—3”
to 2’-6” apart) supporting wood
_ floors was about 43 per cent of the
| wheel load. To be on the safe side

we
_._ (S000

R gto" we will assume 50 per cent. The
—— maximum live load moment will occur

Fig. 369 when one large wheel is in the middle
of the span as shown in Fig. 369.

 

Then we have

MM” = ee x 9x 12=270,000 inch Ibs.

for the maximum bending moment. Now, adding the maximum dead
and live load moments together we have

M’” = 31,000 + 270,000 = 301,000 inch Ibs.

for the total maximum bending moment on a beam. Then we obtain
301,000 + 16,000 = 18.8 for the section modulus. This calls for a 9” x 21#
I-beam. The design of the bridge is shown in Fig. 370.

Complete Design of an 18-Ft. Beam Span with Concrete Floor

226. Data.—

Length of span=18’—0” ¢. to c. of end bearings.
Roadway =16’- 0”.
Specifications, as given in Art. 222.

227. Designing of the Concrete Floor Slab.—The I-beams sup-
porting concrete floor slabs are usually spaced from 2’-3” to 2’-6”
apart and it is usual to limit the minimum thickness of the slab to about
6” and the fill to 3” of gravel. So in this case we will assume the slab
to be 6” thick, the fill 3”, and space the beams as shown at (a), Fig.
371. The weight of the slab per sq. ft. of floor is 150x4$=75#, and
the weight of fill per sq. ft. is 120x4=30#, making in all 75+30=105#
per sq. ft. of floor. Now considering a transverse strip 1 ft. wide as a
continuous beam over eight supports we have

M=- ~ x 105 x 2.25" x 12=- 673.7 inch Ibs.
for maximum bending moment on slab due to dead load.

As is seen, the large wheel of a traction engine will practically
extend from beam to beam and hence the maximum live load on the
transverse strip considered above will really be uniformly distributed.
We know that this load will be distributed along the joists to some
extent. Let us assume that it is distributed for 2 ft. along the joists.
Then we have the live load as shown at (b), Fig. 371. Now consider-
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543
544. STRUCTURAL ENGINEERING

1
ing the transverse strip to be a fixed beam (which is a safe assumption)
we obtain

M’= = x 5,000 x 2.25 x 12 =11,250 inch Ibs.

for the maximum bending moment on the slab due to live load. Then
adding the dead- and live-load moments together we have 11,923”# for
the maximum moment on the slab. This moment requires that the con-
crete slab (6” thick) be reinforced with 4” square rods spaced about 18”

 

 

‘ phy Rods-12 cfs.
ol : = Li: se R aay

    

 

 

 

 

(4)

Fig. 371

centers top and bottom. But this spacing would not be satisfactory as
the rods would be too far apart to properly distribute the stress through
the slab and consequently we will reinforce the slab with 4’ square rods
spaced 12” apart top and bottom in the transverse direction and two }”
square rods at bottom of slab between beams in the longitudinal direc-
tion. The maximum shear on the slab will be about 2,500 +110 =2,610%.
The cross-section of the 1-ft. transverse strip including the reinforcing
is about 795”, So the maximum unit shearing stress on the slab is
2,610+79=33# (about), which is quite satisfactory.

It will be found by calculation that the above slab is safe in case
the beams be spaced as far apart as 3’—3”.

228. Design of Beams.—Assuming a beam to weigh 25% per ft.,
the total dead weight on each intermediate beam will be (75x 2.25) +
(30 x 2.25) +25 =261.25, say 262% per ft. of beam. Then we have

M= ; x 262x18 x 12= 127,333, say 127,000 inch Ibs.

for the maximum dead-load bending moment.

Now, assuming that each intermediate beam supports one-third of
a wheel instead of one-half as in the case of wood floors (see Fig. 369),
we obtain

 

sen x 9x 12=180,000 inch Ibs.

for the maximum moment on each intermediate beam due to live load.
Adding the dead- and live-load moments together we have

127,000 + 180,000 = 307,000 inch lbs.
 

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545
546 STRUCTURAL ENGINEERING

for the total maximum bending moment. Then we have 307,000 + 16,000
=19.2 for the section modulus. This calls for a 9” x21# I-beam, which
is the same as found above for the span with wood floor. The outside
beams will very likely be subjected to about two-thirds of the live load
and one-half of the dead load carried by the intermediate beams. This
requires that the outside beams be 9” x15* channels. The design of
the bridge would be the same as shown in Fig. 370, except the floor
would be of concrete. Fig. 372 shows the design of beam bridges sup-
porting concrete floors as used by the Iowa Highway Commission. It
will be seen that the spans given in Fig. 372 are the clear span in each
case.

PONY TRUSS BRIDGES

Complete Design of a 50-Ft. Pony Truss Bridge with Concrete Floor
and without Joists

229. Data.—

Length =6 panels @ 8’-4’=50’-0” c.c. of end bearings.
Height =6’-6” c.c. chords.

Width =19’-3” c.c. trusses.

Roadway = 18’ — 0”.

Specifications, as given in Art. 222.

230. Design of Concrete Slab.—As the concrete slab is continuous
from one end of the bridge to the other and is supported directly upon
the floor beams (no joists) it is really a continuous beam over seven
supports and there is no objection to considering the slab as an out and
out continuous beam in determining the maximum moments and shears;
but sufficiently accurate results will be obtained by considering the slab
in each end panel as a beam fixed at one end and supported at the other,
and the slab in each of the intermediate panels as fixed beams. In
determining the moments and shears on the slab under either assumption,
a longitudinal strip 1 ft. wide is considered. In the case of live load
only one-fourth of a wheel is considered to be carried by this strip. It
is found from calculation that a slab 74” thick reinforced top and _bot-
tom longitudinally with 3” square rods spaced 9” centers is sufficient
for all panels from 7’ -0” to 8’—6” in length and a slab 8” thick with
the same reinforcing is sufficient for all panels from 8’—6” to 9” —0’ in
length.

231. Design of Intermediate Floor Beams.—Taking the slab as
74” thick and the fill as 3” the dead load on the floor beam from the
slab is 124x8.33=1,033% per ft. of beam and assuming the beam to
weigh 55#, we have 1,033+55=1,088# for the total dead load on the
floor beam per ft. of length. Then we have

M =5 x 1,088 x 19.25” x 12 = 604,753, say 605,000 inch Ibs.

for the maximum bending moment on the floor beam due to dead load.
Placing the wheel loads as shown in Fig. 373, we obtain

M’ =8,437 x 8.12 x 12 = 822,000 inch Ibs.

for the maximum bending moment on the floor beam due to live load.
DESIGN OF SIMPLE HIGHWAY BRIDGES 547

Now adding the above moments together we have
605,000 + 822,000 = 1,427,000 inch Ibs.

for the total maximum bending moment on the floor beam. Then we
Ha oo +16,000=89.2 for the section modulus. This calls for an
— 18” x 55#,

812° 3’ 8/2

3! [sss
C7
9
~~

f

 

 

8,

 

os
pa
40000
a

 

 

*
R=8437,

 

 

 

9.62 9.62"

 

 

 

19°37
Fig. 373

 

232. Design of End Floor Beams.—The maximum bending mo-
ment due to dead load is about one-half of the dead-load moment on the
intermediate beam and the live-load moment is exactly the same. So
for, the total maximum bending moment on an end floor beam we have

318,000 + 822,000 = 1,140,000 inch Ibs.

Then we have 1,140,000+16,000=71.2 for the section modulus.
This also calls for an 1-18” x 55%,
233. Dead-Load Stresses in Trusses.—From Formula (1) of
Art. 221 we obtain
p=4x 504+ 220= 420 lbs.

for the weight of metal per ft. of span and for the dead load from the
floor we have 124 x 18=2,232# per ft. of span. Then for the total dead
load we have 420+2,232=2,652# per ft. of span or 1,326 per ft, of
truss.

Then for a panel load we have 1,326 x8.33=11,045, say 11,-
000#. Practically all of this is at the bottom panel points and hence
we will so consider it.

Ul_+57000 U2+57000 U3
A 8 4)

 

 

 

 

 

 

 

e 09 :
0, @, Oo eo a ©
0° 8 a - x g ¢
ft g a g a
35000 || -35000 -64000 |
Lo (24) Li (22) L2 4e) 3 LO
a= Be (00d) 6 Panels @.6°4"= 50-0"

 

 

Fig. 374

Referring to Fig. 374 it is seen that the stress in the hangers U1-L1
and U3-L3 is a panel load; or 11,000#. The stress in U3-L2 is equal
to one-half a panel load multiplied by secO or (11,000+2) 1.63=8,965,
say 9,000# compression.

The stress in diagonal U1-L2 is equal to one and one-half of a panel
548 STRUCTURAL ENGINEERING

load multiplied by sec or (11,000) 1$x 1.63 = 26,895, say 27,000* tea-
sion.

The stress in the end post LO-U1 is equal to two and one-half of a
panel load multiplied by secO or (11,000) 24 x 1.63 =44,825, say 45,0007
compression. ‘The post U2-L2 does nothing but stiffen the top chord and
hence the only load coming on it is the weight of the metal at joint U2,
which is about 600#. The stress in bottom chord LO-L2 is equal to
Rtan= (11,000) 24x 1.28= 35,200, say 35,000# tension (see Art. 173).

The stress in top chord U1-U3 is equal to (11,000) 4 x 1.28 = 56,320,
say 57,0004 compression. The stress in bottom chord L2-L3 is equal to
(11,000) 44 x 1.28 = 63,360, say 64,000# tension. This completes the de-
termination of the dead-load stresses in the trusses and we will next
take up the determination of the live-load stresses.

234. Determination of Live-Load Stresses in Trusses.—As per
specifications (see 3 of Art. 222) the uniform live load is 100# per sq.
ft. of roadway. Then for the panel load we have

W=100x9x8.33=7,497 Ibs.
We have the following for determining the stresses in the trusses:

Tané = 1.28,
Secd = 1.63,

z Wsecd =2,036 lbs.,

Wtan@ =9,596 lbs.
Loading panel points L5 and L4 (Fig. 375) we have
2,036 x 3=6,108, say 6,000 lbs.

 

 

 

 

 

 

 

 

 

 

ey +38000" U2 +38000" 1/3 U4 Us
+ (4) 14) x =
Oo“*RG NN 0% t~ edo 8
oe aa So, 120 In g Se s &
3 Ry Sy ah +> a ‘
¥ iy Bo ‘9
- 24000" ||-24000" <43000* |!)
Lo (2i) “Ll (28) @2 (48) ZL4 LS 16
Ss + 2? = f
@® © @
S-fanels @ 8-4" = £0*0"
Fig. 875

for the tensile stress in diagonal U3-L4. Loading panel points L5, L4
and L3 we have

2,036 x 6= 12,216, say 12,000 Ibs.
for the maximum compression in diagonal U3-L2. Loading panel points
L5 to L2 we have

2,036 x 10=20,360, say 20,000 lbs.
for the maximum tensile stress in diagonal U1-L2. Loading panel points
L5 to L1 we have

2,036 x 15=80,540, say 30,000 lbs.
for the maximum compression in end post U1-LO.

As is readily seen, the stress in hangers U1-L1, U3-L3, and U5-L5
is tension and equal to a panel load of 7,497 lbs. There is no live-load
DESIGN OF SIMPLE HIGHWAY BRIDGES 549

stress in the posts U2-L2 and U4-L4. Loading all points, L5 to L1,
we have

1
9,596 x2 = =23,990, say 24,000 Ibs.

for the stress in bottom chord LO-L2 and 9,596 x 44 =43,182, say 43,0004
for the stress in bottom chord L2-L4. In a similar manner we obtain

9,596 x 4 = 38,383, say 38,000 Ibs.

for the stress in top chord U1-U3. These chord stresses can be deter-
mined more readily by

direct proportion. For 7
example, the dead-load 19.25
panel load is 11,000# ’

and the live-load panel {4.25 7
load is 1,497#, and ar
hence multiply- Jf

ing (7,497+11,000)

3’ 2
2 i
in any chord member

by the dead-load stress

we would obtain the R=14800"
live-load stress in it. A B

Thus, for the live-load Fig. 876

stress in L2-L3 we have

7,497
11,000

We now have all of the stress due to the uniform live load deter-
mined and next we will consider the stress caused by the engine. Placing
the large wheels transversely over a floor beam as shown in Fig. 376 and
taking moments about A we obtain

(20,000 x 14.25) + 19.25 = 14,800, say 15,000 Ibs.

for the reaction at B, which is the maximum stress in hangers U1-L1,
U3-L3, and U5-L5, This is greater than the stress caused by the uni-
form load and consequently it will be considered in designing these
members.

By placing the wheels as shown in Fig. 377 we obtain a stress of
[ (22,500 x 21.84) +50] 1.63 =15,600, say 16,000# compression in diag-
onal U3-L2, which is greater than the stress caused by the uniform load.
In a similar manner, by placing a large axle at L4 and the smaller one to
the right, we obtain a tensile stress of [ (22,500 x 13) +50] 1.63=9,500,
say 9,000# in diagonal U3-L4, This stress is also larger than that ob-
tained by the use of the uniform live load and hence will be considered
in designing the diagonals U3-L2 and U5-L4.

The hangers U1-L1, ete., and the diagonals U3-L2 and U3-L4 are
the only members that are subjected (as ascertained by trial) to greater
stress from the engine than from the uniform live load and consequently
the stress in the other members due to the engine will not be considered.
In the case of shorter spans the stresses due to the engine will influence
the design of more members. In fact it is always advisable to determine

 

‘

 

 

1000.

 

 

 

 

 

) x 64,000 = 43,650, say 43,000 Ibs.
550 STRUCTURAL ENGINEERING

the stresses due to the engine in most all the members of a truss to make
sure that the absolute maximum stress is obtained in each case. Influence
lines can be used to an advantage in determining the stress in the truss
and especially in the case of the engine load.

The maximum dead-load stresses in the truss are given in Fig. 374

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+
3.66 21.34
Ur ~ U2 Uz
%
SI
‘9
e a 12 WI 14% Le Zé
8 , 5 “ wt
9 yto" 14.20
Wo SP /S@ 84 3"= SOO”
Fig. 377

and the maximum live-load stresses in Fig. 375. Combining these we
obtain the stress diagram shown at the top of Fig. 378.

235. Designing of the Sections of Truss Members.—The work
of designing the truss members is practically the same as previously
given for railroad bridges. The vertical posts and hangers as: far as
stresses are concerned could be very light section, but 4—s 3” x24” x }”
and a plate 7” x4’ is about as light a section as can be used for these
members and obtain good details.

Diagonals U3-L2 must be designed to carry 25,0004 compression or
12,500# tension. The effective length of this member is about 126”,
Then using 2—Ls 33” x23” x4’ =3.760” (34” legs vertically), we have

TL 126
Film

and hence we obtain 16,000—70x113=8,090# for the allowable unit
compression stress. Then 25,000+8,090=3.090’’ is the required area
for compression and 12,500+16,000=0.780” is area required for ten-
sion. If less than 34’ x 24” angles be used L/r would be too great, so we
will use 2—Ls 33” x24” x4” for diagonal U3-L2. The designing of the
other diagonals and the bottom chord is simply a matter of determining
the correct net section in each case as previously explained in the design
of railroad bridges.

The maximum stress in the top chord is 95,000#. Dividing this by,
say 12,000 lbs., we obtain about 84” for the approximate area of cross-
section of the chord. Then assume the top chord to be composed of the
following:

2—[s 6’x8# =4,760"
1—Cov. 12” x4=3.000”

 

7.760”

The radius of gyration of this section with reference to the hori-
zontal axis is 2.38’ and with reference to the vertical axis it is 4.08”,
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553
554 STRUCTURAL ENGINEERING

Considering the chord to fail vertically the length would be taken as
one panel length, or 100”. Then we have
100 -
16,000 - 70 (533) = 13,060 Ibs.
for the allowable unit stress. Dividing this into the stress we have
95,000 +13,060=7.27 sq. ins.

If the chord be considered with reference to the vertical axis the length
should be taken as two panel lengths, or 200”. Then we obtain
200

16,000 - 70 (75) =12,570 Ibs.

for the allowable unit stress. Dividing this into the stress we obtain
95,000 + 12,570 = 7.56 sq. ins.

This shows that the assumed chord section is about correct and hence
will be used. The end posts are designed in a similar manner and the
laterals are designed to transmit a wind load of 300# per ft. of span.
Complete details of the span are shown in Fig. 378.

236. Long Pony Truss Spans.—These spans are designed in the
same manner as shown above for the 50-ft. span. The ends of all spans
65 ft. and over in length should be pin-bearing and have a cast rocker at
one end and a cast fixed shoe at the other and have brackets stiffening
the top chord, as shown in Figs. 379 and 380.

THROUGH TRUSS BRIDGES
Complete Design of a 112-Ft. Through Pratt Truss Span

237. Data.—
Length = Vpanels @ 16’-0”=112’-0” cc. end pins.
Height =20’-0” c.c. chords.
Width =17’-0” c.c. trusses.
Roadway= 16’ - 0”.
Specifications, as per Art. 222.

238. Designing of the Joists and Concrete Slab is the same as
previously shown for beam bridges. The joists will be 9” x21# Is and
9” x 13.25% [s. The slab will be 6” thick reinforced with 4” square rods
spaced 12” centers. The fill will be 3” thick.

239. Designing of the Intermediate Floor Beams.—The weight
of the slab and fill is 105 lbs. per sq. ft. of roadway. The joists can be
taken as weighing 21+24=8.4 lbs. per sq. ft. of roadway. Then from
the slab, fill, and joists we have (105+8.4) 16=1,814# for the load per
ft. of floor beam and assuming the floor beam to weigh 65# per ft. we have
1,814+65=1,879# for the total dead load on the floor beam per ft. of
length. Then for the maximum bending moment on the floor beam due to
dead load we have

poraeJ
M= : x1,879x 17 x12 =814,500 inch lbs.
DESIGN OF SIMPLE HIGHWAY BRIDGES 555

Placing the wheel loads as indicated in Fig. 381 and assuming that

ts of the smaller wheels is transmitted to the floor beam under the larger
wheels, we have

M’ =9,520 x7 x 12 = 800,000 inch Ibs.
for the maximum live-load bending moment on the beam.

”

      

20 M0"

   

Fig. 381

Adding the above dead- and live-load moments together we have
814,500 + 800,000 = 1,614,500 inch Ibs.

for the total maximum bending moment on the beam. Then we have
1,614,500+16,000=101 for the section modulus. This calls for a
20” x 65¥ I.

The designing of the end floor beams is practically the same as for
the intermediate beams. The live load is exactly the same and the dead
load is about one-half as much as for the intermediate beams.

240. Determination of the Dead-Load Stresses in Trusses.—
From (2) of Art. 221 we obtain 2.8 x 112+ 280=594# for the weight of
metal per ft. of span and for the weight of the concrete slab and the
fill we have (75+80) 16=1,680# per ft. of span, making in all 594+
1,680 =2,274# per ft. of span or 1,137, say 1,130 Ibs. per ft. of truss.
Then we have

W=16x1,130=18,080 lbs.

for the panel load. The expressions for the dead-load stresses are given
on the right half of the truss, Fig. 382, and the actual stresses are given
ie le le ay #

Ww
U/t"_+72000_UZV _+87000 U3" +87000 UZ a4 Ttuy
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Fig. 382

on the left half. About one-third of the weight of metal is assumed to be
at the top chord joints. This is about one-twelfth of the total dead load
and hence the distribution shown is assumed. For determining the stresses
we have the following values:
556 STRUCTURAL ENGINEERING

Tand= 0.8.
SecO= 1.28.
W tanO = 14,464 lbs.
W sec = 23,142 lbs.

The stresses are determined as explained in Art. 173.

241. Determination of Live-Load Stresses in Trusses.—We will
first consider the 80-lb. load specified in (3) of Art. 222. For the panel
load we have

P=80x8x 16=10,240 lbs.
For determining the stresses we have the following values:
P tan = 10,240 x 0.8 = 8,192 lbs.
1

 

 

 

 

 

 

 

 

7 P=1,463 lbs.
+ Psecf = 1,872 Ibs.
Loading panel points G and F (Fig. 383) we obtain
4 Cc d f g
£ AS Ne ONE 3
wf | Se pa aS o
wy GF 38a BS x aC . » y EX f
A es TI TA PALA
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6 5 !
@ 6 © © 6
Vig. 383

1,872 x 3 =5,616, say 5,600 lbs.
for the maximum compressive stress in diagonal fE. Loading panel points
G, F, and E we obtain
1,872 x 6 = 11,232, say 11,000 lbs.

for the maximum tensile stress in diagonal dE. Loading panel points
G to D we obtain

1,872 x 10 = 18,720, say 18,800 lbs.

for the maximum tensile stress in diagonal cD. Loading panel points
G to C we obtain

1,872 x 15 = 28,080, say 28,000 lbs.

for the maximum tensile stress in diagonal bC. Loading panel points
G to B we obtain

1,872 x 21= 39,312, say 39,000 lbs.

for the maximum compressive stress in end post bd. Loading panel
points G to E we obtain

1,463 x 6 =8,778, say 8,800 lbs.
DESIGN OF SIMPLE HIGHWAY BRIDGES 557

for the maximum compressive stress in post dD and loading panel points
G to D we obtain
1,463 x 10 = 14,630, say 14,000 lbs.

for the maximum compressive stress in post cC. The stress in hanger bB,
as is evident, is equal to a panel load which is given above as 10,240#.
We now have the stresses in the web members determined and we will
next determine the stresses in the chords. The maximum stress in the
chords occurs when the span is fully loaded. For maximum stress in
bottom chord A-C we have

8,192 x 3 = 24,576, say 24,500 lbs.

For maximum stress in top chord be and bottom CD we have
8,192 x 5 = 40,960, say 41,000 lbs.

For the maximum stress in top chord c-f and bottom chord DE we have
8,192 x 6 = 49,152, say 49,000 lbs.

We now have the stresses due to the uniform load determined and we
-will now consider the stresses due to the engine. Placing the larger wheels
over a floor beam as shown in Fig. 384 5

 

 

 

 

we obtain (using the concentration le" EN
shown in Fig. 381) A 4 ’

12 8 8
(11,560 x 2) iF = 16,320, say 16,300 lbs. x y
for the maximum end reaction on a floor f+16320"
beam which is also the maximum live-

 

; Fig. 384
load stress in each hanger and hence

this will be used in designing the hangers.
Placing the wheel loads longitudinally, as shown in Fig. 385, and
transversely, as shown in Fig. 384, we obtain the concentrations indicated

 

 

 

 

 

 

 

 

 

5 = Jd . f g
‘
& e Ww) 5!
f
:
4 ;
¢ d
a a c D IF la G a]
1e300* = 450*
6. J; a 3 2.
Fig. 385

at E and F. For the shear in panel DE due to these concentrations we
have
3 (ee +2 ee )- 6,985 + 1,385 = 8,370 lbs.
i

This is equal to the maximum stress in post dD due to the engine, and
multiplying the same by sec# we obtain

8,370 x 1.28=10,714, say 10,700 lbs.
for the maximum stress in diagonal dE due to the engine. But as these

 

 
B58 STRUCTURAL ENGINEERING

stresses (in dD and dE) due to the engine are less than produced by the
80-lb. uniform load we need not consider further the stresses produced

by the engine.
As is seen, the hangers are the only truss members receiving maxi-

mum stress from the engine load.

242. Determination of Wind Stresses in the Laterals—Let us
first consider the bottom laterals. The load as given in 4 of Art. 222
is 300# per ft. of span. Then for the panel load we have

P=800 x 16 =4,800 lbs.

For determining the stresses we have the following values:

P=4,800,secew=1.37 and 7 Psecw = 939.

 

 

 

 

 

 

 

 

 

 

 

a Po Y Iarx, ’ x yt se
“
K vv s 4 wo
~~ y Lf @ x . ¥ ie a“
a naw Cc D E F G H
}~G-0

Fig. 386

Loading panels from h to e (Fig. 386) we obtain
939 x 6 =5,634, say 5,600 lbs.

for the maximum stress in lateral dE. Loading panel points h to d we

obtain
939 x 10 =9,390, say 9,400 Ibs.

for the stress in lateral cD. Loading panel points h to c we obtain
939 x 15 = 14,085, say 14,000 lbs.

for the stress in lateral bC. Loading panel points h to b we obtain
939 x 21=19,719, say 20,000 Ibs.

for the stress in diagonal aB. We now have all the stresses necessary
for designing the bottom laterals determined (see Art. 177) and we will
next consider the top laterals. The load as given in 4 of Art. 222 is 100#
per ft. of span. Then for the panel load we have

P’=100x 16 =1,600 lbs. and 4 P’ seew = 313.
Loading panel points g to e (Fig. 387) we obtain
6 x 313 =1,878, say 1,900 lbs.
for the stress in top lateral dE. Loading panel points g to d we obtain
10 x 313 = 3,130, say 3,100 lbs.
DESIGN OF SIMPLE HIGHWAY BRIDGES 559

for the stress in top lateral cD. Loading panel points g to ¢ we obtain
15 x 313 = 4,695, say 4,700 lbs.

for the stress in top lateral bC. The maximum stress will occur in strut
dD when panel points g to d are loaded. The half of the panel load

 

 

 

 

 

 

 

 

6 3 4 | 3
@ @ @ ©

b S d e g

“Dy, J a oe
“ * ge s
7 a Y /
a Cc oa & F G
Fig. 387

applied at D does not affect the strut and hence the maximum stress in
it is equal to

($ P’ +4 P’) =$x 1,600+ 4x 1,600 =2,172, say 2,200 Ibs.

Loading panel points g to c we obtain

(7 Pp’ +3P’) = 3,085, say 3,000 Ibs.

 

 

 

 

 

 

 

 

 

 

 

 

 

7
for the maximum stress in strut cC. a ee < 800%
243. Stress in Portal_—The ays A
maximum stress in the portal mem- Pol [gg eowmene
bers will occur when all of the panel \s “|
points are loaded. The forcesonthe | & Di e
portal will be as shown in Fig. 388: | >) *s
P’ =1,600 Ibs. TY) TS Sle ecm coil
R =2P’=shear in panel bc I v ses v
when all panel points are loaded. u =
, JN tie GS
fe ao abita: |
2 2 7
Fig. 388

Assuming the end posts fixed and taking moments about O we obtain
V = (4,800 x 17.05) +17 =4,814, say 4,800 lbs.
Now assuming =45° we obtain
4,800 x 1.41 = 6,768, say 7,000 lbs.

for the stress in each of the members bD and bE, the same being tension
in one and compression in the other.
Taking moments about D we obtain

(2,400 x 8.55 + 4,000 x 8.5) + 8.5 =6,414, say 6,400 Ibs.

for the compressive stress in member ab. Taking moments about E we
obtain
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560
DESIGN OF SIMPLE HIGHWAY BRIDGES b6L

(2,400 x 8.55 + 800 x 8.5) +8.5 = 3,214, say 3,200 lbs.

for the tensile stress in member bc. This completes the calculations of
the stresses in the structure and the stresses can be written on the stress

sheet Fig. 389.

244. Designing of Sections of the Truss Members.—Intermedi-
ate Post. The length of each of these posts is 20 ft. or 240 ins. L/r
should not exceed 125. Then we have 240-+125=1.92 for the minimum
allowable value of r. Using this value of r we obtain

16,000 —70 x 125 = 7,250 Ibs.

for the allowable unit stress. Dividing this into the stress in post U2-L2
(Fig. 389), we obtain

34,000 +7,250=4.69 sq. ins.

for the required area. It is seen from this that two 6” channels will
be satisfactory for section. The 5” channels can not be used as the r is
too small except in the case of the lightest weight which have not enough
section. Considering 2—[s 0” x 8#=4.760”, we obtain

240 |
16,000 - 0 x <7 =8,821 Ibs.

for the allowable unit stress, and dividing this into the stress in post
U2-L2, we obtain
34,000 + 8,821=3.85 sq. ins.

for the required area of cross-section, which is 0.912” less than contained
in the two channels; but these channels will be used as they are nearest
to the theoretical requirement, and for the same reason these channels
will be used for the other intermediate posts.

Top Chord. Assuming 12,000# as the unit stress and dividing this
into the maximum stress in the top chord, we obtain 136,000+12,000=
11.30” for the approximate area of cross-section.

Let us assume the following section:

1—cov. 12%x}”= 3.000”
2—[s 9” x13.25%= 7.780”

 

10.780”
Taking r as 0.4 of the depth of the chord we obtain
wr (192
p=16,000-70 (F) = 12,267

for the allowable unit stress. Dividing this into the maximum stress in
the top chord we obtain

136,000 + 12,267=11.1 sq. ins. -

for the required area of cross-section, which is only 0.325” more than
the above assumed section. The actual r for the section is practically
3.6, the same as assumed, so the above assumed section will be used for
562 STRUCTURAL ENGINEERING

the top chord throughout. The area required for U1-U2 is less than
this section, but as minimum thickness is used the area can not be changed
without changing the depth of the chord and hence the same section
throughout is used.

End Post. Let us assume the following section for the end post:

1—Cov. 12%x4/= 3.000”
2—-[s 9%x20% =11.760”

14.760”

The radius of gyration about the vertical axis is about 3.8 and
about the horizontal axis it is about 3.5. As a collision strut is used we
need consider the strength only with regard to the vertical axis. Sub-
tracting the depth of the portal we have

25.6-8.5=17.1 ft., or 205 ins.

for the effective length of the end post.
Then we heve

p=16,000—70 (3

for the allowable unit stress for direct compression.
Referring to Fig. 387, it is seen that the maximum moment on the
post occurs at the bottom of the portal and is equal to

2,400 x 8.55 x 12 = 246,240 inch lbs.

The moment of inertia about the vertical axis is about 220 and the dis-
tance to the extreme fiber is 6 ins. Using these values we obtain

246,240
220

for the unit stress due to the cross bending.
For the direct unit stress we have

108,000 + 14.76=7,317 Ibs.
Adding this to the bending stress we have
6,715 + 7,317 = 14,032 Ibs.

for the total combined unit stress. The allowable unit stress given above
(12,220 lbs.) can, according to the specifications (see 16 of Art. 222),
be increased 25 per cent in the case of combined stress. So for the
allowable unit stress we have

12,220 x 1.25 = 15,275 lbs.

x 6=6,715 lbs.

which is 1,243# greater than the actual combined stress, so the assumed
section is a little larger than required, but it will be used as the section
cannot be easily reduced.

The designing of the other members of the structure is mostly a
matter of determining required net areas, which is fully explained pre-
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563
564 STRUCTURAL ENGINEERING

. viously in the design of railroad bridges. After the sections are designed
the stress sheet, Fig. 389, can be completed as shown.

The general details of the span are shown in Fig. 390. These
details will be readily understood if the work on railroad bridges, pre-
viously given, is thoroughly studied.

Complete Design of a 162-Ft. Through Pin-Connected Curved
Chord Pratt Truss Highway Bridge

245. Data—
Length =9 panels @ 18’ —0”=162’ ~0” c.c. end pins.
Height =20’—0” at hip and 27” —-10y4” at center (see Fig. 392)
Width =17’-6” c.c. of trusses.
Roadway = 16’ —0”.
Specifications, as per Art. 222.

246. The Designing of the Floor System is the same as pre-
viously shown (see Arts. 227, 228, and 239).

247. Determination of Dead-Load Stresses in Trusses.—From
(2) of Art. 221 we have

2.8 X 162 + 280 = 734 Ibs.

for weight of metal per ft. of span or 367 lbs. per ft. of truss. For the
weight of slab we have
75x8= 600 lbs.

per ft. of truss, and for the weight of fill we have
30 x 8=240 Ibs.

per ft. of truss. Now, considering one-third of the weight of the metal
at the top chord joints and two-thirds at the bottom chord joints, we have

ce } 18=2,202, say 2,200 lbs.

for the panel load on each top chord joint and
[# (867) + 600 + 240] 18=19,52+, say 19,500 lbs.

 

   

 

 

 

 

 

 

 

 

 

v2 e f
Pe, aaa 1 :
¢ (BB k
c Oy
cA s
as
P] E f G 4 ra ”
8g z é = 4 3 z 7
@ cz) @ @ @ ©
Fig. 391

for the panel load on each bottom chord joint. Now having the panel
‘loads computed the stresses are readily determined either analytically or
graphically in the same manner as previously shown for railroad bridges
(see Arts. 189 and 195).
DESIGN OF SIMPLE HIGHWAY BRIDGES 565

248. Determination of Live-Load Stresses in Trusses.—The
uniform live load produces maximum stress in all members except the
hangers, in which case the engine produces the maximum stress. The
uniform live load as per 3 of Art. 222 is 70 Ibs.eper sq. ft. of roadway.
Then we have

P=%0x8x18=10,080 Ibs.

for the panel load.

The maximum live-load stresses in the chords occur when the span
is fully loaded and hence are directly proportional to the dead-load
stresses in same and can be determined very quickly by multiplying each
dead-load stress by the live-load panel divided by the dead-load panel, or
the live-load stresses in the chords can be determined as explained in
Art. 189.

The maximum stresses in the web members due to the uniform load
are determined in very much the same manner as previously shown for
railroad bridges except that in the latter case wheel loads were used.

The maximum tensile stress in diagonal eF occurs when panel points
K, H, G, and F are loaded. (See Fig. 391.) The shear in panel E-F
ix then equal to 14°P and hence the stress in the diagonal is equal to
1°Psec03. The maximum compressive stress in post eE will occur at the
same time. This stress is equal to+,0 P—V2 where V2 represents the
vertical component of the stress in top chord ed, which is equal to
(42 P x4d/h3) tan¢2 (moments about £).

The maximum tensile stress in diagonal dE will occur when panel
points K to E are loaded and is equal to (48 P—V2) sec 62. The maxi-
mum compressive stress in post dD will occur at the same time and is
equal to48P-—V1, where V1 represents the vertical component of the
stress in top chord cd, which is equal to (4;°P x 3d/h2) tand¢l.

The maximum tensile stress in diagonal cD occurs when panel points
K to D are loaded and is equal to (-3}P—V1) sec.61. The maximum
compressive stress in post cC occurs at the same time and is equal to
21 P—V, where V represents the vertical component of the stress in top
chord cb, which is equal to (4¢Px2d/h1) tan ¢. The maximum tensile
stress in diagonal bC occurs when panel points K to C are loaded and is
equal to (22P-V) sec. 6.

The maximum stress in hanger bB, which is due to the engine load,
is determined as previously explained (see Art. 241).

The maximum tensile stress in counter fG will occur when panel
points K, H, and G are loaded, and is equal to (§P—V3) sec 63, where
V3 represents the vertical component of the stress in top chord fg.

The maximum tension in the posts can be determined as previously
explained for railroad bridges (see Art. 190, pages 422-426).

249. Stresses in Portals and Lateral System.—From (4) ‘of Art.
222 the wind load on the bottom laterals is 300# per ft. of bridge and
100# per ft. of bridge on the top laterals. Then we have 300x18=
5,400# for the panel load on the bottom laterals and 100 x 18=1,800#
for the panel load on the top laterals. The stresses are determined as
explained in Art. 242 (also see Arts. 177-179).
566 STRUCTURAL ENGINEERING

250. Designing of the Sections.—The sections of the truss mem-
bers and lateral. bracing are determined as previously explained (see
Arts. 197, 198, 199, 200, and 244).

After the stresses and sections are determined the stress sheet, Fig.
392, can be made.

251. Details—After the stress sheet is completed, the general
drawing, Fig. 3893, can be made. The calculations for the details are
practically the same as previously shown for railroad bridges (see Art.
203). The student should verify the details shown in Fig. 393.

252. Truss Bridges with Wood Floors are designed in the same
manner as bridges with concrete floors. The total dead load is less than
for bridges with concrete floors, but the live load is the same. The first
cost is less, but in most cases bridges with concrete floors will prove to
be the cheaper in the end.

It is the opinion of the author that all highway bridges should be
designed heavy enough to carry concrete floors even if wood floors are
used, as the county authorities usually insist on putting on the concrete
sooner or later, and if the bridge is not heavy enough to carry a concrete
floor it will have to be replaced by a new bridge, which practically means
the loss of the first bridge.
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569
CHAPTER XIII

SKEW BRIDGES, BRIDGES ON CURVES, ECONOMIC HEIGHT
AND LENGTH OF TRUSSES, AND STRESSES IN PORTALS

253. Skew Bridges.—Bridges crossing over streams, railroads,
streets, and roads at an angle are usually built on the skew. This is

 

 

 

 

 

 

 

(y)

 

 

 

 

 

 

 

 

Nave Girders
7 a
TZ KL
HL VL
y, SHringers-7

 

(2)

 

 

pTruss pSEMIGErS

Y/

 

 

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russ 7

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SL

 

 

 

 

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ONS

 

 

 

 

 

 

 

 

)

Fig. 394

 

done in the case of crossings
over streams in order to ob-
tain piers parallel to the flow
of the stream and in the
other cases to obtain mini-
mum span lengths.

Skew deck plate girder
railroad bridges are usually
constructed as shown at (a),
Fig. 394, through girder
spans as shown at (b), truss
spans as shown at (c)
and (d). The case shown
‘at (c) is where the skew is
a full panel length and the
case shown at (d) is where
the skew is less than a panel.
length.

Skew highway bridges
are constructed very similar
to skew railroad bridges, as
far as the skew is concerned.
End floor beams are, as a
rule, used in the case of
highway bridges, while in
the case of railroad bridges
as a rule they are omitted
and details similar to those
shown in Fig. 305 are used,
except they are constructed
to fit the skew.

The outer ends of the
stringers in railroad bridges
should be square with the
track, as indicated in Fig.
394, so as to avoid having
ties resting upon masonry at
one end and upon a stringer
at the other end. By this
construction we also obtain
SKEW BRIDGES, BRIDGES ON CURVES, TRUSSES AND PORTALS 571

equal stringer concentrations on the first intermediate floor beams, as is
readily seen.

The end posts in each end panel of skew truss bridges should always
be parallel. The stresses in skew bridges are determined practically in
the same manner as previously shown for square spans; the only differ-
ence being that the skew must be taken into account, which is done by
assuming the loads to be applied along the center line of the bridge. Let
the diagram at (m) in Fig. 395 represent the plan of a skew truss bridge.
We would assume the live load
applied along the center line gh.
The influence line for the reac-
tion at A would be as shown at
(0). The influence lines for the
stress in chord cd and diagonal
cD would be as shown at (#)
and (v), respectively. To sat-
isfy the criterion for maximum
live-load moment about, say,
point D, the distance sg would
be taken as & in (5) of Art. 91 pate |
and as gw in case the maximum i er |
moment at B is desired, and so
on. To satisfy the criterion (see
Art. 90) for maximum live-load |
shear in, say, panel BC, the load (t)
would extend from h to and into
panel BC so that the unit load in
that panel (with a load at C)
would be equal (or as nearly hy)
equal as is possible) to the total
load on the span divided by the
distance gh. In determining the Fig. 395
dead-load stress the panel loads
are readily.computed for all panel points. This load will be the same for
all panels except for points a, f, and F, in which case a slight correction
on account of the skew must be made. After the panel loads are computed
the dead-load stresses are readily determined either analytically or graph-
ically. However, it must be borne in mind that the truss is unsym-
metrical with reference to the center of span.

The maximum moment and shears on skew through plate girder
bridges are determined in exactly the same manner as shown above for
truss spans. Skew deck plate girder bridges can be treated the same as
square spans without any appreciable error.

254. Bridges on Curves.—Bridges supporting curved tracks are
subjected to stresses due to centrifugal force and to the loads being
eccentrically applied. These stresses are in addition to dead, live, and
impact stresses.

Stresses Due to Centrifugal Force occur in the laterals system at
the loaded chord. . The laterals and chords constitute the system which
is really a horizontal truss. For centrifugal force in general we have

pd
i

(?)

 

Py

 

 

Q
a
a
a
ty

 

 

 

 

 

fe ee -

 

 

 

 

 

 

 
572 STRUCTURAL ENGINEERING

 

gr

where W represents the weight of the moving body, in pounds, v the
velocity of the body in feet per second, g the acceleration due to gravity,
and r the radius of curvature in feet.

Let D equal degree of curvature. Then we have

 

_ 50
Pr singD,
and hence for a 1° curve we have
50
"= 0.00873 5,730 ft.
Let V equal velocity in miles per hour. Then we have
5,280 22
v3 000" “i5'
Now substituting these values in (1) we have
992
WV? =.0000117WP?. 21.06... eee ee (2)

~ 15? x 32.2 x 5,730

for the centrifugal force for a 1° curve where W equals weight of
the moving body and V equals the velocity in miles per hour. Therefore,
for a 4° curve and 50-mile velocity, we have

F=.0000117 x Wx 50? x4=0.117W.

Then if W be the uniform live load per ft. of span, 0.117 W would
be the centrifugal force per ft. of span which would be used to deter-
mine the stress in the chords and laterals. As an illustration, let Fig.

 

 

 

 

 

 

 

 

 

d
R p— > R
4 a e B | 6Ptanto 'r5Ptany Y3P tan i
Ay x x WT. WS OD
Ss . i \ ey) 9% | ot y
‘ 7 See TW
‘ sg tan tn \25Phantn oP tania JY x
a b c é F g } &
7 2 3 4 5 6

Fig. 396

396 represent the plan of the lateral system at the loaded chord of a
7-panel bridge supporting a track on a 4° curve. Let W represent the
live Joad per ft. of span. Then for a velocity of 50 miles per hour we
have 0.117 for the centrifugal force per ft. of span, and for the panel '
load we have

P=0.117W xd.

Then the maximum stress in the laterals and chords due to this horizontal
SKEW BRIDGES, BRIDGES ON CURVES, TRUSSES AND PORTALS 573

load will be as indicated in Fig. 396. These stresses are determined in
the same manner as previously shown for wind stresses. For example,
to obtain the maximum stress in diagonal eF panel points B, C, and D
would be loaded and in the case of diagonal fF panel points B, C, D,

 

 

| Cs eee QO}
t

Sf

 

a

 

 

 

Fig. 397

and E would be loaded, and so on. The
maximum chord stress would be obtained by
loading all panel points from B to G. The
compression in the floor beams due to centrif-
ugal force is not considered, as the laterals
are practically always connected to the ten-
sion flange of the beams and consequently the

 

 

 

 

 

 

 

 

 

 

 

stress due to centrifugal force only tends to 4 F a a

reverse the dead and live-load tensile stress WY |

in those flanges. r a
The stress in the chords due tc centrif- L_ Sere |

ugal force is always added to the dead, live, 2

and impact stresses in summing up the total Fig. 398

maximum stress.

Stress Due to Eccentricity occurs in the trusses, stringers, and floor
beams. Let the diagram shown in Fig. 397 represent the plan of a
bridge supporting a curved track, where the curve acb represents the
center line of track and the line oo represents the center line or axis of
the bridge which always bisects the middle ordinate (d) of the curve as
indicated. It is readily seen that loads near the center of the span will
load the outer truss and stringer more than the inner truss and stringer
and that just the reverse is true for loads near the ends of the span.

The diagram in Fig. 398 shows the conditions at or near the center
of the span where g represents the center of gravity of the load. Taking
moments above A we obtain

_Wei Fy
r= b a Say oar Was ar es Gea tetar Sek wm) Seiad te emae anes wh uke: Ayan (3)

for the vertical concentration on the truss at B where W represents the
weight of the load and F the centrifugal force. If the super-elevation
of the outer rail be in accord with the speed or velocity of the load the
resultant of W and F will pass through the center of the track, in which
case, taking moments about A, we have

b
roW (Fre) sb=7 4 Woes Ea eh eens eo ear ee ea eg eS (4)

2

for concentration on the outer truss. As is seen, the part We/b is due
to the eccentricity, and as is evident this is greater for a moving load
than for a static load, for in the latter case F would be zero and e would
be less, if not in reality shifted to the other side of the center line of the
574 STRUCTURAL ENGINEERING

floor beam. If the velocity of the load be greater than the super-eleva-
tion of the outer rail provides, e will be increased. However, such a
velocity is not likely to occur for the heavy freight trains which produce
the maximum live-load stress in the trusses. If it were possible to locate
accurately the center of gravity (g) of the load, the concentration could
be determined in any case on the outer truss by taking moments about
A—which would be simply the application of equation -(3)—and like-
wise the concentration on the inner truss could be determined by taking
moments about B. But as it is practically impossible to locate the center
of gravity of the load, as it varies so widely for the different engines and
cars, it is more practical to consider the velocity to accord with the super-
elevation of the outer rail and simply consider the concentration on the
outer truss to be as expressed by equation (+).

As is seen from Fig. 397, e is equal to one-half of the middle ordinate
at the center and ends of the span, and fair results will be obtained by
assuming e equal to one-half of the middle ordinate throughout the entire
length of the span, as the loads near the center of span contribute the
greater part of the chord stresses and the loads near the end contribute
the greater part of the web stresses and hence the actual discrepancy is
slight. Then, to obtain the maximum live-load stresses in the trusses,
we would first calculate the maximum stress in each member the same
as for ordinary bridges, assuming the load applied along the center line
of the bridge, and then increase each of these stresses by the amount
obtained by multiplying the stress in each case by e/b when e is taken
as half the middle ordinate and b as the distance between trusses. That
is, if S be the stress in a member due the load considered applied along
the center line of the bridge, the stress due to the eccentricity e would be
S xe/b, and hence the total stress in the member due to live load would
be $+S8e/b.

In case more accurate results be desired than are obtained by taking
the eccentricity e equal to one-half the middle ordinate, the value: of e
at each floor beam can be computed and the concentration on the trusses
at each of those points determined and the stresses in the trusses due to
same computed. However, in that case an equivalent uniform live load
should be used, as the work would be quite tedious if wheel loads be used.

When the stringers are symmetrically placed with reference to the
center line of the bridge, the maximum moment and shear on the outer
stringers are obtained by first
determining the moment and
shear for symmetrical load
and then increasing each of
these by the amount obtained
by multiplying each by e/b’
where b’ is taken as the dis-

Fig. 399 tance between the stringers.

The moment and shear on the

inner stringers would really be decreased by the same amount; however,

it is usual practice to make the outer and inner stringers of equal
section.

In case the curve is quite sharp the stringers are usually off-set, as
shown in Fig. 399, in which case the stringers are about equally loaded

or

   

Truss or Main
SKEW BRIDGES, BRIDGES ON CURVES, TRUSSES AND PORTALS 575

and they are designed as in ordinary cases. The stringer bracing should
be designed sufficiently heavy to transmit the centrifugal force. The
stress in this bracing is determined in the same manner as shown above
for the lateral system of the structure.

In determining the maximum moment and shear on the floor beams,
the eccentricity of the loading must be taken into account. Otherwise
the work is the same as previously shown.

In the case of double-track bridges the determination of the stresses
due to centrifugal force is the same as for single-track structures, except
that both tracks must be considered.

According to the A.R.E. Specifications, “structures located on curves
shall be designed for the centrifugal force of the live load applied at the
top of the high rail. The centrifugal force shall be considered as live
load and derived from the speed in miles per hour given by the expres-
sion 60-24D, where D equals degree of curve.”

The velocity given by this expression (60—24D) for the different
degrees of curvature and the corresponding centrifugal force are as
follows:

D Vv F D V F

1 ieee 57.5......0.039W AP ee caws 50.0......0.117W
Re eats 55.0......0.071W bee aap ate 47.5......0.132W
Beis wate 52.5......0.097W OL sews 45.0......0.142V

where D represents the degree of curvature, V the velocity (speed) in
miles per hour, F the centrifugal force in pounds, and W the weight of
the load in pounds, which would be the weight of load per ft. of span if
a uniform live load be used.

255. Economic Depth of Trusses.—As is evident, a truss is of
economic depth as regards weight and stiffeners when the weight of the
web members is equal to the weight of the chord members. Of course
the weight of the floor system, which varies with the length of panels, is
part of the total weight of the structure, and hence the panel length is
involved. There are so many variables involved that the theoretical
determination of economic heights and panel lengths is practically impos-
sible. The most satisfactory values for heights and panel lengths are
obtained by actual calculations. The following values are recommended:

————_————For Railroad Bridges —————
Length of Span Length of Panels Height of Truss

 

 

in Feet in Feet in Feet
100 25 30
125 25 30
150 25 31
175 25 33
200 25 39
225 25 45
250 25 50
300 25 55
400 (14 panels) 28.58 (about) 62
500 (16 panels) 31.25 72

Heights for intermediate lengths can be obtained by interpolating.
576

Length of Span
in Feet

105
135
150
162
180
200

STRUCTURAL ENGINEERING

 

For Highway Bridges

Length of Panel Height of Truss
in Feet in Feet
15 20
15 22
16.6 24
- 18 28
20 32
20 36

In order to obtain the required overhead clearance and a satisfac-
tory depth of portal the height of truss is limited to about 30 ft. in the
case of railroad bridges and to about 20 ft. in the case of highway
bridges. The minimum economic panel length for railroad bridges is
about 25 ft. and the maximum about 33 ft., while in the case of highway
bridges the minimum panel length is about 15 ft. and the maximum

about 20 ft.

256. Economic Span Length.—The shortest bridge in the case
of a single span, as a rule, is the cheapest, as the cost of the two abut-
ments in most cases is practically independent of the span length. How-

 

 

 

 

 

 

 

Fig. 400

ever, in a few cases the profile of the crossing is such
that the cost of the abutments is materially affected
by their location. In such cases care must be taken
in locating the abutments so that their cost is a
minimum, the usual limit being when the cost of the
span equals the cost of the two abutments. Often the
required water way or local conditions govern the
length of the span, in which cases economy of span
length can not be considered. In case a bridge is
composed of sevéral spans the spans are practically
of economic length when the cost of superstructure
minus the floor system is equal to the cost of the sub-
structure minus the cost of the end abutments. This
is, of course, assuming that the total length of the
bridge is fixed dither by the required water way or
local conditions. The problem can be solved only by
trial. First a profile of the crossing should be drawn
carefully to scdle and the base of rail or roadway
located thereon; and next the cost of an average pier
computed, and then by using the formule of Art.
124 (Art. 221 in the case of highway bridges) the
weight of metal in the spans of different lengths can
be obtained (the weight of the floor system must be
subtracted in each case) and by careful comparison
of the different bridges possible we can readily as-
certain the economic length of spans.

257. Stresses in Portals—Portals other than the one previously
considered (Art. 179) are sometimes used. The plate girder portal
shown at (a), Fig. 400, is occasionally used in case of bridges having

shallow trusses.
SKEW BRIDGES, BRIDGES ON CURVES, TRUSSES AND PORTALS 577

Assuming the bottom ends of the end posts as hinged, the wind
pressure tends to distort the end posts and portal as indicated at (b).

Considering the forces to the right of section SS and taking moments
about d [see sketch at (c)] we obtain

for the stress in the bottom flange of the portal.

h
But V="(R+P) and H=5(R+P).
Substituting these values in (1) we obtain

h ha
ar (R+P) as (R +P).

When «=0
h

a=, (EEE P Ys eid d tage sier Boga vie ari asd Grea) ede ten warae ote (2)
When ¢=w/2

BHO) ah eek se ed AA AA Se BE eee RA RE (3).
When «=a

h
Bo, eT dehwsienlaad ecbinmsinaare eaba muh nie Fae ee hee (4)

Now it is seen from the above equations that the stress in the bot-
tom flange of the portal is zero at the center of the portal and a maximum
at each end, being tension at one end and compression at the other.

Taking moments about b we obtain

r=H(*) + (x +5) "2 LaumauseMepsiiuaaean (5)

for the stress in the top flange of the portal. Substituting h/w(R+P)
for V and 3(R+P) for H in equation (5) we obtain

F’=3(R+P) (=) + (x +f) = an (R+P)

 

 

e
h R «ah
=p vet Ee = ete) Pees od ane Ses Res es (6).
When x+=0
h R
Pa Se -—-—
F <a. (R+P) ta die Basa eRe eeehaata Savisacatetic eaetea te Suilonaua aes (7).
When #=w/2
R
Fi
F FQ centre meee e eee e eee eeecten eee cen sserneeeeees (8)
When «=
h R
7=_— pie Gap Vesa We Hat Oa eda To AES BuO Beto ate eee 9
F De (R+P)+ 5 (9)
578 STRUCTURAL ENGINEERING

From these equations it is seen that the maximum stress in the top
flange of the portal occurs at the ends, being compression at one end and
tension at the other, and that the minimum stress in the top flange occurs
at the center of the portal where it is equal to #2.

The shear on the portal is constant throughout its length and is
equal to V. In case the bottom ends of the end posts be fixed the above
will apply by substituting }(h—e) for (h—e) and e+h/2 for h (see
Art. 179).

After the stresses in the portal are determined the required sections
are readily computed. Each of the flanges should be such that L/r does
not exceed 120. The portal shown at (a), Fig. 401, is used in case of
limited clearance.

Assuming the bottom ends of the end posts to be hinged, the wind
pressure tends to distort the end posts and the portal as is indicated
at (b). The shear between g and f is constant and is equal to V. The

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x Zz
af]
uv ~~
BA
z
<3
(@@) + |83 | (a)
H
ty
Fig. 401 Fig. 402

moment at g and also at f is equal to Vx b/2 which is obtained by taking
moments about either g or f. The moment at o is zero and at any point
between o and f or between o and g itis Vz. Assuming the moment to be
zero at a and k the strut afgk can be considered as a continuous beam.

Then taking moments at either f or g we have (V xb/2) +d for the
reaction at i or a which is the shear in gk or af. The moment at any
point in gk or af is equal to [(V xb/2)+d]a. If e=d we have Vxb/2
for the moment at g or f which is the same as found above.

The horizontal component of the stress in knee brace gm is obtained
by considering the forces to the right of section 2-2 and taking moments
about k. Then the stress in gm is obtained by multiplying this com-
ponent by secd. The stress in knee brace cf is obtained in the same man-
ner—considering the forces to the left of section 1-1.

In case the bottom ends of the end posts are fixed the above will
apply by substituting $(h-e) for (h-e) and e+4(h-e) for h.

The portal shown at (a), Fig. 402, is known as a lattice portal. It
is distorted by the wind pressure very much the same as the plate girder
portal.

As the shear is constant and equal to V throughout the length of the
portal the diagonals are assumed to be equally stressed. As one system
is in compression and the other in tension the moment of the stresses in
the diagonals about any point on a vertical line through the intersection of
SKEW BRIDGES, BRIDGES ON CURVES, TRUSSES AND PORTALS 579

the diagonals is equal to zero as the moments balance. Then by taking
moments about ¢ and considering the forces to the left of any section
as 2-2, shown at (b), we have

_Hh-Ve
7 e

F

for the stress in the bottom flange at b and taking moments about b we
have

H(h-e) -
_H(h-e)-Va,P

7
= e 2
for the stress in the top flange at c. The flange stress throughout the
portal can be obtained in this manner.

The stress in the diagonals is obtained by dividing the shear (V) by
the number of diagonals cut by a vertical plane, and multiplying this by
the secant of the slope of the diagonals.

The portal shown in Fig. 403 is used to 2 3
some extent, especially for long span bridges. “a__ | I
Let us first assume that the diagonals take ten- ¢
sion only. Then diagonal db would have zero  ,=4 le
stress when the wind pressure is applied as 2 3
indicated and diagonal ac would have zero stress v
if the wind pressure were applied from the s
opposite direction.

Assuming the bottom ends of the end posts 4
hinged, and taking moments about a (ignoring v Vv
diagonal bd) and considering the forces to the Ww
left of section 2-2 we obtain

oS Rt
v

 

 

Q
1
Mm

 

 

 

 

 

 

Fig. 403

for stress (compression) in be, and taking moments about ¢ (ignoring
diagonal bd) and considering the forces to the right of section 3-3 we

obtain
AH P
Pera ack ee
S =A e)+ (x +f)

for the stress (tension) in ad.

The stress in the diagonal ac is equal to Vsecd. If the wind
pressure were applied from the opposite direction to that indicated
diagonal bd would be stressed, and the stress in be and ad would be
determined by taking moments about d and b, respectively. The stress
in diagonal bd would be Vsecd and the stress in diagonal ac would be
zero.

In case the diagonal were considered to take compression as well as
tension we would assume the two equally stressed and then by taking
moments about o and m the stress in be and ad could be determined, as
the sum of the moments of the stresses in the diagonals about these points
would be zero as they would balance. -
580 STRUCTURAL ENGINEERING

If the lower ends of the end posts be considered fixed, the above
would apply if }(h—e) be substituted for (h—e) and e+4(h-e) for h.
The depth of the portal, as a rule, is so shallow, as compared with
the length of the end posts, that the bending of the posts along the ends

of the portal can be ignored, and in case the lower ends of the posts are
fixed the points of contra-

, flexure in the posts can be
es considered as being mid-
D CLA way between the portal

7 and the lower end of the
ee posts, as has been done in
2 : all previous examples. But
e NIB in a few cases the required
| depth of the portal is so

@) . eae that the bending of
4\o _A\| v the end posts along the
° ends of the portal must be
N taken into account in de-
WK Hid termining the points of
e contra-flexure in the posts.
Let the diagram shown,
Fig. 404 (a), Fig. 404, represent
such a case. The distor-
tion of the portal members is so small as compared with the flexure of the
posts that the points C and B can be assumed to remain in a vertical line
and likewise the points D and E. The problem is to determine the value
of z, which is the distance from the lower ends of the end posts up to the
points of contra-flexure.

Let R represent the horizontal component of the forces acting upon
the post at C, and R’ the horizontal component of the forces at B. Then
the post ABC can be considered as a beam fixed at A and acted upon by
the forces R and R’ as indicated at (b).

Then, for the bending moment at any point on this beam between
A and B we have

ay

M=E da? 7 (h-#)-R(c-#)..... {ies ose yaaa Rees (1).

 

 

 

 

 

 

 

 

 

 

 

Integrating, we have

dy x 3 we? :
erft = R(ix-F) -R (co-¥) +C.

But 7 =0 when «=0, therefore C=0, and we have

“
dy _ x? 5 a?
1 (10-5) -z (ce -F) eisemenenanauarene tate ae teronetuens (2).

Integrating again, we obtain
SKEW BRIDGES, BRIDGES ON CURVES, TRUSSES AND PORTALS 581

But y=0 when 2=0, therefore C’=0, and we have

wort §2-2-\—Rf 2-2
Ely=R( 5-2) n(of-=) HuLReemaosses (3).
For the bending moment at any point between C and B we have
rope -
M =EI To SRAM 2)parradarrar sew ewes Gekbeee gerne (4)

 

Integrating, we have

BI Gt=R (ha=F) + 0” ateehecbeetsiede stiles re Nie ieee ae (5).
dz

As is obvious (2) and (5) are equal when x=c in each. So by
substituting c for x in both (2) and (5), equating and reducing, we
obtain

c?
7. Jor ©.
C’=-R 5
Substituting this value of C” in (5) we obtain
2 2
ny 2. B (a0 - =) ah?
dx

Integrating again, we have
2 3 2
Bly=R(h-F)- RE 40" an dechaspinnianeeeuln: (6).

As is obvious, (3) and (6) are equal when x=c in each. So by substi-
tuting c for # in both (3) and (6), equating and reducing, we obtain

3
eee. ¥ ce
C’"=4+R =

Substituting this value of C’’” in (6) we obtain

a? a? , 2 3 ;
Ely-R(AE-T)-P Se DRE cents waceyinctvigdeitavauarso (7).

As points C and B are in the same vertical line (3) and (7) are
equal for x=c in (3) and 2=h in (7). So substituting these values,
equating and reducing, we obtain

R 38 — 3c?b 3c?

= Sr an Se a eet (8).

 

When r=2 we have
M” =R(h-—2)—-R’(c-2)=0.

From this equation we obtain

R _ (c-2)
Fe Gece eh wes eens amen nen rs (9).
582 STRUCTURAL ENGINEERING

Now equating (8) and (9) and reducing we obtain

e(c+2h)
B= 2Qcrh) ee (10).

By applying equation (10) the points of contra-flexure in the posts
can be determined and then the stresses in the portal can be computed
as previously explained.

DRAWING ROOM EXERCISE NO. 9

(a) Design a single-track through pin-connected curve chord Pratt

truss railroad bridge and make a complete stress sheet of same upon
tracing cloth.

Data:
Length of span=9 panels at 26’-0’ =234’-0” e.c. end pins.
Height of truss at center =46’-0”.
Height of truss at hip=31’-0”.
Top chord joints to be on the arc of a parabola.
Live load, Cooper’s E50.
Dead load, to be assumed.
Specifications, A. R. E. Ass’n.

(b) Make a general detail drawing of the above bridge.

(c) Design a single-track through Pettit truss railroad bridge and
make a stress sheet for same.

Data:

Length of span=14 panels at 28’-0’ =392’-0” e.c. end pins.
Live load, Cooper’s £50.
Dead load, to be assumed.
Specifications, A. R. E. Ass’n.

(d) Design a pony truss highway bridge and make a general draw-
ing of same on tracing cloth.

Data:
Length of span=6 panels at 8’-0’’=487-0”.
Height of truss =6’-6”,
Width of roadway =16’-0”.

Concrete floor—?” slab and 3” gravel fill.
Specifications, as per Art. 222.

(e) Design a through pin-connected curve chord highway bridge.
Data:

Length of span=9 panels at 18’-6” =166’-6” c.c. end pins.

Height of truss at center =28’-0”.

Height of truss at hip=20’-0”.

Roadway, 16’-0”.

Concrete floor—6” slab and 3” gravel fill.

Specifications, as per Art. 222.
CHAPTER XIV

DESIGN OF BUILDINGS
MILL BUILDINGS

258. Preliminary.—The simplest type of mill building consists of
a steel roof supported upon brick or concrete walls as shown at (2),
Fig. 405. This type of building is satisfactory for power houses, pumping
stations, and small manufacturing plants where light material is handled.

The type shown at (b) consists of a steel roof supported upon steel
columns. The side and end walls for this type may be brick, hollow
tile, concrete, metal lathing plastered, or corrugated iron. In case the
material handled is heavy, an overhead crane would be installed. This
crane would run lengthwise of the building and be supported upon girders
riveted to the columns.

The type shown at (c) is practically the same as the one shown at
(b) except a shed, known as a lean-to, is attached to each side and a
monitor or ventilator is built onto the top of the roof for the purpose of
ventilating and lighting the building.

The type shown at (d) is simply a double building of the same
type as the one shown at (c) except, as a rule, it would be proportionately
larger.

The type shown at (e) is known as the saw-tooth construction. This
type of construction is used in the case of a very wide building in order
to obtain light in the interior of the building.

All of the types shown above are for steep roof construction where
the roof covering is corrugated iron, slate, concrete, or tile. In case of
flat roof construction, where the roof covering is felt covered with tar
and gravel, the types shown in Fig. 406 are used.

The type shown at (f) consists of a steel roof supported upon brick
or concrete walls.

The type shown at (g) consists of a steel roof supported upon steel
columns. The end and side walls may be corrugated iron, brick, con-
crete, hollow tile, or metal lathing plastered.

The type shown at (h) is the same as the one shown at (9) except
a lean-to is added to each side.

The type shown at (k) is practically the same as the one shown at
(h) except each lean-to is wider and the building is proportionately
larger in every way.

The types of buildings shown in Figs. 405 and 406 are general
types. These are often modified to suit certain special requirements;
in fact it is impossible to indicate types suitable for all cases.

As a rule the nature of the material to be handled and the required
location of the machinery govern the type of building used.

583
 

 

 

 

 

 

LY & NSN
iT Crone I
Column

(2)

Monitor or Ventilator->

: wit
Windows or Louvre: ik] WA

Clear Story

Lean-to

 

 

 

I \Colimn—_|

 

 

Wall-

 

 

(o

Windows Winclows.

Windows

 

 

 

 

 

 

 

 

(2)

Fig. 405

584

 
 

 

 

 

 

(9)

a Cleor Story

 

 

 

TT:

(fr)

 

Clear Story.

Lear-io

 

 

 

 

 

Glass

 

 

 

Gloss —

 

4

 

 

(4).

Fig. 406

585
LOP ‘ST

    

                    

0G tat Opf 'Of-UDe7

.

omopuiy

     

i 109

smopuYy : } Asojgsoyy

SEUNOT JO
FOU

wayrjyuepto seypsaus, oy copy

 

 

586
DESIGN OF BUILDINGS 587

The maximum length of span for the trusses as a rule is limited by
the length of the cranes which have a maximum length of about 80 ft.—
more often 40 or 50 ft. is used.

The names of the different parts of an ordinary steel mill building
are given in Fig. 407. These same names, however, apply in general
to steel mill buildings.

The most common forms of roof trusses are shown in Fig. 408. The
steep pitch roofs are used when the roof covering is corrugated iron,
slate. or tile, and the flat roofs are used when the roof covering is felt

MIR LIM.

 

 

 

 

(Q) Fink Truss (b) fanTruss.
(c) Warren Truss. (A) Pratt Truss.
(e) Flat Warren truss, (f) Flat Frat Truss.
(g) Howe Truss. th) Flat Howe Truss.
Fig. 408

covered with tar and gravel. Concrete roof covering may be used in
either case.

The details shown in Fig. 409 are those of a truss supported upon
columns. The purlins are channels which are supported on the roof
truss at panel points. The roof covering is corrugated iron connected to
the purlins by steel bands.

The details shown in Fig. 410 are those of a roof truss supported
upon walls. The roof covering is slate laid on 2” sheathing. The
purlins are channels which have 6” x3” nailing strips bolted to them.
In this case the purlins are not at panel points and the top chord must
be designed heavy enough to resist, in addition to the direct stress, the
cross bending caused by the purlin loads. Wooden purlins as shown
at (a) are sometimes used, which is very good construction.

The details shown in Fig. 411 are those of a roof without purlins.
The sheathing consists of 2’ x 4” timbers laid on edge, acting as beams
extending from truss to truss. The bay length in the case of such con-
struction should be limited to about 14 or 15 ft. In case the roof load
588 STRUCTURAL ENGINEERING

on the trusses is quite heavy and not supported at panel ‘points, the top
chords of the trusses are constructed of two angles and a plate as shown
in Fig. 411. The plate is principally for transmitting shear.

   

Sliding Windows 3

 

Fig. 409

The details shown in Fig. 412 are those of an ordinary monitor
where louvre bars are used. The details shown in Fig. 413 are those of
an ordinary monitor where pivoted windows are used.

Pitch of a roof truss is the ratio of its height to its length. Thus:
A roof truss 60 ft. long and 20 ft. high would have 4 pitch, and if its
height were 30 ft. the pitch would be 4. The terms “pitch” and “slope”
 

 

 

 

Fig. 410

 

 

 

 

Fig. 411

589
590 STRUCTURAL ENGINEERING

 

 
 

 

   
   
   
   

(| Louvre Bars

\

\s
X

 

 

Fig. 412

are somewhat confusing. The pitch of an ordinary truss is obtained by
dividing its rise or height by its total length, while the slope is obtained
by dividing its height by one-half of its length.

Wil

   

Corrugated Rooting

 

 

 

      

 

 

 

 

 

 

 

 

 

 

 

 

 

Fig. 413

A lean-to truss, as shown at (d), Fig. 405, can be considered as a
half truss.

The pitch of a roof depends upon the roof covering. The following
is recommended :
DESIGN OF BUILDINGS

Roof Covering Pitch
Corrugated Iron...... not less than 4.
Slatéscvesavienwaents « not less than 4, preferably 4 or 4.
MTG cahsueeine 1a So tue wee eh not less than $

Felt, tar, and gravel. ..not more than #5, preferably zy.

591

259. Dead Load.—Except for the roof trusses, the dead load for

mill buildings is readily estimated.
lowing units of weights may be used:

In making these estimates the fol-

 

PmMber «is. ciaewtie wee aexeuias 4.0 lbs. per sq. ft. B. M.
Concrete: gins 64 ose eaten eee Ses 150.0 lbs. per cu. ft.
Corrugated iron, #16............ 3.5 lbs. per sq. ft.
Corrugated iron, #18............ 2.7 Ibs. per sq. ft.
Corrugated iron, #20............ 1.9 Ibs. per sq. ft.
Corrugated iron, #22............ 1.5 lbs. per sq. ft.
Slate, 4” thick, 3’ double lap... 4.5 lbs. per sq. ft.
Slate, 3’; thick, 3” double lap... 6.5 Ibs. per sq. ft.
Tile. (plain) io: ieceas ct eran wet 18.0 lbs. per sq. ft.
Tile (Spanish)................. 8.5 lbs. per sq. ft.
Glass, 4” thick 04 aca vea'ee se 2.0 lbs. per sq. ft.
Gravel, felt, and tar roof covering. 5.0 lbs. per sq. ft.
Hollow tile (gross)............. 40.0 lbs. per cu. ft.
The approximate weight of roof trusses in pounds per sq. ft. of roof
surface is as follows:
Pitch
Flat 4 4 4 4
20-ft. span.......... 2.0 1.9 1.8 1.7 1.5
30-ft. span.......... 2.6 2.5 2.3 2.1 1.8
40-ft. span.......... 3.3 3.2 2.9 2.8 2.4
50-ft. span.......... 3.8 3.6 3.4 3.2 2.7
60-ft. span.......... 4.6 4,4 4.1 3.8 3.2
%0-ft. span.......... 5.2 4.9 4.6 4.3 3.7
80-ft. span.......... 5.7 5.4 5.1 4.8 4.1
90-ft. span.......... 6.4 6.1 5.7 5.3 4.6
100-ft. span.......... 7.0 6.7 6.2 5.8 5.0
120-ft. span.......... 8.0 7.6 7.1 6.7 5.7

260. Snow Load.—The snow load on roofs varies with the locality,
slope of roof, and kind of roof covering. According to Trautwine, fresh
fallen snow weighs from 5 to 12 lbs. per cu. ft., while moistened snow
compacted by rain will weigh from 15 to 50 lbs. per cu. ft.

The following loads in pounds per sq. ft. of roof surface are con-

sidered to be sufficient allowance for snow:

 

Pitch
Locality Flat ¢ 4 4 4 4
Central States.............. 30 380 25 20 15 10
Northwestern States......... 40 40 388 380 20) 12
New England States........ 386 386 32 2% 18 12
Rocky Mountain States..... 35 385 30 26 15 10
592 STRUCTURAL ENGINEERING

These loads should be reduced 20 per cent in case the roof covering
be slate or corrugated iron.

No snow load need be considered south of latitude 35°. However,
in some localities south of this latitude a load of 10 Ibs. per sq. ft. of
roof surface should be included in roof loads to allow for sleet. In fact,
load due to sleet, in all cases, should be taken as 10 lbs. per sq. ft. of roof
surface.

261. Wind Load.—The wind pressure against vertical surfaces
varies with the velocity of the wind and to some extent with the area of
the exposed surface. The wind load (pressure) on vertical surfaces can
be safely considered as 30 lbs. per sq. ft. except for steel towers, where
the exposed surfaces are small, in which case the load should be con-
sidered as 40 lbs. per sq. ft. This 30 lbs. pressure, in the case of mill
buildings, is considered to be applied to the vertical sides and ends of
the buildings.

In the case of wind load on sloping roofs the pressure is assumed to
act perpendicularly or normally to the roof surface. This pressure can be
determined only experimentally. Hutton derived, from his experiments
in 1788, the following formula:

-1
P,,=P (sina) 1,842 cosa
for the normal pressure due to wind upon inclined surfaces, where Py
represents the normal pressure, P the corresponding horizontal pressure,
and a the slope of the surface.
Duchemin, in 1829, derived in the same manner the following

formula:
2 sina
Peat (F25.)

for the normal pressure on inclined surfaces, where the letters signify the
same as in Hutton’s formula.

These two formulas are in general use and the values for the normal
pressures when P equals 30 lbs. are as follows:

Normal Pressure

Slope of Roof Hutton Duchemin
OP sees ala Fay wie uns Goeneleds aie wabdee iar 3.9 5.1
MOE se lela. easidesnins van Ger nantoursta” iehtte 7.2 10.2
tic lantlnceta tars siete So, Sek, ede oe Loch Oe 10.5 14.2
18°-26’ (4 pitch).............. 13.0 17.4
21°-48’ (4 pitch).............. 15.0 19.8
26°-34’ (4 pitch).............. 18.0 22.4
DO) k-tearmauns Rucigeas woeene,eeieisecahate 19.9 24.0
388°-41’ (4 pitch).............. 22.0 25.5
AS re ayns dgule esa dre i teithe Seat at teney aoe 25.1 26.7
‘45° (4 pitch).............004. 27.1 28.2
G0 caceais. tre seeeseies ae nienetee elena ee la ase 30.0 30.0

262. Cranes.—In designing a mill building, it is absolutely neces-
sary that the designer have complete data as regards the type, weight,
DESIGN OF BUILDINGS 593

and capacity of the cranes to be used, and also the clearance required
for their operation.

The data given in Fig. 414 will suffice in most cases for overhead
cranes. However, the general requirements in each case should be

 

 

 

 

 

 

12600
1550
19600 Oo
00 00
Q 900
oO oO
2700 oo
37000 | 507
oo {| 43400
44500 | 58700
oo 900
A6200 |49500
51700 | 66600
700
ioo0 oo
84300
9 |112900
OQ | 7610

 

45900 |i31200

Fig. 414

carefully studied; and often it is advisable to consult the company that
manufactures the crane to be used.

Complete Design of a Roof Supported upon Brick Walls

263. Data.—
Size of building=80 ft. long by 40 ft. wide.
Length of roof trusses=40’-0” c.c. end bearings.
Height of roof trusses=10’-0” (4 pitch).
Length of bay =20’-0”.
594. STRUCTURAL ENGINEERING

Roof covering, slate laid on 2” sheathing.
Purlins, steel channels.

264. Design of Purlins.—As 2” sheathing is specified, the first
thing to do is to determine the distance that the purlins can be apart, so
that the sheathing will not be overstressed or the deflection be too great.

The load on the sheathing includes snow, wind, and the weight of
the slate and sheathing.

The maximum snow and wind load is not likely to occur at the same:
time as the snow would be blown off. Let us assume that the wind pres-
sure is 18 lbs. (Hutton—see Art. 261) and the snow 10 lbs. per sq. ft.
of roof surface. The 10 lbs. snow load provides for sleet, ice, or frozen
snow. Using the weights given in Art. 259, we have the following dead
load per sq. ft. of roof:

vs” slate = 6.5 lbs. per sq. ft. of roof surface
2” sheathing= 8.0 lbs. per sq. ft. of roof surface

14.5 lbs. per sq. ft. of roof surface

Adding this to the snow load we obtain 24.5 Ibs. This load acts
vertically while the wind load acts normally to the roof.

Let ¢ be the slope of the roof. Then for the normal pressure due
to dead and snow load we have

24.5 x cosd = 24.5 x 0.895 = 21.9, say 22 Ibs.
Now adding this to the wind load we have
18 4+22=40 Ibs.

for the total maximum normal
pressure on the roof per sq. ft.
of roof surface. Let x be the
distance between purlins, as in-
dicated in Fig. 415, and consider-
ing the sheathing to be a simple
beam 1 ft. wide, we have

M=%3x40x 2? x 12 = 602? inch lbs.

for the maximum bending mo-
ment. Taking 1,000 lbs. as the
allowable unit stress on the
sheathing and applying Formula
(C) of Art. 53, we have

602* = 1,000 x8,

 

Fig. 415

from which we obtain
w=11.5 ft.

for the distance that the purlins could be apart and the sheathing would
not be overstressed from the 40 Ibs. load. The sheathing should be
capable of supporting a 200-Ib. man in addition to the 12.5 Ibs. dead
load. Considering the man to be midway between purlins (center of
DESIGN OF BUILDINGS 595

span) and resolving the loads normally to the roof, we have
“= (14.5 x 0.895 x 22x 12) + x 0.895 x 7 X12=19.427 + 537%

for the bending moment. Then we. have

19.42? + 53% = 1,000 x 8,
from which:we obtain
2=10.8 ft.

for the distance between the purlins which is only 0.7 ft. less than found
above for the 40 Ibs. load. But we know from experience that such
spacing would not. be practical as the deflection of the sheathing would
be sa great that the slate would be broken. From this it is seen that
the spacing of the purlins depends entirely upon the maximum allowable
deflection of the sheathing: This deflection should not exceed hy of
the distance between the purlins. For the deflection due to the 200-Ib.
man at mid span (see Art. 65) we have the following expression:

ee Px
~48Er°

Placing this equal to 2/800 and taking E as 1,000,000 and’ ;P as 200, we
have
200 0.895 x2
48 x 1,000,000x 8 ~ 800’
from which we obtain
2=51,8” or about 4’-4”.

If the deflection due to the dead load of 14.5 lbs. per sq. ft. of roof
surface be taken into account the value of x will be a little less than 4’-4”,
so the correct distance will pees
be about 4’-0”. The actual
spacing of the purlins de- <b
pends to some extent upon 13
the design of the roof truss.
In this case a Fink truss
will be used and the most
suitable spacing obtainable  ,
of the purlins is shown in
Fig. 416. This gives a dis-
tance of about 3’-8” be-
tween purlins. . Fig. 416

The dead and snow
load as given above is 24.5 Ibs. per sq. ft. of roof surface. Adding to
this the wind load of 18 Ibs., which we will consider to act the same:
as the dead load on the purlins, we have

24.5+ 18 =42.5, say 43 lbs,

for the total load per sq. ft. of roof surface.
Now, as the purlins are 3’-8” apart, we have

43 x 3.66 = 157.38, say 158 lbs.

ae ee ao ees

-0”

2

70

52

 

 

207 Oo”

 
596 STRUCTURAL ENGINEERING

for the roof load per ft. of purlin. Assuming the purlin to weigh 12 lbs.

per ft. we have
158 + 12=170 Ibs.

for the total load per ft. of purlin. Then for the maximum bending
moment on the purlin we have

M’=4x170x20. x12=102,000 inch Ibs.

Dividing this by 16,000 we obtain a section modulus of 6.3 which
calls for a.7” x 124% [ or an 8”x114* [. The 8”x11}# [ will be used
for purlins throughout.

Theoretically the slope of a purlin should be taken into account
in determining the maximum stress on it, but as the above result would
not be changed by ‘so doing (especially if the stiffening effect of the
nailing strips and‘ sheathing be taken into account) we are really justified
in ignoring the slope, as was done in the above calculations.

265. Determination of Stresses in the Trusses Due to Dead
and Snow Load.—There are really two loads to consider: The dead
and snow loads that can be combined with the maximum wind load, and
the dead and maximum snow load that could be combined with one-third
of the maximum wind load. The first is given above as 24.5 Ibs. per sq.
ft. of roof surface where the snow (sleet) is taken as 10 lbs. Combining
this with the maximum wind load (Hutton) we obtained the, 43 Ibs. given
above. For the other load, taking the maximum snow for the Central
States, we have

Snow Dead Load Wind
20 * 14.5 + 6 =40.5 Ibs.

which is less than the combination first considered. So we will use the
24.5 lbs. dead and snow load. The end reaction on each purlin, except
the eave and ridge purlins, including the weight of truss and purlins, is
[(24.54+2.9) 3.66+11.25] 10=1,116 lbs. about, and hence the concentra-
tion on the truss at each of these purlins is 1,116x2=2,232 lbs. Then
the load on the truss wi]l be as shown in Fig. 417.

% 8

 

 

 

 

s Ws &
‘ Ni NEN N ‘
& N N N
N N
® ud N *
N S318 SN a ‘
* nN 1.8 : N
N ql NI N ‘
s ° 3. > N N
s b N 5.9 4 N _
. N N N ®
N N 0
g :
~
40~0"
Fig. 417

The concentration on the panel points, considering the top chord in
each panel as a simple beam, is shown at (a), Fig. 418. The diagram of
the stresses is shown at (b). This diagram is constructed by beginning
at joint A and passing around each.joint clock-wise. Thus, laying off 1-2
DESIGN OF BUILDINGS 597

equal to the reaction and 2-3 equal to the 1,480 Ibs. load and closing on
1 and 3 we obtain 1-2-3-4-1 for the stress diagram for joint A. Starting
at 4 and passing around joint a clock-wise, we obtain 4-3-5-6-4 for the

 

 

Fig. 418

stress diagram for that joint. Starting at 1 and passing around joint B
clock-wise we obtain 1-4-6-7-1 for the stress diagram for that joint.
The stress diagrams for joints b and C can not be drawn as the structure
stands as there are three unknown forces at each joint. We get around
this difficulty by inserting the temporary member Ce and omitting mem-
bers be and ce. Having made this modification we obtain the stress
diagram 7-6-5-9-8-7 for joint b. Beginning at 1 and passing around
joint C clock-wise we obtain 1-7-8-12-s-1 for the stress diagram for that
joint. The only correct stress obtained by drawing the last two diagrams
is the stress in CD which is represented by the line s-1. This stress,
as is readily seen, is not affected by the changing of the web members.
Having determined the stress in CD we can consider members be and ce
in place and Ce omitted, and proceed with the analysis. Considering
joint C and passing around clock-wise we obtain s-1-7-13-s for the stress
diagram for that joint. Having the stress in bC determined, which is
represented by the line 7-13, we can draw the stress diagram for joint b
which is 13-7-6-5-9-10-13.

The remainder of the diagram at (b) is readily constructed. This
diagram represents the stresses in the left half of the truss. The stresses
in the right half are the same, for corresponding members, and hence
the diagram at (b) is sufficient for determining the stress in all the truss
members. By scaling the diagram at (6) the desired stresses in the
truss members are obtained. These are shown on the members at (a).
598 STRUCTURAL ENGINEERING

266. Determination of the Stresses in the Trusses Due to Wind
Load.—The wind load (see table of Art. 261) is 18 Ibs. (Hutton) per
sq. ft. of roof surface. The combined dead and snow load considered
in the last article is 30.5 lbs. per sq. ft. of roof surface. So by multiply-
ing the loads shown in Fig. 418 by g)8s; we obtain the wind loads
shown at (a), Fig. 419, which, of course, act normally to the roof and
upon one side only.

‘ The two ends of the trusses are considered to be equally fixed. Then
the reactions will be parallel to the loads. Laying off the load line MN,

 

 

 

Fig. 419

shown at (b), and drawing the ray diagram MON, and constructing the
corresponding equilibrium polygon ab’ ... g, and drawing the line OE
parallel to the closing line ag, we obtain ME , which represents the reaction
R at A, and EN, which represents the reaction R1 at F.

Then beginning at joint A and passing around each joint clock-wise,
as explained in the last article, the stress'diagram shown at (b) is readily
drawn. By scaling this diagram the desired stresses, which are given at
(4), are obtained.

267. Designing of the Truss Members.—Combining the stresses
shown in Figs. 418 and 419 we obtain the stresses shown in Fig. 420,
which are maximum stresses, and which the truss members must be
designed to transmit.

The members will be designed so that 3” rivets can be used through-
out, which requires that the flanges of the angles through which rivets
pass be not less than 24”,
DESIGN OF BUILDINGS 599

Bottom Chord AB. This is a tension member and the required net
area of cross-section is equal to 28,340+16,000=1.770”". Two angles
will make the most satisfactory section for this member. The least size
permissible is 2} x23’, as 2” rivets are to be used, and besides they

 

 

 

 

. e
on in
45
we
2.38 0
Xt ny
a 6192 ANS \h
WG 2 “So. 4
"a9 3
sb? oh é -4400% = Dd
2 2 een e ae” LK <i
~ 2 t 9
452 TAY y XS
8 oo
+t RA, DAS ALK
b red NO 0 ar
eK ZY eG oh,
50 “ans yy: “ ae iy yy
91 mo 4 = 8
eZ RWS (ad Uh
-28340* — S\ - 26250" -/4400*
A) als 28's 2% $2 5 BS CSB ak ats Z"
40 Lo"
Fig. 420

should be at least this size to secure the necessary rigidity. So we will
use 2—Ls 24x 24” x 4 =2.38-0.42=1.960”, which has 0.195” more
section than required, but the thickness is a minimum and hence this
section is as near the required as is possible to obtain. This same section
will be used for the bottom chord throughout, as it is the minimum.

Members ck and cB. These are tension members and the required
section is only about 0.250”, so we will use 1—L_ 24” x2” x }’=1.06-
0.21=0.855”, which is the smallest angle permissible if 2” rivets are
used.

Members ek and kC. These are tension members and the required
section for ek is 0.94”, and for kC it is 0.614”. We will use 2—Ls 24” x
2” x 4/7 =1.70” net, which are the smallest angles permissible.

Member cC. This is a compression member which has a length of
67”. The value of L/r should not exceed 120. Then for the allow-
able r we have 67/120=0.55. Let us try 2—Ls 247 x2” x} =2.120”.
If the 2” flanges are turned out the least radius of gyration will be 0.79.
Then we have

67
16,000 -'70 x 0.79 = 10,060 Ibs.

for the allowable unit stress. Dividing this into the stress we obtain
8,750 + 10,060 = 0.87 sq. ins.

for the required area of cross-section, which is about one-third of the
area contained in the above angles, but these angles will be used as they
are as small as we can use. The L/r=47/0.79 =0.85, which is compara-
tively low.
600 STRUCTURAL ENGINEERING

Members dk and bB. We will make each of these members of 2—Ls
24 x2”"x4’, which are minimum size angles, the same as used for cC.
The actual requirement, however, can be ascertained in the same manner
as shown above for cC but, as is obvious, this work is not necessary.

Top Chord. This is a compression member. Considered in the
vertical direction the length would be a panel length, and considered in
the transverse direction the length should be taken as two panel lengths,
owing to there being some question as to the actual support given to the
chord by the purlins.

The length, considering the transverse direction, is from A to ce,
which is about 11’-2” or 1384”. Owing to the purlin concentration being
applied between panel points there will be cross bending on the chord
which it must resist in addition to the direct stress. Let us assume
2—Ls 5” x33" x 7G” =7.060". The 5” legs will be placed vertically to
resist cross bending. Assuming that the angles are 2” apart, (b-b), the
least radius of gyration, is 1.50. Then we have

134
16,000 ~ 70 x =, =9,750 Ibs.

for the allowable unit stress in compression. The concentrations on the
top chord due to dead and snow load are given in Fig. 417. The chord
in each panel can be considered as a fixed
beam. The maximum moment will occur
at supports b and d. The loads in panel
Ab (the same as in the other panels) are
shown in Fig. 421. This load (at g) is
obtained by resolving the 1,800 Ibs. purlin
concentration due to dead and snow load
(see Fig. 417) normally to the roof and
adding the concentration due to the 18 lbs.
wind load, which is equal to 18x3.66x
Fig. 421 20=1,317 lbs. Hence, for the concentra-
tion at g we have

1,800 x 0.895 + 18 x 3.66 x 20 = 2,928, say 2,900 Ibs.
Applying (7) of Art. 69, we have

 

Mi=2,900 X60 (2h WI) oa cs anda eaieivttee eeu iiana ves (1)
for the maximum moment which occurs at b.
1.8
= 559 = 0.32.

Substituting ‘this value of k in (1), we obtain
M’ =2,900 x 67 (2 x 0.102 — 0.033 — 0.32) =28,215, say 28,000 inch Ibs.

for the maximum bending moment on the top chord.

The moment of inertia of the 2—Ls 5” x 34’ x 7” about the gravity
axis perpendicular to the long legs is 8.9 x 2=17.8, and the distance from
the top of the chord down to this same axis is 1.63. Then applying
zer ‘Sha

 

 

     

 

     

   

 

 

 

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(LyoT)] "286 fue ry 7 yr. 7 7
Se ML FLOP SO OWES EL ope sexyoy ve00
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ANVSAWOD f %7 ae ash
FSNCL YIMAY SSOLO/
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*58/H # {i i =
oa rk sF
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‘IP=sPIOY] WAyog-L i SS 2
a = : ee

 

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8-77-67 XM 8 PTT}
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601
602 STRUCTURAL ENGINEERING

C of Art 53, we have

 

pes _fx20
M’=31,000= 334
from which we obtain
f =5,177 lbs.

for the maximum compressive unit stress due to cross bending For the
maximum unit stress due to direct compression, we have

35,000 + 80” = 4,375 Ibs,
Now adding this to the stress due to cross bending, found above, we have
5,1774+4,375=l]bs. 9,550 (about).

for the actual maximum unit stress on the top chord. The allowable,
as found above, is 9,750 lbs. So the 2—Ls 5” x 33” x4" =80”, assumed
above, are about the correct size, and hence will be used. ‘

This completes the necessary calculations and next the details can
be drawn. Complete shop drawings for the building are shown in Fig.
422. The details shown here are considered to be self-explanatory.
However, the student should verify the riveting.

 

LULQULLEEEEUEET ET

 

 

TANT PATI

Fig. 423

268. Determination of Stresses in Trusses Supported ' upon
Columns.—Stresses due to dead and snow load are determined just the
DESIGN OF BUILDINGS 603

same as if the trusses were supported upon walls, and are fully treated
in Art. 265. The knee braces are ignored in the analysis.

Stresses Due to Wind Load. The columns are fixed to the masonry
by anchor bolts, which should extend well into the masonry, and by the

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P2
. A FI 2 oo
eh
Ss
1 - .
‘ “ y ee
g Xo or)
a
t ce F S
Se (4)
p Ss
> m
“So.
>
0 p2 k
G h k
(Q)
4 pL 4b 7 ¢
4
owt Cc N oY 4 E
oe v2
eee
‘die Ss & G Sate U iM
3 VIP, Aa 4 2 1 ? —
aan Vv S 4
\
fo
é .
3 H s
. 2 3} \ x :
\ 1 : 7s ' 7 Se
Po \ va
: vo \o xt gf )
tee SS. Nas
i/ x 24
Wana shapee a eed aeeceea ase esee 3S
27 26
Fig. 424

addition of knee braces the columns are quite firmly connected to the
trusses, so that the columns can really be considered as fixed, in which
case the wind pressure will cause them to bend as indicated in Fig. 423,
where O and O’ are the points of contra-flexure.

In determining the stresses in the truss due to wind pressure, we
604 STRUCTURAL ENGINEERING

first locate the points of contra-flexure in the columns by applying (10)
of Art. 257. After the points of contra-flexure are located, we next
consider the ground moved up to that level as indicated at (a) in Fig. 424.

We next compute the value of the wind loads, P1, P2, etc. In com-
puting the horizontal loads (P1 and P2) we take the wind pressure at
30 lbs. per sq. ft. of vertical surface, and in computing the value of the
loads (P3 ... P7) normal to the roof we use the pressure per sq. ft.
of roof surface given in Art. 261, the actual intensity of the pressure
used, of course, being in accord with the slope of the roof.

After the intensities of the loads (P1 ... P7) are computed, we
next determine the horizontal and vertical components of the reactions
on the columns at O and O’. If the columns are of equal moment of
inertia we assume that the horizontal components are equal and that each
is equal to one-half of the horizontal component of all of the loads.
These horizontal components can be graphically determined by laying
off the loads as shown at (b), drawing the vertical AP, and the horizontal
PB. Then the horizontal component in each case is represented by one-
half of the line PB. Let H represent this component. To determine
the vertical components of the reactions at O and O’ we complete the ray
diagram at (b), then construct the corresponding equilibrium polygon
g-n-o ... y as shown at (a). Then by prolonging the segments zn and
uy to intersect at I and from J drawing a line parallel to the line AB,
which represents the resultant F of all of the loads, we obtain the line
of action of the resultant, and prolonging this resultant to intersection at
N with line OO’, and then laying off O’M equal to the vertical component
(=AP) of all of the loads, and drawing the line MO and the ordinates
NT and TU, we have the vertical component V2 of the reaction at O’
represented by the line NT and the vertical component V1 of the reaction
at O by the line TU.

The line MO is really an influence line for the vertical component
of reaction at O’. To show that this is true, let us assume that the re-
sultant F passes through point O’. Then, as is readily seen, the total
vertical component of all of the loads would be taken by column £0’, in
which case the vertical component on column Oc would be zero, and if
the resultant F intersected the line OO’ midway between the two columns
the vertical components of the reactions at O and O’ would be equal and
each equal to one-half of the ordinate O’M. So, evidently, the line MO
is the influence line for the vertical component of the reaction at O’.
That is, if the resultant F intersects the line OO’ at any point C between
O and O’ the vertical component of. the reaction at O’ and O are repre-
sented by the ordinate CD and DG, respectively.

If the resultant F intersects the line OO’ at any point between O
and O’ the vertical components of the reactions will be positive; but if it
intersects at any point EF to the right of O’ the vertical component at O
would be negative, and the vertical component at O’ positive. In that
case the ordinate EK (MO prolonged to K) would represent the positive
vertical component of the reaction at O’, while MS would represent the
negative vertical component of the reaction at O. This last construction
is readily proven: Taking moments about O we have

VxOE=P2x00’,
DESIGN OF BUILDINGS 605

from which we obtain

V2=(VxOE)+00’.
But from similar triangles we have

OE EK _ EK

00’ OM VY’
from which we obtain
EK=(V x OE)+=00’,
and therefore

EK=P2.

After the horizontal and vertical components of the reactions at O
and O’ are thus determined, the actual resultant reaction at each of the
points is readily determined, but it is really just as convenient to use the
components in determining the stresses in the truss.

Having determined the horizontal and vertical components of the
reactions at O and O’ in the above manner, we can readily determine
graphically the stresses in the truss due to the wind loads by first drawing
the auxiliary frames Oac and O’wk and beginning at either O or O’ and
passing around each joint in the same direction. The stress diagram at
(ce) is obtained by starting at O and passing around each joint clock-
wise. The stresses obtained in the auxiliary trusses, including the
stresses in the columns, are to be ignored, but the stresses in the knee
braces and truss proper are not affected by adding the auxiliary trusses
and hence they are correct stresses.

Buildings with Lean-tos. In case the lean-tos are of simple con-
struction, as indicated at (f), Fig. 425, each lean-to column should be
considered as acting only as a simple beam in resisting the wind pressure,
in which case the wind pressure on the side of a lean-to would be trans-
mitted equally to the top and bottom of the columns. Then, considering
the case shown in Fig. 425, the pressure on the side 4B would be trans-
mitted equally to points A and B, producing the loads P1 and P2 (each=
3,000 lbs.). The horizontal pressure of the loads from B to C would be
transmitted directly to the column at C. The horizontal pressure repre-
sented as F (=8,600 lbs.) is obtained by constructing the force diagram
shown at (b). Part of the force F is transmitted to point D and part
to E. The results obtained will be accurate enough if we consider the
part DE of the main column as a simple beam. Then taking moments
about E we have

_Fk
i h

for the part of F transmitted to point D. Then adding this force f to
the 3,200 lbs. at D, we obtain the 9,650 Ibs. force shown at (d). The
part of F transmitted to D can also be determined by constructing the
influence line shown at (c).

After the force or load at D (all other loads are assumed as known)
is determined, the ray diagram shown at (e) can be drawn and the cor-
responding equilibrium polygon a-b-c-d-e-f-g-h .. . . mat (d) constructed.
Then prolonging /m and ab to intersection at J and drawing from I a
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

G
h
} S
e .
e Ce AN
t eee $s
Z 2 eS
S 000
3 = ae
— o 7
‘So
t\Caw
[206
9650" >
bv =
: | ;
‘ ;
o\ Ox lo» 0”
*
*
vi j v8 :
% 8
Wh x\V2
eo #
£
: W/
M N

Fig. 425

606
DESIGN OF BUILDINGS 607

line parallel to the resultant GK we obtain JO” for the line of action of
the resultant of all the wind loads affecting the roof.

By laying off O’P (=V =8,400) equal to the vertical component of
all of these loads, and drawing ON and NO”, we have the vertical com-
ponent V2 of the reaction on the column at O’ represented by the dis-
tance NO”, and the vertical component V1 of the reaction on the column
at O represented by the distance MP. V2 is positive and V’1 negative.
Each of the horizontal components H on each column is equal to one-half
of the distance SK.

Now, having determined all of the loads.and the horizontal and
vertical components of the reactions at O and O’ we can determine the
stresses in the roof truss in exactly the same manner as shown above for
the building without lean-tos.

In case the lean-tos are of the construction shown in Fig. 426, the
wind pressure will cause the columns to bend as indicated. If the shear,
indicated as H1, H2, etc., and the direct stress on the columns at the

 

 

 

Fig. 426

points of contra-flexure—that is, the horizontal and vertical components
of the reactions at these points—were determined, there would be no
difficulty in determining the stresses in the main and lean-to trusses.
The actual values of the horizontal and vertical components of the reac-
tions at the points of contra-flexure depend not only upon the relative
stiffness of the columns but upon the stiffness of the trusses as well. But
as this relative stiffness cannot be determined at all until the structure
is fully designed, and then only to a limited extent, it is evident that the
method used for determining the wind stresses must necessarily be based
to some extent upon practical assumptions.

By considering the main truss and the parts of the columns above
the points of contra-flexure, O and O’, as an independent structure, shown
at (a), Fig. 427, we can determine the horizontal and vertical components
of the reactions at O and O’ by drawing the ray diagram at (b) and the
corresponding equilibrium polygon at (a) and the influence line OJ, all
of which has been previously explained.

Considering the lean-to and the part of the main column below the
608 STRUCTURAL ENGINEERING

point of contra-flexure O as an independent structure, shown at (c), we
know H’ and V’ which are applied at O, as they are equal and opposite,
respectively, to the H’ and V’’ found at (a), and we also know the wind
loads, but the components of the reactions at the points of contra-flexure

 

 

 

 

 

Me = (HE0H6) |

 

Fig. 427

S and S’ and likewise the force H’” are unknown. The force H” is the
reaction from the part of the structure above O due to the wind loads on
the lean-to, and hence is equal to the part of these loads that is trans-
mitted through the main truss to the other side of the building. The
actual value of H” depends upon the rigidity of the columns and trusses.
This rigidity can be ascertained only by tedious analysis of the structure
after it is designed. However, a sufficiently accurate value of H” can
be obtained from the following empirical formula:

nog (2
H =H (5 ia) Pepe ad a eet ite, (1)

where H represents the horizontal component of the wind loads on the
lean-to and L the length of the main roof truss and e the distance from
DESIGN OF BUILDINGS 609

the ground to the top of the lean-to and b the distance from the top of
the lean-to to the knee brace (see Fig. 426).

The values of e, b, and L are known (always) and H can be deter-
mined by constructing the force polygon CDE shown at (e). By sub-
stituting in (1) H” can be determined and then constructing the ray
diagram at (e), and drawing the corresponding equilibrium polygon at
(c) and the influence line ST’, we obtain the vertical components V1 and
V2 of the reactions at § and S’, and then the only unknown forces re-
maining are the horizontal components of these reactions which are rep-
resented at H3 and H4. These forces (H3 and H4) would be consid-
ered to be equal if the two columns were of equal moment of inertia, but
as they never are we must assume relative values. We will assume the
lean-to column to be one-third as rigid as the main column. Then as
H,=H3+H4, we have

HB =i Psand H4=4 Ho,
where H, represents the total horizontal component of all the forces on
the structure shown at (c), including those at O.

Considering the lean-to and part of the main column shown at (d)
as an independent structure, the only applied forces are H”, V’’, and H’
at O’. These forces having been previously determined, we can deter-
mine the vertical component of the reactions at the points of contra-

ee

e?
FE.

 

 

 

 

Fig. 428

flexure, S” and S’’’, by constructing the force polygon at (f) and draw-
ing from O’ the line O’Y parallel to the resultant LM and then the
influence line S”’Z.

Assuming the rigidity of the lean-to column to be one-third of that
of the main column, we have

of , 7 _3 ,
H6=7H'oand H5=7 H

We now have all of the components of the reactions at the points
610 STRUCTURAL ENGINEERING

of contra-flexure determined, and we can now proceed with the determi-
naticn of the stresses in the structure.

By considering the part of the structure above the points of contra-
flexure O and O’ and adding the auxiliary trusses to the sides as indicated
at (a), Fig. 428, the stresses in knee braces and main truss can be graph-
ically determined very readily by beginning at either O or O’ and passing
in the same direction around each joint.

By considering the part of the structure at (c), Fig. 428, as an
independent structure and adding the auxiliary frames, shown dotted,
the stresses in the knee brace and lean-to truss can be graphically deter-
mined very readily by beginning at S and passing in the same direction
around each joint. Likewise the stresses in the knee brace and the
lean-to truss shown at (d), Fig. 428, can be determined.

269. Determination of the Stresses in the Columns.—The col-
umns are subjected to direct stress and cross bending similar to the case
of the end posts in bridges. (See Art. 180.)

After the horizontal components of the reactions at the points of
contra-flexure are determined, the bending moment at any point in a
column is readily computed. For example, the bending moment at any
point in column DE (Fig. 425) is equal to Hz, being a maximum at
D and E.

Complete Design of a Mill Building
270. Data.—

Nature of building, machine shop.
Length of building=11 bays @ 20’-0” =220’-0”,
Width of main shop=60’-0” c.c. of main columns.
Width of lean-tos=24’-0” (lean-to on each side of building).
Roof covering, No. 20 corrugated iron.
Allowable intensities on metal—
Tension...... 16,000 Ibs. per sq. in.
Compression. .16,000-70 L/r where L is the length of
member in ins. and r the least radius of gyration of
“ the cross-section in ins.
Crane a tpi Sieh pei 20-ton overhead in main shop.

271, Preliminary.—In starting the designing the very first thing
to do is to draw a sketch of the cross-section of the building as shown in
Fig. 429, showing the general requirements as to type and pitch of roof,
spacing of purlins, height of lean-to, clearance for crane, height of clear
story, etc. The roof covering is to be of corrugated iron. The pitch of
the main roof will be 4 and the lean-tos about 4 (see Art. 258). The
maximum spacing of the purlins is governed by the kind of roof covering.
In this case No. 20 corrugated iron will be used and, as stated in the
Carnegie handbook, the maximum span must not be more than 6 ft., and
less is preferable. In this case the purlins will be spaced about 4’-6”
centers. The distance from the knee brace to the crane girder, as specified
in Fig. 414, is 6’-2”.

272. Designing of the Purlins.—The weight of the corrugated
iron, as per Art. 259, is 1.9 lbs. per sq. ft. of roof surface. The weight
DESIGN OF BUILDINGS 611

of the purlin will be assumed to be 12 Ibs. per ft. of length. The snow
load (frozen snow and ice) will be taken as 10 lbs. per sq. ft. of roof
surface and the wind load as 22 Ibs. per sq. ft. of roof surface as per

 

 

 

 

 

 

 

 

 

 

 

 

+
a
6
N
e
‘S
;
il
.
>
“I
y
60-9" 2440"
Fig. 429

Art. 261 for roof having 4 pitch. Then for the total load on each inter-
mediate purlin per ft. of length we have the following:

Dead load= 1.9x4.54+12= 20.5 kos 5
Snow load=10 x4.5 = 45.0
Wind load=22 x4.5 = 99.0

164.5, say 165 lbs.

Then for the maximum moment on the purlin we have

M=4x165x20 x12=99,000 inch Ibs.

Dividing this moment by 16,000 we obtain 99,000+16,000=6.2 for the
section modulus, which calls for a 7’ x 12.254 channel or an 8” x 11.254
channel. Use 1—[ 8” x 11.25% for each purlin.

273. Designing of the Lean-to Rafters.—The wind load acts
perpendicularly to the rafter while the dead and snow load acts vertically.
So, for the concentration on the rafter at each intermediate purlin, using
the same unit loads as used in the last article, except the wind load which
is taken as 18 lbs. (the lean-to roof has } pitch), we have:

Dead load=20.5x20x0.908= 372 Ibs.
Snow load =45.0x20x0.908= 817 Ibs.
Wind load=(18x4.5) x20 =1,620 Ibs.

2,809 lbs., say 2,800 lbs.

Then applying this concentration at each purlin and considering
the rafter as a simple beam we have the case fully represented in Fig.
430. The maximum moment on the rafter due to these loads, as is
obvious, will occur at C, the center of span. So taking moments about C
we have
612 STRUCTURAL ENGINEERING

M =[7,000 x 13.2 — (2,800 x 8.8) — 2,800 x 4.4] 12= 665,280 inch Ibs.

for the maximum moment on the rafter due to concentrated loads. As-
suming the rafter to weigh 42 lbs. per ft. of length we have

M’=4x42x264 x12=43,900 inch Ibs.

 

 

 

 

 

 

 

 

 

 

 

# * * #
9 9 iG 9 9
a a a3. 3 &
N NI NC NY N
# 44° +
8 7 9
S 96 8
KR 13.2" s
264"
Fig. 430

for maximum moment oh the rafter due to its own weight. Adding
together the above moments we have

665,280 + 43,900 = 709,180 inch Ibs.

for the total maximum moment on the lean-to rafter. Dividing this by
16,000 we obtain 44.3 for the section modulus which calls for a 12” x 40#
or 15” x42# 1. We will use the 15” x42# I for each lean-to rafter.

274. Determination of Stresses in Trusses Due to Snow and
Dead Load.—The snow and dead load per ft. of purlin as given in Art.
272 is 65.5 lbs. Then for the
concentration on the truss at
each intermediate purlin we
have 65.5x20=1,310 Ibs.
Then, as two of these con-
centrations are transmitted
to each panel point of the top
chord of the truss (see Fig.
429), we obtain 1,310x2=
2,620 Ibs. for the part of the
panel load due to snow, roof
covering, and purlins. The
weight of the roof truss as
per Art. 259 is 3.8 lbs. per
sq. ft. of roof surface. Then,
for the part of the panel load
due to the weight of the roof
truss we have 3.8x9x20=
684 Ibs. Now, adding this
to the part due to the roof
load we have 2,620+684=
3,304, say 3,300 lbs. (1,800
Ibs. of this is due to snow)
for the panel load due to
snow and dead load, and hence the snow and dead load on the truss will
be as shown at (a), Fig. 431. By beginning at A and constructing the

 

Fig. 431
DESIGN OF BUILDINGS 613

stress diagram shown at (b) the stresses shown on the truss are obtained.
Multiplying each of these stresses by 1,500/3,300 the stresses marked D
which are due to dead load alone are obtained.

275. Determination of Stresses in Trusses Due to Wind Load.
—The wind load per sq. ft. of roof surface is 22 lbs. (Hutton), as per
Art. 261. Then for the panel load on the roof we have 22 x 9 x 20 =3,960
lbs. and hence the wind loads on the truss will be as shown in Fig. 432.

 

 

 

 

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A M250 4000
few Oa
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AG BA Bo
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8 gy ( 40" ix 9 6
my = F500 >
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£ gee AIS
oor ce? ie VY AY
GC -2Z000 “400 +7100 #20000 W4000 Me
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8
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on oO”

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

y
9 rm
st
% Ree ler
* N Ln S
4200" Fl= B00” \
: - %
200 H2 oa i
11250 ¥ ee 2 = 1700"
i U =
N RZ \ °
F=8250"
(Cc : aes Ry
Z x anh fe)
eo
'
1 af
t
i oy SS
1 ¢ x
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Bateiiatine soeteate as orice es tauieateasenS N
Fig. 482

The horizontal load at a is equal to 30x 5x 20=3,000 Ibs. The load at b
will be the same as at a plus the horizontal component of the wind load
from the lean-to. The horizontal load at the top and bottom of the
lean-to column is equal to 30x 7x20=4,200 lbs., as shown at (b). The
pitch of the lean-to roof is about 4 (4%), so the wind pressure per
sq. ft. of roof is 18 lbs., as per Art. 261. Then the concentration on the
rafter at each intermediate purlin is equal to 18x4.5 x 20=1,620 lbs.
and hence the wind loads on the lean-to roof will be as shown at (b).
614 STRUCTURAL ENGINEERING

Then by constructing the force polygon at (c) we obtain
F=8,250 lbs.

for the horizontal component of the forces on the lean-to. Then adding
this to the 30x 5x20=3,000 lbs. from the clear story we obtain 3,000+
8,250=11,250 Ibs. for the total wind load at 6. Now having all the
wind loads determined we can proceed with the determination of the
stresses in the knee braces and truss due to same.

The points of contra-flexure in each column will be considered as
being midway between the bottom of the column and knee brace. By
constructing the ray diagram shown at (d), and the corresponding equi-
librium polygon c-d-e . . . n-m at (a), and prolonging the segments cd
and nm to intersection at I, and drawing from I a line parallel to the
resultant AB, we obtain the line of action of the resultant of all of the
forces, which intersects the horizontal line OO’ at N.

Then laying off O’S equal to the vertical component of all of the
forces and drawing the influence line OS and dropping a vertical from
N we have the vertical component of the reaction at O’ given by the
ordinate NT and the vertical component of the reaction at O by the ordi-
nate TU. The horizontal components H1 and H2 of the reactions at O’
and O are equal and each is equal to one-half of the horizontal component
of all the forces, which is represented at (d) by the line CB.

Now having the components of the reactions at the points of contra-
flexure, O and O’, determined, the stresses in the knee braces and truss
due to wind loads, which are given at (a), are determined by adding the
auxiliary trusses, shown dotted, and constructing the stress diagram
shown at (e). The diagram shown at (e) is obtained by beginning at O
and passing around each joint clock-wise. The diagram at (e) should
be verified by the student.

276. Designing of the Truss Members.—The maximum stresses
in each truss member (as given in Figs. 431 and 432), and also the ap-
proximate lengths of the members, are shown on the one diagram in Fig.
433. Having this data we can proceed with the designing of the members.

Member KF. This member is subjected to 35,200 lbs. compression
and 20,000 lbs. tension. Let us assume 2—Ls 5” x 34” xf’ =5.120”
as section. Assuming the 5” legs vertical and 4” apart we have L/r=
177/1.50=118, and hence for the allowable compressive stress we have
16,000 —70 x 118=7,740 Ibs. per sq. in. Then we have 35,200+17,740=
4.50” for the area required for compression, which is 0.614” less than
the assumed section.

For tension the net area of cross-section of the assumed section is
5.12—-0.55=4.570”, The actual area required for tension is equal to
20,000 +16,000=1.250”, From the above it is seen that the assumed
section is somewhat larger than required, but owing to the fact that it
is about as satisfactory as can be obtained (the value of L/r being about
correct) it will be used.

Members AF and FG. The member AF is subjected to 19,300 Ibs.
tension while member FG is subjected to 22,250 Ibs. tension and 13,300
Ibs. compression. The truss is held transversely at G and A by struts,
and as member FG is in compression it is necessary to consider (hori-
DESIGN OF BUILDINGS 615

zontally) AG as one member subjected to the 13,300 Ibs. compression.
As the stresses are small the designing is simply a matter of obtaining
the proper value of L/r. Let us assume 2—Ls 5” x 34” x 38,” =5.120”";
the 5” legs horizontal and the vertical legs 3” apart. Then L/r=
260/2.44= 106 in the horizontal plane and 130/1.03 =126 in the vertical
plane. The first value (106) of L/r is quite satisfactory, but the last

   

 

 

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- 280
x Boh
xg \
2 +7100W,
LeZ§ 2900 OXS.
A S2e55SRE F 2USTRIEKE IL G  ALSKIENE IE.
f BL PERIL KL”

 

Fig. 433

value is a little high (125 being the limit), but as the angles are fixed
more or less at the joints in the vertical plane the above assumed section
will be used. There should be enough rivets in these angles at points 4
and G to develop the angles in tension in order to obtain rigidity.

Members BF and DH. Each of these members is subjected to 6,710
lbs. compression. This stress is so small that the designing is simply a
matter of obtaining the proper value of L/r. Let us assume 2—Ls 24” x
247 x4’ =2.380” for section. Then L/r=72/0.78=92, which is quite
satisfactory, and hence the assumed section will be used.

Member CF. This member is subjected to 20,775 Ibs. tension and
24,700 Ibs. compression. Let us assume 2—Ls 34” x34” x #5” =4.180”
as section L/r=130/1.08=120. Then for the allowable compressive
stress we have 16,000-70x120=7,600 lbs. Dividing this into the com-
pressive stress we obtain 24,700+7,600=3.254” for the area required
for compression. For the net area of cross-section required for tension
we have 20,775+16,000=1.300”. From the above it is seen that the
assumed section is larger for section than required, but as the value of
L/r is about as great as is permissible the assumed section will be used.
616 STRUCTURAL ENGINEERING

Member CG. This member is subjected to 11,800 lbs. tension and
21,600 lbs. compression. Let us assume 2—Ls 4” x4” x #4” =4.800”.
Then L/r=144/1.25=115, and hence for the allowable compressive stress
we have 16,000-—70x115=7,950 Ibs. per sq. in. Dividing this into the
compressive stress we obtain 21,600 + 7,950=2.70” for the area of cross-
section required for compression. For the net area of cross-section
required for tension we have 11,800+16,000=0.730”. From the above
it is seen that the assumed section, as far as area is concerned, is larger
than required, but as L/r is of about the correct value the assumed section
will be used.

Member CH. This member is subjected to 5,975 lbs. tension. The
area of cross-section required is about 0.370”. We will use 1—_ 24” x
24” xf =2.38 —0.44=1.940” net for section. This is about as small an
angle as can be used and the member be in harmony with the other
members of the truss.

Members GH and HE. The member GH is subjected to 19,450 lbs.
tension and 10,650 lbs. compression, while member HE is subjected to
25,525 lbs. tension and 9,520 lbs. compression. As the truss is not sup-
ported transversely at H it is necessary to design EG as one compression
member. Let us assume 2—Ls 5” x 34” x #5’ =5.12U” for section. The
maximum radius of gyration is 2.44 and the least is 1.083. Then L/r=
260/2.44=106 in one case and 130/1.03=126 in the other. For the
maximum allowable compressive unit stress we have 16,000—70x126=
7,180. Dividing this into the maximum compressive stress we obtain
10,650 +7,180=1.480” for the required area of cross-section. For the
required area of cross-section for tension we have 25,525 + 16,000 =1.590”
net. From the above it is seen that the assumed section is larger than it
need be, as far as section is concerned, but that the value of L/r is about
as large as allowed, so the assumed section will be used.

Member GO. This member is subjected to 9,900 lbs. tension and
2,600 Ibs. compression. These stresses are so low that the designing of
the member is simply a matter of obtaining the correct value for L/r.
The length of the member is 200 ins. and hence the radius of gyration of
the member must not be less than 200/125=1.60. By using 2—1s 5”x
34” x75” and 2—Ls 3$”x34$”x}” riveted together as shown at (a),
Fig. 433, we obtain about the correct radius and hence this section will
be used.

Top Chord AB ...E. This member should be of the same section
throughout. It is subjected to a maximum compression of 43,600 Ibs.
and to a maximum tension of 22,100-—8,600=13,500 Ibs. The purlins
hold the truss transversely more or less, but as there is some question as
to the rigidity thus obtained the top chord will be considered unsupported
transversely for half of its length, which is about 18 ft. or 216 ins. The
length as regards the vertical direction is 9 ft. or 108 ins. Then, as
regards the transverse direction, the radius of gyration must not be less
than 216/125=1.73, and as regards the vertical direction the radius of
gyration must not be less than 108/125 =0.86.

There is a purlin concentration midway between panel points of
4 (38,960 + 3,300 x 0.83) =6,700 Ibs. (see Figs. 431 and 432) which causes
a bending moment on the chord (considered as a fixed beam) of 4x
6,700 x 108 =90,450 inch Ibs.
DESIGN OF BUILDINGS 617

Now, it is seen that the section of the top chord must be sufficient to
transmit the direct stress and the cross bending given above and that
the radii must not be less than indicated above.

Let us assume 2—Ls 6” x4” x4”=9.500” for section; the 6” legs
vertical and 4” apart. The radii are about 1.71 and 1.91, which are
satisfactory values. For the allowable compression we have 16,000—70 x
216/1.70=7,100 Ibs. per sq. in. The actual direct stress is 43,600+9.5=
4,600 lbs. per sq. in. and the stress due to cross bending is (90,450 x 1.99)
+34.8=5,180 lbs. per sq. in. Now adding these two stresses together
we obtain 9,780, which is larger than the allowed, so the assumed section is
too small. Let us try 2—Ls 6”x6”x27"=8.720”" The radii are about
2.6 and 1.8. Then L/r in one case is equal to 216/2.6=83, and in the
other 108/1.8=60. Then for the allowable unit stress we have 16,000 -
70 x 83 = 10,190.

For the stress due to direct compression we have 43,600+8.72=
5,000 lbs. per sq. in., and for the stress due to cross bending we have
(90,450 x 1.64) +30.78=4,820. Now adding these two stresses together
we have 5,000+4,820=9,820 lbs., which is only 370 lbs. less than the
allowable, so the 2—Ls 6” x6’ x2” will be used as section for the top
chord. We now have all of the truss members designed and the sections
can be written on the diagram as shown in Fig. 433.

277. Designing of the Lean-to Columns.—We assume that these
columns are not fixed. The load on each purlin due to snow and dead
load, as given in Art. 272, is 65.5 Ibs. per ft. of purlin. The vertical
component of the wind load is equal to (18x4.5) x 0.908=173.5 lbs. per
ft. of purlin. Then for the vertical load on the rafter at each interme-
diate purlin we have (65.5+ 73.5) x20=2,780 Ibs. Then the concentra-
tion on the lean-to column due to the roof load is equal to 2,780 x 3 =8,340
lbs. Adding the weight of one-half of the rafter we have 8,340+ (13.2 x
42) =8,894, say 8,900 lbs., for the total direct load on the column.

The wind load acting perpendicularly to the column is (considering
it uniformly distributed) equal to 30x20=600 lbs. per ft. of column.
Then for the moment on the column due to this load we have $x 600 x 14”
x 12=176,400 inch Ibs. Let us assume an 8” x 34# H-beam (Carnegie)
for section, in which case L/r=168/1.87=89.8. Then for the allowable
unit stress we have 16,000 — 70 x 89.8=9.714 Ibs. and for the actual direct
unit stress we have 8,900+10=890 lbs. For the maximum unit stress
due to cross bending we have

f= (176,400 x 4) +115.4=6,110 Ibs.

Now adding the direct unit stress te this bending stress we have 6,110+
890= 7,000 lbs. for the maximum unit stress on the column, which shows
that the assumed H-beam is larger than required, but as this beam seems
to be the most satisfactory section obtainable it will be used throughout
for the lean-to columns.

278, Designing of the Crane Girders.—The spacing and load on
the wheels of a 20-ton crane having a 60-ft. span are given in Fig. 414.
The maximum moment on the crane girder will occur when the wheels
are in the position shown at (a), Fig. 484. Taking moments about B
618 STRUCTURAL ENGINEERING

the reaction shown at A is obtained. Then taking moments about wheel

C we have
28,046 x 7.58 x 12 =2,551,000 inch Ibs.

for the maximum moment on the crane girder due to the crane. Assum-
ing the crane girder to weigh 80 lbs. per ft. and the rail 60 lbs. per yd.,
making in all 100 lbs. per ft. of girder, we have $x 100 x 20? x 12=60,000

inch Ibs. for the maximum bending moment on the girder due to the weight
of the girder and rail. Adding

this moment to the moment due

* 8 to the crane, we have 2,551,000 +

Qi s 60,000=2,611,000 inch lbs. for
« O

 

 

 

 

 

 

 

 

 

 

 

 

A ©) B the total maximum bending mo-

ea : ment on the crane girder. Di-

y 7.56" pal 7.58! viding this by 16,000 we obtain

e ¢ ag 163 for the section modulus

| sat which calls for a 24” x 80# I,

k which will be used for each
(ats girder.

The maximum reaction on the

Ky, crane girder will occur when the

8 8 crane wheels are in the position

% 5 shown at (b), Fig. 484. Tak-

: Cc ¢ fe ing moments about E we obtain

Si? the reaction 56,100 Ibs. shown

9 2-8” 107 4" at D, which is the maximum end

‘ 20-0" shear on the crane girder due

* to the crane wheels. The end

fa b) shear due to the weight of the

Fig, 434 crane girder and rail is equal to

100+20/2=1,000 Ibs. Then
for the total end shear or reaction on the crane girder we have 56,100+
1,000 = 57,100 lbs.

279. Designing of the Main Columns.—The load applied to the
top of each intermediate column due to snow and dead load is 13,200 lbs.,
as given in Fig. 431. The load from the lean-to rafter due to snow and
dead load is equal to 65.5 x 20x 3=3,930, say 4,000 lbs. The direct stress
due to the wind load is equal to the vertical component of the reaction at
the point of contra-flexure, the maximum of which is given in Fig. 432 as
11,500 Ibs. Then for maximum direct stress on the column above the
crane girder we have 13,200+4,000+11,500=28,700 Ibs. This load
comes on the leeward column. As is seen from Fig. 432, the maximum
moment on the column due to wind load occurs at the knee brace and is
equal to (H2x 12.5) 12=11,500 x 12.5 x 12 =1,725,000 inch lbs. Let us
assume the part of the column above the crane girders to be a built
I-section composed of 4—Ls 6”x34”x 75” and a web 18”x3”. The
distance from the truss down to the crane girder, which is 16 ft. or 192
ins. (see Fig. 429), will be taken as the length of this column. The least
radius of gyration of the section is 2.5. Then we have L/r=192/2.5=
76.8, and hence for the allowable unit stress we have 16,000 — (70 x 76.8)
=10,624 lbs. The moment of inertia of the section with reference to the
DESIGN OF BUILDINGS 619

axis perpendicular to the web is 1,576. Then for the maximum unit
stress due to cross bending we have

f= (1,725,000 x 9) + 1,576 =9,850 lbs.

The area of the cross-section is 26.870”. For the direct stress on the
column we have 28,700 + 26.87 = 1,068 lbs. per sq. in. Adding this to the
bending stress we obtain 9,850 + 1,068 = 10,918 lbs. for the total maximum
unit stress on the column (above crane
girder), which is only 294 lbs. greater
than the 10,624 Ibs. allowable, so the
above assumed section will be used.

The part of the column below the P
crane girder will be at least 11 ins. wider
than the part above (see Fig. 414), so ;
that the crane girder can be connected
directly to the column. This requires é Pp

4
q

H

 

—

that the part below the crane girder be jo
30” wide. So let us assume the part of

the column below the crane girder to P
be composed of 4—Ls 6” x34" x3%"= q
20.120” and a web 30” x,” =9.38,
making in all 29.55” of cross-section. e—68 6
The least radius of gyration is 2.55.
The distance from the crane girder to
the base of the column is 19’-0” or 228”.
So we have L/r=228/2.55=89.4, and
hence for the allowable compression we
have 16,000-—70x89.4=9,742 lbs. per
sq. in. The direct load, considering the
windward column, applied from the part
of the column above the crane girder, is Tip
equal to 13,200+4,000+1,700=18,900
Ibs. This causes a direct unit stress of (A)
18,900 +29.5=641 lbs. The two angles
of the column directly under the crane
girders should be sufficient to carry the
maximum concentration from these gird-
ers. As the crane wheels are equal in
weight the maximum concentration due ( b)
to the wheels is equal to the maximum

end shear on the crane girder, which is Sez Le
given in Art. 272 as 56,100 Ibs. The
concentration due to the weight of the’

lo
crane girders and struts is equal to |
(80+20) 20=2,000 lbs. Then for the (Cc) P
total concentration on the column from -

the crane girders we have 56,100+ fA
2,000=58,100 lbs. The area of cross- Fig. 435

section of the two angles is 10.060”.

Then we have 58,100 +10.06=5,776 lbs. for the direct stress on the two
angles due to the concentration from the crane girders. Adding to this

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

wv.

 

SI
epee

 

 

 
620 STRUCTURAL ENGINEERING

the direct stress transmitted from the part of the column above the crane
girder we obtain 5,776+641=6,417 lbs. for the maximum direct stress
on the two column angles directly under the crane girders.

As is seen from Fig. 432, the maximum bending moment on the part
of the column below the crane girder due to wind is equal to 12.5x
11,500 x 12 =1,725,000 inch lbs., which occurs at the base of the column.
The moment of inertia of the column in reference to the gravity axis
perpendicular to the web is about 4,800. Then we have

f = (1,725,000 x 15) + 4,800 =5,390 lbs. per sq. in.

for the maximum stress due to cross bending. Adding this to the above
direct stress we have 6,417+5,390=11,807 lbs. for the total maximum
unit stress on the two angles directly under the crane girders. The
allowable stress is given above as 9,742 lbs. per sq. in.; but as the maxi-
mum crane load and maximum wind load are not likely to occur at the
same time this stress could safely be increased 25 per cent. So we have
1.25 x 9,742 =12,179 lbs. for the unit stress permissible for the combined
loading. This shows that the assumed section for the part of the column
below the crane girder is about correct and hence will be used. The two
outside angles and web of the column are stressed less than the angles
directly under the crane girders, but it is desirable to have a symmetrical
section and hence the main section of the column as given above is satis-
factory throughout.

280. Designing of Column Base.—Case I. When column is sub-

jected to direct load only. In that case the pressure on the base will be
uniformly distributed as indicated at (a), Fig. 435.

Let P=direct load on column,
p=pressure per square in.,
b=width of base in ins.,
d=\length of base in ins.

Then we have

The value of p should not exceed 600 Ibs., which is the allowable
pressure on concrete masonry.

Case II. When the column is subjected to both direct load and cross
bending but the bending not sufficient to reverse the direct pressure. In
that case the pressure will vary as indicated at (b), Fig. 435.

Let P = direct load on column,
p=Maximum pressure per sq. in.,
p’=minimum pressure per sq. in.,
M = bending moment in in. lbs.,
b= width of base in ins.,
d= length of base in ins.

Then we have
DESIGN OF BUILDINGS \ 621

_P 6M
i eT a (2).
and
P 6M
=a Ea Fee mm ww ee eee we ewe neem reer eee rernsees (38).

Case III. When the column is subjected to both direct load and
cross bending and the bending reverses the direct pressure. In that case,
assuming linear variation, the forces will be as indicated at (c), Fig. 435.

Let P=direct load on column,

p=maximum pressure per sq. in.,

f=stress per sq. in. on the anchor bolt= 10,000 Ibs.,

A =area of cross-section of anchor bolt or bolts on each side of

column,

M=bending moment in in. lbs.,

n=ratio of the modulus of elasticity of steel to masonry =15,
b=width of column base in ins.,

d=length of column base in ins.

The problem is to determine p and A. As is seen from the diagram
at (c)-these could readily be determined if « were known. So we will
first derive an expression for the value of 2.

It is obvious that the moments of the forces about o (center of the

column) must be equal to M.
Then we have

Poe (2 2 gat .
B xb (5 S)tfa $=M Poe) EE has Wale eR EW vetoes (1).

As the summation of the vertical forces must equal 0, we have

From the stress diagram at (c) we have

Se aa

or

 

from which we obtain

 

_f(_2
P=. (75) ce ai athe te ale tareraias ale ie lale yal lab eiata re tay eI SIS eea S ACE (6).
From (5) we obtain '
P
A =$5xb-5 ee Cr ra ec ees ea a are ee eee ee ee ee eae eee era (7)

From (4), (5), - (6) the i expression for 2 can be obtained:
-3da? = = 5 (2M + Pa) oe bf ” (2M +Pd) d= Oiauseass send (8).
622 STRUCTURAL ENGINEERING

The value of # for any case can be determined from (8) and then
the value of p can be obtained from (6) and A from (7). For conven-
ience the anchor bolts are assumed to be exactly at the edge of the base
plate, which of course is not absolutely true, but the error resulting from
the assumption is small.

Now, considering the windward column, the direct load above the
crane girder as previously given is 28,700 lbs. and the load from the
crane girder is 58,100 lbs., and assuming the column to weigh 2,000 Ibs.
we have
P=28,700 + 58,100 + 2,000 = 88,800 Ibs.

for the direct load on the column. As previously found the bending
moment

M =1,725,000 inch lbs.

Let us assume d equals 40” and b equals 16”. Substituting these values in
(8) and assuming f equals 10,000 we obtain

z=20 ins. (about).
Substituting in (6) we obtain
p= 666 lbs.,

which is quite satisfactory, as the pressure is due to direct load and cross
bending which are not likely to occur simultaneously. In fact p could
safely be 800 or 900 Ibs.

Substituting in (7) we obtain

Az (> x 20x 16— 88,800 ) + 10,000 =2.77 sq, ins.

for the area of cross-section of anchor bolts. Use two 13” anchor bolts
on a side.

281. Drawings.—The data given in Articles 270 to 280 is suffi-
cient for making the stress sheet, Fig. 436, except for the gable fram-
ing, girts, and bracing, which can be considered more conveniently
as the work on the stress sheet progresses. The gable framing is de-
signed to resist a wind pressure of 30 lbs. per sq. ft. on the end of the
building. Each gable column can be considered as a vertical beam 35 ft.
long and, as they are about 11 ft. apart, the load on each is 11 x 30=330
lbs. per ft. of length. Then the bending moment on each is 4x330x
35 x 12=606,370"#. Dividing this by 16,000 we obtain 37.9 for the
section modulus. This calls for a 12’ x35# I which will be used for
each gable column.

The bracing between the trusses and main columns in the end bays
should be sufficient to resist the wind pressure on the end of the building.
The stresses in this bracing are determined as indicated in Fig. 437. The
diagram at (a) represents the bracing between the main columns and the
diagram at (b) represents the stresses in the same. The diagram at (c)
represents the bracing in the plane of the roof and the diagram at (d)
‘represents the bracing in the plane of the bottom chord of trusses. The
designing of the bracing in intermediate bays is mostly a matter of judg-
ment as it is intended mostly for rigidity.
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623
624. STRUCTURAL ENGINEERING

The girts are designed to resist a horizontal wind pressure of 30 lbs.
per sq. ft. and in some cases they should act as struts, that is, L/r should
not exceed 125. After the stress sheet, Fig. 436, is completed the shop

 

 

    

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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drawings and bills for the structure can be made. A fair idea of the
details can be obtained from the general drawing, Fig. 438.

HIGH BUILDINGS

282. Preliminary.—In the case of modern high buildings such as
office buildings, hotels, etc., all loads including exterior walls, partitions.
floors, and live loads are practically in every case supported upon a steel:
frame, which consists of columns and beams. While the designing of
        

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625
626 STRUCTURAL ENGINEERING

these buildings as a whole is architectural work the designing of the steel
frame is purely a structural engineering problem and hence will be herein
treated as such.

283. Dead Load.—The dead load includes all permanent parts of
the building and°also all permanent fixtures and méchanisms supported
by the building. In estimating dead loads the following units of weights
may be used:

MSLCEEL sics jet Satanic there ioeiarecor Saetuces 490 Ibs. per cu. ft.
Concrete: 22a ce cee eee ees 150 Ibs. per cu. ft.
Brick walls ...........000005 140 Ibs. per eu. ft.
Hollow tile walls............. 40 Ibs. per cu. ft.
Hollow tile floors (tile alone).. 50 Ibs. per cu. ft.
Wood, 22s0vasivite aed ieans ssw 4.5 Ibs. per sq. ft. B. M.
PIASEER cela casce ey sige aa a ees al 5 Ibs. per sq. ft.

The weights of the other materials can be obtained from handbooks.

284. Floor Load.—The floor load consists of the weight of the
floor proper and the load supported on the floor. The former is usually
referred to as the dead load and the latter as live load. The dead load
is always first assumed and then verified by calculations. The live load,
as a rule, can only be estimated. The building ordinances of the differ-
ent cities specify what these loads shall be in each case. In the case of
office buildings the live load for the first floor is usually taken as 100 to
150 lbs. per sq. ft. of floor; the other floors from 40 to 75 lbs. per sq. ft.

Designing of a Ten-Story Office Building

285. Data:—
Length=7 bays @ 15’-0”
Width =4 bays @ 15’-0” 60’-0”
Basement 107-0”
Height= 5 Ist story 18’-0” + 1367-0”
9 stories @ 12’-0’” = 1087-0”
Dead load to be computed.
{atte floor—100 lbs. per sq. ft. of floor.
Live load

105’-0”

Wout We Ul

All other floors—50 lbs. per sq. ft. of floor.
Roof—25 lbs. per sq. ft. of roof.

Wind load—20 lbs. per sq: ft. of vertical exposed surface.

Type of floor—solid concrete.

All metal to be incased in concrete.

In order to resist the wind pressure on the building the floor
beams will be rigidly connected to the columns. These beams will
be considered as fixed beams for all loads.

The floor framing in general will be as shown in Fig. 439.

Allowable unit stresses will be the same as for highway bridges
(see 12 of Art. 222).

286. Wind Stresses.—As seen from Fig. 439, it is necessary to
compute the wind stresses in bents BB, CC, and EE, at least. The wind
stresses in bents 4A will be one-half as much as in bents BB, and Jike-
wise the stresses in bent DD will be half as much as in bents EE and FF.
DESIGN OF BUILDINGS 627

Let us first consider bents BB, the elevation of which is shown in
Fig. 440. The wind load at the roof is equal to

20x 15 x (6 +4) =3,000 lbs.

At each floor from the tenth down to the third (inclusive) the wind load
is equal to 20x15 x12=3,600 lbs. as shown, and at the second floor it is

to

“I

Light Court

 

‘A
A Q w wy uM
General Plan of Floor Framing.
Fig. 439

equal to 20x 15x (6+9) =4,500 lbs., and at the first floor it is equal to ©
20x 15x9=2,700 Ibs. Having the wind loads determined we can pro-
ceed with the determination of the stresses due to same by beginning at
the roof and working downward story after story.

The columns and the floor beams connecting to them will be con-
sidered fixed and the point of contra-flexure in each column of each story
will be considered to be at mid-story, while the point of contra-flexure in
each floor beam will be considered to be at mid-span. Any part of the
structure between points of contra-flexure can be considered as an inde-
pendent structure.

In determining the stresses the building as a whole will be considered
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Fig. 440

628
DESIGN OF BUILDINGS. 629

to act as a vertical cantilever beam, in which case the direct stress on
the columns will vary directly as their distance from the neutral axis.
Considering the part of the structure above the points of contra-
flexure of the columns in the tenth story we have the independent struc-
ture shown at (a), Fig. 441. As is obvious, the neutral axis will be at

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Fig. 441

O,, tlie center of the building. Then the vertical reaction on the columns
will vary directly as their distance out from O, as indicated at (b).
For the moment of the wind load about O, we have

M=3,000x 6=18,000 ft. Ibs.

This moment must be balanced by the moment of the vertical reactions on
the columns. Let R represent the reaction at F. Then the reaction at
a point one foot (unit) out from O, would be R/30, and hence the reac-
tion at G and H will be (R/30) 15 and similarly at F and K it will be
(2/30) 30. Then taking moments about O, we have

a fh a
2 | (5) + ) 15 | ~18,000=0,

from which we obtain

R= — = 240 Ibs.

The number 75 is a constant that can be used in determining the FR in all
the other stories,

 
630 STRUCTURAL ENGINEERING

Now having the vertical reaction R at F determined, we can readily
determine the reactions on the other columns as the intensity in each case
is directly proportionate to the distance of the column from O,. Then,
for the reaction at G and H, we have (15/30) x 240=120 lbs. and for
the reaction at K we have (30/30) x 240=240 lbs., which of course is
the same as at F.

Having the vertical reactions on the columns determined we can next
determine the vertical shear on the floor beams by simply adding up the
vertical forces beginning at either A or E. For example, beginning at A
we have

V1=240
for the shear on girder AB. For the shear on girder BC we have

V2 =2404+ 120=360.

The structure being symmetrical about O, the shears and reactions on
one side of O, will be equal and opposite to those on the other side, and
hence only one-half of the structure need be considered in determining
these stresses.

Having the vertical reactions on the columns and the vertical shear
on the floor beams determined, the horizontal shears H1, H2, and H3 on
the columns can be computed. Considering the part of the structure
shown at (c) as an independent structure and taking moments about A
we have

240x7.5-H1x6=0,
from which we obtain
H1=300 lbs.

Then summing up the horizontal forces, we have
3,000 - 300 -F=0,

from which we obtain
F =2,700 lbs.

for the direct compression in beam AB. Considering the part of the
structure shown at (d) as an independent structure and taking moments
about B, we have

240 x 7.5 + 360 x 7.5 -H2x6=0,

from which we obtain

H2='750 Ibs.
Then summing up the horizontal forces we have
2,700 - 750 -F1=0,

from which we obtain

F1=1,950 Ibs.

for the direct compression in floor beam BC. Likewise, considering the
part of the structure shown at (e) as an independent structure and taking
moments about C we have

360x15-H3x6=0,
DESIGN OF BUILDINGS 631

from which we obtain
H3 =900 lbs.

Then adding up the horizontal forces we have

1,950 - 900 - F2=0,
from which we obtain
F2=1,050 lbs.

for the direct stress in beam CD. The direct stress in beam DE is equal
to 1,050 -7%50=300 lbs., which is and should be equal to H1.

These shears and stresses can now be written on the diagram in
Fig. 440.

Next considering the part of the structure between the points of
contra-flexure in the columns of the tenth and ninth stories we obtain the
independent structure shown at (a), Fig. 442. Taking moments about

3O’

 

 

 

 

 

 

 

 

 

 

S04"

Fig. 442

O, of the wind forces on the building above this point (see Fig. 440), we
have
3 M =3,000 x 18 + 3,600 x 6 = 75,600 ft. lbs.

Letting R represent the vertical reaction on the column at F we have

R\—2 R\—2
2 | (55) 0 + (3) 15 |= 15.600

from which we obtain

_ 15,600

R= 15 = 1,008 Ibs.
682 STRUCTURAL ENGINEERING

Then the reaction on the column at G is equal to 1,008 (15/30) = 504 Ibs.

Beginning at A and adding up the vertical forces (algebraically)
we have 1,008-240='768 lbs. for the shear on beam AB, 1,008 — 240+
504.—120=1,152 lbs. for the shear on beam BC. Now considering the
part of the structure shown at (c) as an independent structure and taking
moments about A, we have

H1x6+300x6-%68x7.5=0,

from which we obtain

H1=660 lbs.

for the horizontal shear on the column at F,
Similarly, considering the part of the structure shown at (d) as an
independent structure and taking moments about B, we have

H2x64750x 6~768 x 7.5-1,152 x 7.5=0,

from which we obtain
H2=1,650 lbs.

Considering the part of the structure shown at (b) as an independent
structure and taking moments about C, we have

H3x6+900x6-1,152 x 15 =0,

from which we obtain
H3=1,980 lbs.

‘

for the shear on the column at O,. Now adding up the horizontal forces
beginning at A we have 3,600 + 300 — 660 =3,240 Ibs. for the direct com-
pression in beam AB, 3,600 + 300 — 660 —-1,650 + 750 =2,340 lbs. for the
direct compression in beam BC, 2,340+900-1,980=1,260 Ibs. for the
direct compression in beam CD, and 1,260+750-1,650=360 Ibs. for
the direct compression in beam DE.

Continuing in the manner shown above, considering the parts of the
structure between points of contra-flexure in the column of consecutive
stories on down to the base of the building, the direct stresses and shears
on the columns and floor beams, as shown in Fig. 440, can be determined.

It will be seen from Fig. 440 that the horizontal shear on each
column varies from the ninth down to the second story by a constant and
that the vertical shear on the floor beams in each bay varies from the
tenth down to the third floor by a constant, and hence these stresses can
be quickly computed by the use of the constants, which can be determined
after the stresses are computed in the three top stories. Also, it will be
seen (see Fig. 440) that the direct stress in the floor beams.in each bay
is the same from the tenth down to the third floor. After the shears on
the columns and floor beams are computed by the use of the constants,
the direct stress on the columns can be determined by beginning at the
ninth story and adding up the vertical forces included between the points
of contra-flexure. As an example, for the direct stress in the outer left-
hand column of the sixth story, we have 4,272 +2,496=6,768 Ibs. For
the next column to the right we have—2,496 + 2,136 + 3,744=3,384 lbs.,
and so on. To obtain the stresses in the second story floor beams and
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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633
634 STRUCTURAL ENGINEERING

first story columns we first take moments about O,, of the wind forces
above that point. For this moment (see Fig. 440), we have

M=3,000 x 117 + 28,800 x 63 + 4,500 x 9 = 2,205,900 ft. lbs.

Then dividing this by the constant 75 (previously determined) we obtain
29,412 Ibs. for the reaction or direct stress in the outer left-hand column.
The reactions on the other columns can then be quickly determined by
direct proportion, and the remainder of the work of determining the stress
is the same as previously explained.

To obtain the stresses in the floor beams of the first floor and in the
basement columns we would first take moments about O,, (see Fig. 440)
of the wind forces above this point. For this moment, we have

M = 3,000 x 131 + 28,800 x 77 + 4,500 x 23 + 2,700 x 5 = 2,727,600 ft. Ibs.

Then dividing this by the constant 75 we obtain 36,368 lbs. for the reac-
‘tion or direct stress on the outer left-hand column. The reactions on the
other columns are readily determined by direct proportion, and the re-
mainder of the work of de-
termining the stresses is
the same as previously ex-
plained.

From the above it is
seen that all of the direct
stresses and shears in the
columns and floor beams
of the bent are obtained by
fully analyzing only the
three top stories and the
first floor and basement.

We will next consider
the bents CC (through the
light court), the elevation
of which is shown’ in Fig.
' 443. Considering the part
of the structure above the
1

 

 

 

m |9 i :
¥ ms" | 25" Rs points of contra-flexure of
bof = ole > eH a
i > ¥ i the columns in the tenth
'B FL Fe 1D story we obtain the inde-

 

 

l i
ee ‘
% 7 * | endent structure which is
: sa" a , et 7) : Hidwn at (a), Fig. 444.
As these bents are unsym-
metrical we have to deter-
Fig. 444 mine the location of the
neutral axis OO. The re-
actions on the columns will vary directly as their distance from this
neutral axis, as indicated at (b), and the sum of the reactions on one
side must equal the sum of those on the other.
Let z=the distance from A to the neutral axis, and for the sake of
simplicity let a, b, and c represent the bay lengths as indicated, and let R
represent the reaction at E.

 

 

-/22*|
DESIGN OF BUILDINGS 635

Then we have # for the reaction at E, (R/z) (s—a) for the reaction
at G, (R/z) (a+b-—z) for the reaction at H, and (R/z) (a+b+c-z)
for the reaction at K. Then, as the sum of the reactions on one side of
the neutral axis must be equal to the sum of those on the other side of the
neutral axis, we have

R+ (2) (g-a)= (Z) (a+b-2)+(2) (at+b+e-3z),

from which we obtain

sont te easiest eal 2s aces aaa (1)

for the distance from A to the neutral axis. Now, substituting the
numerical values of the bay lengths in (1) we obtain

ae 45+30+410
= 4

which gives the location of the neutral axis.
Taking moments about O, we have

R ——__2 R — R ——2 R sone?
(sia) 21.25 + (sis) 85 + (si55) 878 +(spss) 18.75 -

= 21.25 ft.

3,000 x 6=0,
from which we obtain
18,000 _
R = 43.23. = 416 lbs.

for the reaction on the column at E. The number 43.23 is the constant
for determining the reaction R for all the other stories. Having the
reaction at E determined the reaction on the other columns is readily
determined by direct proportion. For example, the reaction at H is;
(416/21.25) 8.75=171 lbs. After the reactions on the columns are dete
mined the shears on the columns and the shears and direct stresses o1
floor beams can be determined, as previously explained, by consid
the independent portions shown at (c), (d), (e), and (f). In fa-
remainder of the work of determining the direct stresses and ~
the columns and floor beams shown in Fig. 443 is the same -
explained for bents BB.

The shears and direct stresses in the columns and floor
bents EE and FF are given in Fig. 445. These are determin,
same manner as previously shown for bents BB. It will be seen
neutral axis for bents EE and FF is midway between the central -
After the shears and direct stresses on the columns and floor be.
given in Figs. 440, 443, and 445, are determined the stresses at the
connections are readily obtained, as will be seen later.

287. Designing of the Roof Framing.—The columns and
will be spaced as shown in Fig. 446.

The roof covering will be a 4” solid concrete slab.

The assumed loads on the roof will be as follows:
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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lo

Bents -EE & FF.

Pig. 445

636
DESIGN OF BUILDINGS

Live load (Chicago)............ 25 Ibs. per sq. ft. of roof.
Concrete slab....... 50 Ibs. per sq. ft. of roof.
eerie tL Sraeuied ceiling.... 10 lbs. per sq. ft. of roof.

60 Ibs. per sq. ft. of roof,

 

9.

Roof Framing
Fig. 446

637

Beams 1. The dead load on each of these beams is as follows:

Concrete slab and ceiling=60 x 5=300 lbs. per ft. of beam.
Concrete around beam 75 Ibs. per ft. of beam.
Beam (assumed) _15 Ibs. per ft. of beam.

390 lbs. per ft. of beam.

Then, for the maximum bending moment due to dead load we have
638 STRUCTURAL ENGINEERING

4x390x 15 x 12= 181,625 inch Ibs.

The live load is 25x 5=125 lbs. per ft. of beam. Then, for the live load
moment on each beam we have

4x125x 15 x 12=42,187 inch Ibs.

Adding the two moments together we obtain 131,625 +42,187=173,812
inch lbs. for the total maximum moment on the beam. Dividing this by
16,000 we obtain 10.8 for the section modulus which calls for a 7” x 157
I-beam, which will be used.

Dead load =390 x 7.5 =2,925 Ibs.
Live load=125x7.5= 937 Ibs.

3,862 lbs.

End shear

 

Beams 2. These beams will be connected rigidly to the columns to
resist wind pressure and consequently will be fixed beams for all loads.
The dead load on each of these beams is as follows:

Fire wall [see sketch at (b) ] =1x 140 x 4=560 lbs. per ft. of beam.
Concrete slab and ceiling =60%x2.5 150 Ibs. per ft. of beam.
Concrete around beam 120 lbs. per ft. of beam.
Beam (2—[s) 30 Ibs. per ft. of beam.

860 Ibs. per ft. of beam.

Then for the maximum bending moment at the ends of the beam due
to dead load, considering each bracket to be 18 ins. long, we have
7x x 860 x 12? x 12 = 123,840 inch lbs., and one-half of this, or 61,920 inch
Ibs., will be the moment at the center of span (see Art. 69). The live
load on the beam is 2.5x25=62.5 lbs. per ft. of beam. Then for the
bending moment at the ends of the beam due to live load we have 7; x
62.5 x 12x 12=9,000 inch Ibs., and one-half of this, or 4,500 inch Ibs.,
will be the moment at the center of span.

The maximum shear on these girders due to wind pressure is one-half
of the maximum given in Fig. 440, or 360+2=180 lbs. Then for the

pment at the end of the beam (at end of bracket) we have 180x72=
125960 inch Ibs. ;

w combining the maximum dead and live. moments we have
123,840 +-%,000 = 132,840 inch Ibs. Dividing this by 16,000 we obtain
8.3 for section modulus, which calls for 2—[s 6”x8#, but to obtain
rigidity 2—[s ‘K’ x 9.754 will be used. The stress due to wind load is
not considered as ft does not exceed 50 per cent of the live- and dead-load
stresses. This is in xecordance with practice.

For the maximum tnd shear on the beam we have

Dead load=860 7.5=6,450 lbs.
Live load= 62.5x7.5= 469 lbs,
6,919 lbs.

Beams 8. For the load on each of these beams we have the fol-
lowing:

 
DESIGN OF BUILDINGS 639

Concrete slab=50x5 =250 Ibs. per ft. of beam.
Concrete around beam =120 lbs. per ft. of beam.

Beam = 21 Ibs. per ft. of beam.
Ceiling =10x5 = 50 lbs. per ft. of beam.
Live load=25x5 = 125 Ibs. per ft. of beam.

Total = 566 lbs. per ft. of beam.

Considering the beams fixed to the columns by brackets we have 7s x 566 x
12? 12=81,504 inch lbs. for the maximum moment at the ends of the
beam due to dead and live load. The moment due to wind load is 539 x
72 = 38,808 inch lbs., which is less than 50 per cent of the live- and dead-
load moment and hence will not be considered.

Dividing the dead- and live-load moment by 16,000, we obtain
81,504 + 16,000 =5.1 for the section modulus, which calls for a 6” x 12.254
I, but for the sake of rigidity and also to allow for rivet holes at the
brackets we will use 1—I 7” x 15# for each beam.

For the maximum end shear on the beam we have

Dead load =441 x 7.5 =3,307 lbs.
Live load=125x%7.5= 937 Ibs.

4,244 lbs.

Beams 4. There will be two concentrations on each of these beams,
each of which we will assume is equal to twice the maximum end shear on
beams 1. The concrete around the beam will be assumed to weigh 120
Ibs. per ft. of beam and the beam itself will be assumed to weigh 21 Ibs.
The loading will then be as indicated in Fig. 447. These loads include
both live and dead load. ’

Considering the beam to be fixed we have 75 x 141 x12" x 12 = 20,304
inch lbs. for the maximum moment due to the uniform load. &=3.5/12=
0.29 for one load and 8.5/12=

 

 

 

 

 

 

 

 

0.71 for the other. Then substi- Ry i

tuting in (7) of Art. 69 we obtain N N

229,120 inch lbs. for the maxi- mi i

mum moment due to the concen- Ler IE.

trations. This moment occurs at 35’ | 5’ | 35’

the end of the beam. 15! 12° 135
Adding together the two mo- 15/

ments we obtain 20,304+229,- wie. 407

120 =249,424 inch lbs. for the
total maximum bending moment on the beam due to dead and live load.
Dividing this by 16,000 Ibs. we obtain 15.5 for the section modulus which
calls for a 9” x 21% I, which will be used for each of these beams.

For the maximum end shear on these beams we have 7,724+141x
7.5=8,781 lbs. The wind stress is so small that it is negligible.

Beams 5. The dead load on each of these beams is as follows:

Fire wall (same as 2) =560 Ibs. per ft. of beam.
Concrete around beam= 150 lbs. per ft. of beam.
Beam (2 channels) = 30 Ibs. per ft. of beam.

740 Ibs. per ft. of beam.
640 STRUCTURAL ENGINEERING

For the maximum moment on the beam due to this load, considering
the beam fixed, we have 75x 740 x 12’x 12=106,560 inch Ibs.

The moment due to the two concentrated loads will be one-half of
that found for beams 4 or 229,120 +2=114,560 inch lbs. Adding the two
above moments together we have

‘ 106,560 + 114,560 = 221,120 inch Ibs.

for the total maximum moment on the beam. Dividing this by 16,000 we
obtain 13.8, which calls for 2—8” x 11.25# [s, which will be used.

Beams 6 to 12. For the sake of uniformity beams 6 will be the same
size as beams 4, although the moment does not require such a large sec-
tion. Likewise, beams 7 will be the same size as beams 3; beams 8, 9,
and 10 will be the same size as beams 1; and beams 11 and 12 will be the
same size as beams 5.

288. Designing of the Tenth-Story Columns.—
Load on Column C1

Dead load
From beams 2 6,450x 2 =12,900 Ibs.
From beams 1 2,925x2 = 5,850 lbs.
From beams 4 141x7$= 1,058 Ibs.
19,708 lbs.
Live load =25x15x 74 ; = 2,812 Ibs.

Total = 22,520 lbs.
Load on Column C2

Dead load
From beams 2 = 6,450 Ibs.
From beams 1 = 2,925 Ibs.
From beams 5=40 x 74 = 6,450 Ibs.
15,825 lbs.
Live load=25x 74x 74 = 1,406 lbs.

Total=17,231 Ibs.

Load on Column C3
Dead load

From beams 5=6,450 x 2 = 12,900 Ibs.
From beams 1=2,925x2 = 5,850 lbs.
From beams 3 = 3,307 Ibs.

22,057 Ibs.
Live load =25 x 15 x 7 = 2,812: lbs.

Total = 24,869 lbs.

Load on Column Ch ;
Dead load

From beams 1=2,925x4 =11,700 Ibs.
From beams 4=1,058 x2 = 2,116 lbs,
From beams 3=3,307 x2 = 6,614 Ibs.
20,430 Ibs.

Live load =25 15x15 = 5,625 Ibs.

Total = 26,055 Ibs.
DESIGN OF BUILDINGS 641

foad on Column C5
Dead load

From beams 1 = 5,850 lbs.
From beams 3 = 3,307 Ibs.
From beams 7 = 2,200 Ibs.
From beams 8, 9 and 10 = 38,900 Ibs.
From beams 6 = 1,000 Ibs.
16,257 Ibs.

Live load =25 x 124x 15 = 4,687 lbs.

Total = 20,944 Ibs.

Load on Column C6. Total=about 3 of load on column C3 = 16,600
Ibs.
Load on Column C7
Dead load

From beams 2 = 6,450 Ibs
From beams 11 = 6,020 Ibs
Fyom beams 3 = 38,307 lbs
From beams 1 = 5,850 Ibs.
From beams 8 and 9 = 2,600 lbs.
24,227 Ibs.

Live load = 4,200 Ibs.

Total = 28,427 lbs.

Load on Column C7’. The load on this column is about equal to
the load on ‘column C? plus 10,000 lbs. for elevator, which makes a total
load of about 38,000 lbs.

Load on Column C2’. The load on this column is about equal to the
load on column C2 plus 10,000 lbs. for elevator, which makes a total
load of about 27,000 Ibs.

The direct wind loads on these columns are negligible, the maximum
being only about 367 lbs.

The maximum shear on these columns due to wind is 1,194 lbs. in
one direction and 55+ lbs. in the other direction (see Figs. 443 and 445).
Then for the maximum bending moments on the columns, we have 1,194 x
G6 x 12 =85,968 inch lbs. and 554 x 6 x 12 = 39,888 inch lbs., respectively.

The columns should be designed to resist this bending in addition
to the direct load on them. However, the unit stress for the combined
loading can be safely increased 50 per cent over that obtained by the
formula, 16,000-70 L/r.

Let us assume the following section:

41s 47x3"x—"= 8.360”
1—web 12’x8;"= 3.750”
12.110”

‘The Max. [=294.6 and the Min. I=30.2
The Max. r= 4.91 and the Min. r= 1.58

Columns C4 in the center of the building will be subjected to the
642 STRUCTURAL ENGINEERING

greatest bending in the two directions. For the fiber stresses we have
f =39,888 x 4.1 + 30.2 =5,415 lbs.

and
f’ = 85,968 x 6.2 + 294.6 =1,809 lbs.

For the allowable direct stress, we have 16,000-70x144+1.58=
9,620 lbs. per sq. in.

Dividing the area of cross-section of the column into the load on the
column we have 26,055 + 12.11=2,150 lbs. per sq. in. for the direct stress.
Adding this to the maximum wind stress we have 5,415 + 2,150 =7,565 Ibs.
per sq. in., which is quite low, so we will assume a lighter section.

Let us try
4—ls 347x247’ x 4’ =5.760”
1—web 12” x47” =3.000”

8.760”

Max. 2=213.5, Min. 2=15.9
Max.r= 4.94, Min. r= 1.35 i

For the allowable direct stress we have 16,000—70x144+1.35=
8,530 lbs. per sq. in. For the unit stress due to cross bending we have
39,888 x 3.62 + 15.9 =9,080 Ibs.

For the stress due to direct load we have 26,055 +8.76=2,970 Ibs.
per sq. in., which is low. Adding together the stresses due to direct load
and cross bending we have 2,970+9,080=12,050 Ibs. per sq. in. The
allowable is 8,530 1.5=12,795 Ibs. per sq. in. for combined loading,
which is very near the actual stress and hence the last assumed section
can be used. This section is about the minimum that would be used, so
all columns in the tenth story will have this section.

289. Designing of the Tenth-Floor Framing.—The framing in
this floor will be as indicated in Fig. 448.

Beams 1. For the dead load on each of these beams we have the
following:

4” concrete slab =50 x 5=250 lbs. per ft. of beam.
Plaster = 5x5= 26 lbs. per ft. of beam.
Concrete around beam 75 lbs. per ft. of beam.
Beam 18 lbs. per ft. of beam.

Total = 368 lbs. per ft. of beam.

ol

For the maximum moment due to dead load we have 4x 368 x15 x12=
124,200 inch lbs. For the maximum moment due to live load we have

4 (50x 5) x 15°x 12=84,375 inch Ibs. Then for the total maximum mo-
ment we have 124,200 +84,375 = 208,575 inch lbs. Dividing this by 16,000
we obtain 13.04, which calls for an 8” x 18# I, which will be used.

For maximum end shear on the beam we have

Dead load =368 x 74=2,760 Ibs.
Live load =250x74=1,875 Ibs.

Total =4,635 Ibs.

Beams 2. The dead load on each of these beams consists of a story
DESIGN OF BUILDINGS 643

of 12-in. curtain wall, a strip of concrete floor 2.5 ft. wide, and the weight
of the beam and concrete encasing it.

For the weight of the wall we have

8” hollow tile= 40x 0.66=26.4 Ibs. per sq. ft. of wall.
4” prick face =140x 0.33 =46.2 lbs. per sq. ft. of wall.

Total = 72.6 lbs. per sq. ft. of wall.

The curtain wall in each bay will have two openings for windows,
as shown at (a), Fig. 449. The load on the beam from this wall will be

NS
3
S
<S
S
NW

 

cl Gr

/o® Floor Framing.

Fig. 448

as shown at (b), but, as the load near the ends of the beam is supported
directly upon the brackets, the load for computing the stress on the beam
can be taken as indicated at (c),
644 STRUCTURAL ENGINEERING

Then for the maximum moment on the beam due to the curtain wall,
we have

a 18x11 x12= 22,022 inch lbs,

ax 2,504x11 x12=41, 318 inch lbs.
Total = 63,340 inch lbs.

 

 

5°

 

 

2’ 4! Be 4! 2’

 

2!"
ae

 

 

 

 

 

 

 

 

 

‘ (Q)

1g"

 

 

 

2

KR (2)

7]

 

 

 

 

ess*
Sa

 

 

-——_—

 

 

 

 

Tig. 449

For the remainder of the dead load we have

Concrete slab=50 x 2.5=125 Ibs. per ft. of beam.
Beam (2—[s) = 80 lbs. per ft. of beam.
Concrete around beam =150 lbs. per ft. of beam.

Total = 305 lbs. per ft. of beam.
DESIGN OF BUILDINGS 645

For the moment on the beam due to this load we have

5 x805x 11 x 12=36,905 inch Ibs.

For the live load on the beam we have 50x 2.5=125 Ibs. per ft. of
beam. Then for the live-load moment on the beam we have

a 125x 11 x 12=15,125 inch Ibs.

Adding together all of the above moments we have

63,340 + 36,905+ 15,125 = 115,370 inch lbs.

for the total maximum bending moment on the beam due to dead and live
load.

The maximum shear on the beam due to wind load is 1,152 +2=576
Ibs. (see Fig. 440). Then for the moment on the beam due to wind load
we have 576 x 66 = 38,016 inch lbs., which is less than 50 per cent of the
live- and dead-load moment and hence it will be ignored. Dividing the
dead- and live-load moment by 16,000 we have 115,370+16,000=7.2 for
the section modulus, which calls for 2—[s 6” x 8#, but for the sake of
rigidity we will use 2—[s 7” x 9.75#.

Aiasanniek cea Dead load =6,504 Ibs.
. Live load= 937 Ibs.

7,441 Ibs.

 

Beams 8. For the dead load on each of these beams we have the
following: ;
Partition = 40 x 0.33 x 12 = 160 lbs. per ft. of beam.
Concrete slab=50 x5 = 250 lbs. per ft. of beam.
Beam = 21 lbs. per ft. of beam.
Concrete around beam =150 Ibs. per ft. of beam.
Plaster =5x5 25 Ibs. per ft. of beam.

636 lbs. per ft. of beam.

Then for the maximum moment on the beam due to dead load, we

have 7 x 636 x 11° x 12=16,956 inch lbs. The live load is 50x 5=250
Ibs. per ft. of beam. Then for the moment on the beam due to live load,
we have jy x 250 x 11° x 12 =30,250 inch Ibs. Adding together the dead-
and live-load moments we obtain 107,206 inch lbs. for the total maximum
moment due to dead and live load. The maximum shear on these beams
due to wind load, as given in Fig. 440, is 1,152 lbs. So, for the maximum
bending moment on the beams due to wind, we have 1,152 x 66= 76,032
inch lbs. Now subtracting from this 50 per cent of the live- and dead-
load moment we obtain 76,032 — 53,603 = 22,429 inch Ibs., which must be
added to the maximum dead- and live-load moment. So we have

107,205 + 22,429 = 129,635 inch lbs.

bending moment which the beam must be designed to resist. Dividing
646 STRUCTURAL ENGINEERING

this by 16,000 we obtain 8.1 for the section modulus which calls for a
¥’ x 15# I, but as beams 1 are 8-in. beams we will make these the same
size; that is, for each we will use 1—I 8”x18#. This excess size also
provides for rivet holes cut out at the bracket.

Dead load = 636 x 7.5=4,770 Ibs.
Live load =1,875 lbs.

6,645 Ibs.

Maximum end shear

Beams 4. The maximum dead load consists of a 4-in. partition,
two concentrations from beams 1, the weight of the beam, and the concrete
encasing it. For the uniform dead load on the beam we have the
following:

Partition (same as 3) =160 lbs. per ft. of beam.
Beam = 25 lbs. per ft. of beam.
Concrete around beam= 150 Ibs. per ft. of beam.

Total = 335 Ibs. per ft. of beam.

For each concentration due to dead load we have 2,760 x 2=5,520
lbs. For each live-load con-

 

 

 

 

 

 

 

 

 

 

4%, % centration we have 1,875 x 2=

f 3,750 Ibs. Then for the to-

3’ © 5s 3’ tal of each concentration we

have 5,520+ 3,750 =9,270 Ibs.

BIOL EEL The loading will then be as

Z we | 2° indicated in Fig. 450. For
2 the maximum moment (con-

2 sidering the beam fixed) due

Fig. 450 to the uniform load we have

3 x885x 11 x 12=40,535 inch lbs.

For the maximum moment due to the concentrated loads we have

[9,270x 11 (2x0.27 —0.27 —0.27)+9,270x11 (2x0.73 -0.73-
— 0.73) ] 12=240,800 inch Ibs.

Now adding together the above moments we obtain

40,535 + 240,800 = 281,335 inch lbs.

for the total maximum dead- and live-load moment. The maximum shear
on these beams due to wind load, as seen from Fig. 445, is 731 lbs. So,
for the maximum moment on the beams due to wind load we have %731x
66 =48,246 inch lbs., which is less than 50 per cent of the combined dead-
and live-load moment and hence will be ignored.

Then for the required section modulus we have 281,335 +16,000=
17.57, which calls for 1—I 9” x 21#, which will be used.

For the maximum end shear on’ the beam we have

Dead load= 8,032 lbs.
Live load= 3,750 lbs.
Total =11,782 Ibs.

 
DESIGN OF BUILDINGS 647

Beams 5. The dead load supported by each of these beams consists
of a story of 12-in. curtain wall, two concentrations from beams 1, and a
uniform load which consists of the weight of the beam and the concrete
encasing it.

The bending moment due to the curtain wall=63,340 inch Ibs., the
same as for beams 2.

The bending moment due to the concentrations = 120,400 inch lbs.,
one-half of that for beams 4.

For the uniform dead load we have the following:

Beam (2—[s) = 25 Ibs. per ft. of beam.
Concrete around beam= 150 lbs. per ft. of beam.

Total = 175 lbs. per ft. of beam.

For the maximum bending moment due to this load we have
= x1%Bx 11 x 12=21,175 inch Ibs.
aL

Adding together all of the above moments we obtain
63,340 + 120,400 + 21,175 =204,915 inch lbs.

for the tetal maximum moment due to dead and live load. Dividing this
by 16,000 we obtain 12.8 for the section modulus which calls for 2—[s
8” x 11.25#, which will be used.

The wind stress is ignored as it is less than 50 per cent of the live-
and dead-load stresses combined.

For the maximum end shear on the beam we have the following:

Dead load= 8,290 Ibs.
Live lJoad= 1,875 lbs.

10,165 lbs.

Beams 5’. The loading on these’ beams is exactly the same as on
beams 5, except for the wall, which is solid, no openings for windows at
all. This wall, as determined for beams 2, weighs 72.6 lbs. per sq. ft.
Then, ignoring arch action, we have 72.6 x 12=871 lIbs., per ft. of beam.
For the moment due to this load we have

a x8Y1x 11 x 12=105,391 inch Ibs.

Now, adding this to the other moments found for beams 5, we obtain
105,391 + 120,400 + 21,175 = 246,966 inch lbs.

for the total maximum bending moment on the beam. Dividing this by
16,000 we obtain 15.4 for the section modulus, which calls for 2—[s 8” x
11.25%—the same as used for beams 5.

For the maximum end shear we have the following:

Dead load=10,604 Ibs.
Live load= 1,875 lbs.
12,479 lbs.

 
648 STRUCTURAL ENGINEERING

Beams 6, 7, 8, and 9. For the sake of uniformity beams 6 will be
the same size as beams 3, although the moment does not require such a
heavy section. Likewise, beams 7, 8, and 9 will be the same size as
beams 1.

Beams 10. For the dead load on these beams we have the following:

Partition = 160 lbs. per ft. of beam.
Plaster = 25 lbs. per ft. of beam.
Concrete slab = 250 lbs. per ft. of beam.
Beam 25 Ibs. per ft. of beam.

al

Concrete around beam= 15 Ibs. per ft. of beam.

‘535 Ibs. per ft. of beam.

Then for the dead-load moment we have
i xB35x15 x 12=180,562 inch Ibs.

The maximum moment due to live load is 84,375 inch Ibs. the same as
for beams 1. Adding, we have

180,562 + 84,375 = 264,937 inch Ibs.

for the total maximum bending moment on the beam. Dividing this by
16,000 we obtain 16.56 for the required section modulus, which calls for
1—I 8” x23#, Metal would be saved by using the 9” x 21# I but it is
desirable to have beams 10 of the same depth as beams 1.

Maximum end shear is as follows:

Dead load =3,960 Ibs.
Live load=1,875 Ibs.

5,835 Ibs.

Beams 1’. The loading on these beams is exactly the same as that
on beams 1, except a partition is supported at mid-span. For this con-
centration we have 160x5=800 lbs. This produces a moment of 400 x
7.5 x 12=36,000 inch lbs. Adding this to the maximum moment on beams
1 we have

208,575 + 36,000 = 244,575 inch lbs.

for the total maximum moment on beams 1’. Dividing this by 16,000 we
obtain 15.3 for the required section modulus, which calls for an 8” x 20.5#
I, which will be used.

Beams 3’ and 4’. Beams 3’ support less load from partitions than
beams 3, but the difference is not sufficient to change the section, so beams
3’ will be the same size as beams 3. Beams 4’ support no partitions and
consequently could be lighter than beams 4, but for the sake of uniformity
they will be the same size as beams 4.

The floor framing for the other floors down to the first floor would
be the same as the tenth floor if it were not for the wind stresses. How-
ever, the work of designing the framing in these floors is exactly the same
as shown above for the tenth floor, and hence the method of procedure is
fully shown above.

As an example, let us consider beams 4 in the fourth floor. The
DESIGN OF BUILDINGS 649

sum of the dead- and live-load moments is 281,335 inch Ibs., as found
above. The maximum shear on the beams due to wind, as given in Fig.
445, is 4,023 lbs. Then for the maximum moment due to wind load we
have

4,023 x 5.5 x 12 =265,518 inch Ibs.

Subtracting 50 per cent of the dead- and live-load moment from this, we
have 265,518 — 140,667 =124,851 inch lbs. So we have

281,335 + 124,851 =406,186 inch Ibs.

for the moment which the beam must be designed to resist. Dividing this
by 16,000 we obtain 25.4 for the section modulus, which calls for a 10” x
30# I, which will be used.

The floor framing on the first floor is designed to support a live load
of 100 lbs. per sq. ft. of floor. The work of designing this framing is
the same as for the other floors.

290. Designing of the Ninth-Story Columns.—As an example,
let us consider column C4. (See Fig. 448.)
For the maximum direct load on this column we have the following:

From roof (constant) = 26,055 lbs.
From beams 3=4,770 x2 = 9,540 Ibs.
From beams 4= 8,032 x 2 = 16,064 Ibs. ¢36,854 Ibs.
Live load =50x15x15 =11,250 Ibs.
Weight of column and concrete
encasing it = 2,400 lbs.

Total = 65,309 lbs.

The maximum direct wind load equals 514 \bs. (negligible) (see Fig.
443). The maximum shears on the column equal 2,626 lbs. and 1,218
Ibs. (see Figs. 443 and 445). Then for the maximum moments on the
column at the brackets we have 2,626 x (6-2) 12=126,048 inch lbs., and
1,218 x 48 = 58,464 inch lbs.

Let us assume the following section:

41s 4x3xy= 8.360”
1—-web 12x33, = 3.750”

Total = 12.110”

Max. 1=294.6 and Min, 7=30:2
Max. r= 4.91 and Min. r= 1.58

For the allowable unit stress for dead and live loads we have 16,000 -
%0x144+1.58=9,620 lbs. Dividing this into the load we have 65,309 +
9,620=6.790” for the required area of cross-section. For the actual
direct stress we have 65,309 + 12.11=5,390 lbs. per sq. in.

For the maximum unit stresses due to cross bending (from the wind
load) we have

f =126,048 x 6 + 294.6 = 2,568 lbs.

and
f’ = 58,464 x 4.15 + 30.2 = 8,038 Ibs.
650 STRUCTURAL ENGINEERING

Adding the maximum of these to the direct stress we obtain
8,038 + 5,390 = 13,428 lbs.

for the maximum combined unit stress on the column. For the allowable
unit stress for the combined dead, live, and wind loads we have 9,620 x
1.5=14,430 lbs. From this it is seen that the above assumed section is
about the correct size and hence could be used.

The other columns in the ninth story are designed in the same
manner as shown for C+. Unless the loads are quite different the columns
throughout the story would be of the same section—using the maximum
section required.

After the columns in the tenth and ninth stories are designed, as ex-
plained above, the columns in the eighth can be designed in the same
manner, and then the columns in the seventh story; and so on down to the
basement columns, using in each case the loads found at the stories above.
As an example, let us consider column C4 in the first story.

For the direct load on the column (see loads on C4 given above), we
have the following: :

From the roof : 26,055 Ibs.
From 10th, 9th . . . 2nd floors = 36,854 x 9 = 331,686 lbs.
Weight of column above and concrete encasing it= 37,800 lbs.

395,541 Ibs.

 

For the direct wind load on the column we will take 15,006 lbs. (see
Fig. 443). i

For the maximum moments due to wind loads we have

14,442 x (9 - 3) 12 =1,039,824 inch lbs.

and *

6,700 x 72 =482,400 inch Ibs.

a, Let us assume the section shown in Fig. 451 which
15 B : : Se
is composed of the following section:

2—[s 15” x 40# = 23.520”
2—[s 12” x 254 = 14.700”

Total = 38.220”

Max. T=1,389, Min. T="11
Fig. 451 Max. r= 6.02, Min. r= 4.3

    

coum ANB cose

N
S

For the allowable unit stress for live and dead load we have
16,000 — 70 x 216 +4.3=12,470 lbs.

Dividing this into the sum of the dead and live-load stresses we obtain
395,541 +12,470=31.70” for the required section for dead and live load
alone.

For the total direct stress—dead, live, and wind loads combined—we

have
395,541 + 15,006 = 410,547 Ibs.
 

 

 

 
    

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651
652 STRUCTURAL ENGINEERING

|

Dividing this by the area of the cross-section we obtain 410,547 + 38.22=
10,740 lbs. for the actual direct stress on the column due to dead, live, and
wind loads combined. For the unit stresses due to cross bending we have

f =1,039,824 x 9.4 + 1,389 = 7,035 Ibs.
and
f’ = 482,400 x 7.5 +711=5,090 Ibs.

Then for the maximum unit stress we have
10,740 + 7,035 =17,775 lbs.

For the allowable stress we have 12,470x1.5=18,705 lbs. for combined
dead, live, and wind loads. From this it is seen that the assumed section is
about correct and hence could be used.

The other columns throughout the building are designed in the same
manner.

291. Calculation of Details.—A fair idea of the details of the
building can be obtained from Fig. 452. As is seen, knee braces are used
throughout instead of the ordinary brackets. The connections of the
beams to the columns in all cases are sufficient to develop the bending
moment.

As an example, let us consider the details connecting beam 3 to
column 4, as shown at (a), Fig. 452. The moment at O, as given in Art.
289, is 129,635 inch lbs. and the end shear is 6,645 lbs., which we will
assume applied at O. Then, for the moment about the center of the
column (see Art. 66), we have 129,635+6,645 x 18=249,245 inch lbs.
Then dividing this by the arm b (15) we obtain 249,245~+15=16,616
Ibs. for the stress in the knee brace, provided there were but one brace,
but, as there are two holding the beam, stress in each knee brace will be
16,616+2=8,308 lbs. As is seen, the stress in the knee brace causes
shear and tension on the rivets connecting it to the beam and column,
So, as a compromise, we will compute the rivets to take the total stress
in each knee brace. Then we have 8,308 +4,420=1.88, say 2—3” field
rivets required to connect each knee brace to the beam; also to the column.
As is seen, 4 are used in order to obtain good connections. We will next
determine the number of rivets required to connect the end of the beam
to the column. Taking moments about O we have

Rx 18=129,635
from which we obtain
R=7,200 Ibs.

for the force exerted along the column by the beam. So we have required
7,200+4,420=1.6, say 2—3” field rivets. As is seen, 4 are used. All
other connections of the beams to the columns throughout the building
can be caleulated in the same manner.

The other details throughout the building are readily calculated,
provided the work previously given in this book is understood.
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656
PROPERTIES OF ANGLES.

 

For Unequal Flanges.

Table-4

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658
n
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For Equal Flanges

Table-6.

c2c. vo fifi-sixy

40140349) en ey
207 40 fifi-sixy
snipoy

ocx so Afi-eixy

40g Apavsg
sayouy Quonbs ur
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Dag vay
sSpunog ut
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ssauyxoiy

221 Ss

659

h pound
from which we obtain

= Section Modulus-(m)

mel

ral axis to extreme fiber:

,

f1
g
L
¥

Stress on extreme fiber:
Then, M
M
f

Distance from nev:
Bending moment in

Iz Moment of tnectia;

y
M
f

 
Areas and Weights of Bars and Flares
A= area of cross-section in sg.inches

Table-7  w- weight in pounds per lineal tt at 490 pounds per curr

“a a”

“7

AI|WI|AI|WIAI|WIA

128 192
470
A

2/2 Zi

, 409
BPE

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100\ Z40\ 125
/

180 | $10

178 | $95 \ 219
2

4$0\$/0
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Zz

263\ 893

475
219
263
206
I50
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ASE
441

765 \2E/
2$0| 850 | BF
zZ 4
300} 10.20
25 | 1.06 \4.06
G50 | “90 \45E
469
4360 £00\1700

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4.

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668

750

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27.
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950

Ww
149
228
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267
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660
Areas and Weights of Bars and Flares
A= area of cross-section In sg.inches
Table-& w- weight in pounds per lineal tt at 490pounds per cu.ft
ness

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AIWI|AILWIAILWIAIWI|AIWI|A |W | eer

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661
Areas, Weights and Max.lengths of Widellares,
A=oareg of Cross S€CTIO/? 117 5G.178-

Ws weight inpounds per Fr
Table-9 Z = meuomiencre In Ft obfainab/e.

77eESsS
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A Ww te W Zz A Ww L

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662
 

 

Toble Jo

Cgproxiimatre Fade

of Gyration of Column Sections.

 

 

 

 

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= 248

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BSSEEEESERSEESEERESERRS

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&

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663
Toble lt

wn

0.375
é.
0.625
oO.
Ou 875,
1. 000

Shearing ond Bearing Valves of Rivets

A rea a pare Bearing Volues Plates @24000/bs.
°
Rive Jf 4 ” ” ” Xe a” ” ” Ge ”
A oO

Oo
36

7500 |9000

Z Volues for bifferent Ihicknesses of Flofes@20000/bs.
Rivet 4, |\% |% |% 14 |%|%|% |% |%|%
04\ 1100

1963 |1260 |2500|3130

2068 | 3070

4418

6010

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664
Allowable Bearing and Moments on Pins
Table /2
A, ; Bea

of Momen or. Moment
Pin metal \at 25 Pin |VoFmeral.

Mu"

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9200

754800

 
 

PAGES

Applied forces, definition of..........e. ees eenee Ria See ketal See: Se
Beams, theoretical treatment of........ iid ne PEERS yDIRAEAR AO eas PRES 51 to 105
Beams, shearing stresses ON... 1... cece eee eee treet tenets 51 to 53
Beams, bending stressCS On... .... ccc ee eee ete eee eens 53 to 56
Beams, reactlon: ONws.csicsed dea ci aww actceoneaneses aa ase eer wares eae are meee 56
Beams, deflection of,..........0seeeeeeeeee ages ERG TREE Mees ade ess 72 to 106
Beams, cantilever ........ Reta Ay ee Ae pute ywle ee BNR sadn Re ARS 75 to 80
Beams, simple ..... DOORN CMEC s BE PAK RARE y De wel Hees 80 to 84
Beams, fixed at one end and supported at other... ........ cee eee eee eee 86 to 90
Beams, fixed ............ Gacy ouaueenelaner Somas a atesietee Se phissaienodudy sc Se ne usnacdyannnat tahoe 90 to 94
Beams; Overhang ge sys) inst dosiertenc nites ines 8-4 Wears es ciect tosh Wy anata Wen RRO ahaa na ela 84 to 86
Deki, CONTINOUS 264 e pee chon sed 2 Lb ne wre eee ee Rep esa pes gtay ch Ke ERR 94 to 105
Beam bridges, railroad ... cnc cece erect er tees yee eras +eeee 198 to 214
Beam bridges, highways 2.2: ac0cavsuss sete ewondes eb e dee ee ee eme eects 540 to 546
Bending stfess, increment of ..... CLE REL SETEL EC OW EkR HS RES See Sek 64 to 65
Bending moment, maximum on Simple beaMS ......-.. cece erence eee 128 to 132
Bending moment, maximum on simple trusseS .........ssee erence reee 138 to 139
Bridges on curveS ........0+++- rea ke “aie aoe heey RNR ala Ie Kader Wt Ge Oana Lee Ga 571 to 575
Buloines. sil vsdo needs ore eC Shoes haem Ore cene ED ee Re Wed ay ee 583 to 624
Buildings, OMce: sicnssgersscereakoawis sdakapgeta gues seas pues 5 624 to 652
Camber of truss@S «is ss¢c0esssivewes ods asewasw enere ees de eew a aaeeae es 518 to 532
Cantilever beams .......+...-+.5 GOIN eReey eee eewae eh seen potest tev iee 75 to 80
CASTES cae ncdndeen Se Kove Radda the RESLES EEE RES eee teen eee 2
Centrifugal. force, streises G06 16:.0ce02004 5 beeen ae bee RS 571 to 575
Center of gravity, determination of ........... Pisriy Micon AN AS Kcvcedaertcnteaete 44 to 47
Columns, theoretical treatment Of 0.6... cece ee eee tenons 106 to 114
Column; formulas: asi: 0s congivemimedna gia Magus Des acagws eamines 106 to 109
Continuous beams, theoretical treatment .......-.. cece eee poreeeee 94 to 105
Conventional signs ......cee eae eeee Sug s Dee RASA gA EAA Eek eRe oe 11
Cover plates, theeretical Tengtti a... cnc ce RRR De ET HORE nes +++.» 181 to 183
Dead load, railroad bridges ....... 0.0 c ccc eee terete eee 197 to 198
Dead load, highway bridges ........ 0c: e cece reece nee tenet e eee neee 534
Deflection: of beams. svsicicce ck caw anne ers take eee Be waee B aes parariatiac scars; 72 to 106
Deflection) Of trusses: ssasanerveseaivtiawacnasce cig de nea ps wows wanes 518 to 532
Design of I-beams and plate girders, theoretical .............. Seales 172 to 194
Design of railroad bridges:

BORD: WELAE ES isc toi eed cere oe 8 ca OM OAS BROT PMO S OR teeta cece 198 to 214

Deck PME GUUCs. ocak > cc hwiicsckenscaeeewedamenece bbe OOERGaame 215 to 259

Through plate girders ......cceecer etter eee en cence rare ten neenae 259 to 295

Wiaducts  aoacaccniain ces do sewed d tim ace F Gsaneniiaicd eo aaa waite 6 09 Be nena aie ae 295 to 324

WTS USS: DBL CL GG sk otras cineca Lae ac Ss aac rsd aaa Ges naa nn RR NRA Sa aa rere gh 1+» 324 to 532
Design of highway bridges: i.

Beam bridges scssxsigccaas svawacernie esa ny as een semaine emRE Oe Ra to 546

Pony trusses .........6- to 552

Through Pratt trusses to 564

Curved chord' through trusses ....... 0... cece eee eee eee ee ene 564 to 569
Design of a 10 ft. beam bridge, railroad ...... aa dees oyeeetnantnlia teria een Maou g 199 to 201
Design of a 15 ft. beam bridge, railroad .......... cc eee e eee eee 201 to 214
Design of a 50 ft. deck plate girder, railroad ............ cee eee eens 216 to 244
Design of a 75 ft. deck plate girder, railroad ................. pees 244 to 248
Design of a 90 ft. deck plate girder, railroad .............0.ee eee e een 249 to 253
Design of a 60 ft. through plate girder, railroad ...................05, 260 to 286
Design of a 150 ft. Pratt truss bridge, railroad .......... eee ee ee ee eee 326 to 413

667
668 INDEX

 

PAGES
Design of a 225 ft. curved chord truss bridge, railroad .............. _,. 426 to 473
Design of buiIGINgS 2... ccc cece eee ee teeter eee rene ne eee ee ennaes 583 to 652
Distortion, definition Of 00.0... ccc cece eee tenet een e eee eens gen 19
Distortion, determination of ............- iideidas SUeaharalayia We Ge ie atte bis eeca & 5) dopa es 20 to 22
Drawing instruments ...........+.+- ieericiitete y'® Haleels reisinighgasain en eee Se 9
Drawings, size Of ........ceeereeae PACKED ODE ET DEMR REA BOM Te OR 11
Drawings; Shop siaissewewssceesnss eyneeas ene eeecee ey MOREE YS SoS eAS SND 11
Drawing Room Exercise No. 1.1... ccc cee tee tenet enn eee 13
Drawing Room Exercise No. 2 2.1... cece cee eee tent ee eens 214
Drawing Room Pxercise No. 3 oc... ee eee ee eee eee fe ee Nears sey enenantn 259
Drawing Room Exercise No. 4.1... .-e pees cece eee eee eee tenet eee e ne 259
Drawing Room Exercise No. 5 ....... ee eee eee eee eee ia) be waste ania aad a 295
Drawing Room Wxercise No, 6...... 0. ccc escent eee renee rene teereenes 324
Drawing Room Bxercise No. 7 1.0... ccc eee etter tne ere erenene 413
Drawing Room Exercise No. 8 ........ cece cece eee eee nee eens 415
Drawing Room Exercise No. 9.1... . ccc eee ee teen eee eee 582
Economic depth of plate girders .......... cece cece ccee tee veeeuvetans 180 to 181
Economic depth of trusses ........... cee ue eeeneee Gal de Aico ulnsacsee ey iaaws Vi 575
Sconomic length of spanS ............ cece ee eee cet eee teens 576
Eccentrically loaded columns ........... 0... ccc eee eee ete teen nee 112 to 114
Elasticity, definition: Of} 4.1. ae2cice <ctureti iow aanisicmne saw eeaegueas a aishobepmeen va 19
Elastic limit ............005 Be rsnicae'a ta) Ss seensartier’s seeter elute gm mrcagte ie ae ph aielenetel sare S 19
Equilibrium, definitiow of g201c6.4 cvccsvos pieherek poe ee4 nay eEwEwESE Se ~ 81 to 235
Equilibrium of couples ...... ccc eee cette tenn een ee eeeeas 35
Equilibrium polygon: « sscsusis caseee essa xeweee eee PETES SE ewe eee n 38
Equilibrium polygon through twe and three points ................+.... 159 to 161
Equation of the elastic Curve ...... 0... cece eect treet eee 72 to 75
PabricatiOn, — .sé0 hei k Sense wtsac Gs eae s 28 tpebieabasaniasieh darandy a Neer neers hot 7
BU XEO MOSS) 538, 5552s g's ovewsrsvsieuaipideade Osaeds GS Aina Raed aod Aceaaiein deers oad 90 to 94
Flange stress in plate girders ......- ieee eaten iso Wein: a0) pics ia tons Meleemeahebaetaesee an ar ag 178 to 180
Flange:-rivets; Spacing: friend ic gee ease wikrssos Mies Ba cee Reanim wae 8 as 8 4... 185 to 187
Tange splice icc bias vet ecw ae Hie LEE RETR Em Re AR RRO RAE «, 254
Flange increment. acccccesies hs hes eniee Fa weal GE b's 26 TETAS WSS Rw wd 184 to 185
Force, definition of .........-.-.05 TSB aA UNAS OTD ee etetts dk elhehyted grt ea 14
Worce,, Measure Ol: asians s oyays pera tagewws baw hese de 6 Ve Rkaiees Diaaediten ane 14
Force, direction of action 0.0... . ccc eect eee ete ence eee eneeeaee 16
Force; Mae Of AetiOn: ccec sac ies cia otticeecne da BA eae wee new S ees als ke Rain 16
Force polygon oo... cece eee eee ete eee ais ects Henin ia ee dee sore eka EOD 26
Graphical statics, theoretical ....... cc cee eee eee tet ee nee neenae 144 to 161
Graphical determination of resultants of parallel forces ................ 35 to 43
Graphical determination of reactions and moments .................005 147
Graphical analysis of deck plate girders ............ shai Aecdeotyr aha sei gs unenariehc atone 255 to 259
Graphical analysis of through plate girders ............... ccc eeeeceee 286 to 290
Graphical analysis of curved chord bridges .............. 0.0 ceueuceee 473 to 481
Horizontal shearing stress in beams ........... cece cee cece ce eeenes 65 to 68
Highway bridges ......... 00. ccc cee eee eee Wiel eove any a aba ayenee a Z to 569
Impact, railroad bridges .............. ce ece eve vee eee
impact, bighwWiy Bridges: 14.4 cepa ed cedwn ad pega ee EEN ERM S ERBERE Goes
Increment of flange stress to 185
Increment of bending stress to 65
Inertia, definition of ..........

Inertia, moment of ......... SEAS CO AE AMOI onc, Bey Gl ahace x Ai wlerenesat: to 50
Influence JimeS (2... ccc eee a eect c eter eenene ee RSE Or ae eee to 171
Length of cover plates (plate SUTGOES Jia, seievaicarseen Gren: at dy Soke So Reo geass grea ees 181 to 184
LOtterie os i icicc essere aie Handa Babee a aripptiedeieuaGlayeugersuho a Aas etic 9 to 10
+ Live load for railroad bridges ........ 0... cece cece cence nce teteenues 195 to 197
Live load for highway bridges ............. ccc een eee enc eee enecs 538 to 535
Length of span (economic) .......... ccc eee eee ae AAMT pariduansiotohwassseAwenenass 576
Manufacture of SteGl seis gs wierd sain tens Ce Ges a NES cae w nae e a Gare anaes 1
Maximum shear on trussed ...... cece cece cece eee e eee n eee ee ene tneyus 134 to 188
Wind load, highway bridges ........¢....-. Pcs de ansbmapsedenaus esaed Soreneng voraonnalie 535

INDEX 669
PAGES
Maximum umoment on trusses ........... EREMAG ESET ids ie eee. 188 to 139
Mill buildings, design of ...... ccc ec eeeereeevereee cbs teak Cisse .. 583 to 624
Modulus of elasticity.......... heroine 3 eee recta ociantl ehalalinin arrliol harealatorenstea iva 20
Moment of a force «1... cece eect eee Poa e ean pares 23
Moment: Of @ COUPIG ca-vesccicie wed boo eGR ERR en dees tds Cape tice aay ab Ee 23
MOOELOR: 2:0 A itietisnaiecise teavevsctuiavacsectiiess sroshisisdyiniad DeatrtOhravariariateesudent eutheraueond opavecisietaaesuereeae 22
Office buildings, design of....... dDednyontedassGeadeaeastas oyvevisuVehianta ua'youi so. cauchionandenskaicaai a+» 624 to 652
Parabola, graphical construction of .......... Scns aha ae asab chase at alsa 61
Pettit trusses, determination of stresses in ......... 0.000 eevee sping ana ah 481 to 502
Pins, determination of stress in 20... . ccc cece eee eee eenae 118 to 120
Plate girders, economic depth .......... 00: cece eee creer tena ents 180 to 181
Pony trusses (highway) ........0 cece ewe e eee e eee r en eee (Qae-aa wee: 546 to 554
Portals, stresseS in ......... 0 cece cues elias eso cos NE ore ssehaepiema gap asciisuseMars 576 to 582
Radius of gyration ..... 0... cece eee eee eee adda ssuSupitaumectass diag ted prcieasaevensas 48 to 50
Mailroad bridges ..........0..0eaee Sdhatiriets Brdnaddiasenjeuedtg Sok Srananenace iehatiwes ie aasciye 195 to 533
Rankine’s column formula .....-. 0. cc cece eee e erence ane fe areniarlelianoner 106 to 107
Reactions, definition of ........ 0. cece eee ee eee er rere TT eres OE
Reactions, on simple beams ...... ccc eee eee eee eet cence ete eee eeeene 124
Reactions, on simple trusseS ...... cece eect ee ee eee ete renee eens 132
Relation of stress and distortion ....... Ledatidaes Depp eieen Asewenee 19 to 20
Relation of shear to cross bending ........ 0. cece cece eee cee tenes 61 to 64
Resultant of parallel forces ..... 00.0 cece cece ree eens iste aaa ws as ese 43 to 44
Rivets and riveting ..........0..ee0ee Psy Bestaluccluentasynewete Adan Pal enecoeactnceenvang 8 to 5
RG, C20 08 a caincs'c dade naee en aids ws aera eae Geeeeee 12
Rivets, conventional: Signs 0s weiwccciidd a sate wedperevarsaiavecarisnetsl cowie iad Ghdavaiseerieiee 11
Rivets, stress on .....-.200000- LL MOL feed RPE eeeea gare Atimeuwe 115 to 117
Rivets, in flanges of plate girders 1.1.0... ccc eee e cette teen teens 185 to 187
Rollers, theoretical treatment ........ Teer eT ere PEE DEN ERE ORR MRT 120 to 121
Shafting, theoretical treatment .iscccceiesieccens seein venivees vienna 121 to 123
Shear on beams .......... cece cee eeeee Siete RpRis pees Sida ode Mint esti 51 to 53
Shear, M@xiniihi OF BeARIE «ccc s cee ed ea nn eene ens ee ee ead +++. 124 to 128
Shear, maximum simple trusses ..........eeeeueaee Bea valisioy cab so vaedananiae nebo 134 to 138
Skew bridges ...00cc0ssscesecnencuns (ARERR ERED Ea RNA Le 570 to 571
Specifications, railroad bridges 2.10... . ce cee cece eee eee eee tenes 195
Specifications, highway bridges .......... cee ee eee eee ence neta 534 to 540
Stiaight line column foriiuld csccenssecsvnanvaaeneew sede ven esaeaenws 108 to 109
Stiffeners .......... CRNAs SSSA OEE NERS Ste Mea Se ee eee assis 190 to 191
Stress, definition Of ........ cece eect te eee ete ree tenet ete e eens 17 to 19
Stress, maximum .......... ie igs tasSvudeanis Meacerss tue seun maniac eupiest apietiena detnae: ied anieecsuantagtns 68 to 72
GS Eres OM: DOM AIS > cyan eases avec gs neriertederearh pial A eeu onsen ovate PARR Alcaduc aL NED CO 51
Stress: OM TIVES) gcicsesrdienis eciveniecuniieind slew aan aeteloane eile eenina inte seers ahr 115 to 117
StresS on truSseS .. 0. cece ee eect eee eeaae euuiiuiden AD. cca a ates seein tae wigpetanaip el trate 139 to 143
Stresses in curved chord bridges ........ 0... cece eee c cree eee ceeenes 415 to 426
Structural material 1.2... ... ccc eee ee eee eee ees ay neaensey ae ay pa Carta ae 1 to 2
Structural shapes .............0065 acca et paca sso se anesthe tl aa. rm sine ea here te ._ 2
Structiiral draughiting: .a.26csasasiaeeeseae ake deeeKeae tose eid. 9 to 13
TracingS ....ee eee e eee ais ta MEE ag AED eg aan SOND BOR taeda dn cach coed Meas poria ce SnD ee Reaves evan wine 11
Truss bridges (railroad) 1.0... .c cece cece eee eee eee e tte eee . 824 to 532
Types of railroad Iridges .... 4 20ereri yee Ree Cer Seen nae 195
Types of bridge trusses 00... .. cece eee teeter te teen nents 324
Vindueta, railrdad seeps: cis viwcvwsasacerdacwccnd spin ta apa aia ah an esa eeal 295 to 324
Warren trusseS 6.0... cee eee eee nee tenet tent ee 502 to 513
. Web splice ........ CES WERKAGE REDE DWE DHE CmB Bans wie ys eue ee RY 189 to 190
Whipple trusses ........-..ee0ee VOTE ALANS EEE ROMO F204 ORS OT od 514
Wind load, railroad bridges ....... cc cece c cece ence eee tent eenes 198
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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