5'
Art. 105. — Flanged Beams with Unequal Flanges.
By the common theory of flexure, if the two coefficients
of elasticity are equal, it has been shown that if C, Fig. i,
is the centre of gravity of the j^~~^ "
cross-section, the neutral axis [^
of . the section will pass through
that point. Let it now be sup-
posed that the lower flange is in a__
tension, while the upper is in com-
pression. Also let T represent
the ultimate resistance to tension
in bending, and let C represent the
same quantity for compression in ' Fig. i.
bending. Then s'nce intensities vary directly as distances
from the neutral axis,
.f3
A,
I'
F
r
h
T
T
^1 =^7^=^'^-
(l)
662
ROLLED AND CAST FLANGED BEAMS. [Ch. XIV.
The ratio between ultimate -ntensities is represented by
n'. U d=h-{-h^ is the total depth of the beam, and hence
^d
If, as an example, for cast iron there be taken
r T J I '
n =— = 0.2, hi =-d.
C 6
The relation between h and h^ shown in eq. (2) is en-
tirely independent of the form of cross-section. But
according to the principles just given, in order that economy
of material shall obtain, the cross-section should be so de-
signed that h and h^ shall represent the distances of the centre
of gravity from the exterior fibres.
The analytical expression for the distance of the centre
of gravity from DF is
ib'a' + (b-b^)f(d-if) + i{b,-b%' '
^1 bd-\-{b-b')f-\-{b^-b')t^ ' ' ' ^^^
The meaning of the letters used is fully shown in the
figure. In order that the beam shall be equally strong in
the two flanges, the various dimensions of the beam must
be so designed that
x^=\. ....... (4)
It would probably be found far more convenient to cut
sections out of stiff manila paper and balance them upon
a knife-edge.
Art. 105.] FLANGED BEAMS IVITH UNEQUAL FLANGES. 663
The moment of inertia about the axis AB, thus deter-
mined, is
I =\W +hih,^ -{h-h'){h-ty -{hi-h'){hi-hY] . (4a)
This value is
now changed^ to
kl
This value is to be substituted in the formula M=^-,
di
For various beams whose lengths are / and total load W
the greatest value of AI becomes :
Cantileve uniformly loaded,
M= — .
2
Cantilever loaded at end,
M = Wl. '
Beam supported a' each end and uniformly loaded,
'"^ 8 8 ■
Beam supported a each end and loaded at centre,
M= — .
4
The last two cases combined,
?)■
Sometimes the resistance of the web 's omitted from
consideration. In such a case the intensity of stress in
664 ROLLED AND CAST FLANGED BEAMS. [Ch. XIV.
each flange is assumed to be uniform and equal to either
T or C. At the same time the lever-arms of the different
fibres are taken to be uniform, and equal to h for one flange
and h^ for the other, h and h^ now representing the vertical
distances from the neutral axis to the centres of gravity of
the flanges, while d--^h-\-h^.
On these assumptions, if / is the area of the upper flange
and f that of the lower, there will result
M=fC.h+rT.h, (5)
But since the case is one of pure flexure,
fC=f'T (6)
.: M=fC(h + h;)=fCd^f'Td. ... (7)
Also, from eq. (6),
f'-C
(8)
Or, tne areas of the flanges are inversely as the ultimate
resistances.
Frequently there is no compression flange, the section
being like that shown in Fig. 2. In such
case h is equal to h' , or t' is equal to zero;
hence h =h' in eq. (4a) , but no other change
1 is to be made in the second member of that
Fig. 2. equation. Eq. (46) may then be used pre-
cisely as it stands for the internal resisting
moment of a beam with the section shown in Fig. 2.
Prob. I. It is required to design a cast-iron flanged
beam of 5 feet effective span to carry a load of 1800 pounds
applied at the centre of span, the section of the beam to
be like that shown in Fig. 2, i.e., without upper flange.
The greatest permitted working stress in compression will
Art. io6.] FLANGED BEAMS IVITH EQUAL FLANGES. 66$
be 8000 pounds per square inch, and the total depth of the
beam is to be taken at 9 inches.
Referring to eqs. (4a), (46), and Fig. i for the notation,
the given data and the dimensions to be assumed for trial
will, be as follows: d = g inches; b=b'=l inch; bi=S
inches; ^1 = 1 inch; / = 5 feet; and C = 8ooo. The intro-
duction of these values into eq. (3) will give for the distance
of the centre of gravity above the bottom surface of the
beam
/ji =2.6 inches and h=d— hi =6.4 inches.
The preceding trial dimensions will make the beam
weigh about 50 pounds per lineal foot. If all the preced-
ing values are substituted in eqs. (4a) and (46), remembering
that M = — , there will be found
4
W = i994 — 125 =1869 pounds.
The trial dimensions, therefore, give the centre-load
capacity of the beam 69 pounds greater than required,
which may be considered sufficiently near to show that the
assumed dimensions are satisfactory.
Art. 106. — Flanged Beams with Equal Flanges.
Nearly all the flanged beams used in engineering prac-
tice are composed of a web and two equal flanges. It has
already been seen that the ultimate resistances, T and C,
^ of structural steel and wrought iron to tension and com-
pression are essentially equal to each other ; the same may
be said a''so of their coefficients of elasticity for tension
and compression. These conditions require equal flanges
for both steel and wrought-iron rolled beams.
666
ROLLED AND CAST FLANGED BEAMS. [Ch. XIV.
I
I
I
H— + .
I
I
I
I
— B-
In Fig. I is represented the normal cross-section of an
equal -flanged beam. It also approximately represents
what may be taken as the section of c
any wrought-iron or steel I beam, the ^ —
exact forms with the corresponding
moments of inertia being given in hand- Y
books. Although the thickness f of the ^
flanges of such beams is not uniform,
such a mean value may be taken as
will cause the transformed section of
Fig. I to be of the same area as the
original section.
Unless in exceptional cases where
local circumstances compel otherwise,
the beam is always placed with the web vertical, since the
resistance to bending is much greater in that position.
The neutral axis HB will then be at half the depth of the
beam. Taking the dimensions as shown in Fig. i, the mo-
ment of inertia of the cross-section about the axis HB is
D
Fig. I.
7 =
(b-t)h-
12
(i)
while the moment of inertia about CD has the value
2fb' + ht'
L =
12
(2)
With these values of the moment of inertia, the general
formula, M =-r, becomes (remembering that di=- or -
bd^-(b-t)h^
or
M=k
M'=k
6d
2fb^-\-ht^
6b '
(3)
(4)
I
Art. 106.] FLANGED BEAMS IVITH EQUAL FLANGES, 667
k is written for all extreme fibre stress.
Eq. (3) is the only formula of much real value. It will
be found useful in making comparisons with the results
of a simpler formula to be immediately developed.
Let di=^{d+h). Since t' is small compared with
-, the intensity of stress may be considered constant in
2
each flange without much error. In such a case the total
stress in each flange will be kht' = Tbf, and each of those
forces will act with the lever-arm ^di. Hence the moment
of resistance of both flanges will be
kht'-di.
The moment of inertia of the web will be — . Conse-
12
quently its moment of resistance will have very nearly the
value
6 ■
The resisting moment of the whole beam will then be
M=k(bt'dy^^^ (5)
A further approximation is frequently made by writing
d^h for h^\ then if each flange area ht' =/, eq. (5) takes the
form
M
-kd^(f + f) (6)
Eq. (6) shows that the resistance of the web is equivalent
to that of one sixth the same amount concentrated in each
flange.
668 ROLLED ^ND CAST FLANGED BEAMS. [Ch. XIV
If the web is very thin, so that its resistance may be
neglected,
M^kfdi^kbfdi, (7)
or
/=s <«
Cases in which these formulas are admissible will be
given hereafter. It virtually involves the assumption that
the web is used wholly in resisting the shear, while the
flanges resist the whole bending and nothing else. In
other words, the web is assumed to take the place of the
neutral surface in the solid beam, while the direct resistance
to tension and compression of the longitudinal fibres of the
latter is entirely supplied by the flanges.
Again recapitulating the greatest moments in the more
commonly occurring cases:
Cantilever uniformly loaded,
M = —=^ . (9)
2 2 ^
Cantilever loaded at the end,
M = Wl (10)
Beam supported at each end and uniformly loaded,
.M = -3-=^g-. ...... (II)
Beam supported at each end and loaded at centre,
Wl
M = — (12)
4
Beam supported at each end and loaded both uniformly
and at centre.
Art. 107.] ROLLED 3TEHL FL.4NGED BEAMS. 669
^=1(^ + 7) (^3)
In all cases W is the total load or single load, while p, as
usual, is the intensity of uniform load, and / the length of the
beam.
Art. 107. — Rolled Steel Flanged Beams.
The resisting moments of all rolled steel beams sub-
jected to bending are computed by the exact formula
k being the greatest intensity of stress (i.e., in the extreme
fibres) at the distance d^ from the neutral axis about which
the moment of inertia / is taken. In all ordinary cases
the webs of beams are vertical so that the axis for / is
horizontal; but it sometimes is necessary to use the mo-
ment of inertia / computed about the axis passing through
the centre of gravity of section and parallel to the web.
The latter is frequently employed in considering the lateral
bending effect of the compression in the upper flange.
The upper or compression flange of a rolled beam
under transverse load, unless it is laterally supported, is
somewhat in the condition of a long column and, hence,
tends to bend or deflect in a lateral direction. This ten-
dency depends to some extent on the ratio of the length of
flange (/) to the radius of gyration (r) of the section about
the axis parallel to the web, as will be shown in detail in
a later article. It will be found there that the ultimate
compression flange stress decreases as the ratio l^r in-
creases. Hence in Table I there will be found values of
l-^r for the different beams tested.
670 ROLLED /iND C/fST FLANGED BEAMS. [Ch. XIV.
The results of tests given in Table I were found by
Mr. James Christie, Supt. of the Pencoyd Iron Co., and
they are taken from a paper by him in the " Trans. Am.
Soc. C. E." for 1884. All beams, both I and bulb, were
loaded at the centre of span. Hence the moment of the
centre load, W, and the uniform weight of the beam itself,
pi, will be, as shown in eq. (13) of Art. 106,
MJ-iw+i^)J4 (2)
a\ 2/ di
Hence
4
k
=T/(^+?) ••••••• (3)
The known data of each test will give all the quanti-
ties in the second member of eq. (3). The two columns
of elastic and ultimate values of k in the table were com-
puted by eq. (3). The positions of the bulb beams (i.e.,
the bulb either up or down) in the tests are shown by the
skeleton sections in the second column.
The coefficients of elasticity E were computed from
the data of the tests taken below the elastic limit by the
aid of eq. (21), Art. 28:
w+m, (4)
4SE1
W being the centre load and pi the weight of the beam,
the length of span / being given in inches.
All beams were rolled at the Pencoyd Iron Works.
The ''mild steel" contained from o.ii to 0.15 per cent, of
carbon, and the "high steel" about 0.36 per cent, of carbon.
These steels are the same as those referred to in Art. 60.
No. 14 is the only test of a "high" steel beam; all the
Art. 107.]
ROLLED STEEL FLANGED BEAMS.
671
remaining tests being with mild-steel shapes. Tests 3 to
9 inclusive were of deck or bulb beams, as the skeleton
sections show.
Beams 3 and 4 were rolled from the same ingot, as were
also 6 arid 7, as were also 10, 12, and 13, and as were also
16, 17, 18, and 19. All beams were unsupported laterally
in either flange. The moments of inertia were computed
from the actual beam sections. The length of span is
represented by /, while r is the radius of gyration of each
beam section about an axis through its centre of gravity
and parallel to its web. The values of r were as follows:
5 inch I . . . .r=o.54inch. 3 inch I,
6 " "....r = o.63 " 8 " "
7 " "....r=o.7i " 10 " "....r = o.95
9 " " r = o.83 " 12 " •• r = i.oi
Table I.
TRANSVERSE TESTS OF STEEL BEAMS.
. .r=o.59 inch.
. .r=o.88 "
Final
k in Pounds per
Coefficient of
Kind of
Span
/
Moment
Centre
Square
inch at
Elasticity E,
No.
Beam.
in Ins.
of
Load in
Pounds.
in Pounds per
Square Inch.
r
Inertia.
Elastic.
Ultimate
I M
ild 3" I
59
100
2.76
5,500
41,100
45,200
30,890,000
2
3" ''
39
66
2.76
8,300
40,800
45,100
25,011,000
3
' s"?
108
200
12
8,800
50,000
55,000
27,7 18,000
4
' 5"i
108
200
12
8,400
46,900
52,500
25,489,000
5
' 6"2
96
152
22
14,860
51,200
54,300
23,692,000
6
^""
69
97
37.6
34,000
47,100
59,300
18,765,000
7
' 7':?
69
97
37.6
34,000
47,100
59,300
23,040,000
8
' 9"?
240
290
84.8
14,500
46,000
51,300
29,923,000
9
' 9" I
240
290
82.9
13,500
39,800
48,800
30,209,000
10
' 8" I
240
273
70.2
13,000
37,600
44,400
28,889,000
II
8""
240
273
70.3
12,930
37,500
44,100
29,055,000
12
8" "
144
164
70.2
19,480
32,800
39,900
31,313,000
13
' 8" "
96
109
70.2
31,300
40,300
42,800
23,689,000
14 H
igh 3""
39
2.74
1 1,500
54,300
■
27,515,000
IS M
ild 10" "
IS6
164
150.5
22,500
35,000
• ■
28,414,000
16
10" "
168
177
150.5
21,000
35,200
27,182,000
17
! 10" II
180
189
150.5
19,500
35,000
29,160,000
18
10" "
192
202
150.5
18,000
34,400
29,727,000
19
' 12" II
240
238
264.7
24,500
33,400
30,749,000
20
1 2" "
240
238
267.6
24,200
32,500
■
29,568,000
21
' 12""
228
226
273-8
22,000
27-500
29,164,000
22
' 12""
216
214
263.7
29,000
35,600
• •
30,219,000
23
' 12""
204
202
256.7
27,000
32,100
. •
30,030,000
24
' I2"|'
192
190
257.8
34,000
38,000
•
29,709,000
25
' 12' "
192
190
262.6
34,000
37,300
28,234,000
26
' 12""
180
178
262 . 4
36,700
37,700
•
27,717,000
27
' I 2" "
168
166
264.0
38,000
36,300
28,784,000
28
' 12""
156
154
261.7
43,000
38,400
27,818,000
672 ROLLED AND Cy^ST FLANGED BEAMS. [Ch. XIV.
The values of k both for the elastic limit and the ulti-
mate are erratic, and the range of results in the table is not
sufficient to establish any law, but on the whole the small
ratios l^r accompany the larger values of k. The bulb
or deck beams also appear to give larger values of k than
the I beams.
The results of these tests indicate that the greatest
working intensities of stress in the flanges of rolled steel
beams may be taken from 12,000 to 16,000 pounds per
square inch if the length of unsupported compression
flange does not exceed 15 or to 2 0or.
In the work of design, the quantity 1 -^a'^ used in eq. (2),
called the ''section modulus," is much employed, and it
can be taken directly from the Cambria Steel Company's
tables at the end of the book, as can the moment of inertia /.
Eq. (2) shows that
/ M . .
j;--F- (5)
Hence the moment of the loading in inch-pounds di-
vided by the allowed greatest flange stress in pounds per
square inch must be equal or approximately equal to the
section modulus of the required beam.
There may be found in the Proceedings of the Ameri(^an
Society for Testing Materials, 1909, the results of tests
of rolled I beams and girders produced by the Bethlehem
Steel Company and of standard rolled I beams by Profes-
sor Edgar Marburg. Also Professor H. F. Moore gives
results of his testing of steel I beams of the regular or
standard pattern in Bulletin No. 68 of the University of
Illinois. Professor Marburg's main purpose appears to
have been to make comparative tests of the ordinary
I beam and of the wide-flange Bethlehem shapes, while
the principal object of Professor Moore was to investigate
Art. 107.]
ROLLED STEEL ELAN GEO BEAMS.
673
the influence of lateral deflection on the capacity of the
compressive flange without lateral support. Table II
gives the results of these tests, each of professor Marburg's
results except one being an average of three.
Table II.
TESTS OF ROLLED STEEL BEAMS.
Span, I
Ft.
I
r'
Extreme Fibre Stress
k, 'Lbs. per Sq.in.
Modulus of
Elasticity.
Size.
Type.
Bias.
Limit.
Ultimate.
Beth. I
Std. I
Girder
Beth. I
Std. I
Girder
Beth. I
Std. I
Girder
Beth. I*. . ..
Girder ......
15
15
15
15
15
11
20
' 20
20
20
125
167
75
125
167
75
129
176
90
III
84
31,700
' 20,400
26,700
21,800
20,600
22,500
20,900
19,500
15,400
13,000
11,800
46,100
42,200
53,900
37,900
34,700
41,100
34,600
33,000
34,300
32,300
31,000
26,900,000
26,200,000
26,900,000
26,406,000
26,900,000
27,200,000
26,400,000
25,800,000
25,600,000
29,400,000
24,800,000
15" 38 1b.
15" 42 "
15" 73"
15" 38"
15" 42 "
15" 73 "
24" 72 "
24" 80 "
24" 120 "
30" 120 "
30" 175 "
* One beam only.
Prof. Moore's fifteen tests were with 8-inch, 18-pound and one 25-pound I beams,
the spans being 5. 7-5. 7-92, 10, 15. i5 7 and 20 feet. The ratio l-^r' varied from 71 to
286. The ultimate fibre stress k was Max. 36,600; Mean 32,300; Min. 28,100. The
Modulus E was. Max. 32,300,000; Mean 28,400,000; Min. 25,100,000. The Max. E is
to be regarded with doubt.
As is the case with all tests of full-size rolled beams,
the results are seen to vary quite widely. This is largely due
to the fact that such full -size members are seldom true
in all their parts, i.e., the web may be a little twisted on the
cooling bed and the flange will perhaps n3ver be perfectly
plane, consequently the applied load in the testing machine
will not be received with presupposed exactness. Again,
the work of the rolls and the effects of cooling will not be
uniform. At any rate the most scrupulous care in testing
will not prevent many erratic results, apparently unac-
countable.
In order to show these results graphically they have
674
ROLLED AND CAST FLANGED BEAMS. [Ch. XIV.
been .plotted on Plate I and the explanatory matter on
the Plate will make clear the results belonging to each
investigator. The horizontal ordinate is the ratio — , r'
being the radius of gyration of the normal section of
the column about a vertical axis parallel to the web and
passing through its centre. The vertical ordinate is the
intensity k of the extreme fibre stress produced by the
ultimate load on the beam as shown in Table I.
The equation
^=39,000-44-,
represents the broken line drawn on Plate I. It is a tenta-
-50-000
-
'
^ -
PI
atel
H
•
•
40-000
•1
•
X
:
__
T
TtT
T
30-000
•1
•1
.X
-- ..
x
-5-
T^
20-000
10-000
Bethlehem I ? „ ..
V Girder \ ^^^^"^^
X Standard I
— Moore
•
Chri
stie
000 -
"i
/-VJ-'=40
120
200
220
tive expression, as there are not sufficient tests with the
requisite variation of — to justify more than a trial value of k.
Art. 107.] ROLLED STEEL FLANGED BE/IMS. 675
Professor Marburg made no effort to give lateral sup-
port to his beams under test, nor did he endeavor to give
the compressive flange lateral freedom, as did Professor
Moore for a part of his tests. As, however, the results
appear to be about the same, whether the compressive flange
has complete lateral freedom or not, under ordinary cir-
cumstances of testing, no distinction is made on this account
between the various plottings on Plate I. The extremely
high values on that Plate belong to the first nine tests
by Mr. Christie, as given in Table I. They are abnormally
high and whether such results are characteristic of bulb
sections or due to some other reason is not clear.
Prob. I. It is required to design a rolled steel beam
for an eftective span of 20 ft. to carry a uniform load of
725 lbs. per linear foot in addition to the weight of the beam
itself, the circumstances being such that it is not advis-
able to use a greater total depth of beam than 12 ins.
The greatest permitted extreme fibre stress k will be
taken at 12,000 lbs. per sq. in. It will be assumed for
trial purposes that the beam itself will weigh 35 lbs. per
linear foot, so that the total uniform load will be 760 lbs.
per linear foot. The centre moment in inch-pounds will,
therefore, be
^^ 760X20X20X12 ^ . „
J\I =■ ^ =456,000 m. -lbs.
By eq. (5) the section modulus will be 456,000^-12,000
= 38. By referring to the tables in almost any steel com-
pany's handbook it will be found that this section modulus
belongs to a 12-inch, 35-pound steel rolled beam, and
that beam fulfills the requirements of the problem.
Prob. 2. It is required to design a rolled-steel beam
for a 3 2 -ft. effective span to carry a load of 1280 pounds per
linear foot in addition to the weight of the beam, and a
676 ROLLED /IND CAST FLANGF.D BEAMS. [Ch. XIV.
concentrated load of 1 1 ,000 pounds at a point 1 1 feet distant
from one end of the span. The greatest permitted work-
ing stress in the extreme fibres of the beam is 16,000 lbs.
per sq. in.
It will be assumed for trial purposes that a 24-in. beam
weighing 95 lbs. per linear foot will be required so that
the total uniform load per linear foot will be 1375 pounds.
It will then be necessary to ascertain at what point in the
span the maximum bending moment occurs, i.e., at what
point the transverse shear is equal to zero. Let a be the
distance of the concentrated weight from the nearest end
of the span, i.e., a = 11 ft. Then 'et P be the single weight,
p the total uniform load per linear foot, and / the length
of span. The following equation representing the condi-
tion that the transverse shear must be equal to zero may-
be written
pi Pa
Hence x = - +—7 .
2 pi
In the above equation x is obviously the distance from
that end of the span farthest from P to the section of
greatest bending moment. Substituting the above numeri-
cal values in the equation for x, there will result
:r = i6 + 2.75-i8.75 ft.
Since 32 — 18.75=13.25 the following will be the value
of the greatest bending moment in inch-pounds:
.^ y 1375X18.75 11,000X11 „ ,
M^y-^^ ^X 13.25 + X 18.75 j 12
= 2,900,363 inch-pounds.
Art. io8.] DEFLECTION OF ROLLED STEEL[ BEAMS, 677
The section modulus of the beam required is by eq. (5)
2,900,363^16,000 = 181. The section modulus of a 24. -in.
steel beam weighing 85 lbs. per linear foot is 180.7, a-s will
be found by referring to the tables at the end of the book.
Hence that beam will be assumed for the correct solution
of the problem. The fact that the beam w^eighs 10 lbs.
per linear foot less than the assumed weight has too small
an effect upon the greatest bending moment to call for
any revision.
Prob. 3. A steel tee beam of 8 ft. span is to be used as
a purlin to carry a uniform load of 125 lbs. per linear foot
with the web of the tee in a vertical position. The greatest
permitted intensity of stress in the extreme fibre of the
tee is 14,000 lbs. per sq. in. It is required to find the
dimensions of the tee. By referring to eq. (5) the section
modulus will be written
1000X96 Q..
S = ^- = . 86 m.
0X14,000
By referring again to the steel handbook tables it
will be found that a 3 X3 XA in. steel tee weighing 6.6
lbs. per lin. ft. has just the section modulus required.
That tee therefore fulfils the requirements of the problem.
Prob. 4. It is required to support a single weight of
12,000 lbs. at the centre of a span of 13 ft. 6 ins. on two
rolled steel channels with their webs in a vertical position
and separated back to back by a distance of 3 ins., the
greatest permitted intensity of stress in the extreme fibre
of the flanges being 15,000 lbs. Find the size of channels
required.
Art. 108.— The Deflection of Rolled Steel Beams.
The deflections of rolled steel beams may readily be
computed by the formula of Art. 28. The general pro-
678 ROLLED AND CAST FLANGED BEAMS. [Ch. XIV.
cedure will be illustrated by using the equations for a non-
continuous beam simply supported at each end and loaded
by a weight at the centre of span, or uniformly, or in both
ways concurrently. Eq. (20) will give the deflection at
any point located by the coordinate x, while eq. (21) will
give the centre deflection only. The tangent of the in-
clination of the neutral surface at any point located by x
dzv
will be given by the value of -t~ found in eq. (19).
Prob. I. Let the centre deflection of the rolled-steel
beam of Prob. i of Art. 107 be required. Referring to
eq. (21) of Art. 22,
W = o; /== 20 feet = 240 inches; /? = 760 pounds;
7 = 228.3; and E may be taken at 29,000,000.
Hence the centre deflection is
240X240X240X5X760X20 . .
^^ = 48X8X29,00-0,000X228.3 = • 4'4 mch.
If half the external uniform load of 725 pounds per
linear foot had been concentrated at the centre of span,
^^_ 725X20 ,
W =-^—^ ■ =7250 pounds; p =
35
2
and I = 20 ft. =240 ins. Also pi = 700 pounds.
Hence the centre deflection would be
240 X 240 X 240 X (72 50 + 437 • 5) ..,• ^-u
w, = :^ 7; = . ^ ^ ^ men.
1 48X29,000,000X228.3 ^^^
Prob. 2. In Prob. 2 of Art. 107 place the 11,000-pound
weight at the centre of span, then find the inclination of
the neutral surface and the deflection of the 24-inch 85-
Art. 109.] IVROUGHT-IRON ROLLED BEAMS. 679
pound steel beam at the centre and quarter points of the
32-foot span, taking £^ = 29,000,000 pounds.
Art. 109. — Wrought-iron Rolled Beams.
Although wrought-iron rolled beams are not now manu-
factured, being cofnpletely displaced by steel beams, yet
many are still in use. Hence it is advisable to exhibit
the empirical quantities required to design them and to
determine their safe carrying capacities as well as their
deflections under loading.
It has been observed in Art. 107 that the upper or com-
pression flange of a loaded flanged beam will deflect or
tend to deflect laterally at a lower intensity of compressive
stress as the unsupported length of such a flange is in-
creased. The experimental results given in Table I ex-
hibit the values of the intensity of stress K in the extreme
fibres of the beam both at the elastic and ultimate limits,
the usual formula for bending resistance being used,
«=f <-)
In the autumn of 1883 an extensive series of tests of
wrought-iron rolled beams, subjected to bending by centre
loads, was made by the author, assisted by G. H. Elmore,
C.E., at the mechanical laboratory of the Rensselaer Poly-
technic Institute. The object of these tests was to dis-
cover, if possible, the law connecting the value of K for
this class of beams with the length of span when the beam
is entirely without lateral support. The means by which the
latter end was accomplished, and a full detailed account
of the tests will be found in Vol. I, No. i, " Selected Papers
of the Rensselaer Society of Engineers." The main results
of the tests are given in Table I. All the tests were made
on 6 -inch I beams with the same area of normal cross-
68o
ROLLED AND CAST FL/INGFD BEAMS.
Table I.
[Ch. XIV.
K
Final
Perm'nent
Perm'nent
E
No
Span,
Centre
/.
Vertieal
Lateral
Pounds per
Square Inch.
Feet.
Weight,
Pounds.
r
Elastic
Limit,
Ultimate,
Pounds.
Deflection,
Inches.
Deflection,
Inches.
Pounds.
I
- 20
4,060
400
27,726
31,094
0.14
24,170,000
2
4, 200
400
29,623
32,885
0.30
26,374,000
3
I ^^
4.390
360
28,264
30,791
0.2
0.5
24,520,000
4
4,570
360
28.264
32,020
0.18
0.4
24,313,000
5
[i6
4,770
320
26,564
29,579
0.28
1. 00
25,771,000
6
5,270
320
29,596
32,632
0.48
1.25
25,003,000
7
[14
6,130
280
31,191
33,049
0.30
I .20
26,082,000
8
6,125
280
31,164
33,023
0.30
I .10
23,373,000
9
/■ 1 2
7,161
240
30,221
32,907
0.35
1.08
25,287,000
lO
7,350
240
31,314
33,817
0.33
I .09
24,022,000
II
i ^°
9,255
200
33,082
35,358
0.39
1.08
25,115,000
12
9,655
200
33,082
37,064
0.50
1.50
24,218,000
13
[«
11,485
160
29,736
35,010
0.30
0.90
21,61 1,000
14
11,980
160
31,936
36,527
0.29
I .05
21,987,000
15
[ ^
18,300
120
35,497
41,737
0.605
1-53
23,040,000
i6
18,145
120
36,617
41,396
0.67
1.88
20,935,000
17
• 5
22,870
100
34,136
43,434
0.67
1-75
22,023,000
i8
23,065
100
34,136
43,813
0.67
1-75
25,272,000
19
!-
29,985
80
32,619
45,532
0.96
1.70
24,315,000
20
28,585
80
32,619
44,744
0.60
1.86
21,275,000
section of 4.35 square inches. Actual measurement showed
the depth d of the beams to be 6.16 inches. The moment
of inertia of the beam section about a line through its
centre and normal to the web was 7 = 24.336. The radius
of gyration of the same section in reference to a line through
its centre and parallel to the web was r = o.6 inch. / was
the length of span in inches.
If M is the bending moment in inch-pounds, W the
total centre load (including weight of beam), and K the
stress per square inch in extreme fibre, the following
formulas result:
k^
Md
2I
and M =
Wl
k =
Wld
8/ •
(2)
(3)
Art. 109.]
IVROUGHT-IRON ROLLED REAMS.
681
The experimental values of II \ /, d, and / inserted in
the above formula give the values of k shown in the table.
The coefficient of elasticity, E, was found by the usual
formula,
in which w is the deflection caused by W.
The full line is the graphical representation of the values
of k given in Table I. Since k must clearly decrease with
PidLe I.
1 1 1^ 'i ' . 1 1 ■" ' ! — ■■ ~ ~1 ' ■; rn 1 'T- ■■ ■ III 1
M i M 1 1 1 1 i ' ' i :
1 i t - 1 11 r- -r ]-■■■■ 1 t 1
snnt a 1 V- 1 M ' ! ■ ' '
! ; 1 !>^>^^^' ' ■ ' , , , Ml, I ■ i M i 1 1 M M M ■ I M I 1 11 1 i -y- --1
,,,nfin ■ ' Ml' ' 1 M ' fSt'-i.' ' i ' i 1 II ' ' ' ' ' ' ' 1 1 M ' ! 1 ' M ' ' ' Mi' '
^OOi 1 Mil i , M 1 1 r'^-i>^. Mm ' ' M 1 i M i M 1 1 1 MM ■ M M 1 i
1 M M ' ' ' 1 ' '^~^riS-.' 11 M M 1 1 M ' M 1 1 1 1 1 1 1 1
1 \ \ \\ Ml' ' ' M M M^-^-'t. M 'Ti M ' ' M M 1
i MM Ml. . , 1 1 i ! >>T>>J V 1 1 1 M 1 ■ 1
■ ■ --'■ 1 i 1 , : \U\ i^i'SjS-. 1 Mil'' 1 1 1
. 1 • ' M •■ ' 1 i ' ' ' ' ■ ' rKjJs^i 1 M 1 1 ' : : M 1 1 1 M
-r^^ nVH --^^^U^ ' -M4+ + ^^ H--
\ 'rn h
i 1 1 1 III' (flT^ r^-M-^ ' Wl^jAl ' M ■ ' 1 'ii
cUUUU- ■ III 1
LllJl._--_-_--------l-±-^4---.-^.^^4± im^ irl -:^#F^
- M M M M 1 M 1 LI M 1 " ■'■ ' 1 1 M i ' 1 i M 1 M 1 1 ■ +
t 1 1-L M \\\ M iff> a^n M'tiO 1 \ '"■n "on i M ' ^^p M \m
the length of span, and increase with the radius of gyration
of the section about an axis through its centre and parallel
to the web (the latter, of course, being vertical), k has
been plotted in reference to l^r as shown. No simple
formula \yill closely represent this curve, but the bioken
line covers all lengths of span used in ordinary engineering
practice, and is represented by the formula
^=51,000-75-
(5)
For raiVay structures the greatest allowable stress
per square inch in the extreme fibres of rolled beams may
be taken at
^ = 10,000-15--. (6)
682 ROLLHD AND CAST FLANGED BEAMS. [Ch. XIV.
Values of k taken from a large scale plate, like Plate I,
are, however, far preferable to those given by any formula.
The ultimate values of k given in Table I are fairly
representative of the best wrought-iron I beams. The
coefficients of elasticity E range from about 22,000,000 to
about 25,000,000 pounds; the average may be taken about
24,000,000 pounds.
The deflection of wrought-iron beams may be computed
by the formula
WP . .
^ = ^8E7' ^^)
when the load W is at the centre of the beam. In the
general case of a beam carrying the centre load W and
the uniform oad pi, the quantity {W -^Ipl) must displace
W in eq. (7). If the beam carry only the tiniform load pi,
W in eq. (7) must be displaced by \pl.
If it is desired to apply the law expressed in eqs. (5)
and (6) to mild-steel beams, the second members of those
equations may be multipHed by | to f for close approxi-
mations.
CHAPTER XV.
PLATE GIRDERS.
Art. no. — The Design of a Plate Girder.
A PLATE girder is a flanged girder or beam built usually
of plates and angles, the flanges being secured to the web
by the proper number of rivets suitably distributed. The
flanges, unlike those of rolled beams, are usually of vary-
ing sectional area, although occasionally either flange may
be of uniform section throughout when formed of two
angles, or two angles and a cover-plate. Fig. i is a general
view of a plate girder, while Figs. 2, 3, 4, and 5 show
some of the general features of design.
The total length of a plate girder is materially more
than the length of clear span over which the girder is de-
signed to carry load. Blocks or pedestals of masonry or
metal, as the case may be, support the ends of the girders
and rest on the masonry or other supporting masses or
members carrying the girder and its load. The distance
between the centres of these blocks or pedestals is called
the effective span of the girder, as it is the span length
which must be used in computing bending moments,
shears, or reactions. Plate girders must evidently be
somewhat longer than the effective span. >In the Figs, the
relations of the various parts at the end of the plate girder
are shown in detail. The girder illustrated in Fig. i has
683
684 PLATE GIRDERS. [Ch. XV.
an effective span of 68 ft. with the centre of the pedestal
block 15 inches from the face of the masonry abutment
and 12 inches from the extreme end of the girder. The
effective depth of the girder is the vertical distance or
depth between the centres of gravity of the two flanges.
When the girder has cover-plates this effective depth may
be greater than the depth of web plate at the centre of
span and less than that at the ends, even when the web
plate is of uniform depth. It is always customary, how-
ever, to take the effective depth of a plate girder with
uniform depth of w^eb as constant. Frequently that depth
is taken equal to the depth of the web plate; or, again, it
may be taken equal to the depth between the centres of
gravity of the flanges at mid-span without sensible error.
In case the web plate is not of uniform depth the effective
depth might still be taken as the depth of web plate at
the various sections of the girder, or it may be taken as the
depth between centres of gravity of the flanges at the same
sections.
The plate girder shown in Fig. i and to be assumed for
the purposes of design is of the deck type and has a clear
span of 65 ft. 6 ins., an effective span of 68 ft., and a length
over all of 70 ft. The differences between the effective
span and the clear span and total length are obviously
dependent upon the length of span. For short spans
those differences are relatively small, and relatively large
for long spans. The depth of w^eb plate will be taken as
6 ft. 8 ins., and it will be found later that at and in the vicin-
ity of the centre of span three cover-plates will be needed.
The girder will be assumed to be of mild structural steel
and will be supposed to carry a single-track railroad mov-
ing load with the concentrations and spacings shown in
Table I, Art. 21.
The dead load or own weight of the girder and track
will depend somewhat upon whether the girder is of the
^1-7J<^-*
i-li--ipr-i:i)-
, Sole PI. ll"x ^ X 1 6 PC
•Remainder of Bott. li
as Top Flange excexi
' holes to be shop rivi
,3* . 5K
»l«l*C*|<-*k
-IVA
4%
4_|_l_c4£!f.^,
pQ-ffl-l
'6-4-1
i
-(p-OffiO oo-
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O— 0-
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\-T-T-t-T-9'
H-|-oo-i"o-o6-H-l-
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• " |:^:t-i
{To face page 6'oS-^
Art. no.] THE DESIGN OF A PLATE GIRDER. 685
through or deck type. The only difference in computa-
tion arising in those two types is due to the fact that if the
girders are of the deck class (i.e., carrying the moving load
directly on their upper flanges) the rivets connecting the
upper flanges with the webs must be assumed to carry the
wheel concentrations in addition to their other duties, as
will be shown in the following computations. The total
dead load or own weight will be taken as 1400 lbs. per linear
foot. Inasmuch as there are two girders, each will carry
one half of the moving load and one half of the dead load
or own w^eight. It should be observed that the effective
length of span being 68 ft., the two locomotives at the head
of the train load will more than cover the span, so that the
uniform train load will not appear in the computations.
The design of this plate girder will be made in accord-
ance with the provisions of the American Railway Engi-
neering and Maintenance of Way Association and refer-
ences will be made to those provisions.
Bending Moments.
The first computations necessary are those required
to determine the bending moments, and from them the
flange stresses at different points of the span. Those
points may be taken at 5, 8, or 10 ft. apart as may be
desired for the purpose of design; the closer together the
sections are taken the greater will be the degree of accuracy
attained. In the present instance those sections will be
taken 5 feet apart up to 25 ft. from the end of the span,
but the next or final section will be at the centre of span.
After the bending moments are obtained, the flange
stresses at once result by dividing the former by the efl^ec-
tive depth.
Figs. I and 2 show the complete single-track railway
686 PLATE GIRDERS. [Ch. XV.
deck-plate girder span consisting of two girders with the
requisite bracing connections between them. The total
dead load or own weight is a uniform load and consists of:
Lbs. per L-n. Ft.
Track (ties, rails, etc.) 450
Two girders and bracing 1050
Total 1500
Or for one girder -^ — = 750
2
As each girder will carry 750 lbs. of dead load per linear
foot, and as the effective span is 68 ft., the expression for
the dead-load bending moment in foot-pounds at any
point will be as follows :
M=^^(68x-x2) (i)
2
The application of eq. (i) to the sections of the girder
5, ic, 15, 20, 25, and 34 ft. from the ends will give the
following expressions for the bending moments in foot-
pounds :
D. L. Moment.
X Ft. Lbs.
5 . . . . .' 118,120
10 • . 217,500
15 298,100
20 360,000
25 403,100
34 433,500
The moving-load bending moments are next to be found
by using the concentrations shown in Table i, Art. 21.
For this purpose the criterion for the maximum bending
Art. no.] THE DESIGN OF A PLATE GIRDER. 687
moment, cq. (5), Art. 21, must be applied at the assumed
sections in which V (equal to x in the above dead-load
computations) has the values 5, 10, 15, 20, 25, and 34 ft.
The application of that criterion to the section BO, Fig. i,
5 ft. from the end of the span shows that W2, or the first
driving wheel, must rest at the section in question for the
maximum bending moment, the loads W2 to 1^12 inclusive
resting on the span. Wi will be off the span. By the aid
of Table i, Art. 21, the greatest bending moment desired
is:
Ms =^^(9,030,000 + 2 X 273,000) =704,000 ft. -lbs.
68
Similarly for the section CN, 10 ft. from the end of the
span, the criterion eq. (5) of Art. 21 shows that 1/^3 must
be placed at C with W12 2 ft. from the end of the span
and Wi off the span. By the aid of Table i the desired
moment takes the value:
Mio= — (9,030,000-^-2 X273,ooo) — 150,000= 1,260,000 ft. -lbs.
60
Concisely stating the conditions and results for the
remaining sections shown on Fig. i: For DL, 15 feet
from end of span, two positions of moving load, I/F3 at D
and W12 at D satisfy the criterion, but the latter with
13 feet of uniform train load on the span gives the greatest
moment. Total load on the span is
(1^10+ . . . +VF18+3000X13)
and the moment is:
^15=— (6, 3 10, 000 +2 13, 000X13 +3000 X — j -345,000 =
1,715,000 ft. -lbs.
088 PLATE GIRDERS. [Ch. XV.
For EM, 20 feet from end of span, place 1^12 at E\
M20 ==^(6, 310, 000+8(213, 000 H 3£oo\\ _^^^ QQQ^
2,040,000 ft. -lbs.
For GH, 25 feet from end of span, place W12 at G and
the moment is :
M
25
25/
000+232,500X3+3000—) -755.000 =
2,265,000 ft. -lbs.
The moment at the centre of the span can be computed
in the same manner, but by referring to Table II of Art.
2 1 , it will be seen to be :
M34 = 2,435,400 ft. -lbs.
A reference to the American Railway Engineering
and Maintenance of Way Association specifications, Art.
9, will show that the required allowance for impact is
represented by the factor 7, in which V is the length of
load on the span:
1=^
300
L'+30o*
The positions of loading already found for the greatest
moving load moments give the lengths V in feet in the
following table:
Pt.
Loaded
Length, L '.
Impact
Moving Load
Impact Moment
Ft.
Factor I.
Moment, Ft. -lbs.
Ft.-lbs. ■
5
63
.827
704,000
582,000
10
63
.827
1,260,000
1,040,000
15
66
.820
1,715,000
1,405,000
20
61
.897
2,040,000
1,830,000
25
64
.825
2,265,000
1,866,000
34
68
.815
•
2,435,000
1,985,000
Art. no.] THE DESIGN OE A PLATE GIRDER. 689
By adding the dead load or own weight moments,
already computed, to the moving load and impact moments
in the preceding table, the total or resultant moments
will be:
Table I.
T34- Total Moment
^^- Ft.-lbs.
5 1,404,000
10 2,518,000
15 3,418,000
20 4,230,000
25 4,534,000
34 4,855,000
Shears.
Both dead and moving load shears must be computed.
As the dead load or own weight is a uniform load on the
girder, the shear at any point is simply the load between
that point and the centre of span. Hence indicating the
transverse shear at any section by the figure showing its
distance from the end of the span, there will result the
following values, 5o being the end shear or reaction:
5o =34 X750 =25,000 lbs.
S5 = 29 X750 = 21,750
5io = 24 X750 = 18,000
5"i5 = i9X75o = i4,25o
520 = 14X750 = 10.500
525= 9X750= 6,750
534= 0X750= o
The moving load shears will also be needed. Although
there is no systematic criterion for such shears at different
690 PLATE GIRDERS. [Ch. XV.
points in a span traversed by a train of concentrations,
it is a simple matter to find the greatest moving load shears
at the sections contemplated by inspection and trial. The
greatest end shear, i.e., the greatest reaction, has been
found in Art. 21 and is given in Table II of that Article:
End shear for 68-ft. span = 161,700 lbs.
End impact shear =131,800 ''
The impact factors for the shears are computed by the
same formula already used for impact moments.
For a shear 5 feet from end of span: place W2 at the
5 -foot section, then the greatest shear is
c, 9,030,000 + 2X273,000 .,
55 = ^ ^ 7^ -^ = 141,000 lbs.
Do
By trying other positions it will be found that this
gives the greatest shear. Wi is not on the girder and W12
is 2 feet from the end of the span.
For section 10 feet from end: place Wn at the section.
Hence
6,310,000 + 213,000X13 -1 3000X-—
5 0=- ^^ ^ -150.000 =
122,000 lbs.
For section 15 feet from end: place Wn at the section
r.nd there will result
g2
6,310,000 + 213,000X8 +3000 X —
S , = = 1 so, 000 =
104,300 lbs.
Art. no.
THE DESIGN OF A PLATE GIRDER.
691
For 20-ft. section: place W2 at the section and there
will result
c^ 6,QS0,000 ,,
020 = ^^-^ 150,000 =87,200 lbs.
68
For a 2 5 -ft. section: place W2 at the section and the
greatest shear will be
^ S. 240, 000 + 213, 000 X3 11
525=^^-^*^^ T^ ^-150,000 = 71,500 lbs.
Oo
For the centre of span : place 1^2 at that point and the
greatest shear will be :
^ ^,230,000 + 174,000X5 1,
534=^^^-^^^ t:^ ^-150,000=45,300 lbs.
Oo
The loaded lengths in each of these cases to be used
in computing the impact factors are in the order of the
sections beginning with that at 5 feet from the end, 63,
66,. 61, 56, 51, and 42 feet, the latter belonging to the
centre of span. The following tabular statement repre-
sents the elements of these moving load shears and the
impact allowances:
SHEARS AND IMPACT ALLOWANCES
Section.
Loaded
Impact
Moving Load
Impact Shear.
Length. Ft.
Factor.
Shear. Lbs.
Lbs.
5
63
.8.7
141,000
116,500
10
66
.820
122,000
100,000
15
61
■831
104,300
86,600
20
56
.824
87,200
71,800
25
51
•855
71,500
61,100
34
42
•877
45,300
39,700
692 PLATE GIRDERS. [Ch. XV.
Adding together the dead load, moving load and impact
shears as now determined, the following will be the resultant
or total shears at sections under consideration:
Table II.
RESULTANT OR TOTAL SHEARS.
Section. lotal^hears.
End . 319,000
5 279,300
10 240,500
15 205,100
20 169,500
25 139,400
34 85.000
The preceding results or computations due to the dead
and moving loads are the principal data required in the
design of the girder.
Weh Plate.
The effective depth of the girder will tentatively be
taken as 6 feet 8 inches and the depth from the back of
flange angles in the upper flange to the back of the lower
flange angles will be taken as 6 feet 8| inches. As the
depth of the web plate must be taken a little less ihan the
depth from back to back of angles, in order that the flange
plates may not touch the edges of the web plates when
the different parts of the girder are assembled, that depth
should be taken as 6 feet 8 inches. In fact the effective
depth of a plate girder is sometimes prescribed as the depth
of the wxb plate. This depth of web plate will leave
\ inch clear at the top and bottom flanges, which is sufficient
to insure the flange plates freedom from hitting the edges
of the web.
Art. 18 of the Specifications allows a working stress
in shear of 10,000 pounds per square inch of gross cross-
Art. no.] THE DESIGN OF A PLATE GIRDER. 693
section of the web. As the total end shear has been
found to be 319,000 pounds, the gross web plate section
at the end of span should be 31.9 square inches. The
minimum thickness must then be '-^^7^ = .399 inch.
80
A web plate 80 X-^ inch will be used, giving a gross
16
sectional area of 80 X. 43 75 =35 square inches. The sur-
plus area is small and it is judicious design to have it.
This web plate thickness also satisfies Art. 29 of the Speci-
fications which prescribes that ' ' The thickness of web
plates shall not be less than of the unsupported dis-
160
tance between fiange angles," as 6X6 inch flange angles
will be used,
Flanges.
Art. 29 of the Specifications provides that the design
of the flanges may be based either on the moment of inertia
of the net section of the girder or on the assumption that
the flange stress is of constant intensity with its centre
at the centre of gravity of the flange area, the latter
including one-eighth of the gross section of the web, the
difference between one-sixth and one-eighth of the w^eb
section being supposed to cover the material punched out
in the tension side of the web plate. The latter method
will be employed.
Art. 30 of the Speciflcations provides that '' The gross
section of the compression flanges of plate girders shall
not be less than the gross section of the tension flanges."
It will be best, therefore, to design the tension flange
first.
Using the total or resultant bending moment at the
694 PLATE GIRDERS. [Ch. XV.
centre of the span, the trial effective depth of 6 feet 8 inches
will give the centre flange stress as follows:
4,855,000 „ ,.
- ^^' = 728,000 lbs.
6.07
The specifications permit a working tensile stress in
the net section of the tension flange of 16,000 pounds per
square inch. Hence the required net tension flange area
is
728,000
16,000
= 45.5 sq.ms.
The available flange section due to one-eighth the gross
•7 r
sectional area of the web is ^ =4.375 square inches. The
8
amount of flange area to be supplied by the flange plates
and angles is, therefore,
45-5 -4-4 =41-1 sq.ins. .
In providing 41. i square inches it is necessary to know
what rivet holes are to be deducted from each cover-plate
and each flange angle. It is clear that two rivet holes
only need be deducted from each cover-plate, and it is plain
that at least two rivet holes must be deducted from each
flange angle section. In designing cover-plates for flanges
it must be remembered that no such plate must be thicker
than the one under it, i.e., if these plates are not of the same
thickness, the thickest one must lie on the angles, the
remaining thicknesses -to decrease or be the same in passing
outward from the angles. As a trial section let the follow-
ing be assumed:
I
Art. no.
THE DESIGN OF A PLATE GIRDER
695
Angles or Cover-plates.
Gross Area.
Sq.Ins.
Less Rivet Holes.
Sq.Ins.
Net Section.
Sq.Ins.
2 6"X6"xr'
3 covers 14" Xl". ..
16.88
315
4XiX!=30
6XiXf=4-5
13-88
27.00
48.38 •
40.88
As 40.88 square inches is but ij per cent, less than the
desired area, 41.4 square inches, the former may be accepted
subject to further confirmation.
If the centre of gravity of the gross section of the tenta-
tive flange area consisting of the three plates and two
angles indicated above be determined, it will be found
.11 inch above the back of the angles. This will make the
effective depth
6 ft. 8.5 ins. +.22 in. = 6 ft. 8.72 ins.
This increase in effective depth will correspondingly
decrease the centre flange stress so as to make the total
actual net area of 45.3 square inches a little larger than
required. Hence the trial centre tension flange area as
determined above will be accepted as the actual flange area
to be used, i.e., three i4Xi-inch cover-plates and two
angles 6 X 6 X | inch.
Length of Cover-plates.
In the next Article there will be shown two methods
of determining the lengths of cover-plates after the
sections of those plates have been found for the greatest
bending moment, usually taken as at the centre of span.
These two methods are simply different forms of expres-
sion of the same thing. The following notation will be
•used :
696 PLATE GIRDERS [Ch. XV.
Z= length of span in feet;
L\ = length of outside cover-plate in feet;
L2 = length of second cover-plate in feet;
A = total net flange area, square inches ;
a\ =net area of outside cover-plate, square inches;
a2 =net area of second cover-plate, square inches;
as =net area of third cover-plate, square inches.
It has already been seen that if a beam simply supported
at each end be loaded uniformly throughout the span, the
bending moment at any point will be represented by the
vertical ordinate of a parabola whose vertex is over the
centre of span while the end of each branch is at one end
of the span. It is assumed that the greatest bending
moments in the plate girder, already computed, vary by
the same parabolic law. This is not quite true, but suf-
ficiently near for ordinary purposes.
Then, as will be shown in the next Article,
r 7 pi 7- 7 /«l+«2. J 1 jai
-\-a2-\-a2,
A
In this case / = 68 feet and A =45.3 square inches.
ai=a2=az=g sq. ins.
Making these numerical substitutions, there will result
Li =30.7 feet; L2 =42.9 feet; L3 = 52.5 feet. These lengths
are clearly the minimum permissible. In actual construc-
tion it is desirable to have the end of the plate from i to
1.5 feet further from the centre, making the total length of
the plate 2 to 2.5 feet greater than the length computed
above. This lengthening of the cover-plate is essential
in order that the cover-plate metal may be taking stress
at the point where the plate is computed to begin. Also
Art. no.]
THE DESIGN OF A PLATE GIRDER.
697
as will be seen a little further on, the pitch of rivets in
these ends of the cover-plates is made less than in the
body of the plate for greater effectiveness where the plate
begins to take its stress. The lengths of cover-plates
then, beginning with the shortest, will be 33.2, 45.4, and
5 5, feet.
Another method of procedure, more accurate than the
preceding, is to draw a moment curve on the effective
span, which can readily be done by laying down as vertical
ordinates the resultant or total moments as given in Table I.
These moment ordinates would be 5 feet apart except
at the centre of span. The lengths of cover-plates must
be such as to give resisting moments of the flange stresses
at least equal to the external bending moments shown
on such a diagram. The moments of the flange stresses
will require the centres of gravity of parts of the flange
sections to be computed at each moment point. The
following tabulation shows the elements of this method
of procedure for the centre section of the girder:
Section.
One-eighth web plus flange angles
First cover-plate
Second cover-plate
Top cover-plate
Sq. Ins.
18.3
9
9
9
Stress per
Sq. In.
16,000
16,000
16,000
16,000
Lever Ai
Ft.
6.41
6.77
6.83
6.9
Moment.
Ft.-lbs.
1,875,000
976,000
984,000
994,000
This operation must be repeated at each m^oment section
of the girder, but the numerical work need not be repeated
here, being precisely like that for the centre section.
The net lengths of plates found by this method are 32.8,
42.9 and 53.8 feet, a substantial agreement with the lengths
found by the shorter procedure.
In the compression flange the cover-plate lying on the
angles should run the entire length of the girder, especially
698 PLATE GIRDERS. [Ch. XV.
if the girder be of the deck type, i.e., with ties resting upon
the upper flange. That flange being under compression,
it is advisable that the horizontal legs of the angles be
supported throughout their entire length by riveting
them to a cover-plate. This will add to the stiffness and
carrying capacity of the flange. If ties rest directly upon
the upper flange, their deflection tends to bend one side
of it out of its horizontal position, but this tendency will
be materially lessened by the added stiffness gained in
riveting the horizontal angle legs of the flange to the cover-
plate.
Although this process of design has been used in con-
nection with the tension flange, under the specifications
the compression flange is to be made like the tension flange,
i.e., a duplicate of it.
Pitch oj Rivets in Flanges.
Arts. 5 and 31 of the specifications relate to the rivets
required to join the vertical legs of the flange angles to the
web plate. Art. 31 requires that "The flanges of plate
girders shall be connected to the web with a 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
on the flange. The wheel loads where the ties rest on the
flanges shall be assumed to be distributed over three ties."
The chief function of these rivets is to transfer hori-
zontal shear from the web plate to the flanges, as it is in
this way that the flanges receive their stresses. If the
rivets take the direct load of the locomotive driving wheels,
as in the case of a deck girder like that being designed,
they must resist the resultant stress due to both vertical
and horizontal loads.
Art. no.] THE DESIGN OF A PLATE GIRDER. 699
Strictly speaking the number of rivets required between
two moment sections, as shown in Fig. i, should be just
sufficient to give the increase of flange stress in passing
from one section to the next one toward the centre of span.
Art. 31 of the specifications, therefore, requires more
rivets than are needed except at the end of the span. It
is always necessary, however, to introduce more rivets
near the centre of span than is required by actual computa-
tions, for the general stiffness of the girder. Indeed even
more rivets are generally provided than those prescribed
in Art. 31 of the specifications.
If d is the effective depth of the girder at the end of
the span and if the end shear or reaction is R, and if tA is
the flange stress at the distance d from the end of span,
then will the following equation of moments be found,
neglecting the negative moment of any load within the
distance d from the end of the span:
Rd = tAd,
Hence
R=tA,
This shows that an amount of stress equal to the end
shear must be given to each flange within the distance d
from the end. The number of rivets required by this
computation is a little more than necessary if any load
rests upon the girder between the end and the section at
the distance d from it. It will be clear that the general
provision of Art. 31, quoted above, is based upon this
end shear requirement, and it is analytically incorrect,
but the excess of rivets which it calls for adds to the general
stiffness and capacity of the girder.
The weight of one driving wheel is 30,000 pounds, and
it is to be distributed over three ties or 42 inches. As
700
PLATE GIRDERS.
[Ch. XV.
the prescribed impact is loo per cent., the vertical load
per horizontal inch of girder will be:
Tr 2 X 3.0,000 1,
V = ^-^ = 1430 lbs.
42 ^
It is obvious that the flange stress taken by one-eighth
of the sectional area of the web is received directly by the
latter and does not affect the rivets through the vertical
legs of the flange angles. If Ai is the actual net flange
section of cover-plates and angles and A 2 the total flange
area, including one-eighth of the web section, and if 5
is the total shear at any moment section, while d is the
effective depth of the girder, then the horizontal flange
stress H to be taken up per linear inch by the rivets
between two sections the distance d apart will be H =--r^.
d A2
The values of Ai and A 2, beginning at the end section
of the girder, are as follows :
Section
A
A
End
22.88 sq.ins.
27 .26 sq.ins
sft.
22.88 ''
27.26 "
10 "
22.88 ''
27.26 ''
15 "
31.88 ''
36.26 ''
25 "
40 . 88 ' [
45.26 ''
Centre
40.88 ''
45.26 ''
The unit (inch) increments H of horizontal flange
stress found for the various sections by the preceding
formula are:
End H
5 ft.
10
319,500^22,
80.5 27.26
H =
H =
= 3330 lbs.
= 2920 ^'
=2510 "
Art. no.] THE DESIGN OF A PLATE GIRDER, 701
15 ft. H =
= 2170 lbs
25 '' H =
= 1560 ''
Centre H =
= 954 "
Each of the above results gives the horizontal stress H
in pounds per linear inch, over each 80.5 inches of girder
flange for each moment section and to be taken up by the
rivets.
The rivet pitch p at any section will then be determined
by the following formula if K is the working value of one
rivet in shear or bearing:
PVV^-\-H^=K.
Each rivet bears against the web plate as well as against
each vertical leg of the flange angle, and as the web plate
is much thinner than the sum of the thickness of the two
angle legs, the bearing value against the web plate will
be much less than that against the angle legs. Furthermore
each rivet is subjected to double shear, the two shearing
sections of the rivets coinciding with the two faces of the
web plate. K, therefore, must be taken as the least of the
double shearing value and the bearing value against the
web plate. The rivets to be used are |-inch diameter
before being driven and the bearing value of such a rivet
against a ^^-inch plate at 24,000 pounds per square inch
is 9190 pounds and 14,430 pounds in double shear at 12,000
pounds per square inch, both of these working stresses
being in accord with the specifications.
Applying the numerical results thus established to the
formula for the pitch,
there will result :
702 PLATE GIRDERS. [Ch. XV.
At end p=-^= ^ ^ r =^2.55 ins.
Vi43o2+332o^
5 ft. point p = 9i9 ^_^ ^^ 83 ''
V 1430^ + 2920^
10 " p= =3.18 ''
15 " P- =3-53 "
C i
25 P= =4-34
Centre ^= =6.26 •'
If desired a curve can be drawn at the various points
with the corresponding pitch as a vertical ordinate at each
point. Such a curve will give the rivet pitch at g^ny point
in the span, but such detail is not usually required. The
above values of the pitch may be used, with judgment,
without further computations for any part of the girder.
Fig. I shows the pitch used at the different girder points;
it is frequently adjusted to the position of the intermediate
stiff eners.
Pitch of Rivets in Cover-plates.
The number 0/ rivets required in a cover-plate is at once
determined from its net section. In the present case the
net section of each cover-plate is 9 square inches, which,
at 16,000 pounds, gives 144,000 pounds as the stress value
of the plate. The rivets in the cover-plates are subjected
to single shear and the single-shear value of one J-inch
rivet is 7220 pounds. Hence the number of rivets required
to develop the full value of one cover-plate is -^^ = 20
7220
rivets. Between the end of the cover-plate, therefore,
and the point at which the next cover-plate outside of it
begins, there must be at least 20 rivets. As a matter of
fact considerably more than that number will be found,
Art. no.] THE DESIGN OF A PLATE GIRDER. 703
as the pitch must not exceed 6 inches in any case and it
should not be more than 3 inches for a distance of 12 to
18 inches from the end of the plate. It will be seen upon
examining the drawing that these conditions are fulfilled.
Top Flange.
As this flange is in compression, gross areas may be
used. If the provisions of Art. 30 and other Articles of
the specifications be scrutinized, it will be found that they
are fulfilled by the compression flange made up as shown
in the figures, and they need no further detailed attention.
End Stiffeners.
The end stiffeners must be heavy members of their
class and rigidly riveted to the girder, as they take the severe
impact or pounding at the points of support due to rapidly
moving heavy locomotives and trains. Art. 79 of the
specifications provides that " There shall be web stiffeners
generally in pairs, over bearings, at points of concentrated
loading, and at other points where the thickness of the web
is less than one-sixtieth of the unsupported distance between
flange angles. . . . The stiffeners at the ends and at points of
concentrated loads shall be proportioned by the formula
of paragraph 16, the effective length being assumed as
one-half the depth of girders. ..." This provision
makes it necessary to treat the end stiffeners as a column,
the working stress to be:
^ = 16,000 — 70—.
The column load in this case is the maximum end shear
including impact allowance as given by Table II, i.e.,
319,000 pounds.
704 PLATE GIRDERS. [Ch. XV.
If two pairs of sXslxH-inch angles be assumed for
trial with the 3|-inch legs against the web plate, remem-
bering that they will be separated by the thickness of the
plate, the radius of gyration of their combined section
about an axis lying in the centre of a horizontal web section
and parallel to the web will be 3.13 inches. The length
o ^
of the column is — ^=40.25 inches =/. Hence the pre-
scribed formula will give a working stress of 15,100 pounds
per square inch. On this basis
. . J siQ,ooo
Area required = '^—^ = 2 1 sq.ms.
15,000
The actual sectional area of four of the assumed angles
will be 23.24 square inches, which is sufficiently close to
the area required to be accepted as satisfactory.
The entire load is carried to the end stiffeners by the
I -inch rivets which bind them to the web plate. The rivets
are in double shear and bear on the web plate. It has
already been seen that the bearing value on the wxb plate,
9190 pounds per rivet, is much less than the double shear
value. Hence the number of rivets required is ^—^ =35
9190
rivets. This computed number of rivets distributed
throughout the length of the 3|-inch angle legs would make
the pitch too great. The pitch should not exceed about
4 inches, which would make the number of rivets about
40. It is essential, as already indicated, that the end
stiff eners be made exceptionally stiff and rigid.
End stiff eners are not bent, but are riveted onto filling
plates having the same thickness as the flange angle legs.
These filling plates enhance the stiffness and resisting
capacity of the end stiff eners as they, in fact, form a part
of the latter.
Art. no.] THE DESIGN OF A PLATE GIRDER. 705
Intermediate Stiff eners.
By referring to Art. 79 of the specifications there will
be found an empirical formula giving the maximum dis-
tance between intermediate stiffeners, providing, however,
that that distance in no case shall exceed the clear depth of
the web. Intermediate stiffeners are sometimes regarded
as being equivalent to the vertical compression members
of a Pratt truss, but as a matter of fact there is no rational
system of basing their design on computations. They
are almost invariably made of angles, but sectional areas
are determined by experience. Inasmuch as the total
transverse shear at the centre of span is small, they are
sometimes omitted there. As a rule they are never placed
farther apart than the depth of web plate.
As this girder is to carry a heavy railroad load pre-
sumably at high speed, 5X3^Xf-inch steel angles will
be used with the 3 J inch leg placed against the web
plate. As the transverse shear increases toward the end
of the span, the distance apart of these intermediate
stiffeners will correspondingly be decreased. In the central
part of the span this distance is seen to be 5 feet if inches,
but near the ends it is reduced to 3 feet 5 J inches. The
pitch of the rivets in these intermediate stiffeners may vary
from 3 inches to 5 or 6 inches, the greater pitch being near
the mid depth of the web.
Splices in Flanges.
It will be found that cover-plates and flange angles
may be purchased of full lengths required on this plate
girder. When, in general, the girders are so long as to
require splicing of the parts of flanges, those joints for the
tension flange must be so designed as to leave the net
section as large as practicable, as the entire stress must be
7o6 PLATE GIRDERS. [Ch. XV.
carried by the net section. It is good practice and cus-
tomary not to have two joints in adjacent parts concur,
i.e., there should be breaking of joints so as to have a
joint in one part only of the flange at the same section.
In this manner the net section at each joint may attain
its maximum value. In the splicing of angles both legs
should be spliced. In compression, riveted joints can
scarcely be expected to transfer stresses by abutting sur-
faces in those joints. They should be spliced about as
effectively as tension joints, although the question of net
section does not arise, the gross section being available.
Splices in Web Plates.
As one-eighth of the gross web-plate section is considered
as resisting bending as a part of the flange area, the rivets
at a web-plate splice must be suflicient to resist the cor-
responding bending moment. This web-plate moment is,
therefore,
16,000^, 7 >.yO 9 r • 11
• — - — X-tX8o.5^ = 5,670,000 m.-lbs.
8 , 10
There must be two splice- plates, one on each side of
the web, each of which need not be as thick as the main
plate, but in this case f-incli splice-plates have been used
so that the intermediate stiff ener need not be bent. For
this size of girder there should be three rows of rivets on
each side of the joints. If it be assumed that the pitch
be 4 inches in each row, there will be nine rivets in each of
the three rows between the mid depth of the web and the
back of the flange angles. If the loads carried by these
rivets in resisting bending vary directly as the distance
from the neutral axis at mid depth, their resultant will
act at 1X40 = 26.7 inches from that line. The bearing
Art. iio.l THE DESIGN OF A PLATE GIRDER. 707
value of a |-inch rivet against the i^-inch web is 9190
pounds. Hence the resisting moment of the 54 rivets
on one side of the joint is:
M = ^^^-^X2 X26. 7 =6,600,000 in.-lbs.
As this is greater than 5,670,000 in. -lbs., the proposed
arrangement of the joint is satisfactory. The two splice-
plates will, therefore, each be 19 X f inches by 5 feet 8| inches,
as shown in Fig. i.
In general every joint splicing should be tested for
the transverse shear which it must carry. In this instance
it is clear that the splice-plates will carry more shear than
the web.
General Considerations.
The girder proper with its flanges, web, and stiff eners
has been designed in this article without indicating the
manner of connecting such lateral or cross bracing as
would be required in the complete design of a railroad
plate-girder span. The design of such bracing would be
supplementary to the actual design of the girder as made,
and it is the purpose here to illustrate only those principles
belonging to the design of the girder proper. The design
of the bracing and the details of its connection with the
girder belong rather to bridge construction than to the
subject treated here. Fig. 2 has been introduced, however,
as an illustration to indicate the general features of the
complete structure.
Large plate girders are not always built complete in
the shop, although girders nearly 100 feet in length are
frequently and perhaps usually so completed at the present
time. When it is necessary to build them in portions
7o8 PLATE GIRDERS. [Ch. XV.
and rivet the portions together in the field, the general
principles governing the construction of the necessary
field-joints are precisely the same as those illustrated in
this article. They are simply adjusted or adapted to the
exigencies of each particular case.
The bill of material and estimated weight of a single
girder as designed is as follows :
Pounds.
Two 80" Xi^" web plates, 21' \i\" long 5,236
One 80" Xi^" web plate, 26' |" long ; . . 3,094
Four 6"X6"Xf" angles, 70' long 8,036
One 14" X I" cover-plate, 70' long 2,499
One 14" Xf" cover-plate, 55' 5I" long 1,981
Two 14" Xi" cover-plates, 47' 6^" long 1,696
Two 14" X I" cover- plates, 33' 3" long 1,190
Eight ^"XzV'XW angles, 6' 7" long 1,037
Twenty-eight 5" X3I" X I" angles, 6' 7" long 1,917
Four 10" Xf" filler-plates, 5' 8|" long 581
, Four 19" Xf" splice-plates, 5' 8|" long 1,106
Twenty-four 3^"Xf " filler-plates, 5' 8|" long 1,222
Two 14" XI" sole-plates, i' 6" long 107
Rivets 800
Total for one girder 30,502
The weight of girder per linear foot therefore is:
^-^ — =436 lbs.
70
If the plate girder were of the through type, there
would be no change whatever in the procedures of design
which have been followed, but in order to give a better
appearance to the ends they would be formed as shown in
Fig. 5. The latter figure shows the same end stiffness,
depth of girder and the same flange angles as Fig. i.
Art. III. — ^Length of Cover-plates.
There are various methods of determining the lengths
of cover-plates of plate girders involving simple compu-
Art. III.] LENGTH OF COVFM-P LATHS. 709
tations only, which are well illustrated by the following
procedures :
The first of these procedures is based on the assump-
tion that the depth of the girder is uniform and that the
bending moment throughout the length of girder varies
as the ordinate of a parabola as in the case of uniform
loading. The following notation is required:
/ = effective length of span either in feet or inches ;
L= length of cover-plate required in the same unit as /;
A = total net flange area;
a = net cover-plate area required.
Since the flange and cover-plate areas vary directly
as the flange stresses, and as the latter vary as the ordi-
nates of a parabola when the depth of girder is constant,
the following equation will result:
P~A'
or
, a
(i)
='^g
A
Eq. (i) will give the length of the cover-plate whose
area of section is a. Any convenient unit may be taken
for a and A , but the square inch is ordinarily employed.
If there are several cover-plates, a is to be taken suc-
cessively the area of the first, second, third, etc., cover-
plates in summation, i.e., it will first be taken as the net
sectional area of the top cover, then as the net sectional
area of the top cover added to that of the cover-plate
below it, and so on.
The second method is the following, and is applicable
7IO PLATE GIRDERS. [Ch. XV.
to the case of a girder with varying depth, the notation
being as follows :
Let Z£;= uniform load per linear foot, or "equivalent
uniform load" per linear foot;
d and d' represent the effective depths of girder in
feet at the centre of span and at the end of
the cover-plate respectively;
A — a = a'= area of flange section at the end of
cover-plate ;
r = permissible flange stress per square inch;
the bending moment at the end of the cover-plate will
then be
M=w^-~[~] =AdT-w^=d'a'T. . . (2)
8 2 \2 / 8 ^ ^
By solving the second and third members of the pre-
ceding equation there will result
L = 2\/\
{Ad-a'd')T ^ ^^ [{Ad-a'd')T
^■83\^ " :^ . . (3)
w ^ -^ w
It must be remembered that the application of either
of the two preceding methods will give the net length of
the cover-plate. There must be added 12 to 18 ins. at
each end with rivets closely pitched so that the cover-
plate may certainly take its stress at the points where its
effectiveness should begin.
Art. 112. — Pitch of Rivets.
A simple method of finding the pitch of rivets piercing
the vertical legs of the flange angles and the web plate of a
Art. 112.] PITCH OF RIVETS. 711
plate girder at any section of the beam may readily be found
by using the general but elementary expression for the bend-
ing moment,
IPx=M.
By differentiating this equation,
IP.dx = Sdx=dM;
.:> representing the total transverse shear.
If dM is the change of bending mom.ent for the distance
along the flange represented by the pitch of rivets, p, the
change of flange stress for the same distance will be found
by dividing dM by the effective depth of the girder, d. If
the pitch of rivets, p, be placed in the preceding equation in
place of dx, the corresponding change of flange stress will
represent the amount of stress transferred to the flange by
one rivet. Representing that variation of flange stress by
V, the last of the preceding equations may be written
Sp=dv; :. P=^' :
In this equation v represents either the bearing capacity
of one rivet against the web plate or against the two flange
angles, or the double shearing value of the same rivet, i.e.,
the least of those three values. Ordinarily the bearing of
the rivet against the web plate will be less than either of the
two other quantities; hence that bearing value would then
be substituted for v. In general the least of the three pre-
ceding values for one rivet is to be substituted for ^' in an
actual computation. The total transverse shear 5 is always
known at any section or may readily be determined. The
preceding formula for the pitch, therefore, is a very simple
one and is much employed. .__„__.
CHAPTER XVI.
MISCELLANEOUS SUBJECTS.
Art. 113. — Curved Beams in Flexure.
If beams are sharply curved, i.e., if the radius of curva-
ture of the neutral surface is comparatively small, the for-
mulae expressing the common theory of flexure for such
beams will contain the radius of curvature and corre-
sponding variations from the formulae for straight beams.
Let Fig. I represent part of a curved beam subjected to
flexure, AC representing the radius of curvature at the
Fig.
point A before flexure while CA' represents the radius
of curvature of the same surface after flexure takes place.
OAO^ represents the neutral surface. A^f is the continu-
ation of C'A\ Similarly A'b is the continuation of CA'.
Finally, A'b' is drawn parallel to CA. def represents the
normal section of the beam and AA^ is supposed to be
a differential of the length of the neutral surface.
712
Art. 113.] CURBED BE^MS IN FLEXURE. 713
The ordinate dzy is measured from A as an origin
toward B or D, respectively, z is the varying width of
the normal section of the beam and hence it is measured
normal to y and x, the latter being measured along 0A0\
A differential of the section of the beam is zdy.
As the normal sections of the beam are assumed to
remain plane after flexure, let the rate of strain, i.e., the
strain per unit of length of fibre at any point distant y
from the neutral surface be uy, u being the apparent
rate of strain at unit distance from the neutral surface.
By referring to Fig. i there may at once be written:
■ b'b=da^; hh" = (dx-^do^uyi.
By similarity of triangles,
do(^+dx{i^-'^'
'"^ dx
(i)
This equation gives at once :
r
r-r' 7"' , ,
U=-, r-, = (2)
{r-^yy r+y ^ ^
If the beam were originally straight, in which case the
radius of curvature r = 00 , eq. (2) would take the form
u=-, the usual expression for the rate of strain at unit
distance from the neutral surface of a straight beam.
If again the radius of curvature is sufficiently large, so that
r may be written for r-\-y without sensible error:
"=/-7- ^3)
714 MISCELLANEOUS SUBJECTS. [Ch. XVI.
This expression for m may be used for curved beams if
the curvature is not too sharp.
If the radius r is infinitely great, u = -, which is the
value for a straight beam.
Eq. (2) shows that the rate of strain u at unit distance
from the neutral surface and corresponding to the rate of
strain at any distance y is variable, as y appears in the
denominator in such a way as to make u smaller the
greater the distance of the fibre from the neutral surface.
This is in consequence of the curvature of the beam and
results from the assumption that normal sections plane
before flexure remain plane after flexure. With the
increase of length of fibre due to curvature as its distance
from the neutral axis increases, a less rate of strain is
required to keep the section plane after flexure. This
assumption is not strictly true, and it may be a matter of
doubt whether it is necessary or advisable even in the
interests of correct analysis.
If k is the fibre stress of tension or compression at any
distance y from the neutral axis, there may be at once
written :
k=Euy=E(^-,-ijy^^ (4)
The stress on an element zdy of the section will then be :
«,=£(^,^.)sg (s)
Let k' and k'^ he the intensities of stress at the distances
y' and -y" from the neutral surface. Then by eq. (4):
k' ^ y' r-y"
. k" r+y ~y"'
Art. 113.] CURBED BEAMS IN FLEXURE. 715
From this equation:
liy'=y", eq. (5a) becomes:
k" = -k''-±^, (5fa)
r —y
Eq. (55) shows that the intensity of stress at a given
distance from the neutral axis will be greater on the concave
side of the curve than on the convex, and that this relation
holds until the radius of curvature becomes infinitely
great.
In order to locate the neutral axis the integral of the
two members of eq. (5) between the limits of y and —y
must be placed equal to zero, giving eq. (6) :
X>==^&-C
^^=0. ... (6)
Again, the bending moment formed by the direct
stresses of tension and compression in the section may be
written in the usual manner as follows, M representing the
moment :
-&-)x:
r-\-y
^=^ir,-)i f5f (7)
Eq. (6) shows that the neutral axis will not pass through
the centre of gravity of the section. As the intensity of
stress on the convex side of the curve will be less than if
the beam were straight, the neutral axis will be on that
7i6
MISCELLANEOUS SUBJECTS.
[Ch. XVI.
side of the centre of gravity of the section toward the con-
cave surface of the beam. Eq. (7) shows, again, that the
integral is not the moment of inertia of the section about
the neutral axis, but it will reduce to that if the radius
of curvature r be supposed infinitely great.
The integrations shown in the second members of eqs.
(6) and (7) can at once be made when the form of cross-
section is known. Inasmuch as this analysis for curved
beams finds one of its important applications in connection
with the design and carrying capacity of large hooks, a trape-
zoidal cross-section shown in Fig. 2 will be assumed by
way of illustration, and from
that the rectangle section at
once results. In that figure
the larger end CD of the trap-
ezoid will be considered to
lie in the concave or inner
surface of the hook and at
right angles to the plane of
the hook. As the trapezoid
is symmetrical, a = \FH, and
the angle of inclination of a
sloping side as HD to the
centre line will be taken as a.
Then z will represent one-half of the width of the trapezoid
at any point:
Fig. 2.
z=a + {yi —y) tan a.
(8)
If z be inserted in eqs. (6) and (7) there will be required
the following integrations in which yi-\-yo=d:
/-
'' ydy _^
yo^+y
rlog
r+yi
r-yo
. (9)
/:
Art. 113.] CURBED BEAMS IN FLEXURE. 717
r'^^=d(yi^-r)+rUog'±y-\ . . do)
J-y,r+y \ 2 / r-yo
" t^ =rH-hrd{y, -yo) +id(d' -syiyo)
-^,og(^;). . . („)
If these values of z and the integrals given in eqs. (9),
(10) and (11) be substituted in eq. (6), there will at once
result :
log'-±yi= l^ '^ I. . . . (12)
r — yo r\ ir +yi) tan a -\-a]
As known quantities let r-\-yi=R and r—yo=Ro,
then eq. (12) may take the form:
r log -^(R tan a -\-a) = rd tan a -\-d[ - tan a +a ) .
Kq \2 /
Hence :
J(-tana+a)
(r log — -d\ tan a+a log —
After r is determined by eq. (13) there w^ll at once
result :
yi=R-r and yQ=d-yi. . . . (14)
If the section is rectangular, Q:=tan a=o, hence,
^ = ^ and yi^R- — — . . . (14a)
log^ logf-
7i8 MISCELLANEOUS SUBJECTS. [Ch. XVI.
If the section is triangular, a=o and the second member
of eq. (13) will be correspondingly simplified as follows:
r=-i -5 \ (15)
(i?log|-.)
As this expression is independent of o:, yi and yo remain
unchanged whatever may be the value of that angle.
Having thus found yi and >'o, the position of the neutral
axis of the section is determined and the expression for
the bending moment can now be written by the aid of
eqs. (4) and (7), 'the latter being the general expression
for the bending moment. By the aid of eq. (4) the in-
tensity of stress in the extreme fibre at the distance yo
from the neutral axis may be written as follows :
ybo=-£(-,-i)-^ (16)
\r Jr-yQ
Hence,
4.-) = -^- • ; • • <■"
By introducing the second member of eq. (17) in eq.
(7) as w^ell as the value of z from eq. (8) and the integrals
given in eqs. (10) and (11), the following value of the
moment M will result :
-M=^M!:^:^P'((a+3;,tan«)^^-tana^l , (i8)
I \ \U, -r/1 tdii UI.J ; tail ex ; >
yo J-yoi r+y r+y]
2ko{r-yo){.^.^^,.^_ .(d
yo
I (a + (r -^yi) tan a) (-{yi -yo)
.dr+rnog'-±y^)y-^^{d^-iyryA . . (19)
r-yo/ 3 J
Art. 113.] CURl^ED BEAMS IN FLEXURE. 719
As is evident, the factor 2 appears in the second members
of eqs. (18) and (19), for the reason that the section taken
is symmetrical and the varying ordinate z is half the width
of section at any point. If a were taken as the extreme
width of section on the narrow side instead of half that
width and if a were to be so taken that (yi—y) tan a
added to a represents the full width of the section at the
point located by y, the factor 2 would be omitted from the
second member of the value for M.
If the section is rectangular a = tan a=o and the
expression for the moment M then becomes :
_M ^ ^feo(r-yo) f p(^^ _^^) _^^^^, 1 r±n\] (^^^
yo i \2 r-yo/ J
If the section were triangular a = o in the second member
of eq. (19).
These equations may be employed in the design of curved
beams of any form of cross-section or degree of curvature
when those based on the common theory of flexure for
straight beams are not applicable. As a general statement
it may be said that the formulas for straight beams may be
used without essential error in all cases except those of such
special character as hooks and other structural or machine
members in which the curvature is sharp. The applica-
tion of the preceding formula to the case of hooks will be
illustrated in the next article.
Art. 114. — Stresses in Hooks.
The diagram of a hook shown in Fig. i illustrates the
conditions of loading to which hooks in general are sub-
jected. The material to the right of the point of applica-
tion of the load is subjected to no stress whatever except
in a secondary way near that point. On the left of the
720
MISCELLANEOUS SUBJECTS.
[Ch. XVI.
load, however, the arc of the hook, supposed to be circular
in this case, is subjected to direct stress, shear and bending,
the bending moment increasing as that part of the hook
parallel to the loading is approached, but it decreases in
passing on to the shaft of the hook supposed to be in line
with the load. The section of miaximum bending AB
is subjected to the combined direct pull of the load and
Art. 114.
STRESSES IN HOOKS.
721
the bending moment equal to the load multiplied by the
normal distance from its line of action to the centre of
gravity of the section. This cross-section of greatest
bending moment will first be treated as if subjected to.,
pure flexure. The necessary simple analysis required to
determine the greatest intensity of stress in the section will
then be made. In the section
of greatest bending moment
there is no shear.
The cross-section of the
main part of a hook maybe
taken as .approximately trap-
ezoidal, as shown in Figs, i
and 2. In the present in-
stance the greatest dimension
of this cross-section lying in
the central plane of the hook
will be taken as 5 inches and
the corners will be rounded
approximately as shown.
Obviously the integrations
of eqs. (9), (10) and (11) of the
preceding article do not rep-
resent accurately the approx-
imate trapezoid of Fig. 2 . This
integration or its equivalent,
however, may be accomplished with sufficient accuracy
by a number of approximate processes, i.e., by transformed
figures and by dividing the section into a sufficient
number of small parts. A simpler method and one giving
reasonably accurate results is to draw two lines F^C and
FD in such a way as to make a true trapezoid whose resist-
ing moment will be essentially the same as the approx-
imate trapezoid. This will be accomplished if the two
X 2V4- >
Fig. 2.
722 MISCELLANEOUS SUBJECTS. [Ch. XVI.
lines indicated be drawn in such a way that each area
between a broken line as F'C and the inclined full line
of the actual section be three times the combined area
between CB and the curved end of the section and between
AF' and the other curved end of the section. This rela-
tion results from the fact that the bending stress between
the tw^o lines indicated varies in intensity from zero at the
neutral axis to nearly the maximum in the extreme fibre
of the section and has its centre at two-thirds of the distance
from the neutral axis to the extreme fibre. The relation
indicated, therefore, makes the bending moments of the two
parts inside and outside of the actual section equal. This
construction will give for one-half the modified figure:
AF' =AF = .^^ inch=a; BC = 1.1 inches; a =S° 30';
tan a' = .148; i? = 5+2.2 =7.2 inches; R = 2.2 inches.
d = s inches.
Fig. I shows that R and Rq are the interior and exterior
radii respectively of the arc of the hook where the section
of greatest bending moment exists. By introducing these
numerical quantities in eq. (13) of the preceding article
there will at once result :
r =3.87 inches.
Hence,
yi =R-r = s-33 inches;
yo=d—yi=i.6j inches.
By inserting the same numerical values together with
yi and yo in eq. (19) of the preceding articles, the value of
the bending moment becomes :
M =4. 88^0 (i)
Art. 114.] STRESSES IN HOOKS. 723
This moment obviously can be expressed in terms of the
intensity of stress in the extreme fibres on the opposite
side of the section, i.e., 3.33 inches from the neutral surface.
By eq. (4) of the preceding article:
ki =ko
(r-yo)yi
yoir+yiY
After substituting' the values of the quantities already
determined there will be found ^i=.6i/^o. Or there may
be written from the same eq. ^0 = 1.64^1. The bending
moment expressed in terms of the greatest intensity of
stress in the extreme fibres is obviously the form desired
for practical purposes.
Let the hook shown in Fig. i be supposed to carry
a load of 20,000 pounds. The centre of gravity G of
the actual cross-section is 2.13 inches from the side CD
of the cross-section. Fig. 2. Hence the load assumed
will cause a bending moment about the line GH equal
to 20,000 X(2. 13 -1-2.2 =4.33) =86,600 inch-pounds. It is
to be observed that inasmuch as the 20,000 pounds is taken
as uniformly distributed over the cross-section the lever
arm of the load is the normal distance from its line of
action to the centre of gravity of that section, although the
resisting moment of internal stresses has the axis deter-
mined by eq. (14) of the preceding article, the two axes
being parallel to each other.
The greatest intensity of tensile bending stress in the
section therefore takes the following value:
, 86,600 11 . , X
^0 = 5^- = 17^40 lbs. per sq.m. . . (2)
4.00
The uniformly distributed tensile stress equal to the
load will act upon the entire actual area of section, which
724 MISCELLANEOUS SUBJECTS. [Ch. XVI.
is 7.9 square inches. Hence, that tensile intensity will
be — '■ =2530 pounds per square inch. The resultant
7-9
greatest intensity of stress in the entire section will be:
17,740 + 2530 =20,270 lbs. per sq. in. . . (3)
The resultant intensity on the opposite side of the sec-
tion at A, Fig. 2, will be, since ki =.6iko\
— 17,740 X.61 +2530 = —8291 lbs. per sq.in. .
(4)
The minus sign is used because the bending stress is
compression throughout that part of the section indicated
It is commonly observed in actual experience that
hooks or other similar bent members break at the inside
of the section where the curvature is the sharpest. The
eqs. (4) and {s^) of the preceding article indicate clearly
the reason for such failures as the intensity of stress k in
the extreme fibre is shown to vary inversely with the radius
of curvature r-\-y. When, therefore, the curvature is
sharp, i.e., the radius of curvature is small, the fibre stress
k increases rapidly, especially on the inside of the curve
where the radius of curvature is r—y.
This example shows the general method of treating the
stresses in hooks by the common theory of flexure based
on the assumption that normal sections plane before flexure
remain plane after bending.
It is well known that this assumption is not strictly
correct, and it is further known that the ordinary or com-
mon theory of flexure is not accurately applicable to such
short beams as are contemplated in the theory of hooks.
Art. 114.] STRESSES IN HOOKS. 725
Comparison with the Theory of Flexure for Straight Beams.
It is indicated above that the assumptions on which the
preceding analyses are based are not strictly correct. If
it be assumed that the intensity of stress varies directly
as the distance from a neutral axis passing through the
centre of gravity of the section, as for straight beams, and
if k\ is the greatest intensity of stress in the extreme
fibres {FF', Fig. 2) the bending moment will be:
M=^-^ (5)
yc
In this equation I is the moment of inertia about an axis
through the centre of gravity G, Fig. 2, while yc is the
distance of that axis from the most remote fibre at A.
The moment of inertia I of the actual section shown in
Fig. 2 about a neutral axis through the centre of gravity
G Sit the distance 2.87 inches from A is 14.9. Hence,
the bending moment on the preceding assumption is:
M=^k\=s.2k\ (6)
r 2
As the fraction ^^=1.07 this assumption is seen
4.88
to give a result only 7 per cent, greater than that of the
analysis for curved beams if the extreme fibre stress is the
same in amount in both cases. It is true that the result
has the apparent defect of placing the greatest intensity
of stress on the wrong end of the section.
Art. 115. — Eccentric Loading.
The analysis of stresses produced in a column or other
structural member by eccentric loading has already been
726
MISCELLANEOUS SUBJECTS.
[Ch. XVI.
discussed in preceding articles, but it is desirable to con-
sider some further and more general features of that analysis.
A column or structural member is said to be eccentric-
ally loaded when it carries a force or load acting parallel
to its axis but not along that axis. The perpendicular
Fig. I.
distance between the axis of the piece and the line of action
of the load is called the eccentricity of the latter.
Let Fig. I represent the normal cross-section of such a
member when the load P acts at any point Q in that cross-
section. The load P will then act parallel to the axis of
the piece, but at the distance CQ from it, C being supposed
to be the centre of gravity of the section. The ellipse
Art. 115.] ECCENTRIC LOADING. 727
drawn with C as its centre is the elHpse of inertia, the semi-
axes ri and r2 being the principal radii of gyration of the
normal section. Any semi-diameter as CQ' represents
a radius of gyration r\
If the force or load P acts at any point whatever, as Q,
and parallel to the axis of the piece, it will create a bending
moment equal to PxQC. If x' and y^ are the coordinates
of Q the components of that moment will be Px' and Py\
the former about the axis Y, the latter about the axis X.
1 1 and 1 2 being the principal moments of inertia, as already
indicated, the intensities of bending stresses produced by
these two component moments at any point, whose co-
Px' PV
ordinates are x and y, will be —j—x and ----y, respectively.
i 1 J-2
Furthermore, the load P will produce a uniform normal
stress over the entire cross-section of the member, if A
p
is the area of that cross-section, represented by — . The
resultant intensity of stress k therefore at any point of the
section will be :
At the neutral axis the intensity of stress is equal to zero,
hence,
x'x y'y , s.
— j+^=-i. ...... (2)
Eq. (2) is the equation of a straight line, i.e., the neutral
axis, along which the intensity of stress is zero, x and y
being the variable coordinates. It is obvious from eq. (2)
in connection with the general considerations respecting
728 MISCELLANEOUS SUBJECTS. [Ch. XVI.
the action of the load P that the position of the neutral
axis will depend upon the magnitude of that load and the
distance of its line of action from the axis of the member.
If X and y are zero, there will be no bending, and the section
of the member will be subjected to uniform compression
only.
If the point of application Q of P is on the curve its coor-
dinates x' and y' must satisfy the equation of the ellipse:
^'2 r^f2
^+2^ = 1 (3)
The equation of a straight line tangent to the ellipse
at a point whose coordinates are x' and y' is:
"-^+2^ = 1 (4)
When the point of application of P is on the ellipse, x'
and y' have the same values in eqs. (2) and (4). Hence in
that case -p- also has the same value in the two equations,
ax
showing that the neutral axis is parallel to the tangent to
the ellipse at the point where P acts. If in eq. (4) —x'
and —y' be substituted for -\-x and -\-y, that equation will
become identical with eq. (2), i.e., for this case the neutral
axis is tangent to the ellipse at a point diametrically opposite
to the point of application of the load P ; in other words,
the load is applied at one extremity of a diameter and the
neutral axis is tangent to the curve at the other extremity
of that diameter.
In Fig. I if the load P is applied at Q' (on the curve)
the neutral axis N'B' will be tangent to the ellipse at B' ,
the other extremity of the diameter Q'B'.
If the point of application of the force P moves along
Art. 115.] • ECCENTRIC LOADING. 729
the indefinite straight line BQ, the coordinates x' and y'
x'
will vary in the same proportion, making — a constant.
(s)
From eq. (2) :
dy
dx
x'r2^
y' ri2
X
Hence, as — is constant, all neutral axes will be parallel
y
while the point Q moves along a straight line.
Again, the coordinates x and y of the points of inter-
section of the line QB with the neutral axes must neces-
sarily be opposite in sign from x' and y' , as the origin C
lies between them. If therefore —x and —y be inserted
in eq. (2) :
^^+^^1 (6)
By similarity of triangles, a being a constant:
XX II I / \
— : = -=a .. XX =ayx=ayy. ... (7)
y y y yy n//
Eq. (7) in connection with eq. (6) shows that each of
the quantities x'x and y'y is constant, and that is equivalent
to making the products of the segments of the line QB on
either side of C constant:
QCy.CB=Q'C^CB' ^Y"'=Q"Cy.CB". . . (8)
As the point of application of P will always be given, the
quantity to be found will be the distance from the centre C
to the neutral axis, which may be called v. The semi-
diameter y' =CQ' at once becomes known after the ellipse
of inertia is constructed. In general, therefore:
y'2
'=QC (5)
730 MISCELL/INEOUS SUBJECTS. [Ch. XVI.
In some cases the reverse problem is given, i.e., v is
known and the distance of the point of application of the
load P is required. Hence,
■ QC=- do)
V
Rotation of the Neutral Axis about a Fixed Point in It.
One feature of eq. (2) remains to be considered before
the actual application of the preceding results can be made
to form a complete graphical construction. If the co-
ordinates X and y of the neutral axis be considered constant,
while the coordinates x' and y' of the point of application
of the load Pvary, eq. (2) shows that the path of the move-
ment of the point of application of P will be a straight
line, since the equation is of the first degree in respect
to x' and y' . This is equivalent to a movement of rota-
tion of the neutral axis about the fixed point whose coor-
dinates are x and y, while x' and y' determine the path
through which the line of action of P moves. The same
result can be shown by treating eq. (i) in precisely the same
manner for a fixed or constant value of k, that constant
being zero for the neutral axis.
The preceding procedures may be applied to a number
of problems, one or two of which will be illustrated. It is
sometimes desired to determine that part of the cross-sec-
tion of a member of a structure, or sometimes of the struc-
ture itself, within which a resultant load may be applied
anywhere without any change in the kind of stress induced,
usually compression.
Application of Preceding Procedures to Z-har and Rectangular
Sections.
Let it be required to ascertain within what part of a
Z-bar section an axial compressive force may be applied
without any part of the section being subjected to tensile
Art. 115.] ECCENTRIC LOADING. 731
Stress. The Z-bar section is shown in Fig. 2, the depth
of bar being 6 inches and the thickness of metal f inch.
As this section is unsymmetrical the axes for the principal
moments of inertia passing through the centre of gravity
C of the section will be incHned to the central plane of
the^ web. The ellipse of inertia MVNU has MN for its
major axis and UV for its minor axis, the former representing
a moment of inertia of 52 and the latter a minimum moment
of inertia of 5.7, the corresponding radii of gyration being
ri=2.55 inches and r2=.8i inch.
If no part of the cross-section of the bar is to 'be sub-
jected to tension, the outer Hmits or Hnes of that section
such as TS, SO, OL, etc., may be neutral axes for different
positions of the load^, but in no case must the neutral
axis He in any part of the metal section, even to cut across
a corner of it. This means that TS, SO, OL, LH, HE,
and ET will be successively considered neutral axes.' Let
ET be the first neutral axis considered or, rather, ET and
OL may be considered concurrently, as they are parallel
to each other and at the same distance from the centre of
the ellipse. First draw tangents to the ellipse parallel to
ET and OL as shown in the figure. The points of tangency
will fix the diameter DA, which is then extended to R
and W in the assumed neutral axes. As shown in the
preceding demonstration, the square of half the diameter
represented by AD will be equal to CR multipHed by CA
the distance from the centre of the ellipse to the point
of apphcation A of the force P. The distance CR is the
V of eq. (10), while CA is the distance QC desired, / is
half the diameter determined by the two points of tangency
Dividing the square of half the diameter by CR locates
the point A, one of the points desired. In precisely the
same manner D is located by dividing the square of half
the same diameter by CW = CR.
732 MISCELLANEOUS SUBJECTS. [Ch. XVI.
Tangents to the ellipse parallel to TS and HL are then
drawn as shown, one at A^, as indicated, at the lower
extremity of the ellipse and the other at the upper extremity,
thus locating the diameter NCF. Squaring half the diam-
eter so determined and then dividing by the distance from
the centre of the ellipse to TS or HL along the diameter
NF, the points B and F are found. In a precisely similar
way the vertical tangents indicated are drawn parallel to
SO and EH, determining the corresponding diameter. By
the use of that diameter in the manner already indicated,
the points G and K are located.
The points A and B are points of application of the force
P when ET and TS respectively are neutral axes. In the
preceding sections of this article it has been shown that if
a neutral axis such as ET be revolved about a point in it,
as T, to the position TS, the corresponding path of the point
of application of the load will be a straight line, and in this
case AB will be that straight line, since the two points A
and B correspond to the neutral axes ET and TS. By
similarly connecting the other points, the closed figure
ABKDFG is found. So long as the force P acts within
this area no part of the section can be subjected to tension,
but if the point of application is outside of this figure
some part of the section will be in tension
The closed figure thus established is called the '' core
section." Although it possesses much analytic interest,
the ordinary operations of the engineer are such as to make
it of comparatively little value in actual structural opera-
tions.
If any line such as Z'L parallel to a tangent to the
ellipse at g be drawn through a corner L of the Z-bar sec-
tion, and if a line dgZ' be drawn through the same point
of tangency and the centre C, cutting the side of the core
at d, it is shown in the preceding section of this article
Art. 115.]
ECCENTRIC LOADING.
733
that the product of dC by CZ' is equal to the square of the
semi-diameter Cg of the elHpse of inertia For any other
position of a Hne Z'L the same general observation holds,
the line always being parallel to a tangent to the ellipse
Fig. 2.*
at a point through which is drawn the line extending
through the centre and cutting the side of the core.
Probably the most usual section to which the core
* A number of construction lines shown in this figure are drawn for use
in the next article.
734
MISCELLANEOUS SUBJECTS.
[Ch. XVI.
procedure may be applied is the simple rectangle. A
masonry structure having such a horizontal section must
be designed so that compression only may always be
found in it. A simple diagram of pressures will show that
the resultant force or load must act within the middle
third of the section, but Fig. 3 shows the core procedure
appHed to the same axis. AB \s the length of the section
and BD is the width. AB is usually taken as one unit.
The ellipse OLMN is drawn with its
semi-major axis LC representing the
greatest radius of gyration of the rec-
tangle and the semi-minor axis OC is
laid off equal to the least radius of
gyration. Two lines drawn tangent
to the ellipse at M and N parallel to
BD and ED will determine the axes
of the ellipse, in fact already known,
then dividing the square of each semi-
axis by the normal distance of C from
BD and ED, respectively, the dis-
tances CF and CK will be found, thus
fixing the points F and K. The points
H and G are found in precisely the same manner, using the
sides AE and AB respectively. As already indicated, the
distance of H from BD will be one-third oi AB, while K
will be one-third oi BD from AB.
Fig. 3.
General Observations.
The preceding results show that bending combined with
uniform stress induced by a load normal to the section
will prevent the neutral axis from passing through the centre
of gravity of the cross-section. Furthermore, in this general
case the neutral axis or neutral surface will not be at right
angles to the plane containing the axis of the piece and the
Art. ii6.] GENERAL FLEXURE TREATED BY CORE METHOD. 735
line of action of the force unless that plane contains one
of the principal axes of inertia.
Manifestly the neutral axis for any section will be on
the opposite side of the centre of gravity of that section
from the force P. Eq. (8) shows that if the force acts at
C, making QC equal zero, CB will be infinitely great, which
means that the stress will be uniformly distributed, i.e.,
there will be no bending. On the other hand, if the force
P is at an indefinitely great distance from C, making QC
infinity, then will CB be equal to zero, i.e., the neutral
axis will pass through the centre of gravity. This is the
ordinary case of flexure and it is equivalent to taking all
load on the member at right angles to its axis.
Art. 116. — General Flexure Treated by the Core Method.
The procedures given in the preceding article may be
used for the general problem of flexure for straight beams
of any form of cross-section carrying any parallel loads at
right angles to their axes, the loads supposed to be acting
in a plane which contains the axis of the beam in each case.
Under such conditions there will clearly be no direct uni-
form compression on any normal section of a beam. This
is equivalent to assuming that the flexure is produced by an
indefinitely small force acting parallel to the axis (or at
right angles to a normal section) of the beam and at an
infinite distance from the latter.
It is clear, since the product of the distance of the point
of application of a force normal to the cross-section from
the centre of gravity of the latter multiplied by the dis-
tance of the neutral axis from the same point, but on the
opposite side from the point of application of the loading,
must be equal to the square of half the diameter of the
ellipse of inertia, that if that square be divided by
736 MISCELLANEOUS SUBJECTS, [Ch. XVI.
infinity, the distance of the point of appHcation of the
load from the centre, the quotient will be zero, i.e., the
neutral axis must pass through the centre of gravity of
the section.
This condition is further equivalent to taking any finite
loading at right angles to the axis of the beam, as in the
ordinary cases of engineering practice. The stresses found
in the normal section in such cases will be the direct tension
and compression with intensity varying directly as the
normal distance from the neutral axis with the accompany-
ing shears, as in the common theory of flexure.
The preceding investigations show, however, that with
unsymmetrical sections the neutral axis, while passing
through the centre of gravity of the section, is not at right
angles to the plane of loading, unless that plane happens
to contain one of the two principal axes of inertia of the
section.
Let the Z-bar section shown in Fig. 2 of the preceding
article be considered and suppose that the loading acts in
the vertical plane ZZ' , the latter line passing through the
centre of gravity C of the cross-section. It may be con-
sidered that the Z-bar is supported at each and on the lower
surface HL of the lower flange. Inasmuch as the bending
moment acts in the plane ZCZ' the neutral axis will be
drawn through the centre C parallel to the tangents to the
ellipse "where the line ZZ' cuts the latter, as shown at g
and at the opposite end, not lettered, of the vertical diameter.
The diameter A'B' is then the neutral axis desired. The
line Ch drawn at right angles to ZZ' may be considered
the axis of the external bending moment to which the beam
is subjected. The angle between Ch and the neutral
axis is a, as shown.
If the coordinate x be taken as at right angles to the
neutral axis A'B' , and if dA represent an element of the
Art. 1 1 6.] GENERAL FLEXURE TREATED BY CORE METHOD. 737
normal section of the beam, then the distance of that element
from the neutral axis measured parallel to ZZ' will be
X sec a. If k is the maximum intensity of stress at any
point of the section, that stress will occur at L or T, where
the value of x=n is the greatest for the entire section.
The distance of that point parallel to ZZ' will be n sec a.
If M is the value of the external bending moment acting
in the plane ZZ' , dM may be written :
dM = xsec a-dA'Xsec a. . . . (i)
n sec a .
If / is the moment of inertia of the section about the
neutral axis,
M = CdM =-I sec a- =-/A sec a. . . (2)
J n n
In Fig. 2 the line ZZ' cuts at d the side DF of the core.
Let the distance dC be represented by /. Then, as shown
in the preceding article,
jCZ'^Cg.
But the radius of gyration of the section about the axis
A'B' has been shown in Art. 81 to be equal to the normal
distance r' between the neutral axis and the parallel tangent
to the ellipse drawn at g.
Cg = / sec a.
■ It has already been seen that CZ' is equal to n sec a.
r'^ sec a = nj.
If this value of r''^ sec a be substituted in the third
member of Eq. (2) there will result,
M=kAj (3)
738 MISCELLANEOUS SUBJECTS. [Ch. XVI.
Eq. (3) is the expression for the external bending moment
in terms of the greatest intensity of stress in the section,
the area of that section, and the distance j from the centre
of the section to the side of the core as constructed by the
methods explained in the preceding section. Although the
construction has been made with the Z section the method
of procedure is precisely the same for any form of section
whatever.
Component Moments.
By again referring to Fig. (2) of the preceding article
it will be seen that M cos a is that component of the external
moment whose axis is parallel to the neutral axis, while
the component M sin a has an axis be at right angles to the
neutral axis, but lying in the plane of the normal section
of the beam. The former component produces the bending
stresses about the neutral axis, the maximum intensity of
which is k and a deflection normal to it; the latter compo-
nent moment tends to produce an oblique movement of
the beam in consequence of its unsymmetrical section..
This tendency in oblique flexure, especially marked with
unsymmetrical sections, is always toward that position
in which the least radius of gyration of the section (repre-
sented by the least semi-axis of the ellipse of inertia) is
found in the plane of bending, i.e., that plane in which the
bending moment acts. ,
In Fig. 2 of the preceding article ZZ' is the plane in which
the vertical loading acts, and it is clear that the plane in
which the resultant bending compression on one side of the
neutral axis A'B^ and the resultant bending tension on the
other side act is not the plane in which ZZ' lies, but inclined
somewhat to the right of CZ. Inasmuch as these two
planes are neither the same nor parallel, there must be
combined with the couple producing pure flexure such a
Art. 117.J PLANES OF RESISTANCE IN OBLIQUE FLEXURE. 739
couple as to make the resultant external moment equal
and opposite to the internal resisting moment, and the
component of M represented by be is such a couple, Ce
representing the couple producing pure flexure about
These analytic considerations show how essential it is
to give careful consideration to the principles governing
oblique or general flexure for loads not in a plane of
symmetry of a beam and for unsymmetrical sections.
The method of finding the location of the plane of
resistance of the bending stress existing in any normal
section of the beam will be given in the next article.
Art. 117. — Planes of Resistance in Oblique or General Flexure.
The preceding treatment of general flexure has shown
that the plane of action of the external bending moment
will not in general coincide with the plane in which the
internal resisting couple acts. The plane of the external
bending moment is supposed to pass through the axis of
the beam assumed to be straight. If this external bend-
ing couple is to produce pure flexure it must be in equilib-
rium with the internal moment produced by the stresses
in any normal section, and that requires that the two planes
of action shall either coincide or be parallel.
Let it be supposed that the 6X3jXf-inch steel angle
section shown in Fig. i represent any unsymmetrical sec-
tion, and let it also be supposed that G^F is the neutral
axis of the section, G being the centre of gravity; then let
GX and GY be the axes of rectangular coordinates negative
when measured to the left and downwards. The stresses
above GY will be supposed compressive, and those below,
tensile. The intensities will be assimied to vary directly
as the normal distances from GY as in the ordinary theory
740 MISCELLANEOUS SUBJECTS. [Ch. XVI.
of flexure. The centre of all the compressive stresses will
be taken at C and at T for the tensile stresses. The plane
whose trace is CT will be called the plane of resistance,
while AB^ will be taken as the plane of action of the
external bending moment. In other words, if the angle
were to carry vertical loading as a beam AB should be
vertical with the lines of cross-section correspondingly
inclined.
If xi and a'l are the coordinates of the centre C of the
compressive stresses in the section and if a is the intensity
of stress at a unit's distance from the neutral axis GY,
eqs. (i) and (2) will immediately result:
r Cyaxdxdy j Cxydxdy j^
C ( axdxdy C fxdxdy Qi
( j xaxdxdy \ \ x~dxdy j'^
\\ axdxdy ( j xdxdy Qi
The quantities /i and I'l sue the so-called ''product
of inertia " and the moment of inertia of that part of the
cross-section lying above GY, while Qi is obviously the
statical moment of the same part of the cross-section in
reference to the same axis.
If the subscript 2 be used for the corresponding quanti-
ties relating to that part of the section below GY, eqs. (3)
and (4) will at once result, the negative sign being used
in the second member because the coordinates are negative :
^2=-^, (3)
^2=-^^ (4)
Art. 117.] PLANES OF RESISTANCE IN OBLIQUE FLEXURE. 741
Q, I and / represent quantities belonging to the whole
cross-section, then, since G is the centre of gravity of that
section,
Qi=Q2=Q\
I\+I'2=I\
Jl+j2=J.
Fig. I.
It is desired to find the straight line joining C and T,
and in order to do that the general equation of a straight
line may be written as follows:
x+by—c.
(s)
742 MISCELLANEOUS SUBJECTS. [Ch. XVI.
If yi and Xi taken from eqs. (i) and (2) be first written
in eq. (5) and then >'2 and X2 from eqs. (3) and (4), and if
the second of the equations so formed be subtracted from
the first, there will result: b = — j.
Then eq. (5) will take the form
x=jy-\-c (6)
In Fig. I suppose a line parallel to CT drawn from G
to B. If the ordinate xi be produced upward, the line
BC = Gc' will be determined. If in eq. {6) y =0, x=c = Gc' =
BC. The triangles with the bases yi and >'2 will then be
similar and that similarity will be expressed by the following
equation, remembering that x and y are negative:
X\—C —X2 + C f .
— ^= — - (7)
Substituting the values of x\, yi, X2 and y2 established
above there will result the following value of c:
JQ •
Placing this value of c in eq. (6),
"^'f JQ ' .^^^
This is the equation of the line CT, Fig. i, drawn through
the centres of the tensile and compressive resisting stresses
acting in the normal section, i.e., it is the trace of the plane
in which the resisting couple acts. The tangent of the
Art. 117.] PLANES OF RESISTANCE IN OBLIQUE FLEXURE. 743
dx I
angle which it makes with the neutral axis GY is -r- =-.
dy J
If GY is one of the principal axes of inertia of the section
dx
J =0 and -:- becomes infiniteiy great, i.e., in that case
the line GT is at right angles to 6^y and it will presently
be shown that it will pass through G, the centre of gravity
of the section.
If :v=o in eq. (8),
^0= -^ ^ =Gc (9)
The distance Gc' is on the negative side of G. Again if
x=o, there will result :
y= _jQ =G^ (10)
These coordinates Ge and Gc' shown in Fig. i give two
points e and c' in the desired line CT, which must agree
obviously with the points C and T as found by computa-
tions.
If Ge should be zero, eq. (11) will result:
ja'2-I\j2=o (11)
Inasmuch as the moments of inertia I\ and I' 2 will
always have real values for an actual section, in general
if eq. (11) holds true, then must Ji=j2=o. That condi-
tion will of course exist for the principal axes and for the
case where at least one of the coordinate axes is an axis
of symmetry of the section.
Although the figure used for the establishment of the
preceding formulae is the normal section of a steel angle,
those formulae are completely general and are applicable
744 MISCELLANEOUS SUBJECTS. [Ch. XVL
to any form of cross-section whatever, as indicated by
eqs. (i) and (2) and all the equations following.
It is thus seen that if the plane of action AB oi any
external loading producing flexure of a beam with unsym-
metrical cross-section is parallel to the plane whose trace
is CT, there will be pure bending only as the external bend-
ing moment has the same axis as the couple formed by the
internal stresses. The planes of the external bending
moment and that of the internal resisting stresses may in
some cases coincide.
If the steel angle shown in Fig. i is to act as a beam
under vertical loading in pure flexure, the end supports
should be so formed as to make the lines AB and CT verti-
cal. In general, whatever may be the cross-section of a
beam, the latter should be so held at its points of support
that the loading will produce pure flexure. If the section
of the beam has an axis of symmetry, the plane of loading
may be taken through the axes of symmetry of the cross-
section.
Example. The application of the preceding formulae
may be illustrated by using the 6 X si-inch, 22.4-lb. steel
angle shown in Fig. i. The thickness of each leg is .75
inch. By using eqs. (i) and (2) there will at once result:
7'i=9.4i; r2 = i3.94; Ji=5-47\
72=3-04; 7=8.51; = 5-484
Inserting these values in eqs. (i), (2), (3) and (4) there
will result:
yi=i in.; ^^1=1.72 ins.; j2 = -.555 in.;
:;t:2 = —2.54 ins. ; :ro =— 1.02 ins. ; 3/0 = .372 in.
These coordinates are laid off in Fig. i, as shown, so as
to locate the four points C, e, c' and T. In making these
Art. ii8.] DEFLECTION IN OBLIQUE FLEXURE. 745
computations it should be remembered that I'l and I' 2
are moments of inertia of areas, one of whose sides coin-
cides with the axis of y and that the same observation is
also true of the quantities, /i and J2, as well as Q.
Art. 118. — ^Deflection in Oblique Flexure.
The general case of deflection of a beam with unsym-
metrical cross-section, or of a beam with symmetrical
cross-section but loaded obliquely, may readily be found
by the aid of the ordinary formulas for flexure used in
connection with the preceding investigations. The requisite
treatment may be well illustrated by considering the case
of a 6 X3i Xf-inch steel angle, the section of which is shown
in Fig. I to be same as that used in the preceding article.
Such an angle may be considered to be used as a beam
in roof work or for some other similar purpose with the
6 -inch leg placed in a vertical position. It will be assumed
that the span length is 15 feet = 180 inches and that the
angle is to carry as a beam a uniform load of 200 pounds
per linear foot. The data given in an ordinary handbook
on steel sections will show the position of the centre cf
gravity G of the section and enable the ellipse of inertia to
be constructed as in Fig. i.
The maximum radius of gyration represented by the
greater semi-axis of the ellipse is 1.97 inches, while the least
radius of gyration at right angles to the preceding and
represented by the smaller axis is .75 inch. The load
acts in a vertical plane passing through the axis G. The
various dimensions of the cross-section required in the
computations are" all shown in Fig. i.
By drawing vertical tangents on opposite sides of the
ellipse, the neutral axis A'B' drawn through the points of
tangency and the centre G of the ellipse is determined.
746 MISCELLANEOUS SUBJECTS. [Ch. XVI.
This neutral axis of the section makes the angle, 46° 30',
as carefully measured on the diagram, with the horizontal
axis of Y. By drawing a tangent to the ellipse parallel
to A'B^ the radius of gyration about the neutral axis is
found to be i.i inches, i.e., the normal distance between,
the neutral axis and the parallel tangent to the ellipse.
The greatest deflection of the angle beam will be found
at the centre of span at which the moment of the external
forces is
^^ 200X225 ^^^ =67,500 in.-lbs. . . . (i)
8
The component moment, as shown in the preceding
article, with axis parallel to the neutral surface, is
M cos a = .6884^=46,467 in.-lbs. ... (2)
The component moment having an axis at righ^" angles
to the neutral axis is, similarly,
M sin q: = . 7 2 5471^ = 48,964 in. -lbs. ... (3)
The actual flexure is produced by the first of these com-
ponents M cos a. The deflection produced by it will obvi-
ously be normal to the neutral axis, and it can be computed
by the ordinary formula for the deflection at the centre
of span of a beam simply supported at each end and loaded
uniformly throughout its length, the uniform load to be
taken in this case as 200 cos a = 138 pounds per Hnear foot.
If g is the load per linear foot of span, the usual expres-
sion for the centre deflection is w= ^ -r^-r - Substituting
3S4EI
138X15 for g/ in the form.ula, / = i8o inches, £=30,000,000,
Art. ii8.j
DEFLECTION IN OBLIQUE FLEXURE.
747
and 7 = 7.94 (moment of inertia of section about the neutral
axis) there will result:
w =0.66 inch.
(4)
As- cos a = .6884 and sin a = .7254, the vertical deflec-
tion =.66 X. 6884 =.454 inch; and the horizontal deflec-
tion = .66 X.7254 = .479 inch.
Fig.
It is thus seen that the horizontal deflection slightly
exceeds the vertical, in consequence of the major axis of
the ellipse of inertia being slightly inclined to a vertical
748 MISCELLANEOUS SUBJECTS [Ch. XVI.
line, thus causing the inchnation of the neutral axis of
the section to be relatively large.
Precisely the same general treatment would be followed
for any form of cross-section or any other amount or dis-
position of loading.
In the preceding article where the same angle was so
held as to make the plane of loading parallel to . that of
the resisting couple, the horizontal diameter of the ellipse
drawn through G is the neutral axis corresponding to the
conjugate diameter DF, parallel to the trace of the plane
of the resisting internal couple as determined in that
article. The normal distance, 1.95 inches, between the hori-
zontal diameter through G and the horizontal tangent at
F is the radius of gyration corresponding to the horizontal
neutral axis through G. As the area of cross-section of
the steel angle is 6.56 square inches, the moment of inertia
corresponding to the horizontal neutral axis through
G is J =6.56 X 1.95" =24.93, "the moment of inertia of the
cross-section about the neutral axis A'B\ Fig. i, is
7 = 6.56X1.1^ = 7.94. The distance from the horizontal
neutral axis through G to the extreme fibre is 3.82 inches,
while the corresponding distance of the extreme fibre from
A^B'' is 2.3 inches. Hence, the resisting moment for the
horizontal neutral axis through G is
3.82
. For the neutral axis A'B^:
2.3
Hence —p = i.g. In other words, the same angle placed
Art. 119.] ELASTIC ACTION UNDER DIRECT LOADING. 749
SO as to take the vertical loading in a plane parallel to the
resisting internal couple will offer nearly twice as much
bending resistance with the same extreme fibre stress as
when placed with the longer leg vertical. Economic use
of the metal as well as avoidance of unnecessary deflection,
therefore, requires that the beam of unsymmetrical section
shall be so held at its supports as to make the plane of
loading parallel to the resisting plane and as nearly parallel
to the greater axis of the ellipse of inertia of the cross-
section as possible.
Art. 119. — Elastic Action under Direct Loading of a Composite
Piece of Material.
Let it be supposed that a combined straight or cylin-
drical piece of material with length L is subjected to the
direct stress of either tension or compression. If the total
area of cross-section is A, it may be assumed to be composed
of the following parts :
A I =area of cross-section with modulus of elasticity Ei;
A 2 =area of cross-section with modulus of elasticity E2',
A3 =area of cross-section with modulus of elasticity E3;
etc., etc.
Then will
A=Ai-\-A2-\-A3-\-etc (i)
Let the total load P act parallel to L and let / be the
strain per unit of length of the piece, i.e., the unit strain,
then will IL be the total lengthening or shortening of the
piece. Under these conditions every part of the piece
will be subjected to the same rate of longitudinal strain
and the following equation may be at once written:
7SO MISCELLANEOUS SUBJECTS. [Ch. XVI.
EilAi-\-E2lA2-\-E3lA3-]-etc.=P=ElA. . . (2)
Hence,
1 = CO
Also the first and third members of eq. (2) will give
eq. (4) :
^ £iAi+£:2^2+£3^3+etc. , .
E=- J . . . (4)
Eq- (3) will give the lengthening or shortening of each
unit of length of the piece under any assigned load P, the
moduli of elasticity of the areas of the different parts of
the section being known.
The modulus of elasticity E given by eq. (4) may be
considered a mean or average modulus or an equivalent
value for the actual moduli, as the same longitudinal strain
would be yielded by a piece of uniform material having
that modulus of elasticity and the same area of cross-
section as the composite piece.
Art. 120. — ^Helical Spiral Springs.
A spiral spring like that shown in Fig. i takes its load
at the ends as indicated at A and B. In the general case
there may be applied at each end a single load P and a
couple, or either a force or a couple alone may act. The
analysis will be so written as to include concurrent force
and couple or either one separately. The following nota-
tion will be employed:
R = radius of spiral, Fig. i ;
(j) = pitch angle of spiral, Fig. i ;
z = axial elongation or compression of spring under load-
ing;
Art. 120.] HELICAL SPIRAL SPRINGS. 751
/ = length of spiral ;
r = radius of spiral wire ;
P = axial load, Fig. i ;
M = moment of applied twisting couple or torque, as-
sumed to be a right-hand moment ;
u — unit strain at unit distance from, the neutral axis
in bending or flexure ;
a = angle of torsion (unit strain at unit distance from
axis of piece in torsion) ;
T = total twist or rotation of spring measured on central
cylinder of spiral ;
T
T =—= angle of twist of spring in radians.
The force P will be considered positive when it stretches
the spring as shown in Fig. i. If the force P compresses
the spring it must have the negative sign in all the following
analysis.
The moment M will be considered a right-hand moment
when it twists the spiral so as to bring the helical parts
near together, i.e., tightens the spiral. It should be re-
membered that all parts of the spiral are uniformly stressed
or bent. The cross-section of the spiral rod will be con-
sidered circular, although the general analysis is adapted
to any form of cross-section.
The load P produces a moment Mi about the centre of
any section of the spiral rod given by
Mi=PR (i)
The axis of this moment is a horizontal line through
the centre of the section and tangent to the central cylinder
of the spiral shown by a broken-line circle in the lower
part of Fig. i. li A, Fig. 2, be the centre of the section
752
MISCELLANEOUS SUBJECTS.
[Ch. XVI.
considered, KL may be taken as the axis of the moment
PR. If AK, therefore, represent by a convenient scale, the
moment Mi=PR, AG and GK (drawn perpendicular to
AG) will represent by the same scale the component mo-
ments of Ml about those lines as axes passing through
the centre of the section. As the axis ^G* is the axis of
Fig. I.
Fig. 2.
the -spiral rod, it represents a torsion moment. Similarly
GK represents a bending moment as it lies in the section
and, in fact, is a neutral axis. Hence, if the subscripts
t and h mean torsion and bending,
And,
AG=M\=Mi cos (f>.
GK=M\=-M sin ct>.
(2)
(3)
Art. I20.] HELICAL SPIRAL SPRINGS. 753
The moment — M sin has a negative sign because the
triangle AKG, Fig. 2, shows that it will tend to untwist
the spiral of Fig. i, which is opposite to a positive effect.
The right-hand moment M will act at the centre of
section of the spiral rod about an axis parallel to AC,
Fig. I, i.e., about BD, Fig. 2, and AB may represent that
moment. Its two components will be:
BF = M'\=M sin cf>, .... (4)
AF=M",=M cos d, (5)
The resultant moments of torsion and bending at the
section considered will therefore be:
Mt=Mi cos +M sin (j>, ... (6)
M^=M cos 0-Afi sin 0. ... (7)
By the common theory of torsion (correct for a cir-
cular section only) if G is the modulus of shearing elasticity,
the angle of torsion, or unit strain a, is
moment Mi cos (/)+-M'sin 4) ,„s
«= -j-= ■ y. . ... (8)
Evidently, Q=G — (for circle); and Q=G— (for
2 6
square) .
If the exact theory of torsion is used for other sections
of the spiral rod than circular, the corresponding value of
a must be introduced, but no other change is needed.
In the same manner, if E is the modulus of elasticity
for direct stress, I the moment of inertia of the section
754
MISCELLANEOUS SUBJECTS.
[Ch. XVI.
Hadl cos
about its neutral axis, and if Q' =EI =E^^ (for circular
4
section) or Q' =E — (for square section), the unit strain,
12
Uy for bending is,
moment M cos 0— Mi sin , .
«=^^= Q, .. . . (9)
The quantities a and u are unit motions giving to the
spiral spring corresponding motions of rotation and axial
lengthenings or shortenings.
The torsion moment Mt will cause one end of an indefi-
nitely short length dl of the
Radi sin4> spiral rod to rotate through
the angle adl, inducing a
movement of that end, rela-
tive to the axis of the spiral,
perpendicular to the axis of
the rod, equal to Radl, as
shown by Fig. 3. The hori-
zontal component of this
-Budi cos movement tangent to the-
spiral cylinder is, Radl sin 4>,
or for each unit of length of
the rod, Ra sin 0. As the
state of stress is uniform
throughout the spiral rod, the
total circumferential twist of the spiral spring due to
torsion is
Fig. 3.
•Jtotdi sin '
-Rudl
Fig. 4-
r=Rla sin =Rt
And the angle of twist is
Ml cos (/)+Msin
Q
sm (/>.
(10)
r=^=/ Q sm<#..
(loa)
Art. 120.] HELICAL SPIRAL SPRINGS. 755
The axial component of the same movement, as shown
by Fig. 3 is, Radl cos . Hence the total axial movement
due to torsion is
, „jMi cos .
2 = -Rl ^, sm 4>. . . (13)
756 MISCELLANEOUS SUBJECTS. [Ch. XVI.
The angle of twist of the spring under loading will be
the sum of the second members of eqs. (loa) and (12a):
71/r 7 • , , /i I \ , Ti/Tz/sin^ , cos^ (t>\ f N
T=Mi/sm0cos0(^^-^j+M/(^-^+--Q7^j. (14)
The circumferential motion of the spring will be
T-Rr (15)
The axial extension or compression of the spring will
be found by the aid of eqs. (11) and (13) :
+Msin0cosc/)(^-^M. (16)
Eqs. (6) and (7) will enable any spiral spring to be
designed to perform a given duty such as to carry a pre-
scribed load or serve the purposes of a dynamometer, while
eqs. (14) and (16) will give the distortions of the spring,
either angular or axial.
If 5 is the greatest intensity of torsive shear in a normal
section of the spiral rod at the distance r from the centre,
while /p is the polar moment of inertia of the section,
. M,='-U. ...... (17)
r
irr
For a circular section, 1^ = '^^.
2
54
For a square section, Ip =—(£» = side of square).
6
Eq. (17) gives:
s=^. ■ (x8)
Art. I20.] HELICAL SPIRAL SPRINGS. 757
When 5 is given,
r=\ ^ (circular section). , . . (i8a)
\ tS
In both eqs. (6) and (7), Mi and M are known quanti-
ties, as they are the given loads.
Again if k is the intensity of stress in the most remote
fibre at the distance di from the neutral axis, and if I is
the moment of inertia of the section about the neutral axis,
^=—J- (^9)
When k is given,
c^i =f =x/^-* (circular section). . . (loa)
The two intensities 5 and k exist at the same point,
and they are to be used to determine the greatest intensities
of stress in the cross-section of the spiral rod precisely as was
done in Art. 10.
By eq. (2) of that article, the greatest and least inten-
sities of stress (principal stresses of opposite kinds) will be :
k I k^
max. intensity =-+a| 52 H — (tension)
2^4
min. intensity = — \/^^H — (compression).
2 ' 4
At the opposite end of that diameter of section of the
rod normal to the neutral axis where k is compression,
the above " max. intensity " will be compression also,
and the " min. intensity " will be tension.
758 MISCELLANEOUS SUBJECTS. [Ch. XVI.
The planes on which these principal stresses act are
given by eq. (3) of Art. (10) :
, 25
tan 2a = — —.
k
The greatest shear at the same point is given by eq.
(6) of Art. (9) ; i.e., its intensity is half the difference of the
principal intensities, or,
max. — min.
P,= — - — .
There are a number of special cases which may easily
be developed from the preceding general analysis.
Small Pitch Angle.
' If the pitch angle is so small that sin <}> may be con-
sidered zero without essential error,
sin 0=0 and cos = i.
Eqs. (6) and (7) then give:
Mt=Mi=PR\ ..... (20)
M6=M . (21)
From eqs. (14) and (16):
PRH ( X
Art. 120.] HELICAL SPIRAL SPRINGS. 759
Rotation of Spring Prevented.
In this case twisting of the spring is prevented, or r =0.
Eq. (14) then gives:
T\/r M sin (jy cos (Q' -Q) - ..
^=-^^Q'sin^0+Qcos^0- • • • ^^4)
Substituting this value of M in eq. (16) :
^_7^P2/f cQS^ ^ I si^^ ^ (sin cos cf>Y{Q' -QY \ /a
'"1 Q ^ Q' (0^sin^0+Qcos^c/>)QQt ^ ^^
The torsion moment Mt, eq. (6), and bending moment
Mb, eq. (7), are to be computed by using the value of M
given in eq. (24).
Axial Extension or Compression Prevented.
By making 2 =0 in eq. (16),
■ M.= -Mj^^^|fcg. . . . (.6)
Qcos2 0+Qsm2(/)
The angle of twist then becomes:
sin2 , cos2 4> (sin 4> cos0)2(Q' -Q)2
" ^\ Q ^ Q' (Q^cos2c/>4-Qsin2 0)Q^Qr ^'^^
For circular or square sections Q'— Q = ( 6^)(— or—)
and the square of the latter alternative factor is common
to {Q' —QY and Q'Q in the second number of eq. (27), thus
canceling and simplifying the numerical application of
that equation.
In computing Mt and Mb, eqs. (6) and (7), the value
of Ml given by eq. (26) is to be used.
76o MISCELLANEOUS SUBJECTS. [Ch. XVI.
This form of helical spring is employed in the transmis-
sion dynamometer.
Work Performed in Distorting the Spring.
The work performed in producing the angular and axial
distortions r and z by the moment M and force P is easily
found by the aid of eqs. (14) and (16) or corresponding
equations for special cases. The couple whose moment
is M performs work in twisting the spring through the arc
T (measured at unit distance from the axis of the helix)
expressed by
, w,=Mi. ...... (28)
The force P performs work in extending or compressing
the spring the distance z given by the equation
W^=^ . (29)
2
The total work done in the general case will then be :
W = WtVW^=^{Mr+Pz). . . . (30)
For special cases, as already indicated, the corre-
sponding value of T and z must be used in eq. (30).
In wTiting the preceding equations it has been assumed
that both M and P are gradually applied. If they were
suddenly applied, the distortions would be 2t and 2Z and
oscillations having those amplitudes would be set up.
The periods of the amplitudes would depend upon the'
masses moved. •
Art. 121.] PLANE SPIRAL SPRINGS. 761
Art. 121. — ^Plane Spiral Springs.
A plane spiral spring may be represented by Fig. i.
The outer end is fastened at B, but the inner end is secured
to a rotating post or small shaft at C. The spring or coil
is " wound up " to an increasing number of turns by apply-
ing a couple to the shaft C, as in winding a clock.
As a couple only is applied at C, every section of the
spring is subjected to bending by the same couple, i.e.,
there is a uniform bending moment throughout the entire
spring. This uniform condition of stress makes the analysis
of this spring exceedingly simple if the thickness of the metal
is small. , As the spring is a spiral beam subjected to uni-
form bending, the analysis, to be perfectly correct, should
be based on that for curved beams. That procedure would
introduce much complexit}^, and as the thickness of the strip
of metal constituting the spring is small compared with its
radius, no essential error is committed in neglecting the
effects of curvature. The usual cross-section of this type
of spring is a much elongated or narrow rectangle, the
greater dimension of the rectangle being parallel to the
axis of the couple or perpendicular
to the plane of the spring.
If u is the unit strain at unit
distance from the neutral axis of a
section of the spring, I the moment
of inertia of the same section about
the neutral axis, and E the modulus
of elasticity, while M is the moment
applied at C, Fig. i, W//////////////M
M=EJw= constant. . . (i)
Fig.
If / is the total length of the spring and /S the total
angular distortion for that length, then will udl be the
762 MISCELLANEOUS SUBJECTS. [Ch. XVI.
change of direction or angular distortion for each element
dl. Hence,
Mdl=EIudl=EIdl3. ...... (2)
Integrating :
Ml^EI^; and ^^g. . . . (3)
With the thin metal used / is small and 13 may be a
number of complete circles, perhaps sufficient to wrap the
spring closely around the shaft C.
If the moment M is applied gradually, the work done in
producing the total angular distortion /3 is
This is the same as the expression for the work performed
in bending a beam by a moment uniform throughout its
length. In fact the plane spiral is simply a special case of
flexure, the bending moment being uniform.
If the moment M should be applied suddenly, the total
angular distortion would be 2/3, and oscillations having
that amplitude might be set up.
Art. 122. — ^Problems.
Problem i . — A helical spring having a diameter of helix
of .3 inches and composed of twelve complete turns of a
f-inch round steel rod sustains an axial load of 45 pounds.
Find the axial deflection of the spring and the greatest
intensities of torsive shear and bending tension and com-
pression in the rod.
P = 45 lbs. ; ^ = I • 5 ins. ; = 15°; 1 = 117 ins. ;
£^ = 30,000,000; (7 = 12,000,000; • r=i^in.
Art. 122.
PROBLEMS FOR ARTS. 120 AND 121.
763
Mi=PR=6S,s in.-lbs.;
,7rr
Q=G — = 23,373;
2
M=o]
Q =E — = 29,217.
4
Substituting these quantities in eq. (16):
.=3Xii7(^^^+^^^^)68.s=..746m.
\23, 373 29,217/
By eqs. (6) and (7) :
Mt=Mx cos 0=66.2 in.-lbs.;
and
M6= -Ml sin 0= —17.74 in.-lbs.
Trr
TTf^
Since 1^ = '-^-^ and 7= — , eqs. (18) and (19) give:
2 4
5 =9460 lbs. per sq.in. torsive shear;
k =3432 lbs. per sq.in. greatest bending stress.
Problem 2. — Design a helical spring for a transmission
dynamometer for 8 H.P., at 90 revolutions per minute.
Axial distortion of the spring is prevented, or z=o. Let
low working stresses and other data be taken as follows*
Iz = 16,000 lbs. per sq.in.
i? = 3 ins. ; = 11°
G = 12,000,000;
MX9oX27r=8X33.ooo
Q=G— and
2
5 = 12,000 lbs. per sq.in. ;
.*. sin (/) = .i9i and COS0 = .982;
E =30,000,000.
M=466.8 ft.-lbs. =5602 in.-lbs.
Q'=£^.
Eq. (26) then gives:
Ml = — 212 in.-lbs.
764 MISCELLANEOUS SUBJECTS. [Ch. XVI.
By eqs. (6) and (7) :
M/ = 862 in. -lbs. ; and Ms = 5541 in. -lbs.
Solving eqs. (18) and (19) for the radius of the rod:
By eq. (i8a), r = .s6 in.; and by eq. (19a), r = .'j6 in.
Bending of the rod, therefore, requires the- greater
radius, and r = .'/6 in. will be taken.
Eq. (17) gives the greatest torsive shear in a section:
5 = 1250 lbs. per sq.in.
The equations following eq. (19) now give:
max. intensity = -f 16,097 lbs. per sq.in.;
min. intensity = —97 lbs. per sq.in.
The spring will be assumed to have twelve complete
turns, so that its length will be:
/ = 27r3 X 12 Xsec = 230.5 ins.
The twist r at unit distance from the axis of the helix
now becomes :
r = .i59in.
At the distance of 10 inches from the axis the twist
would be 1.59 inches, but the spring is too stiff to be very
sensitive. A higher working stress k may properly be taken.
If in the same problem there be taken 120 revolutions
per minute and an alloy steel for which the working stresses
/^ =40,000 pounds per square inch and 5=30,000 pounds
per square inch may be prescribed, then by using the
results already established :
•M=-^X 5602 =4200 in. -lbs.;
120
Art. 122.] PROBLEMS FOR ARTS. 120 AND 121. 765
Mi = -JX2i2 = — 159 in.-lbs. ;
Mr = f X862 =647 in.-lbs.;
-^6 = 4X5541 =4156 in.-lbs.
For shear: r =^/-X— X.36 =.67 X.36 =.24 in.
\4 2.5
For bending: r =a/-X— X. 76 =.67 X. 76 =.51 in.
y A 2.^
4 2.5
159
(.67)^
795.
At the distance of 10 inches from the axis of the helix
the twist would be ioX.795=7.95 inches.
Problem 3. — What will be the angular distortion j8
of a plane spiral spring i inch by -^-^ inch in section and
20 inches long if the distorting moment is 10 inch-pounds.
Eq. (3) of Art. 121 gives:
10X20 10 X20 X12 Xi2i;,ooo
^=- — =. ^ =10
30,000,000 Xi 30,000,000
(about 1 1 complete turns).
The fibre stress is
10 X--
k = = 1 50,000 lbs. per sq.m.
12 X 125,000
Art. 123.— Flat Plates.
The correct analysis of stresses in loaded fiat plates even
of the simplest form of outline has not yet been made suf-
ficiently workable for ordinary engineering purposes, either
for plates simply resting on edge supports or with edges of
plates rigidly fixed to their supports. It is necessary,
766 MISCELLANEOUS SUBJECTS. fCh. XVI.
therefore, to combine simple, but approximate analysis
based on reasonable assumptions, with experimental results
so as to obtain workable formulae. The following pro-
cedures, due chiefly to Bach and Grashof, are commonly
employed in treating flat plates:
Square Plates — Uniform Load.
In Fig. I let A BCD represent a square plate simply
resting on the edges of a square opening. Tests of such
y plates by Bach have shown that when
increasingly loaded they will ulti-
mately fail along a diagonal, as AB.
Let the plate be uniformly loaded
with p pounds per square unit, then
let moments be taken about the diag-
onal AB. If & is the side of the
Pjg. I. square, the load on the triangular
half of the square is - — , and the distance of its centre
2
from AB is ^^ sin 45° =.23 6^. The upward supporting
forces or reaction on the sides AD and DB will also be
half the load on the plate, — , and its centre will be at the
2
distance ^=.3546 from AB. Hence the moment
2
about AB will be:
M=^(.SS4b-.236b)=.oS9pb^. . . . (i)
If h be the thickness of the plate, the moment of inertia
I about its neutral axis will be:
J b sec 45° h^ ^..^ , X
/= !t^ =.iiSbh^ (2/
12
Art. 123.I SQUARE PLATES. 767
The ordinary flexure formula then gives for the greatest
intensity of bending stress k, assuming it to be uniform
throughout the diagonal section,
^^~T^IW V ^^^
Or, if the thickness is desired,
k^h^^. ....... (4)
Eq. (4) gives the thickness of plate required to carry
the unit load p when the working stress is fe.
Square Plates — Single Centre Load.
If a single load P rests at the centre of a square plate,
using Fig. i and following the same method as in the
preceding section, the moment about the diagonal AB
will be:
7,, Ph sin 45"^ „, , .
M=- ^ ^^ =.i77Pb. .... (5)
The moment of inertia I is the same as befoie and it
is given by eq. (2). Hence, assuming a uniform intensity
k throughout the extreme fibres :
,, _ .i7yPbh _sP ,.
' ii~-4/? ^^^
Or,
h = .S66yj^. (7)
768
MISCELLANEOUS SUBJECTS.
[Ch. XVI.
Rectangular Plates — Uniform Load.
Fig. 2 shows a rectangular plate with sides a and h.
With a much oblong rectangle the indications of tests are
not so well defined as to the section of failure, but tenta-
tively the diagonal section AB may be taken as a close
approximation for usual proportions. DF is a normal
to AB drawn from D. The uniform load on the triangular
half ABD of the plate is - — and its centre of action is at
2
n
the normal distance - from AB. The centre of the sup-
3
porting forces or reaction along the edges AD and DB is
- from AB. Hence the moment about AB is
M
pah In n
2 \2 3
pabn
12
(8)
Fig. 2.
Referring to Fig. (2) :
n=b sin (f) and AB =b sec <]). . . (9)
Therefore the moment of inertia of the diagonal section
is
J _b sec (t)h^ ^ . pab^ sin _ .b sec (ph^
12
12
Art. 123.]
Hence,
CIRCULAR PLATES.
769
7 ah sin cos
k=p -— -\ or
P
sin cos 0. (10)
2/r ' \2k
As is obvious, P=pab is the total load on the plate.
Rectangular Plate — Centre Load.
If a single load P rests at the centre of the plate, the
moment about the diagonal AB, Fig. 2, is produced by the
reaction, only, of the supporting forces along the edges
AB and BD, and its value is
2 2
Consequently,
(11)
k =
3P sin cos 4>
or, h
-4
iP
sm cos 0.
(12)
Circular Plate — Uniform Load — Centre Load.
The circular plate with radius r is shown in Fig. 3.
The same general assumptions are made as in the preceding
cases, i.e., uniform condition of
bending stress throughout the
section of failure and uniform
support along the edge of the
plate. It is clear that the latter
assumption is strictly correct for
the circular outline. Any diam-
eter, as AB, may be taken for
the section of failure.
It will be convenient to sup-
pose the uniform load to be ap-
plied on a circle of radius ri, as shown in Fig. 3. Then
the load on half of the plate is p and its centre is at
770 MISCELLANEOUS SUBJECTS. [Ch. XVI.
AT 1
the distance - — from AB. The edge-supporting force or
reaction, equal to the half load on the plate, has its centre
2T
at the distance of — from AB. The moment about the
TT
latter diameter is, therefore,
2 \ TT 37r/ \ 3 /
If h is the thickness of the plate, as in the preceding
cases, the moment of inertia 7 is
r 2rh^ rh^ , .
1 = =— -; . . , , o , . (14)
12 o
Hence,
Or,
M=prMr~ri)=k— (15)
\ 3 / 3
k=pri^ U • •••... (16)
h=ri
VK-'t) <■'>
If the load is uniform over the entire circular plate,
r=ri, and
M = i--; ^ = ^J; and, h = r^^. . . (18)
If the load is concentrated at essentially a point, ri =0,
but 2^^Il- must be displaced by — ;
-^ ' i|; and, h=M.
irh^ \ irk
These formulae for circular plates are more nearly
M=P-\ ^=^; and, h=J^. . . (19)
TT irh^ \ irk
Art. 123.]
ELLIPTICAL PLATES.
771
correct in analysis and give results more nearly in agree-
ment with tests than those derived for other cases.
Elliptical Plates — Centre Load — Uniform Load.
An elliptical plate is shown in Fig. 4. The approximate
formulae for this case may be conveniently established by
first considering two axial strips of
the same (unit) width, the length
of AB being 2a and of CD, 2b, a
single load being placed at their
intersection. The centre deflec-
tions of the tw^o strips as parts of
the plates must be the same. Let
Pi be the centre load for the strip
AB, and P2 the centre load for CD.
The desired centre deflection for each strip acting as a beam
is given by eq. (28), Art. 28. The equality of the two de-
fl.ections gives the equation, 2a being one span and 2b the
other :
Pia^ P2b\ _ Pi^b^
P2~a^'
Fig. 4.
6EI 6EI
or
(20)
h^
As each strip is of unit width, I = — , h being the thick-
12
ness of plate. Hence the greatest fibre stresses are
b
. Mh „ a
and, k2=zP'.
h^'
Eqs. (21) and (20) then give:
ki _Pi a
k'2~¥2~b
62
(21)
(22)
Eq. (22) shows that ^2 is the greatest fibre stress and,
hence, that the major axis of the ellipse will be the line of
failure, as would be anticipated without the analysis.
772 MISCELLANEOUS SUBJECTS. [Ch. XVI.
If the ellipse of Fig. 4 be elongated by lengthening
the major axis 2a to infinity, the result will be a corre-
spondingly long rectangular plate of 26 width or span.
Hence, the greatest fibre stress for this case of uniform
load will be for a unit cross strip of plate :
, Mh p{2hY hi2 ^b^
This is the greatest intensity of stress for an ellipse
whose major axis 2a is infinity. The other extreme is the
circle for which the greatest intensity of stress is, eq. (18),
^=^g • • • (24)
For ellipses in general, in the absence of a satisfactory
analysis, it is tentatively proposed to wTite:
-(3-!)f («)
When b=a, eq. (25) gives the correct value for a circle,
and when ^ = o the result is correct for the extreme ellipse.
The thickness of plate for a given uniform load p is
= ^V(3 -._')! (.6)
a/ k
Flat Plates Fixed at Edges.
Grashof and others have partly by analysis and partly
empirically deduced a number of formulae for plates fixed
at their edges, i.e., encastre, instead of simply supported.
The following have been used and may be considered fairly
satisfactory, using the same notation as in the preceding
parts of this article.
I. Circular plate with radius r and uniform load p.
The greatest intensity of stress is, if h is the thickness,
Art. 123.] PLATES WITH FIXED EDGES, 773
^=^Jp; and, ^='-Jj|- • • • (27)
II. Stayed flat surfaces, stay bolts being the distance
c apart in two directions at right angles to each other.
Each stay carries the uniform load pc^. The greatest
intensity of stress may be taken:
k=p-^-, and, h=-J^. . . . (28)
III. Rectangular plate a long, b wide, supporting uniform
unit load p. The greatest intensity of stress may be taken :
k=pTnT-r^^h:^ and, h=a^h^^-j-^y^^. (29)
If the plate is square, a = b'.
k=p^,- and, /.=-^^|. . . . (30)
All these plates with edges either fixed or simply sup-
ported are supposed to be truly fiat, as any arching or
dishing changes materially the conditions of stress.
Problem i. — What thickness of steel plate is required
to carry a load of 200 pounds per square inch over a rect-
angular opening 24 by 36 inches. Eq. (10) gives the
expression for the thickness h of the plate when simpl}^
supported along its edges. The total load isP = 2ooX36x
24 = 168,800 pounds.
tan (/) =11 = .667 .•. =33° 40' and sin cos (^ = .461.
If the working stress ^ = 16,000 pounds per square inch;
h=\— — '- X. 461 =1.56 inches. A plate iye inches
\2 X 16,000
thick, therefore, meets the requirements.
Problem 2. — Design a circular steel plate, simply sup-
ported on its edge, for an opening 30 inches in diameter
774 MISCELLANEOUS SUBJECTS. [Ch. XVI.
to carry a load of loo pounds per square inch, if k =-15,000
pounds per square inch. r = i5 inches and P = iooXTrr^ =
100 X 706.9 = 70,690 pounds.
Eq. (18) then gives: h = is\ = 1.22 inches. A
>i 1500
plate I J inches thick will therefore be satisfactory.
If the plate were rigidly fixed along its edge^ eq. (27)
shows that the thickness would be: h = i. 22V^ = i inch
thick.
Art. 124. Resistance of Flues to Collapse.
If a circular tube or flue be subjected to external normal
pressure, such as that of steam or water, the material of
which it is made will be subjected to compression around
the tube, in a plane normal to its axis. If the following
notation be adopted,
/ = length of tube;
d = diameter of tube ;
t = thickness of wall of the tube ;
p = intensity of excess of external pressure over internal ;
then will any longitudinal section It, of one side of the tube,
pld
be subjected to the pressure — . But. let a unit only of
length of tube be considered. This portion of the tube is
approximately in the condition of a column whose length
and cross-section, respectively, are nd and t.
The ultimate resistance of such a column is (Art. 35)
As this ideal column is of rectangular section,
12
Art. 124.] RESISTANCE OF FLUES TC COLLAPSE. yji
and
But P=pd, hence
i2d''
(i)
is the greatest intensity of external pressure which the tube
can carry. But the formulae of Art. 35 are not strictly
applicable to this ideal column. The curvature on the one
hand and the pressure on the other tend to keep it in position
long after it would fail as a column without lateral support.
Hence p will vary inversely as some power of d much less than
the third.
Again, it is clear that a very long tube will be much more
apt to collapse at its middle portion than a short one, as the
latter will derive more support from the end attachments;
and this result has been established by many experiments.
Hence p must be considered as some inverse function of the
length /.
Eq. (i), therefore, can only be taken as typical in form,
and as showing in a general way, only, how the variable
elements enter the value of p. If x, y, and z, therefore, are
variable exponents to be determined by experiment, there
may be written
f='M^ (^)
in which c is an empirical coefficient.
Sir Wm. Fairbairn (" Useful Information for Engineers,
Second Series") made many experiments on wrought-iron
tubes with lap- and butt-joints single riveted. He inferred
776 MISCELLANEOUS SUBJECTS. [Ch. XVI.
from his tests that y=z = i. Two different experiments
would then give
pld = ct-, (3)
p'Ud'=ct'' (4)
Hence
log {pld) =log c-\-x\ogt,
\og{p'l'd')=\ogc-^x\ogt'\
in which "log" means "logarithm." Subtracting one of
these last equations from the other, the value of x becomes
, pld
log'
_ log ipld) - log {p'Vd') --S \p'Vd',
^ \ogt-\ogf ~ /t^ ' ' ' ^^^
log[j^
As p, I, d, t, p',V,d\ and f are known numerical quantities
in every pair of tests, x can at once be computed by eq. (5) ;
c then immediately results from either eq. (3) or eq. (4).
By the application of these equations to his experimental
data, Fairbairn found for wrought-iron tubes:
^ = 9,675,600-^, (6)
in which p is in pounds per square inch, while t, I, and d are
in inches. Eq. (6) is only to be applied to lengths between 18
and 120 inches.
He also found that the following formula gave results
agreeing more nearly with those of experiment, though it is
less simple:
/^■'9 d
;^ = 9, 675, 600-^-0. oo2y (7)
Art. 124.] RESISTANCE OF FLUES TO COLLAPSE. 777
Fairbairn found that by encircling the tubes with stiff
rings he increased their resistance to collapse. In cases
where stick rings exist, it is only necessary to take for I tlie
distance between two adjacent ones.
In .1875 Prof. Unwin, who was Fairbairn' s assistant in
his experimental work, estabHshed formulae with other
exponents and coefficients (" Proc. Inst, of Civ. Engrs.,"
Vol. XL\^I). He considered x, y, and z variable, and
found for lubes with a longitudinal lap-joint:
t-'
/? = 7,363,000^^^^^6 (8)
From one tvibe with a longitudinal butt-joint, he deduced:
/2.21
^ = 9>6i4,ooo ^o.9^x.x6 .(9)
For five tubes with longitudinal and circumferential joints^
he found:
:^ = 15, 547, 000^57^7776 (10)
By using these same experiments of Fairbairn, other
writers have deduced other formula, which, however, are
of the same general form as those given above. It is proba-
ble that the following, which was deduced by J. W. Nystrom,
will give more satisfactory results than any other:
^.692,800^ (")
At the same time, it has the great merit of more simple
application.
From one experiment on an elliptical tube, by Fairbairn,
it would appear that the formulas just given can be approxi-
778
MISCELLANEOUS SUBJECTS.
[Ch. XVI .
mately applied to such tubes by substituting for d twice
the radius of curvature of the elliptical section at either
extremity of the smaller axis. If the greater diameter or
axis of the ellipse is a and the less 6, then, for d, there is
to be substituted -r.
Art. 125. — Approximate Treatment of Solid Metallic Rollers.
An approximate expression for the resistance of a roller
may easily be written. The approximation may be con-
sidered a loose one, but it furnishes a basis for an accurate
empirical formula.
The following investigation contains the improvements
by Prof. J. B. Johnson and Prof. H. T. Eddy on the
method originally given by the
author.
The roller will be assumed
to be composed of indefinitely
thin vertical slices parallel to
its axis. It will also be as-
sumed that the layers or slices
act independently of each
other.
Let E' be the coefficient of
elasticity of the metal pver the
roller.
Let E be the coefficient of elasticity of the metal of
the roller.
Let R be the radius of the roller and R^ the thickness
of the metal above it. ,
Let ze; = intensity of pressure at A ;
. (( ({ (( a
at /i;
'* any other point
Art. 125.] TREATMENT OF SOLID METALLIC ROLLERS. 779
Let P = total weight which the roller sustains per unit
of length.
X be measured horizontally from A as the origin ;
d=AC;
e = DC,
From Fig. i :
E E
•. d = AC = AB + BC = w{^^-^ . . . (i)
and
A'C=A'B^^B'C=p(^^~}j, ... (2)
Dividing eq. (2) by eq. (i),
But
P = jydxJI^£'A'C'dx.
If the curve BAR be assumed to be a parabola, as may
be done without essential error, there will result:
C^' A'Cdx^
^ed.
Hence
P=-we (3)
78o
But
MISCELLANEOUS SUBJECTS.
[Ch. XVI.
e = V2Rd-d'^ = ViRd, nearly.
By inserting the value of d from eq. (i) in the value of
e, just determined, then placing the result in eq. (3),
liR-^R^
p=-^V-^^
This is the desired equation of the line, in which r is
measured normal to the axis of the cylinder or jet, while y
is measured along that axis from the extremity of the jet.
When the material is wholly expelled,
y = —D, and r = o.
Eq. (2) is applicable to the jet only. For the line hF or
Gk, resort will be had to the equation
d(d) _ 2R,' dr
d ~R'-R^' r'
Again integrating between the limits d and D, or r and
R^, and reducing,
d\ '^^'
This value of r is the radius of that portion of the primi-
tive central cylinder which remains over the orifice when D
is reduced to d.
Art. 136. — Positions in the Jet of Horizontal Sections of the
Primitive Central Cylinder.
That portion of the primitive central cylinder below ab,
in Fig. I of Art. 13 3 will be changed to ahKH in Fig. 2 of
the same article.
8i6 THE FLOIV OF SOLIDS. [Ch. XVIII.
If, in the latter Fig., y is the distance from HK to ab,
measured along the axis, then the volume of HKab will
have the value
. rv
J
If d' is the distance ciF^bG, in Fig. i, the equality of
volumes will give
r r'dy^R^'{D-d').
Eq. (i) of Art. 125 gives
R^-Rx^
R2
If A/" is the number of horizontal layers required to com-
pose the total thickness D, and n the number in the depth d',
Hence
?U"&) J'
y'-^A ^-{m) \d (2)
Art. 137.] FINAL RADIUS OF HORIZONTAL SECTION. 817
Tresca computed values of y' for some of his experiments
and compared the results with actual measurements. The
agreement, though not exact, was very satisfactory. Within
limits not extreme, the longer the jet the more satisfactory
was the agreement.
Art. 137,. — Final Radius of a Horizontal Section of the Primitive
Central Cylinder.
Let it be required to determine what radius the section
situated at the distance d^ from the upper surface of the
primitive central cylinder will possess in the jet.
It will only be necessary to put for y in eq. (i) of Art.
135 the value of y' taken from eq. (i) of Art. 136. This
operation gives
Hence
^/\ .R^
r'=RAD) (i)
If R^ is small, as compared with R, there will result ap-
proximately
/d'\y^
Art. 138. — Path of Any Molecule.
The hypotheses on which the theory of flow is based
enable the hypothetical path of any molecule to be easily
established.
8i8 THE FLOIV OF SOLIDS. [Ch. XVIII.
In consequence of the nature of the motion there will be
three portions of the path, each of which will be represented
by its characteristic equation, as follows:
First, let the molecule lie outside of the primitive central
cylinder.
Let R' and H be the original co-ordinates of the mole-
cule considered, measured normal to and along the axis of
the cylinder, respectively, from the centre of the orifice HK
(Fig. I, Art. 133) as an origin, while r and h are the variable
co-ordinates.
The first hypothesis, by which the density remains con-
stant, then gives the following equation:
7t{R'-R'')H = -{R''-r'')K
or
hR'~hr' = {R'-R")H (i)
This is the equation to the path of the molecule, in
which r must always exceed R^.
As this equation is of the third degree, the curve cannot
be one of the conic sections. ■
Second, let the m^olecide move in the space originally occu-
pied by the central cylinder.
While h and r now vary, the volume 7:r^(D~h) must
remain constant. When r^R^^ let h=h^. Hence
r\D-h)=R^'(P-\), ..... (2)
But if h=\ and r = R^ in eq. (i),
Placing this value in eq. (2).
r\D-h)=R,'[D-H^^-^^,y . . . (3)
Art. 138.] PATH OF ANY MOLECULE, 81Q
Third, let the molecule move in the jet.
After the molecule passes the orifice, its path will evi-
dently be a straight line parallel to the axis of the jet. Its
distance r^ from that axis will be found by putting h=o m
eq. (3). Hence
APPENDIX 1.
ELEMENTS OF THEORY OF ELASTICITY IN
AMORPHOUS SOLID BODIES.
CHAPTER I.
GENERAL EQUATIONS.
Art. I. — Expressions for Tangential and Direct Stresses in Terms
of the Rates of Strains at Any Point of a Homogeneous Body.
Let any portion of material perfectly homogeneous be
subjected to any state of stress whatever. At any point as
Oy Fig. I, let there be assumed any three rectangular co-
ordinate planes; then consider any small rectangular par-
allelopiped whose faces are parallel to those planes. Finally
let the stresses on the three faces nearest the origin be re-
solved into components normal and parallel to their planes
of action, whose directions are parallel to the co-ordinate
axis.
The intensities of these tangential and normal compo-
nents will be represented in the usual manner, i.e., p.,3, signi-
fies a tangential intensity on a plane normal to the axis of
X (plane ZY), whose direction is parallel to the axis of
y, while pxx signifies the intensity of a normal stress on
820
Art. I.]
TANGENTIAL AND DIRECT STRESSES.
821
a plane normal to the axis of X (plane ZY) and in the
direction of the axis of X. Two unlike subscripts, there-
fore, indicate a tangential stress, while two of the same kind
signify a normal stress.
Fig. I.
From eq. (3), Art. 2, and eq. (7), Art. 5, there is at
once deduced
5 =
2(i+r)
^=Gcl>.
(i)
Now when the material is subjected to stress the lines
bounding the faces of the parallelepiped will no longer be
at right angles to each other. It has already been shown
in Art. 2 that the angular changes of the lines from right
angles are the characteristic shearing strains, which, multi-
plied by Gs give the shearing intensities.
Let ^^ be the change of angle of the boundary lines
parallel to X and Y.
Let (^2 ^^ "t^^ change of angle of the boundary lines
parallel to Y and Z.
Let ^3, be the change of angle of the boundary line
parallel to Z and X.
822 ELASTICITY IN AMORPHOUS SOLID BODIES. [Ch. I.
Eq. (i) will then give ^he following three equations:
E ^
^^y^TiTTr)'!'^'^ (2)
E ^
^^^==7(rT7)^2^ ...... (3)
E ^
^-=7(7T7)^3 (4)
In Fig. I let the rectangle agfh represent the right pro-
jection of the indefinitely small parallelepiped dx dy dz. If
u, V, and w are the unit strains parallel to the axes of x, y,
and z of the original point h, the rates of variation of strain
-;-, -r, -T-i etc., may be considered constant throughout
dx dy dz
this parallelopiped ; consequently the rectangular faces will
change to oblique parallelograms. The oblique parallelo-
gram dhck, whose diagonals may or may not coincide with
those of agjh, therefore, may represent the strained con-
dition of the latter figure.
Then, by Art. 2, the difference petween dhc and the right
angle at h will represent the strain ^^. But, from Fig. i, ^^
has the following value:
cl)^=dhe-\-bhc. ....... (5)
But the limiting values of the angles in the second mem-
ber are coincident with their tangents ; hence
de be .^.
Art. I.] STRESSES IN TERMS OF STRy^INS. 823
But, again, de is the distortion parallel to OX found by
moving parallel to OY only; hence it is a partial differential
of w, or it has the value
'^=^'^y (7)
In precisely the same manner be is the partial differential
of V in respect to x, or
bc = ^-dx.
dx
By the aid of these considerations, eq. (6) takes the form
du dv
'i'^-Ty+d^-- • • • ■ ■ (8)
If A^y be changed to YZ, and then to ZX, there may be
at once written by the aid of eq. (8)
dv dw .
'^-=dz"'dy\ (9)
dw du . .
^'=5;f+5?- • (^°>
Eqs. (2), (3), and (4) now take the following form:
^(dii dv\ . .
^/dv dw\ , ^
^/dw du\ . .
^-=^5^+^; <'3)
824 ELASTICITY IN AMORRHOUS SOLID BODIES. [Ch. I.
The direct stresses are next to be given in terms of the
displacements u, v, and w. Again, let the rectangular par-
allelopiped dx dy dz be considered. Eq. (i), on page 3,
shows that the strain per unit of length is found by dividing
the intensity of stress by the coefficient of elasticity, if a sin-
gle stress only exists. But in the present instance, any state
of stress whatever is supposed. Consequently the strain
caused by p^^, for example, acting alone must be combined
with the lateral strains induced by pyy and p^.. Denoting
the actual rates of strain along the axes of X, Y, and Z by
/j, l^, and Z3, therefore, the following equations may be at once
written by the aid of the principles given on pages 9 and 10 :
^'=k+(Pyy + Pj^'^ .... (14)
^=h+(p..+PJ^'^ .... (15)
Eliminating between these three equations,
?»«=rf,[u^(^.+4+^3)]; . • (17)
^w„
yy i+^l 2
r^z I +rL ^ I — 2f^ ^ ^ J
But if ti, V, and w are the actual strains at the point where
these stresses exist, the rates of strain l„ l^, and l^ will evi-
Art. I.] STRESSES IN TERMS OF STRAINS. 825
dently be equal to j^,j-^ and T7, respectively. The volume
of the parallelopiped will be changed by those strains to
dx{i+l^)dy(i+l^)dz{i+l^) =dx dy dz{i +1^ + 1^^-!^)
if powers of l^, l^, and l^ above the first be omitted. The
quantity (l^-hl^ + h) is, then, tJie rate of variation of volume,
or tlie amount of variation of volume for a cubic unit. If
there be put
^ dti dv dzv ^ ^ E
= :t-, + -j-+-j-, and G =
dx ' dy dz' 2(1 +r)'
eqs. (17), (18), and (19) wih take the forms
2Gr . ^du . ..
P^^-T^r^+'^dx' • • • • (2°)
2Gr „ ^dv
2Gr dw
p ,= d^2G-T- (22)
The form in which eqs. (14), (15), and (16) are written
shows that if p^^, pyy, or p^^ is positive, the stress is tension,
and compression if it is negative. Consequently a positive
value for any of the intensities in eqs. (20), (21), or (22) will
indicate a tensile stress, while a negative value will show
the stress to be compressive.
The eqs. (14) to (19), together with the elimination in-
volved, also show that the coefficients of elasticitv for ten-
sion and compression have been taken equal to each other,
and that the ratio r is the same for tensile and compressive
strains.
826 ELASTICITY IN AMORPHOUS SOLID BODIES. [Ch. I.
Further, in eqs. (ii), (12), and (13), it has been assumed
that G is the same for all planes.
Hence eqs. (11,) (12), (13), (20), (21), and (22) apply
only to bodies perfectly homogeneous in all directions.
It is to be observed that the co-ordinate axes have been
taken perfectly arbitrarily.
Art. 2. — General Equations of Internal Motion and Equilibrium.
In establishing the general equations of motion and equi-
librium, the principles of dynamics and statics are to be
applied to the forces which act upon the parallelopiped repre-
sented in Fig. I , the edges of which are dx, dy, and dz. The
notation to be used for the intensities of the stresses acting
on the different faces will be the same as that used in the
preceding article.
Let the stresses which act on the faces nearest the origin
be considered negative, while those which act on the other
three faces are taken as positive.
The stresses which act in the direction of the axis of X
are the following:
On the face normal to X, nearest to 0, — p^^ dy dz ;
" ''isiTthestiTomO,(p^^ + -~^dxjdydz;
* dy dx nesLvest to 0, —p^^dydx;
it (i
farthest from 0, (p.^ + ~T^^^ ]dy dx ;
dz dx nearest to 0, —p y^ dz dx ;
** ** farthest from 0, ipy^ + -j^dyjdzdx.
Art. 2.] EQU/tTIONS IN RECTANGULAR CO-ORDINATES.
827
clz
dx
dy
The differential coefficients of the intensities are the rates
of variation of those intensities for each unit of the variable,
which, multipHed by the
differentials of the varia-
bles, give the amounts of
variation for the different dz
edges of the par allelopiped . |d^
Let Xq be the external dx
force acting in the direc-
tion of X on a unit of vol-
ume at the point consid-
ered ; then X^dxciy dz will
be the amount of external
force acting on the paral- Fig. i.
lelopiped.
These constitute all the forces acting on the parallelo-
piped in the direction of the axis of X, and their sum, if un-
d'^u
balanced, must be equal to m-rr^dx dy dz ; in which m is the
mass or inertia of a unit of volume, and dt the differential
of the time. Forming such an equation, therefore, and drop-
ping the common factor dx dy dz, there will result
dx ^ dy ^ dz ^^'>- "^dt^-
(i)
Changing x to y, y to z, and z to x, eq, (i) will become
+ -zJ7f+-^nr+yo = nt:Tr^' ... (2)
dx ^ dy ^ ^- ^^' "^
d'.
Again, in eq. (i), changing x to z, z to y, and y to x,
dx ^ dy '^
^7 _ ^
dz ^"^'-"^dt''
(3)
828 ELASTICITY IN AMORPHOUS SOLID BODIES. [Ch. I.
The line of action of the resultant of all the forces which
act on the indefinitely small parallelopiped, at its limit,
passes through its centre of gravity, consequently it is sub-
jected to the action of no unbalanced moment. The parallelo-
piped, therefore, can have no rotation about an axis passing
through its centre of gravity, whether it be in motion or
equilibrium. Hence, let an axis passing through its centre
of gravity and parallel to the axis of X, be considered. The
only stresses, which, from their direction can possibly have
moments about that axis, are those with the subscripts (yz),
{zy), {yy), or {zz). But those with the last two subscripts
act directly through the centre of the parallelopiped, conse-
dp
quently their moments are zero. The stresses ~r^^dy .dx dz
dpz
and — T^ dz . dx dy are two of six forces whose resultant is
directly opposed to the resultant of those three forces which
represent the increase of the intensities of the normal, or
direct, stresses on three of the faces of the parallelopiped;
these, therefore, have no moments about the assumed axis.
The only stresses remaining are those whose intensities are
pzy and pys. The resultant moment, which must be equal
to zero, then, has the following value:
py^dx dz.dy + pzydx dy .dz = o\ ... (4)
' .*. Pyz=-Pzy (5)
Hence the two intensities are equal to each other.
The negative sign in eq. (5) simply indicates that their
moments have opposite signs or directions; consequently,
that the shears themselves, on adjacent faces, act toward
or from the edge between those faces. In eqs. (i), (2), and
(3), the tangential stresses, or shears, are all to be affected
Art. 2.] EQUATIONS IN RECTANGULAR CO-ORDINATES. 829
by the same sign, since direct, or normal, stresses only can
have different signs.
The eq. (5) is perfectly general, hence there may be
written :
P.y=Py.^ and p,,=p,, (6)
Adopting the notation of Lame, there may be written:
P..=^\^ Pyy-^\^ P..-^\\
Pzy^^v Pxz = T^2^ Pxy-^z\
by which eqs. (i), (2), and (3) take the following forms:
dN, dT, dT, ,, dhi
dT, dN, , dT, ^^
dT, dT, dN, ^
'd^ + ^^-df+^^-'^
The equations (11), (12), (13), (20), (21), and (22) of the
preceding article are really kinematical in nature ; in order
that the principles of dynamics may hold, they must satisfy
eqs. (7), (8), and (9). As the latter stand, by themselves,
they are applicable to rigid bodies as well as elastic ones;
but when the values of A^ and T, in terms of the strains u, v,
and w, have been inserted, they are restricted, in their use,
to elastic bodies only. With those values so inserted, they
form the equations on which are based the mathematical
theory of sound and light vibrations, as well as those of
elastic rods, membranes, etc. In general, they are the equa-
tions of motion which the different parts of the body can
' dt' ' *
. . (7)
d'v
' df' '
. . (8)
d'w
. . (9)
830 ELASTICITY IN AMORPHOUS SOLID BODIES. [Ch. I.
have in reference to each other, in consequence of the elastic
nature of the material of which the body is composed.
If all parts of the body are in equilibrium under the
action of the internal stresses, the rates of variation of the
d'^u d^v . d^w -11 1 i
strams -77^, -7tf» and -7-^, will each be equal to zero.
Hence, eqs. (7), (8), and (9) will take the forms
dN, dT, dT, ^
-57 + ^ + ^+^0 = 0;. . . . (10)
dTo dN^ dT. ^_ , ^
itt+ify+-dr+^'-°'- • ' ' (")
dT, dT, , dN, ^ , ^
-dt + ^ + lk+^'-° (-)
These are the general equations of equilibrium. As they
stand, they apply to a rigid body. For an elastic body, the
values of N and T from the preceding article, in terms of the
strains n, v, and w, must satisfy these equations.
The eqs. (10), (11), and (12) express the three conditions
of equilibrium that the sums of the forces acting on the
small parallelopiped, taken in three rectangular co-ordinate
directions, must each be equal to zero. The other three con-
ditions, indicating that the three component moments about
the same co-ordinate axes must each be equal to zero, are
fulfilled by eqs. (5) and (6). The latter conditions really
eliminate three of the nine unknown stresses. The remaining
six consequently appear in both the equations of motion
and equilibrium.
The equations (7) to (12), inclusive, belong to the interior
of the body. At the exterior surface, only a portion of the
small parallelopiped will exist, and that portion will be a
Art. 2.] EQUATIONS JN RECTANGULAR CO-ORDINATES. 831
tetrahedron, the base of which forms a part of the exterior
surface of the body, and is acted upon by external forces.
Let — be the area of the base of this tetrahedron, and let
2 .
p, q, and r be the angles which a normal to it forms with
the three axes of X, Y, Z, respectively. Then will
da cos p =dy dz, da cos q=dz dx, and da cos r -=dx dy.
Let P be the known intensity of the external force acting
on da, and let tt, /, and p be the angles which its direction
makes with the co-ordinate axes. Then there will result :
Xq=P da. cos 7z, Yq=P da.cos Xy and Zq=P da. cos p.
The origin is now supposed to be so taken that the apex of
the tetrahedron is located between it and the base; hence
that part of the parallelopiped in which acted the stresses
involving the derivatives, or differential coefficients, is
wanting ; consequently those stresses are also wanting.
The sums of the forces, then, which act on the tetra-
hedron, in the co-ordinate directions, are the following:
— (A\ dy dz -{- T.^ dz dx + T^ dy dx) + Pda cos 7t=o\
— (Tg dz dy + N, dz dx 4- T^ dy dx) -h Pda cos ^ = o ;
— ij^ dz dy + T^ dz dx + A^g dy dx) + Pda cos ^ = o.
Substituting from above,
A^j cos ^+ Tg cos g + 72 cos r = P cos tt; . . (13)
T^cosp\N^cosq^T^co?>r=P cos i\ . . (14)
r^ cos ^ + r^ cos (7 + A^g cos r = P cos ^. , . (15)
These equations must always be satisfied at the exterior
surface of the body; and since the external forces must
always be known, in order that a problem may be determi-
nate, they will serve to determine constants which arise
832
ELASTICITY IN AMORPHOUS SOLID BODIES.
[Ch. I.
from the integration of the general equations of motion and
equihbrium.
Art. 3. — Equations of Motion and Equilibrium in Semi-polar
Co-ordinates.
For many purposes it is convenient to have the condi-
tions of motion and equilibrium expressed in either semi-
polar or polar co-ordinates ; the first form of such expression
will be given in this article.
The general analytical method of transformation of co-
ordinates may be applied to the equations of the preceding
article, but the direct treatment of an indefinitely small
portion of the material, limited by co-ordinate surfaces, pos-
sesses many advantages. In Fig. i are shown both the
#^
K
X ^
dx
h
>--^^
^iv
y
e
; 1 \
1 1
Y
s
N
1
>
'^
nM-
Fig. I.
Y
small portion of material and the co-ordinates, semi-polar
as well as rectangular. The angle made by a plane normal
to ZY, and containing OX, with the plane XY is repre-
sented by (f) ; the distance of any point from OX, measured
parallel to ZY, is called r; the third co-ordinate, normal to
Art. 3.] EQUATIONS IN SEMI-POLAR CO-ORDINATES, 833
r and ^, is the co-ordinate x, as before. It is important to
observe that the co-ordinates x, r, and (f>, at any point, are
rectangular.
The indefinitely small portion of material to be con-
sidered will, as shown in Fig. i , be limited by the edges dx, dr^
and r d(j). The faces dx dr are inclined to each other at the
angle d(f).
The intensities of the normal stresses in the directions of
X and r will be indicated by A^^ and R, respectively. The
remainder of the notation will be of the same general char-
acter as that in the preceding article; i.e., T^^ will represent
a shear on the face dr .r dcj) in the direction of r, while N^ is
a normal stress, in the direction of ^, on the face dx dr.
The strains or displacements, in the directions of x, r, and
(j), will be represented by ti, p, and w ; consequently the
unbalanced forces in those directions, per imit of mass,
will be
d^u d^p ^ d^w , ^
"^w^ "^w^ ^^^ ""'w (')
Those forces acting on the faces hf, fe, and he, will be
considered negative ; those acting on the other faces, posi=
tive.
Forces Acting in the Direction of r.
— R.rdcpdx, and
-\-Rr dcj) dx+l-^ — dr = r-^dr + R drjdcf) dx.
— T^rdr dx, and
+ T^rdrdx + —T^d(l).drdx.
^T^r-'f d(j) dr, and
7'X-'
'\-Txr-fd(l)dr + -j^dx.rd(j)dr,
834 ELASTICITY IN AMORPHOUS SOLID BODIES. [Ch. I.
On the face dr dx, nearest to ZOX, there acts the normal
stress ( N^^dr dx + -j-^d^ .drdx\=N'\ and A^' has a com-
ponent acting parallel to the face fe and toward OX, equal to
N' sin {d(f)) =N'^—^=N'd(j). But the second term of this
product will hold ( dx,
— Tx^.r dcf) dr, and
+ Tx^.rd(f)dr + —-T^dx.rd(f)dr,
As in the case of A/",^, in connection with the forces along
r, so the force T^j. dr dx has a component along ^ (normal
to fe) equal to T^rdrdx. sin {d ^dP ' * ^"■'
dT„dRdT,R--N^ d^p_
dT,^ dTr4, dN^^ Tr^ + Tr^ _,«^
dx '^ dr ^ rd4,- r + '""-^^df
(4)
These are the general equations of motion (vibration) in
terms of semi-polar co-ordinates ; if the second members kre
made equal to zero, they become equations of equilibrium.
Eqs. (2), (3), and (4), are not dependent upon the nature of
the body.
Since x, r, and ^ are rectangular, it at once follows that
^ rx — -^ xrj J- r4> = ^ 4>r^ 2-M i x4>^ J- X' • * (5/
836 ELASTICITY IN AMORPHOUS SOLID BODIES. [Ch. I.
In order that eqs. (2), (3), and (4) may be restricted to
elastic bodies, it is necessary to express the six intensities
of stresses involved, in terms of the rates of variation of the
strains in the rectangular co-ordinate directions of x, r, and
(j). Since these co-ordinates are rectangular, the eqs. (11),
(12), (13), (20), (21), and (22) of Article i, may be made
applicable to the present case by some very simple changes
dependent upon the nature of semi-polar co-ordinates.
For the present purpose the strains in the co-ordinate
directions of x, y, and z will be represented by ti\ v', and
w\ Since the axis of x remains the same in the two systems,
evidently
du^ du
dx dx'
From Fig. i it is clear that the axis of y corresponds
exactly to the co-ordinate direction r; hence
dv^ dp
dy ~dr'
From the same Fig. it is seen that the axis of z corre-
sponds to ^, or r(j). But the total differential, dw\ must be
considered as made up of two parts ; consequently the rate
dvu^
of variation -j- will consist of two parts also. If there is no
distortion in the direction of r, or if the distance of a mole-
cule from the origin remains the same, one part will be
div dw
-TT—TT =~-n' If, however, a unit's length of material be re-
d{r(l)) rdcf)
moved from the distance r to r + ^ from the centre 0, Fig. i,
while (j) remains constant, its length will be changed from
I to (i-f-l, in which p may be implicitly positive or
Art. 3. J EQUATIONS IN SEMI-POLAR CO-ORDINATES,
negative. Consequently there will result
837
dw'
dz
dw p
rdcbr'
For the reason already given, there follow
dii^ du dv^ dp
i—=~r~ and T~^ = -r-.
dy dr ax ax
In Fig. 2 let dc be the side of a distorted small portion
of the material, the original position ^
of which was d'e. Od is the distance
r from the origin, ad=dr and ac =
dw, while dd' = w. The angular
change in position of dc is —, = -,- ;
^ ah w ,
but an amount equal to —3 = - is due to the movement of
r, and is not a movement of dc relatively to the material
immediately adjacent to d.
Hence
dp
Fig. 2.
dw' _dw w dv'
d^^d^~^' ^^^^ di
rd(j)'
There only remain the following two, which may be at
once written
dw' dw
dx dx
. du' du
and -r=—Ti,
dz rd(j>
The rate of variation of volume takes the following form
in terms of the new co-ordinates:
~^^ dy dz ~dx^dr'^rd4~^r'
dx
(6)
838 ELASTICITY IN AMORPHOUS SOLID BODIES, [Ch. I.
Accenting. the intensities which belong to the rectan-
gular system x, y, z, the eqs. (11), (12), (13), (20), (21), and
(0.2^. of Art. I, take the following form:
», -'V,-^,«+.^£: (rt
^ 1-2V dr* ^ ^
2Gr - r^( dw p\
(9)
^-^.'=
'■..=n'=<^+.T-r)^ <■■>
n-'-.'=K£+^*)- • • • • • ■(">
If these values are introduced in eqs. (2), (3), and (4),
those equations will be restricted in application to bodies
of homogeneous elasticity only.
The notation t is used to indicate that the r involved is
the ratio of lateral to direct strain, and that it has no rela-
tion w^hatever to the co-ordinate r.
The limiting equations of condition, (13), (14), and (15)
of Art. 2, remain the same, except for the changes of nota-
tion, shown in eqs. (7) to (12), for the intensities N and T.
Art. 4,J
EQUATIONS IN POLAR CO-ORDINATES.
839
Art. 4. — Equations of Motion and Equilibrium in Polar
Co-ordinates.
The relation, in space, existing between the polar and
rectangular systems of co-ordinates is shown in Fig. i . The
angle is measured in the plane ZY and from that oi XY;
Fig. I.
while (p is measured normal to ZY in a plane which contains
OX. The analytical relation existing between the two sys-
tems is, then, the following:
.x = rsin0, y =r cos ([' cos, 6, and z=--r cos (p sin 6.
The indefinitely small portion of material to be considered
IS ah e d. It is limited by the co-ordinate planes located by
(j) and 0, and concentric spherical surfaces with radii r and
r -h dr. The directions r, (p, and (p, at any point, are rectangu-
lar ; hence the sums of the forces acting on the small portion
of the material, taken in these directions, must be found and
put equal to
m
'dp '
m
dt'
and
m
dt'
840 ELASTICITY IN AMORPHOUS SOLID BODIES. [Ch. I.
in which expressions, p, 7;, and co represent the strains in the
direction of r, ^, and ^ respectively.
Those forces which act on the faces ah, bd, and cd will be
considered negative, and those which act on the other faces
positive.
The notation will remain the same as in the preceding
articles, except that the three normal stresses will be indi-
cated by Nr, N^, and N^.
Forces Acting Along r.
— Nr.r d(ff r cos 4f dcf),
i-Nr.r^ cos (l^dil^dcj)
/d(N r^) dN \
"^ ( dr '^ ^ ^'^^^+ 2rNrdrj cos (p dip d,
-T^r-rd([fdr.
+ T^r .rdilfdri- jt^ d .r dip dr . sin aOc = — A^^ .r dip dr . cos ip d-r'^ COS ip dcp dip
+ ( '^^y^ ^r= r''^dr + 2r Tr^dr) cos .r d(p dr.
+ N4>.rdiljdr + -j^d^rd([fdr.
— T^^.r cos (p dcj) dr,
-{jT^^cos (p.r dcf) dr
\ d'' — ^^^^^^ 4^~^7d4'-T^^s\n (pdipjrd^dr.
+ T^rf' d 4^ dr. cos i[f dcj), on face c^.
— T^^ r dip drl sin akc = — y— ) = — T^^ r dip dr. sin if' d(f)y
on face ce.
The lines ak and ck are drawn normal to Oc and Oa,
Forces Acting Along (p.
— Tr^.r cos ip d(j).r dip.
■\-Tr4,r^ cos ip dcp di[}
'dr+2rTr^drj cos (pdipdcp.
— T^^.rd.i/' dr.
+ T^4. r dip dr + -j-^dcp .r dip dr.
— N^.r cos ip dcj) dr.
■\-N^.r cos ip dcp dr '
./d(N^cosiP),^ dN^ . \
+ I ^. d ip = cos ip-T-fdip — N^sm ip dip j r dcp dr.
\-T^.r cos ip dcp dr .dip, on isice be.
f A^,^ .r dip dr. sin akc = + N^f, r dip dr. sin ip dcp, on face ce.
-^^-^^''^' -dr
842 ELASTICITY IN AMORPHOUS SOLID BODIES. [Ch. I.
The volume of the indefinitely small portion of the
material is (omitting second powers of indefinitely small
quantities)
r cos (Jf d(j).r d(fi.dr = AV ,
and its mass is m multiplied by this small volume. The
latter may be made a common factor in each of the three
sums to be taken.
The external forces acting in the directions R, (j), and (p
, will be represented by
RJV, 0JV, and WJV,
respectively.
Taking each of the three sums, already mentioned, and
dropping the common factor J F, there will result
d^ dT^r ^ dT^,r ^ 2 Nr - N^ - A^^ - T^, tan
dr r cos (p.d^ rdip r .
d^p
'df
+ Ro=^^^M^> (i)
dTrj dN^_ dT^^
dr r cos (p .dcp r dip
,^tan<^-r^^tan(/; ^^^ „..
dt
2Tr^ + T4>r-T^^tEin(l>-T^^tan(p d^-q
+ + <^o = ^«Tri^; (2;
dT,, ^ dT,^ ^ dN^
dr r cos (l^d(\) r d(p
2 Tr^, + T,, - N^ tan y^ + A^^ tan ^^ _ d^
Since r, -/>, and ^ are rectangular at any point,
Art. 4. J EQUATIONS IN POLAR CO-ORDINATES, 843
Hence
r r '
2 Tr^ + T^r - tan yA ( jV^ - A ^^) _ 3 r,^, - tan <^ (.V^ - .V^)
r r '
These relations somewhat simplify the first members of
eqs. (2) and (3).
Eqs. (i), (2), and (3) are entirely independent of the
nature of the material ; also, they apply to the case of equi-
librium, if the second members are made equal to zero.
The rectangular rates of strain, at any point, in terms
of r, (j), and (/^ are next to be found. As in the preceding
article, the rates of strain in the rectangular directions of
r, (j), and (p will be indicated by
dv^ dw' du' dv^ du'
Sy' 57"' d^' d^' 57' ^^^•
Remembering the reasoning in connection with the value
dw^
of ~7— , in the preceding article, and attentively considering
Fig. I, there may at once be written,
dtt^ doj p
dx' r d if) f
In Fig. I, if ac = I and ah = uj, while ak =r cot. (p (ak is
perpendicular to aO), the difference in length between ac
and bh will be
CO CO tan (p
rcot d) r
This expression is negative because a decrease in length takes
place in consequence of a movement in the positive direction 1
of nl).
844
ELASTICITY IN AMORPHOUS SOLID BODIES.
[Ch. I.
Again, a consideration of Fig. i, and the reasoning con-
nected with the equation above, will give
dw' drj _ p coicHKp
rcoSil>d(j) r
Without explanation there may at once be written:
dv^ _dp
d^'~dr'
Fig. I of this, and Fig. 2 of the preceding article, give
du' doj CO . dv' dp
dy' dr r
and
rd(lf'
Precisely
dr r dx'
These are to be vised in the expression for T^
the same figures and method give
dv^ dp dw' df] T)
dz' rcosil^dcj) " dy dr r'
which are to be used in finding T,^;..
The expression for -r-j- will be composed of the sum of
two parts. In Fig. 2, ah is the original position of r d(^>, and
after the strain t] exists it takes the position ec. Consequently
ac (equal and parallel to bd and perpen-
dicular to ak) represents the strain tj,
while ed represents drj. vSince, also, fc is
perpendicular to ck, the strains of the kind
Tj change the right angle fck to the angle
fee; or the angle eck is equal to
dw'
d^=''^-
df)
. - ed ca
dck = -1- + — r
do ak
Fig.
r dil) r cot (^'
In Fig. 2, the points a, 6, and k are
identical with the points similarly lettered in Fig. i. The
Art. 4. J EQUATIONS IN POLAR CO-ORDINATES. 845
expression for tt ^^^Y be at once written from Fig. i. There
may, then, finally be written,
dw^' df] 7) tan ^ . du' _ dco
dx' ~~rd(j) r dz' r cos ^ d(j)'
These equations will give the expression for T^^,
The value of
du^ dv' dw'
^^M^d^^'^W
now takes the following form:
^ do df] dii) 2p 6; tan
l) rj ^"
^** = ^(nft. + r cos "i d^ + ^L^j . .... (8)
( dp df) r)\
^ '*^'^\r cos ^d4>^d~r 7)' ^^°>
846 ELASTICITY OF AMORPHOUS SOLID BODIES. [Ch. I.
If these values are inserted iii eqs. (i), (2), and (3), the
resulting equations will be applicable to isotropic material
only.
As in the preceding article, V is used to express the ratio
between direct and lateral strains, and has no relation what-
ever to the co-ordinate r.
It is interesting and important to observe that the equa-
tions of motion and equilibrium for elastic bodies are only
special cases of equations which are entirely independent of
the nature of the material, of equations, in fact, which
express the most general conditions of motion or equilibrium.
CHAPTER IT.
THICK, HOLLOW CYLINDERS AND SPHERES, AND
TORSION.
Art. 5.— Thick, Hollow Cylinders.
In Fig. I is represented a section, taken normal to its
axis, of a circular cylinder whose walls are of the appreciable
thickness t. Let p and p^^ represent the interior and exterior
intensities of pressures, respectively. The material will not
be stressed with uniform intensity throughout the thickness /.
Yet if that thickness, comparatively
speaking, is small, the variation will
also be small; or, in other words,
the intensity of stress throughout
the thickness t ma}^ be considered
constant. This approximate case
will first be considered.
The interior intensity p will be
considered greater than the exterior
p^, consequently the tendency will
be toward rupture along a diametral plane. If, at the same
time, the ends of the cylinder are taken as closed, as will be
done, a tendency to rupture through the section shown in the
figure will exist.
The force tending to produce rupture of the latter kind
will be
F^7:(pr^'-p,r,') (i)
847
848 THICK, HOLLOIV CYLINDERS. [Ch. II.
If N^ represents the intensity of stress developed by this
force,
If the exterior pressure is zero, and if r' is nearly equal to
r, + r'
2
TV— ^^ = ^ r\
^"^^ 2(r,-/) 2t ^^^
In this same approximate case, the tendency to split the
cylinder along a diametral plane, for unit of length, will iDe
If A/"' is the intensity of stress developed by F\
^ =T^. — t — • (4)
xV is thus seen to be twice as great as N^ when p^ = o. If,
therefore, the material has the same ultimate resistance in
both directions the cylinder will fail longitudinally w^hen the
interior intensity is only half great enough to produce trans-
verse rupture, the thickness being assumed to he very small and
the exterior pressure zero.
N^ and N' are tensile stresses, because the interior pres-
sure w^as assumed to be large compared with the exterior. If
the opposite assumption were made, they would be found to
be compression, while the general forms would remain ex-
actly the same.
AjL 5.] THICK, HOLLOIV CYLINDERS. 849
The preceding formulas are too loosely approximate for
many cases. The exact treatment requires the use of the
general equations of equilibrium, and the forms which they
take m Art. 3 are particularly convenient. As in that article,
the axis of x will be taken as the axis of the cylinder.
Since all external pressure is uniform in intensity and
normal in direction, no shearing stresses will exist in the
material of the cylinder. This condition is expressed in the
notation of Art. 3 by putting
T^x = Tfx = Tf.^ = o.
Again the cylinder will be considered closed at the ends,
and the force F, eq. (i), will be assumed to develop a stress
of uniform intensity throughout the transverse section
shown in Fig. i. This condition, in fact, is involved in that
of making all the tangential stresses equal to zero.
Since this case is that of equilibrium, the equations (2),
(3), and (4) of Art. 3 take the following form, after neglect-
ing Xo, Rq, and 0q'.
■5^ = °^ (5)
f +5^=0. («
-VT4>''° (7)
These equations are next to be expressed in terms of the
strains u, p, and w.
In consequence of the manner of application of the exter-
nal forces, all movements of indefinitely small portions of
850 THICK, HOLLOIV CYLINDERS. [Ch. II.
the material will be along the radii and axis of the cylinder.
Hence
ti will be independent of r and (p;
p ^ ^;
The rate of change, therefore, of volume will be (eq. ^6)
of Art. 3)
du dp p
dx dr r ^ ''
* . • 1 1 ^ dd dhi 1 ' r
As p IS mdependent of x, ~^^ =TT2 i hence if the value of
N^ be taken from eq. (7) of Art. 3 and put in eq. (5) of this
article,
dK\ 2GX dhi ^dhi_
dx ~i-2Vdx'^^^"dx'~^'
d'u
.*. -J— 2 = and ti---^ax + a.
But the transverse section in which the origin is located
may be considered fixedt. Consequently if x-~=q, u=q and
thus a' =0. The expression for u is then u =ax.
The ratio u-'rx is the / of eq. (i), on page 3, while the
p of the same equation is simply N^ of eq. (2), given above.
Hence
Art. 5. J THICK, HOLLO JV CYLINDERS. 851
Again, eq. (8) of Art. 3, in connection with eqs. (8)
and (6) of this, gives
2GV /d'p dp _p\^^Jd^ j^Ap._(!\ =0
i.— 2X\dr'^ r dr r^J ' " \dr^ r dr r'y
d(^'
^ d'p ^ dp _p _d"y V, _^
' * dr^ r dr r^ dr"^ dr
dp p
,\ ~ + -=c, or
dr r
r dp + p dr^dypr) ^-cr dr.
cr"^ ^ cr b , ^
.*. pr=— + b, or P = j+-- . . . (10)
This value of p in eqs. (8) and (9) of Art. 3 will give
iV(a + c) c b)
At the interior surface R must be equal to the internal
pressure, and at tlie exterior surface to the external pressure.
Or since negative signs indicate compression,
If r =/.... , R=~p,
li r=r^ . . . . . R=-p^.
Either of these equations is the simple result of applying
eqs. (13), (14), and (15) to the present case, for which
cos /? = cos r = cos ;r = cos ,0 = o,
cos q = cos X = I , and P -= — p or — p^.
852 THICK, HOLLOIV CYLINDERS. [Ch. II.
Applying eq. (11) to the two surfaces,
Subtracting (14) from (13),
r'^ — r^
Inserting this value in eq. (13),
( I — 2t 2'
The general expressions of R and A^,^^, freed from the
arbitrary constants of integration, can now be easily written
by inserting these last two values in eqs. (11) and (12). By
making the insertions there will result
The stress A^,^<^ is a tension directed around the cylinder,
and has been called "hoop tension." Eq. (16) shows that the
hoop tension will be greatest at the interior of the cylinder.
An expression for the thickness, t, of the annulus in terms of
the greatest hoop tension (which will be called h) can easily
be obtained from eq. (16).
Art. 6.] TORSION IN EQUILIBRIUM, 853
If r =r' in that equation,
h =
•• / \2p,-p + hJ '
.■..-=.H(;?r^)'-!----"
Eq. (17) will enable the thickness to be so determined
that the hoop tension shall not exceed any assigned limit h.
If p^ is so small in comparison with p that it may be neg-
lected, t v/ill become
H(?7^)'-l '■«
If p^ is greater than p, N' becomes compression, but
the equations are in no manner changed.
The values of the constants b and c may easily be found
from the two equations immediately preceding eq. (15).
It is interesting to notice that the rate of change of vol-
ume, 6, is equal to (a + c) and therefore constant for all
points.
Art. 6. — Torsion in Equilibrium,
The formulas to be deduced in this article are those first
given by Saint -Venant, and established in substantially the
same manner.
It will in all cases, except that of the final result for a
rectangular cross-section, be convenient to use those equa-
tions of Art. 3 which are given in terms of semi-polar co-
ordinates.
854
TORSION IN EQUILIBRIUM.
[Ch. II.
Let Fig. I represent a cylindrical piece of material, with
any cross-section, fixed in the plane ZY, and let the origin of
co-ordinates be taken at 0. Let
it be twisted also by a couple
P.ab=Pl,
the plane of which is parallel to
ZY. The material will thus be
subjected to no bending, but to
pure torsion.
The axis of the piece is sup-
posed to be parallel to the axis
of X as well as the axis of the
couple. Normal sections of the
piece, originally parallel to ZOYy
will not remain plane after tor-
sion takes place. But the tendency to twist any elementary
portion of the piece about an axis passing through its centre
and parallel to the axis of X will be very small compared
with the tendency to twist it about either the axis of r or
r
■''* ^rd4> ' dr r
o;
(1)
(2)
(3)
(4)
Art. 6.] TORSION IN EQUILIBRIUM. . 855
After introducing the values of T^x and T^x, from eqs.
(10) and (12) of Art. 3, in eqs. (2), (3), and (4) of the same
article, at the same time making the external forces and
second members of those equations equal to zero, and bear-
ing in mind the conditions given above, there will result
dr rd^ r
/dJu d^p d^w d^u dii ^^p\ _
^^\dr^'^dFdx'^rdJdx'^?d4''^7dr^7^)^'^' *^5^
dTrx / d'u d'p\ /M
dTxd. ^/d\v d^M
^(7u^2 +r;7x^- =0 (7)
dx \dx- rd(j)dx
Also by eq. (6) of Art. 3,
dx dr rd(j)^r ^ ^
The cylindrical piece of material is supposed to be of
such length that the portion to which these equations apply
is not a.ffected by the manner of application of the couple.
This portion is, therefore, twisted uniformly from end to
end; consequently the strain u will not vary with any
change in x. Hence
du
Tx^° (9)
Eq. (i) then shows that ^ = 0. This was to be antici-
pated, since a pure shear cannot change the volume or
856 TORSION IN EQUILIBRIUM. [Ch. 11.
density. Because ^ = 0, eqs. (2) and (3) at once give
dp dw p
-/ = — r7 + -=o (10)
dr rd(j) r ' ^
As the torsion is uniform throughout the portion con-
sidered,
— = =— (11)
Eq. (11), in connection with eq. (10), gives
d'^w , .
, ^, -0 (12)
rdxdcj)
Eqs. (11) and (12), in connection with eq. (10), reduce
eq. (5) to the following form:
% K'5?)
d^u ,d^u du d .. \ "• /
^^^dj''^d?'^7dr^''"^d^''^^ dr ' ' ' ^'^^
Both terms of the second member of eq. (6) reduce to
zero by eqs. (9) and (11), and give no new condition. The
second term of the second member of eq. (7) is zero by
eq. (9) ; the remaining term therefore gives
d^w
As the stress is all shearing, p will not vary with (f).
Hence
dp
Alt. 6.] TORSION IN EQUILIBRIUM. 857
Eqs. (10), (11), and (15) show that ^=0, and reduce
eq. (4) to
div w
' Eq. (10) now becomes ^ =0, and shows that w does
not contain 9^; while eq. (14) shows that w does not con-
tain x' or any higher power of x. The strain w, in connec-
tion with these conditions, is to be so determined as to sat-
isfy eq. (16).
If a is a constant, the following form fulfils all condi-
tions :
w = arx (17)
Eq. (17) shows that the strain w, in the direction of cf),
i.e., the angular strain at any point, varies directly as the dis-
tance from the axis of X, and as the distance from the origin
measured along that axis. This is a direct consequence of
making Tr=o.
The quantit}^ a is evidently the angle of torsion, or the
angle through which one end of a unit of fibre, situated at
unit's distance from the axis, is twisted ; for if
r=x = i, w = a.
An equation of condition relative to the exterior surface
of the twisted piece yet remains to be determined ; and that
is to be based on the supposition that no external force what-
ever acts on the outer surface of the piece. In eqs. (13),
(14), and (15) of Art. 2, consequently, P -=o. The conditions
of the problem also make all the stresses except
T^ = T^r and T^ = T^^
858 TORSION IN EQUILIBRIUM. [Ch. II.
equal to zero, while the cylindrical character of the piece
makes
p = go°; .'. cos p=o.
If cos / be written for cos r,
cos t -= sin q.
Eq. (13), just cited, then gives
Txr cos q + T^^, sin q=o (18)
But since ^ = o and w = arx,
and
Eq. (18) now becomes
du
dr dr
^-=^^57 ^^9)
Tx=Gl::^,+ar) (20)
du --^^^^=-r7^' • • • ^''^
rd(j)
in which r^ is the value of r for the perimeter of any normal
section.
Eqs. (13) and (21) are all that are necessary and all that
exist for the determination of the strain u. Eq. (13) must
be fulfilled at all points in the interior of the twisted piece,
while eq. (21) must at the same time hold true at all points
of the exterior surface.
Art. 6.] TORSION IN EQUILIBRIUM. 859
After u is determined, Txr and Tx4> at once result from
eqs. (19) and (20). The resisting moment of torsion then
becomes
M =ffT-4> ^' ^'t'-^^'^^ff^-^^^ dcj) + GaIp. (22)
In this equation Ip= J J r^ .rdcfidr is the polar moment of
inertia of the normal section of the piece about the axis of
A^, and the dotible integral is to be extended over the whole
section.
According to the old or common theory of torsion
M=GaIp.
The third member of eq. (22) shows, however, that such an
expression is not correct unless tt is equal to zero ; i.e., unless
all normal sections remain plane while the piece is subjected
to torsion. It will be seen that this is true for a circular sec-
tion only.
It may sometimes be convenient to put eq. (22) in the
following form:
M-GJ J rdr.-j^dcp + GaIp = Gj u.rdr + Galp. (23)
In this equation u is to be considered as
/
'f' du
di'^'f''
while the remaining integration in r is to be so made that
the whole section shall be covered.
86o TORSION IN EQUILIBRIUM. [Ch. II.
The preceding analysis shows that the old or common
theory of torsion is correct in its expression for torsive
strain, as it is identical with eq. (17) of Art. 6, i.e.,
w = arx ;
but it will be seen later that the remaining formulae of the
common theory are incorrect for all shapes of cross-section
except the circle. Fortunately the torsion members prin-
cipally used in engineering practice are shafts of circular
section.
Equations of Condition in Rectangular Co-ordinates.
In the case of a rectangular normal section, the analysis
is somewhat simplified by taking some of the quantities
used in terms of rectangular co-ordinates.
In the notation of Art. 2 all stresses will be zero except
T3 and T^. Hence eqs. (10), (11), and (12) of that article
reduce to
dy dz '
^=0;
dT_
dx
dT\
dx
The strains in the directions of x, y, and z are, respec-
tively, n, V, and w. Introducing the values of T^ and T.^
in the equations above, in terms of these strains, from
eqs. (11) and (13) of Art. i, and then doing the sam/j.in
reference to the conditions,
Art. 6.] TORSION IN EQUILIBRIUM. 86 1
the following equations will result:
df^d^^°' ^'^^
dv dw . ^
dz + d^''° (^7)
The operations by which these results are reached are
identical with those used above in connection with semi-
polar co-ordinates, and need not be repeated.
Eq. (27) is satisfied by taking
V = OLXZ ;
w= — axy ;
in which a is the angle of torsion, as before.
Eqs. (11) and (13) of Art. 5 then give
^.-KI+e)-<^")^ ■ ■ • <=»'
The element of a normal section is dz dy. Hence the
moment of torsion is
M = ffiT,z-T^)dydz;
.'. M =GJ {zu dz—yu dy) +Galp (31)
Ip=ff{z'+y')dydz
862 TORSION IN EQUILIBRIUM. [Ch. II.
is the polar moment of inertia of any section about the
axis of A^.
The integrals are to be extended over the whole section ;
hence, in eq. (31), zu dz is to be taken as
z dz. I -rdy
J -yo dy ^
and yn dy as
dy
"°du
dz"^''
in which expressions 3/0 and z^ are general co-ordinates of
the perimeter of the normal section.
Eq. (26) is identical with eq. (13), and can be derived
from it, through a change in the independent variables, by
the aid of the relations
z=r cos (p and ;y=rsin^.
Solutions of Eqs. (13) and (21).
It has been shown that the function u, which represents
the strain parahel to the axis of the piece, must satisfy
eq. (13) [or eq. (26)] for all points of any normal section,
and eq. (21) (or a corresponding one in rectangular co-
ordinates) at all points of the perimeter ; and those two are
the only conditions to be satisfied.
It is shown by the ordinary operations of the calculus
that an indefinite number of functions u, of r and 0, will
satisfy eq. (13) ; and, of these, that some are algebraic and
some transcendental.
It is further shown that the various functions u which
satisfy both eqs. (13) and (21) differ only by constants.
Art. 6.] TORSION IN EQUILIBRIUM. 863
If u is first supposed to be algebraic in character, and if
Cp ^2, ^3, etc., represent constant coefficients, the following
general function will satisfy eq. (13):
.( c.r sin 6 + c^r^ sin 2(h + cs^ sin 26+ . . .) , ,
( +c'jrcos ■}- c/^ cos 2^ + ^3r^ cos 3^+ . . .
— c^rsin — c'/^sin 2(j) — c\r- sin t^cJ)— . . . =C. {^^)
C is a constant which changes only with the form of
section.
If -J- and —1-7 be found from eq. (32), w^hile j, be
taken from eq. (33), and if these quantities be then intro-
duced in eq. (21), it will be found that that equation is
satisfied.
The only form of transcendental function needed,
among those to w^hich the integration of eq. (13) or eq. (26)
leads, will be given in connection with the consideration of
pieces with rectangular section, where it will be used.
Elliptical Section about its Centre.
Let a cylindrical piece of material with elliptical normal
section be taken, and let a be the semi-major and b the
semi-minor axis, while the angle is measured from a
with the centre of the ellipse as the origin of co-ordinates,
since the c^dinder will be twisted about its own axis. The
864 TORSION IN EQUILIBRIUM. [Ch. 11.
polar equation of the elliptical perimeter may take the
following shape:
7+7*;?T^^°'"^=^MT^- • • • (34)
By a comparison of eqs. (33) and (34), it is seen that
c^= / 9 , 7 9x and C =
and that all the other constants are zero. Hence eq. (32)
gives
^ = <^ 2{a' + b'f ^^^ ^"^'^^^ ^^^ ^^' • • (35)
The quantity represented by / is evident.
By eqs. (19) and (20)
r;,;.=6'a: ^2_^^2 ^sin 2^; (36)
Tt.6='
xq>
""^^(omt^^^^^^^'^v* * • • (37)
Since "' ^ =dA, A being the area of the ellipse, or
2
7ra6, the second member of eq. (22), by the aid of eq. (37),
may take the form
M^Gaj dJ^ (^^^,r' cos 2 ct> + r'jdr;
r/b'^ — a^r* r*\
Art. 6.] TORSION IN EQUILIBRIUM. 865
Then using eq. (34),
If Ip is the polar moment of inertia of the elHpse (i.e.,
about an axis normal to its plane and passing through its
centre), so that
nabja^ + b^)
^P- 4 '
then
A*
M=Ga^,-j- (39)
4^'Ip ^-^^^
Using / in the manner shown in eq. (35), the resultant
shear at any point becomes, by eq. (24),
T=Gar\/p + 2f cos 2^ + 1.
dT
gives
sin 29^ = 0, or (56 = 90° or o®.
Since / is negative, T will evidently take its maximum
when (j) has such a value that 2/ cos 2^ is positive, or = 90° and r ^b in the value of T,
[Ch. 11.
Taking Ga from eq. (40) and inserting it in eq. (38),
(40)
in which
m 2 "*
rrab'
4
(41)
or the moment of inertia of the section about the major axis.
Equilateral Triangle about its Centre of Gravity.
This case is that of a cyhndrical piece whose normal cross-
section is an equilateral triangle, and the torsion will be sup-
posed about an axis passing through b
the centres of gravity of the different
normal sections. The cross-section is
represented in Fig. 3, G being the h
centre of gravity as well as the origin
of co-ordinates.
Let GH = l,GD = a. Then from the
known properties of such a triangle,
FD=DB-=BF^2aV~i,.
Hence the equation for DB is ; r sin —
Hence the equation for BF is ;
Fig. 3.
2a — r cos
V3
r cos f a = o
XT 1 . r r-^ • • 2a — r cos 6
Hence the equation for FD is; r sm <^ + ,- =0
Art. 6.] TORSION IN EQUILIBRIUM. 867
Taking the product of these three equations and reduc-
ing, there will result for the equation to the perimeter
— — ^COS^(/)-- — (42)
2 6a ^ ' 3 ^^ ^
Comparing this equation with eq. (33),
I . ^ 20^
Hence
c, = — ^ and C= — ••
3 6a 3
r^ sin S(^
«=-«-^^ (43)
A.nd by eqs. (19) and (20)
r^sin^^
T^r=-CfOL — — ; (44)
, r^cos 3(/)
2 a
') (45)
Eq. (22) then gives
M=GaI,-Gaf p^^^drd^;
^GaU-Gaf'^dr;
= Ga\Ip—aW~^ =0.6 GaIp = i.S Gaa'Vs; . (46)
since 7^= polar moment of inertia = 3a ^^3.
868 TORSION IN EQUILIBRIUM. Ch. II.
By eq. (24)
^ ^ I , r^ cos 30 r*
dT .
or
= 0°, 60°, 120°, 180°, 240°, 300°, or 360°.
The values 0°, 120°, 240°, and 360° make
cos30=+i;
hence, for a given value of r, these make T a minimum. The
values 60°, 180°, and 300° make,
cos 3(/) = -i;
hence, for a given value of r, these make T a maximum.
Putting cos 30 = — I in eq. (47),
{-?)
T=Ga[r + ~j. ..... (48)
This value will be the greatest possible when r is the
greatest. But = 60°, 180°, and 300° correspond to the nor-
mal a dropped on each of the three sides of the triangle
from G. Hence r = a, in eq. (48), gives the greatest intensity
of shear T.^, or
m>
T^--G<^a (49)
Art. 6.] TORSION IN EQUILIBRIUM. 869
Or the greatest intensity of shear exists at the middle point
of each side. Those points are the nearest of all, in the
perimeter, to the axis of torsion.
The value of Ga^ from eq. (49), inserted in eq. (46),
gives
M=o.4-r^= — -. .... (50)
a "* 20 ^
in which 1= side of section = 2a\/3.
Rectangular Section about an Axis passing through its
Centre of Gravity.
In this case it will be necessary to consider one of the
transcendental forms to which the integration of eq. (13)
[or (26)] leads; for if the polar equation to the perimeter be
formed, as was done in the preceding case, it will be found
to contain r*, to which no term in eq. (33) corresponds.
If e is the base of the Napierian system of logarithms
(numerically ^ = 2.71828, nearly) and A any constant what-
ever, it is known that the general integral of the partial
differential eq. (13) may be expressed as follows:
-ii=^4^«''cos^ ^nVsin^^ . . . . . (51)
when w^ + w'^ = o; for
d^u d^u du ^ , , ,,, J. , .L
dr^ r^ d(j>^ rdr - ^
But the second member of this equation is evidently
equal to zero if
(w^4-n'^)=o or n'=V —
870 TORSION IN EQUILIBRIUM. [Ch. 11.
These relations make it necessary that neither n or n' shall
be imaginary.
It will hereafter be convenient to use the following no-
tation for hyperbolic sines, cosines, and tangents :
sih i = ; coh t = ; and tah t = -— :
2 2 e^-\-e~^
By the use of Euler's exponential formula, as is well
known, and remembering that n''^=—n^^ eq. (51) may be
put in the following form:
n = i'e^'^cos.^ |-^4^ sin (nr sin ) . sih (nr cos ) — ar^ sin cj) cos 0. (53)
Fig. 4 represents the cross-section with C as the origin of
co-ordinates and axis. The angle (p is measured positively
>^
^iC
Fig. 4.
from CN toward CH. At the points A^ H, K, and L, in the
872 TORSION IN EQUILIBRIUM. [Ch. II.
equation to the perimeter, dr^ will be zero. Hence at those
points, by eq. (21),
du
-J- = ^[A „ sin (nr sin ^) . n cos . coh (nr cos 0)
4- A „ . w sin <^ . cos {nr sin (j>) . sih (nr cos 9^)]
— 2ar sin ^ cos =0.
At the points under consideration 873
for n in that equation. Eq. (53) will then become
<» ^ . (2n—i . ,\ ., /2W— I \
u = 2A^sm. I nr sm ^ I . sih I nr cos I
— ar- sin cos
2
dr
and — ?i will be the tangent of (—) , or
r^d
dr^ . , .. , . •
— —-77 = — tan (
.-9
)=U
Hence
eq.
(2
i) becomes
du '
dr
= tan
4>,
du
rd^j)
or
du
ar sm 4^ = -r- cos
4>-
du
rdcf>
(56)
sin (j).
Substituting from eq. (55), then making
r cos =-,
2
^^ 2fz-i . (2n—i \ . /2n—i
sin 0j
874 TORSION IN EQUILIBRIUM. [Ch. II.
If r sin m(j) and z=r cos cj)
be introduced, that equation will become
u= — azy +
©
ac^Z ^ , ; . (57)
( 2n— 1 \
{2n-iyQoh \—^-b)
This value of u placed in eq. (31) will enable the moment
of torsion to be at once written.
c c
The limits +3^0 and -y^ are +- and , and the limits
+ ^0 and -Zq are +- and — -. Hence
2 2
abc
.. 271—1.
{2n — \Y coh
=^J
= Q, for brevity;
-H :=
ahc
/ 271 I
2n — I
ny
(2W— ij^eoh
\ 2C
-.R.
J
876 6.] TORSION IN EQUILIBRIUM.
For the next integration
[Ch. II.
f.
Qz dz = abc
12 •
+ 'WFf
2bc , 2W— I , 4C" ., /2n—i .\
_ 7 r-.coh- 7:6—7 ^T~2Sih( — ^ — nb)
(2W— i)^ coh
(^•')
/:
Ry dy = abc
■2 /sVc °°
12 \7tl b
c" (2yc_^{2n-iyn
T-^ sih I 7zb
2C
(2n— i)^ coh
2n— I
2C / ^
Thus the integrations indicated in eq. (31) are com-
pleted. Hence
M=G'^fQzdz + fRydy + aIpi. .
Remembering that
M
(58)
But it is known that
2 n^
I (2W— l)* I . 2 .3 ' 2
5"
Art. 6.] TORSION IN EQUILIBRIUM. 877
Hence eq. (58) becomes
r 64. , tah(i^.6) -[
M=Gabc'\ --^j:I ,!_,,, ' |. . (59)
Since
I — tah TT I— tah ^TT i— tahs^r
tah 7z tah ^tt tah t;;r
■)
and since
64
^-g- = 0.209137,
and remembering that
T\2W-i/ ~^~^3^ 5'"^* • * "V 2V295.1215'
^q- (59) becomes
M =GabA - - 0.2 10083^
/i-tah— i-tah^^— \
+ 0.209137^1 ^ + 5 + . . . / 1. (60)
Eq. (60) gives the value of the moment of torsion of a
rectangular bar of material.
878 TORSION IN EQUILIBRIUM. [Ch. 11.
If z had been taken parallel to b, and y parallel to c, a
moment of equal value would have been found, which can
be at once written from eq. (60) by writing b for c and c for b.
That moment will be
M--Gacb'\ --0.210083^
J. (61
, / 1 — tah^ I — tah^
b\ 20 20
+ 0.209137 -\^ J + ^1 +
Eq. (60) should be used when b is greater than c, and eq.
(61) ivhen c is greater than b, because the series in the paren-
theses are then very rapidly converging, and not diverging.
It will never be necessary to take more than three or four
terms and one, only, will ordinarily be sufficient. The follow-
ing are the values of
I — tah —
for a few values of n:
1 ^^^\
I —tah — 1 =0.083 : 0.00373 : 0.000162 : 0.000007 ;
n= I : 2 : 3 : 4
Square Section.
If c =b either eq. (60) or eq. (61) gives
M^Gab^ I -- 0.2 101+ 0.209(1 -tah- j ;
44
/. M =0.1^06 Gab^-=Ga — ^, . . (62)
Art. 6.] TORSION IN EQUILIBRIUM. 879
in which .4 is the area ( = 6^) and Ip is the polar moment of
mertia i ^^^ )•
Rectangle in which b = 2C.
If b = 2C, eq. (60) gives
M =^Ga. 2C^(- — 0.105 +0.1046 (i — tah k) \;
I
3
A'
:. M^o.AS! Go^c'---Ga-~-^, . . . (63)
in which A is the area ( = 2c'^) and /^ = polar moment of in-
ertia
J)c' + h'c ^Sc'
12 6 ■
Rectangle in which b = 4.C.
If 6 ^ 4c, eq. (60) then gives
M =^Gabc^l — 0.0525 j =1.123 G^ac*;
A'
'. M=Ga ^, (64)
40.2 Ip ■ ^ ^^
in which A =area = 4C^ and Ip = polar moment of inertia
_hc^-\-b^c _i']c^
12 .^
88o TORSION IN EQUILIBRIUM. fCh. 11.
If b is greater than 2C, it will be sufficiently near for all
ordinary purposes to write
M -^Ga—[i -o.6sj^) (65)
Greatest Intensity of Shear.
There yet remains to be determined the greatest inten-
sity of shear at any point in a section, and in searching for
this quantity it will be convenient to use eqs. (28) and (29).
It will also be well to observe that by changing z to y,
yto — z, c to b, and 6 to c, in eq. (57), there may be at once
written
. / 2n—i \ .. /2n— I \
u^azy-[-) .ab-I -~
(2n— i)^coh
26
- ncj
.(66)
This amounts to ttirning the co-ordinate axes 90^
Since the resultant shear at any point is
it will be necessary to seek the miaximum of
du V idu V T'
The two following equations will then give the points
desired :
Art. 6.] TORSION IN EQUILIBRIUM.
i?)
du \d^u (du \/ d^u \ ,. ,
+ a.).-. + (rf,-«3'j(5^-«j=o; (67)
dy \dy jdy
dz
du \ dhi \ (du \d^u ,^„.
It is unnecessary to reproduce the complete substitu-
tions in these two equations, but such operations show that
the points of maximum values of T are at the middle points of
the sides of the rectangular sections, omitting the evident fact
that r = o at the centre. It will also be found that the great-
est intensity of shear will exist at the middle points of the
greater sides.
This result may be reached independent of any analytical
test, by bearing in mind that an elongated ellipse closely
approximates a rectangular section, and it has already been
shown that the greatest intensity in an elliptical section is
foimd at the extremities of the smaller axis.
By the aid of eqs. (28), (29), (57), and (66), it will also
be foimd that T^^o at the extremities of the diameter c,
and T^=o at the extremities of the diameter h. The maxi-
mum value of T will then be
r„ = -r,= -ff(|-ay)^^^. . . . (69)
y=
By the use of eq. (57)
du
n . /2n—i \ ^ (2n—i \
{^-j)n-x_^^^ ( ny\ .coh( tcz\
{2n-iyQ.oh ^-^^^^j
TORSION IN EQUILIBRIUM. [Ch. II.
Putting z=-o and y =~ m. this equation, there will result
T„^Gacr^-^ i_^^__^-|.. (70)
I '' ' (2W— i)"-^ coll ^ -r.b] I
If h is greater than c the series appearing in this equation
is very rapidly convergent, and it will never be necessary to
use more than two or three terms if the section is square, and
if h is four or five times c there may be written
T^=Gac . (71)
Square Section.
Making h=c in eq. (70), and making ;z = i, 2, and 3 (i.e.,
taking three terms of the series), there will result
T
T^^ =0.6"] 6 Gac; .'. 6^0: = 1.48— ^
Inserting this value in eq. (62),
M = o.2ib'T^=^^^^^ (72)
M M
•*• '^rn-O.S-ra^-S^S. .... (73)
in which
J b' be
I = — and a =- =-
12 22
Art. 6.] TORSION IN EQUILIBRIUM. 883
Rectangular Section; h = 2C.
Making h^2C in eq. (70), and making w = i, only, there
will result
T
T,„-= 0.93(70:^; :. Ga^i.o^-^,
Inserting this value in eq. (63),
in which
i¥ = o.49^^r,„ = i.47-^ (74)
M M
.'. r^-=o.68ya-2-3, .... (75)
^ be' e' ^ c
I -= — -=7- and a == -.
126 2
Rectangular Section; h=/[C.
Making h = /[c in eq. (70), and making n=-i, only,
7
^,« =0.997 Gac; :. Ga = i.oos-^.
Inserting this value in eq. (64),
IT
M = 1.126 r^r,„ -= 1.69 — -. . . . (76)
m ^ a '
. ., M M
in which
•. T„,-o.6ya=^o.9^, (77)
^ be" c^ , c
y= — =— and a=-
T2 3 2
884 TORSION IN EQUILIBRIUM. [Ch. II.
Circular Section about its Centre.
The torsion of a circular cylinder furnishes the simplest
example of all.
If ro is the radius of the circular section, the polar equa-
tion of that section is
— = C (constant).
Comparing this equation with eq. (33), it is seen that
^1=^2 = ^3== ••• =^1'=^/= ••• =o-
By eq. (32) this gives u=o. Hence all sections remain
plane during torsion.
Eqs. (19) and (20) then give
Txr=o and T^4,=Gar (78)
Eq. (23) gives for the moment of torsion
M=GaI, (79)
or
A ^G
M =o.K TirQ^ .Ga=- ' 27- Q^> .... (80)
in which equation A is the area of the section and
r _^
Art. 6.] TORSION IN EQUILIBRIUM. 88$
The greatest intensity of shear in the section will be ob-
tained by making r = ro in eq. (78), or
T^=Gar,; .\Ga=^^ (81)
r.
Eq. (80) then becomes
i\/ = o.5;rVr^ = 2^ (82)'
.. ^m =0-64— 3=0. 5-j-ro, (83)
in which / = — ^•
4
It is thus seen that the circular section is the only one
treated which remains plane during torsion.
General Observations.
The preceding examples will sufficiently exemplify the
method to be followed in any case. Some general conclu-
sions, however, may be drawn from a consideration of
eq. (33)-
If the perimeter is symmetrical about the line from
which ^ is measured, then r must be the same for + c6 and
— ^; hence
c/ =c^^ =c/ = ... =0.
If the perimeter is symmetrical about a line at right
angles to the zero position of r, then r must be the same for
0=9o°+0' and go^-gS';
886 TORSIONAL OSCILLATIONS. [Ch. II.
hence
^1=^3=^5... =^2'=^/=^/= . . . =0.
In connection with the first of these sets of results,
eq. (32) shows that every axis of symmetry of sections repre-
sented by eq. {^^) will not be moved from its original position
by torsion.
If the section has two axes of symmetry passing through
the origin of co-ordinates, then will all the above constants
be zero, and its equation will become
J 4-^2^' COS 2c^ + ^/'cos 4 can, then, vary with cj) or ([/.
By the aid of these considerations, and after omitting R^,
^0, ^0, and the second members, the eqs. (i), (2), and (3)
of Art. 4 reduce to
dNr ^ 2Nr-N^-N^
-5r+ r =^', •.•.(!)
-A^^ + A^^=o (2)
By eq. (2)
N^=N^.
Eq. (i) then becomes
dNr Nr-N^
-5/ + ^— r-=- (3)
On account of the existing condition of stress which has
just been indicated it at once results that
lfj = CO=0,
and that ^ is a function of r only.
Eqs. (4) to (10) of Art. 4 then reduce to
'-i-'i-' (4)
"'-^r'-"^!--- • • • . . (5)
Nt=Nt = ^^d + 2G^.
I-2C ■ r (6)
894 THICK, HOLLOW SPHERES. [Ch. II.
After substitution of these quantities, eq. (3) becomes
2Gt ((Pp 2rdp—2pdr\ d^p dp p
T^^ [d^ + ?~dF~) + '^dr^ + ^^'7d~^^? '-^""^
or
d'p K'^'r
— o.
dr^ dr
One integration gives
dp 2p ^
Hence 6, the rate of variation of volume, is a constant
quantity. Eq. (7) may take the form
r dp-{- 2p dr = cr dr.
As it stands, this equation is not integrable, biit, by in-
specting its form, it is seen that r is an integrating factor.
Multiplying both sides of the equation, then, by r,
r^dp+ 2rpdr =d(r^p) =cr'^dr;
T ^ CT b
:. r'p = c-+b; .-. P=- + ~2' ... (8)
Substituting from eqs. (7) and (8) in eq. (5),
__ 2GV 2Gc 4bG , ^
A^r=7z:^^+— - 7f; .... (9)
It is obvious what A represents.
Art. 8.] THICK, HOLLOIV SPHERES. . 895
When / and r^ are put for r, A^^ becomes —p and —p^.
Hence
and
These equations express the conditions involved in eqs.
(13), (14), and (15) of Art. 2.
The last equations give
ip,-p)ry\
r'^ — r^
• • -^ ^/3 ^ 3 •
4^^=~^3_^3 »
These quantities make it possible to express A^^ and .Y^
independently of the constants of integration, c and 6, for
those intensities become
_ P/ ,'-pr'' _ (p,-p)ry' I .
^^r — ^3_^ 3 /3_^ 3 '^31 • V-tO/
Thus it is seen that N^=N4, has its greatest value for
the interior surface ; that intensity will be called h.
It is now required to find r^ — r'=tin terms of h, py and /?j.
If r =r' in eq. (11).
896 THICK, HOLLOfV SPHERES. [Ch. II.
Dividing this equation by r'^ and solving,
/3
2(h + p)
2h-p + 3Pi'
^-
,3 2{h+p)
-/. . . (12)
If the intensities p and p^ are given for any case, eq. (12)
will give such a thickness that the greatest tension h (sup-
posing p^ considerably less than p) shall not exceed any
assigned value. If the external pressure is very small com-
pared with the internal, ^^ may be omitted.
The values of .4 and 4Gb allow the expressions for c and
b to be at once written.
If p^ is greater than p, nothing is changed except that
A^^ =A^^ becomes negative, or compression.
CHAPTER III.
THEORY OF FLEXURE.
Art. 9. — General Formulae.
If a prismatic portion of material is either supported at
both -ends, or fixed at one or both ends, and subjected to
the action of external forces whose directions are normal
to, and cut, the axis of the prismatic piece, that piece is said
to be subjected to " flexure." If these external forces have
lines of action which are oblique to the axis of the piece, it
is subjected to combined flexure and direct stress.
Again, if the piece of material is acted upon by a couple
having the same axis with itself, it will be subjected to '' tor-
sion."
The most general case possible is that which combines
these three, and some general equations relating to it will
first be established.
The co-ordinates axis of X will be taken to coincide with
the axis of the prism, and it will be assumed that all external
forces act upon its ends only. Since no external forces act
upon its lateral surface, there will be taken
retaining the notation of Art. 2. These conditions are not
strictly true for the general case, but the errors are, at most,
excessively small for the cases of direct stress or flexure, or
897
898 THEORY OF FLEXURE. [Ch. III.
for a combination of the two. By the use of eqs. (12), (21),
and (22) of Art. i the conditions just given become
r /du dv dw
i — 2r \dx- dy dz
'- + ^+=^)+| = o; ... (I)
r /du dv dw \ dw
T=^VS + 5^+^/+5F^°' • • • ^^^
dv dw , .
dF + d^^°- (3)
Eqs. (i) and (2) then give
dv dw
d^-d^=° (4)
In consequence of eq. (4) eqs. (i) and (2) give
dv _dw du
dy dz dx ^^^
By the aid of eq. (5) and the use of eqs. (11), (13), and
(20) of Art. I, in eqs. (10), (11), and (12) of Art. 2 (in this
case Xq = Yq=Zq = o), there will result
d'^u d^u d^u .^.
d^u d^v
d^^d^' ^''' ••••••• (7)
d^u d^w .
d^z^di^^"" ^^^
Eqs. (3), (5), (6), (7), and (8) are five equations of
condition by which the strains u, v, and w are to be deter-
mined.
Art. 9.] GENERAL FORMUL/E. 899
Let eq. (6) be differentiated in respect to x:
dhi d^u d^u
dx^ dy^ dx dz^ dx
From this equation let there be subtracted the sum of
the results obtained by differentiating eq. (7) in respect to y
and (8) in respect to z:
d^u d^v dhv
dx^ dx^ dy dx^ dz
In this equation substitute the results obtained by
differentiating eq. (5) twice in respect to x, there will result
.3 HP
d^u \dx^
"3J'=^S^ =°- W
This result, in the equation immediately preceding eq.
(9), by the aid of eq. (5) will give
d'v
o.
dx^ dy
After differentiating eq. (7) in respect to y, and substi-
tuting the value immediately above,
d^u _ \dxi
dy'Tx 57" =° ('°)
Eqs. (9) and (10) enable the second equation preceding
eq. (9) to give
^du"
d^u _ \dx^
dz^x dz' ^° ^^'^
goo THEORY OF FLEXURE, [Ch. III.
Let the results obtained by differentiating eq. (7) in
respect to z and (8) in respect to y be added :
d^u ■ d^v d^w
dx dy dz dx^ dz dx^ dy
The sum of the second and third terms of the first mem-
ber of this equation is zero, as is shown by twice differentiat-
ing eq. (3) in respect to x. Hence
d^u \dx, , .
= 0. .... . (12)
dy dz dx dy dz
Eqs. (9), (10), (11), and (12) are sufficient for the
d'VL
determination of the form of the ftinction -j-, if it be assumed
to be algebraic, for
Eq. (9) shows that x"^ does not appear in it ;
" (10) - - f
" ^,2 << <■<■
(11)
((
<(
2:^
<(
(<
((
('
(12)
(<
((
yz
(<
((
((
ii
The products xz and xy may, however, be foimd in the
fimction. Hence if a, a^, a^, b, h^, and h^ are constants,
there may be written
du
— =a + a^z-\-a^y + xib + b,z + b^y). . . . (13)
"Eq. (5) then gives
dv div
jy = :^ = -r{a + a,z + a^y + x(b + b^z + b^)], . (14)
Art. 9.] GENERAL FORMUL/E. 9°!
Substituting from eq. (13) in eqs. (7) and (8),
^2 ==-a^-h^\ (15)
^ = -a,-b,x (16)
The method of treatment of the various partial deriva-
tives in the search for eqs. (13) and (14) is identical with that
given by Clebsch in his " Theorie der Elasticitat Fester
K or per."
It is to be noticed that the preceding treatment has been
entirely independent of the form of cross-section or direction
of external forces.
It is evident from eqs. (13) and (14) that the constant a
depends upon that component of the external force which
acts parallel to the axis of the piece and produces tension or
compression only. For (pages 9, 10) it is known that
if a piece of material be subjected to direct stress only,
du ^ dv dw
-J- =-a and t~ = j~ = — ^^ ;
dx dy dz
the negative sign showing that ra is opposite in kind to a,
both being constant.
Again, if z and y are each equal to zero, eq. (13) shows
that
du .
-T~ =a-\-bx.
dx
Hence hx is a part of the rate of strain in the direction of x
which is uniform over the whole of any normal section of the
piece of material, and it varies directly with x. But such a
902 THEORY OF FLEXURE. [Ch. III.
portion of the rate of strain can only be produced by an
external force acting parallel to the axis of X, and whose
intensity varies directly as x. But in the present case
such a force does not exist. Hence h must equal zero.
The eqs. (13), (14), (15), and (16) show that a^, h^ and
Oj, 62 are symmetrical, so to speak, in reference to the co-
ordinates z^indy, while eqs. (13) and (14) show that the nor-
mal intensity A^^ is dependent on those, and no other, con-
stants in pure flexure in which a = o. It follows, there-
fore, that those two pairs of constants belong to the two
cases of flexure about the two axes of Z and Y.
No direct stress N^ can exist in torsion, which is simply a
twisting or turning about the axis of X.
Since the generality of the deductions will be in no man-
ner affected, pure flexure about the axis of Y will be con-
sidered. For this case
a=a^ = h^ = o =b.
Making these changes in (13) and (14),
du
^=a,z + b,xz; (17)
dv _ dw du
^~5F = -'•5^ = -''(«.« + V^)- • • • (18)
. . du dv dw
■■ ^^Tx+d^ + d^-'(^^ + b^^^(^-^r)-- ■ (19)
Also,
^ i~2r dx
N, = 2G{r-\- i)(a^ + b^x)z=E(a^+b,x)z, . (20)
Art. 9.] GENERAL FORMUL/E. 903
since
2G{r-^i)=E,
Taking the first derivative of N^^
^'=£(a, + V) (21)
This important equation gives the law of variation of
the intensity of stress acting parallel to the axis of a bent
beam, in the case of pure flexure produced by forces exerted
at its extremity. That equation proves that in a given nor-
mal section of the beam, whatever may be the form of the
sectiCfn, the rate of variation of the normal intensity of stress is
constant ; the rate being taken along the direction of the external
forces.
It follows from this that A^^ must vary directly as the
distance from some particular line in the normal section
considered in which its value is zero. Since the external
forces F are normal to the axis of the beam and direction
of A/'j, and because it is necessary for equilibrium that the
sum of all the forces N^dy dz, for a given section, must be
equal to zero, it follows that on one side of this line tension
must exist, and on the other compression.
Let A^ represent the normal intensity of stress at the
distance unity from the line, h the variable width of the
section parallel to 3/, and let i = hdz. The sum of all the
tensile stress in the section will be
[ NzA=n\
Jo J c
zJ.
The total compressive stress will be
Np zJ.
J -zi
904
THEORY OF FLEXURE.
[Ch. III.
The integrals are taken between the limits o and the greatest
value of z in each direction, so as to extend over the entire
section. In order that equilibrium may exist, therefore,
Fig. I.
n\\ zA^r za\ =0.
■•■1:
zA =0.
(22)
Eq. (22) shows that the line of no stress must pass through
the centre of gravity of the normal section.
This line of no stress is called the neutral axis of the
section. Regarding the whole beam, there will be a sur-
face which will contain all the neutral axes of the different
sections, and it is called the neutral surface of the bent
beam. The neutral axis of any section, therefore, is the
line of intersection of the plane of section and neutral sur-
face.
Hereafter the axis of X will be so taken as to traverse
the centres of gravity of the different normal sections
before flexure. The origin of co-ordinates will then be
Art. 9
GENERAL FORMUL/E.
90^
taken at the centre of gravity of the fixed end of the beam,
as shown in Fig. i.
The value of the expression {a^ + h^x), in terms of the
external bending moment, is yet to be determined. Con-
sider any normal section of the beam located
at the distance x from 0, Fig. i, and let
OA =1. Also, let Fig. 2 represent the sec-
tion considered, in which BC is the neutral
axis and d^ and d^ the distances of the most
remote fibres from BC. Let moments of all
the forces acting upon the portion (/ — x) of
the beam be taken about the neutral axis BC. If, again, h
is the variable width of the beam, the internal resisting
moment will be
Fig.
rd' rd'
NJjzdz=E(a^ + b^x)\ z\hdz,
J —di J —di
But the integral expression in this equation is the moment
of inertia of the normal section about the neutral axis, which
will hereafter be represented by I. The moment of the
external force, or forces, F, will be F{l — x), and it will be
equal, but opposite in sign, to the internal resisting moment
Hence
F l-x)=M=-E(a^ + b^x)L
(23)
(a^ + b^x) =
K
EF
(24)
Substituting this quantity in eq. (16),
dhv_M_
dx' "EF
(25)
9o6 THEORY OF FLEXURE, [Ch. IIL
It has already been seen (page 38) that eq. (25) is one
of the most important equations in the whole subject of the
"Resistance of Materials."
An equation exactly similar to (25) may of course be
written from eq. (15); but in such an expression M will
represent the external bending moment about an axis par-
allel to the axis of Z.
No attempt has hitherto been made to determine the
complete values of ti, v, and w, for the mathematical opera-
tions involved are very extended. If, however, a beam be
considered ^vhose width, parallel to the axis of Y, is indefi-
nitely small, u and w may be determined without difficulty.
The conclusions reached in this manner will be applicable
to any long rectangular beam without essential error.
If y is indefinitely small, all terms involving it as a factor
will disappear in ti and w ; or, the expressions for the strains u
and w will he functions of z and x only. But making u and w
functions of z and x only is equivalent to a restriction of
lateral strains to the direction of z only, or to the reduction
of the direct strains one half, since direct strains and lateral
strains in two directions accompany each other in the un-
restricted case. Now as the lateral strain in one direction
is supposed to retain the same amoimt as before, while the
direct strain is considered only half as great, the value of
their ratio for the present case will be twice as great as that
used on pages 9 to 12. Hence 2r must be written for r, in
order that that letter may represent the ratio for the unre-
stricted case, and this will be done in the following equations.
Since w and u are independent of y,
dw du dv
1— =-3- =0, and I^=G-r-'
dy dy ' ^ dx
But, by eq. (14),
V = — 2r(ai + h^x)zy + f{x, z).
Art. 9.] GENERAL FORMULy^, 907
By eq. (3), since
dw
^ = -2ria^ + \x)y + ^J(x, z) =0.
This equation, however, involves a contradiction, for it
makes f(x, z) equal to a function which involves y, which is
impossible. Hence
f{x,z)^o.
Consequently
^^ / 7 s
which is indejfinitely small compared with
^=-2r{a^ + b,x)z,
and is to be considered zero
Because f(x, z) = o,
dv .
This quantity is indefinitely small; hence
Tg = — 2Grh^zy
is of the same magnitude.
Under the assumption made in reference to y, there may
be written, from eqs. (17) and (18),
u = a^xz + h^-z-\-f{z); .... (26)
w=-r{a^z^-\-h^xz'')-[-f{x), . . . (27)
9o8 THEORY OF FLEXURE. [Ch. III.
Using eq. (26) in connection with eq. (6),
By two integrations,
f'{z)^-^-c'z + c" ' (28)
Using eq. (27) in connection with eq. (8),
By two integrations,
The functions u and w now become
u = a^xz + b^-—z '—-c'z + (/^; . . . (29)
^ 3
w= — ra^z^ — rb^xz^ — b^~ ^ — + c^x + (7^. . (30)
The constants of integration c' , c" , etc., depend upon
the values of u and w, and their derivatives, for certain
reference values of the co-ordinates x and z, and also
upon the manner of application of the external forces, F, at
the end of the beam, Fig. i . The last condition is involved
in the application. of eqs. (13), (14), and (15) of Art. 2.
Art. 9.1 GENERAL FORMULA. 909
In Fig. I let the beam be fixed at 0. There will then
result, f or :^; = o and z=o,
(du \
2 = 0<
x = o
(u = o, and w = o)
In virtue of the last condition,
c =^11=0-
In consequence of the first,
After inserting these values in eqs. (29) and (30),
du ^ x^ . .
dz
dw
= — rh^z^ — 6j — — a^x + c^.
^^=^Gt+S = -^^(^+^>^+^'^.- • (31)
The surface of the end of the beam, on which F is applied,
is at the distance I from the origin and parallel to the
plane ZY . Also, the force F has a direction parallel to the
axis of Z. Using the notation of eqs. (13), (14), and (15) of
Art. 2, these conditions give
cos^ = i, cosg = o, cosr=o,
C0S7r=0, COS/ = 0, COS|0 = I.
9IO THEORY OF FLEXURE, [Ch. III.
Since, for x = l,
M = F(l-x)=o,
eqs. (24) and (20) give N.^=o for all points of the end sur-
face. Eq. (15) is, then, the only one of those equations
which is available for the determination of q.
That equation becomes simply
For a given value of z, therefore, any value may be as-
sumed for 72- For the upper and lower surfaces of the beam
let the intensity of shear be zero; or iov z= ±d let T^ = o.
Hence, by eq. (31),
c,=b,(i+r)d\
Fh
.-. T,^^{d'-z^). ...... (32)
The constants a^ and b^ still remain to be found. The
only forces acting upon the portion (/ — x) of the beam are
F and the sum of all the shears T^ which act in the section x.
Let Jy be the indefinitely small width of the beam, which,
since z is finite, is thus really made constant. The princi-
ples of equilibrium require that
r T,.Jy.dz=Gb,(i+r)r {d\ Ay .dz-z\ Ay .dz) =F.
The first part of the integral will be 2 Ayd^, and the second
part will be the moment of inertia of the cross-section (made
Art. 9-1 GENERAL FORM UL/E. 911
rectangular by taking Ay constant) about the neutral axis.
Hence
77 77
2Gh,{i+r)I=F, or ^ - ,g(i +^)/ °gj- • '^H)
.: T.^Yl^d'-^') (34)
If «; = o in eq. (24),
^^.= -£7 (35)
Thus the two conditions of equilibrium are. involved in
the determination of a^ and h^. The complete values of the
strains u and w are, finally,
F I x^ _z^
2 3
^ = gjUT"T-^^^)' • • (36)
w
F / :r^ Z:r^\ F(Px
= -(^lr,^-rx,^--+-)+-^. . . (37)
These results are strictly true for rectangular beams of
indefinitely small width, but they may be applied to any
rectangular beam fixed at one end and loaded at the other,
with sufficient accuracy for the ordinary purposes of the
civil engineer. It is to be remembered that the load at the
end is supposed to be applied according to the law given
^y ^q.- (34) » ^ condition which is never realized. Hence
these formulas are better applicable to long than short
beams.
912 THEORY OF FLEXURE, [Ch. III.
The greatest value of T^, in eq. (34), is found at the
neutral axis by making z = o\ for which it becomes
^^="2r=f^- • • • • • (38)
—J is the mean intensity of shear in the cross-section;
hence the greatest intensity of shear is once and a half as
great as the mean.
In eq. (36), if 2 = 0, u = o. Hence no point of the neu-
tral surface suffers longitudinal displacement.
In eq. (37) the last term of the second member is that
part of the vertical deflection due to the shear at the neu-
tral surface, as is shown by eq. (38). The first term of
the second member, being independent of