■ • ■ ■ ■ £11$$ SBSKS class _XAia5a Book X\ t'f. Gopght N° : ,1 u, ra COPYRIGHT DEPOSIT; ■ ■ i WORKS OF C. E. GREENE PUBLISHED BY JOHN WILEY & SONS. Graphics for Engineers, Architects, and Builders. A manual for designers, and a text-book for scientific schools. Trusses and Arches : Analyzed and Discussed by Graphical Methods. In three parts — published separately. Part I. Roof Trusses: Diagrams for Steady Load, Snow and Wind. 8vo, 80 pp., 3 folding plates. Revised Edition. $1.25. Part II. Bridge Trusses: Single, Continuous, and Draw Spans; Single and Mul- tiple Systems; Straight and Inclined Chords. 8vo, 190 pp., 10 folding plates. Fifth Edition, Revised. $2.50. Part III. Arches, In Wood, Iron, and Stone, for Roofs, Bridges, and \ v all- openings ; Arched Ribs and Braced Arches; Stresses from Wind and Change of Temperature. 8vo, 194 pp., 8 folding plates. Third Edition, Revised. §2 50. Structural Mechanics. Comprising the Strengtn and Resistance of Materials and Elements of Structural Design. With Examples and Problems. By the late Charles E. Greene, A.M., C.E. New Edition, Revised and Enlarged by A. E. Greene. 8vo, viii -f- 244 pages, 99 figures. $2.50 net. STRUCTURAL MECHANICS COMPRISING THE STRENGTH AND RESISTANCE OF MATERIALS AND ELEMENTS OF STRUCTURAL DESIGN WITH EXAMPLES AND PROBLEMS BY CHARLES E. QREENE, A.M., C.E. LATE PROFESSOR OF CIVIL ENGINEERING, UNIVERSITY OF MICHIGAN REVISED BY A. E. GREENE ASSISTANT PROFESSOR OF CIVIL ENGINEERING, UNIVERSITY OF MICHIGAN SECOND EDITION FIRST THOUSAND \ ■ NEW YORK JOHN WILEY & SONS London: CHAPMAN & HALL, Limited 1905 xveu NUV a / SOX 7V Copyright, 1897, BY Charles E. Greene Copyright, 1905, BY A. E. Greene ■ 1 ROBERT DRUMMOND, PRINTtIR, NEW YORK 6~ 2) is the most important, as it is the principal active element and constitutes the greater part of hydraulic cements. The resulting clinker is ground to an impal- pable powder which forms the Portland cement of commerce. Fine grinding is essential, as it has been shown that the coarser particles of the cement are nearly inert. Upon mixing the cement with water, the soluble salts dis- solve and crystallize, that is, the cement sets. The water does not dry out during hardening as in lime mortar, but combines with some of the salts as water of crystallization. This crystal- lization takes place more slowly in the Portland than in the natural cements, but after the Portland cement has set it is much harder and stronger than natural cement. The slower-setting cement mortars are likely to show a greater strength some months or years after use than do the quick-setting ones, which attain considerable strength very soon, but afterwards gain but little. 48. Cement Specifications. — The following specifications of cement are reasonable: MATERIALS. 35 NATURAL CEMENT. Specific gravity: not less than 2.8. Fineness: 90 per cent, to pass a sieve of 10,000 meshes per square inch. Setting: initial set in not less than fifteen minutes; final set in not more than four hours. Soundness: thin pats of neat cement kept in air or in water shall remain sound and show no cracks. Tensile strength: briquettes one inch square in cross-section shall develop after setting one day in air and the remaining time in water : Neat 7 days 100 lb. 11 28 " 2oolb. 1 cement, 1 sand. . 7 " . 60 lb. ..28 " ..1501b. i l u PORTLAND CEMENT. Specific gravity: not less than 3.10. Fineness: 92 per cent, to pass a sieve of 10,000 meshes per square inch. Setting: initial set in not less than thirty minutes; final set in not more than ten hours. (If a quick-setting cement is desired for special work, the time of setting may be shortened and the requirements for tensile strength reduced.) Soundness: a thin pat of neat cement kept in air 28 days shall not crack; another pat allowed to set and then boiled for five hours shall remain sound. Tensile strength: briquettes, as for natural cement: Neat 7 days . 450 lb. " 28 " 550 lb. 1 cement, 3 sand . . 7 " 1 50 lb. " ..28 " 200lb. 49. Concrete. — Concrete is a mixture of cement mortar (cement and sand) with gravel and broken stone, the materials being so proportioned and thoroughly mixed that the gravel fills the spaces among the broken stone; the sand fills the spaces in the gravel; and the cement is rather more than sufficient to fill the 36 STRUCTURAL MECHANICS. interstices of the sand, coating all, and cementing the mass into a solid which possesses in time as much strength as many rocks. It is used in foundations, floors, walls, and for complete structures. The broken stone is usually required to be small enough to pass through a 2-in. or 2|-in. ring. The stone is sometimes omitted. To ascertain the proportions for mixing, fill a box or barrel with broken stone shaken down, and count the buckets of water required to fill the spaces; then empty the barrel, put in the above number of buckets of gravel, and count the buckets of water needed to fill the interstices of the gravel; repeat the operation with that number of buckets of sand, and use an amount of cement a little more than sufficient to fill the spaces in the sand. If the gravel is sandy, screen it before using, in order to keep the proportions true. A very common rule for mixing is one part cement, three parts sand, and five parts broken stone or pebbles, all by measure. The ingredients are mixed dry, then water is added and the mass is mixed again, after which it is deposited in forms in layers 6 or 8 inches thick. Experience has shown that a mixture wet enough to flow makes a denser concrete than a dry mixture, especially if the mass cannot be thoroughly tamped. 50. Paint. — When a film of linseed- oil, which is pressed from flaxseed, is spread on a surface it slowly becomes solid, tough, and leathery by the absorption of oxygen from the air. In order that the film may solidify more rapidly the raw oil may be pre- pared by heating and adding driers, oxides of lead and man- ganese, which aid the oxidation; oil treated in this way is called boiled oil. Driers should be used sparingly, as they lessen the durability of paint. An oil-film is somewhat porous and rather soft, hence its protective and wearing qualities can be improved by the addition of some finely ground pigment to fill the pores and make the film harder and thicker. Most pigments are inert. Paint, then, consists of linseed-oil, a pigment, and a drier. Varnish is sometimes added to make the paint glossy and harder, or tur- pentine may be used to thin it. Varnishes are made by melting resin (resins are vegetable gums, either fossil or recent), combining it with linseed-oil, and MATERIALS. 37 thinning with turpentine. They harden by the evaporation of the turpentine and the oxidation of the oil and resin. The addition of a pigment to varnish makes enamel- or varnish- paint. For painting on wood white lead, the carbonate, and white zinc, an oxide, are pigments extensively used. Iron oxide is largely used on both wood and steel. Red lead, an oxide, and graphite are pigments used on steel. Red lead acts differently from other pigments in that it unites with the oil, and the mixture hardens even if the air is excluded, so that red-lead paint must be mixed as used. Lampblack is often mixed with other pig- ments to advantage, or it is sometimes used alone. As paint is used to form a protective coating, it should not be brushed out too thin, but as heavy a coat as will dry uniformly should be applied. Wood should be given a priming coat of raw linseed-oil, so that the wood shall not absorb the oil from the first coat of paint and leave the pigment without binder. In applying paint to steel-work it is essential for good work that the paint be spread on the clean, bright metal. Rust and mill- scale must be removed before painting if the coat is expected to last. As mill-scale can be removed only by the sand-blast or by pickling in acid, steel is seldom thoroughly cleaned in practice. If paint is applied to rusty iron, the rusting will go on progressively under the paint. Painting should never be done in wet or frosty weather. CHAPTER III. BEAMS. 51. Beams: Reactions. — A beam may be defined to be a piece of a structure, or the structure itself as a whole, subjected to transverse forces and bent by them. If the given forces do not act at right angles to the axis or centre line of the piece, their components in the direction of the axis cause tension or com- pression, to be found separately and provided for; the normal or transverse components alone produce the beam action or bending. As all trusses are skeleton beams, the same general principles apply to their analysis, and a careful study of beams will throw much light on truss action. Certain forces are usually given in amount and location on a beam or assumed. Such are the loads concentrated at points or distributed over given distances, and due to the action of gravity; the pressure arising from wind, water, or earth; or the action of other abutting pieces. It is necessary, in the first place, to satisfy the requirements of equilibrium, that the sum of the transverse forces shall equal zero and that the sum of their moments about any point shall also equal zero. This result is accomplished by finding the magnitudes and direction of the forces required at certain given points, called the points of support, to produce equilibrium. The supporting forces or reactions, exerted by the points of support against the beam, are two or more, except in the rare case where the beam is exactly balanced on one point of support. For cases where the reactions number more than two, see § 109. 52. Beam Supported at Two Points. Reactions. — The simplest and most generally applicable method for finding one 38 BEAMS. 39 of the two unknown reactions is to find the sum of the moments of the given forces about one of the points of support, and to equate this sum with the moment of the other reaction about the same point of support. Hence, divide the sum of the moments of the given external forces about one of the points of support by the distance between the two points of support, usually called the span, to find the reaction at the other point of support. The direction of this reaction is determined by the sign of its moment, as required for equilibrium. The amount of the other reaction is usually obtained by subtracting the one first found from the total given load. w y°° ,11=750 TV-130 ■-12-- 18 * -600=P„ CV D H ~ 7 7T T D H^-5-^ 20 * K 10 ^ !____£ I B lb ,D A Bl P=300 _. _ 200 =P P=93?K '"'* ** ra , P-750 1 Fig. 3 2 l rig. 4 Fig. 5 ' Thus, in the three cases sketched, Pi = W^-^; P2=W — Pi. Examples.— -Fig. 3. If ^=500 lb., A B = 3 o ft., and B C=i8 ft.; J°i = =300 lb., P 2 = 500— 300=200 lb. Fig. 4. If ir=7 5 olb., AB = 2oft., and AC = s ft., P 1 =I5^1^= 20 937i lb., and P 2 = 750- 937^= -1874 lb. Fig. 5. If W=i$o lb., AC=2oft., and AB = 5 ft., P 1 = I i^l= 750 lb., and P 2 = 150— 750= — 600 lb. Note the magnitude of P 1 and P2 as compared with W when the distance between Pi and P 2 is small. Such is often the case when the beam is built into a wall. Where the load is distributed at a known rate over a certain length of the beam, the resultant load and the distance from its point of application to the point of support may be conveniently used. *»«~ ~-*-4— »* 16 v 100 ,200 150 1300 80] a D C E A r — a^ ~^ 1 - ZK P-1920 1280AP 2 * "16 4 •^ ~* P=665& 214% =pj Fig. 6. Fl '»' 7 Example.— Fig. 6. If AB = 4 o ft., A D = 8 ft., DE=i6 ft., and the load on D E is 200 lb. per ft., JF= 3 , 2 oo lb., and CB = 2 4 ft. 4Q STRUCTURAL MECHANICS. Therefore Pi = — =1,920 lb., and P 2 = 3, 200— 1,920=1,280 lb. 40 If several weights are given in position and magnitude, the same process for finding the reactions, or forces exerted by the points of support against the beam, is applicable. Examples. — In Fig. 7, Pi = (100-18+200- 16+ 150- 13 + 300- 11 + 50-8+ 80 -o)-m 6 = 665! lb. P 2 =88o— 6651=214! lb. The work can ^400 Fig. 9 ^^ ^-JOO be checked by taking moments about A to find P 2 , the moment 100-2 then being negative. If the depth of water against a bulkhead, Fig. 8, is 9 ft., and the distance between A and B, the points of support, is 6 ft., A being at the bottom, the unit water pressure at A will be 9X62.5 = 562.5 lb. which may be represented by A D, and at other points will vary with the depth below the surface, or as the ordinates from E A to the inclined line E D. Hence the total pressure on E A, for a strip 1 ft. in hori- zontal width, will be 562.5X9-^2 = 2,531! lb., and the resultant pres- sure will act at C, distant J A E, or 3 ft. from A. P 2 = 2,531^X3 ^6= 1,265.6 lb., and Pi = 2,531.2 — 1,265.6= 1,265.6 lb., a result that might have been anticipated, from the fact that the resultant pressure here passes midway between A and B. Let 1,000 lb. be the weight of pulley and shaft attached by a hanger to the points D and E, Fig. 9. Let the beam A B= 10 ft., A D = 4 ft., D E=4 ft., E B=2 ft.; and let C be 2 ft. away from the beam. As the beam is horizontal, Pi = 1,000X4-^10=400 lb.; P 2 = 1,000— 400= 600 lb., and both act upwards. The 1,000 lb. at C causes two vertical downward forces on the beam, each 500 lb., at D and E. There is also compression of 500 lb. in D E. When the beam is vertical, Fig. 10, by moments, as before, about B, Pi = 1,000 -2 -mo =200 lb. at A acting to the left, being tension or a negative reaction. By moments about A, P 2 =i,ooo-2-Mo=2oo at B, acting to the right. Or Pi + P 2 =o; . * . P x = — P 2 . By similar moments, the 1,000 lb. at C causes two equal and opposite horizontal forces on the beam at D and E, of 500 lb. each, that at D being ten- sion on the connection, or acting towards the right, and that at E acting in the opposite direction. These two forces make a couple MATERIALS. 41 balanced by the couple P\P2- The weight 1,000 lb. multiplied by its arm 2 ft. is balanced by the opposing horizontal forces at D and E, 4 ft. apart. There remains a vertical force of 1,000 lb. in A B, which may all be resisted by the point B, when the compression in DE= 500 lb. and in E B= 1,000 lb.; or all by the point A, when the tension in D E=5oo lb. and in D A= 1,000 lb.; or part may be resisted at A, and the rest at B, the distribution being uncertain. This longitudinal force may be disregarded in discussing the beam, as may the tension or compression in the hanger arms themselves. 53. Bending Moments. — If an imaginary plane of section is passed through any point in a beam, the sum of the moments of all the external forces on one side of that section, taken about a point in the section, must be exactly equal and opposite to the sum of the moments of all the external forces on the other side of that section, taken about the same point. If not, the beam would revolve in the plane of the forces. The moment on the left side of the section tends to make that portion of the beam rotate in one direction about the point of section, and the equal moment on the right side of the section tends to make the right segment rotate in the opposite direction. These two moments cause resistances in the interior of the beam at the section (which stresses will be discussed under resisting moment), with the result that the beam is bent to a slight degree. Either resultant moment on one side of a plane of section, about the section, is called the bending moment at that point, usually denoted by M, and is con- sidered positive when it makes the beam concave on the upper side. Ordinary beams, supported at the ends and carrying loads, have positive bending moments. If upward reactions are positive, weights must be taken as negative and their sign regarded in writing moments. Examples. — Section at D, Fig. 3, 10 ft. from B. On the left of D, and about D, P x ( = 300) -20— 500-8=2,000 ft. -lb., positive bending moment at D. Or, about D, on the right side of the section; P 2 (= 200) -10=2,000 ft. -lb., positive bending moment at D. Usually compute the simpler one. Section at A, Fig. 4, TF-C A= — 750-5= —3,750 ft.-lb. negative bending moment at A, tending to make the beam convex on the upper side. AtD, 10 ft. from B, M= -P 2 - 10= - 187^-10= -1,875 ft.-lb., negative because P2 is negative. At A, taking moments on the right 42 STRUCTURAL MECHANICS. of and about A, M= — 187!- 20= —3,750 ft. -lb., as first obtained. This beam has negative bending moments at all points. In Fig. 5, M at D is— 150- 10= — 1,500 ft.-lb. It is evident that the bending moments at all points between C and A can be found found without knowing the reactions. If this beam is built into a wall, the points of application of P\ and P2 are uncertain, as the pres- sures at A and B are distributed over more or less of the distance that the beam is embedded. The maximum M is at A, and is —150-20= — 3,000 ft.-lb. It is evident that the longer A B is, the smaller the reac- tions are, and hence the greater the security. In Fig. 6, the bending moment at C will be Pi -AC — weight on D C- JD C=i,92o-i6-2oo-8-4=24,32o ft.-lb. At E, M =1,280-16= 20,480 ft.-lb. In Fig. 7, the bending moments at the several points of application of the weights, taking moments of all the external forces on the left of each section about the section, will be — At C, M= — ioo-o=o. At A, M= — 100-2= — 200 ft.-lb. AtD, ilf=-ioo-5 + (665f-2oo)-3 = 8o6£ft.-lb. At E, M= - ioo- 7 + 465!- 5- 150- 2 = 1,328^ ft.-lb. At F, M= — ioo- 10 + 4651-8— 150-5 — 300-3 = 1,075 ft.-lb. And, at B, M will be zero. M max. occurs at E. Do not assume that the maximum bending moment will be found at the point of application of the resultant of the load. The method for finding the point or points of maximum bending moment will be shown later. The moments on the right portion of the beam may be more easily found by taking moments on the right side of any section. Thus at F, M = (P 2 -8o)-8=(2i4J}-8o)-8=i,o75 ft.-lb. Find the bending moment at the middle of E F. i,20ifs ft.-lb. In Fig. 8, the bending moment at section C of the piece A E may be found by considering the portion above C. As the unit pressure at C is 6X62^ lb. = 375 lb. per sq. ft., M at C= P 2 (=i,265.6)-3 — (375X6-J-2)-6-^3 = 1,546.8 ft.-lb. At the section B, M= -(3X62|X3-2)Xi = -28ii ft.-lb. In Fig. 9, as Pi = 400 lb., P2=6oo lb., vertical forces at D and E are each 500 lb.; M at D= 1,600 ft.-lb.; Mat E= 1,200 ft.-lb. In Fig. 10, as Pi = — 200 lb.= — P2, and the horizontal forces at D and E are ± 500 lb.; M at D=-8oo ft.-lb.; M at E=+4oo ft.-lb. The beam will be concave on the left side at D and convex at E. The curvature must change between D and E, where M=o. Let this point be distant x from B. Then 200 •*— 500(3;— 2) = o; .'. x= 3* ft- BEAMS 43- The curved piece A B, Fig. n, with equal and opposite forces applied in the line connecting its ends, will experience a bending moment at any point D, equal to P-CD, this ordinate being perpendicular to the chord. 54. Shearing Forces. — In Fig. 3, of the 500 lb. at C, 300 lb. goes to A and 200 lb. to B. Any vertical section between A and C must therefore have 300 lb. acting vertically in it. On the lei't of such a section there will be 300 lb. from Pi acting upwards, and on the right of the same section there will be 300 lb., coming from W, acting downwards. These two forces, acting in opposite directions on the two sides of the imaginary section, tend to cut the beam off, as would a pair of shears, and either of these two opposite forces is called the shearing force at the section, or simply the shear. When acting upwards on the left side of the section (and downwards on the right side), it is called positive shear. When the reverse is the case the shear will be negative. Examples. — In Fig. 7, where a number of forces are applied to a beam, there must be found at any section between C and A a shear of —100 lb.; between A and D the shear will be — 100+665!— 200= + 365! lb.; between D and E the shear will decrease to 3651—150= 215! lb.; on passing E the shear will change sign, being 2151—300= — 84! lb.; between F and B it will be — 84! — 50= — 134I lb.; and on passing B, it becomes zero, a check on the accuracy of the several calculations. In Fig. 8, the shear just above the support 6 = 3X62^X3-^-2 = 281J lb.; just below the point B the shear is 281^—1,265.6 = —984.4 lb.; and just above A it is 1,265.6 lb. The signs used imply that the left side of A E corresponds to the upper side of an ordinary beam. As the shear is positive above A and negative below B, it changes sign at some intermediate point. Find that point. In Fig. 9, the shear anywhere between A and D is +400 lb.; at all points between D and E it is 400— 500= — 100 lb.; and between E and B is —600 lb. The shear changes sign at D. In Fig. 10, the shear on any horizontal plane of section between B and E is —200 lb.; betwen E and D is —200+500= +300 lb.; and between D and A is +300— 500= — 200 lb. The shear changes sign at both E and D. 55. Summary. — To repeat: — The shearing force at any normal section of a beam may therefore be defined to be the algebraic sum 0} all the transverse forces on one side of the section. 44 STRUCTURAL MECHANICS. When this sum or resulting force acts upward on the left of the section, call it positive; when downward, negative. The bending moment at any right or normal section of a beam may be stated to be the algebraic sum oj the moments of all the transverse forces on one side oj the section, taken about the centre oj gravity oj the section as axis. When this sum or resulting moment is right-handed or clockwise on the left of and about the section, call it positive. A positive moment tends to make the beam concave on what is usually the upper side. By a proposition in mechanics, any force which acts at a given distance from a given point is equivalent to the same force at the point and a moment made up of the force and the perpen- dicular from the point to the line of action of the force. Then in Fig. 7, if a section plane is passed anywhere, as between D and E, the resultant force on the left, which is the algebraic sum of the given forces on the left of the section, is the shear at the section; and this resultant, multiplied by its arm or distance from the point in D E, giving a moment which is the algebraic sum of the moments of the several forces on the left of and about the point, is the bending moment at the section. It is also evident that the resulting action at any section is the sum of the several component actions; and hence that dif- ferent loads may be discussed separately and their effects at any point added algebraically, if they can occur simultaneously. Thus the shears and bending moments arising from the weight of a beam itself may be determined, and to them may be added the shears and bending moments at the same points from other weights imposed on the beam. The numerous examples already given show that formulas are not needed for solving problems in beams, and the student will do well to accustom himself to using the data directly. Formulas, however, will now be derived, which will sometimes be convenient for use, and from which may be deduced certain serviceable relationships. 56. Bending Moment a Maximum where the Shear Changes Sign. — If a beam weighing w per unit of length is supported at each end and carries a system of loads any one of which is BEAMS. 45 distant a from the left support, the shear at a section distant x from the left support is F x = F x -l\W-wx, and the bending moment at the same section is M x = Pix - 1% W(x -a)- iwx 2 . If the beam is a cantilever, that is, a beam fixed in position at the right end and unsupported at the left, the same equation will apply when P x becomes zero. It is seen by comparing the equations above that F is always the first derivative of M, or dM x dx **' Hence, according to the rule for determining maxima and minima, the bending moment is always a maximum (or minimum) at the place where the shear is zero or changes in sign. This criterion is easily applied to locate the points of M maximum. Pass along the beam from the left (or right) until as much load is on the left (or right) of the section as will neutralize P\ (or P 2 ) and the point of M max. is found. Its value can then be computed. If the weight at a certain point is more than enough to reduce F to zero, F changes sign in passing that point, and hence M max. occurs there. For a beam fixed at one end only, F changes sign in passing Pi, and hence M max. is found at the wall. Examples. — M max. occurs in Fig. 3, at C; in Fig. 4, at A; in Fig. 6, at 17.6 ft. from A; in Fig. 7, at A, and again at E; in Fig. 8, at B, and again at a distance x from E such that 62%x- %x= 1,265!; • * • x =>/4o.5 = 6.36 ft.; in Fig. 9, at D; and in Fig. 10, at D and again at E. The bending moments which may not have been found at some of these points can now be computed. The reader who is familiar with graphics can draw the equi- librium or bending-moment polygons or curves, and the shear diagrams, and notice the same relation in them. 46 STRUCTURAL MECHANICS. The unit load may also be considered as the derivative of the shear; F therefore has maximum (or minimum) values where the external forces change in sign. The origin of coordinates may be arbitrarily taken at any point in the length of the beam and general expressions may be written. If — w is the unit load and is constant, ■wx % F x = — jwdx = F M x = f X F x dx = M + F x - \wx 2 , in which Fo and Mo are the constants of integration, the values of F and M at the origin. Thus in the beam above, the bending moment at a second section distant c from the first is M x+c = M x + F x c-I x x +c W(x-a + c)-iwc 2 . The same expression is easily derived by substituting x+c for x in the original equation. Example.— -In Fig. 7, M at D= +896! ft.-lb. F between D and E= + 215I lb. M at E=+896j+ 2I 5fX2= 1,3284 ft.-lb. as in §53. Afmidway between E and F = 8g6|+ 2151X3!— 300X1!= 1,201x5 ft.-lb. 57. Working Formulas — The bending moments and shears for a number of simple cases of common occurrence are given below. In general the bending moment and the shear vary from point to point along a beam, and they may be conveniently repre- sented on a diagram by ordinates whose lengths represent the values of those quantities. In the accompanying figures the upper diagram is one of bending moments and the lower is one of shears. Positive values are laid off above the base line and negative below. The student who has followed the examples of the preceding sections should have no difficulty in computing the ordinates given in Figs. 12 to 15. Figs. 12 and 13 represent cantilever beams, one carrying a load of W at the end, the other carrying a uniformly distributed BEAMS. 47 load of w per unit of length. The bending moment at any section distant x from the free end of the beam of Fig. 13 is M— — wx-\x = — \wx 2 and the bending- moment diagram is therefore a parabola. w Fig. 12 W !~ ■»-f Fig. 14 liiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiinniiiiimiiiiiimi Fig. 13 Fig. 14 shows a beam on two supports carrying a load W, The bending moment is evidently a maximum under the weight. and is equal to W j — , a quantity easily remembered as the weight into the product of the two segments divided by the span. When the load is at mid-span M = \Wl. In Fig. 15 a load of intensity w per unit of length is distributed over the beam. The left reaction is ^wl, and the bending moment at a section dis- tant x from the left support is M = \wl • x — wx' \x = \wx(l —X s ) . The right-hand member of this equation contains x(l — x), which is the product of two variables whose sum is constant* therefore the bending-moment diagram is a parabola. 4 8 STRUCTURAL MECHANICS. The above diagrams are drawn for the applied loads alone, that is, the beams are considered to be without weight. If a beam weighing w per unit of length carries a load W as in Fig. 12 or 14, the bending moment or shear at any section can be found by adding the ordinates of Figs. 12 and 13 or of Figs. 14 and iv Eig. 17 Fig. 16 At any section distant x from one end of a beam on two sup- ports, the bending moment due to a single moving load is a maximum when the load is at the section, when it has a value of M = Wx(l-x)+l. As this is equation of a parabola of altitude \Wl, the absolute maximum moments which can occur in the beam under a single moving load are as shown in Fig. 16. The greatest positive shear at any section occurs when the load is just to the right of that section ; it is equal to the left reaction and consequently is proportional to the distance of that section from the right support, hence the locus of maximum positive shear is a straight line as shown. If a uniform load advances continuously from the right end of the beam of Fig. 17, the positive shear at any section will increase until the load reaches the section, after which it will decrease as the load extends into the left segment. This is evident from the fact that any load placed in the right segment causes positive shear at the section, while any load in the left segment causes negative shear at the section; see Fig. 14. When the load extends a distance, x, from the right support the shear at the head of the load is F=P\ = \wx 2 +l and the curve of max- BEAMS. 49 imum shears is a parabola. To cause maximum negative shears the load must advance from the left. The maximum moment at any section occurs when the whole span is loaded, in which case the moments are given by Fig. 15. 58. Position of Wheel Concentrations for Maximum Moment at any Given Point. — Where moving loads have definite magni- tudes and spacings, as is the case with the wheel weights of a locomotive, the position of the load, on a beam or girder supported at both ends, to give maximum bending moment at any given A -3t— 4 ooOOpO 0*0 00 t- c 1 Fig. 18 section may be found as follows: — Let the given section be C, at a distance a from the left abutment of a beam A B, of span /, Fig. 18. Let Ri be the resultant of all loads on the left of C, and acting at a distance X\ from the left abutment; let R2 be the resultant of all loads to the right of C, and acting at a distance x 2 from the right abutment. The reaction Pi at A, due to R2 alone, is R2X2 -*-/j and the bending moment at C, due to R2 only, is R2CLX2 -=-/• Similarly P 2 , due to R\, is R\Xi +1, and the bending moment at C, due to Ri only, is Ri(l — a)xi -7-/. Hence the total bending moment at C is M=R2 ^ +Rl y-«^ i 1 If the entire system of loads is advanced a short distance d to the lejt, the bending moment at C becomes .., D a(x 2 +d) {l-a)(x l -d) M =A.2 j +xvi' / / The change of bending moment due to moving the loads to the left is ad (l—a)d M'-M = R 2 -j--Rv—y L . 50 STRUCTURAL MECHANICS. If the loads are moved a distance, d, to the right instead of to the left, the change of bending moment is ad (l—a)d M'-M= -R 2 j+Rv-j J ~. From the two values of M'—M it is seen that the bending moment at C will be increased by moving the loads to the left when R 2 a>Ri(l—a) and by moving to the right when Ri(l—a)>R 2 a. When Ri(l—a)=R 2 a the bending moment can- not be increased by moving either to the right or left and is there- fore a maximum. The condition may be written Ri R 2 Ri+R 2 a I— a I Ri+a is the average load per unit of length on the left segment and (Ri +R 2 ) +1 is the average load on the span, hence The bending moment at any point of a beam carrying a system of moving loads is a maximum when the average load on one segment is equal to the average load on the span. Since, for maximum bending moment at any section, a load must be at that section, place a load W n at the given point and compute the above inequality, first considering W n as being just to the right and then just to the left of the section. If the inequality changes sign, the position with W n at the section is one of M max. The value of M max. can be computed as in § 53. If, however, the inequality does not change sign, move the whole system until the next W comes to the section, and test the inequality again. It sometimes happens that two or more different positions of the load will satisfy the condition just explained, and, to determine the absolute M max., each must be worked out numerically. When there are some W's much heavier than others, M max. is likely to occur under some one of them. When other loads are brought on at the right, or pass off at the left, they must not be overlooked. 59. Position of Wheel Concentrations for Maximum Shear at any Given Point. — The shear at point C in the beam or girder of Fig. 19, as the load comes on at the right end, will increase until the first wheel W\ reaches C. When that wheel passes C, the shear at that point suddenly diminishes by W\, and then again gradually increases, until W 2 reaches C. Let R be the sum of all loads on the span when W\ is at C, and x the distance from the centre of gravity of the loads to the right point of support BEAMS. 51 B. The shear at C will be P\ = Rx+l. If the train moves to the left a distance b, the space between W\ and W 2 , so that W 2 has just reached C, the shear at C will be R(x+b) +1 — W\, plus a small quantity k which is the increase in Pi, due to any additional loads which may have come on the span during this advance of the train. The shear at C will therefore be R d) &) G) o o Fig. 19 increased by moving up W 2 , iiRb+l + ky Wi, or (as k can often be neglected) if J TX, R W l Hence move up the next load when the average load per foot on the span is greater than the load on the left divided by the distance between W\ and W 2 . R'c R r W Similarly, W3 should be moved to C if —j->W2 or — > — , 1/ If c R! being the sum of the loads on the span when W 2 is at C, and c the distance between W 2 and W3. It is not necessary to take account of k unless the two sides of the inequality are nearly equal. Example. — Span 60 ft., weights in units of 1,000 lb. 1 2 3 4 5 Weights = 8 15 15 15 15 6 7 9 9 Spacing =8' 6' 4 i' 4i' 7' 5' 8 9 10 9 9 8 15 6' 5' 8' 8 r To apply test for M max. at 15 ft. from left, load advancing from right. With W 2 at quarter span, load on span=io4. If W 2 is just to the right, ^>^ or ^- 4 >-; if W 2 is just to the left, ^>^- 3 ; 60 15 4 1 J '41' therefore move up W 3 to right of quarter span. W i0 now is on the 52 STRUCTURAL MECHANICS. 112 23 112 "?8 span. >— . Consider W% to be just to the left; then — < — , 4 1 41 or the inequality changes with W 3 . P x = (8- 59+ 15 -51 + 45 -4oi+36- 21 + 8 -5) ^60= 64.26. M max. = (64.26)15 — 8- 14—15-6=761,900 ft. -lb. To test for F max. at same point. Put W\ at quarter span. Load 95 8 _ _ tt7 1 i io 4 ^ J 5 on span=QS. ^ >-. Move up WV, load on span now 104. — - <— . 60 8 00 Inequality changes. P 1 = (8- 53 + 15 -45 + 45 *34i+3 6 " 15)^60=53.2. F. max. = Pi — ^1 = 53.2 — 8 = 45,200 lb. When these locomotive wheel loads are distributed to the panel-joints of a bridge truss through the longitudinal stringers, which span the panel distance between floor-beams, the above rule is modified. A load in a panel being supported directly by the stringers is by that means carried to the joints of the truss. When the train advances from the right end until the forward wheels are in the panel under investigation, the shear in that panel is the left reaction minus such part of the loads in the panel as go to the floor-beam to the left. Let R be the resultant of all the loads on the span applied at a distance x from the right sup- port, and let Ri be the resultant of the loads in the panel, of length p, applied at a distance, x± 9 from the right floor-beam. Then _ ^x n X\ I p If the loads are moved a distance, d, to the left, the change of shear is F>-F-d-R& I p and by an argument similar to that of the preceding section, for a maximum RRi l~ p' Hence the shear in any panel of a truss is a maximum when the average load in the panel is equal to the average load on the span. 60. Absolute Maximum Bending Moment on a Beam under Moving Loads. — When a beam or girder of uniform cross-section, such as a rolled I beam, supported at its ends, is subjected to BEAMS. 53 a system of passing loads, such as an engine, heavy truck, or trolley, it generally suffices to determine that position of the system of weights which causes the absolute maximum bending moment, the section where it is found, and its amount. In Fig. 18 let C be that section. Let R = resultant of all loads on the beam and #=its distance from B; R\ = resultant of the loads to the left of C. The reaction at left, Pi = Rx+l; and, since the bending moment at C is to be a maximum, the shear at C must be zero, or v x t> R R\ R-, — jfvi=o. .*. t = — . / I x But the position of loads must also satisfy the condition of § 58, since there is to be maximum bending moment at C, and Ri R Ri Ri — =— . .\ — = — , or a = x. a I ax The point of absolute maximum bending moment therefore is as far from one end as the centre of gravity of the whole load is from the other. This rule may be written: When the middle of the span bisects the distance between the centre of gravity of the whole load on the span and one of the wheels on either side of the centre of gravity, the desired moment is to be found under one of the two wheels. Example. — Beam of span 24 ft. Two wheels 6 ft. apart ? one carrying 2,000 lb. and one 4,000 lb., pass across. Centre of gravity is 2,000-6^6,000=2 ft. from the heavier wheel. Then this wheel is to be placed 1 ft. from mid-span. Reaction= 6,000-11-^24= 2,750 lb. M. max. = 2,750- 11 = 30,250 ft. -lb. 61. Total Tension Equals Total Compression. — If a beam, loaded in any manner, and in equilibrium under the moments caused by the external forces, is cut perpendicularly across by an imaginary plane of section, while the right-handed and left- handed bending moments already shown to exist, § 53, continue to act, it is evident that the left and right segments of the beam can only be restrained from revolving about this section by the internal stresses exerted between the material particles con- tiguous to the section. These stresses must be of such signs, that is tensile and compressive; of such magnitude, provided 54 STRUCTURAL MECHANICS. the material does not give way; and so distributed over the cross-section, as to make a resisting moment just equal to the bending moment at the section. For the former is caused by the latter and balances it. Since the moment arms of these stresses lie in the perpen- dicular plane of section, the components to be considered now will be normal to the section. The tangential components are caused by and balance the external shear. As the external forces which tend to bend a beam are all transverse to it, and have no horizontal components, the internal stresses of tension and compression which are caused by the bending moment must be equal and opposite, as required for a moment or couple, and hence the total normal internal tension on any section must equal the total normal compression. When any oblique or longitudinal external forces act on a beam, there is always found that resultant normal stress on any right section which is required to give equilibrium. 62. Distribution of Internal Stress on any Cross-section. — It may be convenient in the beginning to consider one segment of the beam removed, and equilibrium to be assured between the external moment tending to rotate the remaining segment and the resisting moment developed in the beam at the section, as shown in Fig. 20. If two parallel lines near together are drawn on the side of a beam, perpendicularly to its length, before it is loaded, these lines, when the beam is loaded to any reasonable amount and bent by that loading, will still be straight, as far as can be observed from most careful examination; but they will now con- verge to a point known as the centre of curvature for that part of the beam. An assumption, then, that any and all right sections of the beam, being plane before flexure, are still plane after the flexure of this beam, is reasonable. If the right sections become warped, that warping would apparently cause a cumulative endwise movement of the particles at successive sections, especially in a beam subjected to a constant maximum bending moment over a considerable portion of its span; and such a movement and BEAMS. 55 resulting distortion of the trace of the sectional plane ought there- fore to become apparent to the eye. Such a warping can be perceived in shafts, other than cylindrical, subjected to a twisting couple, but cannot be found in beams. The lines A C and B D just referred to will be found to be farther apart at the convex side of the beam, and nearer together at the concave side, than they first were; hence a line G H, lying somewhere between A B and C D, is unchanged in length. If, in Fig. 20, a line parallel to A C is drawn through H, the extremity ^LAabf, ft c Fig. 20 of the fibre G H which has not changed in length, K L will represent the shortening which I L has undergone in its reduction to I K, and N O will represent the lengthening which M N has experienced in stretching to M O. The lengthening or shorten- ing of the fibres, whose length was originally GH=^, is directly proportional to the distance of the fibre from G H, the place of no change of length, and hence of no longitudinal or normal stress. The diagram, Fig. 1, representing the elongation or shortening of a bar under increasing stresses, shows that, for stresses within the elastic limit, equal increments of lengthening and shortening are occasioned by equal increments of stress. If this beam has not been loaded so heavily as to produce a unit stress on any particle in excess of the elastic limit (and no working beam, one expected to last permanently, should be loaded to excess), the longitudinal unit stresses between the particles will vary as the lengthening and shortening of these fibres, that is, as the dis- tance from the point of no stress. Hence, at any section, the direct stress is uniformly varying, with a maximum tension on 56 STRUCTURAL MECHANICS. the convex side and a maximum compression on the concave side. The stresses on different forms of cross-section A C are shown in Fig. 21. The total tension on the section is always equal to the total compression. Fig. 21 63. Neutral Axis. — The arrows in Figs. 20, 21 may be taken to represent the unit stress at each point of the cross-section, varying as the distance from the plane of no stress, and constant in the direction z. To locate the point or plane of no stress or neutral axis for successive sections : Let j c and } t be the unit stresses of compression and tension between the particles at the extreme edge of any section, distant y c and y t from the point of no stress. It is plain that } c '-ft == y c 'yt from similar triangles, and that the unit stress p at any point distant y from the point of no stress will be n or — y yt or, in general — y, yi from a similar proportion. If zdy is the area of the strip on which the unit stress p is exerted, z being the variable coordinate at right angles to x and y, the total force on zdy will be pzdy = — yzdy, where — is a con- stant, the unit stress at a unit distance. As the total normal tension on the section is to equal the total compression, or their sum is to be zero, § 61, the condition may be written / r +ye — / yzdy = o. yw -yt Therefore the sum of the moments zdy-y of the strips zdy about the axis of z must balance or be zero. Then the axis of z or BEAMS. 57 neutral axis must pass through the centre oj gravity of a thin plate representing the section, and the neutral axis of any section lies in its plane, and usually in a direction perpendicular to the plane oj the applied external forces. The axes of the successive cross- sections make up what is known as the neutral plane of the beam. Although there is no longitudinal or normal tension or compres- sion at that line of the cross-section, it experiences shear, as will be shown later. 64. Resisting Moment. — The law of the variation of stress on the cross-section and the location of the neutral axis have been established. The resisting moment is caused by and is equal to the bending moment. The moments of all the stresses about the neutral axis Z Z is, since p has the same sign as y, and the moments conspire, M = /( +p)zdy( +y) +f(-p) zdy ( -y) = f ' ' ' pyzdy. As }-s-yi denotes the unit strezs at either extreme fibre divided bv its distance from the neutral axis, and p= —y, y\ j r +: - ji M=- / y 2 zdy = L -, and / yw -yt 3'i My 1 T ' where / represents / y 2 zdy about the axis Z Z, lying in the plane of the section, through the centre of gravity of the same and perpendicular to the plane of the external forces applied to the beam. / is termed in mechanics the moment oj inertia of a plane area, and is usually one of the principal moments of inertia of the area. The integral will be of the fourth power, involving the breadth and the cube of the depth. For moments of inertia of plane sections, see Chap. IV. In the above expression for the resisting moment the quantity I-^yi is known as the section modulus. The section moduli of steel beams, angles, etc., are tabulated in the handbooks published by the various steel manufacturers, so that the resisting moment of a steel beam can be readily found by multiplying the section modulus by the working stress. 5^ STRUCTURAL MECHANICS. As moments of inertia of plane areas are of the fourth power, and can be represented by n'bh 3 , where h is the extreme dimension parallel to y, and b to z, and as y\ may be written m'h, the resisting moment can be represented, if n' ' -±-m' '=n, by M = ! - = njbh 2 , 71 being a fraction. For a rectangular section this becomes bh 3 i i ' 12 2 6' and for a circular section -,74 T - M=}-T- + -d= —fd 3 = 0.0982/J 3 . 64 2 32 ' Examples. — A timber beam 6 // Xi2 // , set on edge, with a safe unit stress of 800 lb. will safely resist a bending moment amounting to 8oo-6- 1 2 2 -^ 6= 115,200 in.-lb. A round shaft, 3 in. in diameter, if /= 12,000 lb. will have a safe resisting moment of 12,000- 22 -3 3 -^7 -32 = 31,820 in.-lb. For rectangular sections, either b or h is usually assumed and h or b then found. If the ratio h-^-l is fixed by the desire to secure a certain degree of stiffness (see "Deflection of Beams," Chap. VI.), the unknown quantity is b. Example. — A wooden beam of 12 ft. span carries 3,600 lb. uniformly distributed. M=\Wl= \ -3,600- 12- 12 = 64,800 in.-lb. If /= 1,000, £=1,400,000, and the deflection v is not to exceed -g-^-g- of the span, , v 5// . 1 5 1,000-2-144 . „ from —= _' is obtained - — = — r ; . . h=i$ in. Then / 48.fc.y1 600 48 1,400,000- li assuming /*=i4 in., a practicable size, 64,800=-^- — 6-i4 2 ; and b = 2 in. Economy of material apparently calls for as large a value of h as possible; but the breadth b must be sufficient to give lateral stiffness to the beam, or it may fail by the buckling or sidewise flexure of the compression edge, between those points where it BEAMS. 59 is stayed laterally. The effect of loading as a beam a thin board set on edge will make clear the tendency. When the plane of the applied forces does not pass through the axis of the beam, a twisting or torsional moment is added, which will be discussed in § 86. 65. Limit of Application of M=f I-^yi. — The expression for the resisting moment at any section of a beam, caused by and always equal to the external bending moment at that section, is applicable only when the maximum unit stress / does not exceed the unit stress at the elastic limit of the material. If / exceeds that limit, a uniformly varying stress over the whole section is not found, and the neutral axis may not remain at the centre of gravity. Hence, also, the substitution of breaking weights, obtained by experiments on beams which fail, in >a bending- moment formula which is then equated with JI +yu results in values of /, the then so-called modulus of rupture, agreeing with neither the tensile nor the compressive strength of the material, and therefore of but limited value. This formula is correct for the purpose of design and construction; but its limitation should be kept in mind. 66. The Smaller Value of f-^yi to be Used. — Since from similar triangles fc-^y c = lt-^yt) it is immaterial which ratio is used for M for a given cross-section. But, in designing a cross- section to resist a given moment, if y t and y c are not to be equal, another consideration has weight. A numerical example will bring out the distinction. A beam of 24 in. span is loaded at the middle with a weight of 500 lb. M max. will be \Wl= $00 -6 = 3,000 in. -lb. If the depth of the beam is 5 in., and its section is of such a form that the distance from its centre of gravity to the lower edge is 2 in., and to the upper edge is 3 in., while 1 = 4, then 3,000 = J/r 4 or J/ c • 4. Hence the maximum unit tension } t = 1 , 500 lb. per sq. in., and the maximum unit compression f e = 2,250 lb. per sq. in. But if the material of the above beam must not be subjected to a unit stress greater than 2,000 lb. per sq. in., that unit stress will be found on the compression side; for 2,000 lb. per sq. in. on the tension side would be accompanied by 3,000 lb. per sq. 60 STRUCTURAL MECHAXICS. in. on the compression side; and a unit stress of 2,000 lb. com- pression is only compatible in this case with 2,000-1=1,333 lb. unit stress tension. The beam will safely carry only a moment of 2,000-4^3 = 2,667 in.-lb. Hence, when designing, with a maximum allowed value of /, and using a form of section where y t and y c differ, take that ratio of f+yi which is the smaller. For a few materials, where j c and }t may be taken as differing in magnitude, as perhaps in cast iron, use that ratio } c +y c or ft+yt which gives the smaller value. As the elastic limit in tension and compression for a given material is usually the same, use in computations the larger value of y\. 67. Curved Beams. — An originally curved beam, at any given cross-section made at right angles to its neutral axis, so far as the resisting stresses to bending moments are concerned, is in the same condition with an originally straight beam at a similar and equal cross-section to which the same bending moment is applied. Any definite thrust or tension at its two ends adds a moment at each right section equal to the product of the force into the perpendicular ordinate from the chord to the centre of the section, and a force, parallel to the chord, which force can be resolved into one normal to the section and a shear. Compare Fig. n. 68. Inclined Beams. — A sloping beam is to be treated like a horizontal beam, so far as resisting stress produced by that component of the load which is normal to the beam is concerned. The component of the load which acts along the beam is to be considered as producing a direct thrust along the beam if taken up at the lower end; or a direct tension, if taken up at the upper end, or as divided somewhat indeterminately, if resisted at both ends. If this longitudinal force is axial, the mean unit stress } e caused by it is to be added to the stress /& of the same kind from bending moment at the section where this sum f e +jb will be a maximum. This point can easily be found graphically. If the section of the piece is the unknown quantity, it will commonly suffice to use the value of M max. to determine an approximation to //„ and to correct the section by the resulting value of f e +fb at the point where the sum is largest. If the direct force at the end or ends is not applied axially, BEAMS. 6 1 its moment at any section may augment or diminish the bending moment of the normal components of the load. Cases of inclined beams, for a given load and inclination, are better solved directly than by the application of formulas. Example. — A wooden rafter, 15 ft. long, has a horizontal pro- jection of 12 ft., and a rise of 9 ft., and it carries a uniformly distributed load of 1,500 lb. The normal component of this load will be 1,200 lb., the component along the roof 900 lb. The maximum bending moment, 1 -tii mi 1 1,200X15X12 . .. Tr . . at the middle, will be -— ^ = 27,000 m.-lb. If the sate 8 11 1 • 1 • iiii i,ooobh 2 stress is 1,000 lb., the section to carry this moment should be 6 27,000, or bh 2 = 162. If b = 2>, h=8 in. If the mean thrust, at the middle of the rafter, is 1,250 lb., the maximum thrust, at the bottom end, will be 1,700 lb., and the minimum thrust, at the top end, will be 800 lb. The section of maximum fibre stress will be a very little below the middle. But, if the rafter is 3"X8", /& from bending ... . 27,000-6 ... .. . 1,250 „ moment will be =844 lb. Also, j c — — — =52 lb. Hence 3- 8-8 24 /c+/& = 896 lb., a satisfactory result, if the rafter is stayed laterally by the roof-covering or otherwise. 69. Movement of Neutral Axis if Yield-point is Exceeded. — If it is assumed that cross-sections of a beam still remain plane after the yield-point is passed at the extreme fibres, the stretch and shortening of the fibres at any cross-section will continue to vary with the distance from the neutral axis or plane. Suppose then that the elongation per unit of length of the outer tension fibre has attained an amount equal to O L, Fig. 1. The unit stress on that fibre will be L N. A fibre lying half-way from that edge to the neutral axis will have a unit stress K M. If the beam is rectangular, the total tension on the cross-section must be the area O M N L, O L now being the distance from the neutral axis of the beam to the tension edge. Since the total compression on the section must equal the total tension, an equal area O L'N' must be cut off by L' N' and the compression curve. The neu- tral axis must then divide the given depth of the beam in the ratio of O L to O I/, shifting in this case towards the compression side. Had the compression curve been below the tension curve, the neutral axis would have shifted towards the convex side of the beam. Since L N is less than L' N r , the unit stress on the extreme fibre on the tension side is the less. Hence this displacement of the neutral axis favors the weaker side. If such action continued 62 STRUCTURAL MECHANICS. to the time of fracture, it would account for the fact that the application of the usual formula, fl+yi, to breaking moments gives a value of / which lies between the ultimate tensile and compressive strengths of the material. It must be borne in mind, however, that the compression portion of the section increases in breadth and the tension portion contracts, quite materially for ductile substances, thus adding to the complication. A soft steel bar cannot be broken by flexure as a beam at a < ingle test. A rectangular cross-section also tends to assume the section shown in Fig. 22. The compressed particles in the middle of the width can move up more readily than they can laterally, making the upper surface convex as well as wider, and the particles below at the edges, being drawn or forced in, are crowded down, making the lower surface concave as well as narrower. ^Fig.22** Hence the position of the neutral axis is un- certain, after the yield-point has been passed on either face; but it is probably moved towards the stronger side. 70. Cross-section of Equal Strength. — When a material will safely resist greater compression than tension, or the reverse, it is sometimes the custom to use such a form of cross-section that the centre of gravity lies nearer the weaker side. Cast iron is properly used in sections of this sort. See Fig. 21, section at right. Wrought-iron or steel sections are occasionally rolled or built up in a similar fashion, but the increase in width of the compression flange is then usually intended to increase its lateral stiffness. If / f =safe unit tensile stress, and / c =safe unit compressive stress, the centre of gravity of the section must be found at such point that yt'.y e —ft'}e» when the given safe stresses will occur simultaneously at the section. By composition, y%\y c '.h — }t'}c : h +/« so that the centre of gravity should be distant from the bottom or top, 7 ft j }c ■yt = h7—T, or y c = h- : it+U Jc /«+/.' Example. — If ^=3,000 lb., and / c =9,ooo lb., ^ = 7* -3,000 -4- 12,000 — \h. If a cast-iron x section is to be used, base 10 in., thickness through- BEAMS. 63 out of 1 in., and height of web h', then, by moments around base, t 10 , i2=; , , 165 1r 3,000-165-2 . . .. /= — I ^ + 10-1 + 5-4= — . M=° 2 5 — =82,500 m.-lb., 12 12 4 4-3 the moment that the section will carry. 71. Beam of Uniform Strength. — As has been shown in § 64, the resisting moment may be put into the form M = njbh 2 , where n is a numerical factor depending on the form of cross- section. If, then, for a given load, bh 2 be varied at successive cross-sections to correspond with the variation of the external bending moment, the unit stress on the extreme fibre will be constant; the beam will be equally strong at all sections, except against shear; and there will be no waste of material for a given type of cross-section, provided material is not wasted in shaping. Suppose, for example, that a beam is to be supported at its ends, to carry W at the middle, and to be rectangular in cross - section. The bending moment at any point between one sup- port and the middle is %Wx. Equate this value with the resist- ing moment. \Wx = \jbh 2 . To make / constant at all cross- sections, bh 2 must vary as x from each end to the middle. If h is constant, b must vary as x, or the beam will be lozenge-shaped in plan and rectangular in elevation. If, on the other hand, b is constant, h 2 must vary as x, and the elevation will consist of two parabolas with vertices at the ends of the beam and axis horizontal, while the plan will be rectangular. The section need not be a rectangle. If the ratio of b to h is not fixed, the treatment will be like the above; but, if that ratio is fixed, as for a circular section, or other regular figure, b = ch, and h s must vary as the external bending moment, or, in the case above, as x. The cross-section of the cast-iron beam in the example of the previous section may be varied in accordance with these principles. The following table gives the shape of beams of rectangular cross-section supported and loaded as stated. When a beam supported at both ends carries a single moving load W, passing across the beam, the bending moment at the 64 STRUCTURAL MECHANICS. M bh 2 h 2 Constant, b Constant, vanes as b varies as h 2 varies as Fixed at one end, -Wx X x, triangular plan, x, parabolic eleva- W at other. Fig. 23. tion. Fig. 24. Fixed at one end, — \wx 2 X 2 x 2 , parabolic plan, x 2 , h varies as x, uniform load. Fig. 25. triangular eleva- tion. Fig. 26. Sup't'dbothends ( W^x X 1 triangular plan, Fig. 27. l—x J x ") parabolic ele- W at a from' end. T0-») l—x |- vation. l-xi Fig. 28. Sup't'd both ends, hwx(l— x) x(l—x) x(l—x) parabolic plan. x(l—x), circular uniform load. Fig. 29. or elliptical eleva- tion. Fig. 30. point x, where the load is at any instant, = Wx(l — x) -±1. Such a beam will therefore fall under the last class of the above table. Fig. 23 Fig. 27 Fig. 24 Fig. 28 Fig. 25 Fig. 29 Fig. 26 Fig. 30 BEAMS. 65 Beams which can be cast in form or built up may be made in the above outlines, if desired. Some common examples, such as brackets, girders of varying depth, walking- beams, cranks, grate-bars, etc., are more or less close approximations to such forms. Enough material must also be found at any section to resist the shear, as at the ends of beams supported at the ends. Where a plate girder is used (see Fig. 95) with a constant depth, the cross-section of the flanges, or their thickness when their breadth is constant, will theoretically and approximately follow the fourth column of the preceding table. If the flange section is to be constant or nearly so, the depth must vary in the same way, and not as in the fifth column. Roof- and bridge-trusses are beams of approximate uniform strength, for the different allowable unit stresses and for chang- ing loads. The principles of this section have an influence on the choice of outline for such trusses, and the shapes of moment diagrams suggest truss forms. 72. Distribution of Shearing Stress in the Section of a Beam, Pin, etc. — It will be proved, in § 154, that, at any point in a body under stress, the unit shear on a pair of planes at right angles must be equal. Whatever can be proved true in regard to the unit shear on a longitudinal plane at any point in a beam must therefore be true of the unit shear on a transverse plane at the same point. Fig. 31 represents a portion of a beam bent under any load. The existence of shear on planes parallel to E F is shown by the tendency of the layers to slide by one another upon flexure. Let the cross- section of the beam be constant. If the bending moment at section H, a point close to G, differs from that at G, there will be a shear on the transverse section, because the shear is the first derivative of the bending moment, § 56. The direct stress, here compression, on the face H F of the solid H F E G, will differ from that on the face G E, since the bending moments are different, and that difference will be balanced by a longitudinal horizontal force, or shear, 66 STRUCTURAL MECHANICS. on the plane F E, to oppose the tendency to displacement. If this force along the plane E F is divided by the area E F over which it is distributed, the longitudinal unit shear will be obtained. It follows from the first paragraph that the unit shear at the point E on the transverse section G A must be the same. It is also evident that the farther E F is taken from H G, the greater will be the difference between the total force on H F and that on G E, until the neutral axis is reached, and that the unit shear on the longitudinal plane E F must increase as E F approaches B, the neutral axis. The same thing is true if the plane is sup- posed to lie at different distances from the edge A. Hence, at any transverse section A G, the unit shear on a longitudinal plane is most intense at the neutral axis; and therefore the unit shear on a transverse section A G is unequally distributed, being greatest at B, the neutral axis, and diminishing to zero at A and G. Pins and keys, and rivets which do not fit tightly in their holes, and hence are exposed to bending, have a maximum unit shear at the centre of any cross-section, and this shear muct therefore be greater than the mean value, and must determine the necessary section. To find the mathematical expression for the variation of shear on the plane A G: O B D C is the trace of the neutral plane. B D = E F sensibly — dx. B E = v, B G=y\. Breadth of beam at any point = 2, at neutral axis = Zo- Normal or direct unit stress at the point E on plane AG = p. Unit shear at E = ^; maximum, at B, = q . M and F= bending moment and shearing force at section A G. By §64, ' = ~7~ P = ~jy* The total direct stress on plane G E is pyx M C* J pzdy = Yj yzdy (1) The difference between M at the section through B and M at the section through D must be Fdx, since M = JFdx, by § 56. BEAMS. 67 The horizontal force on E F is the excess of (1) for G E over Fdx f* its value for H F, or —=— I yzdy. Divide by the area z y dx of F E, over which this horizontal force acts, to find the unit shear. Since the mean unit shear = F+S, the ratio of the maximum unit shear to the mean will be found by dividing 8oo ft. -lb. 12. A floor-beam for a bridge spans the roadway a and projects under each sidewalk b. If dead load per foot is w, live load for road- way w', for sidewalk w", write expressions for -\-M max. and — M max. 13. A vessel is 200 ft. long. It carries 5 tons per ft. uniformly distributed, and a central load of 300 tons. Find M max. when at rest; when supported on a wave crest at bow and stern with each bearing 20 ft. long; and when supported amidships only with bearing 30 ft. long. 14. The end of a beam 6 in. wide is built into a wall 18 in. The bending moment at the wall is 600,000 in.-lb. If the top of the beam bears for 9 in. with a uniformly varying pressure and the bottom the same, what is the maximum unit compression on the bearing surface ? 1,852 lb. CHAPTER IV. MOMENTS OF INERTIA OF PLANE AREAS 74. Definitions. — The rectangular moment of inertia, I, of a plane area about an axis lying in that plane is the sum of the products of each elementary area into the square of its distance from the axis. The plane areas whose moments of inertia are sought are commonly cross-sections of beams, and unless other- wise stated the axis about which moments are taken passes through the centre of gravity of the cross-section. The quotient of the moment of inertia by the area is the square of the radius 0} gyration, r. If the area be referred to rectangular axes Z and Y, and if subscripts denote the axes about which moments are taken, I z = I J y 2 dzdy = / zy 2 dy = Sr g 2 ; I y = J J z 2 dzdy = J z 2 ydz = Sr y 2 . If each elementary area be multiplied by the square of its distance from an axis perpendicular to the plane and passing through the centre of gravity, the summation gives the polar moment of inertia, J. As the distance of each elementary area from the axis is Vs 2 +^ 2 , J = ff(z 2 +y 2 )dzdy = I y +I 2 . When the term ' 'moment of inertia" is used without qualification, the rectangular moment is meant. Moments of inertia, being the product of an area into the square of a distance, are of the fourth power and positive. 7i 72 STRUCTURAL MECHANICS. The values of I, J, and r 2 for some common forms of cross- section follow. I. Rectangle, height h, base b. Fig. 33. Axis through the centre of gravity and parallel to b. r r +h \ * j, r +¥i 2, [ T ^v yi bhs bhs m I z = I y 2 zdy = 01 y z dy = —by 3 = — -\ = — . J-ih ' J -hh l_3 J-ih 24 24 12 r b3h o ^ 3 ,, k? I v = — . r 2 = --bh= — . 12 12 12 For an axis through the centre of gravity and perpendicular to the plane, J = I y +I 2 =—(b 2 +h 2 ), and r 2 = — (b 2 +h 2 ). 12 12 II. Triangle, height h, base b. Fig. 34. Axis as above and parallel to b. h:b = ^h-y:z; z =-(th-yY b I 16 16 2 1 \ bit 3 TV243 324 243 324/ "36' 2 _bh 3 Jbhjf_ r = 36 2 18* III. Isosceles triangle, about axis of symmetry. Fig. 35. Height along axis h, base b. h\\b = z\\b-y\ z = hli- z pj. rib / 2y x r y 3 y 4-lib /b 3 h 3\ J lh Z 24 MOMENTS OF INERTIA OF PLANE AREAS. 73 The sum of II and III will be the polar moment, J, about an axis through the centre of gravity and perpendicular to the plane. bh/h 2 b 2 \ i (h 2 b 2 \ i2\3 4/ 0\3 4/ kit y t. Fig. 33 Fig. 34 Fig. 35 Fig- 36 FU. 37 IV. Circle, radius R, diameter d. Fig. 36. If 6^ = angle between the axis of Z and a radius drawn to the extremity of any element parallel to that axis, y = RsinO; %z = R cos 0; dy=R cos OdO. 4 "64* 7 4 = 4^ 4 /""sin 2 cos 2 0d0 = - 2 ic 4 -l|"- sin +0-0~\ -A- 4 10 The polar moment of inertia, 7, may be easily written if r' = variable radius, 7= / r' 2 .2-/ ( // = --. r 2 = -R 2 . Jo 22 Since I z +I y = J, and I y = I z by symmetry, 7 2 = J-i? 4 as before. V. Ellipse. Diameters d and 6. Fig. 37. As the value of z in the ellipse is to that of z in the circle, as the respective horizontal diameters, or as b to d, and as the moment of the strip zdy varies as the breadth, alone, the ellipse having horizontal diameter b, height d, gives h = ndb* 64 ' z ~ 64 'd~ 64 ' J = ^-(d 2 +b 2 )-, r 2 = ~(d 2 +b 2 ). 64 2_7 _^_A 2 r " * ' 4 ~i6* 16 74 STRUCTURAL MECHANICS. VI. The moment of inertia of a hollow section, when the areas bounded respectively by the exterior and interior perimeters have a common axis through their centres of gravity, can be found by subtracting / for the latter from / for the former. Thus : Hollow rectangle, interior dimensions b' and h', exterior b and u; I z ( = ^bh s -b f h f3 ). Hollow circle, interior radius R', exterior radius R; I 2 = \n(R±-R'*). r 2 = {(R 2 +R' 2 )=Md 2 +d' 2 ). The moment of inertia of a hollow ring of outside diameter d and inside diameter d f , the ratio of d' to d being n, may be written I z = &*(#- or one-third the area multiplied by the square of the ordinate to the extreme end. This expression might be derived from I for a rectangle, taken about one base. If the rod is parallel to the axis, and at a distance y\ from it, I = tL-yi 2 , since all particles are equidistant from the axis. Example. — Find approximate value of I of angle of last example. Using centre line, angle is 5l"X3f"Xj". Distance of centre of gravity from heel is 5.75X2.875-7-9.5 = 1.74 in. 7=JXi(4-oi 3 +i.74 3 ) + iX 3 .75Xi.74 2 =i7-3oin. 4 78. Rotation of Axes. — If the moments of inertia about two axes at right angles are known, the moments of inertia about axes making an angle a with / the first may be found. In the solution the quantity / zyds or J I zydzdy occurs, which is called the product 0} inertia and represented by Z. Prod- ucts of inertia may be either positive or negative according to the position of the axes. Fiom Fig. 39, z' = z cos a+y sin a, v' = v cos a —z sin a: Fig. 39 MOMENTS OF INERTIA OF PLANE AREAS. 77 I g ' = f(y cos a—z sin a) 2 dS = jy 2 cos 2 adS — 2 / zy sin a cos adS + / z 2 sin 2 adS, Iy = J (z cos a +v sin a) 2 J5 = J z 2 cos 2 adS -t-2jzy sin a cos aJS +jy 2 sin 2 atf5 Z' = J (z cos a: +v sin a)(y cos a: —2 sin a)dS = / (y 2 —z 2 ) sin a: cos adS +jzy cos 2 adS — J zy sin 2 a^S; 1 1=1 z cos 2 a+iy sin 2 a— 2Z sin a cos a, Iy =I y cos 2 a +/ 2 sin 2 a +2Z sin a: cos a, Z f = (7 2 —Iy) sin a cos a: +Z (cos 2 a: —sin 2 a). 79. Principal Axes. — To find the maximum and minimum values of I J and Iy differentiate with respect to a : dl ' ——= — 2I sin a cos a+2l v sin a cos a — 2Z cos 2 a +2Z sin 2 a, da z y M,' r*, dl' da da Hence maximum and minimum values of IJ and // occur when Z' = o, and since J = 1 g +I y = I B f +I y ' = & constant, I J is a mini- mum when // is a maximum. The axes for which I J and Iy are maximum and minimum are called principal axes. An axis 0} symmetry is always a principal axis since for such an axis Z must be zero, as for every positive ordinate there is a cor- responding negative ordinate. If a figure has two axes of sym- metry not at right angles, the moment of inertia is the same for all axes through the centre of gravity. The angle, , which the principal axis, A A, makes with the original Z axis may be found by making Z' = o: (I g —Iy) sin $ cos +Z (cos 2 -sin 2 4>) =0, i(I z —Iy) sin 2$+Zcos 2*56 = 0, s 2Z tan 2(p= T T . 78 STRUCTURAL MECHANICS. I A , the moment of inertia about the A axis, is found by substi- tuting the value of in the expression for 7/ : I A =I Z cos 2 (j>+Iy sin 2 — Zsin 20; cos 2 = J(i +cos 2(f)), sin 2 = J(i —cos 2) ; 7^ = J7 s (i +cos 20) +il y (i —cos 20) — Z sin 2$ = %(I y —I z )(i —cos 20) —Zsin 20+7 2 ^ I -COS 20 =Z— ; Z sin 20 +7 . tan 20 z ^COS20 — I _ , I— COS 20 = Z — . y , +7 2 ; tan0 sin 20 7^=7 —Z tan 0. sin 20 Similarly the moment of inertia about axis B B at right angles to axis A A is found to be Ib^Iv+Z tan 0. 80. Oblique Loading. — When the plane of loading on a beam does not coincide with one of the principal axes of the section the beam does not deflect in the plane of the loads and the neutral axis is oblique to that plane. In a cross-section of such a beam the stress at a point whose coordinates are a and b referred to the A and B axes may be found by resolving the Fig. 40 bending moment at the section into its components in the direction of each principal axis, finding the stress at the point due to each MOMENTS OF INERTIA OF PLANE AREAS. 79 component and adding algebraically. Thus the stress at the point D, Fig. 40, due to a bending moment, M, is (M sin 0)b (M cos 6) a l A l B It may sometimes be necessary to find the direction of the neutral axis to determine which is the extreme or most stressed fibre. Since there is no stress at the neutral axis, if the last equation be made equal to zero, a and b will become coordinates of some point on the neutral axis, and the ratio between them will be the tangent of the angle, /?, which the neutral axis makes with the A axis, b I A tan B = — = — 7- ctn Q. Example. — A 6X4X§-in. angle is used as a purlin on a roof of J pitch (tan _1 = §), as shown in Fig. 40. What is the maximum fibre stress due to a bending moment if? Centre of gravity lies 0.99 in. from back of longer leg and 1.99 in. from shorter. 72=17.40 in. 4 , I y =6.2'j in. 4 . Z=(£X4)(2-o.99)(i.99-o.25) + (§X5.5)(-o.99 + o.25)(-3. 25 + 1.99) = +6.08 in. 4 . tan 2= 2X6.08 -f- (6. 27— 17.40)= — 1.092, = — 23 46', tan 0= — 0.440. Ia = 1740— 6.o8(— 0.440) = 20.07, 7s = 6.27 + 6.o8(— 0.440) = 3.60. 0=tan -1 1.5 — =8>o° 04', sin^o.985, cos#=o.i73, ctn#=o.i75. tan/?= —20.07X0.175-^-3.60= —0.978, /?= — 44 02'. Laying off angles and scaling gives a= 1.98 in., 6=3.05 in. /=if(o.985X3.o5^ 20.07) + i7 (0.173 X 1.98-^3.60) = 0.243^. If the load is applied along the Y axis, # = 9 o° -<£=ii3° 4 6', sin #=0.915, cos #=—0.403, ctn #=—0.440. tan/? = 20.07X0.440 -7-3. 60 =2.460, /?=67° 53'. 80 STRUCTURAL MECHANICS. The greatest stress is found on the inside edge of the lower leg, where a— + 1.18 in., b= —3.88 in. /= -if (0.915X3.88 -f-2o.o7)-if (0.403X1. 184-3. 60)= -0.309M. Examples. — 1. Find the moment of inertia of a trapezoid, bases a and b, height h, about one base. 2. A 12-in. joist has two mortises cut through it, each 2 in. square, and 2 in. from edge of joist to edge of mortise. How much is that section of the joist weakened ? 2 ^ or 26%. 3. A bridge floor is made of plates rolled to half -hexagon troughs, 6 in. face, 5.2 in. deep, 12 in. opening, J in. thick. Find the resisting moment of a section 18 in. wide. 20.8/. 4. If that floor is 14 ft. between trusses and carries two rails 5 ft. apart, each loaded with 2,000 lb. per running foot, what will be the unit stress? 7>79° lb. 5. Six thin rolled shapes, web a, make a hexagonal column, radius a, with riveted outside flanges, each b in width. Prove that 12(^+26) CHAPTER V. TORSION. 8 1. Torsional Moment. — If a uniform cylindrical bar is twisted by applying equal and opposite couples or moments at two points of the axis, the planes of the couples being perpen- dicular to that axis, the particles on one side of a cross-section tend to rotate about the axis and past the particles on the other side of the section, thus developing a shearing stress that varies with the tendency to displacement of the particles, that is, directly as the distance of each particle from the centre. The unit shear then is constant on any ring, and the shearing stresses thus set up at any section make up the resisting moment to the torsional moment of the applied couple. As all cross-sections are equal and the torsional moment is constant between the two points first referred to, each longitudinal fibre will take the form of a helix. 82. Torsional Moment of a Cylinder. — If the unit shear at the circumference of the outer circle, Fig. 41, of radius R\ and diameter d is qi, the value at a distance R from the centre will be, by the above statement, q = q i R+R 1 . The total shearing force on the face of an infinitesimal particle whose lever-arm is R, and area RdRdO, will be ~R 2 dRdd, and its moment about the centre will be —R 3 dRdd. Hence the resisting moment for a cylinder is T = %T / l / ^ R 3 dRdO = %- = --q l R l z = o.i 9 6q l d* J\\k/ o xJ o tCi 2 8l 82 STRUCTURAL MECHANICS. Hence the resisting moment against torsion resembles in form the resisting moment against flexure, but differs in using the polar moment of inertia of the cross-section for the rectangular one, and in having qu maximum unit shear, in place of /, maxi- mum unit tension or compression. As the rings of metal situated farthest from the centre of a shaft offer the greatest resistance to torsion, it is economical of metal to make the shaft hollow. If d is the internal and D the external diameter of a shaft, the resisting moment is found by subtracting the moment of a shaft of diameter d from that of a shaft of diameter D, hence T =0.196^1 D±-d± D 83. Torsional Moment of a Square Shaft. — If a square bar s twisted and the shear is assumed to vary on the cross-section Fig, 41 with the distance of the particles from the centre, / - J/i 4 , r\ —h\ and r = i^r 6 -0.236^. This assumption is not correct. The unit shear is actually the greatest at the middle of each side. For rectangular sections the preceding treatment would be seriously in error, but for a square section the error is not important. The last coefficient should be about 0.208. For shafts the cylindrical form is now almost universal. See § 85. TORSION. 83 The exact investigation for sections other than circles is very involved. See "Theory of the Elasticity of Solid Bodies, " Clebsch, translated from the German into French, with Notes by de St. Venant and Flamant; and report of Chief of Engineers, U. S. A., for 1895, p. 3041, Part IV. Example. — Design a round shaft to transmit 500 H.P. at 100 revolutions per minute if !=OM + MR, ^ 2 = OM-MR. MR 2 = MN 2 + NR 2 . Pi=itWif 2 + qi 2 , (1) p2 = hf-^if 2 + qi 2 . By § 152, qmax = i(Pi-p2) =vj/ 2 +^ 2 . Since M=fl+y\ and T = q±J ' +yi and since 2! = J for a square or a circle, (1) may be multiplied by / +yi and transformed to M 1 ^i(M + VM 2 + T 2 ) > (2) in which M = original bending moment at the section, T= original torsional moment, and Mi = equivalent resulting bending moment for which the shaft should be designed so that the unit stress shall not exceed / when both M and T occur at the given section. If (1) is multiplied by J -s-yu there results T 1 =M+VM 2 + T 2 (3) as an equivalent torsional moment in which T\ ^piJ+yi. Although neither equation (2) nor (3) is rational, since pi does not act on a right section, the conception of an equivalent bending moment is less objectionable than the conception of an equivalent torsional moment because pi is a normal stress. But in designing shafting it is usual to employ the same working value for unit fibre stress, /, as for unit shear, qi; consequently the results obtained are the same whether equation (1), (2), or (3) is used. If the deformation of the material is taken into account as in 86 STRUCTURAL MECHANICS. §175, the new principal stresses become pi =pi—\p2 and i>2 /== i>2-ii>i. pi'-Jf + l^ W+tf, t^im+^VW+T 2 . Examples. — 1. If the pulley of the previous example weighs 500 lb. and is 12 in. from the hanger, on a free end of the shaft, and the un- balanced belt-pull of 2,000 lb. is horizontal, the resultant force on the pulley is 5oovi 2 + 4 2 = 2,060 lb., which causes a bending moment of 24,720 in.-lb. at the hanger. Then M\ = J[24,72o+\/(24 J 720 2 + 30,ooo 2 )]=3i,8oo in.-lb., which will cause a fibre stress of 31,800x2 s p\=- 77—^=10,400 lb. per sq. in. 0.196X5^ 2. The wooden roller of a windlass is 4 ft. between bearings. What should be its diameter to safely lift 4,000 lb. with a 2-in. rope and a crank at each end, both cranks being used and / being 800 lb. ? 8Jin. 3. Design a shaft to transmit 500 horse-power at 80 revolutions per min., if #1=9,000 lb. d=6 in. 4. How large a shaft will be required to resist a torsional moment of 1,600 ft. -lb. if #1=7,500 lb.? If the shaft is 75 ft. long and C= 11,200,000, what will be the angle of torsion? 2§ in.; 30J . 5. What torsional moment will a hollow shaft of 5 in. internal and 10 in. external diameter transmit if #1 = 8,000 lb. per sq. in.? What is the size of a solid shaft to transmit the same torsional moment ? What is the difference in weight between the two shafts if a bar of steel of one inch section and one foot long weighs 3.4 lb. ? CHAPTER VI. FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 87. Introduction. — As the stresses of tension and compression which make up the resisting moment at any section of a beam cause elongation and shortening of the respective longitudinal elements or layers on either side of the neutral plane, a curvature of the beam will result. This curvature will be found to depend upon the material used for the beam, upon the magnitude and distribution of the load, the span of the beam and manner of support, and upon the dimensions and form of cross-section. It is at times desirable to ascertain the amount of deflection, or perpendicular displacement from its original position, of any point, or of the most displaced point, of any given beam carrying a given load. Further, the investigation of the forces and moments which act on beams supported in any other than the ways already discussed requires the use of equations that take account of the bending of the beams under these moments. There are too many unknown quantities to admit of a solution by the principles of statics alone. The required equations involve expres- sions for the inclination or slope of the tangent to the curved neutral axis of the bent beam at any point, and its deflection, or perpendicular displacement, at any point from its original straight line, or from a given axis. The curve assumed by the neutral axis of the bent beam is called the elastic curve. 87 A ip Fig. 44 88 STRUCTURAL MECHANICS. 88. Formula for Curvature. — If, through the points A and B, on the elastic curve, Fig. 44, and distant ds apart, normals C D and K G to the curve of this neutral axis are drawn, the distance from A B to their intersection will be the radius of curvature p for that portion of the curve. If through A a plane F H is passed parallel to K G, the distance F C will be the elon- gation, or H D will be the shortening, from the unit stress /, of the extreme fibre which was ds long before flexure. Cross- sections plane before flexure are plane after flexure, § 62. AO=p', AC=Vi; CF = ^, § 10. From similar triangles / E\'i ACF and O A B, p:ds = yi'-ds, or p = ~j-. As, by §64, M = — , — = -f^f, the reciprocal of the radius of curvature, called y\ p & the curvature or the amount of bending at any one point. 89. Slope and Deflection. — If the elastic curve is referred to rectangular coordinates, x being measured parallel to the original straight axis of the beam, and v perpendicular to the same, the calculus gives for the radius of curvature [ \dx/ _ dx 2 For very slight curvature, such as exists in practical, safe beams, — is a very small quality, and in comparison with unity its square may be neglected. Then p dx 2 EI' As M is a function of x, as has been seen already, the first dv definite integral, — , will give the tangent of the inclination or the slope of the tangent to the curve of the neutral axis at any point FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 89 x, and the second integral will give v, the deflection, or perpen- dicular ordinate to the curve from the axis of x. Thus dv ^=slope, »; _ di T,-rd 2 v Ei-r- = moment, M = EI^— 2 \ ^ = shear, F=EI^; dF d±v 3— = load, w = EIj—r. dx dx? While the following applications of the operations indicated in the last paragraph are examples only, the results in most of them will be serviceable for reference. The student must be careful, in solving problems of this class, to use a general value for M, and not M maximum. The origin of coordinates will be taken at a point of support, such a point being definitely located; x is measured horizontally, v vertically, and — v denotes deflection downwards. The greatest deflection for a given load is Vx- The greatest allowable deflection for a fibre stress / is V\. If M +1 is constant, the beam bends to the arc of a circle. This happens where M is constant and / is constant, or where I varies as M. Example. — The middle segment of a timber beam carrying a load, 4EI W, at each quarter-point bends to the arc of a circle. M= \Wl; p=—z-y. If the beam is 6 in.Xi2 in.X2o ft., W=2,ooo lb. and £=1,500,000 1 r * \ , q o,- • a 1 4- 1,500,000-864 lb. per sq. in., then I= T ^-6'i2 6 = 864 m. 4 and p= ^ H l2 r 2,000-20-12 = 10,800 in. = 900 ft. 90. Beam Fixed at One End; Single Load at the other; origin at the wall; length =/. Let the abscissa of any point on the elastic curve be x, and the ordinate v. At that point M x =-W(l-x). 9° Then STRUCTURAL MECHANICS. (Pv_ W dx*~~Er~ x) ' The slope at the point x is found by integrating. dv W i= dx = -EI {lx - ix)+C ' At the point where x = o, the slope is zero and therefore C = o. Integrating again, v=-^(ilx 2 -ix*)+C.' When x = o, v = o, therefore C' = o. The equation just found is the equation of the elastic curve. If x is made equal to / in the — X above equations, the slope and the deflection at the end of the beam are found to be WP ^max. 2EV WP v. 3EI' To determine the maximum allowable deflection of a given beam consistent with a safe unit stress in the extreme fibre at the section of maximum bending moment, substitute, in the expression for the value of W in terms of /. M = Vl -wiJl WP fP $EI~2>Eyi \ W=- FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 9 1 Example. — If /=6o in., 6 = 4 in., h=S in., W = Soolb., £=1,400,000; _ 4-8 3 512 800 -6o 2 - 3 / = = ; max. slope = ° — = 0.006; v max = 12 3 r 2-1,400,000-512 max - 800 -6o 3 - 3 . . = 0.24 in.; and if /= 1,200 lb., maximum safe de- 3-1,400,000-512 1,200X60? flection= 3X1,400,000X4 0.26 in. It will be seen that, for a given weight, the maximum bending moment varies as the length /; the maximum slope varies as / 2 , and the maximum deflection as I 3 . The slope and deflection also vary inversely as /, or inversely as the breadth and the cube of the depth of the beam. The maximum safe deflection, how- ever, consistent with the working unit stress /, varies as I 2 , and inversely as y± } or the depth of the beam. These relationships are true for other cases, as will be seen in what follows. The ease with which problems regarding deflection are solved depends greatly upon the point taken for the origin, as it influences the value of the constants of integration. 91. Beam Fixed at One End; Uniform Load of w per unit over the whole length, /; origin at the wall. M x = -\w(l-x) 2 . d 2 v w dx 2= ~7EI ( - l2 ~ 2lx + x2 ' ); dv w ax 2EP 6 When x=o, i=o; .'. C = o. w 92 STRUCTURAL MECHANICS. When x = o, v = o; .'. C = 0. When x = l, (wl)P "max. RJ?T ' For maximum safe deflection consistent with unit stress, /, in the extreme fibre at the dangerous section, M max , = - (wl) \l = — ; (wl) = 2 — ; yi ytf _ (wl)P ft 2 Vl ~~ SEI ~4Eyi 92. Combination of Uniform Load and Single Load at one end of a beam fixed at the other end. Add the corresponding values of the two cases preceding. M max = -[Wl + h(vl)l]; Note, in the expression for v max ., the relative deflections due to a load at the end and to the same load distributed along the beam; and compare with the respective maximum bending moments. Example. — If the preceding beam weighs 50 lb., the additional .,, , co-6o 3 - 3 . ,, . deflection will be = 0.005 m -> to ° small a quantity to 8- 1,400,000- 512 be of importance. In the majority of cases the weight of the beam itself may be neglected, unless the span is long. 93. Beam Supported at Both Ends; Uniform Load of w per unit over the whole length /; origin at left point of support. M x = %w(l—x)x. d 2 v w dx 2 2EP dv w FLEXURE AND DEFLECTION OF SIMPLE BEAMS 93 From conditions of symmetry it is evident that the slope is zero when x = \l\ hence, to determine the constant of integration, make o=lP-&P+C\ C=- T V 3 ; w As v=o when x = o or /, C'=o. When x = U, ^'max. 5 (wlW 384 EI Fig. 47 For maximum safe deflection consistent with a unit stress, /, in the extreme fibre at the middle section, i(w/)/=-; wl=^; "1=75 5 P 48 Eyi Examples- -A pine floor-joist, uniformly loaded, section 2X12 in., span 14 ft. = 168 in., has deflected f in. at the middle. Is it safe? £=1,500,000. By the last formula, 3 _ 5 I4 2 -I2 2 4 48 1,500,000-6' /= 2,300 lb., and the beam is overloaded. What weight is it carrying ? By formula for v max ., 3 5 7 x 4 3 ' i2 3 - 12 4 384 1,500,000-2 • 12 3; w/= 5,248 lb. 94 STRUCTURAL MECHANICS. 94. Beam Supported at Both Ends ; Single Load W at middle of span I; origin at left point of support. d 2 v W dx 2 ~ 2 EI X ' This expression will apply only from x = o to x = \l\ but, as the two halves of the deflection curve are symmetrical, the discussion of the left half will suffice. l ~dx~2EI^ X +C) ' When x = \h *-a .\ C= -\l 2 . W When #=0, z>=o. .". C'=o. The limit x=l is not applicable. Vmax -~~ 4 SEf For maximum safe deflection consistent with a unit fibre stress,/, \Wl=~\ W = ^; ^i = -C-. 1 )'i y\l i2Ey 1 Example. — A 10-in. steel I beam of 33 lb. per ft. and 7=162; span, 15 ft. ,= 180 in., carries in addition a uniform load of 767 lb. per ft. of span and 6,000 lb. concentrated at the middle. What will be its deflection and the maximum unit fibre stress? /q-8oo-i!j 6,ooo\ 180 3 Vmax={- a + ~V I ^ = °'35 m -*> \ 384 48 / 29,000,000- 162 /•162 /8oo-i5 . 6,ooo\ '-— -^— ^ ? + — J180; /= 16,667 lb. FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 95 95. Single Weight on Beam of Span /, supported at both ends; IF at a given distance a from the origin, which is at the left point of support. Fig- 48 As the load is eccentric, the curve of the beam is unsym- metrical, and equations must be written for each portion, xa. ON LEFT OF WEIGHT. _ .J- a ,. W(l-a) P i== W—; M x =-±- ] — -x. d 2 v W dx-=ETl«- a)X ' v=-~(&x 3 -±ax 3 +Cx+C ff ). (1) v = o, when #=0; C"=o. ON RIGHT OF WEIGHT. Wa Wa P 2 =— ; M x =—{l-x). d 2 v W . , dx- 2 = ETf al - ax) - %=w£ alx -^+ Cf)=i ' (2> v^^-aalxt-baxt+C'x+C"'). Ell v = o, when x = l; .-, c"" = -ia/ 3 -C7. For equations to determine the constants C and C ', use the value x = a, when it will be evident that i for the left segment must give the same value as does i for the right segment, and v at a must be the same when obtained from the left column as when obtained from the right. Therefore, from (1) and (2), \a 2 l - Ja 3 + C = a 2 l - §a 3 + C C = C' + \a 2 l v at a gives \aH - \a± + C a + |a 3 / = \aH -\ a * + C'a- %aP - CI. C=-\a?-\al 2 . C" = \aH. C = \a 2 l - Ja 3 - \al 2 . aH a 3 of 3 W (I— a , . a 2 lx a 3 x al 2 x\ v =eTi\— x+ ——£—T)- i) (l— x)[a 2 — (2/— x)x]. 96 STRUCTURAL MECHANICS. As a is assumed to be less than J/, and the substitution of dv x = a in the value of -=- on the left gives a slope which is negative, the point of v maXi will be found on the right of W, and for that dv value of x which makes -7- on the right zero. Hence \ax 2 — alx + Ja 3 + Ja/ 2 = o. which is the distance from the left-hand support to the point of maximum deflection. Substitute in the expression for v on the right, to obtain the maximum deflection. It should be noticed that, when the weight is eccentric, the point of maximum deflection is found between the weight and the mid-span, and not at the point of maximum bending moment, which latter is under the weight. 96. Two Equal Weights on Beam of Span /, supported at the ends ; each W, symmetrically placed, distant a from one end. This case may be solved by itself, but can be more readily treated by reference to § 95. Thus the maximum deflection will be at the middle, and can be found by making x = \l in the above value of v for the right segment and doubling the result. Then The deflection under a weight will be given by the addition of v at a and v at (/ — a) of the preceding case. Thus Wa 2 Wa 2 Va= -&nP- a * m > Vl ~ a= -6Eii (p - 2a2) ' Wa 2 V2LtW=-^j(sl-4a). FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 97 Example. — A round iron bar, 12 ft. long and 2 in. diameter, carries two weights of 200 lb. each at points 3 ft. distant from either of the two supported ends. The deflection at a weight= 200 • 3 6 2 • 7 • 4 6-28,000,000- 22 (3-i2 2 -4-3-i2) = o.56 in. ''OO '^'T2'1'A. The maximum unit bending; stress is = 9,160 lb„ 22 DEFLECTION OF BEAMS OF UNIFORM STRENGTH. It will be apparent that a beam of uniform strength will not be so stiff as a corresponding beam of uniform section sufficient to carry safely the maximum bending moment; for the stiffness arising from the additional material in the second case is lost. 97. Uniform Strength and Uniform Depth. (See § 71.) — Since M = njbh 2 and varies as bh 2 , and I = n f bh s and varies as bh 3 , M +1 varies as 1 +h. But if h is constant, M +1 is constant d 2 v and -n is constant. Therefore all beams of this class bend to the arc of a circle. I. Beam fixed at one end only, and loaded with W at the other. Fig. 22. If Iq is the moment of inertia at the largest section, which is in this case at the wall, and / is the variable moment of inertia, I b l—x l—x I Io b Q I ' / Io d?v__M_ W Wl_ dx 2 ~EI~~ ED -*/'-- EIq - dv Wl „ ■j-=— -^rrX + C', C = 0. dx EIq Wl v=—=rrx 2 +C; C'=o. WP 'max. 2EU a deflection 50 per cent, in excess of that of the corresponding uniform beam, while the maximum slope is twice as great. 9 8 STRUCTURAL MECHANICS. Examples. — If a triangular sheet of metal, like the dotted triangle in Fig. 49, is cut into strips, as represented by the dotted lines, and these strips are superimposed as shown above, the strips, if fixed at the ends, and subjected to W as shown, will tend to bend in arcs of circles, and will remain approximately in contact. If l=io in., b=4 in., h = J in., JF=4oo lb. and £ = 28,000,000, the deflection will be 400- io 3 '4 3 - 12 = 1.37 in. 2-28,000,000-4 An elliptical steel spring 2 ft. long, of 4 layers as shown, each 2 in. broad and J in. thick, under a load of 100 lb. at its middle, will, 1 00 • 1 2 • 8 3 • 1 2 if JE= 20,000,000, deflect 0=2.3 in. The maximum unit 2-29,000,000-8 fibre stress will be =28,800 lb. Note that one-half of the o weight is found at each hinge, and that the deflection of one arm is doubled by the use of two springs as shown. II. Beam fixed at one end only and uniformly loaded with w per unit. Fig. 25. L h ^ l ~ x)2 (7-*) 2 I 2 h~b Q ~~ P ' / "Jo d 2 v w .. _ wl 2 dx 2 ~ 2EI {1 X)2 ~ dx 2 dv dx wl 2 v = 1) = 2EI wl 2 4EI0 wl* 4£V x + C; x 2 + C; 2EI0' C=o. C'=o. a deflection twice that of the corresponding uniform beam. FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 99 In these two cases there are no constants of integration, since t~ and v=o when x=o. dx III. Beam supported at both ends and carrying W at middle. Fig. 27. I :I = b:b =x:^l. d 2 v Wx Wl m dx 2 2EI 4EI0 ' dv Wl dx 4E/0 Slope=o when x = \l) .*. C= —J/. Wl v=-gj-(ix 2 -±lx) + C; C'=o. WP V *ttnv 32EI0' a deflection 50 per cent, greater than for a corresponding uniform beam. IV. Beam supported at both ends, and uniformly loaded with w per unit of length. Fig. 29. / : I = b :b = x(l — x) : \l 2 . d 2 v w wl 2 ~JZ2 = 'Zkt\' /X ~ x ) = ' dx 2 2ED J 8EI dv wl? dx SEIq (x+C); C = -\l. v = ^{hx 2 - \lx + C) ; C* = o. o£L1q wl* V max. = - 64jE / ' a deflection 20 per cent, greater than for a corresponding uniform beam. 98. Uniform Strength and Uniform Breadth. — In these h? cases, as b is constant, I:Io = h 3 :h 3 or I = Iq t-§. IOO STRUCTURAL MECHANICS. V. Beam fixed at one end only, and loaded with W at the other. Fig. 24. Jl 2 l-X T _r hS _r ( l ~ X )* ho 2 ~~ /-/ V ° l l ' dx 2 ~ Er x) ~ W x) ; dv Wfl N , _ N ^ (-2(/-x)* + C); C = 2#. 2?/ "max, -p j 7i /»/ (Wo 2 IF/ 1 2 IF/ 3 - EIo 1 ^ VJ '"^J- 3 £/ ' or twice the deflection of a corresponding uniform beam. VI. Beam fixed at one end only and uniformly loaded with w per unit of length. Fig. 26. h l—x (l—x) 3 [log (I -x) -log I]. dx 2EI0 wl 3 V max. 2EJ I [log (I -x) -log l]dx* Jo wl 3 I y = I x log (7— a) — x— /log (/—#)— a? log/ 2EI0L- J 7£'/ 3 7t'/ 4 = ^gj"(^ log O -/ -/ log O -/ log / + / log /) = - -gp or four times the deflection of a corresponding uniform beam. dx * Log (/— #)=m; dx=dv; x=v; du=— . /P X X I \og(l—x)dx = x\og{l—x)+ I - 1 — tf#. By division,- = — i+j ; J l — X I x I — X /~—dx= — I dx+l I 7— = — x— I log (I— x). l—x J J l—x FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 101 VII. Beam supported at both ends and carrying W at middle. Fig. 28. h 2 x J8x 3 h 2 2< d 2 v Wx I 3 Wfl dx 2 2 EI 4.x/ 2 EI dv Wl x -* dx 4V2EI0 (2^+C). dv When#=|/, j- = o> an d C = -2V 7 !/. W7* 2\/2EU 2V 2 EI - f\x±-V±l)d § X% — Vi/x J^ 3 /2 /* /* 2\Z2EIq\$ 2V 2 2>/2 IF/ 3 24EI0 or twice the deflection of a corresponding uniform beam. VIII. Beam supported at both ends, and uniformly loaded with w per unit of length. Fig. 30. &2 X (l-X) 8Vx*(l-x)3 h 2 ~ {I 2 * ° P ' d 2 v w wl 3 dx~ 2 = 2~EI (lx ~ X)= 76ET dv wl 3 ( 2X Tx=76ET V eKm T +c,; (lx— x 2 )~*\ dv — = o, when # = hi ; .*. C = —versin" 1 1= — J^. V "'*-=lSrJ (™-sin-iy-W) and W= ^r- 1 2 fbh 2 ft 2 1 f 2 * 23/ 6Eh 18 E The allowable shock, or the resilience, is therefore propor- tioned to j 2 +E, which is known as the modulus 0} resilience of the material, and to the volume of the beam. These relation- ships hold for other sections, and beams loaded and supported differently. The above formula should not be rigorously applied to a drop test, unless / is below the yield-point. Example. — A 2X2 in.-bar of steel, 5 ft. between supports, if /= 16,000 lb., ought not to be subjected, from a central weight, to more . 1 i6,ooo 2 -2 2 -6o ., than — = 118 in. -lb. of energy. 18 29,000,000 If the load is distributed over a similar beam, the deflection at each point will be v, and the total work done will be \ i vwdx. If w is uniform, and the beam is supported at its ends, r l w 2 r l /ix 3 x 4 Px\ w I vdx = —p^h I \—> — )dx = Jo 4E1J Q V 6 12 12 / 24.0EI' As -7r =— and Vi 8 vi l tfEyi the above expression becomes w 2 l 5 8 7N Resilience = j^ = — (wl)v\ . 240EI 25 v 104 STRUCTURAL MECHANICS. Example. — A weighted wheel of 1,000 lb. drops J in. by reason of a pebble in its path at the middle of a beam 3X12 in., 15 ft. span. If E= 1,400,000 to find /: External work= 1,000(1+^)= - — . _ / 2 = / 2 . 18 -E 70,000 I II 2 1 80 ■ 180 t Q . y = = }= Z — f 12 Eyi 6-12-E 28,000 o 18 5ooH — -f= / 2 . /= 2,150 lb. per sq. in. J 28 70,000 D r -1 Resulting deflection=o.69 in. Static unit stress would be 625 lb. and ^=0.2 in. In an actual bridge the shock is distributed more 01 less in the floor and adjacent beams. 10 1. Internal Work. — The internal work done in a beam may be divided into two parts, that of extending and compressing the fibres, and that of distorting them, the first being due to the bending stresses, and the second to the shearing stresses. The work of bending will be found first. Let the cross-section be constant. The unit stress at any point of a cross-section = p; the force on a layer zdy = pzdy. The elongation or shortening of a fibre unity of section and dx long by the unit stress p = pdx+E. The work done in stretching or 1 p 2 } M shortening the volume zdydx = ---^r -zdydx. But p= — y = -jy* The work done on so much of the beam as is included between two cross-sections dx apart will be M 2 f +y * M 2 ydx I y 2 zdy = —^jdx. 2 £/^V_, > m >-2ET Substitute the value of M for a particular case in terms of x and integrate for the whole length of the beam. Thus for a beam supported at ends and loaded with W at the middle, M = %Wx at any point distant x from one end for values of x between o and hi- Then pi W 2 p l - / M 2 dx = -^rp I x 2 dx = J n aEIJ ' W 2 P 2EIJ Q 4EIJ 96EI' FLEXURE AND DEFLECTION OF SIMPLE BEAMS. 105 The internal work due to shear is not readily found unless a simple form of cross-section is chosen. If the section is rect- angular the shear at any point on the cross-section is, by § 72, 6F The distortion in a length dx is qdx-^C and the work done upon q qdx a volume zdydx is — -—^r -zdy. The work done upon the volumes between two sections dx apart is dx f +ih isf 2 r + * h 3 F 2 For a beam carrying a single load at the centre, as above, F = ^W, a constant, and the work done by the shearing forces is 3 W 2 l 2obhC 102. Deflection Due to Shear. — In the cases of flexure so far treated the bending moment only has been taken into account in finding the deflection. The shearing stresses, however, cause an additional deflection which is generally too small to be of practical account. The deflection of a rectangular beam carrying a single load at the centre can be found from the results of the preceding section since the internal work must be equal to the external. W 2J3 3W 2l 2 v 9 6EI^2obhC v = 48EI ' iobhC The first of these terms is the deflection when the shear is not taken into account, and it has the same value as was obtained 6Eh 2 in § 94. The ratio of the second term to the first is r 2 , which shows that in rectangular beams of ordinary proportions the shearing deflection is but a small proportion of the whole. 106 STRUCTURAL MECHANICS. Examples. — i. What is the deflection at the middle of a 2Xi2-in. pine joist of 12 ft. = 144 in. span, supported at ends and uniformly loaded with 3,200 lb.? £=1,600,000. 0.27 in. 2. What is the deflection if the load is at the middle? 0.432 in. 3. Find the stiffest rectangular cross-section, bh, to be obtained from a round log of diameter d. b=\d. 4. A 4X6-in. joist laid flatwise on supports 10 ft. apart is loaded with 1,000 lb. at the middle. The deflection is found to be 0.7 in. What is E? 1,607,000. 5. What is the maximum safe deflection of a 12-in. floor-joist, 14 ft. span, if /= 1,200 lb. and £=1,600,000? 0.37 in. for uniform load; 0.29 in. for load at middle. 6. What is the diameter of the smallest circle into which J-in. steel wire can be coiled without exceeding a fibre stress of 20,000 lb. per sq. in. ? CHAPTER VII. RESTRAINED AND CONTINUOUS BEAMS. 103. Restrained Beams. — When a beam is kept from rotating at one or both points of support, by being built into a wall, or by the application of a moment of such a magnitude that the tangent to the curve of the neutral plane at the point of support is forced to remain in its original direction (commonly horizontal) at such point, the beam is termed fixed at one or both supports. The magnitude of the moment at the point of support depends upon the span, the load and its position. It is the existence of this at present unknown moment which calls for the application of deflection equations to the solution of such problems as those which follow, there being too many unknown quantities to permit the treatment of Chapter III. In applying the results obtained in the following cases to actual problems, one should feel sure that the beam is definitely fixed in direction at the given point. Otherwise the values of M, F> and v will be only approximately true. 104. Beam of Span /, Carrying a Single Weight W in the middle and supported and fixed at both ends. Origin .at left support. Fig. 50. The reactions and end moments are now unknown. The beam may be considered either as built in at its ends (as at the right in above figure), or as having an unknown couple or moment Qb applied at each point of support (as at the left), of a magnitude just sufficient to keep the tangent there horizontal. The reaction at either end will then be jW + Q, while the shear between the points of support will still be±JTF. For values of x < J/, M x = -Q(b+x) + (lW+Q)x=lWx-Qb. 107 no8 STRUCTURAL MECHANICS. If this value is compared with that of M x in § 94, it is seen that a constant subtractive or negative moment is now felt over the whole span in combination with the usual ^Wx. Q^-b Fig ..50 d?v 1 fin " t Slope is zero when # = 0; v. C = o. Also slope is zero when Jv O Cm 0==J i_WP-iQbl. -Qb=-iWl, the negative bending moment at either end. That it is negative appears by making x = o in M x above. If this value of Qb is substituted in the first equation, giving M x = \W{x — \T), the point of contraflexure is located at x = \l, and the bending moment at middle, where x = \l, is M„ wx , = + \Wl, or one-half the amount in § 94. Substituting the value of Qb in the equation for slope, w w 4E1 (X "™ " nvX)» As v = o when x = o, C = o. When x = \l, RESTRAINED AND CONTINUOUS BEAMS. 109 W fP P_\ WP 4£/\24 16/ 192-E/' The beam is therefore jour times as stiff as when only sup- ported at ends. As — = -B~' W = — -r and Vi y\ 8 j\l 2^Eyi so that only one-half the deflection is allowable that is permitted in § 94, but the beam may safely carry twice the load. It is useful to notice that this beam has a bending moment at the middle equal to that which would exist there if the beam were cut at the points of contraflexure and simply supported at those points, and that the two end segments, of length J/, act like two cantilevers each carrying %W, the shear at the point of contraflexure. If the weight were not at the middle the moments at the two ends would differ, equations would be needed for each of the two segments, and the solution, while possible, would be much more complicated. Example. — A wooden beam, 6 in. square and 7^ ft. span, is built into the wall at both ends. A central weight of 3,000 lb. will give a max. fibre stress of R 75 = 937i lb. per sq. in. at the middle and both "? OOO * QO * I 2 ends. The deflection will be -7= 0.07 in. if E= 1 , coo,ooo. 192- 1,500,000 -(r The allowable deflection for /= 1,200 is — = 0.00 in., and 24-1,500,000-3 max. allowable W= 3,000 •■§■= 3,860 lb. 105. Beam of Span /, Uniform Load of w per unit over the whole span, fixed at both ends. Origin at left support. Fig. 51. As in the previous case, the reaction at either end may be represented by \wl-\-Q. The shear at x is \wl— wx, which ex- pression changes sign at the middle and at either point of support; hence at those places will be found M max . M x =(^wl + Q)x — ^wx 2 —Q(b + x) = \wlx — \wx 2 —Qb. HO STRUCTURAL MECHANICS. Compare with § 93. dv dv - — = when * = o; .*. C = o. -r- = o when *=*; .-. — - -Qbl = o. —Qb= , Fig. 51 the negative moment at each point of support. If x = %l, , wl 2 M at middle =w/ 2 (i -J- - T V) = — • Substitute the value of Qb and get w dv dx 2 EI (±x*-ilx 2 + ll 2 x); w (x A lx s l 2 x 2 \ RESTRAINED AND CONTINUOUS BEAMS. Since v = o when x=o, C' = o. Whenx = J/, w I I 4 I* l 4 \ (wl)l 3 3 S 4 EP in "max. (JL J— —\ - \IQ2 48 48/ 2EI\ig2 48 48 which is one-fifth the value of § 93. The points of contraflexure occur where M x =o; .-. x 2 -Ix + \l 2 = o ; x = il± ih/h The second term is the distance from the middle, each way, to the points of contraflexure. If M is calculated for the middle point of a span I+V3, it will prove to be wl 2 +24, as above. Since (wl)l jl 12 f I . W = — , wl = — r, and V\ — — w~ , or 0.3 as much as in § 93. The beam may safely carry, however, 50 per cent, more load. Example. — An 8-in. I beam of 12 ft. span, carrying 1,000 lb. per foot, if firmly fixed at both ends and not to have a larger unit stress than = 48. An I,000 -12- I2-I2-4 12,000 lb., should have a value of /= ' ' 12-12 ,000 8-in. steel beam, 18 lb. to the foot, 7=57-8, will satisfy the requirement, the load then being 1,018 lb. per foot. The deflection will be 0.6 in. Fig. 53 106. Beam of Span /, fixed or horizontal at P2, supported at Pi and carrying a uniform load of w per unit. Fig. 52. Origin at Pi and reaction unknown. H2 STRUCTURAL MECHANICS. It will be seen from the sketch that a beam of length 2/, rest- ing upon three equidistant supports in the same straight line, will come under this case. d 2 v 1 dv 1 dv — =0 when x=l; .'. C = \wl z — \P\l 2 . I /P\X 3 wx 4 wl 3 x PJ 2 x \ v =jrA-ir- — + ~Z — +C). EI\ 6 24 6 2 / v = o when x = o', .*. C f =0. If x=l, v = o, and Pi = (A-i)™l 4 + (i-h)l 3 = >l- F x =§wl-wx. Substitute this value of Pi in the above equations. dv w , / 3/^2 _ 1~3 i J3\ dx~ 2 EP 8 3 2T h w •7/ f±Jr& — lr 4 — 1/3-r^ dv dx "Forv max% make -7- = 0, or §/x 2 -J# 3 — ^y 3 =o. 8^c 3 — 9/x 2 +/ 3 = o. As a minimum value of v, or v = o, occurs for x=l, divide by x— I and obtain Sx 2 — Ix— / 2 = o, or x=l 7 — =0.4215/. (w/)/ 3 Then iw. = -0.0054— — . To find points of M maXm put F x = o, or #=§/. Also, by inspection, M max , when x = l. For * = f /, M max . = (& - T f ? ) w/ 2 = T f ¥ w/ 2 . RESTRAINED AND CONTINUOUS BEAMS. 113 For x = l, M max .=(§-i)wP= -\wl 2 . For the point of contraflexure §/# — Jx 2 =o, or x — \l t as was to be expected from the position of the point of maximum positive 31. Note again that the point of maximum bending moment is not the point of maximum deflection. It will be seen that a continuous beam of two equal spans /, uniformly loaded with w per unit, has end reactions of | wl and a central reaction of 2X%wl=^wl; that points of contraflexure divide each span at \l from the middle pier, and that the bending moment at the middle of the remaining segment of f/ is, as above, f -f •lwl 2 = ji- 8 wl 2 . It will also be seen that, since the bending moment at P> is— Jw/ 2 , a uniform beam, continuous over two equal spans, each /, is no stronger than the same beam of span / with the same uniform load. It is, however, about two and a half times as stiff. Example. — A girder spanning two equal openings of 15 ft. and carrying a 16-in. brick wall 10 ft. high of no lb. per cubic ft. will . . , 5-4- no- 10-15 .. 1 .-Hi throw a load of =27,500 lb. on the middle post and 1 i< .4-iio-io-i^k _ „ must resist a bending moment of - =41,250 ft. -lb. 107. Two-span Beam, with Middle Support Lowered. — A uniform beam, uniformly loaded and supported at its ends, will have a certain deflection at the middle which can be calculated. If the middle point is then lifted by a jack until returned to the straight line through the two end supports, the pressure on the jack, by § 106, will be five-eighths of the load on the beam. Since deflection is proportional to the weight, other things being equal, if the jack is then lowered one-fifth of the first deflection referred to, the pressure on the jack will be reduced one-fifth, or to one-half of the load on the beam. Hence, if a uniformly loaded beam of two equal, continuous spans has its middle support lower than those at its ends by one-fifth of the above deflection, the middle reaction will be one-half the whole weight, the bending moment will be zero at the middle, and the beam may be cut at that point without disturbance of the forces. 108. Beam of Span /, fixed at left and supported at right end, ii4 STRUCTURAL MECHANICS. and carrying a single weight IF at a distance a from the fixed end. Fig- 53- Origin at fixed end. Fig. 53 The reaction at the supported end, being at present unknown, will be denoted by P 2 , and moments will be taken on the right of any section x. From lack of symmetry, separate expressions must be written for segments on either side of W. BETWEEN W AND FIXED END. M x = P 2 (l-x)-W(a-x) dv i 7x = ~EI iP2{lX ~ ^ 2) ~ W{ax ~ ** 2 ) + C] i r llx 2 x\ dv -— =o when x=o; dx v=o when x=o; (ax 2 x 3 \ t-jt) C=o. C'=o. BETWEEN W AND SUPPORTED END. M x =P 2 {l-x) i r /^ 2 ^ 3 \ n ^sLMt— ) +c "* +c "'] dv , r dv If # = a, — on left = — on right. If # = a, v on left=z/ on right. C" = TF( - Ja 3 + Ja 3 + i^ 3 ) = \Wa*. If *=/, ?;atP 2 = o; .'. P 2 (J/ 3 -^ 3 ) -W°< 2 l + Wa z =°> or P 2 = TTfl 2 (J/-Ja)^J/3 PFa 2 2/ 3 (3*-<0- RESTRAINED AND CONTINUOUS BEAMS. 115 If this value of P2 is substituted in the above equations the desired expressions are obtained. Thus — ( 3 l-a)(l-x)-W(a-x). M * = —ji II 11 VV (L Mx = -~s{3l-a)(l-x)-W(a-x). M x = — r ( 3 J-a)(/-*). M max . , by inspection, when x = o, or x = a. Wa 2 W M maXt =— w($l~ a)— Wa— — j^a(l—a)(2l—a) at fixed end, TTT 9 and =—p-(l — a)(3l — a) at the weight. The point of contraflexure occurs between W and the fixed end, where M = o, or a 2/ — a -(^l — a)(l—x)—(a—x) = o. .'. x = al- 2 / 3VO 2/(7 + aj -a 2 ' 6?X The maximum deflection will be found where t~ = o, on the right or left, according to the value of a. The above beam may be regarded in the light of two equal continuous spans with W on each, distant a each side of the middle point of support. In solving the more intricate problems in the flexure of beams, as well as those just treated, each equation of condition can be used but once in the same problem, and as many unknown quantities can be determined as there are independent equations of condition. The reactions and moments at the points of support are usually unknown, and must be found by the aid of such flexure equations as have just been used. Example. — A bridge stringer which is continuous over two successive openings of 12 ft. each and carries a weight from the wheels of a wagon of 3,000 lb. at each side of and 3 ft. from the middle support will be horizontal over that support. Then —M maXm = - 3 ^-3-9" 2 i = - 5,9o6. 2 5 ft-lb. + M max =^21.f. 33 . 9 = 2,320.3 ft.-lb. P 2 =- o-3 2 -33=258 lb. Reaction at middle sup- port from both spans = 2 (3,000— 258)= 5,484 lb. n6 STRUCTURAL MECHANICS. 109. Clapeyron's Formula, or the Three-moment Theorem for Continuous Loading. — To find the reactions, shears, and bending moments for a horizontal uniform continuous beam loaded with Wi, w 2 , w 3 , etc., loads per running unit over the successive spans h, l 2 , h> etc. Fig. 54. P , Pi, P 2 , etc., denote the unknown reactions; M , Mi, M 2 , etc., the unknown bend- ing moments at points of support, o, 1, 2, etc.; F , F i} F 2 , etc., denote the shears immediately to the right of o, 1, 2, etc.; while Fi, F 2 , etc., denote the shears immediately to the left of the points of support 1, 2, etc. The origin of coordinates is first taken at O and the sup- ports are on a level. +M makes the beam concave on the upper side. As positive shear acts upward at the left of any cross- section, w is negative. Consider the condition of equilibrium of the first span 0-1, or h, loaded throughout with Wi per unit of length. Take moments on the left side of and about a section S, distant x from the origin O. The bending moment at S is RESTRAINED AND CONTINUOUS BEAMS. 117 d 2 v M = EI-p^ = M +F x—%WiX 2 (1) Let to, i\, t2-'-in — tangent of inclination of the neutral axis at o, 1, 2, . . . N. Integrate (1) and transpose the con- stant of integration, to, to the left-hand member and thus obtain an expression for the difference in slope or inclination of the two tangents to the bent beam at O and S. fdv \ EIl-j--io)==MoX + %FoX 2 -iw 1 x 3 . ... (2) dv When x=lu t~=*i> an d hence ax E/ft-^^Mo/i + IW-MW • • • (3) Integrate (2) and determine constant as zero, because v—o when #=0. EI(v-iox) = iMoX 2 + %F x s -? l T WiX 4: . . . . (4) Make x=h, then •z/=i>i=o, and -Elioh = Woh 2 + W1 3 -?Wi 4 , or -£/*WWi + Wi 2 -?Wi 3 (5) Eliminate z by subtracting (5) from (3). Elh^iMoh+iFol^-lw^ (6) If the origin is taken at 1 instead of o an equation like (5) is obtained for the second span l 2j or -EHi^Wih+Wihf-ThwM* (7) Add (6) and (7), obtaining o = iMoh + \Mih + Woh 2 + JiW - W1 3 -AW2 3 . (8) Il8 STRUCTURAL MECHANICS. The unknown slopes have thus been eliminated. The next step is to remove either M or F. Equation (i) must equal M\, for x = /i . Therefore Mi = M o + F h - iwih 2 , or F = j + \w x l x , Mi-_M Q t i t j r- M 2 -M\ , . In the same way, tor second span, F\ = ; 1- \w 2 l 2 * *2 Substitute the values in (8) and obtain M h Mih Mi-M 0l u\l^ M 2 -Mi t o = 1 1 h-\ — - — V 2 h 223 66 w 2 h 3 wJi 3 w 2 h 3 12 8 24 .'. Af o*i + 2M1 (h + h) + M 2 l 2 = - JOi^i 3 + w 2h 3 ), . . (9) which is Clapeyron's formula for pier moments for a continuous beam, with continuous load, uniform per span. Notice the symmetry of the expression. The negative sign to the second member indicates that the bending moments at points of sup- port are usually negative. Example. — Three spans, 30 ft., 60 ft., and 30 ft. in succession. Load on first and last 500 lb. per ft., on middle span 300 lb. per ft. No moment at either outer end. Then Mo=o. M\ = M 2 by sym- metry. 2M 1 • 90+ M 2 - 60= — i(5oo • 3o 3 + 300 • 60 3 ). Mi= -81,562^ ft.-lb. F = - 2,7181+7,500= +4,78ii lb. F\= +4,7811—30-500= — io,2i8f lb. F\ = 300 -30= 9,000 lb. .*. Pq^^^ 4,78i:| lb.; Pi = P2=io,2i8f+ 9,000= 19, 2i8f lb. The bending moment and shear at any point can now be readily determined. If the two adjacent spans are equal and have the same load, M + 4Mi+M 2 = -\wl 2 (10) If there are n spans, n — i equations can be written between n + 1 quantities Mo, Mi . . . M n . But if the beam is simply placed on the points of support, the extremities being unre- strained, Mo=o and M n =o, and there remain n — 1 equations RESTRAINED AND CONTIGUOUS BEAMS. 119 to determine Mi . . . M n -i. If the beam is fixed at the ends, the equations io = o and i n = o will complete the required number no. Shears and Reactions. — As the shear is the first derivative of the bending moment, § 56, from (1) is obtained -^=F = Fo-WiX, (11) as was to be expected, +F acting upwards on the left of the section. A similar equation can be written for each span. The reaction at any point of support will be equal to the shear on its right plus that on its left with the sign reversed. As the shear on its left is usually negative, the arithmetical sum of F n and F n ' commonly gives the reaction. A simple example may make the application plainer. Given two equal spans on three supports, Wi=w 2 =w. M =o, M 2 =o. (10) gives Mi= —\wl 2 . F = — \wl + \wl = f wl; F\ = \wl —wl=— f w/. Fi = \wl + \wl = \wl ; F-2 = %wl —wl=— fwl. Po = iwl ; Pi = (f + f )wl = \wl ; P 2 = f wl 1 ii'I 3 (5) gives i = -—(o + T \-^)wP=--^j. (6) gives ii=^j(o + -^-i)wl 3 =o, and the analogous equation for the second span is which differs from to only in direction of slope. , . . „ T dv wl 3 wx 3 (2) gives EI T% = -— + T 3 jrrote 2__. zul 3 IVX^ (4) gives EIv= —x + -Azl'Ix 3 — — — • v 48 48 24 120 STRUCTURAL MECHANICS. These equations determine the slope and deflection at each dv point. Putting -j~ = o, there results I 3 — glx 2 + 8x 3 =o, containing the root x=l, already known. Therefore divide by l—x and obtain l 2 +lx—8x 2 =o, which is satisfied for # = 0.4215/, the point of maximum deflection. The substitution of this value in the equation for v will yield v max . From (1) M = %wlx — \wx 2 . If M =0, §/ — |# = o, or x = \l, the point of contraflexure. Differentiate M and get F = %wl— wx. If F=o, # = -§/, the point of + M maXm /. M max , = Iwl • f / -\w • -fcl = T | ¥ w/ 2 . Example. — If a uniformly loaded continuous beam covers five equal spans, M + 4M1 + M 2 =- %wl 2 = Mi + 4M2 + M 3 =M 2 + 4 J^j + ^4 = M 3 + 4 M 4 +M 5 . M =o\ M 5 =o. Then M x = -fawP=M±\ M 2 = -^wl 2 =M 3 . F =Hwl; F^=-Uwl; F^^wl; F 2 '=-^wl; F 2 =^wl, etc. P =Uwl=P 5 : Pi=f|^=P 4 ; Fo=Hwl=P 3 . The sum of the reactions must equal $wl. in. Coefficients for Moments and Shears. — It has been found that the numerical coefficients for moments and shears at the points of support, when all spans are equal and the load is uniform throughout, may be tabulated easily for reference and use. Thus the values of M and F just obtained for the five equal spans can be selected from the lines marked V. The reactions are given by the arithmetical addition of the shears. The sum of the reactions must equal the total load. The shears at the two ends of any span differ by the whole load on the span, the shear at the right end being negative. The dashes represent the spans. RESTRAINED AND CONTINUOUS BEAMS. 12 I SHEAR AND REACTION COEFFICIENTS. I. i-iwl; TT 1 5 5 3L W J. 11. g g g g«", m4 6 5 5 6 4 . • TO" TO" TO TO TO TO J TV 11 17 IS 13 13 15 17 11. 1V - 28 28 28 28 28 28 28 28) V 15 2.1 2 18 JL9 19 18. 2 0. 2.3 15. v • 3 8 38 38 18" 3 8 "5838 3838 3 8 > etc., etc. PIER MOMENT COEFFICIENTS. II. — J— wl 2 ; in. -^v-rV-^ 2 ; _3 2 _3_ I*' " 28 28 28 j \T 4 3 3 4 . v • 38 38 38 38 > etc., etc. The rule for writing either table is as follows: For an even number of spans, the numbers in any horizontal line are obtained by multiplying the fraction above, in any diagonal row, both numerator and denominator, by two, and adding the numerator and denominator of the preceding fraction. Thus, in the first 2X6+ 5 17 . . .,12X1 + 13 table, 1 o = ~77> an d m the second table, — — rs =_ ^> 2X10 + 8 28 2X10 + 8 28 2X 3+ 2 8 or w , — a = • For an odd number of spans, add the two 2X38 + 28 104 r preceding fractions in the same diagonal row, numerator to numerator and denominator to denominator. Thus, -^ -=—?;• 28 + 10 ^8 The denominators agree in both tables. A recollection of two or three quantities will enable one to write all the others. Example. — Continuous beam of five equal spans, each /, carrying w per foot. Where and what is the max. + M in second span? Shear changes sign at f|/ from left end of span. If this span were inde- pendent, + J/ at that point would be \tv1 2 - 2 ° T =£-^. — . The 19-19 361 8 negative or subtractive moment is (3 3 -§-+ -sVif)^ 2 - The difference between these values is +M max - 122 STRUCTURAL MECHANICS. A more general investigation will produce equations which are of great practical value in the solution of problems concerning continuous bridges, swing-bridges, etc., as follows: 112. Three-moment Theorem for a Single Weight. — O is the origin, Fig. 55; the supports are at distances h below the axis of X. A single weight W n is distant kl n from O on the span /„, k being a fraction, less than unity, of the span in which W is situated. The moment at section S beyond W n will be, as in the former discussion, M x = M n + F n x-W n (x-kl n ) (1) If x=l m M x = M n+ i, and from (1) M n+l -M n , F n = 1 + W„(i-k) (ia) In For an unloaded span, W=o, and F m = - ? . For the shear on the left of a section at the right end of the nth span, F n+1 '=F n -W n = -^ ±} 1 ~-W n k. For an unloaded span, W=o, and , M m -M m -i I m—l As F m r is the shear at left of support m, and F m is the shear at right of the same support; the reaction there will be the sum of F m and —F m ' or Mm+i -Mm M m - X -M m ±m — ? 1 7 hn t'm—l To get values of i and v for span l n it is necessary to write separate equations for the two segments into which W n divides RESTRAINED AND CONTINUOUS BEAMS. 123 the span. Equation (1) applies to the right segment, and by omitting the last term the equation for the left segment is obtained. Equate -^y to j— 2 and integrate between the limits of o and kl n on the left and between kl n and x on the right. Then on the left /kln rkln dx + F n I xdx, . . (2) and on the right /dv %J kln The sum of (2) and (3) gives the change of slope between the support n and the section S. El(-^-i n )= M n I dx + F n I xdx - W n I (x - kl n )dx =M n x+iF n x 2 -iW n (x-kl n ) 2 (4) 124 STRUCTURAL MECHANICS. Integrate the last equation between the same limits as before. EI(v -i n x - h n ) = \M n x 2 + \F n x 3 - iW n (x-kl n ) 3 , . (5) which is the general equation of the curve of the neutral axis, the term in W disappearing for values of x less than kl n . If x=l n , v = h n+ i. If the value of F n from (ia) is inserted in (5), the slope at support n is i w= ^~^ -^[ 2 Kj n+ M w+1 / w + TFJ w 2 (2i ! -3^ + ^)]. (6 ) The equation of the curve is therefore completely deter- mi ed when M n and M n +\ are known. The equation of this curve, between W n and the « + ith support is given by (5), and the tangent of its angle with the axis of X by (4). If the value of F n from (ia) and of i n from (6) are substituted in (4), and dv x = l ny -j- will be the tangent i n+ i at n + i 3 Or i n+1 = ^^—^^jiMnln + 2M n+l l n + WJ n 2 (k -&)]. Remove the origin from O to N, and derive an expression for i n by diminishing the indices. Equate with (6) and transpose. M n -lln-l + 2M n Qn-\ +ln) +M n+ il n -W J ,?{2k - 3 F + ^ 3 ), (7) which is the most general form of the three-moment theorem for a girder of constant cross-section. In using the theorem the following factors may be of service: k-k 3 = k(l-k)(l+k)] 2k-7,k 2 + k 3 = k(l-k)(2-k). RESTRAINED AND CONTINUOUS BEAMS. 125 Pier moments are usually negative and the end moments zero. When the supports are on a level, hi=h 2 , etc., and the term containing EI disappears. Any reaction P n = F n —F n '\ _. Pn= M^M_n + M n ^-M n + Wn{i _ k) + Wn _ ih It should be borne in mind that all the preceding deflection formulas are derived on the assumption that the deflection is due entirely to bending moment, the deformation from shear being neglected. In a solid beam the amount of deflection due to shear is very small, but such is not the case with a truss. In a truss the deflection due to the deformation of the web may in some cases amount to half the total deflection. For this reason the deflection formulas and the three-moment theorem should not be applied to trusses. Example. — Three-span continuous girder carrying loads as shown. Supports on a level. 000 000 O 30 (n I 2Q M 40' co 40 2 3 A A A A 5°' I0 °' 75' 2 • 150!/"! + IOoif 2= — 2,000 • 2,500 • f • I • I — 1,000 ■ 10,000 • to"* To - ' TO - 3 ,000 • 10,000 • T 6 o • J * I 0- iooM 1 + 2.i75M 2 =-i,ooo-io,ooo- T %- T 8 o4t-3' 00 °- IO > ooo-T ( o-fV-rf 30olfi + iooiW"2 = — 14,880,000, T00M1 + 350^/2= — 13,440,000. M x = -40,670 ft.-lb. M 2 = - 26,780 ft.-lb. -40,670 2,0 00 • 20 P = 1 ==_I 3 lb. — 26,780+40,670 40,670 i,ooo-8 3,000-4 2,000*3 "1 = 1 r — 1 1 = + 4, 152 lb. 100 50 10 10 5 J + 26,780 -40,670+26,780 1,000-2 3,000-6 „ J J 2= = — 1 1 ' i =+2,215 ID. 75 100 10 10 -26,780 -^3= — =—357 lb. 75 — 13 + 4,152 + 2,218 — 357 = 2,000+1,000+3,000=6,000. 126 STRUCTURAL MECHANICS. Examples. — i. A brick wall 16 in. thick, 12 ft. high, and 32 ft. long, weighing 108 lbs. per cubic ft., is carried on a beam supported by four columns, one at each end, and one 8 ft. from each end. Find M at the two middle columns and the reactions. M= — 31,104 ft. -lb.; P\ = 3,024 lb.; P2= 24,624 lb. 2. Two successive openings of 8 ft. each are to be spanned. Which will be stronger for a uniform load, two 8 ft. joists end to end or one 16 ft. long? Find their relative stiffness. 3. A beam of three equal spans carries a single weight. What will be the reactions and their signs at the third and fourth points of support when W is in the middle of the first span? -A IF; +W. ,V 4. A beam loaded with 50 lb. per foot rests on two supports 15 ft. apart and projects 5 ft. beyond at one end. W T hat additional weight must be applied to that end to make the beam horizontal at the nearer point of support? 156 J lb. at the end, or 312^ lb. distributed. 5. A beam of two equal spans on level supports carries a single load, W, in the left span. Prove that iWPF( 4 -5£+£ 3 ); P2=W{2,k-k*)\ P 3 =-iW(k-k 3 ). CHAPTER VIII. PIECES UNDER TENSION. 113. Central Pull. — If the resultant tension P acts along the axis of the piece, the stress may be considered as uniformly dis- tributed on the cross-section S. If, then, / is the maximum safe working stress per square inch for the kind of load which causes P, Fig. 56, P = jS, or S=j for the necessary section which need not be exceeded throughout that portion of the piece where the above conditions apply. Changes due to connections will require a larger section. If owing to lack of uniformity in the material or the direct application of P at the end of a wide bar to a limited portion only of the width the stress may not be considered as uniformly distributed at a particular cross-section, injurious stress may be prevented by taking the mean stress / at a smaller value and obtaining a larger cross-section. If there is lack of homogeneity, or two materials are used together, or two or more bars work side by side, those fibres which offer the greatest resistance to stretching will be subject to the greatest stress. Fortunately the slight yielding and bending of connecting parts tend to restore equality of action. A long tension member has a much greater resisting power against shock than a short one of equal strength per square inch. See § 20. 114. Eccentric Pull. — If the variation of stress on a cross- section is due to the fact that the line of action of the applied 127 128 STRUCTURAL MECHANICS. force does not traverse the centre of figure of the cross-section S, the force P f that can be imposed without causing a unit stress greater than / at any point in the section is less than P of the pre- ceding formula, and depends upon the perpendicular distance yo of the action line of P from the centre of S. For safe stresses, which must lie well within the elastic limit, the unit stress is proportional to the stretch, and plane cross- sections of the bar before the force is applied are assumed to re- main plane after the bar is stretched. It is impossible to detect experimentally that this assumption is not true. Were the plane sections to become even slightly warped, the cumulative warpings of successive sections in a long bar ought to become apparent to the eye. No reference is intended here to local distortion preceding failure. If the stress on any section is not uniform and the successive sections remain plane, they must be a little inclined to one another. The stress on any cross-section S must therefore vary uniformly in the direction of the deviation of the action line of P' from the centre, (Fig. 56), and be constant on lines at right angles to that deviation. The force, P', acting at a distance, y , from the axis of the tie is equivalent to the same force, P', at the centre and a moment, P'yo- The stress on each particle of any section may therefore be divided into two parts, that due to direct tension and that due to the moment or to bending. The former is uniform across the section and is }t- s - The latter is zero at the centre of gravity of the section and a maximum at the edge toward which the load deviates, where it is PIECES UNDER TENSION. 1 29 My 1 P'yoyi I Sr 2 Note that /* is the mean stress, which is always the existing stress at the centre of gravity of the cross-section. Example. — A square bar 1 in. in section carries 6,000 lb. ten- sion. The centre of the eve at the end is J in. out of line. Then /= 6,000(1 + J- J- 1 2) = 15,000 lb. per sq. in., 2\ times the mean and probably the intended stress. A bar which is not perfectly straight before tension is applied to it tends to straighten itself under a pull, but the stress will not become uniform on a cross-section. The bar is weaker in the ratio of f t to /, as it might carry fS if the force were cen- tral, but now can safely carry only f t S. If a thrust is applied to a bent bar, there is a tendency to increased deviation from a straight line, and to an increase in the variation of stress. It is seen from the example above that a small deviation yo will have a decided effect in increasing / for a given P', or in diminishing the allowable load for a given unit stress. Herein may be the explanation of some considerable variations of the strength of apparently similar pieces under test; and, on account of such effect, added to other reasons, allowable working stresses may well be and are reduced below what otherwise might be used. 115. Hooks. — The bending action on and the strength of a hook are given by the same formulas. Here yi will be the dis- tance from the inside edge to the centre of the cross-section, and yo the distance from the action line of the load to the same centre. Then the maximum unit tension i-U^-f) Example. — A hook, the section of which in the bend is elliptical, iXf in., carries the link of a chain at a distance of J in. horizontally 2 2 I • C Jl 2 t from the inside of the bend. Then S= — ?r~^ = i sq. in.; r 2 = — = — 7 8-4 H ' 16 16' 130 STRUCTUR.iL MECHANICS. §74, V.; >' =i in. Then /=2P(i + i-J-i6) = i8P. If /= 8,000, P=45o lb.; if /= 12,000, ^=650 lb. Compare with the given section. The ordinate of the bend should be reduced as much as possible. 116. Combined Tie and Beam. — If to a tension member transverse forces are applied, or if it is horizontal and its weight is of importance, the unit tensile stress on the convex edge, due to the maximum bending moment, must be added to the unit stress at that point due to the direct pull. As in § 114, 1 (M max .)yi P Jb = j , and tt = y But / = /& + /* must not exceed the safe unit tension, and the needed section is, since I=Sr 2 , In this case the sections may vary, since the external bending moment M varies from point to point. If the piece is rectangular in section, as with, timber, the formula may be written bh 2 ' bh' fh\ h In practical calculation of such a rectangular section, if h is assumed, it is sufficient to compute the breadth to carry M and add enough breadth to carry P, when the combined section will have exactly / at the edge. Example. — A rectangular wooden beam of 12 ft. span carries a single weight of 3,000 lb. at the quarter span, and, as part of a truss, resists a pull of 20,000 lb. If /= 1,000 lb., what should be ihe 1 i • i -> if- ^,000 • 3«o- 12 section under the weight? M max = = 81,000 m.-lb. 12 81,000 -6 , f9 _, Tr . , .. 20,000 f _ — 1 =^2 =4 86 # if h=i2, 6 = 3.37. Also, ■ =1.67. En- 1 ,000 1 ,000 '12 PIECES UNDER TENSION. I3 1 tire breadth = 3.37 + 1.67 = 5.04. Section=5Xi2 in. The same result is obtained by the formula , 1 /6- 81,000 , \ 0= h 20,000 ]. 12- 1,000 \ 12 / If the tensile stress in a combined tie and beam is small as compared with the stress due to flexure the above solution is accurate enough, but when the tensile stress is large the error is considerable. The transverse forces acting on the beam cause it to deflect and consequently the line in which the direct pull acts does not pass through the centre of gravity of each section of the beam and a bending moment is produced which partially counteracts the bending moment due to the transverse forces. The bending moment at any sect'on then is M-PvJ—, y.i in which v is the deflection of the tie-beam at the section where }b is found. By referring to Chapter VI it is seen that the deflec- tion of a beam may be written Eyi where k is a constant depending upon the way in which the beam is loaded. If it is assumed that v is proportional to /& in the case under consideration as well as in the case of a beam not subjected to tension, there results Myi Pl 2 ' I + k— E For a beam supported at both ends and uniformly loaded k = ^; carrying a single load at the centre, k = -}%\ carrying a single load at the quarter-point, k= ^V If the ends are free to turn and the transverse load is uniformly distributed or is a single load near the centre, k may be taken as T V. An expression for the fibre stress in a horizontal steel eye-bar bending under its own weight is readily derived from the last equation. Since a bar of steel one square inch in area and one 13 2 STRUCTURAL MECHANICS. foot long weighs 3.4 lb. and E is 29,000,000, there results when all dimensions are in inches: 4, 900,000 h Jb = ft + 23,000,000 ( h\ 2 ' I Example. — An eye-bar 8X1 in., 25 ft. long, carries 100,000 lb. ten- sion. /$= 12,500 lb. per sq. in. If the effect of the deflection is neg- lected, /&= 2,390 lb. Taking the deflection into account by the last formula, /&= 1,360 lb. 117. Action Line of P Moved towards the Concave Side. — It will be economical, if it can be done, in a member having such compound action, to move the line on which P acts towards the concave side. If there are bending moments of opposite signs at different points of the length, or at the same point at different times, such adjustment cannot be made. If vo is made equal to \{M max ) t-P, one-half of the bending moment will be annulled at the point where M max . exists, and at the point of no bending moment from transverse forces an equal amount of bending moment will be introduced. The unit stresses on the extreme fibres at the two sections will be the same, but reversed one for the other. Example. — A horizontal bar, 6X1 in. section and 15 ft. long, has a tension of 33,750 lb. It carries 100 lb. per foot uniformly dis- nr IOO-I5-I5-I2 ... tnbuted. M max = =33, 750 m.-lb. .'. y may be made o ii, Then/f r omM_=3M^ =±5)62slb . oneitheredge . But / from P=^ 2 |^(i±|^) = 5,625(i±J) = 5,625±2,8i2.5. Stress at top at ends and at bottom at middle 8,43 7 J lb.; at bottom at ends and at top at middle 2,812^ lb. tension. The extreme fibre stress from bending moment of the load varies as the ordinates to a parabola; that from Pv is constant. A rectangle superimposed on a parabolic segment will show the resultant fibre stress at each section. 118. Connecting-rod. — If a bar oscillates laterally rapidly, as does a connecting-rod on an engine, or a parallel rod on loco- motive drivers, there are forces developed due to the acceleration, and at certain positions of the bar these forces are transverse and PIECES UXDER TENSION. 133 cause bending. When a parallel rod is in its highest or lowest position the centrifugal force, due to the circular motion of every part of the rod, is acting at right angles to the bar, which is then subjected to a uniformly distributed transverse load of wv 2 w =— A-n 2 R, gr g in which n is the number of revolutions per second, R the radius of the crank, w the weight of unit length of bar and g the acceler- ation due to gravity. The rotating end of a connecting-rod at two points in its path is under the influence of a transverse force whose intensity is obtained by the above formula, but as there can be no transverse force at the sliding end due to acceleration, the rod as a whole may be considered to be acted upon by transverse forces varying uniformly from wv 2 +gr at one end to zero at the other. The maximum fibre stress due to such loading may be found and added to the tensile or compressive stress caused by the pull or thrust along the rod. An I-shaped section is suitable for such members. Owing to the rapid variations and alternation of stress, the maximum unit stress should be small. Mass is disadvan- tageous in such rods. 119. Tension and Torsion. — A tension bar may be subjected to torsion when it is adjusted by a nut at the end, or by a turn- buckle. The moment of torsion will give rise to a unit shear at the extreme fibre, for a round rod, of q\ = T -^-o.ig6d 3 by § 82, or at the middle of the side, for a square rod, of qi = T^-o.2o8h 3 by § 83, either of which, combined with j = P+-S, the tensile stress, will give p x = J/+\/(i/ 2 + ? 2 ). See § 86. Example. — A round bar, 2 in. diameter, to be adjusted to a pull of 10,000 lb. per square inch, calls for the application to the turnbuckle of 200 lb. with an arm of 30 in., one-half of which moment may be sup- posed to affect either half of the rod. If the turnbuckle is near one end, the shorter piece will experience the greater part of the moment. 7. 000*2-7 11 mi • • • , a=- 5—*-= 1,910 lb. Ine maximum unit tension on the outside 22 • i d fibres of the rod will be 5,ooo-fv(5> 00 ° 2 + i j9 i ° 2 ) = 10,350 lb. 120. Tension Connections. — If a tension member is spliced, or is connected at its ends to other members by rivets, the splice should be so made or the rivets should be so distributed across 134 STRUCTURAL MECHANICS. the section as to secure a uniform distribution of stress. An angle-iron used in tension should be connected by both flanges if the whole section is considered to be efficient. One or more rivet-holes must be deducted in calculating the effective section, depending on the spacing of the rivets. See § 185. If the stress is not uniformly distributed on the cross-section, the required size will be found by § 114, •S=-t( i +~~ J~ )• Transverse bolts and bolt-holes are similar to rivets and rivet- holes. Timbers may be spliced by clamps with indents, and by scarfed joints, in which cases the net section is much reduced; so that timber, while resisting tension well, is not economical for ties, on account of the great waste by cutting away. How- ever, where the tie serves also as a beam, timber .may be very suitable. 121. Screw-threads and Nuts. — If a metal tie is secured by screw-threads and nuts, the section at the bottom of the thread should be some 15 per cent, larger than the given tension would require, to allow for the local weakening caused by cutting the threads. Bars are often upset or enlarged at the thread to give the necessary net section and thus save the material which would be needed for an increase of diameter throughout the length of the rod. • To avoid stripping the thread, the cylindrical surface, whose area is the circumference at the bottom of the thread multiplied by the effective thickness of the nut, should be, when multiplied by the safe unit shear, at least equal to the net cross-section of the rod multiplied by the safe unit tension. 2r.Ri-t-qi=xRi 2 j i or }Ri = 2qiL As q\ is usually taken less than /; as with a square thread only half of the thickness is effective, and with a standard V thread quite a portion of the thickness must be deducted, nuts are usually given a thickness nearly or quite equal to the net diameter of the rod. Heads of bolts may be materially thinner. 122. Eye-bars. — The eye for the connection of a tension bar to a pin is seen in Fig. 57. The pin is turned and the eye bored to a reasonably close fit. Since bearing first takes place at the back of the pin, the most intense pressure will be found there, and it will probably diminish at different points of the semi- circumference until the horizontal diameter is reached. The pressure on the pin may be found to extend slightly below that diameter. If these pressures are assumed to be normal, and PIECES UNDER TENSION. 135 are laid off in succession at 2 r -2, 2-3, 3-4, . . . 5-6, and closed with 6-6', the pull on the bar, a pole can be assumed at o and an equilibrium polygon or curve drawn, cutting the centre line of the material of the eye about as shown. By moving o hori- zontally and changing the point of beginning near A vertically, this equilibrium curve can be brought to the right position to satisfy the requirement that the sum of the products, from A to D, of each ordinate normal to the equilibrium curve multiplied by the force 0-2, 0-3, etc., there acting, shall equal zero. This require- ment means that the tangents at A and at D to the centre line shall make the same angle with each other, before and while the pull is applied. (See Greene's Graphics, Part II., Chap. VI.) It will be seen that the resultant force 0-1 at the section at A is smaller than 0-4 or 0-6, the pull at B or C. Hence considerable deviation of the resultant force from the centre line at that section is not a serious matter. The eye was formerly made with an un- necessary enlargement at A, but is now made commonly circular through more than half of its perim- eter. The edges of greatest stress are at A on the outside, B on the inside, and C on the outside. This neck should be wide. The material in front of the pin within the dotted triangular area is of no service. In the looped eye of Fig. 57, made by bending the bar around the pin, that space is empty. Experiment has shown that for strength this loop should be long, from two to two and one-half diameters of the pin. If it were not for the weld and the excess of metal on either side of the pin, such a form of eye would be a satisfactory one. If jo is the deviation of the force from the centre of section at A or B, Fig. 57, the half width being )'i and the pull 0-1 or 0-4 being P, the unit stress at the extreme edge, by § 114, will be Fig. 57 where h= width of eye on one side. It is necessary to make the pin from § to f the width of the bar, in order to develop the strength of the latter, that is, to give sufficient 136 STRUCTURAL MECHANICS. bearing or compression area back of the pin. The right section through the eye exceeds that of the bar from 33 to 50 per cent. Examples. — 1. A round bolt 1} in. in diameter carries a load of 20,000 lb. As its head is not square to its length, the centre of resist- ance is probably J in. from the axis of the bolt. What is / in this case, and how much greater than the mean stress? /= 41,500 lb. Since the elastic limit has been passed the actual maximum stress is probably less. 2. Find the maximum stress and the mean stress for a pull of 20,000 lb. on a square eye-bar 1JX1J in. if the pin is J in. out of cen- tre. 26,670 lb. 3. A rectangular bar, section 2X1 in., has a central pull of 8,000 lb. Then /= 4,000 lb. If the bar is widened on one side to 3 in. with- out change of force and its point of application, what is /? 5,333 lb. CHAPTER IX. COMPRESSION PIECES— COLUMNS, POSTS, AND STRUTS. 123. Blocks in Compression. — If the height of the piece is quite small as compared with either of its transverse dimensions, and the load upon it is centrally imposed, the load or force P may reasonably be considered as uniformly distributed over the cross - section S, and the unit stress / upon each square inch of section will be given by the formula P=/5, or /=|, as is the case with any tension member when the force is centrally applied. 124. Load not Central. — So also when the action line of the resultant load cannot be considered as central, but deviates from the axis of the piece a distance y , the force P can be replaced by the same force acting in the axis and a couple or moment Py 0> which moment must be resisted at every cross-section by a uniformly varying stress, forming a resisting moment exactly like that found at a section of a beam. Compare Fig. 56, and change tension to compression. If j c is the uniform unit stress due to a central load of P, and if }b is the unit stress on the extreme fibre lying in the direction in which v is measured, the latter stress being due to the moment Pyo, 1 =JL 1 _Pyoyi_P yovi Jc s , lb j s r 2 ' 137 I3 8 STRUCTURAL MECHANICS. The greatest unit stress on the section is The load that such a piece will carry is r 2 By comparison with the formula of the preceding section it will be seen that the piece, when the load is eccentric, is weaker in the ratio The values of y^S-^I or vi^r 2 are given below for some of the common sections of columns, Vi being measured in the direction h. I y x S y x +r 2 _ fr// 3 , 6 Rectangle — hh oh — 12 A Square — hh h 2 — 12 & Circle — Jd J^ 2 — 64 d bh*-b'h* 6h(bh- b'h') Hollow rectangle. . . \h oh — b'h —^ — ,,,,„ ■ b 12 bh° — b'h' 3 Hollow circle *(<* 4 - or double the ordinate at the point of contraflexure, F. Extracting the square root of the above equation and separating the variables gives \EI dv dx P \/ 2VqV — V 2 ' \ EI . ,v ™ #= vhft versm 1 — + C. As i>=o when # = o, C'=o. P v=Vq versm [ x\\—j = v [ i -cos (x^jff). As i —cos 6 = 2 sin 2 J#, v = 2^o sm 2 ( ix>y. If, in this equation, x = \l a value of *Vz*. is obtained to be equated with the previous value, 2V§. *W. = 2^0 = 2^0 sin 2 ( }/\J-^7 ) . sin (iN £/) = i ; IN £/ * **• 4tz 2 EI 144 STRUCTURAL MECHANICS. which is commonly known as Eulefs jormula. P is the ultimate strength of the column and is independent of the deflection; that is, within certain limits P is the same, no matter what small deflection is given the column. Under a load less than P the column would straighten ; under a greater load it would fail. The equation of the curve of the column is v = 2^0 sin 2 \y x )' To find the points of contraflexure, make v=Vq. sin \-j-x)=\/\=sm 45° = sin— . •"• T = i or x = \l from either end. Hence the curve is made of four equal portions, A F, F D, D E, and E II. A column hinged, pin-ended, or free to turn at its ends, and of length represented by EF = J/, will have the same portion of stress at D that is due to bending as does a column of length AH=/, which is fixed at its ends. If, in actual cases, F is considered to be practically in the same position horizontally as before loading, it may be said that a column fixed at one end and hinged at the other, of length F H = j/, will also have the same portion of stress arising from bending. The maximum deflection will then occur at one-third of its length from the hinged end. This result has been verified by direct experiment on a full-sized steel bridge member. 130. Rankine's Column Formula. — Euler's formula can be put in the form P 4* 2 E l\ 2 ' (7) P I This curve, when plotted with -~- and — as variables, has the form of the curve B C, .Fig. 59, where the greatest mean unit COLUMNS, POSTS, AND STRUTS. 145 stress which any ideal column of a given ratio of / to r can endure is given by the ordinate to the curve. But as the yield-point marks the ultimate strength of metal in compression, for short columns the line A B, whose ordinate is the yield-point, takes the place of the curve. For long, slender columns (when l+r exceeds 200, say) Euler's formula gives results close to the ultimate strength found by loading actual columns to destruction; but for shorter columns the formula gives results much too large. It must be remembered that actual columns do not satisfy the conditions from which Euler's formula was derived. No real column is perfectly straight A \b oWUU 20000 yO P s 1,144,001 8 ),000 2 , 100OO P _ s , 40,000 Z2 C_ 28,600 r" 100 200 300 400 Fig. 59 and homogeneous, nor can the load be applied exactly along the axis, and as no two columns are exactly alike no theoretical formula can exactly fit all cases. For these reasons Euler's formula is never used in designing. When the results of actual column tests are plotted, they follow in a general way some such curve as ADC. Rankine's P p jormula, -=- = -^, is the equation of such a curve, in which i+a— p is the yield-point of the material and a is a numerical coefficient. By modifying the value of a the curve can be made to agree fairly well with the results of tests. For a short block when 146 STRUCTURAL MECHANICS. l+r approaches zero the formula becomes P-^S = p. If a is made equal to p +4- 2 E, the curve of Rankine's formula approaches Euler's for large values of l-^-r, where Euler's agrees well with the results of tests. This value of a has been used by some authorities and is believed to give conservative values. The curve of Fig. 59 was so plotted. However, the value of any column formula depends upon its agreement with the results of tests, and in practice a is generally an empirical factor. In order that P of Rankine's formula may be the load which an actual column can safely carry instead of its ultimate strength, the greatest allowable unit stress on the material, /, is to be sub- stituted for the yield-point, p. The formula then becomes P I I 2 ' 1 + fl- it is to be regarded as giving the mean unit stress on any cross- section when the stress on the extreme fibre at the point of greatest deflection is /. In designing columns the load, P, and the length, /, are usu- ally known. The form of cross-section is then assumed which fixes 5 and r. If the assumed values satisfy the equation the section chosen is satisfactory; if not, another trial must be made. The radius of gyration, r, to be used is the one perpendicular to the neutral plane, that is, it is measured in the direction in which the column is likely to deflect. Unless the column is restrained in some way it will, of course, deflect in the direction of the least radius of gyration of the cross-section. Example. — What load will a 12-in. 31 J-lb. I beam 10 ft. long carry I 2 as a column if the mean unit stress is 1 2,000 -f- ( 1+ — ^ ) ? r about 30,ooor z • tu u 1 • o a • /2 T 4,4oo axis through web=i.oi in. = 9.26 sq. in. -5= = 14,100. r z 1.02 ^^ = 0.392. ^^=8,620 lb. ^=8,620X9.26= 79,800 lb. 36,000 ^ 1.392 131. Multipliers of a. — As seen above, 2/ should theoretically be substituted for / when the column is hinged or free to turn COLUMNS, POSTS, AND STRUTS. 147 at its ends, in order to obtain the equivalent length of a column which is fixed at the ends; and for a column fixed at one end and hinged at the other -f/ should be substituted for I for the same reason. Or, more conveniently, a may be used for a column fixed at the ends and of length /; 4a for a column hinged at both ends and of length /; and -^-a for a column hinged at one end and fixed at the other, length /. Actual tests, carried, how- ever, to the extreme of bending or crippling, appear to show that a column bearing on a pin at each end is not hinged or per- fectly free to turn; hence the multipliers of a more commonly used, instead of 1, J -g 6 , an d 4, are 1, f, and 2. 132. Pin Friction. — Some regard columns as neither per- fectly fixed nor perfectly hinged, and use but one value of a for all, which might perhaps then be taken as a mean value. The moment of friction on a pin is considerable. If P is the load on a post or strut, d the diameter of the pin, and tan the coefficient of friction of the post on the pin, the moment of fric- tion at the pin will be P-J^-tan <£; and this moment, if greater than Mo = Pv , will keep the post restrained at the end, so that the tangent there to the curve remains in its original direction. As P and Vq increase, Mq will become the greater when v exceeds \& tan ; the column will then be imperfectly restrained at its ends, and the inclination will change. As the friction of motion is less than that of rest, such movement when started may be rapid. Some tested columns, showing at first the curve of Fig. 58 known as triple flexure, have suddenly sprung into a single curve and at once offered less resistance. It may be doubted whether ordinary columns under working loads ever develop a value of Vq sufficient to overcome the pin friction, unless the column is very slender and the diameter of the pin small. The custom is quite general of using the same column formula for struts with fixed and with hinged ends. 133. Straight-line Column Formula. — In engineering struc- tures columns which have a greater ratio of length to radius of gyration than 100 or 150 are very rarely used. Within this limit a straight line can be drawn which will represent the aver- age results of a series of column tests, plotted as described in § 130, as well as Rankine's formula does. As the equation of a straight line is easier to solve than the equation of a curve, the 148 STRUCTURAL MECHANICS. straight-line formula is much used. For working values it has the form P ! l in which / is the greatest allowable unit stress on the material and c is a numerical coefficient determined empirically. In this country both Rankine's and the straight-line formulas are exten- sively used in designing columns. Examples of each will be found in Chapter X. 134. Swelled Columns. — Some posts and struts, especially such as are built up of angles connected with lacing, are swelled or made of greater depth in the middle. If the strut is perfectly free to turn at the ends, such increase in the value of r 2 may be quite effective, and r 2 for the middle section may be used in determining the value of P, provided the latter does not too closely approach the uniform compression value at the narrower ends. But if the strut is fixed at the ends, or is attached by a pin whose diameter is large enough to make a considerable friction- moment, such enlargement at the middle is useless ; for an equally large value of r 2 ought to be found at the ends also. Hence swelled columns and struts are but little used. 135. Column Eccentrically Loaded. — When the load is applied eccen- trically to a long column, the maxi- mum unit stress found in the extreme fibre on the concave side must be due to three combined effects : 1st. The stress due to the load P, or p = P+S. (Fig. 60.) 2d. The stress due to the resisting moment set up by Pyo. 3d. The stress due to the resisting moment set up by Pvo. yoyi Fig. 60 From § 124, f = j[i+- r r) = Po + Po P I l 2 \ I 2 From § 130, / = £ ^i+a^j =po + poay 2 . COLUMNS, POSTS, AND STRUTS. 149 If then the column is long and the line of action of P deviates from the original axis of the column by a distance y , the three expressions, po, poyoyi+r 2 , and p al 2 +r 2 should be added, giving I 2 , Wi\ P / i = po[i+a-i+- r r). s i 2 , yoyi i+a- y— y- This formula may be put in a form more convenient for designing : P , Pyoyi S= j T /r I 2 The first term gives the area of cross-section necessary to resist the direct thrust and the second term the area to resist the bending moment due to eccentric loading. Since y will determine the direction of flexure, r must be taken in this case in the direction yo\ that is, the moment of inertia and r must be obtained about the axis through the centre of gravity and lying in the plane of the section, perpendicular to y . That the moment Py , although small, has a decided weaken- ing effect on a column is proved by experiment, and its unintended presence may explain some anomalies in tests. 136. Transverse Force on a Column. — The resisting power of a column or strut to which a transverse force is applied in ad- dition to the load in the direction of the axis, and the proper dimensions of the strut, are involved in some doubt. Theo- retically, the formula is deduced as follows: From the formula for the resisting moment of a beam, M = }I -z-yi, the stress on the outer layer from such bending moment is My\ -v-J. Hence if M is that particular value of the bending moment (from the transverse load or force) which exists at the section of maximum strut deflection, where the column stress is greatest (that is, at the middle for a column with hinged ends, but perhaps at the ends for a column with fixed ends, since M may then be greater at the ends), the maximum unit stress on the concave side 150 STRUCTURAL MECHANICS. P\ My ] S ' P S = p < My j ' jr 2 * P If the straight-line formula is used instead of Rankine's, the last equation has the form P Myi S = r ^ • Or, again, it may be said that at the section in question P is moved laterally by the moment M a distance y = M+P. Then, by § 135, l = / 5 P Myi which reduces to the forms given above. The value of r 2 to be used in these formulas is that for flexure in the plane of M. Example. — A deck railway-bridge with 20-ft. panels has ties laid directly on the top chord, thus imposing a load of 2,500 lb. per foot. If the direct thrust is 249,000 lb., what should be the size of chord for a working stress on columns of 10,000— 45/^- r and on beams of 9,000 lb. per sq. in. ? Try a section composed of two 20-in. 65-lb. I beams and one 24Xi-in. cover-plate. 5=50.16 sq. in. The centre of gravity of the section lies 2.45 in. above middle of I beams, yi to upper fibre = 8.05 in. 7=3,298 in. 4 r about horizontal axis=8.n in. M= J -2, 500- 20-20-12= 1, 500,000 in.-lb. Working column stress = 10,000 -45X240-^8.10=8,670 lb. per sq. in. 249,000^-8,670=28.7 sq. in. 1,^00,000X8.05 . . A required for column action. J Qt ^q, t = 2 °-4 sq. in. required n 9,000X0.11X0.11 for beam action. 28.7 + 20.4=49.1 sq. in., total. Assumed section is sufficient. COLUMNS, POSTS, AND STRUTS. 15 1 137. Lacing-bars. — The parts of built-up posts are usually connected with lacing straps or bars. These bars carry the shear due to the bending moment arising from the tendency of the post to bend and should be able to stand the tension and compression induced by the shear. At the ends and where other menibers are connected, in order to insure a distribution of load over both members, batten or connection plates at least as long as the transverse distance between rivet rows are used in good practice. Each piece should be of equal strength throughout all its details. A post or strut composed of two channels connected by lacing-bars and tie-plates is proportioned for a certain load, the mean unit stress being reduced in accordance with the formula in which the variable is the ratio of the length to the least radius of gyration of the whole section. In the lengths between the lacing-bars, this ratio for one channel with its own least radius should not be greater than for the entire post. Nor should the flange of the channel have any greater tendency to buckle than should one channel by itself. The same thing applies to the ends of posts, where flanges are sometimes cut away to admit other members. Quite a large bending moment may be thrown on such ends, when the plane of a lateral system of bracing does not pass through the pins or points of connection of the main truss system. The usual bridge specification reads : Single lacing-bars shall have a thickness of not less than ^, and double bars, connected by a rivet at their intersection, of not less than -fo of the distance between the rivets connecting them to the member. Their width shall be: For 15-in. channels, or built sections with 3 J- and 4-in. angles, 2 J in., with f-in. rivets; for 12- or 10-in. channels, or built sections with 3-in. angles, 2j-in., with f-in. rivets; for 9- or 8-in. channels, or built sections with 2j-in. angles, 2 in., with f-in. rivets. All segments of compression members, connected by lacing only, shall have ties or batten plates placed as near the ends as practicable. These plates shall have a length of not less than the greatest depth or width of the member and shall have a thick- ness of not less than j-q of the distance between the rivets con- necting them to the compression member. Examples. — 1. A single angle-iron 6X4Xf in. and 6 ft. 8 in. long is in compression. Use r=o.g or obtain it from § 76. 5=3.61 sq. in. If P + S= 12,000 — 34/4- r, what will it carry? 32,400 lb. 2. A square timber post 16 ft. long is expected to support 80,000 lb. IS 2 STRUCTURAL MECHANICS. If /=i,ooo and the subtractive term is 2/4-r, what is the size? 10X10 in. 3. What is the safe load on a hollow cylindrical cast-iron column 13 ft. 6 in. long, 6 in. external diameter, and 1 in. thick, if it has a broad, flat base, but is not restrained at its upper end? (J= 9,000 lb., E= 17,000,000, S= 15.7 sq. in.) 124,000 lb. 4. If a short wooden post 12 in. square carries 28,800 lb. load, and the centre of pressure is 3 in. perpendicularly from the middle of one edge, what will be the maximum and the mean unit pressure, and the maximum unit tension, if any? 500 lb.; 200 lb.; —100 lb. CHAPTER X. SAFE WORKING STRESSES. 138. Endurance of Metals Under Stress. — It is important to determine how long a piece may be expected to endure stress when constant, when repeated, or when varied and perhaps reversed; and it is still more important to find what working stresses may be allowed upon a given material in order that rupture by the stresses may be postponed indefinitely. The experiments carried on by Wohler and Spangenberg, and afterwards continued by Bauschinger, show the action of iron and steel under repeated stress. Tests were made on speci- mens in tension, bending, and torsion. * A number of bars of wrought iron and steel were subjected repeatedly to a load which varied between zero and an amount somewhat less than the ulti- mate strength. When the bar broke under the load a similar bar was tested under a reduced load and was found to resist a greater number of applications before rupture. Finally a load was reached which did not cause rupture after many million repetitions. The stress caused by such load was taken as the safe strength of the material and was called the primitive saje strength The following table is an illustration : WOHLER' S TESTS OF PHCENIX IRON IN TENSION. Stress varying between Number of Applications. o and 46,500 lb. per sq. in 800 o 43,000 IO/,000 o " 39,000 " " " " 341,000 o " 35,000 " " il " 481,000 O " 31,000 " " " " IO,I 2,000 * For a description of machines and tests, see Unwin, Testing of Materials of Construction. 153 154 STRUCTURAL MECHANICS. As a result of Wohler's and Spangenberg's experiments the latter states that alternations of stress may take place between the following limits of tensile (— ) and compressive ( + ) stress in pounds per square inch with equal security against rupture: PHCENIX IRON AXLE. — 15,500 +15,500 Difference 31,000 — 29,000 o " 29,000 -43,000 -23,500 " i9o°o KRUPP CAST-STEEL AXLE. — 27,000 +27,000 Difference 54,000 —47,000 o " 47,000 — 78,000 —34,000 44,000 UNHARDENED SPRING STEEL. Difference 48,500 44,000 39,000 29,000 —48,500 o — 68,000 —24,000 — 78,000 —39,000 -87,500 -58,500 The figures are approximate. The results illustrate Wohler's law, that rupture oj material may be caused by repeated alternations 0) stress none 0) which attains the absolute breaking limit. The greater the range oj stress the smaller the stress which will cause rupture. The term jatigue oj metals is often applied to the phenomena just described. 139. Bauschinger's Experiments. — Professor Bauschinger, besides making tests similar to Wohler's, determined the effect which repeated stresses had upon the position of the elastic limit and the yield -point. (§13, 14, and 15.) He derived the following results from tensile tests: Wrought-iron plate. . . . Bessemer soft-steel plate, Iron, flat Iron, flat Thomas steel axle , Primitive Safe Stress, u. 28,000 34,000 31,000 34.QCO 43,000 Original Elastic Limit. 14,800 33,900 25,700 3 2 >3°° 38,100 Original Yield- I 29,700 41,800 32,600 35.100 Ultimate Strength, t. 54,600 62,000 57,600 57,200 87,000 SAFE WORKING STRESSES 155 From an inspection of the table it is seen thai the primitive safe stress on sound bars of iron and sled for tensions alternating from zero to that stress is a little less than the original yield-point of the metal. Although the safe stress was found to be greater than the original elastic limit, yet when bars which remained unbroken after several million applications of stress were tested statically it was found that the clastic limit had risen above the stress applied. The inference is that the primitive safe stress is below the elastic limit although the clastic limit may be an artificially raised one. Pulling a bar with loads above the original elastic limit and below the yield-point raises the elastic limit in tension but lowers it in compression and vice versa. These artificially produced elastic limits are very unstable. When a bar is subjected to a few alternations of equal and opposite stresses which are equal to or somewhat exceed the original elastic limits, the latter tend toward fixed positions (ailed by Ba use hinder natural clastic limits. He advanced the view that the original elastic limits of rolled steel and iron are artificially raised by the stresses set up during manufacture. These natural clastic limits seem to correspond with Wohler's range of stress for unlimited alternating stresses. 140. Alteration of Structure.- If a bar has been subjected to repetitions oi stress somewhat below the yield -point, any genera] weakening or deterioration in the quality of the bar ought to show itself in some way; but bars which have endured millions of repeated stresses or bars which have broken under repeated stresses give no indication of any weakening when tested statically. Instead of being weakened by repeated stresses well below the primitive safe strength, a bar is really improved in tenacity. Bars fractured in the Wohler machines did not draw out, no matter how ductile the material mighl be, but broke as if the material were brittle. From these facts wc arc forced to conclude that whatever weakness there may be is confined to the surface of rupture. Such breaks appear to be detail breaks. Continuity is first destroyed at a Haw or overstressed spot, and from that point the fracture spreads gradually until the section 156 STRUCTURAL MECHANICS. is so weakened that the load is eccentric and bending stresses .are set up which cause the piece to break by flexure. From the results of Wohler's and Bauschinger's tests it would appear that steel might undergo an indefinite number of repe- titions of stress within the elastic limit without rupture, or that within that limit the elasticity w r as perfect. However, certain experiments on the thermoelectric and magnetic properties of iron under stress and on the molecular friction of torsional pen- dulums give results inconsistent with the theory of perfect elas- ticity. Engineers differ more or less in their interpretations of the experimental results but agree in using lower working stresses for varying than for static loads. 141. Launhardt- Weyrauch Formula. — A number of formulas based on Wohler's experiments have been advanced for the determination of the allowable unit stress on iron and steel when the range, as well as the magnitude, of the stress is considered. Launhardt proposed the following formula for the breaking load of a member which is subjected to stresses varying between a maximum and a minimum stress of the same kind: a=u\ 1 t—u min. stress\ u max. stress )• a = breaking load under the given conditions. u — primitive safe stress. t = ultimate strength, static. v = greatest stress piece will bear if repeated from +v to — v indefinitely. Weyrauch extended Launhardt's formula to cover cases of alternate tension and compression, in which case the minimum stress is to be considered as negative. / u—v min. stress\ \ u max. stress/ As the results of the different experiments vary more or less, the different authorities do not agree as to the values of /, u, and v to be substituted in the formulas. In some of Wohler's experiments u appears to be greater than J/ and to approxi- mate to §/. This value is assumed by Weyrauch and gives (t—u)+u=i. Likewise if v = hi the coefficient of the second formula becomes h and both formulas reduce to SAFE WORKING STRESSES. 1 57 / , min. stressN a=u[i+l ), \ max. stress/ the sign of the second term changing for reversed stresses. The choice of values of u and v to use in the equations seems to have been determined largely by the result sought. The formulas are of rather doubtful value, but they are more or less used for determining working stresses in steel bridges. The following relationships of /, u, and v are given by some authorities, and they conform more nearly with the results of the experiments quoted in this chapter: Steady load a = t Load varying from o to u a =u ■ = \t " " " +i/to-i' a=v = \t 142. Reduction of Unit Stresses. — The safe working stress to be used for any material will depend upon several considerations : Whether the structure is to be temporary or permanent; whether the load is stationary or variable and moving ; if moving, whether its application is accompanied by severe dynamic shock and perhaps pounding; whether the load assumed for calculation is the absolute maximum; whether such maximum is applied rarely or is likely to occur frequently; whether the stresses obtained are exact or approximate; whether there are likely to be secondary stresses due to moments arising from changes of the assumed frame; what the importance of the piece is in the structure, and the possible damage that might be caused by its failure. The allowable unit stresses of different kinds, and for greater or less change of load, will be further reduced to provide against: Distribution of stress on any cross-section somewhat different from that assumed; variations in quality of material; imperfec- tions of workmanship, causing unequal distribution of stress; scantness of dimensions; corrosion, wear, or other deterioration from lapse of time or neglect; lack of exactness of calculation. The allowable unit stresses so determined will be but a small fraction of the ultimate or breaking strength of the material; and it is evident that the idea that it will require several times the allowable maximum working load to cause a structure to fail is seriously in error. 158 STRUCTURAL MECHANICS. Overconfidence of the in experienced designer in the correct- ness of his design may be checked by a study of this section. 143. Load and Impact. — The design should be completely carried out, both in the principal parts and in the details. The latter require the most careful study, that they may be at once effective, simple, and practical.* All the exterior forces which may possibly act upon a structure should be considered, and due provision should be made for resisting them. The static load, the live load, pressure from wind and snow, vibration, shock, and centrifugal force should be provided for, as should also deteriora- tion from time, neglect, or probable abuse. A truss over a machine- shop may at some time be used for supporting a heavy weight by block and tackle, or a line of shafting may be added; a small surplus of material in the original design will then prove of value. Light, slender members in a bridge truss, while theoretically able to resist the load shown by the strain sheet, are of small value in time of accident. The tendency from year to year is towards heavier construction. Secondary stresses, as they are called, are due — first, to the moments at intersections or joints, when the axes of the mem- bers coming together at a connection do not intersect at a common point ; and second, to the moments set up at joints by the resistance to rotation experienced by the several parts when the frame or truss is deflected by a moving load. If symmetrical sections are used for the members, if the connecting rivets are symmetrically placed, and if the axes of the intersecting members meet at one point, secondary stresses will be much reduced. All members of a structure should be of equal strengh, and the connections should develop the full strength of the body of the members connected. The connections should be as direct as possible. When a live load is joined to a static load, the judgment of the designer, or of the one who prepares the specifi- cations for the designer, must be exercised. A warehouse lloor to be loaded with a certain class of goods has maximum stresses from a static load. The floor of a drill-room, ball-room, or highway bridge receives maximum loading from a crowd of people * A portion of these paragraphs is extracted from a lecture by Mr. C. C. Schneider. SAFE WORKING STRESSES. 1 59 the possible density of which can be ascertained. But if these masses of people keep moving, and more particularly if they keep step, the effort of their weight will be increased by the vibrations resulting therefrom. This action is generally called impact. In the case of a building, the floor-joists, receiving the impact directly, will be most affected; the girders which carry the joists will be less affected, and the columns which support the girders will receive a smaller percentage of the impact, the proportionate effect growing less as the number of stories below the given floor increases. In the absence of trustworthy data from which to determine this impact, it is left to the judgment of the engineer to increase the live load by a certain percentage, or to decrease the allowable unit stress, for each case, to provide for the effect, as will be seen in the values given later. For economy, make designs which will simplify the shop work, reduce the cost, and insure ease of fitting and erection. Avoid an excess of blacksmith work and much use of bent pieces. 144. Dead and Live Load. — From what has been said it is readily seen that a moving or live load has a much more serious effect on a truss than a static or dead load, and in the case of a railroad bridge the stresses in the members due to a rapidly moving train are much greater than the stresses would be under such a train at rest. In designing the members of steel bridges this increase of stress is usually provided for by one of three ways : Using different working stresses for dead and live load, as a unit stress of 18,000 lb. per sq. in. for dead load and 9,000 lb. for live. When this method is employed the ratio is commonly 2 to 1. Using a varying unit stress found from a modified form of the (min. stressx 1 4 — — '■ — 7 ) . If max. stress/ the stresses reverse the fraction is negative. The coefficient \ in the original formula is dropped arbitrarily. Adding a certain percentage of the live load, called impact, to the sum of the dead- and live-load stresses, and using a constant working stress as 18,000 lb., one impact formula often used is Impact = \ 7- X live-load stress. By "length " is meant r 300 + length J ° the distance in feet through which the load must pass to produce maximum stress in the member in question. Such experiments as have been made on bridges under moving load indicate that i6o STRUCTURAL MECHANICS. the impact is not as great as the formula gives. The third method is sometimes combined with the first or second, but with the use of a different impact formula. While these three ways of designing are quite different in theory and method, the resulting sizes of members are much the same. 145, Working Stresses for Timber. — The following table is from Bulletin No. 12, U. S. Dept. of Agriculture, Div. of Forestry: Safe Unit Stresses at 18% Moisture. Structures freely exposed to the weather. Compression. Bending. Shear. With Grain. Across Grain. Extreme iubre. With Grain. Tension. l,O0O 84c 700 760 880 650 800 215 215 15° 140 170 115 400 1,550 1,300 880 1,090 1,320 1,200 125 IOO 75 200 12,000 9,000 7,000 Redwood White oak IO OOO Factor of safety 5 3 5 4 I The values given were found by taking the average of the lowest 10 per cent, of the results of tests on non-defective timber and dividing by the factor of safety given in the last row. As timber never fails in tension the safe tensile stress is not given; the last column shows the ultimate strength. For structures protected from the weather the compressive and bending values may be increased 20 per cent. The shearing value should not be increased. The table was constructed for designing railway trestles. The following safe working stresses have been recommended by the Committee on Strength of Bridge and Trestle Timbers, Am. Assn. of Ry. Supts. of Bridges and Buildings : SAFE WORKING STRESSES. 161 Tension. Compression. Bend- ing. Shear. With Grain. Across Grain. With Grain. Across Grain. Ex- treme Fibre. With Grain. End Bear- ing. Cols. under is Diams. Across Grain. Vv hite pine 1,200 700 Sco 1,200 Soo 6c 7co i,oco 60 5° 5° 1,600 1,100 1,200 1,600 1,200 1,000 700 800 1,200 800 800 Sco 900 35° 200 2CO SOO 2CO I50 2CO "CO 1200 700 700 1,100 700 600 75° 1,000 15° ICO 150 IOO IOO IOO 200 1,250 rnn Norway pine Spruce and Eastern fir. Hemlock 750 Redwood White oak 2CO 1,400 I, COO 10 IO 5 5 4 6 4 4 Messrs. Kidwell and Moore deduced the following column formulas based on the values given in the preceding table. Least cross-dimension of column = h; length =/. l + h<2$ /-=-/*> 25 Long-leaf pine 1,000— 8.5/ -z-h 1,000— 11/ + h White pine 700—6. l-^-h 700— Sl-^-h Norway pine, spruce, and Eastern fir 800— 6.5/ s-h 800— 8/ -r-h Douglas spruce 1,200—9. l^-h 1,200— 12/ -i-h Hemlock 800 — 7.6/-^ /z- 800—9. 5/ -Wz Redwood 800 — 8.6/ s-h 800 — iil+h Oak goo—g.l+h goo— ul~Ji For buildings Mr. C. C. Schneider recommends the following unit stresses: Compression. Bending. With Grain. Across Grain. Extreme Fibre. Shear with End Bearing. Columns under 10 Diameters. White pine, spruce. .. Hemlock 1,500 1,000 800 1,200 I, COO 600 500 1,000 350 200 200 500 1,500 1,000 800 1,200 IOO IOO IOO 200 White oak 1 62 STRUCTURAL MECHANICS. P / I For columns when l+h>io, -77 = / — t, in which / is o 100 h the compressive stress for a short column taken from the table. The strength of columns built up by bolting two or more pieces of timber together is no greater than that of the individual pieces acting independently. The names of woods vary in different localities; long-leaf, short-leaf, and one or two other Southern species of pine are not distinguished in the market and are called Southern, Georgia, or yellow pine. Douglas spruce is also known as Oregon fir, spruce, or pine. 146. Working Stresses for Railroad Bridges. — Mr. Theodore Cooper's Specifications, edition of 1901, give the following unit stresses in pounds per square inch. MEDIUM STEEL IN TENSION. Main truss members — dead load 20,000 live " 10,000 Floor-beams and stringers 10,000 Lateral and sway bracing — wind load. . . . . i8,cco " " " live " 12,000 Floor-beam hangers and members liable to sudden loading 6,000 For soft steel in tension reduce the unit stresses 10 per cent. MEDIUM STEEL IN COMPRESSION. Chord segments — dead load 20,000 —90/ ■*■ r " " live " 10,000— 45/ -r- r Posts of through-bridges — dead load 1 7,000 —90/ -=-r " " " live " 8,500-45/^r Posts of deck-bridges and trestles — dead load 18,000 —80/ -=-r 11 " " " " live " . ... 9,000-40/^r Lateral and sway bracing — wind load 13,000 —6ol-±r " " " live " 8,700 -40/^r For sojt steel in compression reduce the unit stresses 15 per cent. The ratio of / to r shall not exceed 100 for main members and 120 for laterals. When a member is subjected to alternating stresses, each stress shall be increased by eight-tenths of the lesser and the member designed for that stress which gives the larger section. SAFE WORKING STRESSES. 1 63 PINS AND RIVETS. Shear 9,000 Bearing (thickness Xdiameter) 15,000 Bending (pins) 18,000 For floor rivets increase number of rivets 25%. " field " " " " " 50%. ' lateral " decrease " " " 2>3%- PEDESTALS. Pressure on rollers, pounds per lineal inch, 30oXdiam. in inches. Pressure of bedplates on masonry, pounds per square inch, 250. The specifications of the American Bridge Co., 1900, give the following working stresses: The maximum stress in a member is found by adding the dead 300 load, live load, and impact. Impact = — — -yX live-load stress, r L 300 + L in which L is length in feet of distance to be loaded to produce maximum stress in member. Soft Steel. Medium Steel. Tension 15*000 17,000 „ . ivooo 17,000 Compression. ^— —- I 2 I 2 1 + — — o" x + 2" 13,500^ iijooor- Shear on web-plates 9,000 10,000 The ratio of / to r shall not exceed 100 for main compression members and 120 for laterals. When a member is subjected to alternating stresses the total section shall be equal to the sum of the areas required for each stress. PINS AND RIVETS. Shear 11,000 12,000 Bearing (thickness Xdiameter) 22,000 24,000 Bending (pins) 22,000 25,000 For hand-driven field rivets increase number 25%. 11 power-driven " " " " 10%. 1 64 STRUCTURAL MECHANICS. PEDESTALS. Pressure on rollers, pounds per lineal inch, 12,000V diam. in inches. Pressure of bedplates on masonry, pounds per square inch, 400. 147. Working Stresses for Highway Bridges. — Mr. Cooper's Specifications, 1901, give the following stresses: MEDIUM STEEL IN TENSION. Main truss members — dead load 25,000 live 1 2, 500 Floor-beams, stringers, and riveted girders 13,000 Lateral and sway bracing — wind and live load 18,000 Floor-beam hangers and members liable to sudden loading 8,000 For sojt steel in tension reduce the unit stresses 10 per cent. MEDIUM STEEL IN COMPRESSION. Chord segments — dead load 24,000 — 1 10/ -f-r " " live " 12,500-- 55/^-r Posts of through -bridges — dead load 20,000 — 90/ -±r " " " live " 10,000- 45/-^ Posts of deck-bridges — dead load 22,000 — 80/ -r-r " " " live " 11,000- 40/ ^r Lateral and sway bracing — wind load 13,000 — 60/ -*-r " " * " live " 8,700- 40/^r For soft steel in compression reduce the unit stresses 15 per cent. The ratio of / to r shall not exceed 100 for main members and 120 for laterals. When a member is subjected to alternating stresses, each stress shall be increased by eight-tenths of the lesser and the member designed for that stress which gives the larger section. PINS AND RIVETS. Shear 10,000 Bearing (diameter X thickness) 18,000 Bending (pins) 20,000 For floor rivets increase number 25%. " field " " " 50%. " lateral " decrease " 3°%» SAFE WORKING STRESSES. 165 PEDESTALS Pressure on rollers, pounds per lineal inch, 3ooXdiam. in inches. Pressure of bed-plates on masonry, pounds per square inch, 250. TIMBER FLOOR-JOISTS. Fibre stress on yellow pine or white oak 1,200 ' ' white pine or spruce 1,000 1 1 t i The working stresses given by the American Bridge Co. Specifications for Highway Bridges, 1901, are the same as for Railway Bridges, with the following exceptions: The sum of dead and live load is to be increased by 25 per cent, of the live load to compensate for the effect of impact and vibration. The ratio of / to r shall not exceed 120 for main members and 140 for laterals. When a member is subjected to alternating stresses, design for that stress which gives the larger section. Fibre stresses on floor-joists same as in preceding specification. 148. Working Stresses for Buildings. — A set of building specifi- cations by Mr. C. C. Schneider may be found in the Transactions of the American Society of Civil Engineers, Vol. LIV, June, 1905. Some of the working stresses recommended are abstracted below. Structural steel of ultimate strength between 55,000 and 65,000 lb. per sq. in. is to be used. Tension 16,000 Compression 16,000 — 70/ -s-r Shear on web plates (gross section) 10,000 For wind bracing and the combined stresses due to wind and the other loading the permissible stresses may be increased 25 per cent. The ratio of / to r shall not exceed 125 for main compression members and 150 for wind bracing. Members subjected to alternating stresses shall be propor- tioned for the stress giving the larger section. 1 66 STRUCTURAL MECHANICS. PINS AND RIVETS. Shear on pins and rivets 1 2,000 " " bolts 9,000 Bearing on pins and rivets 24,000 " bolts 18,000 Bending on pins 24,000 For field rivets increase number 33 per cent. WALL PLATES AND PEDESTALS. Bearing pressures in pounds per square inch on Brickwork, cement mortar 200 Rubble masonry, cement mortar 200 Portland cement concrete 350 First-class masonry, sandstone 400 " " " granite 600 MASONRY. Permissible pressure in masonry, tons per square foot : Common brick, natural cement mortar 10 " ' * Portland cement mortar 12 Hard-burned brick, Portland cement mortar 15 Rubble masonry, Portland cement mortar 10 Coursed rubble " " ' ' 12 Portland cement concrete, 1-2-5 2 ° 149. Machinery, etc. — The designing of machinery has not been systematized as has the designing of structural steel work, and the choice of working stresses is to a large extent a matter of individual judgment. The following values are taken from a table in Unwin's Machine Design and are for unvarying stress; if the stress varies, but does not reverse, multiply by two-thirds; if the stress reverses, multiply by one-third. Tension. Cast iron 4,200 Iron forging 15,000 Mild steel 20,000 Cast steel 24,000 For shafting of marine engines 9,000 lb. per sq. in., on wrought iron and 12,000 lb. on steel is commonly used, the shaft being designed for the maximum and not the mean twisting moment. These values are about those used in other machinery. The working stress in boilers made of 60,000 lb. steel may be as high as 13,000 lb. per sq. in. with first-class workmanship, but ordinarily 10,000 to 11,000 lb. is employed. Dmpression. I2,6oO Torsion 2,IOO 15,000 6,000 20,000 8,000 24,000 10,000 CHAPTER XI. INTERNAL STRESS: CHANGE OF FORM. 150. Introduction. — Let any body to which external forces are applied be cut by a plane of section. The force with which one part of the body acts against the other part of the body is the stress on the plane of section. This stress is distributed over the section and may be uniform or may vary. Its intensity, that is, the unit stress, at any point is found by dividing the stress on an indefinitely small area surrounding the point by the area. Stresses may be normal, tangential or oblique to the surface on which they act, and they are completely determined only when both intensity and direction are known. It is desirable to know the magnitude and kind of the unit stress at each point in order to be sure that the material can safely resist it; or to determine the required cross-section to reduce the existing stresses to safe values. A unit stress is expressed as a certain number of pounds of tension, compression, or shear on a square inch of section. If the plane of section is changed in direction, the force on the section may be changed and the area of section may also be changed so that the unit stress on the new section is altered from that on ihe old in two ways. Stresses per square inch, or unit stresses, therefore cannot be resolved and compounded as can forces, unless they happen to act on the same plane. Generally, each unit stress may be multiplied by the area over which it acts, and the several forces so obtained may be compounded or resolved as desired ; the final force or forces divided by the areas on which they act will give the desired unit stresses. Where the stress on a plane varies from point to point, as does the direct stress on the right section of the team, and as does the shearing stress also in the same case, the investigation is supposed to be confined to so small a portion of the body that the stress over any plane may be considered to be sensibly constant. 167 i68 STRUCTURAL MECHANICS 151. Stress on a Section Oblique to a Given Force. — Suppose a short column or bar to carry a force of direct compression or tension, of magnitude P, cen- trally applied and uniformly distributed over the cross- section, 5. The unit stress on and perpendicular to the right section will be pi = P+S. On an oblique section CD, Fig. 2, making an angle with the right section A B, the unit stress will be P+S sec 6= pi cos d, making an angle of with the normal to the oblique section on which it acts. If this oblique unit stress is resolved normally and tangentially to C D, the Normal unit stress = # n = ^i cos 6 -cos 6 = pi cos 2 Q\ Tangential do. = q —p\ cos 6 sin 0. The normal unit stress on the oblique plane is of the same kind as P, tension or compression; the tangential unit stress, or shear, tends to make one part of the prism slide down and the other part slide up the plane. The largest normal unit stress for different planes is found when 6 = 0, which defines the fracturing plane for tension; the minimum normal unit stress occurs for # = 90°; and the greatest unit shear is found for # = 45°, when we have q max . = 2pi- 152. Combined Stresses. — The action line of P may be taken for the axis of X. Two equal and opposite forces, pull or thrust, may then be applied along the axis of Y, and the normal and tangential unit stresses found on the plane just discussed; and similarly for the direction Z. The normal unit stresses, since they act on the same area, may then be added algebraically, and the shearing stresses may be combined; finally a resultant ob- lique unit stress may be found on the given plane. Thus, if pi is the unit stress on a section normal to the X axis, and p2 the unit stress on a section normal to the Y axis, INTERNAL STRESS: CHANGE OF FORM. 169 pn = pi cos 2 d + p 2 cos 2 (qo°-#) = p x cos 2 d + p 2 sin 2 0, q =pi sin cos — p 2 cos sin = (pi —pz) sin (9 cos 0. _/> n is a maximum when # = or 90 , and q is a maximum when 0=45°. Pnmax=P\ Or p 2 ', 1max. = i(Pl-p2)' A more convenient method will, however, be developed and used in the following sections. As most of the forces which act on engineering structures lie in one plane or parallel planes, such cases chiefly will be considered. 153. Unit Shears on Planes at Right Angles. — If, in the pre- ceding illustration, the unit stresses, both normal and tangential, are found on another plane N which makes an angle 0' = 90°— with the right section, there will result p n ' = pi cos 2 e r =p x sin 2 0; q f = q. Hence, on a pair of planes of section, both of which are at right angles to the plane of external forces and to each other, the unit shears are 0] equal magnitude, and, since p n + pn=piy the unit normal stresses are together equal to the original normal unit stress. It is further evident that one normal unit stress p,/ may be found by subtraction as soon as the other is known, and that ordinary resolution on these two planes of the original unit stress would be erroneous. 154. Unit Shears on Planes at Right Angles Always Equal. — Since, as before stated, other forces, in other directions, may be simultaneously applied to the given body, and their effects found on the same two planes, it follows that, in any body under stress, equal unit shearing stresses will exist on two planes each of which is perpendicular to the direction of the shear on the other. Example. — A closed cylindrical re- Fig. 61 ceiver, \ in. thick, has a spiral riveted joint making an angle of 30 with the axis of the cylinder, and 170 STRUCTURAL MECHANICS. a portion 2 in.X4 in. of the cylinder, Fig. 61, has the given tensions of 2,500 lb. acting upon it. Then ^ = 2,500 4- 2-^=5,000 lb. per sq. in., and />2= 2,500-7-4- J= 2,500 lb. per sq. in. p n = 5,ooo-f+2, 5oo-J=4,375 lb. per sq. in., 9=5,000-0.433-2,500-0.433 = 1,082 lb. per sq. in., />=\/(4>375 2 + I > 82 2 ) = 4 J 5 7 lb. per sq. in., or 4,507-^=1,127 lb. per linear inch of joint, which value will deter- mine the necessary pitch of the rivets for strength. The stress on a joint at right angles to the above can be similarly found. An easier process will be given in § 161. 155. Compound Stress is the internal state of stress in a body caused by the combined action of two or more simple stresses (or balanced sets of external forces) in different directions, as in the above example. The investigations which follow are those of compound stress, but they will, as above stated, be chiefly confined to stresses in or parallel to one plane. 156. Conjugate Stresses: Principal Stresses. — If the stress on a given plane in a body is in a given direction, the stress on a plane parallel to that direction must be parallel to the first-mentioned plane. For the equal re- sultant forces exerted by the other parts of the body on the faces A B and C D of the pris- matic particle, Fig. 62, are directly opposed to one another, their common line of action trav- ersing the axis of X through O; and they are therefore independently balanced. Therefore the forces exerted by the other parts of the body on the faces A D and B C of this prism must be independently balanced and have their resulcants directly opposed; which cannot be unless their direction is parallel to the plane Y O Y. A pair of stresses, each acting on a plane parallel to the direc- tion of the other, is said to be conjugate. Their unit values are independent of each other, and they may be of the same or oppo- site kinds. If they are normal to their planes, and hence at right angles to each other, they are called principal stresses. Examples. — The unit stress found in § 154 makes an angle with the plane on which it acts whose tangent is 4,375^1,082 = 4.04. Upon a INTERNAL STRESS: CHANGE OF FORM. 171 new plane cutting the metal in this direction the stress must act in a direction parallel to the joint referred to. If a plane be conceived parallel to a side-hill surface, at a given vertical distance below the same, the pressure at all points of that plane, being due to the weight of the prism of earth above any square foot of the plane, will be vertical and uniform. Then must the pressure on a vertical plane transverse to the slope be parallel to the surface of the ground. That the pressure against the vertical plane is not hori- zontal, but inclined in the direction stated, is shown by the movement of sewer trench sheeting and braces, when the braces are not inclined up hill, but are put in horizontally. 157. Shearing Stress. — If the stresses on a pair of planes are entirely tangential to those planes, the unit shears must be equal. Consider them as acting along the faces of a small prismatic particle ABCD, which lies at O. The moment of the total shear on the two faces A B and C D must bal- ance the moment for the faces A D and B C, for equilibrium. Fig. 63 ?ABEF = ?'ADHG. But the area of ABCD, AB-EF^AD-HG; .*. q' = q. This construction shows further that a shear cannot act alone as a simple stress, but must be combined with an equal unit shear on a different plane. 158. Two Equal and Like Principal Stresses. — If a pair of principal stresses, § 156, are equal unit stresses of the same kind, pi and p2, Fig. 64, the stress on every plane is normal to that plane, and of the same kind and magnitude. 1 w p. "P% A -i> 2 " kit K Fig. 64 Let pi act in the direction O X on the plane O' B' of the prismatic particle O' A' B' which lies at O, and p 2 act in the direc- 172 STRUCTURAL MECHANICS. tion Y on the plane O' A', pi and p 2 being equal unit stresses of the same kind. Make OD=^-0'B / , the total force on O'B' , and O E=p 2 -O f A', the total force on O' A', both being positive. Complete the rectangle O D R E. Then must R O, applied to the plane A'B', be necessary to insure equilibrium of the prism O' A' B\ Hence p' = R O +A' B'. Since p 1 =p 2 , OP OE OR m ,_ O' B'~0'A' _ A'B' ; ''' p -Pi-? 2 ' Because of similarity of triangles A' O' B' and O E R, R O is perpendicular to A B, or ft to A' B', and is of the same kind as pi and p2> Example. — Fluid pressure is normal to every plane passing through a given point, and equal to the pressure per square inch on the horizontal plane traversing the point. Here manifestly the three coordinate axes of X, Y, and Z might be taken in any position, as all stresses are princi- pal stresses. 159. Two Equal and Unlike Principal Stresses. — If a pair of principal stresses are equal unit stresses of opposite kinds, as pi and - p2, Fig. 64, the unit stress on every plane will be the same in magnitude, but the angle which its direction makes with the normal to its plane will be bisected by the axis of principal stress, and its kind will agree with that of the principal stress to which it lies the nearer. In this case lay off Oe in the negative direction, to represent —p2-0" A"; construct the rectangle O Dre, and draw rO which will be the required force distributed over A" B" to balance the forces O" B" and O" A". This force rO will be the same in magnitude as R O, making ft = pi = p2 and rO will make the Bame angle with O X or O Y as R O does. As R O lies on the normal to A B, and O X bisects R Or, the statement as to position is proved. The stress ft agrees in kind with that one of the principal stresses to which its direction is nearer. 160. Two Shears at Right Angles Equivalent to an Equal Pull and Thrust.— If the plane A" B" is at an angle of 45 with O X, rO will coincide with A B and becomes a shear. Therefore two INTERNAL STRESS: CHANGE OF FORM. 173 equal unit stresses of opposite kinds, that is a pull and a thrust and normal to planes at right angles to one another, are equivalent to, and give rise to equal unit shears on planes making 45 with the first planes and hence at right angles to each other; and vice versa. \ < Example. — If, at a point in the web of a plate girder, Fig. 65, there is a unit shear, and nothing but shear, on a vertical plane, of 4,000 ^~^~ ~~~^ 65 lb., there must be a unit shear of 4,000 lb., and nothing but shear, on the horizontal plane at that point and on the two planes inclined at 45 to the vertical through the same point there will be, on one, only a normal unit tension of 4,000 lb., and on the other an equal normal unit thrust. 161. Stress on any Plane, when the Principal Stresses are Given. — This general problem is solved by dividing it into two special cases ; the one that of two equal and like principal stresses, the other that of two equal and unlike principal stresses. Let the two principal unit stresses be p\=0 D, and ^ = OF, of any magnitude, and of the same kind or opposite kinds. Fig. 66. The direction, magnitude and kind of the unit stress on any plane A B is desired. Let pi be the greater. Divide pi and p2 into their half sum and difference as follows : Pi = i(pi+p2)+i(pi-p2), and p2 = KPi+p2)-i(Pi~p2). The distance O C or O E will represent the half sum i(pi+p2), and C D or E F the half difference \(p\-p2)- If pi and p2 are the same sign the right-hand figure will result; if of opposite signs, the left-hand figure will be obtained. By the principle of § 158, when the two equal principal unit stresses O C and O E are considered, lay off O M on the normal to the plane whose trace is A B, for the direction and magnitude of the unit stress on A B due to ^(pi+p2)- There remain C D and E F representing +\(pi—p2) on the vertical axis, and —\{pi—p2) on the horizontal axis respectively. 174 STRUCTURAL MECHANICS. By § 159, lay off O Q, making the same angle with O X as does O M, but on the opposite side, or in the contrary direction, for the magnitude and direction of stress on plane A B due to ±h(Pi—p2)- As O M and O Q both act on the same unit of area of A B, R O, in the opposite direction to their resultant O R, will give the direction and magnitude of the unit stress on A' B' to balance pi on O' B' and p 2 on O' A'. In the figure on the right R O is positive, or compression. If, in the figure on the left, where pi is + and p 2 is -, R O falls so far to the left as to come on the other side of A B, it will agree with p 2 , and Fig. 66 be negative or tension. If A B is taken much more steeply inclined, such will be the case. The small prisms illustrate the constructions. If R O falls on A B, it will be shear. Some constructions for different inclinations of plane A B will be help- ful to an understanding of the matter. A much abbreviated construction is as follows: — Strike a semicircle from M on the normal, with a radius M = %(pi +P2), and draw M R through the points where the semicircle cuts the axes of pi and p 2 . The angle N M R is thus double the angle MOD, since it is an exterior angle at the vertex of an isosceles triangle. Lay off MR = |(/i-^ 2 ) in the direction of the axis of greatest stress, and R O will be the desired unit stress on A B. If p 2 is opposite in kind to pi, M R will be greater than M O, and R will go beyond P. INTERNAL STRESS: CHANGE OF FORM. 175 162. Ellipse of Stress. — For different planes A B through O, pi and p2 being given, the locus of M is a circle of radius %(pi+p2), and the locus of R is an ellipse (as will be proved below), with major and minor semidiameters pi and p2- Hence it is seen why pi and p2, normal to the respective planes, and at right angles, are called principal stresses. If three principal stresses, coinciding in direction with the rectangular axes of X, Y and Z, simultaneously act on a given point, an ellipsoid constructed on them as semidiameters will limit and determine all possible stresses on the various planes which can be passed through that point in the body. That the locus of R is an ellipse may be proved as follows : Drop a perpendicular R S from R on O X. P M O and O M G are isosceles triangles. 2 sm#, S = GRcosMPO=^cos^ O R = ^ r =\/(S0 2 + RS)=v / (^i 2 cos 2 ^ + ^2 2 sin 2 ^), . (1) which is the value of the radius vector of an ellipse in terms of the eccentric angle, the origin being at the centre. As If one of these normals were revolved around O to coincide with the other, the point M' would fall on M, but M' R' would diverge from M R, while equal in length to it. Hence, when p and p r are the given quantities, let A B, Fig. 68, represent the trace of the plane on which p acts, O N the Fig. 67 normal to that plane, and O R the unit stress p in magnitude and actual direction of action on A B. OR represents either tension or compression, as the case may be. Now let the plane on which ft acts, together with its normal and ft itself in its relative position, be revolved about O until it coincides with A B. Its normal will fall on O N and ft will be found at O R', on one side or the other of O N, if it is of the same kind with p; or it is to be laid off on the dotted line below, if of the opposite sign. In other words, lay off ft from O, at the same angle with O N which it makes with the normal to its own plane. It is well, for accuracy of construction, to draw it on the same side of the normal as p, the result being the same as if it were drawn on the other side. (The change from one side of the normal to the other simply consists in using the corresponding line on the 1/8 STRUCTURAL MECHANICS . other side of the main axis of the ellipse of stress). Thus is found O R' or -O R' as the case may be. Draw R R' and, since R M R' must be an isosceles triangle, bisect R R' at T and drop a perpendicular to R R' from T on to the normal, cutting the latter at M. Then since, as previously pointed out, O M = HP1+P2) and MR = MR' = }(^i-^ 2 ) — with M as a centre, and radius M R, describe a semicirc e ; OX will be pi and O S will be po. Since p is in its true position, and the angle N M R = 2 M O D of Fig. 66 or 2 M O X of Fig. 67, the direction of the axis X along which p\ acts will bisect N M R, and the axis along which p2 acts will be perpendicular to axis X. They may be now at once drawn through O, if desired. 165. From any Two Stresses to Find Other Stresses. — From the preceding construction, § 164, the stress on any other plane may now be found. All possible values of p consistent with the two, O R and O R', first given, will terminate, in Fig. 68, on the semicircle just drawn, as at R", and the greatest possible ob- liquity to the normal to any plane through O will be found by drawing from O a tangent to this semicircle. 166. When Shearing Planes are Possible. — In case the lower end of the semicircle cuts below O, Fig. 69, p\ and p2 are of opposite signs, all obliquities of stress are possible, and the dis- tance from O to the point where the semicircle cuts A B, being perpendicular to the normal O X, gives the unit shear on the shearing planes. If p\ and p2 are drawn through O in position, and the ellipse of stress is then constructed on them as semi- diameters (as can be readily done by drawing two concentric circles with pi and p2 respectively as radii, and projecting at right angles, parallel to pi and p2, to an intersection, the two points where any radius cuts both circles), an arc described from O, with a radius equal to this unit shear and cutting the ellipse, will locate a point in the shearing plane which may then be drawn through that point and O. Two shearing planes are thus given, as was proved to be necessary, § 157. The above solution may be considered the general case. INTERNAL STRESS: CHANGE OF FORM. 179 167. From Conjugate Stresses to Find Principal Stresses. — If p and p' are conjugate stresses, it is evident, from definition, and from Fig. 62, that they are equally inclined to their respective normals. Hence O R' will fall on O R, when revolved, both O R and O R' lying above O when of the same sign, and on opposite sides of A B when of opposite signs. The rest of the construction follows as before, being somewhat simplified. It may be noted that, when p\ = ± p2, the propositions of §§ 158 and 159 are again illustrated. One who is interested in a mathematical discussion of this subject is referred to Rankine's "Applied Mechanics," where it is treated at considerable length. This graphical discussion is much simpler, less liable to error, and determines the stresses in their true places. i68„ Stresses in a Beam — The varying tensile and compres- sive stresses on any section of a beam are accompanied by vary- ing shearing stresses on that section and by equal shears on a longitudinal section. The direct or normal stresses due to bend- ing moment vary uniformly from the neutral axis either way ? § 62; the shears are most intense at the neutral axis, § 72. The normal and shearing stresses on the cross-section of a rectangular beam and also the resultant stresses on the section are shown in 0: g Pn } f ■ n Fig. 70 Fig. 70, a and b. The maximum and minimum unit stresses (that is, the principal stresses) at any point, with their directions, may be found by a slight modification of § 164 as follows : Let Fig. 71 represent the stresses acting on a particle, as the i8o STRUCTURAL MECHANICS. one at O, Fig. 70. Lay off ON = /> tt on the normal to the plane A B, on which it acts in its true position. As q acts on the same y b 1/ y k d >l ^ o ■^o- a b Fig. 71 area of A B as does p n , lay off R N = . Then sinROM=^4 2 pi 1 +sin (j> 182 STRUCTURAL MECHANICS. In the figure, pi is taken larger than p2', if p2 is larger than pi, the triangle will lie on the other side of the normal and the ratio becomes P2 i+sin pi~i —sin ' If pi is the principal pressure acting vertically and due to the weight of the earth itself, its conjugate pressure, p2, must lie between the values given by the two preceding equations. Its exact amount can be determined by the principal of least resist- ance, for pi, which is caused by the action of gravity, produces a tendency for the earth to spread laterally, which tendency gives rise to the pressure, p2. As p2 is caused by pi, p2 will. not increase beyond the least amount sufficient to balance pi, hence p2 i —sin 4> pi i +sin (j> and in Fig. 73 the angle R O M is equal to <£. In the preceding discussion pi has been considered to act on a horizontal, and p2 on a vertical plane, and such will be the case when the surface of the ground is level, but if the ground slopes, the pressure on a plane parallel to the surface is vertical and the direction of pi inclines slightly from the vertical, as shown in the following example: Example. — Find the pressure on the back of the retaining-wall shown in Fig. 74, and also the resultant pressure on the joint C O'. The pressure on the plane O O' passed parallel to the surface of the ground is vertical and due to the weight of the earth upon it of depth K O. But the prism of earth resting on one square foot of the plane has a smaller horizontal section than one square foot, and the ratio of the unit vertical pressure on the plane through O to the weight of a verti- cal column of earth one square foot in cross-section will be that of the normal O H to the vertical O K. Hence O R(= O H) represents in feet of earth the pressure per square foot on the plane O O'. Draw a line, O S, making the angle of repose, cj>, with the normal and by trial find on O H a centre, M, from which may be drawn a semicircle tangent to the line, O S, and passing through R. Then O M = %(pi + p 2 ) and INTERNAL STRESS: CHANGE OF FORM. 183 M R=i(pi — p 2 ) f and M X gives the direction of p u the unit of pressure being the weight of one cubic foot of earth. The principal stresses are now known and the pressure on the back of the wall can be found. To find the pressure on O' B at O', draw O' A parallel to M X and O' M' normal to the back of the wall. Lay off O' M' equal to O M, and M' R' equal to M R and making the same angle with O' A as O' M' does. Then R' O' gives the direction of pres- sure on the back of the wall and, when measured by the scale of the drawing and multiplied by the weight of a cubic foot of earth, gives the pressure in pounds per square foot at O'. As the pressure on the back increases regularly with the distance below the surface of the ground, the centre of pressure will be at D, one-third of the slant height from O', and the total earth pressure against one foot in length of the wall will be P= JX O' B X O' R'X weight of a cubic foot of earth. If W is the weight of one foot length of wall applied at the centre of gravity, G, the resultant of P and W is the resultant pressure on the bed-joint, O' C, and the point where it cuts the joint should lie within the middle third. 170. Change of Volume. — If ^ = unit stress per square inch on the cross-section of a prism, and X is the resulting stretch or 184 STRUCTURAL MECHANICS. shortening per unit oj length, then by definition E = p+X, if p does not exceed the elastic limit. When a prism is extended or compressed by a simple longi- tudinal stress, it contracts or expands laterally, Fig. 75. This contraction or expansion per unit of breadth may be written TX-^-m in which m, the ratio of longitudinal extension to lateral contraction, is a constant for a given material, and for most solids lies between 2 and 4. A simple longitudinal tension p then accompanies a Longitudinal stretch = X = p+E per unit of length, and a Transverse contraction = — X + m= — p-r-mE per unit of breadth. For ordinary solids X is so small that it makes no difference whether it is measured per unit of original or per unit of stretched length. The original length will be used here. The new length of the prism is l(i+X) and the cross-section is S(i— X-^-m) 2 . The volume has changed from SI to . Sl(i+X — 2X+m) nearly, if higher powers of X p\ than the first are dropped, since the unit deforma- tions are very small. The change of unit volume is therefore Xl 1— — J. Thus, if m is nearly 4, for metals, the change of volume of one cubic unit is \X nearly, the volume being increased for longitudinal tension. If there were no change of volume, m would be 2, as is the case for india rubber for small deformations. Similarly, for compression the change of the unit volume is nearly— \X for metals, the volume being diminished. Example. — Steel, £=29,000,000; ^=20,000 lb. per sq. in. tension; the extension will be of its initial length, the lateral contraction i,45° will be about of its initial width, and its increase of volume about 5,800 1 2,900' Fig. 75 INTERNAL STRESS: CHANGE OF FORM. 185 171. Effect of Two Principal Stresses. — Denote the stresses by p\ and p 2 , treated as tensile. If they are compressive, reverse the signs. Under the action of pi there will be the following stretch of the sides per unit of length: Fig. 76, parallel to O C, h E' parallel to OB and O A, --%. r mE Under the action of p2 there will be parallel to O B, P2 E' Fig. 76 parallel to O A and O C, mE' Adding the parallel changes or stretches parallel to O C, h=\\p\-^A\ parallel to O B, X 2 =^(p 2 -— ) ; parallel to O A, A 3 mE (P1+P2). If pi and p 2 are equal unit stresses, but of opposite signs, the changes of length become P 1 E [ 1+ m' p 1 ~E\ 1+ m' ; and zero or, putting either of these two changes equal to X, the lengths to the sides of the cube originally unity per edge will be i+A, 1 —A, and 1, and the volume, neglecting A 2 , is unchanged. i86 STRUCTURAL MECHANICS. 172. Effect of Two Shears. — The cube shown in Fig. 77 has been deformed by the action of two equal and opposite prin- cipal stresses, and a square, traced on one side of the original cube, has been distorted to a rhombus, the angles of which are greater and less than a right angle by the amount, <£. By § 169, two equal principal stresses of opposite sign are equivalent to two unit shears of the same amount per square inch on planes at 45 with the principal axes; hence the distortion of the prism, whose face is the rhombus, results from two equal shears at right angles. Now one-half the angle \- — 6 has for its tangent J(i — X) -s- J(i-M); hence 1 -A t/ , t . 1 —tan h , 7TI = tanJ(^-^)= i+tan ^; or A=tan #. But as cj> is small, X = %, or = 2X. Therefore a stretch and an equal shortening, along a pair of rectangular axes, are equivalent to a simple distortion relatively to a pair of axes making angles of 45 with the original axes ; and the amount of distortion is double that of either of the direct changes of length which compose it. This fact also appears from the consid- eration that a distortion of a square is equivalent to an elon- gation of one diagonal and a shortening of the other in equal proportions. Example. — For steel, as before, X= , d> = = 4' 4 c", if j>\ — * 1,450' r 725 * ^ ' tl — /> 2 = 20,000 lb. 173. Modulus of Shearing Elasticity. — Similarly, equal shear- ing stresses q on two pairs of faces of a cube, in directions par- allel to the third face, will distort that third face into a rhombus, each angle being altered an amount <£, there being distortion of shape only, and not change of volume, Fig. 78. INTERNAL STRESS: CHANGE OF FORM. 187 Under the law which has been proved true within the elastic limit, and the definition of the modulus of elasticity, § 10, a modulus of transverse (or shearing) elasticity, C, also called coefficient of rigidity, as E may be called coefficient of stiffness, may be written, C = q + . As these two unit shears are equivalent to a unit pull and thrust of the same magnitude Fig> r8 per square inch, at right angles with each other and at 45 with these shears, the case is identical with the preceding one. Then 4> = 2X, and UU+ 1 -). .-. E\ m] 4> = 2p m + i E m p 1 mE (f> 2 m-\-i But, as p = q = C6, For iron and steel m is nearly 4, which gives C = \E. For wrought iron and steel, C is one or two one-hundredths less than 0.4E. Some use §£. C — 11,000,000 is a fair value. 174. Stress on One Plane the Cause of Other Stresses. — The elongation produced by a pull, the shortening produced by' a thrust, and the distortion due to a shear can be laid off as graphi- cal quantities and discussed as were unit stresses themselves. All the deductions as to stresses have their counterparts in regard to changes of form. There has been found an ellipse of stress for forces in one plane, when two stresses are given. Also, when three stresses not in one plane are given, there is an ellipsoid of stress which includes all possible unit stresses that can act on planes in different directions through any point in a body. So there is an ellipse or ellipsoid that governs change of form. Whether the movement of one particle towards, from or by its neighbor sets up a resisting thrust, pull or shear, or the appli- cation of a pressure, tension, or shear is considered to cause a corresponding compression, extension, or distortion, the stresses and the elastic change of form coexist. Hence it follows that, l88 STRUCTURAL MECHANICS. when a bar is extended under a pull and is diminished in lateral dimensions, a compressive stress acting at right angles to the pull must be aroused between the particles, and measured per unit of area of longitudinal planes, together with shears on some inclined sections. That such a state of things can exist may be seen from the following suggestions. It may be conceived that the particles of a body are not in absolute contact, but are in a state of equilibrium from mutual actions on one another. They resist with increasing stress all attempts to make them approach or recede from each other, and, if the elastic limit has not been exceeded, they return to their normal positions when the external forces cease to act. The par- ticles in a body under no stress may then be con- E 1 iff. 79 ceived to be equidistant from each other. The smallest applied external force will probably cause change in their positions. If, in the bar to which tension is to be applied, a circle is drawn about any point, experiment and what has been stated about change of form in different directions show, that the diam- eter in the direction of the pull will be lengthened when the force is applied, the diameter at right angles will be shortened, and the circle will become an ellipse. In Fig. 79, particle 1 moves to 1', 2 to 2', 4 to 4 r , and 7 to 7'. As they were all equidistant from o in the beginning, 1 in moving to i' offers a tensile resistance, 7 resists the tendency to approach o, while a particle near 4, mov- ing to 4', does not change its distance from o, but moves laterally setting up a shearing stress. A sphere will similarly become an ellipsoid. 175. Actual Resulting Stresses. — Let —pi be the unit tension in the direction 0-1, and + pi-^-m the accompanying unit thrust in the direction 7-0. If a pull is applied to the solid in the direc- tion 0-7, which develops —po tension in that direction on the plane 0-1, and +p2+m thrust in the direction 1-0, the resultant unit tension along 0-1 on the plane 0-7 will be — pi + p2+m, and alone: 0-7 on the plane 0-1 will be —po + pi +tn- It follows that INTERNAL STRESS: CHANGE OF FORM. I?9 the tension in the direction o-i will be less than when the first pull was acting alone. Hence a plate is stronger to resist two pulls at right angles than when subjected to one only. The opposite deduction can be drawn if one principal stress is of opposite sign to the other. A boiler plate has a tension in a tangential direction, that is, on a linear inch of longitudinal element, of pr, or of pr-^t per square inch, where p = steam pressure per square inch, r = radius in inches, and t = thickness of plate. On a circumferential inch the pull is one-half as much. Then, by § 170, and what has been stated above, if m = 4, pi = — pr pi v?» = — \pr p2 = — \py p2 -*■ m = — \ pr pi=pi-p2+m=-lpr p2=p2—pi+m=-\pr Hence the true unit tension is less than the apparent tension by 12 \ per cent, and the boiler is stronger than it would be if the longitudinal tension from the steam pressure on the heads did not exist. If tension is applied to the ends of a rectangular prism, and external compre sion is added to all four sides, the true unit ten- sion is much increased, or the piece is decidedly weaker in resist- ing the pull. Example. — At a certain point in a conical steel piston there exist principal stresses of 3,160 and 1,570 lb., of opposite signs. Then ^ 1 , = 3,i6o + i-i,57o = 3,55o lb.; p 2 '= 1,570+^3,160=2,360 lb. Since test experiments to determine tensile and compressive strength are made by the application of a single direct force, the values so determined are compatible with the existence of the opposite stress on planes at right angles with the cross-section. Hence the working stresses for any material may fairly be con- sidered to be a little higher than ordinary experiments show, 190 STRUCTURAL MECHANICS. provided account is taken at the same time of all the stresses which act on a particle. 176. Cooper's Lines. — Steel plate as it comes from the mill has a firmly adhering but very brittle film of oxide of iron on the surface. This film is dislodged by the extension of a test speci- men in tension when the yield-point is passed. If a hole is punched at moderate speed in a steel plate, so that the particles under the punch have some opportunity to flow laterally under the compression, there will be a radial compressive stress in all directions outwardly from the circumference of the hole, and a tensile stress circumferentially. These opposite principal stresses cause shearing planes to exist whose obliquity depends upon the relative magnitudes of the principal stresses by § 163. The scale breaks on these lines of shear and there result curves where the bright metal shows through, branching out from the hole, intersecting and fading away. The process of shearing a bar will develop the same curves from the flow of the metal on the face at the cut end. They are known as Cooper's lines. These lines show that deformation takes place at considerable distances from the immediate point of shearing or punching. Examples. — 1. A pull of 1,000 lb. per sq. in. and a thrust of 3,000 lb. .per sq. in. are principal stresses. Find the kind, direction, and magnitude of the stress on a plane at 45 with either principal plane. 2. Find the stress per running unit of length of joint for a spiral riveted pipe when the line of rivets makes an angle of 45 with the axis of the pipe and when it makes an angle of 6o°. 0.707 p; 0.5 p. 3. A rivet is under the action of a shearing stress of 8,000 lb. per sq. in. and a tensile stress, due to the contraction of the rivet in cooling, of 6,000 lb. per sq. in. Find pi and p2- P\= -n,54o lb.; ^ 2 = + 5,54olb. 4. A connecting plate to which several members are attached has a unit tension on a certain section of 6,500 lb. at an angle of 30 with the normal. On a plane at 6o° with the first plane the unit stress of 5,000 lb. compression is found at 45 with its plane. Find the principal unit stresses and the shear. —6,600; +4,800; 5,700. 5. Assuming the weight of earth to be 105 lb. per cu. ft. and the horizontal pressure to be one-third the vertical, what is the direction and unit pressure per square foot on a plane making an angle of 15 with the vertical at a point 12 ft. under ground, if the surface is level? 515 lb.; 39^° with the horizon. 6. A stand-pipe, 25 ft. diam., 100 ft. high. The tension in lowest ring, if J in. thick, is 7,440 lb. per sq. in. If plates range regularly from J in. thick at base to j in. at top, neglecting lap, the compression at base will be about 215 lb. per sq. in. For a wind pressure of 40 INTERNAL STRESS: CHANGE OF FORM. 19 1 lb. per sq. ft., reduced 50% for cylindrical surface, and treated as if acting on a vertical section, M at base =2,500,000 ft. -lb. Compression on leeward side at base = 485 lb. per sq. in. If p\= — 7,440 lb., p2= 215 + 485 = 700 lb., find the stress and its inclination for a plane at 30 to the vertical? —6,193 Mb.; 34° 4°'- Prove that the shearing plane is 17 04' from horizontal, and that the shear is 2,284 lt>- per sq. in. CHAPTER XII. RIVETS: PINS. 177. Riveted Joints. — There are four different ways in which riveted joints and connections may fail. The rivets may shear off; the hole may elongate and the plate cripple in the line of stress; the plate may tear along a series of rivet-holes, more or less at right angles to the line of stress ; or the metal may fracture between the rivet-hole and the edge of the plate in the line of stress. From the consideration that a perfect joint is one offering equal resistance to each of these modes of failure, the proper pro- portions for the various riveted connections are deduced. 178. Resistance to Shear. — The safe resistance of a rivet to shearing off depends upon the safe unit shear and the area of the rivet cross-section, which varies as the square of the diameter of the rivet. When one plate is drawn out from between two others, a rivet is sheared at two cross-sections at once, and is twice as effective in resisting any such action. Rivets so circumstanced are said to be in double shear, and their number is determined on that basis. 179. Bearing Resistance. — The resistance against elongation of the hole or crippling the plate depends on the safe unit com- pression and what is known as the bearing area, the thickness of the plate multiplied by the semicircumference of the hole. As the semicircumference varies as the diameter, it is more con- venient, and sufficiently accurate, to use the product of the thick- ness of the plate and the diameter of the rivet with a value of allowable unit compression about fifty per cent greater than usual. In practice the bearing values is always given in terms of the diameter. 180. Resistance of Plate. — The resistance to tearing across the plate through a line of holes, or in a zigzag through two lines 192 RIVETS: PINS. 1 93 of holes in the same approximate direction, depends on the safe unit tensile stress multiplied by the cross-section of the plate after deducting the holes. If the transverse pitch, or distance between centres of rivets, is considerable, an assumption of uniform distribution of tension on that cross-section is not likely to be true. The resistance of the metal between the rivet-hole and the edge of the plate in the line of stress is usually taken as the safe unit shear for the plate multiplied by the thickness and twice the distance from the rivet-hole to the edge. Some, however, consider that the resisting moment of the strip of metal in front of the rivet-holes is called into action. 181. Bending: Friction. — There are those who advise the com- puting of a rivet shank as if it were subjected to a bending moment. If the rivet fills the hole and is well driven, there is no bending moment exerted on it, unless it passes through several plates. As practical tests have shown that rivets cannot surely be made to fill the holes, if the combined thickness of the plates exceeds five diameters of the rivet, this limitation will diminish the im- portance of the question of bending. No account is taken of the friction induced in the joint bv its compression and the cooling of the rivet, and such friction gives added strength. As the rivet is closed up hot, the shank is under more or less tension when cold. Moreover, the head is not given the thickness required in the head of a bolt under tension, and therefore rivets are not available for any more ten- rion, and should not be used for that purpose. Tight-fitting turned bolts are required in such a case. 182. Spacing. — The rivets should be well placed in a joint or connection, in order to insure a nearly uniform distribution of stress in the piece; they should be symmetrically arranged, be placed where they can be conveniently driven, and be spaced so that the holes can be definitely and easily located in laying out the work. 183. Minimum Diameter of Rivets. — The punch must have a little clearance in the die. The wad of metal shears out below the punch with more ease and with less effect on the surrounding 194 STRUCTURAL MECHANICS. metal when it can flow, as it were, a little laterally, and it then comes out as a smooth frustum of a cone with hollowed sides, reminding one of the vena contracta. The punch must also be a little larger than the rivet, to permit the ready entrance of the rivet-shank at a high heat. The diameter of the hole is com- monly computed at \ inch in excess of the nominal diameter of the rivet; but the rivet is treated as if of its nominal diameter. One other consideration has weight in determining the minimum diameter of the rivet. If the diameter of the rivet is less than the thickness of the plate, the punch will not be likely to endure the work of punching. A diameter one and a half times the thickness of the plate is often thought desirable. 184. Number and Size of Rivets. — Formulas are of little or no value in designing ordinary joints and connections. Boiler joints and similar work can be computed' by formulas, but to no great advantage. Tables are used which give what is termed the shearing value of different rivet cross-sections in pounds for a certain allowable unit shear, and the bearing or compression value of different thicknesses of plate and diameters of rivet for a certain allowable unit compression. For a given thickness of plate, that diameter of rivet is the best whose two values, as above, most nearly agree. The quotient of the force to be trans- mitted through the connection or through a running foot of a boiler joint, divided by the less of the two practicable values, will give the minimum number of rivets. Their distribution is governed by the considerations previously referred to. Whether a joint in a boiler requires one, two, or three rows of rivets depends upon the number needed per foot. Example. — Two tension bars, 6 in. by £ in., carrying 30,000 lb., are to be connected by a short plate on each side. Let unit shear be 7,500 lb. per sq„ in., unit bearing 15,000 lb. on the diameter, and unit tension 12,000 lb. The bearing value of a f-in. rivet in a J-in. plate is 5,625 lb., its shearing value in double shear is 2X3,310=6,620 lb. Hence 30,000-^5,625 = 6 rivets necessary. If these rivets can be so arranged that a deduction of but one rivet-hole is necessary from the cross-section of the tie, (6— (f +i))£= 2.56 sq. in. net section, which will carry 30,750 lb. at 12,000 lb. unit tension. Each cover-plate can- not be less than \ in. thick, and, as will be seen presently, should be RIVETS: PINS. 195 made a little more. The length will depend on the distribution of the rivets, to be taken up next. 185. Arrangement of Rivets. — Long joints, under tension, like those of boilers, are connected by one or more rows of rivets, I to t r 1 1 000 1 000 1 000 K o U o \ O] O ! O / O! io O O jo O jo O O O Io O jo O \ O io L&4 E^l O O Io o o o J ojo o\ j 000 H 000 IOOOO 0000 00 0/ Fig. 80 as shown at A and B, Fig. 80. If more than one row is needed, the rivets are staggered, as at B, and the rows should be separated such a distance that fracture by tension is no more likely on a 196 STRUCTURAL MECHANICS. zigzag line than across a row. Experiments have shown that a plate will break along a zigzag line such as is shown at B unless the net length of that line exceeds by about one-third the net length of a line through a row. To prevent tearing out at the edge of the plate, the usual specification of at least one and a half rivet diameters from centre of hole to edge of plate will suffice. The tendency of a lap-joint to cause an uneven distribution of stress by reason of bending, and the same tendency when a single cover is used, is shown at C. The increase of stress thus caused should be offset by increased thickness of plate. A cover strip on each side is preferable if not objectionable for other reasons. In splicing ties, D shows a bad arrangement, the upper plan failing to distribute the stress evenly across the tie, and the lower plan wasting the section by excessive cutting away. The rivets at E are well distributed across the breadth, and weaken the tie by but one hole, as only two-thirds of the stress passes the section reduced by two holes; and, unless the net section at this place is less than two-thirds of the section reduced by one hole, it is equally strong. Thus b—d=§(b — 2d), or a breadth equal to or greater than four diameters will satisfy this requirement. The covers, however, will be weakened by two holes, and hence their combined thickness, when two are used, should exceed the thickness of the tie. F similarly is better than G, and the tie at F is again weakened by but one hole. The sectional area of the plate shown at H is diminished by two holes at m and four at n, but the stress on section n is less than that on m by the stress the rivets at m transmit to the splice plate. Consequently, in designing a splice, if the area cut out by the two extra rivets at n multiplied by the working stress in the plate does not exceed the working value of the two rivets at m, the plate will be weakened by two holes only. In the splice shown the sections at m and n will be equally strong when b - 2d = \%(b - +d) or when b = iSd. As it is desirable to transmit all but the proper fraction of the tension past the first rivet, the corners of the cover F or H are RIVETS: PINS. *97 clipped off, thus increasing the unit tension in the reduced section and increasing its stretch to correspond more nearly with the unit tension and elongation of the tie beneath. The appearance of the connection is also improved. It is not desirable to make splice plates as short as possible, because a short splice is likely to be weak. In splicing members built up of shapes this is especially the case, as a uniform dis- tribution of stress over the whole cross-section is not easily secured when the stress passes into the member within a length less than its width. Short splice plates may be lengthened without increasing the number of rivets by using a greater pitch. 186. Remarks. — If the member is in compression, the holes are not deducted, since the rivets completely fill the holes; and the strength is computed on the gross section. Unless special care is exercised in bringing two connected compression pieces into close contact at their ends, good practice requires the use of a sufficient number of rivets at the connection to transmit the given force. Rivet-heads in boiler work are flat cones. In bridge and structural work they are segments of spheres, known as button heads, and are finished neatly by means of a die. These heads may be flattened when room is wanting, and countersunk heads are used where it is necessary to have a finished flat surface. Members of a truss which meet at an angle are connected by plates and rivets. The axes of the several members should if possible intersect in a common point. If they do not, moments are introduced which give rise to what are known as secondary stresses, as distinguished from the primary stresses due to the direct forces in the pieces of the frame. Such secondary stresses may be of considerable magnitude in an ill-designed joint. It is desirable to arrange the rivets in rows which can be easily laid out in the shop, and to make the spacing regular, avoiding the use of awkward fractions as much as possible. Commercial rivet diameters vary by eighths of an inch, §-, §-, and f-in. rivets being the ones frequently used. As much uniformity as possible in the size of rivets will tend tc economy in cost. 198 STRUCTURAL MECHANICS. 187. Structural Riveting. — The following rules for structural work are in harmony with good practice: Holes in steel f in. thick or less may be punched; when steel of greater thickness is used, the holes shall be subpunched and reamed or drilled from the solid. The diameter of the die shall not exceed that of the punch by more than ts of an inch, and all rivet-holes shall be so accu- rately spaced and punched that, when the several parts are assembled together, a rivet & in. less in diameter than the hole can generally be entered hot into any hole. The pitch of rivets, in the direction of the stress, shall never exceed 6 in., nor 16 times the thickness of the thinnest plate connected, and not more than 30 times that thickness at right angles to the stress. At the ends of built compression members the pitch shall not exceed 4 diameters of the rivet for a length equal to twice the width of the member. The distance from the edge of any piece to the centre of a rivet-hole must not be less than ij times the diameter of the rivet, nor exceed 8 times the thickness of the plate; and the distance between centres of rivet-holes shall not be less than 3 diameters of the rivet. The effective diameter of a driven rivet will be assumed to be the same as its diameter before driving; but the rivet-hole will be assumed to be one-eighth inch diameter greater than the un- driven rivet. In structural riveting these relationships between unit working stresses are very commonly used : Bearing stress = 1 J X tensile stress ; Shearing stress = J X bearing stress = f X tensile stress. The shearing area of the rivets, therefore, should exceed by one-third the net area of the tension member they connect. See §§ 146, 147, 148. 188. Boiler-riveting. — Boiler work admits of standardization much more readily than structural work and standard boiler joints, which make the tensile strength of the net plate equal to the strength of the rivets, have been very generally adopted. A triple-riveted boiler joint is shown in Fig. 80, L. The most notable point of difference between boiler and structural riveting is that it is not customary to consider the bearing of rivets in boiler RIVETS: PINS. 199 work. Rivets are figured for shear only and in double shear are considered to have but one and three-quarters time the value of rivets in single shear, instead of twice the value, as in structural work. Custom in this country makes the allowable unit shear on rivets approximately two-thirds the unit tensile stress in the plate, but the British Board of Trade rule gives about four-fifths. The diameter of the rivets used should be about twice the thick- ness of the plate. In ordinary cases there is no danger that the rivets will be too far apart to render the joint water- or steam- tight, when the edge of the plate is properly closed down with a calking-tool. 189. Pins: Reinforcing Plates. — The pieces of a frame are frequently connected by pins instead of rivets. The axes of the several pieces are thus made to meet in a common point, if the pin-hole is central in each member. Pins are subjected to com- pression on their cylindrical surfaces, to shear on the cross-section, and to bending moments. The compression on the pin-hole is reduced to the proper unit stress, if necessary, by riveting re- inforcing plates to the sides of the members, as shown at K, Fig. 80. A sufficient number of rivets to transmit the proper proportion of the force must be used, with a due consideration of the shearing value of a rivet and its bearing value in the re- inforcing plate or the member itself, whichever gives the less value. No more rivets should be considered as efficient behind the pin than the section of the reinforcing plate each side of the pin-hole will be equivalent to. When the pin passes through the web of a large built member, such as a post or a top chord of a bridge, the web is often so thin that more than one reinforcing plate on either side is needed. It is then economical to make the several plates of increasing length, the shortest on the outside, and determine the number of rivets in each portion accordingly. Pin-plates should be made long for the same reason as given for making long splice plates. The longest plate is sometimes required to extend 6 in. inside the tie-plates so that the stress may be transferred to the flanges and not overtax the web. See K, Fig. 80. 200 STRUCTURAL MECHANICS. 190. Shear and Bearing. — The shear at any section of the pin is found from the given forces in the pieces connected. The resultant of the forces in the pieces on one side of any pin section will be the shear at that section. As the pin will probably not fit the hole tightly (a difference of diameter of one-fiftieth of an inch being usually permitted), the maximum unit shear will be four-thirds of the mean (§ 72). Specifications frequently give a reduced value for mean unit shear, which provides for this un- equal distribution. Bearing area is figured as if projected on the diameter. 191. Bending Moments on Pins. — At a joint where several pieces are assembled, the resisting moment required to balance the maximum bending moment on the pin caused by the forces in those pieces will generally determine the diameter of the pin. In computing the bending moments, the centre line of each piece or bearing is considered the point of application of the force which it carries. This assumption is likely to give a result some- what in excess of the truth, as any yielding tends to diminish the arm of each force. The process of finding the bending moments will be made clear by an illustration. Fig. 81 shows the plan and elevation Fier. 81 of the pieces on a pin, with the forces and directions marked. The thickness of the pieces, which are supposed to be in contact, is also shown. The joint must be symmetrically arranged, to avoid torsion, and simultaneous forces must be used, which reduce RIVETS: PINS. 201 to zero for equilibrium. As the joint is symmetrical, the com- putation is carried no farther than the piece adjoining the middle. Resolve the given forces on two convenient rectangular axes, here horizontal and vertical. Set the horizontal components in order in the column marked H, the vertical ones in the column marked V. Their addition in succession gives the shears, marked F. The next column shows the distance from centre to centre of each piece. Fdx is then the increment of bending moment; and the summation of increments gives, in the column M, the bending moment at the middle of each piece, from the horizontal and from the vertical components respectively. The square root of the sum of the squares of any pair of component bending moments will be the resultant bending moment at that section. It is comparatively easy to pick out the pair of components which will give a maximum bending moment on the pin. Equate this value with the resisting moment of a circular section and find the necessary diameter. H. F. dx. Fdx. M. A +10,000 +10,000 B —40,000 —30,000 +11,250 +11,250 2. C o —30,000 —22,500 —11,250 D +15,000 —15,000 —18,750 —30,000 7 8 E +15,000 -i3> 12 5 -43» I2 5 V. A - 5.780 - 5,78o T l B C D — 2,890 - 5.780 - 8,670 ~ 8,670 f 8 7 - 6,503 ~ 4,335 - 5,419 ~ 6,503 -10,838 -16,257 E + 8,670 8 - 7,586 -23,843 202 STRUCTURAL MECHANICS. MatD = \/(3o,ooo 2 + i6,257 2 ); M at E= V (43, 12 52 + 23,8432). The latter is plainly the larger, and is 49,210 in. -lb. The pieces can be rearranged on this pin to give a smaller moment. The maximum moment is not always found at the middle. The bending moment at any point of the beam or shaft, when the forces do not lie in one plane, can be found in the same way. A solution of the above problem by graphics can be found in the author's Graphics, Part II, Bridge Trusses. Examples. — 1. A tie-bar J in. thick and carrying 24,000 lb. is spliced with a butt-joint and two covers. If unit shear is 7.500 lb., unit bearing on diameter is 15,000 lb., and unit tension is 10,000 lb., find the number, pitch, and arrangement of j-in. rivets needed, and the width of the bar. 2. The longitudinal lap-joint of a boiler must resist 52,000 lb. ten- sion per linear foot. If the unit working stress for the shell is 12,000 lb. and the other stresses as above, what size of rivet is best, for triple riveting, what the pitch, and the thickness of the shell? CHAPTER XIII. ENVELOPES. 192. Stress in a Thin Cylinder. — Boilers, tanks, and pipes under uniform internal normal pressure of p per square inch. Conceive a thin cylinder, of radius r, to be cut by any diame- tral plane, such as the one represented in Fig. 82, and consider the equilibrium of the half cylinder, which is illustrated on the left. It is evident that, for unity of distance along the cylinder, the total pressure on the diameter, 2pr, must balance the sum of the components of the pressure on the semi-circumference in a direction perpendicular to the diameter. This pressure, 2pr, uniformly distributed over the diameter, must cause a tension T in the material at each end to hold the diameter in place. Hence T = pr. As all points of the circle are similarly situated, the tension in the ring at all points is constant and equal to pr. If the thick- ness is multiplied by the safe working tension / per square inch, it may be equated with pr, giving Required net thickness = pr+f. In a boiler or similar cylinder made up of plates an increase of thickness will be required to compensate for the rivet-holes. If a is the pitch, or distance from centre to centre, of consecu- 203 204 STRUCTURAL MECHANICS. tive rivets in one row along a joint, and d the diameter of the rivet-hole, the effective length a to carry the tension is reduced to a—d, and the gross thickness of plate must not be less than pr a j a—d' Example. — The circumferential tension in a boiler, 4 ft. diameter, carrying 120 lb. steam pressure is 120-24=2,880 lb. per linear inch of 00 _ length of shell, which will require a plate — in. thick (net), if / is 10,000 not to exceed 10,000 lb. per sq. in. Net thickness = ^ in. If a longi- tudinal joint has J-in. rivet-holes, at 2J in. pitch, in two rows, the i-i r 1 ill 2,880* 2j . . thickness of plate must not be less than r= T V in. 10,000 -i+ 193. Another Proof of the value of T may be obtained as follows: The small force on arc ds = pds. The vertical compo- nent of this force = pds sin = pdx. The entire component on pdx = 2pr, which must be resisted by the tension in the material at the two ends of the diameter. The same result will be obtained graphically by laying off a load line = 2 pds, which become a regular polygon of an infinite number of sides, i.e., a circle, with the lines to the pole making the radii of the length pr. The cylinder, under these circumstances, is in stable equi- librium. If not perfectly circular, it tends to become so, small bending moments arising where deviation from the circle exists. Hence a lap-joint in the boiler shell causes a stress from the resisting moment to be combined with the tension at the joint. The above investigation applies only to cylinders so thin that the tension may be considered as distributed uniformly over the secton of the plate. For riveting see Chapter XII. 194. Stress in a Right Section. — The total pressure from p on a right section of the cylinder is r.r 2 p, which will also be the resultant pressure on the head in the direction of the axis of the cylinder, whether the head is flat or not. This pressure causes tension in every longitudinal element of the cylinder, or in every cross-section. As this cross-section is 27rrXthickness, the longi- tudinal tension per linear inch of a circumferential joint is ENVELOPES. 205 Kr 2 p-±27zr = \pr, or one-half the amount per linear inch of a longitudinal joint. Hence a boiler is twice as strong against rupture on a circumferential joint as on a longitudinal joint, and hence the longitudinal seams are often double riveted while the circumferential ones are single riveted. 195. Stress in any Curved Ring under Normal Pressure. — The stress in a circular ring of radius r, under internal or ex- ternal normal unit pressure, p, is pr per linear unit of section of the ring, being tension in the first case and compression in the second. Similarly, in single curved envelopes in equilibrium under normal pressure (that is, envelopes in which the stress acts in the direction of the shell) the stress at any point per linear unit along an element is equal to pp, in which p is the radius of curvature of the cu ve cut out by a plane normal to the element and passing through the given point. Fig. 83 shows the trace of an arc of shell of width unity and of length pdd acted upon from within by a normal pressure, p. Then P = ppdd and for equilibrium the sum of the components acting in the direction of P must be zero or P = 2 T sin \dd = Tdd. Hence T = pp. 196. Thin Spherical Shell : Segmental Head. — If a thin hollow sphere of radius / has a uniform normal unit pressure p applied to it within, the total interior pressure on a meridian plane will be -r' 2 p, and the tension per linear inch of shell will be -r f2 p^2-r' = hpr f . If p is applied externally, the stress in the material will be com- pression. It may be noted that the double curvature of the sphere is associated with half the stress which is found in the cylinder of single curvature having the same radius. If a segment of a sphere is used to close or cap the end of a cylinder or boiler, the same value will hold good. In this case the radius r f is greater than r for the cylinder. If the segmental end is fastened to the cylinder by a bolted 206 STRUCTURAL MECHANICS. flange, the combined tension on the bolts will be 7ir 2 p, as this is the total force on a right section of the cylinder. The flange itself will be in compression. The pressure p from* below, in Fig. 84, causes a pull per circumferential unit in the direction of a tangent at B, which pull has just been shown to be equal to \pr' . It may be resolved into vertical and hori- zontal components. The vertical component B C is, by § 194, \pr. The horizontal component h must be proportioned to the vertical component as A O to A B, the sides of the right-angled triangle to which they are respec- C Y~^f^* ^?7/\. tively perpendicular. As AO = V(r' 2 -r 2 ), h : \pr = \/{r' 2 —r 2 ) : r, or h = \p\/ (V 2 — r 2 ) . As h is a uniform normal pressure Fig. 84 r applied from without (or tension applied from within) in the plane of the flange, the compression on the cross-section of the latter will be hr or ^pr\/(r f2 —r 2 ), to be divided by that cross-section for finding the unit compression. Segmental bottoms of cylinders are sometimes turned inward. The principles are the same. Example. — A segmental spherical top to a cylinder of 24 in. diam- eter, under 100 lb. steam pressure, has a radius of 15 in. with a versed sine of 6 in. The tension in top = J- 100-15 = 750 lb. per linear inch. If its thickness is i in., the stress per sq. in. is 3,000 lb. The total pull on the flange bolts is 100-144-22^7 = 45,260 lb. A f-in. bolt has about 0.3 sq. in. section at bottom of thread, giving a tension value of about 3,000 lb. if /= 10,000 lb. There would be needed some 15 bolts, about 5i in. centre to centre on a circumference of 26 in. diameter. The compression in the flange is ^-100-12-9=5,400 lb. A 2X|-in. flange with a f-in. hole has a section ^-ii=| sq. in., giving a unit compression in the flange of 1-5,400=8,600 lb. per sq. in. A similar compression acts in the connecting circle between a water-tank and the conical or spherical bottom sometimes built. See §§ 204, 205. 197. Thick Hollow Cylinder. — If the walls of a hollow cylin- der or sphere are comparatively thick, it will not be sufficiently accurate to assume that the stress in any section is uniformly distributed throughout it. If the material were perfectly rigid, ENVELOPES. 207 the internal or external pressure would be resisted by the imme- diate layer against which the pressure was exerted, and the re- mainder of the material would be useless. As, however, the substance of which the wall is composed yields under the force applied, the pressure is transmitted from particle to particle, decreasing as it is transmitted, since each layer resists or neu- tralizes a portion of the normal pressure and undergoes ex- tension or compression in so doing. 198. Greater Pressure on Inside. — Let Fig. 85 represent the right section of a thick hollow cylinder, such as that of a hydraulic press. Let r± and r<2. be the internal and external radii in inches ; pi and p2 the internal and external normal unit pressures in pounds per square inch, pi being the greater; and p the unit normal pressure on any ring whose radius is r. If a hoop is shrunk on to the cylinder, p2 will be the unit normal pressure thus applied to the exterior of the cylinder. The unit tensile stress found in a thin layer of radius r and thickness dr will be denoted by /, and will be due to that portion 0} p which is resisted by the layer and not transmitted to the next exterior layer. The total tension on the radial section of a ring lying between ri and r is piri—pr, since the pressure, pi, on the inside sets up a tension of piri in the ring and the pressure, p, acting on the outside of the ring sets up a compression of pr. This total tension* may also be expressed by /: tdr. As p and r are variables, there is obtained by differentiating the equation pr= I tdr, or pin —d(pr) =tdr, pdr + rdp + tdr = o. (1) Another equation can be deduced from the enlargement of the cylinder. The fibres or layers between the limits r\ and r, being com- pressed, will be diminished in thickness. The compression of a piece an inch in thickness by a unit stress p Tig. 85 208 STRUCTURAL MECHANICS. will be p+E, § 10, and of one dr thick will be pdr-^-E. The total diminution of thickness between r x and r, from what it was at first, will therefore be \L ***• But the annular fibre or ring whose radius is r and length 27tr has been elongated t-^-E per inch of length. Its length will now be 2xrl i +t<) an d its radius rl i +tJ- The internal radius must similarly have become rA i + e-L where / is the value of t for radius Y\. The thickness r— r\ has now become rli+- — rA i + p), and, by subtracting this value from r— r^ there is found the diminution of thickness, rip- r-p. This expression may be equated with the previous one for decrease of thickness, or / j \fj dr - E E HnJ r x Since the first term is constant, there is now obtained by differentiating this equation, —d(tr)=pdr, or tdr + rdt + pdr=o. . . . (2) Add (1) and (2), and multiply by r to make a complete differ- ential. Then integrate 2(t+p)rdr+r 2 (dt+dp) =0; r 2 (t + p)= constant; .-. =fi 2 (/ + ^i) = r 2 \? + p 2 ). . (3) Again, subtract (1) from (2), and then integrate dt—dp=o. /—^constant; .*. =}—pi=f'—p2* . (4) From (3) and (4) are obtained, by addition and subtraction, t= +—5 p = -\ — « . . (5) 2 r 2, 2 2 p- 2 ENVELOPES. 209 If the internal radius is given, the external radius, and hence the required thickness, r^ — fi, is found by eliminating f from (3) and (4), ^ ) (6) -P1 + 2P2/ v J }~pl+2p If p2 is atmospheric pressure, it may be neglected when pi is large. In that case As r 2 becomes infinite when the denominator of (6) is zero, it appears that no thickness will suffice to bring / within the safe unit stress, if pi exceeds } + 2p2. These formulas do not apply to bursting pressures, nor to those which bring / above the elastic limit; for E will not then be constant. They serve for designing or testing safe construc- tion. Examples. — Cylinders of the hydraulic jacks, for forcing forward the shield used in constructing the Port Huron tunnel, were of cast steel, 12 in. outside diameter, 8 in. diameter of piston, with J in. clear- ance around same; pressure 2,000 lb. per sq. in. T2 2 36-16 /+ 2,000 ^ = ^s^ = F^S' / '° 3 °' A cast-iron water-pipe at the Comstock mine was 6 in. bore, i\ in. thick, and was under a water pressure of 1,500 lb. on the sq. in., or about 3,400 ft. of water. Here 7=2,770 lb. per sq. in. for static pressure, while the formula for a thin cylinder gives 1,800 lb. 199. Greater Pressure on Outside. — In this case the direc- tion or sign of / will be reversed, it being compression in place of tension. From the preceding equations, without independent analysis, by making / negative, there result: — d(pr) = —tdr; d(tr) = pdr. pdr + rdp—tdr=o; tdr+rdt—pdr=o. r\p-t)=ri\pi-])=r 2 \p2-f). t + p = l + pi=f' + p2- 2IO STRUCTURAL MECHANICS. The outer radius and pressure will now be taken as given quantities, and the unit compression in the ring at any point will be 2 r z 2 2 r z 2 which becomes, if pi is neglected as small, The external pressure p 2 must be less than i(f + pi), if t\ is to have any value. It will be seen from /in (7) that the com- pression is greatest at the interior. Example. — An iron cylinder 3 ft. internal diameter resists 1,150 lb. per sq. in. external pressure. The required thickness, if /= 9,000 lb., is given by l8= l( I _^ =0 . 86 , 2 . N \ 9>oo°/ 72=20.9 in. Thickness=3 in. 200. Action of Hoops. — To counteract in a greater or less degree the unequal distribution of the tension in thick hollow cylinders for withstanding great internal pressures, hoops are shrunk on to the cylinders, sometimes one on another, so that before the internal pressure is applied, the internal cylinder is in a state of circumferential compression, and the exterior hoop in a state of tension. If the internal pressure on the hoop is computed, for a given value of / in the hoop, and this pressure is then used for p 2 on the cylinder, the allowable internal pressure pi on the cylinder consistent with a permissible / in this cylinder can be found. There is, however, an uncertainty as to the pressure p 2 exerted by the hoop. ENVELOPES. 211 Examples. — A hoop i in. thick is shrunk on a cylinder of 6 in. external radius and 3 in. internal radius, so that the maximum unit tension in the hoop is 10,000 lb. per sq. in. This stress, by § 198, will be due to an internal pressure on the hoop of 1,530 lb. per sq. in. F °"=\lte 10,000+ p\\ 40 10,000+^1 or 000 — pi/ 36 10,000—^1 This external pressure p2 on the cylinder will cause a compressive unit stress in the interior circumference of the cylinder when empty, after the hoop is shrunk on, of 4,080 lb., and will permit an internal pressure in the bore of 8,448 lbs. per sq. in., consistent with /= 10,000 lb. For — = - — — — . The cylinder alone, without the hoop, would 9 10,000—^1 + 3,000 allow a value of p\ given by — = — '- — , or £1 = 6,000 lb. If the 9 10,000 — ^1 cylinder had been 4 in. thick, the internal pressure might have been 6,900 lb. The gain with the hoop, for the same quantity of material, is 1,548 lb., or some 22%. Hydraulic cylinder for a canal lift at La Louviere, Belgium, 6 ft. 9 in. interior diameter, 4 in. thick, of cast iron, hooped with steel. Hoops 2 in. thick and continuous. When tested, before hooping, one burst with an internal pressure of 2,175 lb. per sq. in., one at 2,280 lb., and a third at 2,190 lb. These results, if the formula is supposed to apply at rupture, give an average tensile strength of 23,400 lb. per sq. in. The hoops were supposed to have such shrinkage that an internal pressure of 540 lb. per sq. in. would give a tension on the cast iron of 1,400 lb., and on the steel of 10,600 lb. per sq. in. The ram is 6 ft. 6} in. diam. and 3 in. thick, of cast iron, an example of the greater pressure outside. 201. Thick Hollow Sphere. — Greater pressure on inside. Let Fig. 85 represent a meridian section of the sphere. Suppose /, t, etc., to be perpendicular to the plane of the paper. The entire normal pressure on the circle of radius Y\ will be pinri 2 , and the tension on the ring between radii Y\ and r will be n(piri 2 —pr 2 ). Any ring of radius r and thickness dr will carry 27trtdr, and hence is derived the first equation 7i(piri 2 — pr 2 ) =27r / rtdr, or —d(pr 2 )=2rtdr. .*. r 2 dp-\-2pr dr + 2rt dr = o. The second equation will be the same as obtained for the cylinder. —d(tr) = pdr, or rdt + tdr + pdr = o. 212 STRUCTURAL MECHANICS. Strike out the common factor r from the first equation, multiply the second by 2, and subtract. 2rdt—rdp=o, or 2dt—dp=o. .\ 2t— p = constant; .'. z =2J — pi = 2J f —p2- • • (9) Again, add the first to the second and multiply by r 2 . r 3 (dp+dt) +y 2 dr(p + t) =0. /. r 3 (p + i) = constant; .'. =r x \f + p l ) = r 2 3 (f , + p 2 )- • (10) From (9) and (10), 2} -pi r x 3 t + pi 2f-p i ri 3 f + pi 3 ^ 3 3 3 r 3 3 3// 2 (/ + />i) ^=vh — x - ^ — r) ( I2 ) \\ 2/ -P1 + 3P2/ These formulas are not applicable to bursting pressures for the reason given before. For a finite value of r 2 , pi must be less than 2/ + 3^2- If ^2 is atmospheric pressure, it may be neglected, and 202. Sphere: Greater Pressure on Outside. — Here again t changes to compression or reverses in sign, yielding . 2/' + p 2 , r 2 3 f-p 2 A 2f + ^ 2 f 2 3 /' -#2 3 r 3 3 r 3 3 That fi shall be greater than zero requires that p2 The error which would arise from con- sidering t as uniformly distributed is manifest. The dotted s Fig. 87 Fig. 88 \Fig. 89 circles show the respective cylinders or spheres. Fig. 86 gives the external and internal tensile stress for p\ in the interior of a thick cylinder. Fig. 87 shows the distribution of compression when the greater pressure is from without. Figs. ZZ and 89 represent thick spheres under similar pressures. 204. Tank with Conical Bottom. — A water-tank of radius r may be built with a conical bottom and be supported only at the perimeter by a circular girder. The pressure of the water in pounds per square inch at any point is p =0.434 X depth of point below surface. In the cylinder the stress per unit of length on a vertical joint is pr; the stress on a horizontal joint is equal to the weight of the sides lying above the joint and is generally insignificant. 214 STRUCTURAL MECHANICS. Any horizontal joint in the cone such as A A of Fig. 90 must carry a load, W, composed of the weight of water in the cylinder whose base is the circle A A, the water in the cone ABA and the weight of the metal shell of the same cone. The last item is comparatively insignificant. As the cone, like the sides, is built of thin plates, the stresses in the cone must always act tangentially to the shell, so W ' ■±2izr\ is the vertical component of Tij the stress per unit of length on the horizontal joint, and „ Wsecd 1 27^ ' which varies from zero at B to a maximum at C. To find the stress on a joint along an element of the cone imagine a ring of slant height unity to be cut out by two hori- zontal planes as shown in the figure. Substitute for the pressure, p, which acts normally around the ring, the two components, p tan 6 and p sec 6. Of these the former acting along the ele- ENVELOPES. 215 ments causes no stress on an elemental joint while the latter causes a tension in the ring of T2 = pr\ sec = pp. It can be proved that n sec 6 is the radius of curvature of the conic section cut out by a plane normal to the element, B C, hence T2 is equal to the pressure into the radius of curvature as shown in § 195. This tensile stress varies from zero at B to a maxi- mum at C. The total weight of the tank and contents is carried by a circular girder, which in turn is supported by three or more posts, consequently the girder is subjected to both bending and torsional moments. At C the stress, T\, is resolved into vertical and horizontal components, the former of which is W -T- 27ir and is the vertical load per unit of length of girder ; the . W tan d , . , . latter is — - — which causes a compressive ring stress in the 2.7CV p W tan girder of . ■ & 27T Example.— A circular tank, 40 ft. in diameter, has a conical bottom for which #=45°. The depth of water above the apex is 60 ft. Weight of cu. ft. of water, 62.5 lb. Sec 0=1.414. Tension in lowest vertical ring of sides is 40X0.434X20X12 = 4,170 lb. per lin. in. Tension in radial joint of cone at A, half way up, is 50X0.434X120X1.414=3,680 lb. per lin. in. of joint. Same at C is 40X0.434X 240 X 1.414= 5,900 lb. per lin. in. For tension on horizontal joint at A, half way up, W=7rX 100(50+^X10)62.5. ^ = 1X10X53.33X62.5X1.414=23,600 lb. per lin. ft.= 1,960 lb. per lin. in. of joint. At C, r 1 = JX 20X46.67 X62.5X 1.414 = 41,250 lb. per ft. = 3,440 lb. per lin. in. The vertical com- ponent of Tx at C is 41,250-7-1.414=29,200 lb. per ft. of girder. As the horizontal and vertical components are equal, the compression in the girder is 29,200X20=584,000 lb. 205. Tank with Spherical Bottom.— The stresses in the spheri- cal bottom are found in the same way as in a conical bottom. Any horizontal joint as A A carries the cylinder of water whose base is the circle A A, the segment of water ABA, and that part 2l6 STRUCTURAL MECHANICS. of the shell below the joint. The volume of the segment is 7ia 2 {r r — \d) if a = r'{i — cos 6). If the weight on the whole joint is W, the vertical component of the stress per linear unit of joint is W -T- 2~r' sin and the stress is T W 1 2tt/ sin 2 #" The tension per unit of length on any meridian joint is, by § 196, T 2 = hpr' The vertical load on the girder is W-±2~r per unit of length, and the horizontal force applied to the girder W bv the bottom of the tank is J 2nr tan a per unit of length, which causes a com- in the girder. ^ W\ Fig. 91 w pressive stress of — A r 2~ tan a The stresses in the spherical bottom are smaller than those in the conical bottom. Example. — A circular tank, 40 ft. diameter and 40 ft. high, has a spherical bottom for which « = 45°. Then ^'=28.3 ft. and the ex- treme height = 48.3 ft. Tension in radial joint at bottom = ^X62.5X 28.3X48.3 = 42,645 lb. per linear ft. = 3, 554 lb. per in. of joint. Ten- sion in radial joint at A, half way up, where 6=22%°, is 40,770 lb. per ft. or 3,397 lb. per in. of joint. At C, tension is 35,350 lb. per ft., or 2,946 lb. per in. of radial joint. Tension in horizontal joint at A is 41,760 lb. per ft., or 3,480 lb. per in., and at C is 39,220 lb. per ft., or 3,268 lb. per in. of joint. Compression in circular girder= 554,700 lb. 206. Conical Piston. — The cone C B C of Fig. 92 represents a conical piston of radius r, subtending an angle 2d, with a normal steam pressure, p, per unit of area applied over its exterior or interior, the supporting force being supplied by the piston-rod at B. The force in the direction of the rod on any section A A of radius r\ is pn(r 2 —r\ 2 ) which becomes at the vertex pzr 2 , the force on the piston-rod. This force will be compression on the rod and tension in the cone, if p acts on the exterior of the cone, ENVELOPES. 217 and the reverse if p acts within the cone. The unit stress in the metal of the cone at this section will be found by multiplying this force by sec d, and then dividing by the cross-section, 2tzTi/, in which t is the thickness of the metal. The unit stress on the circumferential section is then pi = — (r 2 —Vi 2 ) sec 6, Fig. 92 which is a maximum at the piston-rod if / is constant. The unit stress at A on a radial section is, by § 204, p2=—=—sec d. When pi is compressive, p 2 is tensile and vice versa. Example— -Conical piston, Fig. 92. ^=24 in., radius of rod=3 in., 0=6q°. Thickness for 7-1=17 in. is 1.5 in.; for ^ = 8 in. is 1.9 in. Steam pressure= 100 lb. per sq. in. Sec #=2.79. For ^1=17 in. pi== ioo(2 4 2 -i7 2 ) 2.79^(2X17X1. 5) = 1,570 lb. per sq. in.; *' p 2 = 100X17X2.79-^15 = 3,160 lb. per sq. in. For n = 8 in., ^ = 4,700 lb. and ^2=1,175. lb - For alternating stresses on steel castings these stresses are satisfactory. See example § 175. Fig. 93 207. Dome.— A dome subjected to vertical forces symmetri- cally placed around its axis, such as its own weight, may be treated as follows : C B C of Fig. 93 represents a meridian section of a 2i8 STRLCTURAL MECHANICS. dome, a hemisphere as shown, but the results to be deduced are true for any surface of revolution about a vertical axis. If a horizontal plane, A A, is passed through the dome to cut out a circle of radius r x and all the weight from the crown to that sec- tion is denoted by W, the stress in the shell per linear unit of circumferential joint is Wsecd T i — . 2xri i is always compressive and is a maximum at the base. To find the stress on a meridian section pass two horizontal planes through the dome so as to cut out a thin ring of mean radius t\. If the total load above the ring is W\ and the total vertical force supporting the ring is W 2 , the weight of the ring is W2—W1. As the shell of the dome is thin the stresses in the shell are tangential to the surface and the ring is acted upon by a system of forces around its circumference as shown on the right side of the figure. Resolve the forces into vertical and horizontal components as shown on the left. Acting upon the upper edge per unit of mean length of ring is the vertical com- pel , , . . . Wi ctn 0i „ ponent and the horizontal component ■ . By sub- r 27ZTi r 2WT\ J stituting W 2 and 6 2 for W\ and 6 y the components acting on the lower edge are found. The vertical components, together with the weight of the ring balance among themselves, but there is an unbalanced horizontal force of H= (W\ ctn 0i — W 2 ctn 6 2 ) 27zr\ which causes tension or compression in the ring depending on whether it acts outward or inward. The stress in the ring is, therefore, Hr\ and its intensity per linear unit of joint, T 2 , is found by dividing by the width of the ring measured along the meridian. At the crown T 2 is compressive and equal to 7\, but it diminishes as A is taken lower and lower down and becomes tensile in the lower part of the structure. 208. Resistance of a Ring to a Single Load. — When a ring is acted upon by two equal and opposite forces as shown in Fig. 94, the curve becomes flatter at A and sharper at B, showing that ENVELOPES. 219 bending moments of opposite sign are set up at these points. From conditions of symmetry it is seen that each quadrant is acted upon by vertical forces of \W together with an unknown moment at each end. Imagine the quadrant to be removed from mv Fig. 94 the circle and horizontal levers to be attached at A and B so that the forces can be moved horizontally such a distance as to cause the actual moments existing at A and B. For equilibrium the forces, when moved, must be applied in the same line, and the moment at any section of the ring will be determined when the line of application of the forces is fixed. To fix that line the deformation of the arc must be considered. 1 M M Bv § S8, —=-^-p which becomes da=-^ ds if the angle between 3 p EI EI to the two radii of Fig. 44 is da. This equation gives the change in the angle between two right sections ds apart, caused by the bending moment, and is true for curved beams as well as straight. To find the change between the two sections a distance s apart inte- grate from zero to s. In the ring under consideration the tan- gents at A and B remain horizontal and vertical respectively, as seen from the condition of symmetry, hence the change in the angle between right sections at A and B is zero and as E and i" -r.r WXDC and are constant / Mds=o. M at any point C is 2 */ o must be expressed in terms of r and 6 to be integrated. Then hWr £ (a—rcoz 6)d0 = o 220 STRUCTURAL MECHANICS. [ {ad — r sin 0) =o = \ Fig. ro Fix. 97 required flange areas vary as the o dinates to a parabola (Fig. 96) and the length of any flange-plate is given by in which / is the length of the girder from centre to centre of base-plates, S the net flange area required at the centre, and s the net area of the plate whose length is desired plus the area of such plates as may lie outside it. If the girder is so long that the plates or angles must be spliced, additional cross-section must be supplied by covers at the splices, with lengths permitting sufficient rivets to transmit the force. Even compression joints, though milled and butted together, are spliced in good practice. The net area of the cover- plate and splice angles should be equal to that of the largest piece spliced. Only one piece should be cut at any one section, and enough lap should be given for the use of sufficient rivets to carry the stress the piece would have carried if uncut. 214. Rivet Pitch. — If a strip of flange (Fig. 97) of a width equal to the pitch of the rivets connecting the flanges to the web is cut out by two imaginary planes, shearing forces, F, act on the two sides forming a couple with an arm equal to the pitch. Under the usual assumption that the web carries shear only, there are 226 STRUCTURAL MECHANICS. no other forces acting on the planes of section, and the only forces which can keep the strip from rotating are those supplied by the two flange rivets. Hence by equating the two couples the pitch can be found. _. , rivet value X depth Pitch = r £— . shear As the flange-angles are supposed to be fully stressed at the point where the first flange-plate begins, the increments of flange stress coming out of the web must pass through the flange-angles and into the flange-plate. The rivets connecting the flange-angles and the flange-plates must therefore resist the same stress as do the rivets connecting the web plate and the flange-angles, and the above formula applies to both, although the rivet values in the two cases will be different. The rivets through the web are in bearing (or double shear if the web is thick), while those in the flange-plate are in single shear and occur in pairs. But practi- cally the pitch in one leg of an angle must be the same or an even multiple of the pitch in the other, so that the rivets may be stag- gered. Make the pitch of rivets in inches and eighths, not decimals; do not vary the pitch frequently, and do not exceed a six-inch pitch, so that the parts may be kept in contact. If flange-plates are wide, and two or more are superimposed, another row of rivets on each side, with long pitch, may be required to insure contact at edges. Care must be taken that a local heavy load at any point on the flange does not bring more shear or bearing stress on rivets in the vertical legs of the flange-angles than allowed in combination with the existing stress from the web at that place and time. Webs are occasionally doubled, making box girders, suitable for extremely heavy loads. The interior, if not then accessible for painting, should be thoroughly coated before assembling. If the web must be spliced, use a splice plate on each side for that purpose, having the proper thickness for rivet bearing and enough rivets to carry all the shear at that section; there should be two rows of rivets on each side of the joint. PLATE GIRDERS. 227 215. Stiffeners. — At points where a heavy load is concentrated on the girder, stiffeners, C, consisting of an angle on each side, should be riveted to throw the load into the web and to prevent the crushing of the girder. They should, for a similar reason, be used at both points of support, D. Such stiffeners act as columns and may be so figured, but as the stress in them varies from a maximum at one end to zero at the other, a good rule is to consider the length of the column equal to half the depth of the girder. Since the thrust at 45 to the horizontal tends to buckle the web, and the equal tension at right angles to the thrust opposes the buckling, it is conceivable that a deep, thin web, while it has more ability to carry such thrust as a column or strut than it would have if the tension were not restraining it, may still buckle under the compressive stress ; and it is a question whether stiffeners may not be needed to counteract such tendency. They might be placed in the line of thrust, sloping up at 45 from either abutment, but such an arrangement is never used. They are placed vertically, as at C, and spaced by a more or less arbitrary rule. A common formula is : The web of the girder must be stiffened if the shear per square inch exceeds d 10,000 — 75—, where d= clear distance between flange-angles, or between stiffeners if needed, and t= thickness of web. Another rule calls for stiffeners at distances apart not greater than the depth of the girder, when the thickness of the web is less than one-sixtieth of the unsupported distance between flange-angles. There is no rational method of determining the size of stiffeners used only to keep the web from buckling. The usual practice is about this : make the outstanding leg of the angle 3 J inches for girders less than 4 feet deep, 4 inches for girders 5 feet deep, and 5 inches for girders over 7 feet deep. Experience appears to show that stiffeners are not needed at such frequent intervals as the formula would demand. An 228 STRUCTURAL MECHANICS. insufficient allowance for the action of tension in the web in keeping the compression from buckling it, is probably the cause of the disagreement. Interior stiffeners may be crimped at the ends, or fillers may be used under them to avoid the offset. End stiffeners and stiffeners under concentrated loads should not be crimped; they should fit tightly under the flange that the load may pass in at the ends. Example. — A plate girder of 30 ft. span, load 3,000 lb. per ft., /= 15,000 lb. per sq. in., PF=oo,ooo lb., and M max =\Wl = 4,050,000 in.-lb. Assume extreme depth as 42 in., effective depth, 39 in. Net flange section at middle =4,050,000 -5- (39- 15,000) = 7 sq. in. A f-in. web 42 in. deep will have 15 J sq. in. area. Two flanges, each 7 sq. in. net + allowance for rivet-holes, will fairly equal the web. Use f-in. rivets. Let the flange-angles be 2 — 4X3X1 in. = 4.96 sq. in. Deduct 2 holes, |X J=o.66. Net plate= 7 — 4.3 = 2.7 sq. in. A plate 9Xf=3§ sq. in.; deduct two holes = 0.66, leaving 2.71 sq. in. Two angles and plate, gross sec ion=4.96+3-37 = 8. 33 sq. in. Resisting moment of net section of angles = 4-3X 15,000X39 = 2,515,500 in.-lb. Such a bending moment will be found at a distance x from either end, given by P\x— J- 3,ooox 2 = 2,515,500. #=5.9 ft. .'. Cut off the plate 5 ft. from each end. Shearing value of one f-in. rivet at 10,000 lb. per sq. in. = 4,400 lb. Bearing value in f-in. plate at 20,000 lb. = 5,600 lb. Max. shear in web= 45,000 lb. Pitch for flange-angles, since bearing resistance is less than double shear, =5,600 -39-^-45,000= 4.85 = 4! in. Make 3-in. pitch for 2 ft., then 4f-in. for 6 ft., then 6-in. pitch to middle. Rivets in end web stiffeners, 45,000-7-5,600=9. Max. shear in web = 45,000 j -7-(42-f) = 2,78o; 10,000— 75— = 2,800, since ^=42 — 6. No other stiffeners needed. By the other rule 36-7-1=96, and stiffeners are needed. CHAPTER XV. SPRINGS AND PLATES. 216. Elliptic Spring. — The elliptic spring is treated in § 97. If the load on the spring is W and the span is /, the deflection is The fibre stress is Wl 3 v = 16 Eld /= 6Mp 3 Wl b h 2 ~2 b h 2 ' and the work done upon the spring is %Wv, which is equal to the resilience of the spring. Hence n n . 1 W 2 P 1 j 2 , Resihence = 7^7— =-? -=; • volume. 32 £/ 6 £ 217. Straight Spring. — If a beam of uniform section, fixed at one end, has a couple or moment applied to it (Fig. 98) in place of a single transverse force, it will, as shown in § 89, bend to the arc of a circle. The deflection will be, if I is the length of the beam, Eig. 98 V = Ml 2 jl 2 2 EI 2Eyi ' The work done by the rotation of a couple is the product of its moment by the angle through which it turns. For a deflec- 229 230 STRUCTURAL MECHANICS. tion, dv, the free end of the spring turns through an angle 2dv +1] hence t, .,. rw , 4E1 r , 2E1 // p Resilience = 2 / —rdv = — tt- / vdv = —pr~v 2 = 5 • -7;. ,/ / / 3 J P 2yi 2 E For a rectangular section &&, these quantities become 6Ml 2 _ }P V ~Ebh*~Eh' 1 P 1 P Resilience = 7- • -ft • ^ = f * "F • volume. For a circular section the number 6 in the last expression will be replaced by 8. 218. Coiled Spring. — In practice the rectangular or cylin- drical bar is bent into a spiral and subjected to a couple which, as a couple can be rotated in its plane without change, acts equally at all sections of the spring. The developed length of the spiral is/. 219. Helical Spring. — A cylindrical bar whose length is / and diameter d, when fixed at one end and subjected to a twisting moment T = Pa at the other, if the elastic limit is not exceeded, 32 77 by § 84, is twisted through an angle # = -7^-74. The work ex- pended in the torsion is z 64/ From § 84, # = 7^r> and therefore I <7i 2 Tld 2 1 <7i 2 Resilience = — 77 -/ = ~ • volume. 4 C 4 4 C 2 4 2 P If C=-E and Q\=-U work= ^-volume, while for flexure, 5 5 5 E 1 / 2 as shown in the preceding section, work = 7- --tv volume, a smaller SPRINGS AND PLATES. 231 quantity; hence a spring of given weight can store more energy if the stresses are torsional than if they are bending. If this bar is bent into a helix of radius, a, and the force, P, is applied at the centre in the direction of the axis of the cylinder, the moment, Pa, will twist the bar throughout its length. Then i6Pa nd 3 ft "ST. or P =Wa qi - The deflection of the spring is v = ad, since, as the force P de- scends, the spring decends, and the action is the same as if the spring remained in place and the arm revolved through an angle 6. The force P is too small to cause any appreciable compression (or extension) of the material in the direction of its length. 32 Pa 2 1 2al q x /\.7rna 2 q 1 if n= number of turns of the helix and l = 27tan. If the section of the spring is not circular, substitute the proper value of qi or the resisting moment from § 85. If the rod is hollow, multiply the exterior volume by (1— -^-J. For a square section and a given deflection, P will be about 65 per cent, of the load for an equal circular section. C for steel is from 10,500,000 to 12,000,000. Example. — A helical spring, of round steel rod, 1 in. diameter, making 8 turns of 3-in. radius, carries 1,000 lb. i6-i,ooo«v7 4- 22-8-0- iq, 273 ?i = ^ = 15,273. v=~ v * =1-15 in. 22 J '° 7-1-12,000,000 ° 220. Circular Plates. — The analysis of plates supported or built in and restrained at their edges, and loaded centrally or over the entire surface, is extremely difficult. The following formulas from Grashof's "Theorie der Elasticitat und Festig- keit " may be used. The coefficient of lateral contraction is taken as \, or m = 4. 232 STRUCTURAL MECHANICS. I. Circular plate of radius r and thickness /, supported around its perimeter and loaded with w per square inch. / x =unit stress on extreme fibre in the direction of the radius, at a distance x from the centre; / 2/ = unit stress perpendicular to the radius, in the plane of the plate, at the same distance x from the centre. h ~I2St 2{jr 3 ^ h ~I2St 2{ ' Wr } ' 117 wr 2 i8q7cT 4 fx = fy max. (for X = 0) =—^ ~y> V = J^^p' For the same value of /, the maximum stress is independent of r, provided the total load WTir 2 is constant. II Same plate, built in or fixed at the perimeter. j = Al™ (2 _ 2) . / = ^-(r 2 -x 2 ) At the centre, j x = fy At the circumference j y is zero, and } x is maximum. 45 wr 2 45 wr 4 Umax. = ^~~fi' V °=~^6W III. Circular plate supported at the perimeter and carrying a single weight W at the centre. Loaded portion has a radius r . 4 q W(. r i\ ,45 Wf r 4 \ These expressions become maxima for x = Vq, and the second is the greater. 117 IF> 2 i>o = 64- £/ 3 For values of r-^r = 10, 20, 30, 40, 50, 60, Lax. =14 1-7 J '9 2.0 2.1 2.2 IF-f/ 2 . SPRINGS AND PLATES. 233 If r =o, the stress becomes infinite, as is to be expected, since W will then be concentrated at a point, and the unit load becomes infinitely great. It is not well to make r very small. IV. Same plate, built in or fixed at the perimeter. . 45 Wf r \ 45 W r The maximum value of / is f yj for x = r . 45 Wr 2 Vq 64- Efi ' For values of r+r = 10, 20, 30, 40, 50, 60, fmax. =LO 1.3 1.5 1-6 1.7 1.8 JF-^ 2 . 221. Rectangular Plates. — The problem of the resistance of rectangular plates is more complex than that of circular plates. Grashof gives the following results : V. Rectangular plate of length a, breadth b, and thickness t,. a>b, built in or fixed at edges and carrying a uniform load of w per square inch. b*-wa 2 a±-wb 2 l a = „(r.\ I 7viw2 ' l b ~ 2(> 4 + 6 4 J/ 2 ' /0 2( 4 W The deflection at the centre is Vq= . , ,, — =-=. and for a a 4 + 4 32E/ 3 cc-a 4 square plate, ^-^. VI. Plate carrying a uniform load of w per square inch and supported at rows of points making squares of side a. Fire- box sheet with staybolts. / = r 1 k wa 4 64 i 2 ' ^ S12 £/ 3 ' 234 STRUCTURAL MECHANICS. Navier gives formulas for rectangular plates which are sup- posed to be very thin. Approximate values from those formulas are as follows : VII. Rectangular plate, as in V, but supported around the edges. a 4 b 2 w a 4 b 4 w ^ = a92 ^T^ 2 ^ ; Vo= °' 19 {oJTW 2 Ef^ VIII. Rectangular plate, supported at edges and carrying a single weight W at centre. a*b W aW W / = 2 - 2 VW 2 ^ ; Vo = °' 46 W+V) 2 Ef'' For the same total load, / is independent of the size of the plate, provided the ratio a to b and the thickness are unchanged. Example. — A steel plate 36 in. square and } in. thick, supported at edges, carries 430 lb. per sq.ft., or 3 lb. per sq. in. /=o.Q2- 4-36- 36-3- 16 = 14,3°° lb - o.iq -?6 4 -V4 3 1 . v= — -•- — — — = £ in. 4 30,000,000 CHAPTER XVI. REINFORCED CONCRETE. 222. Reinforced Concrete. — As concrete has small tensile strength and is likely to crack when built in large masses, it can be reinforced to advantage by steel bars or wire netting imbedded within it. This form of construction is much used and is espe- cially applicable to beams and slabs, in which the steel is placed near the tension edge. By its use a great saving of material in masonry structures can often be effected, since the strength of the structure can be depended upon, rather than its weight. Among its advantages for buildings may be mentioned the fact that it is fireproof and that the metal is protected from rust. The expansion and contraction of the two materials from changes of temperature are so nearly alike that heat and cold produce no ill effects. When used in beams of any considerable span,, there is the disadvantage that the dead load is a large proportion of the total load. To compute the strength of a structure composed of two materials which act together, the modulus of elasticity of each material must be known, and it is here that the chief difficulty in computing the strength of reinforced concrete members lies. The modulus of elasticity of concrete is uncertain; it not only depends upon the composition of the concrete, but, for a given concrete, varies with the age, while for a particular specimen the stress deformation diagram is not a straight line as for steel, but the ratio of the stress to the deformation decreases with the load. The concrete takes a permanent set, even for small loads, in consequence of which a reinforced-concrete beam, after being released from a load, is in a condition of internal stress. 235 236 STRUCTURAL MECHANICS. Considering, then, our lack of knowledge of the exact stresses to be expected in the concrete, it is best to adopt as simple a method of computation as is reasonable. 223. Beams. — Beams may be figured according to the common theory of flexure, and the following assumptions will be made: 1. The steel and the surrounding concrete stretch equally. 2. Cross-sections, plane before bending, are plane after bending, 3. The modulus of elasticity is constant w T ithin the working stress. 4. The tension is borne entirely by the steel. As the tensile strength of concrete is low and as a crack in the beam would entirely prevent the concrete from resisting tension, this is a proper assumption to make and is on the side of safety. 5. There is no initial stress on the section. Neutral Axis L Fig- 90 From these assumptions it follows that the stresses acting on the cross-section are as shown in Fig. 99. Let E s = modulus of elasticity of steel; E c = " " " " concrete; e = E s +E c ; j s = unit stress in steel ; } c = " " " concrete at extreme compression fibre; X = deformation of fibre at unit distance from the neutral axis, between two sections originally unit distance apart ; kbh = sectional area of steel; y = distance from neutral axis to extreme compression fibre ; c=h-r-y; w=a numerical coefficient. REINFORCED CONCRETE. 237 By § 10, } s = lE s (k-y), f c = XE c y y or the ratio between the unit stresses is j s E s h —y __ h—y Ic~E c y ~ e ~ (1) As the total tension on the cross-section must equal the total compression, for a rectangular cross-section \byf c = kbhf 8 ( 2 ) Combining (1) and (2) gives fa y h—y } c 2kh y Solving this equation for y gives the location of the neutral axis, y = (-ek+Ve 2 k* + 2ek)h=ch (3) This equation shows that for beams made of a given quality of concrete the location of the neutral axis depends only upon the percentage of reinforcement. The resultant of the compressive stresses on the cross-section is applied at a point \y from the top of the beam and together with the tensile stress in the steel forms a couple whose arm is h -iy. The moment of this couple, which is the resisting moment of the beam, is M = ibyJ c (h-iy) (4 ) = kbh/ s (h-iy), the first expression giving the moment in terms of the stress in the concrete and the second in terms of the stress in the steel. If the value of y from (3) is substituted in (4) there results M = ic(i-lc)l c bh 2 = n c f c bh 2 (5) = k(i-ic)} s bh 2 =n 8 f 8 bh 2 . 23 8 STRUCTURAL MECHANICS. As long as j c and j s are the unit stresses in the concrete and in the steel, the two forms of (5) must be equal, but if f c and f s are arbitrarily chosen working stresses, that form of (5) must be used in designing which will give the smaller value of M . Therefore use n c when n c f c ~y The actual fibre stresses, Jc 2 & however, can be made to assume any given ratio by changing the proportion of steel reinforcement, and that proportion can be found by eliminating y from (1) and (2) with the result k = efc 2 2 ts 2 + ejc)s faffs . ) . . . • (6) Equation (5) can be readily applied to the design of beams by tabulating the values of n c and n s for different values of k. k e=S. e= 10. e=i2. c n c ns c tie n s c n c n s .003 .004 .005 .006 .007 .008 .010 .012 .OI4 .Ol6 .Ol8 .020 196 223 246 266 284 300 328 353 375 394 411 428 .092 .103 •113 . 121 . 128 •135 . 146 ■155 .164 .171 .177 .183 .00280 .00370 .00459 .00546 .00633 .00720 .00891 .217 .246 .270 .292 ■3ii .328 •358 .384 .407 .428 • 447 .463 . IOI •113 .123 .132 .140 . I46 •157 .167 .176 .183 . I9O . I96 .00278 .00367 .00455 .00542 .00628 .00713 .00880 235 265 291 314 334 353 384 411 435 457 476 493 .108 . 121 •131 .140 .148 •!55 .167 .177 .186 .194 .200 .206 .00277 .00368 .00452 ■00537 .00622 .00706 .00872 When reinforced concrete beams are tested to destruction, they sometimes fail by the opening of diagonal cracks which seem to follow in a general way the lines of principal stress. To prevent that mode of failure most designers provide some form of rein- forcement in a vertical plane. Stirrups or loops of wire lying in planes of section and spaced at intervals less than the depth of the beam are often employed, but no satisfactory method of determining their size and spacing has been proposed. REINFORCED CONCRETE. 239 Another mode of failure to be guarded against is the slipping of the steel in the surrounding concrete. That the beam may not be weak in this respect, the change of stress in the bars, between two sections a short distance apart (say one inch), must not exceed the area of the surface of the bars between the two sections multiplied by the safe unit adhesive stress of concrete to steel. This change of stress between two sections one inch apart is F-^-h(i-^c), since the shear in the beam measures the change of bending moment at any section, as is shown in § 56. Rods which have been roughened or corrugated may be used to diminish the likelihood of slipping. Example. — Design a beam of 16 ft. span to carry an external load of 500 lb. per ft. if / c =5oo lb., f s = 16,000 lb., and e=io, assuming the beam to weigh 300 lb. per ft. Af=J-8oo- 16-16-12 = 307,200 in.-lb. If each material is to be stressed to its limit, the proportion of reinforcement needed is 10-^-2-32-42 = 0.00372, or say 0.4%. Then n c =o.iiT, and bh 2 = 307,200 -3- (o. 113X500) = 5,430. b=i2j in., h=2i in., and the area of steel is 0.004X121X21 = 1.03 sq. in., or say 4— T 9 g- in. rounds. If there are 2 in. of concrete below the rods, the beam weighs 150X12^X23-^144=284 lb. per ft. If 0.8% of steel is used, instead of 0.4%, bh 2 = 42 10. 6=io|in., h=2o in., and 1.68 sq. in. of steel is needed ; use 1 — f in. and 2 — J in. rounds. Adhesion between concrete and steel. F wa *- = 800X8 = 6,400 lb. For first beam 6,400-^21(1 — 4X0.246)= 33 2 lb. Superficial area of 4— A rounds = 4- T V- 2 T 2 -=7.o7 sq. in. 332-^7.07 = 47 lb. per sq. in. For second beam 359-^7.86=46 lb. per sq. in. 224. Columns. — Reinforced concrete columns are built with steel rods embedded parallel to and spaced symmetrically about the axis. The rods should be tied together by wire or bands at intervals not greater than the diameter of the column. A common design is a square section with the rods near the corners. As the ratio of length to breadth is generally small, it is usual to design such columns as short blocks. The ratio between the intensity of the stress in the steel and in the concrete is IC &C 240 STRUCTURAL MECHANICS. and the total load the column may carry is P=} c S(i-k)+j s kS=} c S[i+k(e-i)], in which 5 is the total area of cross-section of the column. With the usual working stresses and ratio of E s to E c employed in designing, the greatest allowed unit stress in the reinforcement is so small that economy requires the amount of steel to be reduced as much as reasonable, hence the load P will be but slightly greater than f c S. 225. Safe Working Stresses. — The following safe unit stress e. may be used in buildings: Concrete, bending 500 lb. per sq. in. ' ' direct compression 350 ' ' ' ' " " Steel, tension 16,000 " " " '• Adhesion of concrete to steel 50 ' ' " " ' E s +E c = 10 is a fair average value; the ratio may vary con- siderably without materially affecting the proportions of the beam. The weight of reinforced concrete is about 150 lb. per cu. ft. The concrete should be rich; one part cement, two, sand; four, broken stone is a good mixture, the stone being broken to pass through a f-inch ring. It should be mixed wet and placed with great care to insure the proper bedding of the steel. The proportion of steel reinforcement to use in beams is a question of economy, which is most easily solved by designing a number of sections of the same ratio of b to h, but with different values of k, and figuring the cost per foot of each. A considerable variation in k affects the cost but slightly, for ordinary prices of concrete and of steel. The percentage of reinforcement is usually between J and ij; £=0.007 * s an average value. The concrete lying below the reinforcing rods serves merely to protect the steel. For fireproofing two inches is sufficient. Steel thoroughly covered with concrete does not rust, and bars, which were covered with rust when placed in concrete, have been found to be bright when removed after some time. INDEX. Angle of repose, 181 Annealing, 29 Ashlar, 30 Axis, neutral, 56, 61, 79 Bauschinger's experiments, 154 Beams, 2, 51, 87, 107 bending moments, 41, 44 cantilever, 45 Clapeyron's formula, 1 18 column and beam, 149 continuous, 116 curved, 60 deflection, 87 elastic curve, 87 fixed, 107 flexure of, 87 flitched, 102 I beams, 69, 173, 221 impact, 103 inclined, 60 modulus of rupture, 59 moving loads on, 48 neutral axis, 56 oblique loading, 78 reactions, 38 reinforced concrete, 236 resilience of, 103 restrained, 107 sandwich, 102 shaft and beam, 84 shear, external, 43 shear, internal, 65 slope, 88 stiffness, 89 stresses in, 54, 65, 179 three-moment theorem, 1 16, 122 tie and beam, 130 timber, 68 torsion on, 84 uniform strength, 63, 97 work, internal, 104 Bending and compression, 149 and tension, 130 and torsion, 84 Bending moment, see Moment, 4 1 Bessemer process, 25 Blocks in compression, 15, 137 Boilers, 189. 203 rivets, 198 working stresses, 166 Bricks, 31 Bridges, shear in panel, 52 working stresses, 162, 164 Buildings, working stresses, 165 Burnettizing, 21 Cantilevers, 45, 47 Carbon and iron, 22 Cast iron, 22 properties of, 23 working stresses, 166 Cement, 33 Centrifugal force, 133 Clapeyron's formula, 118 Clay, 31 Coefficient, see Modulus. Columns, 2, 137 beam and column, 149 deflection, 142 designing, 146 eccentric load, 148 ends, fixed or hinged, 146 Euler's formula, 142 flexure, direction of, 141, 146 Gordon's formula, 144 ideal column, 142 lacing-bars, 151 pin ends, 147 radius of gyration, 146 Rankine's formula, 144 reinforced concrete, 239 short. 137, 145 straightdine formula, 147 241 242 INDEX, Columns, swelled, 148 timber, 162 transverse force on, 149 working stresses, 162 yield-point, 141 Combined stresses, 3 bending and compression, 149 " " tension, 130 " " torsion, 84 tension and torsion 133 Compression, 3, 15, 137 bending and, 149 eccentric load, 137 granular materials under, 15 Concrete, 35, 235 Concrete-steel, 235 Cone, stresses in, 213, 216 Connecting-rod, 132 Continuous beams, 116 Cooper's lines, 190 Crank, 84 Curvature of beams, 88 Curve, elastic, 87 stress-deformation, 9 Cylinders, thick, 206 thin, 203 Deflection of beams, 87 simple, 89 restrained, 107 uniform strength, 97 Deflection of columns, 142 Deformation, 6, 184 Distortion, 7, 1S6 Dome, 217 Ductility, 6, 17, 26, 29 Earth pressure, 181 Eccentric load, 127, 137, 148 Elastic curve, 87 Elastic limit, 9, 154 Elasticity, modulus of, £, 6 cast iron, 23 concrete, 235, 240 steel, 26 stone, 30 timber, 22 wrought iron, 24 Elasticity, shearing modulus of, C, 7, 186 Ellipse of stress, 174 Elongation, work of, 8, 10, 13, 27 Envelopes, 203 Equilibrium, conditions of, I Euler's formula, 142 Eyebars, 131, 134 Fatigue of metals, 154 Flexure, common theory of, 54 Girders, see Beams. Girder, plate, see Plate girder, 221 Gyration, radius of, 71 Hooks, 129 Hoops, 210 I beam, 69, 173, 221 Impact, 159 Inertia, moment of, 57. 71 product of, 76 Iron, cast, 22 malleable, 28 wrought, 24 Joints, masonry, 138 riveted, 195 Lattice bars, 151 Launhardt-Weyrauch formula, 156 Lime, 32 Lines of principal stress, 180 Linseed oil, 36 Loads, dead and live, 159 eccentric, 127, 137, 148 sudden application, 14 wheel loads on beam, 49 Machinery, working stresses, 166 Manganese in steel, 26 Masonry, 30 working stresses, 166 Materials, 19 Middle third, 138 Modulus of elasticity, see Elasticity, 6 of resilience, 13 of rigidity, 186 of rupture, 59 section, 57 Moment, bending, 41 maximum, 44 on p'ns, 200 position of load for maximum, 49 sign of, 41 Moment of inertia, 71 Moment of resistance, 57 oblique loading, 78 Moment, torsional, 81 Mortar, 32 Neutral axis, 56, 61, 79 Nuts, 134 Oblique load on beam, 78 Open-hearth process, 25 Paint, 36 Parallel rod, 133 Pedestals, 166 Permanent set, 8 INDEX. ^43 Phosphorus in steel. 26 Pier-moment coefficients, 121 Pig iron, 23 Pins, 109, 200 distribution of shear in, 67 friction, 147 working stresses, 163 Pipes, 203 Piston, conical, 189, 216 Plaster, 33 Plate girder, 221 stresses in web. 69, 173 Plates, resistance of, 231 Polar moment of inertia, 71 Portland cement, 34 Posts, see Columns, 137 Power, shaft to transmit, 83 Principal stresses, 170 lines of, 180 Product of inertia, 76 Puddling furnace, 24 Pull and thrust at right angles, 172 Punching steel, effect of, 28 Radius of gyration, 71 Rafter, 60 Rankine's column formula, 144 Rankine's theory of earth pressure, 181 Reactions of beams, 38 Rectangular beams, 58 Repose, angle of, 181 Resilience, definition, 13 modulus of, 13, 103 of bar, 13 of beam, 103 of springs, 229 Resisting moment, see Moment, 57 Restrained beams, 107 Retaining wall, 182 Rigidity, modulus of, 187 Ring under normal pressure, 205 under single load, 218 Rivets, 192 plate girder, 225 steel for, 26 working stresses, 163 Rollers, 163 Rubble. 31 Rupture, modulus of, 59 Safe working stresses, 153 Sandwich beams. 102 Screw threads, 134 Secondary stresses, 158, 197 Section modulus, 57 Set, permanent, 8 Setting of cement, 34 Shafts, 81 working stresses, 166 Shear, 3, 43 Shear, deflection due to. 105 derivative of bending monu-nt, 45 distribution on section of beam, 65 modulus of elasticity. C, 186 position of load for maximum, 48, 50 sign of, 43 timber beams, 68 two shears at right angles, 169, 172, 186 work of, 105 .Shearing planes, 176 Sign of bending moment, 41 compression and tension, 4 shear, 43 .Silicon in iron, 23 Slope of beam, 88 Spangenberg's experiments, 153 Sphere, stresses in, 205, 211 Splices, 195 in plate girder, 223, 225, 226 Springs, 98, 229 Steel, 25 shearing and punching, 28 structural, 26 tool, 28 working stresses, 162 Steel concrete, 235 Stiffness of beams, 89 Sti (fliers, 227 Stirrups, 238 Stone, 29 Straight line formula, 147 Strain, see Deformation. Strength, beams of uniform, 63, 97 cross-sections of equal, 62 ultimate, 1 1 Strength of cast iron, 23 steel, 26 timber, 22, 160 wrought iron, 24 Stresses, 2, 167 alternating, 154 conjugate, 170 distribution on section of beam, 54. 65, 179 ellipse of, 174 internal, 2, 167 lines of principal, 180 principal, 170 reversal of, 155, 162 secondary, 158, 197 sign of, 3 unit, 4, 167 working, 160. 240 Stress-deformation diagram, 9 Struts, see Columns, 2, 137 Sudden loading, effect of, 14 Sulphur in steel, 26 Tanks, 213 244 INDEX. Tempering, 26, 28 Tension, 3, 127 bending and, 130 connections, 133, 196 eccentric load, 127 torsion and, 128, 133 Three-moment theorem, 1 16, 122 Tie, 2, 127 and beam, 130 Timber, 19 column formulas, 16 1 modulus of elasticity, 22 shear in beams, 68 strength of, 22 working stresses, 160 Torsion, 81 bending and, 84 resilience of, 230 tension and, 133 twist of shaft, 83 Trees, growth of, 19 Twist of shaft, 83 Ultimate strength, II Uniform strength, beams of, 63, 97 Unit stresses, 4 Varnish, 36 Varying cross-section, 12 Volume, change of, 183 Wall, retaining, 182 middle third, 138 Web of plate girder, 223 stresses in, 69, 173 Welding, 26 Wheel loads. 49 Wohler's experiments, 153 Wood, 19 Work of elongation, 8, IO, 13, 27 flexure, 104 shear, 105 springs, 229 Working stresses, 153 reinforced concrete, 240 Wrought iron, 24 Yield-point, 10 SHORT-TITLE CATALOGUE OF THE PUBLICATIONS OP JOHN WILEY & SONS, New York. Londoh: CHAPMAN & HALL, Limited. ARRANGED UNDER SUBJECTS. 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(Herrmann — Klein.) 8vo, Machinery of Transmission and Governors. (Herrmann — Klein.). .8vo, Wolff's Windmill as a Prime Mover 8vo, Wood's Turbines 8vo, MATERIALS OF ENGINEERING. Bovey's Strength of Materials and Theory of Structures 8vo, 7 50 Burr's Elasticity and Resistance of the Materials of Engineering! 6th Edition. Reset 8vo, Church's Mechanics of Engineering 8vo, Johnson's Materials of Construction 8vo, Keep's Cast Iron 8vo, Lanza's Applied Mechanics 8vo, Martens's Handbook on Testing Materials. (Henning.) 8vo, Merriman's Mechanics of Materials. 8vo, Strength of Materials nmo, Metcalf's Steel. A manual for Steel-users nmo. Sabin's Industrial and Artistic Technology of Paints and Varnish 8vo, Smith's Materials of Machines nmo, Thurston's Materials of Engineering 3 vols., 8vo, Part II. Iron and Steel 8vo, Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their Constituents 8vo, Text-book of the Materials of Construction 8vo, 12 I 50 I 50 ■i OO 3 00 2 OO 1 25 1 OO 2 50 5 OO 2 50 1 50 3 OO 5 OO 2 OO 4 OO 4 OO 5 OO 4 OO 1 50 3 50 3 OO 2 OO 3 OO 1 50 3 OO 3 OO 3 OO 3 OO 1 OO 7 50 5 OO S OO 3 OO 2 50 7 50 6 00 6 00 2 50 7 50 7 50 5 00 1 00 2 00 3 00 I 00 S 00 3 50 2 50 5 00 Wood's (De V.) Treatise on the Resistance of Materials and an Appendix on the Preservation of Timber 8vo, 2 00 Wood's (De V.) Elements of Analytical Mechanics 8vo, 3 00 Wood's (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and Ste»L 8vo, 4 00 STEAM-ENGINES AND BOILERS. Berry's Temperature-entropy Diagram nmo Carnot's Reflections on the Motive Power of Heat. (Thurston.) nmo Dawson's "Engineering" and Electric Traction Pocket-book. . . .i6mo, mor. Ford's Boiler Making for Boiler Makers i8mo Goss's Locomotive Sparks 8vo Hemcnway's Indicator Practice and Steam-engine Economy i2mo Hutton's Mechanical Engineering of Power Plants 8vo Heat and Heat-engines 8vo Kent's Steam boiler Economy 8vo Kneass's Practice and Theory of the Injector 8vo MacCord's Slide-valves „ 8vo Meyer's Modern Locomotive Construction 4to Peabody's Manual of the Steam-engine Indicator i2mo Tables of the Properties of Saturated Steam and Other Vapors 8vo Thermodynamics of the Steam-engine and Other Heat-engines 8vo Valve-gears for Steam-engines 8vo Peabody and Miller's Steam-boilers 8vo Pray's Twenty Years with the Indicator Large 8vo Pupin's Thermodynamics of Reversible Cycles in Gases and Saturated Vapors (Osterberg. ) nmo Reagan's Locomotives: Simple Compound, and Electric i2mo Rontgen's Principles of Thermodynamics. (Du Bois.) 8vo Sinclair's Locomotive Engine Running and Management nmo Smart's Handbook of Engineering Laboratory Practice nmo Snow's Steam-boiler Practice 8vo Spangler's Valve-gears .' 8vo Notes on Thermodynamics . . i2mo Spangler, Greene, and Marshall's Elements of Steam-engineering 8vo Thurston's Handy Tables 8vo Manual of the Steam-engine 2 vols., 8vo Part I. History, Structure, and Theory 8vo Part II. Design, Construction, and Operation 8vo Handbook of Engine and Boiler Trials, and the Use of the Indicator and the Prony Brake 8vo Stationary Steam-engines 8vo Steam-boiler Explosions in Theory and in Practice nmo Manual of Steam-boilers, their Designs, Construction, and Operation 8vo Weisbach's Heat. Steam, and Steam-engines. (Du Bois.) 8vo Whitham's Steam-engine Design 8vo Wilson's Treatise on Steam-boilers. (Flather.) i6mo Wood's Thermodynamics, Heat Motors, and Refrigerating Machines. ..8vo , I 25 , I 50 . 5 00 , 1 00 , 2 00 t 2 00 , 5 00 , 5 00 , 4 00 , 1 50 , 2 00 , 10 00 . 1 50 , 1 00 » 5 00 , 2 50 . 4 00 1 2 50 , I 25 , 2 50 , 5 00 , 2 00 , 2 50 . 3 CO , 2 50 , I 00 , 3 00 , I 50 , 10 00 , 6 00 , 6 00 . 5 00 , 2 50 >» 1 50 . 5 00 . 5 00 . 5 00 , 2 50 , 4 00 MECHANICS AND MACHINERY. Barr's Kinematics of Machinery 8vo, 2 50 Bovey's Strength of Materials and Theory of Structures 8vo, 7 50 Chase's The Art of Pattern-making i2mo, 2 50 Church!s Mechanics of Engineering „ 8vo, 6 00 13 Church's Notes and Examples in Mechanics 8vo» 2 00 Compton's First Lessons in Metal-working nmo, 1 50 Compton and De Groodt's The Speed Lathe nmo, 1 50 Cromwell's Treatise on Toothed Gearing i2mo, 1 50 Treatise on Belts and Pulleys nmo, 1 50 Dana's Text-book of Elementary Mechanics for Colleges and Schools, .nmo, 1 50 Dingey's Machinery Pattern Making i2mo, 2 00 Dredge's Record of the Transportation Exhibits Building of the World's Columbian Exposition of 1893 4to half mcrocco, 5 00 Du Bois's Elementary Principles of Mechanics: Vol. I. Kinematics 8vo, Vol. II. Statics 8vo, Vol. III. Kinetics 8vo, Mechanics of Engineering. Vol. I Small 4to, Vol. II Small 4to 8 Durley's Kinematics of Machines 8vo, Fitzgerald's Boston Machinist i6mo, Flather's Dynamometers, and the Measurement of Power nmo, Rope Driving nmo, Goss's Locomotive Sparks 8vo, Hall's Car Lubrication nmo, Holly's Art of Saw Filing i8mo, James's Kinematics of a Point and the Rational Mechanics of a Particle. Sm.8vo,2 00 * Johnson's (W. W.) Theoretical Mechanics nmo, Johnson's (L. J.) Statics by Graphic and Algebraic Methods 8vo, Jones's Machine Design: Part I. Kinematics of Machinery 8vo, Part II. Form, Strength, and Proportions of Parts 8vc, Kerr's Power and Power Transmission 8vo, Lanza's Applied Mechanics 8vo, Leonard's Machine Shop, Tools, and Methods 8vo, *Lorenz's Modern Refrigerating Machinery. (Pope, Haven, and Dcan.).8vo, MacCord's Kinematics; or, Practical Mechanism 8vo, Velocity Diagrams 8vo, Maurer's Technical Mechanics .. 8vo, Merriman's Mechanics of Materials 8vo, * Elements of Mechanics nmo, * Michie's Elements of Analytical Mechanics 8vo, Reagan's Locomotives: Simple, Compound, and Electric nmo, Reid's Course in Mechanical Drawing 8vo, Text-book of Mechanical Drawing and Elementary Machine Design. 8vo, Richards's Compressed Air nmo, Robinson's Principles of Mechanism 8vo, Ryan, Norris, and Hoxie's Electrical Machinery. Vol. 1 8vo, Schwamb and Merrill's Elements of Mechanism 8vo, Sinclair's Locomotive-engine Running and Management nmo, Smith's (0.) Press-working of Metals 8vo, Smith's (A. W.) Materials of Machines nmo, Spangler, Greene, and Marshall's Elements of Steam-engineering 8vo, Thurston's Treatise on Friction and Lost Yfork in Machinery and Mill Work 8vo, Animal as a Machine and Prime Motor, and the Laws of Energetics. i2mo, Warren's Elements of Machine Construction and Drawing 8vo, Weisbach's Kinematics and Power of Transmission. (Herrmann — Klein.) . 8vo, Machinery of Transmission and Governors. (Herrmann — Klem.).8vo, Wood's Elements of Analytical Mechanics 8vo, Principles of Elementary Mechanics nmo, Turbines 8vo. The World's Columbian Exposition of 1893 4to, 14 3 50 4 OO 3 50 7 50 10 00 4 00 1 00 3 00 t 00 2 00 I 00 75 0,2 00 3 00 2 00 I 50 3 00 2 00 7 50 4 00 4 00 5 00 1 50 4 00 5 00 I 00 4 00 2 50 2 00 3 00 I 50 3 00 2 50 3 00 2 00 3 00 I 00 3 00 3 00 1 00 7 50 5 00 5 00 3 CO 1 25 2 50 I 00 METALLURGY. Egleston's Metallurgy of Silver, Gold, and Mercury: Vol. L Silver 8vo v 7 50 Vol. II. Gold and Mercury 8vo, 7 50 ** Iles's Lead-smelting. (Postage 9 cents additional.) i2mo, 2 50 Keep's Cast Iron 8vo, 2 50 Kunhardt's Practice of Ore Dressing in Europe 8vo, 1 50 LeChatelier's High-temperature Measurements. (Boudouard — Burgess. )i2mo, 3 00 Metcalf's Steel. A Manual for Steel-users nmo, 2 00 Smith's Materials of Machines nmo, 1 00 Thurston's Materials of Engineering. In Three Parts 8vo 8 00 Part II. Iron and Steel 8vo, 3 50 Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their Constituents 8vo, 2 50 Ulke's Modern Electrolytic Copper Refining 8vo, 3 00 MINERALOGY. Barringer's Description of Minerals of Commercial Value. Oblong, morocco, 2 50 Boyd's Resources of Southwest Virginia 8vo, 3 00 Map of Southwest Virignia Pocket-book form. 2 00 Brush's Manual of Determinative Mineralogy. (Penfield.) 8vo, 4 00 Chester's Catalogue of Minerals 8vo, paper, 1 00 Cloth, 1 25 Dictionary of the Names of Minerals 8vo, 3 50 Dana's System of Mineralogy Large 8vo, half leather, 12 50 First Appendix to Dana's New " System of Mineralogy." Large 8vo, 1 00 Text-book of Mineralogy 8vo, 4 00 Minerals and How to Study Them nmo, 1 50 Catalogue of American Localities of Minerals Large 8vo, 1 00 Manual of Mineralogy and Petrography i2mo , 2 00 Douglas's Untechnical Addresses on Technical Subjects nmo, 1 00 Eakle's Mineral Tables 8vo, 1 25 Egleston's Catalogue of Minerals and Synonyms 8vo ( 2 50 Hussak's The Determination of Rock-forming Minerals, (Smith.) .Small 8vo, 2 00 Merrill's Non-metallic Minerals: Their Occurrence and Uses 8vo, 4 00 * Penfield's Notes on Determinative Mineralogy and Record of Mineral Tests. 8vo paper, o 50 Rosenbusch's Microscopical Physiography of the Rock-making Minerals (Iddings.) 8vo. * Tillman's Text-book of Important Minerals and Rocks .8vo. Williams's Manual of Lithology 8vo, MINING. Beard's Ventilation of Mines i2mo, Boyd's Resources of Southwest Virginia 8vo, Map of Southwest Virginia Pocket book form, Douglas's Untechnical Addresses on Technical Subjects i2mo, * Drinker's Tunneling, Explosive Compounds, and Rock Drills ,4to.hf mor Eissler's Modern High Explosives 8vo. Fowler's Sewage Works Analyses .-. nmo Goodyear's Coal-mines of the Western Coast of the United States nmo. Ihlseng's Manual of Mining '. . , 8vcv. ** Iles's Lead-smelting. (Postage ox. additional.). , nmo, Kunhardt's Practice of Ore Dressing in Europe 8vo v O'Driscoll's Notes on the Treatment of Gold Ores 8vo. * Walke's Lectures on Explosives Svo, Wilson's Cyanide Processes. . . . : , nmo., Chlorination Process nmo, 15 5 OO 2 OO 3 OO 2 50 3 00 2 OO I OO 15 OO 4 OO 2 OO 2 50 5 OO 2 50 I 50 2 OO 4 OO 1 50 1 50 Wilson's Hydraulic and Placer Mining i2mo Treatise on Practical and Theoretical Mine Ventilation nmo, SANITARY SCIENCE. Bashore's Sanitation of a Country House i2mo Folwell's Sewerage. (Designing, Construction, and Maintenance.) 8vo, Water-supply Engineering 8 V0> Fuertes's Water and Public Health i2mo, Water-filtration Works i2mo, Gerhard's Guide to Sanitary House-inspection i6mo, Goodrich's Economic Disposal of Town's Refuse Demy 8vo, Hazen's Filtration of Public Water-supplies 8vo, Leach's The Inspection and Analysis of Food with Special Reference to State Control 8vo, Mason's Water-supply. (Considered principally from a Sanitary Standpoint) 8vo, Examination of Water. (Chemical and Bacteriological.) i2mo, Merriman's Elements of Sanitary Engineering 8vo, Ogden's Sewer Design i2mo, Prescott and Winslow's Elements of Water Bacteriology, with Special Refer- ence to Sanitary Water Analysis i2mo, * Price's Handbook on Sanitation i2mo, Richards's Cost of Food. A Study in Dietaries i2mo, Cost of Living as Modified by Sanitaiy Science nmo, Richards and Woodman's Air, Water, and Food from a Sanitary Stand- point 8vo, * Richards and Williams's The Dietary Computer 8vo, Rideal's Sewage and Bacterial Purification of Sewage 8vo, Turueaure and Russell's Public Water-supplies 8vo, Von Behring's Suppression of Tuberculosis. (Bolduan.) nmo, Whipple's Microscopy of Drinking-water 8vo, Woodhull's Notes on Military Hygiene i6mo, MISCELLANEOUS. De Fursac's Manual of Psychiatry. (Rosanoff and Collins.). . . .Large nmo, 2 50 Emmons's Geological Guide-book of the Rocky Mountain Excursion of the International Congress of Geologists Large 8vo, 1 50 Ferrel's Popular Treatise on the Winds 8vo. 4 00 Haines's American Railway Management nmo, 2 50 Mott's Composition, Digestibility, and Nutritive Value of Food. Mounted chart, 1 25 Fallacy of the Present Theory of Sound i6mo, 1 00 Ricketts's History of Rensselaer Polytechnic Institute, 1824-1894. . 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Small 4to, half morocco, 5 00 Letteras's Hebrew Bible 8vo, 2 23 16 2 00 I 25 I 00 3 oe 4 00 1 50 2 50 1 00 3 50 3 00 7 50 4 00 1 25 2 00 2 00 1 25 1 50 1 00 1 00 2 00 1 50 3 50 5 00 1 00 3 50 1 50 NOV 6 I « HHfl , '\ ■ ■ ' vU \ ■ i I I ■ ■ 1 1 ■ man* # o ** " i <•. i Km ml H kQOO Hi i ■ VV. ■ ■ I I m UK 39*i HUM TOcjj NB ul&8 hj£b I ■ m .;■-«»■