§^ ©hcowtical and §<:a(tiat Statist ON THE STRENGTH OF BEAMS AND COLUMNS; IX WHICH THE ULTIMATE A^B THE ELAS- TIC LIMIT STRENGTH OF BEAMS AND COL- UMNS IS COMPUTED FROM THE ULTIMATE AND ELASTIC LIMIT COMPRESSIVE AND TENSILE STRENGTH OF THE MATERIAL, BY MEANS OF FORMULAS DEDUCED FROM THE CORRECT AND NEW THEORY OF THE TRANSVERSE STRENGTH OF MATERIALS. BY / EOBEET H. COUSI]Srs, Civil Engineer, Formerly Assistant Pi'ofessor of Mathematics at the Virginia Military Institute, Lexington, Ya. E. & F. N. SPON, NEW YORK : 12 CORTLANDT STREET. LONDON : 12 5 Strand, 1889, {All tights 7'eserved.] .t** Copyright, 1889, BY R. H. COUSINS. ■ e B6 INTEODUCTIOK. For more than two centuries the mathematical and me- chanical laws that govern the transverse strength of Beams and Columns have received the attention of the most expert mathematicians of all countries. Galileo, in 1()38, formulated and published the first theory on the subject. He was fol- lowed bj such philosophers as Mariotte, Leilmitz, Bernouilli, Coulomb, and others, each amending and extending the work of his predecessor, until the year 1824:, when Navier succinctly stated the theory that is recognized to be correct at the present day, and to which subsequent writers and investigators have added but little. This theory has neither received the endorsement of the experimenters nor of some of the theoretical writers. *' Ex- cepting as exhibiting approximately the laws of the phenom- ena, the theory of the strength of materials has many prac- tical defects" (Wiesbach). " It has long been known that under the existing theory of beams, which recognizes only two ele- ments of strength — namely, the resistance to direct compression and extension — the strength of a bar of iron subjected to a transverse strain cannot be reconciled with the results obtained from experiments on direct tension, if the neutral axis is in the centre of the bar" (Barlow). During the present century much time and means have been expended in attempts to solve, experimentally, the prob- lems that have eno^aofed the attention of tlie mathematicians, and as the result of their labors we find such experimenters as Ilodgkinson, Fairbairn, and others, whose names are house- IV INTRODUCTION hold words in tlie literature of the suhject, adopting empirical rules for the strength of Beams and Columns rather than the rational formulas deduced bj the scientists. " For no theory of the rupture of a simple beam has jet been proposed which fully satisfies the critical experimenter" (De Yolson Wood). That we should be able to deduce the strength of Beams and Columns from the known teiisile and compressive strength of the material composing them, has been apparent to many writers and experimenters on the subject, but to the present time no theory has been advanced that embodies the mathe- matical and mechanical principles necessary to its accomplish- ment. The theory herein advanced and the formulas resulting therefrom deduce the strength of Beams and Colunms from tlie direct crushing and tensile strength of the material com- posing them, without the aid of that coefficient that has no place in ^nature, the Modulus of Rupture. The theory and the formulas deduced therefrom are in strict accord with cor- rect mechanical and mathematical principles, and the writer believes that they will fully satisfy the results obtained by the experimenter. The great practical benefits to be derived from the correct theory of the strength of Beams and Columns will be evident, when we consider the countless tons of metal that have been made into railroad rails, rolled beams, and the other various shapes, and that the manufacturers were without knowledge of the work to be performed by the diiferent parts of the beam or column in sustaining the load that it was in- tended to carry. The best that they have been able to do is to compute the strength by the aid of an empirical quantity deduced from experiments on '' similar beams." The correct theory will enable them to foretell the strength of any untried shape, and the reason for the strength of those that have been long m use, which is the " true object of theory." R 11 Dallas, Texas, March 13, 1889. * * " CONTENTS. CHAPTER I. 1. 2. 3. 4 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. FORCES DEFINED AND CLASSED. Force Defined, Stress or Strain, The Load, Equilibrium and Resultant Bending Moment— Concentrated Forces, General Formulas for Bending Moments, Uniformly Varying Forces— Rectangular Areas, Resultant, ....... Moment of Uniformly Varying Forces, Uniformly Varying Forces — Circular Segment Areas, Moment of Uniformly Varying Forces — Circular Segment Areas, Resultant, Moment of Uniformly Varying Forces — Circular Arcs, Resultant, " " " . " " . . 3 6 8 8 9 11 13 13 14 16 17 18 19 CHAPTER II. RESISTANCE OF CROSS SECTIONS TO RUPTURE. 18. Moment of Resistance, ...... 19. Neutral Line, 20. Bending A and Moment of Resistance, . 21. Equilibrium, 22. Position of the Neutral Line, ..... 23. Neutral Line at the Transverse Elastic Limit, 24. Movement of the Neutral Line, with the Deflection, 25. Neutral Lines of Rupture in a Rectangular Section, 26. Relative Value of the Compressive and Tensile Strains, 27. Summary of the Tlieory, 4il . 21 22 . 23 24 . 24 26 . 28 s, . 28 . 31 vi CONTENTS. CHAPTER III. TRANSVERSE STRENGTH. ART, PAGE 28. Coefficients of Strength, 38 29. Elasticity of Materials, 38 30. Elastic Limits, 34 31. Elastic Limit of Beams, 34 32. Working Load and Factors of Safety, .34 33. General Formula — Transverse Strength, 35 34. Relative Transverse Strength of a Beam, 35 35. Moment of Resistance — Rectangular Sections, .... 37 36. The Xeutral Line " " .... 38 37. Transverse Strength " " .... 38 38. Designing a Beam " " .... 39 39. To Compute the Compressive Strain, 39 39. " " " Tensile " 39 40. Moment of Resistance — Hodgkinson Sections, . . . .40 4L Neutral Line " " .... 42 42. Transverse Strength " " .... 44 43. To Design a Hodgkinson Section, 44 44. Moment of Resistance — Dbl. T and Hollow Rectangular Sections, 47 45. Neutral Line " " " " " 48 46. Transverse Strength .* «< << " .< 4^ 47. To Design a " " " " " 48 48. Moment of Resistance — Invtd T, Dbl. Invtd. T and U Sections, 52 49. Neutral Line " " " " " " 58 50. Transverse Strength " " " " " " 53 51. To Design an " " " " " " 53 52. Moment of Resistance — Circular Sections, . . . . .55 53. Neutral Line. " " .... 56 54. Transverse Strength " " 56 55. To Compute the Compressive Strain of Circular Sections, . 58 56. " " " Tensile .<.<.< << ^ 5^ 57. Relative Strength of Circular and Square Sections, . . . 59 58. Moment of Resistance — Hollow Circular Sections, . . .60 59. Neutral Line " " " ... 60 60. Transverse Strength " " " ... 60 CHAPTER IV. CAST- 1 RON BEAMS. 61. Compressive Strength of Cast-Iron, ...,,. 64 62. Tensile " " " 64 CONTENTS. VU ABT. PAGE 68. Ratio of the Compressive to tlie Tensile Strength, ... 65 64. Transverse Strength of Cast-iron, 65 65. To Compute the Compressive Strength of Cast-Iron, . . 67 65. " " " Tensile ....... ... 67 66. Neutral Line— Rectangular Cast-Iron Beams, .... 71 67. Transverse Strength " *' " .... 71 68. To Design a Rectangular Cast-iron Beam, .... 73 69. Hodgkinson Beams, 74 70. Mr. Hodgkinson 's Experiments, 74 71. Neutral Line — Hodgkinson Cast-Iron Beams, . . . .75 72. Transverse Strength " " " . , . . 75 73. To Design a Hodgkinson Cast-Iron Beam, 78 74. Neutral Line — Double T and Box Cast-Iron Beams, . . 78 75. Transverse Strength " " " " " . . .78 76. To Design a Double T and Box Cast-Iron Beam, ... 79 77. Neutral Line— Cast-Iron Circular Beams, 80 78. Transverse Strength " " *' .... 81 79. Movement of the Neutral Line 82 80. Relative Strength of Circular and Square Cast-Iron Beams, . 83 81. Neutral Line — Hollow Circular Cast-Iron Beams, . . .85. 82. Transverse Strength " " " " . . . 86 CHAPTER V. WROUGHT-IRON AND STEEL BEAMS. 88. Compressive Strength of Wrought-Iron , . . . . 87 84. Tensile .... .. g;. 85. Compressive " " Steel, 88 86. Tensile ...... gg 87. To Compute Compressive Strength of Wrought-Iron and Steel, 88 88. " " Tensile " " " " " 90 89. Transverse Strength of Wrought-Iron Beams, .... 91 90. " " " Steel Beams, 92 91. Neutral Line — Rectangular W^rought-Iron and Steel Beams, . 92 92. Transverse Strength " " - .« «« , 92 98. Neutral Line— W^rt. Iron and Sfl Dbl. T and Rolled Eye-Beams, 96 94. Transverse Strength " " " " " " " " 96 95. Transverse Strength — Wrought-Iron and Steel Double T Beams, Flanges Unequ _,. ^ f^r Tx' Tbx^ The moment = J hdx = 7, — , . The moment = Tbd' 8 Dividing the moment by the resultant we obtain the lever arm = frfx. 9. Problem I. Required the Moment of an uniformly VARYING compressive FOKCE OF MAXIMUM INTENSITY, C, WITH RESPECT TO AN AXIS, ^, IN THE BACK OF THE PRESSURE WEDGE. Let ABD (Fig. 5) represent a section through the pressure wedge. MM The resultant = c 2 -5 10 STRENGTH OF BEAMS AND COLUMNS. The lever-arm = ^d^ , • R = — — • (11) Problem II. Hequired the mortient of only a jportion of the force distributed over a depth EB = d^ and width Z*, — the force heing zero in intensity at A ; the axis heing at B (Fig. 5). d — d -^— — ^ • C = the intensity of the force at E^ cCq d — ' = the lever-arm of this force at E, considered to be constant over the distance EB^ M — - " ' }^ — *^^ intensity of the force at B less that d. 2X E, - = its lever-arm. Having considered the force to be divided into two portions, the first, GE^ constant in intensity over the depth EB^ and the second increasing in intensity from zero at G to ( 1 — -^-^ — ■- j (7 at i^, by adding the moments of these lyarts we obtain R. Qdr {Sd, - 2d,) a (12) UNIFORMLY VARYING FORCES. 11 By the Calculus : Let a-c = c?c = the depth AB (Fig. 5), ' X = any distance from A, C = the intensity of the force at the depth x, Xc bdx = a small area of the base of the wedge, Jq X, \ 2xc dXc/ .-. Bc = -^, or Eq. 11. Integrating the above expression for Be, between the limits x = d^ and x = di, we obtain ^'^'' {Mc - 2di) C, or Eq. 12. i?e = 6c?c 10, Problem III. Required the Moment, ^„ of an UNIFORMLY VARYING TENSILE FORCE, WITH RESPECT TO AN AXIS, (9, PARALLEL TO THE EDGE, A^ OF THE TENSION WEDGE, AND AT THE DISTANCE OA = d^ FROM IT. Fig. 6 Let ABD (Fig. 6) represent a section through the tension wedge. hcl^. T . , — 7^ — = the resultant. or. f/e -}- i^h = its lever-arm, Jx^ hcL {U + d,) T. (13). 12 STKENGTH OF BEAMS AND COLUMNS. Problem TV. Iiequired the moment of only a lym^t of the above force distributed over the depth EB = d^ and the width h^. Consider tlie force to be divided into two portions. The iirst, with the intensity at ^ constant over the distance EB ; the second increasing in intensity from zero 2ii G to -^ T atE. d^ (1 j-jT^ the intensity of the force at E, ("-'!')=' its lever-arm, d -^ T = the intensity of the force at B, less that E, (d — — M = its lever-arm. By adding the moments we obtain B, = ^\^SdA i^d - d,) - d: (3rZ - 2«f,)]. (14) By the Calculus : Let a*! = ^T = the distance AB, X = any distance from A, icT — =: the intensity of the force at any distance x, (dc -\-x) = its lever-arm, bdx = the differential of the area of the base, . • . i?^ = -|-' {2d + dc) T, or Eci. 13. Integrating the above expression for i?T, between the limits x = dr — d-2 and X = d/t, and placing h^ for b, we have UIS^IFORMLY VARYING FORCES. 13 6fZ, D i?^ = J^ r^drd^ {d - d,) - dj" {ddc - 2^2)."] T. This is not identical in form with Eq. 14, but gives the same numerical result. Section TV. — Uniformly Varying Forces — Circular Segment Areas. 11. Resultant and Lever-arm. The resultant and the moment of an uniformly varying force, distributed ov^er circular segment areas, cannot be readily deduced by means of the elementary methods , used when the areas were rect- angular ; but this can be accomplished with the aid of the Calculus. 12. Problem. Kequieed the moment of an uniformly VARYING FORCE DISTRIBUTED OVER A CIRCULAR SEGMENT, WITH RESPECT TO AN AXIS THAT IS A TANGENT TO THE BASE OF THE PRESSURE WEDGE AND PARALLEL TO ITS EDGE, WHICH IS A CHORD OF THE CIRCLE. Let, in Fig. 7, OBS represent the axis, MNB the base of the pressure wedge, and ABD a section through AB, Notation. — Adopting the notation defined in Art. 7 with the following in addition : 14 STRENGTH OF BEAMS AND COLUMNS. Let r = the radius of the circle, d = d^ -\- dj: = the diameter, a?c = d^ = the versine of one half the arc MBN^ X := any distance from the axis OBS^ — C = the intensity of the pressure at any distance, a?, from the axis OBS, 2 {2?'x — x^)^dx = the differential of the area of the base, Integrating, we obtain Substituting the following equalities : r versin. — = one half of the arc of the segment MBN Xq := CCq y reducing, we obtain ^^'"^ 24dc\ ^^c^T l^(l\{dc-r) + r'^(30?'-14^e)] -^Comp.arc (12(?e- 15r)?'n(t5) from which the moment of any compressed wedge may be computed. The factor Comp. Arc is the arc MBN. 13. Problem. Eequired the moment of an uniformly VARYING TENSILE FORCE, DISTRIBUTED OVER THE SEGMENT OF A CIRCLE, WITH RESPECT TO AN AXIS THAT IS A TANGENT TO THE CIRCLE, PARALLEL TO THE EDGE OF THE TENSION WEDGE, AND AT THE DISTANCE d^ FROM IT. Let, in Fig. 8, FOS represent the axis, MNB the base of the tension wedge, MN the edge, and OABD a section through OB. UNIFORMLY VARYING FORCES. 15 Let x^ — d^ — tlie distance AB, X = any distance from B toward A. ^T — ^ . T — the intensity of the force at any distance x^ — x from A^ d — X — \i^ lever-arm. 2 {^rx — x")^ dx — the differential of the area of the base of the wedge. For notation, refer to Arts. 7 and 12. E^ = f'^-^IiII^T{d -x)2 {2rx - x'fdx, «/ x^ r,E^ = ^T[{2rx.-x\)[^-^^ ^'"^ M~ "^ 24^/ ^ r V 6 24 V J Substituting the following values : (2a^', - Q?\) =: 4/^/. 7' — 1 Tversin. — =: one half the tension arc MBN. tC"J» ■ " rp • reducing, we obtain 16 STRENGTH OF BEAMS AND COLUMNS. R^ ^ 24^/ L '^^c'^T [4<^\ (or — d^) + ISr' (r — d^)] + Tension arc {12d^ — 9r)r^) 1 n g\ from wliicli the required moment may be computed. 14. Problem. Kequired the re- sultant OR AMOUNT OF DIRECT FORCE OF AN UNIFORMLY VARYING FORCE DISTRIB- UTED OVER A CIRCULAR SEGMENT AREA. Let Fig. 9 represent the pressure wedge, MBN the base or circular seg- ment, and B the origin of co-ordinates. Let a?c = AB the versine of one half of the arc MBN^ X = any distance from the origin B^ k 2 {^rx — x'') dx — the differential of tlie segment, T^= the resultant or volume of the pressure wedge, 6^— the greatest height, DB^ '.^ca?. ^ Qy^ 2[^rx-x')dx, vbf. . • . \ =2C\ {^rx — x^) ( -^-! — \-{-rversin.-^[ - — - 1 • Substituting the equalities given in Art. 12 and reducing, we have -\- Segment arc {d^ — r) Zr • (17) The resultant of a tension force may be obtained from the above equation by placing T for C. The cubic contents of any cylindrical wedge may be com- puted from Eq. 17 by substituting for 6^, expressed in pounds, A, the greatest height of the wedge in inches. UNIFORMLY VARYING FORCES. 17 The area of tlie base of the pressure wedge or that of any segment of a circle may be computed from the following : fl Area = 2 \/cl^dj; (^c— ^) "h ^eg77ie7it arc Y.-f-, in which ^4 = versine of one half the arc of the segment, r = the radius, d^ :=. the diameter, less d^. {\nA) Section N .— Uniformly Yarying Forces — Circular Arcs. 15. Problem, Kequired the moment of an uniformly VARYING COMPRESSIVE FORCE DISTRIBUTED OVER THE ARC OF A CIRCLE, WITH RESPECT TO AN AXIS THAT IS A TANGENT TO THE CIRCLE AND PARALLEL TO THE CHORD JOINING THE EXTREMI- TIES OF THE ARC. Let, in Fig. 10, OBS represent the axis, MBN the arc, ABD a section through AB^ B the origin of co-ordinates, C the intensity of the force at B, and zero that at M and li. J) Fig. 1 d, A For flotation refer to Arts. T and 12. Let Xc = dc — the versine AB oi one half the arc JJTBJV X = any variable distance from B or the axis. 18 STRENGTH OF BEAMS AND C0LU3INS. C — the intensity of the force at x distance, Xq r = the radius of the circle, = twice the difierential of the arc, V2rx — X r> /"'^c 2/ 'ax ( a?c — x) p r Integrating, we obtain E, = 2Cr{2rx - x'jf^^^sjilM r - r) y ^Xq/ -J Substituting the vahies given in Art. 12, we have ^„ = ^ [|/4^[2(3r - d^)] + Co?,>.p.arc{2d, - 3r)] - (18) 16. Problem. Hequired the moment of an uniformly VARYING FORCE DISTRIBUTED OVER THE ARC OF A CIRCLE, WITH RESPECT TO AN AXIS THAT IS A TANGENT TO THE CIRCLE, PARAL- LEL TO THE CHORD CONNECTING THE EXTREMITIES OF THE ARC, AND AT THE DISTANCE cl^ FROM IT. Let, in Fig. 11, I^OS represent the axis, MBY the arc. rxgn Or 4-' rt T I B OABD a section tlirougli OB, 7'the intensity of the force at B, and zero at Jfand iT. UNIFORMLY VARYING FORCES. 19 For notation refer to Arts. 7 and 12. Let a?T = cIt. = the versine of one half the arc MBN^ X = any distance from the origin B^ tJCr, iV T = the intensity of the force at any distance w, d — X := its lever-arm '^rdx — the differential of the arc, V2rx — xi 2rdx i?, = f^^ _^^iL-_ (^^-ZJ^\ {d - X) JO y'^rx — x' V ^'x / Integrating, we obtain ^T = -^ I Vd,d^ [2 {r + d,)] + Tension arc (2r7^- r) J, (19) from which the required moment may be computed. 17. Problem. Hequiked the resultant or amount of DIRECT FORCE OF AN UNIFORMLY VARYING FORCE DISTRIBUTED OVER AN ARC OF A CIRCLE. Let, in Fig. 9, page 16, J/J5 3^ represent the arc of the circle, MNBD a wedge whose cylindrical surface is equal to the re- quired resultant, the force being C in intensity at B and zero at M and N. Let x^ = 6?c = the distance AB^ X — any distance from the origin of co-ordinates B^ J^ C = the intensity of the force at the point a?, Xq ''Irdx ./ o ^= the differential of the arc, Y := the resultant or volume of the pressure wedge, y _ rx^ X, — X ^^ "Irdx Jo X. ^c V2rx — i X 20 STRENGTH OF BEAMS AND COLUMNS. . • • ^=^i\j^ VdA+ ^V^^^^^ ^^^ K - r)!^, (20) from which the required resultant may be computed. The curved surface of a cylindrical wedge may be com- puted from the above formula by substituting for C^ expressed in pounds, A, the greatest height of the wedge in inches. CHAPTER II. RESISTANCE OF CROSS-SECTIONS TO RUPTURE. 18. Moment of Resistance. The cross-section is the shape of the figure and the area that any material, such as a beam, would show, should it be cut into two pieces by a plane perpendicular to its length, and its resistance to rujpUire at this plane or section is the number of inch-pounds that its fibres Avill oifer to forces tendino; to cross-hrecik the beam or material of which it is a section : this is called the Moraent of Resistance of the cross-section. The Moment of Resistance varies in amount with the material and the shape of the cross-section, but it is entirely independent of the length of the beam and of the manner in which the load may be applied, in each case ; the same cross- section and material will offer the same number of inch-pounds of resistance when broken across. 19. Neutral Line. AVhen a beam is broken across, or is acted upon by forces that bend or tend to break it, we know from observation that its fibres on the lower or convex side, AB (Fig. 12), are in a state of tension, and that those on the 22 STRENGTH OF BEAMS AND COLUMNS. upper or concave side, CD^ are compressed ; but our knowl- edge obtained from observation is limited to what takes place on the surface of the beam. We can only know what takes ])lace within such a beam by reasoning from analogy ; there is a tensile strain in the lower side of the beam, AB^ and just the reverse character of strain in the upper side, CD. In order that these two directly opposite strains may exist in the same beam at the same time, both strains must decrease from the surface toward a common point within tlie beam, where both strains become zero in intensity, and they may be classed and treated as uniformly varying forces. A line, 7^Z, for the longitudinal section of the beam, or a plane for the beam, is called the neutral line or line of no Htrain y its position in a beam having a cross-section of a given shape, at the instant of rapture, depends upon the material alone, or upon the ratio existing between the hreaJc- Ing compressive and tensile fibre strains. No line in this plane lias any of the properties of an axis that are usually assigned to it by writers on this subject. 20. Bending Moment and Moment of Resist- ance. In order to obtain the relation existing between the Bending Moment of the applied load and the Moment of Resistance of the cross-section of the beam, conceive one half of the beam, ABCD (Fig. 12), to be removed and the bent-lever, og f ^\^. 13), to be substituted for it, and that the same cohesion to exist between the fibres of the bent-lever and those of the beam along the line, fg., that originally existed between the fibres of the two halves of the beam along the same line, the hent-lever^ however, preserving its distinctive (character of a hent-lever. The applied load., B, causes the l)ent-lever, og fi and the half of the beam, A gfD., to move downward in the direction of the lower arrow of the figure, and the end of the lever, 0, and that of the half of the beam, /), to revolve around f in the direction of the upper arrows, RESISTANCE OF CROSS-SECTIONS TO RUPTURE. 23 /h T -^ FlQ 13 the effect being identical with tliat produced by conceiving tlie fulcrum^ f^ to remain stationary and a moving force, Z -f- 2, to be applied to the end of the bent-lever, 0, and at the same time to be pressed in the direction of its length by a force, C^ equal in magnitude to the amount of direct com- pression along the line, fn^ above the neutral line^ nl. When the bent-lever is made to revolve around the fulcrum, y, it meets with an opposition of compression to its motion, de- creasing in intensity from ih^ fulcrum^ f^ to the neutral line^ n^ when it becomes zero in value ; at this point the opposition changes to a tensile resistance which increases in intensity from zero at n to its maximum, T^ at (j. The hent-lever and the original vertical section of the beam, fg^ are pressed closest together at \hQ fulcrum^ f -, from this point they continue to separate, by virtue of the ductility and compressibility of the material composing the beam, until rupture takes place, either in its upper or lower fibres ; the action of the bent-lever being identical with that of the half of the beam for wdiich it was substituted. 21o Equilibrium. The Bending Moment of the applied load, Z, and the Moment of Resistance of the tensile and 24 STRENGTH OF BEAMS AND COLUMNS. compressive fibre strains must be taken or computed witli reference to the fulcrum^ f^ and in order that equilibrium shall exist, the Bending Moment of the applied load, Z, must be equal to the Moment of Resistance, or the sum of the moments of the tensile and compressive fibre strains, and that the latter must be equal to each other in magnitude. 22; Position of the Neutral Line. By deducing general formulas for the onoments of the tensile and compi^es- sive fibre strain in a cross-section of a given shape, placing them equal to each other and deducing from the equation so formed a general formula for either gn or fn, tlie position of the neutral line may be found in any section of the same shape by substituting in the formula its dimensions and any Jcnoivn values for the tensile and compressive fibre strains. In sections of an uniform shape, such as the rectangle and the circle, the depth of the neutral line below the compressed side of the beam may be obtained by multiplying the depth of the rectangle and the diameter of the circle by a deter- mined quantity that is constant for each shape and material ; this constant multiplier being its ]30sition when the depth of the rectangle and diameter of the circle is unity. But in irregular shaped sections, such as the T, Double T, Box, Rolled-eyebeain and other sliapes, in which the metal is not continuous from the neutral line to the top and bottom, a special solution must be made for eacli case to determine the position of the neutral line from which to compute the Moment of Resistance of the section. Before beams of this character are manufactured, an economical 2:)osition should be assumed and a sufficient area of metal placed above and below the neutral line to furnish the required Moment of Resistance. 23. Neutral Line at the Transverse Elastic Limit. Our object in testing, to destruction, the strength of any piece of construction material is to obtain information RESISTANCE OF CROSS-SECTIONS TO RUPTURE. 25 that will guide us to a correct knowledge of its use, when safety to life a.nd property is demanded, whether this destruc- tion be by means of extension or compression / as we apply the load in small instalments there are only two points in its intensity at which we can record the knowledge thus gained for future intelligent, comparative use when it has reached the elastic Umit, and when it is sufficiently intense to rupture or destroy the piece of material tested. When a beam is broken by a transverse load that has been applied in small instalments, we have two similar points, the elastic liirhit load and the rupturing load, and these are the only points at which our information can be used to compute the strength of similar material when used in structures. We know that in an ordinary beam without a load its com- pressive and tensile fibre strain is zero, and that the breaking transverse load produces the hreahing compressive and tensile strain in the fibres. Kow, does the transverse elastic limit load produce the elastic fibre strain limits % Is the neutral surface the same as that for rupture ? When the beam is un- loaded each plane of fibres is a neutral su'rface ; as the load is applied the compressive and tensile strain penetrates the beam from the top and bottom respectively ; theoretically we know they must meet at a common point within the beam at the instant of rupture^ and that this is fully 'sustained by ex- periments will be shown in the sequel. Our theory demands that when the same ratio exists between the tensile and com- pressive elastic fibre strain limits that does between the ulti- mate or breaking strains, in order that equilibrium shall exist, the neutral surfaces must be identical, but the theory does not require that the transverse elastic limit load shall j^roduce the elastic fibre strain limits ; we can only gain the desired informa- tion from discussing a numerical example. The mean compressive and tensile elastic fibre strain limits for good wrought-iron is C = 30000 pounds and T = 30000 pounds per square inch. With these strains, from formulas de- 26 STRENGTH OF BEAMS AND COLUMNS. duced in the sequel, the centre elastic limit transverse load of a bar of wrouglit-iron six inches square and ten feet span is 44100 pounds. The centre elastic limit transverse load of a bar of wrought-iron, one inch square and one foot span, is 2250 pounds, and that of the above beam from the well- known formula is, ' J , hd"^' 6 X 36 X 2250 ,_^^ . Load = —J— = - — — TTj = 48600 pounds. From this practical identity of results, as we have only used average values, and other special tests given in the sequel, we are authorized to conclude that the transverse elastic limit load produces the tensile and compressive fibre strain elastic limits. 24. Movement of the Neutral Line with the De- flection. Having established the fact that the transverse elastic limit load produces the elastic fibre strain limits — and our theory requires that the elastic limit neutral line and the neutral line of rupture shall coincide only when their ratios are the same, but should they be unequal they must occupy diiferent positions — in the sequel it will be shown that where these ratios are unequal the elastic limit neutral line is situated hettoeen the neutral line of rupture and the bottom or ex- tended side of the beam, and that as the loading advanced from the elastic limit load to the rupturing load, the neutral surface must have moved upward or toward the compressed surface of the beam. From the above we conclude that as there was no change in the condition of the loading that could have reversed the direction in which the neutral line moved from its position at the elastic limit to that at rupture^ the neutral Ihie at the in- ception of the loading was at the bottom or extended side of the beam, and that as the loading progressed it 'moved up- ward or toward the compressed side of the beam— the tension area, to avoid rupture in its fibres, continues to encroach upon RESISTANCE OF CROSS-SECTIONS TO RUPTURE. 27 the compressed area until the rupturing strain is produced in botli the top and the bottom of the beam. The neiitral line^ at the inception of the loading, being at the bottom or extended side of the beam, it can only be moved upward by reason of the deflection and equally with it. If, from the dimensions of the beam, it should not be able to deflect sufficiently to move the neutral line to the position required for equilibriuin between moments of resistance of the 'idtimate fibre strains, the true breaking strength will not be obtained for the beam. When this is the case the observed breaking load will be too large for wooden, wrought-iron, steel and tough cast-iron beams, and too small for the more fractious varieties of material ; for should the compressive strain reach its ultimate limit before the tensile strain an in- crease of the load will develop a crushing strain in excess of the true crushing intensity, as is frequently done in crushing short blocks; the beam will, however, continue to deflect under tliese increased loads, and will finally develop the full tensile strength, wdien the beam will be broken by a load much in excess of its true breaking load. From the above, the reason for the variation in the modulus of Txiptiire that is required in the '' common theory of flexure " is apparent, as the shorter beams in most series of experiments, especially of cast-iron, did not deflect sufficiently to break with the true breaking load, and, therefore, it re- quired a larger inodidus or empirical coefficient for the shorter beams than for the longer beams of the same series of tests. At the instant of deflection the hending ^noment of the applied load is held in equilibrium by a jpurely compressive resistance, distributed over the section as an uniformly vary- ing force, being zero in intensity at the bottom or extended side of the beam and greatest in intensity on the opposite side. This is a very important principle, as from it we shall, in tlie sequel, deduce the correct theory of the strength of columns. 28 STRENGTH OF BEAMS AND COLUMNS. 25. Neutral Lines of Rupture in Rectangular Sections. It maj not be out of place here to anticipate the resuks obtained in the sequel, by stating the positions of the neutral lines of rupture in a rectangular section when composed of the different kinds of material used in construc- tion. Let the section be six inches deep, q the ratio of the com- pressive to the tensile strains of rupture, or G ^ T. r depth of neutral line -^ ^ ^ . r, K below the com- \ *^''*-"'°" ? = ^-5 1 pressed side of the \ 2-45 ins. I section J " q = 6. " " 3.00 " " q =z 4:. " " 3.24 " Steel ^ = 1.5 " " 4.29 " Wrought-iron q = 1. " " 4.68 " Beech, English....^ = 0.7Y5 " " 4.90 " " American q = 0.383 " " 5.36 " From this comparative statement we observe the order in which the neutral lines of rupture are arranged in rectangular sections ; the same order of arrangement exists in all other uniform sections. For different kinds of wood and cast-iron the neutral line of rupture lies between the extremes given in the above table of comparison. 26. Relative Value of the Compressive and Tensile Strains. Experiments have fully shown that the compressive and tensile strains do not possess equal values as factors in determining the transverse load that a beam will bear, and that the influence of the tensile strain predomi- nates. The elastic limit or technical hreaMng load of a wrought- iron beam one inch square and twelve inches span, loaded in the middle, is 2000 pounds when HESISTAl^CE OF CROSS-SECTIOT^S TO RUPTURE. 29 C = 30000, T = 30000 pounds, and the Moment of Ee- sistance = 6084 inch-pounds from our formulas. We will now endeavor to trace the effect that will be pro- duced upon the amount of the transverse breaking load from varying the exact and relative values of C and T from those in the above-described wrought-iron beam, which will be taken as our standard of comparison. Since the crushing and tensile strength of the material composing any beam must each sustain one half of the breaking load of the beam, the per cent of loss or gain in the transverse strength should be, approximately^ one half the sum of the per cents of the losses or gains in the values of C and J', but this will be found to be true for only the smaller ratios oi C -^ T that exist in materials of construction. The centre breaking load of a white pine beam one inch square and twelve inches span is 450 pounds when C = 5000, T = 10000 pounds, and the Moment of Ke- sistance = 1260 inch-pounds. In passing from the standard wrought-iron beam where C ^=^ T to the white pine beam where (7 = 0.5 T^ the following changes take place : Loss in the value of C 83.4 per cent. " T. m.Q '' Apparent loss to the transverse load 75.0 " Actual '' " " 77.5 " Loss to the Moment of Resistance 79.2 " results, practically, identical for this ratio. From Mr. Kirkaldy's experiments a bar of steel one inch square and twelve inches span will break with a centre load of 6400 pounds when C — 160000, T = 70000 pounds. Moment of Eesistance = 19200 inch-pounds. In passing from our standard wrought- iron beam to the steel beam, the following changes take place : 30 STRENGTH OF BEAMS AND COLUMNS. Gain in the value of C .433.3 per cent. " " T 133.3 " Apparent gain to the transverse load 283.3 " Actual " " '^ '' 220.0 " Gain to the Moment of Resistance 215.0 " Mr. Hodgkinson found the centre breaking load of a certain cast-iron beam, one inch square and twelve inches span, to be 2000 pounds when C = 115000, T = 14200 pounds, and the Moment of Ee- sistance = 6600 inch-pounds. In passing from our standard wrought-iron to the cast-iron beam of the same size, the following changes take place : Gain in the value of C -|- 283.4 per cent. Loss " " " '' T - 52.6 " " Apparent gain to the transverse load .... 115.4 " " Actual " '^ " " 0.0 " " Gain to the Moment of Resistance 8.4 " " In this experiment it required 283.4 per cent gain in the value of C to offset a loss of 52.6 per cent in the value of T^ or that the compressive strength does not sustain its proper proportion of tlie load. This great discrepancy between the legitimate theoretical deductions and the results obtained from experiments cannot be reconciled on the hypothesis that the forces are in equi- librium with respect to an axis that lies within the beam, the moment in each case, for rectangular sections, being the resultants of the tensile and compressive forces, multiplied by two thirds of the depth of their respective areas, showing that the compressive strain works under no disadvantages ; but on our theory this discrepancy is fully accounted for. The lever- arm of the crusJung resultant is one third the depth of the RESISTANCE OF CKOSS-SECTIONS TO RUPTURE 31 compressed area, while that of the tensile resultant is only one third less than the total depth of rectangular beams. 27. Suiiimary of the Theory. The theory herein advanced to explain the relation that exists between the Bend- ing Moment of the applied load and the Moment of Resist- ance of the material composing the beam, may be expressed by the following hypotheses : 1st. Tlie fibres of the beam on its convex side are extended and those on the concave side are compressed in the direction of the length of the beam, and there are no strains but those of extension and compression. 2d. There is a layer or plane of fibres between the extended and compressed sides of the beam that is neither extended nor compressed, w^hich is called the neutral surface or neutral line for any line in this plane. 3d. The strains of compression and extension in the fibres of the beam are, in intensity, directly proportional to their dis- tance from the neutral surface. 4th. The axis or origin of moments for the tensile and com- pressive resistance of the fibres of any section at right angles to length of the beam, is a line of the section at its intersection with the top or compressed side of the beam. 5th. The fibres of a beam wall be ruptured by either the tensile or compressive strains in its concave and convex sur- faces, whenever they reach in intensity those found by experi- ment to be the direct breaking tensile and compressive fibre strains for the material composing the beam. 6th. The Bending Moment of the load at any section is equal to the sum of the moments of resistance to compression and extension of the fibres, or to the Moment of Resistance of the secti/)n of the beam. 7th. The sum of the moments of resistance of the fibres to compression is equal to the sum of the moments of resistance of the fibres to extension. 32 STRENGTH OF BEAMS AND COLUMNS. 8tli. The algebraic sum of tlie direct forces of compression and extension can never become zero. 9th. The Moment of Resistance of the section is equal to the sum of the moments of resistance to the compression and extension of its fibres. 10th. The transverse elastic limit load produces the tensile and compressive fibre strain elastic limits. CHAPTER III. TRANSVERSE STRENGTH. Section I. — General Conditions. 28. Coefficients or Modvili of Streng^th are quanti- ties expressing the intensity of the strain under which a piece of a given material gives way when strained in a given manner, such intensity being expressed in units of weight for each unit of sectional area of the material over wliich the strain is distributed. The unit of weight ordinarily employed in expressing the strength of materials is the nuinber ofjjonnds a/voirduj)ois on the square inch. Coefficients of Strength are of as many different kinds as there are different ways of breaking a piece of material. Coefficients of Tensile Strength or Tenacity is the strain necessary to rupture or pull apart a prismatic bar of any given material whose section is one square inch, w^hen acting in the direction of the length of the bar. This strain is the T oi our formulas. Coefficient of Crushing Strength or Compression is the pressure required to crush a prism of a given material whose section is one square inch, and whose length does not exceed from one to Jive times its diameter, in order that there may be no tendency to give way by bending sideways. This pressure is the C of our formulas. 29. Elasticity of Materials. It is found by experi- ment that if the load necessary to produce a strain and fracture of a given kind is applied in small instalments, that before the load becomes sufficiently intense to produce rupture, it will 34 strength: of beams at^d colum:n^s. cause a change to take place in the form of the material, and if the load is removed before this intensity of the fibre strain passes certain limits, the material possesses the power of re- turning to its original form. This is called its elasticity. 30. Elastic Limits. When the material possesses the power of recovering its exact original form without " set" on the removal of a load of a given intensity, the greatest load under which it will do this is called the limit of perfect elas- ticity. The limit of elasticity as ordinarily defined and used by experimenters is that point or intensity of strain where equal instalments or increments of the applied load cease to pro- duce equal changes of form, or where the change in form in- creases more rapidly than the load. 31. The Elastic Limit of Beams may be deter- mined by applying small equal parts of the load and noting the increase in deflection after each increase of the load, al- lowing sufficient time for each increase of the load to pro- duce its full effect. When it is found that the deflections increase more rapidly than the load, its elastic limit has been reached and passed. The relation between the elastic limit load of the beam and the elastic limit of the tensile and com- pressive fibre strains of the material composing the beam will be shown in the sequel, or that the elastic limit load of the beam produces the elastic limit strain for the fibres. 32. Working Load and Factor of Safety. The greatest load that any piece of material, used in a struc- ture, is expected to bear is called the worMng load. The hreaking load to be provided for in designing a piece of material to be used in a structure is made greater than the worMng load in a certain ratio that is detennined from ex- perience, in order to provide for unforeseen defects in the material and a possible increase in the magnitude of the ex- pected working load. TRANSVERSE STRENGTH. 85 The factor of safety is the ratio or quotient obtained by dividing the hreahing load by the worthing load required. 33. General Formula. In our first chapter we deduced rules or formulas, from which can be computed the Greatest Bending Moment that a load applied to a beam in a given manner will produce without reference to the shape of its cross-section, or to the material composing the beam. In our second chapter i)rinciijles are deduced from which can be computed the Greatest Moment of Resistance cross- sections of the various shapes and material will exert at the instant of rupture, without reference to the length of the beam or to the manner in which the load may be applied. To avoid repetition, the formulas for the Moments of Re- sistance are deduced in this chapter. Our knowledge of the transverse strength of beams will now be complete if we com- pute and make the Greatest Moment of Resistance of the cross-section of the beam equal to the Greatest Bending Moment of the applied load. Let i? = the Greatest Moment of Resistance of the l)eam, L = the total applied load in pounds, s = the span, the distance between the supports in inches, n = the factor defined in Art. 6, page 7, M = the Greatest Bending Moment of the applied load; (21) from which the breaking load of the beam may be computed. 34. Relative Transverse Strength of a Beam. Referring to the values of the factor, n, of Eq. 21, as given Eq. 8, page 7, we have . Z = nLs R ns 36 STRENGTH OF BEAMS AND COLUMNS. in Art. 0, page 7, it is found in each ease to be either unity or a fraction^ and when these values are introduced for n^ in Eq. 21, it is equivalent to multiplying the numerator, R^ by the denominator of the fraction, n ; hence if we make 1 — = m, n Eq. 21 will become Z = — . (22) s Computing and tabulating the values of m from those of /i, Art. 6, we obtain the following relation between the breaking loads of a beam, the span, material and cross-section remain- ing the same or constant ; the breaking load of a beam, fixed at one end and loaded at the other, being the unit or standard of strength. m Beam fixed at one end and loaded at the other 1 Beam '' " " " uniformly... 2 Beam supported at its ends and loaded at its middle 4 Beam " " " " " uniformly 8 Beam fixed at both ends and loaded at its middle 8 Beam " " " " " uniformly 12 Eq. 22 is the general formula from which will be de- duced the transverse strength of beams of all sections by giving to the factor, R, its proper value. TRANSVERSE STRENGTH. 37 Section II. — Transverse Strength — Rectangular Sections. 35. Moment of Resistance. In Fig. 14, let ^^i>X represent the section, nl the neutral line, AnlX the compressed area, BnlD the extended area, AnBMnNA any section through the strain wedges at right angles to the neutral line, and AX Adopting the notation heretofore used in the axis or origin. Art. 7, we have C = the greatest intensity, per square inch, in peunds of the compressive strain, T = the greatest intensity, per square inch, in pounds of the tensile strain, H^ =: the moment of the compressive strain in inch-pounds, R^ = the moment of the tensile strain in inch-pounds, R =^ R^-\- Rj, = the Moment of Resistance of the section in inch pounds, r/c = the depth of the compressed area. An, in inches, d.^, =: the depth of the extended area, Rn, in inches, d = 6?c + ^^4 = tl^e depth of the section AR in inches, h = the width of the section RR in inches, q =^ the quotient arising from dividing O by T, L = the total applied load in pounds, s = the span, the distance between the supports in inches, m = the factor, defined in Art. 34, page 35. 38 STRENGTH OF BEAMS AND COLUMNS. From Eq. 11, page 10, we have, for the moment of com- pressive resistance, i?c = *^, (23) and for the moment of the tensile resistance from Eq. 13, -R^^^-pi^d+cQT. . (24) 36. The Neutral Line, From the 7th hypothesis, Art. 23, page 31, we have, from equating the second members of Eqs. 23 and 24, 6 6 ^ 1 c; J from which and c\ = ^(-0>5 + 1/2^+2.25) ^26) The position of the neutral line in any rectangular sec- tion may be computed from either of these formulas ; it is independent of the absolute intensity of the maximum tensile and compressive iibre strains, but depends upon their ratio C-^ T— q for its position. The application of these formulas is illustrated in Examples 5, 6 and 7 of the sequel. 37, Transverse Strength, General formulas for the transverse strength may be obtained by substituting for 7? the Moment of Eesistance in Eq. 22, page 36, its values 2i?c == 27?^ from Eqs. 23 and 24. • '^ 3^-' ^^'^ . • . Z = '"^^L {2d + d.) T, (28) OS TRANSVERSE STRENGTH. 39 From eitlier of these equations, either the breaking load or the elastic limit load may be computed by giving to C and T the breaking or elastic limit values of the material composing the beam. The application of Eq. 27 is illustrated in Exam- ples 7, 26, 27, 28, 36, 37, 38, 39 and 40, and Eq. 28 in Ex- amples 5, 6, 7, 8, 25 and 30 of the sequel. 38. To Design a Beam. In designing a beam of a rectangular section that shall break with a given total applied load, distributed over the span of the beam in a given manner, the span of the beam will be determined from the position in which it is to be used, and the depth, cZ, in inches will be as- sumed / the criLsliimj and tensile strength of the material composing the beam must be hiown / it will then only be nec- essary to determine the width^ Z>, in order that the beam shall break with the required load. From Eq. 25 deduce the position of the neutral line^ which is independent of the width, 5, then from Eq. 27 deduce the value of h, the width •••^=^„ (29) which gives the required width, since all of the factors in the second member of this equation are known quantities. Should the assumed (iQ^t\i and the computed ^Y\di\\ not give an economical section for the beam, a second depth must be as- sumed from this information and a second width computed ; this process should be repeated until a satisfactory result is secured. 39. To Compute the Compressive and Tensile Strains. Pkoblem I. — The position of the neutral line and the crushing strain of the material of a rectangidar section may he computed from the knotvn transverse hreaking load and tensile strength. 40 STRENGTH OF BEAMS AND COLUMNS. From Eq. 28, page 38, deduce tlie value of r/„ and which gives the required position of the neutral line, illus- trated in Examples 1, 2, 21, 22 and 35. The crusJiing strength can be computed by deducing its value, 6', from Eq. 27, mbd: ^ ^ The application is illustrated in Exaniples 1, 2 and 21. Problem II. — The position of the neutral line and the tensile strength of the material of a recto/ngular section may he com^puted from the known transverse hreaking load of the heam and the crushing strength of the matei'ial. From Eq. 27 we have a, ^ Z-^. (32) mo (J for the position of the neutral line, which is illustrated in Ex- ample 4. From Eq. 28 we have for the required tensile strength rp O l^S /"QQX "" mhd^ {M -f d)' ^ ^ illustrated in Example 4. TRANSVERSE STRENGTH. 41 Section 111. — Transverse Strength — Ilodghlnson Section. 40. Moment of Resistance. In Fig. 15 let ABDX represent the section, AOX tlie axis or origin of moments. Ji y^ I 1 -1- , 6. t ^C B D nl the neutral line, the area above it being convpressed and that below it extended. OnEMnNO^ a section through the strain wedges on the line OE. d = the depth of the web and section, OE, in inches, d^ =z the " " npper flanges in inches, d^ — the " " lower a u h = the width of the web in inches, h^ = the sum of the widths of the upper flanges in inches, or AX - 5, ^2 = the sum of the widths of the lower flanges in inches, or BI) - h. For other notation refer to Art. 35. From Eqs. 11 and 12, page 10, we have for the moments of com^pressive resistance, Web, E, = 6 ' 7 7 2 Upper flanges, E^ = -^-j- {^d — 2d^) C. 42 STRENGTH OF BEAMS AND COLUMNS. Adding, we obtain for the section, and for the ?no?nents of tensile resistance from Eqs. 13 and 14, page 11, Web, E, = ^-^ {2d + d,) r, Lower flanges, ^. = ^ \j>dA {U - d,) — d,' (Sd - 2^7,) 1 T. Adding the above equations we obtain for the moinent of tensile resistance of the section, 7?,= VM^\U-\-d^ ■^llM.dJ^U-d^-dX^d-M,)]]^ (35) and the Moment of Resistance of the Hodgkinson section is 2^e = 2/4 from Eqs. 34 and 35. 41. The Nevitral Line. By equating the vakies of R^ and 7?T given l)y Eqs. 34 and 35, and deducing from the equation so formed the vahie of d^^ the position of the neutral line is determined, but as this will involve the solution of a biquadratic equation, the general solution will be too complex for ordinary use. The following approximate formula obtained by neglecting certain quantities that do not materially affect the result in ordinary cases, will give its position sufficiently near for all practical purposes when the beam is made of cast-iron : , ^ F+^' + V^- - l//^' + p% (A 36) p and k must be made equal to one third and one half of the numerical quantities, 3/> and 2k, respectively, and their algebraic sign must be that of the term in which tliey appear in the reduced equation. From the above equation one value of d^ will be determined, and with it, Eq. (A 36) may be reduced to an equation of the second degree by dividing the cubic equation by d^ plus or minus the numerical value deduced above, giving the numer- ical value in the division the reverse sign to that computed. From this resulting equation of the second degree, the other two values of d^ may be computed ; the value that represents the position of the neutral Hue may be generally determined from inspection. Problem II. — Required the position of the neutral line in any flanged heam when the transverse load and tensile strength of the material have heen obtained from experiments. In Eq. 38, wdiich gives the relation between the Moment of 44 STRENGTH OF BEAMS AND COLUMNS. the applied load and the Moment of Resistance of the section, all of the terms become known quantities except d^ ; substitut- ing for the letters their values in any given case, it will always reduce to the following general form : d^ — ad^ ± ind^ ± ^^ = 0, in which a^ in^ and n are numerical quantities. By substitut- ing for 6?T a new unknown quantity, (Ij. — X — |— — , the equation will reduce to the following general form : x' ± Zpx ± 2A: = 0, (B 36) which can be solved by Cardan's Rule, as in the preceding Problem. 42. Transverse Strength, To obtain a general formula from which the transverse strength of a Hodgkinson beam may be computed, place for R^ the Moment of Resist- ance in Eq. 22, page 36, its values, 2^^ = 2^^? from Eqs. 34 and 35, and we have and L = |j^.[w; + h,d; {3d, - 2fZ,)], (37) Z=~\'hd^'' {2d-\-d,)+b,[3d,d^{2d-d,)-d,' {Zd-M,)]]{SS) oaT^sL. J From either of these formulas, either the breaking or elastic limit load may be computed. Their application is illustrated in Examples 8, 9, 10, 22, 29, 30 and 31. 43. To Design a Hodgkinson Beam. An econom- ical position for the neutral line must be assumed and a suffi- cient area of the metal arranged above and below it, to furnish the required cotnjpressive and tensile resistance / the depth of TRANSVERSE STRENGTH. 45 tlie beam, d^ and the thickness of the web, J, should also be assumed. Let L = the required breaking load in pounds, s = the span in inches, TYh =: the factor defined in Art. 35. Step I. — Assume the thickness of the web, J, and its depth, d^ which is also the depth of the beam, and an economi- cal position for tlie neutral line. The compressed area above the neutral line must furnish one half and the extended area below it the other half of the required strength of the beam. For the compressed area : Step II, — Place the value of the compressive resistance given by Eq. 11, page 10, for 7? in Eq. 22, page 36, and we shall have for the load, Z„ sustained by the compressed area of the web^ A - —^^' (39) Step III. — For the top flanges deduct the load, Zj, ob- tained in Step XL, from one half of the required breaking load, Z, of the beam ; the balance is the breaking load, Z^, for the top flanges, or, A - f - A- m In order to design an area of section that will sustain this load, we must assume a convenient depth, c?,, for the top flanges and obtain the sum of their widths, 5,, by placing for R^ in Eq. 22, page 36, its value for this case, from Eq. 12, page 10, and we will have J _ l,d^d^ (2^7 - d,) - d: (Sd - 2d,)] mT ^ ' One half the width computed from this formula must l>e arranged on each side of the web. Step VI. — Should the computed dimensions from those assumed produce a badly designed section, new dimensions must be assumed from the knowledge thus obtained, and a second computation made as in the first instance. TRANSVEESE STRENGTH. 47 Section IV. — Transverse Strength — Double T and Hollow Rectangle or Box Section. 44. Moinent of Resistance. Let Fig. 16 represent the sections of tlie Double T and Box Sections respectively ; the same letters in the text apply to both sections. A ^ . h I I- ^ ^ ^ T At ->E B. aj ... h. J^ D Let ABDX represent the sections, OnEMnNO a section through the strain wedge on the line OE^ A OX the axes, and nl the neutral lines. Notation for the Double T : 1) = the width of the web in inches, l)^ = 1).^ = the sum of the widths of the flanges of tlie double T in inches, or ^^ — h, d^ = the depth of the top flanges of the double T, d, = the " " " bottom " " '' Notation for the Box Section : 1) =z the sum of the widths of the sides of the box in inches, h^ =z h^ =z the inside width of the box, rZj =r the depth of the top of the box, d„ = the " " " bottom of the box. Giving the letters the above definition the moments oi tensile 48 STRENGTH OF BEAMS AND COLUMNS. and compressive resistance may be obtained from the following formulas : B, from Eq. 34, (47) E^ '' " 35. (48) 45. Neutral Line. The position of the neutral line in the Double T and Box Sections, when constructed of cast-iron, may be computed approximately from the formula given for the neutral line in the Hodgkinson beam, or < from Eq. 36. (49) When beams of these sections are made of wrought-iron and steel, the neutral line is so close to the bottom of the section that its position cannot be determined from an approximate formula. An exact solution of a long equation of the fourth degree must be made to determine its position. 46. Transverse Strength. The breaking strength, Z, may be computed from the formula given for the trans- verse strength of the Hodgkinson beam, by noting the defini- tion of the letters given in Arts. 35 and 44. L from Eq. 37, (50) L " " 38. (51) 47. To Design a Double T and Box Section, Step I. Assume the depths <:Z, of the Box or Double T and the position of the neutral line, also the width, h, of the web of the Double T, or the sum of the equal widths of the sides of the box. The compressed area above the assumed position of the neutral line must sustain one half and the extended area below it the other half of the Bending Moment of the applied load, Z, that the beam is required- to break with. For the compressed area : Step II. — The moment of compressive resistance that the sides of the Box or the web of the Double T will oifer to the TEANSVERSE STRENGTH. 49 Bending Moment of the applied load from Eq. 11, page 10, being substituted for B, in Eq. 22, page 36, will give for its proportion of the load Z. = "^^. (52) Step III. — For the to]) flanges of the Double T or the tojp of the Box Section deduct the load, Zj, found by Step II. from one half of the total ap]3lied load, Z ; the remainder will be the breaking load, Z^, for the top flanges or the top of the box, as the case may require, or A = I - L,. (53) To design an area of section that will resist this load, Z^, assume a width, ^j, for the sum of the widths of the top flanges or the top of the Box, and compute therefrom their depth, d„ by substituting for the unoment of C07npressive resistance^ R^ in Eq. 22, page 36, its value in this case from Eq. 12, page 10, and we will have A = 14: {^^^ must be placed be- tween the sides of the Box and the balance in two equal project- ing flanges on the outside of the Box at the top and bottom ; the total width of the rolled plate that forms tli« top and bottom of the Box is equal to (J -f- Jj), and the depth of the plates that form its sides is equal to cZ — (d^ -\- d^ as found by Steps III. and y. For the Double T : Riveted plate sections. The width of the top and bottom plates is equal to {b -\- b^), and the depth of the web plate is d — {d, + d,). Rolled Eyebeams are arranged like that for a cast-iron Double T beam. 52 STRENGTH OF BEAMS AND COLUMNS. Section Y. — Transverse Strength. The Inverted T, Double Inverted T and |_J Sections. 48. Moment of Resistance. Let Figs. 17, 18 and 19 respectively represent the three sections, nl the neutral lines, and AXt]iQ agues or origin of moments of resistance for each section. ^_X A X Tx^u \ '1^ .1 I, . • - -1 ^ ^ ■^ J. a Nl/ ^l. ^ c Fi^J 8 t J\ X .1 "M. V \ Fig. 19 D M N In addition to the notation given in Art. 35, observe the following for nse in the formulas given in this Section : The Inverted T, Fig. 17. h = the width of the web in inches, h.^ — the sum of the widths of tlie flanges-, or BD — J, d^ = the depth of the flanges in inches. The Donhle Inverted T, Fig. 18. h z=z tlie sum of the widths of the webs in inches. h d„ flanges, or MJV — h, the depth of the flanges in inches. The LI Section, Fig. 19. h = tlie sum of the widths of the webs in inches, h^ — the widtli of the bottom in inches, or OP — h, d^ = the depth of the bottom in inches. Noting tlie above definitions, the moment of compressive resistance for each section will be from Eq. 11, page 10, rCo = — 7. — , (60) TRAI^SVERSE STRENGTH. 53 and for the mom£7it of tensile resistance tlie formula given for the Hodgkinson beam will apply, or R^ from Eq. 35, page 42. (61) The Moment of Resistance of the section will be B^ jR,-{-B., = 27?, = 27?„ (62) from the above equations. 49. The Neutral Line. In cast-iron beams of these sections the neutral line will be given approximately by the following formula : . _ dh + VmdbA (g + 1) + f^'d' (4g + 5) 50. Transverse Strength. Place for E in Eq. 22 the values for the Moment of Resistance of the sections, and we will have L = -3^-. (64) and L from Eq. 38, page 44. (65) From either of these formulas the transverse breaking load may be computed. 51. To Design an Inverted T, Double In- verted T and i_| Sections. Stei> I. — Assume a value for the depth, f7, and the width or sum of the widths of the webs, J, and an economical posi- tion for the neutral line^ and from these compute the othei- dimensions. Step II. — The compressed area above the neutral line must have a moment of resistance equal to one half the Bend- ing Moment of the applied load, Z, or hd^^ G Ls rr^c^. 54 STREI^GTH OF BEAMS AND COLUMNS. If our assumed values for h and d^ do not satisfy this equa- tion, other values must be assumed until the two members give identical numerical quantities. For the extended area : Step III. — The load, Z, sustained by the extended area of the web or webs, ^, may be obtained by substituting their mo- 7nents of resistance for H in Eq. 22, page 36, its value from Eq. 13, page 11, L, = 'bd,{U + dy'^. (67) Step IV, — For the flanges and hottom deduct the load Z, found by Step III., from one half of the applied load Z, the remainder, L^^ is the load that must be sustained by the flanges and bottom. A = ^ - A- (68) In order to design an area that will sustain this portion Z, of the applied Z, assume a convenient depth, d^^ and from this compute the width h^^ by substituting for B in Eq. 22, page 36, its value, the moment of tensile resistance for this case from Eq. 14, page 12, Z, = \UA {^d - d,) - d: {M - U:)~\ ^^M^, (69) from which I — M^L^s .^^x ' M4^ {2d - ^,) - d,' (Sd - M,) mT' ^ ^ which gives the required width. TRANSVERSE STRENGTH. 55 Section YI. — Transverse Strength — Circular Sections. 52. 3Ioment of Resistance. In Fig. 20, let BnOl represent the section, BiiOMnNB a section tlirongli the strain wedges on the line BO^ B the origin of co-ordinates, DBS the axis or origin of moments, nl the neutral line^ nBl the compressed area, and n 01 the extended area Adopting the notation heretofore used and defined, we have C = the greatest intensity of the compressive strain in pounds, per square inch, T = the greatest intensity of the tensile strain in pounds, per square inch, B^ = the moment of the tensile resistance in inch-pounds, 7?c = the moment of the compressive resistance in inch- pounds, B =z the Moment of Kesistance of the section, d^ = the versine Bn of one half of the arc nBl in inches, d^, = the versine On of one half of the arc nOl in inches, d =: df,-{- d^, = the diameter of the circle in inches, r = the radius of the circle in inches, q -— the quotient arising from dividing by T^ 56 STRENGTH OF BEAMS AND COLUMNS. L — tlie total applied load in pounds, s = tlie span, the distance between the supports in inches, m — the factor defined in Art. 34, page 35, Comp. arc = the arc bounding the compressed area in inches. Tension arc = the arc bounding the extended area in inches. For the moment of coinjpressive resistance from Eq. 15, page 14, Ave have K^ = C 21^Z, r Vd.d,, [4.d: {d, - r) + r' (30r - Ud,)] + Comp. arc (12<^c — 15r)/ J, (71) and for the moment of tensile resistance B^ = T 24:d. r Vd,d^ [UJ" (5r - d^) + ISr' {r - d,)] + Tension arc {12d^ — 9r) 7-0, (72) and for the Moment of Resistance of the circular section, B = B, + B, = 2B, = 2E,. (73) 53, The Neutral Line, The position of the neutral line in circular wooden, cast-iron, wrought-iron, and steel beams will be found tabulated on pages 80, 100 and 112. 54, Transverse Strength. Substituting for i?, in Eq. 22, page 36, its value 21?, ^ 2R^ from Eqs. 71 and 72, we have ^ = 1^, [ ^^- [^^c^ (^^c - r) + r' {30r - 1^)] + Comj). arc {\2d, - l^r)r'~^, (74) TRANSVERSE STRENGTH. 57 and L = riiT 12r4s _ [ |/.///, \\d: {^or - cJ;) + 18r^ (r - <,)] + Tension cu^c {V^d^ — 9r) r"" , (75) from which the breaking load of a circular beam may be com- puted. By observing the following equalities, much time and labor will be economized in comparing the results computed from these formulas : \/d,d^ ]Jtd; (d, - 7') + r' (SOr - Ud,)] = VdAl'^dT" (5^^ - d,) + ISr' {r - d^)], =F (12fZe - 1-5;") r' = ± (12^, - 9r) r% Comp. arc — 27rr — Tension arc. Another Method, The Moments of Kesistance of circular sections are to each other as the cubes of their rcidii / hence by computing and tabulating the Moments of Resist- ance for all required jiositions of the neutral line, in a circle whose radius is unity ^ those for any other circle composed of the same material may be computed by multiplying the tabular number by the cube of the radius. Let f^ = the second member of Eq. 74, when r = 1^ except the factor - — - s f^. = the second member of Eq. 75, when /» =: 1, except m J. the factor . s . • . Z = "^f^, (76) and L ^ !!?:!5zr. (77) 58 STRENGTH OF BEAMS AND COLUMNS. The application of Eq. 76 is illustrated in Examples 11, 32 and 33, and Eq. 77 in Examples 12, 13 and 14. Tables giving the computed values of /. and /,, for the diiferent positions of the neutral line in cast-iron, wrought- iron and steel, circular sections, whose radius is unity, are given on pages 80 and 100. 55. To Compute the Compressive Strain. Problem. — T/ie position of the neutral line and the com- pressive strain of the material of a circular section may he computed from the Jcnown transverse hreaking load and tensile strength of the material. Deducing the value of f from Eq. 77, we have from the proper table in the sequel take the value of q cor- responding to this value of f ; then C=qT. (79) Illustrated by Examj)les 3 and 23. 56. To Compute tlie Tensile Strain, Problem. — The tensile strain of the material of a circular section may he computed from the known transverse hreak- ing load and compressive strength of the material. From Eq. 76 deduce the value of /„ • • • /c = -— ^7>- (80) mr (J ^ From the Table given in the sequel for the material take the value of q^ corresponding to the computed value of /c^ then ^ = f (81) Illustrated by Example 24. TRANSVERSE STRENGTH. 59 57. Relative Strength of Circular and Square Beams, Constant ratios exist between tlie Moments of Ee- sistance of the circle and its inscribed, circumscribed and equal area square, when the material composing the square and the circle has the same tensile and compressive strength, and consequently the same ratios exist between the transverse strength of the beams of which they are sections, the sj)an and manner of loading the circular and square beam being the same. General Formula : Let f = the position of the neutral line in the square when the side is unity, and d^ = fd^, h = d and r^ =: -—, 8 hence, hd ^ O f^d^ C Strength of the Square = — ^ — z=''-^- — ■, from Eq. 23. Strength of the Circle ■= rfC = ^f^^ f roj^ Eq. 76, 8 from which we have Square : Circle : : ^^ : ^^, ^ 3 8' . • . Circle = Square X — - - . (82^) ^ 8^/y'^ ^ ^ Case I. — Wlien the circle is inscribed loithin the square^ d = d, and Eq. 82 becomes Circle = Square X ^^^ (83) Illustrated by Examples 15 and 16 of the sequel. Case II. — When the square is inscribed within the circle side of the square, d = diameter of the circle d X 0.707, and Eq. 82 becomes Circle = Sonar e X • — (84) ^ 2.8271/2 ^ ^ 60 STRENGTH OF BEAMS AND COLUIINS. Case III. — WTien the square and circle are equal in area side of the square^ d =■ diameter of the circle d X 0.886, and Eq. 82 becomes *" /^ (85) Circle = Square X 5.564 f Illustrated by Examples 18 and 19. Section YII. — Transverse Strength — Hollow Circular Sections. 58. Moment of Resistance, In Fig. 21, let BnOl represent the section, BnOMnNB a section through the strain wedges on the line BO, B the origin of co-ordinates, DBS the axis or origin of moments, and nl the neutral line. %-2i a. _ ^./ _ — ^^ A t T J ^ -> O Let t = the thickness of the metal ring in inches, r = the radius of the outer circumference. r, — r — ^ = a a a inner u tC= CsiTid tT= T. For other notation refer to Art. 52. When t, the thickness of the metal ring, is very small, the entire strain distributed over the metal of the section may be TRANSVERSE STRENGTH. 61 treated as if it was all concentrated in the outer surface of the cylindrical beam ; if ^ is not very thin, r — ^ or the mean ra- dius must be used instead of r in the following formulas. Making the substitution of tC for C and tT ior T in Eqs. 18 and 19, the moments of coTripressive and tensile resistance become ^c = ^[ Vd^, X 2 (3/- - d,) + Comp. arc {2d, - 3r)1, (86) and J^T = ^1 ^d4^ x2{r^d^)-\- Te7ision arc {2d^— ?') J- ^^'^) For the Moment of Resistance of the section, E^B,^B, = 2E, = 2^,. (88) 59, The Neutral Line, The position of the neutral line in hollow circular sections of cast-iron, wrought-iron and steel, when the radius of the outer circumference is unity, will be found tabulated in the sequel. 60, Transverse Streng^th. Substituting for the Moment of Resistance B, in Eq. 22, page 36, its values in this case, 2^c = 2^^^ from Eqs. 86 and 87, we have Z = mrt C s dc and Z = mrtT S Orr [ Vd,d^ X 2 {2>r - d,)+Comp. arc {2d, - 3r) J, (89) [^ VdA X2{r + d^)+ Tension arc {2d^— r)\ (90) From either of the above formulas the transverse strength of hollow cylindrical beams may be computed. 62 STRENGTH OF BEAMS AND COLUMNS. In these formulas the following equalities exist : Vd^X 2 (3/' — d^) = Vd^d^ X 2 (/^ + d^\ ± {2d, -Sr)= T {2d, - r\ Oomp. arc = 27rr — Tension arc. When the metal ring is very thin, r in the above formulas is the radius of the outer circle, otherwise it is the radius of the mean circle, r — — . Another Method, The Moments of Eesistance of thin hollow circular sections are to each other as the squares of their radii ; hence, bj computing and tabulating the Moments of Resistance for all required positions of the neutral line, in a thin hollow circle whose radius is unity, those for any other thin hollow circle composed of the same material may be computed by multipljdng the tabular number by the square of the radius. Lety^ = the second member of Eq. 89, when radius = 1, ex- cept the factor , s f^ — the second member of Eq. 90, when radius = 1, ex- cept the factor . With this notation Eqs. 89 and 90 become L = ^^^^, (91) s and 4 L = 'Hi'^fil. (92) TRANSVERSE STRENGTH. 63 Tables giving the computed values oif^ and f^. for the posi- tions of the neutral line in thin hollow circles, whose radius is unity, will be found in the sequel. The application of the above formulas is illustrated in Examples 20 and 34. CHAPTEK lY. CAST-IRON BEAMS. Section I. — General Conditions. 61. Compressive Strength, CruMng. — The crush- ing strength of cast-iron that is nsually obtained by experi- menters and recorded for use in designing structures, is the number of pounds avoirdupois that it requires to crush a prism of the material whose sectional area is one square inch and length from one and a half to three times the diameter, under which condition it is found to be more nearly constant in value for the same material than when the height bears a greater or less ratio to tlie least diameter of the prism tested. Yalue. — The range of values for the crushing strength of cast-iron may be taken as being from 85000 pounds to 125000 pounds per square inch ; the mean is al)0ut 100000 pounds. Elastic Limit. — The compressive elasticity of cast-iron as recorded by the earlier experimenters appears to be very defective, but improved methods of manufacture have pro- duced a cast-iron from which modern experimenters find the increase in the amount of the compression to be, practically, in direct proportion to the increase in the load, within the elastic limit of the cast-iron. The compressive elastic limit varies from three fifths to nearly the crushing strength. 63. Tensile Streiijj^th. T'^3J:_X_60 ^ pounds. 4 X 1.02 (0.4968)^ ^ The following Example is taken from Major Wade's Ex- periments on the " Strength and other Properties of Metals for Cannon," made for the United States Government : Example 2.— Kequired the Crushing Strength of Major Wade's cast-iron in third fusion, from the transverse strength of a rectangular beam, when The depth. . .d = 2.01 inches, T = 26569 pounds by test, '^ breadth h = 2.008 " Z = 16172 '' " " " span 6' == 20. " m = 4 for a centre load. * Barlow's " Strensrth of Materials," p. 152. GENERAL CONDITIONS. 69 For the position of the neutral line, we have from Eq. 30, _ IX?:^ _ /9 X 4x 2.008(2.0iF26569^ 12 X 16172 X 20 _ ^„ ^^^^ ^-2 y 4 X 4 X 2.008 X 2¥569 " ^ '^^^^' . • . ^e = 2.01 - 0.8835 = 1.1265 inches. Then we can compute the crushing strength, O, from Eq. 31, „ 3 X 16172 X 20 ^._^ , Example 3. — Required the Crushing Strength of certain ,st-ir when cast-iron from the transverse strength of a circular beam The diameter d = l'M29, T= 29400 pounds mean of tests, '• span s = 20'^0, Z = 2118 " by test, 771 = 4. From Eq. Y8 we have ^ 2118 X 20 ^ -^^ 4 (0.5645/ 29400 From the Table, page 80, for the above value of /t> we have q = 3.275, and from Eq. 79, . • . 0= 29400 X 3.275 = 96285 pounds. From the United States Government Report of the Tests of Iron and Steel for 1884,"^ the mean crushing strength of this cast-iron was C — 100700 pounds. The C^mshing Strength may be computed from the Trans- verse Strength of the Hodgkinson or any other flanged beam. When the neutral line is not situated above the top line of the tension flange, its position may be computed from Eq. 30, page 40, as if the beam were rectangular. When the neutral * Senate Ex. Doc. No. 35— 49th Congress, 1st Session, p. 284. 70 STRENGTH OF BEAMS AND COLUMNS. line is located above the top line of the tension flange, its position can be determined by the method given in Problem I., page 43, but in either case the value of C must be computed bj formula 37, page 4-1:, C being then the only unknown quantity that it will contain. Tensile Strength. — This may be computed from the known Crushing Strength of the material and the Transverse Strength of any rectangular, circular or flanged beam. Example 4. — Required the Tensile Strength from the Trans- verse Strength of a rectangular beam made of a certain kind of cast-iron that was tested by Captain James, of England,* when The depth d = 0.75 inches, C= 60233 pounds by test, " breadth h = 0.75 " Z = 134 " " • " " span s = 54.0 " fn = 4: The position of the neutral line is to be computed from Eq. 32, page 40, . ,/ 3X134X54 r. ^ ' I ^^ = ^' 4-^0:75-^^60233 = '-^ ^^^^^^^^' hence d^, = 0.75 — 0.4 = 0.35 inches. Then we compute the required value of the tensile strength, Ty from Eq. 33, page 40, ^ = 1 TTl^ TTTi^VT^ r. ^K ■ ^ -T^ = 14509 DOUuds. 4 X 0.75 X 0.35 (2 X 0.75 + 0.4) ^ Mr. Hodgkinson determined for this iron, Clyde, ISTo. 3, O = 106039 and T = 23468 pounds. * British " Report on the Application of Iron to Railway Structures," p. 257. EECTANGULAR CAST-IRON BEAMS. 71 Section II. — Rectangular Cast-iron Beams. 66. The Neutral Line. The position of the neutral hne in rectangular cast-iron beams, for the different required ratios of C ^ T — q^ have been computed from Eq. 2(3, page 38, and tabulated below for reference. Table of positions of the neutral line in rectangidar cast- iron heams. Ratio of Crushing 1 I Depth of Neutral Line Ratio of Crushing Depth of Neutral Line to Tenacity Below the Crushed Side to Tenacity, Below the Crushed Side or of the Beam, or or of the Beam, or C-5- T^ q. dc. C-f- T=q. dc. 8.0 0.4191 d 5.5 0.4831 d 7.875 0.4127 d 5 . 375 0.4871 d 7.75 0.4243 d 5.25 0.4913 d 7.625 0.4226 d 5.125 0.4956 d 7.5 0.4298 d 5.0 0.5000 d 7 375 4326 d 4.875 0.5045 d 7.25 0.4354 d 4.75 0.5092 d 7.125 0.4384 d 4.625 5139 d 7.0 0.4414 d 4.5 5189 d 6.875 0.4444 d 4.375 5240 d 6.75 0.4475 d 4.25 0.5293 d 6.625 0.4507 d 4.125 0.5347 d 6.5 0.4540 d 4.0 0.5403 d 6.375 0.4573 d 3.875 0.5461 d 6.25 0.4608 d 3.75 0.5521 d 6.125 0.4642 d 3.625 0.5583 d 6.0 0.4678 d 3.5 5647 d 5.875 0.4715 d 3.375 0.5714 d 5.75 0.4754 (/ 3.25 0.5784 d 5.625 0.4791 d 3.125 0.5855 d 3.0 0.5930 d Should lower numerical values of q be required, refer to the table of neutral lines in rectangular Sections of Steel. 67. Transverse Streng-th, From either Eq. 27 or 28, as may be most convenient, the transverse strength of rect- angular cast-iron beams may be computed. 72 STRENGTH OF BEAMS AND COLUMNS. Example 5. — Required tlie centre breaking load of a rect- angular cast-iron beam when Tlie dej^tli d — 1 incli, C = 10243i pounds by test, " breadth l =. 1 '^ 7^=10724: " " " " span 6' == 54: inches, ^ = 6 . 125 and ??^ = 4. The position of the neutral line may be computed from Eq. 26, . 1(- 0.5-fi/2x 6.125+2.25) ,, ,^,,, . -, < ^ r, ,0- I 1 -^- = 0.4642 mches, or its value may be taken from the Table. The transverse breaking load, Z, from Eq. 28, becomes ^ ^ 4 X 1(0.4642)- X102J34 ^ ^^, ^^^^^^ 3 X 54 ^ Mr. Hodgkinson, in his experiments, gives the above values of C and T for Blaen-Avon No. 2 iron, and 447 pounds as the mean transverse streno^th of 4 beams tested, wliile Captain James found 556 pounds to be the mean of 3 beams of the above dimensions for this brand of iron. Example 6.— Required the centre breaking load of the fol- lowing : 1 inch square cast-iron beams, the span being 54 inches, and the values of C^ 7" and the experimental break- ing loads being from Mr. Hodgkinson' s experiments. It is very probable tlmt tliese bars did not deflect sufficiently to de- velop the true transverse strength.'" * Barlow's " Strengtli of Muteriuls," p. 163. RECTANGULAR CAST-IROJS^ BEAMS. 73 Tested Strength per Transverse Square Inch. Neutral Line, or dc. Strength. Crushing C. Tensile T. Computed, lbs. Tested. lbs. CaiTon No 2, C B.... 106375 16683 0.457 548 476 ( ( ( < 2, H. B. . . 108540 13505 0.418 468 463 i( te 3, C. B.... 115442 14200 0.416 493 446 " " 3, H. B. . . 133440 17755 0.427 606 527 Buffery " 1, C. B.... 93366 17466 0.488 549 463 ii ( ( 1, H. B. . . 86397 13434 0.456 429 436 Mean . . . 515 468 Example T. — Required tlie centre breaking load of a rect- angular cast-iron beam, from Mr. Barlow's experiments, when The depth ..d= 2.00 inches, C = 95462 lbs., computed, Ex. 1, " breadth h = 1.99 " T= 18750 " by test, " span. . . ..9 = 60.00 " ^ = 5 . 091 and m = 4. For the neutral line we have, from Eq. 26, d^ = 0''. 993-1, Computed transverse strength from Eq. 27 or 28 . . . .4175 lbs., Tested " " 3863 " 68. To Design a Rectangular Cast-iron Beam. The principles and formulas required are given in Art. 38, page 39. 74 STRENGTH OF BEAMS AND COLUMNS. Section III. — Ilodghinson Cast-iron Beains. 69. In this section we will consider the principles as if applied to that class of cast-iron beams whose bottom or tension flange is larger than its top or compressed flange, though they are applicable to all other forms of flanged beams. 70. Mr. Hodgkinson's Experiments, Previous to the year 181:0, when Mr. Hodgkinson commenced his ex- perimental investigation into the strength of materials, the only metal beam that had been used in structures to any ex- tent was the Inverted T, Fig. IT, and the Double T, Fig. 1(), in which the top and bottom flanges Avere equal in area. Commencing his experiments with the equal flanged Double T, he found that, by increasing the area of the lower or tension flange by small amounts, he continued to increase the transverse strength of the beam, his unit of measure and standard of comparison being the quotient obtained from di- viding the breaking load by the area of the section of the beam expressed in inches, and that this continued to increase until he had reached the point where the tension flange was about six times the area of the compressed flange ; increasing- it to a greater ratio he found that the transverse strength j^er square inch of section began to decrease. Experimenting with cast-iron, in which the ratio of C to jT, or the crushing to the tensile strength, was about six^ he recommended that the ten- sion flange should have six times the area of the compressed flange, in order to obtain the greatest transverse strength per square inch of section. But subsequent investigators, experi- menting with cast-iron possessing greater tenacity^ or a less ratio of (7 to 7^ than that used bj^ Mr. Hodgkinson, found the greatest strength with a lower ratio of extended to compressed flange than that recommended by him. The reason for this will be apparent when the formulas for the strength of these beams are examined. 1 ' 1 .\/ . . _ b. _■ b ; 2 Z 1 HODGKINSON CAST-IRON BEAMS. 75 71. Neutral Line. When the neutral Kne is not situ- ated witliin the tension flanges, its position may be ajjjjwxi- mately computed from Eq. 3G, page 42. Should the neutral line lie witliin the tension flange, its position may IJe accu- rately computed from Eq. 30, or from the methods given in Probs. 1 and 2, page 43, when the transverse load and either the compressive or tensile strength has been determined from experiments. AFT 1 73. Transverse Strength. The transverse strength of the Hodgkinson or of any beam having flanges at Xh^ top and the bottom may be computed from either Eq. 37 or 38, as may be most con- venient. ExA^MPLE 8. — Required the centre breaking load of a Hodg- kinson cast-iron beam when The depth of the beam ^ = 5.125 ins., C = 96000 lbs., " breadth of the Aveb J = 0.34 '' T = 16000 " " " " top flanges, h, = 1.22 " q= 6 " depth '' " "' d, = 0.315 " s = U ins. " '' bottom " d,= 0.56 '' m = 4. '' breadth " " '' h, — 4.83 " Nexdral Line. — Our formula 36, giving the position of the neutral line approximately only, places it in each case a little too near the tension flange. In this example it locates it within the tension flange, and we will assume that it coin- cides with its upper line, or d^ = 4^565 and d^ = 0.56. Erom Eq. 37 the transverse breaking load becomes J- _ 0.34 (4.565)3+1.22 (0.315)^ (3X4.565 -2X0.B15^ .vQfinon i-«ifiiy.. J^ - 3X4565X54 ' ^X^^OOO = 1,616 lbs. Tlie tension area beiner continuous from the neutral line to the 76 STRENGTH OF BEAMS AND COLUMNS. bottom of the beam, Eq. 28, page 38, can be used to compute the load from the tensile strength T, in wliich h ^ h -\- h^ oi this example. -r 4x5.17x0.56^2x5.125+4.565)16000 _,^^,,, „ L = • ^ — z~, ' — 16944 lbs. 3Xo4 These two values of Z, not being identical in numerical value, our assumed position for the neutral line must liave been a little too low down in the section for exact equilibrium. The above described beam was one of the series of beams that was used by Mr. Hodgkinson to determine the " section of greatest strength." Its transverse breaking load was, by test, 16730 pounds." The value for T that we use was also determined from experiment ; the ratio was supposed to be 6^ -^ T = 6. In this series of beams the neutral line of rupture "was continuously lowered in the section, from pro- gressively increasing the area of the tension flange until it finally reached the upper surface of this flange. Example 9. — Required the centre breaking load of a Hodg- kinson Cast-iron Beam (Fig. 15), when The depth of the beam d = 14.0 ins., C = 75983 lbs. by test, " breadth of the web b= 1.0" 7=13815" " " top flano:es di= 1.0" ^ = 5.5, "depth " " " b, = 2.5" ,s = 16 feet, " " " bottom flanges . ..^2 = 1.75" m = 4. " breadth " " " ...M= 11.00" From Eq. 36 the position of the neutral line becomes , _ 1X14+ Vf2xl4xllXl.7 5(5.5 + l)4-(lyl4)M4vo 5+5) _,.,_. "^^ - 2x1(5.5+1) - ^'^'^ '"'" which is within the tension flanges, assuming d^ := 14 — 1.75 == 12.25. * Barlow's " Strength of Materials," p. 177. HODGKINSON CAST-IROI^^ BEAMS. 77 From Eq. 37 the breaking load, Z, becomes _ 1(12.25)3 + 2.5 (1 )^3x12.25-2x1) _ ^-^ - 3 X 12725 X 12 X 16 ^ ^ ^ '^^ ~ ^'^^ " • From the tensile strength, T^ and Eq. 28, L becomes ^ 4 X 12 X 1.75 (2 X 14 + 12.25) 13815 ^^^. ^„ ,, X = — ^ — , — sUlOy lbs. 3 X 12 X 16 From this near identity of values for L we conclude that our assumed position for the neutral line was very near the correct position. Two of these beams w^ere constructed of Calder No. 1 cast- iron, and broken with centre loads of 73920 and 76160 pounds respectively by Mr. Owens, Inspector of Metals for the British Government."^' The values of G and 7^ were determined by Mr. Hodgkinson for this brand of iron. The bottom table contained six times the area of the top tahle^ as recommended by Mr. Hodgkinson. These beams broke with the deflections 1^^87 and 2''.0 re- spectively, and the full strength of the section was thus developed by the transverse load. The following statement gives the proportion of the load sustained by each uneinher of the transverse section and the proportion of the area that it occupied : Load. Area, Compressed Flanges 2.3 7.0 per cent. " Web 47.7 34.3 " Extended '" : 4.2 5.0 '' '' Flanges 45.8 53.7 " 100.0 100.0 The web extends through the total depth of the beam. * Box's " Strength of Materials," p. 203. 4 78 STRENGTH OF BEAMS AND COLUMNS. 73. To Design a Hodgkiiison Beam. The formula required in designing a IJodgkinson beam will be found in Art. 42, page 44. Section IV. — Douhle T and Box Cast-iron Beams. 74. The Neutral Line. The position of the neutral line may be approximately computed from Eq. 36, page 42, by giving the letters of tlie formula their definitions in Art. 44. 75, Transverse Strength. The Transverse Strength ^ , ^-vj. may be' computed from Eqs. 37 and 38, I ' 1 • ^ I giving the letters their definition in Art. (-- j( i,_ "& 44 I X Example 10. — Required the centre — "|i— , ^ breaking load of a Hollow Rectangular "^ I Cast-iron Beam whose outside dimension is 3';i25 X 3".125, inside 2".375 X 2".375, thickness of metal all around 0.375, when The depth of the beam. .....d = 3'M25 C = 84000 lbs. " breadth '' sides b = 2 X 0".375 T= 14000 '' " top b, = 2^375 q =Q, '' depth '' " ^. =:. 0'^375 ^ == 6 f eet, " " " bottom .. .r/, = 0".375 m == 4. " breadth '' " . ..^, = 2^^375 The position of the neutral line d^ from Eq. 36 becomes . dc = 3.125X0.754- 4/12x3.125 X 0-75 X 2.37 5 (6+1)4- (3.125+0.75)' (4x6+5 )= ' ' 2X0.75 (6 + 1) 1".965. DOUBLE T AND BOX CAST-IRON BEAMS. 79 The breaking load from Eq. 37 becomes , 0. 75 (1.965)» +2.375 (0.375)^3X1.965-2X0.375) , L= ^3^ 965X12X6 X4x84000=5760 lbs., and from the tensile strength 7", and Eq. 38, L becomes L = 5178 pounds. Tn the investigations preliminary to the construction of the Menai Straits Tubular Bridge, Mr. Stephenson broke this beam with 5387 pounds.^ 76. To Design a Double T and Hollow Rectangular Cast- iron Beam. This may be done by using the formula and directions given in Art. 47, page 48. * " Britannia and Conway Tubular Bridges," p. 429. 80 STRENGTH OF BEAMS AND COLUMNS. Section Y. — Circular Cast-iron Beams. 7 7 . Neutral Line. Tahle of positions of the neutral line in Circular Cast-Iron Beams and factors for use in Eqs. 76 and 77. Ratio of Crushing to Tenacity, or Depth of Nentral Line Below the Crushed Side of the Beam, or dc. 8. 7.875 7.75 •7.625 7.5 7.375 7.25 7.125 7.0. 6.875 6.75 6.625 6.5 6.375 6.25 • 6.125 6.0 5.875 5.75 5.625 5.5 5.375 5.25 5.125 5.0 4.875 4.75 4.625 4.5 4.375 4.25 4.125 4.0 3.886 3.75 3.625 3.5 3.375 3.25 3.125 3.0 0.3984 d 0.4004 d 0.4022 0.4041 4062 0.4087 0.4112 0.4137 0.4161 „ 0.418^ d 0.4201 d 0.4236 0.4262 0.4289 0.4316 0.4344 0.4373 0.4402 0.4432 0.4463 0.4494 0.4526 0.4559 0.4594 0.4629 0.4665 0.4702 0.4740 0.4780 d 0.4821 d 0.4863 d 0.4906 d 0.4951 0.5000 0.5045 0.5095 0.5146 0.5200 d 0.5256 d 5314 d 0.5375 d Factors for Computing the Moment OF Resistance, r = 1. fc for Crushing Strain, C. 0.3226 0.3260 0.3298 0.3336 0.3380 0.3426 0.3472 0.3518 0.3568 0.3620 0.3672 0.3726 0.3780 3836 0.3892 0.3952 0.4014 0.4078 0.4142 0.4210 0.4278 0.4350 0.4426 0.4504 0.4584 0.4668 0.4752 0.4842 0.4938 0.5036 0.5138 0.5244 0.5356 0.5478 5593 0.5722 0.5855 0.5996 6145 0.6302 6468 /V for Tensile Strain, T. 2.5804 2.5722 2.5640 2.5556 2.5466 2.5356 2 5250 2 5138 2.5030 2.4920 2.4808 2.4694 2.4578 2.4458 2.4346 2.4208 2.4080 2.3952 2.3812 2.3664 2.3536 2.3392 2.3240 2.3080 2.2916 2 2750 2 2580 2.2400 2.2218 2.2028 2.1832 2.1630 2.1424 2.1186 2.0974 2.0739 2.0493 2.0237 1.9970 1.9691 1.9399 CIRCULAR CAST-IRO]^^■ BEAMS. 81 From this Table the position of the neutral line is obtained by placing for cl^ the diameter in inches and computing the product indicated. •78. The Transverse Streiig:tli of Circular Cast-iron Beams may be ac- ^* curately computed from either Eq. 74 or 75, but as this involves a tedious calcu- lation Eq. 76 or 77 will give results, practi.'.allv exact, with much less labor. Example 11.— Eequired the centre breaking load of a Cir- cular Cast-iron Beam, when The diameter .7 = 2^0, 6^ = 95200 pounds computed, Ex 2 " ^Pa« '^ = 20^0, T = 26569 " by test, m =4, q = 3.58. The values of the factors /, and/,, obtained by interpolation Irom the Table, page 80, and substituted in Eqs. 76 and 77, give ^_4(l)n).577x 95200 ,_ ' ^ "20 "" ^^^^^' pounds. Major Wade, in his experiments, broke this beam with 11112 pounds. Example 12.— Eequired the centre breaking load of a Cir- cular Cast-Iron Beam when The diameter.. .,/ = 2^42, C = 95200 lbs. computed, Ex. 2, ^P^^ -^ = 20^0, T = 26569 - bv test, m = 4, q = 3.58. /. = 2.065 from the Table, which, substituted in Ecp 77, gives ^_4(1.21)='2.065x 26569 , ^ o?) = 1^^^^ pounds. 82 STRENGTH OF BEAMS AND COLUMNS. Major Wade broke four beams of the above dimensions with 18141, 20419, 19997 and 18225 pounds respectively, the mean strength being 19198 pounds. Example 13. — Required the centre transverse elastic liinit load of a Circular Cast-Iron Beam, when The diameter d ■= 1M29, C = 20000 pounds by test, " span s = 20".0, T = 17000 m = 4, q = 1.111. The value of ^/t = 1.1818 from the Table, page- 80, substi- tuted in Eq. 77, gives ^ 4 (0.5645y 1.1818 X 17000 . ^^^ , X = — ■= IOC) 2 pounds. The elastic limit load, of the two of these beams that were tested, was 1130 pounds each, the values of 6'' and T being the mean of three tests, as given in the United States Government Eeport of the Tests of Iron and Steel for 1884."^ Example 14. — Kequired the centre breaking load of the Circular Cast-Iron Beam whose elastic Ihnit load was com- puted in Example 13, when > C = 100700 pounds by test, /,. = 2.0333 from the Table, T= 29400 " " ^=3.422. Erom Eq. 77 we have J- 4 (0.5645)' 2.0333 X 29400 X = — ^ ~ = 2131 pounds. 20 ^ The breaking load of the two beams described in Exampk^ 13 was 2118 and 1795 pounds respectively. 79. Movement of the Neutral Line. In the Cir- cular Cast-iron Beam, Example 13, the depth of the neutral * Senate Ex. Doc. No. 35 -49th Congress, 1st Session, p. 284. CIRCULAR CAST-IRON BEAMS. 83 line below tlie compressed surface at the elastic limit was, from the Table, page 80, d^ = 0".7853 in Example 14, at the instant of rupture, d^ — 0".5846 ; hence, as the loading pro- gressed, the neutral line moved upvmrd, or from the tension side toward the compressed side of the beam. 80. Relative Strength of Circular and Square Cast-iron Beams. Case I. — When the circle is inscribed within the square. The relation required is given in Eq. 83. Example 15. — Required the relation between the transverse strength of a square and the inscribed circular cast-iron beam when 6^ -=- 7^ = 5. f = 0.5 from Table, page 71, /"c = 0.4581: from Table, page 80. 3 X /c 3 X 0.4584 ^ „„_ . • . Strength of Circle = Strength of the Square X 0.6876. Example 16. — Required the centre breaking transverse load of the circular beam inscribed within the square, from the tested strength of the square beam, wdien The side of the square d = 2". 01, q = 3.58, " diameter of the circle fZ= 2".01,/e = 0.577 from the Table, "span ^ =: 20^0,/ .=: 0.561 " " With these values Eq. 83 becomes 3 X/e ^ 3 X 0.577 ^ ^ 8 X /' 8 (0.561)"^ Actual b'k'g load square beam, Maj. AVade's tests, 16172 lbs. " " '' circular " " " " 11112 " Comp'd " " '' " =16172X0.687=11110 " Case H. — When the square is inscribed tvlthin the circle. The relation will be given by Eq. 84. Example 17. — Required the relation between the strength 84 STRENGTH OF BEAMS AND COLUMNS. of the circular and its inscribed square beam, when C -^ T r= ^. f = 0.5 from the Table, page 71, /c = 0.4584 from the Table, page 80. . 3 /; ^ 3 X 0.4584 ^ * * 2.8271 /•■' 2.8271 (0.5/ * ' . • . Sti'ength of the Circle — Strength of the Square X 1.945. Case III. — When the circular is equal in area to that of the square heam. The relation will be given bj Eq. 85. Example 18. — Required the relation between the transverse strength of the circular and the square beam, when their areas are equal and 6^ -^ 7^ = 5. / = 0.5 from the Table, page 71,/e = 0.4584 from the Table, page 80. , 3/e ^ 3X0.4584 _^ ' ' 5.564/^ " 5.564 X (0.5/ " ' . • . Strength of the Circle = Strength of the Square X 0.987. Example 19. — Required the centre breaking load of a Circu- lar Cast-Iron Beam from that of the square beam of equal area, when The side of the square d- 1".01, g = 5.091, " diameter of the circle <^ = l".145,/e= 0.4562, from the Table, page 80, " span.. s=60".0 /= 0.496 " " " " 71. With these values Eq. 85 becomes 3 X 0.4562 = 1. 5.564 (0.496/ Mean b'k'g I'd of 4 of these square beams from Mr. Barlow's tests,* 519 lbs., " " of the circular beam from Mr. Barlow's tests, 519 pounds, Comp'd " " " " " = 519 X 1 = 519 pounds. * Barlow's "Strength of Materials," p. 153. HOLLOW CIRCULAR CAST-IRON BEAMS. 85 Section YI. — Hollow Circular Cast-iron Beams. 81. Neutral Line. Factors for Computing the Moment Ratio of Crushing to Tenacity, or Depth of Neutral Line Below the Crushed Side of the Beam, or rfc- op Resistance, r = 1. V-^ T^q. /c for Crushing /t for Tensile Strain, C. Strain, T. 8.319 0.5000 d 0.8584 7.1414 8.25 0.5020 d 0.8640 7.1282 8.125 0.5057 d 0.8742 7.1104 8.0 0.5093 d 0.8852 7.1820 7.875 0.5131 d 0.8962 7.0580 7.75 0.5170 d 0.9076 7.0352 7.625 0.52U9 d 0.9190 7.0078 7.5 0.5249 d 0.9304 6.9818 7.375 0.5290 d 0.9430 6.9550 7.25 0.5331 d 0.9556 6.9278 7.125 0.5374 d 0.9684 6.8996 7.0 0.5418 d 0.9816 6.8704 6.875 0.5462 d 0.9950 6.8410 6.75 0.5507 d 1.0090 6.8104 6.625 0.5553 d 1.0232 6.7790 6.5 0.5600 d 1.0380 6.7470 6.375 0.5647 d 1.0534 6.7174 6.25 0.5697 d 1.0688 6.6800 6.125 0.5748 d 1.0848 6.6442 6.0 0.5800 d 1.1008 6.6092 5.875 0.5852 d 1.1184 6.5694 5 75 0..5907 d 1.1356 6.5296 5.625 0.5962 d 1.1540 6.4918 5.5 0.6018 d 1.1730 6.4492 5.375 0.6076 d 1.1924 6.4062 5.25 0.6134 d 1.2124 6.3640 5.125 0.6193 d 1.2330 6.3184 5.0 0.6255 c? 1.2542 6.2702 4.875 0.6314 d 1.2748 6.2266 4.75 0.6384 <^ 1.2994 6.1718 4.625 0.6451 d 1.3230 6.1192 4.5 0.6520 ^ 1.3482 6.0638 4.375 0.6590 <^ 1.3740 6.0064 4.25 0.6662 d 1.4000 5.9492 4.125 0.6735 ri'2~>ri0 = ^'^^^^^ ^' ■ •' ' ^ — 003111 ^ S^"^^^ pounds. Example 23. — Required the Crushing Strength of certain steel made by the Otis Iron and Steel Co., of Cleveland, Ohio, from the centre breaking load of a circular beam when The diameter = l'M29, T = 83500 pounds mean of 10 tests, " span = 20'^0, L = 3860 '' one test. From E(j. 78 we have . _ _^^_^^__ ^ 1 2S49 •''' 4: (0.5645/ 83500 TJ From the Table, page 100, q fractured / this breaking load may be computed from the tensile and compressive elastic Ihnit coefiicients of wrought-iron. The hreahing-dowji load. — While wrought-iron in beams cannot be broken transversely by rupturing its fibres, yet there is an intensity of the transverse load at which it does not appear to be able to offer any further resistance to the action of the bending load ; this load may be called the hreaMny- down load ; in wrought-iron beams it is about twice the technical breaking or elastic limit load. Wrought-iron when rolled into T beams may become sufficiently hard and unyielding from the intense pressure re- quired to make it fill the rolls, that it will offer sufficient resistance to actually fracture the fibres by the tensile strain, as in Example 22. 92 STRENGTH OF BEAMS AND COLUMNS. 90. Transverse Streng^th of Steel. As in the case of wrought-iron beams the elastic hmit load of steel beams is technically called the breaking load. The hreahing-down load of steel beams is about f tlis of the elastic limit load. Section II. — Rectangular Wrought-Iron om^d Steel Beams. 91. Neutral Line. The position of the neutral line may be computed from either formula 25 or 26. From Eq. 25 the position has been computed for the different values of C ^ T — q^ that are required in wrought-iron and steel rect- angular beams, and the results tabulated below for reference. TaUe of Positions of the Neutral Line in Eectangidar Wrought-Iron and Steel Beams : For ratios greater than C -^ T — Z refer to the Table, page 80. 93. Transverse Streng^th. Either Eq. 27 or 28 may be used to compute the transverse strength of rectangular wrought-iron and steel beams, as may be most convenient. RECTANGULAR WROUGHT-IROI^ AND STEEL BEAMS. 93 ExAMPr.E 25. — Required the centre breaking or elastic limit load of a Rectangular Wrouglit-Iron Beam, when The depth tZ = 3".0, C = 30000 pounds mean value, '' breadth l = 1".5, T = 30000 " '' " span s = 33".0, q — 1, and m =- 4. The position of the neutral line may be computed or taken directly from the Table for q = 1. d, = 0.Y805 X 3 = 2^3415, d^ = S - 2.3415 = 0,6585. From Eq. 28 the value of the load, Z, required becomes ^ 4 X 1.5 X 0.6585 (2 X 3 -f 2.3415) 30000 _^^ , X = -r TTT^ = 9681 pounds. 3 X 33 ^ Mr. Barlow tested this iron and found its tensile elasticity perfect with. 22400 pounds ; the transverse elastic limit load was, by experiment, be- tween 9520 and 10080 pounds. A num- ber of beams tested gave similar re- sults. The deflection with 10080 pounds was 0".963inone bar and 0".624 in another; hence the elastic fibre strain limits were fully developed."' Example 26. — Required the centre breaking or elastic limit load of a Rectangular Wrought-Iron Beam of Swedish iron, when The depth d = 2".0, C ^ 2263T pounds by test, " breadth ?> = 2".0, T = 24052 " '' span s= 25\0, q =i 0.94, and m = 4. The position of the neutral line may be computed from Ecp 25, or taken from the Table, by proportion between the posi- * Barlow's " Strength of Materials," p. 278. 94 STRENGTH OF BEAMS AND COLUMNS. tions for q = 1 and q = 0.9, for which d^ = 1.58. With these values the required load, Z, becomes from Eq. 27, ^ 4 X 2 ( 1.58/22 637 ...^ , X = ---^ =. 6003 pounds. Mr. Kirkaldy determined, experimental!}^, that the centre breaking load of this beam was between 6000 and 6500 pounds, and that the elastic limit values of C and T were as given above. The position of the neutral line for the hrealii/ng-doivn load of tliis beam (Exariple 21) was 1".446 below the top or compressed side of the beam, and as the loading progressed it must have moved upward from its position at the elastic limit, 1".58, to 1".446, its position with the hreaJcing-doivn load. The deflection with 6000 pounds was 0".378, and Avitli 6500 0''.506 ; it only required 0".46 to fully develop the elastic iibre strain limits.* Example 27. — Required the centre transverse elastic limit load of a Rectangular Steel Beam, when The 'depth ....d^ r.75, C = 48000 pounds by test, " breadth h = 1''.75, C = 52000 " '' '' span s = 25".0, q = 0.923, and m = 4. The position of the neutral line, computed from Eq. 25, or taken from the Table by proportion, is d, = 1".3867 and d, ^ 0.3633. The required load, Z, from Eq. 27, becomes J. 4X1.75(1.3867)^8000 _^,^ , L = -3^25"^ ^ pounds. * Barlow's " Strength of Materials," p. 256. RECTANGULAR WROUGHT-IRON AND STEEL BEAMS. 95 Mr. Kirkaldy determined by experiment that tlie elastic limit load of each of 4 of these beams was between 8000 and 9000 pounds, and that the values of C and T were as given. The deflection with 8000 pounds was 0".476, and it only required 0".3633 to develop the elastic compressive and tensile flbre strain limits at the same instant.^ Example 28. — Required the hreahing-down load of the beam described in Example 27, when C = 1.59582 pounds by test, q = 2.3, T = 69336 " " m = 4. For the position of the neutral line from Eq. 25 or from the Table, d, = V'AVl and d^ = 0^^629. The breaking-down load, Z, becomes from Eq. 27, ^ 4 X 1.75(1.12iy 159582 _..^. , Z= ^ — 3'^ ^^ ' = 18 ab pounds. The mean hreaking-down load of 4 of these beams was 16477 pounds from Mr. Kirkaldy's experiments. The upward movement of the neutral line is seen in these Examples, in Example 27, d^ — 1^^3867 and in Example 28, d^ = l'M21. * Barlow's " Strength of Materials," p. 253. 96 STRENGTH OF BEAMS AND COLUMNS. Section III. — Doiihh? T, Rolled Eye-Beams and IIollow Rectangle or Box, Wrought- Iron and Steel Beams. 93. Nevitral Line. The neutral lines in these wrought- iron and steel beams are so near the bottom or tension side of the beam that their positions cannot be computed from ap- proximate formulas, and the exact formula is too complicated to be useful in practice. Before beams of these sections are maimfactured, its position should be assumed and a sufficient area of the metal placed above and below the assumed neutral line to insure equilibrium from the known values of the crush- ing and tensile strength of the material. In most practical examples of these beams the neutral line is located within the tension flange, and its position can be computed from Eq. 30, page 40, when L the load and T the tensile strength have been determined experimentallv, or bv the methods given in Problems I. and II., page 43. 94. Transverse Strength. AVhen the position of the neutral line is known the transverse strength may be com- puted from either Eq. 37 or 38, that were deduced for the transverse strength of the Ilodgkinson beam, giving the letters of the formulas the meaning deflned in Art. 44, page 47. The transverse strength of both the Tee and Double- Headed Railroad Rails may be very accurately computed from the above formulas, although they are partly bounded in outline by curved lines. The web must extend from the bottom to the top of the rail, and an equivalent in area right line section must be formed for the metal that remains in both the base and head of the i-ail, which will enable the formulas to be applied to these forms of beams. 95. Double T heams with tension and compression flanges uneq^ial m area. Example 29. — Required the centre elastic limit load of a Wrought-Iron Rolled Double T Beam, when DOUBLE T, ROLLED EYE-BEAMS, ETC. 97 Depth of beam d — 8". 38, G — 36320 lbs. required for eq'brium, Breadth of web h - 0".325, T = 30000 pounds mean value, " " top flanges 6] = 3". 175, m = 4, Depth " " " d^ = 1".0, s = 11 feet, "bottom'' (?2 = 0".38, Breadth " !' " 5o = 3". 675. The neutral line will be assumed to be near tlie top line of the tension flanges, or ^4 == 8';05 and d^ = 0.33. The elastic limit load, Z, becomes from Eq. 37, IfE _ b _ _ b. _• _ I) ^ — -b, L = 0.325 (8.05)'^ 4-2.175 (1)^ (3 X 8.05— 2X1) 3 X 8.05 X 132 X 4 X 36320 = 9924 lbs., and from Eq. 36, ^_ 4 X 4 (3 X 8.38 + 8.05) 30000 "" 3 X 132 = 9924 pounds. The elastic limit load was between the applied loads 9493 and 11253 pounds from Mr. William Fairbairn's experiments, and their deflections were 0'^46 and 0^^60 ; the deflection re- quired was 0''.33.* Example 30. — Required the centre elastic limit transverse load of a Rolled Wrought-Iron Double T Beam when Depth of T beam d = 9". 44, C = 37165 lbs. required for eq'brium, Breadth of web b = 0".35, T = 30000 pounds mean value, " top flanges &i = 2". 4, s = 10 feet. Depth " " " d, = l".0, m=4. " " bottom " d-2=Q"A4:, f?c = 9.06 assumed, Breadth " " " h^ = 3". 95, d^ = 0.38 '' *Fairbairn, " On Cast and Wrought-Iron," p. 102. 98 STRENGTH OF BEAMS AND COLUMNS. With these values the load, Z, becomes from Eq. 28, J _ 4 X 4.3 X 0.38 (2 X 9.4 4 + 9.06) 30000 _ ., koi o ik X _ 3 X 12 X 10 - ^^^^^ ^^^•' and from Eq. 38, ^ 0.35 (9.06f + 2.4 (If (8 X 9.06 - 2 X 1) 3 X 9.06 X 12 X 10 X 4 X 37165 = 15218 lbs. The elastic limit load was between the applied loads 14693 and 16373 pounds from Mr. William Fairbairn's experiments, and the deflection was 0''.35 and 0'^45 with these loads re- spectively.* In Examples 29 and 30 the tensile elastic limit has been assumed to be 30000 pounds, the mean value for wrought- iron, and from it the position of the neutral line has been computed from Eq. 30, and then the required value of C to produce equilibrium. rc A. 1 1 \ rxT 96. Double T beams with tension and compression flanges equal in area. Example 31. — Required the centre transverse elastic limit load of a Rolled Wrousrht-Iron Evebeam when Depth of beam d = 9".0, (J = 63170 lbs. required for eq'brium. Breadth of web b = 0".6, T = 30000 " mean value, " top flanges bi = 4". 775, m = 4, Depth " " " di = l".0, 5 =14 feet, " bottom'' d.2 = 1".0, do = 8".2, Breadth " " " b-i = 4".775, d.^ = 0".8. With these values the load, Z, becomes from Eq. 37, r - 0> 6(8.2f + 4. 775(1)^3X8.2-2 X 1) ^ , ^ .o... ^' - 3 X 8.2 X 12 X 14 ^^^ ^^^'^ 26823 lbs., «*, * Fairbairn, "^On Cast and Wrought-Iron," p 103. ^ , DOUBLE T, ROLLED EYE-BEAMS, ETC. 99 and from Eq. 28, in which ^ = 5 -j- ^2 = 5.375, _ 5.375x0.8(2x9 + 8.2) ^^^^^ ^ ^^^^3 ^ - 3 X 12 X U The vahie of T is taken at the mean, from which d^ must be C.8 and 6^= 63170, that equilibrium shall exist between the moments of the tensile and compressive resistances. The rolled eyebeam, of which the data given in Example 31 is the equivalent right line section, was tested by the Phoe- nix Iron Co. of Pennsylvania,* and its centre elastic limit load was found to be between the applied loads 26880 pounds and 28000 pounds ; the deflection was 0".572 and 0''.600 with these loads respectively. The proportion of the load and area of the section that is sustained by each member is given in the following table : Load. Area. Compression Flanges 12.3 31.9 per cent. Web 37.7 32.9 " Tension " 0.55 3.2 '' '' Flanges 49.45 24.1 " " " (practically lost) 7.9 " 100.00 100.0 The transverse elastic liimt load of the rectangular wrought- iron beam 5".375 X 9".0, from which the above described eyeheam may be supposed to have been cut, is from Eq. 27, when C = 30000 and T — 30000 pounds, the mean values for bar-iron, ^ 4 (7.02)'' 30000 ^^.^- , ^ ^ 3 X 12 X 14 ^ P""'"' From which it will be observed that the eyebeam, while con- taining only 31 per cent of the area of the rectangular beam, * Phoenix Iron Company's " Handbook of Useful Information." 100 STRENGTH OF BEAMS AND 'COLUMNS. is able to sustain 42.5 per cent of its load, which is supposed to be due to the elevatioji of the elastic limit during the pro- cess of rolling, or the top of the beam must have been laid with steel. Section \Y.— Circular Wrought-Iron and Steel Beams. 97. Neutral I^iiie. The position of the neutral line in Circular Wrought-Iron and Steel Beams for the different ratios of C -^ T — q required has been computed, also the factors /e and/, for use in Eqs. 76 and 77, and tabulated be- low^ for reference. Factors for Computing the Moment of Eatio of Crushing Depth of Neutral Line Below the Crushed Side of the Resistance, Radius = 1. to Tenacity, or C -4- T — q. Beam, or dc. /c for Crushing T for Tensile Strain C. Strain T. 3.0 0.5375 d 0.6468 1.9399 2.875 0.5438 d 0.6642 1.9093 3.75 0.5505 d 0.6827 1.8766 2.625 0.5574 d 0.7024 1.8434 2.5 0.5646 d 0.7230 1.8084 3.375 0.5724 d 0.7458 1.7718 2.25 0.5804 d 0.7696 1.7314 2.125 0.5890 d 0.7952 1.6896 2.0 0.5980 d 0.8228 1.6454 1.875 0.6076 d 0.8526 1.5984 1.75 0.6178 d 0.8848 1.5482 1.625 0.6277 d 0.9198 1.4944 1.5 0.6404 d 0.9580 1.4368 1.375 0.6525 d 1.0000 1.3748 1.25 0.6666 d 1.0462 1.3074 1.125 0.6816 d 1.0976 1.2328 1.0 0.6976 d 1.1548 1.1548 From this Table the position of the neutral line in any wrought-iron or steel beam may be obtained by multiplying the diameter d, expressed in inches, by the decimal factor CIRCULAR WROUGHT-IRON AND STEEL BEAMS. 101 corresponding to the ratio C -^ T =^ q. For ratios intermedi- ate in value the position may be obtained by proportion. 98. Transverse Strength. This may be computed from either Eq. 71 or 75, but with much less labor from Eq. 76 or 77, using the values of f^ and f^ given in the above Table. Example 32. — Required the centre breaking or elastic limit load of a Circu- lar Steel Beam when The diameter d = l'M29, C = 41160 lbs. mean of four tests, " span s =z 20'^0, T = 39200 " " '' ten " " f actor ... ?y2. — 1, q ^ 1.05. The position of the neutral line and factor, f^, becomes, by proportion from the Table, d, = 0.6912 X l'M29 = 0.780, f, = 1.132. With these values the load, Z, required becomes from Eq. 76, ^ 4 (0.5645y 1.182 X 41160 _,^^^ , J^ = -^ — ^ = 16^6 pounds. From the United States Government Report of the Tests of Iron and Steel for the year 1885,^ the elastic limit load of two of these beams that were tested was 1638 pounds witli tlie values of O and T as given in the Example. Example 33. — Required the centre elastic limit load of a cylindrical Phoenix pin supported at both ends when * Senate Ex. Doc. No. 36-49tli Congress, 1st Session, p. 690. 102 STRENGTH OF BEAMS AND COLUMNS. The diameter. ... d— 2".5, C = 30000 pounds from tests, " span 6' = 24".0, T ^ 30000 " " factor m = 4:, f^= 1.1548 from the Table. Then from Eq. 76 we have for the required load, _ 40:25)^1^54^^^^0000 ^ ^^3^^ ^^^^^^^ 24 ^ The elastic limit strength of this pin was 11000 pounds. (Watertown Arsenal Report of Tests of Iron and Steel for 1881.)^ The computed ultimate strength is 18795 pounds, with C and T == 50000 pounds, the recorded ultimate strength is 20000 pounds. In order that these wrought-iron pins should truly cross- Ireak, their deflection should not be less than 0.3024 d\ in the above examples the ultimate deflection was 1''.278 ; while that required for true cross-breaking was 0''.75, the recorded de- flection with 18000 pounds was 0''.78. A number of these pins, made of Phoenix and Pencoyd iron, were tested with diameters ranging from 2^ to 5 inches, but the span was so short, in nearly all cases, that the tensile flbre strain was not fully developed, causing the transverse elastic limit load to be not well defined and the ultimate load to be larger than it should be. Though the computed elastic limit loads do not differ very greatly from those determined by tests, there is a very great difference between the computed and experimental ultimate loads, the latter being the greater. This series of tests illustrates the correctness of the state- ment made in Art. 24, that the hend'mg moment of the ap- plied load at the i7icej)tio?i of the deflection of a beam is held in equilibrium by tlie moment of a purely compressive resist- ance that is distributed over the section in an uniformly vary- * House of Representatives Ex. Doc. No. 12, 1st Session, 47th Congress, p. 171. CIRCULAR WROIJGHT-IROIS^ AND STEEL BEAMS. 103 iiig strain. Test E'o. Y41, page ISO, of the Eeport, was of Plioeiiix iron 4^' in diameter and 24'^ span. The bending moment of tlie centre applied load is 6 Z from Eq. 3, page 5, and the moment of compressive resistance is the product of the resultant (the area of the section by one half, (7, the elastic limit compressive strength), by its lever-arm, %d, the distance of the centre of gravity of the pressure wedge below the axis ; hence 3d ,, d' € -8-><"4 X 2' ^L = -;-x TT -- X ,,, L — — - — ^ ~ = 47124 pounds. 384 384 ^ The observed elastic limit load was 48000 pounds with a deflection of 0^^0585, but as no correction appears to have been made for the settling of the beam on its bearings, we may conclude that the beam was on the verge of true deflec- tion with its transverse elastic limit load. 104: STRENGTH OF BEAMS AND COLUMNS. Section Y.— Hollow Circular Wrought-Iron and Steel Seams. 99. Neutral Line. The position of the neutral line in Hollow Circular Wrought-Iron and Steel Beams for the different ratios of C ^ T = q required has been computed, also the factors f^ and f^, for use in Eqs. 91 and 92, and tabulated below. Factors for Computing the Moment or Ratio of Crushing Depth of Neutral Line Below the Crushed Side of the Beam, or dc. Resistance, Rabius = 1. to Tenacity, or C+ T=q. /c for Crushing /t for Tensile Strain C. Strain T. 3.0 0.7500 d 1.7320 5.1961 2.875 0.7597 d 1.7737 5.0992 2.75 0.7697 d 1.8172 5.0000 2.625 0.7800 d 1.8628 4.8898 2.5 0.7906 d 1.9105 4.7760 2.375 0.8014 d 1.9601 4.6566 2.25 0.8122 d 2.0108 4.5334 2.125 0.8241 d 2.0645 4.3S28 2.0 0.8358 d 2.1251 4.2490 1.875 0.8478 d 2.1853 4.0966 1.75 0.8600 d 2.2479 3.9338 1.625 0.8727 d 2.3135 3.7586 1.5 0.8851 d 2.3823 3.5722 1.375 0.8979 d 2.4534 3.3728 1.25 0.9107 d 2.5267 3.1588 1.125 0.9233 d 2.6024 2.9296 1.0 0.9360 d 3.6806 2.6806 100. Transverse Streng'tli. This may be com- puted from either Eq. 89, 90, 91 or 92. Example 34. — Required the centre breaking transverse load of a Hollow Wrought-Iron Cylindrical Beam, supported at both ends, when The outer diameter. . .d = 4'^0, 6^ and T = 45000 pounds, ^' thickness of metals = 0'M875, /; = 2.6806 from Table, " sj3an s = 6'.0 m = 4. WROUGHT-IRON AND STEEL BEAMS. 105 Then from Eq. 91 we have for the load, ^ 4 (2y 2.6806 X 0.1875 x 45000 ^,_^ L = — -^ a— = 50 /^6Vcr;??'.v gives, practically, accurate results, or such as are within the limits of the varia- tion in strength of the material, while those from the more recent experimenters make their computed transverse strength from twenty-five to one hundred per cent greater than their experiments gave, which indicates that with improved testing machines the more recent experimenters, such as Professor Thurston, Laslet and Hatfield, obtained their crushing strength at a point nearer the total destruction of the wood than the older with machines having less power. 102. To Compute tlie Comi>ressive and Ten- sile Strength. Crushing. — From the fibrous charactei* of timber and its weak lateral adhesion it is difficult to de- TIMBER BEAMS. 107 termine its crushing strength from direct pressure on small specimens, or that intensity of crushing strain that holds its tensile strength in equiUbrium : this can be accurately com- puted from Eqs. 31 and 79, when the breaking, transverse and tensile strength have been determined from experiments. Example 35. — Eequired the Crushing Strength of Teak- wood from the known tensile strength and the centre trans- verse breaking load of a rectangular beam when Depth. . .<:Z=2.0 ins., 7^=15000 lbs. mean of Mr. Barlow's tests, Breadth .^=2.0 " Z=938 " " " " " Span .... ^=7.0 feet, 7/?. =4. The position of the neutral line becomes from Eq. 30, d = ^ X ^ - , /9 X 4 X 2 (2)^ 15 000 — 12 X 938 X 12^~7 ^ q 35^^^ 2 r 4X4X2X 15000 . ' . d, = 2.0 - 0.35 = 1.65 inches, and the crushing value of O from Eq. 31, ^ 3 X 938 X 12 X 7 ^^._ , 6 = -. ?rm^^V2 — = IO80O pounds. 4x2 (1.65/ ^ From Mr. Hodgkinson's experiments O —- 12100 pounds for Teak. Tensile /Strength. — This may be computed from Eqs. 32 and 33, and from Eqs. 80 and 81, when the transverse and crushing strength have been ascertained from experiments. Rectangular Wooden Beams. 103. Neutral Line. Table of positions of the neutral line in rectangular sections of wood, and the mean ultimate crushing and tensile strength of the different varieties of timber. 108 STRENGTH OF BEAMS AND COLUMNS. Ash, American " English Beech, American . . . " English Birch, American Black Birch, English Cedar, American Red Elm Fir, White Spruce . . " " Christiana Deal.... Oak, American Red " White English , Pine, Am. South-) em Long-leaf. . . i" Pine, Am. White. . . " Am. Yellow. . " Red, European ** " Dantzic... " Riua .... Poplar Walnut, Black, Breaking Strength PER Square Inch IN Pounds. Eatio of Crushing to Tenacity, C Crushing, C. Tensile, T. f=Q. 4400 11000 0.4 5800 14000 0.4142 8600 H 12000 31 0.7166 5800 15000 0.3866 6900 18000 0.3833 7700 JI 9000 i? 0.4142 9300 H 12000 B 0.7750 7000 11600 0.6034 4530 H 11700 i? 0.3872 6400 H 15000 31 0.4266 6000 10300 0.5825 6830 14000 3/ 0.4878 6500 31 10000 31 0.6500 5850 M 12000 J/ 0.4875 6000 10000 0.6 7200 18000 0.4 9100 18000 0.5055 6500 H 10000 31 0.65 9500 H 19000 31 0.5 8000 12600 0.6349 8000 19200 0.4166 5000 10000 0.5 5400 M 12000 31 0.*45 * 7500 3f 14000 M 0.5357 5400 31 8000 31 0.575 574^ H 11549 B 0.4977 6586 // 12857 B 0.5122 5100 T 7000 T 0.73 Depth of Neutral Line Be- low the Crushed Side of the Beam, or 6?c. Authorities. 0.8900 dB., Barlow. 0.8877 d H., Hodgkinson. 0.8267 ^^M., Molesworth. 0.8933 rZR., Rankine. 0.8941 d 0.8878 d 0.8165 d 0.8477 d 0.8939 d 0.8843 d 0.8518 d 0.8706 d 0.8376 d 0.8711 d 0.8495 d 8900 d 0.8673 d 0.8376 d 0.8685 d 0.8410 d 0.8525 d 0.8685 d 0.8792 d 0.8559 d 0.8533 d 0.8690 «Z 0.8660 d 0.8243 d Tlie position of the neutral lines in the above Table was computed from Eq. 25, d being unity. 104. Transverse Strength. Tim- ber beams breaking with a well-defined fracture in its fibres, the transverse strength may therefore be computed from the crushing and tensile strength by TIMBER BEAMS. 109 means of either Eq. 27 or 28, as the rectangular is the form that is principally used in wooden beams. Example 36. — Required the uniformly distributed breaking load of an American White Pine Rectangular Beam, when The depth d = 14'^0, C = 5000 pounds mean of tests, " breadth.... J = 6^0,^=10000 '' " '' " span s — 28'. 0, m = 8, Art. 31, and q = 0.5. The position of the neutral line from the Table is r/e = 0.8685 X H = 12.16 inches, and the breaking load from Eq. 27 becomes ^ 8X6 (12.16)' 6000 _„„^ . . ^ - 3X12X28 ' = ^'^^^ P""^^'- The Table in Trautwine's Engineers' Pocket-Book gives 37800 pounds as the breaking load of this beam. Example 37. — Required the breaking load of a Rectangular English Oak Beam fixed at one end and loaded at the other, when The depth d = 2".0, = 6500 pounds mean by test, " breadth b = 2'^0, T = 10000 " '' " " span c9 = I'.O, q = 0.65 and m = 1, Art. 34. The position of the neutral line is, from the Table, d^ = 0.8376 X 2 = 1.6752 inches, the load, Z, from Eq. 27, becomes Z = lX2(m52y65O0 ^ 253 3 X 12 X 1 ^ 110 STRENGTH OF BEAMS AND COLUMNS. From Colonel Beaiifoy's experiments the mean breaking load of 6 of these beams was 258 pounds each.* Example 38. — Required the centre breaking load of the beam described in Example 37, when The span = 7.0 feet, m = 4. The required load, Z, from Eq. 27, becomes L ^ iX_2_(M752):65^a0 ^ ^^^ 3X 12X 7 ^ From Mr. Barlow's experiments the mean breaking load of three of these beams was 637 pounds each. Example 39. — Required the centre breaking loads of the following Rectangular Wooden Beams, when. The depth ^ = 2".0, C = values from the Table. " breadth h =z 2'\0, T = " " " '' span 5 = 50".0, m, =^ 4:. The position of the neutral line is obtained by midtiplying the depth, >/, by the factor corresponding to the ratio C ^ T given in the Table, for each case. Computed. 2 tests. Christiana Deal. .Z = o T ;^n -= 969 940 and 1052 lbs. d X 50 English A«h ^^ 4x2a.6534y 8600^ 1304 an^5 (6.374)- 6000 ^ ^3,^3 3 X 12 X 15 ^ M. Buff on, in experiments made for the French Govern - ment,f broke two of these beams with 13828 and 14634 pounds respectively ; from Rondelet's experiments the values of C and T are supposed to be equal to those given in the Table for American Red Oak. Circular Wooden Beams. 105. Neutral Line. The position of the neutral line and the value of the factors y^ and/!r for use in Eqs. 76 and 77 may be obtained from the following Table, by proportion when necessary : ♦Barlow's " Strength of Materials," p. 86. f Ibid., p. 56. 112 STKENGTH OF BEAMS AND COLUMISTS. Ratio of Cruehing to Tenacity, Depth of Neutral Line Below the Crushed Side of the Beam, or Factors for Computing the Moment OF Resistakce, Radius = 1. or C-i- T-q, da. /c for Crushing Si rain, C. /r for Tensile Strain, T. 1.0 0.6976 d 1.1548 1.1548 0.9829 0.7000 d 1.1632 1.1433 0.9473 0.7050 d 1.1812 1.1189 0.9126 0.7100 6^ 1.1991 1.0943 0.8790 0.7150 d 1.2174 1.0708 0.8463 0.7200 d 1.2357 1.0458 0.8133 0.7250 d 1.2540 1.0200 0.7837 0.7300 d 1.2726 0.9974 0.7541 0.7350 d 1.2912 0.9732 0.7245 0.7400 d 1.3098 0.9490 0.6965 0.7450 d 1.3287 0.9251 0.6686 0.7500 (^ 1.3476 0.9013 0.6423 0.7550 d 1.3666 0.8775 0.6161 0.7600 d 1.3856 0.8538 0.5912 0.7650 d 1.4048 0.8304 0.5664 0.7700 d 1.4240 0.8066 0.5429 0.7750^ 1.4434 0.7832 0.5194 0.7800 d 1.4628 0.7598 0.4971 0.7850 d 1.4828 0.7365 0.4749 0.7900 d 1.5020 0.7133 0.4534 0.7950^ 1.5216 0.6895 0.4320 0.8000 d 1.5412 0.6658 0.4129 0.8050 d 1.5611 0.6442 0.3938 0.8100 6? 1.5810 0.6226 106. Transverse Strength, The transverse strength of circular wooden beams may be com- puted from either Eq. 74 or 75, also from Eq. 76 or 77, with the aid of the values of the factors f^ and f^. deduced from the above Table. Example 41. — Required the centre Deal Beam, when breaking load of a Circular Christiana The diameter d = 2".0, C ~ .5850 pounds mean of tests, " span ... s = 48".0, T =: 12000 " " " " m =4, q= 0.4875. • TIMBER BEAMS. 113 The required load, Z, becomes from Eq. Y7, L = ill )' 0-T26 X 12 000 ^ ^^g ^^^^^^^_ Mr. Barlow broke tliree of tliese beams with 710, 796 and 780 pounds respectively, the mean being 772 pounds."^ 107. Relative Strength of Square and Circu- lar Timber Beams. The required relation will be ob- tained from either Eq. 83, 81 or 85, as the case may require. Example 12. — Required the centre breaking load of a Cir- cular Christiana Deal Beam from that of the circumscribed square beam of the same span and material, when Side of the square ... 6? = 2".0, (7= 5850 lbs. mean of tests, Diameterof the circle 6?= 2".0, 7"=: 12000 " " " " Span s = 18^0, ^ = 0.1875, y, from the Table, page 108, becomes 0.8711 when d = 1, and/*c, from the Table, page 112, becomes 1.1912 when ^ = 1. With these values Eq. 83 becomes 3/e _ 3 X 1.1912 ^ ^ ^3^. 8/ ~ 8 X (6.8711)' ' ' hence. Circle = Square X 0.737. Breaking strength of 2" sq. beam, Mr. Barlow's tests, 1117 lbs. " " " 2" circ. " 1117 X 0.737 823 " Mean breaking strength of 3 beams, Example 39 ... . 772 " Mr. Barlow says that the 2" square and the 2" diameter cir- cular beams were cut from the same plank, " which was a very fine specimen of Cliristiana deal." The breaking strength of the 2" square and 48'' span beam, with the above values of T and G, should have been 986 pounds, and the circular beam 986 X 0.737 = 726 pounds, as in Example 11. * Barlow's " Strength of Materials," p. 78. CHAPTER YII. STRENGTH OF COLUMNS. 108. General Conditions of Failnre of Col- nniiis. Enler and Tredgold are credited with the only well- recognized attempts that have been made to deduce rational formulas for the strength of columns, but the basis of each of their theories involves the assumption of the existence of conditions that render their application to the actual pheno- mena observed in practice inapplicable without the aid of empirical factors determined from experiments. The law^s that govern the breaking strength of pillars were investigated, experimentally, by Mr. Eaton Hodgkinson in 1S40, but no rational theory has ever been advanced that will explain the phenomena observed by him and subsequent in- vestigators ; and our constmctors are to this time using the empirical rules of either Hodgkinson or Gordon, that were deduced from the former's experiments, to compute the re- quisite dimensions of their pillars, or as they mny have been modified to conform to the results obtained by subsequent in- vestigators from new material or that produced from improved methods of manufacture. The effect that will be produced upon a given piece of material, when subjected to a strain in the direction of ita length, depends entirely upon its deflection. The defiection varies with the material and with the ratio of the length to the least diameter. It has been found, experimentally, accord- ing to Mr. Tredgold, '' that when a piece of timber is com- pressed in the direction of its length, it yields to the force in a different manner according to the proportion between its STRENGTH OF COLUMNS. 115 length and the area of its cross-section," which is according to the amount it is able to bend or deflect laterally. The material of a pillar when subjected to an applied load in the direction of its length, in order to avoid the strain thus brought upon its flbres, deflects laterally and assumes a form composed of one or more curves, as its ends may be round or flat ; a round end pillar assumes the one curve form and the flat end pillar takes a form made up of three or four curves. The former is represented in Fig. 34, and will be called the Triple Flexure form ; it is that in which the pillar offers the least resistance to breaking by an applied load. When a pillar of the Triple Flexure form fails with deflec- tion, tension exists in its fibres at g (Fig. 3-1), and compression at m ; hence there must be a point in the line gam.^ at which there is no strain ; likewise there must be a point without strain in the line gds ; and compression existing in the flbres at the points <9, f and Z", the entire side, kef ho, must be in like condition. The fibre strains at two sections of the pillar, such as ao and db, must therefore increase uniformly in inten- sity from zero at a and d, to its greatest at c {ind Z>, respec- tively. The failure of a few columns along the lines etc and dh indicates that they join the points of 'reverse curvature of each side of the pillar, though a fracture beginning at c may follow the line of less resistance, cm. There is also a point, n, in the line g y, at which the strain is zero in intensity, and a curved line, a n d, connecting these zero points is the neutral surface of the column. Since the column must bend symmetrically there is a point in each of the lines ac and dh that is one fourth the length of the column from each of its ends, nik and so, respectively, thus making the middle curve of the triple flexure form one half of the length of the column. Tlie angle that the lines ac and dh make with the plane of the ends varies with the material, but in all pillars it is supposed to approximate the angle of 45 degrees. 116 STRENGTH OF BEAMS AND COLUMNS. However short a pillar may be when loaded, it will attempt to assmne the triple flexure form. The curves above and below the lines ac and clh being the first to take their shape, thus give rise to the various phenomena that are observed when short blocks of granular textured material, such as cast-iron, are crushed. When the points c, f and h coincide, the block is sheared across at one plane ; when a coincides with m and d with 8 the block splits up into four or more wedges. The strength of a j^illar generally dimin- ishes as it becomes longer in proportion to its least diameter. Should a series of pil- lars of an uniform section be constructed with progressively increasing lengths and the strength of each be obtained by ex- periment, there would be found one length whose strength per square inch of section would be greatest ; the strain must be uniformly distributed over the section of the pillar and is the crushing value of C for the material. As the length increases the strength decreases, until a length of pillar is reached whose strength is just one half of the greatest strength above described ; the strain must be vmformly varying in intensity — zero at g and great- Fxg. J4 est at f^ where its intensity per square inch is the crushing value of the material. The pillar of our series, with this ratio of length to its least diameter, will fail by crushing without deflection when it is on the verge of fail- ing by crushing with deflection ; twice the mean intensity of its strength per square inch of section is the crushing strength of the material, or that criishing strength that equilibrates the tenacity when beams and columns fail by cross-hreahing . STRENGTH OF COLUMNS. 117 At the inception of the deflection of the pillars of our series that fail with deflection, the neutral line coincides" with the side, 7ngs^ of the pillar ; as its length continues to increase it will now be able to deflect, all compression having been re- moved from one side, and the neutral line will move from , the observed deflection will be too small by its distance from the. line Tio^ which must be parallel to the line of direction of the resultant of the applied load. The deflection is an inci- dental quantity ; the distance required is the lever-arm of the resultant of the load, which is the perpendicular distance from the fulcrum, y, to its line of direction. I z=z the length of the pillar in feet and decimals, d^ =r the depth of the compressed area in inches, c = a constant quantity determined from experiments, S = the breaking deflection in inches. Then '-'i (93) The factor, c, must be determined for the different material used in structures, from experiments on fixed, round, two pin, and on one pin and one fixed end columns ; it has a different value for each material and style of end connections, and probably for each different shape of cross-section. Experiments. The experiments required to determine the laws that govern the deflection of colunms of a given section and material may l)e very much simplified by determining defi- nitely the following points : STRENGTH OF COLU.AINS. 123 1st. The ratio of I ^ d that gives the true crushing vahie of the materiah 2d. The largest ratio oi I -^ d that sustains its greatest load without deflection. 3d. The smallest ratio of I ^ d that eross-hreaks. 4th. The largest ratio of / -^ d tliat cross-hreaks without leverage of the appKed load. The law governing the change in the strength from the 1st to the 2d and from the 2d to the 3d ratio would then be re- quired ; from the 3d to the 4th ratio the strength would be constant, and for larger ratios than the 4th the deflection, or lever-arm of the load, would be the only varying element, and only for these pillars will it be necessary to determine the values of the factor c, in Eq. 93, for the various kinds of material used in structures. JSTo attempt will be made to determine, for any given material, the limits above defined, from the incomplete record of experiments at present existing. 112. Classification of Pillars. From the descrip- tion of the manner of " failure " of columns of various lengths, it is evident that they may be divided into the follow- ing general classes : 1st. Pillars tHxVt fail by crushing. 2d. Pillars that fail by cross-breaking. These classes may be again divided into Cases, which, for convenience of reference, will be numbered consecutively. Pillars that fail hy crushing. Case I. — Pillars that fail with the full crushing STRENGTH OF THE MATERIAL. Case II. — Pillars that fail with less than the full CRUSHING strength OF THE MATERIAL AND WITHOUT DEFLEC- TION. 124 STRENGTH OF BEAMS AND COLUMNS. Case III. — Pillars that fail with less than the FULL CRUSHING STRENGTH OF THE MATERIAL AND WITH DE- FLECTION. Pillars that fail hy cross-hreaking. Case IV. — Pillars that cross-break from compression. Case V. — Pillars that cross-break from compression AND CANTILEVERAGE. End Connections. Columns are again classed from the form or shape of their end connections, while thej all fall under one or the other of the iive cases given above. In columns of a given material and dimensions, the only element of variation in their strength aris- ing from differences in its end connections is the deflecticm. A flat or fixed end column is one whose ends are planes at right angles to its length ; for a given material and dimen- sions it deflects less, and is, therefore, the form of greatest strength. A round end column is one whose ends are spherical ; con- sequently, being free to move laterally, it deflects most and is the form of least strength. Pin end is a class of columns used in bridge construction in which the load is applied to pins passing through holes in its head and foot, and by them transmitted to the column. The smaller the pin and the less frictional resistance that may be developed between the ^nn and its soffit^ the nearer it ap- proaches the condition of a true round end column, and the less strength it will exhibit, while the larger the pin and the greater frictional resistance that may be developed tlie nearer it approaches the condition of a flat end pillar, and the greater strength it will exhibit. One pin and one flat end^ in its strength, follows laws similar to the above and a mean between flat and round end pillars. STRENGTH OF COLUMT^S. 125 Mateeial and Section. In practice, columns are again classed as wooden, cast-iron, wrouglit-iron and steel, from the material of which thej are made, and from the shape of their cross-sections into rect- angnlar, circular, hollow circular, angle-iron, box, channel, eje- beam, tee, etc. From the incomjDlete tests at present made the lengths of columns of a given section and material that belong to each of the five cases of failure given above cannot be determined. In the examples quoted in the sequel these limits are, however, well defined in some special cases of material. 113. Case I. — Columns that fail with the full crush- INO strength of the material. Let A represent the area of the section of the pillar in square inches, then as the load must be umforinly distributed for this case, the crushing load will be L^A-xC. ■ (94) 114." Case II. — Columns that fail with less than the FULL CRUSHING STRENGTH OF THE MATERIAL AND WITHOUT DE- FLECTION. In cdl pillars of this class, at failure, the strain at the surface of one side is constant and equal to the erusliing value of C for the material ; the strain at the surface of the opposite side varies with the length, from the crushing value of C to zero in intensity, at which length the strain is uniformly varying, the mean intensity being 6^ -^ 2. l^ = the length of pillar that fails with full crushing strength, I = the " " '^ " " " a mean intensity, (7^2, X = the " " " whose strength is required, A = the area of the section of the pillar in square inches. The values of l^ and I vary with the material and are de- termined experimentally ; assuming that the intensity of the 12Q STRENGTH OF BEAMS AND COLUMNS. strain on one side of the pillar varies with the length, then for the length x it will he C ( -, V I and the mean for the sec- (^3' tion will be (^+^(7 v )) "^ 2? and the load that the pillar of the length x will fail with is For the longest column to which this formula applies, Z = a?, and Eq. 95 becomes A y C L = ^t (95A) To Compute the Crushing Str'mgth. — The above described manner of failure of pillars and the formulas resulting may be used very advantageously to compute the crushing strength of any material, which, in this case, is not affected by the tensile strength. And we are thus furnished w^th a valuable means of testing the correctness of the computed crushing strength for the same material when broken in a beam by the methods given in Articles 39 and 55 ; also that obtained from direct experiments with short blocks. Deducing the value of C from Eq. 95A, we have 9 V 7^ C = ^-AA, . (^^5B) which is the formula required from which to compute the crushing strength of the material in a pillar that fails or sus- tains its greatest load 'without deflection wdien it is on the verge of failing with deflection. Example 43. — Required the greatest strength of a Rect- angular Wrought-lron Column, tested with two l"h pin ends, when the deflection (^ = 0, area A — 8.85 square inches, and STRENGTH OF COLUMNS. 127 — 50000, assumed to be equal to the tensile strength ob- tained from direct tests. From Eq. 95 A we have ^ ^ aS^X^OC) ^ 231250 pounds. From the United States Government Watertown Arsenal Tests, 1882-1883,^ the greatest strength of nine of these col- umns was found by experiments, the mean was 234850 pounds, the deflection varied from 0''.02 to 0'M2, the mean was0".07; the lengths varied from 5-1: to 78 inches, and they were all approximately three inches square. The mean strength of four of these columns, tested with one flat and one \"\ pin end was 210800 pounds, the mean deflection, 0".12 ; the length of two was 90 inches and that of the other two columns, 120 inches. The mean strength of four of these columns, tested ^\\\\flat ends, was 217875 pounds, the mean deflection was 0''.13, the lengths were 90 and 120 inches. 115. Case III. — Columns that fail with less than the FULL CRUSHING STRENGTH OF THE MATERIAL AND WITH DEFLEC- TION. The pillars of this class vary in length for any material and section, between the limits of I of Case XL, or that length at which the mean intensity of the crushing load is 6^ -^ 2, and that length of Case lY. that flrst fails by cross-breaking, both limits being determined by experiment. T' = the greatest tensile strain at the middle section. C = the " compressive strain required to balance T\ Then assuming that the intensity of the tensile strain in the convex side of the pillar varies with the deflection, ^, from * Senate Ex. Doc. No. 5— 48th Congress, 1st Session, pp. 60-67. 128 STRENGTH OF BEAMS AND COLUMNS. to T^ the tenacity of tlie material, the following formulas are deduced for the various sections : .'. C = q- -J-. By giving to q and ^ the proper value for the section, the following formulas have been deduced : Rectangular Pillars. From Eqs. 27 and 28, page 38, we have ^ _ T K -f- — cj ^^ which cL = ^, d, Then the expression for C ' becomes, from substituting these values, ^, ^ S' iZd - 6) T d, {d - df ' The sum of the greatest intensities of the two pressure wedges at '^ failure" is O -\- C . The value of d^. in this formula is that required for rupture of the column when broken as a beam. Example 44. — Required the greatest strength of a Rect- angular AVrought-Iron Column, tested with two 1''^ pin ends, when The diameter ^ = S'^OO, C = 50000 pounds assumed, " breadth h = 3".00, T = 50000 '' by test, '' deflection d = 0".34, d^ = 0''.66 for rupture. STRENGTH OF COLUMNS. 129 From Eq. 96 we have T^-f^nnnn i (0-34)2 (3x3 - 34) .^,^„„~| 3 (3 - 0.34) ^,-^.„ , L —\ 50000 A TTTTT-r. 7rn-.-T- -^0000 = 2422d8 pounds. L O.bb 1,3 — 0.34)- J 2 Tlie greatest strengtli was 284000 pounds,* tlie length was 80 inches. Circular Pillars. Bj substituting C -\-C' for 2(7 in Eq. 100, page 131, whicli gives the resultant of the load for this case, we have for the load that the pillar will fail with, Z = [('+?f].-/ (97) The numerical values required for the factors q and f are to be taken from the Table, page 131, for that position of the neutral line that corresponds to d^ = d —6. The value of d^. is that re- quired for rupture of the material in a circular beam, with its full compressive and tensile strength. Example 45. — Required the greatest load sustained by a Cylindrical Column of Midvale Steel, when The diameter d = 1^M29, C = " length.... Z = 8".96, T = " deflection S = 0''.25, o = 152000 pounds mean of tests, 112285 " " " " 1.353. From the Table, page 131, y = 1.1 839, for the neutral line at failure, d, = l'M29 - 0.25, q = 0.5278, and d^ = 0.38, the position of the neutral line of rupttire. * From the United States Government Watertown Arsenal Teste, 1883, Senate Ex, Doc. No. 5— 48tb Congress, 1st Session, page 56. 130 STRENGTJI OF BEAMS AND COLUMNS. From Eq. 97 we have L = (152000 + ^ ^5 X Q-^J^^^^ X ^^^^^'^ -^ (0.5645)-^ 1.1839 = 72056 pounds. The greatest strength was 81250 pounds."^ Hollow Cikculak Pillars. By siibstitutnig O -\- C for 2(7 in Eq. 101, page 135, which gives the resultant of the load for this case, w^e have Z=[C+2JI ]/■//„ (98) in which the numerical values required for q and ^/^ ^re to be taken from the Table, page 136, for that ~P^^^^^^ position of the neutral line at failure that //^'"^^"^^yX corresponds to d^ =: d — ^, the radius / / \ \ of the outer circle being r and t the hL -I - - . . .v|( j. ji thickness of the metal, both to be ex- \ >v ^ yy pressed in inches. The numerical value \^^^J^^/ of d.j, is that required for rupture of the material in a hollow circular beam, with its full compressive and tensile strength. 116. Case IV. —Columns that cross-bkeak from com- pression. The shortest pillar of this class, for a given section and material, is that (Fig. 31) in w4iich gn — 6 = d^, or that depth of extended area required for equilibrium when a beam of this section and material is broken by a transverse load, and the longest is that that just deflects sufficiently to allow the re- sultant of the applied load, Z, to pass through the fulcrum,/ ; this varies with the section, for a given material. * From the United States Government Watertown Arsenal Tests for 1883-84, Senate Ex. Doc. No. 35— 49tU Congress, 1st Session, p. 369. STRENGTH OF COLUMNS. 131 The load is the siom of the two pressure wedges. In rect- angular areas or those that may be divided into rectangular areas, the load is the compressed area, multiplied by the crush- ing strength of the material. Rectangular Pillars. The position of the neutral line must be computed from either Eq. 25 or 26, page 38. Then, in order that the breaking load of a given pillar may be deduced from the principles applicable to Case lY., (5", the deflection, must be equal to or greater than ^4, and equal to or less than \d. Then L = M,C, (99) from which the required load may be computed, being the sum of the two pressure wedges required for rupture. Example 46. — Required the breaking load of a Rectangular Yellow Pine Column, tested with flat ends, when The diameter.. cZ = 5".5, C = 5230 pounds mean of four tests, " breadth.. J =5".5, r = 15178 " " "tests, " deflection (^ = 0".66, 17 r:^ 0.337. The position of the neutral line, computed from Eq. 26, page 38, is d^ = 5". 39, and d^ = 0.11, being less than the de- flection, the column failed by cross-breaking ; the deflection being less than one third of the least diameter, it failed with- out cantileverage. . • . Z = 5.5 X 5.39 X 5230 = 155043 pounds. In a series of experiments to test the strength of yellow pine columns with flat ends, conducted on the United States Gov- 132 STRENGTH OF BEAMS AND COLUMNS. ernment testing macliine, at Watertown Arsenal, 1881-2,* thirty columns were broken, whose lengths varied from I = ^1d to ^ = 45<;/, which appears to be the limits of cross-hreaking^ without leverage for the yellow pine tested. The crushing strength obtained from short blocks differed so greatly that a msan was not admissible, though two consistent classes can be made of the material from these tests. The mean of four tests was C = 5230 pounds, from which the required load in the above example was computed ; the mean load from three tests was 154000 pounds. The mean value of C from three other tests was 3600 pounds, from which the computed load in the following Table was obtained. The tensile strength was obtained on the same machine the previous year, and not from the material composing the columns tested. The " experimental '' load in the Table is a mean of the number of tests given in the last column ; from " failure at knots and diagonally," the remainder of the thirty experiments mentioned above were rejected. Dimensions in Inches. Load in Pounds. I -*• d. Deflection in Inctiep. No. of Tests. I. b. d. Computed. Experiment'] 27.0 180 15.6 6.6 .5 to .62 344822 409000 3 30.8 180 12.0 5.8 .73 " 1.26 235000 250000 1 31.2 240 9.67 7.7 .82 " 1.6 249253 281000 2 31.2 180 15.5 5.6 .52 290718 344000 2 32.8 180 5.5 5.5 .52 101574 129500 2 36.0 180 12.1 5.0 1.3 " 1.8 202554 230000 1 38.2 210 5.5 5.4 1.03 '' 1.95 100584 97330 3 40.0 180 11.6 4.4 .9 " 1.30 170800 126350 2 43.0 320 9.28 7.4 .99 " 2.55 231516 199830 3 45.0 240 5.4 5.4 1.28 98776 84900 2 45.0 180 11.35 4.1 .95 166394 142000 1 In the above Table the second and sixth tests carried the maximum load with deflections that varied a half an inch in amount before failure. * Senate Ex. Doc. No. 1 — 47th Congress, 2d Session, p. 321. STRENGTH OF COLUMN'S. 183 Example 47. — Required the breaking loads of the series of White Pine Cohinins tested with flat ends, on the United States Government machine, at Watertown Arsenal, 1881-2,* when T = 10000 pounds mean of other tests, C — 2500 pounds mean of five of these tests. The position of the neutral line from Eq. 26 for «/ = 0.25 is d^ = 0.92rZ. The breaking loads in the following Table were computed from Eq. 99, as in Example 46 ; the first pillar of the Table did not deflect quite sufficiently to belong to this class of columns, but as the error is small, the lower limit is taken to be, as in yel- low pine, Z ~ 27<:Z ; the greater limit was not determined by the tests. The " experimental loads " and the dimensions are the means of those given for three tests. DiMEN SIGNS IN Inches. Load in Pounds. Deflection in Inches. /-»- d. I. h. d. Computed. Experimental. 27 180 15.6 6.66 .4 to .3 239400 227300 32 180 15.6 5.62 .37 " .83 202800 164100 32 240 9.4 7.45 .7 " 1.2 161200 170300 33 180 11.3 5.4 .6 165400 156200 36 280 9.6 7.69 .66 " 1.2 169900 153300 40 180 11.6 4.48 .43 " 1.25 119700 130700 43 320 9.3 7.47 .7 " 1.67 159700 147600 CiRCULAK Pillars. For this Case the deflection, (5^, must be equal to or greater than d^^ the depth of the tension area given by the Table be- low for the neutral line, when C -^ T = 6-d) hd: ' Then we will have for the applied load, Z, Z = hd,C-hd,C\ * Ex. Doc. No. 23, House of Representatives, 46tli Congress, 2d Session, p. 278. 138 STRENGTH OF BEAMS AND COLUMNS. from which, hy substituting the above value of C\ we ob- tain ^ ;^3-:r7^- (102) When SS becomes equal to or less than d, the depth, the for- mula becomes that for rectangular pillars of Case lY., the cantilever effect of the load having disappeared. As before stated, the formulas deduced for the strength of columns that fail by cross-breaking with cantileverage are iden- tical in form with those given by Mr. Lewis Gordon, but, unlike his, they give exact values and show when the formula does not apply. This uncertainty as to the length of pillars to which Gordon's for- nmlas did not apply has been the chief objection to their use in practice. Adapting Gordon's formula for the strength of rectangu- lar pillars to our notation, we have J. AC 1 + cI From making A the full area of the column, the factor C could never be the crushing strength of the material, but is always something less than it in value. The second member of the denominator makes the formula true for all lengths of pillars, which experiment does not confirm, ^nd is of the same general form as our formula for the deflection given by Eq. 93. ^o successful effort has ever been luade to so modify Gor- don's formula that the computed and experimental strength STRENGTH OF COLUMNS. 139 of wooden columns would be the same in value ; and for the reason that in all published results of experiments on wooden pillars, except for a few on yellow pine made at the Watertown Arsenal, they all failed without can ti lev- erage of the applied load, and the deflection did not enter as a factor to decrease the strength as contemplated by Gor- don. In the tests of white and yellow pine columns made at the Watertown Arsenal, and described in Examples 46 and 47, the strength was not decreased from cantileverage. For any given rectangular section of these pillars, the strength is the same for all lengths from twenty-seven to forty-five times the least diameter or least side of the rectangle, which includes the lengths of all yellow and white pine pillars that are used in structures. Example 50. — Required the breaking load of a Rectangu- lar Wrought-Iron CJolumn, tested with two V'^ pin ends, when The diameter d = 3".0, O = 50000 pounds assumed. " l)readth h = 3'.0, T = 50000 pounds mean tests. " deflection 6 = 1".61, q = 1. The position of the neutral line from the Table, page 92, is d^ = 0.7Sd; from Eq. 102 we have ax 1.61 -3.0 1 + 0.783 ^ "^ 2.34 This example and those in the following Table are from tlie series of tests described in Example 44, page 128, as given in " Senate Ex. Doc. Ko. 5— 48th Congress, 1st Session," pp. 68 to 102. 140 STRENGTH OF BEAMS AND COLUMNS. Dimensions 3" X 3". Deflection IN Inches. Load in Pounds. I -^d. Length in Tiif'hpR 1. mJ.\j Lx\^D , From. To. Computed. Experimental. 28 84 0.5 216000 30 90 0.5 218975 32 96 0.5 'I'.ei" 197666* 219000 34 102 0.8 1.61 197000 208500 36 108 0.8 1.80 178760 190500 38 114 0.5 1.80 178760 181125 40 120 0.6 1.9 168000 181750 42 126 0.5 1.67 188800 169700 44 132 0.4 2.30 131700 178950 46 188 0.4 1.98 155600 157825 48 144 45 2.16 141100 155790 50 150 0.45 2.30 131700 158065 52 156 0.35 2.50 120088 155000 54 162 0.30 2.50 120088 147750 56 168 0.30 2.80 106000 149875 58 174 0.85 2.50 120088 128775 60 180 0.30 2.50 120088 126875 When under the greatest load sustained by the above col- umns, they "suddenly sprung" to a cross-hreaking deflection with cantilever age. The " Deflection from " in the Table is that at which the " sudden spring " began, and " Deflection to " is that to whicli it sprung / the deflection in the Table is the mean of those of two tests, in most cases. '^ The " Experimental " load is the mean of two tests, in each case, and is that sustained by the column at the beginning of the " sudden spring." From the " Computed " loads it will be seen that the momentum of the " sudden spring " caused the observed deflections to exceed the trne deflection in many examples of this series of tests. Circular Pillars. 6 — \d =^ the lever-arm of the load, Z, f = the factor in Table, page 134, /, = the " " " pages 80, 100 and 112. STRENGTH OF COLUMNS. 141 From Eq. 76, page 57, the moment of the applied load, Z, becomes , _ Z{8d - Sd) . C = r% and for the load from the pressure wedges given by Eq. 100 we have L ^ 2rYC - 2rYC'. Substituting for C its value given above, we have L=. 1 + 2/(8()^ — Zd), ■rf. When 8(^ becomes equal to or less than Zd the load must be computed from the formula for Circular Pillars, Case lY., as in this position of the pillar there will J be no cantilever effect of the applied load to be deducted from the sum of the two pressure wedges. (103) Hollow Circular Pillars. d = the outer diameter in inches, r = the " radius " " {6 — \d) = the lever-arm of the load, Z, /J) = the factor given in Table, page 136, /„ = the " " " pages 85 and 104. t = the thickness of the metal ring in inches. From Eq. 91, page 62, the moment of the applied load be- comes 142 STRENGTH OF BEAMS AND COLUMNS. _ A^S-d) For the load, Z, we have Substituting the value of C ^ we obtain When ^S is equal to or less than the outer diameter, d, the strength of the column must be com- puted from Eq. 101, Case lY., as there will then be no cantilever effect to de- duct. Example 51. — Required the Breaking Strength of a Phoenix Column tested witli flat ends ; C = T = 60000 is the assumed strength of the iron. Outer diameter d — 8".0, q = C -^ T = 1, Thickness of metal, t = 0".85, /o = 2.9732 from Table, page 136, for g = 1^ Deflection 5 = 2".47,/e = 2.6806 " " " 104, " q = U From Eq. 104 we have ^ _ 2 X 4 X 0.35 X 2.9732 X 60000 _ 49949 7 ^ 39^426 ~ ^ , 2.9732 (4 X 2.47 - 8) ~ 1+0.26 "^ 2 X 4 X 2.6806 This example is taken from the series of experiments^' de- scribed in Example 49, page 136 ; the tested breaking load was 416000 pounds ; the length was 28 feet. * Report of the United States Board appointed to test iron and steel, Ex. Doc. No. 23, House of Representatives, 46th Congress, 2d Session, page 270. STREIN^GTH OF COLUMNS. 143 Angle-Iron, Box, Channel, Eye-Beam and Tee Pillars. In pillars of the above sections, the distance of the neutral line from that side of the pillar that will most likelj be its concave side, when broken, must be computed by the rules heretofore given for it in beams of the section of the pillar. The sum of the two pressure wedges will then be the product of the compressed area, A^^ by the crushing strength of the material. The distance, g, of the centre of gravity of the applied load wedge from the point k (Fig. 34), must be determined, which is obtained by computing liF, the moment of the load wedge with respect to the axis, A', in which F is the greatest intensity of the load at ^', it being zero at m., and 7? a factor that depends for its value upon the section ; for any given dimensions it reduces to a numerical quantity without assign- ing any value to F\ and dividing this moment, ^7% by the volume of the load wedge, AF -^- 2, in which A is the area of the section, we have BF--- J: 9 and iF _2B ^ __L{S-g) Then we have Substituting the above value of C, we deduce . , AAS^-j). (105) B^ is a factor of the mom,ent of resistance that the section of the pillar offers to the cantilever bending of the load ; for a given section it becomes a numerical quantity without assign- 144 STRENGTH OF BEAMS AND COLUMNS. ing any value to C wlien computed by tlie rules given for the mouient of resistance of the section when strained in a beam. When S becomes equal to or less than g in value the canti- lever strain ceases to exist, and the pillar belongs to Case lY., .-. L^ A,C. (106) When d becomes less than r/^, the depth of the extended area required for rupture, the pillar belongs to Case III. in which d^ gives the position of the neutral line of rupture, and ^ the ratio of the compressive strain that will be required to hold in equilibrium the tensile strain developed by the bending of the pillar as a cantilever. Equation 105 is the general formula for the strength of pillars of all lengths, sections and material of which the formulas heretofore deduced in this chapter are only the forms it will assume for special cases. The denominator of the second member of the formula must never be less than unity. CHAPTER YIII. COMBINED BEAMS AND COLUMNS. 118^ General Statement. In roof -trusses, cranes, derricks, platforms supported by cantilevers, trussed beams and other structures, there is used a class of pieces of material that, from the manner in which they are loaded, do not belong ex- clusively to either horizontal beams or columns, but partake of the nature of both, in the manner in which they support the load to which they are subjected. The theory of the transverse strength of these Combined Beams and Columns gives the solution of the general problem of the transverse strength of all beams, without regard to the special angle that the axis of the beam makes with the line of direction of the loading and supporting forces. Horizontal beams acting under vertical loads and columns are only special cases of tlie general problem, in which certain factors, that cause the strength of the same piece of material to vary with its angle of inclination to the horizon, disappear, from the general rale for these cases, by becoming zero in value. But on account of their great importance, we have, in the preceding chapters, deduced separately the principles and rules from which the strength of these special cases may be com- puted. Should one end of a horizontal beam, such as Ije (Fig. 44), be fixed in a vertical wall, and a load be attached to its free end, not suspended, as in the figure, none of this load will di- rectly compress or rest upon the cross-section of tlie beam, he. Now let the wall be revolved around the point 7^, to a horizon- tal position, thus bringing the beam, he^ to the vertical, then the 146 STRENGTH OF BEAMS AND COLUMNS. entire load will rest upon or directly compress the cross-section of the beam, now converted into a pillar ; the load has thus been gradually converted from a.non-compressing to a compressing load with its full weight. The bending moment of the attached load is greatest when the beam is in the horizontal position, and it gradually diminishes as its lever-arm, s^ becomes less in value with the revolution of the wall, and becomes zero in value when the beam occupies the vertical position. On the other hand, should the wall be revolved to the horizontal position around the point P^ the bending moment will gradually de- crease and become zero, while the tensile strain w^ill increase from zero to that of the full w^eight of the load, when the beam becomes vertical. From the above illustration the origin of the special cases of horizontal beams and colicmns is apparent, and the reason for the special rules for their strength. A similar illustration could be deduced from a beam supported at both ends, by con- ceiving it to occupy all positions from the horizontal to the vertical. In order to deduce the relation that exists between the applied load and the resistance of the combined beam and column at the instant of rupture^ it is assumed., in the analy- sis, that they only deflect enough to admit of their cross-hr cak- ing as a column, without cantileverage of the applied load, as in Class lY., Chapter YII. In very long beams, how^ever, the compression resulting from pressure applied to its ends will act with the leverage explained in Case Y., page 137, and the strength must be computed from tiie formulas there given, but the compression resulting from the transverse bending of the load will be the same in each case, and will compress the beam without cantileverage. 119. Notation. In addition to the notation heretofore used and defined in Art. 35, page 37, the following will be used in this Chapter : COMBINED BEAMS AND COLUMNS. 147 (7 = 0'-{- C" ^=^ tlie greatest intensity of the compressive strain in pounds per square inch, C = the greatest intensity of tlie compressive strain in pounds per square inch, arising from the bending component of the load, C"^= the greatest intensity of the compressive strain in pounds per square inch, arising from the compressing compo- nent of the apphed load, Zb = the bending component of the applied load in pounds, Zc == the compressing component of the applied load in pounds, I = the unsupported length of the beam in inches, s = the span, the horizontal distance between the supports in inches, h — the difference between the heights of the ends of the beam in inches, a = the angle that the beam makes with the horizon. In many of the different methods of loading and support- ing beams given in this Chapter, the greatest resulting com- pressive strain at any section is the sinn of two or more dis- tinct pressures, which will be represented by the letter O, with corresponding accents, such as C\ C\ C"\ C"'\ etc. Inclined Beams. 120. General Conditions. The effect produced by a load when applied to this class of beams will manifest itself in two distinctly different ways ; each separate effect must be computed, and their sum will be the total effect produced by the load upon the beam. This is accomplished by decompos- ing the applied load into two components^ by the well-known theorem of the parallelogram of forces. One component, Z^, must be at right angles to the axis of the beam, and the other, Zc, parallel with it ; the first component will bend the piece as a heam, while the second will compress it as a column. The parallelogram of forces for inclined beams is a rect- 148 STRENGTH OF BEAMS AND COLUMNS. angle, and the relation between the components and the load is obtained from that of the three sides in a right-angle tri- angle, in which COS. a =^ — h sin. a = — V , h tan. a=^ -. s 121. Inclined Beam Fixed and Supported at One End. This method of " fixing " and loading beams is illustrated by two different positions of the beam, he^ in Fig. 44. Two Cases will be considered. Case I. — When the load is applied at the fi^ee end of the heam. Let the right-angle triangle, eot (Fig. 44), represent the parallelogram of forces, in which et represents the applied load, drawn to any given scale, then ot will be the hending and oe the compressing component of the applied load, Z, from which ot =: et COS. a and oe = et sin. a, Ls T' L. (108) COMBINED BEAMS AND COLUMNS. 149 and A = ~-^ (109) The bending moment of the component, Z^, its lever-arm being /, will be Bending Moment, L^ = -^Xl = Ls. (HO) Having decomposed the applied load, Z, into two compo- nents, one perpendicular and the other parallel to the inclined beam, we can now ascertain the effect that will be produced by the original load upon it, by a combination of the methods used to compute the effect produced upon horizontal beams and columns by vertical applied loads. Rectangular Beams. The transverse strength of a rectangular beam loaded and fixed as in this Case will now be deduced from the foregoing formulas. The bending moment from Eq. 110 must be made equal to the moment of resistance from Eq. 23, page 38, in which r.Ls='^, (111) 3 ZLs d. hd,C'. (112) The compressing component, Z^, from Eq. 109, must be re- sisted by an equal compression produced in the section at the face of the wall, as given by Eq. 99, page 131, in which C = C" and L = ^^ ^=:hd,G", (113) 150 STRENGTH OF BEAMS AND COLUMNS. Adding Eqs. 112 and 113 we have, by making C ' -{- C " ^=. C^ the crushing strength of the material, from which L = _J^^_. (115) When the beam is horizontal, h^^O and I = s, the formula reduces under this hypothesis to that given in Eq. 27, page 38, in which m = 1. In a vertical beam, ^ = and I = h, the fornmla reduces to that given for the strength of rectangular columns, Eq. 99. To Design a Rectangidar Seam. The length I and height h will be controlled by the position in which the beam is to be used. A convenient depth, d, must then be assumed, and the value of d^ computed from Eq. 26, page 38 ; then from Eq. 114 we obtain from which the required breadth, h, of a beam that will break with a given load, Z, will be obtained by giving to O the value of the crushing strength of the material of which the beam is to be constructed. Case II. — When the Load is uniformly distributed OVER THE unsupported LENGTH OF THE BEAM. The resultant of the applied load will pass through the middle of the length of the beam and the triangle, eot (Fig. 44), will, as in Case I., give the relation between the load and its two components ; hence A = ^, (in) Ze = ^- (118) COMBINED BEAMS AND COLUMNS. 151 The bending moment of the component, Z^, will be, Bending Moment =^ — - x ^- — --5 (^l^) its lever-arm being Z -^ 2. Rectangular Beams. The transverse strength of a rectangular beam loaded and fixed as in this Case may be obtained by the same process heretofore used in Case I., • /- ^i^^ (120-, To Design a Rectangular Beain. Assume a depth, r7, and compute J^ from Eq. 26, page 38, then from Eq. 120 we have , _ {Zh + 2r4A) L 133. Inclined Beam, supported at one end and stayed or heUl in position at the otlier with- out vertical support. This class of Inclined Beams is illustrated in Figs. 45, 46 and 47. Five different cases of loading will be considered. Case I. — When the Inclined Beam is loaded at its STAYED END. The effect produced by the load, Z, upon the beam, he (Fig. 45), is simply to compress it as a colunm, the load being held in equilibrium by a pnll along the tie ec^ and a thrust along the inclined beam. In the right-angle triangle, toe^ let et represent the load 152 STRENGTH OF BEAMS AND COLUMNS. drawn to any given scale, then to gives the pull and oe the thrust. et te 01 = -: • 5 oe = -. — tan. a sin. a Ls The pull = -J-, and The thrust L, n h (122) (123) from which the compressing effect of the applied load, Z, can be ascertained. Rectangular Beams. The direct compression given bj Eq. 123 must be made equal to the resistance offered bj the section of the beam as a column, then from Eq. 99, page 131, h ' L = hhd.C I (124) COMBINED BEAMS AND COLUMNS. 153 Case II, — When the Inclined Beam is loaded at its MIDDLE. Let he (Fig. 46) represent the beam loaded at its middle, t^ witli the load, Z, then in the right-angle triangle, toe^ et rep- resents the load drawn to any scale, ot^ perpendienlar to the beam, hc^ the hending component, Zg, and ot parallel to he, the directly compressing component, L^. _ Lh A- -J- (125) (126) The hending moment produced by the component, Zg, is identical with that produced by conceiving the beam to be " fixed " at its middle, t, and loaded at the free end with . • . Bending Moment, Zb = — y X ^ == ~T~' (1^^) In order that the loading and supporting forces shall be in equilihrium, one half of the hending component, L^, must be supported at each end of the beam h and c. The half of L^ at G cannot be directly supported at that point, but must be 154 STRENGTH OF BEAMS AND COLUMNS. carried to the ground bj some means. Tlie method of stay- ing the end of the beam, he, represented in the hgure is tliat used in roof -trusses ; hence the load, Z3 -^ 2 at c', must be de- composed into two components, one in the direction of each beam or rafter ; if the angle, hck, is a right angle the entire load, Zb -^ 2, compresses the rafter, ch, as a pillar, thus reach- ing the support, Jc. In roof-trusses the inclined beams, he and ck, are usually loaded in the same manner, and eaeJi beam will, therefore, carry to e a component, L^ -^ 2, of its load for support, Avhich will be equivalent to the component, L^ -^ 2, compressing each rafter to which it is applied, but much increased in amount from the manner in which it is converted from a load that is perpendicular to the beam into a compressing load that is parallel to its axis. In the two equal triangles, cmn, let em in each represent the load, Ljj -^ 2, €)i the equal components that the rafters exchange w^ith each other, and mn the equal components that the loads, Z3 -^ 2, produce that strain the rafters to which they are applied. When the angle, 5c^, is greater than 90° the sum of the components {inn -f ei%) will be greater than L^ -.- 2, when hek = 90°, {?nn + en) = Z^ -^ 2, nm becoming Izero. When hek is less than 90° the component, mn, is a tensile- strain ; the total compression {cm — m?i) wdll then be less than Z«-2. In roof -trusses the angle, hek, is generally greater than 90°, and the total compression upon each rafter resulting from the component of the load that it carries to c is {mn -\- en). mn = me . tan. (90° — 2<^), 7n.e en = eos. (90° - 2a)' Addlns: and substitutino: me — -—r, we have Ls^ 21 COMBINED BEAMS AND COLUMNS. 155 (™. + ..) = ^ = |(.«».(90»-2.)+— ^J.-^,). (128) The factor, (?, in this equation, for convenience is . = .«n,.(90»-2.) + --^_^^. (129) Wlien the angle a ^ 45°, tan. (90°— 2a) := 0, — 4 — ^--^ = 1, COS. ( t/U JiCtj and the amount of compression will become — ? =: —^ , the angle, hck^ being a right-angle. The total compression produced by the single load, Z, at the middle of each rafter will be the sura of the three distinct parts, represented by Eqs. 126, 127 and 128. Rectangulak Beams. The amount of direct compression from Eqs. 126 and 128 must be equal to the resistance of the beam as a column, from Eq. 99, page 131, in which C — C and C— C" respectively. .-. ^=:hd,C\ (130) and :^^ := M,C'". (131) The hending Tuionient from Eq. 127 must be equal to the moment of resistance of the section as a beam from Eq. 23, page 38, in which C ^ G" . • • -T'"~3~' ^^^^^ .-. ?I:'L = hd,0'\ (133) 4:d^ Adding these separate components, as given by Eqs. 130, 156 STRPJNGTH OF BEAMS AND COLUMNS. 131 and 133, and making C'+ C" + C" = C, the greatest compressive strength of the material, we have .-. L = iii^ Q35X 35^ + (4A+2c6')6^e from which the required breaking load may be computed. Case III, — When the Load is uniformly distributed OVER THE LENGTH OF THE INCLINED BeAM. The bending moment and the component L^ will be identi- cal with those produced by conceiving one half of the total load, Z, to be concentrated at its middle section, t (Fig. 46) ; then, from Eqs. 125, 126 and 127, Case 11., by making Z = Z -i- 2, we have L. = §, (136) A = ~, (137) Ls Bending Moment, L^^^ ^-' (1^^) o The bending moment is produced by a component of the load, Z, that is perpendicular to the length of the beam, and uniformly distributed over its length, he. In order that equili- brium shall exist, one half of this uniformly distributed com- ponent, — -^ must be supported at each end of the beam, and as it cannot be directly supported at the stayed end, c, it must be carried to the supports h and k, as described in Case II., Eq. 128. The method of analysis of the strains in a simple roof-truss, given in this and the preceding Case, differs from that usually COMBINED BEAMS AND COLUMNS. 157 pursued by writers. One half of the load on each rafter is usually considered to be supported by h and h^ and the other half of eacli rafter load to be supported at c, by reacting against each other. The transverse strength is then computed as if the rafter were a horizontal beam, which is equivalent to saying, that for a given material, section and span of roof, as in Fig. 46, the transverse strength would be the same for the infinite number of roof -trusses that could be constructed between these points of support by varying t[\Q jpitch or angle that the rafter makes with the horizon, which, of course, is not the case. Rectangular Beams. From Eqs. 128, 137 and 138, ,we obtain, by a process similar to that used in Case III., Z (--\- - 4- — _) = hd,C, (139) . • . Z = Sj^^ , Q40) 3,9^+(4A + 2t'^)^e ^ Example 52. — Required the uniformly distributed breaking load of a White Pine Rafter, as in Fig. 46, when The span 5 = 15'. 0, 6^ = 5000 pounds mean of tests, " length Z = 16'. 8, Z = 10000 " " " " " rise A = 7'. 5, ^ = 0.5, " depth d= 9''.0, d^ = OMSd, from Table, page 108, " breadth... &= 5^0, ^ ^ 26°.34'. From Eq. 129, , and longitudinally with either a compressive or tensile load, L^. COMBINED BEAMS AND COLUMNS. 165 From Eq. 27, page 38, in which 6^ = C\ we have L = ^Ls Zs tncL hcLC\ (168) (169) And from Eq. 99, page 131, we have L, = M,C'\ (170) Adding Eqs. 169 and 170, we have, bv making C'-\-C"^^C^ tncL + A = hd,a (171) From which either Z, Zc, or h maj be computed when the other two are known. « OS z.= h = mM;C - ZLs (172) (173) (174) Hollow Circular Beams. The relation between the transverse load, Z, and the result- ing moment of resistmice of the beam is given by Eq. 91, page 62, in which C = C. 166 STRENGTH OF BEAMS AND COLUMNS. /c =: the factor from the Table, pages 81 and 104, /; = the " " " " page 136. X = mrtjcC — ^L±i/-f — • 2/„Z mr/^ (175) = S/Y/^^'^ (176) The compression from the longitudinal strain, Zc, must be equal to the resistance of the beam as a column from Eq. 101, page 136, in which C = C" . L, = ^rtf,C". ,(177) Adding Eqs. 176 and 177 we have + Ze = ^rtf.C (178) 2/oZ . . X- 2^. - • (l^y) The compressing load, Z^ in the foregoing formulas, in practice, generally results from the transverse load, Z, its effect being transmitted to the ends of the beam by a vertical post placed under the centre of the beam and connected with the ends by inclined truss-rods. For a single load, Z, at the middle it is usually the practice to consider one half of it as producing a tensile strain, on each of the inclined truss-rods ; this is only strictly correct when the beam is cut into two pieces at its centre. When the beam is continuous it can only strain the vertical post after it l)egins to deflect ; therefore the load that would have caused the untrussed beam to deflect from the horizontal position could bring no strain upon the post. I ]sr D E X. PAGE Beams. See Transverse Strength, Moment of Resistance and Neutral Line Beams, Cast-Iron, 64 <( Wrought- Iron, 87 < I Steel, 87 i t Timber, 106 (( Horizontal, 3 " Inclined, 148 " Trussed, 164 << Elastic Limit, 34 " and Columns, Combined, 145 Bendin g Moments, 3 " General Formula, 6 ( < Moment and Moment of Resistance, .... 22 Coefficients of Strength, 33 Columns, Box Section, 163 " Classed,, 123 < ( Deflection of, 121 < ( Experiments, 122 I i End Connections, 124 " Failure of, without Deflection, 125 << " with " 127 " " without Cantileverage, , 130 ' ' " with 137 " General Conditions of Failure, 114 < ( Gordon's Formula, 138 < c Pin End, 118 " Resistance of, 119 < < Rectangular Sections, 128, 131, 137 " Circular Sections, 129, 133, 140 << Hollow Circular Sections, 130, 135, 141 <( Angle-Iron, Eye-Beam, Channel, etc.. 143 Combined Beams and Columns, 145 168 I^^DEX. PAGE Crushing Strength, To Compute 39, 58, 126 of Cast-Iron, 64 To Compute, 67 " Wrought- Iron, 87 To Compute, .... 88 Designing a Rectangular Beam, 39 " a Hodgkinson " 44 a Double T " 48 a Box " 48 an Inverted T "......• • 53 a Double T " o3 Elastic Limits Defined, 33 " Limit of Cast-Iron, 65 " *' Wrought-Iron, 87 " " Steel, 88 " " Materials, 33 Equilibrium, 3 " Moment of Resistance and Bending Moment, . . 23 Factors of Safety, 34 Forces Defined, 1 " Uniform Intensity, • • 2 " Uniformly Varying Intensity, 2 Gordon's Formula, 1^38 Inclined Beams Supported at One End, 148 " Both Ends, 160 Load, • • • 2 Moment of Resistance Defined, 21 " Box Sections, General Formula, ... 47 " Circular Sections," " . . .55 Double T Section,'' " ... 47 Hodgkinson " " " ... 41 " " Hollow Circular Sections, General Formula, 60 '' " " Eye-Beam Section, General Formula, . . 47 Inverted T, " " . . 52 *' " Rectangular Sections, " *' . . .37 " " Concentrated Parallel Forces, 3 " " Uniformly Varying Forces, 8 Neutral Line Defined, 21 " " Box Sections, General Formula, 48 " " Cast-Iron Sections, 78 " " " Wrought-Iron and Steel Sections, .... 96 " " Circular Sections, 56 " •' " Cast-Iron Sections, 80 INDEX. 169 Neutral Line Circular Cast-iron Sections, Table of, " " " "Wrouglit-Iron and Steel Sections, " " " " Table of, " " " Wooden Sections, Table of, . " Double T Section, General Formula, " " Wrouglit-Iron and Steel Sections, " " " " Cast-Iron Sections, ..... " " Eye-Beam Sections, " " Hodg'kinson Section, General Formula, . " " " Cast-Iron Section, .... " " Wrought-Iron Section, " Hollow Circular Sections, " Cast-Iron Sections, Table of, . " " " " Wrought-Iron Sections, Table of , " " Inverted T, General Formula, .... " " Movement of, " " " " from Experiments, .... " Rectangular Section, General Formula, " " " Cast-Iron Sections, .... Table of, . " " " Wrought-Iron and Steel Sections, " " " Tableof, Wooden Sections, Table of, " " Position of, " "of Rupture in a Rectangular Section, " Transverse Elastic Limit, Notation for Rectangular Sections, " " Hodgkinson Beam, " Double T and Box Beams, " " Inverted T and Double Inverted T, " Circular Beams " " Hollow Circular Beams, " " Columns, ..... " " Inclined Beams, .... " " Uniformly Varying Forces, Pin End Columns, ...... Relative Transverse Strength of Beams, " Strength of Square and Circular Beams, Cast- Iron Beams Timber I'AGE 80 100 100 112 112 48 96 78 96 42 75 96 61 85 85 104 104 58 26 82 , 38 ' 71 71 92 92 108 24 28 24 37 41 47 52 55 60 121 146 8 118 35 59 83 113 170 INDEX. 8, 16 89 Relative Value of the Crushing and Tensile Strength, . Resultant Defined . " Uniformly Varying Forces, Strain, Stress, Tensile Strength. To Compute, of Cast-Iron, «* «' " " To Compute, " " " Wrought-Iron and Steel, . '. . . " '< " " To Compute, " " " Timber, «« " " " To Compute, Transverse Strength, General Conditions, .... Box Beams, General Formulas, . " Cast Iron Beams, .... " Wrought-Iron and Steel Beams, " Cast-iron Beams, Circular Beams, General Formula, . Cast- Iron Beams, Wrought-Iron and -Steel Beams, ^" " " Wooden Beams, Double T Beam, General Formula, . '< " " " Wrought-Iron and Steel Beams, " " " Cast- Iron Beams, Eye-Beams, Wrought Iron and Steel, . Hodgkinson Beam, General Formula, '« " " Cast-iron Beam, . Wrought-Iron and Steel Beams, " Hollow Circular Beams, General Formula, '* " " Cast-iron Beams, " " " Wrought Iron and Steel Beams Inverted T, General Formula, . Rectangular Beams, General Formula. . Cast-Iron Beams, . " " " Wrought-Iron and Steel Beams, " " " Wooden Beams, . Steel Beams, " " Wrought-Iron Beams, .... " " Timber Beams, Trussed Beams, Working Load, 28 2 19 1 1 , 58 64 67 87 90 106 106 21 48 78 96 65 56 81 101 113 48 96 78 96 44 75 96 61 86 104 53 38 71 92 108 92 91 106 164 34 LIBRARY OF CONGRESS 019 449 670 9