A TREATISE ON THE STRENGTH OF BRIDGES AND ROOFS. ■,v w PREFACE. A T t"-ca."t\ s>s. on thfi t Ss^r^e, \~\ aVK o C BRIDGES AND ROOFS, WITH PRACTICAL APPLICATIONS AND EXAMPLES, FOR THE USE OP ENGINEERS AND STUDENTS. SAMUEL H. SIIREVE, A. M., Civ. Era NEW YORK: D. VAN NOSTEAND, PUBLISHER, 23 Murray St. and 27 Warren St. 1873. V>VK * o-- Entered according to Act of Congress, in the year 1873, by D. VAN NOSTRAND, in the Office of the Librarian of Congress at Washington. PREFACE. In the following treatise I have applied the simpler pro- cesses of Algebra to the discussion of the subject of strains in single span trusses, and have obtained many formulae for practical application, sufficiently elucidating this method, I hope, to render easy the determination of other formulas adapted to any form of truss that ingenuity may suggest. Only algebraic processes have been employed, because they are simpler, more comprehensible, more practical, and more accurate in practice, than those of the higher mathematics. Uniformly distributed loads alone have been considered,, whether full or partial, since it may be possible to load each part with the load which may be brought upon any one part; and if we consider the density of the whole load to equal the maximum density at any point, any other case can only produce a less strain. No comparison is made between the different systems which are given, many of them for the first time, since practical details of construction affect the theoretical economy, and to fully consider these would be beyond the scope of the present volume; there are many cases, however, where the practical difficulties may be ignored, since they are so nearly equal and a comparison readily made. The changes IV PKEFACE. in the form of a truss caused by a want of rigidity, or temperature, will not affect the values of the formulae. The discussion of the strains affecting drawbridges, leading directly to the subject of cantilever trusses, which includes continu- ous trusses, the Sedley system, and many other important forms, for which we have no practical formulae, must, with the subject of arched trusses, be reserved for a future volume. S. H. S. New York, January, 1878. CONTENTS. At:T. CHAPTER I. INTRODUCTORY. PAGE 1. Truss ...... 1 2. Strains .... . 1 3. Chords . , 1 4. Braces . . . < . 1 5. Ties , 1 6. Struts . , 1 7. Panels . , 1 8. The principles of the lever 2 9. The resolution of forces . 3 10. The equality of moments - 5 11. Each member of a truss may represent the line of a strain or force • 5 CHAPTER II. SIMPLE TRUSSES. Case I. — A Simple Truss supported at the ends and loaded at the centre only. Art. 12. Description . . • • 7 " 13. The horizontal strain • . . .7 " 16. The vertical strain • • . .9 v i CONTENTS. PAGB 14: 14 15 17 Art. 18. The strain in the struts . . . • 1° " 20. Example . . . • • • 11 Case II. — A Simple Truss supported at the ends and loaded at a point between the centre and one abutment. Art. 21. The horizontal strain . . .12 " 22. The vertical strain " 24. The longitudinal strain in the braces " 25. Example .... " 26. The ratio of horizontal strains . •" 27. Amount of horizontal strain in the braces . 17 Case III. — A Truss uniformly loaded throughout its ^length. Art. 28. Description . . • • .18 " 29. The horizontal strain . . . .19 « 30. Example . . . • • • 21 " 31. Graphic representation of the horizontal strain . 21 11 32. The vertical strain . . • .23 " 33. The longitudinal strains in the struts . . 24 " 34. The vertical strain uniform in the braces be- tween the weights . • • .25 « 36. The abutment bearing any part of the weight determined by the sign of the equation " 37. Graphic representation of the vertical strain . 26 Case IV.— A truss loaded from one abutment only a por- tion of its length. Art. 39. Description . • • • • 28 " 40. The horizontal strain in the unloaded part . 28 26 CONTENTS. Vll PAGE Art. 41. The vertical strain in the unloaded part . 29 " 43. The horizontal strain in the loaded part . 29 " 44. The vertical strain in the loaded part . . 30 4i 45. The point of no vertical strain in every fully or partially loaded truss coincides with the point of greatest horizontal strain . .31 Example . . ■ . . . .32 Each abutment's share of the weight from the nearest portion of the load . . .32 « 46. Example 33 Chord strains in the unloaded part . .33 Chord strains in the loaded part . .34 Vertical strains in the unloaded part . . 34 Tension in the ties in the loaded part . 35 Compression in the struts in the loaded part . 35 Case V. — A Truss subject to a uniform constant load throughout its length, and a uniform inovable load. Art. 47. Description . . . . .36 " 48. Maximum strains only needed . . .36 " 49. Horizontal strains . . . .36 u 50. Yertical strains . . . . .37 The vertical strain from a partial load, passing in one direction at any point, is greatest when the load extending from the abutment reaches that point . . . .38 Yertical strain at any point greater from a partial than from a full load . . .38 * 51. Point of no vertical strain where there are a uniform load and a partial load . . 39 Example . . . . . . 39 " 52. Counterbracing . . . . .41 yiii CONTENTS. Art. 53. Graphic representation of the vertical strain from the moving load . . . .41 " 54. Practical form of the moving load equation . 44 Example . . . • .48 CHAPTER III. A SIMPLE TRUSS WITH INCLINED STRUTS AND VERTICAL TIES, SUBJECT TO THE ACTION OF A CONSTANT AND A VERTICAL LOAD. Art. 55. Description . • . • .52 Horizontal strains . . . .53 Tension in the ties . . . .54 Compression in the struts . . .55 CHAPTER IV. TRUSSES WITH VERTICAL STRUTS AND INCLINED TIES SUBJECT TO CONSTANT AND TO MOVING LOADS. Case I. — A Simple Truss. Art. 57. Description . . . . .56 " 58. The horizontal strains . . . .56 " 59. The vertical' strains . . . .57 Case II. — A Double Truss with an even number of panels. Art. 60. Description . . . . .59 " 61. Horizontal strains . . . .60 " 62. Vertical strains . . . . .63 . " 63. Vertical strains from the moving load . . 63 CONTENTS. . ix PAGE Art. 64. Example . . . . , .65 il 66. Chord strains equal in amount between the same inclined braces . . . .68 Case III. — A Double Truss containing an odd number of panels. Art. 67. Description . . . . . .68 " 69. Horizontal strains . , . .69 " 70. The end panels do not affect the horizontal strains . . . . . .72 rt 71. Equations with distances from the centre . 73 " 72. Vertical strains from the constant load . . 74 " 74. Vertical strains from the moving load . . 76 " 75. Example of the application of the vertical equa- tions . . . . . ,79 " 76. The Quincy Eailroad Bridge . . .82 Description . . . . .83 Chord strains . . . . .85 Counterbraces • . . . .86 Vertical strains . . . . .86 Case IV. — A Triple Truss containing an even number of panels. Art. 77. Description . . . . .90 " 78. Horizontal strains . . . .90 " 79. Strain in the lower chord members . . 95 " 80. Vertical strains from the constant load . . 97 " 81. Vertical strains from the moving load . . 99 " 82. Horizontal equations, with the variable measured from the centre of the truss . . .102 * 83. Example . . . . . .103 X CONTENTS. Art. 84. Horizontal strains " 85. Vertical strains . u 86. Griethausen bridge Description " 87. Horizontal strains <{ 88. Vertical strains from the constant load " 89. Vertical strains from the moving load Strains in the braces PAGR . 103 . 105 . 109 . 110 .111 . 114 . 114 . 1W Case V. — A Triple Truss containing an odd number of panels. Art. 90. Description . . . . .119 " 91. Horizontal strains . . . .119 " 92. Vertical strains from the constant load . 122 " 93. Vertical strains from the moving load . 123 " 94. Example . . '. . . , 123 " 95. Horizontal strains . . . .124 iC 96.. Vertical strains . . . . . 125- " 97. A simple form of the moving load equation for general use .. . . .129 Case VI. — A Quadruple Truss containing an even num- ber of panels. Art. 98. Description . . . . # jgQ " 99. Horizontal strains .... 130 " . 100. Vertical strains from a constant load . . 135 " 101. Horizontal tension in the lower chord end members . . . . . 13& " 102. Vertical strains from a moving load . . 138 1 " 103. Example U1 " 104. Horizontal strains .... 142 CONTENTS. xi PAGB Art. 105. Yertical strains • . 143 Case VII. — A Quadruple Truss containing an odd num- ber of panels. Art. 106. Description . . . . . 144 107. Horizontal strains . . . . 145 108. Vertical strains from a constant load . . 147 109. Horizontal tensions in lower chord end mem- bers . . . . . 148 110. Vertical strains from a moving load . . 149 111. Example . . . . . .149 112. Horizontal strains . . . .151 113. Vertical strains • 151 114. Other compound trusses . . .153 115. Simple forms of equations . . . 153 CHAPTER V. TRUSSES WITH HORIZONTAL CHORDS, WITH STRUTS AND TIES OF EQUAL INCLINATIONS, SUBJECT TO CONSTANT AND TO MOVING LOADS'. Case I. — A Simple Truss. Art. 116. "Warren girder, horizontal strains " 117. Vertical strains from a constant load " 118. Vertical strains from a moving load " 119. Example Chord strains Brace strains . 154 . 155 . 156 . 157 . 157 . 158 Xll CONTENTS. Case II. — A Double Truss containing an even number of panels. Art. 120. Description " 121. Horizontal strains u 122. Vertical strains from a constant load " 123. Vertical strains from a moving load " 124. Example .... Chord strains Brace strains PAGE 159 160 162 163 163 164 165 Case III. — A Double Truss containing an odd number of panels. Art. 125. Horizontal strains .... 166 " 126. Vertical strains from a constant load . .167 " 127. Vertical strains from a moving load . .167 Case IV. — A Quadruple Truss containing an even num- ber of panels . Art. 128. Description . . . . . .168 " 129. Horizontal strains . . . .168 " 130. Vertical strains under a full load . .170 " 131. Horizontal strains in the end chord members . 171 " 132. Vertical strains from a moving load . .172 " 133. Example . . . . .172 Chord strains . . . . .173 Brace strains . . . . .174 CONTENTS. X1U CHAPTER VI. TRUSSES WITH HORIZONTAL CHORDS AND INCLINED BRACES, THE TIES HAVING A DIFFERENT INCLINATION FROM THE STRUTS, SUBJECT TO CONSTANT AND MOVING LOADS. Case I. — A Double Truss, whose upper chord is divided into an even number of panels or members. Art. 135. Post Truss PAGE . 176 " 136. Horizontal strains . . , . 177 " 137. Vertical strains from a full load . 180 " 138. Vertical strains from a moving load , . 180 " 139. Example Chord strains . 185 . 186 Brace strains . . 187 Case II. — A Double Truss, whose upper chord is divided into an odd number of panels or members. Art. 140. Description . . . . .188 141. Horizontal strains . . . .188 142. Vertical strains from a constant load . 190 143. Vertical strains from a moving load . . 191 144. Example 194 145. Horizontal strains . . . 194 146. Vertical strains .... 195 147. This system capable of extended application . 196 XIV CONTENTS. CHAPTER VII. Art. 148. " 149. « 150. " 151. 152. 153. 154. 155. 156. 158. 159. 160. 161. INCLINED TRUSSES OR RAFTERS. PAOB Description . . . . 197 Horizontal reaction of the abutments . .197 Ambiguity of strains to be avoided . . 198 The proportion between the horizontal and vertical reactions of the wall . .199 Longitudinal reaction . 199 Transverse reaction .... 200 Truss strains . . . . 201 Longitudinal strains .... 202 Transverse strains .... 205 Example . . . , . 207 Longitudinal strains .... 207 Transverse strains . . . . 209 An inolined truss whose struts and ties make the same angle with the chord . . 209 CHAPTER VIII. TRIANGULAR TRUSSES. Art. 162. Description " 164. Longitudinal strains " 165. Horizontal strains " 166. Vertical strains . •« 167. Example . 211 212 213 214 216 CONTENTS. XV Art. 167. Chord strains . Brace strains " 168. Inclined struts and ties Chord strains Brace strains " 169. The triangular truss subject to movable loads When the load covers less than half the truss When the load covers more than half the truss Greatest strains from a full load PAGK 216 217 217 218 218 218 220 223 224 Case I. — A Simple Triangular Truss with vertical struts A B C D®E Y aGbHcTh K L M M N O P QeRf SgT U V Fig. 6. 21. — Let w = the weight upon the truss, I = the length of the truss, d = the depth of the truss, x = the distance of one of the abutments to any one of the vertical braces or end of a panel, p = the horizontal length of a panel, m = the distance of the weight from the left abutment, n = the distance of the weight from the right abutment, H and H' = the horizontal strains in the chords, Y = the vertical strain. By the principles of the lever (8), the reaction of the IVTYL right abutment is — j- . The segment to the right of a vertical section through any panel end 6, is held in equi- librium by the reaction of the right abutment and the strains at b and/. Taking moments around/, distant x THE STRENGTH OF BRIDGES AND ROOFS. 13 from the right abutment, we obtain, by the same reason- ing as in the previous case, wmx H -ir ••••(c) for the horizontal strain at b, or the amount of compres- sion in be. The value of H in this equation varies directly as x, and is greatest when x is greatest or equal to n, that is under the weight. If the moments to the left of w and the reaction of the left abutment be taken we shall have, H:--gp .... (7) x being measured from the left abutment. These equa- tions also give the tension in the lower chord. If an- other point be taken in the section through eg, one panel length nearer the abutment, we shall have, _ wm(x — p) H ^ . ... (8) A value evidently less than the value of H in Eq. (6), showing that there is in be an excess of compres- sion over the amount in eh. This excess may, as in the previous case (15), be considered as the sole horizontal force at c, and is equal to the horizontal component of the strain in c/J which strain is (9) tension. The strain in eg is consequently compression and is equal to the vertical component of the strain in ef. Subtracting Eq. (8), from Eq. (6), we have, H " H T~M~ • • • < 9 > for the horizontal component of the strain in the tie ef % 14 A TREATISE ON and as it is a constant and independent of any value of #, for the horizontal component of the strain in any tie between the weight and the right abutment. Similarly, wnp „ is the horizontal component of the strain in the ties between the weight and the left abutment. 22. Vertical strain. — The vertical component may be obtained from the horizontal component of the strain in the ties to the right of the weight, by the proportion used before : wmp ivm p : d :: __ : __ Whence, V = -j- ; (10) is the vertical component of the strain in each of the ties, and the total compression in each of the struts in the right segment ; And similarly, V = -p (11) is the vertical component of the strain in each of the ties and the total compression in each of the struts in the left segment. 23. — Hence we obtain this law, equally applicable to the previous case : The vertical strain in either segment of a truss loaded at one point only is equal to the reaction of the abutment on ivhich the segment rests. 24. Longitudinal strain. — The longitudinal strain in the inclined braces may be, as before, obtained from this proportion ; as the depth of the truss is to the length of THE STRENGTH OF BRIDGES AND ROOFS. 15 the brace, so is the vertical reaction of the abutment to the strain ; or, . _ wm wm (be) the tension in the ties to the right of the weight. And similarly the tension in the ties to the left of the weight may be found. 25. — Example: Let Fig. 6 represent a truss sixty feet long and four feet deep, divided into twelve panels of uniform length and supporting a load of sixty tons at the distance of twenty feet from the left abutment. Here — w = 60 Tons, 1= 60 Feet,* d= 4 " m = 20 " n = 40 " wmx 60x20X0? H = 5 x. dl 4x60 Substituting the different values of x or the distances from the right abutment to the panel ends, we obtain the following strains in the right segment. This table is arranged as was the previous one. Values of x. 5 10 15 20 25 30 35 40 Strains in Tons. 25 50 75 100 125 150 175 200 Compression in. M L K I H G F E Tension in. W V U T S R Q For the left segment, H = to' wnx 60x40Xff 4X60 = 10 x, 16 A TREATISE ON substituting the different values of x, or the distances from the left abutments to the panel ends, we have the following : — Value of x. 5 10 15 20 Strains in Tons. 50 100 150 200 Compression in. A B C D Tension in. N P The length of the ties is V 4 J + 5" wm(be) 6.4 feet, cil = 128 tons, & d£ i? H THE STRENGTH OF BEIDGES AND EOOFS. 17 for the horizontal strain under the weight ; and if x in Eq. (2) be made equal to -3- we have H = ^ .... (13) Fig. 7. In Fig. 7. upon ab % made equal to I, describe a sem- icircle with a radius equal to — ; then the vertical cd will be a mean proportional between m and n, or m : cd :: cd : n Whence mn — (cdf Consequently, in a truss, loaded with a single weight, if the iveight he placed at different points, the horizontal strains resulting therefrom under the load, are to each other as the squares of the vertical distances at the points where the weight is placed, between a horizontal line equal in length to the truss and a semicircle inscribed thereon with a radius equal to half the length of the truss. 27. — The horizontal strain in either of the above cases between the weight and the abutment decreases at the panel ends in uniform quantities, and at these points passes through the inclined braces from one chord to the other where it neutralizes an equal amount of the oppo- site strain; or the braces contain all the vertical strain 18 A TREATISE ON and an amount of horizontal strain equal to the total st ram in one chord. In the first example above, the compression passes down the struts to the lower chord; in the second, the tension passes up the ties to the upper chord ; hence, in a truss there is no neutral axis or line in which no hori- zontal force exists. CASE III.— A TRUSS UNIFORMLY LOADED THROUGHOUT UTS LENGTH. Fig. 8. 28. — Let w = the whole weight upon the truss. I = the length of the truss, d = the depth of the truss, x = the distance to the end of a panel from one abutment. p = the length of a panel. u = the distance from the same abutment from which x is measured, equal to x — —-, being the distance to the cen- it tre of a panel. H = the horizontal strain. Y = the vertical strain. THE STKENGTH OF BKIDGES AND ROOFS. 19 In a truss the load is to be considered as concen- trated at the ends of the panels, for it is at these points that the connections are made between the truss and the members of the bridge which receive the load. A half panel load is therefore borne by either abutment. 29. Horizontal strain. — The weight upon either abut- w ment is — . The segment to the right of any vertical Z section, HA, is held in equilibrium by the reaction of the right abutment, the load on Am, and the strains at H and A. Taking moments around h distant x from the right abutment, we have the moment of the right abut- wx ment, equal to -~-, in one direction, and the load on the Z segment to the right of HA multiplied by the distance of its centre of gravity from A, for the moment in the opposite direction ; their difference is the horizontal strain at H multiplied by its distance d. The moment of the load on x may be found as follows : The weight coming directly upon the abutment is half a panel load, -—-, its moment is therefore -~~ ; the weight upon A has no moment, and the weight upon the remain- w der of the segment is evidently y multiplied by x — p, V and the distance of its centre of gravity is — ; whence its Z L . w . v x wx wpx . moment is y (^ — p) — = ——r *—, or the moment r. x , ill i ' • wx wpx , WpX WX* of the whole load on a? is — — — -f~- + -f-~ — -—— ; we J.L Zl Zo Zt 20 A TEEATISE ON can therefore form, for the horizontal strain, the follow- ing equation: — -ct 7 wx wx 7 Wl,e„ce H-5-g, . • • (U) (The half panel load resting directly upon the abut- ment has been generally disregarded in calculating the strains in trusses as it does not affect the results ; but it has here been introduced because the equations are ren- dered simpler, and w represents thereby the whole load upon the truss). Eq. (14) gives the compression in members of the upper chord, in the form of truss shown on the centre side of the panel end to which x is measured and the tension in the lower chord members on the abutment side of the same point, and is not confined as in the last preceding case to points between the centre of gravity of the load and the abutment, but is true for any value of #, which may be measured from either end. Differentiating Eq. (14), we shall find that H attains its maximum value when x — — , or at the centre; where, substituting — for x, we have, H = || • • • • (15) Comparing this with Eq. (1 3) we see, that the horizontal strain at the centre of a uniformly loaded truss is one half what it would oe if the same load were concentrated at the centre. THE STEENGTH OF BEIDGES AKD EOOFS. 21 At the abutment H becomes zero. 30. — Example: Let Fig. 8 represent a truss 110 feet long, 12.5 feet deep, divided into eleven panels of uniform length, loaded on the lower chord with a load of fifteen tons per running foot, or a total load of 165 tons. Here I = 110 feet, d = 12.5 " w = 165 tons. Substituting the values of these constants in Eq. (14), we have, 165 a 165 x % = 6.6x — .06 x 9 _ wx wx n ~Jd~~2dl 2 X 12.5 - 2X12.5X110 Whence the horizontal strains in the chords are as fol- lows : — Values of x 10 or 100 20 or 90 30 or 80 40 or 70 50 or 60 Strains in Tons. 60 108 144 168 180 Compression in. KL&BC IK&CD HI&DE GH & EF FG Tension in. Im & ah kl& he ik & cd hi & de ef,fg&gh 31. — Eq. 14, is the equation of a segment of a para- bola whose diameter is equal to — referred to rectangu- lar axes, whose origin is at the intersection of the diam- eter with the curve. Let AX and AY, Fig 9, represent the axes, II be- ing measured on AX and x on AY. Make x — 0, and H become zero, or the curve passes through the origin A ; make x = I its maximum value and again H = 0, 22 A TKEATISE ON or the curve intersects AY at the distance I from A ; make x = -5- and H = -^-r, its maximum value at the Fig. 9. vertex of the curve ; and at any point the distance be- tween AY and the curve represents the strain in the truss at that distance from the abutment ; practically this may be done as follows : — Let AB, Fig. 10, be equal to Z, the length of the truss, and upon its centre, C, erect a perpendicular whose height is equal to -^-, by any scale. Then through ADB construct a parabola, and the length of any vertical be- tween AB and ADB by the same scale will give the- horizontal strain in either chord at the distance from either abutment that the vertical is from A or B. THE STRENGTH OF BRIDGES AND ROOFS. 23 32. The Vertical Strain. — Taking Eq. (14), -p. _ WX WX* Jd~~ "M' at the end of any panel distant x from the abutment, the horizontal strain at the next panel end towards the abut- ment is, since the distance is x — p, w(x- p) __ w(x-pY U 2dl ' ' ' K ' Subtracting Eq. (16) from Eq. (14) we have, •a- tt' — wx w ( x — P) wx _l w ( x — P)" ~~U U ¥^ Ul » _ wp wp _ p " U dl K 2'' j • P and since x — ~ = u 9 '2d dl ' ' * ( U) The excess of horizontal strain in the upper chord thrusting towards the abutment beyond that thrusting towards the centre and which (9) consequently causes in a truss of this form compression in the inclined and ten- sion in the vertical braces. It is therefore at any point equal to the horizontal component of the strain in the strut at the same point, whose vertical component is equal to the tension in the tie which meets it at the up- per chord. This vertical strain can, as before, be ob- tained from the proportion, (18) vjp wpw . w Jd W : 2 " wu _- w wu 2 ~ T' • • 24 A TREATISE ON "Whence — The vertical strain at any point in a uni- formly loaded truss is equal to the weight borne by one abutment, less the weight between that point and the abut- ment : the point being measured from the nearest abut- ment. Eq. (18) varies inversely with the different val- r i • i I i i W 1 ues oi u, being zero when u = — or at the centre, and — when u = 0, or at the abutment. Substituting the values of the constants in the above example, Eq. (18) becomes, V = 82.5 - 15. u, and the different values of u are, 5, 15, 25, 35, 45, whence we obtain the following: table: — Values of u. 5 15 25 35 45 Strains in Tons. 75 60 45 30 15 Tension in. Bb&Ll Cc&Kk Dd&Ii Ee&Hh Ff &Gg When u = 55, V = 0, or there is no strain at the centre. 33. Longitudinal Strains in the Struts. — The lonffitu- dinal strain in any strut is the vertical strain divided by the depth of the truss and multiplied by the length of the strut. Performing this operation with Eq. (18) we have, since the length of a strut is 16 feet, L = 105.6 — 1.92 u (19) whence we can form the following table of compression in the struts : THE STRENGTH OF BRIDGES AND ROOFS. 25 Values of u. 5 15 25 35 45 Strains in Tons. 96 76.8 57.6 38.4 19.2 Compression in. Lm & Ba Kl&Cb Ik&Dc Hi & Ed Gh&Fe 34. — In Fig. 8, at the centre, there is, as is shown by the Eq. (18), no vertical strain ; then taking the weight on the next panel point g, 15 tons, we find it produces a tension equal to its amount in Gg, and a vertical strain, or vertical component of a strain, of the same amount in Gh ; there the weight on the next panel point, h, is added to it, and the vertical strain in Hh and also in Hi, is equal to the sum of the two weights, and so on to the end of the truss. Now, suppose the load to be placed on the upper chord ; then the weight would be at G, and there would be no strain in Gg ; but the vertical strain, or the ver- tical component of the strain in Gh, would be the same as the weight on G, and would equal the tension in Hh. c d SI*. 11. If the braces were inclined the opposite way, as in Fig. 11, representing a panel to the right of the centre, and the load upon the lower chord, there would be no strain upon ac, but the vertical strain in cb would equal that in bd ; if the load was upon the upper chord, the 26 A TREATISE ON vertical strain in ac would equal that in be ; or, in any case, the vertical strain is constant in the braces between the weights. The vertical strain, or the vertical com- ponent of the strain, in the inclined braces remains the same, whether the truss be loaded upon the upper or upon the lower chord, and whatever may be the inclina- tion of the brace. 35. — Therefore, Eq. (18) gives the vertical strain, or the vertical component of the strain, in t/ie inclined brace, ivhose centre is distant u from the abutment, and the ver- tical or total strain in the vertical brace attached to the unloaded end of the inclined brace. 36. — At the centre, as stated before, there is no ver- tical strain ; or, the vertical strain on either side of the centre of a uniformly loaded truss, passes to the abut- ment on that side, and if u in Eq. (18) be made greater than -~-, V will have a minus value, whence this rule : In a vertical equation, when V has a plus value, the ver- tical strain given thereby is passing to the abutment from ivhich u is measured, when a minus value, to the opposite abutment 37.— Eq. (18.) w wu V = 2 ~P is the equation of a straight line referred to rectangular axes. Let AX, Fig. 12, on which u is measured, and AY on which the values of V are measured, repre- w sent the axes; when u = 0, then V = -~ ; and make THE STRENGTH OF BRIDGES AND ROOFS. 27 w AB to any scale equal to -& when u I v 2' v o, therefore, lay off AC on AX equal to 5-, AD equal to I. Draw DE parallel and equal to AB and join BE. On AD, at the distances from A represented by the values :. 12. § of w, erect perpendiculars meeting the line BE, the lengths of these lines will give, by the same scale by which AB was measured, the values of V at the differ- ent points. The lines above AB showing the strains passing to the left abutment, those below, the strains- passing to the right abutment. 38. — The horizontal and vertical equations given above show, that where the horizontal strain is greatest there is no vertical strain, and where the vertical strain is greatest the horizontal strain is least. 28 A TREATISE ON CASE IV.— A TRUSS LOADED FROM ONE ABUTMENT ONLY A PORTION OF THE LENGTH. •#— aftMMteftt y •m"" «/ .C ^K.- Fig. 13. 39. ■Let w = the whole weight upon the truss uni- formly distributed, extending from one abutment a distance equal to 2 ra., I = the length of the truss, d = the depth " " p = the length of a panel, m = the distance of the centre of gravity of the load from the loaded abut- ment, n = the distance of the centre of gravity of the load from the unloaded abut- ment, y = the length of the unloaded part, x = the distance of any panel end from the unloaded abutment, H = the horizontal strain, V = the vertical strain, L = the longitudinal inclined brace strain. 40. Horizontal Strain in the Unloaded Part. — By the IDYYh principles of the lever, (8), —r- is the reaction of the right abutment, and the segment to the right of any THE STRENGTH OF BRIDGES AND ROOFS. 29 vertical section at a panel end, gk, in the unloaded part, is held in equilibrium by the reaction of the abutment, and the strains at g and k. Taking moments around k distant x from the unloaded abutment, we have, H = ~ » (20) for the strain at g, or similar to the strain in the case of the truss loaded at one point only, (21). In the lower chord the tension at k is evidently the same. 41. Vertical Strain in the Unloaded Part. — The ver- tical strain is also plainly — -- or the reaction of the un- loaded abutment, as in the similar case, (21). 42. — In the above operations w expresses the weight of the load upon the truss. It is more convenient and sometimes necessary that the equation should express the weight of a full load of uniform density with the partial load. Let w f — the weight of a full uniform load of equal vf density with the partial load. Then w — y- (I — y) and I — y wm w f (l — y)' 2 m = -j~, and .-. ~p= 2r ■■ 43. Horizontal Strain in the Loaded Part. — The seg- ment to the right of any panel end, be, in the loaded part is held in equilibrium by the reaction of the right abutment, the load on x — y, which is j-(# — 2/)? an ^ tne strains at b and c. Taking moments around c distant 30 A TREATISE ON x from the right abutment, we have therefore, ^~. being the distance from c of the centre of gravity of the load on x — y, B Whence _ ™\l-yYx w' , . x-y w^-yYx w'(x- y y " 2dr 2dT~* ' ' ( } is the compression in the upper chord at b. The same result may be obtained for the tension in the lower chord, at c. At a point one panel length nearer the right abut- ment, _ ™\l-yY(x-p) w'(x- p-yY idF ~2dl * ' " {22) 44. Vertical Strain In the Loaded Part. — The vertical strain may be obtained, as in the previous cases, from the difference in the horizontal strains at the different ends of the same panel. Subtracting Eq. (22) from Eq. (21) we have, tt tt, _ w'p w'py w'px w'p 7 ^ M : : 2dF - dl + ~dT — ~2dT And ^:cZ::H-H^.^ a _^(,_|_., ) . And since x — ~ — u w f (l—yY vJ V = -^ _ -j(u-y). . - . (23) In this equation, the less u—y becomes, the greater is the value of V, and the latter is greatest when u—y be- THE STRENGTH OF BRIDGES AND ROOFS. 31 comes zero, or in the brace at the end of the load, where it equals — \nf or tne reaction of the abutment. Again, V decreases with any increase of u — y, and there may be a point where V = ; to find this point, make V of Eq. (23) equal to zero, •■ ~~2F~ ~ r y) F+y* and u = g, • is the distance from the unloaded abut- ment to the point where there is no vertical strain. To the right of this point V has a positive value and the vertical strain passes to the right abutment ; to the left, a negative value, and the vertical strain passes to the other abutment. If Eq. (21) be differentiated to find the maximum value of II, it will be found to be when V+if x == ai or m the same panel, in which there is no vertical strain. Another proof of the rule in (38). 45. — In Eq. (23) u cannot equal ; ,, until — s« — w' — -j-(u — y), or until we have passed an amount of the V load equal to that borne by the abutment ; hence this important rule : There is a point in every fully or par- tially loaded truss where there is no vertical strain, but where the horizontal strain is greatest, which divides the load into the tivo parts borne by the two abutments, the part on either side of this point being borne by the abut- ment on that side. 32 A TKEATISE ON Therefore, knowing the load borne by either abut- ment, we have only to pass from the abutment along the load, full or partial, until we measure an equal weight, and we reach the point of no vertical strain. Example. — If a truss be 50 feet long, and is loaded from one abutment a distance of 30 feet, at the rate of 1.5 ton per foot, where is the point of no vertical strain ? Here, w' — 75 tons, I = 50 feet, y = 20 feet, iv'{l— y y 75 (50— 20) 2 " " ~W 2X50* = 13 - 5tons > the reaction of the unloaded abutment. Beginning at the end of the load towards the un- loaded abutment, which is 20 feet from that abutment, we must go from the end of the load 9 feet towards the other end of the truss before we have passed 13.5 tons and reached the point of no vertical strain. And by the formula found above l'+y : 2500+400 ~~2~f~ — iqq — 29 feet from the abutment. — A?is. JVo vertical strain can pass this point which can exist at one place only in any truss, no matter how loaded. Each, abutment's share of the weight comes directly from THE STRENGTH OF BRIDGES AND ROOFS. 33 that part of the load nearest to it ; and vertical strains in the same truss cannot pass each other. ABCDEFGHI KLMNOPQRST T — wmm NNwxNNm pabcde fghikl mnopqrs tuvj Fig. 14. 46. — Let Fig. 14 represent a truss 80 feet long, 6 feet deep, divided into 20 uniform panels, and support- ing on the lower chord a load of 25 tons, extending from the left abutment to the centre of the truss. Here, w = 25 tons, /. w = 50 tons, I = 80 feet, d = 6 feet, p = 4 feet, y = 40 feet, Length of inclined braces = 7.2 feet. The equation for the horizontal strains in the unloaded part is, H = ^=yy = im=££ = i.04i7,. 2dr 2x6x80 s Whence the following table of the strains in the chords: Values of x 4 8 12 16 20 24 28 32 36 40 Strains in Tons. 4.2 8.3 12.5 16.6 20.8 25 29.2 33.3 37.5 41.7 Compres- sion in. ST RS QR PQ OP NO MN LM KL IK Tension in. uv tu St rs qr pq op no mn lm 34 A TREATISE ON The equation for the horizontal strains in the loaded part is v/x(l-yY w'(x—yY 50ft(80— 40) a H " UV ~ 2dl ~ 2x60x80 Whence the following table: Values of x. 44 48 52 56 60 64 68 72 76 Strains in Tons. 45 46.7 46.7 45 41.7 36.7 30 21.7 11.7. Compres- sion in. HI GH GH FG EF DE CD BC AB Tension in. kl ik hi&gh *g ef de cd be ab The equation for the vertical strain in the unloaded part is iv'(l— y y 50(80— 40) 3 V = -\/L = - 2 ^py^ = 6.25 tons, tension in all the vertical ties from K£ to Tu inclusive, and the vertical component of the strain, which is from their inclination, compression, in the inclined braces of the unloaded part. 7.2 V L = V X length of strut divided by d = ' = 7.5 tons, compression in the struts from Km to Tu inclusive. The equation for the vertical strain in the loaded part is, w '(Z-y)' w'(u-y) 21' I = 31.25 — .625m. THE STRENGTH OF BRIDGES AND ROOFS. 35 Whence we form the following table for the tensions in the ties in the loaded part : Values of u 42 46 50 54 58 62 66 70 74 78 Strains in Tons. 5 2.5 —2.5 —5 —7.5 —10 -12.5 —15 -17.5 Tension in Ik Hi Gh Fg- Ef De Cd Be Ab The equation applies to the ties at the ends of the inclined braces nearest the centre of the truss, for the reasons given in (35). From the above table it will be seen that there is no vertical strain in the panel GH fti, and that to the left of this panel the strain has the minus sign, showing that the weight is now passing to the abutment opposite that from which u is measured. From the previous table it is also seen that the horizontal strain is greatest in the same panel, hence it is from this point that the braces must incline in opposite directions to the abutments. Multiplying the vertical strain by the length of strut, and dividing it by the depth of the truss, as before, we have the longitudinal strains in the struts in the loaded part as follows : Values of u 42 46 50 54 58 62 66 70 74 78 Strains in Tons. 6 3 —3 —6 —9 —12 —15 —18 —21 Compression in. 11 Hk Qg Ff Ee Dd Cc Bb Aa 36 A TREATISE ON CASE V. A TRUSS SUBJECT TO A UNIFORM CONSTANT LOAD THROUGHOUT ITS LENGTH, AND A UNIFORM MOVABLE LOAD. 47. — Let w = the weight of the full constant load, w = the weight of the full load whose *t> weight per lineal foot is the same as that of the partial load, I = the length of the truss, d = the depth " p = the length of a panel, x = the distance from one abutment to a panel end, P y = the length of the unloaded part, H = the horizontal strain, V = the vertical strain. 48. — In the previous cases the weight of the truss itself has been entirely disregarded ; but this is the case of a truss the weight of which is considered, subject to the action of a rolling load, and is to a certain extent a combination of the two previous cases. Here it is neces- sary to obtain the maximum strains only to which each member of the truss is subject, whether from a full or partial load. 49. Horizontal Strains. — In Eq. (14) we have, TT _ W'X w'x " '2d ~ ~UV THE STRENGTH OF BRIDGES AND ROOFS. 37 for the horizontal strain under a full load in either chord. In Eq. (21), _ w'x(l— y) % w'jx—y) 9 a T UV ~ Ul ' for the horizontal strain at the same point from a load of equal density but covering only a part of the truss, U-y. Eq. (21) will reduce to this form, H - 2d — 2dl — 2dl V— lh ■ ■ < 24 > which is less than Eq. (14) hy the quantity or the horizontal strain at any point is greatest under a full load, no matter how small y may be, or how large a portion of the truss may be loaded. Hence, where w is the constant truss load, and w' the weight of a full rolling load, the equation for the greatest horizontal strain is ^ = —u~ - ui ; " " ' " (25) SO. Vertical Strain. — In Eq. (23) we have, for the vertical strain at u, from a partial load reaching from one abutment a distance equal to I — y. Let y and u be measured from the right abutment and confining this equation to the right of the point of no vertical 38 A TREATISE OlS" strain, or to its positive values, it is evident that, consid- ering y for the moment constant, V is greater as u — y is less, and is greatest when u — y is least, or when u = y or at the end of the load; where, therefore, Y = ^=yl. ------- (26) Or, the vertical strain from a partial load passing in one direction at any point is greatest when the load extending from the abutment reaches to that point. Eq. (26) will reduce to this form, w f w'u w'u* /9 v v = T~-r + -2r (27) The vertical strain at the same point from a full load of equal density is, Eq. (18), w' w'u Y = t r If u in this equation be less than 3- or when V has a positive value, it follows, That when a truss is par- tially but more than half loaded, the load extending from one abutment, the vertical strain at any point at the end w'u* of the load is greater by -gjr than the vertical strain at the same point from a full load of equal density. The greatest vertical strain, therefore, in a truss sub- ject to a rolling load, is the strain at any point from the constant truss weight added to the strain from the roll- ing load when it reaches that point and covers the THE STEENGTH OF BEIDGES AND KOOFS. 39 greater segment of the truss; hence, adding Eq. (18) and Eq. (26), w wu v/( l—uy , 9ft . y being here equal to u, is the vertical strain from the rolling load w'/and constant load w. 51. — In a truss uniformly loaded there is a point of no vertical strain at the centre, and in a truss partially loaded, where the truss weight is disregarded, there is also a point of no vertical strain, distant „, from the unloaded abutment (44), but in no case can there be two points of no vertical strain, for a section of the truss cannot support a portion of the weight without trans- mitting it entirely to one or partially to either abut- ment. This point of no vertical strain can be found by making Eq. (28) = 0. w wu u/(l — v)* Whence V = -j — -j-+ g f 7 = Let w f = aw, eliminate w and we have "Whence, I u aV — 'Zalu+au* 2~7 + 2? ~ ° Example. — In a truss 200 feet long, whose perma- nent uniform load is 75 tons, and the weight of its full load of equal density with the partial load, 150 tons; how far from the unloaded abutment is the end of the 40 A. TKEATISE ON partial load when it is at the point of no vertical strain, or what is the value of u ? w' Here, — = a = 2. I = 200 Whence, I + - — Z i/ — + - 200 = 200 + 2 200 4/ £ + i = 126.8 ft.— Ans. That is, when the partial load covers 73.2 feet of the truss, the end of it is at the point of no vertical strain. The weight borne by the farther abutment is equal to the weight of 126.8 feet of the truss, or 47.5 tons. When the truss is unloaded the weight on the abutment is 37.5 tons. When the load of .75 ton per foot covers 73.2 feet, by the principles of the lever, 10 tons are added to the weight on the farther abutment. While the end of the movable load rests at the point of no vertical strain the weight upon the farther abutment comes solely from the weight of the truss, and none from the load upon it. When the end of the load approaches the unloaded abutment, passing the point referred to of no vertical strain from the fixed and movable loads, the weight upon this abutment is increased, but as the load still covers less than half the truss, the larger portion of the increased load is borne by the nearer abutment ; the point of no vertical strain, therefore, does not remain stationary, but moves after the end of the load, and reaches the centre when the load covers the truss. THE STRENGTH OF BRIDGES AND ROOFS. 41 52. Counterbracing. — It is therefore evident that on the loaded abutment side of this point there can be no vertical strain passing to the farther abutment ; conse- quently, it is only from this point that it is necessary to arrange braces to carry the vertical strain to the farther abutment. The braces between these points on either side of the centre, and the centre itself, are termed coun- terbraces, and come into use only under the action of moving loads. 53.— Eq. (18) added to Eq. (26), or w'(l-yY w wu v= 2--r + 2? (30) may be represented as in Fig. 15. \ I M %/\ / / \ A I g/k l, Fig. 15. Let AB equal the length of the truss ; AD and BC w each equal to a scale, -^; then (37) the perpendicular dis- tance at any point between EB and EC is the amount of vertical strain from the permanent load at that point passing to the abutment B, and the perpendicular distance between AE and DE at any point represents the amount of vertical strain from the same load, at that point passing 42 A TREATISE ON to the abutment A ; these distances being the values of V in Eq. (18) ; the positive and negative values being on the opposite sides of AB. Next, let the load be brought on at A and extend from A to I, and let BH, to the same scale as before, y/H yV represent ~r a — , the reaction of the abutment B, and as this is a constant between I and B, draw I G equal and parallel to BH and join GH. The vertical distance at any point between LH and EC will give the vertical strain from the constant and the movable loads at that point passing to the abutment B and the values I of V in Eq. (30) where y is greater and u less than

y» 1S represented by the vertical distances w wu between GH and IB ; and as -^ — -y is a minus quan- tity to be subtracted from — \r > tnen tne vertical dis- THE STRENGTH OF BRIDGES AND ROOFS. 43 tance between KE and ME is to be deducted from the vertical distance between KL and ME. Consequently the vertical distance at any point between KC and KH rep- resents the vertical strain at that point passing to the abutment B ; and by similar reasoning it may be shown that the vertical strain passing to the abutment A is represented by the vertical distances between DK and the two lines EG and GK. K is the point of no ver- tical strain. An examination of Eq. (23) added to Eq. (18), or \r - w '( l —yy _ w '( u —y) i 2F . w wu will give the vertical strains under the truss, and is easily made, but has no practical value. Eq. (26), V = zip — "i is the equation of a pa- rabola, and Eq. (27) may be shown as in Fig. 16. X \ X a L / / y / K \ Fig. 16. Let AB represent the length of the truss, BC and 44 A TREATISE ON W , . _ . . w' AD each ^ and AF and BE each 5- ; then the vertical distances between DC and FE represent the strains from the weight of the truss and of a full load. Through AE draw a parabola ; the vertical distance to any point in the parabola may be found from the value of V in Eq. (26) at that point. Similarly draw the parabola BF. The vertical distances between HC and the curve HE represent the vertical strains from the constant load and the end of the load passing to the abutment B. When the load passes to the left, the vertical strains to A are represented by the vertical distances between ID and the curve IF. Where the curves intersect the line DC, H and I are the points of no vertical strain. w f n u y 54. — The equation of the moving load, — c*» — > requires some change before it can be practically applied to determining strains ; because, in its present form, it assumes that no weight comes directly, that is, without passing through the braces, upon the next unloaded panel point. If the load were suspended at each panel point or end, the equation could be applied without change ; but as it comes first upon a girder, or string- piece, resting upon the panel points, any load in a panel must affect both ends of the panel. Let AI (Fig. 17) represent a truss, and A B C D, &c, the different panel ends ; and let the load extend from the abutment A to midway between B and C, then there are one and a half panel loads upon the truss ; but B does not bear a full panel load, and cannot until the THE STRENGTH OF BRIDGES AND ROOFS. 45 load extends to C, or B cannot have a full load until C has a half load ; so that «« is greater than the ver- tical strain in the brace from B towards the farther abutment, because a part of the strain is in the brace from C. ABC DEFGH I Fig. 17. Let the vertical distance at any point between AI and the parabola Abe - - - i represent, as shown above, the weight borne by the unloaded, or, in this figure, the right abutment, when the load extends from the left or loaded abutment to that point, and the vertical dis- tances between AI and the smaller similar parabolas Ab f Be' CcZ', &c, represent the weights coming upon the farther panel points B, C, &c. ? as the load extending from the left abutment traverses the girders resting upon these points. When the load extends from A to B, B6 is the weight borne by I, but &b' is the weight upon B, which is greater than the weight upon the abut- 46 A TEEATISE ON merit I ; continuing the load from B towards C, the vertical distance from the line BC to the parabola Be' at any point, will represent the strain upon C, when the load reaches that point, and the vertical distance at the same point between BC and the parabola be represents the strain at the same time upon the farther abutment, I. When the former equals the latter, or when the curve Be' intersects the curve £c, all the strain upon I comes from that portion of the load bearing upon C, and none from that portion upon B, and the vertical distance between the two curves, before their intersec- tion, represents the vertical strain upon I that comes from the load upon B ; consequently the greatest strain upon the brace at B towards I is when the vertical dis- tance between the two parabolas, between A and B, is greatest. This occurs where a vertical line will inter- sect the two curves at points where their tangents are parallel ; and since the weight on AB is to the whole load upon AI as the distance AB is to the distance AI, the horizontal distances of the tangent points referred to from the panel end and the abutment bear the same pro- portion to each other. Let V represent the length of the partial load, and let it extend from one abutment beyond one panel, to where the tangents to the two curves at points vertically above each other are parallel to each other; letp be the length of a panel, and I the length of the truss ; then pV I :p::l' : -j , the distance the load extends into the • THE STRENGTH OF BRIDGES AND ROOFS. 47 second panel ; and the length of the load V will there- fore be »-,+$ whence, VI = pi + pi'; and v - f l-p ( 31 ) When the load covers any number, n, of panels, the value of V will be *.-.'£ « Dividing the weight of the whole movable load by the length of the truss, and multiplying by j , we have the weight of the load on 7^7", whence (8) the reaction of the unloaded abutment is, in this case, v = ltef m But, as explained before, this is greater than the strain in the braces from the last loaded panel point, because a certain amount of the load is borne by the panel point beyond the load. Hence this last weight is to be de- ducted from Eq. (33). The distance from the loaded abutment to the far- thest panel end or point which is under the load is np, therefore, npl j=p- n p 48 A TEEATISE ON is the distance the load extends beyond on to the partly loaded panel, and the weight on this distance is, t(S - *) _ v/_ i up* \ Of this weight, the part which is borne by the end be- yond the load, or the end farthest from the loaded abut- ment, of that panel into which the load extends, is, by the principles of the lever, WAi-p) < 34 > This amount is to be deducted from Eq. (33) to ob- tain the correct strain upon the brace from the last panel end under the load, towards the unloaded abutment, whence, vJ / npl \* _ w r i up* \ " {i- P y w f inpV —7)?pl\ / 3 37 w n p o = 2?(l-~p) * • v - 2i(i- P y (6b) is the greatest vertical strain from a partial load, where n represents the number of panels loaded, towards the unloaded abutment upon the brace at the last loaded panel point. Example. — Let a truss be 80 feet long, divided into 8 panels, and loaded at the rate of one ton per foot: THE STKENGTH OF BBtDGES AND KOOFS. 49 What is the greatest vertical strain upon the brace from the last loaded panel end, when six panels are loaded ? Here, vf = 80 tons, ^ I = 80 feet, p = 10 " n= 6 " Substituting these values in Eq. (35) we have v/nY 80x6 a Xl0 a V ~~ 21(1— p) ~ 2x80(80-10) ~ 2 °- 71 tons ' v/ If in ot»(£— u )% (l— u ) na d De en made equal to 65, 80 we should have 9 nAi (65) 3 = 26.41 tons for the reaction of the unloaded abutment. But as the load extends half way from one panel end to the other, the unloaded panel end would support of £ of £ panel load, or 1.25 ton ; hence 26.41 — 1.25 = 25.16 tons, would be the greatest strain upon the brace from the sixth panel end. t a i n P l 6X10 X80 OQ ^ In the case supposed, 7— = an— 10 ~ "°-^ feet, is the length of the load, and consequently, 29.38 tons is the reaction of the unloaded abutment; and 3.67 tons the weight on the first unloaded panel end, whence, 29.38 - 3.67 = 25.71 tons, as before, for the greatest strain on the brace. When np is greater than —„ or when the load covers more than half the truss, 9 ,, , v is passing in the same 4 50 A TKEATISE ON direction as the vertical strain at the same point from the constant truss load, and consequently, Eq. (35) may be added to Eq. (18), giving w wu w'n'p* Y = 2~ir + wp^) - ■ • • <- 36 > for the maximum vertical strain, from both the perma- nent and the movable loads, in the braces of that panel whose centre is distant u from the unloaded abutment, and where n is the number of panel points in I — u. A l — u — ~ 2 reference to Fig. (14) will show that n = ; substituting this value in Eq. (36) we obtain, ''(Z-w-f)' w V -2~T"+ 2l(l-j>) ' " ' (37 > an equation giving the same results and containing only one variable quantity, u. But when np of Eq. (35) is less than -~, then the / 2 2 ID Vb U vertical strain, 9/ ,, , , is passing towards the cen- w wu tre or m an opposite direction to -^ — -y-, the vertical strain from the constant truss weight at the same point, and the difference between the two is the total vertical strain at u. The less of these may be considered as iv ivu neutralizing its amount in the greater; but -^ — ~i~ ? the constant truss strain, also meets the vertical strain from the first panel point outside the load, or the quan- THE STRENGTH OF BRIDGES AND ROOFS. 51 W f I Tip* \ * tity, Yl[j~Z z — )i ^t (**)i an( ^ ^as keen l essene( i D y this amount outside the load. Hence, since o7t~ \ a W 110 u Eq. (33) and -~- -- —r are each diminished by the same amount, their difference will remain the same, and we have for vertical strains to the farther abutment when the truss is less than half loaded. Substituting for n its value 4- v / comes V 2 I T 2(l-pY ' W the vertical strain to the unloaded abutment when the load covers less than half the truss affecting the brace at the last loaded panel end. 52 A TREATISE ON CHAPTER III. TIES SUBJECT TO THE ACTION OF A CONSTANT AND A MOVING LOAD. ABC DEFGHIKLMNOPQR ,^/Vl^/b This equation will not apply to the other simple truss, on account of its half panel at the ends. The uniform simple truss has a full panel load at each panel point, Off 1 THE STRENGTH OF BRIDGES AND ROOFS. X\*. 61 and a half panel load upon each abutment ; the other simple truss has a full panel load at each panel point, and none upon the abutment. To obtain an equation w for the latter, we have 7- X x 1 for the moment of the re- action of the abutment, at any panel point distant x' from that abutment ; the load upon the truss between w this point and the abutment is ^i + (^—p)* P being the length of the end panel ; and the distance of its centre of gravity is -5 + ^ ; this is apparent from the figure of the truss. Whence, __, wot/ w . i N /x'+p\ ivx' wx f% H = id - iiP-ti rrJ = ij—ur +^ - - ; - • <«) is the compression in the upper chord, and tension in the lower chord of the truss of Fig. 22 at any point distant x' from the abutment. If x in Eq. (40) be equal to ME in Fig. 22, H will be the horizontal strain in MO and km ; and if x' of Eq. (41) be equal to LR in Fig. 21, H' will be the horizon- tal strain in LN and il. Hence, the horizontal strain in the upper chord of the double truss at any panel point, is equal to the hori- zontal strain in that one of the simple trusses whose panel end is at the same point, added to the strain at that panel end of the other simple truss which comes next towards the centre ; or H at x in Fig. 20 is equal 62 A TEEATISE ON to H at a? in one of the simple trusses, added to H' at x + p in the other simple truss. Making, therefore, xf of Eq. (41) equal to x +p, and adding the equation so changed to Eq. (40), we have,* wx wx* pwx pw H = Yd ~~ ~2dl ~~ Jdl + Id ' " 2d v* + 2 / 2dZ \ x ' 2 j + Ul ^' For the panel points of the double truss, common also to simple truss, Fig. 22, the strains at x' are equal to the strains at x' in one simple truss added to the strains in the other simple truss at a/+ p, making, there- fore, x of Eq. (40) equal to x'-\- p, and adding the equa- tion so changed to Eq. (41), we obtain the same result as before ; or Eq. (42) gives the strains in all the mem- bers of the upper chord on the nearest abutment side of the panel points to which x is measured. In the lower chord, the strain at any point of the double truss is equal to the strain in one simple truss at the same point added to the strain in the other simple truss at the next panel point nearer the abutment; making, therefore, x f of Eq. (41) = x r — p, or x of Eq. /40) = x — p and adding the equation so changed to the other unchanged, the result from either addition is, wx wx* pwx piv H = Jd~~~~2dl + Jdl~ Id' ~ 2cT 2 j Ul\ x 2 j + Ul ^ 6) * The algebraic process, being very simple, is omitted. THE STRENGTH OF BRIDGES AND ROOFS. 63 an equation giving the tensions in the lower-chord members, on the centre sides of the points to which x is measured. 62. Vertical strains. — The simple trusses being en- tirely independent of each other in their vertical actions, the equations of their vertical strains are to be deduced from the simple truss horizontal strains of Eqs. (40) and (41) as in (32), whence we obtain, from either Eq. T-f-5; (") for the vertical strain from a full load in either simple truss, u being the distance to the centre of a panel of a simple truss, and not to the centre of a panel of the com- pound truss. In the simple truss of Fig. 22, the centre of the end panel is considered as at the abutment, and the first value of u for that truss is therefore zero. 63. Vertical Strains from the Moving Load. — The effect of the moving load upon the panel points of a compound truss differs from that upon the points of a single truss, because, in the former case, a panel point or end of one simple truss can be fully loaded without the next panel point belonging to the same simple truss being affected by any portion of the load. So that the effects of the load in a compound truss are the same as if the different portions of it, or panel loads, were sus- pended at the ends of the panels. Therefore, w' representing the weight of the full movable load, 64 A TREATISE O^ will give the greatest vertical strain from the moving load, w', upon the simple truss of Fig. 22. Hence, adding this to Eq. (44), w ivu iv f „ , v is the equation of, the vertical strain from the constant load, w, and the moving load, w\ in simple truss Fig. 22. In the simple truss of Fig. 21, where u' is the distance from the unloaded abutment to the centre of one of the panels, "kt(1 — u' — p), as an inspection of the fig- ure will show, is the load on I — v! ; dividing by I and multiplying by ~ , the distance of its centre of gravity from the abutment from which the load extends, and we have, w' V= 11 (Q-U')-P), - - - - (46) for the vertical strain upon the truss from the moving load. Adding this to the equation for the constant load, we have, w wu ' u/ Y = 4 ~ "21 + -U^Q-^y-P") ' ' ( 47 ) for the equation of the vertical strain from the constant and the moving loads in simple truss Fig. 21. THE STRENGTH OF BEIDGES AND EOOFS. 65 64. Example. — In Fig. 20, Let w' = 160 tons, the weight of the full moving load, w = 80 tons, the weight of the truss, I = 160 feet, the length of the truss, d = 20 feet, the depth of the truss, p = 10 feet, the length of a panel, x = the distance from the abutment to the end of a panel, u == distance from the abutment to the centre of a panel of either of the simple trusses. The load is upon the lower chord. Substituting the values of these constants in Eq. (42), we have, H = (w'+w ) 2d (160+80) %+f) fw Wf + (w'+w)p' 2dl v ~ T 2 1 ' Ml «n (160+80) 1(V 2x20 v" + 2J 2X20X160 ^ + 2 j + (160+80)10' 8x20x160 = 6(a+5) - .0375 (x+5) 2 + .9375. from which we can form the following table of com- pressions in the members of the upper chord : Values of x. 10 20 30 40 50 60 70 80 Strains in Tons. 82.5 127.5 165 195 217.5 232.5 240 240 Compression in AB& QR BC& PQ CD& OP DE& NO EF& MN FG& LM GH& KL HI& IK 06 A TEEATISE ON Substituting the values of the constants in Eq. (43), we have, H = 6(a>— 5) - .0375 0-5) 3 + .9375, whence we can form the following table of tensions in the lower chord : Values of X. 10 20 30 40 50 60 70 Strains in Tons. 30 82.5 127.5 165 195 217.5 232.5 Tension in be & pq cd & op de & no ef &mn fg&lm gh&kl lii&ik There are no strains in ab and qr. Substituting the values of the constants in Eq. (45), we have, .r 80 SOu 16 ° /,™ V ^ T - 2X160 + 4x(160)-< 160 - tt >' (160 — u) M 20 — .25^ + 640 "We can form the following table of compressions in the struts of simple truss of Fig. 22 : Values of u. 20 40 60 80 Strains in Tons. 60 45.6 32.5 20.6 10 Compression in Aa&Rr Bb&Qq Dd&Oo Ff&Mm Hh&Kk 1 Dividing the same Eq. by d and multiplying by the length of the tie, or V X 1-414, for all ties except the end ones, where it is, V X 1.118, and we obtain the fol- lowing table of tensions in the ties of Fig. 22. (The THE STRENGTH OF BRIDGES AND ROOFS. 67 quantities by which V is multiplied are the secants of the angles made by the ties with a vertical line.) Values of u. 20 40 60 80 100 Strains in Tons. 67.1 64.5 46 29.1 14.1 0.9 Tension in Ab&Rq Bd&Qo Df & Om Fh & Mk Hk&Kk Km&Hf 1 V has a negative value when u == 110. Substituting constants in Eq. (47), we have, V = 20 - .2hu + (160 — u'Y~ 100 640 Whence the following table of compressions in the struts of simple truss of Fig. 21 : Values of u'. 10 30 50 70 15 Strains in Tons. 52.5 38.8 26.3 Compression in Aa&Rr Cc&Pp Ee&Nn Gg&Ll The total compression in the end struts, Aa and Rr, from the two simple trusses is 52.5 + 60 = 112.5 tons. Multiplying V of Eq. (47) by 1 % 414, as before, we have the following table of tensions in the ties of truss Fig. 21 : Values of u \ 10 30 50 70 90 Strains in Tons. 74.2 54.9 37.2 21.2 7.1 Tension in Ac& Rp Ce&Pn Eg& Nl Gi& Li Il&Ig 68 A TEEATISE ON 65. — The double truss shown in Fig. 20 has an even number of panels, and each half of it has also an even number. If two panels be added, each half of the truss will then contain an odd number of panels, and simple truss of Fig. 22 will be uniformly divided, while the ends of simple truss of Fig. 21 will become similar to the ends of the other simple truss in the example given. Each truss, however, will still support one-half the full load, and the horizontal equations will remain unchanged. The vertical equations for the moving load, however, are entirely dependent upon the end panels of the simple trusses, and equation (45) will apply to that truss which is divided into uniform panels, that is, terminates with a panel equal to two panels of the compound truss ; while vertical equation (47) will always apply to that simple truss whose end panel is equal to one panel of the compound truss. 60. — By referring to the examples given, it will be seen that the strains in the two chords are equal in amount between the same inclined braces. CASE III. A DOUBLE TRUSS CONTAINING AN ODD NUMBER OF PANELS. 1212121212112121212121 ABCDEPGHIKLMNOPQRSTUVW Fie:. 23. 67. — Let Fig. 23 represent a double truss, the coun- terbracing shown by the dotted lines, containing an odd number of panels. THE STRENGTH OF BRIDGES AND ROOFS. 69 This truss is also composed of two simple trusses, the panel points of one being shown by the figures 1,1,1, and the panel points of the other by the figures 2,2,2. The two simple trusses will be distinguished as truss No. 1 and truss No. 2, as they are numbered in the Figure. Simple truss No. 1, having full end panels, may be considered as bearing the half panel loads resting directly on the abutments, and has consequently one panel load more than the other ; or it supports half the whole weight and half a panel load ; while Truss No. 2 supports half the whole weight, less half a panel load. This is when the truss is fully loaded. 68 Let I = the length of the truss, d = the depth of the truss, p = the length of a panel of the double truss, w = the weight upon the truss, uniformly distributed, x = the distances to the panel ends from one abutment, H = the horizontal strain, V = the vertical strain. 69. Horizontal strain The weight on simple truss w wp No. 1 being -q + -^f , the reaction of each abutment is therefore J Ur + -4r )? and as the truss is divided into uniform panels, the moment of the load on any segment 70 A TREATISE ON (VJX* \ -aTji from which we readily obtain the equation, TT 7 . iw ivp \ WX* =»"*(! + ■«?)— IT whence, .p. _ wx WX* wpx ti ~Jd-~idl+-IdP ( 48 > is the compression in the upper chord of Simple Truss No. 1 on the abutment side, from which x is measured, of the points 1,1,1, &c.,and the tension in the lower chord of the same truss on the centre side of the same points. The reaction of each abutment upon Simple Truss JNo. 2 is i \-^ — ~2l)i aT1 d tQe moment of the load on any segment whose length is x, is, as in the Simple Truss of Fig. 22, j lwx n ivp\ * \~2r~lrJ' from which we obtain the equation, ] whence, 2 l2 21 1* ? [ 21 21 P wx wx" wpx' ivp* Y1 -u--m--uT + idi - • < 49 > is the compression in the upper chord of Simple Truss No. 2 on the side towards that abutment from which of is measured, of the points 2,2,2, &c, and the tension in the lower chord on the centre side of the same points. It is evident that, here, as in the previous case, in the THE STRENGTH OF BRIDGES AND ROOFS. 71 upper chord of the double truss, the compression at any panel point is the compression at the same point of one of the simple trusses added to the compression in the other simple truss at one panel length nearer the centre ; and similarly in the lower chord of the double truss, the tension at any panel point is equal to the tension at the same point of one of the simple trusses, added to the tension in the other simple truss at one panel length nearer the abutment. Therefore, making x' of Eq. (49) equal to x + p, and adding the equation so changed to Eq. (48), we have, _ wx wx* wp ivpx wp 2 ' Jd ~~ 2dT + Id — ~2dl ~ ~4dl for the compression in the members of the upper chord on the abutment side of the points 1, 1, &c. And making x of Eq. (48) equal to x f + p, and add- ing to (Eq. (49), we have, wx f tvp wx'* wpot' wp it ~"~ t ~-jr _____ __£__, 2d ^ Ad ~ 2dl 2dl " Ul W s . V\ W / / P\* 3 W P* , V for the compression in the members of the upper chord on the abutment side of the points 2, 2, 2, &c. In these and the subsequent equations x cannot have a value greater than 5-, because the simple trusses are not symmetrical, as in the previous cases, beyond the centre. 72 A TREATISE ON Making x f of Eq. (49) equal to x — p of Eq. (48), and adding the two, we have, wx wp wx wpx ( wp* 2d U 2dl 2 2dl + ±dV 3wp* o p\ w p\ dwp< d( X -2)-2d^-2)+-Sdf' ' < 52 > for the tension in the members of the lower chord of the double truss on the centre side of the points of Simple Truss No. 1. And making x of Eq. (48) equal to x'— p of Eq. (49) and adding the two, we have, wx' wp WX' H = ~2d~~~U~"2dl wpx wp ~ 2d^ x ~ 2/ 2S X ~ 2 1 ~ Sdl (53) for the tension in the members of the lower chord of the double truss on the centre side of the points of Simple Truss No. 2. 70. — Let there be an odd number of panels on either side of the centre panel as in Fig. 24, 2 12121121212 Fig. 24. and numbering the trusses as before, Simple Truss No. 1 has now a half panel at either end, while the end panels of Simple Truss No. 2 are uniform with the THE STRENGTH OF BRIDGES AND ROOFS. 73 others. Truss No. 1 still supports one panel load more than Truss No. 2, but the moment of the load on the segment x is wx* wp whence, wx H ~ Id + wpx wx* wp" Adl " 4dl + ±dV - (54) as the horizontal strain in Simple Truss No. 1. The moment of the load on segment x' of Simple Truss No. 2 is -j-jTi whence, _ wx' wx' a wpx' . H -¥""¥~"M ...... (55) Following the same process as before, substituting and adding, we obtain the same results, that is, Eq. (50) for compression in the upper chord at the points 1, 1, &c. ; Eq. (51) for compression in the same chord at the points 2, 2, &c. ; Eq. (52) for tension in the lower chord at the points 1, 1, &c, and Eq. (53) for tension in the same chord at the points 2, 2, &c. Hence we see that these equations are not affected by the panels at the ends of the simple trusses, but that Eqs. (50) and (52) belong to the panel points of that simple truss whose braces form the centre panel, and Eqs. (51) and (53) belong to the points of the other simple truss 71. — It will lead to less confusion, therefore, to measure the points from the centre of the truss, as well 7-i A TREATISE ON as render the equations simpler ; putting in Eq. (50), -x — z f or x ; z being the distance from the centre of the truss to the same point to which x is measured, we have, wl w p\* wp* for upper-chord compressions at the panel points of Simple Truss No. 1. In Eq. (51). putting r z' for x\ we have, for upper-chord compressions at the panel points of Simple Truss No. 2. In Eq. (52), putting ~ z for a?, we have, H = Wl W / j9 8^ — 2^ 3wp j^+fj+w' - - - ( 58 > for lower-chord tensions at the panel points of simplt Truss No. 1. In Eq. (50), putting ^ z' for x\ we have, wl w_ ,, p\ wf_ n Rrl 9/7/^+9 Q.77' " " " " V Dt V Sell for lower-chord tensions at the panel points of Simple Truss No. 2. 72. Vertical Strain from the Constant Load. — Under a uniformly distributed load, the vertical strains in one simple truss are unaffected by those in the other, and THE STRENGTH OF BRIDGES AND ROOFS. 75 the equations are therefore to be deduced, as before, from the simple- truss horizontal equations. (32) From Eq. (48) or Eq. (54) we obtain for Simple Truss No. 1, w wu wp V = T— 2T+4T ( 60 > From Eq. (49) or Eq. (55), for Simple Truss No. 2, w wu' wp Y = T — u—£> ( 61 > u and u' being the distances to the centres of the panels of the simple trusses, and whenever either simple truss begins with a half panel, u or u' in the equation which belongs to that truss must be made equal to zero. It will be noticed that the constant-load vertical equations, like the compound-truss horizontal equations, are un- affected by the terminations of the simple trusses ; but are determined in their application to either simple truss by the position of the braces of that truss at the centre. 73. — If the difference in the successive values of u and u' be constant when these quantities exceed ^-, each will then represent the distances to the centres of the panels of the other simple truss than that for which the equa- tion in which it is found was obtained. That is, \ ,j of (u ) Eq. j«- I , which, when less than ^-, is the distance from one abutment to the centre of any panel point of Simple Truss ] 2 [ becomes, when greater than ~-, the distance 76 A TREATISE ON from the same abutment to the centre of any panel of Simple Truss \A- Further, when u or v! becomes greater than — , the equation to which it belongs gives the vertical strain in the other simple truss, or that one to the centres of whose panels it now represents the distances; that is Eq. \„+ [ when W i s greater than 5-1 gives the vertical strain in Simple Truss No. \ . [passing to the abutment opposite that from which < ,[■ is measured. For Eq. (60), becomes, when u is greater than ~-, and consequently equal to I — u\ or the same as Eq. (61) when u' is measured, as indi- cated by the minus sign, from the opposite abutment ; and similarly is Eq. (61) changed. 74. Vertical Strains from the Moving Load. — It will be seen from the plan of the truss, Fig. 23, that the passing load, before reaching the centre, transmits that portion of its weight which is borne by the farther abut- ment through the counterbraces from one of the simple trusses to the other. One half of one simple truss being THE STRENGTH OF BBLDGES AND ROOFS. 77 thus connected with the opposite half of the other, we have two other simple trusses in this same double truss, different from the former simple trusses, and in their ver- tical action under a moving load entirely independent of each other; they are shown in Figs. 25 and 26. A B D F H K M V w Fig. 26. Either of these is simply the other reversed. In this case, as well as in any combination of simple trusses, any number of panels, in either simple truss, extending from one end may be considered as fully loaded without throwing any weight upon the panel point outside the load as in the case of the Simple Truss (54). Let the moving load extend any distance from the j • .t t [ end of truss of Fig. )nn[, or that abutment on which a half panel end of the simple truss rests, then, V = ^ l p? is the reaction of the opposite abutment, and the great- est vertical strain at any point from the moving load passing to the unloaded abutment, w' being the weight of the full movable load. 78 A TKEATISE ON But if the load extend from the opposite abutment, or that on which a full panel end cf one of these trusses rests; then Eq. (45), w' will give the reaction of the opposite abutment and the greatest vertical strain. These equations depend for their application upon the length of the end panel of the simple truss upon which the load enters, and are not affected by the length of the panel at the other end of this simple truss. And since either simple truss may have a full or half panel, as the double truss contains more or less, either of the moving-load equations may be added to either of the simple-truss constant-load equations. There is no dif- ficulty, however, in determining how the addition is to be made in any case. If the moving load extend from the half panel end of the trusses of Fig. 25 and Fig. 26, and cover more than half the truss, then it is plain that j^ must be added to the equation of that simple truss which has a full panel end, since it is the braces of this truss , . . w'(l — uf that transmit the strain v „ — '- to the unloaded abut- ment ; and if the load covers less than half the truss, W '(l — u)\ then — -j= is acting upon the braces of the other simple truss between the end of the load and the centre of the truss. Hence we have this simple rule : THE STRENGTH OF BRIDGES AND ROOFS. 79 W ' * is to be added to the equation of that sim- ple truss which has full panel ends. Similarly it may be shown that j-p W ~~ U Y~ ' P^ 1S to be added to the equation of that simple truss which has half panel ends. f5. Example of tlie Application of the Vertical Equa- tions. — In Fig. 23, let I = 210 feet, the length of the truss, d = 20 feet, the depth of the truss, p = 10 feet, the length of a panel, w = 105 tons, the weight of the truss, w' = 210 tons, the weight of a full movable load. Since Simple Truss No. 1 has full panel ends, we have, V = w'(l — u)* — U) WW. p\ , .-. % for the maximum vertical strains in the braces of this simple truss, and in the counterbraces of Simple Truss No. 2. Substituting the values of the constants in Eq. (62) we have, 210(210 — m)' u 105 105 10\ 4x(210) a + 4 ~~ 2X210^ ~~~2J _ (210 — u) 840 + 26.25 - 0.25 (w — 5). 80 A TREATISE ON This gives the maximum strains in the struts. For the tension in the ties, Eq. (62), must be multiplied by 1.414, the secant of the angle made by the ties with the ver- ticals ; whence we can form the following table : Values of u. 10 30 50 70 90 110 130 Strains in Tons. 75 58.6 45.5 33.3 22.1 11.9 2.6 Compres- sion in Ww& Aa Uu& Cc Ss& Ee Qq & Oo & Ii Mm & LI Kk & Nr Strains in Tons. 106.1 82.9 64.3 47.1 31.2 16.8 3.7 Tension in Wu& Ac Us& Ce Sq & Eg Qo & Gi Om & 11 Mk & Ln Kh & Np The strains in the end braces, that is, the end strut and the ties attached to it, in this and in the next table, are determined as follows : It will be seen by referring to the double truss with an even number of panels, that the strain on the end ties of either simple truss is the same as if it had been obtained from the constant-load vertical equation, with w f + w substituted for w, or the same as where the truss is fully loaded. This is because the simple trusses are entirely independent of each other, in their vertical action either under a full or a partial load. But in the present case, when the truss is partially load- ed, one half of one simple truss is connected by the counterbraces to the opposite half of the other simple THE STRENGTH OF BRIDGES AND EOOFS. 81 truss, and strains come, more or less, upon these counter- braces, until the opposite halves of the same simple truss are fully loaded, then the counterbraces are released, and the moving-load equation is no longer applicable. Hence, in this case, the greatest strains upon the end braces? which are when the truss is fully loaded, are to be deter- mined from the simple-truss constant-load vertical Eqs. (60) and (61), w of these equations being changed to w'+ w. Since Simple Truss No. 2 has half panel ends, we have, V = ^l(l-uJ-f]+^-^ r (w'+ F ), . (63) for the maximum vertical strains in all the braces of this simple truss, and in the counterbraces of Simple Truss No. 1. Substituting the constants in Eq. (63) we have, 105 . , 10\ 2X210^"*" 2 J' ^(2io-,T-ioo +2625 _ 025K+5> This gives the maximum strains in the struts ; for the strains in the ties, we must multiply Eq. (63) by 1.414, as before, except for the end ties, when it is multiplied by 1.118 (the secant of their angle); whence we can form the following table: 82 A TREATISE ON Values of u'. 20 40 60 80 100 120 Strains in Tons. 75 62.9 49.3 36.7 25 14.4 4.5 Compres- sion in Ww & Aa Vv& Bb Tt & Dd Br & Ff Pp & Hh Nn & Kk LI & Mm Strains in Tons. 83.9 88.9 67.7 51.9 35.4 20.2 6.4 Tension in Wv& Ab Vt & 13a Tr & Df Bp & Fh Pn & Hk Nl & Km Li & Mo. It will be observed that the struts LI and Mm are common to both trusses, and are subjected to a greater strain by Eq. (62), and that Kk and Nn are also com- mon, but subjected to greater by Eq. (63), or when the longer segment is loaded. The compression upon the end struts Aa and Ww, from the passing load, 75+75 = 150 tons, the sum of the strains in the simple trusses. 76. The Quincy Railroad Bridge. — The longest span of this bridge offers an excellent opportunity for the application of the preceding formula?, as it is divided into an odd number of panels, with end panels differing from the others, and probably presents as complicated an example of the double truss as is likely to occur. THE STRENGTH OF BRIDGES AND ROOFS. 83 AB CDEFGHIKLMNOPQRS QUINCY RAILROAD BRIDGE. Fig. 27. Let I = 247 feet, the length of the truss, d = 26 feet, the depth of the truss, p = 13 feet, the length of a panel, w = 198,150 lbs., the weight of the truss, uniformly distributed. w r = 328,750 lbs., the weight of the uniform full movable load. In this case, the panel points of Simple Truss No. 1, or the simple truss whose braces form the centre panel, are K, 1, M, n, O, p, Q, r, S and t, to the right of the centre ; and to the left, I, k, G, h. E, f, C, d, A and b ; the re- maining points are those of Simple Truss No. 2. In the upper chord, the uniformity of the double truss ex- tends from A to S, and in the lower chord, from c to s ; Eqs. (56, 57, 58, 59) will therefore apply between those points, for the horizontal strains ; from a to c and from s to u, the horizontal tension is readily found from the moment of the reaction of the abutment around A or S. This tension being greatest under a full load, where — 5 — is the reaction of the abutment, whence we have, deducting half panel load on the abutment, * These dimensions and weights are from the" description of the Quincy Bridge by Mr. T. C. Clarke, C. E. 84 A TREATISE ON wp wp 7 _ (w'+w)p __ {v/+ w)p U ~- ~2d~~2dl' ■ '2d 2dl for the strain in a c and s u. Substituting the values of the constants in Eq. (56), w being equal to w'+ w, we have, (328,750+198,150) X 2 47 328,750+198,150 H_ " 8X26 2X26X247 13\ 9 (328,750+198,150)13' ^"~*T/ ~ 8X26X247 = 623,961-41.023(2 — 6.5) 2 , Here, z is the distance from the centre to the panel points, and as this equation belongs to Simple Truss No. 1, the different values of z are 6.5, 32.5, 58.5, 84.5, and the amount of compression, given by the substitu- tion of these different values of z in the equation, is con- tained in that member of the upper chord on the abut- ment side of the points to which z is measured. Substituting the values of the constants in Eq. (57) we have, H = 630,894 - 41.023(2 - 6.5) 2 : z is here the distance from the centre to the panel points of Simple Truss No. 2, and its values are consequently 19.5, 45.5, 71.5, and 97.5, by the substitution of which, in the equation, we obtain the compression in the upper- chord members on the abutment side of the point to which z is measured. Eq. (58) becomes H = 630,894 — 41.023 (z + 6.5)\ tension in the lower chord on the centre side of panel THE STRENGTH OF BEIDGES AND EOOFS. 85 points of Simple Truss No. 1, values of z being 6.5, 32.5, 58.5, and 84.5 ; and Eq. (59) becomes H = 653,961 — 41.023 (z + 6.5)\ tension in the lower chord on the centre side of panel points of Simple Truss No. 2, the values of z being 19.5 45.5, 71.5, and 97.5. The strains in the lower chord are the same in amount as those in the upper chord between the same inclined braces, and consequently the lower-cord equa- tions are only needed in this case to obtain the strain in c d and r s. From these equations, by the substitution of the dif- ferent values of z, we can form the following table of strains in the chords Values of 8. 6.5 19.5 32.5 45.5 58.5 71.5 84.5 97.5 Strains in lbs. 6239G1 623961 596229 568498 513035 457571 374377 291182 Compres- sion in HI, IK &KL GH& LM FG& MN EF& NO DE& OP CD& PQ BC& QR AB& RS Strains in lbs. 623961 596229 568498 513035 457571 374377 291182 180256 124792 Tension in kl ik& lm hi& mn gli& no fg& op ef & pq de& qr cd & 1 ab, bc.j rs st&-tu The load is upon the lower chord, consequently the struts have the same vertical strain as those ties to the upper ends of which they are attached. 86 A TREATISE ON In the figure, the counterbraces needed under the effects of the moving load are shown by the dotted lines. In the bridge itself there are counterbraces from every point in the lower chord, except the points b and t ; why the reason which compels the insertion of these superfluous counterbraces is not applicable to these two points it is impossible to say. When the moving load covers part of the truss, band t may be considered as belonging to Simple Truss No. 1, for the other might be removed and this one would support these points. In which case this simple truss, then, has end panels of half the length of the others; that is, it has a panel point distant from the abutment half the length of one of its panels ; and being connected with Simple Truss No. 2 at the centre by the counterbraces, the maximum vertical I strain in the braces of the latter, when u is less than ^-j and in the counterbraces of No. 1, when a is greater than ~-, is, from the moving load, u, as before, being the distance from the abutment to the centre of the panel in which are the braces whose strains are to be determined ; and adding to Eq. (61), we have for the total vertical strain, The part of the moving load borne by the points b and THE STEENGTH OF BEIDGES AND ROOFS. 87 t may be considered as belonging to either simple truss ; since the distance of one of these points is p from the nearest abutment, when it is loaded, the reaction of w p the farther abutment is plainly (8) ~w~- When the load enters Simple Truss No. 2, which has uniform end panels, the vertical moving-load strain from it isEq. (45), / r 3 W W T> ■jp \_(l — u'Y — p 2 ] ; to which add — *- and constant-load strain, Eq. (60), and we have, V = 1^-^-^ + f-fjK-f), - (65) for the total maximum vertical strains in the braces of No. 1 when u' is less than ~", and in the counterbraces of No. 2 when greater than ~- The ambiguity in regard to the load on the points b and t renders it necessary to provide in one simple truss for a slight excess of strain. This arises from the fact that the symmetry of the truss is broken at these points. Substituting the values of the constants in Eq. (64) we have, • 328,750 t v 198,150 198,150, V = i^m < 247 - ->'+ ~i 2xW (-+ 6 - 5 >' = 1.347(247 - ^) 2 +49,537.5 — 401.1(^+6.5), compression in the struts of Simple Truss No. 2, and vertical component of the tension in the ties. V X 1.414 88 A TREATISE ON gives the longitudinal strain of the latter. Whence we have the following table of strains in the braces of this truss Values of u. 13 39 65 91 117 143 Strains in lbs. 124,016 89,564 65,477 43,211 22,765 Compression in Su & Aa Rs&Bc Pq&De No&Eg Lm&Hi Strains in lbs. 124,016 126,643 95,584 61,100 32,189 5,858 Tension in Ss & Ac Rq & Be Po&Dg Nm & Fi Lk&Hi Ih&Kn The compression in Aa and Su, and the tension in Ac & Ss is obtained from constant-load equation, w made equal to w'+ w, and multiplied by 1.118, the secant of the angle of these two braces. When u = 1 30, the strain in Ik and Kl is less than when the opposite segment of the truss is loaded. Substituting the values of the constants in Eq. (65), we have, v=™C(^- M y + 3(i3')] + ^ 198,150 ~"2X247 (M_6 ' 5) = 1.347 (I— uy+ 50,220 - 401.1 (u - 6.5), compression in the struts of Simple Truss No. 1, and vertical component of the tension in the ties ; VX 1.414 gives the longitudinal strain in the latter, the strain in the end braces from the constant-load equations, as be- THE STRENGTH OF BRIDGES AND ROOFS. 89 fore ; whence we have the following table of strains in the braces of this truss : Values of u. 26 52 78 104 130 Strains in lbs. 124,016 83,190 60,013 38,658 19,122 Compression in Su& Aa Qr&Cd Op&Ef Mn&Gh Kl&Ik Strains in lbs. 156,850 117,631 84,858 54,662 26,028 Tension in Sr&Ad Qp&Cf On&Eh Gk&Ml Ki&Im As the compression upon Su and Aa comes from both simple trusses, we have for its total amount, 124,016+124,016+27,732X1.118 = 279,035 lbs. The tension upon Ab and St is the weight of a full panel load, 27,732 lbs. All the count erbraces needed in this truss are, Ih, Im, Ki, Kn, Lk, £1, Hi, Ik, Kl, and Lm. 90 A TREATISE ON CASE IV. — A TRIPLE TRUSS CONTAINING AN EVEN NUMBER OF PANELS. 77.— Let Fig. 28 represent a triple truss divided into an even number of panels. 132132132132123123123123 1 132132132132123123 1231231 Fig. 28. -L. Let I — the length of the truss, d = the depth of the truss, p = the length of a panel, w = the weight supported by the truss, uni- formly distributed, x, x' x" = the distances from one abutment to the panel ends, u &c. = x +~"> &c - H & V = the horizontal and vertical strains. 78. Horizontal Strains. — Under the maximum uni- form load, in which case the horizontal strains are the greatest, this truss may be considered as divided into three simple trusses whose vertical strains, or strains having vertical components, do not in any way affect each other. THE STRENGTH OF BRIDGES AND R These simple trusses, with the counterbrac^ by the dotted lines in Fig. 28 removed, are, Truss No. 1, Fig. 29, whose braces meet at the centre ; Truss j \ Fiff. CO. No. 2, Fig. 30, whose braces come next ; and Truss No. 3, Fig. 31. Fig. 31. The regular panels of these simple trusses are equal to three panels of the triple truss in length ; Simple Truss No. 1 is uniform throughout, and having a full panel length at the end may be . considered as sup- porting the half panel loads resting upon the abut- ments. The end panels of Simple Truss No. 2 are two-thirds, and the end panels of Simple Truss No. 3 one-third the length of their other panels ; and the sym- metry of both is broken at the centre. Under the uniform full load each simple truss bears w one-third the weight, or -~ ; the reaction of each abut- ment upon Simple Truss No. 1, is therefore -r, the 92 A TREATISE ON IVX weight upon a segment of the truss, \ of -57- x being the distance from the abutment to the panel points of the truss hence, wx wx H = ~M ~ Ul ( 66 ) is the horizontal strain in the upper and lower chords of Simple Truss No. 1. In simple Truss No. 2, the reaction of the abut- w ment is 77, the weight upon any segment of the truss is w \ of -j(x' — 2p), (x' being the distance from the abutment to a panel point of this truss, and confined to values les3 than -£ I, and the distance of the centre of gravity of this X' 7) weight from the point to which 2/ is measured, is "9+4 ; x\ whence we have, for the moment of the load on vjx'' wpx' wp a "W - ~W ~ "37" ; therefor ^ _ ivx' wx' 2 wpa/ wp* n ~ 6d ~~ Ml + ~6dl+ Ml - ' ' ( G7 ) is the horizontal strain in the upper and lower chords of Simple Truss No. 2. Similarly, in Simple Truss No. 3, the reaction of each . w IV abutment is g-, the load on x" is \ of ~j(x ,f - p),(x"' being the distance from the abutment to a panel point of this truss, and confined to values less than ^), and the dis- THE STRENGTH OF BRIDGES AND ROOFS. 93 tance of its centre of gravity from the point to which x" x" is measured is -5- + p % whence we have for the moment . _ _ . „ wx"* wpx" top* of the load on x , ~^y~ + /»? — -«j- > Therefore, is the horizontal strain in the upper and lower chords of Simple Truss No. 3. These equations give the strains in the simple trusses on the abutment sides in the upper, and on the centre sides in the lower chord, of the points to which x, x' and x" are measured ; the strain in the upper chord of the triple truss, at any panel point, is the strain in that sim- ple truss whose panel point is at the same place, added to the simple-truss strains at the next two panel points towards the centre ; and, in like manner, the strain in the lower chord of the triple truss, at any panel point, is the strain in one simple truss at the same point, added to the simple-truss strains at the next two panel points towards the abutment. Hence, to find the compression in the triple truss at the panel points of Simple Truss No. 1, we must make x" of Eq. (68) equal to x + p, x! of Eq. (67) equal to x+ 2p, and add the equations so changed to Eq. (QQ) m , performing this operation, we obtain, wx wx* wp wpx . H== "2rf- ~2dl + ~2d-~dT t " " " " (69) For the compression in the triple truss, at the panel 94 A TEEATISE ON points of Simple Truss No. 2, make x of Eq. (66) equal to x'+p, x ! " of Eq. (68) equal to %'+ 2p, and then, add- ing to Eq. (67), we have, wx' wx'* wp wpx' wp* H = U~~M + ~2d~~~dl~ Ml " ' (7 °' For the compression at the panel points of Simple Truss No. 3, in the triple truss, make x f of Eq. (67) equal to x''-\-p, x of Eq. (66) equal to x + 2p, and, adding to Eq. (68), we have, _ wx" wx" 2 wp wpx" the same as Eq. (69) ; or, but two equations are re- quired for the upper chord, one, Eq. (69), for the points of Simple Trusses No. 1 and No. 3, and one, Eq. (70), for the points of Simple Truss No. 2. In the lower chord, by a similar process, and substi- tuting — p and — 2p for p and 2p, we obtain one equa- tion, w __ wx wx* vjp wpx for the tension in the triple truss at the panel points of both Simple Trusses, Nos. 1 and 2, and, for the tension in the triple truss at the panel points of Simple Truss No. 3. These equations, giving the maximum strains in the chords of the truss of Fig. 28, will remain the same so THE STRENGTH OF BRIDGES AND ROOFS. 95 long as the truss contains an even number of panels, and are not affected by the end panels of the simple trusses. That is to say, if, in the case supposed, another panel be added at either end, though Simple truss No. 1 will then support \(w ~-\ + -j-; Simple Truss No. 2, the same, and Simple Truss No. 3, \(iv vM , the same equations will still apply ; the trusses being numbered, as in this case, from the centre. It must be remembered that the moment of the load on any simple truss is not affected by the proportion of the full load which it bears, but depends upon the length of the end panel of that simple truss. For example, if this truss be lengthened two panels, Simple Truss No. 1 bears ~-, but the o o moment of the load upon it is, because its end panel is one-third the length of its other panels, the same as the moment of the load upon Simple Truss No. 3, in the example above, which has a panel of equal length, or, wx wpx vrp Ml + ~Ml ~ ~Ml This is because the centre of gravity is affected by the length of the end panel. 79. Strain in the Second Lower-Chord Members.— In the lower chord, as the equation applies to the abut- ment end of any member, it is evident that, to obtain the strain in the second member from the abutment, we require the simple-truss strain for the point next the abutment, added only to the simple-truss strain at the 96 A TEEATISE ON panel point on the abutment, as the latter is the only panel point . on the abutment side which may be added to the former. But making x, in the equation of the latter simple truss, zero, this strain becomes zero, and we need, consequently, only the simple-truss strain at the point one panel length from the abutment. This strain will differ as the point may belong to the different simple trusses, and is simply the reaction of the abut- ment upon that simple truss to which the point referred to belongs, multiplied by^ and divided by d. If the point belongs to Simple Truss No. 1, the re- action of the abutment is always, wp whence we have, \( w + H = !+S • - • <»> for the strain in the second or next to end member of the lower chord. If the point belongs to Simple Trnss No. 2, the re- action of the abutment is always, i wp) -.0* — fj> whence we have, „ wp wp' ^~Td~'m^ ' ' < 75) for the strain in the second member. And if the point belongs to Simple Truss No. 3, the reaction of the abut- _ w ment is always g-, whence we have, THE STRENGTH OF BRIDGES AND ROOFS. 97 for the strain in the second member. There is no strain in the end members of the lower chord, and the strains in all the other members may be obtained from the regular equations given above. If the inclined braces were struts, instead of ties, a similar process would determine the strains in the cor- responding members of the upper chord. 80. Vertical Strains from the Constant Load. — The vertical equations are obtained from the simple-truss horizontal equations, as in the previous cases, because, under a full load, the simple-truss braces are uncon- nected, and act independently of each other, and the difference in the horizontal strains in any simple truss at x and at x +3p, or at the two ends of a panel, is the horizontal component of the strain in the inclined brace connecting these two points, and, as before, the ver- tical component may be obtained from the proportion, Zp:d. In Simple Truss No. 1, if the end panel be of uni- form length with the others, or equal to 3p, TT _ wx wx* 6d 6dl is the horizontal strain ; if the end panel = 2p, it be- comes .p. __ wx ivx* top 7 = 13 ~ ~Mi + ~MP 98 A TKEATISE ON and if the end panel = jp, it is _ wx wx* wp* 11 = ~U ~ Ml + ~2>dl - from either of which equations we obtain, as before, V = f-f, - - - - (77) for the vertical strain from a constant load in Simple Truss No. 1, u being equal to x + -y, or the distance to the centre of a panel of the simple truss. Whence we see that the vertical equation is independent of the pro- portion of the whole weight which the simple truss may sustain. From Eq. (67) we obtain, ▼-S-S«-f). <"> for the vertical strain ia Simple Tj;uss No. 2. And from Eq. (68) we obtain, v =!->•+!). w for the vertical strain in Simple Truss No. 3. In these equations, t/., ^', and w" represent the distances from one abutment to the centres of the panels of the three simple trusses, and in Simple Truss No. 1, the difference in the different consecutive values of u is uniform, or equal to 3p ; this is true of u' and u" only to the centre of the truss, because the simple trusses to which they be- long are not uniform beyond that point. If, in Eq. (79). THE STRENGTH OF BRIDGES AND ROOFS. 99 u" be made greater than ~-, and the regular increment in its values be kept the same, 3p, it becomes the dis- tance to the centre of a panel of Simple Truss No. 2, and the values of V will then give the vertical strains in the latter truss passing to or sustained by the abutment opposite to that from which u" is measured. This may be proved by making u" = I — v! whence we shall have the equation of Simple Truss No. 2, with the minus sign. Similarly, if u' be made greater than ~-, we ob- tain the vertical strain in Simple Truss No. 3 beyond the centre. 81. Vertical Strains from the moving Load. — Disre- garding the constant load, if we suppose one end of the truss to be loaded, and trace the course of the vertical strain to the other end, we find, in Simple Truss No. 1, that it follows the braces and counterbraces of that truss throughout ; but in the other simple trusses, the vertical strain at the centre passes through the counter- braces from one to the other ; or, if we consider a part less than one-half of either Simple Truss No. 2 or No. 3 as loaded, the farther abutment reacts upon it through the other simple truss. Hence we have, under the moving load, two other simple trusses, composed of the opposite halves of Simple Trusses Nos. 2 and 3, as shown in Figs. 32 and 33— one being the other reversed. 100 A TREATISE ON A\ A \ Wis. 32. The dotted lines represent the counterbraces. In the figure, Simple Truss No. 1 has a panel end equal to 3p. Let u be the distance from one abutment to the centre of any panel, then \ oi-j-(l — u |- ] , (w' being the weight of the full uniform moving load), will be the weight upon the panel points of this truss within the space I — • u, and u being unloaded, \ (I — u + -|-j is the distance of the centre of gravity of this weight from the loaded abutment, or the abutment from which the load extends ; dividing by Z, and multiplying the above quantities (8) we have, _«7 -uy--f -(80) for the reaction of the unloaded abutment and the ver- tical strain throughout the unloaded part. Adding Eq. (80) to Eq. (77), we obtain the max- imum vertical strain in Simple Truss No. 1, from the moving and the constant loads. In Fig. 32, which has the left end panel equal to p y and the right end panel equal to 2jo, let u" be the dis- THE STRENGTH OF BRIDGES AND ROOFS. 101 tance from the right abutment to the centre of any panel, and the length of the unloaded portion of the truss. Then, if the left segment of this truss be loaded, % of ~f~(l — u" + & ] will be the weight upon the panel points within the distance I — ?j", and the distance of the centre of gravity of this load from the left abutment will be \ (I — u" — — ), whence we have V w \n ..„\. P 6? ((J-«")'-£Ji • - ( 81 ) for the equation of the moving load when the load ex- tends from the left end of Fig. 32, or the right end of Fig. 33, or that end of either truss which has an end panel equal to p, and is to be added, in this case, to the equation of the simple truss at the unloaded end of this truss, or to Eq. (79). In Fig. 33, which has the left end panel equal to 2^>, and the right end panel equal to p, let u' be the distance from the right abutment to the centre of any panel, and the length of the unloaded portion of the truss ; then, if w' the left segment of the truss be loaded, \ of -j-(l u 2 J will be the weight upon the panel points, within the distance I — u\ and the distance of the centre of gravity of this load from the left abutment will be \ of (I — u' + «- ) , whence we have, V=^((Z- M 0'-f-], - - (82) 102 A TREATISE ON same as Eq. (81), for the equation of the moving load, when the load extends from left end of Fig. 33 or right end of Fig. 32, or that end of either truss which has a panel equal to 2p, and is to be added to the equation of the simple truss at the unloaded abutment. It will be seen that, as the triple truss may vary in the number of its panels, any one may have an end panel equal to p, 2p, or 3p, and consequently, either of the equations of the moving load may, in different ex- amples, apply to either of the simple trusses; the appli- cation depending solely upon the length of the end panel. The moving-load equation, in any case, is to be added to the constant-load equation of that simple truss through whose braces the reaction of the unloaded abutment acts, to obtain the maximum vertical strain. 82. Horizontal Equations, with the variable meas- ured from the Centre of the Truss. — In Eq. (69) make x = -x 2, and we have, TT wl w . . wp , v H =M-Wl( z -Pr+2TV ■ ■ ■ < 83 > for the compression in the upper chord at the panel points of Simple Trusses Nos. 1 and 3. In Eq. 70, make x' = ^ — z', and we have, wl w , V ■ H = «-iaP-*)' < 84 > for the compression in the upper chord at the panel THE STRENGTH OF BRIDGES AND ROOFS. 103 points of Simple Truss No. 2, z and z' being measured from the centre of the truss. By measuring from the centre of the truss, instead of from the ends, the equa- tions are simpler and more readily applied. §3. — Example. 213213213213212312 3 12312312 ABCDEFGHIKLMNO N'M'L'K' I' H' G' F E' D'C'B' A' / ab c d e f g li i k 1 mnon'm'l'k'i'h'g' fe' d' c' b'a' ^pj Fig. 34. Let Fig. 34 represent a triple truss with an even number of panels. Let I = 24:7 feet, the length of the truss, d — 26 feet, the depth of the truss, p = 9.5 feet, the length of a panel, to == 100 tons, the constant truss weight, w' = 160 tons, the full movable weight. The load is upon the lower chord. The simple- truss panel points are distinguished in the upper chord by their numbers. §4. Horizontal strains. — For the upper-chord com- pressions we have Eqs. (83 and 84), which, with the values of the constants given above, are, 260x247 260 H 8x26 310.58 ■ 2x26x247 (z>- P y ( V_l 26 °X( 9 - 5 ) 8 \fT#)± 2X26X247' 49.4 for the points 1 and 3. And 104 A TEEATISE ON H 49.4 ' 260x247 260 8x26 2X26X — (z?-py= 308.75 Whence we can form the following table of strains in the upper chord. Values of 2 and z f . 9.5 19 28.5 38 47.5 57 Strains in Tons. 308.75 308.75 308.75 303.27 292.31 281.35 264.91 Compres- sion in NO& ON' MN & N'M' LM& M'L' KL & L'K' IK& KT HI & I'H' GH& H'G' {Continued) Values of z and z' 66.5 76 85.5 95 104.5 114 Strains in Tons. 242.98 221.06 195.66 160.78 127.89 89.53 Compres- sion in FG& G'F EF& F'E' DE& E'D' CD& D'C BC& C'B' AB& BA' In the lower chord, as the tensions are equal in amount to the compressions in the upper chord, between the same inclined braces, we have onlv the second and third members from the abutments to determine, the first member being subject to no strain. Since the point one panel length from the abutment belongs to Simple Truss No. 1, we have for the strain in be and c'b', Eq. (74), H = 260x9.5 260x(9.5) 5 6x26 6X26X247 = 16.44. THE STPwENGTH OF BEIDGES AND ROOFS. 105 and from Eq. (73), making %"= 19, we have the tension in cd and dV. We can consequently form the fol- lowing table without further calculation for the lower- chord tensions : Strains in Tons. 303.27 292.31 281.35 264.91 242.98 22-1.06 Tension in no & o'n' mn & n'm' lm& ml' kl& l'k' ik& k'i' hi& i'h' (Continued.) Strains in Tons. 195.66 160.78 127.89 89.53 45.67 16.44 Tension in gli & h'g' fg& ef & fe' de & e'd' cd & d'c' be & c'b' 85. Vertical strains. — Since Simple Truss No. 1 has a panel point distant^) from the abutment, Eq. (81) will give the strains from the moving load, and adding this to Eq. (77), we have W V v = TT A(i-uy-j- + -_ er\ v - J 4; ' 6 u ' w w u (85) Substituting the values of the constants, we have, (8 „ _„..)._ s^+»? V = 160 6x(247) 2 100t*" 3x(247) 8 = "1830^7 K 247 ~ U "^— 22 ' 5625 1 + 16 ' 67 u" ~7zr 10G A TREATISE ON The first value of v! f in this equation is — 4.75, be- cause the centre of the first panel of this truss, if it were uniform with the others, would be 4.75 feet back from the abutment ; the other values of u" are the successive distances from the abutment to the centres of the panels ; whence we can form the following table of compression in the struts of this simple truss : Values of u". Strains in Tons. —4.75 23.75 52.25 80.75 109.25 137.75 45.00 35.23 26.18 17.84 10.20 3.28 Compres- sion in Aa & A'a' Bb & B'b' Ee& E'e' Hb& H'h' LI & LI' Oo The next value of u" will give V a negative value. The tension in the ties is found by multiplying Eq. (85) by the secants of their angles, which, for the end ties, is 1.065, and for the others, 1.48 ; whence we can form the following table Values of u. — 4.75 23.75 52.25 80.75 109.25 137.74 Strains in Tons. 47.92 52.14 38.75 ... 26.4 15.1 4.85 Tension in Ab& A'b' Be & Be' Eh& E'k' HI & HT Lo& l'o' 01 & .01' The strains in the end braces are given by Eq. (85) as well as by the constant-load equations, since this sim- ple truss is the same under the moving as under the con- THE STRENGTH OF BRIDGES AND ROOFS. 107 stant load, but in the other simple trusses it is different ; there, as before explained, the strains in the end braces are to be obtained from the constant-load equations alone. Since Simple Truss No. 2 has an end panel equal to 3p, we must add Eq. (80) to Simple Truss No. 3 con- stant-load equation, Eq, (79), for the vertical strains in all the braces of Simple Truss No. 3, and in the counter- braces of Simple Truss No. 2, when u" is greater than ~- ; whence we have, V = ^f/7_^-^U^_ (I — u") w \l'( U " + 2 ] (86) Substituting the values of the constants, 8 V = r 18302.7 u"+ 4.75 7.41 [(247 - u'J— 203.0625] + 16.67 The first value of u" is 4.75, whence, and the other values of u'\ we can form the following table : Values of u". 4.75 33.25" 61.75 90.25 118.75 147.25 Strains in Tons. 40.00 31.5 22.68 14.57 7.18 0.5 Compres- sion in. Aa & A'a' Cc& CV Ff & F'f Ii & IT Mm & M'm' Nn& N'n' 108 A TREATISE ON The end braces from the constant-load equations as before, w'+ w being put for w. Multiplying Eq. (86) by the secants of the angles made by the ties, which for the end tie is 1.24, and for the others 1.48, we can form the following table : 1 Values of u". 4.75 33.25 61.75 90.25 118.75 147.25 Strains in Tons. 49.60 46.62 33.57 21.56 10.63 0.74 Tension in Ac& A'c' Cf & C'f Fi& F'i' Im & I'm' Mn'& M'n N'k'& Nk Finally, since Simple Truss No. 3 has an end panel equal to 2p, we must add Eq. (82) to Simple Truss No. 2 constant-load equation, Eq. (78) for the vertical strains in all the braces of the latter, and in the counter- braces of the former, when v! is greater than this divided by the moment ol its sup- / w wp \ port is I — —-) (x + a), the excess from the lengthened _ wa ^;a ., .. end panel, —7- — -77" J adding these two quantities, and dividing by a 7 , we have, wa wpa ~2d + ~6dl < 88 > for the increase in the compression in the upper chord, between the points B and B', and the increase in the tension in the lower chord, between the points e and e'. Adding this quantity to Eqs. (83 and 84), and substi- tuting the values of the constants given above, we have for the upper-chord strains, H = 651.14 — .02474(z — 8) 2 , for the panel points of Simple Trusses Nos.l and 3, and H = 649.56 - .02474(2 - 8)\ for the panel points of Simple Truss No. 2. The strain in A B and A' B' is most readily obtained by multiplying the reaction of the support upon Simple Truss No. 2, given above, by 12, the distance of the last panel point, and dividing by 24, the depth of the truss. The strains in the lower chord are taken from the strains in the upper chord between the same inclined braces. Hence we form the following table of strains in THE STRENGTH OF BRIDGES AND ROOFS. 113 the chords, w + w' being substituted for w in the equa- tions : Values of z. 8 16 24 32 40 48 56 Strains in Tons. 649.56 649.56 649.56 644.81 635.31 625.81 611.56 592.56 Compression in WX& XW VW& WT UV& V'U' TU& U'T' ST& T'S' RS& S'R' QR& RQ' PQ& Q'P' Tension in wx & x'w' vw & w'v' uv & v'u' tu & u't' st & tV (Continued.) Values of 2. 64 72 80 88 96 104 112 Strains in Tons. 573.56 549.81 521.31 592.81 459.56 421.56 383.56 Compres- sion in 0P& P'O' N0& ON' MN& N'M' LM& M'L' KL & L'K' IK& KT HI& I'H' Tension in rs & s'r' qr & r'q' pq& q'p' op & po' no & o'n' mn & n'm' lm & m'l' Continued. Values of z. 120 128 136 144 152 160 Strains in Tons. 340.81 293.31 241.81 193.56 136.56 79.56 33.25 Compres- sion in GH& H'G' FG& G'F' EF& F'E' DE& E'D' CD& DC BC& C'B' AB& B'A' Tension in kl & ik & k'i' hi& i'h' gh & fg& ef & f'e' de& e'd' 114 A TREATISE ON §§. Vertical Strains from the Constant Load. — As the Eqs. (77, 78, and 79) represent the weight at any point borne by the simple trusses between that point and the point of no vertical strain, it is evident they are un- affected by the peculiarity of this case, and apply with- out change. 89. Vertical Strains from tlie Moving Load. — The ver- tical strain from the moving load is affected by this change or elongation of the end panels, as is evident when we consider the effect upon a lever, loaded at any other point than the centre, of increasing its length at either or both ends. The equation of the moving load, being the reaction of the unloaded abutment, depends entirely upon the principles of the lever, and is affected by any arrange- ment of the simple trusses that alters the position of the centre of gravity of the load on the segment I — w, as well as by any elongation of the trusses. Let the load be supposed to enter Simple Truss No. 1, at the end, then the weight on the segment I — u, u being the distance from the abutment beyond the load w f p\ to the end of the load, is ^ of -j-(l — u -~- -£) ; the length of the lever, or Simple Truss No. 1, is I -\ • 2#, and the distance of the centre of gravity from the sup- port on which the truss rests at the loaded end is i (I — u + ~ J + a ; whence we have for the equation of the moving load on this truss, THE STRENGTH OF BRIDGES AND ROOFS. 115 V =6^T^)^-«r-^+2a( Z - M -f]].- ( 89) It must be remembered that I is the distance between the abutments, or points c and c', and not the outside length of the truss ; and u the distance from the abut- ment and not from the end of the truss. Adding Eq. (88) to Eq. (77), we have the greatest vertical strain in the truss. Next, let the load be supposed to enter Simple Truss No. 2, at the end ; the load on the segment I — u' is \ of -j-(l — u + y- ; the length of this truss, since Simple Truss No. 2 is added to Simple Truss No. 3, at the centre, is I + 2a, and the distance of the centre of gravity from the end of the truss which is loaded, is \ (I — u — -- + 2a, therefore, v = w^ (l -^-^ +4a{l - u ' + i)^ ■ < 90 > which is to be added to Eq. (79) for the total vertical strain. And, lastly, the load on I — u" in Simple Truss w' &P^\ No. 3, is ^ of -j- (I — u" — ~ ; length of the truss I + 2a ; and the distance of the centre of gravity, \ (I — u'+ —-] ; therefore, Y -WT^) Kl - u '' y -T] ; * * (91) to be added to Eq. (78) for the total vertical strain. 116 A TREATISE ON Substituting the values of the constants in Eq. (89), and Eq. (77) added, we have, (320 -t*y-4tt+ 1248 fi7 _H„ 2488 -r^.oi 96 • whence we can form the following table of strains in the struts of Simple Truss No. 1 ; the load being upon the lower chord : Values of u. 4 28 52 76 100 124 148 172 Strains in Tons. 61.75 52.61 43.92 35.69 27.92 20.62 13.78 7.4 Compression in Bb& B'b' Ee& E'e' H'li' L1& LT Oo& O'o' Rr& RY Uu& U'u' Xx. Multiplying the vertical strain by 1.414, the secant of the tie angle for all the ties except the end ones, which are to be multiplied by 1.667, we have the following table of tensions in the ties : Values of u. 4 28 52 76 100 124 148 172 196 Strains in Tons. 102.94 74.39 62.10 50.47 39.48 29.16 19.48 10.46 2.09 Tension in Be & BE' Eh& Eh' H1& HT Lo & L'o' Or& Or' Ru& RV Ux& U'x Xu'& X'u U'r' & Ur Substituting the values of the constants in Eqs. (90) and (79) added, we have, (320- W y-8«'+2560 l \ ,. ^ 2488 + 2L67 ~ 96<" + 4 > THE STRENGTH OF BRIDGES AND ROOFS. 117 whence we can form the following table of compression in the struts of Simple Truss No. 3, and counterstruts of Simple Truss No. 2. The end braces from the constant- ^ad equation. Values of u'. 12 36 60 84 108 132 156 180 Strains in Tons. 57.00 49.57 41.00 32.89 25.24 17.56 11.34 5.08 Compression in Cc& C'c' Ff & ft Ii& IT Mm & M'm' Pp& P'p' Ss& S's' Vv& VV W'w' & Ww Multiplying the vertical strain by 1.414, we have the following table of tension in the ties of Simple Truss No. 3, and counterties of Simple Truss No. 2 : Values of u f . 12 36 60 84 108 132 156 180 . L Strains in Tons. 80.60 70.09 59.97 46.51 34.29 24.83 16.03 7.8 i Tension in Cf & C'f Fi& FT Im & I'm' Mp& M'p' Ps& P's' Sv& SV Vw'& V'w WT& Wl. Substituting the values of the constants in Eqs. (91) and (78) added, we have, (320-^) -144 13 2488 + Z * b ' W V ' whence we can form the following table of compression in the struts of Simple Truss No. 2, and counterstruts of Simple Truss No. 3. The end braces from the constant- load equations. 118 A TKEATISE ON Values of u". —4 20 44 68 92 116 140 164 188 Strains in Tons. 66.5 55.61 46.81 38.47 30.59 23.17 16.21 9.72 3.69 Compres- sion in Aa & A'a' Dd& D'd' Kk& K'k' Nn & N'n' Qq& Q'q' Tt& T't' Ww& W'w' W &Vv Multiplying the vertical strain for the end ties by 1.118, and for the others by 1.414, we have the fol- lowing table of tension in the ties of Simple Truss No. 2, and counterties of Simple Truss No. 3 : Values of u". —4 20 44 68 92 116 140 164 188 Strains in Tons. 74.35 78.63 66.19 54.4 43.25 32.76 22.92 13.74 5.22 Tension in Ad& Ad' D'g' Gk& G'k' Kn & K'n' Nq& N'q' Qt& Q't' Tw& T'w' Wv'& W'v V's'& Vs THE STRENGTH OF BRIDGES AND ROOFS. 119 CASE V.— A TRIPLE TRUSS CONTAINING AN ODD NUMBER OF PANELS. 90.— Let Fig. 36 represent a triple truss composed of three simple trusses, and containing an odd number of panels, the counterbraces being shown by the dotted lines. 1321321321123 1241231 132 132132112312312 31j Fig. 36. Let I = the length of the truss, d = the depth of the truss, p = the length of a panel, w == the weight supported by the truss, #, x' & x !f = the distances from one abutment to the panel ends, u, u' & u" =a the distances from one abutment to the centres of the panels of the simple trusses. H & V = the horizontal and vertical strains. The simple trusses composing the triple truss are numbered, as in the previous case, from the centre, the panel points belonging to each being shown in the figure by the numbers. 91. Horizontal strains. — Under the maximum uni- form load, w, the weight on simple Truss No. 1, is 120 A TEEATISE ON £ (w — ~T~)-\ — p? an d the moment around any panel point of this simple truss of the weight on the segment x is, as in the previous case where the simple truss had the same end panel, J of tt™ TT 7 ! . wp \ wpx wx % whence, was wx* wpx H= 6J"~" m + ~MT- (92 ) The weight on simple Truss No. 2 is \(w ~~\ and the moment around any panel point of this simple (wx'* pwx' wp*\ 2T~ "21"/' whence, _ was' wx' % wp* H ' - 12~ ~ isr + isr ( 93 > The weight on Simple Truss No. 3 is \(w — -j-\ , and the moment around any panel point of this simple truss, of the weight on the segment x", is, as before, 1 r wx"* wpx" wp*' I 21 + 21 I \ » whence, „„ wa?" wx" % wpx 6d Ml Ml +!••-- w THE STRENGTH OF BRIDGES AND ROOFS. 121 These are the horizontal strains in the upper and lower chords of the simple trusses at their, different panel points. Following the same process which was pursued in the previous case (77), in adding the simple- trass equations, we obtain, wx wx* wp wpx wp H = ~2d ~~ Ml + 23 ~~ ~~df ~~ Ml " " (95 ' Making x = ~ z, we have, H -^-5-^-^ - - - < 96 > for the compression in the upper chord of the triple truss at the panel points of the Simple Trusses No. 1 and No. 2. For the panel points of Simple Truss No. 3 we ob- tain, by the same process, Sd 2dT p) dl J (J ' } z and z! being the distances from the centre of the truss to the panel points of the respective simple trusses. For the tension in the lower chord of the compound truss, in like manner, we obtain, for panel points of Simple Truss No. 1, and TT wl w . for the panel points of Simple Trusses No. 2 and No. 3. 6 122 A TREATISE Otf These lower-chord equations will not apply to the second panel from the abutment, as that member is only within the action of two of the simple trusses, when x in one of them equals zero, as explained in (78.) If the point in the lower chord distant p from the abutment belongs to Simple Truss No. 1, the reaction of the abutment upon this truss is always, w wp _ wp wp* whence, 6d T 6dl If the point belong to Simple Truss No. 2, the re- action of the abutment upon this truss is always, whence, _ wp wp* H -6d + "6dT- " " " " (101) And if the point belong to Simple Truss No. 3, the re- action of the abutment upon this truss is always, wp) whence, n-^2L ^ lion 92. Vertical Strain from a Constant Load.— Deducing the equations of the constant vertical strain from the horizontal equations for the simple trusses, as the ver- THE STRENGTH OF BEIDGES AND ROOFS. 123 tical strains in the different simple trusses are independ- ent of each other, we have, for Simple Truss No. 1, w wu' v = 6 - "ST " for Simple Truss No. 2, and w w for Simple Truss No. 3. (103) (101) (105) 93. Vertical Strains from the Moving Load. — In this case, Simple Truss No. 2 is always symmetrical at the centre, or its opposite halves are united by the counter- bracing, while one half of Simple Truss No. 1 is con- nected by the counterbraces with the opposite half of Simple Truss No. 3. This arrangement, however, does not affect the equations for the moving load, and Eqs. (80 and 81) apply in this case as well as in the pre- vious ; their application being governed by the same principles. 94. — Example — Let Fig. 37 represent a triple truss- composed of three simple trusses, and containing an odd number of panels ; loaded on the lower chord. A BCDE PGH IKLMNOPQRST U V "W X Y bcde fgh i k lmnopq rst u vwxy Fig. 37. 124 A TREATISE ON Let I = 230 feet, the length of the truss, d = 20 feet, the depth of the truss, p = 10 feet, the length of a panel, w' = 230 tons, the weight of the full moving load, w = 115 tons, the constant truss weight. 95. Horizontal strains. — Substituting these values in Eqs. (96) and (97), we have, for upper-chord com- pressions, w being (w'-f-w), H = 495.9375 -.0375(2-10)% where z is the distance from the centre of the truss to the panel points of Simple Trusses Nos. 1 and 2, and H = 503.4375 — .0375(2' -10) 3 , where z' is the distance from the centre of the truss to the panel point of Simple Truss No. 3. The lower-chord tensions are the same as the upper- chord strains between the same inclined braces, arfd we shall not need the lower-chord equations except for the third member from the abutment, cd and vw. The strain in these members is the strain at c or w, which are points in Simple Truss No. 1, and consequently given by Eq. (98). As b and x are points in Simple Truss No. 2, the strain in be and wx is given by Eq. <101.) Whence we have the following table of strains in the upper and lower chords : THE STRENGTH OF BRIDGES AND ROOFS. 125 Values of 2. 5 15 35 45 65 Values of z.' 25 55 Strains in Tons. 495 495 495 472.5 450 427.5 382.5 Compres- sion in LM, MN &NO KL& OP IK& PQ HI& QR GH& RS FG& ST EF& TU Tension in mn lm & no kl& op ik& pq M& qr {Continued.) Values of z. 75 95 105 95 Values of z'. 85 Strains in Tons. 337.5 292.5 225 157.5 90 30 Compres- sion in DE& UV ■ CD& VW BC& WX AB& XY Tension in gh& rs St ef & tu de& uv cd& VW be & WX There is no strain in ab and xy. 96. Vertical strains.— The moving-load vertical equa- tion depends, as before stated, upon the end panel of the simple truss; and the constant-load vertical equation for any simple truss is to be added to that moving-load 126 A TREATISE ON equation which applies to that simple truss to which the first is united by the counterbraces at the centre. Hence, Eq. (103) for Simple Truss No. 1 is to be added to Eq. (80), whence we have, v = %{Q - uy -¥]+?- £(« -p). - - (ice) Substituting constants, (230 -„y- 225 19 ; 1G7 ,!Lzio 1380 r 6 Whence we can form the following table of compressions in the struts. The strains in the end braces from the constant-load equations, as explained before. Values of u. 5 35 65 95 125 Strains in Tons. 60.00 42.39 29.57 18.04 7.82 Compression in Aa& Yy C!c& Ww Ff & Tt Ii& Qq Mm& Nn Multiplying V by 1.414 for the end tie, and by 1.803 for the others, these being the secants of the tie angles, we obtain the following table of tensions in the ties : Values of u. 5 35 65 95 125 Strains in Tons. 84.84 76.43 53.31 32.53 14.10 Mp& Nk Tensiou in Ac & 1 Cf & Yw Wt Fi & Tg Im & Qn THE STRENGTH OF BRIDGES AND ROOFS. 127 For Simple Truss No. 2 we must add Eq. (104) to Eq. (81) and we have, Substituting constants, &m,i^ (230 — w') — 25 , J> „„„ u' V=i 1880 + m67 -6' whence we can form the following table of compression in the struts : Values of u '. • —5 25 55 85 115 Strains in Tons. 60 45.43 32.17 20.22 9.57 Compression in Aa & Yy Bb& Xx Ee & Uu Hh& Rr L1& Oo Multiplying V by the secants of the tie angles, 1.118 for the first, and 1.803 for the others, we can form the folio win s table of strains in the ties : Values of u'. — 5 25 55 85 115 145 Strains in Tons. 67.08 81.99 58.01 36.46 17.26 0.40 Tension in Ab& Yx Be & Xu Eh& Ur HI & Ro Lo& 01 Or & Lh Simple Truss No. 2 is the same under the moving and under the constant load, hence Eq. (107) gives the strains in the end braces. 128 A TREATISE ON For Simple Truss No. 3, Eq. (103) is to be added to E m the extreme case, only by the quantity 225 r-QoTT > consequently, if this quantity be omitted, the dif- ference in the result will be immaterial and upon the safe side, and we shall have but the simple form w' "jnr$ — U Y *° ^ e added to the simple- truss constant-load equations. Moreover, all confusion will be avoided. 130 A TREATISE ON CASE VI. A QUADRUPLE TRUSS CONTAINING AN EVEN NUM- BER OF PANELS. 9§. — Let Fig. (38) represent a truss composed of four simple trusses, and containing an even number of panels. 2143214321-432123412341234123 The simple trusses are numbered from the centre, as in the previous cases, and their different panel points are shown by the numbers in the figure. Let I — the length of the truss, d = the depth of the truss, p = the length of a panel, w = the weight, #, a/, x f \ & x'" = distances from one abutment to the panel ends, u, u\ u", & u'" = the distances to the centres of the panels of the simple trusses. H & V = the horizontal and vertical strains. 99. Horizontal strains. — Under the full uniform load, the horizontal strains are the greatest, and are deter- mined as follows : the result would be the same if the quadruple truss contained any even number of panels. • THE STRENGTH OF BRIDGES AND ROOFS. 131 In this figure, Simple Truss No. 1 bears 7- w "1 2 " -4-^, U U W 3 u V 4 U w wp 4 r Taking moments around the panel points of Simple Truss No. 1, the moment of the abutment reaction is -— , the load on x is -j(x— 2p), the distance of its cen- tre of gravity from the point to which a? is measured, — +i>? whence we have __ wx wx wp - ,„. ^ for the horizontal strains in the upper and lower chords of this truss. Taking moments around the panel points of Simple Truss No. 2, the moment of the abutment reaction is WX WpX ' • , W.j nit -g- + "or - ? the load on x' is *t(« — J9), the distance of its centre of gravity from the point to which x f is meas- , . of 3» . ured iSg- + "9", whence we have, _ waf wx ri pwx' %p*w /iia\ K -~87~w + ~4sr + w' - " m (110) for the horizontal strains in the upper and lower chord of this truss. 132 A TREATISE ON Taking moments around the panel points of Simple Truss No. 3, the moment of the abutment reaction is -~n-~, and the moment of the load on x" is, since this wx" % truss has a full panel end, -tt, whence we have, _„ wx" wx" % w ' = -8T—m> - - - - cm) for the horizontal strains in the upper and lower chords of this truss. And taking moments around the panel points of Simple Truss No. 4, the moment of the abutment reac- . wx'" ivpx'" .W, '_ tion is j, — - — 7tj— , the load on x'"- is -j(x'"— 3p), the distance of its centre of gravity from the point to which x'"-\- p x'" is measured is — « — > whence we have, ; for the compression at the panel points of Simple Trusses No. 2 and No. 3. Making x = -r — 2;, and a/ = | 2', Eq. (113) be- comes where z is the distance from the centre of the truss to the panel points of Simple Trusses No. 1 and No. 4 ; and Eq. (114) becomes wl w f 3p pw where z f is the distance from the centre to the panel points of Simple Trusses No. 2 and No. 3. In the lower chord of the compound truss the tension at any point, x, is the simple-truss strain at a?, added to H' when x f == x — p, H" " x" =x-2 Pl andH'" « x m = x-Sp, and similarly for the points x\ x", and x'". Whence we obtain, changing x and x' to 5- — 2; and £ — 2' 134 A TEEATISE OK wl w 8p\' 9p'w H -M-M^ + TJ +_ M' (117) for the panel points of Simple Trusses No. 1 and No 2: and tot w 3p\* />'«> H ~ M^2JA 2 + TJ + W * * V < 118 > for the panel points of Simple Trusses No. 3 and No. 4 ; z and 2/ beinsr measured from the centre of the truss. It will be seen from these equations, that in this case also the horizontal strain in the two chords is the same between the same inclined members. Hence, in prac- tice, we shall not need the last two equations. The three end members of the lower chord are not subject to the action or the strains of all the four simple trusses, and consequently Eqs. (117 and 118) are not applicable to* them. The first, or end member of the lower chord is subject to the strain of only one simple truss when x is zero, and consequently the strain is zero ; the second member is subject to the strain of the first member, and the strain of that simple truss whose point is one panel length from the abutment, and the third member has the strain of the second member added to the strain of that simple truss whose panel point is two panel lengths from the abutment. . These strains can be most readily determined from the vertical equations for the constant load, and are given below. THE STEENGTH OF BEIDGES AND KOOFS. 135 100. Vertical Strain* from a Constant Load. — In this case, as in the others, the simple trusses are entirely in- dependent of each other in their vertical action under a uniform constant load; hence the equations are to be obtained from the simple-truss horizontal equations ; whence, by the process previously described, we have, from Eqs. (109, 110, 111 and 112), *- : f-3f i (119 » ' for Simple Truss No. 1, V '=l-ik«'-PV- - - - "(120) for Simple Truss No. 2, V-M?. <*"> for Simple Truss No. 3, and for Simple Truss No. 4. In which w, u\ vJ\ and u'" are the distances from the abut- ment to the centres of the panels of the respective simple trusses. 101. Horizontal Tension in the Lower-Chord End Members. — The strains in these, referred to above, may be found as follows : 136 A TKEATISE ON Let Fig. 39 represent the four end panels in a quad- ruple truss, and let d be a panel point in Simple Truss ABODE Fig. 39. No. 4, and c a panel point in Simple Truss No. 3. The strain in be is the strain in the latter truss at c added to that in the former at d. The reaction of the abutment, or the vertical strain in Ee, from Simple Truss No. 4 is Eq. (122), where w u'"= — p, V = 7T ; and from moments around D, H wp (123) w is the tension in the second member, where Simple Truss No. 4 has a panel point distant p from the abutment. The vertical reaction of the abutment from Simple Truss No. 3 is Eq. (121), where u" = 0, whence V = and from moments around C, This added to Eq. (120) gives _ Spw H - 13" * • (124) tension in be, or third member of the lower chord, where THE STRENGTH OF BRIDGES AND ROOFS. 137 Simple Truss No. 4 has a panel point distant p from the abutment. Next, let d be a panel point in Simple Truss No. 3, and c consequently a panel point in Simple Truss No. 2. The reaction of the abutment from Simple Truss No. 3 IV WD is Eq. (121), where u" = — p, whence V = -s + -jj, and 11 = ib- + it™ - - ( 125 ) strain in the second member ; and the reaction of the abutment from Simple Truss No. 2 is Eq. (120), where . - T w wp r _ pw pw . . _ u'= 0, whence V = tt + -n, and li — j~j + Vy?? which added to Eq. (125) gives for the tension in the third member of the lower chord where Simple Truss No. 3 has a panel point distant p from the abutment. Next, let d be a panel point of Simple Truss No. 2, and c consequently a panel point of Simple Truss No. 1. The reaction of the abutment from the former is Eq. w pw (120), where v! = — p, whence V = jr -f- -~i, therefore, pw tfw H -83" + w * ; (127) is the tension in the second member ; and the reaction from Simple Truss No. 1 is Eq. (119), where u = 0, 138 A TREATISE ON w pw whence V = rr and H = -j-t, which, added to Eq. (127), gives _Spw p'w H ~ Sd + Ut " ' ' < 128 > for the tension in the third member when Simple Truss No. 2 has a panel point distant p from the abutment. The remaining case is where d is a panel point of Simple Truss No. 1, and c of Simple Truss No. 4. The reaction of the abutment from the former is Eq. (119), where u = — p, whence V = -~ + ^y, and __ pw pw is the tension in the second member ; and the reaction of the abutment from Simple Truss No. 4 is Eq. (122), where «"'= 0, whence V = f _ f , and H = *g -||, which, added to Eq. (129), gives xx Bpw p^w for the tension in the third member of the lower chord when Simple Truss No. 1 has a panel point distant p from the abutment. 102. Vertical Strains from a Moving Load. — Let Figs. 40, 41, 42, and 43 represent the different ends of the four simple trusses. Let the vertical lines show the THE STRENGTH OF BRIDGES AND ROOFS. 139 divisions into simple-truss panels, and the points in the lower chord the panel points of the compound truss. -k Fig. 40. In Fig. 40 u is the distance from the right abut- ment to the centres of the simple-truss panels or the points a, b, c, d, &c. ; w f being the weight of the full w' movable load, J of -j-(l—u+p) is the load on the panel points of this truss within the distance I — u ; and the distance of its centre of gravity from the left abut- l — u — p ment is g — — , whence the reaction of the other abutment, or the greatest vertical strain from the mov- able load upon the simple truss whose loaded end has a panel = p, is, V = w' I(i-»Y-Pl. (131)1 u- Fig. 41. In Fig. 41 u' is the distance from the right abut- ment to the centre of the simple-truss panels, or the w points a, b, c, d, &c. ; whence jp(l — u'Y is the load on 140 A TKEATISE ON I — u\ distance of its centre of gravity from the loaded I u end, — ~ — , and the greatest vertical strain upon the simple truss whose panel at the loaded end = 2p, is w Y = w( i- U y. (132) 1 * t t 1 < T t 1 : U— - 1 1 1 b C Fig a 42. w In Fig. 42, similarly, the load on I— u" is -rj- (I — u" — p), the distance of its centre of gravity from the . _ . I — u ff -\- p loaded end, ^ > whence the greatest vertical strain upon the simple truss whose panel at the loaded end = 3p, is w V=8f[(Z-«")-^] (133) -Ur- Fig. 43. v/ In Figure 43, similarly, the load on I — v!" is ~4j-(l — u"'— 2p), the distance of its centre of gravity from the loaded end, -- ? f whence the greatest THE STRENGTH OF BRIDGES AND ROOFS. 141 vertical strain upon the simple truss whose panel end at the loaded end «*= 4p is w *Pl (134) In this case, Simple Trusses No. 1 and No. 3 are con- tinuous, and their panels uniform throughout, but one half of Simple Truss No. 2 is united to the opposite half of Simple Truss No. 4. It has been fully explained how to determine which moving-load vertical equation is to be added to the constant-load vertical equation belonging to any simple truss. In the truss shown in Fig. 38, Eq. (131) is to be added to Eq. (122), Eq. (132) to Eq. (119), Eq. (133) to Eq. (120), and Eq. (134) to Eq. (121.) 103. Example. — Let Fig. 44 represent a quadruple truss containing an even number of panels, and loaded upon the lower chord. A B C D E F G H I K L M N O P Q R Q' P' O'n' m' I/k' i' H' g' F' E' d' C' B' A' |H| ab c d e f g h i k 1 m n o p q r q'p'o'n'm'l'k'i'h'g'f' e'd'c' b'a' |i|§| 1111 14 321432143214321234123412341234 1 Ml Fig. 44. Let I = 320 feet, the length of the truss, d = 27.5 the depth of the truss, p = 10 feet, the length of a panel, v? = 208 tons, the weight of the full uniform movable load, • w = 144 tons, the weight of the truss. 142 A TREATISE ON Values of z. Values of z' Strains in Tons. Compression in Tension in 512 QR and RQ' 10 512 PQ and Q'P' 20 512 OP and P O' 30 512 NO and O'N' 40 504 MN and N'M'. qr and r'q' 50 488 LM and ML' pq and q'p' 60 472 KL and L'K' op and p'o' 70 456 IK and KT no and o'n' 80 432 HI and I'H' mn and n'm' 90 400 GH and H'G' lm and ml' 100 368 FG and G'F kl and l'k' 110 336 EF and F'E' ik and k'i' 120 296 DE and ED' lii and ill' 130 248 CD and D'C gh and h'g' 140 200 BCandC'B' fg and g'f ' 150 152 AB & B A' ef and f e' 130 96 de and e'd' 48 cd and d'c' 16 be and c'b' 104. Horizontal strains. — Substituting these values in Eqs. (115) and (116) for the upper-chord strains, in Eq. (117) for the fourth lower-chord member, and in Eqs. (123) and (124) for the second and third lower- chord members, since Simple Truss No. 4 has a* panel THE STKENGTH OF BKIDGES AND EOOFS. 143 point distant p from the abutment; the other lower- chord strains being taken from the upper-chord strains between the same inclined braces ; w in these equa- tions being w'+ u\ or 352 tons ; we have the preceding table ; there is no strain in ab and l/a'. 105. Vertical strain*. — Since Simple Truss No. 1 has a panel end = 4j9, Eq. (134) is to be added to Eq. (119) to obtain the maximum vertical strains in this truss; Simple Truss No. 2 has a panel end = 3p, therefore Eq. (133) is to be added to Eq. (122) to obtain the greatest vertical straihs in No. 4 ; No. 3 has a panel end = 2p, therefore Eq. (132) is to be added to Eq. (121); and No. 4 has a panel end = jp, therefore Eq. (131) is to be added to Eq. (120) to obtain the maxi- mum vertical strain in Simple Truss No. 2. Making the variable, in the first case, w, in the second, u\ in the third, u'\ and in the fourth, u'", substituting the values above, w of the constant-load equations being 144 tons, and multiplying the vertical strain by secants of the tie angles, we can form the following table of strains in the braces. The secants are 1.765, 1.48, 1.236, and 1.063. Total compression on Aa and AV is 170.5 tons. The strains in the end braces of those simple trusses which are uniform beyond the centre, No. 1 and No. 3, are found from the same equations which give the strains in the other braces ; but in the end braces of No. 2 and No. 4 the strains are determined from the constant-load portion of the equations, u/+ w being put for iv. 144 A TREATISE ON Values of u. Values, of u' Values of u." Values of u'" Strains in Tons. Compres- sion in Strains in Tons. Tension in —10 44.00 44.00 Aa & A'a' 46.77 Ab&A'b' Aa & A'a' 54.38 Ac & A'c' 10 44.00 Aa & A'a' 65.12 Ad&A'd 20 38.50 Aa & A'a' 67.95 Ae&A'e' 30 34.83 Bb & B'b' 61.47 Bf & B'f 40 33.41 Cc & C'c' 58.97 Cg&C'g' 50 31.98 Dd & D'd' 56.53 Dh&D'h' 60 28.31 Ee & E'e' 49.97 Ei & E'i' 70 24.84 Ff & F'f 43.84 Fk&F'k' 80 23.62 Gg&G'g' 41.69 Gl & GT 90 22.41 Hh & H'h' 39.55 Hm & H'm' 100 18.94 Ii & IT 33.43 In & In' 110 15.67 Kk & K'k' 27.66 Ko&K'o' 120 14.66 LI & LT 25.87 Lp & L'p' 140 130 13.64 Mm&M'm' 24.07 Mq&M'q' 10.36 Nn&NV 18.29 Nr&NV 150 7.31 Oo & O'o' 12.90 Oq'&O'q 160 6.50 Pp & P'p' 11.47 Pp' & P'p 170 5.71 Qq & Q'q' 10.01 Qo' & Q'o 180 2.62 Rr' 4.62 Rn' & R'n CASE VII.— A QUADRUPLE TRUSS, CONTAINING AN ODD NUMBER OF PANELS. 106.— In this quadruple truss, as in the others, the equa- tions are not affected by the number of panels the truss THE STRENGTH OF BRIDGES AND ROOFS. U5 contains, but by the condition whether the number be odd or even. Hence, it is necessary to determine equa- tions for this case. It should be remembered that, in determining equations for any case, the simplest form of truss may be taken, provided it fulfils the necessary con- ditions. Let Fig. 45 represent a quadruple truss containing an odd number of panels. This, like the other, is divisible into four simple trusses, numbered, as in the 14 3 2143211 4 12 8 4 1 MS \ Si XIN- 143214a 2 112341 3 4 1; Fig. 45. previous cases, from the centre towards the abutments, and whose panel points are shown by the numbers. Let I — the length of the truss, d = the depth of the truss, p = the length of a panel of the compound truss, w — the maximum weight uniformly dis- tributed, #, ccf #", & x" f = the distances from one abutment to the panel points of the simple trusses, u, u f ,v! f &u" f — the distances from the same abutment to the centres of the simple-truss panels, H & V = the horizontal and vertical strains. 107. Horizontal strain*. — The moment of the load on a segment of any simple truss in this case is the same as 10 146 A TREATISE OK the moment of the load on the same segment of that sim- ple truss in the previous case whose end panel is similar. In this figure, wp \ wp Simple Truss No. 1 bears £ (w — -j- J + -y-* u u 3 « *<—?), 4 « i(— T), whence, 11 - Sd~ Sdl^ Sdl' ' " ( i6b ) for Simple Truss No. 1 ; 11 ~ 8<* "" Sdl + 8ett + 8^ ' " " V ob ' for Simple Truss No. 2 ; 11 ~ Sd ~ Sdl ""Sdl "*" 2<*P " " ^ i} for Simple Truss No. 3 ; and, n/// _^ wa ///a 3/W , 3p'tfl -"- ~~8cZ ~~ 8<« ~ Sdl + 8cZP " ^ 5 ' for Simple Truss No. 4. These are the horizontal strains in the upper and lower chords of the simple trusses. Adding them in the same manner as in the previous cases we have, wl w SpV 3p'u> THE STRENGTH OF BRIDGES AND ROOFS. 147 z being the distance from the centre of the truss, for the compressions in the upper chord at the panel points of Simple Trusses No. 1 and No. 3. z 7 being the distance from the centre of the truss, for the compressions in the upper chord at the panel points of Simple Truss No. 2, and, H=—- — (z"~ -^ \'+ -^-i - -1141) "■ Sd 2dl {z 2 j + 8dl {lil) z" being the distance from the centre of the truss, for the compressions in the upper chord at the panel points of Simple Truss No. 4. For the tension in the lower chord we obtain, TT - Wl W It A. 8j> \'l 15j ^° 11491 at the panel points of No. 1 ; wl w , 3^\ 2 Sp'w s+ MM at the panel points of No. 2 and No. 4 ; and H - 8^ - iar + Ti ~ sir ■ ' * (144) at the panel points of No. 3. 108. Vertical Strains from a Constant L.oad- — From Eqs. (135, 136, 137, and 138) we have, for No. 1 ; V = 8-lT( M '-2)' "• - " < 146 > for No. 2 ; 148 A TREATISE ON v=f-fk«:+f). - - • o« for No. 3 ; and for No. 4. 109. Horizontal Tensions in Lower-Chord End Hem- bers.— The strains are obtained as in (99). See Fig. 39, whence, __ pw p*w H -"8rf — 8d? " " " (U9) in the second member ; and h = ^-^ - • • a»> in the third member, when a panel point of No. 4 is distant^ from the abutment. pw phv U ~U + Mt " " " " < 151) in the second member ; and _ 3pw Sp*w H -~8J + W ' * " < 152) in the third member, when a panel point of No. 3 is distant p from the abutment. „ _pw Sp'w h -m + w - - - - < 153) in the second member, and THE STRENGTH OF BRIDGES AND ROOFS. 149 in the third member, when a panel point of No. 2 is distant J? from the abutment. And in the second member ; and *-&-£•• • - • < iM > in the third member, when a panel point of No. 1 is distant p from the abutment. 110. Vertical Strains from a Moving: Load. — Eqs. (131), (132), (133) and (134) apply in this case. None of the simple trusses are continuous, but one half of No. 1 is connected by the counterbracing with the opposite half of No. 4, and one half of No. 2 is .similarly connected with the opposite half of No. 3 ; hence care must be exercised in the addition of the vertical equation, and it is always safest to make a diagram of the simple trusses, as connected under the moving load. ill. Example. — Let Fig. 46 represent a quadruple truss containing an odd number of panels. Let I = 328 feet, the length of the truss, d = 32 feet, the depth of the truss, p = 8 feet, the length of the panel, w f = 328 tons, the weight of the full uniform movable load, w = 164 tons, the constant truss load. The load is upon the lower chord. 150 A TEEATISE 01ST Values of z. Values of z'. Values of z". Strains in Tons. Compression in Tension in 4 630 UV VV & V'U' 12 630 TU & U'T' 20 630 ST & ST' 28 630 RS & S'R' vv' 36 618 QR & R'Q' uv & v'u' 44 606 PQ & Q'P' tu & u't' 52 594 OP & P'O' st & t's' 60 582 NO&O'N' ,rs & s'r' 68 558 MN & N'M' qr & r'q' 76 534 LM & ML' pq & q'p' 84 510 KL & L'K' op & p'o' 100 92 486 IK & KT no & o'n' . 450 HI & I'H' mn & n'm' 108 414 GH & H'G' lm & m'l' 116 378 FG & G'F' kl & IV 124 342 EF & F'E' ik & k'i' 132 294 DE & E'D' hi & i'h' 140 246 CD & DC gh & h'g' 148 198 BC & C'B' fg&g'f 156 150 AB & B'A' ef & f e' 140 90 de & e'd' 45 cd & d'c' 15 be & c'b' THE STRENGTH OF BRIDGES AND ROOFS. 151 112. Horizontal strains.— For the upper chord, use Eqs. (139), (140) and (141) ; for the lower chord, take the strains from the upper chord between the same in- clined braces, and beyond these, for the fourth member, use Eq. (143), for the third member Eq. (150), and for the second, Eq. (149) ; w of these equations being made equal to vf + w A or 492 tons. Hence we can form the preceding table of strains in the chords. 113. Vertical strains. — For the strains in the struts of Simple Truss No. 1, add Eq. (131) to Eq. (145). In No. 2 add Eq. (132) to Eq. (146). In No. 3 add Eq. (133) to Eq. (147), and in No. 4 add Eq. (134) to Eq. (148). For the ties multiply the vertical strain so obtained by the secants of the tie angles, which are 1.414, 1.25, 1.118 and 1.0308. The following gives the strains in the braces : CDKFQH IKLMNOPQBBTUVVUT8RQPO «' i/k' i' h'o' p'k'p'c' b' a' Fig. 46. 152 A TREATISE ON Values of u. Values of u'. Values of u". Values of u'". Strains in Tons. Compres- sion in Strains in Tons. Tension in —8 60.00 Aa & A'a' 63.24 Ab & A'b' 60.00 Aa & A'a' 72.11 Ac & A'c' 8 60.00 Aa & A'a' 84.84 Ad & A'd' 16 60.00 Aa & A'a' 100.00 Ae & A'e' 24 51.12 Bb & B'b' 72.28 Bf & BT 32 49.36 Cc & C'c' 69.80 Cg & Cg' 40 47.61 Dd & D'd' 67.32 Dh & D'h' 48 45.85 Ee & E'e' 64.83 Ei&E'i' 56 40.10 Ff & FT 56.70 Fk & F'k' 64 38.53 Qg & G'g' 54.48 Gl & GT 72 36.97 Hh & H'h' 52.28 Hm & H'ni/ 80 35.41 Ii & IT 49.69 In & IV 88 29.85 Kk & K'k' 42.21 Ko & K'o' 96 28.49 LI & LT 40.28 Lp & L'p' 104 27.12 Mm& M'm' 38.35 Mq&M'q' 112 25.76 Nn & N'n' 36.42 Xr & NY 120 20.39 Oo & O'o' 28.83 Os & O's' 128 19.22 Pp & P'p' 27.18 Pt & Ft' 136 18.05 Qq & Q'q' 25.52 Qu & Q'u' 144 16.88 Rr & R'r' 23.87 Rv & R'v' 152 11.71 Ss & S's' 16.71 Sv' & S'v 160 10.73 Tt & T't' 15.17 Tu' & T'u 168 9.75 Uu & U'u' 13.79 Ut' & U't 176 8.78 Vv & VV 12.41 Vs' & V'f 184' 5.37 V'r' & Vi 192 4.27 U'q' & Uq 200 3.17 T'p' & Tp 208 2.06 S'o' & So THE STRENGTH OF BRIDGES AND ROOFS. 153 Total compression in Aa and A'a' is 240 tons ; the strains in the end braces determined as in the pre- vious cases. 114. Other Compound Trus§es.— In the same manner equations may be deduced for any compound truss with vertical struts and inclined ties composed of any number of simple trusses, and so many examples have been given that the process can present no difficulties. 115. Simple Forms of Equations.— The difference be- tween the other moving-load equations in the last case and -^—(l — uf is very slight, and being always on the safe side, the latter form may be used in practice for all the simple trusses in this case, and all difficulty in the addition of equations thereby avoided. 154 A TREATISE ON CHAPTER V. TRUSSES WITH HORIZONTAL CHORDS, WITH STRUTS AND TIES OF EQUAL INCLINATIONS, SUBJECT TO CONSTANT AND TO MOVING LOADS. CASE I. — A SIMPLE TRUSS. A D G II c d e Fiff. 47. 116. Horizontal strains.— What is generally termed a Warren Girder, shown in Fig. 47, is an example of this description of truss, supporting the load upon the upper chord. In this case moments may be taken around the panel points of the upper chord to obtain the tension in « the members of the lower chord vertically beneath these points, and since the load may be considered as concen- trated at these points, we have the same conditions which in (28) gave us Eq. (14), H wx wx for the maximum horizontal strain. This equation can be applied only to the lower chord. For the strain in the upper chord, moments must be taken THE STRENGTH OF BRIDGES AND ROOFS. 155 around the panel points in the lower chord, since it is only at these points that a vertical section can be made which does not cut more than one- member (having a moment) subject to a horizontal strain. Let x' be the horizontal distance from one abutment to any one of the lower-chord panel points, w the full uniform load, and p the distance between the points ; and we can form the equation, tj/ __ wtf __ wx ! * _ wf n . -v U M sdV ' K } for the compressions in the members of the upper chords opposite the lower-chord panel points, distant x f from the abutment. 117. Vertical Strains from a Constant Load.— Eq. (14), which gives the lower-chord tension, gives also the com- pression in the upper chord at the point to which x is measured ; that is, the resultant compression from the strains in the braces and the chord-member on one side of the point. Similarly, Eq. (157) gives the tension at the points in the lower chord. Therefore, if the first be the strain at F, as it is at the same time in ef, and the second the strain at f, as it is at the same time in EG, the difference between the two will be the horizontal component of the strain in the brace Ff, and its vertical component may be obtained from the proportion of x— x' : d. Performing this operation, we have for the difference 156 A TREATISE ON between Eqs. (14) and (157), or the horizontal compo- nent of the strain in the brace, H - H = ld {x ~ x) ~ W^ x - x) HH + U Since p = 2(x — x'), we have from the proportion above, and as x = x' + — . V = ^-^?' ... (159) is the vertical component of the strain in Ff. If Eq. (14) next represent the strain at G, and be subtracted from Eq. (157), representing the strain at f, we shall again obtain V — w - wx ' 2 T' for the vertical component of the strain in fG. This equation is the same as Eq. (18), x 1 here being the same as w, that is the distance from the abutment to midway between the loaded points. 118. Vertical Strains from the Moving Load.— As this case comprises but a single system, one panel point can- not be fully loaded without a portion of the weight coming upon the next point, and thus rendering the conditions similar to those of the case in (54), and con- THE STRENGTH OF BRIDGES AND R sequently Eq. (37) applies wheji the truss is halt or more than half loaded, and Eq. (39) when less than half loaded. 119.— Example.— In Fig. 47, Let I = 80 feet, the length of the truss, d = 5 feet, the depth of the truss, jp = 10 feet, the distance between the panel points, in the same chord. w' = 80 tons, the weight of the full uniform movable load, w — 40 tons, the weight of the constant load. Substituting the values of these constants in Eqs. (14) and (157), we can form the following table of horizontal strains : Values of x. 10 20 30 40 Val ues of x'. 5 15 25 35 Strains in Tons. 105 180 225 240 52.5 142.5 202.5 232.5 Compression in AB& HI BC& GH CD& FG DE& EF Tension in ab & bc& cd & ef de For the maximum vertical strain we have Eq. (37), see (54), -r 7 to wu v =2— r+ "<* W Kb U 2/ 21(1 —p) ' 158 A TREATISE ON when the load covers l\alf and more than half the truss ; and Eq. (39), V w ~2 wu w V- U 7Z -T + 2(i - F y when the load covers less than half the truss. In these equations u is the distance from the abut- ment to a point midway between two panel points in the loaded chord, and the equations give the vertical components of the strains in the braces between these two points. As long as V has a plus value, that brace whose loaded chord end is nearer to the abutment from which u is measured is a tie, while that brace whose unloaded chord end is nearer is a strut. As the equations are applied to the different ends of the truss, it will be seen that some of the braces near the centre act sometimes as struts and sometimes as ties; these are termed counterbraces. * Substituting the values of the constants, and multiplying by 1.414 the secant of the angle of all the braces, Values of u. 5 15 25 35 45 Strains in Tons. 74.24 54.03 35.86 19.70 6.85 Compression in Ba&Hh Cb & Gg Dc&Ff Ed&Ee Fe & Dd Tension in Aa&Ih Bb&Hg Cc&Gf Dd&Fe Ee & Ed. THE STRENGTH OF BRIDGES AND EOOFS. 159 CASE II.— A DOUBLE TRUSS CONTAINING AN EVEN NUMBER OP PANELS. AB CDEFGH IK LMNOPQ R 120. — Let Fig. 48 represent a truss composed of the two simple trusses of Figs. (49) and (50). The first of which will be termed Simple Truss No. 1, since its braces meet at the centre, and the other Simple Truss No. 2. A B H KM Q R e i i Fig. 49. In these figures those braces which act solely as coun- terbraces are removed. E G N R a b d f h km o q r Fig. 50. In these trusses the vertical strains, or those strains having vertical components, are entirely independent of each other; and, as in the previous cases of com- 160 A TREATISE ON pound trusses, the horizontal strain in the compound truss is the sum of the strains in the simple trusses. 121. Horizontal Strain*. Let I = the length of the truss, d = the depth of the truss, %> = the length of a panel or chord-member. w = the full weight, uniformly distributed, a?, x' &c. = the distance from the abutment to any panel point. Each simple truss bears half the load which is upon the lower chord. ••■ H = = -=p • • • (16») is the compression in the upper chord opposite to the loaded points of Simple Truss No. 1, to which x is the distance, and tvx> wx^ w£_ m m m (161) is the compression in the upper chord opposite to the loaded points of Simple Truss No. 2, to which x' is the distance. The value of H in one simple truss at x, added to the value of IT in the other simple truss at a point nearer the centre by the value of p, will give the amount of horizontal strain in the upper chord of the double truss between x and x +J9, or in that member of the upper chord which is on the centre side of the point to which x is measured. THE STRENGTH OF BRIDGES AND ROOFS. 161 Making x f of Eq. (161) equal to x +p of Eq. (160), and adding the two equations, we have, and making x = — — z, we have, Here z is the distance from the centre of the truss to the abutment end, and z — S- consequently the distance to 25 the centre end of that upper-chord member whose strain is given, by H. If x of Eq. (160) be made equal to a' + p of Eq. (161), and the two equations added, we shall have the same result, or Eq. (163) will give the compression in all the members of the upper chord. In Fig. (49), tt WX WX* WV* /i/?i\ H= 4j-M-ii' • ■ (164) . is the tension in the lower chord opposite the upper- chord panel points to which x is measured. In Fig. 50, -rr, WX' WX ,% n aK \ H = 4rf-15T' " ' * (165) is the tension in the lower chord opposite the upper- chord panel points to which x' is measured. 11 162 A TREATISE ON The tension in this case, in one simple truss at x, added to the tension in the other simple truss, when x' = x +J9, will give the tension in that member of the double truss on the centre side of the point to which x is measured. Making x' of Eq. (165) equal to x +JP, and adding to Eq. (164), or making x of Eq. (164) equal to x' +p, and adding to Eq. (165), we obtain, H wl w / p\* Zwp ~Sd -m^ z —2) -mt> ' ■ • (166) for the tension in any member of the lower chord, where z — S- is the distance from the centre of the truss to the centre of the member. 122. Vertical SI rains from a Constant L.oad. — It will be seen from the last case that the constant-load vertical equations may be obtained from the simple-truss hori- zontal equations, as in the previous cases ; that is, from the difference in these horizontal equations applying to the same chord. In the last case the vertical equation was obtained from the difference in the horizontal strains at the two ends of the same brace ; but it would be the same had it been obtained from the difference in the upper-chord equation alone, at two ends of the same panel. The proportion was S- : d ; but if we take the horizontal strains at the two ends of a panel of the simple truss, their difference is just double the difference THE STRENGTH OF BRIDGES AND ROOFS. 163 at the two ends of a brace, because, between two weights or loaded panel points, the horizontal strains or hori- zontal components of the strains in the two braces are equal ; the proportion will therefore be p : d, whence we have for either simple truss, V-|-^, - - - - (167) in which u is the distance from the abutment to the centre of any loaded chord-member of the simple trusses. 123. Vertical Strains from a Moving Load. — Under a moving load the simple trusses present the same cases as Fig. 21 and Fig. 22 ; and Eq. (46) applies to that simple truss having a full end panel, or in this case to Fig. 49 ; and Eq. 45 to that simple truss having an end panel of half the length of its other panels, in this case to Fig. 50. Each of these is to be added to constant load equation (167). 124. Example. — Let Fig. 48 represent a truss in which I =160 feet, the length of the truss, d = 20 feet, the depth of the truss,

UV V wp (169) where z' is the distance from the centre to the panel points of Simple Truss No. 2 in the upper chord, and H' is the strain in that member on the abutment side of the point to which t! is measured. THE STRENGTH OF BRIDGES AND ROOFS. 167 In the lower chord, z being the distance to the panel points of No. 1 in the lower chord, and H the tension in the member on the centre side of the point to which z is measured; and z being the distance to the panel points of No. 2, and H' the tension in the member on the centre side of the point to which z* is measured. In all these equations w is the maximum uniform load, or equal to w' -\-w of the examples previously given. 126. Vertical Strains from a Constant Load. — The vertical equation for a constant load upon No. 1 is, v -f-»-f- • • ■ < 1!2 > and for the constant load upon No. 2, 7-t-wP-iY ■ ■ ■ (m > 127. Vertical Strains from a Moving Load. — The ver- tical equations for the moving load are the same as in the last case, and their application is governed by the same principles which have been explained in the previous examples of double trusses. An example of the application of the equations belonging to this case is unnecessary. 168 A TREATISE ON CASE IV. — A QUADRUPLE TRUSS CONTAINING AN EVEN NUMBER OF PANELS. ABODE FGH I KLMNOPQ, RSTUVWXYZ Fig. 52. 128. — Omitting the braces which are used solely as counterbraces, and numbering as before, we have the four simple trusses shown in Fig. (53). 32143214321 12341234123 14321432143212341234123 41 Fig. 53. 129. Horizontal Strains. — In this figure the weight borne by each truss is --, the load being upon the lower chord. By reasoning as in the previous case of a quadruple truss, we obtain, for the upper chord of Simple Truss No. 1, TT _ WX WX Mm'- ■ • < 174 > x being the distance to a point in the lower chord. For No. 2, TT, _ wx' wx" pwx' Sp'w n „ K , THE STRENGTH OF BRIDGES AND ROOFS. 169 For No. 3, H --W m + M m (176) For No. 4, 8d r " 8<£ " 4dZ 8cB'" V ; An inspection of this compound truss will show that the compression in any member of the upper chord, QR for example, is the compression in that simple truss, to whose point in the lower chord R is opposite, or Simple Truss No. 1 at a?, added to the strains in No. 4 at x +■ p, in No. 3 at x + 2p, and in No. 2 at x — p. Making these changes in the values of a/, #", and x 1 ", and then adding the equations, we obtain, after making x= l --z, H 2 is here the distance from the centre to the panel points in the upper chord of Simple Truss No. 3, and H is the strain in the member on the centre side of that point ; and, similarly, the same equation will apply when z is made the distance to any panel point in the upper chord of No. 4. By a similar process we obtain for the remaining members of the upper chord, %' being the distance from the centre to any upper- 170 A TREATISE ON chord point of No. 1 and No. 2 ; H' being the compres- sion on the centre side of that point. In the lower chord, T T_ivl w t ' p) 7 15p*iv nftm H -8d-2dl {Z --2l— 8dT " " (180) is the tension in those members on the centre side of the points in the same chord of Simple Trusses No. 3 and No. 4, and Sd 2dl K 21 8 in the lower chord ; and, H _ Spw p 7 w m - - (189) ■i i. ~~~8d idV in the upper chord, when No. 1 has a l lower-chord point distant p from the abutment. H _ Spw Sp'w - - (190) in the lower chord ; and, H _ Spw . = ~8d~~ Y p'tv • • - (191) 172. A TKEATISE ON in the upper chord, when No. 2 has a lower-chord point distant p from the abutment. 3^;_7pV m m m (192) 8d 4dl J y ' in the lower chord ; and, H = 8g! + 2g» . . . (193) Sd Adl in the upper chord, when No. 3 has a lower-chord point distant p from the abutment. 132. Vertical Straine from a Moving Load. — The equa- tions determined in the case of the quadruple truss with vertical struts and inclined ties are equally applicable to this truss, and their application is governed by the same principles. 133. Example. In Fig. 52, Let I = 288 feet, the length of the truss, d — 24 feet, the depth of the truss, p— 12 feet, the length of a panel or chord-mem- ber, v/= 288 tons, the weight of a full movable load, w = 144 tons, the constant-truss weight. The load is upon the lower chord. For the chord-members we use Eqs. (178), (179), (180), and (181), except for the ends, where Eqs. (186) and (187) ; w of these equations being equal to uf + w, THE STEENGTH OF BRIDGES AND ROOFS. 173 or 432 tons ; whence we can form the following table : Values of 2 of Eqs. (179) & (180). Values of 2 of Eqs. (178) & (181). Strains in Tons. Compression in Strains in Tons. Tension in 12 639 mn & no 24 648 LM, MN, NO &0P 621 lm & op 36 630 KL&PQ 603 kl & pq 48 594 IK&QR 585 ik & qr 60 558 HI&RS 549 hi & rs 72 522 GH&ST 495 gh & st 84 468 FG&TU 441 fg & tu • 96 396 EF&UV 387 ef & uv 108 324 DE&VW 315 de & vw 120 252 CD&WX 225 cd & wx 132 162 BC&XY 135 be & xy 81 AB&YZ 72 ab &yz For the strains in the braces we have, for Simple Truss No. 4, Eq. (131) + Eq. (183), For No. 3, Eq. (132) + Eq. (184), For No. 2, Eq. (133) + Eq. (185), and For No. 1, Eq. (134) + Eq. (1§2). 174 A TREATISE ON Substituting the values of the constants, and multi- plying for the end braces by 1.118, and for all the others by 1.414, we have the following table: Values of u in Eq. (133) & Eq. (185). Values of u in Eq. (132) & Eq. (184). Values of u in Eq. (131) & Eq. (184). Values of u in Eq. (134) & Eq. (182). Strains in Tons. Com- pression in Tension in —12 60.38 Ab&Zy Ba &Yz 76.36 Ac&Zx 12 76.36 Bd&Yw 24 62.98 Ce & Xv Ca&Xz 36 53.03 Df&Wu Db&Wy 48 52.32 Eg&Vt Ec & Vx 60 48.78 Fh &Us Fd&Uw 72 40.57 Gi&Tr Ge & Tv 84 33.94 Hk& Sq Hf &Su 96 81.11 Il&Rp Ig&Rt 108 . 28.28 Km&Qo Kh&Qs 120 20.77 Ln& Pn Li&Pr 132 14.85 Mo.& Om Mk&Oq 144 12.73 Np&Nl Nl&Np 156 10.61 Oq&Mk Om&Mo 168 3.8 Pr&Li Pn & Ln Here Ln, Mo, Np, Oq, Pr, Pn, Om, Nl, Mk, and Li are subject to both tension and compression, or act as counterbraces. The compression in the end posts is the vertical strain in zy and zx, or 108 tons. THE STEENGTH OF BRIDGES AND ROOFS. 175 134. — It is not deemed necessary to give further ex- amples of compound trusses with equally inclined struts and ties, the principles governing the determination of equations in any case having been sufficiently illustrated. 176 A TREATISE ON CHAPTER VI. TRUSSES WITH HORIZONTAL CHORDS AND INCLINED BRACES, THE TIES HAVING A DIFFERENT INCLINATION FROM THE CASE I.— A DOUBLE TRUSS WHOSE UPPER CHORD IS DIVIDED INTO AN EVEN NUMBER OF PANELS OR MEMBERS. 135. — There is but one truss of the character described in the heading of this chapter in general use, the Post Truss, Fig. 54, where the struts have a horizontal extent of half that of the ties, the extent of the latter being equal to the depth of the truss. 212121212121212121212121212 A B C D E F G H I KLMNON'M'L'K'l'H'G'FE'D'C'B'A' / b c d e f g li i k 1 m n o o' n' m' 1' k' i' h' s' f e' d' c' b Wk y l 2 1 2 12 1 2 1 2 1 2 1 1 2 1 21 2 12 12 1 2 1 I Fig. 54. Under the action of the full load the truss may be resolved into two simple trusses, one of which is shown in Fig. 55, and termed, as in the previous cases, since its braces meet at the centre, No. 1. A B D F H K M M' K' H' F' D' B' A' V\AA/\/\/\ AAAAAM |] b d f h £ m o~? m' k' h' f d' b'a' l| Fig. 55. THE STRENGTH OF BRIDGES AND ROOFS. 177 The other, shown in Fig. 56, is termed No. 2. A C E G I L L' I' G' E' C A ' K/\/\/\/\/\ AA/W\/I a c e g i 1 n n' 1' i' g' e' c' Fig.. 56. In both the counterbraces are removed. 136. Horizontal strains. — Under a full load upon the lower chord, Simple Truss No. 1, having the panel point next the abutment, will be considered as bearing the load which comes directly upon the abutment, and which, since the point is only half a panel length from the abutment, is one-fourth a panel load. On the next to the abutment panel point rests a half panel load from one side, and a fourth panel load from the other, making upon it in all three-fourths a panel load. Let I = the length of the truss, d == the depth of the truss, p = the length of a panel or any upper-chord mem- ber. The load, w, is upon the lower-chord. w wp Simple Truss No. 1 bears — + -^, and No. 2, — — -j- in this example. Let x be the distance from the abutment to any point in the lower chord of No. 1 ; then the moment of the reaction of the abutment is — £-, the moment 4 21 \ of the load directly upon the abutment, ^-, of the load 12 178 A TREATISE ON on the next panel point, -~f(? — %\ and on tne re " mainder of the truss i of ™(x — J~\, (x— £ ), whence we have, 2 u= ^_wx pwx ,P™ . . . . (194) for the compressions in the upper-chord members oppo- site the panel points in the lower chord to which x is measured. Taking moments around the panel points in the upper chord of No. 1, distant x f from the abutment, we obtain for the tensions in the lower-chord members of this simple truss opposite these points, -TT, _ wx' wx'* ptvx 1 p'w (AQr\ z ~Id~~W~ i ~~IdT~Ml' " { ° ; From moments around the lower-chord points of Simple Truss No. 2, distant x" from the abutment, the compressions in the upper-chord members opposite these points are given by ' Ad"~Ul ~4dl + 16dl' ' " ^ J And for the lower-chord members of the same truss, x'" being the distance to the upper-chord panel points, ~^d~~idr + idf ' (m) The strain in the members of the upper chord of the double truss is found by making K'i of Eq. jjj^j THE STRENGTH OF BRIDGES AND ROOFS. 179 equal to •] ,/7 ; [, and adding the equation so changed to Eq. lioftN whence we have for the compressions in those members opposite the lower-chord points of No. I 1, x having been made equal to H -T^~M ( ^2J~8jr " (198) And for the compressions in those members opposite the lower-chord points of No. 2, x' f having been made equal to -- — z\ % and z' being respectively the distances from the centre of the truss to the lower-chord panel points of Simple Trusses No. 1 and No. 2. The tension in the lower chord of the double truss is found by making j£ I of Eq. j x J j equal to j^, + £ j , (195) and adding the equation so changed to Eq. jiq^r; whence the strain opposite the upper-chord point of No. 1 is, making x f = -- — z, B. = ™ l -™(z+£)\ - - (200) 180 A TREATISE ON and opposite the upper-chord points of No. 2, making H = ^-^+£)'-^. - (201) 8J 2dP ^2 J 2dl K } 137. Vertical Strains from a full Load. — As in the Other compound trusses under a full load, the vertical strains here, in the simple trusses, are independent of each other, and may be obtained from the upper-chord sim- ple-truss horizontal equations, and are, v = f-I-(-f). ■ • • ' + f )• < 206 > is, as long as it has a plus value, a strain passing to the abutment from which u is measured. When it has a minus value it passes to the other abutment and ceases to be of use, for it indicates a less strain than when the load extends to the same point from the other abut- ment. The value of v! differs from that of u in this man- ner : Suppose the segment a to m, inclusive, to be loaded as before, then v! of the latter part of the equa- tion is the distance from the right abutment to n, or the centre of a panel of a simple truss, but u is the distance from the same abutment to the centre of a panel of the double truss, in this case, to midway between m and n ; hence u = v! + ^, and the equation becomes V = g< <*-«)<-£) +f-^. - (207) This equation extends from that point where it first has a plus value to the centre, but no farther. When the load covers half the truss, the vertical strain to the unloaded abutment may take one of three courses, as explained before. First, suppose it all to pass from the point o to 184 A TREATISE ON N', or o' to N, then since ^ = — , Eq. (204) becomes Y=^-^, --- - (208) and this is the vertical strain throughout the braces, of Simple Truss No. 2 ; then, as the load passes on, and the successive points of this simple truss become -Loaded, we have, for the load upon them, within the space I — u, £ «'-*>• -£-£)' - - (209) adding this to Eq. (208), we have, w 4? Again, adding to Eq. (203), we have, for the total vertical strain, or vertical component of the strain, in the braces of Simple Truss No. 2, when the load covers more than half the truss. When the truss is half loaded, the strain from all except the point next the centre may pass through the braces of No. 1 to the unloaded abutment, or all of Eq. (204), when u = _ + p, which is therefore, 2 ^#,/ v _w' pw'pw . y ~T + W + ~2T ' ' (212) THE STRENGTH OF BRIDGES AND ROOFS. 185 The load on the panel points of No. 1, in the space I 2" — ^ 1S > J /!«'/># n,*nn' nntnn' v -i6~ir + ^r _ i6?-" " (213) Adding these and Eq. (202), we have, for the maximum vertical strains in the braces of No. 1. 139. Example. — In Fig. 54, Let I =312 feet, the length of the truss, d = 24 feet, the depth of the truss, p = 12 feet, the length of a panel, **/= 312 tons, the weight of a full movable load, w = 156 tons, the weight of the truss. For the horizontal chord strains, we use Eqs. (198), <199), (200), and (201). Substituting the values of the above constants, we can form the following table of strains : 186 A TREATISE ON" Values of z in Eq. (198). Values of z in Eq. (199). Strains in Tons. Compres- sion in Values of z in Eq. (200). Values of z in Eq. (201). Strains in Tons. Tension in 6 759.375 XO & ON' 759.375 oo' 18 759.375 MN& N'M' 12 745.875 no & o'n' 30 741.375 LM & M'L' 24 732.375 ran & n'm' 42 723.375 KL & L'K 36 700.875 lm & ml' 54 687.375 IK & KT 48 669.375 kl & l'k' 66 651.375 HI & I'H' 60 619.875 ik & k'i' 78 597.375 GH & H'G' 72 570.375 hi & i'h' 90 543.375 FG & G'F 84 502.875 gh & h'g' 102 471.375 EF & F'E' 96 435.375 tg & g'i' 114 399.375 DE & E'D' 108 349.875 ef & f e' 12G 309.375 CD & D'C 120 264.375 de & e'd" 138 219.375 BC & C'B' 132 160.875 cd & d'c' 150 111.375 AB & BA' 144 57.375 be & c'b' For the vertical strains, or strains having vertical components, we use Eq. (20G) for the counterbraces when the load covers less than half the truss; Eq. (214) for the braces of No. 1, and Eq. (210) for the braces of No. 2, when the truss is more than half loaded. Each THE STRENGTH OF BRIDGES AND ROOFS. 187 brace must be proportioned for the greatest strain which can come upon it in any case. The vertical strain is to be multiplied by 1.031 for the struts, and 1.25 for the ties, these being the secants of their angles. The follow- ing is a table of the strains in the braces : Values of u in Eq. (214). Values of u in Eq. (211). Values of u in Eq. (206). Strains in Tons. Compres- sion in Strains in Tons. Tension in 125.27 Ab & A'b' 6 164.93 Ac & A'c' 18 124.70 Bb & B'b' 151.19 Bd & B'd' 30 124.37 Cc & C'c' 150.79 Ce & C'e' 42 107.33 Dd & D'd' 125.13 Df & DT 54 98.20 Ee & E'e' 119.06 Eg & E'g' 66 90.91 Ff & F'f 110.22 Fh & F'h' 78 82.26 Qg & G'g' 99.74 Gi & G'i' 90 75.37 Hh & H'h' 91.38 Hk & H'k' 102 67.27 Ii & IT 81.56 11 & IT 114 60.93 Kk & K'k' 73.88 Km&K'm' 126 53.23 LI & LT 64.54 Ln & L'n' 138 47.37 Mm&M'm 57.44 Mo & M'o' 156 40.16 Nn & N'n' 48.71 No' & N'o 168 31.44 On' & On 180 19.83 N'm' & Nm 192 6.27 Ml' & Ml The compression in the end struts is 229.5 tons. 188 A TREATISE ON The ambiguity caused by the arrangement of the struts at the centre and the counterbraces may be en- tirely avoided, and the weight divided between the two simple trusses, by the arrangement shown in the follow- ing case : CASE • II.— A DOUBLE TRUSS WHOSE UPPER CHORD IS DIVIDED INTO AN ODD NUMBER OF PANELS OR MEM- BERS. 140.— Let the Post Truss, described in the previous case, be opened at the centre, so that the upper ends of the centre struts are one panel length apart, and we have, adding the ties, the truss shown in Fig. (57). A BODKFOHTKLMNOo' n'm'i/ k' i'h'q' t' e'd' c' m'a' 1 212121212121221212121212121 T7T 212121212121212 12 12 12 12 1 21 2 [H bode f gb. ik lrnnop o'n'iriT k'i'k'g'f e'd/c'b' , Fig. 57. The struts and ties have the same inclinations as in the previous case. 141. Horizontal Strains.— Under the action of a full uniform load this truss may be resolved, as in previous cases, into two simple trusses, whose separate panel points are shown in the figure by the numbers 1, 1, etc., and 2, 2, etc. Let I = the length of the truss, d = the depth of the truss, p = the length of a panel, w = the weight of a full uniform load. THE STRENGTH OF BRIDGES AND ROOFS. 189 The load is upon the lower chord. As before, the quarter panel load, directly upon the abutment, is con- sidered as belonging to that simple truss which has the panel point next the abutment. Therefore, in this example No. 1 bears — — — v, and No. 2 bears — + -£-. Hence, taking moments around any points of No. 1 in the lower chord, distant x from the abutment, we obtain for the strain in the upper chord opposite this point, wx wx* Zp'w and for the lower chord, opposite any upper-chord panel point distant x' from the abutment, wa/ wx" H -1J-"I5T- ■ ■ ■ (216 > From moments around the points of No. 2 we have for the upper-chord member opposite any lower-chord point, distant x" from the abutment, H -^--m + TtdV ' ' ' (217) and for the lower-chord member, opposite any upper- chord point distant x'" from the abutment, w „OsC ™*!1 &*'.'.-: (218) These are the strains in the chords of the simple trusses under their separate loads. The strain in the upper chord of the double truss, opposite any lower- 190 A TREATISE OK chord panel point, is the strain in the upper chord of the simple truss at the same point added to the strain in the other simple truss at the next panel point towards the centre ; whence we obtain but one equation for the upper chord : Hss -sz-iaC— ») + w - - ( 219 > z being = — — #, or the distance from the centre of the truss to the lower-chord panel point, opposite to which is the member whose strain is given by the equation, and is, therefore, the distance to the centre of that member. In the lower chord of the double truss the strain at any point is the simple-truss strain at that point added to the other simple-truss strain at the next panel to- wards the abutment, whence we obtair for any member of the lower chord, H - sd-m( z+ i)- jip (220) 2/ being = — — a?, or the distance from the centre of the truss to the centre of any lower-chord member whose strain is given by the equation. 142. Vertical Strain from a Constant Load.— This, as before, is obtained from the simple-truss horizontal equations, and is for Simple Truss No. 1, TT _ w wu 4 'IP ( 221 > where u is the distance from the abutment to any point THE STKENGTH OF BKIDGES AND ROOFS. 191 midway between the lower-chord points of No. 1 ; and for No. 2 it is, V = j— W .... (222) where vl is the distance from the abutment to any point midway between the lower-chord points of No. 2. 143. Vertical Strains from a Moving Load. — If we consider the truss as half loaded, it will be seen from the figure that the portion of the weight borne by the unloaded abutment, excepting that upon the central panel point, passes from the end of the load to that abutment solely through the braces of Simple Truss No. 2. If less than half the truss be loaded, the load ex- tending from one end, the weight borne by the unloaded abutment passes by means of the counterbraces from one simple truss to the other until it reaches the centre, beyond which it is confined to No. 2, as before. In the case of a less than half load, the conditions are similar to the previously- described Post Truss, and Eq. (204), _. w' v» ->•-?) gives the vertical strain passing to the abutment from which u is measured, the minimum value of u being ._ + P and w f being the weight of the full load. To this strain must be added the equation of the con- 192 A TEEATISE ON (221) (222) stant-truss strain. This is Eq. j %<%* f a( ^ed to -^ / 221 1 when j M 'i is made equal to -K i^|- Whence, V = |-|«)' + |); - -_.- (223) and since w' + -2. of this equation is the distance to the same point as u of Eq. (204), we have, adding these equations, v =S(('-«)'-f)+l-x- " < 224) It must be borne in mind that u, in this equation, is the distance from the abutment to the centre of a panel of the double truss, and that its least value is — + — , and that the equation is only needed as long as V has a plus value. When the load extends from one abutment beyond the centre, we have for the strains on the braces of No. 2, the strain of Eq. (204), when u =- + -£-, or, Then, as the successive points of No. 2 become loaded, we have, for the proportion of weight passing to the unloaded abutment from which u' is measured, THE STRENGTH OF BRIDGES AND EOOFS. 193 which is to be added to Eq. (225), and to the constant- truss strain of No. 2, Eq. (222)j and we have, V - w ' a _ tt V + w ' _ W 'P 4. w L «** . f2 2 to V-_(< "O+jg -^ + r "2!", (226) for the maximum strain upon the braces of No. 2, in that panel whose centre is distant u' from the abutment. Until the moving load, extending from one abut- ment, reaches the central panel point, no vertical strain from it comes upon the braces of No. 1 in the unloaded half ; the greatest strain upon these braces from the cen- tral point is when it has a panel load upon it. As the successive points towards the unloaded abutment come under the load, we have for the proportion of weight borne by this abutment and the greatest vertical strain from the movable load in the panel of No. 1 to which u is measured, and adding this to Eq. (221), we have, for the maximum strain in No. 1. There is a portion of the load on No. 1, between the centre and the end of the load, that passes to the other abutment, or the one from which the load extends, the strain of which is taken back to the centre by the coun- terbraces. 13 194 A TREATISE ON 144. Example. — In Fig. (57), Let I = 324 feet, the length of the truss, d = 24 feet, the depth of the truss, p = 12 feet, the length of a panel, w' = 324 tons, the weight of a full movable load, w = 162 tons, the weight of the uniform permanent load. 145. Horizontal strains. — Substitute the values of the constants in Eqs. (219) and (220), w of these equations being w' + w, or 486 tons, and we have the following table of strains : Values of 2 in Eq. (219). Strains in Tons. Compres- sion in Values of z' in Eq. (220). Strains in Tons. Tension in 820.125 00' 6 813.375 op & po' 12 820.125 NO & NO' 18 779.875 no & o'n' 24 811.125 MN & M'N' 30 777.375 mn & n'm' 86 793.125 LM & L'M' 42 745.875 lm & ml' 48 766.125 KL & K'L' 54 705.375 kl & l'k' GO 730.125 IK & IK' 66 655.875 ik & k i' 72 685.125 HI & HI' 78 596.375 hi & ill' 84 631.125 GH & G'H' 10 529.875 gh & h'g' 06 568.125 FG & F G' 102 453.375 fg & g'f ' 108 496.125 EF & E'F 114 367.875 ef & f'e' 120 415.125 DE & D E' 126 273.375 de & e'd' 133 325.125 CD & CD' 138 171.875 cd & d'e' 144 226.125 BC & B'C 150 57.375 be & c'b' 156 118.125 AB & A'B' STBENGTE ES AND KOOflfe, Ca 195 146. Vertical strains— Substitute the values of the constants in Eqs. (228), (226), and (224), and multiplying the results by 1.25 for the ties, and 1.031 for the struts, we have the following table of strains in the braces : Values of u in Eq. (224). 168 180 192 204 Strains in Tons. » 23.67 14.36 1.20 Compres- sion in Oo& O'o' Nn& N'n' Mm& M'm' Strains in Tons. 43.13 28.70 17.41 1.45 Tension in oO & o'O nO& n'O' mN& m'V 1M& I'M' Values of v in Eq. (226). 150 126 102 78 54 30 6 Strains in Tons. 9.28 22.57 36.78 51.89 67.93 84.88 Compres- sion in Nn& N'n' L1& 1/1' Ii & 11' Gg& G'g' Ee& EV Cc& C'c' Strains in Tons. 11.25 27.36 44.59 62.91 82.36 102.91 124.59 Tension in Np& Np Ln& L'n' 11 & IT Gi& GT Eg & E'g' Ce& C'e' Ac & A'c' Values of v in Eq. (228). 138 114 90 66 42 18 Strains in Tons. 51.49 65.24 79.90 95.47 111.98 129.39 Compres- sion in Mm & M'm' Kk& K'k' Hh& Hh' Ff& F'f Dd& D'd' Bb& B'b' : Strains in Tons. 62.43 79.10 96.88 115.75 135.76 156.88 125.27 Tension in Mo& M'o' Km& K'm' Hk& H'k' Fh& F'h' Df & BY Bd & B'd' Ab& A'b' 196 A TREATISE ON Total compression on the end struts, 238.5 tons. The compression on Nn and N'n' is greater in the second equation than in the first. 147.— This system of bracing, like the others, may be extended to compound trusses containing any number of simple trusses, for which equations may be similarly found. THE STRENGTH OP BRIDGES AND ROOFS. 197 CHAPTER VII. INCLINED TRUSSES OR RAFTERS. 148. — The same principles govern the investigation of strains in inclined as in horizontal trusses, and mo- ments are to be taken and equations formed in the same manner. The inclined truss thrusts horizontally as well as vertically against its abutments or walls, or the latter may be considered as having a horizontal reaction in- ward as well as a vertical reaction upward. Fig. 58. 149. Horizontal Reaction. — Let AB and AC, Fig. 58, represent a pair of inclined trusses or rafters, joined together at the apex and to the walls by one chord only. Let I = the horizontal distance, BC, between the walls, V = the length of each rafter, h = the height of the apex, A, above the walls, w=the whole weight, uniformly distributed over the two rafters. 198 A TKEATISE ON The rafter to the right of A is kept in equilibrium by the vertical reaction of the wall, — , the weight on the rafter, — , and the inward reaction of the wall ; hence, taking moments around A, we have for the moment of the latter, the difference between the moments of the other forces, or, 2^2 24' vH =f^ (229) is the horizontal reaction of either wall. Taking moments around D, in the line of the tops of the walls, it will be found that the horizontal thrust at A is the same. 150— Had the rafters been joined by both chords at A, it would be impossible to determine the proportion of strain to which each chord would then become sub- ject. In such a case one rafter could not bear an equal or certain fixed proportion of strain to the other unless the mechanical construction possessed a theoretical pre- cision almost unattainable in practice; and even were this done, the different expansions and contractions of the two chords would immediately affect the amount of strain upon each. All ambiguity of strains in trusses must, where prac- ticable, be avoided, and we shall therefore consider the rafters as joined together by one chord only. THE STRENGTH OF BRIDGES AND EOOFS. 199 151. The proportion between the Horizontal and Ver- tical Reactions of the Wall. — To find the proportion between the horizontal and vertical reactions of the wall, we have, — : •__ :: h : — 8A 2 4 Draw, in Fig. 58, the vertical line BE = h, and the horizontal line EF = — . If BE represent the amount of vertical reaction, or — , then EF will represent the horizontal reaction, and this proportion will hold good for any value of w, great or small. The resultant force of BE and EF is represented by the line BF. This line gives the direction and amount of thrust of the rafter, and BE and EF are the vertical and horizontal components of this thrust. It will be noticed that the resultant thrust is not in the line of the rafter, and that its direction depends on the inclination of the rafter and not on w. 152. Longitudinal Reaction. — Instead of horizontal and vertical components, this thrust of the rafter, repre- sented by the line BF, may be resolved into two other components, one shown by BA in the line of the rafter, and the other by GF at right angles to it. The first of these we shall term the Longitudinal Reaction, and the other the Transverse Reaction, the value of the former being found as follows : Produce the lines EF and BG 200 A TEEATISE ON until they meet at I, and from F draw FG parallel to BI ; therefore FI = FE = -, and the triangle FGI is 4 similar to the triangle ADB, and L ■■!• L 2 *" 4 ' 81 V :I ::i: JL = the length of GI. Then, 8Z 7 "! ' 2A 16M 7 ' and wl' _ wV == wZ' w ,,„_ -f^^wV wh 2h 16A? " 2A 4AZ' 1 ' 4F" 1 " IF or, L = ^ + ^, . . . /230^ 4A 4f * ^ is the longitudinal reaction of the wall or abutment. 153. Transverse Reaction.— The transverse reaction is found by a similar process. l ' : h :: T : TjJ - tlie distance GF, 4 4r ' and ^ . M .. w 9 wl % ' 4f " 2" " 8F ; or, rry _ wl 8P ----- (231) is the transverse reaction of the wall. THE STKENGTH OF BKIDGES AND HOOFS. 201 154. Truss strains.— Let Fig. 59 represent a pair of rafters, Fig. 59. or inclined trusses, joined to each other and to the wall by the upper chord, and divided into uniform panels by the braces. Let w = the weight, uniformly distributed over the two rafters. I = the distance between the walls or lower ends of the rafters. I' = the length of each rafter between the points of its connection. h = the height of the apex above the supports, or the height of one end of a rafter above the other. d = the depth of the rafters, or the distance between the chords at right angles to the lengths. x = the distance along the upper chord to any panel point from the point where it is joined to the abutment. L = the longitudinal chord strain. T == the transverse strain. The weight is considered as concentrated at the 202 A TREATISE ON upper-chord panel points— the abutment bearing half a panel load. 155. Longitudinal Strains. — If we take moments around any panel point in the upper chord distant x from the wall, the moment of the longitudinal strain in the lower chord, in the same transverse section, is Lx^ ; and it is equal to the moment of the transverse reaction of the abutment, less the moment of the transverse com- ponent of the load on x : the longitudinal reaction of the abutment, and the longitudinal component of the weight on a?, passing through the point about which mo- ments are taken, have no moment. The weight on x is - - X os; and if it be represented Jib by the line ah in Fig. 59, be will be its transverse, and ac its longitudinal component, and the triangle abc will be similar to the triangle ACD ; hence we obtain the former component from the proportion I .. wx . wlx V : 2 " 21' ' 4 206 A TREATISE ON V This strain becomes zero when u = — , and increases towards either end ; but when u becomes greater than — , T has a minus value, indicating a strain passing to the opposite end to that from which u is measured. 157.— At either end of the rafter, T = W' <»!) At the upper end, where x = l\ Eq. (232) becomes •zero, and Eq. (234), putting V for x — -2., becomes L -£-£- ■ ■ ■ -m Hence we have two strains at the apex, one, Eq. <238), thrusting upward and against the other rafter ; the other thrusting downward and also against the other rafter. The horizontal component of Eq. (238) is found from the proportion, l> . l_ . . wV __ wh t wl __ whl # ' 2 " 4A " IF" ! 8l 87 1 and of Eq. (237) from the proportion, V • h ' ' w ^ • w ^ adding, we have, u= id_wM whl_wl 8h 8/" ^ 8F ~ 81' ' ' ( ' total horizontal strain at the apex. THE STRENGTH OF BRIDGES AND ROOFS. 207 The vertical component of Eq. (238) is from the pro- 7 , 7 wl' wh w ivli* xl T portion, v \ h \\ — - — — - : — — — -, a thrust upward. The vertical component of Eq. (237) is from the pro- -,, I wl wV portion, V : — : : — - : — _ , and since — = o — h*. = — — -— -, 4 ' 16f 4 IP a thrust downward, equal in amount to the other and neutralizing it, leaving only a horizontal strain at the apex. 158. Example. — Let Fig. 59 represent a pair of rafter trusses ; Let Z = 150 feet, the distance between the lower ends of the rafters, V = 80 feet, the length of either rafter, h = 27.84 feet, the height of the apex above the wall, p == 5 feet, the length of a panel, d— 2.5 feet, the transverse depth of either truss, w— 20 tons, the weight, distributed uniformly over the trusses. 159. Longitudinal Strains. — For the upper-chord strains substitute values in Eq. (233), the first value of z being zero, and in Eq. (234), and add the strains so found ; for the lower chord use only Eq. (233), the first value of % being 5. In the table of chord-strains below, the strains for the upper chord, from the two equations, are given separately and then added. 208 A TREATISE ON Values of 2 in Eq. (233). Strains in Tons. Tension in Com- pression in Values of x in Eq. (234). Strains in Tons. Total upper- chord com- pression. 35 8.79 St TC 5 16.00 24.79 30 16.41 rs ST 10 15.78 32.19 25 22.86 qr RS 15 15.57 38.43 20 28.12 pq QR 20 15.35 43.47 15 32.20 op PQ 25 15.13 47.33 10 35.16 no OP 30 14.91 50.07 5 36.91 mn NO 35 14.70 50.61 37.5 MN 40 14.48 51.98 37.5 LM 45 14.26 51.76 5 36.91 Ira KL 50 14.04 50.95 10 35.16 kl IK 55 13.82 48.98 15 32.20 ik HI 60 13.60 45.80 20 28.12 hi GH 65 13.39 41.51 25 22.86 ga FG 70 18.17 36.03 30 16.41 tg EP 75 12.95 29.36 35 8.79 ef AE 80 12.73 21.52 THE STRENGTH OF BRIDGES AND ROOFS. 209 160. Transverse strains. — The load being concentrated upon the upper-chord points, the strut belongs to that point to which u is measured ; Eq. (236) gives the strains in the struts, and multiplied by 2.236, the secant of the angle made by the ties with the struts, gives the strains in the ties ; whence, by substituting the values given above, we can form the following table : Values of u. 2.5 7.5 12.5 17.5 22.5 27.5 32.5 37.5 Strains in Tons. 4.39 3.81 3.22 2.64 2.05 1.46 .88 .58 Compression in Tt& Ee Ss& Ff Rr& Qq & Hh Pp& Ii Oo& Kk Nn& LI Mm Strains in Tons. 9.82 8.52 7.20 5.90 4.58 3.26 1.97 .65 Tension in Ae & tc Ef & sT Fg& rS Gh& qR Hi& pQ Ik& oP K1& nO Lm & mN 161. An Inclined Truss whose Struts and Ties make the same Angle with the Chord.— If the truss shown in Fig. (48) be used as a rafter, the longitudinal and transverse reactions of the wall remain the same ; the longitudinal component of the weight on any segment remains the same, and Eq. (234) will apply without change; the moments for the lower chord are taken similarly, so that Eq. (232) holds for that also. But Eq. (232) is adapted, in this case, to the upper chord, because x must be meas- ured to the lower-chord panel points, instead of to the loaded points. 14 210 A TREATISE ON The moment of the weight on the segment x' is in this case ^H-^ and its transverse component is given by the proportion, ;/ . I . . wx' _ wp 1 . ivlx* . wlp* m 'I'' Id? 4dF ' 8dT "*" 8dF ] therefore Eq. (232) for the upper chord becomes T whf wlx'* wlp /o^a\ x' being the distances to the lower-chord points. In this case vertical equation (236) applies without change, u being the distance to the centre of any upper- chord panel member. The struts and ties are distin- guished by their inclinations, and are the same as in the same truss placed horizontally. THE STRENGTH OF BRIDGES AND ROOFS. 211 CHAPTER VIII. TRIANGULAR TRUSSES. 162. — Triangular Trusses are those in which one horizontal chord forms the base of an isosceles triangle, the remaining two sides acting as the other chord, with braces between. They are generally used to support roofs, but also possess many advantages for bridge pur- poses. 163. — Let AB and BC, Fig. 60, represent two rafters whose thrust at their supports is taken by the hori- zontal chord BC, and connected to the chord by braces, omitted in the figure. ABC is a triangular truss, AB and AC acting as one chord, the strains in which will be termed the longi- tudinal strains. 212 A TKKATISE ON 164. Longitudinal Strains.— Let w = the weight, uniformly distributed over the chord struts, AB and BC, I = the horizontal distance between the abutments, B and C, V = the length of either chord strut, d — the height of the apex above the abutments, or the depth of the truss at the centre, x — the horizontal distance from one of the abut- ments to any point between that abutment and the centre, H = the horizontal strain in the horizontal chord, L = the longitudinal strain in the chord struts, V = the vertical strain, or vertical component of the strain affecting the braces. Disregarding for the present the braces, and taking moments around a, in a vertical section cutting no in- clined brace, distant x from the abutment C, that part of AC to the right of ab is kept in equilibrium by the reaction of the right abutment, — , the load on be, — — , and the longitudinal strain at /. Hence L, the longitudinal strain at fr, multiplied by ac, the distance from a at right angles to its direction; T w wx wx % -L x ac = 2 2 From similar triangles, V : d :: x : ac, or ac — ' , V and THE STRENGTH OF BRIDGES AND ROOFS. 213 Substituting, we have, T dx _ wx wx* j^^wtwVx . . . . (241) From the shape of this truss it is evident that the maximum value of x is — , and that its limits are zero 2' and — ; when it is zero, L =^> W the longitudinal strain at the abutment, whose vertical component, found from the proportion V : d, is — ; or at the abutment the chord strut contains the whole verti- cal strain. "When x — --, L=|* - - - - - - (243) and its horizontal component, from the proportion 165. Horizontal strains. — Taking moments around Z>, those of the load on be and the reaction of the abutment remain the same as above, and we have for the strain at a, Hx(«&)= T - Tr and — : d :: x : — — , the distance ah. 2 I ' 214 A TEEATISE ON Substituting, we have, and tt 2dx _wx wx % H = ^-^f. (244) This strain, like the longitudinal, is greatest at the abutment, where x = 0, and H = *L - - ... (245) Ad and least at the centre, where x = — , and, J* H = S : (246) or it equals the strain in a horizontal truss at the centre, and increases from that point to the abutment, where it is double that amount. 166. Vertical strains. — H of Eq. (244) being the hor- izontal strain at a ; tt/_ wl wa/ ~U~ Id' will be the horizontal strain at the point d, distant x' from the same abutment. The strain at d is evidently greater than that at a ; subtracting, is the difference ; this may be caused by a tie from d to 6, with a strut from b to a, which would add to the ten- THE STRENGTH OF BRIDGES AND ROOFS. 215 sion at d ; or a strut from a to c, with a vertical tie cd, which would neutralize a certain amount of tension at a. Let the latter case be supposed ; then, since — -(x — x') is the horizontal component of the compression in ac, its vertical component is found from the proportion ad : dc. , , 2dx' .. w / ,v . wx' x — x : — — :: —Ax — x') : — — . I 4d K ; 21 Hence, V = *|p (247) is the vertical component of the strain in the brace ac. Now, if the truss be divided into panels of uniform horizontal length, braced as in Fig. (62), with inclined struts and vertical ties, and the weight considered as concentrated at the upper panel points, it will still per- fectly fulfil the conditions on which the above equations have been based ; and Eq. (247) is the vertical compo- nent of the strain in that strut whose upper end is dis- tant x horizontally from the abutment. It will be noticed that Eq. (247), or the vertical strain in the braces, increases from zero at the abutment, where x = 0, to — at the centre, where x = — '4 ' 2 Attention must be directed to the difference between the equations of the triangular and of the horizontal trusses. The tension in any vertical tie is plainly the vertical strain in that strut to whose lower end it is attached, 216 A TREATISE OTX except the centre tie, which receives the vertical strain from two struts. 167. Example— In Fig. (61), Let iv — 50 tons, whole weight, I = 100 feet, the length of the truss, I' = 51.4 feet, the length of the chord struts, d— 12.5 feet, the height of the truss at the centre, p = 10 feet, the horizontal length of a panel. From Eqs. (241) and (244) we have the following table of the longitudinal and horizontal strains : Values of x. 10 20 30 40 50 Strains in Tons. 92.56 82.27 71.99 61.70 51.49 Compression in AB & KL BC & IK CD&HI DE&GH EF&FG Strains in Tons. 90 80 70 60 Tension in Ab&hL be & gli cd & fg de& ef From Eq. (247) we obtain the vertical strains and the compression in the struts from the proportion of the vertical height of the strut to its length, or what is the THE STRENGTH OF BRIDGES AND ROOFS. 217 same, multiplying the vertical strain by the secant of the angle of the strut and a vertical. Values of x. 10 20 30 40 Strains in Tons. 10.31 11.18 12.5 14.14 Compression in Bb&Kk Cc&Ig Dd&Hf Ee&Ge Strains in Tons. 2.5 5 7.5 20 Tension in Cb&Ih Dc& Hg Ed&Gf Fe 168. Inclined Struts and Tie§. — If a triangular truss be braced as in Fig. (62), the lower 3 -^aAAAx^l b c d e f g h Fig. 62. chord points are not vertically beneath the upper points, and the longitudinal equation becomes changed ; and j dx wx wx 9 . wp* whence, j _ wV __ wl'x , wl'p* ~M ~Ul' USE (248) w v The vertical equation, — — , remains the same, x being — i still the horizontal distance to the upper end of any strut ; the tie connected to the lower end of any strut 218 A TREATISE ON havio o- the same vertical strain as that strut. If the weight and dimensions of Fig. (62) be assumed the same as those of Fig. (61), we have the following table of longitudinal and horizontal strains : Values of X. 5 15 25 35 45 Strains in Tons 92.56 85.71 76.11 66.13 56.01 Compres- sion in AB&KL BC&IK CD & HI DE & GH EF&FG Values of X. 10 20 30 40 50 Strains in Tons. 90 80 70 60 50 Tension in Ab&iL be & hi cd & gh de&fg ef And the following table of strains in the braces Values of x. 10 20 30 40 Strains in Tons. 5.59 7.07 9.01 11.18 Compression in Ki&Bb Cc&Ili Dd&Hg Ee&Gf Strains in Tons. 3.53 6.00 8.39 10.76 Tension in Cb&Ii Dc&Hh Ed&Gg Fe&Ff 169. The Triangular Truss subject to Movable Loads. — When used for bridge purposes the load will be upon THE STRENGTH OP BRIDGES AND ROOFS. 219 the horizontal chord, and generally the truss will be in- verted, in which case the longitudinal and horizontal strains will remain the same in amount, though changed in character ; the vertical equation will also remain the same. We shall now consider the effects of movable loads to which the truss is subject when used as a bridge. Let the truss, Fig. (62) inverted, be partially, but more than half loaded, let the load extend from the left abutment to a point distant u from the right abutment, and let w f — the weight of a full load of uniform density with the partial load. By the principles of the lever, — (I — uf is the reaction of the right abutment. Taking moments around a point in the chord tie, or the lower chord, distant x horizontally from the right abutment, x being less than u, we have for the strain in the horizontal chord, tt %dx w'x n v, and, H "&- u)\ - - (249) a constant for th( 3 entire distance i u. In the same manner for the tension in the lower chord, T dx wx n ■u)\ and L id' n tt)«, • m » (250) 220 A TEEATISE ON also a constant, or, the compression in the upper chord and the tension in the lower chord are the same from the end of the partial load to the abutment. Conse- quently no strain comes upon the braces between the end of the load and the abutment, when the partial load cov- ers more than half the truss. Eq. (249) may be put in this form : H wl tvu^ U 2d { u 2iP Under a full load, the least horizontal strain through- out the above distance, zz, would be at the end of u ; therefore, putting x=u, Eq. (244) becomes, H wl id wu 4d' evidently greater than the strain from a partial load at the same place. Similarly, it may be shown that the longitudinal strain is greatest under a full load. The above Eqs., (249) and (250), are true for any value of «, while x does not exceed -; that is, when the 2 ' load covers less than half the truss, Eq. (249) is the horizontal, and Eq. (250) the longitudinal strain from the abutment to the centre, and being constant, do not cause any strain in the braces between those points. Let Fig. (63) represent a truss less than half loaded. Fiff. 63. THE STRENGTH OF BRIDGES AND ROOFS. 221 Let x represent the distance from the right abutment to a point between the centre of the truss and the end of the load, being therefore greater than—, but less than u ; then, 1 -d-l x- 2d v- x ) 2 "' * 1 ' the vertical distance between the chords! And from moments around 6, H 2d ( l ~ x ) - wx (l v y and B _ = M j~^Y (251) 4dl(l — x is the horizontal strain in the upper chord between the centre and the end of the load, and increases from the centre, where it is the same as Eq. (249), as x increases, to the end of the load, where x == w, and it becomes, TT WU n v a strain very evidently, by comparison with Eq. (244), less than the strain at the same point under a full load. This decrease in the value of H, as we pass from the centre, requires ties inclined as in full load, that is, as in Fig. (64), or struts with the opposite inclination. The less than half load requires the same braces as a full load. Let H wx(l — u) 7 4dl(l — x) ' 12 222 A TREATISE ON be the compression at a, and tt, _ wx'j l — u) 2 == 4cli(i _ X ') be the compression at c, then IT' H = w $ — u )( x ~ x ') 4:d(l - X)(l - X') is the horizontal component of the strain in Jc, and this component is to the vertical component as ac is to ah ; or, . 2d(l — x) .. H _ H / . w(Z - w) 1 I 21(1 — x') greatest when x' is greatest, or when it equals u, in which case we have, V = |(Z- W ), (252) for the vertical strain in the braces at the end of a less than half load. The equation of the vertical strain at the same point from a full load is, Eq. (247), V — wx but in this case x is the distance from the left abutment, or I — u ; that is, the vertical strain in the braces from the less than half load is equal to the vertical strain at the same point from a full load. Again, ^ — L, the reaction of the right abut- Ah ment, is less than — ' T" ^ ; or the vertical strain in the THE STRENGTH OF BRIDGES AND ROOFS. 223 braces at the end of the load is greater than the reaction of the unloaded abutment. This seeming contradiction, that the braces between the centre and the end of the load bear a greater weight than the abutment, can be readily explained. In this case, and as H was determined, L = wV u 2dP (I - „), the tension in the chord tie under the end of the load, and l':h Ul Q v . wu n «), the vertical component of this longitudinal strain. This strain is passing towards the loaded end of the truss, or, in the present example, towards the left abutment. Sub- tracting this from the vertical strain in the brace, since it passes in the opposite direction, w n v wu /7 W 21 'ir »)-5C*—y, the reaction of the right abutment, or the excess of ver- tical strain in the braces is neutralized by an excess of equal amount in the chord tie. It is next necessary to ascertain the bracing required *9 Fig. 64. under a partial but greater than half load. In Fig. 64, 224 A TREATISE ON let u equal the length of the unloaded part, as before, and x the distance from the abutment to a point under the load ; then from moments around d we have, and W /7 V2 w S = ^- M >'-4i>- M >'' • ' < 253) in which H increases as x decreases ; this increase re- quires ties inclined as in a full load, or as in Fig. (65) ; that is, as far as the centre, where H reaches its least value; beyond, H increases to the farther abutment, requiring ties with the opposite inclination. At any point, 5, from moments around c, 4dl K ' Ux' K J Whence, U v ; U [ xx' r and, as before, x — x'' ?^L •• —(x — cv') -J^L i x — x ' \ % > ^ V ^L a strain passing in the same direction, and requiring the same kind of brace, but less by ~ than the vertical 2lx strain from a full load of equal density. Hence we see that in every case the strains from the full load are the greatest, and that no braces are required for a partial load except those needed under a full load. THE STRENGTH OF BRIDGES AND ROOFS. 225 CASE I.— A SIMPLE TRIANGULAR TRUSS WITH VERTICAL STRUTS AND INCLINED TIES. 170. Example In Fig. 65, ABCDEFGH I KLMNOPQR Fig. 65. Let I = 200 feet, the length of the truss, d = 20 feet, the depth of the truss at the centre, p = 12.5 the length of a panel, w = 200 tons, the weight supported by the truss when fully loaded. The horizontal strains in the upper chord are ob- tained from Eq. (244), and the longitudinal strains in the lower-chord ties, from Eq. (241), whence the follow- ing table : Values of x. 12.5 25 37.5 50 62.5 75 87.5 100 Strains in Tons, from Eq. (244). 375 350 325 300 275 250 225 Compression in AB,BC PQ&QR CD& OP DE& NO EF& MN FG& LM GH& KL HI& IK Strains in Tons, from Eq. (241). 386.6 360.9 335 309.3 283.5 257.8 231.9 206.2 Tension in Ab& qR be & pq cd & op de & no ef & mn fg& lm gh& kl hi& ik 15 226 A TREATISE ON Eq. (247) gives the vertical strain in the ties, a being the distance to the abutment end of any tie ; whence, from the proportion of the vertical to the longitudinal, we obtain the tensions in the ties. The vertical strain in any strut distant x from the abutment, is a panel load, or ^, added to the vertical It strain in the tie on the abutment side of the strut, or — , when xf = x — p ; substituting and adding, we have, V = y +!<*-*) =f +f, - -(254) is the compression in any strut distant x from the abut- ment. Whence the following table of strains in the braces : Valuee of x. 12.5 25 37.5 50 62.5 75 87.5 100 Strains in Tons. 25.8 27.9 31.2 35.3 40.1 45.1 50.4 Tension in bc& Pq cD& Op dE& No eF& Mn fG& Lm Kl Ik Strains in Tons. 12.5 18.75 25 31.25 37.5 43.75 50 100 Compression in Bb& Qq Cc& Pp Dd& Oo Ee& Nn Ff & Mm Gg& LI Hh& Kk Ii The strain in Ii is double Eq. (254), as it receives the vertical strain from both sides. THE STRENGTH OF BRIDGES AND ROOFS. 227 CASE II. A DOUBLE TRUSS WITH VERTICAL STRUTS AND INCLINED TIES. 1212121212121212121212121 ABCDBFGHI RLXNM' V K' I' H' G' F'E' W C B' A' Fig. 66. in.— This truss, like the preceding double trusses, may be divided into two simple trusses whose braces act independently of each other. The simple truss which has a full panel at the end will be designated as No. 1, the other as No. 2 ; their separate panel points are num- bered in the figure. Each truss bears half the load. Let I = the length of the truss, d = the depth at the centre, p e= the length of a panel measured horizontally, w = the weight, uniformly distributed. irs. Horizontal strains. — Simple Truss No. 1 has the half-panel load resting directly upon the abutment, and has also a full panel at the end, hence the horizontal strain in the upper chord from this simple truss is H = rl-S- < 255 > Simple Truss No. 2 has a full panel load a half (simple truss) panel length from the abutment, and its horizontal strain is similar to Eq. (162), top* -TT/ __ Wl WX' , ~Sd ~Sd 8dx r (256) 228 A TREATISE ON In the compound truss, the strain at any point in the upper or horizontal chord is the strain in one simple truss at the same point added to the strain in the other simple truss at the next panel towards the abutment. Hence, making x' of Eq. (256) = x — p, and adding to Eq. (255), we have for the horizontal chord strain at any panel point of Simple Truss No. 1, jr __wl _wx , wpx (257) Id Id^ M(x-p)' X And making x of Eq. (255) = x' — p, and adding to Eq. (256), we have for the horizontal chord strain at any panel point of Simple Truss No. 2, H=— — — + *?P + i^- - - (258) U Ad^Sd^ $dx' } 173. Vertical Strains in the Braces.— From Eq. (255) the vertical component of the strain in any tie of Simple Truss No. 1 is, where x is the horizontal distance to the lower end of the tie, V = — . (259) The vertical strain in any strut is the vertical strain in that tie to which the upper end of the strut is at- tached, and one panel load, or JB. ; if x be the horizontal distance of the strut from the abutment, the strain in it is -B + Eq. (259) at x — 2p, or 4 At, THE STRENGTH OF BRIDGES AND ROOF& From Eq. (256) the vertical component of the strain in any tie of Simple Truss No. 2 is, where x' is the hor- izontal distance from the abutment to the lower end of the tie, V = ^+ w ? . - - - (261) The vertical strain in any strut is, as before, the vertical strain in that tie to which the upper end of the strut is attached and one panel load ; or when xf is the horizontal distance of the strut from the abutment, the compression in it is, v - 5 (a> m + 4Z(aj , _ 2i? + 2j)) + *f* 174. Longitudinal strains. — Taking moments around any panel point in the horizontal chord of Simple Truss No. 1, we obtain, L = t*_t2fr . . . . . (263 ) 4cZ 4 ' K J 175. Example. — Let Fig. (66) represent a triangular truss of the character described, in which I = 240 feet, the length of the truss, V = 123.69 feet, the length of each section of the lower chord, d = 30 feet, the depth of the truss at the centre, p = 10 feet, the horizontal length of a panel, w = 300 tons, the full uniform load. The equations obtained above will apply to and give the strains in every member of the truss, except the two central horizontal chord-members. A part of the hori- zontal strain which the equation gives as belonging to them is taken by the inclined strut. The amount of horizontal strain in this strut may be obtained from its vertical strain by the proportion of the vertical to the horizontal extent of the strut. It is then to be deducted from the amount of strain obtained from the equation applicable to the horizontal member. This change is owing to the uniformity of the truss (on which the equations were based) being broken by the inclination of the strut. The following table gives the strains in the horizon- tal chord members : THE STRENGTH OF BRIDGES AND ROOFS. 231 Values of x in Eq. (257). Values of x' in Eq. (258). Strains in Tons. Compression in 20 575. AB, BC, CD & DC, CB', BA' 30 541.67 DE & ED' 40 516.67 EF & F'E' 50 490. FG & G'F' 60 465 GH & H'G' 70 439.14 HI & I'H' 80 414.29 IK & K T 90 388.89 KL & L'K' 100 363.89 LM & M'L' 110 325.01 MN & NM' The following table gives the tension in the lower- chord members : Values of a;' in Eq. (266). Values of x' . in Eq. (265). Strains in Tons. Tension in 10 592.65 Ab & b'A' 20 558.26 be & c'b' 30 532.46 cd & d'e' 40 504.95 de & e'd' 50 479.15 ef & f V 60 452.63 tg & g'f 70 427.81 gh & h'g' 80 400.60 hi & i'h' 90 374.80 ik & k'i' 100 348.44 kl & l'k' 110 322.94 lm & ml' 232 A TREATISE ON The following table gives the strains in the braces Values of x in Eq. (259). 20 40 60 80 100 Strains in Tons. 25.78 27.95 29.47 35.35 40.02 Tension in cE& E.'c' eG& G'e' gl& I'g' iL& L'i' 1N& NT Values of x in Eq. (260). 20 40 60 80 100 120 Strains in Tons. 12.5 18.75 25 31.25 37.5 87.5 Compression in Cc & C'c' Ee& E'e' Gg& G'g' Ii& l'i' L1& L'I' Nn& N'n' Values of x in Eq. (261). 10 3,0 50 70 90 Strains in Tons. 33.58 28.45 30.32 33.74 38.00 Tension in D'b' dF& F'd' fH& H'f hK& K'k' Km& M'k' Values of x in Eq. (262). 10 30 50 70 90 110 Strains in Tons. 12.5 16.67 22.5 28.57 34.61 43.12 Compression in Bb& B'b' Dd& D'd' Ff & FT Hli& H'h' Kk& K'k' Mn& Mn The strain determined from the equations being multiplied by the secant of the angle of the brace to which it is applied. THE STRENGTH OF BRIDGES AND ROOFS. 233 CASE III,— A DOUBLE TRUSS WITH INCLINED STRUTS AND TIES. A B CDEFGH I KLMNM'L'K'I'H'6'FE'D'C'B'A' 12121212 12121212121212121 r^^ Fig. 67. 176. — This truss may be divided and numbered as in the last case. Let I = the length of the truss, V — the length of each half of the lower chord, d = the depth at the centre, p = the horizontal length of a panel, w = the maximum weight, uniformly distributed. 177. Horizontal Strains. — The moments of the load on any segment of the simple trusses are the same as in Case II., Chap. V. Hence, taking moments around any panel point of Simple Truss No. 1, in the lower chord, distant horizontally x from the nearest abutment, we have TT %dx _ wx wx 7 ivp* t ~"r~id ir whence rr _ wl wx wp /*a*\ Similarly, from moments of Simple Truss No. 2, we have W _WI_W'X /MR v H 83 "83' " " " " (268) 234 A TREATISE ON The strain in the upper or horizontal chord on the abutment side of the point to which x is measured equals the strain at #, added to the strain at x\ when x' = x — p ; therefore, making x' = x — p, and adding Eqs. (268) and (267), we have H=^~^+^~^\ - - (269) Ad 4d Sd Sdx v ; Again, making x = x' — p, and then adding the equations, we obtain for the horizontal strain in the members on the abutment side of the point to which a/ is measured, w-= w ^ — wx ' -L W P — w fi men U ~" 4d Sd "" Sd{x' -p)' " { " } 178. Longitudinal strains. — Taking moments around any panel point of No. 1, in the horizontal chord, we have dx wx wx* L , V 4 U whence T Wl' Wl'X /nrr-ix h = Td—ur {271) the tension in the lower chord. And for No. 2, U Ul ^idlx' K ' Make of = x — p, and adding, we have 2d " 2dl ^ 4dl(x --p? ' ' K ' THE STRENGTH OF BRIDGES AND ROOFS. 235 for the strain in any lower chord member whose centre end is opposite any upper-chord panel point of No. 1, distant x from the abutment. Make x = x' — jp, add, and we have T — *^ w ^ x ' 4- W ^P 4- W ^'P* C274) 2d ~2dT~ t ldl~ Jr Idlx~'' " V ; for any lower-chord member whose centre end is oppo- site, or vertically beneath, a panel point of No. 2 distant x' from the abutment. 179. Vertical strains. — Let Fig. (68) represent a seg- ment of Simple Truss No. 1 ; Fig. 68. then from moments around d, TT __ wl won _^ wp 1 % "83 Sd - Sdx ; and from moments around e, TT, _ wl __ wx f ~8d ~8d' w 2 ( - x ~ x ' ) + i% (275) (276) Their difference, H-H' is the horizontal component of the compression in bd. 236 A TREATISE ON Its vertical component, from the proportion p or , 2dx . x — x : —j- , is for the vertical component of the strain in the struts of No. 1, x being the horizontal distance to their lower ends. From moments around e, H== «rf_««? (278) From moments around f, H' = — — — '— W P (279) 8cZ U Ux r K ' Subtracting, we obtain, as before, ■y- _ wa/ wp IT TV for the vertical component of the strain in the ties of No. 1, xf being the horizontal distance to their lower ends. For Simple Truss No. 2, by a similar process we ob- tain v =T«+a^y " • • < 280) for the vertical component of the strain in the struts, x being the horizontal distance to their lower ends ; and V =!£?.'- W P X ' /sail THE STRENGTH OF BRIDGES AND ROOFS. 237 for the vertical component of the strain in the ties, a/ being the horizontal distance of their lower ends. 180. Example. — Let Fig. 67 represent a trass in which I = 240 feet, the length of the truss, V = 123.69 feet, the length of each section of the lower chord, (I S3 30 feet, the depth of the truss at the centre, p = 10 feet, the horizontal extent of a panel, w = 300 tons, the full uniform load. Substituting these values in the above equations, we obtain the following table of strains for the different values of x and x f : Strains in Tons. Compression in Strains in Tons. Tension in 575. AB & A'B' 592.69 Ab & A'b' 550 BC & B'C 558.32 be & b'c' 533.33 CD & CD' 532.46 cd & c'd' 508.33 DE & D'E' 505.06 de & d'e' 485 EF & E'F 479.14 ef & e'f 460 FG & F'G' 452.78 fg & f'g' 435.73 GH & G'H' 426.8 gh&g'h' 410.72 HI & HT 400.94 hi & h'i' 386.21 IK & I'K' 374.8 ik & i'k' 361.11 KL & K'L' 349.03 kl & k'l' 336.36 LM & L'M' 322.94 lm & I'm' 311.36 MN&M'N' - 238 A TEEATISE ON The following is a table of the strains in the braces Strains in Tons. Compression in Strains in Tons. Tension in 27.95 Be & BV 20.83 Cd & C'd' 9.32 Dc & D'c' 23.57 De & D'e' 10.41 Ed & E'd' 24.01 Ef & E'f 14.14 Fe & F'e' 27.04 Fg & E'g' 16.01 Gf & G'f 28.79 Gh & G'li' 19.32 Hg & H'g 31.94 Hi & Hi' 21.68 Hi & I'll' 34.19 Ik & Ik' 24.84 Ki & KT 36.90 Kl & KT 27.35 Lk & L'k' 39.90 Lm & L'm' 30.60 Ml & Ml' 43.12 Mn & M'n 33.25 Nm & Nm' 75 Nn THE STRENGTH OF BRIDGES AND ROOFS. 239 CHAPTER IX. BOW-STRING TRUSSES. 181.-— A Bow-String Truss has one curved and one straight chord, the former meeting the latter at the abutments, and conveying to it the whole of its hori- zontal thrust, so that the reaction of the abutment is entirely vertical. The curved chord may be of any curvature, and on this depends to a great extent the amount of strain to which it will be subject. The chords are also connected by braces. 182. Parabolic Bow-String Trusses. — Let Fig. 69, j5 ADB represent the chords of a Parabolic Bow-String Truss, the braces being removed, whose arc, ADB, is the segment of a parabola, with the vertex at D. Let the chord AB receive the whole horizontal thrust of the 240 A TREATISE ON arc, so that the abutments, A and B, receive only a downward vertical thrust from the truss and its load. Let I = the length of the horizontal chord, AB, d = the depth of the truss at the centre, DC, being always the versed sine of the arc, w = the weight, uniformly distributed per unit of the horizontal chord, AB, x = the horizontal distance of any point from one abutment, y = the vertical distance between the chord and arc at a?, L == the longitudinal strain in the arc, H = the horizontal strain in the chord, V = the vertical strain or vertical component of a strain. 183. Longitudinal strains. — Disregarding the braces, and taking moments around any point, a, distant x from the abutment B, in a vertical plane cutting no inclined braces, we have L(ac)=™-^= WX V- X \ . - (281) 2 21 21 wx — being the moment of the reaction of the abutment, and — - the moment of the load on x. The strain L at Jib ' b is a thrust in the line of the tangent at that point; let bf be that tangent, then ac is its perpendicular distance from a, and L X (ac) is the moment of the longitudinal strain at b. THE STRENGTH OF BRIDGES AND ROOFS. 241 Let D be the origin of the axes to which the parabola is referred, and we have from the equation of the para- bola, - = 2nd, I* or 2/z, the parameter, — -7-7. Again, from the equation of the parabola, . :.^ = x{l-x), - - - - (282) and y ,-: W-s) .... ( 283) The angle abc being equal to the angle bfe, in the similar triangles abc and bef, fb : be :: ah : ac, or let t = tangent fb, and we have whence , . I ... t :- — %:: y :ac; ac = « Substituting this, and- the value of x (l—x) of Eq (282) in Eq. (281), we have x) y < 2A4ii' 16 242 A TREATISE ON whence, L = _^L_, .... (284) Ml a-* for the longitudinal strain in the arc at any point distant x horizontally from the abutment. Moving from any point towards the nearest abutment, t, the tangent, in- creases more rapidly than — — #, and consequently the longitudinal strain becomes the greatest at the abut- ment ; moving from the abutment the strain decreases until we reach the centre, where t = - x, when L =S < 285 > In the case of a truss of this form, the limits of x are and Z, since the uniformity of the truss is not broken as in the triangular truss. 184. Horizontal strains. — Taking moments around the point 6, we have for the strain in the horizontal chord, -rr, rvx wx* wx n , Substituting the value of x(l — x) in Eq. (282) we obtain y 21 wr and H-g, ..... (2 86) THE STRENGTH OF BRIDGES AND ROOFS. 243 a uniform strain throughout the chord and equal to the longitudinal strain in the arc at the centre ; and since it is subject to no increase nor decrease throughout its lengthy there can be no horizontal strain in the braces, or no inclined braces are required under a uniform load. The load may rest upon the arc, or be suspended from it by ties, and still fulfil the conditions upon which the equations are based. 185. — In practice the arc is composed of a number of straight members, whose points of intersection are in the curve. In Fig. 70, let the arc be divided into a number Fisf. 70. of short straight members by the division of the truss into an even number of panels of equal horizontal length, and let the load be considered as concentrated in equal amounts at the panel points. Then the moments of the abutment reaction, and of the load on the segment x remain the same ; but L, the longitudinal strain, is now in the line of the arc member, instead of in the line of the tangent to the curve at the point x. The value of y, in Eq. (283), remains the same, because x is always measured to the panel ends, at which points the truss intersects the curve of the parabola. 244 A TREATISE OX In Fig. 70 let moments be taken around a ; then, as before, L X ^ = lfiV~' 00 ^ and from similar triangles, bg : gh :: y : ac. Let gh =p, the horizontal length of a panel, and p' equal the length of the chord member hg ; whence, sub- stituting the value of y, Eq. (283), we have whence, then, whence L , . Adxil — x) p':p:: L ±: Adpxll — x) ac = — ± — i L p'V ' 4:dpx(l — x) wx(l — x) An equation in which we have but one variable p\ and entirely independent of the values of a?, and which shows that the longitudinal strain in any arc member is to the horizontal chord strain as the length of that mem- ber is to its horizontal extent The horizontal chord strain is not affected by the division of the truss into panels. In Fig. (70) the truss is divided into an even num- ber of panels, and d, the depth of the truss at the centre, THE STRENGTH OF BRIDGES AND ROOFS. 245 is also the distance of the vertex of the parabola above the horizontal chord, or the versed size of the arc. If, however, the truss be divided into an odd number of panels, then d, in all the previous equations, is no longer the depth of the truss at the centre, because, at that point, the curve passes outside the truss. Hence it will be seen that d is the versed sine of the arc, and not always the depth of the truss. 186. The Horizontal Component of the Longitudinal or Arc strain.— The horizontal component of the longi tudinal strain in any member is from the proportion r * Up ' 8d' a constant throughout the arc, and equal to the horizon- tal strain in the lower chord. 187. The Vertical Component of the Longitudinal strain. — Let y be the vertical distance of the upper end of an arc member, and y' the vertical distance of the lower end of the same member above the lower chord. Then subtracting from the value of ?/, Eq. (283), at a?, the value of y' at a? — jp, we have, after reduction, y _ y , ^dp{l-2X+ p)^ m _ (28g) the vertical extent of any arc member whose centre end is distant x from the nearest abutment, and from the proportion of its length to this vertical extent, we have z . 4dp(l — 2x+p) . . wlp' . y — w wx \ W P (289") the vertical component of the strain in the arc member. 246 A TREATISE ON Whence it will be seen that the strain in each arc member is the resultant of all the vertical strain or weight that comes upon the truss between it and the centre, the whole weight concentrated upon its own cen- tre end included, and the constant horizontal chord strain. 188. Strains from a Partial Load — Longitudinal strains. — Let vf represent the weight of a full uniform load of equal density with the partial load, and let the load extend from one abutment a distance equal to I — u; (let the truss be divided as before into panels of equal horizontal length, and let u be measured to the centre of a panel). From the principles of the lever, . — ' — '. is the vertical reaction of the unloaded abutment ; then, taking moments around a point in the horizontal chord in the unloaded part, x being less than w, we have, But _ Adpxjl — w) an equation evidently greatest when x is greatest, or ap- proaches i/, or at the end of the load. If I — x be made = I — u, THE STRENGTH OF BRIDGES AND ROOFS. 247 less than Eq. (285), or at any point the longitudinal strain from a partial load is less than the strain from the full load. 1§9. Horizontal Strains from a Partial Load. — If mo- ments be taken around any panel point in the arc in the unloaded part, x being less than w, Hy = w f x(l — u)* 2? ' and by substituting value of y, Eq. (283), (292) which is also greatest at the end of the load, and, at any point, less than the strain from a full load. If either this strain, or L, the longitudinal strain, be taken at any point under the load, it will be found to be less than the strain at the same point from a full load. Eq. (292) varies with the different values of #, and consequently its changes must be taken or caused by inclined braces. Let abed, Fig. (71), represent a panel in the un- 248 A TKEATISE ON loaded part of a parabolic bow-string truss, and let x be the distance from the right or unloaded abutment to c?, and x' the distance from the same abutment to c ; then, rr _. w'(l — u)* ~~ Sd(l — x) is the horizontal strain at d, and i^t _ w f (i—uy Sd(l — x') is the strain at c ; now, since the horizontal chord is subject to tension, and the strain at c is greater than the strain at d, cb is a tie, and ac and bd struts. 190. Vertical Strains in the Braces from the Partial Load.— The horizontal component of the strain in be is consequently H' — H, and its vertical component may be found from the proportion cd : bd, or, y : x' — x. Subtracting IF— H, and putting for y its value, Eq. (283), we have, x' — x' * dx ( l — x ) .. w '( l — U Y(®' — x ) . y r " u{i — x'){i-w) w'x(l — u) 7 (293) 2F(l-x') In the case of a simple truss, or truss with but one system of braces, it was shown (54) that the reaction of the unloaded abutment does not equal the true amount of vertical strain in the brace at the end of the load, but that it is given by (Eq. 35), .f~*~» Y _ wnp "21(1— p)' THE STRENGTH OF BRIDGES AND ROOFS. 249 In Fig. (71) x' is the horizontal distance from the abutment towards which the tie leans to the horizontal chord end of the tie, and x the horizontal distance from the same abutment to the arc end of the same tie. Therefore w, in Eq. (35), that is, the number of loaded panel points in (I — u) of Eq. (293), is equal to ±— ; and substituting — - — £— - tor — ±—— — '- in p ' °2l(l—2>) 2? Eq. (293), in place of n\ its value just given, and for x\ its value, x + _p, we have, V = w'(l — x— pfx _ w\l — x—p)x , 2q4 v 2l(l—p)(l-x-p) ' 21(1— p) "' ' V ' for the greatest vertical strain in any tie whose arc end is distant a?, measured horizontally, from the abutment towards which the tie leans ; which strain results solely from the moving load. Under the effects of the moving load the vertical braces act as struts, and the greatest strain upon any one — bd, for example, in Fig. (71) — is when the panel point d is outside of the load, or when the tie cb is sub- ject to its greatest tension. Then the strain in bd is equal to the vertical component of the strain in de, which may be determined from Eq. (293), x' and x being the distances to d said/. But (54) the panel point c cannot be fully loaded without a certain portion of the load coming upon the point d, which can cause no strain in / a a bd; hence, as before, we must substitute _ J vn P for 21(1 — p) — ^-^ — -, and x r being the distance of the vertical brace 250 A TREATISE ON from the abutment, n = — ^, x = «/ — jp, whence v _ w'x(l — uY _ w'(l — x'—p)\x f —p) m m , 295 v V , "2P(i— (C) 2l(l—p)(l — x / ) ' V ' is the compression in the vertical brace distant x' from the abutment, from the moving load w'. From this is to be deducted the tension caused by the permanent load. CASE I.— A SINGLE TRUSS, WITH VERTICAL AND INCLINED BRACES. 191. Example I.— Upper Chord Arched.— In Fig. (72), F G HIKLMN o P Let I = 200 feet, the length of the truss, d = 25 feet, the depth of the truss at the centre, p = 10 feet, the length of a panel, w' = 200 tons, the weight of the full movable load, w = 100 tons, the permanent truss weight. Eq. (286), by substituting the values of the con- stants, becomes = («/+ J£ )Z (200 + 100) x 200 _ 8<2 "200 ' the maximum tension throughout the length of the horizontal chord. THE STRENGTH OF BRIDGES AND ROOFS. 251 Similarly Eq. (285) becomes L = 30p'. Substituting the different values of p\ or the lengths of the different arc-members, we can form the following table of strains in the upper or arched chord. The length of any arc-member is evidently the square root of the sum of the squares of the horizontal length of the panel and of the difference in height of the two ends of the arc-member. Values of p'. Strains in Tons. Compression in 11.07 332.1 AB&UV 10.91 327.3 BC&TU 10.68 320.4 CD&ST 10.51 315.3 DE&RS 10.37 311.1 EF&QR 10.25 307.5 FG&PQ 10.15 304.5 GH&OP 10.07 302.1 HI&NO 10.03 300.9 IK&MN 10.00 300 KL&LM 252 A TKEATISE ON i When the truss is fully loaded, all the weight, except that of the arc, is borne by the vertical braces ; if we assume that the weight of the arc, or the weight imme- diately upon it, is one-third of the permanent truss load, then, when the truss is fully loaded, ^-S + J^L is the amount of tension upon each vertical brace. Substitu- ting the values above, we have, !^X^ + SXIOOXIO = 13<3 3 t 200 __ 3 X 200 for the maximum tension in each vertical brace. The tension from the permanent truss load is — f-, or 3.33 tons, which is to be deducted from the compres- sion caused by the moving load. The vertical differ from the inclined braces in one respect : those of the latter, leaning in one direction, are affected only when the load is moving in the direction in which their arc ends are inclined ; while the former are subject to strains from a load in either direction, and the greatest compression upon them is when the load covers more than half the truss ; or when x\ in Eq. (295), does not exceed —. Substituting the values of the constants in Eq. (295) it becomes v== (190-fl/) , (a/-10) 380(200 - x') ' whence, deducting the constant tension of 3.33 tons, we THE STRENGTH OF BRIDGES AND EO have the following table of maximum compressions in the vertical braces : Values of x'. Strains in Tons. Compression in 20 1.19 Cc&Tt 30 4.60 Dd&Ss 40 7.77 Ee &Rr 50 10.42 Ff &Qq 60 12.55 Gg &Pp 70 14.16 Hh&Oo 80 15.22 Ii&Nn 90 15.83 Kk & Mm 100 15.85 LI There is no compression in Bb and Uu. Substituting values in Eq. (294) it becomes (90 — xYx V 16000 — 160^' whence, multiplying the vertical strains obtained there- from by the secants of the tie angles, we have the fol- lowing table of tensions in ties from the moving load 254 A TREATISE OJS" Values of x. Strains in Tons. Tension in 10 11.05 tTJ&Bc 20 13.37 sT&Cd 30 16.05 rS&De 40 18.62 qR& Ef 50 20.87 pQ& Fg 60 22.74 oP&Gn 70 24.15 nO&Hi 80 25.09 mN&Ik 90 25.54 1M&K1 100 25.51 kL&Lm 110 24.97 iK&Mn 120 23.95 hl&No 130 22.43 gH&Op 140 20.40 fG&Pq 150 17.89 eF&Qr 160 14.89 dE& Rs 170 11.37 cD&St 180 7.08 bC& Tu THE STRENGTH OF BRIDGES AtfD ROOFS. 255 192. Example II. -The Lower Chord Arched —In Fig. (73), ABCDEFGH IK f S Fig. 73. Let I — 90 feet, the length of the truss, d = 8.1 feet, the versed sine of the arc, p = 10 feet, the length of a panel, w' = 90 tons, the weight of the full uniform movable load. w = 45 tons, the weight of the permanent truss load. The inversion of the arc results in many practical advantages ; the longer chord becomes subject to ten- sion, the shorter to compression, and the vertical braces to only one kind of strain, while the transverse bracing, to prevent flexure in the truss, does not interfere with the necessary headway. For the compression in the horizontal chord we have Eq. (285), H = -1= 187.5 tons. Sd For the tension in the arc we have Eq. (286), L = 18.75/, whence the following table for the different values of p 1 : Values of p'. 10 10.01 10.03 10.05 Strains in Tons. 187.5 187.8 188.1 188.4 Tension in de, ef & fg cd & gh be & hi Ab&ik 256 A TKEATISE ON It will be readily seen that Eq. (293) becomes in this case, ' n\i—x) ' (296) where x f is the distance, measured horizontally, from the abutment to the arc, or farthest end of the inclined tie. Making the change required, as explained in (54) for simple trusses, n of Eq. (35), the number of loaded panel points in I — u, is equal to - -, and#=a?'— p, whence P Eq. (296) becomes v _ w'(i—x'yx' (297) 2l(l—p)(l — x'+p) Substituting the values given above, we have, V - (90 - xjx' 16000 — 160 #'' Multiplying the vertical strains obtained for the dif- ferent values of x' by the secants of the angles, we form the following table of maximum tension in the ties. Values of x. 20 30 40 50 60 70 80 Strains in Tons. 15.73 16.40 16.67 16.00 12.09 11.99 7.98 Tension in cB gH& dC f G & eD eF& fE dE& cD & liG bC& iH Assuming, as in the last example, that one-third of the permanent truss load is borne directly by the arc, THE STRENGTH OF BRIDGES AND ROOFS. 257 we have for the compression upon the vertical struts from the truss fully loaded, —f- + — l -S- = 13.33 tons. The greatest compression upon these struts from the maximum moving load is when the ties to whose upper ends «they are attached are subject to the greatest ten- sion. This is given by Eq. (297), but the difference be- tween this equation and Eq. (296) is also borne by the struts ; hence Eq. (296) gives the total compression in the struts from the moving load. In this equation x is the distance from the abutment to the strut, x' = x + J9, and u = x + ^. Substituting these values, we have, V = w\x+ P )(l-x—^ _ {x + 10){s& _ w y Zl\l — x) 180(90 -tf) The greatest compression upon any strut is when x of this equation is less than — : that is, when the truss is more than half loaded. Adding 3.33 tons to the above, we have the following table : Values of x. 20 30 40 Strains in Tons. 13.33 13.39 14.53 14.78 Compression in Bb & Ii Cc & Hli Dd&Gg Ee&Ff The greatest compression in Bb and Ii is from the full load. 17 258 A TREATISE OTX CASE II.— A COMPOUND TRUSS, WITH VERTICAL AND IN- CLINED BRACES. 193.-In Fig. (74), b cd ef eh ik lmnopqr s tut's'r'q'p' o'n'm' 1' k' i' h' g' f ' e' d' C b 1231231231231 2 3 12 3 13 2 13 213 2 13 2 13 2 1 8 2 1 Fig. 74. Let I = 152 metres, the length of the truss, d = 20 metres, the depth of the truss at the centre, p = 4 metres, the length of a panel, or distance between the vertical braces, w' = 500 tons, the maximum movable load, w = 500 tons, the permanent truss weight. Under the full load this truss does not differ from the previous cases, and Eqs. (286) and (287) give the chord strains-. Therefore, -rr wl 1000x152 Q _ n = M- 8X20 - 95Qt0 ^ is the tension throughout the horizontal chord ; and wlp' 237.5p', Multiplying by the different values of p\ or lengths of the arc-members, we have the following table of com- pressions in the arc : THE STRENGTH OF BRIDGES AND ROOFS. 259 Values of p'. Strains in Tons. Compression in 4.49 1066.38 AB & B'A' , 4.44 1054.5 BC & C'B' 4.39 1042.63 CD & DC 4.34 1030.75 DE & ED' 4.30 1021.25 EF & E'F' 4.27 1014.13 FG & F'G' 4.23 1004.63 GH & G'H' 4.19 995.13 HI & I'H' 4.16 998 IK & KT 4.14 983.25 KL & L'K' 4.11 976.13 LM & M'L' 4.08 969 MN&N'M' 4.06 964.25 NO & ON' 4.05 961.88 OP & P'O' 4.03 957.13 PQ & Q'P' 4.02 954.75 QR & R'Q' 4.01 952.38 RS & S'R' 4.00 950 ST & T'S' 4.00 950 TU & U'T' Under the moving load we have three systems of braces acting independently of each other, as in the cases of the horizontal triple trusses. We will number these systems, or simple trusses, as Nos. 1, 2, and 3, their horizontal chord panel points being numbered in the figure. Eq. (293) gives the vertical component of the 260 A TREATISE ON tensions in the inclined ties in either of these simple trusses when w '" ~~, — of that equation representing the unloaded abutment reaction in a simple truss is changed to the amount of the abutment reaction upon each of these simple trusses. In Simple Truss No. 1, which has a panel point next the abutment, the reaction of the farther abutment is f as explained before. Eq. (293) therefore becomes w'((7— uY--p\x V = ii__. - - (298) 6P(Z — a/) In this equation x is the distance to the arc end of the tie (from the abutment towards which the tie in- clines), x' to the chord end, and equal to x + 3p, and u is equal to x + -J-. Substituting these values, Eq. (298) 2 may be reduced to this form : V = ^(l- x+j W \ . . (299) Inserting the constants in this equation we obtain for the different values of x the vertical components of the tensions in the inclined braces, which, multiplied by their secants, give the following table : THE STRENGTH OF BRIDGES AND EOOFS. 261 Values of x. Strains in Tons. Tension in 4 16.75 Be & B'e' 16 20.12 Eh & E'h' 28 20.88 HI & HT 40 22.20 Lo & L'o' 52 25.15 Or & Or' 64 26.39 Ru & R'u 76 26.84 Ur' & Ur 88 26.46 R'o' & Ro 100 25.47 OT & 01 112 23.81 L'h' & Lb. 124 22.36 H'e' & He 136 27.51 E'b' & Eb If we assume that one-half of the permanent truss weight is borne directly by the arc, we shall have for the maximum tension upon each vertical brace, occurring when the truss is fully loaded, ^+^ = 19.74 tons; and for the tension from the permanent truss weight, * !££ = 6.58 tons. Lb 262 A TREATISE ON T?he greatest compression upon these braces is, as explained before, when they are outside the load ex : tending from one abutment and covering the next panel point of the simple truss to which they belong. This strain is given by Eq. (298), where, if x' be the distance from the unloaded abutment to any vertical brace, x = x'— 3p, and u = x'+ -£-. Substituting these values, Eq. (298) becomes, Y - 2l 4 ] (300) 2^-00 Subtracting the tension from the permanent weight from this equation, we obtain the table below of the maxi- mum compressions. This equation is greater when x' is greater than — , or when the load covers less than half the truss, and for any value of x' greater than 124, or less than 28, the compression from the moving load is less than the constant tension. Values of x. 124 112 100 88 76 Strains in Tons. 0.65 3.81 6.38f 7.74 8.30 Compression in Hh & H'h' LI & LT Oo & O'o' Rr & R'r' Uu Eq. (299) will also give the vertical strain in the in- clined braces of Simple Truss No. 2. Simple Truss No. 1, in this example, is the same if the load enter at either end of the truss ; the other simple trusses differ. As the equations depend upon the simple truss panel at 13 THE STRENGTH OF BRIDGES AND ROOFS. 263 that abutment from which the load extends, it will be seen that that equation which applies to the inclined braces from the points numbered 2 in the figure when the load is moving from the left to the right, also applies to the inclined braces from the points numbered 3 when the load moves in the opposite direction. Substituting values in Eq. (299), and multiplying by the secants of the tie angles, we have the following table : Values of x. Strains in Tons. Tension in 132 , 24.34 Fc & F'c' 120 24.71 If & IT 108 . 24.46 Mi & M'i' 96 25.89 Pm & P'm' 84 26.71 Sp & S'p' 72 26.79 T's & Ts' 60 25.02 Q't' & Qt 48 25.09 N'q' & Nq 36 22.51 K'n' & Kn 24 20.05 G'k' & Gk 12 17.76 D'g' & Dg In Simple Truss No. 3, when the load is passing to 21 264 A TREATISE ON the right, or that simple truss whose panel point is 3p from the abutment, Eq. (293) becomes, v = (301) 6F(Z— a;') As before, a; is the distance from the abutment to the arc end of the tie, u — % + -S and x' — x + &p ; whence Eq. (301) reduces to, Y _ w'x(l-x) . . . . (302) 61* When the load is moving to the right this equation applies to the inclined braces from the points numbered 3, and to the inclined braces from the points numbered 2 when the load moves in the opposite direction. Substituting values in Eq. (302), and multiplying by the secants of the tie angles, we have the following table : Values of x. Strains in Tons. Tension in 128 20.01 Gd & G'd' 116 22.47 Kg & K'g' 104 23.58 Nk & N'k' 92 25.49 Qn & Q'n' 80 26.55 Tq & T'q' 68 26.50 S't & St' 56 25.57 P's' & Ps 44 24.01 M'p' & Mp 32 21.67 I'm' & Irn 20 19.18 FT & Fi 8 16.83 C'f & Cf THE STRENGTH OF BRIDGES AND ROOFS. 265 The vertical braces at the points numbered 2 are subject to different strains as the load moves to .the right or left. The same is true of the vertical braces at the points numbered 3. The maximum compression upon one system of points, when the load is moving in one direction, is the same as that upon the other when it is moving in the opposite direction. If the load be mov- ing to the right, the maximum compression is upon the verticals at the points 2, 2, etc. ; and if to the left, upon the verticals at the points 3, 3, etc., by Eq. (300), from which we form the following table, the permanent load upon the horizontal chord being deducted. Val ues of x. 36 48 60 72 84 ' 96 108 120 Strains in Tons. 2.45 5.32 7.33 8.25 8.08 6.92 4.58 1.83 Compression in Kk& K'k' Nn& Qq& Q'q' Tt& Tt' SV& Ss P'p'& Pp M'm'& Mm I'i'& Ii The compression in Dd, Gg, G'g', and D'd' is less than the tension from the permanent load. The com- pression upon the verticals at the points 3, 3, etc., when the load moves to the right, or upon the verticals at 2, 2, etc., when to the left, may be determined from Eq. (302), as Eq. (300) was obtained from Eq. (298). and will be found to be less than that given above. 266 A TREATISE ON CASE III. — A SIMPLE TRUSS, WITH ALL THE BRACES IN- CLINED, AND HAVING EQUAL HORIZONTAL EXTENT. 194. — It is only when the intensity of the load is uniform horizontally upon the arc, or concentrated, or suspended by vertical ties in equal amounts, at points equidistant horizontally, that the strain in the hori- zontal chord is uniform throughout its length, and the strain in any arc-member is equal to the hori- zontal strain divided by the uniform panel length and multiplied by the length of that arc-member. This is evident when we consider that a brace can transmit only a longitudinal strain, and that consequently the strain in an inclined brace must have a horizontal as well as a vertical component. The proportion of these components to each other depends on the inclination of the brace. If each brace in Fig. (75) be assumed to support a half panel load, then the horizontal component of its strain varies with • the inclination of the brace, and the strain in the hori- zontal chord cannot remain uniform. If the horizontal components be equal in all the braces, then each brace cannot support the same amount of weight. A BCDEFGHIKLMN , . c 7^ULOX^T m e * g fa i k Fig. 75. 195. — Let Fig. (75) represent a Bow-String Truss, THE STRENGTH OF BRIDGES AND ROOFS. 267 with a single system of inclined braces of uniform hori- zontal extent, the lower chord-points being in the curve of a parabola, as before ; Let I = the horizontal length of the truss, d = the versed sine of the curve at the centre, p = the length of the panels on the horizontal chord, p' = the length of an arc-member, w = the maximum load, permanent and movable, x = the horizontal distance of any panel point from one abutment ; y = the vertical distance of any lower or arc chord panel point from the horizontal chord. In the investigation of the chord strains w will be considered as concentrated at the horizontal chord panel points. 196. Horizontal Chord Strains. — Taking moments around any panel point in the arc, distant x from the abutment, we have, -tt __ wx wx 7 _ wp* V == 2 ~W ~8P h^- + -£- being the moment of the load on the segment j . 4dx(l — x) i #, and since y = ^— -, we have, i H = ^-_^L_, . . (303) 268 A TREATISE ON for the strain in any horizontal chord-member whose centre is distant x from the abutment. 197. Longitudinal Arc Strains. — If moments be taken around any point in the horizontal chord, distant x from the abutment, L/I^V^ ----—, - - (304) I 2 )rf 2 2V v J y' and y being the vertical distances of the two ends of the arc-member opposite the point x from the horizontal chord. Since y f is measured at x + i, and y at x — -±-, y'+y Ad{x+ 1 ) ( j_0_ |) +4 <%_ f ) (Z-a+f ) 2 ~ 2Z' Uilx — d—^ Substituting this in Eq. (304) we obtain, (wl , w/;? 2 Z2d(lx+x'— ^1 < - - (305) for the strain in any arc-member whose centre is distant horizontally x from the abutment. 198. Vertical Strains in the Braces from a Full Load. — Let Eq. (303) represent the strain in any horizontal chord-member, and let H' = w ^ - W P^ M 3'2djc'(l — x f ) THE STRENGTH OF BEIDGES AND ROOFS. 269 represent the strain in the next member towards the centre of the* truss, x' being equal to x + p, and confined to value less than — . 2 Since this strain increases from the abutment to- wards the centre, the difference must cause the same kind of strain in the braces whose arc ends are inclined towards that abutment from which x is measured, that there is in the horizontal chord ; in this case compres- sion, since this chord is uppermost. The difference between the two strains is mdx(i—x)x\i-x) ' v J Its vertical component is, from the proportion, p . 4dx(l — x) 2 I wp\_x' (I — x')—(l — a?)] Ux\l — x') ivl H— H : V - - - (307) a quantity always less than—-, or one-fourth of a panel load ; hence the other brace is also subject to compres- sion. Had the vertical component of the difference in the horizontal strains been a panel load, it would prove the other brace to be subject to tension, or to no strain. 270 ^ TREATISE ON In Fig. 76, representing a segment from the right of the centre of a truss similar to Fig. (75), let the vertical section db be distant x from the right abutment, and the vertical section ca be distant x from the same abutment ; Fig. 76. then Eq. (306) will represent the excess of horizontal strain in eb over that in ea, and Eq. (307) will represent the vertical strain, or the vertical component of the strain, of which Eq. (306) is the horizontal component. But this does not prove that there is a greater amount of vertical strain in eh than in ea, for they have different inclinations. From this vertical strain, how- ever, may be determined the proportion of the panel load upon e, borne by ea or eb. Deduct from —£-, or the load upon the point e, the weight, Eq. (307), borne by the excess of horizontal strain in eb, and, of the remainder, let V = the amount borne by ea ; and V, that borne by eb ; then the hori- zontal components of the compressions in ea and eb, of which V and V are the vertical components, are equal. Let H equal this horizontal component, uniting with V or V, then H: V'::|:y,andH:V::|:-y, and THE STRENGTH OF BRIDGES AND ROOFS. 271 ft-S xVwSE, - - -(308) and since V'+V= S? - ^l^fc^fc^ Z 4fo?'(Z — #') Substituting the value of V, Eq. (308), y/ + y- V / (yM-y\ Substituting the values of y* and y, y/ ( y'+ y) - y; gg - «Q+ gg ~ g) 2/ x'(l — x') m y, x'(l — x')+ x(l—x) _wp wp[x'(l— x')— x(l— x) ) " 'x(l - x') I ilx (l-x') \ which may be reduced to, y/ '_ up , tvpx'jl — gj / 309 x 4Z 2Z[ff'(Z— a^+^Z— a>)]' V ' The vertical component of the strain in that brace whose arc end is farthest from the abutment from which x is measured, a/ being the distance to the arc end and x = x' -p. Since, in Eq. (309), „_ ^fe^ \ 1 * ' X f (l-rX')+X(l—X) is less than — , V is greater than -^y-, or, under a full load, the weight upon that brace whose arc end is nearer the centre is greater than the weight upon the other brace from the same horizontal chord point. This equa- tion is necessary for the strain upon the braces from the permanent load. 272 A TREATISE ON The vertical strain upon the other brace from the same point in the horizontal chord is most readily deter- mined by subtracting Eq. (309) from ^. 199. Vertical Strains in the Braces from a Movable Load.— In Fig. (76) let the load extend from the left abut- ment, so that the point e is not loaded. Then, since the chord can take no portion of the vertical strain, it fol- lows that the vertical component of the strain in one of the braces, ae or be, is equal to that in the other, or V in ae — V in be ; but since the inclinations of the braces differ, the horizontal components of their strains must differ also, and the sum of these horizontal components equals the difference in the horizontal strains in the chord -members fe and eg. If a be distant x\ and b be distant x from the right and unloaded abutment, then 9/a being the reaction of that abutment, w f being weight of the full movable load, it, _ w'x'Q — u) 2 Wy< is the strain in ye, and xt __ wx(l — u) % 2iy the strain in eg. The vertical components of the strains in ae and he are to their horizontal components as the vertical extents and THE STRENGTH OF BRIDGES AND ROOFS. 273 of these braces are to their horizontal extents, the latter being E ; hence, y' : E ;: Y : —E, the horizontal com- if ponent of the strain in ae, and y : E : : V : -— 2, the hori- zontal component of the strain in he. Therefore, Ye + Ye = Y p(y+y r ) = H' - h 2^/ 2y 2y'y _ w'x'(l — uY wx(l — u,y = Wj/ Wy ' Vjp(y + 3/0 = ^"^ (*V - «rf) Substitute for y' and y their values as given in Eq. (283), and we have, omitting — , since it is common to t all the members of the equation, Yp\x(l-x) +x'(l - w>y\ = w '< 1 ~ i££$h gl ~ wx '( lx ~ ') I iv'(l — uyxx\x! — a?), f ; and since p — x f — x • v=- w'(! — uyxa/ mm llx(l-x)+x'(l-x')] J K ' is the vertical component of the strain in either ae or Z>e, and is greatest when (I — y) % is greatest, or when the load extends from the abutment opposite to that from which x 1 x, and u are measured, and fullv loads the 274 A TREATISE ON panel point next these two braces ; in this case the panel pointy! The longitudinal strains in ab and be have a result- ant at b, which passes in the direction of the end of the truss resting on the (in this case right) abutment. For let 7/, the vertical depth at b, represent the vertical abut- ment reaction, then V : H :: y : x; or, w(l-+-uy . tv(l — uY .. 4dx(l — x) W~ ' Sdif^x) f : X ' Therefore the amount only, and not the direction of this resultant, is affected by the weight of the load. The relative proportions of the components of this strain borne by ab and be consequently remain the same, even when the point e is loaded. Hence Eq. (310) will deter- mine the strain in fa when /is loaded, as well as in ae and be; and the greatest compression in fa, and tension in ae, are when the point / is loaded. To apply the equation, since it is a single truss, .'^'^i 21 wnj> . i gnhatitntfifl (ZA\ fnr^""^' n r N must be substituted (54) for " ~~ U) ; and 2 since x' = x + p ,n = > Eq. (310) becomes = P w>(l-~-x-lXx(x + p) y— £/ (311) for the vertical component of the greatest f ~ when the arc is above > I from a movable (compression when the are is below, j * movable THE STRENGTH OF BRIDGES AND ROOFS. 275 load in any brace whose arc and nearest end is distant x from the abutment. To this strain must be added the strain from the permanent load, which is-y^ — Eq. (309), •y- _ 3wp _ wpx"(l — x') "IT 2l[x'(l — x f )+x(l — x)y x' being equal to x +p, this becomes V= ™£ 4. wpx(l—w) f312 s U \ 2l[2x(l-p - x)+p(l -p)]' V ' In this equation w represents the weight of the per- manent truss load, and this equation, added to Eq. (311), gives the total maximum vertical strain in those braces to which Eq. (311) applies. l-x'-P 2 Making x = x' — jp, n = > Eq. (310), be- comes w'(l-x'-¥Yx'(x f -p) V = i ?i , (313) 1(1 -p)\2x* (l+p- x') —p(l +jp)] for the vertical component of the greatest \ . [ *■ ° (tension ) when the arc is I , [■ , from a movable load in any (below)' J brace whose arc and farthest end is distant x' from the abutment. From this must be subtracted Eq. (309), giving the strain from the permanent load in which x is to be made equal to x' — p. 276 A TREATISE" ON 200. Example. — In Fig. (75), Let 1=120 feet, the length of the truss, d = 14.4 feet, the versed sine of the curve of the lower chord, p = 10 feet, the horizontal distance between the panel points, w' = 120 tons, the weight of the full movable load, w = 48 tons, the permanent truss weight. Substituting values in Eq. (303), w being changed to w' + w, we can form the following table of strains in the horizontal chord : Values of x. 5 15 25 35 45 55 Strains in Tons. 167.4 172.2 173.2 173.5 173.7 173.8 Compression in AB& MN BC& LM CD& KL DE& IK EF& HI FG& GH For the strains in the arc, we obtain from Eq. (305), w being changed to w' + w, the following table : Values of X. 10 20 30 40 50 60 Values of p'. 10.77 10.50 10.28 10.03 10.02 10 Strains in Tons. 184.1 192.9 186.1 181.6 176.9 176.6 176.2 Tension in Ab & nN be & mn cd &lm de&kl ef &ik fg&hi gn THE STRENGTH OF BRIDGES AND ROOFS. 277 The strain in the first, or end member, is found from the strain in the horizontal end member, dividing by 5 its horizontal extent, and multiplying by 5.5 its length. Supposing one-half of the permanent truss weight to be borne directly by the arc, w of Eq. (309) is equal to 24 tons ; we have from Eqs. (311) and (312) the follow- ing table of compression in the braces, the vertical strain being multiplied by the secant of the brace angle. Values of x. Strains in Tons. Compression in 5 4.60 Mn&Bb 15 9.52 Lm & Cc 25 12.99 Kl&Dd 35 15.40 Ik&Ee 45 16.85 Hi&Ff 55 17.36 Gh& Qg . 65 16.96 Fg&Hh 75 16.63 Ef&Ii 85 13.41 De&Kk 95 10.28 Cd&Ll 105 6.34 Bc&Mm 278 A TREATISE ON Similarly from Eqs. (313) and (309), we obtain the folio win £ table of tensions in the braces ; Values of x. Strains in Tons. Tension in 15 2.97 Mm&Bc 25 6.78 Ll&Cd 35 9.29 Kk&De 45 10.75 Ii&Ef 55 11.28 Hh&Fg 65 10.99 Qg&Qh. 75 9.91 Ff &Hi 85 7.93 Ee&Ik 95 5.25 Dd&Kl 105 2.15 Cc & Lm THE STRENGTH OF BRIDGES AND ROOFS. 279 CASE IV. — A DOUBLE TRUSS, WITH ALL THE BRACES IN- CLINED, AND HAVING EQUAL HORIZONTAL EXTENT. 201 — Let Fig. (77) represent a bow-string parabolic Truss, containing two systems of braces, which permit H bcdefgliiklninopqrstu r. 77. the division of the truss into two simple trusses, whose chords are common, but whose braces act independently of each. Let Fig. (78) represent one of these trusses, designated as Simple Truss No. 1, and Fig. (79) the other, designated as Simple Truss No. 2. I L N Fig. 78. Fig. 79. 202. — Let I = the horizontal length of the truss, d = the versed sine of the curve, 280 A TREATISE ON p = the horizontal length of the panels of the double truss, jp' = the length of an arc member, w = the maximum load, uniformly distrib- uted, x = the horizontal distance of any panel point from one abutment, y = fjg(? — #) ? t he vertical distance be- V tween the chords at any panel point x. 203. Horizontal Chord Strains. — Each truss bears one-half the load, and the half panel load at each abut- ment is taken as belonging to No. 2. Taking moments around a point in the arc chord of No. 1. Fig. (80), distant x from the abutment. •pr 4:dx(l— x) wx wx* P 4 4J ' H = wl 1ST " - - - " < 314) is the tension in the lower, or horizontal chord, between the points b and u. But outside of these points, or be- tween them and the abutment, this equation will not apply, because there is no panel point in the upper chord of this truss. The manner of determining the strain in the end member will be shown hereafter. THE STRENGTH OF BRIDGES AND ROOFS. 281 Taking moments around the upper chord panel points of No. 2, we have, XT/ Adx(l — x)_ wx _ wx* wp* f 4" U IP whence, M ~ 163 16:y , i:h:v':.h = ^, Then *=•, and v= v X y y y' v+ y--T' and ^_ wpy' ky'+yY Substituting for y[ and y their values, 4cfa ' ^ ~ X "> and fM=f ), we havej V- wpx'(l—x>) , ,. l(x>(l-~x>)+x(l-x) ' ' ^ 4) for the vertical component of the strain in that brace whose arc end leans from the abutment from which a/ is measured, x being = a/ — 2p. The vertical component of the strain in the other brace from the same point is WD -j Eq. (324). It will be observed that, since THE STRENGTH OF BRIDGES AND ROOFS. 285 x'(l — x') + x(l — x) is less than 2x'(l — x') when x' is not greater than — ; Eq. (324) is greater than — £, or, the brace leaning towards the centre supports more than half a panel load. In Simple Truss No. 2 we have the same case which was explained in (193), and to which, consequently, Eq. (309) will apply. 206. Tertical Strains in the Braces from a Moving Load. — A comparison of these Simple Trusses with Case (III) will show that the vertical strains, or vertical components of the strains in the braces, are similar; and that Eq. (310) representing twice the reaction of the abutment from which x is measured, multiplied by —2- — , (x being the distance from the x'(l-x') + x(l-x)' V & abutment whose reaction is taken to the arc end of any brace leaning towards the same abutment, x' being = a? + 2p, and u being = a?,) will give the maximum tension in that brace. Hence, in Simple Truss No. 1, the reaction of the abutment from a partial load, found as in the horizontal trusses, being, — ,T ? Eq. (310) becomes, 4& v= w'(l — xYx(v + 2p) 2r[x(l — x) + ( + 2j?)(l — xx— 2p)J w'(l — xyx(x + 2p) W[x(l— 2p — x) +p(l — 2p)J (325) 286 A TREATISE ON This being tension, the tension from the constant truss weight is to be added to it. The greatest compression in the braces of the same simple truss is, as has been explained in the previous case, given by Eq. (310), u being = x 1 and x = x — 2p, whence Eq. (310) becomes V = w'(l — xyx'(x' — 2p) 2F[(rf — 2p) (l — a/ + 2p) + x' (I — a?')] u/(l — afyaf(ti—$p) ( . ' U\x\l + 2p — x) —jp(l + 2p)] ' V ; x' being the horizontal distance from the unloaded abut- ment to the arc end of the brace leaning from that abut- ment ; and since the strain is compression, it is necessary to deduct from it the strain from the constant load. In Simple Truss No. 2, by a similar process of rea- soning, Eq. (310) becomes changed to __ w f [_(l — xY — />(> + 2p) m -v U\x(l — x — 2p)+jp(l — 2p)]' V ; for the tension in those braces whose arc ends are dis- tant x horizontally from the abutment towards which they lean. And for the compression in those braces whose arc ends are distant x' horizontally from the abutment from which they lean, Eq. (310) becomes changed to U'\x\l + ip - x)-p(l + 2p)] ' K ' To Eq. (327) is to be added the constant truss load, THE STRENGTH OF BRIDGES AND ROOFS. 287 Eq. (324) ; and from Eq. (328) is to be subtracted the constant truss load to attain the maximum result. 207. Example.— In Fig. (77), Let I = 200 feet, the length of the truss, d = 25 feet, the depth of the truss at the cen- tre, and the versed sine of the curve, jp = 10 feet, the horizontal distance between the panel points, w = 100 tons, the permanent truss weight, two-thirds of which is supported by the braces, u/ = 200 tons, the weight of the full movable load. From Eq. (318) for the end member, and Eq. (316) for the other members, we obtain the following table of strains in the horizontal chord, w of these equations being the maximum load, 300 tons : Values of x. 10 30 50 70 90 Strains in Tons. 308.77 292.11 297.06 298.00 .298.35 298.49 Tension in Ab& uV be & tu cd, de, rs, & st ef, tg, pq. & qr gn. hi, no & op ik,kl,lm & mn From Eqs. (321), (322), and (323), we may form the following table of compressions in the arc chord ; w being here also the maximum load, 300 tons : 288 A TREATISE ON Values of x. Strains in Tons. Compression in 341.81 AB&UV 334.37 BC&TU 30 323.54 CD&ST 30 318.39 DE&RS 50 313.17 EF & QR 50 309.55 FG&PQ 70 306.17 GH&OP 70 303.76 HI&NO 90 302.41 IK&MN 90 301.81 KL&LM From Eq. [325), added to two-thirds of -f- — Eq. (324), we obtain the tensions; and from Eq- (326), less two-thirds of Eq. (324), the compressions in the braces of Simple Truss No. 1, as given in the following table, the vertical strains being multiplied by the secants of the angles of the braces. THE STRENGTH OF BRIDGES AND ROOFS. 289 Values of x&x'. 20 40 60 80 100 120 140 160 180 Strains in Tons. 11.48 13.94 16.18 16.60 18.32 18.09 17.18 16.07 19.92 Tension in Cd& Ts Ef & Rq Gh& Po Ik& Nm Lm& Lk No& Ih Pq& Gf Rs& Ed Tu& Cb Strains in Tons. 3.53 6.71 8.48 9.17 8.85 7.58 5.44 2.51 Compres- . sion in Rs& Ed Pq& Gf No& Ih Lm& Lk Ik& Nm Gli& Po Ef & Rq Cd# Ts Again, from Eq. (327), added to two-thirds of -2— — V Eq. (309), we obtain the tensions, and from Eq. (328), less two-thirds of Eq. (309), the compressions in the braces of Simple Truss No. 2, as given in the following table ; the vertical strain being multiplied by the secants of the angles of the braces : Values of x & x'. 10 30 50 70 90 110 130 150 170 Strains in Tons. 12.40 12.81 15.18 17.00 17.97 18.06 17.79 16.41 15.80 Tension in Bc& Ut De& Sr Fg& Hi& On K1& Ml Mn& Ki Op& Hg Qr& Fe St& Dc Strains in Tons. 1.84 5.47 7.76 8.89 8.98 8.52 6.15 3.24 Compres- sion in St& Dc Qr& Fe Op& Hg Mn& Ki K1& Ml Hi& On Fg& Qp De & Sr 19 990 A TREATISE (XN" 208. — By similar processes equations may be prepared for Parabolic Bow String Trusses containing any num- ber of Simple Trusses, or single systems of bracing. 209. Bow String Trusses, with Arcs of any Curva- ture. — It is only when y, or the vertical distance between the arc and the horizontal chord at any point, bears some ratio to the horizontal distance of that point from the abutment, that we can express the former in terms of the latter, as in the Parabolic Bow String. This can be done when the curve can be referred to rectangular axes, as is the case with the circle, ellipse, and many others. But in these cases the equations become more intricate, and it is better that thev should contain two variables than to be rendered liable to error by their prolixity. The following equations will apply to any Bow String Truss, as they are entirely independent of the curvature of the arc. 210 — Let I = the length of the truss, p = the horizontal length of a panel. j/ = the length of any arc member, w = the weight, maximum and uniform, upon the truss, x = the horizontal distance of any panel point from one abutment, y = the vertical distance between the chords at x THE STRENGTH OF BRIDGES AND ROOFS. 291 ail. Horizontal strains. — From moments around any panel point in the arc of Fig. (80), we Fig. 80. have for the tension in the horizontal chord, wx wx ny = whence, XT _ wx wx B _ = wx__wxl . . . . (329) 2y 2ly' K J and at any other point x\ H' = ^-^ - - - (330) 2y' 2ly' K ' If H' be greater than H, then the horizontal chord end of the inclined brace between the two points x' and x must be distant x\ x f and the arc end distant x from the abutment. That is, if the horizontal strain at c be greater than the horizontal strain at <#, a brace is neces- sary from c to b. If, on the contrary, the strain be greater at d, then the brace must be from a to d. The vertical components of the strains in these braces, neces- sary to ascertain the effects of the constant load, may be obtained, as has been shown before, from the difference in the horizontal strains. 212. Longitudinal Arc Strains. — From moments around any point in the horizontal chord we have, L=£(^-«^l > . , (331) 292 A TREATISE ON Eqs. (329) and (331) apply to the members of the chords on that side of any point to which x is measured, on which, under a full load, no brace is in action. 213. Vertical Strains in the Braces from a Moving Load.-— Under a partial load, extending from one abut- ment a distance I — u, at any point, #, outside the load, x and u being measured from the same abutment, and w' being the weight of a full load of equal density with the partial load, the strain in the lower chord will be, TT __ W'(l — UfX u 2ty and at the next panel point x\ (332) H/ = y'(E-tt)V .... (333) Subtracting n TT_TT' w'jl-uy jx y'-x'y) _ ,„„.. Wyy> ■ K > Let x' = x — p, then H - H' = w \ l - u V(®y' — x y+py) / 335 \ WW ' v is the difference in the horizontal strains at two consecu- tive panel points outside, and consequently the horizon- tal component of the strain in the brace. If y' be the vertical extent of this brace between x and x — p y p being its horizontal extent, we have from the propor- tion p : y>::Tt-W : V = "'(*-")'[% '-y)+gy] (336) IVpy THE STRENGTH OF BRIDGES AND ROOFS. 293 for the vertical component of the strain in any inclined brace between the panel points x and x — p. This equation must be changed, as has been done in the previous cases to suit different systems of bracing ; — : ~~ a ■- be altered to represent the reaction of the un- loaded abutment of the simple truss to which it is to be applied, and p is always to represent the horizontal length of a simple truss panel. Examples of the appli- cation of these equations are considered unnecessary. 294 A TREATISE OS CHAPTER X. LENTICULAR TRUSSES. 214.— The form of this peculiar truss, known also as the Pauli System, is shown in the following figure : D E F G H I K L M X It is composed of two equal parabolic arcs for chords meeting at the ends, and braced with vertical and in- clined braces. It is not capable of supporting any greater weight than a Bow String Truss of equal depth and length, and practically possesses many disad- vantages. 215. Longitudinal Chord Strain§.— In Fig. (82) Let I = the length of the truss, d = the depth of the truss at the centre, p = the horizontal length of a panel, THE STRENGTH OF BRIDGES AND ROOES. 295 p f = the length of any arc member, x = the horizontal distance of any vertical brace from the abutment, y = the depth of the truss at any point x, w = the weight upon the truss uniformly distributed. Taking moments around a in the lower chord distant horizontal x from the abutment B, we have, wx vox XJ) 2 21 (337) The versed sine of either arc is — , therefore Eq. (283) It the vertical height of any panel point in the upper chord above the abutments is, 2dx(l — x) f : (338) and the vertical depth of any panel point m the lower chord below the abutment in the same vertical section is also 1— ?., whence the vertical depth of the truss 296 A TREATISE ON at any panel point distant x from the abutment is, as in £(ZX (l •*■— X) the case of the Parabolic Bow String Truss, ^ L From similar triangles, af : ah : : gc : 6c, or substituting the values of these quantities, whence, af:^!tl^:p-.p', r_ ±dx{l — x)p Substituting this value in Eq. (337), we obtain wlp' (339) Sdp Equally true for either chord, and containing but one variable p', 216. Vertical Strain in the Vertical Braces. — The depth of the truss at any panel end, ab, distant x from the abutment, is — ~\~ — - ; at the next panel end to- wards this abutment, ce, where the distance is x — p, it 4d(x -p)(l - x +p) is . Subtracting these quantities, we have 4dp(l-2x + p)^ .... {340) It for the difference in the depths of the truss at the two THE STRENGTH OF BRIDGES AND ROOFS. 297 ends of a panel, or in the lengths of ab and ce. Half of this quantity is consequently the vertical extent of each chord member of this panel^ or b is vertically one- half of this distance above c, and a vertically one-half the same beneath e. Since be : by, so is the longitudinal strain in be to its vertical component, we have, , . 2dp(l — 2% + p) .. wlp' . y _ w __ wx wp /ojn p * f sdp ' i ~2i T ir { } in I Its horizontal component is a constant •— the lower chord member in the same panel has the same vertical component. Hence each arc or chord member supports one-half the weight that comes upon the truss between its end towards the centre and the centre, and one-fourth of a panel load ; and the horizontal component remain- ing uniform, there is no strain upon the inclined braces. If a?, in Eq. (341), be made =% — p, we shall obtain the vertical component of the strain in the next chord mem- ber towards the abutment, and if this be subtracted from Eq. (341), with x unchanged, the remainder is ^y, a con- stant. Therefore, at each panel point, as we pass from the centre towards the abutment, there is an increase in the weight borne by each chord of half a panel load ; or each chord bears half the load ; and if the whole load come first upon one chord, only one-half of it is taken by that chord, while the other half is transmitted by the vertical braces to the other chord ; hence, when the truss is fully loaded, each brace is subject to a strain of -2L 298 A TKEATISE ON 217. Vertical Si rain* In tlie Braces from a moving i.oad. — It is evident that as the two chords have, at the two ends of any panel, the safne difference in their verti- cal distances, as there is in the case of the Parabolic Bow String, they can take no more of the vertical strain from the moving load. Taking moments around any panel point in either chord distant x from the abutment, x and u being measured from the same abutment and x being less than u ; I — u being the length of a partial load, and w' being the weight of a full load of equal density with the partial load, the strain in either chord is, j- 4:dx(l—x)p__ w'ty—ufx ~ly ~2? ' whence L= w l~ vy f < 342 > 8d\l — x)p For simplicity take the horizontal component of this strain, which is H =!|3r • • • - m A strain increasing with x or greatest at the end of the load. At x' Subtracting, H — H' = w 'Q — u y( x — jg O 8tf(Z_ x)(l— off 1 is the horizontal component of the strain in the inclined THE STRENGTH OF BRIDGES AND ROOFS. 299 brace whose ends are distant x and x f from the abut- ment towards which the brace leans ( ), or in ac. The vertical extent of this brace is one- half of ce, the depth of the truss at x' and one-half of ba the depth at a?, its horizontal extent is p ; hence . 2dx(l — x) . 2dx (I — x) .. w'(l — u)\x — x) . v 9 : f "*" t 8d(l—x)(l — x') '' _ w'{l—uy j 2x _ pQ+p) \ (345 . 4Z a (l-x (l-x){l-x+p)y K > x — p being substituted for x\ for the vertical compo- nent of the strain in any inclined brace from the moving load, when the lower end of the brace is distant x hori- zontally from the abutment towards which it leans. This strain is evidently greatest when the load extends from the opposite abutment to that from which x is measured, and covers the panel point, distant x from the abutment. This being a simple truss, — ~ , the re- Tub action of the unloaded abutment must be changed for the reasons explained in (54) to - F ; and since n r — l\L p) the number of loaded panels in I — u, equals , this P quantity becomes —j- U Substituting this in Eq. (345) we obtain V=^±4J2»-M=C:}, - (346) for the maximum vertical component of the strain in 300 A TREATISE ON any inclined brace whose lower end is distant x from the abutment towards which it leans. This strain is evidently always tension, and it pro- duces compression in the vertical brace to whose upper end it is attached. This compression is greatest when the tension in the inclined brace is greatest, or when the load extends from one abutment and covers the next panel point to this brace. Referring to Fig. (82), let the load extend from the abutment A so that the point h is loaded and a outside the load. It is evident that the vertical strains in the chord member ib and the inclined brace hb, compression in the former and tension in the latter, produce all the vertical strain that may exist in the chord member be and the vertical brace ba ; or the vertical strain in ib and lib equals the vertical strain in ba and be ; hence, if we subtract the vertical strain in be from the sum of the strains in ib and bh< we have for the remainder the ver- tical strain in ba. Let V, Eq. (345), be the vertical strain in ib, x being the horizontal distance from the abutment B to point h. From moments around A, sd(i^p ' ■ (347) the reaction of the abutment bein«- w 'v / — u ) . t -u e ver tical component of this from the proportion of its length to its vertical extent, , . 2dp(J--2x+p) .. W 'Q- U y p > _ l' ' " Hd(l~x)p ' = v'(l-u)'(l-2x + p) THE STRENGTH OF BRIDGES AND ROOFS. 301 Adding Eqs. (345) and (348), we have, _ u /(l— u ) % i 2x _ p(l + p) t l-Vx+ PlftAQ i? \i-x (i-x)(i-x+py " i-x y which readily reduces to Y= w'(!-uy(!+p) . . . (350) U\l-x+p) ' for the total vertical strain coming upon the point b y and taken by the members be and ba. If x of Eq. (348) be changed to x — jp, we shall have for the vertical strain in &c, Y _ w'(l-u)Xl-2x + 3p) _ m „ 51 U\l-x+p) v ; Subtract Eq. (351) from Eq. (350) and w'(l-u)'2(x-p) _ _ _ (352) 4F{l — w+p) is the remainder and the strain in ba. In this equation the reaction of the abutment W ' ~ U * must be changed LI l j wnp an ^ s i nC e x is the distance to the panel 21(1— p)' l point beyond the vertical brace to which the equation refers, let it be changed to x' + p, so that x f shall be the distance to the brace. Then, since n = . %—, mak- P ing these changes, Eq. (352) becomes W '(i~x'- P yx' _ m _ (53 2l(l~p)(l-x') ' The vertical braces are subject to the action of the 302 A TREATISE ON load moving in either direction, or a strain may come upon them from either of the ties attached to their upper ends. The greatest strain is when the load covers more than half the truss, or when x' does not exceed — . If the roadway be upon the upper chord the proportion of the compression from the constant load must be added to Eq. (353). If upon the lower chord the tension from the constant load must be deducted. 218. Example. — In Fig. (81), Let I = 350 feet, the length of the truss, d = 49 feet, the depth of the truss at the centre, p = 25 feet, the horizontal length of a panel, w' = 350 tons, the weight of the full movable load, w = 175 tons, the constant truss weight. Substituting values in Eq. (339) we have the follow- ing table of chord strains, w of this equation being the maximum load, 525 tons : Values of p.' 25.83 25.60 25.40 25.24 25.12 25.04 25.01 Strains in Tons. 484.31 480.00 476.25 473.25 471.00 469.50 468.94 Compression in AB & OP BC& NO CD& MN DE& LM EF& KL FG& IK GH& HI Tension in Ab& oP be & no cd& mn de & lm ef & kl fg& ik gh& hi inn THE STKENGTH OF BKIDGES AND KOOFS> 303 Substituting values in Eq. (346), and multiplying the vertical strain by the secants of the inclined brace angles, we may form the following table : Values of x. Strains in Tons. Tension in 50 30.96 Bc&nO 75 33.41 Cd&mN 100 38.68 De&lM 125 40.32 Ef &kL 150 45.13 Fg&iK 175 45.94 Gh & hi 200 44.99 Hi & gH 225 42.31 Ik&fG 250 37.93 Kl&eF 275 31.90 Lm&dE 300 24.30 Mn&cD 325 14.95 No&bC If three-fourths of the constant load be considered as concentrated upon the lower chord, which also receives the moving load (the upper chord bearing directly the other fourth), each vertical brace will be subject to a 304 A TREATISE ON" tension of one-fourth of a panel load from the constant load, or ^2, and one-half of a panel load of the moving 41 load, or !££; hence, -^ + ~f = 15.625 tons is the maximum tension in the vertical braces. Substituting values in Eq. (353), and deducting from the results the constant load tension of —|-= 3.125, the 41 following table of maximum compressions in the verti- cal braces may be formed : Values of x. 25 50 75 100 125 150 175 Strains in Tons. 7.53 16.27 23.10 28.03 31.07 32.21 31.49 Compression in Bb& Oo Cc& Nn Dd& Mm Ee& LI Ff & Kk Gj* Hh THE STRENGTH OF BRIDGES AND ROOFS. 305 CHAPTER XI. THE KUILENBERG TRUSS. 219. —The plan of this truss is shown in Fig. (83). 1 « * 1 ? ! J »ls £ \\VZ°^.Z **~"l?. 1 2 3 , M AA cd e f g h i k lmnopqr stuvwv'u't's'r'q'p' o'n'mTk' i'h'g'P e'd'c' 1*18313313213213 2 1321 231231231 2312312312 Fig. 83. For trusses of this character, having the upper chord arched, but not meeting the lower and horizontal chord at the abutments, equations may be prepared, containing but one variable, as in the previous cases ; but in this case to express the depth of the truss at any point in terms of the distance of that point from the abutment, would give a long and somewhat intricate equation. For this reason, and to give an example of the manner in which any form of truss may be analyzed, and its strains determined, only the simplest forms of equations will be used. 220. ^Let I d 20 152 metres, the distance between the abutments, 20 metres, the depth of the truss at the centre, 306 A TREATISE ON p = 4 metres, the length of a panel, n = 1.2 metres, the distance on the abut- ment between the end posts, *w' = 500 tons, the full movable load, w = 500 tons, the constant truss weight, uniformly distributed, x = the horizontal distance of a panel point from the abutment not from the end of the truss, y = the vertical depth of the truss at x. The truss extends beyond the abutment at either end a certain distance = 2n, ah = hc— n. The upper chord is the arc of a circle whose diameter = 493.33 metres, and the length of the end post — 8 metres. 221. Horizontal strains.-H-Under the full load, in which case the chord strains are greatest, this truss, like the trusses with horizontal chords, may be divided into three simple trusses, No. 1.— Fig. (84). These weights are assumed. THE STRENGTH OF BRIDGES AND ROOFS. 307 No. 2— Fig. (85). P s v V s No. 3.-Fig. {86). o R v_ 5* » o' Simple Truss No. 1 bears — (w -r- _£.), (w, for con- venience, expressing the full load and the truss weight), therefore the moment of the reaction of the abutment upon it is at any point a?, since its end post is 2n beyond the abutment, ff+ $)♦:+■* and the moment of the load on x is, llwx ' wpx wp*\ 3V 21 21 I Hence H, the horizontal component of the compression in the upper chord, or the tension in the lower chord is, wpn' # rr _ wx wx , wp wn V ~~T '6^' + ""3I + "3"" h ~3^ , whence, H = w % (Ix — x? + 2tf + 2ln + 2pn). (354) 308 A TEEATISE ON Substituting values given above, w being 1000 tons, w _ 125#[ (152 -a;) +406.4] Whence the following table of chord strains in Simple Truss No. 1 : Values of x. Values of y. Horizontal Strains in Tons. Horizontal Component of Compression in Tension in 4 9.92 110.34 AD & AD' dg & d'g' 16 13.04 217.14 DG & DG' gk & g'k' 28 15.57 273.13 GK & G'K' kn & k'n' 40 17.52 305.81 KN & K'N' nq & n'q' 52 18.9 325.25 NQ & N'Q' qt & q't' T 64 19.72 335.75 QT & Q'T' tw & t'w' 76 20 338.95 TW & T'W There being no tension in ad. Simple Truss No. 2 bears — (w + -j-) , and end at the abutment, the moment of the abutment reaction is, mg wx wpx ~6~ "6T wx % ithe load on x is — -, hence 61 THE STRENGTH OF BRIDGES AND ROOFS. 309 H IV 6ly (to — x* +jpx). (355) Substituting values, w being put for (w' + w,) H 250 203i/ jar(156— a?)| whence the following table of chord strains in Simple Truss No. 2 : Values of x. Values of y. Strains in Tons. Horizontal Component of Compression in Tension in 12 12.05 157.24 CF & C'F' fi & f 'i' 24 14.79 234.86 FI & FT im & i'm' 36 16.93 279.79 IM & I'M' mp & m'p' 48 18.5 307.25 MP & M'P' ps & p's' 60 19.51 323.72 PS & PS' sv & sV 72 19.97 332.07 SW & S'W No strain in cf. Simple Truss No. 3 bears — (w + —f-j — -—-, and since its end post is distant n beyond the abutment, the moment of the abutment reaction is (— — -JL )(x+n), the moment oi the load on x is — - — -J— — -±L ; 61 61 U ' hence, H _(w _ wp\ ( v _ m? , wpx wp\ My If ~ui {x +n) w + ~6T + -3i' 310 A TREATISE ON ... B. = ~\lx-x , -px + 2p' + nl^-2 I )n); (356) 6ly substituting values, 250 H = 203y L(U$-cc) + 204.8 |, whence the following table of chord strains in Simple Truss No. 3 : Values of x. Values of y. Strains in Tons. Horizontal Component of Compression in Tension in 8 11.03 131.7 BE & B'E' eh & e'h' 20 13.95 217.25 EH & EH' hi & hi' 32 16.28 263.80 HL & H'L' lo & l'o' 44 18.04 290.57 LO & L'O' or & o'r' 5G 19.24 305.29 OR & OR' ru & r'u' 68 19.88 311.34 RW & RW No strain in be. These are the horizontal strains in the chords of the Simple Trusses. The strain in the Compound Truss in any member is the sum of the strains in the Simple Trusses affecting that member; for example, in the upper chord the strain in HI is the sum of the strains in HL in No. 3, in GK in No. 1, and in FI in No. 2. In the lower chord the strain in hi is the sum of the strains in hi, gk, and/z. Consequently we may arrange a table as follows : put the panel points, beginning at THE STRENGTH OF BRIDGES AND ROOFS. 311 the end of the truss, in a vertical column ; opposite to and in line with each point, place the strain extending from that point to the next point towards the centre of the same simple truss ; then at any point the sum of the simple truss strain at that point, and the two strains immediately above it, is the total strain in that member of the upper chord whose abutment end is at the point referred to. Similarly of the lower chord tensions, thus : Upper Chord Panel Points. Lower Chord Panel Points. Simple Truss Strains. Compound Truss Strains. Horizontal Component of Compression in Tension in A d 110.34 110.34 AB & A'B' de & d'e' B e 131.70 242.14 BC & B'C ef & e'f C f 157.24 399.28 CD & CD' ig & f 'g' D g 217.14 506.08 DE & D'E' gh & g'h' E h 217.25 „ 591.63 EF & E'F' hi & hi' F i 234.86 669.25 FG & F'G' ik & i'k' G k 273.13 725.24 GH & G'H' kl & kT H 1 2G3.80 771.79 HI & HI' lm & I'm' I m 279.79 816.72 IK & I'K' mn & m'n K n 305.81 849.40 KL & K'L' no & n'o' L 290.57 876.17 LM & L'M' op & o'p' M P 307.25 903.63 MN & M'N pq & p'q' N q 325.25 923.07 NO & N'O' qr & q'r' r 305.29 937.79 OP & OP' rs & r's' P s 323.72 954.26 PQ & P'Q' st & s't' Q t 335.75 964.76 QR & Q'R' tu & t'u R u 311.34 970.81 RS & R'S'* uv & uV S V 332.07 979.16 ST & ST' vw & v'w T 338.95 982.36 TU, UV, VW, T'UMJ'V & V'W 312 A TREATISE ON Multiplying the horizontal components of the com- pressions in the upper chord by the length of the mem- ber, and dividing by the horizontal length of the panels, we obtain the following table of upper chord compres- sions AB& AB BC& B'C CD& CD' DE& D'E' EF& E'F F'GK G'H' HI& HI' IK& IK' KL& K'L' LM& L'M' 119.54 262.32 413.25 523.79 610.87 691.60 747.00 793.01 837.12 868.51 893.69 {Continued) MN& M'N' NO& NO' 0P& OP' PQ& P'Q' QR& Q'R' RS& R'S' ST& S'T' TU& T'U' UV& U'V VW& V'W 921.70 939.22 951.86 966.19 974.41 980.52 986.50 987.77 984.82 982.36 The excess of horizontal strain in both chords of No. 1, caused by this objectionable extension of the truss w beyond the abutments, is the quantity —-(In + pn), Sly amounting to 20.53 tons at the centre, increasing to 41.38 tons at the abutments. In No. 3 this excess, w (nl — 2pn), is 10. tons at the centre and 18.45 tons at the abutments, making a total unnecessary excess of strain of 30.53 tons a tons at the abutments. strain of 30.53 tons at the centre, increasing to 59.83 THE STRENGTH OF BRIDGES AND ROOFS. 313 222. Vertical Strains from the Permanent Load.— The above tables show that under a full load the chord strains increase from the abutments to the centre, and consequently the vertical braces are struts and the in- clined braces ties. The figures of the Simple Trusses, (84), (85), and (86), show all the braces needed under a full or constant uniform load. If the truss supports only its own weight, w, the strains in the chords will be only one-half those given in the tables, since w is one- half of w f + w ; and the horizontal component of the strain in any brace (Kn, of Simple Truss No. 1, for example), from the constant load, w, will be the difference between the horizontal components of the strains in GK and KN, or, what is the same, between the strains in kn and nq ; hence Sp : y :: H — H' : V, or the horizontal extent of any brace is to its vertical extent as the hori- zontal component of the strain in that brace is to its ver- tical component. And the compression from the same load in any vertical brace is equal to the vertical strain in the tie which meets that brace at the lower chord, less a panel weight of that load, —S-, the load being con- side red as concentrated at the lower chord panel points. In this manner the following tables of the strains in the vertical struts and of the vertical components of the tensions in the inclined ties from the permanent load w are formed : 314 A TEEATISE ON Simple Truss No. 1. Strains in Tons. 68.96 44.14 30.42 21.20 14.19 8.27 2.63 4.89 7.90 Tension in Ad& A'd' Dg& D'g' Gk& G'k' Kn & K'n Nq& N'q' Qt & Qt' Tw& T'w Tt& T't' VVw Strains in Tons. 85.51 55.8 80.98 17.26 8.04 1.03 Compres- sion in Aa & A'a' Dd& D'd' Kk& K'k' Nn& N'n' Qq& Q'q' As the vertical strain in Qt is 8.27, and the load upon point t is —±- — 13.16, the strain in Tt is tension, L and its amount is 13.16 — 8.27 = 4.89. The vertical strains in Tw and T'w = 5.26 ; the load at w = 13.16 ; hence 15.16 — 5.26 = 7.90 tension in Ww. Simple Truss No. 2. Strains in Tons. 57.32 38.97 27.68 19.37 12.69 6.78 0.47 6.37 Tension in Cf & Of Fi& FT Im& I'm' Mp& M'p' Ps& P's' Sv& S'v' Ss& S's' Vv& V'v' Strains in Tons. 78.94 44.16 25.81 14.52 6.21 Compression in Cc& C'c' Ff & FT Ii& IT Mm& M'm' Pp& P'p' There is tension in Ss and Vv, as explained after the previous table, because the inclined ties from their lower chord points do not support a panel load. the strength of bridges and roofs. 315 Simple Truss No. 3. Strains in Tons. 59.91 39.32 27.02 18.16 11.06 4.85 2.10 8.31 Tension in Be& Be' Eh& E'h' H1& HT Lo & L'o' Or& Or' Ru& RV Rr & R'r' Uu & U'u' Strains in Tons. 78.94 46.75 26.16 13.86 5.00 Compression in Bb& B'b' Ee& E'e' Hh& H'h' L1& LT Oo & O'o' The cause of the tension in Rr and Uu was explained before ; the compression in each end post of any simple truss is obviously half the weight borne by that truss. 223. Vertical Strains from a Moving Load. — Let I — U represent the length of the moving load as before, ex- tending from one abutment to the centre of a panel of Simple Truss No. 1 ; then the load on I — u is —(I — u + £ ] ; divide this by I + 4^, which is the length of the leverage of this simple truss, and multiply by — (Z — u — ■£- J + 2n, the distance of the centre of gravity of the load from the loaded abutment, and we have for the reaction of the unloaded abutment upon this simple truss. 316 A TREATISE ON Taking moments around any panel point of this truss outside the load, x and u being measured from the same abutment, we have, H = — (358) y for the tension in the lower chord and the horizontal component of the compression in the upper chord. And at the next panel point of the same truss towards the abutment from which x is measured, H / = V(g-8p) .... (359) Subtracting Eq. (359) from Eq. (358; H - H' = \yW-y) + *Py\ . (360) yy is the horizontal component of the brace which extends from x to x — 3p, and which, since the horizontal strain increases towards the end of the load, is a tie, if the upper end inclines towards the abutment from which x is measured. The vertical extent of this tie is y\ its horizontal extent is 3^>; hence, multiplying Eq. (360) by y\ the vertical extent of this tie, and dividing by its horizontal extent, 3p, we obtain Y[ a V-ti + i\ - - - (361) for the vertical component of the strain in any inclined brace, outside the load, from a moving load, whose lower chord end is distant x from the unloaded abutment, whose vertical extent is y\ y being the vertical height THE STRENGTH OF BRIDGES AND ROOFS. 317 of the truss at x. Substituting for V its value given by Eq (357), we have Substituting the values given above, this becomes (152 -mV- 4.8 1*4-735.2) W-y) , j ) (363) 286.0032 ( 12j ) The greatest strain from this equation upon any in- clined tie is when the load covers the point to which x is measured, or when u = x — -|-, and this strain is greater than the strain from the full load. To this strain from the moving load is to be added the strain from the constant load so long as the two are acting upon the same abutment ; but when x of Eq. (363) becomes greater than — , then the strains in the same panel are acting in opposite directions, and one neutralizes its amount in the other. While Eq. (363), with x greater than — , is greater than the constant load vertical strain, in the same panel in which the brace from a? to a? — 3p is, a tie whose upper end inclines to- wards the abutment from which x is measured is needed to support the difference in these vertical strains. In the following table the constant load vertical strains taken from the table for Simple Truss No. 1 are placed with the plus sign, as we proceed from either end towards the centre and beyond with the minus sign. 318 A TREATISE ON This is upon the same reasoning given before. The sum is the strain in the inclined brace. Values of u. Values of X. Values oiy. Values oiy'. Strains in Tons from Moving Load. Constant Load Vertical Strains. Total Strains in Tons. Tension in 10 16 13.04 9.92 +49.65 +44.14 +93.79 Dg& D'g' 22 28 15.57 13.04 + 38.06 +30.42 + 68.48 Gk& G'k' 34 40 17.52 15.57 +31.88 +21.20 + 53.08 Kn& K'n' 46 52 18.9 17.52 +28.11 + 14.19 -^ 42.30 Nq& N'q' 58 64 19.72 18.9 + 25.28 + 8.27 + 33.55 Qt & Q't' 70 76 20. 19.72 +22.72 + 2.63 +25.35 T\v& T'w 82 88 19.72 20 +20.16 - 2.63 + 17.53 Wt'& Wt 94 100 18.9 19.72 +17.83 - 8.27 + 9.56 There is no inclined brace from T' to q' or from T to q, consequently the strain, 9.56 tons, is compression in the braces with the opposite inclinations, Q't r or Qt, with the moving load assumed, which is certainly not too great. "When u =106, the strain from the moving load to the abutment from which it is measured, is less than the permanent strain in the same panel to the opposite abutment, and consequently no further counter- bracing is necessary. Subtracting from these strains the weight of a panel THE STKENGTH OF BRIDGES AND EOOFS. 319 load permanent and movable, A — ~ — '-±- == 26.32 tons, we obtain the compression in the vertical braces con- nected with the lower ends of the ties, as follows : Strains in Tons. 67.46 42.16 26.76 15.98 7.23 Compression in Gg & G'g' Kk & K'k' Nn & N'n' Qq & Q'q' Tt & T't' No compression in ¥w. Under the moving load one-half of Simple Truss No. 2 is united by the counterbraces with the opposite half of Simple Truss No. 3. Let the load be considered as extending from the abutment end of Simple Truss No. 3, or from left end of Fig. (87). The load on I — u is, u being distances from the right abutment to the centres of the panels of Simple Truss No. 2 while less than — , and when greater than , to the centres of the panels of Simple Truss No. 3, 1 w' /r p\ 3 r % Dividing by the length of the leverage of the halves 320 A TREATISE ON of these two trusses, I + n, and multiplying by L(7 — u + — )+ n, the distance of the centre of gravity of the load from the loaded abutment, we have for the reaction of the unloaded abutment upon Simple Truss No. 2 when u is less than — , and upon No. 3 mi when u is greater than — . L Substituting values in Eq. (364), we obtain by the same process of reasoning as in the case of Simple Truss No. 1, (156- M )'-2.4 M + 356 j%'-y) | V_ ~ 279.4368 "l-lty - >' (365) for the vertical component of the strain, from the mov- ing load, in any brace of Simple Truss No. 2 whose lower chord end is distant x from the abutment towards which the brace leans, while x does not exceed — , and 2 beyond that point in the counterbraces of Simple Truss No. 3 having similar inclinations. In the following table the strains from the moving load are from Eq. (365) ; the strains from the constant load are from the table given before for Simple Truss No. 2 ; to the centre and beyond that, where they be- come minus, from Simple Truss No. 3. THE STRENGTH OF BRIDGES AND ROOFS. 321 9 O ■ a > o ■ > O 02 d 2- ■ P "a Strain in Tons from Moving Load. Constant Load Vertical Strains. d d ±[ 3 m° J o H & d .2 *S © 18 24 14.79 12.07 +41.33 +38.97 +80.30 Fi & FT 30 36 16.93 14.79 + 33.61 +27.68 +61.29 Im & I'm' 42 48 18.5 16.93 +29.21 + 19.37 +48.58 Mp & M'p' 54 60 19.51 18.50 + 26.08 + 12.69 +38.77 Ps & P's' 66 72 19.97 19.51 +23.42 + 6.78 +30.20 Sv & SV 78 84 19.88 19.97 +20.84 + 0.00 +20.84 Vu' & V'u 90 96 19.24 19.88 + 18.04 - 4.85 + 13.19 UY & Ur 102 118 18.04 19.24 + 14.93 -11.06 + 3.87 The compression in Or and O'r', since there are no ties, Ro and B/o' is 3.87 tons. Beyond u = 102 no counterbracing is necessary. Subtracting from the strains a panel load, * — ~Xr — '-L I = 26.32 tons, we may form the following table of com- pressions in the vertical braces : Strains in Tons. 53.98 34.97 22.26 12.45 3.88 7.68 Compression in Ii & IT Mm & M'm' Pp& P'p' Ss & S's' Vv& vv UV& Uu If the load be considered as extending from the 21 322 A TREATISE ON abutment end of Simple Truss No. 2, or from right end of Fig. (87), by similar reasoning we obtain for the reaction of the unloaded abutment, and substi- tuting values, (152-tt)'-86 j s(y'-y) + 1 } . . (367) 279.4368 I 12y ) x for the vertical component of the strain from the moving load in any inclined brace of Simple Truss No. 3 whose lower chord end is distant x from the abutment towards which the brace leans, while x does not exceed —, and beyond that point in the counterbraces of Simple Truss No. 2, having similar inclinations. The following table is prepared as were the previous two of the inclined brace strains : Values of u. 8 «M o DQ o tn > Strains in Tons from Moving Load. Constant Load Vertical Strains. X/l o & a .2 14 20 13.95 11.03 + 44.21 + 39.32 +83.53 Eli & Eh' 26 32 16.28 13.95 + 35.09 +27.02 +62.11 HI & HT 38 44 18.04 16.28 + 29.78 + 18.16 +47.94 Lo & L'o' 50 56 19.24 18.04 +26.30 +11.06 +37.36 Or & Or' 62 68 19.88 19.24 +23.61 + 14.85 +28.46 Ru&R'u' 74 80 19.97 19.88 +20.97 +20.97 Uv' & U'v 86 92 19.51 19.97 + 18.26 - 6.78 + 12.48 V's' & Vs 98 104 18.5 19.51 + 15.16 12.69 + 2.47 , THE STRENGTH OF BRIDGES AND ROOFS. 323 The compression in Ps and P's', since there are no ties, Sp and S'p' is 2.47 tons. Beyond w — 98, counter- bracing is unnecessary. Whence the compression in the vertical braces is as follows : Strains in Tons. 57.21 35.79 21.62 11.04 Tension in Hh & H'h' LI & LT Oo & O'o' Rr & R'r' The maximum vertical compressions upon the end struts and ties of each truss, are when the truss is fully loaded, and are consequently double the strains given in the tables for the permanent load strains ; the second struts receive the maximum strains at the same time. The maximum tensions in the central vertical braces also occur under the maximum load ; hence the following table : Strains in Tons. 171.02 157.88 157.88 111.60 93.50 88.32 Compres- sion in Aa & A'a' Bb& B'b' Cc& C'c' Dd& D'd' Ee & Ee' Ff & FT Strains in Tons. 137.92 119.82 114.64 4.20 0.94 9.78 16.62 12.74 15.80 Tension in Ad& Ad' Be& Be' Cf & C'f Rr& R'r' Ss& S's' Tt& Tt' Uu& U'u' Vv& V'v' Ww The strains which have been determined for the in- clined braces are the vertical components of the strains to which they are subject ; hence it is necessary to divide them by the vertical extents and multiply them by the lengths of the braces, which is the same as multiplying 324 A TREATISE ON them by the secants of the angles the braces make with a vertical line, to obtain the longitudinal strains. Per- forming this operation we may form the following table : Strains in Tons. Tension in Strains in Tons. — i Tension in 176.71 Ad & Ad' 51.28 Nq & N'q' 178.08 Be & Be' 44.88 Or & Or' 198.59 Cf & C'f ' 46.21 Ps & Ps' 144.09 Bg & D'g' 39.74 Qt & Q't' 126.39 Eh & E'h' 33.55 Ru & RV 113.23 Fi & FT 35.45 Sv & SV 93.06 Gk & G'k' 29.18 Tw & T'w 81.70 HI & HT 24.49 Uv' & U'v 85.03 Im & I'm' 24.31 Vu' & V'u . 67.02 Kn & K'n' 20.49 Wt' & Wt 59.54 Lo & L'o' 13.94 V's' & Vs 59.54 Mp & M'p' 15:41 UY & Ur Strain in Tons. 4.65 2.94 11.32 Compression in Or & Or' Ps & Ps' Qt & Q't' THE STRENGTH OF BRIDGES AND ROOFS. 325 CHAPTER XII THE BOLLMAN AND FINK TRUSSES. 224. — These trusses take their names from the design- ers, and, strictly speaking, do not come within the definition of a truss, as given in the opening chapter. Each may be considered, however, as composed of a number of simple triangular trusses, and the determina- tion of the strains affecting them presents no difficulty. CASE I. THE BOLLMAN TRUSS. 225. The Hoi i man Truss. — Let Fig. (88) represent a truss in which the only chord, AB, is horizontal, and in b c d e fghi k 1 mno p q Fig. 88. which the inclined braces meet the chord at its ends only. It is evident that the strain affecting this chord is compression, and that it is uniform throughout the length of the chord, for no brace whose strain has a 326 A TREATISE ON horizontal component meets the chord except at its ends, and that it is greatest when the truss is fully loaded. 226. Horizontal Strains. Let I = the length of the truss, d = the depth of the truss, or the length of the vertical braces, p = the length of a panel, or the distance between the vertical braces, w = the maximum uniform weight, both fixed and movable. Taking moments around a point in the centre of truss, and in the line of the lower ends of the vertical braces, the strains in the inclined braces crossing the centre may be disregarded, as their amounts in opposite directions exactly balance each other, and we have, therefore, H =;4 (m is the strain in the horizontal chord. 227. Vertical Strains in the Vertical Braces. — The vertical braces are struts, and each one can only support the maximum panel load, for no brace connects with its upper end to bring any greater vertical strain upon it ; hence its greatest strain is when it is fully loaded either from a full truss load or from a moving load ; therefore V= T ( 367 > is the maximum strain in each vertical brace. THE STRENGTH OF BRIDGES AND ROOFS. 327 228. Vertical Strains in the Inclined Braces. — Each vertical brace, with the inclined braces attached to its lower end, is, with the horizontal chord, a simple trian- gular truss. Let x be the distance of a vertical strut from one abutment ; then by the principles of the lever, since -J- is the load upon the strut, -j- _i- I X (I — x) Wp WpX /Q£Q\ = _-___, .... (368) is the reaction of that abutment from which x is mea- sured upon the inclined brace from the lower end of the strut to which x is measured, and consequently the ver- tical component of the strain, which is tension, in that inclined brace. Similarly, V = ^ (369) V is the vertical component of the strain in the other in- clined brace from the same strut. One equation only, however, is needed. Dividing these equations by d, the depth of the truss, or the vertical extent of the inclined tie, and mul- tiplying by the length of the tie itself, we obtain the longitudinal strain. 229. Example.— In Fig. (88), Let I = 160 feet, the length of the truss or chord, d = 15 feet, the depth of the truss, p = 10 feet, the length of a panel, w = 240 tons, the full weight of the truss and of the load. 328 A TEEATISE ON Therefore — _ = 2 40Xl60 = 320 tons, compression , , , A , wp _ 240X10 throughout the horizontal chord. And -j- j^ — = 15 tons, compression in each vertical strut; and wp wpx _ -i e 3# X T" " 32* Substituting values of x, dividing by d, and multi- plying by the lengths of the ties, we have the following table of tensions in the ties : Values of x. Strains in Tons. Tension in 10 16.09 bA& qB 20 21.36 cA& pB 30 27.25 dA & oB 40 32.04 eA & nB 50 35.89 f A & mB 60 38.65 gA &1B 70 40.26 hA&kB 80 40.71 iA&iB 90 39.92 kA&hB 100 37.92 lA&gB 110 34.69 mA&fB 120 28.23 nA & eB 130 24.54 oA & dB 140 17.6 pA & cB 150 9.42 qA&bB THE STRENGTH OF BRIDGES AND ROOEi IVEESI 4gR 230. A Truss Containing an odd number of Panels.— This truss may be divided into an odd number of panels or chord members, in which case, by taking moments as before, we obtain for the chord strain, 011*3 H = wl wp Sdl (37-0) The strains in the braces are found from the same equations as in the previous case. 231.— In practice, in the Bolhnan Truss, there are two light rods from the bottom of each truss to the tops of the next on either side, and a horizontal rod connect- ing the lower ends of the struts. CASE II. THE FINK TRUSS. 232. The Fink Truss. — In this truss, shown in Fig. (89), as in the Bollman, the single chord is subject to compression, the vertical braces are struts, and the in- clined braces ties. ABCDEFGHI L M N O P Q B Pig. 89. 233. Vertical strains. — The struts attached to the chord at the points B, D, F, H, K, M, O, and Q, or the alternate struts beginning with those next the abutments, each bear one panel load, since, as no brace meets them 330 A TREATISE ON or is connected with them at the chord, they can receive no weight or strain except from the load immediately upon them. The braces from their lower ends having equal inclinations receive equal amounts of vertical strain ; or Let I = the length of the truss, p = the length of a panel, w = the uniform maximum load, then ^ is the weight upon each of these struts, and V ^2 the vertical component of the tension in each of the ties from their lower ends, or in all the ties in the truss having a horizontal extent of p, or a panel length. The struts from the points C, G, L, and P, have, in in addition to the panel load upon them, a half panel load from each of the inclined braces which meet them at the chord, and consequently are subject to a vertical strain of — £- ; the inclined braces from their lower ends c therefore have each a vertical strain ^, and a horizon- tal extent of two panel lengths, or 2p ; whence we form this rule. The vertical component of the strain in any tie equals half the panel loads in its horizontal extent ; thus nR has a horizontal extent of four panels, the vertical com- ponent of its strain is therefore <2 ^P- ; iR has a horizon- tal extent of eight panels, the vertical component of its THE STRENGTH OF BRIDGES AND ROOFS. 331 strain is therefore — -±- ; or let n == number of panels in l the horizontal extent of a tie, then — ±~- is the vertical component of the strain in that tie, and this divided by the depth of the truss, and multiplied by the length of the tie, will give the longitudinal strain. Evidently -2~- is the compression in the strut to whose lower end the tie, having a horizontal extent of n r is attached. 234. Horizontal Strains. — The strain in the chord from any tie having a horizontal extent of p, or the horizontal component of the strain in that tie, is From the ties whose horizontal extent = 2p, From the ties whose horizontal extent = 4p, j . i n .. 2wp. 8wp*. and from those whose horizontal extent = &£>, b &L It is evident that all these strains affect each member of the chord, and that, therefore, W P* i % W P* i % W P* i 3% W P* _ 4:2iwp* 332 A TREATISE ON is the uniform compression throughout the chord of a Fink Truss, containing sixteen panels. It will be noticed that general equations are not o-iven for this truss, but special equations are determined for a truss divided, as shown in the figure ; and for a truss containing a different number of panels a different horizontal equation must be prepared. It will also be noticed that the greatest strains are from the full load. 235. Example.— In Fig. (89), Let I = 160 feet, the length of the truss, d — 15 feet, the depth of the truss, p = 10 feet, the length of a panel, w = 240 tons, the maximum uniform load. Hence 42 W_ 42£x 240x10x10 _ dl 15x160 compression throughout the upper chord. For the struts, -^— = 15 n, whence, V 425 tons, Values of n. 1 2 4 8 Strains in Tons. 15 30 60 120 Compression in Bb,Dd,Ef, Hh, Kk/Mm, Oo, & Qq Cc, G£, LI, &Pp Ee&Nn Ii THE STRENGTH OF BRIDGES AND ROOFS. 333 For the tension in the ties, — |~, divided by d and Jib multiplied by their lengths, we have, Values of n. 1 2 4 8 Strains in Tons. 9.01 24.41 85.44 325.66 Tension in Ab, Cb, Cd, Ed, Ef,Gf,Gh,Ih, Ik, Lk, Lm, Nm, No, Po, Pq, & Rq. Ac, Ec, Eg, Ig, 11, Nl, Np, & Rp Ae, Ie, In, & Rn Ai&Ri 334 A TEEATISE ON CHAPTER XIII. RESISTANCE OF MATERIALS TO COMPRESSION AND TENSION. I. — COMPRESSION. 236. — Mr. Eaton Hodgkinson, to whom we are almost wholly indebted for our knowledge of the strength of pillars and struts, divides them into three classes : 1. Short pillars, whose lengths, compared with their diameters, is so short that they yield by crushing alone. 2. Medium, or short flexible pillars, whose lengths, compared with their diameters, is such that they fail partly by flexure or bending, and partly by crushing. 3. Long flexible pillars, whose lengths are so great, compared with their diameters, that they fail by flexure or bending, like a girder subject to a transverse strain, the breaking weight being much less than the crushing weight. 237. Formulae for the Strength of Pillars Empirical. — The formulae for the resistance of long pillars to flexure or bending are generally based upon the formula? for the resistance of a girder supported at both ends and subject to a transverse weight or pressure, although the THE STRENGTH OF BRIDGES AND ROOFS. 335 latter are not altogether satisfactory, nor do they entirely conform to the results of experiment. The girder and the long pillar both yield from flex- ure, but there is a difference in the conditions of the two which will materially affect the formulas for the resist- ance of the latter to the flexure derived from their resemblance. A girder, supported at both ends, and subject to a transverse pressure or weight, has its great- est longitudinal strain concentrated at one point, the compression on one side being equal to the tension on the opposite, and has no longitudinal strain at the ends ; but a loaded pillar or strut must sustain, in some part of any transverse section, throughout its whole length, an amount of compression equal to the weight imposed upon it. If the line of this weight coincide with the axis of the pillar, failure will result from crushing, unaffected by the length, and there is no resemblance to the girder ; and if, as is generally true of long pillars, this coincidence does not exist, the conditions of a girder are not immediately fulfilled, for the line of pressure may be nearer one side, so that the compression upon that side may be greater than upon the other, where the compression may be slight, or where there may actually be no strain. This may be true even with a slight bend- ing of the pillar. But flexure soon produces tension in the convex side of the pillar, and this tension may resemble that in a loaded girder ; that is, it may be greatest at one point and diminish in a certain ratio from that point towards the ends. It must diminish or 336 A TREATISE ON vanish at some point, because it cannot exist at the ends, and it cannot vanish without neutralizing, or being neu- tralized by, an equal amount of compression ; hence it either lessens the compression in the concave side, or when tension exists in a pillar, there is present an amount of compression equal to the weight imposed and to the tension. Therefore it would seem that a long pillar, yielding by flexure, would more nearly resemble a girder bending under a uniform load, and subject to two equal and opposite thrusts at its ends. In the absence of sufficient experiment our only re- course is to such formulae, and their modifications or adaptations to different forms, empirical though they be, as have generally been adopted with safety, using such as are adapted to truss struts and chord sections, each section being considered as a separate pillar. 23§. Short Wooden i»iiiur». — The crushing weight of American yellow pine is given by Prof. Rankine as 5,400 pounds per square inch, and of oak as 6,700. These quantities seem small, but there is no doubt that the crushing weight of white pine, given by some American authors as 10,000 pounds per square inch, is too great. The extreme safe working load for wooden pillars may be put at 800 pounds per square inch for permanent structures ; for temporary purposes this load may be exceeded. Rondolet found that wooden pillars do not yield by flexure until their length exceeds ten times their smallest diameter. The load of 800 pounds per square inch, however, may be carried until the THE STKENGTH OF BEIDGES AND ROOFS. 337 length so exceeds the diameter that the formula for flexure, given beyond, gives a less weight, which is when the length is about 29 J times the diameter. 239. Long Wooden Pillars. — Prof. Rankine gives this formula for the working strength of long rectangu- lar wooden pillars : w _ 300,000 fr? .... (371) where Wis the safe working strain in pounds, b is the breadth in inches, d is the depth, or least lateral dimension, in inches, I is the length in inches. Example. — What is the safe working strain for a wooden pillar 10 feet long, 4 inches deep, and 5 inches in breadth ? tt w 300,000x5x4 3 apP - ; Here, W = 2 = 6667 pounds. ' 14400 F 240. Square Wooden Pillars. — u Of rectangular wooden pillars, it was proved experimentally, that the pillar of greatest strength, where the length and quantity of material is the same, is a square." — Hodghinson. 241. Solid Pillars of different Transverse Sections.— " It appears that the strength of (long) circular, square, and triangular pillars of the same quality, weight, and length vary as 55.299, 51.537, and 61.056, the last being the strongest." — Hodghinson. 22 338 A TREATISE ON Or if the strength of a long, round pillar be 100, that of a square solid pillar will be 93, and that of a triangular solid pillar 110, the pillars being of equal length and transverse section. 241. — The crushing strain of cast-iron is 85,000 pounds per square inch. 242. The Load for Ca§t-Iron Solid Pillars. — The formula for solid round cast-iron pillars, whose lengths exceeds 30 times their diameters, deduced from Hodg- kinson, is W=^?T, .... (372) in which Wis the safe working load in pounds, d, the diameter in inches, Z, the length in feet. Tables of the powers of d 35 and I t63 are given at the end of the chapter. Example.— What is the safe weight for a solid pillar of cast-iron 10 feet long and 3 inches in diameter? Taking the values of 10 163 and 3 35 from the tables, we have w _ 6500X46.765 • _- Q , W ~ 42.658 -=' 078 P QU » ds - 243. The Load for Cast-iron Hollow Pillars. w= 6500(^_-^) _ _ _ (378) THE STRENGTH OF BEIDGES AND ROOFS. 339 in which W, 746 THE STKENGTH OF BRIDGES AND EOOFS. 341 248. Equality in the Thickness of Hollow Pillars not important. — Mr. Hodgkinson remarks that, " where there is an inequality in the thickness of hollow pillars it does not produce much diminution of strength.' 1 Mr. Stoney adds : " In practice, neither the excess nor the want of thickness should exceed twenty-five per cent, of the average thickness. If, for instance, a hollow pillar is specified to be one inch in thickness, then in no place should the metal be less than three-quarters of an inch, nor more than one and a quarter inch thick. 249. — " In all the pillars with rounded ends, those with increased middles were stronger than uniform pillars of the same weight, the increase being about one- seventh of the weight borne by the former." — Hodg- kinson. They were solid cast-iron pillars. 250. Formulae for Medium Pillars. — The above for- mulae apply to all pillars whose lengths exceed thirty times their external diameters (for cast-iron and timber, and sixty diameters for wrought iron). For pillars shorter than this they give too great a weight, and for such Mr. Hodgkinson has modified his formulae, from which we have, W -60WT8? - - - - < 379 > where W == the safe working load in pounds, W = the safe working load in pounds, derived from the formulae given above for long pillars, 342 A TREATISE ON c = the crushing weight of the pillar ; that is, the crushing weight per square inch of the material, multiplied by the sectional area in inches. 251. — In hollow cast-iron pillars Mr. Hodgkinson found that no additional strength was obtained by en- larging the diameter at the middle. Solid square pillars " do not break in a direction parallel to their sides, but to their diagonals nearly." — Hodgkinson. 252. wrought iron. — Mr. Hodgkinson gives the crushing weight of wrought iron as 35,800 pounds per square inch, being much less than that of cast-iron; hence a short cast-iron pillar can resist a much greater crushing strain than wrought-iron. But, as Mr. Stoney remarks, the strength of very long pillars depends not on the strength of the material, but on its stiffness and capability of resisting flexure ; hence, " although a short pillar of cast-iron will bear a much greater weight than a similar pillar of wrought-iron, yet a very long wrought- iron pillar will support a greater weight than a similar one of cast-iron, as the co-efficiency of elasticity is con- siderably higher than that of cast-iron." 253. Long Solid Wrought Iron Pillars.— The formula for the safe working load of solid wrought-iron pillars is, W = 24967 d™ (380) THE STRENGTH OF BRIDGES AND ROOFS. 343 The working load being put at one-fourth the breaking weight ; W, d, and I having the same meaning as before. 254. Hollow Wrought Iron Pillars. — The strength per square inch of hollow wrought-iron pillars seem to de- pend greatly upon a ratio between the thickness of the plate and the diameter of the pillar, though this ratio is not satisfactorily known. Prof. Rankine says : " The ultimate resistance of a square wrought-iron pillar, when the thickness of the plate is not less than one-thirtieth the diameter or side of the pillar, is 27,000 pounds per square inch.'" The strongest form of a hollow rectangular pillar is where the chief part of the material is concen- trated at the angles. 255. Relative Strength of Long Pillars. — Putting the strength of a long cast-iron at 1,000, the strength of similar pillars of other materials was found by Mr. Hodgkinson to be : Cast-iron, - "Wrought-iron, Cast-steel, Oak, - - Pine, 1,000 1,745 2,518 109 79 Hence, from the formulae for cast-iron, the strength of similar pillars of other materials may be found. 344 A TREATISE ON II. TENSION". 256. — No formula is required for tension, since the strength of any member of a truss subject to a tensile strain varies directly as its weakest transverse section, and is unaffected by the length ; hence we need only to know the tearing weight of different materials and the safe working load. The safe working tension for timber may be put at 900 pounds per square inch of the smallest transverse section. Cast-iron is not suited for tension ; its tearing weight is 15,680 pounds per square inch, and its safe working load should not exceed one-sixth of this. The tensile strength of wrought-iron is 54,000 pounds per square inch ; its safe working load should not exceed one-fourth of this. THE STRENGTH OF BRIDGES AND EOOFS. 345 Table I. — Powers of Lengths, or Z 1 63 iMP— 1. 9 ms _ 32.927 17 !- 63 — 101.305 2 i-63 _ 3095 10 »*■ _ 42.658 18 1>63 mm 111.197 3 I-**— 5.994 11 M3_ 49.828 19 Ml M 121.442 4"»— 9.58 12 ms _ 57.42 20 >•«- 132.032 5 mi — 13.782 13 Mf = 65.423 21 ms = 142.961 6 > 63 = 18.552 14 1.63 = 73823 22 ms — 154.223 7 »•# — 23.851 15 mi — 8 2.609 23 V- 63 = 165.812 8 mi = 29.651 16 mi = 91.773 24 MJ = 177.723 Table II. — Powers of Diameters, or d 3.5 1 8 - 5 — 1. 4.75 s ' 6 = 233.58 8.5 3 - 5 = 1790.47 1.25 3 - 5 — 2.184 5 3 -s = 279.51 8.75 3 ' 5 = 1981.66 1.5 3 -» — 4.134 5.25 3 - 5 = 331.56 93.5 = 2187.00 1.75 »- 5 — 7.09 5.5 3 ' 5 =390.18 9.25 3 - 6 = 2407.11 2 3 - 5 =11.314 5.75 3 - 6 = 455.87 9.5 3S = 2642.61 2.25 3 - 5 =17.086 6 8 ' 5 = 529.09 9.75 3 - 5 = 2894.12 2.5 »- 5 =24.705 6.25 3 - 5 = 610.35 10 3.5 = 3162.28 2.75 3 ' 5 — 34.488 6.5 3 - 5 =700.16 10.25 3 - 5 = 3447.73 3 3 - 5 — 46.765 6.75 3 * 6 — 799.03 10.5 3 ' 5 = 3751.13 3.25 3 - 5 = 61.886 7 3 - 6 = 907.49 10.75 3 - 5 = 4073.14 3.5 3 ' 5 =80.212 7.25 3 ' 5 = 1026.08 H3.5 = 4414.43 3.75 3 ' 5 = 102.12 7.53.5 M 1155.35 11.25 3 - 5 = 4775.66 4 3 - 5 = 128.00 7.75 3 ' 5 = 1295.85 11.5 3 6 = 5157.54 4.25 3>6 = 158.26 8 3 ' 5 = 1448.15 11.75 35 — 49.174 8.5 2 - 5 —210.63 1.25 2 - s — 1.747 5*« — 54.90 8.75 s - 5 — 226.47 1.5 s * — 2.756 5.25 S5 — 63.153 9 2 ' 6 — 243.00 1.75 *- 5 — 4.052 5.5 s ' 5 — 70.943 9.25 2 - 6 — 260.23 2 s - 5 — 5.657 5.75 2 - 6 — 79.281 9.5 2 -» —278.17 2.25 *- 5 — 7.594 6 2.5 _ 88>181 9.75 2 - 8 — 296.83 2.5 2 - 5 — 9.882 6.25 25 — 97.654 10 2 -» — 316.23 2.75 85 — 12.541 6.5 »- 6 —107.716 10.25 25 — 336.36 3 «•* — 15.588 6.75 »- 5 — 118.375 10.5 2 - 5 —357.25 3-25 25 — 19.042 7 2 - 5 —129.642 10.75 2 ' 6 — 378.99 3.5 95 —22.917 7.25 s - 5 — 141.53 11 2 - 5 — 401.31 3.75 « 5 — 27.232 7.5 2 - 5 —154.05 11.25 2 - 5 — 424.50 4 2 * — 32.00 7.75 2 * — 167.21 11.5 *•» —448.48 4.25 ! - 5 — 37.237 8 2 ' 6 —181.02 11.75 2 -« — 473.26 4.5 25 —42.967 8.25 25 — 195.37 12 2 - 5 — 498.83 THE END. ¥ THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 50 CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. AP * 1Q mm T#N tffftl 4 LD 21-100m-7,'40(6936s) *'i. '74-? TC THE UNIVERSITY OF CALIFORNIA LIBRARY