REESE LIBRARY OF THK UNIVERSITY OF CALIFORNIA. Received ^^Jf^**^ - A ccessions No.4*3 ^ 33 ^ he y No ^ -so V B.VV? : n : m MECHANICS OF THE GIRDER: Treatise on ^Bridges and T^oofs, IN WHICH THE NECESSARY AND SUFFICIENT WEIGHT OF THE STRUCTURE IS CALCULATED, NOT ASSUMED; THE NUMBER OF PANELS AND HEIGHT OF GIRDER THAT RENDER THE BRIDGE WEIGHT LEAST, FOR A GIVEN SPAN, LIVE LOAD, AND WIND PRESSURE, ARE DETERMINED. BY JOHN DAVENPORT CREHORE, C.E. 'Inveniam viam out faciam." NEW YORK : JOHN WILEY AND SONS. 1886. COPYRIGHT, 1886, BY MRS. J. D. CREHORE. JJJJ RAND, AVERY, AND CO., ELECTROTYPERS AND PRINTERS, BOSTON. PREFACE. THE "Mechanics of the Girder" is presented to the public in an unfinished condition, just as it was left at the author's ' death, in October, 1884. All that then re- mained to be done was to carry out an example in each of the twelve classes of girders in a manner similar to that of the Brunei Girder in Class I. (Sections 2 and 3, Chapter X.), and the Double Parabolic Bow and Post Truss in Class II. (Chapter XL). Of all these, the Post Truss promised to yield the most prolific results ; and it may be possible, before another edition is published, to complete this calculation at least, if not to introduce other examples from the later classes. However, the a priori method of the author is fully set forth previous to the tenth chapter ; and it is believed that no one else has as yet published any so satisfactory results from this method, if, indeed, the method has been hitherto attempted with any degree of success. The author's family feel deeply grateful to Professor John N. Stockwell for his kindness in devoting much of his valuable time to the supervision of the proof-reading, for the many suggestions he has given during the publi- iv PREFACE. cation, and particularly for his offer to conduct the work of completing the remaining examples. At his own sug- gestion, however, it has been thought expedient to delay no longer the publication of the completed portion of the book, and to leave any additional matter to be inserted later. WILLIAM W. CREHORE. July 29, 1886. ANALYTICAL TABLE OF CONTENTS. CHAPTER I. PRESSURES IN ONE PLANE. ART. PAGE 1. Force or pressure defined I 2. Resultant, component, equilibrium, defined 2 3. The parallelogram of forces 2 4. The triangle of forces 3 5. The resolution of a force 4 6. The resolution of many forces acting in one plane at a given point ... 4 7. The composition of forces 5 8. Polygon of forces 6 CHAPTER II. MOMENT OF A FORCE. 9. Definition and measurement of moment 9 10. Couples 9 11. Resultant of many co-axal couples 12 12. Arm of the resultant couple 14 13. Direction of the resultant couple 14 CHAPTER III. MOMENTS OF THE EXTERNAL FORCES APPLIED TO A BEAM OR GIRDER. SECTION i. The Semi-Beam, or Girder fixed at One End and free at the Other. 14. Formulae for concentrated and for uniformly distributed loads 17 15. Moments due many equal weights placed at equal intervals along the beam. Examples 19 1 6. Moments due many unequal weights placed at irregular intervals ... 22 v VI ANALYTICAL TABLE OF CONTENTS. SECTION 2. Horizontal Girder supported only at its Ends, which are not fixed. ART. PAGE 17. Data for required formulae 23 18. Re-actions at the piers 24 19. Tabulated moments for any load or pressure on the girder 25 20. Moments due uniform discontinuous load on any part of the beam. Examples 27 21. Moment at the foremost end of a uniform load advancing by panel weights. Example 33 22. Moment at the r th weight due n I equal weights uniformly distributed . 34 23. Moment at foremost end of advancing uniform load when the two end intervals are different from the common interval. Example .... 34 24. Moment at any weight due n equal weights applied at equal intervals, end intervals different from common interval. Example 35 25. Difference of simultaneous moments at consecutive points of division. First and second differences of these simultaneous moments. Exam- ples. Simple computation for maxima moments and differences . . 36 26. Point of greatest moment due uniform discontinuous load. Verification . 40 27. Greatest moment at foremost end of advancing uniform load. Example. Tabulated results 41 28. General expression for point of greatest moment due uniform dead and advancing uniform live load 44 29. Moment at foremost end of continuously distributed uniform load, advan- cing by continuous increments, and not by leaps. Example .... 45 30. Moment due both dead and live loads one interval beyond the foremost end of advancing load. Convenient method of computation .... 46 31. Position of foremost end of uniform continuous live load when moment at that end is a maximum 48 32. Position of foremost end of same when moment there is maximum for combined dead and live loads 48 33. Maximum moment due any uniform partial or complete continuous load . 48 34. Point of greatest moment due full continuous uniform load 49 35. Point of greatest moment due both dead and live loads 49 36. Moments when the panels are equal, but the weights are unequal, and are not applied at equal intervals. Example. Two locomotives . . 49 37. Moments due the same locomotives when their weight is uniformly dis- tributed throughout their length 57 SECTION 3. Horizontal Girder of One Span, with Fixed Ends. Effects of End Moments. 38. Formula for computing the influence of the end moments on the normal moments 59 39. Formula for the correction of the normal differences of moment for influ- ence of end moments. Examples illustrative 60 ANALYTICAL TABLE OF CONTENTS, vii CHAPTER IV. STRAINS IN FRAMED OR BUILT GIRDERS, DEDUCED FROM THE MOMENTS OF THE EXTERNAL FORCES AND FROM THE SHEARING-FORCES, AND FROM THESE COMBINED. ART. PAGE 40. Statement of a case 64 41. Strains found 66 42. Shearing-forces and shearing-strains. Definitions, with Table 1 67 43. Expressions for shearing force and strain 69 44. Formulae for increments of shearing-strains due to end moments .... 73 45. Shearing-strain at any vertical section of a girder in terms of the vertical components of the forces which are impressed upon the shearing-plane through the members of the girder cut by that plane 74 46. Strains in all members of a girder determined from the given shearing- forces . . . . i 75 47. Examples illustrative of two methods of determining strains in open girders . 76 48. Maxima strains in the web members of an open girder, deduced from the moments and shearing-forces combined, for uniform discontinuous dead and live loads 80 49. Classification of girders 86 CHAPTER V. MOMENTS OF RESISTANCE OF THE INTERNAL FORCES OF A BEAM OR GIRDER HAVING A CONTINUOUS WEB. SECTION i. General Formula found and applied to Particular Cross-Sections of^ Beams with Continuous Web. 50. Mode of estimating the moment of resistance . 134 51. Beam of rectangular cross-section 136 52. Hollow beam of rectangular cross-section 137 53. Beam composed of two vertical plates and two horizontal channels . . .138 54. Beam composed of two vertical I-beams and two equal horizontal plates . 139 55. Beam composed of two vertical channels and two horizontal plates . . . 140 56. Beam of T-shaped section 140 57. Solid or hollow square beam with diagonal vertical compared with the same beam when a side is vertical .'..... 142 58. Solid or hollow beam of circular cross-section < 144 59. Solid or hollow beam of elliptical cross-section 145 60. Relation among the ultimate resistances of material to tension, to com- pression, and to cross-breaking ; with Table II 146 viii ANALYTICAL TABLE OF CONTENTS. SECTION 2. Moment of Inertia and Radius of Gyration of a Given Cross-Section. ART. PAGE 61. Definitions 154 62. Eprmulae for moment of inertia and the square of radius of gyration. Table III 154 CHAPTER VI. DEFLECTION, END MOMENTS, AND POINTS OF CONTRARY FLEXURE FOUND. CAMBER. SECTION i. Deflection of the Semi- Beam having a Uniform Cross-Section. 63. Equation of the elastic curve as applied to a beam or pillar 157 64. Deflection of the semi-beam under one weight 158 65. Deflection of the semi-beam under uniform load per unit 159 66. Semi-girder, partial uniform load on each unit of length at any distance from fixed end 160 67. Total deflection of semi-beam at its free extremity when supporting a load not reaching that extremity. Example 162 68. Total deflection of semi-beam at its free extremity when supporting equal weights at equal intervals, and last weight is at the free end . . . .164 69. Total deflection of semi-beam at its free extremity when weight at free end is one-half every other weight 166 70. Deflection at a given interval due to all the equally distributed equal weights 166 71. Deflection at a given interval when weight at the free end is w th part of every other. Example 167 SECTION 2. Deflection of a Beam of Uniform Cross-Section, supported at its Free Unflxed Ends. 72. Deflection due the beam's own weight supposed to be uniform. Example . 169 73. Deflection due a concentrated load placed at a given horizontal distance from the end of the beam. Example 170 74. Deflection due a uniformly distributed partial load 174 75. Application to the special case of an advancing continuous load. Examples. 178 76. Deflection at any point due any number of equal weights placed at equal intervals along the beam. Examples 183 ANALYTICAL TABLE OF CONTENTS. ix SECTION 3. The Influence of Fixed Ends upon the Deflection of a Beam of Uniform Cross-Section, supported at its Two Extremities, which are assumed to be Level, and One or Both of Them fixed Horizontally or Otherwise. Determination of the End Moments and Points of Contrary Flexure. ART. PAGE 77. Deflection due end couples 189 78. Deflection at any point of beam, with load continuous and uniform throughout '. 190 79. Deflection of beam horizontally fixed at one end, and simply supported at the other. Example 192 80. Deflection of beam fixed horizontally at both ends, due to a concentrated load placed at a given distance from the left end of beam 194 81. Points of maximum deflection and points of contrary flexure. Examples . 197 82. Deflection due any number of equal weights placed at equal intervals along beam fixed horizontally at both ends 201 83- Deflection of beam fixed at right end, and simply supported at the left, uniformly loaded 207 84. Deflection, end moments, and points of contrary flexure due a partial uni- form load continuously distributed when both ends of the beam are fixed horizontally 209 85. Partial or full continuous uniform load on any portion of a beam fixed horizontally at the right end, but simply supported at the left . . . . 213 86. Examples illustrative 214 87. Established formulae applied when intervals are fractional, not integral. Examples 219 88. Continuous uniform load on beam fixed horizontally at both ends . . .221 SECTION 4. Deflection of a Girder of Variable Cross-Section in Terms of the Constant Unit Strain upon the Extreme Fibres of the Section ; that is, Deflection of a Beam of Uniform Strength. End Moments for Fixed Beams. 89. How cross-sections of members should be proportioned. Equation used . 224 90. Deflection of semi-girder of uniform height and strength. Example . . 224 91. Deflection of the semi-girder of uniform strength but of variable height. Examples . 226 92. Additional applications 235 93. Deflection of the girder of uniform strength, supported at both ends, either fixed or free, and the height of the girder being either uniform or vari- able. Examples 235 94. Preceding results arranged according to amount of deflection 240 95. Total deflection nearly in the inverse ratio of the areas of the figures of the girders i 241 ANALYTICAL TABLE OF CONTENTS. 96. These formulae may be employed for girders of continuous web. Examples . 243 97. Thickness of continuous webbed girder. Example 246 98. Thickness of the beam at any point. Example 248 99. Determination of any dimension of cross-section. Example 249 100. Beam of uniform strength fixed horizontally at both ends 251 101. Beam of uniform strength fixed horizontally at both ends; points of con- trary flexure 252 102. Beam of uniform strength and height fixed at both ends, and bearing a concentrated weight. Examples 252 103. Beam of uniform strength, height, and load, fixed horizontally at both ends. Rectangular cross-section. Example 256 104. Concentrated weight at given distance from the left end of a beam of uni- form strength, fixed at the right end only. Example 258 105. Continuous uniform load on a beam of uniform strength, fixed at the right end only. Example , 261 106. Beam of uniform strength and uniformly varying height, fixed at both ends. Examples 263 .107. Beam of uniform strength and uniformly varying height, fixed at one end. Examples 273 SECTION 5. Camber. 108. Definition. Modes of giving camber 277 109. Change of length calculated from the unit strain 278 no. Elongation and contraction calculated from deflection . 279 in. Application to Classes II., IV., VII., IX., X., and XII. of article 49. Example 281 112. Applied to the straight upper horizontal chord of Classes III., IV., VIII., IX., XI., XII 284 Ji3. Treatment for change of length in chords and verticals. Length of diag- onals changed. Example 286 CHAPTER VII. THE STRENGTH OF PILLARS, COLUMNS, POSTS, OR STRUTS. SECTldN I. Strength of Pillars by Rational Formula. 114. Definitions 289 115. Pillars of uniform cross-section ; three cases 290 ANALYTICAL TABLE OF CONTENTS. xi SECTION 2. Hodgkinsorfs Empirical Formula: for the Strength of Cast-Iron and Timber Pillars. ART. PAGE 116. Formulae for cast-iron pillars 296 117. Formulae for timber pillars 298 SECTION 3. Gordon's Empirical Formula, with Rankine^s Modification. 118. Gordon's formula deduced 298 119. Rankine's modification of same, introducing the least radius of gyration of the cross-section 300 1 20. Explanation of variations. Table IV 300 SECTION 4. Strength of Pillars compiited by the Preceding Formula^ and compared with the Strength experimentally determined. 121. Tables V., VI., VII., VIIL, IX., X., XL, XII 303 CHAPTER VIII. PROPORTIONS AND WEIGHTS OF ALL THE MEMBERS OF A BRIDGE EXCEPTING THE GIRDERS PROPER. 122. The floor 312 123. The joists, longitudinal 312 124. The wrought-iron I floor beams, transverse 314 125. The system of lateral support 316 126. Additional weight 317 127. Example 317 CHAPTER IX. OPEN GIRDERS WITH EQUAL AND PARALLEL STRAIGHT CHORDS. CLASS IX. SECTION i. The Pratt Truss of Single System and Uniform Live Load. Wind Pressure. 128. Strains in terms of the structure's unknown weight 318 129. Weight of the structure determined 324 130. Bridge of two equal girders with additional permanent weight. Strain sheet 330 Xll ANALYTICAL TABLE OF CONTENTS. ART. PAGE 131. Discussion of preceding strain sheet 333 132. Changes to be made in strain sheet 334 133. Lateral sway-bracing 337 134. Best number of panels, and best height. Preliminary operations . . . .337 135. Strains due to wind. Weight of beams 338 136. Strains due to wind. Diagonal strains 341 137. Lateral strengthening of end posts 343 138. Additional strains in end members required to resist wind pressure . . . 343 139. Formulae for weight to be added to resist adjustment and distortion strains . 348 140. Exemplification of method of article 138 351 141. Exemplification of method of article 139 365 142. Exemplification of method of article 139 with double the load 373 143. Another example 382 144. Deductions and comparisons 391 SECTION 2. The Pratt Truss of Single System under Varying Live Load, without taking Account of Wind Pressure. 145. Example of the two locomotives again 392 146. For deck bridge 398 147. Col. Merrill's example for Pratt Truss 399 CHAPTER X. CALCULATION OF THE WEIGHT OF BRIDGES HAVING GIRDERS OF CLASS I., AND DETERMINATION OF THE NUMBER OF PANELS AND THE HEIGHT OF GIRDER, WHICH RENDER THE BRIDGE WEIGHT LEAST FOR A GIVEN SPAN AND UNIFORM LIVE LOAD. LIMITING SPAN FOUND. SECTION i. General Specifications for Iron Bridges, issued in 1879 ty the New York, Lake Erie, and Western Railroad Company. O. Chanute, Chief Engineer. 148. General specifications for iron bridges ,....413 SECTION 2. The Brunei Girder of Single System. 149. Dimensions computed 423 150. Moments and strains in chords due to the total panel weight 427 ANALYTICAL TABLE OF CONTENTS. Xlll ART. PAGE 151. Weights of these wrought-iron chords 429 152. Greatest strains and weights in the web system 433 153. Weight of floor 440 154. Weight of joists 441 155. I floor beams 441 156. More convenient method for weight of joists 442 157. Weight of wrought-iron I-beams 443 158. Transverse I-beams 444 159. Horizontal wind pressure '. . . . . 446 160. The horizontal diagonals 446 161. The horizontal struts; that is, in this case, the quantity of iron to be added to the transverse I-beams by reason of wind pressure .... 448 162. The chords required in the horizontal system to resist wind force . . . 450 163. Vertical supports 451 164. Lateral head bracing as additional security against deflection 455 165. Necessary amount of material for the triangular web system of latticed struts or columns 459 166. Weight of the bridge 462 167. Table showing the number of panels and height which simultaneously ren- der the total bridge weight a minimum 466 168. Inferences from table 471 169. Example. Strain sheet 472 SECTION 3. The Brunei Girder of Double System. 170. Expression for height 485 171. Weight of chords 486 172. Weight of girder diagonals 488 173. Whole weight of girders due to loads 490 174. Weight of floor 490 175. Weight of longitudinal I-beams 491 176. Weight of transverse I-beams 492 177. Weight added to transverse I-beams for wind 492 178. Weight of horizontal diagonals 493 179. Weight of wind chords . 493 180. Weight of all verticals 494 181. Weight of head system 496 182. Equations and table giving best height for least bridge weight. Two Brunei Girders of double system 501 183. Inferences from table 512 184. Example. Strain sheet 512 xiv ANALYTICAL TABLE OF CONTENTS. CHAPTER XL BRIDGES OF CLASS II. BEST NUMBER OF PANELS AND BEST HEIGHT DETERMINED FOR A GIVEN SPAN UNDER A GIVEN UNIFORM LIVE LOAD, LEAST BRIDGE WEIGHT AND LIMITING SPAN FOUND. SECTION i. The Parabolic Bowstring Girder of Double Triangular System, with the Extreme Diagonals omitted^ and a Vertical Suspender at Extreme Panel Point. ART. PAGE 185. Dimensions 5 2 3 186. Weight of chords 524 187. The girder diagonals 526 188. Floor 532 189. Weight of longitudinal I-beams 532 190. Weight of transverse I-beams 532 191. Weight of horizontal diagonals 532 192. Weight of wind chords 532 193. Weight of verticals 534 194. The head lateral system 535 195. Equations and table giving best height and best number of panels for least bridge weight 535 196. Inferences from table 540 197. Example. Strain sheet 540 SECTION 2. The Post Truss with Parabolic Top Chord. 198. Dimensions and expressions for height 546 199. Moments due a total dead load of uniform panel weight 549 200. Weights of top and bottom chords due a total dead load of uniform panel weight . 558 201. Greatest strains in the girder diagonals, and their weights 563 MECHANICS OF THE GIRDER, CHAPTER I. PRESSURES IN ONE PLANE. i. FORCE is a cause which changes, or tends to change, the condition of matter as to rest or motion. Whether there is or is not, in fact, any difference between force and pressure, it is sufficient for the purposes of this volume to treat them as identical, since it is with their measurable effects alone that we are here concerned. A force is said to be given when its point of application, its direction, its line of action, and its intensity are known. Two pressures are equal which, acting on the same point, along the same line, and in opposite directions, neutralize each other ; and, if two equal pressures act at the same point in the same direction, the result of their combined action is twice that of each separate pressure. Pressures, therefore, may be compared by means of numbers expressing their intensities. Since the intensity of any one of the pressures to be compared may be taken as the standard, it follows that the unit pressure is entirely arbitrary, and may be a finite or an infinitesimal pressure. When pressures are expressed by symbols, such as P, Q, R> etc., it is to be understood that these letters stand for num- MECHANICS OF THE GIRDER. bers denoting the number of times the concrete unit is taken. Otherwise, such an expression as P 2 , being the square of a concrete pressure, would be unintelligible. A force or pressure may be conveniently represented by a geometrical straight line ; one end of the line denoting the point of application of the force, the direction of the line being coincident with the direction of the force, and the number of linear units in the line being equal to the number of force units to be represented. 2. When many pressures act at the same time on a mate- rial particle, the result of their combined action is generally a definite pressure in a definite direction. This definite press- ure is called the resultant of the acting or impressed pressures ; and these latter, with reference to the resultant, are styled com- ponents. When the resultant is zero, the pressures are said to be in equilibrium ; when the resultant of the given pressures is not zero, equilibrium may evidently be produced by intro- ducing a new force which shall neutralize this resultant. 3. Parallelogram of Forces It is shown in elementary works on mechanics, that if two forces act upon a single point, and their intensities and direc- A ^> tions be represented by two ad- jacent sides of a parallelogram, then the diagonal of the paral- lelogram drawn to the intersec- tion of those two sides will represent, both in magnitude and direction, the resultant of the two given forces. If P t and P 2 , Fig. i, are two forces acting at the point <9, represented in magnitude and direction by the lines OA and OB, then, completing the parallelogram, AOBC, the resultant will be represented, in magnitude and direction, by the diagonal OC R. When, therefore, a force is applied at O equal in intensity to R, and acting in the same line but in the opposite PRESSURES IN ONE PLANE. direction, it will balance the given forces P lt P 2 , and the three forces will be in equilibrium. 4. Triangle of Forces. Since, in Fig. i, AO = BC = P iy the three sides of the triangle BOC (or A OC) represent, in mag- nitude and direction, three forces, P u P 2 , and R, which, acting in one plane on a given point, are in equilibrium ; the direction of the forces being that of a point traversing the perimeter of the triangle. In this manner the value of the resultant may be constructed. A formula for the value of R is found by solving the tri- angle of forces, where two sides and the angle included between them are given. Thus, if c = AOB = the angle between the given lines of action of P l and P 2 , we have, from the geometry of the figure, putting BOC 6 (tketa), R 2 = P* + P 2 2 + 2P 1 P 2 cos;-. (i) Sin = sin <:. (2) EXAMPLE. Let P, = 8, P 2 = 12, c = 75. Then R* = 8 2 + i2 2 + 2 x 8 x 12 cos 75 = 208 + 192 X 0.25882 = 2 S 7. R = 16.053. Sin# = - x 0.96593 = 0.48471. 16.053 e = 28 59 ' 4 o". If the lines of action of the two forces, P 19 P 2 , are at right angles to each other, cos^ becomes zero, and equation (i) re- duces to R 2 = P 2 4- P 2 2 , where R is the hypothenuse, and P l and P 2 are the other sides of a right-angled triangle. MECHANICS OF THE GIRDER. EXAMPLE. When P x = 8, and P 2 = 12, R 2 = 8 2 + i2 2 = 208. R = 14.422. In this case, Fig. 2, we have > = R sin = P 2 tan (9. 2 R cos = />, cot (9. = jP, -*- sin = P t cosec 0. ? = P 2 -5- cos = P 2 sec ^. (3) FIG. 2. 5. Resolution of a Force. Conversely, any force, R, acting at a given point with known intensity and direction, may be resolved into two component forces acting at the same point, having definite intensities and directions. Manifestly also may each one of the two components be resolved into two components, and so on without limit. EXAMPLE. Resolve the force R = 100 tons, acting at the point O y Fig. 2, in the direction OC, into its horizontal and vertical components ; 6 being equal to 28 59' 40". From (3), /! = R sin = 100 x 0.48471 = 48.471 tons. P 2 = R cos0 = 100 X 0.87467 = 87.467 tons. 6. Resolution of Many Forces acting in One Plane at a Given Point. Let there be any number of forces, P a P 2 , P 3 , etc., Fig. 3, acting in the plane of the axes XX', YY', at their point of intersection, O ; and let a (alpha) symbolize the angle between the line of action of any force and the axis of x. PRESSURES IN ONE PLANE. Resolving each force into its horizontal and vertical com- ponents, and calling the sum of the horizontal components X> and the sum of the vertical components F, these results : X = P l cos <* r 4- P 2 cos a 2 + P 3 cos 3 + . . . = 2/>cos , (4) Y P I sin x + P 2 sin 2 +- /> 3 sin a 3 + . . . = SPsin a ; (5) the symbol 2 (sigmd) denoting the sum of the terms having the form P cos a or P sin a. Y 7- Thus, for all the given forces acting in their various directions on the point O have been substituted two other forces, X and F, acting at the same point ; the one horizon- tally, the other vertically, and in the plane of the original forces. Now, if R is the resultant of the two forces X and F, it must also be the resultant of the forces P P 2 , P y etc. ; and, being the angle between the resultant and the axis of x, we shall have Rcos9 = X = IPcosa, (6) R sin 6 = F = XPsin , (7) .-. fr = X 2 + Y*, (8) = -V (9) 6 MECHANICS OF THE GIRDER. When the given forces are in equilibrium, the resultant vanishes, and X = ZPcosa = o. (10) Y = 2/>sin a = o. (u) EXAMPLE. Let P I = 10 tons, a, = 40. P 2 = 20 tons, a 2 r= 150. P 3 = 30 tons, a 3 = 250 1 10. P 4 = 40 tons, a 4 = 300 = 60. Required the intensity, R> and the direction, 0, of the re- sultant. X = 10 cos 40 + 20 cos 150 + 30 cos 250 4- 40 cos 300 = 10 cos 40 20 COS 30 30 COS 70 4- 40 COS 60 = 10 X 0.76604 20 X 0.86603 30 x 0.34202 4- 40 x 0.5 = 0.0792 tons. Y = 10 sin 40 4- 20 sin 150 4- 30 sin 250 -f- 40 sin 300 = 10 sin 40 + 20 sin 30 30 sin 70 40 sin 60 = 10 x 0.64279 + 20 X 0.5 30 x 0.93969 40 x 0.86603 = 46.404 tons. 0.0792 = [(o.o792) 2 4- ( 46.404) 2 ]s = Y -t- sin0 = 46.404065 tons. The resultant is therefore in the fourth quadrant, and makes an angle of 5' 52" with the axis of y. This substitution of one force for many others is called the composition of forces. 8. Polygon of Forces. Let S t , S 2 , 5 3 , etc., in Fig. 4, be the five sides of a closed polygon. Measure the inclination of each side to the horizon, as indicated in the figure, for c iy c 2) etc. ; then the sum of the horizontal projections of all the PRESSURES IN ONE PLANE. sides is, in accordance with the trigonometrical signs of the cosine, found to be Since cos c = Si cos = + 10 x 4 -f ioo x 12 20 x 8 + 30 x 10 200 x 6 = 40 -f- 1200 1 60 + 300 1200 = +180 foot- tons, ' r - = ' 95493 foot ' Since the product Rr is positive, and the direction of the line of R, when drawn through the fixed, point, is into the third quadrant, it follows that r must be laid off below the fixed point on the perpendicular to the line of R> as shown by OE in the figure ; E being supposed rigidly connected with the solid AOCB. MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 1 7 CHAPTER III. MOMENTS OF THE EXTERNAL FORCES APPLIED TO A BEAM OR GIRDER. SECTION i. The Semi-Beam, or Girder fixed at One End and free at the Other. 14. We can now find expressions for the moments devel- oped in any section of a beam or girder, by the action of any forces in the plane of the beam, in whatsoever manner applied Let us first take a beam fixed at one end and free at the other, or a semi-beam as it is called. Let EOAB, Fig. 8, represent a beam fixed to a wall along the line AB h. Sup- pose the weight of the beam to be w pounds for every unit of its length / = AO. Assume, also, that the length b = DC has an additional uniform load of in/ pounds per linear unit, both w and w' being continuously distributed throughout their re- spective lengths. Also let W be a concentrated weight or pressure at the distance a' BJ from the fixed end of the beam. Let BC a = the distance from the wall to the nearer end of the uniform load ufb. Let P be any pressure acting at any point, G, with any inclination, a, to the arm FG ; and call the horizontal distance of the point G from the wall a" AK. 18 MECHANICS OF THE GIRDER. Suppose the beam horizontal, and all the applied pressures, except P t vertical. Let VS be any vertical section of the beam at the distance x from the fixed end AB. It is required to find the moment of the applied forces which must be resisted by the internal forces of the beam at the section VS. W J D a B S FIG. 8. Manifestly only the pressures at the left of VS affect that section. Taking the moments of these sinister pressures about any point, F, in the vertical section VS, and remembering that downward pressures on the left of VS give negative moments, we have the following equations for the required moment MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 19 SEMI-GIRDER. LENGTH = /. (See Fig. 8.) Load. Conditions. Force left of VS. Arm. Moments about F. w x a' M=o. (19) w x =. o W a' M = Wa'. (20) w a' = 1 W 1 X M = W(l x). (21) w x = o, a' = / W I M= _ 0V (max.). (22) wl xaznd<(a+t>) w'(a+b - x) \(a+b-x) M = W (a + b x) 2 . (26) w'b a = o w 1 ^ x) W - *) M = \w' (b x} 2 . (27) w'b a = o, b = I w'(l - x) i(/ - x) M = -\w' (1 x) 2 . (28) w'b x = or < a w'b \b + a x M = w'b (\b + a x). (29) w'b x = o,a=o,l> = t w'l i/ M = \w'l 2 (max.). (30) P x* } (34) The moment at the interval r due to these remaining (n r) equal weights is M= - n n n (35) EXAMPLE I. A semi-girder projects 50 feet, and is loaded at intervals of 10 feet with a weight of 10 tons ; required the moment at the fixed end due to the 5 equal weights. From (32), M= Wl 10 x 50 x - = 2 -1500 foot-tons. EXAMPLE 2. The same conditions continuing, required the moment at the fixed end due to the first 3 of the weights. From (33), M = - Wl rr -10 X 50 x -600 foot-tons. 2X5 MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 21 EXAMPLE 3. With same conditions of beam and load, re- quired the moment due, at the fixed end, from the remaining 2 weights. From (34), M = - * - 3(3_jO\ _ 00 foot-tons. 2 x 5 / EXAMPLE 4. At 10 feet from the fixed end of the beam, what is the moment due to the 4 weights beyond ? From (35), M = - Wl^ ~ T + l)(5 ~ T) = -looo foot-tons. 2 X 5 EXAMPLE 5. If the given semi-beam weighs 0.8 ton to the linear foot, what is the moment at its centre and at its fixed end ? From (23), if x = /, M= \w(\lY \ X 0.8 X so 2 = 250 foot- tons. From (25), M= % X 0^8 x 50 2 = looo foot- tons. EXAMPLE 6. Suppose the same beam to be covered with the uniform load 0.6 ton for the space of 15 feet, beginning 25 feet from the fixed end ; required the moment due to this load at 30 feet from the fixed end. Here a/ = 0.6, b = 15, a = 25, x = 30. From (26), M= | x 0.6(25 + 15 30)* = 30 foot- tons. 22 MECHANICS OF THE GIRDER. EXAMPLE 7. If the load 0.6 ton per foot covers the first 35 feet of the beam, and the moment at 10 feet is required, we have b = 35, a o, x = 10; and, from (27), M \ X 0.6(35 I0 ) 2 = 18 7-5 foot-tons. EXAMPLE 8. If the load 0.6 ton per foot covers the entire beam, the moment at the centre is, from (28), M = | X 0.6 X (50 25) 2 = 187.5 foot-tons, and at the fixed end M | X 0.6 X 5o 2 = 750 foot-tons. EXAMPLE 9. If the uniform load 0.6 ton covers 40 feet of the beam, beginning at the free end, then the moment at 5 feet from the fixed end is, from (29), M= 0.6 X 4o(J X 40 + 10 5) = 600 foot-tons. EXAMPLE 10. If the force P, Fig. 8, is 4 tons, and its line of action makes an angle of 30 with the line GF P = 20 feet, then the moment due to P at the point Fis, from (31), M 4 X 0.5 X 20 = 40 foot-tons. 1 6. If there are several concentrated weights, W W 2 , W y etc., or pressures, P lt P 2 , P y etc., or detached uniform loads, bjW T r t b 2 w 21 b^w^ etc., at different points on the left of the section VS, we must evidently sum the moments due to the separate pressures for the total moment. MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 2$ Thus we may write M x = -^W(a' - x) - \w(l- x) 2 - %uf(a + b - x) 2 + a x)b + S/'sina/, (36) where M x is the moment, with reference to any point of any vertical section of a semi-beam, due to all the forces applied to the beam between its free end and the given vertical section. It should be observed, that, for all pressures whose lines of action are vertical, the moments will be the same, whatever point of reference is taken in the vertical section VS \ for such pressures have no horizontal component. SECTION 2. 17. We next take a beam or girder, horizontal, supported at its ends, and loaded in any manner whatsoever. Such a girder is also said to have its ends free; since they simply rest upon two level supports, and are fixed in no other manner. x FIG. 9. Let the beam OABE, Fig. 9, be supported at the two points O and A, and be subjected to the following pressures : 24 MECHANICS OF THE GIRDER. w = weight of beam per linear unit. w' = uniform load per linear unit of the length b = CD. W = a concentrated weight or vertical pressure at any point, K. P = any force at distance a" = OF from O. F x = vertical re-action of the left support. V* = vertical re-action of the right support. / = length of girder. h = height of girder. a = OD = the horizontal distance from the origin to the nearer end of the uniform load bud. a' = the horizontal distance from O to the weight W. .x =. the horizontal distance of any vertical section, VS, from O, the origin of co-ordinates. y =, the vertical distance of any point in the section VS from the horizontal line AO. The vertical section VS is made by any plane cutting the 'beam perpendicular to the line AO. It is required to find the moment generated at any vertical .section, VS, by the action of each of the given pressures. Since at any given cross-section, there can be but one mo- .ment due to the given simultaneous pressures, it follows that we may determine this moment, either by using the pressures applied upon the left side of the given section, or by using the applied pressures on the right side of the same section. In the following table we use the pressures that act on the left of the section VS ; and consequently downward pressures give negative moments, and upward pressures give positive moments, in accordance with our previous notation. 18. The sum of the re-actions V T and V 2 for the simple girder with free ends is equal to the total weight of the girder and its load. MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 2$ The resistances V t and V 2 due to any concentrated weight, W, are, since there can be but one moment for the vertical section through W y inversely proportional to the horizontal distances of W from the points of support ; and we have, from equation (13), M = VJ = V 2 (l- of), ... Fi = F 2 ^= W- F 2 , (37) K 2 = W a T (38) V t = W 1 ^-. (39) Or, by proportion, F, : F 2 : : / - a' : J, Similarly, for the uniform load bit/, the re-actions F x and Z^ will be inversely proportional to the distances of the centre of gravity of the uniform load from the points of support. 19. In the following table we have, First column, load whose moment is sought. Second column, re-action at left support, giving -\-M. Third column, conditions of load and plane VS. Fourth column, part of load on left of VS, giving M. - Fifth column, arm of V v Sixth column, arm of load on left of VS. 26 MECHANICS OF THE GIRDER. S S 'ft II II d d II II v > > v g : ^ * x i v i H H ^ 5 Q O ^ * J II II II V II J H * * * * H 5R .9 . * MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 2J 20. Moments due Uniform Discontinuous Load on any Part of the Beam. Let r t r 2 denote the number of equal weights, W, at equal intervals, c, between any two consecutive weights on the whole or any part of the girder. We may shorten the numerical computation of moments, as in case of the semi-girder, by first summing the series that arises in the expression for M. For this purpose let r r 2 = the number of equal weights, W, on the length x\ (r 2 + i)c = the distance from the left end of the beam to the nearest weight. If this distance is less than c, that is, not a full interval, it follows that r 2 will be a negative proper fraction. Now (r, r 2 ) (r r 2 ) = r, r = the num- ber of equal weights between the point x and the right end of the beam. The three differences, r x r 2 , r r 2 , and r l r, must be integers, since each denotes a number of equal weights. If one of the three quantities r, r r 2 , is not an integer, neither of the other two is an integer, and the decimal part of each is the same, except that, when r 2 is negative, its value is less by unity than the common decimal part of r and r,. Let us first find the moment due r r 2 equal weights, W, at any point, x, between the last weight and right-hand end of the girder. We use equation (43), giving to a f the successive values c(r 2 + i), c(r 2 + 2), c(r 2 + 3), . . . cr, and taking the sum ; thus, ^a' = c\_(r z + i) 4- (r a + 2) + (r a + 3) + - - r] = &(r-r 2 )(r + r 2 + i), ... M x = ^(r Ir 2 )(r + r 2 4- (/-*), (60) where x cannot be less than cr. EXAMPLE i. Length of beam = / = 100 feet = ioc,r =. 6J, r 2 = 2\ ; what is the moment at the fourth weight, W=8 tons, due the 4 weights = 32 tons ? 28 MECHANICS OF THE GIRDER. Here x = re = 65, .-. M r = x 4 X 10 x 35 = 16 X 35 = 560 foot-tons. 2 x 100 If x = (r + i)c = 75, .'. M r + I = 1 6 X 25 s= 400 foot-tons. If * = (r + 2)* = 85, .*. M r + * = 16 X 15 = 240 foot-tons. If * = (r + 3)* = 95, .*. -^- + 3 = 16 x 5 = 80 foot-tons. This shows a uniform decrease of moment for each interval beyond the given load. Equation (40) gives for a single weight, W, applied at any point, a', the moment at any distance, x, between the weight and the left end of the beam. By giving to a' the successive values c(r + i), c(r + 2), c(r + 3), ... cr lt and summing for aT and of, we find - r, - r, (61) which is the moment at any point, x, between the left end of the girder and the nearest weight, which is at the (r -\- i) ih point of division ; the number of weights being r, r t and x not being greater than c(r + i). MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 2Q EXAMPLE 2. Let /= 100 feet = loc, W= 8 tons, r r = 6J, r = 2j; what is the moment at the first weight due the r t r = 4 equal weights ? Here x (r + i)c = 35, o (2 X 4 X 100 10 X 4 X 10)35 = 560 foot-to 2 X 100 x = re = 25, .*. J/ r = 1 6 X 25 = 400 foot-tons. x = (r i)c = 15, .*. M r _i = 16 X 15 = 240 foot-tons. # = (r - 2) of the beam due r I r 2 equal weights, W, placed at equal and consecutive intervals, c, over the whole or any part of its length. 3O MECHANICS OF THE GIRDER. Here r cannot be less than r 2 nor greater than r n and x lies between re and (r + i)c for the loaded part of the beam, but may have any value between o and r 2 c where r r 2 , and any value between r,c and / where r = TV EXAMPLE 3. Let / = 100 feet = 10 c, W = 8 tons, r 2 = 2j, r, = 6J ; what is the moment at the fourth weight ? Here x = r^c = 65 feet, and (62) gives, If r = r iy M ri = | (o 10x4X10)65-1-10x100x4X101 = 560 foot-tons. 200 \ ) Or, if r = r l i, g f -v M ri = - j (2X i X 100 10X4 X 10)65 + 10 X 100X3X9 t = 560 foot- tons. What is the moment one interval beyond the last weight ? Here x = (r, -f- i)c = 75, and r = r, = 6J, in (62) ; 8 C * ) .*. J/ ri + I = -- \(o 10x4X10)75 + 10x100X4X101 200 ( j = 400 foot-tons. If n = the whole number of intervals in the girder's length, we have c = -, and (62) becomes M x = [2(r s - r)-(r s - r a )(r x + r 2 + i)] / , (63) from which we may find the simultaneous moments at all points throughout the girder due to any uniform discontinuous par- tial or full load. MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 31 EXAMPLE 4. Let a uniform load of 4 weights, each = W = 8 tons, spaced c = 10 feet from weight to weight, come upon a girder loofeet long, and move forward to the centre ; required the simultaneous moments throughout the girder as the fore- most end of the load passes the points^ = 5, 15, 25, 35, etc., n = 10. Owing to the important applications of this formula which are to follow, we add the complete solution of this problem, and may remark that by giving to x the values 10, 20, 30, 40, etc., we can find the simultaneous moments at the full intervals as the foremost end of the load passes the successive points of division. Or we may give x any value we please between o and /, and so suit the equal or unequal panel lengths of any girder. As the load, now central, passes off to the right, it is evident these moments will be reversed. UNIVERSITY MECHANICS OF THE GIRDER. *f P 00 00 co ^ 3 * VO Tf Tj- Pl VO ON *^t" ON "^ * 00 M Tf O\ O 'S pt co 3- $ **t* vo vO M VO PJ r}- VO OO O M CO ^ VO af OO N PI OO P> VO O * M CO OO N 00 ON P> OO Tf O O VO C-l OO M ^ > *- 1 LO LO vo vo LO vo vo O O M O -4 PJ O O co T)- vovo O vovo r^ 00 O t^oo ON O + XXXX + XXXX + XXXX ' *P) "cO 1 ^- VOM PI CO^vOi-! P) CO^i-vOi-i M CO^J- O O * O LO LO LOVO" LO vo vo VOQO* vo vo vo vo'o^ vo vo vo vo HHNCO MMCOTf MCOTJ-vOi-cCO-rJ- VOVO * g d d d d d d d 666X6666X6666^6666 rt 1 I M 1 1 1 1 >-> PI M CO X XX X O O O O O 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ! 1 1 WHH "tCOPI"-' Tj-COPli-H rJ-COPlM XX X X X X X X X X X X X X OOOOOOOOOOOOOOOOOO PIP) PIPINP) P1P1P4P) PJP1MPJ g| O O O O OOOOOOOOOOOOOOOOOO t Tf rj- rj- m*. IO to to to to to to Q o HH O t 1 N O LOLOLOLOLOLOVOVOvOvovOVOVOvovovovovo-i-' MPiroO'-HPJCOTj-OPJcO'tf-voocoTi- vovo C d d d d d d d ddddddddddddddddddg ^ IO to to to to to to voLOVovovovovovovovovOLOvovovovovovog *~ d ~ M pi pi p3 co COCOCO^Tj-TJ-Ti-^-vovovovo VOVO VO VO VO VO In IO to to to to ^O to d d M d .M PJ d L0vovovovovovovovovovovovovovovovovo rt Mpicod>-M 3 4, ^4 6, M 6 7,M 7 9>M g = i 2 3 4 5 6 7 8 9 X X X X X X X X X 2 3 4 5 6 7 8 9 10 X X X X X X X X X 9 8 7 6 5 4 3 2 I X X X X X X X X X 4 4 4 4 4 4 4 4 4 - 72 192 336 480 600 672 672 576 360 foot-tons, foot- tons, foot-tons, foot-tons, foot-tons, foot-tons, foot-tons, foot-tons, foot-tons. 34 MECHANICS OF THE GIRDER. 22. From equation (63), by putting r 2 = o, r l = n i, and r/ = ?r , we derive n (65) -which is the moment at the r th weight due i equal weights, W, placed at equal intervals, -, throughout the girder. EXAMPLE. Uniform discontinuous load ; W = 8 tons, / = 100 feet, n 10. If r = i, M I = 40 x 9 X i = 360 foot-tons. 2, M 2 = 40 X 8 X 2 = 640 foot-tons. 3, J/ 3 = 40 X 7X3= 840 foot-tons. 4, J/ 4 = 40 x 6 X 4 = 960 foot- tons. 5,^ = 40x5x5 = 1000 foot- tons. And these moments are to be reversed for the other half- span. 23. Suppose that the first and last intervals into which the beam is divided are each = \c = , while every other is = c, 2n and that we wish to find the moment at the foremost end of a uniform load of equal intervals, c, as that end passes each point of division of the beam. For this object we employ equation (60), makings = re = , and r 2 = J, and have n " r) - (66) MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 35 EXAMPLE. Let / = : ioo feet , w=s tons, n =. 10. For r = 0.5, M t = 4 x I X 9i = 3 8 foot-tons. i-5, M 2 = 4 X 4 X 8* = 136 foot-tons. 2.5, M 3 = 4 X 9 X 7i = 270 foot-tons. 7.S, M. \J ^ = 4 x 36 X 4i = 6 4 8 foot-tons. 6.5, J/ 7 = 4 X 49 X 3j = 686 foot-tons. 7-5, ^s = 4 X 64 X 2^ = 640 foot-tons. 8.5, J/ 9 = 4 X 81 X ij = 486 foot-tons. 9-5> M w = 4 X IOO X ^ = 200 foot-tons. 24. From equation (63), by making r 2 = J, r t = n and x = , we also obtain n (67) which is the moment at any weight due n equal weights, W f applied at equal intervals, -, except the interval at each end, which is = . Here r takes the values }, f , f , |, . . . 2n "" l ' 21 1 2 and n denotes the whole number of full intervals in the length, /, which in this case is also the whole number of weights. EXAMPLE. Let, as before, / = ioo feet, n = 10, W=8 tons ; then For r = 0.5, M! = 10(1 x 19 + i) = 200 foot-tons. 1.5, M 2 = 10(3 x 17 -h i) = 520 foot-tons. 2.5, M 3 = 10(5 X 15 + i) = 760 foot-tons. 3.5, M 4 = 10(7 x 13 + 1) = 920 foot-tons. 4.5, M 5 = 10(9 x ii 4- i) = I0 foot-tons. The same to be reversed for the other half. 36 MECHANICS OF THE GIRDER. 25. Difference of Simultaneous Moments at Consecu- tive Points of Division. By making r 2 ~ o, and x = (r + i)c t in equation (60), we have 2H 2 -r- i), (68) which is the moment one interval, c, beyond the foremost end of a uniform load consisting of r equal weights, W y at the dis- tances c, 2c, 3^, ... re, respectively, from the left end of the beam. Subtracting (64) from (68), we have = M r + I -M r = -r(r + i), (69) 272 which is the difference between the simultaneous moments at any two consecutive points of division on the unloaded end of a beam, which has r equal weights at full intervals on the other end. For finding the difference of simultaneous moments at con- secutive points of division on the loaded part of the beam, we use (62), making r 2 = o, and x re, (r + i)c, (r -\- 2)c, etc., in succession. Thus, + clr(r + i = J2(r I -r)-r 1 (r 1 + i)l, (70) which is the first order of differences for the loaded part or end of the beam : and &M is an increasing function of r, for a given value of r, and will be greatest when r, is greatest ; MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 37 that is, when r, = n I, its limit. But at this limit of r x the girder is loaded, and the positive differences on the left half will equal the negative differences for the corresponding inter- vals on the right half of the girder. Putting n i for r I in (70), we have (71) which gives the difference of simultaneous moments for each interval, c, of the beam fully loaded with n I weights, W, applied at equal and all full intervals, c, or with n weights, W, when each end interval = \c. Subtract (69), which is negative, from (71), whose positive values in one half-span are equal to its corresponding negative values in the other half-span, and the remainder is which is positive, since n > r, and both n and r are integers. Therefore the greatest negative difference computed by (71) for any interval is numerically less than the difference computed by (69) for the same interval in the second half- span; that is, both half-spans, if the uniform load is to travel either way. Consequently we use (69) in finding the greatest difference of simultaneous moments for any interval due a uni- form discontinuous moving load. It may be observed here that (69), for the unloaded end of the beam, gives a constant first difference, while (70), for the loaded end, gives a first difference which is not constant. By putting r -f- i for r in (70), and subtracting (70) from the re- sulting equation, we find the second difference, A(A^f) = -Wf, (72) which is constant and negative, and may be conveniently em- ployed in some computations. MECHANICS OF THE GIRDER. EXAMPLE i. Let a girder of 10 panels, each 10 feet, be laden with a permanent load of 4 tons at each panel point, and a discontinuous uniform rolling load of 8 tons to be applied at the same points as the load advances ; required the greatest moments at these panel points, and the greatest difference of simultaneous moments at any two consecutive panel points, due to both these loads. The greatest moments will occur when both loads cover the beam. We have, then, in equation (65), W= 12, /= 100, n = 10, == 60, and r = i, 2, 3, etc., in succession, for the greatest 2n moments. The difference of moments at consecutive panel points due dead load is to be computed by (71), making W = 4, c = 10, n = 10, = 20, and r = o, i, 2, 3, 4, etc., in succession. 2 And the greatest difference of simultaneous moments for each interval due live load is found by using equation (69), We when W = 8, c = 10, n = 10, = 4, and r= i, 2, 3, 4, etc., 2n in succession. COMPUTATION FOR GREATEST MOMENTS AND DIFFERENCES. No. of the Panel Point, r. 1 2 3 4 5 6 7 8 9 Greatest moments = 60(10 r)r 540 960 1260 1440 1500 1440 1260 960 S4o Differences, dead load = 20(9 2r) 1 80 140 100 60 20 20 60 IOO 140 180 Greatest differences, live load = 4/-(r + i) 8 24 -48 80 120 168 224 288 -360 Total differences for both loads 1 80 TO2 76 12 60 IAO 228 '32J. 428 54 Differences, load moving ( X 3* yu 60 A T I 4 228 O^T- 324 -428 54 either way ( 540 428 324 228 140 60 MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 39 If in equation (66), instead of the factor (/ re], we write [/ (r + i)c], and then subtract (66) from the resulting equa- tion, we shall have AJ/= --(r-M) 2 , (73) which gives the greatest difference of simultaneous moments at any two consecutive points of division, due live load advan- cing by equal panel weights, when the two extreme panels have each but half the length of every intervening panel. Here ob- serve that r takes the successive values J, J, f , f , . . . and that c = for the two extreme panels, but c = - for all 2H n others. EXAMPLE 2. Given the same loads and length of girder as in example I, but the panel points being now at the distances 5, 15, 25, 35, etc., from either end ; required the greatest mo- ment at each of these points, and the greatest difference of simultaneous moments at any two consecutive panel points, for both live and dead loads. Equation (67) gives the greatest moments if we make / = Wl 100, n = 10, W = 12, = 15, and r = |, f, f, |, etc., in suc- oH cession. The differences for dead load are computed from (71) by putting W = 4, c = 5 in two-end panels, c = 10 in all others, n 10, and r = J, J, f, f , etc., - = 10 or 20. The greatest differences for live load are found from (73), where W = 8, c = 5 or 10, n = 10, and r = J, J, f, f, etc., 2n 4 o MECHANICS OF THE GIRDER. COMPUTATION FOR GREATEST MOMENTS AND GREATEST SIMULTANEOUS DIFFERENCES FOR EACH INTERVAL. r. -* i 1 I I f * Greatest moments, U/-7I i K-'H 300 780 1140 1380 1500 1500 1380 1140 780 300 Differences, dead load, ^(_i_ 2 r) 100 160 120 80 40 40 80 120 160 TOO Differences, live load, ~ (r + %) 2 4 16 -36 -64 100 144 196 -256 324 2OO Total differences . . 100 156 104 44 24 100 -184 -276 -376 -484 300 Differences to be used j 300 484 376 276 24 184 100 100 -184 24 -276 -376 -484 300 26. To determine the Point in any Girder simply sup- ported at its Two Ends, and carrying any Partial or Com- plete Uniform Discontinuous Load, where the Moment due that Load is Greatest. The required greatest moment will occur at a point within the loaded part of the girder, since for any partial load the simultaneous moments decrease from either end of the load to the corresponding end of the girder. If, therefore, we put x = in (63), and call the result M r , n then in M r thus found put (r + i) for r, giving M r + TJ and equate &M r = M r + x M r to zero, we shall find r = r, (r t - 2H (74) and the panel point of greatest moment lies between re and (r + i)c, except when re and (r + i)c are panel points. Let us verify this statement by referring to example 4, MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 41 article 20. Taking r lt r 2 , r, and the greatest moment, from that example, we compute r by (74), and write as below : ft. r 2 . r. Mmax. r by (74). 6.5 2-5 4-5 or 5-5 640 4-5 5-5 i-5 4-5 624 3-9 4-5 -5 3-5 544 3-3 3-5 --5 3-5 416 2.7 2-5 -o-5 2-5 270 2.05 i-5 --5 t-5 136 i-3 0.5 --5 -5 38 045 r I 27. If in equation (63) we make r = r,, x , r 2 = r, e, e being the number of equal weights on the beam, we shall find, after putting &M fi = M fi + x M fi = o, r.-2ip* (75) But when the advancing load reaches back to the left end of the girder, we may not know how many weights will give a maximum moment at the foremost weight. In that case we deduce AJ/ r = M r + x M r o from (64), and find r = r, = \(n - i) (76) for a girder of equal panels to receive an advancing load of equal weights applied at successive panel points. And for a girder each of whose two extreme panels is one- half of any other, the advancing load to be applied at panel points, we derive Aj/ r M r + x M T o, from (66), and get r = r, = \(2n - 5 + 4 - 2)- (77) In all these cases the panel point at foremost end, having the greatest moment as the load advances, lies between r^ and (r, + i)c, except when rj and (r, + i);: are panel points. 42 MECHANICS OF THE GIRDER. s 00 S- 3- % g * - ^ * * 1 * Jf ** ^t- VO CO O M N M CO * to ON ON oo oo oo * O O O O O O O N vo M r^. ON "* 1 + oo -3- oo M vo o M M tf "Si" - ^ i *JI_+ * VO OO VO rl- N O i-O VO to d CO VO N Tf Tj- CO N tj- 1 5. * M tO CO N N ^ ? ? * ^> > * J S ON ON hH HH i wwtoMto^ vovo t^oo ONO XXXX X X XX >. lx. X. Second Term. M x . O 10 0.004 90 O 0.2 90 10 180 180 10 20 0.004 IOO 80 32 0.2 80 20 3 20 352 20 30 0.004 400 7 112 O.2 70 30 420 532 30 40 0.004 900 60 216 O.2 60 40 480 696 40 50 0.004 1600 50 3 20 0.2 SO 50 5 00 820 50 60 0.004 2500 40 400 0.2 40 60 480 880 60 70 0.004 3600 30 432 O.2 30 7 420 852 70 80 0.004 4900 20 392 0.2 20 80 3 20 712 80 9 0.004 6400 IO 2 5 6 0.2 10 9 1 80 436 90 IOO 0.004 8100 0.2 O IOO O Second computation, using (85), r. trtl 23' r 2 . r i. First Term. w/ 2 M* r + i. r i. Second Term. ^r + i. 4 9 20 I 9 180 180 I 4 i 8 32 2O 2 8 320 352 2 4 4 7 112 20 3 7 420 532 3 4 9 6 216 20 4 6 480 696 4 4 16 5 320 2O 5 5 500 820 5 4 25 4 400 2O 6 4 480 880 6 4 36 3 432 20 7 3 420 852 7 4 49 2 392 2O 8 2 320 712 8 4 64 I 2 S 6 20 9 I 180 43 6 9 4 81 20 IO o This second computation will, in general, be found more convenient than the first, since n and r are usually integers, and not very large. 48 MECHANICS OF THE GIRDER. 31. When a uniform continuous load is coming upon one end of a girder, we may find the position of the foremost end of the load at the instant the moment at that end reaches its maximum value, by differentiating (81) with respect to b, and putting - o. Thus, do do .'. b = \L (86) 32. The position of the foremost end of the live load when the moment is a maximum there for combined dead and live loads, is determined by differentiating (83), and making = o. do Thus, dM = (2rfJ _ ,. + /2 _ 2/ ^ = do 2.1 .'. J ^.Z,- i (< + >l 8 3 .2 * rt O" 1 V> 1 vovovovosovo^oo 777 a ' ^ 1 1 1 \ * \ * ' ^ o a* 1 1 1 1 1 1 I 1 * 1 5SS-5^55^ ^\% 3. 3. 4t 3 "018 | vol a" a" a* a* a a IS yi a'a"^^^^^^ s In To Tol "vo Co"! . K^ i I 1. II* 1 Ma * t/3 S ^ a* TTTTT^T'T 888 8 5 1 3, 3 3, * T3 Hoe*ee^.e + * 5M Spr] ^ lir^riirb^'^bftrbr SSSSSS. 1 - 1 1 o Q-^^^^S^J , 7 a"" ' 8 T T T ?l * * ^ s Z ^ J3 r a* IIMIM^' 8 88 sE -S v5 ^ \ 3 "*i S o rt ^ 2 ssstss^^s . ] l Z , (92) which is a constant, while the point B is anywhere in the space 5. If both drivers are between two consecutive joints, as AB, we have c c c which is not a constant, but reaches its greatest value within the prescribed limits when x == c S ; that is, when Therefore D is the greatest pressure that can come upon any joint from the drivers of one engine. Now, when the foremost driver, D lt is at B, the second truck wheel is between B and C; and when the second driver is at MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 53 B> the first tender wheel is between A and B. In the former case the increment of weight at B from the truck would be - l -k ; in the latter case the increment at B from the tender c c S would be - 2 -t. Therefore the first or second driver at B c S, gives the greatest pressure at that point according as - k c-S 2 is greater or less than - /. In the present example, c S l 10 8.5 f-k = x 5-75 = 0.86250, c S 2 10 7.25 -' = - " - X 5.25 = 1.44375- We will, therefore, find the pressures at points whose intervals are equal to c, when the second driver of the foremost engine is at one of these points. Let this point be the third, counting from the right-hand pier. Then Fig. 1 1 shows the positions of all the wheels with ref- erence to the joints at the equal intervals ; and a simple calcu- fc fc t it r( p\ k k B Q B.TJ Q 8 Q 4 Q 7M Q*O 7.H Q *> C) ""SO Q ^8'Q _ A I 3.75+5.75+.60 | 7.60 + 2.60 | 1.504 7.25*1i I 2.75+A26 \ 7.7B--2.Uft | 8. 26 1 3.75 | 2^8 | a 7 a s a s a 4 a, a 2 a, a t t t t Ji X n C _ _ _ Q 4 Q T.25 Q 4 O 7.2; ( ) T.75 {) B 1 6+4.1-1 \ 6- 26."* 3.75 | .25*7.25 + 4.68 | 6.25 + 4.75 | a.,, a to a 9 a. a* FIG. ii. lation according to the principle involved in (38) and (39) gives the total pressure at each joint, which pressure is to be substi- tuted for W in the equation (91). 54 MECHANICS OF THE GIRDER. In this position of the locomotives the pressures at the equal intervals due to the weights on the adjacent panels are 1. At A, o.20o = 1.15000 tons. 2. Ata,, 1.425^ = 8.19375 tons. 3. At a 2 , 0.375/6 + 0.775/7 = 10.29375 tons. 4. At a 3 , 1.225/7 + 0.275* = 14.30625 tons. 5. At # 4 , 1.750* = 9.18750 tons. 6. At # s , 1.725* = 9.05625 tons. 7. At #6, 0.250* -h 1.325^ = 8.93125 tons. 8. At0 7 , 0.675^ + o-5 2 5^ = 9-39375 tons - 9. At 8 , 1.225/2 + 0.025* = I2 -99375 tons * 10. At 9 , 0.250/7+ i .600* = 11.02500 tons. 11. At IO , 1.775* = 9-3 l8 75 tons - 12. At , 0.600* = 3.15000 tons. Total, 107.00000 tons. Stopping with the second driver at a y we see that the hind- most tender truck has not yet come upon the girder. We com- pute, however, the moments due this load as it advances panel by panel, till the twelfth weight is upon the girder, and the first, second, and third have passed off; using indices M u M t _ 2 , . . . M 3 _ II} M 4 _ 12 , to denote, inclusively, what panel weights produce the simultaneous moments opposite M. MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 55 -t^ -p 4x 4*. 4* O ON *^J ON HI p ON ON p 4^ CO OO Co OJ VO u U> M HI ON tO ^-J ON OO HI tO 4^ Ul ON 10 O r^ u f\\ K^ VQ fjj ^b O ON 4^ 4^ Ul to -a to 4- 4>> O O W to M OOUr VOto OO^J^I 4^VOU> W 1-14^ ONOl4^. U)Ln O 4^ ON^J 4^t>JO\ OOJ ON LnC^J tX(Jl Un OO OOVO tO4^On HI (0 4- Lrr M4:i.4i. O to tO^J OO OVO4^ to >H tri t/i U) OO^J O Ln vb Q\ OSU> " 60 ^J^TC*i-l OOOJ tOLriLn Ui NO O tOt/r - ONLn VO O 04 t/T OOtOUivb ^ItOVOtO^I ONONVO ONLn 4>- OOi-i w OOOO ONU) OJU) "^J >H i-iVO ^JLn4- OJ -'-i H 00 4^ fO 4^ HI ON Co HI O HI 4. ^J OJ vO Ln M 4*- MECHANICS OF THE GIRDER. & 5 voONCOr^O ONH-I w cow ONCO i-iri-i-iqwcpvoqqvocooN Miirrrrrrrr ^2 l! i i i i i i i i i i r r CO 1-^ M q W CO vo vo ON MOO ^J" r^ ON O ON O M w CO O r^ M co MOO ^ r^ f^ r^>. Tj" vo t^>. M co M covO O rj-w ONCOCOVOVO M i r TT r i 'j i M VO O i r rt rt r vo ON co i r r i a i i i N O O vo MOO O O OVC)MD OO >-< ro M M M fOWVQVO I vococor^sOwcovo M W M CO CO CO CO T)-TJ->O OOO ONM COM -^- ON fO t^OO ONO M W NO M CM m ro ro di t^ c? O O tf- w t-s I s * II 1 1 3 ? 7 T T wo sT""** o o ON o> * 1 o' to 1 in 1 in < 2 $. 3- 1 1 1 1 q 8 o q 3 ' |T K ' of ' ? M ( M M ? 1 P-l 2 OO O OO t^ OO t t^ in ' w d pi M- M t^ ^ t>. 7 i 7+7 g2 S2_2 j; 2 ^ S JToS g. g R. I 1 1 M t S Jo 8 S , , q 09 vo 1 tn t^ f oo" H | N M TJ- w O ^ c<") j CO \O OO | | C> in co u-> o 10 o 10 H 10 | | ro K ^ tQ Tt- \O R 8 R 8 * vo in H r^> ^ w g. g. , - +| I + 1 s-si i i h 4i J5 .* .1 z ^^139 v j ^3 ft S S P MOMENTS OF FORCES APPLIED TO BEAM OR GIRDER. 57 The moments for dead load, given opposite Mw, have been computed by (65), whilst (91) has been used in finding the moments due the advancing load. The differences are taken directly from the computed mo- ments ; and we must evidently use for each half-span the greatest difference due any interval, the load being supposed to travel either way. 37. Let us now suppose that this same live load of 107 tons is distributed uniformly over the 10 panels, so that W panel weight = 10.7 tons. We then find by means of (65) the great- est moments due live load, and by means of (69) the greatest differences of moment due live load. Taking th,e moments due dead load as found above, we write : WEIGHT OF Two LOCOMOTIVES UNIFORMLY DISTRIBUTED. DISTANCE FROM PIER. 10 20 30 40 50 r. - I 2 3 4 ^(r + i)r 64 20 2 2 Full live load, difference .... 481.50 225.00 374.50 175.00 267.50 125.00 160.50 75.00 53-50 25.00 4 5 6 Maximum difference + . . . Maximum difference .... 706.50 481.50 549-50 856.00 392.50 1123.50 235-50 1284.00 78.50 -82.00 1337.50 7 Dead load, M 225.00 400.00 525.00 600.00 625.00 8 Total M maximum 706.50 1256.00 1648.50 1884.00 1962.50 DISTANCE FROM PIER. 60 70 80 90 100 r. 5 6 7 8 9 Wc (r + ^)r 224.7O 385.20 481.50 2 2 Full live load, difference .... 53-5 I60.50 -267.50 -374-50 481.50 4 5 6 Maximum difference + . . . . Maximum difference .... -185.50 -299.70 424.60 856 oo 560.20 48l.50 -706.50 Dead load, M 225 oo 3 1884 oo 164.8 c;o MECHANICS OF THE GIRDER. A comparison of these maxima moments and differences with those just found for the natural distribution of the weights of these two locomotives, shows but one moment and one differ- ence to be less for uniform load than for naturally distributed load. The extreme length of wheel base of these two united locomotives is 2 X 44.5 + 8 = 97 feet. It will be observed, that, when the second driver of the sec- ond engine is at a joint, the weight at that joint is greater than the weight we have used in the calculation of moments. But, when the second driver of the second engine is at a joint, the second driver of the first engine is 2.5 feet (see Fig. n) from a joint ; so that, assuming the coupled locomotives to travel either way, our calculation is correct. We will close this section with an example including every kind of loading contemplated herein. EXAMPLE. Let W =. 20.0 tons, a' = 50 ft. w = 0.4 tons, / = 100 ft. w' = 0.8 tons, b = 20 ft., a = 40 ft. P = 10.0 tons, a" 50 ft., a = 30. Find the moments and differences of moment for every ten feet throughout the girder. In this calculation we use equations (40), (43), (49), (53), (55) (57)> (47) > and (4 8 )> witn tne following result : DISTANCE. 10 20 30 40 50 60 70 80 90 100 For W, M IOO 200 300 400 5o 400 300 200 IOO Forw, M 1 80 320 420 480 500 480 420 320 180 o ?OT'Z G, etc., be horizon- tally projected at the centres of the horizontal projections of STRAINS IN FRAMED OR BUILT GIRDERS. 65 BD, DF, FH, etc., respectively. Let the inclinations of the segments of the top chord, AC, CE, EG, etc., to the horizon be 2 , 3 , etc. ; those of the segments of the bottom chord, BD, DF, FH, etc., to the horizon be ft, ft, ft, etc. ; the inclination to the horizon of a Y web member, as BA, DC, FE, etc., be < 2 > $ 3 > etc. ; and the inclination to the horizon of a Z web mem- ber, as AD, CF, EH, etc., be 6 a 2 , 3 , etc. : each angle of incli- nation of chord, a, ft to be measured from the horizontal drawn through the left end of the chord segment, and each angle, <, 0, to be measured from the horizontal through the lower extrem- ity of the web member, as in trigonometrical notation. FIG. 13. Assume, further, that each member of the structure is capa- ble of resisting the strain that may come upon it, either of tension or compression ; and, for distinction, call strains in compression positive, and tensile strains negative. Also, the simultaneous forces acting at each apex are sup- posed to be in equilibrium, and the structure at rest. All the dimensions of the skeleton girder, as Fig. 13, are given, and may be varied so as to represent all the usual forms of girder, as illustrated below. Let P symbolize the strain along any segment of the top chord ; U the strain along any segment of the bottom chord ; Y the strain along a Y web member whose slope is , as defined 66 MECHANICS OF THE GIRDER. above, and length v ; and Z the strain along a Z web member whose slope is 0, and length z. Therefore / = v sin ( 0), and q -srsin (180 (9 + a) = ^sin (B a), where / is nega- tive, and represents the line drawn from any upper apex per- pendicular to the chord opposite it, and q is positive, and denotes the length of the perpendicular drawn from any lower .apex to the chord opposite. Let H r and H r+l denote the two simultaneous horizontal strains at consecutive apices, whose difference, A//, is the great- est of all differences of simultaneous horizontal strains for that interval ; and let M n M r+l , be the corresponding moments, and 7/ r , // r+I , the heights or vertical distances from those apices to the axis of each chord opposite. Then H M h, and lH is the horizontal component of the strain developed in the diag- onal or web member for the interval to which A// belongs. 41. Suppose that the greatest moments due the given load- ing have been computed for the vertical section at right angles to the plane of the girder through each point, B, A, D, C, F, etc., Fig. 13 ; that is, at intervals each equal to c : and call these moments M lt M 2 , M y etc., at these consecutive intervals. Also, if the simultaneous moments which yield the greatest difference of horizontal strains at any two consecutive apices are different from these greatest moments, as may be the case for rolling loads, suppose such moments known. We then have, from the figure and from the principles of articles 10 and 3, Strain along BD, U^ = M t -r p l ; along AC, P l = M 3 DF, U 2 = M 4 -r / 2 ; CE, P 2 = M s Fff,Us = M 6 -f/ 3 ; EG, P 3 = M 7 Generally U - M r -f /; P =M r + l -g. Strain along AB, K x = ^ff -f cos ^, ; along AD, Z x = CD, Y 2 = ^ff -T- cos^ 2 ; CF, Z 2 = Generally Y = L r H -f cos ^ ; Z - A r + x H (96) cos0 2 . (97) cos 0. \ STRAWS IN FRAMED OR BUILT GIRDERS. 6/ 42. Shearing Forces and Strains. If through any mate- rial body or structure we conceive a plane to pass, dividing the body into any two parts whatsoever, and assume either one of the two parts to be fixed in position, while the other part slides or tends to slide in any direction along this plane, then the force acting in a line parallel to the dividing-plane, and causing this sliding or tendency to slide, is called a shearing-force, and the strain on the particles of the body lying in this plane, resisting or tending to resist the shearing-force, is called the shearing-strain. The amount of shearing-strain per unit of the shearing-surface is its intensity ; and the intensity at the in- stant of rupture (that is, at the beginning of actual sliding of one part of the body over the other along the shearing-plane) is the breaking shearing-strain, and is, in general, peculiar to each kind of material, and must be determined by experiment. The published results of trustworthy experiments for deter- mining the ultimate resistance of materials to shearing, are very meagre ; and the following table, compiled from the works of two of the best authorities I know, viz., Professor Rankine and Mr. Bindon B. Stoney, is probably as worthy of confidence as any published records of the kind. 68 MECHANICS OF THE GIRDER. TABLE I. ULTIMATE RESISTANCE OF MATERIALS TO SHEARING, IN POUNDS, PER SQUARE INCH. MATERIAL. Resistance to Shearing. Remarks. Metals. Cast-iron . . 27700 R. Tensile strength ranges from 1 3400 to 29000. R. Cast-iron . . ( 17920 to ) ( 20160 S. ) " Substantially its tensile strength." S. Wrought-iron . 55059 s. Mean of 5 tests by Mr. Jones, punching-plates. Wrought-iron . 50400 S. Mean of 2 tests, Mr. Little, hammered scrap, inch punch. Wrought-iron . 43456 s. Mean of 4 tests, Mr. Little, hammered scrap, two-inch punch. Wrought-iron . 50848 S. Bar, 0.5 x 3 inches, punched both ways, Mr. Little, mean. Wrought-iron . 48160 S. 2 bars, 1x3 inches, punched both ways, Mr. Little, mean. Wrought-iron . 46144 S. Flanged tire, 1.8 x 5 inches, edgewise, by Mr. Little. Wrought-iron . 52192 S. Rivet, I inch, Mr. Clark. Tensile strength 53760. Wrought-iron . 45696 S. Rivet, I inch, 2 plates, Mr. Clark. Wrought-iron . 49952 S. Rivet, I inch, 3 plates, Mr. Clark. Wrought-iron . 50000 R. Steel .... 63796 s. Kirkaldy, rivet steel, tensile strength 86450. Timber. Fir . . . S I d' f "R 1 Fir, red pine . ( 500 to ) 1 800 R. ) Fir, spruce . . 600 R. Fir, larch . . ( 970 to ) i 1700 R. ) Oak .... 2300 R. Oak .... 4000 S. Across grain, Rankine's deduction from Par- sons's tests of English oak treenails. Ash and elm . 1400 R. Abbreviations : R., Rankine ; S., Stoney. STRAINS IN FRAMED OR BUILT GIRDERS. 69 Instead of a plane cutting the body into two parts, we may conceive it cut into two separate parts by any cylindrical sur- face, and may suppose the sliding, or tendency to slide, to be in the direction of the generating line of the cylindrical surface, as in the case of a cylindrical punch. 43. From the definition of shearing-force, it follows, that if any girder, as Fig. 13, be cut by a vertical plane at right angles to its own plane, then the shearing force or strain at this ver- tical section is equal to the algebraic sum of the vertical com- ponents of all the forces impressed upon either side of this vertical plane. And these two algebraic sums of the vertical components of the forces impressed upon the opposite sides of this vertical plane will have contrary signs, and be numerically equal, except when a vertical force or weight is applied in the vertical plane itself, in which case the shearing-strains on oppo- site sides of the shearing-plane will differ by the value of this weight applied in the vertical plane. Since the resultant of parallel forces is simply their algebraic sum, if the external forces applied to a girder are all vertical (that _ is, made up of the applied weights and the consequent vertical resistances of the piers), the shearing-force on either side of a vertical shearing-plane is merely the difference be- tween the sum of the weights and the re-action of the pier on that side. If, therefore, S denotes the shearing-force on either side of the shearing-plane, W being positive and denoting any weight, and V being the vertical re-action and negative, on the side chosen, we then have S=V + ^ W, (98) where 2* W is the sum of all the weights between the shearing- plane and the point of support having the re-action V\ that is, of all the weights on the length x. MECHANICS OF THE GIRDER. In case of the semi-beam for the free end, V = o, and Sum of weights on / x, S = J-X W. Sum of weights on /, S = (99) ^ being measured from the fixed end. When the girder is supported at both ends, the re-actions due to a single weight, W, applied at the distance a f . Fig. 9, from the left support, are, by equations (38) and (39), At left support, V^ W I -a' At right support, V 2 = W ; calling them negative. And for any number of different weights applied at different points, (100) Therefore for this case the shearing-strain at a vertical section distant x from the left support is or (101) If upon the girder supported at both ends there are i STRAINS IN FRAMED OR BUILT GIRDERS. equal weights, W, at equal intervals, c = -, and J W upon each end of the girder on a pier, we have I= --W = r a , 2 = -w + (r + \W = \W(2r - n + i) ; (102) the shearing-plane being at the r th point of division counted from the left. For a uniform continuous load, !w, upon a girder supported at its extremities, r, = -\iw = r a . | At any point, x, ( I0 3) *S = \lw -f- wx. \ For a uniform continuous load, Iw, upon a semi-girder at any point, x, measured from the fixed end, the shearing-strain is S = (/-*),;] and when x = o, (104) For any partial uniform continuous load, bw r , Fig. 9, on a beam simply supported at its two ends, the re-actions of the piers are (105) which re-actions are identical with the shearing-strains for the unloaded parts of the beam. MECHANICS OF THE GIRDER. But for the loaded part b, the shearing-strain is V) + it/(x a), or S = - (106) If a = o, and x = , (106) becomes (io 7 ) 24 which is the shearing-strain at the foremost end of a uniform continuous load reaching to the left end of the beam supported at both ends. And equation (107) gives the greatest positive and the greatest negative value of 5 for this kind of load ; since, in equations (106), x cannot be greater than b, and in the first of those equations 5 is an increasing function of x, while in the second 5 is a decreasing function of x. The shearing-strain at any point, x, of the partial uniform continuous load on a semi-beam, Fig. 8; is S = it/ (a + & x); (108) x being measured from the fixed end, and not being greater than a + b, nor less than a. In order to simplify the application of equations (101) for the important case of a partial or complete uniform discontinu- ous moving-load, Z, to be applied at equal intervals, c -, n along the girder, we proceed as in article 20, where we found the moments due such a load on a beam supported at both ends. Let r, r 2 = the number of weights, Z, on the beam STRAINS IN FRAMED OR BUILT GIRDERS. 73 at any instant. Take r not less than r 2 nor greater than r,. Then, in the first of equations (101), we have 2a' = r l -r 2 and V = <-[(r a + i) + (r a + 2) + (r a + 3) + r,] = ^.(r, - r a )(r, + r 2 + i) for the first term. But in the second term, 2*L, we must take L no times for the left unloaded end of the beam, r r 2 times for the loaded part, and r t r z times for the unloaded part on the right end. Therefore S = -y[(r r - r a )/ - f-(r x - r 2 )(r, + r 2 + i)] (109) for the shearing-strain left of the load. S = - j[(r f - r a )/ - ^(r, - r a )(r f + r 2 + i) -(r-r a )/J, (no) which is the shearing-strain between the points re and (r + i)^. S = (r, - r a )(r x + r 2 + i) = -^(r, + (m) 2/ 2W if r 2 = o, and c = - ; and this is the shearing-strain at and n beyond the foremost end of a uniform discontinuous load reach- ing back to the left end of the beam. 44. The influence of end moments on the normal shearing- strains may be regarded as operating upon that term only of the shearing-strain which expresses the re-action of the pier. Now, the pier moment M M acting at the right-hand pier, will M affect the re-action V^ of the left pier by the amount -* ; and the pier moment M u acting at the left pier, will affect the 74 MECHANICS OF THE GIRDER. M reaction V 2 of the right pier by the amount -- - 1 . But, by the principles of article 10, the force -- ', acting at the right end JI/T of the lever arm, /, induces a re-action, + - *, at the left end of that arm, that is, in this case, at the left pier, where, conse- quently, Similarly (112) which are the increments of the shearing-strains due to the end moments, and are to be added algebraically to the shear- ing-strains found for the given load on the same beam simply supported at its two ends. The values of M^ and M 2 are here arbitrary, but will be determined for particular cases in subsequent chapters of this work. 45. To find the Shearing-Strain at any Vertical Section of a Girder (Fig. 13) in Terms of the Vertical Components of the Forces which are impressed upon the Shearing- Plane through the Members of the Girder cut by that Plane. Using the notation already given in article 40, Fig. 13, and equations (3), we have, as the vertical component resulting from all the pressures on the left of each odd vertical plane, or the planes through the lower apices, B, D, F y etc., Left of B, S = V lt Left of D, S 2 = P l sin a, + Z x sin 0, - U, sin ft ; Left of F 9 S 4 = -P 2 sin 2 + Z 2 sin 2 - U 2 sin ft ; Left of (27-+ i) th apex, S 2r = -P r sma r + Z r sm0 r - / r sinft; (113) counting r on P, Z, and U. STRAINS IN FRAMED OR BUILT GIRDERS. 75 And on the left of each even vertical plane, or those through the upper apices, A, C, E, etc., Left of A, S t = P sin F, sin < T /, sin /? T ; Left of C, S 3 = /> sintf, - F 2 sin< 2 - 4 sin ft ; Left of , S s = P 2 sin 2 F 3 sin 3 / 3 sin @ 3 ; Left of (2r) th apex, 1 F r sin< r 7 r sin/? r ; (114) counting r on F and U. These values of 5 may be used to verify solutions by equa- tions (96) and (97), as will be illustrated in some of the exam- ples below. 46. Strains in all Members of a Girder determined from the Given Shearing-Forces. Equilibrium of the system re- quires that at each apex, Fig. 13, the sum of the horizontal forces, as well as the sum of the vertical forces, shall vanish. Therefore at any lower apex we have C7 r _ l cosft r _ I Z r _ 1 cosO r _ I Y r cos , U r cos(3 r = o, (115) and ! + Z r cos6 r + F r cos< r P r cosa r = 6 (116) at any upper apex. The four equations, (113), (114), (115), (u6), enable us to determine the four quantities, U r , P r , Y r) Z r , in terms of / > r _ J , and the given shearing-strains, S 2r _^ and .S 2r , if we use the auxiliary equation, P r -i cosa r _ l = [7 r _ I cosfi r _ I Z r _ I cosQ r _ 1 , (117) expressing the equality of the horizontal strains at any lower apex, and at the point directly above it in the top chord. 76 MECHANICS OF THE GIRDER. Therefore, after solving and reducing, P r _ i sin ( may be continued through the girder. 47. We will now give examples illustrating the determina- tion of strains in open girders, first by the method of moments, and second by the method of shearing-strains, and will verify the solutions by equations (113) and (114). EXAMPLE i. Let B, Fig. 13, represent the unsupported end of a semi-girder, whose fixed end coincides with the vertical plane passing through E, and at right angles to the plane of the girder. Let the horizontal distance between consecutive apices, B, A, D, C, etc., = c 10 feet, and the elevation of the apices, in feet, above the point B be as shown in the first line of the solution below. These elevations, with the horizontal distance c=io feet, furnish all the angles and lines required. If any apex is below B, its elevation is negative ; and all angles and trigonometric functions follow the ordinary trigonometrical laws. At each apex, top and bottom, of this semi-beam, let a weight, W i ton, be applied. Required the strains due this load in every member of the girder. The moment M is given by (35), where ;/ = 5, ^ 5> ^= ! and r takes the values 4, 3, 2, i, o, in succession. STRAINS IN FRAMED OR BUILT GIRDERS. *JJ W ^OIIIOIIIIIIIIIIIIIIIIIIIll I^V gg|^ V g| w f-L v?"!^ -n % C | < 5^ >) . n o* 1 S^ ^^ 1 1 1 1 *> 1 1 1 1 &Jg ^lo 1 "K 1 1 1 1 1 1 fa 1 ^ 1 - 5 9 1 ^J ^^o % CO CO >(. O ON C7NJ2 ^O M fc tfl < u ON 06 L/-J O ON ON ^ f^s ^ i^ ON ^ 1/3 ' rt 3 jg W S 0* c? * 5 O N *iO CO ^- >-0%t ij Q 1 f^"bi ''''^''''^So 1 ^'*^" '"'I' i coicon, NONVO O O COONM CJ ON oN^g ^ - 1 ON M S C 3 H S J e o ' ^~ * j. - + g 5 o 1 . . *** CC CC ^So ^"'olcnlo C 1 02. O^tn *^3w '--^w^o "5tnu o o w Ji Vj ts\ l || ^CCbObO .fcObO . CCfcJDbXlW) .WJbO M O 1 O || to II 60 U) " W^2 o co oo a*2 ^--^^^iS-e-i2u o ti oo ^\ MECHANICS OF THE GIRDER. K 1 , , 1 1 I 1 1 I 1 l I I 1 1 1 ' 1 ' ' ' w >0 I a Concluded. M ^ d od M d 06 cji d d m <-5 c^ 1 ^i- 1 1 PQ *. ^ J ^ o TO % ro g N vo rT . ^co vo O M !?dNM N d^^d^NO d4 S3 1 , to 1 r/) r> i u ro vO ON "^t" ^O O H- w w oo -o ON 1 1 1 1 i^^^^vg^i i^i^s&^o 1 ' ' S 1 5>8* ^Jo^vSg? ^- M o to MOfo^ddfo Shearing-*. Cfi* 1 N ii II 1 i Q oo t^ o ro oo 1000 * N ^r S * "- 1 VOMOO M vo ^loocol^Ssr^l 1 Tj-Tfi-ni-ivO'-oONl 1 "^ O Q\ LO i-O "^ , t^ W t^>. ^O C^ VO ON ^O to vO > uT M I 1 M o\ i-. 6 M c5 H- d ON 1 1 1 rf * * 2 - 1 1 1 1 " r?0 &^3 1 1 SvB" ? 5vf vV |"vi 1 1 Nl "? <-or^ rj--nr^r^ 1 -or^r^ H^ ON O ^ ON O O ON O w 1 1 1 H P " \* ^S.s * !~i 1 . C/) -. rt LS g ~, \ ^(^ w S 1 s ; ; ^. ; * ; i ; ^ ; ^ ; ; ^ ; ; ; !> ; ^ ; !* 3 X . . + . * . 7 . .|- - .|. . u> . . 11 . ii . * 1 + s SB . -2 ^ ^ ^ c I't Q 1 * ' o *T I' K K H f^ 1 :o :a ;JL:I '.^ : > * + . r ^ -f ' " -Ti . .' . ' '^ :^ : .=. 9 * a 1 M t'< 1 1 STRAINS IN FRAMED OR BUILT GIRDERS. 79 LOGARITHMIC SOLUTION BY EQUATIONS (118) TO (121). Method of Shearing-Strains. r. 1 2 3
7' V i 5 log^W-i log cos
r . ... 0.48 1;64 1 . 7Q7 2 2.7776 log/V-i . . o. 2 240067 n o.74526i9 Q.Q2i2t;8t; Q 07180^8 log/V-isin(r 0r) 9.6863166 Q Q288l IQ 0.4462948 Q.Q28 IQ 1 ^ 0.8532540 Q 0447777 log/ r 0.7 C7 CQ47 O.^lSoQQ 1 ^ 0.008016"? Ur O t\72I 3.2060 8 1081 S 2r 2 4 log^r O.3OIO3OO 0.6020600 0.70^02 <2n 96oc;6oi8 log Szr cos Or 0.00605 5 2W o 2976618;* S*r cos Or 1.01404 1.084 q log sin (Or @r) Q Q4.7 C2I4. Q Q C C6478 log Ur sin (Or Pr) Q 7OW26I 0.4777477 Ur sin(0 r ftr) o. ^0702 2.0768 1.52106 4.Q6l 7 lognum 0.1821464;* 0.69 5 w c qw O.Qc8l'JQ7 O.QCO7776 log/V o 2 240067 n O 74 ^26lQW Pr 1,674.0 c c6^C log sin OT -i 8.QQ78007 8.6984422 8o MECHANICS OF THE GIRDER. LOGARITHMIC SOLUTION BY EQUATIONS (118) TO (121). Concluded. r. 1 2 3 9.22i9o64 0.4477041 n 0.16666 O 27777 8.6984422 8.8778446 log UT sin ftr 8.4.CCQ46Q Q.7QIQ44I o 0080167 Ur sin /3 r 0.028^7 0.246^7 o 1.028^7 7.O7QO 4 72^2 log num . 0.012277972 o 488538072 Q.Q4I "$Q I Q Q46QO4O Q Q447777 logFr. . 0.0706448;* o ^41674072 o 7 298087 w Kr I 1766 7 480? From P r - 1 sin a r - 1 comes P r sin or, Numerator of (121) . . . . -0.16666 I 8619 J-4 UU -0.27777 7 0688 O'jWy o 2600^87 j.yuoo o 59861581 log sin 0r Q Q7 ^4741 Q Q7864O8 logZr O 77A/4.8/l6 Z r 2 1601 4o/ A1 N - B - The method of shearing-strains, though applicable, is not conveniently used for live loads, since every change of load requires recomputation from the beginning. 48. Maxima Strains in the Web Members of an Open Girder, deduced from the Moments and Shearing-Forces combined, for Uniform Discontinuous Dead and Live Loads. Let W panel weight of dead load at (n i) equidistant points, L = panel weight of live load to be applied at the same points, M w = moment at each panel point due dead load, by equation (65), M L = moment due live load at its foremost end, by equation (64), STRAINS IN FRAMED OR BUILT GIRDERS. 8 1 Sw shearing-force at each panel point due dead load, by (102), S L = shearing-force at foremost end of live load due live load, by (in) ; .*. Sw + SL = greatest shearing-force simultaneous with MW -f- ML, w - = H = simultaneous horizontal chord strain. h Now, we have on the immediate right of any vertical plane- through an upper apex (Fig. 1 3), C/cosfi = H r = -Pcosa -f Zcos0, Psina + ZsmO 7 sin ft = S r . Whence, after eliminating U and P, there results, cos/3 where H and 5 belong to the vertical section through the upper extremity of the ZQ member, which joins the left end of the P a chord segment, to the right end of the Up chord segment. Similarly, on the immediate right of the vertical plane through the consecutive lower apex, cos a ., cos where ff t and 5 X belong to the given vertical plane through the lower extremity of the Y member, which joins the left 82 MECHANICS OF THE GIRDER. . end of the U lft chord segment, to the right end of the P a chord segment of equation (122). It may here be observed, that according to our notation '(article 40, Fig. 13), in any symmetrical girder, in either half- span is the supplement of < in the corresponding panel of the other half-span ; also a and ft in either half-span are respec- tively equal to a and ft of the corresponding panel of the other half-span. So z of the first half-span equals the corre- sponding v of the second. EXAMPLE. Uniform discontinuous dead and live loads. Let all that part of Fig. 13 which is on the left of the vertical line through E represent one of the equal half-spans of a girder supported at its two ends, B, and L not shown in the figure. Take the dimensions for each half-span the same as those already given in the example of article 47. Let the dead load, W^ = 4 tons, be applied at each apex, top and bottom ; and the live load, L t = 8 tons, at the same points progressively. Each extreme apex may be supposed to bear J( W + L) when fully loaded ; but this will only affect the resistances V v and V 2 , so far as the present method of com- puting strains is concerned. We may find greatest strains as .follows : STRAINS IN FRAMED OR BUILT GIRDERS. ~ ^~ % J' ' ' '!>' ' 'S^S^ 00 ?"?;? & 1 OOv vWf^ff 1 OO OO * , * 5 s 5 s 4 ff i? M' N N OD ^ M M' 1 1 + - O o. 3 K. O O>OvO-*-oONOOro 4. N + H S>S IO x OO 1O M CO * - T " I d 1 * * 1 1 J ION OoMt^. -^-rOOO r^ M N^OM"^" iOOO M V M * M d> M" g * fe 6 O ONNVOvO0-LMNrHMiri 5- O>--VOH^ -jLvom M CM ON 0? 2 * S | 1 - 1 1 1 1 1 1 1 M' N M' I 5 H 5 m 5 N tx r% QO"M" O^ QOONM-^-odrOM" >^ ' 1^ ' ' S,vS 2 + M + M 3 a v 3- vS 1 s; S "* -^f \O ^ M ^ iO S ' a ? ^f^SH 2 ^ % ? * . * 1 " * * 01 fe S 5 2 o" 2 J? \O ro ro t^ t- O<^ t^ MD OOQNOOvOOONN CO M M ^0 N M tN. l/> t^* - % : OOO-*1-fO OOWN^ NVC "o^'csS 11 3-sl:^ Ti '1 ie2 oo ;v CO M M N M H o e OOOO*^ 1 OO*O fc O Q W 2 ^ ^ ^-vo 8 J? vBS'co 1 'So> 8&22 c3c; - c s v S I >a 'd Q oo ro io 10 o m ir, OO O 00 ff 2 C ff 1* 'in %, S I! K%2 1 r!n ^-00 N 7 | t N . T 2 C ^O^i 2 ?M . 8 . . 1 8 .j ^1. .^.^i ^ . Sj -?!li| f-f fliiljlj H;4|V}.s bO*'ltObfiW) MMM 84 MECHANICS OF THE GIRDER. s f$ 2 f % ' ' ' * ' ' v & i? ' ' ' ' $ 00^ I 10 o\ H g 8ei vc | vO "0 ^> ^> vS S? ^ S ^ ii ( v iiii V vS* c ^ "r v8 1? ' ' ^ ro ' ' 1 1 M ^ VO^ Q (T) *** OS VO* ^ W VO OS VO ^" O N N CO' O 0* OS M ^. 5- H Os w ^ VO + I" | | | *& ^ w In * ^ d & 8; . ' a !> , , , , , , , i i . C C ' ' 3- P g R s. % J? H a v? ' OO M to 0> M vO % co ^ vo vo , % a v g N ^^ s^ ooS V ^^SI^^IH^^ vtB> , , M 0? 0' M l\ H Jf H OS H VO vO I 1 vO ^O H 1 5W N 5 CO 00 VON fe W V M-VOOS^ M f^OO rO IO i , W 00 0' N d O" VO 00 o 0^ O TS vg | 1 1 1 1 4 v J> ' * 5 J iooO(' fO roio ^v $ gj L i v s ^i ff 11111 ''"''v^'''^ 1 1 S 7 O' so 1 Os N f^ vo" II 1 vo ^ ^ 1 V 'gf^J^S vS SL? Q i 1^1 i . i 1 1 g i 5 | | ' 'fl-y 1 ' 1 1 "o, oroOootT-Osoo^oooo osos" lf > loddosOosd^vdo" oso' ^ osf" * ^^ S f i ' V % a S 11111 r t t f i i i .1 i i o> i H . "* O H Os M VO 10 1 5 ro N M e^ (j, ' ro r^ 10 7 7 vo 7 T a H ' i ' "' . o -^ . ? X 8 Jr * l?l :^f ; : : T::Jrl :?K' : :f : :fi t^. IH rt 11 . rt >ii> i^_, i ^^ >-i rt ou ? r rS r"^'o2u ' -e- ' 6 S 111*7.3,: ' .IS3I 8^gl 'i^ -H ae c oo N . i a o Zi a o 4>gc . , *i - x'x ximum Z + ximum Z -a^^foooo" 1 " 1 'taobobn-M bobo^3oo Ibobo, N^;^ ^ ^^^^ < ^ ^ ^ 2 ,2 'go-e.-e.oo^,^^ ct rt h I gj CJ K STKAINS IN FRAMED OR BUILT GIRDERS. I- ::: * f s -; 1- : t! > \ M * i | $ i i i i *& i " i^.vco |jj 8? f C/3 00 O VO O<*>M CO M M H (0 tO ^ va M CO OO CO 4*- O votOOJl04-t04*-tOO ^ "ON OO ^J N M O WON vo vjoauitOO * ! K> O VO O H "w "M M H M tO JO p i s 1 j, P VO P i. "tO a H H M N W 'lOOv CO vjvjtntOM s S^vOO ONfO vo M w !^ WW r H ON^J ON^ ON^ltOVOO H COM^ICO K)-^ <-" OO ^1 ON IA4 VOM^ONOI III wO^"^> ON o v> -^^i 4^M QQQ50 OMM BON o\ O MM IOCO OOOOO , ! I I & M VO M in ^. H M M 10 W VO-*>-ON S St. -iiootovbo M jr & 1 w ^ifois^ K S 1 CO O\ ^n OvOtOtdtO 4^-O\P IHVJ\OOON p 5 1 9 4fcON*^j WQON"^! ON ONtobovb OJQ OJ^ON 00 O OMCoOJOOONOOONO |^ | -fisvSSo 3 - J H VO M ooo - 86 MECHANICS OF THE GIRDER. The chord strains, U and P, are to be found as before ; their values being greatest when the two uniform loads cover the beam. In the second line of this last solution, M L is the moment due live load at its foremost end as that end passes the suc- cessive apices. In the third line, M+^L is the moment one interval beyond the foremost end, and simultaneous with M L . It is manifest, from what precedes, that we need compute the moments M Wt M L , and M +lL , only when h is not constant ; as, when h does not vary, we may find &M W and &M L by (71) and (69), whence A// = &M -H h. 49. We now proceed to classify girders according to the form which the general equations assume when particular values are assigned to one or more of their variables ; first, recapitulating the general equations of the method of moments, and of the method of moments and shearing-forces. From equations (95), (96), (97), (122), (123), we arrange GENERAL FORMULAE. METHOD OF Moments. Moments and Shearing-Forces. P = n(,-0 = -*rcoB0. p = v sin(0 ft) = hr cos/3. 9 = z sin(0 a) = hr+i cos a. q = z sin(0 a) = hr +i cos a. H = M h. H = (M w +Mj-) +A. &H = Hr+i. H r , or S Inclined . d Horizontal, o III.. . Horizontal, Inclined . Inclined . e Inclined IV. . . Horizontal, Inclined . Inclined . d Horizontal, V. . . Inclined . a Inclined . Vertical . 90 Inclined . (8 VI. . . Inclined . a Vertical . 90 Inclined . e Inclined . f VII. . Inclined . a Inclined . * Vertical . 90 Horizontal, o VIII. . Horizontal, Inclined . f Vertical . 90 Inclined a IX.. . Horizontal, Inclined . $ Vertical . 90 Horizontal, X. . . Inclined . a Vertical . 90 Inclined . e Horizontal, XL . . Horizontal, Vertical . 90 Inclined . e Inclined . XII. . Horizontal, o Vertical . 90 Inclined . e Horizontal, o The conditions yielding the twelve classes may be briefly stated thus : neither, 1 With regard to a and /3 we may have the one, the other, = ; 4 conditions. both, (neither, 1 the one, I = 90 3 conditions, the other, J Combining these conditions gives twelve classes and no more. STRAINS IN FRAMED OR BUILT GIRDERS. 8 9 CLASS I. ALL MEMBERS BUT ONE INCLINED. Use General Formulae. P 3 FIG. 14. THE CRESCENT GIRDER. P 2 FIG. 15. THE TRUNCATED CRESCENT. Pa FIG. 16. THE DOUBLE Bow, OR BRUNEL GIRDER. FIG. 17. INVERTED TRUNCATED CRESCENT. 9 o MECHANICS OF THE GIRDER. FIG. 18. INVERTED OR SUSPENDED CRESCENT. FIG. 19. ROOF PRINCIPAL. FIG. 20. ROOF PRINCIPAL. FIG. 2i. BENT GIRDER OF FOUR SYSTEMS. STRAIN'S IN FRAMED OR BUILT GIRDERS. 9 1 FIG. 210. FIRST SYSTEM. FIG. 2ib. SECOND SYSTEM. FIG. 2ic. THIRD SYSTEM. FIG. zid. FOURTH SYSTEM. 9 2 MECHANICS OF THE GIRDER. FIG. 22. DOUBLE Bow OF Two SYSTEMS. FIG. 220. FIRST SYSTEM. FIG. 16 SHOWS THE SECOND SYSTEM. FIG. 23. SEGMENT OF ROOF PRINCIPAL. (SEE FIG. 71.) FIG. 230. DIAGONALS IN COMPRESSION. STRAINS IN FRAMED OR BUILT GIRDERS. 93 FIG. 23^. DIAGONALS IN TENSION. FIG. 24. PARALLEL CHORDS. TRIANGULAR. FIG. 25. THE BRACED ARCH. ST. Louis BRIDGE SYSTEM. (SEE ARTICLE.) 94 MECHANICS OF THE GIRDER. V Although we have supposed the linear dimensions of the girder known, we will now give a mode of finding them from the known length /, and central height //, in case of girders having either chord, or both chords, circular or parabolic. FIG. 26. ist, Lower chord horizontal, and upper chord circular, as ABCED, Fig. 26. Let / = 2DC length of bottom chord, h = AD =. central height of girder. Then the radius (124) Take D, the centre of the chord of any arc, as ABC or AJ$ t for the origin of rectangular co-ordinates ; the axis of x being horizontal, and that of y being vertical. Then the equa- tion to the curve ABC is and (y + R - hY = R 2 - x 2 , = y = h- R x)(R - x), (125) which is the height of the bowstring girder at any point, E, whose distance DE from the origin is x. STRAINS IN FRAMED OR BUILT GIRDERS. 95 Thus, if / = ioo, and h = 10, we have R = IO 2 + 4 X io 2 130; and from (125) we find, When x = o, y = 10 = h. x = 10, y = 9.6148. x = 20, y = 8.4523. x = 3> J = 6.4912. x = 40, ^ = 3-693 1 - * = 45> J = I -9 6 3 I - x = 5 = V> y = - The length of any diagonal, DB y whose inclination to the horizon = <, is BE DE V = -: 7 = 1 ' sin cos <^ or, in general, v = y r -r- sin<^ = &x + cos<, (126) &c being one panel length. The length of any diagonal, AE, whose inclination to the horizon = 0, not acute, is z = = DE . sin cos ' or, in general, .-jV-. + stotf--^. (I27 ) The length of any chord, AB, whose inclination to the hori- zon = a, is a= DE = ^ cos a sin a ' or, in general, a = -^- = -^L. (128) cos sin a 96 MECHANICS OF THE GIRDER. From (126), From (127), ^L = ->_, + A* = tan 0. (130) cos From (128), HL^L AJF + A* = tana. (131) cos a Whence , 0, and a, and their sines and cosines, may be taken from a table of natural or logarithmic circular functions. To determine the length of any part of the circular arc ABC, or of the whole arc, find the angle at the centre corre- sponding to the given chord of the required arc ; then the required length of arc is to the whole circumference as the angle at the centre is to four right angles. Thus, if C denotes the angle at the centre whose chord is the span / = 100, we have .-. C = 45 14' 23" = 45 .2 3 9722. Circumference = 360 degrees. Length of circumference = 2-n-R = 2 X 3.14159 X 130 = 816.8134, /. Required arc 2ABC = < x 816.8134 = 102.645. Or, the length of the arc subtended by a given chord may be taken from a table constructed for the purpose. 2d, Both chords or flanges circular, as ABCB.A,, Fig. 26, where the chords meet at the ends of the girder ; or, as Figs. 15 and 17, where the cjiords do not meet at the ends. STRAINS IN FRAMED OR BUILT GIRDERS. 97 If the curves meet, as in Figs. 14, 16, 18, and 26, then / in equation (124) will be common to both arcs, ABC, A 1 B 1 C\ but the central heights of the two arcs will be h = AD, and hi = A^D. Hence, for the upper curve, the radius If, as in Fig. 15, the curves do not intersect at the ends,, then, for each curve, / will be the chord subtended by the arc, h will be the central height of each arc above its chord, and the origin of co-ordinates for each curve will be at the centre of its own chord. The ordinates y, corresponding to the same values of x, are to be found for each curve by (125) ; and if y t = an ordinate to the upper curve, and y == the corresponding ordinate to the lower curve, and e = the difference in height of the two ori- gins, then j, -f- e y = the height of girder at any point, x. And when e = o, as in Fig. 26, the height of the girder ABCB.A, at any point, B, is BB, 7, y. FB = BE - AD = Ay, in general. F,B, = B,E - A,D = &y s , in general. DE = AF = A,F t = A#, in general. Ay A_y, tan = -, tan ai = ~ AA, + BF = y lr - y r+I , y, r - y r tan(9 = -- - - , tan A = -~ -- 98 MECHANICS OF THE GIRDER. From these tangents, a, 0, <, and their sines and cosines, are to be found as before. We then have y l y r &x - 4* - "S* = c^' . = A,B = ^^p = - ti , (133) sm 6 cos & a = AB = L _-*-, (134) sm a cos x T- Ay, A^ / x a, A^Bi = -r^- = . (135) cosaj 3d, When the curvature of one or both chords of the girder is parabolic, we proceed as in case of the circular chords just discussed, except in finding the ordinates and length of the curve, which only, therefore, we need now determine. Let the curves, Fig. 26, now be parabolas, whose vertices are at A and A l respectively. Take the origin of rectangular co-ordinates, as before, at D ; the axis of x being horizontal, and that of y vertical. Then the equation to the curve ABC is y =(i-4|!^ ; (I36 ) to the curve A^B.C, (I37) For the same value of x, (136) and (137) give -i)i (138) which is the height of the girder at any point whose distance is x from the centre or origin. STRAINS IN FRAMED OR BUILT GIRDERS. 99 Thus, if / = ioo, and h = 10, (136) gives, When x = o, y = 10 = h. x = 10, y = 9.6. x = 20, y = 8.4. x = 30, y = 6.4. x = 40, ^ = 3.6. * = 45> ^ = i-9- x = 5 = K y = o- And if /*, = ^,Z> = 20, A = ^Z> = 10, and / = 2DC = ioo, equation (138) gives the heights of girder ACA t as below : When x = o, h = 10 = /t t h\ x = 10, h m = 9.6 ; x = 20, ^ 20 = 8.4 ; x = 3> ^30 = 6.4 j x = 40, /; 40 = 3.6 ; X = 45. ' ^45 = I-9> ^ = 50 = -|/, 7z so = o ; which are the same as the heights of the girder ADC just found, since /i t = 2/1, and, from (136) and (137), that is, the ordinates to the two curves, for the same value of x, are proportional to their central heights. The length of the parabolic arc, in terms of the chord / and the central height //, is (140) where log means the common logarithm. IOO MECHANICS OF THE GIRDER. If / = loo feet = span, and h 10 feet = central height, of parabolic arc, then (140) gives 5 = 102.606 feet. From these examples it appears, that, when the curvature is small, there is but little difference between the ordinates and arc of the circular and the ordinates and arc of the parabolic girder of the same central height and span. Instead of these exact determinations of the linear dimen- sions of a girder, the figure may be drawn to a scale, and the length of each member measured, where greater accuracy is not required. It is proper to observe here, that, in all cases of curved flange, the line of action of the flange strain, P or U, is the chord of the arc between adjacent apices, and not the arc itself. When, therefore, either flange of a girder is curved, and not polygonal, there is developed midway between adjacent apices in the same flange a deflecting force tending to increase the curvature of a compressed flange, and to dimmish the curvature of a flange in tension. For the amount of this deflecting force F we have, if P is the strain along the chord BIC, Fig. 26, F = 2P\a&HCL (141) Or, if C is the angle at the centre of the circle whose chord is BC, then i - cos /. F = 2/>(cosecC - cot|C) ; (142) and the strain along the chord HC of each half of the arc BC is P'= P H- cosZfCY. (143) STRAfJVS IN FRAMED OR BUILT GIRDERS. IOI Similarly, in cases like P v Fig. 19, there is a deflecting force generated at the ridge equal to F = 2/> 3 tan a; and the strain along the upper segment of each rafter is, as indicated, P 3 -r- cos a. The bending-moment due to this deflecting force is given by equation (46), M = Fa, (144) where a is the length of the chord BC. The amount of material required to neutralize this moment will be determined in the sequel. But it is already manifest that the least amount of resisting material will be required when the line of pressure coincides with the axis of the resist- ing member. Multiple or compound web systems, as those represented in Figs. 21 and 22, may be separated into the single systems of which they are composed, when the sum of all the strains found for the same member will be the strain sought for that member. CLASS II. BOTTOM CHORD HORIZONTAL, OTHER MEMBERS INCLINED. 3 = o. U, u 2 u 3 u 4 FIG. 27. THE PARABOLIC OR CIRCULAR BOWSTRING. 102 MECHANICS OF THE GIRDER. U 2 U 3 FIG. 28. THE TRUNCATED BOWSTRING. U 2 U 3 FIG. 29. ROOF PRINCIPAL. U 2 U 3 FIG. 30. ROOF PRINCIPAL. FIG. 31. TIED RAFTERS. STRAINS IN FRAMED OR BUILT GIRDERS. 103 U 2 FIG. 32. ROOF PRINCIPAL. FIG. 33. ROOF PRINCIPAL OF Two SYSTEMS. FIG. ^a. FIRST SYSTEM. FIG. & SECOND SYSTEM. 104 MECHANICS OF THE GIRDER. u, FIG. 34. ROOF MAIN, COMPOUND SYSTEM. (SEE FIG. 70, CLASS VIII.) FIG. 35. BOWSTRING OF Two SYSTEMS. TRIANGULAR. U 2 , U 3 U 4 U 6 FIG. 350. FIRST SYSTEM. FIG. 27 SHOWS THE SECOND SYSTEM. FIG. 36. THE POST TRUSS WITH CURVED TOP. STRAINS IN FRAMED OR BUILT GIRDERS. 105 As Counters. As Mains. FIG. 360. FIRST SYSTEM. 1 2 3 As Counters. As Mains. FIG. 36^. SECOND SYSTEM. FIG. 37. DOUBLE TRIANGULAR TRUNCATED BOWSTRING. SYSTEM OF KANSAS-CITY BRIDGE. FIG. 370. FIRST SYSTEM. io6 MECHANICS OF THE GIRDER. U 2 U 3 FIG. yjb. SECOND SYSTEM. FORMULA FOR CLASS II. /3 = o. Method of Moments. H = M -r- h. XT' cosoc U = -^,. cos z _ A r + 1 ^ In case of vertical end posts, as in Fig. 37^, where Fand Z become indeterminate by the above equations, we have x = Z^ sin I -f P l sin = F 5 sin P 4 sin 4 . Where a vertical section through any apex cuts a web member, as in Fig. 33$ for Z l = F 3 , and in Figs. 36^ and 36^ for P 4 and the counters F and Z, we do not have H, the hori- zontal component of chord strain, equal to M -h //, but may proceed as follows : STRAINS IN FRAMED OR BUILT GIRDERS. ID/ Find the moments M iy M 2 , M 3 , etc., at vertical planes through consecutive apices. Call the heights above the bottom .chord at which the cut diagonal meets these vertical planes in each panel, a and b> as in Fig. 38. FIG. 38. Then, taking moments about A and B, we have Mj. h^ cos a, + a, Y l cos < M 2 = /i 2 P l cos ! -h b l Y l cos p _ -L\ (146) (148) (aji 2 ( AI)COSJ Taking moments about A^ and B there results M 2 = h^ /, cos ft (/^2 ^) ^i cos(/>,, ( I 5) ... ^/; ii-i ^ ^-. (151) I0 8 MECHANICS OF THE GIRDER. Similarly, or by increasing the indices of a, b, a, and by i, and those of M and h by 2, we find (aji- bji^ cos p 2 Then, taking moments about C gives (152) Now, in case of the Post Truss, we need only the counter- strains F, since Z, P, and U have their greatest values as main strains. And when both chords are horizontal, the horizontal pro- jection of the Z member, or strut, is one-third of the horizontal projection of the Y member, or tie, as usually built ; hence a = \b = \h, * - - Sf A TT which, it will be seen, is the same thing as F x = - , pro- cos vided A// is taken for the interval equal to the horizontal projection of the F member, while ^H belongs to one-third of that interval ; since the foremost end of the live load, at the instant the value of F is here sought, is at the foot of the F member, being applied either directly at the lower apex, or STRAINS IN FRAMED OR BUILT GIRDERS. 109 indirectly at the upper apex, and reaching the bottom chord through the Z member terminating there. In other words, when the counter-strain F, due live load, is greatest, there is no part of the live load applied on the right of the foot of the Y member, or of the top of the ^member above. In multiple systems, where the chords are not straight lines, in finding total chord strains, care should be taken to reduce all strains that are to be added, so that their lines of action will be parallel ; horizontal, for instance. CLASS III. TOP CHORD HORIZONTAL, OTHER MEMBERS INCLINED, a o. FIG. 39. TRUSSED BEAM. FIG. 40. TRUSSED BEAM. P, Pi FIG. 41. INVERTED BOWSTRING. no MECHANICS OF THE GIRDER. FIG. 42. INVERTED TRUNCATED BOWSTRING. FORMULA FOR CLASS III. a =. o. Method of Moments. H = M -h- h. P = H r U = r + cos /. f- cos . f ^- cos CLASS IV. BOTH CHORDS HORIZONTAL, WEB MEMBERS INCLINED, a o, (3 = o. Pi P 2 P 3 u, U 2 u 3 U 4 FIG. 43. THE TRIANGULAR GIRDER. ERECT, OR "THROUGH." STRAINS IN FRAMED OR BUILT GIRDERS. Ill U 2 U 3 FIG. 44. TRIANGULAR GIRDER. SUSPENDED, OR "DECK." a o, /? = o. FIG. 45. DOUBLE TRIANGULAR GIRDER. FIGS. 43 AND 44 COMBINED. FIG. 46. QUADRUPLE TRIANGULAR SYSTEM. EACH SYSTEM INDEPENDENT. U 3 FiG.'4&*. FIRST SYSTEM. 112 MECHANICS OF THE GIRDER. FIG. 46^. SECOND SYSTEM. U, FIG. 46^. TiiRD SYSTEM. FIG. 46d. FOURTH SYSTEM. FIG. 47. THE POST TRUSS. Two SYSTEMS. STRAINS IN FRAMED OR BUILT GIRDERS. 113 U 7 Z 7 As Counters. As Mains. FIG. 470. FIRST SYSTEM. U, U 2 U U 4 U 5 U 6 As Counters. As Mains. FIG. 47^. SECOND SYSTEM. FIG. 47. Z = A r + iH -r- cos 0. 114 MECHANICS OF THE GIRDER. CLASS V. ALTERNATE WEB MEMBERS VERTICAL. BOTH CHORDS INCLINED. Generally { Verticals in compression, \ Q = ( Diagonals in tension. - j Only one set of diagonals shown in figures. These are counters in first half-span, mains in second half-span. Ui U 2 Us U 4 U 5 U a U 7 U 8 FIG. 48. THE CRESCENT. FIG. 49. TRUNCATED CRESCENT. FIG. 50. DOUBLE Bow, OR BRUNEL GIRDER. STRAINS IN FRAMED OR BUILT GIRDERS. 115 FIG. 51. TRUNCATED CRESCENT INVERTED. U 4 U 6 FIG. 52. CRESCENT SUSPENDED. VJi FIG. 53. ROOF PRINCIPAL. FIG. 54. TRUSSED RIB. MECHANICS OF THE GIRDER. FIG, 55. DOME PRINCIPAL. PRIMARY SYSTEM. SECONDARY SYSTEM SAME AS IN FIG. 54. Z = FIG. 56. BENT TRUSS. DOUBLE SYSTEM. FORMULA FOR CLASS V. = 90. Method of Moments. H = M + h. M fJ . -Mr. h r + T h r P = H -r- cos a. TT TT /D T ^ j-f P r sin a r Y r sin r = H r + ! tan a r + I H r tan a r (Load applied at bottom.) H r tan & + //r + 1 tan /J r + x A r ZTtan < r . (Load applied at top.) STRAINS IN FRAMED OR BUILT GIRDERS. 117 The value of Z in equation Z = kH H- cos 0, here becomes indeterminate, since &H = o for the horizontal projection of the Z member, and cos 90 o. CLASS VI. BOTH CHORDS INCLINED. ALTERNATE WEB MEMBERS VERTICAL. ( Verticals in tension, ) Generally J . \ $ = ( Diagonals in compression. ) 90' Only one set of diagonals shown in figures. These are counters in first half-span, mains in second half-span. FIG. 57. THE CRESCENT. FIG. 58. TRUNCATED CRESCENT. U 4 U 5 FIG. 59. THE BRUNEL GIRDER. MECHANICS OF THE GIRDER. FIG. 60. TRUNCATED CRESCENT SUSPENDED. FIG. 61. SUSPENDED CRESCENT. Y = P r + 1 sin a r FORMULA FOR CLASS VI. 90. Method of Moments. H = M -j- h. ^r +I _ Mr^ h r+ i h r P H -r- cos a. #_, = -H + cos (3. P r sin a r Z r + x sin r + 1 = Z^. + ! tan r + ! ZT r tan a r ^ r (Load applied at bottom.) ff tan O r + z . (Load applied at top.) Z = & r H + COS ^. Multiple systems of this class are seldom built, since long struts are not economical. STRAIA r S IN FRAMED OR BUILT GIRDERS. 1 19 CLASS VII. BOTTOM CHORD HORIZONTAL. fi = o. ALTERNATE WEB MEMBERS VERTICAL. 90. In general, j Verticals * ( Diagonals i compression, in tension. But one set of diagonals shown in figures. These are counters in first half-span, mains in second half-span. FIG. 62. THE BOWSTRING. U 2 U 3 U, U 5 U ff U 7 U 8 U 9 FIG. 63. TRUNCATED BOWSTRING. FIG. 64. RAFTERS AND TIE. I2O MECHANICS OF THE GIRDER. ~OT~ U 3 U 4 U 5 U 6 U, U 8 U 9 FIG. 65. TOP CHORD UNIFORMLY SLOPED. a x a z = a 3 . U, U 2 U 3 U 4 U 5 U 9 FIG. 66. POLYGONAL TOP CHORD. U 7 U 8 U 2 U 3 U 4 U 5 U 6 U 7 FIG. 67. ROOF PRINCIPAL. a^ a 2 a 3 . U 2 U 3 U 4 U 5 U e U 7 U 8 U fl U IO U tl FIG. 68. PARALLEL BRACING. STRAINS IN FRAMED OR BUILT GIRDERS. 121 FIG. 69. UNIFORM SLOPE. DOUBLE SYSTEM. FIG. 70. CURVED TOP. DOUBLE SYSTEM. FIG. 71. ROOF PRINCIPAL. (SEE FIG. 23.) 122 MECHANICS OF THE GIRDER. simultaneous. FIG. jia. THE TWIN FISHES, LONG SPAN. (SEE FIG. 22.) FORMULAE FOR CLASS VII. ft = o, = 90. Method of Moments* H = M -f- h. H = Mr + * - ^ h r + i h r P = H r -r- cos a. U = -Zf r + I . y = z = Foremost end of live load at Zr i for maximum Y and Zr> When it is desired to have the diagonals in each half-span parallel for a given number of panels, as in Fig. 68, the lengths of the panels and the inclination of the diagonals may be found as follows : Let / = length of span. 7z = height at each end. h = central height. m = number of panels in each half-span. < = inclination to horizon of counters in first half-span, and of mains in second half-span. a = inclination of top chord. A/= the variable panel length. STRAINS IN FRAMED OR BUILT GIRDERS. Then = , and AA = <_. tan < tan , = h - A* = h - tan \ tan < = Aol 14- , , tand)/ tan< = Generally, h r = A c = h ( i tana> \ = // 1 _ tan V " z \ tan / ' Ao = ^ _ ssjy, tan a / x .-. tan = r (JSS) '-(if for the first half-span. Similarly, for the second half-span i * + MA'. (I5 8) \ tan ^>/ 124 MECHANICS OF THE GIRDER. CLASS VIII. TOP CHORD HORIZONTAL, a = o. STRUTS VERTICAL. = 90. FIG. 72. THE FINK TRUSS. U 1 FIG. 720. MAIN SUSPENDERS. P P U u. FIG. "j2b. SECONDARIES. TERTIARIES. FIG. 73. THE BOLLMAN TRUSS. STRAINS IN FRAMED OR BUILT GIRDERS. 125 FIG. 74. TRUSSED RAFTER OF FIG. 34. FIG. 75. SPANDREL FILLED. U 2 FIG. 76. TRUSSED BEAM, THREE LINKS. R P 2 P* PS Pa P 7 P 8 FIG. 77. CATENARIAN LINKS. 126 MECHANICS OF THE GIRDER. FIG. 78. TRUSSED BEAM. FIG. 79. THE POINT SUSPENSION, STIFFENED CATENARY. FORMULAE FOR CLASS VIII. a = o, 6 = 90. 4 Method of Moments. H = M -T- h. h r + i h r -H r -f- COS f3. - cos. P U Y Z When the vertical member has no diagonal attached at its top, then, of course, the strain upon the vertical is, for Class VIIL, equal to the load applied at the upper apex. P, STRAINS IN FRAMED OR BUILT GIRDERS. I2/ CLASS IX. BOTH CHORDS HORIZONTAL, a = o. Verticals in compression. f$ = o. Diagonals in tension. = 90. P 2 P 3 P 4 P 5 P 8 Pr U, U 2 u 3 "07 U 8 FIG. 80. THE PRATT TRUSS. FIG. Sew. THE LINVILLE, OR PRATT OF Two SYSTEMS. FIG. 8o. PRATT TRUSS OF THREE SYSTEMS. 128 MECHANICS OF THE GIRDER. For end struts, tan = 2 tan <. FlG. SOC. LlNVILLE, WITH INCLINED END-POSTS. FIG. Sod. THREE SYSTEMS, INCLINED END POSTS. / 4 FIG. 8o = 90. Diagonals in compression. U 2 U 3 U 4 U 5 U 6 U 7 U 8 FIG. 81. THE POLYGONAL BOWSTRING. U, U 2 U 3 U 4 U 5 U e U 7 U 8 FIG. 82. THE HOWE TRUSS, WITH CURVED TOP. 2 U 3 U 4 U 5 U 6 U 7 U 8 FIG. 83. HOWE TRUSS, INCLINED TOP CHORD. STRAINS IN FRAMED OR BUILT GIRDERS. 131 FIG. 84. RAFTERS, WITH VERTICAL TIE. FIG. 84*2. SYSTEMS OF THE SCHAFFHAUSEN TRUSS. FORMULA FOR CLASS X. ft = o, < = 90. Method of Moments. If = M -f- h. . h r + i. h r Hr + T -5- COS . -H r . P = U = Y = Z = 132 MECHANICS OF THE GIRDER. CLASS XI. TOP CHORD HORIZONTAL, a = o. STRUTS INCLINED. TIES VERTICAL. = 90. R P 2 Pa Pe U 6 FIG. 85. SUSPENDED Bow. FIG. 86. FILLED SPANDRELS. FORMULAE FOR CLASS XI. a = o, < = 90 Method of Moments. H = M -*- h. M r ir + i U = Y = Z = cos ^. STRAIN'S IN FRAMED OR BUILT GIRDERS. 133 CLASS XII. BOTH CHORDS HORIZONTAL, a = o, STRUTS INCLINED. TIES VERTICAL. = 90. = o. R P 5 P 6 P 7 U 2 U 3 U 4 , U 5 U 6 FIG. 87. THE HOWE TRUSS. U 8 Y 3 U, U 2 U 3 U* U 6 U 6 U 7 U 8 FIG. 88. HOWE TRUSS, INCLINED END POSTS. FORMULAE FOR CLASS XII. a = o, ft = o, < = 90. Method of Moments. Moments and Shearing-Forces. H M -T- h. -5- h. P U Y H = U = -H W + L . Y = -S +I . Z = 6* -T- sin 0. 134 MECHANICS OF THE GIRDER. CHAPTER V. MOMENTS OF RESISTANCE OF THE INTERNAL FORCES OF A BEAM OR GIRDER HAVING A CONTINUOUS WEB. SECTION i. General Formula found and applied to Particular Cross-Sections of Beams with Continuous Web. 50. The mode of estimating the moment of resistance offered by the cohe- sion of the particles of the material com- posing a beam, we now proceed to illus- trate. Let ABCD 9 Fig. 89, be the vertical longitudinal central section of a beam of any cross-section whatever, under the influence of given applied forces or pressures. It is required to find the moment of resistance offered by FIG. 89. beam at any normal section, OTQ. the material of the MOMENTS OF RESISTANCE OF INTERNAL FORCES. 135 Let NS be the intersection of the neutral surface of the beam with the plane of the paper. The neutral surface of a beam coincides with the position of that longitudinal lamina which, for a given strain, is neither compressed nor elongated. All fibres not in the neutral surface are assumed to be increased or diminished in length by a quantity in direct pro- portion to their distance from the neutral surface, and also in direct proportion to the intensity of the force acting on the fibres. Let / = the force acting on a unit of area of any given normal section, at right angles to the section, and at the unit's distance from the neutral sur- face, either above or below. dz = an element of the thickness of the beam.^ dy = an element of its depth. y = the distance of the fibre whose area is dz dy from the neutral surface. Then the pressure upon the area dz dy is fy dy dz, and the elementary moment due this stress is dM = ffdydz; whence the total moment for the cross-section is (159) which expression is to be integrated between limits depending on the form of the given cross-section whose centre of gravity may be taken as the origin of co-ordinates. 136 MECHANICS OF THE GIRDER. 51. We will now apply (159) to the determination of the moment of resistance, R, for various cross-sections occurring in practice. Let the beam have a rectangular cross-section of the breadth b y and the height h, as in Fig. 90. FIG. 90. Then equation (159) becomes = f y*dydz = fdy, (160) where / is the unit strain at the unit's distance from the neu- tral surface, and B = \hf = the unit strain at the distance \h from the neutral surface, or at the upper and lower surfaces of the beam, since the neutral surface is here assumed to be in its centre. The quantity B is the unit strain which, at the instant of rupture, would be developed at the upper and lower surfaces of a beam having its neutral surface midway between those outer surfaces. B is called the modulus of rupture, or the ulti- mate unit resistance of the material to cross-breaking. A table giving values of B is inserted in article 60. MOMENTS OF RESISTANCE OF INTERNAL FORCES. 137 52. Beam of Hollow Rectangular Section, or Beam of Equal Flanges, Fig. 91. b b FIG. 91. Let ^ = height of beam. hi = height of cavity or web. b = breadth of beam or flange. # x = breadth of cavity. Then, as in article 51, we shall have, for the whole area b X h, and for the area of the cavity #, X // or 2 X J# x X ^,, Whence, for net area of cross-section, R = (161) where ^ = J/z/" = unit strain at the upper and lower surfaces of the beam. If the beam is square and hollow, so that h = b, and h* = b u we have, from equation (161), R = (162) 138 MECHANICS OF THE GIRDER. 53. Beam composed of Two Vertical Plates and Two Horizontal Channels. FIG. 92. \ Let the two plates and the two channels, Fig. 92, have equal cross-sections. b = entire breadth of beam. h = entire height of beam. bi = width of channel. b 2 = width of its web. h T = distance between channels. h 2 = depth of channel cavity. The neutral axis (that is, the line of intersection of the neutral surface with the normal section) is here central. Whence, as in article 51, we have, for the area (b b^ X h, for the area b 2 X (h 2// 2 ), for the area , X h n MOMENTS OF RESISTANCE OF INTERNAL FORCES. 139 Whence the total moment of resistance, R = [(j - b 6/1 - w], (163) where B =. \hf = unit strain in extreme top and bottom fibres. 54. Beam composed of Two Vertical I-Beams and Two Equal Horizontal Plates. _ i FIG. 93. In Fig. 93, let / =: height of beam. hi = height of I-beams. // 2 = height of their webs. b = width of plates. bi = width between beams. b 2 = width of cavities of beams. Then, proceeding as in the last article, we find R = .(40 - bji? - b (164) 140 MECHANICS OF THE GIRDER. 55- Beam composed of Two Vertical Channels and Two Horizontal Plates. 6 A. FIG. 94. In Fig. 94, let h = height of beam. ^j = height of channels. h 2 = height of their webs. b =. width of plates. bi = width between channels. b 2 = width of cavities. Then we have 56. Beam with but One Flange. '6 N i&i A fc, Ti. B C (165) FIG. 95. Let the cross-section, Fig. 95, have the form of the letter T. MOMENTS OF RESISTANCE OF INTERNAL FORCES. 141 b = width of flange. h = whole' height of beam. b b l = thickness of web. h^ = height of web. g = distance of neutral axis NA from any line, BC, parallel to NA, in the plane of the given cross- section. ist, To find , and determine the position of the neutral axis. Take the moment of the area of the section about BC as an axis, and divide this moment by the area. The quotient will be the value of . Moment of flange about BC = b(h - h,) X \(h + h,) = \b(h* - h*). Moment of web about BC = h*(b d t ) X JA, = \h?(b - h). Total moment of area about BC is Total area = bh bji* ; therefore " 2d, By means of (159) we find, For area b X (h s), for area (^ ^,) X , I 4 2 MECHANICS OF THE GIRDER. for area b^ X (/*i ), * 3 = ^ f y3y = i/& x (*i - ) 3 - t/o Whence the moment of resistance due to the net cross-section is R = [*(* - 8 ) 3 + (* - *) 3 - M*x - *) 3 ]> ( I66 > where B = ef = unit strain at the extreme edge of the beam. In a similar manner may all other beams be treated whose cross-sections are composed of rectangles having two sides par- allel to the neutral axis. 57. Solid or Hollow Beam of Square Cross-Section and Diagonal Vertical. FIG. 96. Let b, Fig. 96, be a side of the beam's cross-section, and b l a side of the square concavity whose centre coincides with the beam's centre. Then the diagonals are b^2 and b^2 ; and (159) becomes, since z = Jfl/2 y, For solid beam, R = 12 where ^ = \fb2 = intensity of stress at extreme upper or lower edge of the beam whose diagonal is vertical. MOMENTS OF RESISTANCE OF INTERNAL FORCES. 143 If in (160) we make h = b, then R = where B = \fb = intensity of stress at upper or lower surface of a square beam whose side is vertical. Hence, although the identity of the middle members of equations (160) and (167) shows that the total moment of resist- ance, R, is the same for a given solid square beam whether its side or its diagonal be vertical, yet the extreme fibres for these two positions of the beam are strained in the ratio of their distances from the neutral axes; that is, in the ratio of i to ^2. If, therefore, B expresses the ultimate strength of the material, when in equation (167) it is equal to J//H/2, we may evidently give to B the same extreme value in equation (160), and thus make the beam V/2 times stronger when its side is vertical than when its diagonal is vertical. Again, for the vacant square whose side is b lt since z = y, we have r 4*i/ r- =2/1 f(i4Vfl - Jo And therefore, for a hollow square beam with diagonal vertical, the moment of resistance is X = J*B - 6 '\ (168) 12 b where B \fb^2 = unit strain at extreme edge of beam when the diagonal is vertical. 144 MECHANICS OF THE GIRDER. 58. Solid or Hollow Beam of Circular Cross-Section. FIG. 97. Let r = radius of the outer circle, Fig. 97, and r, = radius of the inner circle. The equation of the outer circle is /. z = (r* - and equation (159) becomes T 4 (169) which is the moment of resistance for a solid beam of circular cross-section with the radius r, and where B = /r = the unit strain on the highest and lowest fibres. If the beam is hollow, the inner and outer circles being concentric, we manifestly have R = 0.7854^'^-^, where B = unit strain on highest and lowest fibres. (170) MOMENTS OF RESISTANCE OF INTERNAL FORCES. 145 59. Beam of Elliptical Cross-Section, Solid or Hollow ; Longer Axis vertical ; Axes of Outer and Inner Ellipses coincident. FIG. 98. Let k, Fig. 98, be the length of the semi-transverse axis of the outer ellipse, and h l that of the inner ellipse ; b = length of semi-conjugate axis of outer ellipse, and , = the same for the inner ellipse. The equation of the outer ellipse is z 2 y 2 Whence (159) becomes *"*lp 7> T jK. =^ 0.7854^^, which is the moment of resistance for the solid elliptical beam, where B = /if = unit strain on highest and lowest fibres. 146 MECHANICS OF THE GIRDER. Similarly, for the area of the inner ellipse, and therefore, for the hollow elliptical' beam, the moment of resistance is R = 0.7854***' ~ W , (i 70 where B = hf = unit strain on highest and lowest fibres. 60. These illustrations may suffice for girders of continuous web. We close this section with a table giving the limiting value of B, in pounds avoirdupois to the square inch, for the ordinary materials used in beams ; that is, the values of B in this table are values which cannot be exceeded in the equations of this section, and represent the ultimate resistance of the material to cross-breaking. It should, however, be borne in mind, that B may not repre- sent the actual unit strain which the material is capable of resisting, either in tension or compression ; but that it, in gen- eral, has some mean value between the ultimate resistance of material to crushing, and the ultimate resistance of the same material to tearing by direct pull. Continuing the suppositions made in article 50, we may find the relation existing among these three ultimate unit strains in rectangular beams as follows : Let us take h depth of beam. b = breadth of beam. / = length of beam. x = distance of the neutral surface from the compressed side of the beam. MOMENTS OF RESISTANCE OF INTERNAL FORCES. 147 C ultimate resistance of the material to crushing by direct thrust, in pounds, per square inch. T = ultimate resistance of the material to extension, in pounds, per square inch. B = the unit strain which, at the instant of rupture, would be developed at the upper and lower surfaces of a beam having its neutral surface midway between these outer surfaces ; that is, B = the modulus of rupture, also in pounds per square inch. Then, using (159), we find for the compressed part of any cross-section, moment of internal forces, (173) and for the extended part of the same cross-section, But, if the neutral axis bisected the given cross-section, we should have moment of internal forces on either side of this axis, R = \Bb(&hy. (175) Now, since experimental researches show, that, for many materials used in construction, these three expressions are nearly equal to one another, we have, approximately \Cb& = \Tb(h - *Y = \Bb(\K)*; ( l l 6 ) whence J T r= 9 (177) (178) C ^T \/C + \/T B 148 MECHANICS OF THE GIRDER. (179) (180) When, therefore, any two of the three quantities, C, T, and B, are given, the third may be found, and also the position of the neutral surface. It is probable, that after the elastic limit of the material is passed, and rupture is about to take place, the expressions in (176) do not represent the actual moments, but are similar functions of C, T, and B, and are proportional to the forces then developed. For within the elastic limits the forces are \CJ>x = \TJ>(h - x) = \BJ>(&K), (181) C,, T I9 and B T being the limited unit strains at the surfaces. But when the strain on the extreme fibres passes the elastic limit, and the fibres expand or contract more rapidly than the strain increases, then an increment is given to all the previous inner strains proportional to their distances from the neutral surface, which is equivalent to introducing a factor of the form kx into the expressions (181), whereby, at the instant of rup- ture, they become \Cbkx* = \Tbk(h - x} 2 = \Bbk(\hY, (182) which is identical with (176). Experiments indicate, that in the case of cast-iron, owing to the superior hardness of the outer over the inner portions of the metal, an increment should be given to B in (180) equal to one-ninth of itself. MOMENTS OF RESISTANCE OF INTERNAL FORCES. 149 TABLE II. MATERIAL. Authority. Ultimate Resistance, in Lbs., per Square Inch, to Modulus of Elasticity, E. Com- pression, C. Tension, T. Cross- Breaking, B. Cast-Iron. Stoney. Rankine. Thurston. Stoney. it Rankine. Lovett. Stoney. 105945 H. 86284 H. II2000 87429 127323 40320 36000 40000 225568 F. 16720 H. 15298 H. 16500 20500 3447 57555 K. 60000 70000 51000 357oo 28600 64000 70000 100000 90000 50915 83391 F. 71658 K. 37695 H.F. 45696 C. 30240 C. 26880 C. 38304 C. 45965 c. 38250 45760 67035 51341 c. 74995 C. 30240 C. 70291 C. 43814 C. 61824 53760 51475 52567 C. 42000 128083 K. 115181 K. 12090000 17000000 "450754 15968254 24000000 29000000 25300000 15000000 27311111 31000000 Means of 16 irons . . . Bars not > i inch wide . . . Bars 3 inches wide .... Bars, small round .... Wrought-Iron. Bars, previously strained . . Bars, new round Boiler tubes, welded .... Circular tubes, riveted . . . Rolled I-beams, about . . . T-iron, flange up about . . . T-iron, flange down about . . Average Bars and bolts . . ... Plates Plates, double-riveted . . . Plates, single-riveted . . . Wire Wire Mean of 113 tests .... Mean of 27 tests ..... Steel. Bessemer, hammered . Bessemer, rolled ISO MECHANICS OF THE GIRDER. TABLE II. Continued. MATERIAL. Authority. Ultimate Resistance, in Lbs., per Square Inch, to Modulus of Elasticity, E. Com- pression, C. Tension, T. Cross- Breaking, B. Steel (continued). Crucible, hammered .... Crucible, rolled ..... Cast, not hardened .... Cast, low temper ..... Stoney. M Rankine. < Stoney. Rankine. Stoney. M ( Rankine. Stoney. Rankine. Stoney. Rankine. Haswell. Stoney. Rankine. Stoney. 198944 Wd. 354544 Wd. 391985 Wd. 372598 Wd. 6831 H. 9363 H. 9000 11500 9363 H. 11663 H. 6402 H. 10300 5863 H. 5860 5350 6586 H. 7293 H. 10300 10331 H. 85546 K. 68589 K. 1 00000 130000 80000 13900 Mu. 16700 Bv. 170006. 9360 115006. 17300 Mu. 15000 Bv. 20000 B. 20000 11400 Bv. 11400 13300 Ro. 10500 Bv. 12100 Bv. 11500 6000 Mu. 12900 Bv. 14000 14400 Bv. 147840 K. 118272 K. "7344 12156 B. 13000 10500 9336 B. 12366 B.D. 11568 B. 14670 T. 4596 D. 8958 D. 7400 10660 9372 B. 7850 4692 B.D. 11820 D.N. 29000000 42000000 1600000 1350000 1645000 486000 1140000 1020000 Cast, mean temper .... Cast, high temper .... Bars .... Bars Plates, average Average Wood. Alder Ash Ash Beech Beech Beech Birch, American Box Box Cedar, American white . . . Cedar of Lebanon .... Cedar of Lebanon .... Chestnut, Spanish .... Chestnut, horse . . . Deal, Christiana . . Deal, red Deal, white Elm Elm Elm, English ... Elm, Canada Rock .... MOMENTS OF RESISTANCE OF INTERNAL FORCES. 151 TABLE II. Continued. MATERIAL. Authority. Ultimate Resistance, in Lbs., per Square Inch, to Modulus of Elasticity, E. Com- pression, C. Tension, T. Cross- Breaking, B. Wood (continued) . Fir, spruce Fir, spruce, American black . Stoney. Rankine. Stoney. N Rankine. Stoney. Haswell. Rankine. Stoney. Rankine. Haswell. Rankine. Stoney. < a a 6819 H. 5375 6200 5400 5570 5568 H. 9900 9"3 8200 8198 H. 8150 7700 10000 6000 10058 H. 5982 H. 7518 H. 6790 H. I2OOO B. I2OOO I4OOO 12400 9OOO IOOOO 10220 Ro. 8900 Bv. 11800 Bv. 11800 16000 20100 Mu. 8000 21800 8000 B. 16500 Bv. 17400 Bv. 10600 IOOOO 19800 10250 IOOOO B. 19800 Bv. 13950 Ro. 7650 Mu. 8076 M. 6216 D. 7392 B. 7100 954 9900 12300 5000 IOOOO 6852 D. 12774 D.N. 8790 D. 8010 B.D. 5466 D. 12078 N. I2OOO II2OO 20580 B. 7600 II500 10314 M.N. 10164 D. 8700 13600 10600 10164 B.D. 8898 M. IOI22D.N. 10458 B.D 9162 B.D. 10362 B.D. 7374 D.N. 1460000 1900000 1400000 1800000 900000 1360000 1255000 1200000 1750000 2I5OOOO Fir, yellow pine, American . Fir, spruce Fir larch Hemlock Hickory, American .... Hickory, bitter-nut ..... Larch Larch .... Larch, American Lignum-vitae Lignum-vitae Locust Locust . . Locust . ... Maple Maple Maple Oak, European . Oak, European .... Oak, American red .... Oak, English Oak, English Oak, French Oak, Quebec Oak, American red .... Oak, American white . . . Pine, American red .... Pine, American pitch . . . Pine, American white . . 152 MECHANICS OF THE GIRDER. TABLE II. Continued. MATERIAL. Authority. Ultimate Resistance, in Lbs., per Square Inch, to Modulus of Elasticity, E. Com- pression, C. Tension, T. Cross- Breaking, B. Wood (continued). Pine, American yellow . . . Pine, Norway Stoney. Rankine. Stoney. Rankine. M Stoney. Rankine. Stoney. Rankine. Stoney. Haswell. Stoney. , Rankine. Moseley. Stoney. Rankine. Stoney. 5445 H. 7082 H. I2IOI H. 12000 7227 H. 6128 H. 3i73 Wi. i344oWi. 55o IIOOO 4000 4500 3050 F. 18043 Wi. 3216 Re. 20160 Wi. 5500 2185 7884 55oo 2200 17344 14300 Bv. 7287 Bv. 13000 13000 Bv. 15000 B. 15000 8130 Mu. 7800 Bv. 14000 Bv. 670 2800 55i H. 722 Bu. 1054 Bu. 1261 Bu. 9600 12800 7110 B.D. 9600 12648 B.M. I2OOO igOOO 14980 6600 456 Wi. 2442 Wi. 1698 Wi. 2484 Wi. 1252 H. 2697 H. 2664 2010 Re. 5142 Re. 2360 IIOO 5000 7370 1040000 2400000 13000000 16000000 Pine, Norway Sycamore Teak Teak, Indian Teak, Indian Teak African Walnut Walnut Willow Willow Stone. Granite Limestone Limestone . Limestone . Limestone . Limestone . Limestone .... Marble Marble > . . . . Marble Marble, white .... Marble, black Marble, black Sandstone Slate Slate Slate MOMENTS OF RESISTANCE OF INTERNAL FORCES. 153 TABLE II. Concluded. MATERIAL. Authority. Ultimate Resistance, in Lbs., per Square Inch, to Modulus of Elasticity, E. Com- pression, C. Tension, T. Cross- Breaking, B. Bricks, etc. Pale red Stoney. Rankine. Stoney. ft 562 Re. 808 Re. 1717 Re. 2240 Gr. 618 Ro. 5984 Gr. 280 300 51 358 Gr. 71 Ro. 546 Gr. 604 Gr. 632 Gr. 627 Gr. 666 Gr. 709 Gr. - ~ Red Fire Gault clay . Lime mortar, average . . . Portland cement Plaster of Paris ROMAN CEMENT: The value of B, the modulus of rupture in Table II., is that due to a rectangular cross-section, unless otherwise specified. The works from which this Table is made up are the fol- lowing well-known authorities : ist, "A Manual of Civil Engineering," by William John Macquorn Rankine. 2d, " The Theory of Strains in Girders and Similar Struc- tures," by Bindon B. Stoney. 3d, " The Mechanical Principles of Engineering and Archi- tecture," by Henry Moseley. 4th, " Engineers' and Mechanics' Pocket-Book," by Charles H. Haswell. 5th, " Report on the Progress of Work, etc., of the Cincin- nati Southern Railway," by Thomas D. Lovett. 154 MECHANICS OF THE GIRDER. 6th, " Report on Tests of Salisbury Cast-Iron," in " Railroad Gazette" of Nov. 30, 1877, by Robert H. Thurston. Names of the experimenters cited are thus abbreviated : viz., H., Hodgkinson ; F., Fairbairn ; Bv., Bevan ; Bu., Buchanan ; B., Barlow ; D., Denison ; N., Nelson ; M., Moore ; K., Kir- kaldy ; Ro., Rondelet ; Re., Rennie ; C, Clark ; Gr., Grant ; Wi., Wilkinson ; Wd.,Wade; Mu., Musschenbroek; T., Trickett. SECTION 2. Moment of Inertia and Radius of Gyration of a Given Cross-Section. 61. In equation (159) the factor fffdydz^I (183) is called the moment of inertia of the surface of the cross-sec- tion, relatively to the axis of z, the factor being analogous to the real moment of inertia of a material plate whose thickness is unity. The moment of inertia divided by the area of the section gives the square of the radius of gyration, which we will call r 2 . We then have, if 5 is that area, Square of radius of gyration = r 2 = -~ = rr ' . (184) 62. From the moments of resistance already found, equa- tions (160) to (171), and from similar applications of (183), we derive values of /and of r 2 as given below in Table III., where the axes of gyration are assumed to pass through the centre of gravity of the cross-section. MOMENTS OF RESISTANCE OF INTERNAL FORCES. 155 TABLE III. FORM OF CROSS-SECTION. Moment of Inertia of Section, Square of Radius of Gyration, r 2 i. Rectangle (Fig. 90). About least axis b Max. Min. Max. Min. | I Yqbh* \$b h -fa(bh? Mi 3 ) bh* Mi 3 About greater axis h . . . . 2. Square. 3. Hollow Rectangle (Fig. 91). About greater axis h . . . . 4. Hollow Square. \2(bh l>iAi) ^(h h 2 )b 3 \2(bh MX) &(h z + hf). b 2 A 5. I-Section (Fig. 91). About vertical axis h . . . . About horizontal axis b . . . . A =. area of flanges. B = area of web. 6. Plates and Channels (Fig. 92). About axis 6, normal to plates . About axis l>, parallel to plates ) (Fig. 04.) . j 12(A + B) 12(A + B) 7-T6 1 . /^S 1 . 7-=-^. 7. Plates and I-Beams (Fig. 93). About axis b, parallel to plates . About axis /;, normal to plates . i 5 6 MECHANICS OF THE GIRDER. TABLE III. Concluded. FORM OF CROSS-SECTION. Moment of Inertia of Section, 7. Square of Radius of Gyration, r 2 . 8. T-Section, erect (Fig. 95). About axis b, parallel to flange bh z Mi 2 _ hej-ht of I \b(h - e) 3 I S 2(bh normal to flange . 9. Angle Iron; equal ribs 3, thick 1 Min. j I '-r S. Of unequal ribs h and b . . . Min Ph z (b + h}t Ph* 10. Channel Iron; h = depth of flanges -f thickness of web, A = area of flanges, B =. area of web Min. 4*4) ^(^,+^D- n. Star Iron, or cross of equal arms h Min JLji 12. Ellipse (Fig. 98). About minor axis 2b .... About major axis 2h .... 13. Circle ; radius h (Fig. 97) . . . 14. Hollow Ellipse (Fig. 98). About minor axis 2b Max. Min. if About major axis 2h Min 4 15. Hollow Circle (Fig. 97). Radii, h, h* 4 4 #w DEFLECTION, END MOMENTS, ETC., FOUND. 157 CHAPTER VI. DEFLECTION, END MOMENTS, AND POINTS OF CONTRARY FLEXURE FOUND. CAMBER. SECTION i. Deflection of the Semi-Beam having a Uniform Cross-Section. 63. Equation of the Elastic Curve as applied to a Beam or Pillar. Let N, Fig. 89, be the origin, and x and y the current co-ordinates, of the neutral line NTS of any beam or column under a given load ; TU = a unit of the length of the beam ; VT VU = the radius of curvature at any point = P=" ^T^=-^ ('85) when the deflection of the beam or pillar is small compared with its length. (See Differential Calculus.) PP^ = incre- ment of unit on extended side due to flexure ; RR^ = decre- ment of unit on compressed side due to flexure ; a = the angle included between the tangent to the curve at any point and the axis of x, so that tan a =. -^-. dx j Suppose the unit strain required to extend a unit of length by the space PP^ to be T iy and that required to compress a 158 MECHANICS OF THE GIRDER. unit of length by the space RR^ to be C u and that required to extend or compress a unit of length by its own length, TU, to be E = the modulus of transverse elasticity ; and put C x + T t = 2Bi = total unit strain at surfaces. We then have, if h is the depth of the beam, and if the dis- placing forces E and zB^ are proportional to the displacements they cause, P,P ^R,R = TU PU RU " UV 9 f - Multiplying (186) by / = moment of inertia of cross-section, we find which is the moment of resistance of the cross-section, since B l -T- \h = f t and / -=ffy 2 dydz of equation (159). If, therefore, we put EI^ equal to the moment at the section due the external forces acting on a beam or pillar of uniform cross-section, and perform two successive integrations, we shall have an equation in which y is the deflection of the neutral line at the distance x from the origin of co-ordinates. 64. Deflection of the Semi-Beam under One Weight. Let L } Fig. 8, the point where the neutral line of the semi- beam meets the wall, be the origin of co-ordinates, and call x positive to the left, and y positive downward, in accordance with the notation in article 14. Take semi-beam of length /, with concentrated load W at distance a' from fixed end. DEFLECTION, END MOMENTS, ETC., FOUND. 159 From equation (18), M x = - W(a' - x) = -EI^Z by (187), dx 2 Integrating, with the condition that -2- = o when x o, 6?Jv we have EI& = wit ' dx \ Integrating again, with the condition that y = o when x o, /. ^"Ty = i Deflection at any point, x> is M7 (188) And when x = ^' = /, we have Maximum deflection Z> = -, (189) where the proper values of E and I are to be taken from Tables II. and III., according to the material used, and the form of cross-section. 65. Semi-Girder, Length /, Uniform Load w per Unit. From equations (23) and (187), = %w(l - *)*. i6o MECHANICS OF THE GIRDER. When = o,* = o; dx dx When y = o, x o ; .-. Ely = Deflection anywhere, Maximum deflection, (I90) (191) 66. Semi-Girder, Partial Uniform Load, /, on each Unit of Length, b, at Distance a from Fixed End. Fig. 8. When x a, or x < a, equations (29) and (187) give - x) dy -f. = o when ^r = o, dx y = o when ^r = o, DEFLECTION-, END MOMENTS, ETC., FOUND. l6l Deflection x not > a, * (I92) Again, -^ tan a when x = (a + 3), we have, from (26) and (187), = tan a when x = a, ! I ) 1 62 MECHANICS OF THE GIRDER. y =. y^ when x = a, .-. EI(y - y t ) - (x a) iz.na.EI - a}\. ) 4 Whence, after eliminating y t and tan a, we find the deflection + 6(a + bY(o? - a 2 ) - ^(x - a) + 6a*b 2 + 803J], (193) And if ^ = a + ^, we have Maximum deflection D ^,[3(0 + <^) 4 3^ 4 4 3 ^]. (194) 67. If it be required to find the total deflection of a semi- beam at its free extremity when it supports partial uniform or concentrated loads not reaching that extremity, we may proceed as follows : Find the deflection at the free end due the beam's own weight, Iw ; then find the deflection, D, and the inclination, a, of the beam at any point bearing a concentrated load, W, or at the outer end of any partial uniform load, bid ; using tan a, compute the end deflection due W or bw' by the formula + (/ a )tan a for bw r , > 2 = D w + (/ - and we have deflection at L1 \1l / 3 free end due the r th weight, <-> Hence the end deflection due the first r weights is + i) -r^r+i)']. (197) When r = ;/, (197) becomes /> = .ZL( + 'Kv + O t ( , 9 8) 24^7 n which is the deflection at the free end due n equal weights at equal intervals, -. n 166 MECHANICS OF THE GIRDER. 69. If the weight at the free end is but \W, while every other weight is W, as is generally the case with a uniform load of panel weights, we must diminish the deflection last found by the deflection due J W at the free end, which, according to (189), is This quantity taken from the value of D in (198) leaves W> j(,, + I )(y, + i) 1 S "~ "" " 4 ' (I99) which is the deflection at the free end of a semi-girder of uni- form cross-section when the load is distributed in equal panel weights ; there being but the half panel weight at the free end. 70. To find the Deflection at the r th Interval due to all the n Equally Distributed Equal Weights, W. For the first r weights, (198) applies if we put r for n, and /, for /; /j being the distance of the r ih weight from the fixed end. That is, Wl? (r+ i)( 3 r+ i) For the (n r) weights beyond the r th , we have, from (188), since x now becomes / y = the deflection at the r th weight due to any one weight at the distance a' from the fixed end, DEFLECTION, END MOMENTS, ETC., FOUND. Therefore, by summing (201), we find _ {] ~ EI2 the deflection at the r th weight due all the weights beyond. Adding (200) and (202), there results D = -r(6 2 + r* - 4 rn - 2 r + 6n + i), (203) which is the deflection at the r th weight due to all the n given weights, W. 71. If the weight at the free end is the m^ part of W t instead of W y the deflection due at the r th point of divis- m ion is, by putting a' /, and / z = , in (201), = _ _ " 2 i " mEI2n 2 6ni 2EIn m Subtracting this value of y from the deflection in (203), we have, finally, 1 M + r*- 4 rn-2r + 6n + i- I2n " * r \ \ m I - r 2 + i) (204) if m = 2 ; st> that (204) is the deflection at the r th panel point due to a full load of equal panel weights. 1 68 MECHANICS OF THE GIRDER. EXAMPLE. Take the oak semi-beam of the example in article 67, and suppose it loaded with n 5 equal weights, of 100 pounds each, at intervals of 3 feet. El = 12,000,000. Then the deflection at the free end due the 5 weights is, by equation (198), _ IPO X is 3 (5 + i)(3 X 5 + i) 24 x 12000000 5 = 0.0225 foot = 0.27 inch; to which add 0.17086, the deflection due the beam's own weight, for the total deflection at the free end = 0.44086 inch. If the fifth or end weight is but J X 100 = 50 pounds, then, by (199), IPO x 153 | (5 + i)( 3 x 5 + i) ) 24 x 12000000) 5 4 ) = 0.0178125 foot = 0.21375 i nc h; D = and the total deflection =0.21375 + 0.17086 = 0.38461 inch. To find the deflection at the third loaded point due the 5 equal weights in their positions, we use (203), taking r = 3 ; thus, = 0.0104625 foot = 0.12555 i And the deflection at this third interval, due the beam's own weight of 27 pounds per linear foot, is, by equation (190), put- ting x = 9, w 27, 27 y = - (6 X is 2 X o 2 4 X 15 X o 3 + O 4 ) 24 x i 2000000 v = 0.006766 foot = 0.081192 inch. Therefore the total deflection at the third interval is 0.125550 -f- 0.081192 = 0.206742 inch. DEFLECTION, END MOMENTS, ETC., FOUND. 169 SECTION 2. Deflection of a Beam of Uniform Cross-Section, supported at its Free Unfixed Ends. 72. Deflection due the Beam's own Weight, supposed to be Uniform. For the cases in this section we employ the same notation as that given in article 17, Fig. 9, excepting that we take the origin of co-ordinates at O lt a point in the neutral surface, instead of using O as before, in order that y may be the deflection of the neutral line, as it is in the expression for dy the moment of the internal forces, R El-r-; = M, of dx* article 63. We now have x positive to the right, and y posi- tive downwards. From equations (49) and (187), dy Integrating, with the condition that -j- = o when x = Integrating again, with the condition that jj/ o when x = o, we have, after reducing, ( 2 5) which is the deflection of the uniformly loaded beam at any point, w being the load per unit of the beam's length, /. MECHANICS OF THE GIRDER. If in (205) we put x \l, we have (206) the deflection at the centre due a continuous uniform load, Iw. EXAMPLE. Beam of oak, 54 pounds per cubic foot. Length 15 feet = /. Rectangular cross-section, depth i foot = h ; breadth \ foot = b. E 2,000,000 pounds per inch. .'. E = 288000000 pounds per square foot, / = -fabh* = ^, El = 1 2000000; all dimensions in feet. Weight per foot of length = i X J X i X 54 = 27 pounds. Deflection due beam's own weight at a point 5 feet from either end, by (205), is y = - 2J - - (5 4 - 2 X 15 X 5 3 + i5 3 X 5) 24 X 12000000 = 0.001289 foot. From (206), the central deflection is = 5 X 27 X 15* = foot _ 384 x 12000000 73. Deflection due a Concentrated Load, W t placed at the Horizontal Distance a' from the Origin or End of the Beam. When x < a', equations (40) and (187) apply; that is, dx* DEFLECTION, END MOMENTS, ETC., FOUND. Let a be the angle of inclination, or slope, of the beam at point of a when x = a', the point of application of W. Then, integrating, -r = tan a Again, y = o when x = o, But when # > ', use equations (43) and (187), giving dy Integrating between the limits tan a and -j-, and a' and Integrating again between the limits o and y, and / and x, there results - (x - /)tan] By putting ^ = a and j/ = j x in equations (207) and (208), we find MECHANICS OF THE GIRDER. Putting this value of tan a in (207), we get, after reducing, y = W( *~\(2l - W* - *], (209) which is the deflection at any point between the origin and the weight W. If x = a', we have the deflection at the loaded point, And if x = a' = \l, tan = o ; and Deflection at centre = D = -~, (211) which is a maximum, since F is now at the centre. Comparing (211) and (206), where wl = W = entire load on the beam, we see that the deflection at the centre due the load, Iw, uniformly distributed continuously, is five-eighths of the deflection at the centre due the same amount of load con- centrated at that point. By putting / d for a' y and I x for x t in (209), or by substituting the value of tana = "Wj*{ -- a' I + ftf' 2 J in (208), we have /2 ~' z ' z)(/ ~* ) ~ (/ ~* )3] ' (2I2) which is the deflection when x > a' \ that is, at any point between the weight W and the right-hand support. I EXAMPLE. Take a beam of pine weighing 40 pounds per cubic foot, of rectangular cross-section. Depth h = i8j inches, breadth = b = 15 inches, length = 12 J feet = 150 DEFLECTION, END MOMENTS, ETC., FOUND. 173 inches. Call E = 1,680,000 pounds per square inch, I = -i2^ 3 = ' ~1T~ ~ = 7>9 l 4.$3i2$, El 168 X 79,145,312.5 ; f , 18.5 X 15 X 40 beam s own weight per inch of length = w = - 3 - = 6.4236^- pounds. Deflection due beam's own weight, Iw, at a point 48 inches from one end is, by (205), - 6.4236$ - ( 4 84-2X 150x48' +150* X 48) 24 X i68x 79145312.5 = 0.0027 i ncn - Deflection at centre, from beam's weight, by (206), is 5 x 6.4236^ x 150* D - 384 X 168 X 79 i4S3".S = - 32 mCh ' which is a maximum. Deflection at the point of application, due weight W = 17,935 pounds placed a r 48 inches from end of beam, is, by (210), D = - o - X (150 - 48)* = 0.07185 inch. 3 X 168 X 79145312.5 Deflection at the centre when W^ = 17,935 pounds is placed 48 inches from one end, is found by equation (212), making x 17935 X 48 / I S 2 \ v = -, 1 iqo 2 48 2 = 0.078017 i 12 x 168 X 79i453i2.5\ 3 4 / And when W is at the centre, the central deflection is, from (211), D -r J -^~ = 0.0948 inch. 48 X 168 X 79145312.5 Add deflection due beam's own weight for total maximum deflection = 0.098 inch. 1/4 MECHANICS OF THE GIRDER. 74. Deflection due a Partial Load, ix/b t Uniformly Dis- tributed Continuously over the Length, b, beginning at the Horizontal Distance,, #, from the Origin, O lt or Left End of the Beam, Fig. 9. To find this deflection, we use, when x < a, equations (53) and (187), giving d* y / _ ( a + U) Jjg = -w'b- ^-f- -4 = -* (say). Let a be the angle of inclination, or slope, of the beam at the distance a from the left end ; then integrating, with the condi- tion that - = tan a when x = a, dx dx Again, y = o when x = o, .-. EI(y - x tana) = -*J?- - a *x\ (213) 2 \3 / Let y =. y T when x =. a, - a -^). (214) But when x > a and < (a + 6), equations (55) and (187) are to be employed, yielding DEFLECTION, END MOMENTS, ETC,, FOUND. 175 And, if J3 is the angle of inclination at the distance (a + b) from the origin, we may integrate as follows : -^ tan a when x a, = gy* - A _ L_^(. _ .) + ^(* _ a). y y^ when ;r = a, - /|> -* -(*-) tana] = ^^-^ - a *(x - a) | _^^_ a2(x _ a) ^^^ Let j = y 2 when ^ = a + #, Fig. 9 ; so that, after redu- cing, (215) becomes t -y,-b tan ) = - + + __. ( 2l6 ) / \ 2 12 12 2 8 / Or, we may integrate in a different manner ; first, with the con- dition -^ = tan /? when x = ^ + ^, dx = g/^3_ (a + ^3) 2 ( 3 ) 2 ( Also j = j/ 2 when ^r = + #, i;6 MECHANICS OF THE GIRDER. But in (217) y = y^ when x = a ; therefore, after reducing, we have EI(y. -y,+ Jtan/Q = + + - - ( 2 i8) / 2 12 6 2 2 12 2 24 For the remaining part of the beam, Fig. 9, that is, when x > (a + )> equations (57) and (187) give x - a - ty) = (w'b - e)x - w'b(a = tan 6 when ^r = a + <^, dx - (a zr o when ^r = /, _(*_/) tan )8] _ /3 + 6) (a + tf) ( X - /), (219) which becomes, if we put y 2 for y, and a + for ;r, and reduce, ^[>2- (a + l>- /)tan^8] = (l-a - &)*(a + ^). (220) From equations (214), (216), (218), and (220), we may now determine the four quantities, tan a, tan /?, y lt y 2 , so that they can be eliminated from (213), (215), (217), and (219). _-- \ 3 26 24 2636 DEFLECTION, END MOMENTS, ETC., FOUND. 177 = 6 8 3aW lafrl n&l .a*l* .abl* .Pl*\ , N - ---- 1 --- 1 --- 1 ). ^224) 2 6 24 3 2 6 / We have, then, from (213), where x is not greater than a y , (225) which is the deflection due w'b at any point between the origin and the beginning of the partial continuous uniform load it/b, Fig. 9. For the uniformly loaded part, &, of the beam, we find, from (215), which is the deflection due / at any point of the loaded por- tion b, since x is here not less than a, nor greater than a + b, Fig. 9. Equation (219) gives the deflection for the remaining part of the beam, that is, where x is not less than a + b ; and we find + (*-/) tan & (227) which is the deflection due ii/b at any point between the right- hand end of the beam and the load w'b, Fig. 9. 1 78 MECHANICS OF THE GIRDER. If x = a b \l in (225) or (226), we have the central deflection when one-half of the beam is uniformly loaded con- tinuously ; viz., y = 5*"' " 2 X which, if ze/ = w, is one-half the deflection found by (206) for the fully loaded beam. The same result may be obtained from (226) or (227) by putting x = b = /, and a = o ; for then the one-half load is upon the other end of the beam. The greatest deflection due a partial uniform load evidently occurs when the centre of the load and centre of the beam are in the same vertical line ; that is, when a + \b = J/, a = \(l b\ and b = I 2a. Then, putting x = \l in (226), we may find the greatest deflection a partial uniform load can produce. But if it is required to find the maximum deflection of the beam when a given partial uniform continuous load has any given position upon it, we may differentiate (225), (226), or (227), put -^ = o, and so find a value of x that shallr ender^ a cJv maximum. If we then add the deflection at the point so found, due the beam's own weight, we have the total deflection. 75. An important application of (226) and (227) may be made if we take a = o ; for in that case the partial uniform continuous load begins at the left end of the beam, so that, by assigning successively increasing values to b> we may find the deflection at any point due an advancing continuous uniform load w'b. If a = o, jr = o, tana = -^- TJ (P - tfl + 4/ 2 ), and (226) becomes w ' ( DEFLECTION, END MOMENTS, ETC., FOUND. 179 which is the deflection at any point of the loaded part of the beam, where x is not greater than b. Also, if a = o, and (227) becomes which is the deflection at any point of the unloaded part of the beam, where x is not less than b. EXAMPLES. Partial uniform continuous load, it/b. Wrought- iron 15-inch I-beam. Length 30 feet = 360 inches = /. Mo- ment of inertia / = -^(bjij bjif), by Table III. 5. Let h 2 = 15- inches, b 2 = 5| inches, h^ = I2| inches, b l = 4| inches, Fig. 91 ; putting /i 2 for k y and b 2 for b, to avoid confusion here. Beam supported at ends. Load w' = 75 pounds per inch of the length b> .'. I = T V(5-375 X 153 - 4.75 x 12.753) = 691. Take E = 24,000,000, Table II., /. El = 16584000000, all dimensions to be in inches. Let the load cover the first 10 feet of the beam. ist, What is the deflection at the end of the load ? We have x = b = -J/ = 120 inches; and (228) applies, giving 75 X 36o 4 /i 1511 i i i6 5 8 4 oo 0.23444 inch. y _ I T * vv \^f _ \s 24 X 16584000000^81 180 MECHANICS OF THE GIRDER. 2d, What is the deflection at the centre of the beam ? We have from (229), if x J/, and b = J/, 75 X X 36o 4 (73 24 x 16584000000! 8 4 = 0.24421 inch. 3d, What is the deflection 10 feet from the unloaded end of the beam ? Here we use (229) also, putting x |7, and b = J/, . 75 X I X 360* j 19 5 24 X 1 65 84000000 ( 27 n 9 = 0.19176 inch. 4th, Suppose it is now required to find the point of greatest 'deflection due this same load of 75 pounds to the inch on 10 feet of one end of the beam. Differentiating (229), we find, since b = 4/, omitting constants. Putting this value of -^ equal to zero, we MM? at once have x = 0.43892^ which is the point of greatest deflection ; and, by placing this value of x in (229), there results y = 0.24847 inch, which is the greatest deflection of the beam due this load along one end. 5th, But if this same load be moved along to the centre, so that we have a = \l = b, we find the greatest deflection the DEFLECTION, END MOMENTS, ETC., FOUND. l8l load can produce, by putting x = \l in equation (226), where y t becomes = IIW * and tan a = / n * from (223) and 1944^7 648^7 (221). Thus (226) becomes 75 X 36Q 4 ) i / i _I\_, _!/ _ 2 x 16584000000(1216 _I\_, _!/ _ \ 8i/ 8i\2 37 Deflection at centre = y = 0.50063 inch, \w'l central ; deflection at centre = y = 0.24421 inch, Jo// at either end ; /. y = 0.50063 4- 2 x 0.24421 = 0.98905 inch, equals deflection at centre when the given 15 -inch I-beam of 30 feet between supports is loaded with 13.5 tons, uniformly dis- tributed continuously. And this result accords exactly with that given by (206) ; thus, 5 x 75 x 360* = 384 X 16584000000 = '9 8 95 mch; where, as in other values of the deflection, we have retained several unnecessary decimals, in order to test the accuracy of the solutions. 6th, If this beam is half loaded with 75 pounds to the inch, we have in (228), for the deflection at the centre, x = b = \l = 1 80 inches ; and _ 75><36o 4 I i 131/1 i i\i "24X16584000000(16 24 8 \i6 8 4/2 = 0.494525 inch, which is half the deflection due the fully loaded beam, as just found. 1 82 MECHANICS OF THE GIRDER. 7th, The maximum deflection due this half-load on one end of the beam is found, both in position and magnitude, by differ- entiating (228), putting-^ = o, and solving the resulting cubic clx equation, putting b = J/, / = 360. Thus, omitting constant factors, = 4 *3 _ x 3 C ( r 2 + f r 2 )> and we have, by summing, ' = c(r 2 + i + r 2 + 2 + r 2 + 3 + . . . + r) Hence (212) becomes i) 2 - r 2 2 (r 2 + i (r - r 2 )(r + r 2 + i)(/- ^) 3 , (230) which is the deflection due r r 2 equal weights, W, at any point, x f between the r th weight and the right end of the beam ; r 2 being the number of full intervals vacant at the left end, and x being not less than cr. For the r l r equal weights between the point x and the right end of the beam, we employ (209), which reduces to 1 84 MECHANICS OF THE GIRDER. If in this equation a' takes the successive values c(r + i), c(r + 2), c(r + 3), c(r + 4), . . . c(r + r, - r), then, by sum- ming, we find ^a r = <-[(r + i) + (r + 2) + (r + 3) + + r,] = \c(r, - r)(r t + r + i), + i) 2 + (r + 2) 2 + (r + 3) 2 + n 2 ] (r, + i)(2r, + i) - r(r + i)( 2 r + i)], + i)s + (r + 2)3 + (r + 3 )3 + . . . r x a] = i* 3 [>, a (r s + i) 2 - 7*(r + i) 2 ], Sa /0 = r, - r. Hence, for this case, (209) becomes W y 2.^22,11 { + IXr, - r)(r + r + i) - 4(r, - r)/]^J, (231) which is the deflection due the r, r equal weights on the beam at any point, x, between the left end and the (r + i) th weight ; x not being greater than c(r + i). Adding the deflections given by (230) and (231), and calling their sum also y, we have W ( r , y = rrSwU^fr - r )< r + r ' + - 4(^i - 6c(r r 2 )(r + r 2 + i -- r a )(r, + r 2 + i) + ^[^(r, + i) 2 - r 2 2 (r 2 + i) 2 ] i)(2r r + i) r(r + i)(2r + i)]^ + i) 2 - ^ 2 (^ 2 + i) 2 ]l, (232) DEFLECTION, END MOMENTS, ETC., FOUND. I8 5 which is the deflection at any point, x, due the r, r 2 equal weights, W\ where r, denotes the number of intervals between the last weight and the left end of the beam, r a number of full intervals not less than r 2 , the number of unloaded intervals at the left end of the beam, nor greater than TV If in (232) we put c = / -f- , x = -J/, r 2 = o, r, = ;z I, and r = # when** is even, but r = J(w i) when n is odd, we shall find D (5 2 - 4), n even, Wt* 4 2 i), # odd, (233) which is the deflection at the centre due the r, = i equal weights, H 7 , covering the beam of ^ equal intervals (/ -f- ). EXAMPLES. Let us take the same 1 5-inch I-beam we employed in the examples of article 75, for which / = 691, E = 24,000,000, / = 360 inches. Take 3 weights of 4,500 pounds each, placed at intervals of 60 inches, beginning at one end of the beam ; then the deflection at the centre is given by (230) if we put W 4,500, / = 360, c \l 60, r 2 = o, r 3, x = |/, El 16,584,000,000. Thus, = 4 X 36o3__/ 24 X 16584000000^ 6 r* \/ XX " N/ >l \/ 2A-X3X4X 2X- 26 X3X4X - j = 0.6154 inch. 8 / If 2 more equal weights are added at the same interval, so as to cover the beam, the central deflection due these last 2 is, by (231), where r, = 5, r = 3, 1 86 MECHANICS OF THE GIRDER. 4=500 X 36o 3 (if \i - 24 X i6 5 8 4 ooooool 4 X 6( 25 - 9 + 5 - 3 )i - 2 X -If 5 X 6 X ii - 3 X 4 X 7\- + r I 2 5 x 36 - 9 x i6V 36\ /2 63\ / 2 + 2 X 1(25 - 9 + 5 - 3 V -4 X 2 x * j = 0.3517 inch. If we compute the central deflection due these 5 equal weights by (233), we have n 6, and _ 45 X 3603 /5 x 36 - 4\ _ = 384 x 16584000000^ 6 ~ / '9 6 7i cn > which is the sum of the deflections found by (230) and (231). Again, if there are 8 weights upon the beam, each equal to W .= 3,000 pounds, at intervals of 40 inches, we have n = 9, / = 360 inches ; and (233) gives the central deflection, 3000 x 3602 / 5 x 94 - 4 x 9 2 - A D = 384 X 16584000000^- -JT- -) = '97926 mch. In these examples the weight has been purposely chosen equal to 75 pounds to the inch for the entire length of the beam, except a half-interval, (/ -f- 2w), at each end ; so that we may compare the results with the central deflection of the same beam, computed by (206) for the continuous uniform load of 75 pounds to the inch, which deflection we have found to be 0.98905 inch. Now it will be found that the central deflection due the dis- continuous load, (n i) W, at equal intervals, (/ -=- n), will be less than that due the continuous uniform load, /w, until n becomes infinite, and W= infinitesimal, when (233) becomes n identical with (206). DEFLECTION, END MOMENTS, ETC., POUND. l8/ The greatest difference between the central deflections of these two loads, (;/ i) W and Iw, manifestly occurs when n = 2 ; that is, when there is but one weight, and that at the centre, and equal to W = . Equation (233) then becomes D = -4 , which is four-fifths of the deflection due Iw con- 384^7 tinuously distributed uniformly, as shown by (206). From these considerations it appears, that, in practice, the Nformulse found in article 75, for a uniform continuous load, are applicable to a uniform load distributed, as above, discontinu- ously, or by panel weights, each equal to (Iw -f- ) provided n is large. But in any case, whether there be many or few intervals, we may find, by means of equation (232), the greatest deflection due any partial or complete load of equal panel weights, W t and the point where it occurs. For this purpose, differentiate (232) with respect to x, and put -f = o. This gives ax [6f(r t - r a )(r x + r 2 + i) - 12 (r, - r)/> 2 i2/ 2 . (246) Hence, from (245), y = (2 ^ ~ 3/ ^ 3 + /3 ^ )j (247) which is the deflection at any point of a beam horizontally fixed at one end, and simply supported at the other, under a uniform load w per unit of length, /; x to be measured from the unfixed end. Since, now, M t = o, the total moment due Iw at any point is, from (49) and (93), M x = \w(l - *)* DEFLECTION, END MOMENTS, ETC., FOUND. 1 93 If M x = O, x = & (248) which is the distance of the point of contrary flexure from the free end of the beam, under the load Iw uniformly distributed continuously. EXAMPLES. Suppose the wrought-iron 15 -inch I-beam 3cr feet in length, of the examples in article 75, to be fixed hori- zontally at both ends, and loaded uniformly with 75 pounds to each inch of its length ; what is the deflection 10 feet from either end ? We now have / = 691, E = 24,000,000, / = 360,, x \l or |/. Hence (242) gives the deflection y = 75 X 360* X 4 = 6 inch . 24 X 16584000000 X 8 I At the centre, where x = |-/, (242) gives = 75 X 360* x i = gl inch ^ 24 X 16584000000 X 16 which is one-fifth of that given by (206) for beam with free ends. If only one end of the beam is fixed, (247) gives, When* = \l, y= 7$ X 36o 4 X 10 =O . 39O74 inch. 24 X 16584000000 X 8 i * = i/, 75X3604X1 =0 . 39563 inch . 24 X 16584000000 X 8 * = f/, y= 75 X 360* X 7 = o.2 735 2inch. 24 X 16584000000 X 8 i x= 151.7846 inches, ^=0.41 14 1 inch, a maximum. IQ4 MECHANICS OF THE GIRDER. 80. Deflection of a Beam fixed Horizontally at Both Ends, due to a Concentrated Load, W, placed at the Hori- zontal Distance a from the Left End of the Beam. From equations (40), (187), and (93), we have the total moment due W when x is not greater than a' y M x = -*/ = W-* - ^L + M l} (a49 ) Integrating, as in article 73, -^ = tan a when x a', -^) -*('-'). (.50) Again, ^ = o when ^r = o, /. ^"/(^ x tan a) #V _ /) + M, - MJx* , 2 \ ^/jc 2 , \ , , = i - 'L - \ - !(__ - + llf. - MAxt _ , _ I 3 ^^(*--/) (254) Now 7 in (251) is equal to j/ in (254) when x = #'; there fore, from (251) and (254), we find [WV(f^ + |/ 2 - aft) Ell + Jlf,(il> + & - a'l) - Jf,W - |/0]. But in (250) we now have = o when x = o, dx ... tana = | Wa'*?-^- 1 + M/ - a'l) - M 2 \. (256) -Y/ ( 2 \ 2 / 2 ) Also, in (253) -j- o when # = /, /. tana = -\_W(W* + J7> - *V) j^y/ + tK^f x - J/ 2 ) (^ - / 2 ) + MJ(l - a')]. (257) From (255), (256), and (257), we find M* = -^(l-a'Ya', (258) M 2 = -^(l-a')a'*, (259) which are the end moments developed by the weight W in any position, a'. 196 MECHANICS OF THE GIRDER. If the weight W is at the centre, a' J/, and M, = M 2 = -\WL (260) Eliminating J/ n M 2 , and tana from equation (251), we find, x not being greater than a', (261) which is the deflection at any point between the weight W and the left end of the fixed beam with ends horizontal. Again, eliminating M I} M 2) and tan 192^7 which is the central deflection when the weight W is at the centre of the fixed beam, and is one-fourth of that due the same load on the same beam with its ends not fixed, as seen by equation (211). To find one point of contrary flexure, we put M x = o in equation (249), and, after eliminating M, and M 2> have ( 264 ) DEFLECTION, END MOMENTS, ETC., FOUND. 197 If a' = V, * = I/. (265) For the other point of contrary flexure, put M x = o in (252), and the result is /( 2 / - a') * = y^y (266) If a' = \l, x = f/. (267) From (265) and (267) it appears that when the concentrated load, W y is at the centre of the fixed beam, the points of con- trary flexure are each midway between the centre and end of the beam. 81. If the beam is fixed horizontally at the right-hand end, but only supported at the left end, we have M t = o ; while M 2 may be found from (255) and (257), since the condition that = o when* = o, on which (256) depends, does not now exist. . M = = 2 (a' 2 - I 2 )a f . (268) This value of M 2 placed in either (255) or (257), while M l = o, gives tana = -(^3/a _ ^'2/3 4. iS (275) DEFLECTION, END MOMENTS, ETC., FOUND. 199 the point of maximum deflection between W and fixed end, where x cannot be less than a' ; that is, a' in this formula cannot be greater than x. Putting at- for x in (275), we may find easily, by trial, that ci 0.414213/13 the greatest value a' can have in this case of a maximum value of y between the weight W and the horizon- tally fixed end of the beam. The point of contrary flexure may be found from (252) by putting M x = o, M! = o, and M 2 as in (268). This substitu- tion gives If the weight Wis at the centre, a' = ^/, and x = T 8 T /, (277) which is the distance of the point of contrary flexure from the free end of the beam. If a? =z o, x = |/; and if #' = /,.#=:/: which are the limits to the range of the point of contrary flexure, for a con- centrated load W, on a beam fixed horizontally at one end, and free at the other ; x being measured from the free end. EXAMPLES. Take the 1 5-inch I-beam of article 75, and suppose it bears a concentrated load W = 27,000 pounds, and that both ends are fixed horizontally. We have, as before, / = 691, E = 24,000,000, / = 360 inches. When W is at the centre, what is the deflection halfway between the centre and either end of the beam ? Put a' = l/ and x = \l in (261), or a' = \l and x = \l in (262), and find = 27000 X r/ a __ _ 2N 1 6 x i 65 84000000 LV + (t - i + f)A] = o-i978i inch. 2OO MECHANICS OF THE GIRDER. At the centre the deflection is, by (263), D = *7ooX36o3 = 6 192 X 16584000000 which is one-fourth of 1.58248 = the deflection due the same load on the same beam with free ends. And this 1.58248 is, again, eight-fifths of 0.98905, the deflection found by (206) for the same load continuously distributed uniformly over the same beam with free ends. The points of contrary flexure are given, by (265) and (267), at 90 inches and 270 inches from either end. Now, since the deflection at the quarter-points is just one-half that at the cen- tre, it follows that, in this case, the end of the neutral line, the point of contrary flexure, and the centre are in the same straight line. When H^is at the distance ct |/from the left end of the beam, what is the maximum deflection ? Differentiating (262), and putting ? = o, we find x = o.4/, /. y = 0.2136 inch. Or, if a f = |/, we find in the same way, from (261), x = o.6/, /. y = 0.2136 inch. If a' = |/, the points of contrary flexure are, by (264) and (266), x = T y, * = f /. But if a' = /, Let us now suppose that this beam is fixed horizontally at the right-hand end, but is simply supported at the left end. DEFLECTION, END MOMENTS, ETC., FOUND. 2OI When W = 27,000 pounds is at the centre, what is the deflection at the quarter-points ? Putting a' = J/, and x = J/, we find, from (270), y = 0.5316 inch. But if a' = J/, and ;tr = f /, (271) gives y = 0.3091 inch. H 7 remaining at the centre, the central deflection is, from (272), D 0.69234 inch. Also, from (274) and (270), the maximum deflection due W at the centre is y = 0.70769 inch. If we place the weight W = 27,000 pounds at the distance a' = 0.4142 1 3/ from the free end for the maximum value of the deflection y, we shall find, by (275), x a' = 0.414213/5 and from (270) or (271),^ = 0.74534 inch, which is the greatest deflection W can produce on this beam, since it is at the point of maximum deflection. Putting a' O.4i42i3/in (276), we find x = 0.7071064 the point of contrary flexure when W is at the lowest point of the beam fixed horizontally at one end ; x to be measured from the free end. 82. Any Number, r l r 2 , Equal Weights, W, placed at Equal Intervals, c, along the Beam ; the First Weight being (r 2 + i) Intervals from the Left End, and the Beam being fixed Horizontally at Both Ends. Let r r 2 denote the number of equal weights, and r equal the number of full in- tervals, between the point x and the origin or left end of the beam, Fig. \2 ; then r t r = the number of weights between the point x and the right end, if any. 2O2 MECHANICS OF THE GIRDER. The deflection at the point x due any one of the r r 2 equal weights, W y is given by equation (262). Let a' in that equation take the successive values c(r 2 -\- i), c(r 2 + 2), c( r 2 + 3), . . . c(r 2 + r r 2 ) ; then, by summing, we have r 2 [(r 2 + i) 2 + (r a + 2 ) 2 + (r a + s) 2 + . . . + r 2 ] [r(r + i)(2r + i) - r a (r a + i)( 2 r a + i)], + i) 3 + (r a + 2)3 + (r a + 3)2 +...+: - !r- 2 ( -i. v _ 2 C- 4- i VI 4 which values, put in the place of a? 2 and a'* in (262), give (278) which is the deflection due r r 2 equal weights at any point, Xj between the r th interval and the right end of the beam having both ends horizontally fixed ; x being not less than cr. If in (278) we make x cr, and r 2 = o, then + s)/ - 2c*r*(r + i)], (279) which is the deflection, at the r th weight, due r equal weights, W, along the left end of the beam at equal intervals, c. Again, the deflection at the point x due any one of the r, r equal weights beyond the point x y is given by equation (261). DEFLECTION, END MOMENTS, ETC., FOUND. 2O3 Let a' in that equation take the successive values c(r -\- i), c(r + 2), c(r -f- 3), . . . cr l ; then summing as in article 76, and putting the values of S#', %a', ^ /2 , 2 2 4- (r a 4- 3) 2 + - - - + ? O'i + i)(2r z + i) - r 2 (r 2 + i)( 2 r a 4- i)], DEFLECTION, END MOMENTS, ETC., FOUND. 20$ +. !)3 4. ( r2 + 2 )s + ( TZ + 3)3 = M^O, + i) 2 - r 2 2 (r 2 + i) 2 ], OO-, + r 2 + i)/ 2 i)(2r x + i) - r 2 (r 2 + i)( 2 r 2 + i)]/ + i^O, 2 ^ + i) 2 - r 2 2 (r 2 -f i) 2 ]J, (283) - Mr, 2 (r z + i) 2 - 9 2 2 (r 2 + i)^]J, (284) which are the end moments due r z r 2 equal weights, W\ both ends of beams fixed horizontally. The greatest deflection due r, r 2 equal weights, W y placed at equal consecutive intervals anywhere along the beam, may be found by the following method : If in equation (281) we provisionally make x = cr, we shall have y r . Then, putting r + i for r in this value of y rt we find y r + ! ; and therefore Ay = > + I - y r . Now, by making by = o, we obtain a value of r the inte- gral part of which, not less than r 2 nor greater than r x , will be the value of r in (281) when 7 is a maximum. Then, differen- tiating (281), and putting -^ = o, we find a value of x which renders y a maximum. Although this solution is rigorous, it need not often be employed, since (281) gives the deflection at as many points as we please, and a close approximation to the greatest value of y may be found by a few trials. An example will be given. 2O6 MECHANICS OF THE GIRDER. For finding the points of contrary flexure, we have, from (6 1) and (93), M x = [(r, - r)/ - \c(r, - r) (r, + r + i)> which is the moment due the r t r equal weights at any point, x, between the (r + i) th weight and the left end of the beam ; x being not greater than c(r + i). Also, from (60) and (93), M x = c(r - r a ) (r + r 2 + i) (/ - *) - ^ ^ M *x + M t , (286) which is the moment due the r r 2 equal weights at any point, x, between the r th weight and the right end of the beam ; x not being less than cr. If we now add equations (285) and (286), representing the three resulting moments still by the symbols M XJ M iy M 2 , we shall have - r - r 2 ) + M 19 (287) which is the moment due all the r, r 2 equal weights at any point, x, between the r th and the (r + i) th weights ; x being not less than cr, nor greater than c(r + i), between r 2 and r x . Now, at the points of contrary flexure, we have, in (287), M x = o ; and therefore oc -- ' , - r)/ - ^(r x - r a ) (r, + r 2 + i)] But in this expression for x, whose value lies somewhere between cr and c(r + i), we cannot tell what to call r. Let us, DEFLECTION, END MOMENTS, ETC., FOUND. 2O? therefore, in (288), put cr in the place of x, and determine the values of r, which must be integers ; we can then find x, since the other quantities in (288) are given. By putting cr for x in (288) we obtain r = - y/^:' - r,' - r, + e*, (289) where = l -jj^ - r t % + -(r, r 2 ) (r, + r 2 + i). If (289) gives values of r not integral, the decimals must be rejected, and the integers retained. Equation (289) will give r an integer only when there happens to be a point of contrary flexure at the r ih interval ; that is, where x really equals cr. Having r 2 equal intervals, c, without weights at the left end of the beam fixed horizontally at both ends, succeeded by r l r 2 equal weights, we have found the corresponding end moments in equations (283) and (284). By making r 2 = o in those equations, there results i)/ - 6/ 2 - tfr^rt + i)], (290) i) - 2(2r, + i)/], (291) which are the end moments due r x equal weights, W, placed at equal intervals, c, along the beam fixed horizontally at both ends ; the first weight being at the distance c from the left end. 83. Beam fixed Horizontally at the Right End, and simply supported at the Left, uniformly loaded for a Part or All of its Length with Equal Weights, W, at Equal Intervals, c. If the first weight is r 2 + i intervals from the free end, and if there are r r 2 equal weights, then the deflection at 208 MECHANICS OF THE GIRDER. any point, x, between the r th interval and the horizontally fixed end of the beam is given by (271), provided we put therein For a', \c(r - r a )(r + r a + i). For a'\ (r + i) 2 - r 2 2 (r 2 + i) 2 ]. But if the first weight is at the distance c(r + i) from the free end, and if there are r l r equal weights at equal inter- vals, c, beyond, then the deflection at any point, x, between the (r + i) th weight and the free end of the beam is given by equation (270) if there we substitute For a', r, r. For a', \c(r* - r )(r s + r+ i). For a' 2 , ^[r x (r x + iX2r x + i) - r(r + i)(2r + i)]. For of 3 , (r + i) 2 ]. If, then, we add the two deflections thus derived from (271) and (270), we shall have + i) 2 - T 2 2 (r 2 + i) 2 ] - 2(r x - - 3t(r - r a )(r + r 2 i) 2 - (r a + i) 2 r 2 2 ]/ 2 + f^(r x - r a ) (r, + r 2 + i)/ 4 r x + i)(2r x + i) -(r + i)(2r + i)r]/^}* + !) - (r a + i) 2 r 2 2 ]/3J, (292) DEFLECTION, END MOMENTS, ETC., FOUND. 2OQ which is the deflection due all the r l r 2 equal weights at any point, x, between the r th and the (r -f- i) th points of division ; x being not less than cr nor greater than c(r -+- i), but r from r 2 to TV In this case, where M t o, M 2 is derived from (268), which reduces to M 2 ^ - a' I 2 ). For a' put \c(r, - r 2 )(r T + r 2 + i). For of* put J* 3 [>i a (ri + i) 2 - r 2 2 (r 2 + i) 2 ]. We then have - Mr x - r 2 )(r I + r 2 + i)/^ (293) which is the end moment due r^ r 2 equal weights, W, uni- formly distributed at equal intervals, c, on any part of the beam fixed horizontally at one end and simply supported at the other ; r l and r 2 to be counted from the free end. 84. Deflection, End Moments, and Points of Contrary Flexure, due a Partial Uniform Load continuously distrib- uted, when Both Ends of the Beam are fixed Horizontally. - We might proceed in this case as in article 74, using equa- tions (53), (187), and (93); but, as the process is tedious, we employ the following method instead, utilizing results already obtained. Let n denote, as heretofore, the whole number of intervals, each equal to (/ -f- n}. Let r 2 denote a certain part of n, which we will call jn ; let r x = a "j" n, where neither a nor a + b can exceed /. 2IO MECHANICS OF THE GIRDER. Now, for a uniform continuous load we must have in the values of M I and M 2 , equations (290) and (291), n, r 2 , and r 1 infinite, and W infinitesimal ; so that we must put n W w'l if w' = the weight per unit of the length. Making these substitutions in (290) and (291), they become - <*"% ( 2 94) (295) which are the end moments due the uniform continuous load it/ per unit on the length b, measured to the right from a point at the distance a from the left end of the beam fixed horizon- tally at both extremities. Now, if a o (that is, if the continuous uniform load begins at the left end, and extends over the length b), equations (294) .and (295) reduce to (296) - 4*0, ( 2 97) which are the end moments due the continuous uniform load w' per unit, on the length b> measured from the left end. It may be noted here, that if in (296) and (297), while a = o, we suppose b = /, these values of M l and M 2 become each equal to ^w'l 2 , which accords with equation (241) for the fully loaded beam. DEFLECTION, END MOMENTS, ETC., FOUND. 211 Let us now put the values of M t and M 2 , as given by (294) and (295), into equation (236) ; we shall then have w f > (298) which is that part of the deflection due to the influence of the end moments, the beam horizontally fixed at both ends being loaded with w' per unit for any part, b t of the beam's length, /; x varying from o to /. If x be now restricted so as not to exceed a, and we add y in (298) toy in (225), the sum will be the deflection due w'b at any point between the origin and the beginning of the partial continuous uniform load w'b. If again we limit x between the values a and a + b, and add the values of y in equations (298) and (226), the sum will be the deflection due in w'b at any point of the loaded portion b. Finally, by making x not less than a + b in (298), and add- ing that equation to (227), the sum of the second members will be the deflection due w'b at any point between the right-hand end of the beam and the load w'b. It is evident, that, by assigning the proper values to a and b, we may place the load anywhere upon the beam, and give it any magnitude not exceeding w'l. Also, we may put many partial uniform continuous loads, wj> u wj)^ w^by etc., upon the beam, by so choosing the values of a,, a 2 , a y etc., b u b 2 , b z , etc., that the partial loads shall take desired positions, whether they are required to be equal to each other, or to overlap, or to have intervals between them. But it is not necessary to formulate the deflection for such totals here. 212 MECHANICS OF THE GIRDER. It remains to find the points of contrary flexure for partial continuous uniform loads, ix/b, when the beam is fixed horizon- tally at both ends. If there is a point of contrary flexure between the left end of the beam and the beginning of the partial load (that is, within the length a), we use equations (53) and (93), giving n/r ,jl a *b Mr M 2 M x = wb *-# -x If M x = O, ' , - M 2 - w'b(l - a - where the values of M l and M 2 are to be taken from (294) and (295), and x cannot be greater than a. Should (299) yield a value of x either negative or greater than a, there is no point of contrary flexure in the part a. For the loaded part of the beam b, we have equations (55) and (93), giving ** IT a M x = wb \, * 2 ,,- x a} 2 -- - - -x + M l = o, .-. x = e L - a* + ?, (300) in _ a _ U} M, M 2 . where = - *- 7- - + a. / wl M! and M 2 are given by (294) and (295). When, in (300), either value of x is less than a or greater than a + b, it must be rejected ; and when both values of x are in this condition, there is no point of contrary flexure in the loaded part b. DEFLECTION, END MOMENTS, ETC., FOUND. 213 In finding the point of contrary flexure between the right end of the beam and the load w'b, we employ equations (57) and (93) ; taking, as before, the values of M x and M 2 from (294) and (295). Thus, M x = v/b(a + \b) 1 - - Ml ~ Mz x + M t = o, M t - M 2 + z^O 4- ty) Equations (299) and (301) show that there can be but one point of contrary flexure between either end of the beam and the adjacent end of the load, while (300) indicates that there may be two such points within the length b covered by the uniform load w'b. 85. Partial or Full Continuous Uniform Load, /, on any Portion of a Beam fixed Horizontally at the Right End, but simply Supported at the Left. Proceeding as in article 84, we make c = -, r = ^n, r l = a *" n, and substi- n / / tute these values in (293), which, when n is infinite, and W infinitesimal and = , becomes n M * = &3 (a + ^ )4 ~ a " ~ 2l ^ (a + b ^ ~ ^' (3 2) which is the moment at the fixed end due the uniform continu- ous load, w f b, anywhere on the beam. Here, if a = o, and b /, the beam is fully covered by the load, and M 2 = \w'l 2 , in agreement with equation (246). If in (302) a =. o, we have as the moment at t^e fixed end, when the partial load w'b, begins at the free end, - (3 3) 214 MECHANICS OF THE GIRDER. Substituting the value of M 2 as given by (302), in equation (237), we obtain (304) which is the deflection due the end moment M 2 when M^ = o, and the load is w'b in any position ; x varying from o to /. If, as in article 84, x be now limited so as not to exceed a, and we add y in (304) to y in (225), the algebraic sum will be the deflection due w'b at any point, x> between the free end of the beam and the beginning of the load w'b. If, again, x be limited between the values a and a -f- b y and we add algebraically the values of y in equations (304) and (226), the result will be the deflection due w'b at any point, x, of the loaded portion b. Also, by making x not less than a + b in (304), and adding that equation to (227), the sum of the second members will be the deflection due w'b at any point, x, between the right or fixed end of the beam and the load w'b ; x measured, as usual, from the free end of the beam. The point of contrary flexure for the beam fixed horizontally at one end and simply supported at the other, which is taken as the origin, is found for a partial continuous uniform load, w'by by means of equations (299), (300), and (301) for their respective cases, by putting M, = o, and taking M 2 from (302). 86. Examples illustrating Articles 82, 83. For the sake of comparing the deflection of the same beam when one or both its ends are fixed, with its deflection when both ends are simply supported, we further consider the 1 5-inch rolled wrought-iron I-beam of 30 feet clear span, whose moment of inertia I = 691, and whose modulus of elasticity E = 24,000,000, as given in article 75. DEFLECTION, END MOMENTS, ETC., FOUND. 21 5 ist, Take 3 weights, of 4,500 pounds each, placed at inter- vals of 60 inches, beginning at the left end of the beam fixed horizontally at both ends ; then the deflection at the centre is given by (278) if we put W = 4,500 pounds, / = 360 inches, c = i/ = 60, r = 3, and x \l\ El being 16,584,000,000. Thus, 4500x36* |/I X _L X84 _I X _^ XI44 V 6 X 16584000000(^2 36 2 216 /8 + (3 x 144 _ 8_4\i + i x 84 x i _ i x 144) = 8 inch> \4 216 36/4 2 36 2 4 216) 2d, If 2 other equal weights, 4,500 pounds each, be added at the same interval of 60 inches, so as to cover the beam with concentrated loads, the central deflection due these last 2 is, by (280), where r = 3, and r, 5, or by (278), making r = 2, inch. 3d, For the 5 equal weights now on this beam, (281) gives the deflection = 4500 x 360* l/i x i x _ 2 _ 2oo\i 6 X 16584000000(^2 36 432/8 = 0.19781 inch; or, (282) gives the same much more simply. This value is, as it should be, the sum of the two deflections last found. 2l6 MECHANICS OF THE GIRDER. 4th, Suppose the fifth weight removed from the beam, what is the deflection at the fourth weight ? Use (279), making r = 4, c \l = 60 inches, W = 4,500 pounds ; ._ = 45 X 36* X 16 x 5 / _ i x 6 24 X 16584000000 X 2i6\ 6 16 X 33 2 X 64 X 5\ -36" ~2l6 1 = ' 13748 mch ' 5th, These 4 equal weights remaining on the beam, what is the deflection at the third weight, or centre ? In equation (281), put x \l = cr, r = 3, r x = 4, c = ^/; 4500 X 36o 3 j /i8o _ _ 400X1 6 x 16584000000(^72 432/8 A x 400 180 24^1 84 i V"; r^ rr^ ~rz" *T~ 24V 12/4 X 216 36 12/4 7 2 6th, The same 4 weights remaining, what is the deflection at the second weight ? Use (281), calling r, = 4, r = 2, x = r^: = -|/, ^ = ^/; 4500 x 3603 |/i8o 4oo\ i 6 x 16584000000 ( ^^72" ' ~ 4/32/ 2~7 400 i 80 /th, What are the end moments due these 4 weights in the same position as above ? Use (283) and (284), making r = 4, c = \l 60, W = 4,500; = 750000 inch-pounds. 45 X 3 ^Y3 x 20 2 x 9) = 600000 inch-pounds. 12 36^6 / DEFLECTION, END MOMENTS, ETC., FOUND. 21 7 8th, When all the 5 weights are on the beam uniformly dis- tributed as above, r = 5, c \l = 60, W = 4,500. Then, by (283) and (284), 4500 x 360 30/44 9 o\ M, = -- x -g-^-g- -- 6 - -^J = -78 7500 inch-pounds. 4500 X 360 3/3 \ M 2 = - I2 - X -gU X 30 2 x II ) = 78 75 oo inch-pounds. 9th, The 4 equal weights of 4,500 pounds still occupying the first 4 intervals on this beam, where are the points of contrary flexure ? Here we have M^ = 750,000, M 2 = 600,000, W = 4,500, c \l = 60, r, = 4, r 2 = d. These values put in (289) give r = 4-6595 or 1-1923. We have then, rejecting the decimals, r = i or 4. Hence (288) becomes -750000 + j. x 45 x *|fl x 2 = -* - 45oo( 3 - i X * X 4 X 5) = 74.806 for r= i, 750000 4- | x 4500 x s-jp x 20 = - 4500(0 - i X i X 4 X 5 ) ' " ioth, If these 4 weights occupy the last 4 intervals, leaving the first vacant, we shall have M 1 = 600,000, J/ 2 = 750,000, r, = 5, r 2 = i, c = / = 60, PF = 4,500 ; so that, from (289), we find r = 4.80372 or 1.34442, that is, 4 or i. These values placed in (288) give x = 84.706 for r = i, x = 285.194 for r = 4, which accords with example 9th, since 360 84.706 = 275.294, and 360 285.194 = 74.806. 2l8 MECHANICS OF THE GIRDER. nth, When all 5 weights are on the beam at equal inter- vals, M! = M 2 787,500 by example 8th. Also, c \l = 60, r t 5, r 2 o, W = 4,500. From (289), we find that r must be i or 4 ; and therefore (288) gives, as the points of con- trary flexure, x = y6f for r = i , x = 283^ for r = 4. The sum of these values of x is 360, as it should be, since the load is symmetrical. 1 2th, Let there be on this beam weights at the end of the second and third intervals, and find the end moments and points of contrary flexure. We now have W 4,500, c = \l = 60 inches, r I = 3, r 2 = i ; so that (290) and (291) become M 4500 x 360(6 i 4f ' = -75 -jgx 2x5-4x^(3x4x7-1x2x3) 4. -3-(i44 _ 4) j = -442500, 210 ) -4500 _ 4) J = _ Using these moments in (289), we find r =. 1.2460 or 4.2354; we use r = i or 4. Therefore, from (288), x = 79.254 or 2 5 I -53> which are the points of contrary flexure sought. 1 3th, When these 2 equal weights are at the second and third points of division, as in the twelfth example, what is the maxi- mum deflection of the beam, and at what point does it occur ? Using (281), where now W = 4,500, c = 60 = J/, r t = 3, r 2 = i, and provisionally putting x = cr, we find y r ; then, putting r + i for r in the value of y n we find jv + i J and then, making Aj/ = jj/r + r y r o, we obtain r 3 6.722r 2 + 9.1117* -f 2.676 = o, DEFLECTION, END MOMENTS, ETC., FOUND. 2IQ from which we easily see, as was suspected, that a positive value of r lies between 2 and 3 for a maximum y. Making, therefore, r =: 2 in equation (281), and differentiat- ing with respect to x y then putting -^- o, we find x = 0.473914 which, substituted in (281), r being 2, gives y = 0.11553 inch, a maximum. At centre, y 0.11478 inch, at second weight. At \l t y = 0.09514 inch, at first weight. 87. When the uniform discontinuous load is applied at equal consecutive intervals, the first weight being placed at no inte- gral number of times the common interval from the left end of the beam, we may proceed in finding the deflection, end moments, and points of contrary flexure as in article 20, where r, / and r 2 need not be integral, but where the differences, r l r 2 , r r 2) r l r, each denoting a number of weights, must be integral. In this way the deflection formulae already established in this chapter for full intervals, r, r lt r 2 , being whole numbers, also apply to the case now under consideration, where r, r,, and r 2 have the same fractional part, except that, when r 2 is negative, its value is less, by unity, than the common decimal part of r and / as before shown. EXAMPLE i. Beam fixed horizontally at right-hand end, simply supported at left end. Length = 360 inches = /, c = l/ =r 60 inches ; 6 weights, each = W = 4,500 pounds, applied at the intervals \c> c, c, c, c, c, \c\ depth and cross-section of I-beam as in the example of article 75, where the moment of inertia of section I =691, and E = 24,000,000. What is the deflection at the centre under this load ? 22O MECHANICS OF THE GIRDER. Using equation (292), where now r z J, r z 5^-, r = 2^, and x == -|/ = 1 80, we find central deflection = 45Q X 360* ( i /3 . I . 6 a 12 x 16584000000(8^2 ' 6 ' + ^(44375 + 9 - 10.4583) - 0.1771! = 0.3993 i And the greatest deflection due this full load on the beam fixed horizontally at the right-hand end is found by putting x = cr provisionally in (292), and making y r + l y r = o Ajj/. This equation indicates a value of r between f and f. Calling r = J in (292), and putting -2- = o, we find x = 0.420777= 151.477 inches, which is greater than c(r + i) = 6o(f -j- i) = 150 inches, an inadmissible result. Hence we see that the approximate equation AJJ/ = o gave r too small. Now, calling r = f in (292), and making -^- = o, we find .ar = 0.4174047 = 150.265 inches, which is between cr and ^(r + i), as it should be. With r = |, and x = 0.4174047, (292) gives greatest deflec- tion jj/ = 0.41463 inch ; while at the centre it was 0.3993 inch. The end moment in this case removes the point of greatest deflection 180 150.265 = 29.735 inches from the centre. The end moment due this load is given by (293), where r l = -y-, r z = , = / = 60, and W = 4,500 pounds ; and it is, in inch-pounds, 4500 x 360(1 i M 2 = ^~ = -1231875. DEFLECTION, END MOMENTS, ETC., FOUND. 221 The point of contrary flexure is found by adding equations (62) and (93), and equating the sum of the second members to zero. Thus, since M l = o, we have r)/- c(r, - r 2 ) (r t + r 2 + i)] + - r a ) (r + r 2 + i) = o, (305) -c(r-r a )(r + r 2 + i)/ _ - JJf- (306) - r) - <-(r, - r a ) (r, + r 2 + i) + Making x = re provisionally in (306), we find r = 4.5353. Calling r = |, and J/ 2 = 1,231,875, (306) yields the point of contrary flexure x = 0.7547174 which is between re and (r + i)c (that is, between O./5/ and J-J/), though very close to the former. If both ends of this beam are free under this load of 6 .equal weights, we find by (232), at the point x = 0.41 74O4/, y =. 0.9676 inch. And, by (237), the deflection due M 2 = 1,231,875 is y = 0.5530 inch, which added to 0.9676 gives y 0.4146 inch, as found by (292) above. 88. Continuous Uniform Load, wfb, on Beam fixed Hori- zontally at Both Ends. Take the examples of article 75, and apply to the deflections there found the effects of the end moments as given by equation (298). ist, In the first example of article 75, for beam with free ends, the deflection, when x = b = -J/ := 120 inches, and a = o, was found to be y = 0.23/1/1/1 inch. 222 MECHANICS OF THE GIRDER. Now, by (298), the effect of end moments on the deflection in this case is _ 75 X 360* | /j. _ 2 _ _V_i_ _ i\ 24 x 16584000000(^27 9 8i/\27 3/ /8 6 3 \/i A) Q- )(- -)> = 0.19392 inch. ^27 9 8i/\9 3/j Therefore the deflection sought is y = 0.23444 0.19392 = 0.04052 inch; the left third of the 1 5-inch I-beam bearing 75 pounds to the inch, both ends being fixed horizontally. 2d, Again, in the second example of article 75 the central deflection = 0.24421 inch, under the same conditions. If, now, in (298) we make a = o, b \l = 1 20 inches, x /, we get the effect of end moments on deflection = 75 * 36* | _ j*_ /i __ i\ 24 x 16584000000! 81 \8 2/ ^ -f- jr~( )\~ ~ 0.20514 inch. Therefore the required deflection is y = 0.24421 0.20514 = 0.03907 inch at the centre of the beam fixed horizontally at both ends. 3d, Applying the value of y in (298) to the deflection found in the third example of article 75, where a = o, b = ^/, x = f /, we find, for the beam with fixed ends, y = 0.19176 0.17077 = 0.02099 inch. DEFLECTION, END MOMENTS, ETC., FOUND. 22$ 4th, The greatest deflection due 75 pounds per inch on the left third of this I-beam fixed at both ends, is found by adding equations (229) and (298), and in the resulting equation making -2- = o when a = o, and b = A/. dx This gives x |/, whence the greatest deflection y = 0.042199 inch at |/ from the left end of the beam, which is (| |)/ = T y beyond the end of the load. 5th, The end moments for this load of 75 pounds per inch on the left third of this 1 5-inch I-beam 30 feet long, where / rr 691, E = 24,000,000, a = o, b = ^l == 1 20 inches, are given by equations (296) and (297), as follows : 2 9 f - - 6 } = 330000 inch-pounds, M 2 = 75 - - - = ~ 8ooo inch-pounds. 6th, With these values of M l and M 2 , equation (300) gives the first point of contrary flexure, x = 100.926 37.229 = 63.697 inches, since in (300) x cannot be greater than (a + b) = (o + J/) = 1 20 inches. The second point of contrary flexure is derived from (301), where we find x == 260.69 inches. The mode of procedure when only one end of the beam is fixed horizontally is so similar to that just exemplified for two fixed ends, that further examples seem to be unnecessary. 224 MECHANICS OF THE GIRDER. SECTION 4. Deflection of a Girder of Variable Cross-Section in Terms of the Con- stant Unit Strain upon the Extreme Fibres of the Section ; that is, Deflection of a Beam of Uniform Strength. End Moments for Fixed Beams. 89. Economy in the construction of built beams or framed girders requires that the cross-sections of the various members, as well as that of the whole structure, should be proportioned to the greatest strains allowed upon the sections ; and, when the dimensions of parts are so adjusted, it is clear that the unit strain of tension, compression, or bending will be constant throughout the girder. The complete realization of this condition is, for obvious con- siderations, probably seldom attained ; but it is a condition so yearly approximated in practice as to require examination here. For this case we employ equation (186) ; viz., \ 17^, which is independent of /, the moment of inertia of the cross- section, 4Jfid in which B, is constant for a given load, and equal to the mean of the unit strains upon the fibres at the upper and lower surfaces of the beam, and h height of cross-section. 90. Deflection of Semi-Girder of Uniform Height, k, and Uniform Strength. Using the notation of article 64, as illustrated by Fig. 8, and integrating (186), with the sign of E positive for the semi-girder, first, with the condition dx* that = o when x = o, dx h DEFLECTION, END MOMENTS, ETC., FOUND. 22$ secondly, 7 = when x = o, h ' y = -^- (307) which is the deflection at any point, x y of the semi-girder of uniform height and strength. If x /, D = ^m' (3 8) which is the deflection at the free end of the semi-girder of uniform height and strength. It may be observed that (307) is the equation of a parabola with its vertex at the origin of co-ordinates. EXAMPLE. Take an open-webbed semi-girder of wrought- iron whose effective height, h, is 20 feet = 240 inches, length, /, = 50 feet = 600 inches ; and suppose the allowed unit strain in the top chord is C t 8,000 pounds per square inch, and in the bottom chord TI = 10,000 pounds per square inch. Then calling, as we may do without sensible error, the top and bot- tom chords extreme fibres of the cross-section, we have Bi = \(Ci + T,} = 9000. (309) Take E = 25,000,000, then 9000 x 6oo 2 Deflection at free end = D = 240 x ' a5O ooooo = ' 54 mch ' 9000 x 300 2 Deflection at centre = - - - - = 0.135 mc h. 240 x 25000000 226 MECHANICS OF THE GIRDER. It should be remembered that the deflection of a framed girder due to its first full load is likely to be greater than that computed by these formulas, by reason of the yielding of the joints and probable straightening of some of the parts in tension. It is customary, therefore, in computing the deflec- tion of a girder under its first loads until the frame becomes "set," to take E ranging from 15,000,000 to 20,000,000 for wrought-iron, according to the accuracy of the joint fittings and general workmanship ; afterwards the ordinary value of E may be used. 91. Deflection of the Semi-Girder of Uniform Strength but of Variable Height. (a) Let the semi-girder be like either half of Fig. 64, 65, 67, 33, 34, 39, or 83 ; that is, let it slope uniformly from the fixed end, whose height we will call // to the free end, whose height is h v Then the height at any point, x, is (310) .and Hence (186) becomes, for this semi-girder, E(h Q - h,Y # = i 2B* "' dh 2 k' Integrating, with the condition that -^ = o when h . 2HJ* dh where log e denotes the Napierian logarithm. DEFLECTION, END MOMENTS, ETC., FOUND. 22/ Again, y = o when h = h 1t h \ *; - v + *" y - which is the deflection of the uniformly sloping semi-girder at any point where the height is h ; log denoting the common logarithm, and the girder being of uniform strength. Putting for // in (311) its value as taken from (310), we have y in terms of x ; thus, /&i/+(/&o hi}x which is the same as (311). If the semi-girder of uniform slope and strength comes to a point at the free end, we have at that end h = o = h ; and therefore (311) becomes 7" 7>% (313) which is twice the deflection given by (308) for semi-beam of the same length but of the uniform height h v When //o = hi = //, the value of y in (311) and (312) is in- determinate, but is given by (307). EXAMPLE.. Length of semi-girder / = 50 feet ; height at fixed end = 20 feet, at free end 10 feet ; B^ = 9,000; E = 25,000,000 pounds per square inch. What is the deflection at the free end ? //, = 240 inches, h = k = 120 inches, / = 600 inches. 228 MECHANICS OF THE GIRDER. By (3 1 1), = 0.6628 inch. (b) Semi-Girder with Either or Both Chords Parabolic. Open Frame. First, take a case like the half of Fig. 63, supposing the top chord parabolic, and, as in all these cases, the members formed as for a semi-beam. Let / = length of semi-girder, // r = its height at the fixed end, h Q = height at free end, and h = variable height. Then, by equation (136), putting for the h in that equation h* 7z , and adding 7z to the second member for our present case, we have, / also being put for J/, This value of h placed in (186) gives, after reducing, and making m? = - - , -.^ = _^ . (3:5) dx 2 m 2 l 2 x 2 Integrating (315), first with the condition -J- = o when * = q, BJ dx * ml x where log e means Napierian logarithm. Integrating again, with the condition^ = o when x = o, x)log(m/ + x) (ml x}\og(ml x) 2ml\ogml~\, DEFLECTION, END MOMENTS, ETC., FOUND. 22Q '- *) (317) where log denotes common logarithm, andjp is the deflection at any point, x, of the semi-girder of uniform strength, and of the form of one-half of Fig. 63, when the top chord is parabolic. EXAMPLE. Let B^ = 9,000, E = 25,000,000, / = 600 inches, //, = 240 inches, k = 120 inches. If, now, x = /, we have the deflection at the free end of the girder, from (317), 2.302585 x 9000 x 600 y = ~ /- ( 2 3-9) = -59757 25000000 x which is, as it manifestly should be, less than the deflection just found by (311) for the semi-beam of equal length and depth of ends, but of uniform slope, and greater than the deflection of semi-beam of same length and uniform depth = h u found by equation (308). If this girder comes to a point at the free end (that is, if it is the half of the parabolic bowstring), we have, in (3 1 7), k =. o, m i ; + (I ~ *)lQg(/ - X) - 2/bg/], ( 3 l8) which is the deflection at any point, x. 230 MECHANICS OF THE GIRDER. When, in (318),^ = /, we have the deflection at the free end of the parabolic semi-bowstring ; thus, ~ I. D = I.3862Q5 which, according to (313), is - - - of the deflection at the free end of the semi-girder of same length and height at fixed end, but sloping uniformly to a point. From the identity in the form of equations (136), (137), and (138), and from the manner in which (317), (318), and (319) have been derived from (136), it follows that the deflection of any parabolic semi-girder of uniform strength, whether the half- crescent, or the half double bowstring, may be found from (317), (318), and (319), provided we make /^ the height of girder at fixed end, and h Q = its height at the free end. (c) Semi-Girder with Circular Arc for Top Chord. Uniform Strength. Let, as before, //, = height at fixed end, h Q = height at free end, for a girder like the right half of Fig. 63, fixed at the vertical plane through the centre ; the top chord being now supposed circular. If ft is the radius of the circle, the height of the semi-girder at any point, x y is given by equation (125), h = h, - h Q + h Q - (320) h hi R -f /?cos0; (321) being the arc between the point (x> y) of equation (125) and the fixed end of the girder. DEFLECTION, END MOMENTS, ETC., FOUND. Therefore (186) becomes E d z y i , v 7* (3 22 ) 2-Z?! dx z hi R -j- R cos But x = R sin 0, cos dB ( v Integrating first with the condition that -^ = o when ^ = o, dx we have (for this first integration, see Price, "Infinitesimal Calculus," vol. ii. p. 85), after reducing, and putting a = cos a, COS where, as usual, log e means Napierian logarithm. For the second integration, between the limits o and 7, o and 0, (324) takes the form 2J3.R sin a. CQC ^ + 6> 2 The first term is easily integrated thus : / OcosOdQ = 0sm0 - \ sm0J0 Jo Jo . (325) 0sm0 4- cos^ - i 232 MECHANICS OF THE GIRDER. Integrate the second term also by parts, according to the form fudv uv fvdu. (326) Take u log e - = log e cos log e cos a ' . CC -f- 2 2 2 dfr = cos //0, /. 27 = sin 0. sin 4 sm 2 tan cos cos - 2 2 Therefore the second term of the second member of (325) becomes a - 9 cos a sin log e ff tan ?2 + tan ^1\ s i n o^ + 2j v 2 2 y 4 2 But 2 2 COS C6 + COS and the second term reduces to since CQS <*__0 J cosa + cos0' 2 Now a r sing^ = _ a p(cog 8 + cogg) = _ atoge(cosa J cos a 4- cos J cos a + cos DEFLECTION, END MOMENTS, ETC., FOUND. 233 Whence, finally, the integral of (325) is a - cos E sin a a + !0 , o ... y _ s in0 + cos0 - i - COS where m^ = 0.4342945, the modulus of common logarithms, log ; and 7 is the deflection at any point, x =. R sin 0, of the semi-girder having its top chord circular and bottom chord straight, like the truncated bowstring. When 7/o o (that is, when the semi-girder is half of the common bowstring girder), the last term of (327) becomes infi- nite for a = cos 6, which is the case if x = I length of semi-girder, and sin = I -r- R. But in this case sin0 = since; and (327) is easily reduced to , (328) which is the deflection at the free end of the circular semi- bow-string of uniform strength. EXAMPLE i. Semi-bowstring. / = 600 inches, h^ = 240 inches, B l = 9,000, E = 25,000,000, wrought-iron. ... R = /2 + h * = 870 inches =72.5 feet, 234 MECHANICS OF THE GIRDER. sin(9 = /-r-^ = = 0.689655, 6 = 43 36' io".i5, 87 \ C o S = ^I_A = = 0.724138. In arc, = 43 ' 6 282 7r = 0.761013, 1 80 a = -^-^ = -- & = 0.724138 = cos a = cos0, fl 87 /. a = 180 - 43 36' io".is = 136 2/49^5, Therefore the deflection at the free end of this semi-bow- string of uniform strength is, by (328), 2 x 9000 x 870 y = - 25000000 ('5 2 4836 + 0.724138 - i + 1.145371) = 0.71745 inch, which is a little less than 2 X 1.08 = 0.7486 inch = deflection at free end of parabolic semi-bowstring, by (319). And this should be so, since the top chord of the parabolic girder lies just below that of the circular bowstring of the same central height and same span. EXAMPLE 2. Semi-girder, truncated bowstring, circular. / = 600 inches, //, = 240, h Q = 120, B* 9,000, wrought-iron ; E = 25,000,000 ; .-. R = /2 + ^' ~ h >* = 130 feet =1560 inches, 2 A x - Ao DEFLECTION, END MOMENTS, ETC., FOUND. 235 Use equation (327). sin 6 = I -i- R -fa, = 22 3 / 1 1"-5 = 22>6l g 986 ^ = 0.39479 m arc. IoO 2O 130 II 0050 = 0.923077, = cos= I3Q - 5 sina = 0.532939, a = 180 - 32 i2'is".3 = 147 47' 44^7 ; L^= 8 S i2'28".i, LH_? = 62 35 'i6".6; 2 2 2 X QOOO X 1560 ' > = - 25000000 (0.151842+0.923077 - 1+0.455710) = 0.596003 inch, which is the deflection at the free end, and is, as was to be expected, a little less than that found by (317) for the parabolic semi-girder of the same length and end heights. 92. Equations (327) and (328) apply also to the double cir- cular bowstring, truncated or otherwise, provided the radii of the two curves are the same. But when these radii are differ- ent, we may, without sensible error, employ the equations (317), (318), and (319), deduced for the deflection of the para- bolic semi-girder of uniform strength, and applicable to all the cases, including the crescent and the double bow; the computed deflection being always a little greater than that due the cir- cular semi-girder of the same end heights and span. 93. Deflection of the Girder of Uniform Strength sup- ported at Both Ends, either Fixed or Free, and the Height of the Girder being either Uniform or Variable. Since the deflection of a girder may be defined as the difference of level between the position of any one of its points before bend- ing and the position 'of the same point after bending, under 236 MECHANICS OF THE GIRDER. the given load, it follows that the formulae already established for the deflection of the semi-girder of uniform strength also apply to the present case, provided we take the origin of co-or- dinates in the neutral axis at the centre of the span, and call y positive upward, and write \l for /; / being the length of the girder in all cases, and the neutral axis taken horizontal. EXAMPLES. Take an open webbed girder of wrought-iron, height at the centre 25 feet = 300 inches, span = 200 feet = 2,400 inches ; therefore k t = 300, \l 1,200. Let B^ %(d + T t ) 9,000, E = 25,000,000. What is the deflection at the centre ? EXAMPLE i. Height uniform = k = /i, 300; therefore central deflection is, by (308), OOOO X I2OO 2 D = -- = 1.728 inches. 25000000 X 300 EXAMPLE 2. Truncated circular bowstring, h^ = 300, k = 1 80 = end height, /2 r h = 120. R = (VY + (*. " h Y = I200 2 + 120- Use equation (327). = 2 X 120 sin = ^ = = 0.198020, 6 = 1 1 25' i6".3 ; R 55 ^0.980:98. In arc, = TT = 0.199338. 1 80 7 7? cos a = a -^ - = 0.950495. K DEFLECTION, END MOMENTS, ETC., FOUND. 237 + 0) = 8639' 3 i".i5, a = 180 - 18 6' 14" = 161 53' 46". a - 0) = 75 14' 14". 85. = 0.039473, \og = -0.2218488. cosM^l sinj? cosK-0) = -0.408267. & cosj(a + 6) sin a cosi(a + 0) (0.408267 0.2218488) = 0.407994. 25000000 = 1.86169 inches. ( EXAMPLE 3. Truncated parabolic bowstring, equation (317). 300, 7/o = 1 80, x -|-/ = 1,200, m* - IJL = 2.5. fil "O = 1897.366, m(\l) + x 3097.366, w(-|/) x = 697.366. log|w/= 3.2781512, log (-|^/+ JP) = 3.4909925, log (^ml x) 2.8434608. Therefore (317) becomes 2.302585 x 9000 x 1200 , y = 5 2_2 - Z - (10812.88 -|- 1982.93 12439.70) 7 1.58114 X 120 X 25000000 v = 1.86695 inches. EXAMPLE 4. Chords of uniform slope, h^ = 300, h Q = 1 80 //, \l = 1,200. Use equation (311). y = ~ ' ' 7 ~~~ " """ < 3 i8o[ 2.302585 log - " ---> 2 ( \ _ 2 X 9OOO X I2OO 2 25000000 x i20 2 \^ ~ \ ^ u ' u ~ 180 = 2.0197 inches. 238 MECHANICS OF THE GIRDER. EXAMPLE 5. Circular bowstring. \l = 1,200, /^ = 300, h Q = o. Use equation (328). - = 212.5 f eet = 2 55 inches. sin0 = = 0470588, = 284'2i". R 7? Jt cos0 = - = 0.882353. In arc, = 0.501346. I? cos a = cos# = a = 0.882353. a= 180- 2 84'2i"= 151 55' 39". *( - <0 = 6I-SS' 39", ^ 2[cos : ( t:, )]2 = -0-57573x8- = 0.23593. 2 X QOOO X - , ,- y = (0.2^91 + 0.882353 J 25000000 + 2.302585 x 0.882353 X 0.5757318) = 2.36477 inches. EXAMPLE 6. Parabolic bowstring. \l 1,200, h* = 300, By equation (319), 1.^8620^ X QOOO X I2OO 2 D = = 2.39552 inches. 25000000 X 300 EXAMPLE 7. Girder sloping uniformly from centre to ends. \l = 1,200, h r 300, k = o. By equation (313), 2 X 9000 X I200 2 D = = 3.4^6 inches. 25000000 X 300 DEFLECTION, END MOMENTS, ETC., FOUND. 239 EXAMPLE 8. Parabolic crescent. //, = 300, h = o, \l = 1,200. The deflection in this case must be the same as that in the sixth example, for the parabolic bowstring. /. D = 2.39552 inches. EXAMPLE 9. Girder like Fig. 5 3, sloping uniformly from centre to ends. // x 300, 7z = o, \l 1,200. Deflection the same as in example 7, viz., D 3.456 inches. EXAMPLE 10. Girder like Fig. 66, polygonal. Find the deflection for each part having a uniform slope, separately, and add the results for the total central deflection, after correcting. Take /i t = 300 at Z 4 , and h 240 at Z 6 , the quarter-section. Then \l = 600, and equation^ 11) gives the deflection at Z 6 , thus, 2 x 9000 x 600% y = 25000000 x 60* l>o - 240(2.302585 log fJ + O] = 0.46408 inch. Similarly, for the end quarter, equation (313) gives 2 x 9000 x 6oo 2 = 25000000 x 240 ' = i. 08 inches. But, before adding these results, we must find, as in article 67, how much the free end of the semi-beam is deflected by reason of the bending of the part between Z" 4 and ^ 6 ; that is, we must add to 0.46408 the quantity \l X tan a = 600 tan a & dx 240 MECHANICS OF THE GIRDER. From (311), = tana = _. 1 , .log.|. = 0.0016066, *i) "4 600 X 0.0016066 = 0.96398 inch, /. Total deflection = 0.46408 -f 0.96398 -f- 1.08 = 2.50806 inches, which is greater than the deflection found in example 6, for the parabolic bowstring ; and it will be found that although the girder, Fig. 66, is deeper at the quarter-points than the para- bolic bow of example 6, yet at the } and f points the latter is the deeper. In like manner may we proceed in all cases of irregular forms, whether there be two or more changes of slope ; but, in general, we may use the formulae already found for regular forms, with sufficient accuracy, always choosing the one most fitting for the case in hand. f 94. We may arrange the results found in these examples according to the amount of the deflection, and thus the more clearly perceive the effect of form upon the bending of girders of uniform strength. All the girders here represented are 200 feet in length if supported at both ends, or 100 feet long if semi-girders ; the deflection being the same in either case. Since in all the formulas the deflection varies directly as 2* K^i + T *\ we may fi n d the deflection of girders of the E E same dimensions, but of other material than wrought-iron, by substituting for E the proper value taken from Table II., and for Ct and 2" x the allowed unit strain. If, for pine, T t = 1,200, C t 552, E 1,460,000, then B^ 7? /? = 876, and - = 0.0006, but for wrought-iron -r = 0.00036 ; hence, for a girder of uniform strength and of given span and DEFLECTION, END MOMENTS, ETC., FOUND. 24! height, the deflection, if the material is pine, will be five-thirds of the deflection were the material wrought-iron ; that is, allow- ing Ct and Ji the above values. If the compressed chord be of pine, C l = 552, and the other of wrought-iron, T t = 10,000, and if E = 13,230,000 = ry 4(25,000,000 + 1,460,000), we have B^ = 5,276, - = 0.0004.. R Hence a combination of pine and wrought-iron gives a deflec- tion -4- = times that due wrought-iron alone, with these 3-6 9 unit strains. Were the compressed chord of cast-iron, for which C t 15,000, while the other chord is of wrought-iron, T l = 10,000, and E = \(2 5, 000,000 + 12,000,000) = 18,500,000, we should n have BI 12,500, 0.00067567, and the deflection would E be 1.877 times that of the girder of same size in wrought- iron. 95. By inspecting the following table, we see that for open girders of the same central height, same length, and of uniform strength, the total deflection is NEARLY in the inverse ratio of the areas of the figures of the girders. This is exactly the ratio of the deflections in case of the girder of uniform height and of that sloping uniformly to a point : viz., ratio of areas, \ ; ratio of deflections, \. We may, therefore, without appreciable error, employ this principle in finding the total deflection of open girders of uniform strength and variable height. 2 4 2 MECHANICS OF THE GIRDER. EXAMPLES. DEFLECTION OF OPEN WEBBED GIRDERS OF UNIFORM STRENGTH. Length = / = 200 feet, central height = /Zj = 25 feet. MATERIAL. Wrought- Iron. Pine. Wrt.-Iron and Pine. Wrt. and Cast Iron. Equa- tion. E 9,000 25,000,000 876 1,460,000 5,276 13,230,000 12,500 18,500,000 Form. Description. Def., ins. Def., ins. Def., ins. Def., ins. 100 \ 25 25 25 I 1. S Uniform ) height J 1.728 2.88 1.92 3-24 (308) 2. 25 ^ ^"""^ i One chord | circular f 1.86169 3.1028 2.0685 3494 (327) ^===^ 3. ~^ === ^ "751 One chord ) parabolic ) 1.86695 3.III6 2.0744 3.504 (317) ^ s====== ^ 4. 25 15] Uniform ) slope ) 2.0197 3.366 2.2441 3-791 (311) 25 IB] 5. Tf^ <^" Circular \ Bowstrfhg ) 2.36477 3.9413 2.6275 4438 (328) 25 ^ . 25 -^^^ Parabolic ) curves ) 2.39552 3.9925 2.6617 4.496 (319) 25 ^ 7. 25 ~^-~^^^ Uniform ) slope } 3456 5.76 3-84 6.48 (313) 2T^ ^_^ 25 ^ -**^" 25 ~^^Z=~ 8. ^S^^ Polygonal ) chord ) 2.50806 4.1801 2.7867 4703 (3") DEFLECTION, END MOMENTS, ETC., FOUND. 243 We give below, the deflections of girders of wrought-iron for the eight cases just tabulated, but now computed by this Method of Areas : No. Area of One-Half Girder. Deflection. Deflection by Formulae. I 2500 square feet. 1.7280 inches. 1.72800 inches. 2 2167 " 1.9931 1.86169 " 3 2167 " 1.9938 1.86695 " 4 2000 2.1600 " 2.01970 " 5 1737 " 2.4871 2.36477 " 6 1667 " 2.5920 " 2.39552 " 7 1250 " " 3.4560 3.45600 " 8 1625 " 2.6585 2.50806 " The deflections of such girders as those shown in Figs. 19, 20, 29, 30, 31, 32, 33, 34, 39, 40, 53, 54, 55, etc., are therefore easily found by the method of areas. It should be noticed that in the preceding table of deflec- tions of the same girder in different materials, a factor of safety equal to 10 has been allowed for pine, while 5 is the factor allowed for wrought and for cast iron. The modulus of elasticity for cast-iron, E = 12,000,000, is so small, that, in spite of its large resistance to compression, d = 15,000, the open beam made of wrought and cast iron, and of uniform strength, has greater deflection than that of wrought-iron alone, and, indeed, greater than that of pine alone, with the low unit strain here allowed. 96. Finally, if the beam be of uniform strength, but have a continuous web, the formulae already deduced for girders of uniform strength and of open web may be employed by assign- ing to BI its proper value derived from Table II. 244 MECHANICS OF THE GIRDER. EXAMPLE i. Plate girder of uniform strength and uniform height, wrought-iron. Take the length / = 50 feet = 600 inches, height h = 5 feet = 60 inches ; the girder being sup- ported at both ends. By Table II., B = 42,000 breaking unit strain for plate beams. Allowing a safety factor of 5, we have B l = 8,400 ; and calling E = 25,000,000, and putting / for / in equation (308), there results the central deflection, D = < 30' = . 504 inch . 25000000 X 60 EXAMPLE 2. Take a plate girder of the same length, 50 feet, and same central height, 5 feet, but sloping uniformly from centre to ends, where the height is 2 feet. Then, if the girder is of uniform strength, we have, from equation (311), , x 8400 x [6o _ ( , 25000000 X (60 24)* L = 0.65376 inch. EXAMPLE 3. Cast-iron beam of uniform strength, and height h = 3 feet = 36 inches, -|-/ = 6 feet = 72 inches. Take B t *&%& = 7,650, E = 17,000,000 (Table IL). Then, by (308), 7650 x 72 2 D = - --- - - 7 = 0.0648 inch. 17000000 X 36 EXAMPLE 4. If this cast-iron beam of uniform strength slope uniformly from centre to ends, where h = 12 inches, then, by (311), - 7 ^6 - 12(2.302585 logff + i)] = 0.0876 inch. DEFLECTION, END MOMENTS, ETC., FOUND. 24$ EXAMPLE 5. Oak beam of uniform strength and height. Take \l 1 20 inches, h = 1 8 inches, B^ = - L ^ , E 2, 1 50,000 ; then (308) gives 1060 X 120* D = 2150000 x 18 EXAMPLE 6. If this oak beam of uniform strength slope uniformly from centre to ends, where h = 12 inches, then, by (311), 2 X 1060 X I20 2 ^ 18 - ^(2.302585 log it + 1)] 2150000 x (12 - ts) = 0.4474 inch. EXAMPLE 7. Beam of Bessemer hammered steel, uniform strength. \l =. 72 inches ; height at centre, /i t = 20 inches, at ends, // = 10 inches. Take B, = l^d&l = 25,616, E = 31,000,000 (Table II.). Then deflection at centre is, from (311), 2 X 25616 X 72 2 ' = 31000000(10 - 2o)'t 30 - IO ( 2 -3 2 S85 logM + i)] = 1.1196 inches. For same beam of wrought-iron, 2 x 9000 x 72 2 y = 25000000(10 - 2o)l> ~ ^(2-302585 log fg + i)] = 0.4878 inch, which is less than half the deflection of the same beam in steel. But if we suppose this beam to be of rectangular cross- section, and to bear a concentrated weight, W, at its centre, where the height is //, = 20 inches, and the thickness b = 2 246 MECHANICS OF THE GIRDER. inches, the length being / = 144 inches, then, from equations (46) and (160), we have moment at centre, M = \Wl = %Bbh 2 = %M 2 for safety, 3 M4 ' W = 3.7037 x 25616 = 94874 pounds for steel, W 3.7037 X 9000 = 33333 pounds for wrought-iron. Hence, under the assumed unit strains, the steel beam bears = 2.8462 times the weight at the centre of the wrought- 9000 iron beam of the same dimensions, while the deflection of the steel beam is ^-^ - = 2.2953 times that of the wrought-iron 0.4878 beam ; that is, what is shown in all the formulae, the weight W varies directly with the unit strain B lt while for the same unit strain the deflection varies inversely as the modulus of elasticity, E. Therefore in the present case, so far as deflection is con- cerned, the advantage of steel over wrought-iron, under same load, is - p = , which is the simple ratio of the moduli of elasticity. 97. The thickness, b, of a continuous webbed girder of uniform strength at any rectangular section of given height, //, may be found, in general, by equating the moment, M, due the external forces, to the moment of resistance, R, of the internal forces of the beam at the given section, and solving with respect to b. DEFLECTION, END MOMENTS, ETC., FOUND. 247 For a beam of rectangular cross-section, bearing a con- centrated load at its centre, equations (45) and (160) give M = Wx = IBM 2 - BI allowed unit strain. (329) where, if the height, //, be uniform, b varies as x\ making the horizontal projection or ground plan of each half of the beam a triangle with a vertex at the end of the beam, where b = x = o, and a base at the beam's centre, where b = - - . EXAMPLE. Oak beam of uniform strength, and height h = 15 inches, length = 15 feet, weight applied at centre W = 4,000 pounds allowed unit strain = - = 1,060 10 pounds per square inch. What must be the thickness of this beam at the centre ? Here x = \l 90 inches, /. b = 3 x 4 X 90 = inches< 1060 x i 2 It must be remembered that wherever the moment becomes zero, causing b, the thickness of the beam, to vanish by the formulae, we must, nevertheless, have at all such points suffi- cient material to resist, with the proper margin of safety, the shearing-strain which may there be developed, and the re-action of the supports. In this example the shearing-strain at each end of this beam 4000 is = 2,000 pounds. Now, by Table L, the ultimate resist- ance to shearing is, for oak, across the grain, 4,000 pounds ; 248 MECHANICS OF THE GIRDER. one-tenth of which is 400 pounds, to be allowed to each square inch of the vertical section at each end. Therefore - = 5 square inches of section at least the 400 beam must have at each end; that is, the depth being 15 inches, the thickness is \ inch. But there is another consid- eration to be attended to ; viz., the bearing-surface at the ends must be sufficient to resist with safety and permanence the pressure coming upon it. This beam as now estimated is \ inch thick at each end, and 4.53 inches at its centre. Hence it must have 8.903 inches of its length at each end upon the support, in order to secure a bearing of 3^ square inches, required for 2,000 pounds with an allowed unit strain of 600 pounds to the square inch, in com- pression. Again, a beam so thin at the ends would lack lateral stiff- ness unless it were walled in. In practice, therefore, even when it is desired to use the least material possible, it is customary to make those parts of a beam which theoretically, or rather, by formula, are almost nothing, of such size as a just regard to all these require- ments, as well as to the good appearance of the structure, may demand. Let it not be inferred that theory and practice are at vari- ance here, for such is not the case. The equations which determine the thickness of the beam do not pretend to take into the account all the conditions affecting the sufficiency of the beam for its purpose. And hence the theory is not complete till the modifying conditions are introduced. 98. If the beam of uniform strength be loaded uniformly with w units of weight to the unit of length, we have, from equations (49) and (160), DEFLECTION, END MOMENTS, ETC., FOUND. 249 putting BI for B, and the cross-section being rectangular ; which is the thickness of the beam at any point, x, measured from the end. When h is constant, (330) is the equation of a parabola ; the vertex being at the end of the beam. Thickness at end = b = o. x = o. Thickness at centre = b = ^ W . x \L Horizontal projection, two parabolas. EXAMPLE. Oak beam, uniform strength. Height uniform = h 15 inches, length / = 180 inches, B, = I)O 6o, 10 8OOO A ' 1 w = - - = 44! pounds per inch. 1 80 Then thickness at centre is x 400 X i8o 2 * x 400 x iocr . , 2 2 = 4.53 inches. 4 X 9 X 1060 x is 2 99. If the cross-section of the beam of uniform strength be of either form, Fig. 91, then, by assigning values to three of the dimensions, //, // b, b iy we may, from equation (161) and the equation expressing the moment due the given load, find the fourth dimension of the cross-section, which, therefore, becomes known at every point. In like manner may we determine any one dimension of any cross-section whose moment of resistance, R, is known. 250 MECHANICS OF THE GIRDER. EXAMPLE. Take a tubular plate girder of the dimensions given in example i, article 96 ; viz., / = 50 feet, h = 5 feet, B t = 8,400, uniform strength and height. Cross-section as in Fig. 91, where let b = 12 inches, x = \\\ inches; the side plates being J inch thick each, h = 60 inches. From (49) and (161), we have equal to 58 inches if w! = 123,508 pounds, the total uniform load on beam, and x = ^/ = 300 inches. At the centre, therefore, the top and bottom plates must have the thickness of i inch each ; while at the ends, where x o, (331) gives hi = hi ) 3 = 1.02174^ = 61.3044 inches, \-W which renders h h^ 1.3044 inches negative, showing that the cross-section of the side plates is more than sufficient at the ends to resist the moment. We may find at what distance from either end of this beam the top and bottom plates begin to be needed, by putting /^ = h = 12 in (331), and finding x. This gives x = 69.19 inches, for which the side plates alone are sufficient if properly braced laterally. Now, the shearing-strain at each end of the beam supporting this load is J X 123,508 61,754 pounds; and, calling the allowed shearing-strain 8,000 pounds to the square inch, we require -o = 7.72 inches in cross-section of the two plates, whereas we have 2 X f X 60 = 45 square inches. But DEFLECTION, END MOMENTS, ETC., FOUND. in order to have sufficient bearing-surface on the abutments, allowing the iron to bear 8,000 pounds to the square inch in compression also, the beam must be supported for at least 7 72 ' 3 = 10.3 inches of its length at each end. 2 X g^ The semi-girder of uniform strength and continuous web is to be treated in the same manner as the girder just considered when we seek its variable cross-section. 100. Beam of Uniform Strength fixed Horizontally at Both Ends. By definition the beam of uniform strength is equally efficient at all sections to resist the strains generated by the external forces. Hence, when this beam is horizontally fixed at both ends, and loaded with a concentrated or with a continuous load, the points of contrary flexure are, for any style of beam or girder, practically midway between the centre of gravity of the load and the ends of the girder ; since there is as much reason for their being on one side of this midway point as there is for their being upon the other side of it, and no more. And the beam of uniform strength is such only with reference to a particular mode of loading. That is, if the unit strain is uniform throughout the girder for a given position of the load, a change in the position of the load causes a change in the relative values of the total strains in the members or parts of the girder, and therefore a change in the unit strain on each member, if, as is assumed, the cross-sections of the members be not changed. In general, we have for any girder, from equations (184) and (187), Moment due internal forces, M x = ^^ = 2jgltS>2 , (332) h h where B v = allowed unit strain in bending, 5 = area of any cross-section, r radius of gyration of the section about its neutral axis, h = height of section. 252 MECHANICS OF THE GIRDER. By equating the last member of (332) to the known moment due the external forces applied to the girder, any one of the four quantities B iy S, r, h t may be found. But, when the girder is fixed at one or both ends, we need to know the point or points of contrary flexure, in order to determine the end moments. 101. For the girder of uniform height and strength, fixed at both ends, it follows from the uniformity of the unit strain and height, which causes a uniformity of curvature, that, as already stated, each point of contrary flexure is sensibly midway between the centre of gravity of the load and the correspond- ing end of the girder. Assuming that the height and strength are uniform, and that, for any required form of cross-section, the necessary variation in its area is attained by varying the thickness of the beam only, we shall have, in (332), r, k, and B, constant, so that the variable area, S, may be found at once for any section of the beam ; and from 5 the thickness is to be determined. 102. Beam of Uniform Strength and Height fixed at Both Ends, and bearing a Concentrated "Weight, IV, at the Distance a' from the Left End. The moment at any point between the weight and left end of the beam, that is, when x is not greater than a' y is given by equations (40), (93), and (332), thus, ( 333 ) Now, M x = o when x = \a' y ... o = W l ~ 7 a a' - -(M, - M 2 ) + M t . (334) DEFLECTION, END MOMENTS, ETC., FOUND. 253 Also, when x is not less than a', we have, from (43), (93), and (332), . (335) M x = o when x = \(l + ^0> o = W l -=-^a' - \(M, - M 2 ) - ^(M, - M 2 ) + M,. (336) From (334) and (336), we find a >). ( 337 ) That is, the end moments are equal and negative for any given position of the load. Eliminating M l and M t from (333) and (335), we obtain = ^f-, (338) = mL^l (x _ V ). (339) 2 ' h **, (340) (34i) 254 MECHANICS OF THE GIRDER. EXAMPLE i. If the varying cross-section is a rectangle of the breadth b, and constant height //, we have r 2 = y^ 2 , and (339) and (341) become bh = - (x - X)> (342) + a' \ ~ x } (345) If, further, the weight, W, is at the centre of the girder, a' = /, and when - I/). ( 34 6) ~ X) - (347) i WJ In (346), for x = o, b = b, = - ^_^-, at left end. x = /, ^ = o, at quarter point. # = i/, ^ = 1^?-, at centre. ^BJi 2 ~ M7/ In (347), for x = J/, ^ = 2-, at centre. x = |/, b o, at quarter point. - ; a t right end. DEFLECTION. END MOMENTS, ETC., FOUND. 255 If the beam is of oak, and B t = ^B = 1,060 pounds, E = 2,150,000, / = i So inches, h = 15 inches, W = 4,000 pounds, 3 X 4000 X 1 80 tb^A=r^= -*,= ~ 4 x I0 6o X IS 2 " -2-264 inches; the algebraic sign only indicating the direction of the inclina- tion of the vertical planes forming the sides, to the vertical longitudinal plane of the r3am. Fig. 99 shows this beam thus loaded, in plan and elevation. It is evident that the deflection of the part BD = |7, or of the part AB = \l, as a semi-beam, is equal to the deflection of a beam of uniform strength and height supported but not fixed at the points B and D, and bearing the concentrated weight W. But, by equation (307), the deflection of the part AB or BD is, since for x we must put \l, D FIG. 99. Therefore the total deflection at C y the centre of AE t is equal to - IO - = 0.1331 inch in the present case. 8 X 2150000 X 15 256 MECHANICS OF THE GIRDER. At the points of contrary flexure, where b = o, the beam, of course, must be enlarged, to resist with safety the shearing- strains. The shearing-strain at each of these points is now J W = 2,000 pounds. By Table I., article 42, the ultimate shearing- strength of oak across the grain is 4,000 pounds to the inch ; or the working-strength is 400 potfnds to the square inch of cross-section. We require, therefore, at least %$ 5 square inches of area at each point of contrary flexure ; that is, the beam, being 15 inches deep, must be at least J inch thick at these points, even when restrained from moving laterally. 103. Beam of Uniform Strength, Height, and Load, fixed Horizontally at Both Ends. Rectangular Cross-Section. Equations (49) and (93) give M x = >(/ - *)* - l ~ Z * + M l = \BJh\ (349) Make x = o, then M = M 2 = \BJ>Ji\ But, as in article 102, b o when x = \l or f /, ... % w (i _ i/)i/ + M, = o, Mt = M 2 = --&0// 2 , M c = \w(l - V)V ~ -h 1 * = ?W 2 > M x = \w(l - oc)x - &wl* = \B,bh\ (350) DEFLECTION, END MOMENTS, ETC., FOUND. 2$? When x = o, or x = /, (351) gives which is the width of beam at either fixed end ; which is the width of the beam at centre, being one-third of the width at either end. Equation (351) being that of a parabola with respect to the variables b and x, the plan of this uniformly loaded beam, of uniform strength and height, fixed at its ends, is shown in Fig. 100. I ELEVATION. PLAN. FIG. 100. EXAMPLE. Oak beam. / = 180 inches, h = 15 inches, ends fixed. Take B l = 1,060, E = 2,150,000, w = &$$ = 44$ pounds to the linear inch. Then j. = M 2 = _ x x i8o 2 = 135000 inch-pounds. 3 2 9 M c = 45000 inch-pounds. 258 MECHANICS OF THE GIRDER. b, = b 2 = - 9 * -F * i8o 2 = _ . n> = thicknesg at 16 X 1060 x is 2 b c = 1.1321 inches, at centre. The maximum deflection is evidently given, as in article 102, by equation (348), and is D = 0.1331 inch. 104. Concentrated Weight, W, at the distance a' from the Unfixed End of a Beam of Uniform Strength, fixed at the Right End, but simply supported at the Left. The point of contrary flexure must be at the distance ;F O = l(/ -f- cf) from the unfixed end, in order that the greatest positive ;moment may be equal to the greatest negative moment. Equations (43) and (93) apply, giving, since M, = o, M x = W-=- a f + M = \BJ>h* (352) .if the cross-section be rectangular, and x 'y a'. For x = /, M 2 = \BJ)Ji\ .For * = (/ + a'), M 2 = - W a '^ ~~ a when b = o. ~ DEFLECTION, END MOMENTS, ETC., FOUND. When***', = *>%'(' ~ f > ^V b+* I Which shows that the width at Z>, Fig. 101, is the same as the width at C for h constant. When x ^ a', use (40) and (93), giving M x = W*-=- x + ^ = i^M*. (354) To find the lowest point, E, in the curve, Fig. 101, we equate the deflection, D between the lowest point and left end of the beam, to the total deflection, D 2 + Z? 3 , between the same point and the right end of the beam, and solve the equa- tion Z> x = A + D y For the length AE = z, (307) gives EB = l + a ' - z BD = - JkLfi which is the deflection at the lowest point, E. 26O MECHANICS OF THE GIRDER. EXAMPLE. Given W = 4,000 pounds at the distance a' \l from the unfixed end, A, Fig. 101 ; / = 180 inches; h = 15 inches = uniform height of beam ; B^ =. -^B = 1,060 pounds = working inch strain for oak ; E = 2,150,000. Cross- section rectangular. Then width of beam is, At left end, (354), x = o, b = o. At the weight, (353), , 4000 x 1 80 * = * = IQ 6o x 15- = 3 ' 19 (353), * = |/, b = o. At fixed end, (353), j = _ 4000 X 180 = _ 1060 x i5 2 the negative sign simply showing that the lines cd, Cjd u have crossed somewhere between x = \l and x = /. Moment at fixed end, Jlf 2 = 4Q00 3 ^- = 120000. 1 + Moment at the weight, = 120000 inch-pounds. DEFLECTION, END MOMENTS, ETC., FOUND. 26 1 The deflection at the lowest point, E y is given by (355), 1060 x ~ D l = 4 x 2150000 x 15 X (|) = 0.1849 C E FIG. 101. 105. Continuous Uniform Load, wl y on a Beam of Uni- form Strength, fixed at the Right End, and simply sup- ported at the Left, Fig. 101. The figure shows the curvature and the plan, when the section is rectangular ; and equations (49) and (93) give, since M I = o, M x = for these conditions. For x = /, (356) wl 2 For x = M 2 = - b = o. 262 MECHANICS OF THE GIRDER. For * = M = *>/ 2 b = b = f^(2l - Z x)x. (357) Hence the breadth at 7 is one-third of that at Z> when the height is uniform, as seen in the parabolic plan, Fig. 101, derived from equation (357). The deflection at E is given, by (355), for this case also, provided we put \l for a'. EXAMPLE. Let / 180 inches, h = 15, B^ = ^B = 1,060 pounds = working unit strain for bending oak, E = 2,150,000. Cross-section rectangular ; load wl = 8,000 pounds uniformly distributed continuously, 44$ pounds to the inch. From (356) and (357), x = o, M l = o. x \l, M = T ^ X ^ X i8o 2 = 80000 inch-pounds. x |/, M = o. # = /, J/ 2 = | X ^ X i8o 2 = 240000 inch-pounds. Width at left end, ^ = o, by (357)- Width at i/, x = .oi 26 inches. 3 X 1060 x Width at |/, J = o. Width at right end, b 2 = 6.0378 inches. DEFLECTION, END MOMENTS, ETC,, FOUND. 263 Put -J/for tf'in (355), and find the deflection D = 0.1849 at lowest point. 106. Beam of Uniform Strength and Uniformly Varying Height, fixed at Both Ends. The end moments M l =. M 2 are determined for this case as in articles 102 and 103, for the same kind of load. To find the deflection of this beam, we may regard it as composed of four semi-beams, Fig. 102. ist, AJ3, fixed at A deflection A- 2d, BC, fixed at C, the lowest point; deflection Z> 2 . 3d, CE, fixed at C, the lowest point ; deflection D y 4th, EF, fixed at F; deflection Z> 4 . Now we must have D* + D 2 = D 3 + D# from which the lowest point and its deflection may be found. EXAMPLE i. Take one-half of the girder shown in Fig. 65, and suppose the ends of this half to be immovably fixed. Call the length / = 100 feet = 1,200 inches ; and height at left end, h Q 1 80 inches ; height at right end, k, 300 inches. Let the girder be of wrought-iron, and, as in article 93, take B t = i(i ~H ^i) 9>ooo pounds, E = 25,000,000 ; and suppose the load to be a concentrated weight, W = 200,000 pounds at the centre, no account being here taken of the girder's own weight. By article 100, the points of contrary flexure are ^^ = 2$ feet from the centre of the beam ; and from equation (337), since a' /, M, = M 2 = M c = \ x 200000 x X \ X 1200, = 30000000 inch-pounds. 264 MECHANICS OF THE GIRDER. The area 5 of any cross-section on the left of the weight is given by (339), and on the right by (341). But these equations suppose the section of the top chord to be equal to that of the bottom chord in the same vertical plane of section, and at the centre give Sc = 200QOO X 240 X 600 X j X 1200 = inches 2 X QOOO X 1200 X J X 240 2 at left end, (339), = 200000 x 180 X 600 X -j X 1200 = incheg 2 x 9000 X 1200 X J X i8o 2 at right end, (341), S 2 = 20000Q X 300 X 600 x I X 1200 = 22>222 inches . 2 X 9000 X 1200 X J X 300 2 the negative sign indicating only a difference in the direction of the lateral faces of the chords, that is, change of slope laterally. But if S' = area of section of chord in compression, S" = area of section of chord in tension, we have C,S' = P = --, (358) cos a TVS" = U = -* (359) cos/5 H = M + h, according to the notation and equations of article 49. DEFLECTION, END MOMENTS, ETC., FOUND. 265 From which, S' = , (160) Cih cos a S "=T^- < 36I) Calling Ci = 8,000 pounds, 7", = 10,000 pounds, a being the inclination of the top chord for all parts between the points of contrary flexure, while (3 = o, and ft being the slope of top chord for the remainder of the beam, while a =. o, we have, at either end, tan = 25 IQQ 15 = o.i, cos/3 = 0.99503 ; tana = o, cos a = i. At centre, tana = o.i, cos a = 0.99503; tan/3 = o, cos/3 = i. At left end, (361), area of top section, s ,, = - 30000000 - = i 10000 x 180 x 0.99503 At left end, (360), area of bottom section, s , = = 8000 X 180 X i At centre, (360), area of top section, 5 , = - 30000000 - = I5 inches . 8000 X 240 X 0.99503 266 MECHANICS OF THE GIRDER. At centre, (361), area of bottom section, 30000000 S " = loooo x 240 X i = I2 '5 inches " At right end, (361), area of top section, 30000000 = 10000 x 300 x 0.99503 = Ia 5 mches - At right end, (360), area of bottom section, s , = - 3oooooo = QO inches _ 8000 X 300 X I The totals are : At left end, S* = 37-5 82 inches; at centre, S c = 28.203 inches ; at right end, S 2 = 22.550 inches; differing somewhat, as was to be expected, from the areas com- puted on the supposition of equal top and bottom sections. The deflection for each part of this girder is given by (311). See Fig. 102. ist, AB t fixed at A ; /^ = 180 inches, h =. h 210, -i -, 7z hi = 30, and we have h 7 2 X QOOO X 300 2 ( / 7 { I8 - 2IO ( I - ^302585 log g 25000000 x 30* = 125 x 2 ' 37IS = ' I?I inch- DEFLECTION, END MOMENTS, ETC., FOUND. 267 4th, EF y fixed at F\ //, = 300, h h Q = 270, j hi 10 A, - //, - -30, -^ = -, 2X 9000x300' I _ 270 / 2 . 302585 Iog I9 + ,\\ 25000000 x (-3H V 9 /) o = -y- X 1.5528 = o.i 12 inch. Now, in equation (311), - - is constant, since the length, //o //! /, varies as the height, h. Therefore, * 2d, BC, fixed at C\ h 210, y = A = yf-O^ - 210 x 2.302585 log ^ -f 210 X 2.302585 log 210 210). 3d, CE, fixed at C ; /* = 270, 2.302585 log A, H- 270 x 2.302585 log 270 - 270), and A + A = A + A- After making the substitutions, and reducing, we find hi = 236.15 inches, which is the depth of the girder at lowest point, C. Hence dis- tance of lowest point from left end is 10(236.15 180) = 561.5 inches. A = o.i 08 inch, > 3 = 0.167 inch. A + A = A + A = 0.279 inch at C 268 MECHANICS OF THE GIRDER. EXAMPLE 2. Take the same girder as in the preceding example, but let the load, W = 200,000 pounds, be 75 feet from the left end. Then the points of contrary flexure are, where x = |/, and # = !/, and by (337), since now a' = |/, MI M 2 \ X 200000 X I X J X 1200 = 22500000 inch-pounds. At the weight, x = a' = f /, (338) gives M = 200000 X J X f X 1200 = 22500000 inch-pounds. At the centre, x = |7, M c = 200000 x \ X \ X 1200 = 7500000 inch-pounds. At left end, by (361), area of top section, S = 22500000 = 12.56 inches. 10000 X 180 X 0.99503 At left end, by (360), area of bottom section, S' = 22 5 000 = 15.63 inches. 8000 x 180 X i At the weight, (360), area of top section, S' = 2250000 = 10.47 inches. 8000 X 270 X 0.99503 At the weight, (361), area of bottom section, S" = 22 5 0000 = 8.33 inches. 10000 x 270 X i DEFLECTION, END MOMENTS, ETC., FOUND. 269 At right end, (361), area of top section, 221500000 S " = 10000 x 300 x 0.99503 = 7-54 mches. At right end, (360), area of bottom section, s , = 22500000 = g inches _ 8000 X 300 X I Since equations (338) and (340) are of the first degree with respect to M and x, and since h varies uniformly with x, we PLAN OF TOP CHORD. PLAN OF BOTTOM CHORD. FlG. 102. have M and 5 in (360) and (361), varying uniformly from the ends of the girder and from the weight to the points of con- trary flexure, as shown in plan of chords, Fig. 102, where the 270 MECHANICS OF THE GIRDER. depth of each chord is supposed to be uniform, and the varia- tion in size of chord attained by varying the thickness only. The deflection for each of the four semi-beams into which the girder now becomes divided, is to be found as in the preced- ing example, where the load was at the centre. We now have, for the part AB = f/, fixed at A, /i t = 180, h = h Q 22$, k, -j- k f, k /i, = 45 ; and (311) becomes 2 x 9000 x y == Dr = 25000000 x 45o 2 j / 5\) -^ \ 1 80 - 225 i - 2.302585 log- > 45 2 ( \ 4/) = - X 5.206 = 0.37486 inch. For the fourth part, EF \l, fixed at F] 7/ x = 300, h = h = 285, h, -f- h = f^, //o ?h 15 ; y = A = iFs^ 00 - 28 5 T + 2.302585 log = x 0.382 = 0.0275 inch. For the second part, BC y h = h = 225, A = -dhK 7 *' - 22 5 x 2.302585 log^ x + 225 x 2.302585^225 - 225). For the third part, CE, h k = 285, y = A = T*(* - 285 X 2.302585 log A x H- 285 x 2.302585^285 - 285). And since A + D 2 = D z + > 4 , we find h^ = 234.761 inches; that is, the lowest point, C, is now at the distance DEFLECTION, END MOMENTS, ETC., FOUND. 2/1 10(234.761 1 80) = 547.61 inches from the left end of the beam. Using this value of h u we find D 2 and D y and have finally, A = o-3749> A = 0-3622, D 2 = 0.0148, D 4 = 0.0275, Deflection at C = 0.3897 inch = 0.3897 inch; an apparently paradoxical result, since, when the same load, W = 200,006 pounds, was at the centre of the girder of uni- form strength, and having the same varying height and same length, / = 100 feet, the deflection was only 0.279 inch at the lowest point. The paradox vanishes, however, when we take into account the difference in the length of the component semi-beams for the two cases. Indeed, it may be easily shown that a girder of uniform height and strength, bearing a concen- trated load, both ends being fixed, deflects least when that load is at the centre, and the four component semi-beams are of equal length. Suppose that, in (307), we have, for First semi-beam, x = \a', according to article 100; second semi-beam, third semi-beam, fourth semi-beam, 2/2 MECHANICS OF THE GIRDER. Then, if u is half the sum of the four deflections, that is, if u = the total deflection of the beam, we have + W - a ) 2 ]. (362) XJJsfl, Put Therefore a' \l renders u a minimum, since 2a r is positive and / constant. In a similar manner, from (311), may the position of the load be found on the beam of uniform strength and uniformly vary- ing height, the ends being fixed, when it is required to know what position of a given load gives the least deflection. EXAMPLE 3. Continuous uniform load wl = 400,000 pounds upon the same girder, Fig. 102. Since the moments of the external forces are independent of the height, equation (350) applies here, giving for x = o, M! = -/$ X 400000 x 1200 = 45000000 inch-pounds; x = \l t M c = J- X 400000 x 1200 + M! = 15000000 inch-pounds; x = /, M 2 = MI. By equations (360) and (361), we find At left end, section of top chord, 415000000 S =10000 x 180x0.99503 = 2 S' I2 S mches ' At left end, section of bottom chord, = 3I . 2SO inches . 8000 X 180 X I DEFLECTION, END MOMENTS, ETC., FOUND. 2/3 At centre, section of top chord, S' = __ 15000000 - = 7 . 8S1 inches. 8000 x 240 x 0-99503 At centre, section of bottom chord, 15000000 S = xoooo x 240 X i = 6 ' 2 5 mches - At right end, section of top chord, 45000000 S = zoooo x 300 X 0.99503 = '5-075 At right end, section of bottom chord, 45000000 5 = 8000 x 300 x i = I inches - The deflection must be the same as in example i ; viz., D l + D 2 0.279 inch, since the centre of gravity of each of the two loads is at the same point, and the unit strain the same. 107. Beam of Uniform Strength and Uniformly Varying Height, fixed at One End, and simply supported at the Other. Since the position of the point of contrary flexure depends upon the moments due the external forces, which moments are independent of the height of the girder, we already have, in articles 104 and 105, the point of contrary flexure, and the moment at the fixed end, M 2 , for the present case of uniformly varying height, if the load be either concen- trated or uniform and continuous. The cross-section at any point is given generally by equation (332), the deflection of each of the three component semi-beams 274 MECHANICS OF THE GIRDER. by (311), and the equation D l = D 2 + D 3 fixes the lowest point. EXAMPLE i. Take a girder of the same varying height and same length .as in the examples of article 106, Fig. 102, of wrought-iron, but now fixed at the right end and simply sup- ported at the left ; that is, let E = 25,000,000, B^ = 9,000, C t = 8,000, T t 10,000, / = 1,200 inches, h Q = 180 inches = height at left end, //, = 300 inches = height at fixed end, tana = = o. i = tan of slope of top chord, cos a = 0.99503, tan/3 = o, cos (3 I, since bottom chord is horizontal. Let the load W = 200,000 pounds be at the distance a' = |-/ from the unfixed end, the point of contrary flexure .being at x = (/ + a f ) = \L Then, from (352), Moment at fixed end = M 2 = 200000- | = 32000000 inch-pounds. Moment at weight, M a f = 32000000 inch-pounds. Moment at left end, M I = o. At the left end, (360) and (361) give the chord cross-sec- tions = o ; but, of course, as before shown and exemplified for all such cases, the end must be enlarged to bear the shearing and crushing strains with permanent safety. At the load, (360) gives S' = 15.46 inches = section at top. At the load, (361) gives S" = 12.31 inches = section at bottom. DEFLECTION, END MOMENTS, ETC., FOUND. 2?$ At fixed end, (361) gives S" = 10.72 inches = section at top. At fixed end, (360) gives S' = 13.33 inches = section at bottom. Applying equation (311) to the three parts of this beam, AB, BD, DE, we find the deflection, Fig. 103, AB, fixed at B\ Ji Q 180, y = D, = T M/^ - 180 x 2.302585 (log^, - log 1 80) - 180]. BD, fixed at B\ h Q 280, y = D 2 = -rf^/*! 280 x 2.302585 (log hi log 280) 280]. DE, fixed at E\ h Q = 280, //, = 300, 7/ //, = 20, & = IS " 14 y = D z = yMsoo - 280(2.302585 log if + i)] = yfj X 0.681 = 0.049 inch. From the equation D 1 = Z7 2 + Z) 3 we have ^! = 229.731 inches, /. DI = 0.419 inch = deflection at lowest point, B. EXAMPLE 2. Take the same girder, with the same condi- tions, as in example i, except that the load is now wl = 400,000 pounds, uniformly distributed. 2 7 6 MECHANICS OF THE GIRDER. The moments are found by (356) ; thus, x = o, M = MI = o. 17 n/r 400000 X 1200 /:/-/-/-/-/: X = -|/, M = " = 26666666. Io x = /, J^ = 80000000 inch-pounds. PLAN, UNIFORM LOAD. FIG. 103. Using these moments, we find the required cross-sections, by means of equations (360) and (361), as follows : x = %l, section of top chord, S' 15.23 square inches. x = |/, section of bottom chord, S" = 12.12 square inches. x = /, section of top chord, S" = 26.80 square inches. x = /, section of bottom chord, S' 33.33 square inches. DEFLECTION, END MOMENTS, ETC., FOUND. 2JJ Deflection for the three parts, by (311) : FG, fixed at G ; // = 1 80 inches, y = D, = TtbC/Zx - 1 80 x 2.302585 (log ^ - logiSo) 180]. GH, fixed at G ; /i = 260 inches, y = Z> 2 = yf^-^1 260 X 2.302585 (log hi log 260) 260]. HI, fixed at /; h Q = 260, //, = 300, h /*, = 40, & = iS 4, 13' ^ = A = rMs 00 - 260(2.302585 log ^f + i)]. From A = A + Z> 3 , we find /z x = 226.547 at the point x = 10(226.547 1 80) = 465.47 inches, .*. Z^, = 0.37 inch at G, Fig. 103. SECTION 5. Camber. 108. Camber is the slight upward curving or crowning that is sometimes given to a girder, in order to obviate the sagging which would otherwise result from the deflection of the same girder made without this slight arching. The effect of camber is, therefore, to keep the track line straight under the working- load, and thereby prevent that increase of stress which would otherwise be developed by the falling and rising of loads mov- ing rapidly along a line originally straight. In no other respect 2/8 MECHANICS OF THE GIRDER. does camber augment the efficiency of the structure. Some- times, however, a greater upward curvature than that here con- templated is given to the floor line of highway bridges, as being more pleasing to the eye ; but so large a convexity, if effected in the girder itself, is always at the expense of material or of efficiency, as will appear from a comparison of the capabilities of two girders shaped like Figs. 23 and 80, of equal length and equal height between axes of chords. It is evident that camber may be given to the floor or track line in three ways : ist, The girders may be made in normal shape, and the floor or track line be raised sufficiently to counteract the deflec- tion due the total load. In this case the two chords of each girder will sag, while the cambered floor line becomes straight under load. 2d, The chord which carries the floor line may be cambered, while the other is built in normal shape. In this case the uncambered chord will sag, while the other assumes its normal shape under load. 3d, The girder may be so built, that, before the load is imposed, its proper floor line will have a deflection equal and opposite to the deflection due the total load, and that the whole girder will assume its normal shape under load. We need examine and exemplify only the second and third cases. 109. Change of Length calculated from the Unit Strain. If A z = total contraction for the original length / and X 2 = total elongation for the original length / 2 , of any strained member, we have, within the elastic limit where the amount of displacement per unit of length varies as the stress, DEFLECTION, END MOMENTS, ETC., FOUND. 279 For compressed member, XV T /T X X \ i = -jr j (364) for extended member, ^ = ^r 2 ; (365) / x and 7^ being the allowed unit strains, and E e and ^ the moduli of compressive and of tensile elasticity respectively. The total difference between the lengths of the two chords of a girder after deflection is, therefore, A = \ v + A, = (% + fiV (366) provided the chords were of equal length, /, before deflection, and of uniform strength. If an originally straight girder of equal and parallel chords take the circular form, Fig. 104, after deflection, the neutral line being midway between the chords, we must have for it, ... X = X, + A 2 = + = (368) E c E t E if E = \(E C -J- E}) = modulus of transverse elasticity, and BI = ^(C l + 7 1 ,) = bending unit strain. no. Elongation and Contraction calculated from Deflec- tion. Let ABCD, Fig. 104, represent an open-built semi- girder fixed horizontally at A and C, and having its deflection, D = NH, greatly exaggerated in the figure ; the actual lines NFM and HM being sensibly equal each to /, the original 280 MECHANICS OF THE GIRDER. length of the parallel chords AB and CD of the semigirder whose height is AC = h. FIG. 104. We may without appreciable error, for the present purpose, regard the deflection curve as circular. Take, as radii of curvature, r for neutral line NM, r + \h for the extended chord, r J/z for the contracted chord. Then, from the geometry of the figure, we have /* = D(2r - D), r = (369) since D is very small compared with r. DEFLECTION, END MOMENTS, ETC., FOUND. 28 1 Also, from the figure, r / r + \h I + A/ ( 37 o) which is the length to be given to the chord in compression, the figure being inverted. Again, r -\h I - A,' ... /-A, = /(y ;^, (37-) the length required for the chord in tension, the figure being inverted. Subtracting (371) from (370), we find the total difference in length required, X = X, + X 2 = ^=^, (37*) after eliminating r by means of (369). B I 2 It may be noted that (368) and (372) give us D = -~ k,h , which is equation (308) for the semi-girder of the length /. In (369) and (372), we must, of course, put J/ for /, when we apply these equations to a girder of the length / supported at both ends. in. Suppose, now, that it is required to camber only the straight, horizontal bottom chord, to which the moving-load is to be applied. This supposition includes Classes II., IV., VII., IX., X., and XII. of article 49. 282 MECHANICS OF THE GIRDER. We may, by the formula proper for the given girder, find the deflection at each panel point, or apex, of the bottom chord. If we now assume that no apices of the bottom chord are to be moved horizontally, by reason of the adopted camber, we must theoretically increase each normal panel length, c, of this chord, in the ratio = ; 8 being the inclination to the c cos/3 horizon of any panel length, c + A of the bottom chord when cambered, and A being the change of length in the bottom chord for any panel, by reason of the adopted camber. Also, tan p -, D! and D being the deflections at any two consecutive apices. FIG. 105. Then each vertical member, as FB, Fig. 105, coming down to a lower apex, B, must be shortened by the amount of deflec- tion, D lt computed for the apex ; and each diagonal or brace, BE v, terminating at the same apex, must be shortened in the ratio - = l v \ + h 2 (373) where h is the normal difference of level between the ends of any diagonal. DEFLECTION, END MOMENTS, ETC., FOUND. 283 In this case the effective depth of the girder at the centre has been lessened by D. EXAMPLE. Let us find the changes of length required to effect camber in the bottom chord alone of a wrought-iron para- bolic bowstring, where / = 2,400 inches, h = 225, n 16 panels of c ^^ = 150 inches each, C, = 6,698 pounds, T, = 10,000 pounds, B* = 8,349 pounds, and E 24,000,000 after the frame has taken its permanent " set ; " but, as ex- plained in article 90, we will, for the present purpose, take E = 16,000,000, to provide for any sagging that might other- wise be caused by the first full load, beyond what the elasticity of the frame can recover. By equation (319), putting \l = 1,200 for /, we have deflec- tion at centre, 1.386295 x 8349 x 1200* D = 16000000 x 225 - = 4.62 9 7 inches. Now we may, by using equation (31 8),* find the deflection at each panel point ; but it will be practically accurate, and more simple, to regard the cambered bottom chord as a parabola, having the central height D =. 4.6297 inches, and then find, by equations (136) and (137), both the normal heights, /i, and the height of each lower apex after camber is effected. Thus, (136) now becomes y - /i _ -^-\ x 4*6297 V 2 4 oo 2 / for bottom chord, and y [ i .4_ ) x 225 V 24007 for top chord ; the origin being at the middle of the bottom 284 MECHANICS OF THE GIRDER. chord in its normal shape. From these equations and (373) we compute D t z/, /, h, in inches. X h D AZ? h-D v-*w, VHT*r i/ z' 225.00 4-63 0.07 220.37 220.44 216.85 266.64 263.67 ISO 221.48 4.56 0.22 216.92 217.14 206.38 263.91 258.14 300 210.94 4-34 0.36 206.60 206.96 189.09 255.60 241.36 45 19343 3.98 -0. 5 I 189.45 189.96 164.77 242.04 222.82 600 168.75 3-47 0.65 165.28 165.93 I33-64 223.68 200.90 750 I37.II 2.82 -0.79 134.29 135-08 95.62 201.86 177.88 900 98.44 2.03 -0-95 96.41 97.36 5o-7i 174.76 158.34 1050 5 2 -74 1.08 -1. 08 51.66 52-74 Theoretically, the end panel lengths of bottom chord, where the inclination, /?, is greatest, would become (i5o 2 + 1.0852)5 = 150.0039 inches. But this is practically equal to the normal length, 150 inches ; hence we will not change the panel lengths of the bottom chord. It will be perceived that the girder thus cambered becomes the parabolic crescent. Instead of computing dimensions as above, it is evident that the elevation may be drawn accurately, on a large scale, from the central deflection Z>, and h and /; and then all desired lengths can be taken off as accurately as the work will be " laid out " in the shop. The camber curve may always be drawn circular for an originally straight chord. 112. Similarly, if we would camber only the straight upper horizontal chord of Classes III., IV., VIIL, IX., XL, XII., of DEFLECTION, END MOMENTS, ETC., FOUND. 28 5 article 49, without moving appreciably the upper apices hori- zontally, we must increase the normal length of each vertical FIG. 106. member by the deflection due at its place, and the normal length of each diagonal in the ratio See Fig. 106, where the efficient depth of the girder has been increased at the centre by the value of the deflection D. EXAMPLE. Let us invert the uncambered girder of article in, and effect the same amount of camber, D = 4.6297 inches, in the straight top chord alone. We have the same values of D and h as before, and readily find the required lengths of ver- ticals and diagonals, in inches, numbering from the centre. (374) . (375) 286 MECHANICS OF THE GIRDER. X h D h + D V^r+x i/ W-^r z' O 225.00 4-63 229.63 229.56 274.22 226.11 271.34 J 5 221.48 4-56 226.04 225.82 271.09 215-50 262.57* 300 210.94 4-34 215.28 214.92 262.08 197.77 248.22 45 I 9343 3.98 197.41 196.90 247-53 17273 228.77 600 168.75 3-47 172.22 I7I-57 227.90 140.58 205.58 750 i37-ii 2.82 139-93 I39-I4 204.59 101.26 180.98 900 98.44 2.03 100.47 99-52 180.01 54-77 159.69 1050 S 2 -74 i. 08 53-82 In like manner may we effect camber in a straight chord of any one of the classes cited in this and the preceding article. And, if it is required to preserve the normal height between chords after camber, we must change both. 113. When it is desired that the effective depth of the girder be not altered by the camber, then both chords must be displaced vertically by the amount of the deflection at the sev- eral apices, and in the opposite direction. In articles in and 112 we have made no appreciable change in the length of either chord by reason of camber ; and, of course, the length of each chord will be changed as the load takes out the camber. Strictly, regarding camber as the inverse of the operation performed by deflection, we should increase the normal length of the compressed chord by A x , equation (364), and diminish that of the stretched chord by X,, equation (365) ; but, since the whole change of length required in either chord is very small for each panel, we shall distribute, in the present case, the whole difference, X X x + A^, of length due to deflection additively among the panel lengths of the compressed chord. Hence camber thus produced will require no change in the normal panel lengths of the chord in tension, which, under the load, will resume its normal line, but increased in length by A 2 . DEFLECTION, END MOMENTS, ETC., FOUND. 287 At the same time, the compressed chord loses A, of its incre- ment, and retains X 2 . This change of length in either chord which rests upon the points of support, may, if necessary, be provided for in the same manner that provision is made for the effect of change of tem- perature. The length of any vertical member is not to be altered appreciably for camber in this case, since the vertical displacement of each chord is assumed to be the same for any given value of x ; and the slight change in their length caused by the spreading of the verticals to fit the change in the com- pressed chord, is hardly measurable. But the length of any diagonal member will be changed as follows : Let ABCE, Fig. 107, represent any one of the n normal panels of a girder, and A'B'C'E' the same panel when cambered B' FIG. 107. by adding - to - =1 c, the horizontal projection of the chord n n AB in compression. The panel points in both chords are dis- placed vertically by the deflections D r CC r == BB', and D r + x = EE f AA f , appreciably ; and the points A and B 288 MECHANICS OF THE GIRDER. are removed horizontally by the space -. Hence practically we have E'C' = EC, A'B' = n A'E' = AE, '= BC, E'F = -, ^C'=- n n /3 being now the inclination of EC or E'C' to the horizon. = |/ + AY + /^. _ ^tanjSVj 4 , (376) - (377) Instead of -, we may evidently employ equations (364) and n (365) in finding the proper increment for any panel length of compressed chord. EXAMPLE. Take BC k r = 240 inches, AE = h r + x 200 inches, ^/^ = - = c = 150 inches, - = ^A 0.12 inch, ?2 A = 0.06 inch, tan ft = l - X 24 ~ 2 - - Then 2;z 2 150 15 ^'^= AB + 0.12, 0'= [(i5o.o6) 2 -f (240 - 150 x A) 2 ]* = 266.304 = if = [(i5o.o6) 2 + (200 + 2o) 2 ]5 inches. PILLARS. 289 v CHAPTER VII. / PILLARS. SECTION i. Strength of Pillars, by Rational Formula. 114. Under the general term pillars we shall include col- umns, posts, struts, props, braces in compression, and, in a word, every member in a structure whose function it is to resist compressive force applied at its end, and, in general, in the line of the longitudinal axis of the member. It is assumed that a pillar has no lateral support or pressure applied between its ends, except when, owing to an unavoidable existing lateral force (as, for example, the weight of a horizontal strut), a counter-force is applied as a balance. But a pillar may have its ends in the conditions known as round, hinged, flat, imbedded, fixed ; the two ends being in the same or in differ- ent conditions. Pillars may be long or short, solid or hollow ; may have a uniform or variable cross-section of any desired form. Long pillars yield chiefly by bending and breaking across ; short blocks of ordinary building material yield by being crushed without bending, properly so called. At what exact ratio of length to diameter bending first takes place in a given material, is not at? present very definitely ascertained ; but it will be safe to assume, in the present state of our knowledge, that bending will occur when this ratio is as low as three for such 290 MECHANICS OF THE GIRDER. material as wrought-iron. Experiment has shown what, per- haps, we might have inferred from a stalk of wheat, that material is saved by using hollow instead of solid pillars to .-support a given load. 115. Pillars of Uniform Cross-Section. Let / = length of pillar, h = least diameter, r = least radius of gyration of cross-section, 5 area of cross-section, D greatest deflection of pillar ; ;all in inches. E = modulus of transverse elasticity, C = crushing-strength of standard specimen of the ma- terial, P = breaking-weight applied at the end of the pillar and in the line of its axis before deflection, p Q = = breaking-weight per square inch of cross-section ; vj ;all in pounds per square inch. ./ = Sr 2 = least moment of inertia (so called) of cross-section. M x = moment of forces developed in any normal cross- section by the given load P. M l = the end moment at the lower end when that end is fixed. M 2 = the end moment at the top when the upper end is fixed. Suppose the pillar vertical, Figs. 108, 109, no, and take the origin of rectangular co-ordinates at the lowest point of the pillar's axis, which call also the axis of x ; that of y being hori- zontal. Then, from equations (15), (93), and (187), we have the moment at any height, x, It, = -EI = ' ~ >x -M I + fy, (378) PILLARS. 291 wherein no account is taken of the modified condition of every cross-section due to the longitudinal pressure, Q, per unit. Now, since the full unit strength of the cross-section of the unloaded pillar is C, and the remaining unit strength of the loaded pillar's cross-section is (C 0, it follows that the expression for the moment of the internal forces developed in any cross-section must be diminished in the ratio 7^- We then have M x = -P?- = ^ -x - Mi + Py (379) ctoc^ c if 2 = EI(C - = Ef*(C - PC QC There will be three cases, according as we consider neither, both, or one only, of the ends fixed. CASE I. If neither end can produce any moment, M I = M 2 = o ; and we have, from (379), e ^ = _y. ( 3 8o) Multiplying by 2dy, dvd Integrating this equation, and putting a 2 for the arbitrary constant of integration, from which 292 MECHANICS OF THE GIRDER. Integrating again between the limits, for x y o and /; and for y, o and o ; - = :r, where n may be any whole number ; but, in order that P may have the least value it can have, consistent with the bending of the pillar necessarily assumed in establishing equations (378) and (379), n must be equal to unity. (See Rankine's "Applied Mechanics," p. 352.) - Q Cl 2 (381) FIG. 1 08. which is the formula for pillars with rounded ends that can generate no end moments, Fig. 108. The curved line shows the deflected axis. CASE II. If both ends of the pillar are equally fixed, Fig. 109, so that the elastic curve at each end, after flexure, has for its tangent the original undeflected axis, th'en, in equa- tion (379), M, = M 2 , whence *.-*-* (382) Multiplying by 2cfy, equation (382) becomes PILLARS. 293 Integrating, we find dx 2 = 2M,y - Py* + a, (383) where a, the arbitrary constant, must vanish, since -f- = o when dx y = o. Hence, from (383), Integrating again, with the condition that y = o when x o, there results, after can- cellin ! ^r-> (393) k, being the pillar's internal diameter, and A " representing the strength of a pillar I foot long, and I inch in diameter, and being a constant for a given quality of iron, but ranging in value, for different irons, from 75,000 to 112,000." The mean values of A adopted by Professor Rankine are, Solid pillars with rounded ends, A = 14.9 tons = 33376 pounds; solid pillars with flat ends, A = 44.16 tons = 98918 pounds; PILLARS. 297 hollow pillars with rounded ends, A 13 tons = 29120 pounds; hollow pillars with flat ends, A = 44.3 tons = 99232 pounds. It hence results experimentally that "fixing" both ends of a pillar, Fig. 109, enables it to support about three times the load which would break it were the ends unfixed, Fig. 108, and incapable of developing moment. For a pillar fixed at one end and rounded at the other, Fig. no, Mr. Hodgkinson found the strength to be a mean between the two strengths of the same pillar when both ends are rounded and when both ends are flat. We then have, for cast-iron pillars, Solid, one flat and one round end, A = 66147 pounds; hollow, one flat and one round end, A 64176 pounds. When the length is less than 30 or 15 times the diameter respectively, Mr. Hodgkinson first finds P by equations (392) and (393), and then corrects P by means of this supplementary formula ; P l being the corrected value sought. _ PCS CS (395) 298 MECHANICS OF THE GIRDER. which is an empirical equation identical in form with (381), (385), and (391), analytically established. 117. The Hodgkinson formula for the ultimate resistance of pillars of oak and of red pine to crushing by bending, as adopted by Professor Rankine, "Applied Mechanics," p. 365, is, with our notation, article 115, <2 = | = so cj, ( 39 6) a formula to be used only when Q < C, the crushing-strength of the material, Table II., article 60. Applications of the Hodgkinson formulae are given in tables of article 121. SECTION 3. Gordon's Empirical Formula, with Rankings Modification. p 118. We have, in article 115, Q = the direct unit o pressure of the load upon every cross-section of the pillar. Now, if B l is the additional unit pressure due to bending- moment upon those fibres where the bending-moment is great- est, and if f denote the greatest intensity of unit pressure, we have / = <2 + ^. (397) Regarding, with reference to the central moment, a loaded pillar of uniform cross-section as in the condition of a beam supported at both ends, and carrying the central weight W = -j> since equations (15), (46), and (187) give us M = PD = \Wl = 4 (398) PILLARS. 299 we find Wl = 8^,7 = 4&g/P h / 2 ' from (211); ~ ~ULti T> From which, for a given value of :, But (398) gives Bi - (399) y o/& if / = kk 2 S, k being a constant depending upon the form of the pillar's cross-section (see Table III., article 62) ; /. Bi ~ ^ Whence (397) becomes f and a being constants to be determined by experiment ; /. - = L , (400) which is of the form "proposed by Tredgold," and is now known as the " Gordon Formula," having been, after some "disuse, revived by Mr. Lewis Gordon, who determined the values " of a and /, for certain materials, from the results of Mr. Hodgkinson's experiments. 3OO MECHANICS OF THE GIRDER. 1 19. If, in equation (399), we put Sr 2 for /, using still the notation of article 115, we find PDh P/ 2 '~ (40I) Therefore, from (397), (402) which is Professor Rankine's modification of the Gordon for mula ; r being the least radius of gyration of the cross-section. The Gordon (400) and the Rankine (402) formulae are iden tical if we make i2o. Supposing f to be constant for varying conditions of the pillar, both a and a l will be found to require different co- efficients, according as the pillar has neither, one, or both, of its ends fixed. Assuming that equations (400) and (402) apply to a pillar fixed in direction at both ends, Fig. 109, we see that the length, /, between the points of contrary flexure, is in the con- dition of a pillar not fixed at its ends, and has only the strength of a pillar of twice its length, 2/, fixed at both ends ; that is, for a pillar rounded at both ends, we have, Gordon's formula, -r-k y (404) PILLARS. 301 Rankine's formula, P = ^ . (405) S i + Similarly, in Fig. 109, the length, /, between either point of contrary flexure and the remoter end is in the condition of a pillar with one fixed and one rounded end, and has only the strength of a pillar f / in length. We have, then, for a pillar fixed at one end and rounded at the other, Gordon's formula, _ L _ (406) S Rankine's formula, (407) This is Mr. Hodgkinson's ingenious explanation of the vari- ation among the strengths of these three classes of pillars, a variation which he discovered by a comparison of the results of his experiments. If we invert the three numerical co-efficients of the fractions in the denominators of (400), (404), (406), or (402), (405), (407) (viz., i, 4, J^-), and multiply the inverted numbers by 4, we have the relation, 4, i, 2.25 ; while 4, i, 2.28, is the relation of the corresponding constants in equations (385), (381), (391), deter- mined analytically. We may hence infer that the degree of approximation in (391) is close to the true value. Especially, since we can seldom tell the exact amount of influence which given end bearings exert, may we regard (391) practically cor- rect. 302 MECHANICS OF THE GIRDER. VALUES OF / AND a A., American Bridge Company TABLE IV. OF THE GORDON, AND OF/ X AND a l OF THE RANKINE FORMULA. , Chicago, 111.; K., Keystone Bridge Company, Pittsburgh, Penn. Material. Form of Section. Experi- menters. Authority. Gordon Formula. Rankine Formula. / a A * Iron, Cast . . Hodgkinson. Gordon. 80000 400 - - Iron, Cast . . O Hodgkinson. Gordon. 80000 267 - - Iron, Cast cjja Hodgkinson. Gordon. 80000 133 - - Iron, Wrought . Hodgkinson. Gordon. 36000 3000 - - Iron, Wrought . ft Hodgkinson. Rankine. 36000 3000 36000 36000 Iron, Wrought . ei Hodgkinson. Stoney. 35840 3000 - - Iron, Wrought . Hodgkinson. Stoney. 30660 3000 - - Iron, Wrought . Hodgkinson. Stoney. 40032 3000 - - ' Iron, Wrought . LT E^ Davies. Unwin. 42560 900 - - Iron, Wrought . = O A. K. Lovett. 49580 3000 42980 36000 Iron, Wrought . n A. K. Lovett. 43725 3000 38650 36000 Iron, Wrought . 3C A. Lovett. 38271 3000 37029 36000 Iron, Wrought . !3 K. Lovett. 36523 3000 33531 36000 Steel, Mild . . - Baker. 67200 1400 - - Steel, Strong . - Baker. 114240 900 - - Steel, Mild . . ' - Baker. 67200 2480 - - Steel, Strong . (C - Baker. 114240 1600 - - Timber . . . tt Hodgkinson. Rankine. 7200 250 - - Oak and Fir . Rondelet, Stoney. 1.5 C of Table II. 250 1 . Stone and Brick, (I Rankine. Cof Table II. 600 - - PILLARS. 303 SECTION 4. Strength of Pillars computed by the Preceding Formula, and compared with the Strength experimentally determined. 121. The following tables, V., VL, VII., VIIL, IX., X., XII., contain data derived from experiments on the strength of pil- lars, probably as trustworthy as any yet made and published. To these tests the appropriate formulae, either direct or in- verted, have been applied ; and the values of /, C, or Q for a given pillar, computed by different formulae, have been tabu- lated in the same horizontal line. In Table XI. no experimental values are given, but the assumed values of E and C are within the limits fixed by experiments upon steel. In Table VII., when the thickness, /, of the metal is less than a fifty-fifth part of the least diameter, k, of the pillar, the computed value of Q, the breaking-weight, in pounds, per square inch, has been diminished in the ratio -7-, as seems to be required by the test^. For columns having rounded or hinged ends, in Table V, the formulae for those having one flat and one round end have been used, as more in harmony with the tests than the formulae for columns having no end moments. It must be confessed that there are anomalies of consider- able magnitude in the experiments themselves ; and, of course, there appear corresponding variations from the test values in the numbers computed according to the laws of the applied formulae. It is to be regretted that we have not, accompanying these tests for Q, also experimental determinations of C and of E, for each pillar tabulated, but have been obliged to use probable mean values of C in all the calculations of Q, and probable mean values of E in all but Table V. 304 MECHANICS OF THE GIRDER. OT CINCINNATI >f an I-beam; :d. - Hf " M t^O N -*t-00 00 er> ID * Oi * O O\\O 00 IDOO ID O> t~.\O x-> "^ ^ el 3jji| ! ^, Modulus of Transverse Elasticity. .2 H D tx H -^- O ONVO t^ ID H CDvO ID t^CO VOVO COMO^ O^HTf'Ot-'iDOfD "(MNWNMrONN NMMMNWNMrO cTrON Sc?rO(McTN^f')S S ui J3 d iit ^ s ll >- 03 3 " J, ID ID IDVO OO ID fOOO MOO -^ -* -h H mvQ CD ^- -4- tv O^OO O< fO f o d\ o\ c>i d>cd oo d< c> c> t>- c>od co o o o oo & 6 a u 2 y ^ 111 J^fc?!f88^ W ?!?l! W *^n WgjrsJ! , o | I W J a W Q < 1 7 + *" rv Tf-vo vo N ^mwco t^r^^j-o c^ow tx o O t^woo t^ H ** ooo mo OON WiOd ron O iO moo moo lOtx.rOiOtOt-x^HVO "<* rovO IN O f^> oT & H* Oi I ng- Weight, in Lb 'H H rt 0* fl II S 1 PQ 4 OOHtOONHHCOOHlOHNvOrOO --O>OO O O O>OO O O O c/2 .^OiOOOOOtO^OOO "HHNINHI-iTj-lOtOeMHMMl-IHl-irOrO-^CgHMl-IHCSMqHM i .3 ^- ^- ,j- ^-00 00 00 00 OO * 00 00 00 CO Tj- CO CO 00 CO Tj-CO Tt-00 * -*OO OO 00 00 ; l yj ^j-o ^hioiow OvO txiON IOO O ^"OvO t^vo OOtoO^"ONO*oO C O CNI OO ONOO CO ON W -^- ON rOCO IO t^ O *O CM rovO CM f^OO IO O CM rO O CO lO 10 o * rooo toco ro ro ro oco 100 ioioONt>.-.o6 cJ H M ro ci 6 ro ^- c^.od M ei M ro d H M d M" ro i M CM ro TJ- iovo r^oo ON O H CM ro <*- ovo t^-oo o- O n N ro iovO txoo ON 1010101010101010 *OvO vOvOvOvOvONOvOvOvO t^txt^C>Cxt>.txtxC^t^ PILLARS. 307 TABLE VIII. HOLLOW CYLINDRICAL PILLARS OF WROUGHT- IRON. Ends flat and well bedded. Hodgkinson's Experiments. DATA FROM BINDON B. STONEY'S "THEORY OF STRAINS IN GIRDERS AND. SIMILAR STRUCTURES." h z 8r 2 . No. 5 h l + h l + r h + t Q = Breaking- Wt., in Lbs., per Sq. Inch, Sec- tional Area. Diame- ter. Ratio of Length to Diame- ter. Ratio of Length to Radius of Gyration. Ratio of Diame- ter to Thick- ness of Metal. * By - Experi- ment. Gordon Formula, (?= _ J 00^0_ Equation (385), 1 + - 30oo 2 I4 V*H sq. ins. ins. 80 0.444 1-495 , 80.00 226.274 15.00 M673 12782 12873 81 0.610 1.964 60.00 172.816 18.80 23206 18204 18127 82 1-435 2.340 51.28 145.042 10.80 22179 21342 21810 83 1.605 2.350 51.00 144.250 9.70 21572 21451 21926 84 0.804 2.490 47.80 I35.I99 23.27 29798 22735 23289 85 0.444 1-495 40.00 "3-137 15.00 31180 26120 26914' 86 i-35 3.000 40.00 112.877 20.00 27671 26120 26959 87 0.610 1.964 30-50 86.267 18.80 33299 30571 3^745 88 1.414 3-035 29.60 83-721 18.00 29789 30997 32212 89 1.707 4-050 29.60 83.721 29.00 27657 30997 32212 90 1.900 4.060 29.60 83.721 26.10 26263 30997 32212 9* i-37i 2-335 25.70 72.690 11.40 29998 32823 34219 92 1.472 | 2.350 25-50 72.125 10.60 29330 33024 34321 93 0.804 2.490 24.10 68.165 23-27 35ioo 33554 35025 94 1.613 4.052 22.20 62.791 30.90 33331 34399 3596i 95 2.879 4.000 22.20 62.791 16.50 26046 34399 3596i 96 2.897 4.000 22.20 62.791 16.00 26503 34399 3596i 97 2-837 4.000 22.2O 62.791 16.50 27816 34399 3596i 98 0.804 2.490 21. OO 59-397 23-27 36489 34917 36536 99 0.444 1-495 20.00 56.569 15.00 34220 35338 37004 100 1.800 6.180 1940 54-87I 65.00 33375 35586 37280 101 2.540 6.360 18.90 53-334 49.00 35985 35789 37524 1 02 0.610 1.964 I5.30 43-275 1 8.80 36980 37 I 5 I 39028 103 2.895 3-995 15.00 42.426 16.30 30024 37256 39*45 104 2.848 3-995 15.00 42.426 16.50 34453 37256 39M5 105 2-547 6.366 14.10 39-88i 48.90 41664 3756i 39487 106 1.407 2-343 12.80 36.204 II. 10 38214 37976 39953 107 1-435 2-335 12.80 36.204 11.40 36639 37976 39953 108 1-435 2-335 12.80 36-204 11.40 35389 37976 39953 109 1.651 2.383 12.50 35-355 9-70 33107 38067 40055 no 1-358 2-343 12.30 34-790 11.60 39569 38127 . 40030 III 1-554 2-373 12.20 34-507 10.27 36906 38i57 40155 112 1.799 6-175 9-70 28.075 61.10 38355 38832 40853 "3 1.414 3.000 9-30 26.305 19.60 37392 38928 41023 114 2.845 4.000 7.00 19.799 16.00 47844 39406 41182 "5 2.850 4.026 6-95 19-657 16.00 48567 39415 4 J 573 116 1.799 6.125 4.90 13-859 62.50 41361 39732 4i93i Assume E = 24,000,000. C = 42,290, f = 40,050, Means. 308 MECHANICS OF THE GIRDER. TABLE IX. SOLID CYLINDRICAL PILLARS OF CAST-IRON. Ends flat and well bedded. Hodgkinson's Experiments. DATA, AND PER CENT OF VARIATION FROM Q, BY THE HODGKINSON AND GORDON FORMUUE, TAKEN FROM WlLLIAM E. MERRILL'S " IRON TRUSS BRIDGES FOR RAILROADS." h z = i6r 2 , E = 12,215,000, C = 109,801, y= 80,000. /+* h Q Variation from Q, per cent, by Gordon's For- No. Ratio of Length Diameter. Breaking- Weight, by Hodgkinson's Formulae, Equations (392), mula, Equation (400), 80000 Equation (385). to Diameter. Experiment. (394). ins. Ibs. per sq. in. 117 4 0.520 107674 - 6 29.000 - 3-641 n8 8 0.500 88964 - 3 21.000 + 0.085 119 10 0.777 67502 +13 4-000 +19.227 120 13 0.768 55959 +13 O.OOI +21.444 121 IS 0.500 57321 + 2 II.OOO + 5-267 122 15 0.785 50182 +U + Won +20.243 I2 3 15 I. 000 51248 + 9 + 7.000 +I7-74I 124 20 0.500 45485 i 13.000 - 1.758 125 20 0.775 45596 i lo.ooo - 1.998 126 20 1.022 38770 +12 + 5-000 +I5-257 I2 7 24 0.500 36644 + 2 II.OOO - 3-291 128 26 0.777 32860 + \ 9.000 - 3-503 I2 9 30 O.5IO 33"i 10 25.000 -22.497 I 3 30 1. 010 25350 + 5 3.000 + 1.231 131 39 0.770 18921 - 8 13.000 11.284 132 39 1.560 I5I53 + 6 +11.000 +10.777 133 40 0.510 18749 2 13.000 -14.241 134 47 1.290 12291 - 3 + \ 1.261 135 61 0.500 8464 + 6 7.000 10.881 136 61 0.997 7990 + \ 2.000 - 5-581 137 79 0.770 5274 + 2 8.000 -12.286 138 119 0.510 2384 +19 7.000 12.416 Equation (392), as used in Table IX., is P = 98922. See " Iron Truss Bridges for Railroads," p. 26. For E = 12,215,000, see Stoney's "Theory of Strains," p. 180. PILLARS. 309 TABLE X. SOLID CYLINDRICAL PILLARS OF CAST-IRON. Ends rounded. Hodgkinson's Experiments. DATA, AND PER CENT OF VARIATION FROM Q, BY THE HODGKINSON AND GORDON FORMUUE, TAKEN FROM WlLLIAM E. MERRILL'S " IRON TRUSS BRIDGES FOR RAILROADS." h z = i6r 2 , E = 15,268,750, C = 109,801, / = 80,000. M h Q Variation from Q, per cent, by No. Ratio of Breaking- Hodgkinson's Formulae, Gordon's For- mula, Equation, Length to Diameter. Weight, by Equation (392), 2 = _8oooo_ Diameter. Experiment. P *m7Q T J_ I+ ^ 3379 dV)' 7 looA* ins. Ibs. per sq. in. i39 8 0.500 76939 -25 -34 18.269 140 10 0.770 49280 - 7 -i 8 + 2.8 77 141 13 0.760 38590 -IS -25 4-203 142 15 0.497 27124 + ii +"735 i43 15 0.990 25660 + 10 - 6 +18.110 144 20 0.760 20331 -13 21 4-633 i45 20 1. 010 19642 - 9 _I 9 - 1.288 146 20 1.520 17928 + 3 10 + 8.149 i47 23 1.290 13187 + 5 - 7 +16.175 148 26 0.767 14289 -23 -29 -I3-472 149 3 0.500 9697 -13 -19 1.464 15 30 0.990 7931 + 9 2 +20.477 151 31 1.940 7717 +13 - 3 +16.599 152 31 1.960 8051 +14 - 3 +"763 i53 34 1-765 6360 + 5 10 +19.262 i54 34 1.780 7058 + 6 IO + 7-467 i55 39 0.770 5854 - 5 -17 + 0.137 156 39 1-535 5755 + * -16 + 1-859 i57 40 1.520 5985 - i 14 6.650 158 47 1.290 4367 _ -18 6.000 i59 47 1-295 4149 ~~ 3 -18 i. 060 160 61 0.500 2745 - 5 -23 - 9.872 161 61 0.990 2471 + 8 -16 + O.I2I 162 79 0.770 1675 + 2 -24 II.I05 163 121 0.500 728 + 10 -25 -12.083 3io MECHANICS OF THE GIRDER. E C Eq NIOOO- O M rotxosvo r-NwO>-4U">O rOHNVOtxCO 1000 ^-miH t^ Wior^ot-tvo 000 Ooo rooo O ow oiotx 8s ? M OJ s S s llsf g ' *rl a 8 33333 ooooo K ff ff H M 333333333 bJOMMMbCbOMMbJ) CCGCCCCCC 22SS52252 OO VO t^ PILLARS. TABLE XII. . SOLID SQUARE PILLARS OF PINE. DATA FROM BINDON B. STONEY'S "THEORY OF STRAINS IN GIRDERS AND SIMILAR STRUCTURES." k z = i2T 2 . Take E = 1460000, C = f = 5000. l + k Q = Breaking- Weight, in Lbs., per Square Inch. No. Ratio of Length to Rondelet's Brereton's Tests. Gordon Formula, Hodgkinson's Formula, Equation Equation (391). Equation (385). Least Diam- eter. Propor- tionals. Flat Ends. Ends in Ordinary Manner. I+ 2loA2 SooC/2 No End Moment. One End fully fixed. Both Ends fully fixed. h z 184 i 5000 _ _ 5000 _ _ _ 185 12 4167 - 3176 5000 3126 3959 4349 1 86 24 2500 - 1513 4940 1471 2437 3126 187 36 1667 - 809 1929 782 1485 2135 188 48 833 - 489 1085 462 960 1471 189 60 4 J 7 - 325 693 313 660 1076 190 7 2 209 - 230 483 221 478 642 191 10 - 1867 3571 5000 3530 4228 4529 192 20 - 1789 1923 5000 l8 7 6 2889 3530 193 30 - 1400 1087 2777 1053 1891 2581 194 40 I 1244 676 1563 653 1273 1875 312 MECHANICS OF THE GIRDER. CHAPTER VIII. PROPORTIONS AND WEIGHTS OF ALL THE MEMBERS OF A BRIDGE EXCEPTING THE GIRDERS PROPER. 122. The Floor. Let / = length of floor, in feet. q = breadth of floor, in feet. t = thickness of floor, in feet. u weight of one cubic foot of the material, in pounds. .'. Volume of floor = Iqt cubic feet = o.o 1 2lqt thousand feet, board measure. F weight of floor = ulqt pounds. (408) 123. The Joists, Longitudinal. / -7- n = length of joist in each panel, in feet. d = depth of joist, in inches. b thickness of joist, in inches. n number of equal panels. g = distance between centres of joists, in feet. q -r- g = number of joists in any panel, each of the two out- side ones having the thickness \b, and being counted as one-half a joist. nc l ~^~ g whole number of joists in the n panels. L = panel weight of uniform load, in tons. x = weight of one cubic foot of the material, in pounds. PROPORTIOA T S y ETC., OF MEMBERS OF A BRIDGE. 313 Weight upon the joists of one panel = h 20ooZ pounds, n * tf ultg 2 ooogL Uniformly distributed load on one joist = 1 pounds. Add weight of joist itself = - pounds. Total uniform load for each joist is, therefore, ultg 2OoogL bdfai I where w is the number of pounds per linear foot to be sup- ported by one joist. Now, by equation (52), we have for the external forces, greatest moment at centre, //\ 2 Iw I ul 2 tg i M = = X =-+-- + foot-pounds ; and for the internal forces of a rectangular beam, equation (160), the moment of resistance is R = \Bbd z inch-pounds = -faBbd 2 foot-pounds. Introducing /, the factor of safety, and equating M and R -7- f, we find ul z tg (409) Taking the value of B from Table II., and assigning a value to b or d, we may find, from (409), the required depth or thick- ness of each joist. 3 14 MECHANICS OF THE GIRDER. If we neglect the weight of the joist itself, which omission the factor of safety may well warrant, the last term in (409) vanishes, and we have at once Thickness of joist = b 2 (uqlt -f 2OooZ). Depth of joist = d = \ -^-(uqlt -f 2OOO*)l*. ( tfqbB j / = weight of (nq -*- g) joists = i ' pounds. (410) In a similar manner may the dimensions and weight of any other joist or beam or stringer be found ; that is, by equating the greatest moment due the external forces acting on the beam, to the greatest allowable moment due the internal forces resisting. 124. The Wrought-Iron I Floor Beams, Transverse, supporting the Joists, Floor, and Load. Let d z = depth of beam, in inches. dj. depth of web, in inches. d 2 di = depth of two flanges, in inches. b 2 = breadth of 'one flange. b 2 bi = thickness of web. q T = entire length of beam, in feet. 5 = cross-section of beam, in square inches. n i = number of beams in bridge. m = weight of one cubic inch of wrought-iron, in pounds. F + J + 20OonL D = = uniform load on any beam, in pounds. Then, by equation (52), Moment of external forces = M = \Dq^ inch-pounds. PROPORTIONS, ETC., OF MEMBERS OF A BRIDGE. 315 And, from equation (161), Moment of internal forces = R = B ^ b * d * ~ b * d ^ inch-pounds. 6d 2 Whence, introducing / as the factor of safety, Let us take now the dimensions of the cross-section of a well-proportioned I-beam, as, for instance, d 2 15, d* = I2j, bi 5|, bi = 4|, and express the relation Therefore (411) becomes /. 4 = 3.80122 Area of section = S = bA , or inches, J computed. m = weight of one cubic inch of wrought- iron. U = weight of horizontal struts = \2qrn (n + i)S t . (415) X = weight of horizontal diagonals = ^SmnS 2 i/ q 2 -\ -. (416) 126. Finally, there should be added whatever weight of wood or iron is not included in the foregoing specifications, but is employed in the actual completion and equipment of the struc- ture. Call this weight p pounds to the panel ; then we have Y = weight of residue = np pounds. (4 1 ?) 127. Take K = weight of bridge exclusive of the girders, in pounds ; then K = F+J + P + U + X + Y pounds. (418) And if G weight of girders, in pounds, Weight of bridge = 2OQQnW = K + G pounds. (419) MECHANICS OF THE GIRDER. CHAPTER IX. OPEN GIRDERS WITH EQUAL AND PARALLEL STRAIGHT CHORDS. CLASS IX. SECTION i. The Pratt Truss of Single System and Uniform Live Load. Wind Pressure. 128. Strains in Terms of the Structure's Unknown Weight. Let Fig. 1 1 1 represent a girder, or built beam, having a discontinuous or open web, and its flanges or chords A abcdefahB = inclination of diagonals to chords. nh I ' \lr 2 -4- /7 2 \//-2 -f k 2 = length of a diagonal. sin cos < cos 'c 2 -f ^ 2 = 1 / + h 2 = -V/ 2 +- n 2 h 2 length of a diagonal. Assume the entire weight of the structure supported by the girder, including the girder's own weight, to be uniform through- out, and equal to n W tons applied at the lower joints ; viz., \ W at C, \ W at D, and W at each of the other n i joints, apices, or panel points, a,, b c etc. W is called the panel weight, or apex load, due to the permanent weight of the struc- ture. Total pressure at C or D = \n W = resistance of pier to the permanent load. Assume also a uniform moving-load, nL, advancing by apex loads, Z, from left to right, upon and over the girder. We then have total weight = n( W + L) tons ; weight at each apex = W + L tons when fully loaded. With these data, we proceed to find the greatest strains developed in each member of the girder by the permanent load n W, and the uniform moving-load nL. 32O MECHANICS OF THE GIRDER. _ I _ (a) To find the moment at each joint due the entire weight n(W -f- L)> and thence the horizontal strain in chords by equation (95). H M -f- h = moment divided by height. Equation (65) applies here if for W we put W -f- L, and we have (W , - ^ -(n - i) X i, (W -f L)l, .'. H a = - - (n i) x i = strain on Aa, a A ; , Mb = tz~(* - 2 ) x 2 > .'. Z?3 = - ^^ - ( 2) X 2 = strain on ab, b l c l ; (W +L)l ^c = - - ^-^( n - 3) X 3, '* Hc = - ~ - { n ~ 3) X 3 = strain on be, c,d, ; Md = - - ^ n ~ 4) X 4, (W + Z)/ (n 4) X 4 = strain on cd, d 1 e 1 \ M h = ( - V -^\n -(-!)]( -i), (W + Z)/ .*. Zfo = ^^ [ (# i)] ( i) = strain on hB, g^h^ ; where H is the greatest horizontal strain, in tons, at the suc- cessive joints ; the strain on each chord being assumed to act at the centre or axis of the chord, whose depth is small com- pared with h. GIRDERS WITH EQUAL AND PARALLEL CHORDS. $21 (b) To find the compression on verticals, and the tension on diagonals, due to permanent load, nW, alone. From equation (65), dividing by h, and from the formulae for Class IX., article 49, H A = o; W7 H a = L(n - i) X i, 2/2 n Wl = H a HA : (n i) = hor. component of Aa^^ ; H a = - (n 3) = hor. component of ab l ;, .'. H c Hb = - (n 5) = hor. component of bc l ; 2.nn TT rrt' r / \ -i / \ * HB Hh - (n i) I = hor. component Therefore, from the triangle of forces, equations (3), the vertical components are n h I .'. ZA = \W(n i) = compression on AC or BD, Z a = \W(n 3) = compression on aa t or hh lt Zb \W(n 5) = compression on bb^ or gg lt ZB = \W(n i) = compression on BD or 322 MECHANICS OF THE GIRDER. And the strain Y along any diagonal is &H -4- COS <, or ~ - ^ * . + 72 2 ^ 2 Z -i- sin 6 = - nh = (n i) = tension on Aa l or .%, W Y a = : (n 3) = tension on ab l or hg l9 2 sin d> = : -(n 5) = tension on bc^ or gf t , 2 sm (w i) = tension on ^ x or (c) Maximum strain on verticals and diagonals from moving- load, nL, alone. To find this strain Z L , we subtract equation (64) from (68), divide remainder by h for greatest difference of horizontal strains at adjacent joints, and multiply the quotient by tan < ; thus, after putting Z for W, the difference between (68) and (64) is X r(r H- i) = maximum &Hh (say), x 2. x r(r + i) = ^ X lill^L), ( 422 ) h 2n 2 n 2 where r is the number of apex loads on the girder as the mov- ing-load advances, and Z L is the compression on the (r + i) th vertical ; GIRDERS WITH EQUAL AND PARALLEL CHORDS. 323 Z& = - X i = compression on bb lt r i ; n T Z c X 3 = compression on cc ly r 2 ; n Zd = X 6 = compression on */ is y = Z L -v- sin<; (423) Yb = - X i = tension on aj, r = i ; Y c = - X 3 = tension on bj, r = 2 ; X 6 = tension on c^d, r = 3 ; x 10 = tension on d t e, r 4 ; L (n i)n YB = . , x - = tension on h^B, r = n i. (d) Combining the strains due n W and nL, and, for conven- ience, writing N for -^ Z)/ , we find, for any number of panels : 324 MECHANICS OF THE GIRDER. MAXIMA STRAINS IN PRATT TRUSS. (a 1) N. 2 (n-2) N. 3 (n-8) N. i (n 4) N. (u-1) N. 2 (n-2) N. FIG. 112. 3 (n-3) N. UNIFORM DEAD AND LIVE LOADS. Loads applied at lower joints : W =. panel weight of dead load. L = panel weight of live load. / length of truss from centre to centre of end pins. h height of truss from centre to centre of pins. n = number of panels. nh V/ 2 -f n 2 h 2 N = W 2nh 2 tan 129. Weight of the Structure determined. (a) To find the weight of the top chord. Suppose <2 -f- / to be the greatest allowable pressure to the square inch of section of top chord, and Q to be of the same denomination with W and L ; and suppose /to be a number called the factor of safety. Q is known as the breaking-weight GIRDERS WITH EQUAL AND PARALLEL CHORDS. 325 of a column of the given material, having the length of one panel, and the cross-section of the top chord for any given panel. Let m = weight of one cubic inch of the material, in pounds. We then have, for each one of the equal panel lengths of the top chord, Area of section = J~jL square inches, Volume of one panel length = ^ cubic inches, Qn Weight of one panel length = I2m f l H pounds, Qn Weight of top chord = ^H pounds. Qn From (a} of the preceding article we find [i + 2 -I- 3 -f 4 -f . . . (n i) terms] - [i 2 + 2* + 3* + 4* + . . . (rc - i) terms] -f- i# 2 for n even, + J(# 2 ~~ i) for w odd. j T\7 znh (W + L)l * - ^ ' X (2n 2 + 3 2), n even; 12 X A(23 + 3^ _ 2 _ 3), Odd. Substituting these values for ^H, we have Weight of top chord = m f l *( w + L \ 2n * + - _ 2 ) 2Qnh (n even), (424) (wodd). 326 MECHANICS OF THE GIRDER. (b) Similarly, if T -f- / the greatest allowable tensile strain, we find Weight of bottom chord = -^H pounds. 4- 2(n i) for two end panels, n\_i + 2 + 3 + 4 + . . . (n i) terms] - [i 2 + 2 2 4- 3 2 + ..-(- i) terms] J 2 for even, %(n 2 i) for n odd. znh = i- x -rV( 2w3 "~ 3^ 2 + 22 24) when n is even, 21) when n is odd. /. Weight of bottom chord even), - 3^ 22 - 21) (odd). (425) (c) In finding the weight of the verticals, let Q t -f- / be the allowed working unit strain in compression ; then . Area of section = W Q square inches, Volume of one strut = - ^ - cubic inches, \2mfh(Zw 4- ZL) Weight of one strut = -Q pounds, Weight of all verticals = - -Q pounds. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 327 Hence, from the strain sheet, Fig. 112, using the proper limits of summation, we derive, when n is even, = 2 x 2 _ Wn 2 ) 4 --- (J) terms] \ = 2 x ! (J) - [i + 3 + 5 + 7 . . . (%n) terms] [~i + 3 + 6 + 10 + . . . f- - i\ terms! + if n - -}[n - ( - + i Y] for middle strut I S>-)} But when is odd, we thus sum, = 2X :j^ / I + 3 2 x N 2 2 \ 2 + 2 ( i + 3 + 6 + 10 + . . . '^-5 terms = X ^V\7^ 3 ~~ 3^ 2 ~ 7 W + 3). ^ Wherefore 3;/% ^F 2 w/7?Z Weight of verticals = ~^7\ H -J 6 therefore weight of diagonals due uniform moving-load, nL, alone is * - '> <**> The weight of the diagonals due to the dead load, n W, is manifestly to be derived from the weight of the verticals due dead load if for h we put h -f- sin 2 0, and for Q l we put T. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 329 .-. Weight of diagonals = ^f hL (" - i) + 2 Tsm 2 (n even), = j> (*odd). (e) Taking the sum of the weights thus found, we have, when n is even, total weight of girder, in pounds, _ mfl z (W 4- Z)/2/Z 2 + 3^ 2 , Q Tn Atell (429) But when n is 0dk/, total weight of girder, in pounds, is 2 Q 3?* 2 zn 3 2^ 3^ + 22^ 2i + ,nfhL\ V ~ & '- V* + 3 + A*gL=j) ). ( 43 o) ( 2 ^ 2^sm 2 ) EXAMPLE i. Wrought-iron girder of 6 equal panels. Take w = 6, / = 60 feet, // = 10 feet, / = 4, ?/z = y 5 ^ pound, L = 8 tons, 7" = 24 tons, g = 16 tons, (2i = I2 tons, tan = I, sin <^> = ^2 = 0.70711. Therefore, from (429), = 483-333^ + 4267 pounds, equal to 2000/2 W if n W is the girder's own weight in tons. /. Panel weight of girder = W 0.3704775 ton, Total weight of girder = nW = 2.2228650 tons. 330 MECHANICS OF THE GIRDER. 130. But if the structure is a bridge having two equal gir- ders whose combined weight is G, and an additional permanent weight of K pounds, then the weight of the bridge is 20ocm W = K + G pounds, as shown by equation (419). Continuing the first example of article 129, we compute K as follows : For the floor, we have / = 60 feet = length. Take q 18 feet = breadth. t = feet = thickness. u = 54 pounds = weight of one cubic foot of oak. From (408), Weight of floor = 54 X 60 x 18 X -jj = 12150 pounds = F. The joists : / -5- n = 10 feet = panel length of joist. Take b = 3 inches = thickness. g = 2 feet space between centres. q 4- g =. 9 = number of joists in each panel. nq -T- g = 54 = number of joists in bridge. , = 54 = . B = 10,600. / = <* Then, by article 123, we have d=\ 9X9X60x2 / x I8 x 6o x M+ 200 o X 6 X (6 2 X 18 X 3 X io6oo\ 12 = 7.1424 inches. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 331 Call d = 8 inches, 3 X 8 X 60 X 18 X 54 .-. Weight of joists = - = 4860 pounds = J. For the iron I-beams, we have, from article 124, F 4- / H- 2QoonL D = 18835 pounds. Take q t = 19 feet = entire length of beam. f 4 = factor of safety. B = 52,567, from Table II. Whence, by equation (412), Required depth of beam = d 2 = 3.8oi22f 7 ) = 11.435 inches. Call d 2 = 12 inches ; then, by (413), Area of section = S = r.M 3 r/ l88 35 X 19 X 4 V x /_IL_Y \ 5 2 5 6 7 / \"-435/ = 12.84 square inches, since similar sections are to each other as the squares of their like dimensions. Now this cross-section, 12.84, agrees very nearly with that of the "12-inch light I-beam" of the Union Iron Mills, Pitts- burgh, Penn., whose weight is 42 pounds to the foot, and area = 42 X -^ 12.6 square inches. Using this beam, we then have Weight of 5 floor beams = 5x19X42 = 3990 pounds = P. Use full head trussing ; the struts to be composed of two T-bars, each 5} pounds to the foot, latticed with i| X J inch 332 MECHANICS OF THE GIRDER. braces, at 45 degrees, the whole weighing 12 \ pounds to the running foot ; length = 18 feet. Weight of (n + i) horizontal struts = 7X18X12^ = 1575 pounds = U. Let the horizontal diagonal ties be ij inches in diameter, weighing 3.359 pounds to the foot. Then Weight of 24 horizontal ties = 24 x 3-35 9V io 2 + i8 2 = 1660 pounds = X. Call the residue 100 pounds to the panel ; that is, in all = 600 pounds Y. .\ K=F + J + P + U + X + Y= 24835 pounds, G = weight of girders = 4267 + 483.333^, K + G = weight of bridge = 29102 + 483.333^ = 1 2000 W pounds ; /. Panel weight of bridge = W = 2.526947 tons, Total weight of bridge = nW = 15.161682 tons. Panel weight of dead load on each girder = 1.26347 tons, Panel weight of live load on each girder = 4 tons. + L) = 5.26347 tons = total panel weight for one girder. Putting this value, 5.26347 tons, for W + L, in the expres- sion for N, article 128, (d\ we find c. 26347 X 60 N = - -T- = J* J v - = 2.63174 tons. 2nh 2 X 6 X io And from the strain sheet, Fig. 112, the greatest chord strains are ff t = 2.63174 x 5 X i = 13.15870 tons, H 2 = 2.63174 x 4 X 2 = 21.05392 tons, H z = 2.63174 X 3 X 3 = 23.68566 tons. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 333 Putting | W = 1.26347 for W, and \L 4 for L, the same strain sheet gives, for each of two girders : Greatest compression on verticals : Z, 0.63174 x 5 -h 0.33333 x 5 X 6 = 13.15870 tons, Z 2 = 0.63174 x 3 H- 0.33333 x 4 X 5 = 8.56188 tons, Z 3 = 0.63174 x i + 0.33333 x 3 X 4 = 4-63174 tons, Z 4 = 0.33333 x 2 x 3 = 2.00000 tons. Also, for the diagonals : L becomes = 0.94281 ton, 2 sin < becomes - 6 3 T 74 = 0.89342 ton. sin And, from Fig. 112 : Greatest strain on diagonals : Y lt counter, = o Y 2) counter, = 0.94281 X Y 3 , counter, = 0.94281 x Y 4 , main, = 0.94281 x Y s , main, = 0.94281 X Y 6) main, = 0.94281 x GREATEST STRAINS, IN TONS. - 0.89342 x 5 < i 0.89342 x 3 < 3 - 0.89342 x i = 6 H- 0.89342 x i = 10 -f 0.89342 x 3 = 12.10836 tons; 15 -f- 0.89342 x 5 = 18.60925 tons. o ton; o ton; 1.93501 tons; 6.55028 tons; REQUIRED SECTIONS, IN SQUARE INCHES. LOAD APPLIED AT BOTTOM JOINTS. DIAGONALS IN TENSION. CLASS IX. FIG. 113. 131. Now, it is very evident that the bridge would depend entirely upon the floor system for stability in a longitudinal direction if we should omit the bottom chords and the counter ties, which are marked as receiving no strain in the end panels. 334 MECHANICS OF THE GIRDER. It is therefore usual to insert these members in the end panels, and also, in this style of girder, to stiffen the bottom chords by cross-bracing in the end panels, so that each bottom chord may there act as a strut. Some builders place counters in all panels, even where the assumed behavior of the given load does not require them. By so doing they provide for concentrated loads greater than the assumed uniform apex load, as well as enhance the symmetry of the structure. In short-span bridges, as in the present example, some of the vertical struts require a greater cross-section than the actual downward pressure upon them would indicate ; for, besides this pressure along the axis of the strut, it should be able to resist probable lateral blows from the travel of the road, even though every strut be protected from ordinary collision with hubs. To provide for the increase of bridge weight from these sources, above the weight computed from the given load nL y we have added the last term, F, of K, which term should be large enough to cover every thing not otherwise included. No absolutely definite rule can be given for the size of these parts, but the smallest counter ties should be so large as not to look wiry, say not less than one inch in cross-section ; the bottom chord in the first panel may equal in size that of the second panel ; and the size of any vertical strut should enable it to resist such lateral shocks as are probable in the situa- tion. 132. Thus far the panel weights, W and L, have been assumed to be applied at the lower joints only. In the nature of the case, however, it is plain that the weight of the top chords and the system of head-bracing, as also the weight of the girder diagonals, can only reach the bottom joints through the vertical struts. But as the weight of these members is small compared with the whole weight of the bridge, and as GIRDERS WITH EQUAL AND PARALLEL CHORDS. 335 the calculation is a little more simple when W is applied at one point instead of two, it is usual to make the above assumption. It is proper, however, in this place, to indicate the changes to be made in the strain sheet, Fig. 112, by changes in the dis- tribution of the loads. ist, Suppose the panel weight, L, of live load, and the panel weight of the floor system, and half the panel weight of the girders, to be applied at each lower joint, and the other half of the girders' weight, and the system of head-trussing, to be uniformly distributed at the upper joints. We have G = weight of girders, in pounds. G -T- 2.11 = one-half panel weight of girders, in pounds. Take A = panel weight of head-bracing, X* /. Load at each lower joint = L + W A -- , 2H f Load at each upper joint = A -\ -- pounds. 2H The strain sheet, Fig. 112, applies to this case if to the Q compression on each vertical we add A -\ - pounds, but to each end post \( A + --Y And the additional weight of the verticals due to this change of loading is ^ V 2nf which is to be added to the weight of verticals in equation (426), and, of course, to the second member of (429) and (430), and thence a new expression for G be found. 2d, Suppose we have a "deck" bridge, and that both W and L are applied at the upper joints. 336 MECHANICS OF THE GIRDER. Then to each vertical compression given in strain sheet, Fig. 112, we must add W -\- L ; and the additional weight of the .verticals is, with J( W -f- L) on each end post, 1L(W 4- L)n pounds, to be placed in the second member of (429) and (430), provided the bridge has its points of support at the bottom, as in the figure ; but if the points of support are at the ends of the upper chords, then no end posts are required, and their weight may be deducted from the second member of (429) and (430), and be added. 3d, In case of the deck bridge, if we suppose half the weight of the girders, and the weight, nA lt of the bottom hori- zontal bracing, to be applied uniformly at the bottom joints, while the remainder of the loading is applied at the upper joints, we must then add to the pressure on each vertical, Fig. 112, ,, L -f W A t pounds, 2H instead of W + L. And the additional weight of girders from this source is + W _ Ai _ n ds> . or W . _ Ai _ and put them into suitable form for general application. The value of F, the weight of floor, (408), is independent of ;/ and //, and requires no change. t 338 MECHANICS OF THE GIRDER. If for joists we call d = b\ (431) we shall have a good ratio of breadth, b, to depth, d\ and, in (410), bd = fo, and (432) 2000;zZ) ' (433) which is the weight of the joists, in pounds. Restoring the value of D, we write, for (414), JCF+ /+ 2000Z) ?,/)! />= 15.460680*7, (-i) - ^g -f pounds, (434) 'equal to weight of (n i) wrought-iron I-beams having the proportions assumed in deriving equation (412). 135. If we take into account the greatest probable pressure of wind horizontally against the side of each open girder and its moving-load, or against the entire side of each wholly cov- ered structure, we find the strains due to wind, in the chords and entire lateral system, by making the proper changes in the .strain sheet, Fig. 112. For any through bridge of Class IX., let the uniform wind pressure to be resisted by the top or bottom lateral system be W l = \hw~ tons per panel ; w being the horizontal pressure of wind per square foot, in tons. And for the bottom lateral system, which alone is affected by the wind pressure against the moving-load, let the uniform moving wind pressure per panel be L, = sw~ tons ; c being the height of train or other moving-load, in feet. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 339 From (424) we derive the additional weight of top chords due to wind pressure by substituting 2 W t = - - for ( W -\- L) ; since, in order to provide for the wind coming either way, we must increase each chord for increased compression, and by putting q for /i, and formulating thus, Weight of top chords) mffihw, 2 . v j j t 7T~^ V T o n 2 ) due to wind ) 2 (V;z 2 ^ ( even), = mffihw^ H 2 Qn 3 g (n odd). 2 2n 3) flDjJ (435) Similarly, from (425), putting (2 W t -f 2L,) = (h + 2s) for ( W + L), and ^ for //, Weight of bottom ) _ mfl*w(h +2g) , chords due to wind J 2 Tr$q (n even), x 24) ( odd). (436) And, from (426), we derive the weight of the horizontal struts between the top chords by putting W t = \hw- for W, o for L y 11 q for //, Q, for Q a and adding, by reason of the load being applied to the compressed chord, as explained in article 132. 340 MECHANICS OF THE GIRDER. Weight of top horizontal ) ynfqwlh struts due to wind ) = ^ = U. (437) The floor beams which carry the moving load generally act as the horizontal struts between the loaded chords ; and they are usually so large, in comparison with the struts actually required to resist the wind pressure, that we may with little error make no further allowance for these beams acting as hori- zontal struts than that already suggested in article 124. But, if it is required, we can find the additional metal to compensate the floor beams for this end pressure by treating each beam as a pillar whose least diameter is its depth, since the longitudinal joists or stringers prevent deflection sideways. Thus, q l being the length, d the depth, of the wrought-iron I floor beams, and 5 the cross-section due to the total effect of wind pressure, P, in tons, applied longitudinally at the end of a beam, we have, from equation (400), square inches, to be added to section of each beam, in order to neutralize effect of wind upon the loaded horizontal system of struts. -flP equals total additional section of I-beams ; / being the factor of safety. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 34! Now, in this case, 2

14 28.5! + 3f^ 15 1(28-51 + 3! W) (28.5! + 3W)sec0 15 32.71* + At\W 16 1(32.71*. +4**n (32.71* +4*W)sec0 16 37.134 + 5 W 17 1(37-134 + 5 ^) (37.134 + 5 ^)sec0 In the first 6 panels we shall introduce counters of I square inch cross-section, and therefore capable of resisting safely 5 tons where theoretically no strain appears. Also, we shall call the strain on the bottom chords in the first panel equal to that of the second panel ; viz., $W + 37.134 tons. For each panel length of top chord, take ratio of length to least diameter =15. Then the Gordon formula becomes, equa- tion (400), P 18 _ = (/ = = 16.744 tons per square inch. T I 5 3000 = 3.3488 tons = allowed inch strain on top chords. 408 MECHANICS OF THE GIRDER. For each vertical strut, take ratio of length to least diam eter = 37.5. Then P 18 = Q" = - - = 12.255 tons per square inch. 3000 I 2.2 ^ 15 ' 2.451 tons = allowed inch strain on verticals. In tension, 5 tons allowed. Factor of safety / = 5 for all parts of girder; panel length of chords = 12.5 feet; length of verticals = 18.75 f eet ; length of diagonals = 23 feet ; wrought- iron, -f^ pound per cubic inch. From these data we find, Weight of top chords = 4(124^' + 872.244) X I2 ' 5 X I0 = 6171.36 fT+ 43411 pounds. 3.3488 x 3 Weight of bottom chords = 4(107$ W + 763-064) X "' 5 * I0 = 3588.89 W + 25435 pounds. Weight of verticals = 4 (2iJfF + 193-3571) Xf 18.75 X I0 k 2.451 x 3 Weight of diagonals = 3238.47^" + 29585 pounds. 23 X io sec X - = 221 1.38 JF + 24539 pounds. 5 * 3 Weight of 2 girders G = 15210.10^ + 122970 pounds. W is in tons. GIRDERS WITH EQUAL AffD PARALLEL CHORDS. 409 To find the constant part of the bridge weight = K. 2 rails, 200 feet, 56 pounds per yard, = 7,467 pounds. Rails 5 feet between centres. Ties 6X8 inches, 16 inches between centres. 12 x 200 = jtj o t i eSj j f ee j- long, 40 pounds per cubic foot. 16 Weight of ties 150 X X 7 X 40 = 14,000 pounds. 144 2 track stringers, each 15 X i8J inches, 40 pounds per cubic V T C V T 8 1- foot, - - ?" X 200 X 40 = 30,833 pounds. 144 Depth of stringer is thus found : Length between bearings =12.5 feet. Uniform load = -^ of rails and ties = ^ffi 3 - = 1,342 pounds. Concentrated load = panel weight of engine = 2 X 17,600 = 35,200 pounds, which is equivalent to uniform load of 70,400 pounds. .-. Uniform load on 2 stringers 30 inches wide = 71742 pounds. For pine, take factor of safety = 10, and the ultimate inch resistance to cross-breaking B = 8,000 pounds ; .-. Moment due external forces = J X 71742 x 12 X 12.5, from equation (52). Moment of resistance due internal forces, equation (160), 8oo x 3V, 6 which becomes, after introducing the factor of safety, 8000 X OO 410 MECHANICS OF THE GIRDER. Equating moments of external and allowable internal forces, we find J x 71742 x 12 x 12.5 = /. Depth = d = 18.338 inches, called i8J. 15 pairs I-beams, heavy 15-inch, 67 pounds, 16 feet, = 32,160 pounds. Depth of beam is thus determined : Panel weight of rails = 467 pounds, Panel weight of ties == 875 pounds, Panel weight of stringers = 1927 pounds, Panel weight of engine = 35200 pounds. Weight on each pair of I-beams = 38469 pounds. Now, this weight is actually concentrated at two points 5 feet apart, under the rails, each point being 5 J feet from the nearer end of the I-beams. The moment at the centre of the double beam is therefore, according to equation (43), M c = 2 x 19234.5 x T %- x sJ x 12 = \W x 16 x 12, by equation (46), if W r is the weight at centre producing an equivalent moment ; .-. W = 26447.4, and 52,895 pounds is the equivalent load uniformly distributed for 2 beams. Therefore uniform load on I beam is 26,447 pounds. And, if 6 is the factor of safety for beams, equation (412) gives, B being = 50,000, 726447 X 16 X 6\i Depth of beam = d 2 = 3.80122! - - ) = 14.076 inches. \ 50000 / GIRDERS WITH EQUAL AND PARALLEL CHORDS. 4! I From (413), 726447 X 16 X 6\f Area = 1.288391 r 7: ) = 17.667 square inches. \ ^ oooo / 1 1 \ 2 Area of similar beam iq inches deep = 17.667 x 14.0762 = 20.062 square inches. Area of heavy 15 -inch 67-pound beam of the) Union Iron Mills, Pittsburgh, Penn., f = 2=! P= i 4 -P=the L = the ^=the L 2 for square end compression members. for compression members with one pin and one square end. for compression members with pin bearings. 20000^? 2 allowed compression per square inch of cross-section. length of compression member, in inches. least radius of gyration of the section, in inches. 3. The lateral struts shall be proportioned by the above formulae to resist the resultant due to an assumed initial strain of 10,000 pounds per square inch upon all the rods attaching to them, produced by adjusting the bridge. 4. In beams and girders, compression shall be limited, as follows : Pounds per Square Inch. In rolled beams used as cross floor beams and stringers . . In riveted plate girders used as cross floor beams, gross section, In riveted longitudinal plate girders over 20 feet long, gross 1 0000 6000 6000 In riveted longitudinal plate girders under 20 feet long, gross section .*. COOO 5. Members subjected to alternate strains of tension and compres- sion shall be proportioned to resist each of them. The strains, however, shall be assumed to be increased by an amount equal to eight-tenths of the least strain. 6. The rivets and bolts connecting all parts of the girders must be so spaced that the shearing-strain per square inch shall not exceed 6,000 pounds, nor the pressure upon the bearing-surface exceed 12,000 pounds CALCULATION OF THE WEIGHT OF BRIDGES. 417 per square inch of the projected semi-intrados (diameter x thickness of piece) of the rivet or bolt hole. 7. Pins shall be so proportioned that the shearing-strain shall not Bending- exceed 7,500 pounds per square inch, nor the crushing-strain upon the strains, projected area of the semi-intrados (diameter x thickness of piece) of any member connected to the pin be greater than 12,000 pounds per square inch, nor the bending-strain exceed 15,000 pounds per square inch, when the centres of bearings of the strained members are taken as the points of application of the strains. 8. In case any member is subjected to a bending-strain from local loadings (such as distributed floors on deck bridges), in addition to the strain produced by its position as a member of the structure, it must be proportioned to resist the combined strains. 9. Plate girders shall be proportioned upon the supposition that the Plate girders., bending or chord strains are resisted entirely by the upper and lower flanges, and that the shearing or web strains are resisted entirely by the web plate. 10. The compression flanges of beams and girders shall be stayed against transverse crippling when their length is more than thirty times their width. 11. The unsupported width of any plate subjected to compression shall never exceed thirty times its thickness. 12. In members subject to tensile strains, full allowance shall be made for reduction of section by rivet holes, screw threads, etc. 13. The iron in the web plates shall not have a shearing-strain greater than 4,000 pounds per square inch, and no web plate shall be less than { inch in thickness. 14. No wrought-iron shall be used less than -^ inch thick, except in places where both sides are always accessible for cleaning and painting. DETAILS OF CONSTRUCTION. 1. All the connections and details of the several parts of the structure shall be of such strength, that, upon testing, rupture shall occur in the body of the members rather than in any of their details or connections. 2. Preference will be had for such details as will be most accessible for inspection, cleaning, and painting. 3. The web of plate girders must be spliced at all joints by a plate on each side of the web. T-iron must not be used for splices. 4. When the least thickness of the web is less than one-eightieth of the depth of a girder, the web shall be stiffened at intervals not over twice the depth of the g'rder. 5. The pitch of rivets in all classes of work shall never exceed 6 inches, nor sixteen times the thinnest outside plate, nor be less than three diameters of the rivet. MECHANICS OF THE GIRDER. 6. The rivets used will generally be \ and inch diameter. 7. The distance between the edge of any piece and the centre of a rivet hole must never be less than 1-4- inches, except for bars less than 2-^ inches wide ; when practicable, it shall be at least two diameters of rivets. 8. When plates more than 12 inches wide are used in the flanges of plate or lattice girders, an extra line of rivets, with a pitch of not over 9 inches, shall be driven along each edge, to draw the plates together, and prevent the entrance of water 9. In punching plate or other iron, the diameter of the dye shall in no case exceed the diameter of the punch by more than -^ of an inch. 10. All rivet holes must be so accurately punched, that, when the several parts forming one member are assembled together, a rivet ^ inch less in diameter than the hole can be entered, hot, into any hole, without reaming or straining the iron by " drifts." 11. The rivets, when driven, must completely fill the holes. 12. The rivet heads must be hemispherical, and of a uniform size for the same-sized rivets throughout the work. They must be full and neatly made, and be concentric to the rivet hole. 13. Whenever possible, all rivets must be machine-driven. 14. The several pieces forming one built member must fit closely together, and, when riveted, shall be free from twists, bends, or open joints. 15. All joints in riveted work, whether in tension or compression members, must be fully spliced, as no reliance will be placed upon abutting joints. The ends, however, must be dressed straight and true, so that there shall be no open joints. 1 6. The heads of eye-bars shall be so proportioned that the bar will break in the body instead of in the eye. The form of the head and the mode of manufacture shall be subject to the approval of the chief engineer of the railroad company. 17. The bars must be free from flaws, and of full thickness in the necks. They shall be perfectly straight before boring. The holes shall be in the centre of the head, and on the centre line of the bar. 1 8. The bars must be bored of exact lengths, and the pin hole -fa inch larger than the diameter of the pin. 19. The lower chord shall be packed as narrow as possible. 20. The pins shall be turned straight and smooth, and shall fit the pin holes within -fa of an inch. 21. The diameter of the pin shall not be less than two-thirds the largest dimension of any tension member attached to it. Its effective length shall not be greater than the breadth of the foot of the post plus four times the diameter of the pin. The several members attaching to the pin shall be packed close together, and all vacant spaces between the chords and posts must be filled with wrought-iron filling-rings. Lower chords and suspension bars. Pins. CALCULATION OF THE WEIGHT OF BRIDGES. 419 22. All rods and hangers with screw ends shall be upset at the ends, Upset screw so that the diameter at the bottom of the threads shall be ^ mcn larger ends, than any part of the body of the bar. 23. All threads must be of the United States standard, except at the ends of the pins. 24. Floor beam hangers shall be so placed that they can be readily Floor beam examined at all times. When fitted with screw ends, they shall be pro- hangers, vided with check nuts. 25. When bent loops are used, they must fit perfectly around the pin throughout its semi-circumference. 26. Compression members shall be of wrought-iron of approved Compression forms. members. 27. The pitch of rivets, for a length of two diameters at the ends, - shall not be over four times the diameter of the rivets. 28. The open sides of all trough-shaped sections shall be stayed by diagonal lattice work at distances not exceeding the width of the member. The size of bars shall be duly proportioned to the width. 29. All pin holes shall be re-enforced by additional material, so as not to exceed the allowed pressure on the pins. These re-enforcing plates must contain enough rivets to transfer the proportion of pressure which comes upon them. 30. Pin holes shall be bored exactly perpendicular to a vertical plane passing through the centre line of each member, when placed in a posi- tion similar to that it is to occupy in the finished structure. 31. The ends of all square-ended members shall be planed smooth, Abutting and exactly square to the centre line of strain. joints. 32. All members must be free from twists or bends. Portions exposed to view shall be neatly finished. 33. The sections of the top chord shall be connected at the abutting Splicing of ends by splices sufficient to hold them truly in position. top chord. 34. In no case shall any lateral or diagonal rod have a less area than Lateral of a square inch. bracing. 35. The attachment of the lateral system to the chords shall be thor- oughly efficient. If connected to suspended floor beams, the latter shall be stayed against all motion. 36. All through bridges with top lateral bracing shall have wrought- Transverse iron portals of approved design at each end of the span, connected diagonal rigidly to the end posts. bracing. 37. When the height of the trusses exceeds 25 feet, overhead diagonal bracing shall be attached to each post and to the top lateral struts. 38. Pony trusses and through-plate or lattice girders shall be stayed by knee braces or gusset plates attached to the top chords, at the ends, and at intermediate points not more than 10 feet apart, and attached below to the cross floor beams or to the transverse struts. 420 MECHANICS OF THE GIRDER. In all deck bridges, diagonal bracing shall be provided at each panel. In double-track bridges, this bracing shall be proportioned to resist the unequal loading of the trusses. The diagonal bracing at the ends shall be of the same equivalent strength as the end top lateral bracing. Bed plates. 39. All bed plates must be of such dimensions that the greatest press- ure upon the masonry shall not exceed 250 pounds per square inch. Friction . 40. All bridges over 50 feet span shall have at one end nests of rollers. turned friction rollers formed of wrought-iron, running between planed surfaces. The rollers shall not be less than 2 inches diameter, and shall be so proportioned that the pressure per lineal inch of rollers shall not exceed the product of the square root of the diameter of the roller, in inches, multiplied by 500 pounds (500^). 41. Bridges less than 50 feet span will be secured at one end to the masonry, and the other end shall be free to move by sliding upon planed surfaces. Camber. 42. All bridges will be given a camber by making the panel lengths of the top chord longer than those of the bottom chord in the proportion of \ of an inch to every 10 feet. Tension tests. QUALITY OF IRON. 1. All wrought-iron must be tough, fibrous, and uniform in character. It shall have a limit of elasticity of not less than 26,000 pounds per square inch. Finished bars must be thoroughly welded during the rolling, and free from injurious seams, blisters, buckles, cinder spots, or imperfect edges. 2. For all tension members, the muck bars shall be rolled into flats, and again cut, piled, and rolled into finished sizes. They shall stand the following tests : 3. Full-sized pieces of flat, round, or square iron, not over 4^ inches in sectional area, shall have an ultimate strength of 50,000 pounds per square inch, and stretch 12^ per cent in their whole length. Bars of a larger sectional area than 4^ square inches, when tested in the usual way, will be allowed a reduction of 1,000 pounds per square inch for each additional square inch of section, down to a minimum of 46,000 pounds per square inch. 4. When tested in specimens of uniform sectional area of at least square inch for a distance of 10 inches taken from tension members which have been rolled to a section not more than 4^ square inches, the iron shall show an ultimate strength of 52,000 pounds per square inch, and stretch 18 per cent in a distance of 8 inches. Specimens taken from bars of a larger cross-section than 4^ inches will be allowed a reduction of 500 pounds for each additional square inch of section, down to a minimum of 50,000 pounds. CALCULATION OF THE WEIGHT OF BRIDGES. 421 5. The same-sized specimen taken from angle and other shaped iron shall have an ultimate strength of 50,000 pounds per square inch, and elongate 15 per cent in 8 inches. 6. The same-sized specimen taken from plate iron shall have an ulti- mate strength of 48,000 pounds, and elongate 15 per cent in 8 inches. 7. All iron for tension members must bend cold, for about 90 degrees, Bending- to a curve whose diameter is not over twice the thickness of the piece, tests, without cracking. At least one sample in three must bend 180 degrees to this curve without cracking. When nicked on one side, and bent by a blow from a sledge, the fracture must be nearly all fibrous, showing but few crystalline specks. 8. Specimens from angle, plate, and shaped iron must stand bending cold through 90 degrees, and to a curve whose diameter is not over three times its thickness, without cracking. When nicked or bent, its fracture must be mostly fibrous. 9. Rivets and pins shall be made from the best double-refined iron. 10. The cast-iron must be of the best quality of soft gray iron. Cast-iron. n. All facilities for inspection of iron and workmanship shall be Tests. furnished by the contractor. He shall furnish, without charge, such specimens (prepared) of the several kinds of iron to be used as may be required to determine their character. 12. Full-sized parts of the structure maybe tested at the option of the chief engineer of the railroad company ; but, if tested to destruction, such material shall be paid for at cost, less its scrap value, to the con- tractor, if it proves satisfactory. If it does not stand the specified tests, it will be considered rejected material, and be solely at the cost of the contractor. WORKMANSHIP. 1. All workmanship shall be first-class in every particular. 2. Abutting joints in truss bridges shall be in exact contact through- out. 3. Bars which are to be placed side by side in the structure shall be bored at the same temperature, and of such equal length, that, upon being piled on each other, the pins shall pass through the holes at both ends without driving. 4. Whenever necessary for the protection of the thread, provision shall be made for the use of pilot nuts, in erection. PAINTING. 1. All work shall be painted at the shop with one good coat of selected iron-ore paint and pure linseed-oil. 2. In riveted work, all surfaces coming in contact shall be painted before being riveted together. 422 MECHANICS OF THE GIRDER. Bed plates, the inside of closed sections, and all parts of the work which will not be accessible for painting after erection, shall have two coats of paint. 3. Pins, bored pin holes, and turned friction rollers shall be coated with white lead and tallow before being shipped from the shop. 4. After the structure is erected, the ironwork shall be thoroughly and evenly painted with two additional coats of paint mixed with pure linseed-oil, of such color as may be directed ; the tension members being, however, generally of lighter color than the compression members. ERECTION. 1. The railroad company will take down the old bridge if any exists. It will furnish the lower falseworks, or supporting-trestles, only. The use of these falseworks by the contractor shall be construed as his approval of them. 2. The contractor shall furnish all other staging (the plan and con- struction of which must be approved by the chief engineer), and shall erect and adjust all the ironwork complete. 3. The contractor shall so conduct all his operations as not to impede the running of the trains or the operations of the road. 4. The contractor shall assume all risks of accidents to men or material during the manufacture and erection of the bridge. ADDITIONAL STIPULATIONS. The above specifications are approved. Chief Engineer N. K, L. E. t & W. R.R. The above specifications are accepted. Contractor. These specifications will be modified to suit the character of the bridges here considered. CALCULATION OF THE WEIGHT OF BRIDGES. 423 SECTION 2. The Brunei Girder of Single System. 149. Take the form of girder shown in Fig. 16, Class I., article 49, where the end lengths of top chord are shorter than other segments. Bridge to have 2 equal parabolic double-bow girders, top and bottom chords of the same curvature. Floor carrying load attached by vertical struts and suspenders to bottom and top panel points, and in the plane of the axes of the two girders. To compute the dimensions, let / = span in feet, h = height of the two equal parabolas composing each girder at the centre of span. Take the axis of x horizontal, that of y vertical ; then the equation to the upper parabola, origin at top, is x 2 = ay. But, with origin at left end, the equation is 2hf x 2 \ , . = y(- - 7} (47*) If there are n equal panels, then the value of y at the panel point, where x = , becomes y = ^ r (* = r ); . (473) and the whole height is 2y = ^* ~ r) * (474) By (473), For r = o, y = o ; r = 2, y = 2/i 2n ~ 4 ; ^^ 2 424 MECHANICS OF THE GIRDER, Twice these values of y will be the heights between curves of parabolas ; but, as we shall here assume each chord to be polygonal, that is, to be straight from apex to apex, the actual heights of girder at these points are manifestly 4- ( v 2/1 n 2 2h ,, \ h r = y r _i 4- v 2h = -T - (475) VALUES OF s IN (475). r = 4 6 8 10 12 14 16 18 20 22 24 I 5 9 13 17 21 25 29 33 37 41 45 2 7 IS 23 3 1 39 47 55 63 7i 79 87 3 17 29 41 53 65 77 89 101 "3 125 4 31 47 63 79 95 in 127 H3 159 5 49 69 89 109 129 149 169 189 6 7i 95 119 M3 167 191 215 7 97 125 153 181 209 237 8 127 159 191 223 255 9 161 197 233 269 10 199 239 279 ii 241 285 12 287 Also we have cos 2 a, 2 / cos 2 3 2 / COS 2 2 COS 2 4 CALCULATION OF THE WEIGHT OF BRIDGES. 42$ and generally cos 2 a r n 2 / 2 VALUES OF s, IN (476). (476) r =:4 G 8 10 12 14 16 18 20 22 24 I 3 5 7 9 II 13 IS I? 19 21 2 3 2 2 4 6 8 10 12 M 16 18 20 3 2 4 6 8 10 I ^ 14 16 4 o 2 4 6 8 10 12 5 2 4 6 8 6 o 2 4 In the same manner, we derive ih* COS i_ = , + &L (n _ 2 y, _L_ =I+ g ( _ IO ). cos 2 /? r / 2 2 VALUES OF e 2 IN (477). (477) r = 4 6 8 10 12 14 16 18 20 22 24 I 2 4 6 8 10 12 14 16 10 2O 22 2 o 2 4 6 8 10 12 M 16 18 3 o 2 4 6 8 10 12 M 4 o 2 4 6 8 10 5 o 2 4 6 6 o 2 426 MECHANICS OF THE GIRDER. I COS 2 COS 2 r VALUES OF c 3 IN (478). r = 4 6 8 10 12 14 16 18 20 22 24 I 7 13 19 25 31 37 43 49 55 61 67 2 3 17 31 45 59 73 87 101 "5 129 H3 3 5 27 49 7i 93 "5 i37 i59 181 203 4 7 37 67 97 127 i57 187 217 247 5 9 47 85 123 161 199 237 275 6 ii 57 103 149 i95 241 287 7 U 67 121 i75 229 283 8 15 77 '39 2OI 263 9 17 87 157 227 10 19 97 T 75 ii 21 107 12 23 a denoting slope of any segment of top chord, /? 'denoting slope of any segment of bottom chord, denoting slope of any Z web member, as shown in Fig. 16. = length of end segment of top chord. = length of any other segment of top chord. length of any segment of bottom chord. = length of any Z member. 72 COS n cos CALCULATION OF THE WEIGHT OF BRIDGES. 427 Take g t = 18 feet = extreme width of bridge. q = 1 6 feet = width of floor. 0, = inclination to plane of girder, of any horizontal diagonal. Then sn (479) 18 = length of horizontal diagonal supposed to reach from end to end of the transverse I floor beams. 150. To compute the Moments and Strains in Chords due to the Total Panel Weight, W + Z, applied at Each Apex, Top and Bottom. We have, from equation (65), moments at the vertical sections through these apices, W l(n - r)r = W (480) VALUES OF s 4 IN (480). r ,. t 6 8 10 12 14 16 18 20 22 24 , 3 5 7 9 II 13 IS 17 19 21 23 2 4 8 12 16 20 24 28 3 2 36 40 44 3 9 15 21 27 33 39 45 5 1 57 63 4 16 24 32 40 48 56 64 72 80 5 25 35 45 55 65 . 75 85 95 6 36 48 60 72 84 96 108 7 49 63 77 91 105 119 8 64 80 96 112 128 9 81 99 117 135 10 100 120 140 ii 121 143 12 144 428 MECHANICS OF THE GIRDER. Therefore, since Hr = ?, equations (475) and (480) give H = which is the horizontal component of the greatest strain in each chord, at a point directly above or below an opposite apex. Hence, in the present case : STRAINS IN TOP CHORD. STRAINS IN BOTTOM CHORD. VALUES OF - = 0.5 -\ , IN (481). = 4 6 8 10 12 14 16 18 20 22 24 0.600000 -S55555 0.538462 0.529412 0.523810 0.520000 0.517241 0.515152 0-513514 0.512195 0.511111 0.571429 -533333 0.521739 0.516129 0.512821 0.510638 0.509091 - 507937 0.507042 0.5063290.505747 0.529412 0.51*241 0.512195 0.509434 0.507692 0.506494 0.505618 0.504950 0.504425 0.504000 0.516129 0.510638 0.507937 0.5063290.505263 0.504505 0.503937 o- 5 3497 0.503145 0.510204 0.507246 0.505618 0.504587 0.503876 0.503356 0.5029590.502646 0.507042 0.5052630.504202 0-503497 0.502994 o. 502618 0.502326 0.505155 0.504000 0.503268 0.502762 0.502392 0.502110 0-503937 0.503145 0.502618 o. 502242 o. 501961 0.5031150.502538 0.502146 0.501859 0.502513 0.502092 o. 501792 0.502075 -5 OI 754 0.501742 CALCULATION OF THE WEIGHT OF BRIDGES. 429 Owing to the peculiar form (0.5 -f- Y these values are easily written from a table of reciprocals. 151. Weights of these Wrought-Iron Chords. W = unknown panel weight of bridge. = given panel weight of live load. Q = allowed inch strain in top chord. T = allowed inch strain in bottom chord, all in tons, say. p z= cross-section of top chord, square inches. TJ = cross-section of bottom chord, square inches. VOLUME OF SEGMENTS OF TOP CHORD, CUBIC INCHES. 24/^6 Qn cos 2 / Qn cos 2 2 ' (?# cos 2 3 (? cos 2 4 VOLUME OF SEGMENTS OF BOTTOM CHORD, CUBIC INCHES. 7 cos 2 /? 7>/ cos 2 ft 7 cos 2 Tn cos 2 ) _ j Weight of top chords, ) _ jj_ 24^ ZT in pounds, j 18 ;z^ cos 2 a 18 <2 A 4 e ^ x A l* I by reason of (476) and (481). In summing (484), it will be seen that only one of the two extreme panels is to be counted. 43 MECHANICS OF THE GIRDER. M f , ^ l ^ . H $ "8 vS ON S O o o O in t^. T}- co ON m m in tv I 1 N co R Q w 1 ^8 c? o^ O g w \r> in m u-j m 10 o O O O O O* 6 *o 4 ; j i - A .- j. P5 e GO to t^ ^ ro ro OO QO ^ oo m o ^ oi co ffl v) 9| v) 0) xn ^x co o O oo c* o> o o o o o <*- ^ ^^ K M pa i _^\_ | H H CO N ON 10 -^t" ro oo H O O O O 10 10 m in \o 10 o B rH 1 I'l I 1 O rh H ^^ CPi (T) !z g INCLUDI S 3 vo ro N T|- OO^OlO Tf- M H O O VD 10 10 10 10 10 0* 0* 6 ro Q P i w iH O CO CO M * co o IN o m t>. M co oJ d oo in M ro B pq J ^_._ j> _ S O ro 8 O 10 O O CO W W t^ rH ^ Jn , H n & S, S" N rH ro o co m H H o o' o' o' ro rf H *\ " . ! oo m in in ^ 1 s CO CO -^ N oq in q H 6 6 6 3 vo ~ 00 % N ^ ^ OO oo -^t* Q oo in in in o vo" 00 ro t> d o' o* H PQ B CD in ro N S S 1 o? S JS CO ro u"> CO M H *"! h ^ O -* 8 5 S" .(, Q M O w H" 80 8 td s v fn Er 5 S a M Ik HNTj-vOOOON^I k H * VO 00 OJ CALCULATION OF THE WEIGHT OF BRIDGES. 431 We shall here assume that the ratio of the panel length of top chord to its least radius of gyration is 100, and that sizes of iron can be exactly fitted to meet the required strains. We have, then, for the segments of the top chord, by our specifica- tions for columns with flat ends, IQO 2 40OOO = 3.2 tons per square inch of section. These values of 2- 4 , 2-V, and Q, put in (484), give, Weight of top chords, ) in pounds, f n W + L (o.6ioi2O/ 2 -f- 0.703 1 3/* 2 ) 4 h (o.8449o8/ 2 + 1.050677^) 6 (1.092744/2 + 1.40235^) 8 (I.345284/ 2 + 1.75268^) 10 (i.6ooi92/ 2 -f- 2.ioi79/ 2 ) 12 (i.85649O/ 2 -f- 2.45oio^ 2 ) 14 (2.H3692/ 2 4- 2.79790^) 16 (2.37i52i/ 2 -h S-MSST^ 2 ) 18 (2.629797/ 2 -f 3.49263^) 20 (2.888 4 i 3 / 2 + 3.83976^) 22 (3.147290^ + 4.18679^) 24 Similarly we find, from (477) and (481), Weight of bottom chords,) _ jj_ x 24/^ H in pounds, j 18 nT cos 2 /? Th n 2 e to be summed as follows : 432 MECHANICS OF THE GIRDER. p; m co M s VO S S vg (N !n i m m > ft "" O o' o' vo S ir ON M vO in ON 91 S fc? ^J O co ** , U n in O O in in In m in m N z ?o r^ vO 01 CO ^ * i in in O in o m 1 sT CO -" v vO 00 m co ss 1 1 oo" "S o 1 o 1 CO 00 O m O o o -a- i ^ TJ- oo~ 8 in ,n o O O '* /' ^- 1 vo w* m iH in S S g CO S I -X. 1 1 i ! 6 d CO PI 5- ^ Oi | 00 ^ in VO in S5 O M s*+~> ^7' VO 00 9 c ^ 2" i M S H 1 m 61 ? II M *. j H CO in tx M ^ co 00 vo ON Cv ON 00 H \O f~. 8 t^ CO OO O OO N * vo o> in M ? fl 05^ ^ vo CO ? 00 M -^- HI Q 1 O j j g. a S ct , m N vo * H oo o m o> c o oo' o oo' ci vO ON in H I H rN vO OO FH 8 $ S OO 00 N O 1 O H co t^ co 1 50 iH N TJ- i- O ON OS m vS CO vo H O 8 & <5 N " 3; CO ^t" H JW ON 0> 00 ^ O O * N OO" 230,9244 2 ^ ON M | a >> 5 is u S s SS ^I HIU SF3 SI? 5 &l >& IIS s n-r ^ & 438 MECHANICS OF THE GIRDER. VALUES OF 5 s 6 IN (490). = 4 6 8 10 12 14 16 18 20 22 24 0.62857 0.57778 0.555*8 0.54269 0.53480 0.52936 0.52539 0.52236 0.51999 0.51806 0.51647 o-S 1 ?^ 0.52058 0.51686 0.51393 0.51179 0.51018 0.50896 0.50799 0.50721 0.50657 Q-49475 0.50803 0.50827 0-50739 0.50651 0.50578 0.50516 0.50466 0.50424 0.48466 0.50242 0.50440 0.50444 0.50413 0.50378 0.50344 0.50315 0.47895 0.49923 0.50218 0.50276 0.50277 0.50263 0.50246 0.47529 0.49716 0.50076 0.50167 0.50189 0.50189 0.47273 o.4957o 0-49975 0.50089 0.50127 0.47084 0.49463 0.49901 0.50033 0.46939 0.49381 0.49844 0.46825 0.49317 0.46731 0.62857 1-09543 I-5705I 2.05224 2-53837 3-02746 3-51859 4.01129 4-50511 4-99985 5-49530 Since the girder diagonals are to have their end bearings on the chord pins, and are to have their centres attached to the floor system, we shall take the ratio of whole length of diagonal to least radius of gyration = 100, and the allowed inch strain + TOO 2 2OOOO = - tons. We have, then, after multiplying by 1.8, as above explained, Q A /T" Cross-section of i diagonal = ' ' Volume of i diagonal Weight of i diagonal Weight of all girder diag- ) _ onals, in pounds, from (478) and (490). <2i cos k I2/ J I >C[UcUC I .8A^ llCllCb, cubic HH inches. pounds. iff X 72 18 Ts 'x lj n x I2/ x 3 -^ < cos 2 1.8 5 (J 3 ( n :os 2 <9 CALCULATION OF THE WEIGHT OF BRIDGES. 439 ^ llh to 4 to i 1 + 1 i OJ OJ ON U GC \ ? 1 1 1 + I | 01 01 H 00 s ,7; to ON 1 + + 1 1 1 ON VI OJ M tn o to fe S 8 H 01 to ON M 8' SO * 1 + + + 1 1 1 t 01 M b b 8. bo s ON ON ~O (0 4>- 4- ON 1 + + + 1 1 | I OO M VI OJ M 01 vC Ol 1 Ol VI Ol M O 4^ 1 I* + + + + 1 1 H f 1 S " "0 "ON U 5 1 1 CO 1 + + + + 1 1 1 1 | Ol M vj OJ O 01 VO UJ >0 t Oi 00 to M O O O M OJ to O 4*- O OJ Ol s s 8 a ^ 1 U 1 + + + + + i i i | 1 to ^i oj vO ui O OJ ^j M 01 to vO 10 H b *o o b H bo i r S $ 3 S^fSsS I 8 1 + + + + + i i i i 1 i OJ O 01 OJ b io ff 2 g? 10 4- VO M VI OJ p 01 vp OJ O 01 OJ CO ON O O' >-J 00 w vO 24.0790 sr r !' II I ft. . ON o o o 00 O to 4. o OJ ^ O U) I Ol ^ O 10 01 VI 0\ Oi Oi ON 00 M CO Oi to OO OJ M % i O OJ H VO NO ON 00 vg 4>- O f M | 4*. ON Ol CO VI 4*. M W? ^ M NO \O 4" OJ ON IH OJ O M OJ VO vi vi 5 ON o 4v I <* i tO 01 vj CO ON OJ ^ % ON t OJ 4v M Ol Ol 4w 00 VI 00 H O 01 4>- 01 O M OJ to ON 4? ^O S^ 3> (OVI MOJ MVOO1 H V)IOHO4>-4>.>-i|0 vOOl MOJ MVOvboi MV1V) K) ONIO4>. M ^s i ONOIOJ OOJOvO w M4-V1^ (2 RN H M M M H f vO 10 ON4- ONO OvO OOto OJ OOVi OOO H ONO MO "S H4W4*. OO OOOl ONON OJ ON4- MOlOJ O OOVO 01 H to to to to to M 4k 10 O ONO OOOJ ONOOH 4h.tHMtOMtOVlOl4k.O Olt-OViOOONOJOV| " t I VI O Oi H tO t04t M 4v VI ONM Ovjvl to OOONCOO 4-OJOJ to OJ 4>- ONONIO M ? H to OJ * ^ OJ OJ M H ^ L O OOO4*4^OOV|O1 M 5OJ4^VJ ONO OOONO OO CO M M 1 JOJ MOJ 10 MUH ONIO OO4v 440 MECHANICS OF THE GIRDER. Placing the values of 2( 5 &)> 3 3 2 ( 5 e 6 ), 2 A- 4 , and * 2 3 2 A- 4 , in (491), we find e Weight of girder diagonals, in pounds, M _ W h L (o.O32i42/ 2 4~ o.393?5/z 2 ) 4 (o, 7 6 7 S 5 /> + 2.16563^) ii' 4 (0.020589/ 2 4- 0.32779^) (0.205393/ 2 + 5.15096^) 6 (0.015004/2 4- 0.26906^) (0.220852/ 2 4- 9.32845^) 8 (o.OII78l/ 2 4- O.22622/J 2 ) (o.2 3 o877/ 2 + I4.7IO96/? 2 ) 10 (o.oo9688/ 2 4- 0.19425^) (0.237972/ 2 + 21. 29814^) 12 (o.oo8229/ 2 4- O.I7048/! 2 ) (o.243278/ 2 4- 29.08846/^2) 14 (o.oo7i48/ 2 4- 0.15133/0 (o.2474Oi/ 2 4- 38-.o8o42// 2 ) 16 (0.006326^ 4- 0.13892^) (o.250705/ 2 4- 48.27465^) 18 (o.oo566o/ 2 4- o.i23i3/? 2 ) (o.2534i2/ 2 + 59.66964^) 20 (0.005 1 28/ 2 + - II 35 2 ^ 2 ) (0.255654^ 4- 72.26527^) 22 (o.oo4686/ 2 4- 0.10456^) (o.2 5 7 59 2/ 2 + 86.06221^) 2 4 Weight of girders without head system, in pounds, n _ w h (I.042262/ 2 4- 1.49688^) + ^ (i. I 86 9 o 5 / 2 + 3.26876^) 4 (1.41 2338/2 + 2.03690^) (1.597142/2 4- 6.86oo7/z 2 ) 6 (i.8ii55o/ 2 4- 2.56531^) (2.oi 7 3 9 8/ 2 + ii.6247o/^ 2 ) 8 (2.22I543/ 2 4- 3.10097/z 2 ) (2.440634/ 2 -f 17.58571^) 10 (2.6 3 68 7 3/ 2 4- 3-64326^) (2.865157/24- 24-747 I 5 /z2 ) 12 (3.05531 1/ 2 4- 4.19149^) (3.290360/2 4- 33.10947/22) r 4 (3-475 72 1/ 2 + 4.7438^ 2 ) (3-7i5974^ 2 + 42.67217^) 16 (3.897495^ + 5.30065^) ( 4 .i4i8 7 4/ 2 + 53-43638^ 2 ) 18 (4.32O2O3/ 2 4- 5- 8 5439^ 2 ) (4-567955 /2 + 654OO9O/2 2 ) 20 (4.743644^ + 6. 41402/22) (4.994 1 7o/ 2 4- 78.565 77/2 2 ) 22 (5. 167629^ 4- 6.97399^) (5-420535/2 4- 92.93164^) 24 153. The floor ; to be made of 2^-inch oak planks weighing 52 pounds per cubic foot. Width = q 16 feet in (432). CALCULATION OF THE WEIGHT OF BRIDGES. 44! .'. Weight of floor = F = ^| X 16 X 52/ = ^/pounds. (492) 154. The joists of oak; longitudinal, spaced 2 feet between centres, and consequently 9 in number, all of equal size. By (431) and (432), we have Depth of joists = b* = (9 X 10 X 2/^ + 20OO ^ finches, (493) ( 16 x 10000/2 ) calling B = 10,000 pounds = breaking inch strain for oak, f = 10 = factor of safety, g = 2 feet = distance between centres of joists. And, from (433), Weight of 9 joists = J - -j 0.001125-^(7^ + 200o/zZ) \ {I ) f 0.001125 (F + 2QooZ) \ pounds. (494) n 2 } ' 155. The wrought-iron I floor beams, ;/ I in number (single or in pairs, according to live load, as shown below), each bearing I panel weight of live load, of floor, and of joists, besides its own weight, which is provided for as explained in article 124. Also, the I-beams must bear the longitudinal strain due to wind pressure, and to initial strain on the horizontal diagonals inserted between them in each panel. Taking the proportions of the beam's section as in article D 124, and calling = 10,000 pounds = allowed inch strain for wrought-iron beams, (412) gives ( (F 4- J -f- 2ooo//Z) 1 8 ) ir Depth of I-beam = bd*, (497) by (161) ; B T being the allowed inch strain. . Now we will take bd 2 ='-S\ n (d in inches, / in feet), CALCULATION OF THE WEIGHT OF BRIDGES. 443 S = -bd 2 square inches. (498) That is, the cross-section, S, of a joist is taken equal to - times its breadth, b> multiplied by the square of its depth, d. Then, calling g = 2 feet, q = 16 feet, w = 52 pounds, t = - 5. feet, Z?, = 1,000 pounds per square inch for oak, equation (497) becomes, after reducing, /520/ \ = o.oo 1125! - + 2000 L J square inches, (499) (500) = 7-3125^ 25! - 4- 2OooZ )/ pounds. \ O 1 ' 63375 pound n n pounu, 0.1584375 pound, 4 0.1056250 pound, 6 0.0792187 pound, 8 - o6 3375 pound, 10 0.0528125 pound, 12 0.0452679 pound, 14 0.0396094 pound, 16 0.0352083 pound, 18 0.0316875 pound, 20 0.0288068 pound, 22 0.0264063 pound, 24 which are the weights of the 9 joists for oak. 157. If, instead of wood, we use wrought-iron I-beams to support the floor and load, we may assume the beam's cross- section to have such proportions (see manufacturers' tables) that / ion 444 MECHANICS OF THE GIRDER. where / = moment of inertia of section ; 5 = area of section, in square inches ; d = depth of beam, in inches ; / = length of bridge, in feet. Then, from equations (52) and (187), we have moments of external and internal forces, M = i x glF + 20ooZ\ x iz/ = *BJ = ls l_ s % q\ n J n d n ' /. - = , to be used with makers' tables, d zBi S = o.oooi5f - h 2OooZJ square inches, if g =. 3.2 feet, q = 16 feet, B l = 10,000 pounds per square inch. \ (502) Weight ight of 6 weight-iron longi-) = ^ = 6 tudinal I-beams, in pounds, ) x x I2/A (5<>3) / x = 6Z/ + - , -^- pound, 0.1300000 pound, 0.0866666 pound, 0.0650000 pound, 0.0520000 pound, o-433333 pound, 0.0371429 pound, 0.0325000 pound, 0.0288889 pound, 0.0260000 pound, 0.0236364 pound, 0.0216667 pound. 4 6 8 10 12 H 16 18 20 22 2 4 158. For the transverse I-beams supporting the longitudinal I-beams, the floor, and load, we have on each, exclusive of its own weight, F + L + 200o;zZ pounds. CALCULATION OF THE WEIGHT OF BRIDGES. 445 From (52) and (187), Take B^ = 10,000 pounds per square inch. q l = 1 8 feet = entire length of beam. We shall assume the cross-section of each transverse I-beam to be such that 7 = 2$d (55) whether the beam be rolled or made up of plate and angle iron. Therefore, from (504) and (505), / S = 0.000675 520/ 6Z/ - - 4- + + 2oooZ. (506) Weight of (n i) transverse I-beams due load, in pounds, = (n i) x -f s X 12 x 186* /520/ 0.5 2/ 2 6LI \ = 0.0405 ( - i)l H ^ h + 20ooZ ) (507) = / n 5.2650 + / 2 0.003949 + Z/ 0.18225 4- Z 243 4 5.8500 0.002925 0.20250 405 6 6.1425 0.002303 0.21262 567 8 6.3180 0.001895 0.21870 729 10 6.435 0.001609 0.22275 891 12 6.5186 0.001397 0.22564 i53 14 6.5812 0.001234 0.22781 1215 16 6.6300 0.001105 0.22950 1377 18 6.6690 O.OOIOOO 0.23085 1539 20 6.7009 0.000914 0.23195 1701 22 6.7275 0.000841 0.23287 1863 24 446 MECHANICS OF THE GIRDER. 159. In order that the wind pressure may be a function of the height, /i, of the girders, as it manifestly ought to be, we will assume, for wind pressure against these highway bridges, 50 pounds per square foot of actual vertical surface presented by both girders, estimated at X - square feet, to the run- ning-foot of bridge. Therefore Wind pressure per linear foot = 2500- pounds. Wind pressure per panel length = W t = 2500- pounds. No account is here taken of vertical wind force. Let the strains due this horizontal force of wind be provided for along the floor system which is placed midway between the top and bottom chords (that is, insert horizontal diagonals in each panel between the transverse I-beams), and increase the cross-section already found for these transverse beams by an amount required by the wind force ; and let the longitudinal strain due wind be taken up by 2 longitudinal wrought-iron bars, or channels, extending the entire length of bridge, securely attached to the outside of the outer longitudinal floor beams, to the top of every transverse beam, and to each girder diagonal, as already explained. These two longitudinal bars or channels, and the two out- side floor beams, are to be made continuous throughout, and capable of resisting either tension or compression. It will be noticed, that the floor and load use only one-half the capacity of these two outside longitudinal floor beams ; also, that all floor beams, longitudinal and transverse, are, from the manner of their loading, unable to deflect horizontally. 160. The Horizontal Diagonals, 2*1 in Number. To provide for travelling gusts of wind, we shall here assume that CALCULATION OF THE WEIGHT OF BRIDGES. 447 this panel pressure, W t = 2,500- pounds, is a uniform live load. Therefore the strains upon the diagonals are given in Fig. 112 if we put W l in the place of Z, and make W = o. But, since we must provide for this wind pressure coming upon either side of the bridge, it is plain that all horizontal diagonals must be "mains," and the two in any panel equal in size. We must, therefore, take - terms of the following series four times : Strain on horizontal diagonals due wind = l ^\n(n-i}, (_i)(- 2 ), (-2)(-3),... -terms}, (508) 2sin etc.]. (510) 3 These I-beams being unable to deflect horizontally, and having considerable depth, we may take for them, under this wind pressure, the unit strain, CALCULATION- OF THE WEIGHT OF BRIDGES. 449 8000 8000 pounds per square inch, 20000 x l I 4- 0. 933 I2 7 where -^ is put for the square of the radius of gyration about an axis normal to web of beam. Hence the areas of sections to be added to the I-beams, by reason of wind, are S=-^[(- i) 2 , ( ~2) 2 > ( - 3)% etc.]. (511) Taking m T \, q, = 18, JF Z = 2,500-, we find Weight of iron to be added to transverse I-beams, on account of wind, in pounds, _ 1\2 _^ (n - i) 2 4- ( - 2) 2 4- = 4 X 12 ^7 V ..(-- 1} terms /I ^ ' \ z / 2mfj,W,, -. , -. = - ' L (7 i2 4- 2) (;z even), 3<2a = M% + 0.972^7^ - 12 -f ^X \ 2 4 /A */ n = ^ 18.188 4- - 64.15 4 31-597 176.91 6 46,111 344.09 8 60.625 565-71 10 - 75-173 841-75 12 89-733 1172.18 14 104.297 I 557'i5 16 118.866 1996.49 18 133-438 2490.27 20 148.008 3038.47 22 162.587 3641.11 24 (5") 450 MECHANICS OF THE GIRDER. 162. The Chords required in the Horizontal System to resist Wind Force. The strains generated in these chords by the panel pressure of wind, W lt are given in Fig. 1.12 if, for N = W + L l, we put N = ^, thus : 211/1 2Jiq l .Maxima chord strains in horizontal system W I = ^-C(" ~ *)> 2 ( H ~ 2 )> 3( ~ 3), etc.], (513) for each chord, since the wind may act on either side of the bridge. Now, since these strains will be sometimes in tension and sometimes in compression, these wind chords must be con- structed so as to resist either kind of strain. Then, of course, a cross-section sufficient for the greatest compressive strain will be ample for the maximum tensile strain. And, because these chords are to be securely attached to the girder diagonals, to the outside longitudinal I-beams whose strength is only one-half taxed in supporting the floor and live load, and to the transverse I-beams, we shall take unsupported length r: ^ ; IOO, radius of gyration as in case of the top chords, article 151 ; also, call the ends fixed. The axis of gyration is normal to the plane of girder, since the floor prevents these struts from deflecting horizon- tally. Then Q = 3.2 tons = 6400 pounds = allowed inch strain. Cross-section of wind chords for each panel = JZ!\n - i, ,(* - ), 3 ( - 3), -.. ?( - *-)} (5M) 2nq l Q\ 2\ 2/J CALCULATION OF THE WEIGHT OF BRIDGES. 451 Weight of wind chords, in pounds, = 4 X X -- 1 8 n 0.006028164/2 -f ^ 4 - a),--- terms (5*5) = /z/ 2 0.0158239 0.0147353 0.0141285 0.0137442 0.0134800 0.0132866 0.0131395 0.0130238 0.0129304 0.0128534 0.0127889 n = 4 6 8 10 12 14 16 18 20 22 24 163. The vertical supports for the transverse floor beams and their load must also resist the moment due that part of the wind pressure which acts upon the chords and girder diagonals. ist, The total weight to be upheld by each of these vertical struts and hangers is the (2;/) th part of the sum of the weights of the live load, the floor, the longitudinal I-beams, the trans- verse I-beams taken times, the horizontal diagonals in 11 i the floor system, and the wind chords ; all of which have now been found in terms of / and L. Call this weight g x pounds on each strut -tmd on each hanger, ;/ being the number of panels. Each strut, of course, transmits the load, e M , to the panel point below ; while each hanger or suspender transmits the load, e w , to the alternate panel points above. 452 MECHANICS OF THE GIRDER. For the struts, we may take 8000 " = 5333 P unds 2OOOO as the allowed inch strain in compression. .*. Cross-section of a strut due) ? .... > = o = - square inches. (516) vertical forces ) Q 3 Weight of all struts due ) = 2 x J. x ^ x I22j; vertical forces ) 18 Q 3 ( 4^ = o.ooo2o8|f n \/is n pounds, (517) since, from (473), for lower parabola, zh ( fn \ ) 2v = < (211 4) + (AH 16) + (6/2 36) . . . ( i | terms i n 2 \ f Similarly, for the suspenders, take T^ =. 6,000 pounds = the allowed inch strain in tension. Cross-section of suspender) = s i Q ^^ due vertical forces ) T t Weight of all suspenders due) 2X j[_ x x I22 vertical forces ) 18 T t = 0.000185 f* 4- -V^ pounds, (519) \ n l since, for the upper parabola, -i)+ (3 - 9) + (5 - 25) f CALCULATION OF THE WEIGHT OF BRIDGES. 453 2cl, We shall assume, that, of the wind pressure, 125 pounds act upon each running-foot of the two top chords and of the two bottom chords, tending to rotate each girder about its longitu- dinal axis. We may note, that, in general, these forces act- ing upon one chord will be nearly balanced by the forces acting upon the other ; but in certain cases a gale may strike one chord and not the other. Acting, then, in lines normal to the plane of each girder, at 2/ / each panel point or apex, is the pressure of 62.5 X -- ='125- // // pounds, with the lever arm, y, causing a moment, at the wide end of the strut or suspender, of 125- y foot-pounds. We take n no account here of the fact that each end segment of the top chord is only about one-half the length of any other. Let these struts and suspenders, acting also as lateral braces to the chords where there is no lateral head system, have a breadth of effective base equal to -$y. The broad end of the suspender is to be attached to the top chord and head lateral strut whenever it would obstruct unduly the roadway below. Otherwise, and in all cases where head laterals are wanting, the suspender has its broad end securely bolted to the transverse I-beam in the floor system. Then, if S 2 = cross-section of the two members or flanges of each strut or suspender, we have, at the broad end of each, this equality of moments, 75 / = 7^ n square mches > ( 520) if B t = 1(5,333 + 6,000) 5,667 pounds = allowed inch strain in bending. 454 MECHANICS OF THE GIRDER. It will be noticed that the cross-section, S 2 , is uniform throughout the member if the two flanges meet at one end, as we shall assume they do, and shall illustrate in specifications. We have, then, Weight of verticals required to resist bending-moment due wind, in pounds, = 2 X 78 X 1^'n X I2 ^ = -93o 3 922(i - }U t (521) since now ^y = -[( - i) 4- 2(n - 2) + 3( 3) . . . ( - i) terms] = ^ - ^. From (517), (519), and (521), we find Weight of all vertical supports, in pounds, ^(0.000393518* - a0 f 2963 ) + 0.9803922(1-^/1 (522) (o.ooi i88Z7+2.3765Z+o.oooooo6737 3 +o.ooooi7i67 2 + 0.990247 +0.0081) (o.ooi2o6Z7+3.2i54Z + 0.0000006327 3 + o.ooooi3o67 2 + 1.003737 + 0.0152) (o.ooi2i4Z7+4.o464Z+o.oooooo6o47 3 +o.ooooio527 2 + 1.010357+0.0244) (o.ooi2i8Z7+4.8733Z+o.oooooo5847 3 +o.ooooo88o7 2 + 1.014707 + 0.0358) (o.oo 1 22 1 Z7 + 5-698oZ + 0.0000005707 3 + 0.000007 5^7 2 + i .01 7 587 + 0.0494) (o.ooi223Z7 +6.521 iZ+o.oooooo5597 3 + o.ooooo6627 2 + 1.019887 + 0.0650) (o.ooi 224Z7+7-3434Z+ 0.0000005 507 3 + o.ooooo5897 2 + 1.021727 + 0.0828) (o.ooi225Z7+8.i65oZ+o.oooooo54o7 3 +o.ooooo53i7 2 + 1.023427 + 0.1028) (o.ooi225Z7+8.986iZ+o.oooooo5377 3 +o.ooooo4837 2 + 1.024877 + 0.1249) (o.ooi226Z7+9.8o69Z+ o.oooooo53i7 3 +o.ooooo4437 2 + 1.026267 + 0.1491) 24 CALCULATION OF THE WEIGHT OF BRIDGES. 455 since, for s n in (22), we have 4 = o.7So37Z7 + I040-5Z -f o.oi69o8/ 2 -f 22.54417 -f 0.00252 1/// 2 -f 5.843/5 + 10.69-, e 6 = 0.5202 5Z7 + 1040. 5^ -f 0.00751 5/ 2 -f 15.02907 -f- O.OOI474/// 2 + 6.030/2 + 17.69-^, 8 0.390^7 + 1040.5^ -f 0.004227/ 2 + 1 1.27207 + o.ooiO22/z7 2 -f 6.i84// + 24.58-, f IO = o.3i2i5Z7+ 1040.5^, + o.oo27057 2 + 9.01767 + 0.0007 7 7 M 2 + 6.268^ + 31.43-, e I2 = o.26oi3Z7 + IO4O.5Z + o.ooiSjgt 2 -f- 7.51477 -f- o.ooo624//7 2 -)- 6.322/2 -f 38.26 *, f u = 0.22297^7 + 1040.5^ + o.ooi38o7 2 + 6.44127 + 0.000520//7 2 + 6.359/2 + 45.08-, I6 0.19509^7 -f- IO4O.5/, + O.OOIO577 2 + 5.63607 + o.ooo446//7 2 + 6.387^ + 51.90-, e I8 = O.I7342Z7 + 1040. $L -f o.ooo8357 2 -(- 5.00987 -f o.ooo389/^7 2 + 6.407/2 -f- 58.71-, e 2Q 0.15608^7 -f- 1040. 5^ -f o.ooo6767 2 + 4.50887 + o.ooo346//7 2 + 6.424/5 -(- 65.53-, e 22 o.i4i89Z7 + IO4O-5Z -f o.ooo5597 2 -f 4.09897 + 0.00031 1//7 2 -f 6.437/2 -j- 72.34-, e 24 = o.i3oo6Z7 + I040-5Z + o.ooo47o7 2 -f 3.75747 + o.ooo282/27 2 + 6.448^ + 79.15-, which, as above defined, is the weight upheld vertically by each support. It will be observed, that, in all terms involving h z in the expression for weight of vertical supports, equation (522), // 2 has been replaced by \hl. This substitution is simply for conven- ience, and, being made in these small terms only, does not practically affect the accuracy of our resulting equations, while we are hereby relieved of higher powers of k than the second, in the value of W. 164. As additional security against deflection, out of the plane of the girder, by the top chord, we shall insert a system of head lateral bracing between the two top chords where the MECHANICS OF THE GIRDER. height is sufficient. For this purpose, the two top chords are the flanges of a great longitudinal strut or column, whose tendency to deflect laterally must be overcome by this head web system of diagonals and struts. Suppose the moment at the centre of this system be that due to \ W* acting upon each panel length, or to \ W l acting upon the windward side at each joint of the windward top chord ; that is, by (480), where now we must put \n for ;/, and \n for r, and 2,500- pounds for W u we have n Moment at centre = M = -faWJn = 78.125/^7; and the longitudinal horizontal strain, H = 78.125 pounds at centre, ?i ?i which in each flange may be considered to decrease uniformly to the ends, as. is practically the case with that part of the strain due to bending-moment in a pillar. Therefore, for each double panel length, *ff = H X ~ = 312.5; \n nq, requiring each diagonal tie to resist 312.5 /// and to have a cross-section 312.5^7 . S - - - - - r - = - ^r square inches, (523) 15000 cos a cos 2 nq l CALCULATION OF THE WEIGHT OF BRIDGES. 457 since cos a may, for these central panels receiving the head system, be put = i without practical error. Length of each head diagonal = practically. n cos

2 , we have, by our specifications, the end pressure brought by each pair of diagonals upon the end of each head strut ; and, calling the inch strain in compression on these struts 2,500 pounds, wo have the cross-section of each head strut, in square inches, 453 MECHANICS OF THE GIRDER. 312.5 X 2 X ioooo///tan 2 15000 X 2500/2^, " 1 ^ h being in feet, and tan < 2 z= ^. Weight of I - 2 J head struts, in pounds, = h n o 4 5 6 10 8 15 v 10 20 12 25 14 30 16 35 18 40 20 45 22 5 24 (sa6) h in feet. Since we have already provided, in the floor system, for the whole bending-force of the wind, and are now simply stiffening the head system laterally as a column, or to meet adjustment strains, and strains due to imperfections in workmanship, it will manifestly suffice if we call the additional chord strain in each segment of top chord within the head system equal to = 312.5 = 17.36$ pounds. And, as 6,400 pounds is the allowed inch strain in top chords, we have CALCULATION OF THE WEIGHT OF BRIDGES. 459 Cross-section to be added to each segment of top chord due to strain on head diagonals, square inches, = S = 0.00271267 . (527) Weight of added iron in ( - - 3 J of the central double panels, for top chords, in pounds, (5=8) n - 4 6 0.0005651 8 0.0007234 10 0-0007535 12 0.0007381 14 0.0007064 16 0.0006698 18 0.0006329 20 0.0005978 22 0.0005651 2 4 165. To find the Necessary Amount of Material for the Triangular Web System of Latticed Struts or Columns. Let /' = length of strut. d = effective width of strut. A = area of both flanges in section normal to axis of strut, in square inches. A l = area of diagonals in the same section normal to axis of strut, and not to its own axis, in square inches. 45 inclination of diagonal to axis of strut. BI allowed inch strain, both in flanges and diagonals, in the present case. 460 MECHANICS OF THE GIRDER. Then, moment at centre, M = \AdB M Longitudinal flange strain at centre = H = = a Now, since H decreases uniformly from the centre to the ends, at least practically, which is the longitudinal component of diagonal strain. And s- cos L = strain on diagonal ; / COS0 Ad , v * (529) = A if// -^- ^ = 20 > = T Vif/ A -^ = 30, = JQ If /' -5- + & Tf DC 3 464 MECHANICS OF THE GIRDER. 37 o LO " ft vo vo ON O i^, ^ fO OO vy-i CO + < - r-* w 8 w 1 is I n \\ + ."S o 4- CALCULATION OF THE WEIGHT OF BRIDGES. 465 a ft 5 + 8, + ^ 8 o + s? l-O iy-) u-) q ~ 0.0245504/ 2 + 1 + 96.30706) + o. + i VO CO 3 ic io - co 00 s o c? O 6 + CJ CD 10 C\ CO VO fl C XI XI = c s ^ V i> O CN VO V O ^J" ^^ CO 10 To i- CN CO O C i ro co ^ s s II . i 466 MECHANICS OF THE GIRDER. In all these cases, the value of h which renders W a mini- mum has been found by the simple method of article 140, equa- tions (469) and (470). The limiting spans just given have been determined by putting the denominator of (543) equal to zero, and substituting the assigned values of h. It will be seen that these limiting spans are independent of the live load, nL, and therefore represent the limit to the length of each girder imposed by its own weight. The effect of live load on the limiting span will be considered below. 167. Having found W and //, it is easy to compute the weights of all parts of the bridge from the expressions for weights in terms of W, /i, L, and /. The computation affords a perfect verification of the accuracy of the work. We give below a table showing the number of panels and the height, which simultane- ously render the total bridge weight, nW, a minimum for various spans ranging from 50 to 1,000 feet, and have thus probably ex- tended the table far beyond any economical use of this girder. Of course, we find great heights ; but it should be remem- b**ed that one-half of this central height, h, is below the plane of the floor system, where the points of support are situated. Also the width, 18 feet, becomes too small for the highest girders ; but it has been retained in this set of examples, to pre- serve uniformity in data. To illustrate the change in central height and bridge weight, as the number of panels varies for the same span, we have given the solutions corresponding to three values of n> including that one which renders n W least. Also bridge weights and central heights are given for 2,000, 3,000, and 4,000 pounds of live load to the running-foot ; the weights being minima values. The 3Oth line of this table exhibits the effect of a small live load upon the length of the limiting span, as resulting from the substitution of \W for L in the equations for weight. Of course, we do not mean that the live load is small near the limit when W is infinite. The reader cannot fail to notice how prolific in useful and in- teresting results these general equations for bridge weight are. CALCULATION OF THE WEIGHT OF BRIDGES. 467 N -^00 \O OOO O "t- t^OO OO 04 N O _ . rOvO M \O O N **)-CO ? S S S ^ S 00 So? "M S; M - "^ lOroiow fxTf-pioo i^-^-r^ot Q o O O<^O ^O PI O m ro M i-^ ro^O O N M ro ooo t^ m 1000 V - ^^^^ ?* * <8 o^. ^4 ?^> M ^oo" 21 o' d" t^ H-*MMfOOt^OM *vO t^ N ro N 10 P) Ml H^ M oo^oo^ ro tovq^ t^. M^ ON 10 roMONt^.roONOm-*ONNinc)-* N ^^ * C'J ) MH IOC>1 H' IOC ON VO 99 ^Zo co ro H ^ m O co co ffi 8 . 8g\8?8;ra8raR*,*fr%aa . ^ ^- CM CM ON -^-NO CMCMHCMCMt^CMHCMinnt^CM t^OO CM in K e 3. 8> ~* n m i & 1 2 "s a 8 '5 " 'oc'c"3s - sSt? .S 3 - .S 3 .3 . rt ,-" ID o Q 5 H = 1 J-g 7 ^ ^ "S s T g. M " ^ -S, 0* "."fil 1 |l^ w . . ;|S,;2,;|M .jr.'j-s -s s jv| i i jr^ ^ M -I -gag ^||>^ .2 i ^^^ ^ 111 rffM -III lilll'B 1 ^ i 1^ I fl 1l ?g r lllllJ!l -1 -ii^ 1 ^a^MfSfS &< cS s HHBOSi-!HH3J?i>3}WESpQS HH HlnHj H CM rO^-^NOt^OOON SHHH"H^H^?S5Sc?ctw'N C Sc?^> : CALCULATION OF THE WEIGHT OF BRIDGES. 469 .2 03 P M O 00 O 2 - vo m t> v 10 t^ VO ll Illl 10 moo O m M \O 1000 -j- t>. o^ m ^ N - ^ N - ocT cT ^? cT t-T ON rC m t^vo tC m -*& t^co oo ^f O 10 m O moo cj *-* IX.M--^-O t^w IOIOM m miOM tCn t^iOO 4 ONVO IOOO VO OO t^ ON N &*8 m 10 mvcj 'S^SL- 00 j - L ^ ? ^ H: ^ i^ 4vi M -^-M ro-^-W -^-O cow ci moo ^ O 10 ov Ov -i-vO O ON M 10 TJ-O w * 10 m 10 * M m M * O * t^NO NCIO M->J- MP^MO 00 to N M M VO M 00^ M - N " c^ t^. M mvo O ^- xovo O WVONO t^t^t^^- 1^.00 CONO 10 o N oo O co moo <*- ^L -4- o" t^co'vo'vo" "" o" t^ N"O" -^oo" d> m M oo N ON 00 M O OxOO OOvOmiOON f) MOMCV1 ^ m M m ^ M" H* 10 10 10 m t*^ 1000 MOO O M ^mm-^-o -^-M ^-oo -I-NO_O 1000 mvo IN 10 M cv 10 ^ moo O vo"o6" 10 M*O~ O"oo" ON ON tC -^OO'ND" m M t~. N OO r^MOoocooNiomioov d MQMCNI - -> w O ONOO NO t^ ON m m cvi m^t-M 10 mvo H oo - ^00 N_ m o, cvi^ i-. c^vo^ + H - ^.^ og- sg mONONmcj P..N t-.. |,JJJJJ|jJ^4JJJ ll Heigh ds pe above 'oo'vo OOCIOOONONQ-OMCNICVI ^lO-^-'^-'^-Lrjtoioioioio ilii! in* 2-8 a t -5 arts, us Thousan Foot of S rm Live -I -I TQfi 'il'l'lll 'Is, 'I'll - 5 ^Ja3(2d d &< S * E^c^'::-,j.K^>ffiffi 4/0 MECHANICS OF THE GIRDER. vo M o tf 1 2 en ps w Q X ?1 t oo c? ,/, R. co co" 38 u t "? H- - 8.SS. ~ 5 o vq q ^ o q^ . C^ rOMD 00 "^" CO ^ O Jill ro t^ -^- o tx o : 8S-S?S?S < ? ONroO > CN?c?Incoq' s ~ OO ON ? o^r^ q, O H ON i-^OO^ ON CO O^ ,-;-, ^ Jj" ^ ON M CM ON O m COVO O inM ino >H cv t^ u^ ,0 ON O irjco O >o M \o \O HlOOHW^OliO 2 STlr, O 8 vo Jn S. co O 1 m !? CM coot? *- 8 int^inq^^f'inw min^j-co ONOO ; tvT o" cTvo" co in covo" ON ONOO" ON 6" m H vo co co m M in ON M CM r^vo in co w cvi t^ CM in OO ** CO M t^\O ON\D COCOwvO CO-^J-O t^ON-^* W in CM -rt-vD CO N 10 Tj- CO N r>. rt- P) ON -* CO N ON TJ- in Ovov Pioo co in tC cT Tl- r^ M (M COOO O CO^O O covo HMWHVO cT H" H" in r in ONOc M ON CM NHCOm WONCS^ CO vo ^ " O oo oo t^ t^ ON r^oo < ^- OO ONOO O vO CO ~ ~ t~- *OO I VO_VO K VO, M vp ON invj vo" CM" -f tCco"oo" CM" o" inoo"oo"vo" ^"00" M t^ CM i cooo O t^ ro o O -*-oo IN co in CM ON 1?^ q_^; co ^- cooo O cococoo m 2& S^ici. CJ - 2 J 2 2 .a I I 11 1 ' ? 'S 'o rt ^ 5 -1 I 1^3 il 'So^. t^ OO ON O H (M CO * invO t^OO ON O M ^ CO * invO t^OO ON O i-c CALCULATION OF THE WEIGHT OF BRIDGES. 4/1 168. Among the inferences which may be legitimately drawn from the table of article 167 in regard to bridges having two lenticular Brunei girders of single system of the same width but of varying span, height, and uniform live load, are the following ; viz. (see Fig. 116), ist, The best number of panels varies approximately as the cube root of the span, n <* fi nearly. (544) 2d, The best central height for a uniform live load of 2,000 pounds per linear foot is about X span, 4.2 h =l nearly; (545) and for different loads the best height for same span and same number of panels varies very nearly as the sixth root of the live load, h oc (nL)~* nearly. (546) 3d, For spans less than 500 feet, the least bridge weight varies approximately with the product of the best central height of girder multiplied by the span (that is, with the geo- metrical area of the girder, since the parabolic area is propor- tional to this product) ; / < 500, nW c< lh nearly. (54?) 4th, For the same span and same number of panels, and best central height of girder, the least bridge weight varies approximately with the square root of the uniform live load ; nW 'oc (nZ,y* nearly. (548) Or, by reason of (546), the least bridge weight for same span 472 MECHANICS OF THE GIRDER. and best central height of girder varies nearly as the cube of this best central height ; nW oc (nL)* oc fa nearly. (549) 5th, For each span of 500 feet and over, a large increase of live load, and consequently of best height, causes a diminution in the number of panels corresponding to minimum bridge weight. 6th, Small deviations from the best height and best number of panels, or from either of them, do not greatly affect the bridge weight ; but large deviations either way, in this respect, cause a great increase in bridge weight, as shown in sixth line of table, h = -j 1 ^/, thus rendering the girders only about one-half as high as minimum bridge weight requires. 7th, The limiting span increases slowly with the number of panels, till a maximum value depending upon - and - is JLf It reached. 8th, We cannot, for a given span, assign the best height and the best number of panels till we know the live load which is to be imposed. 169. EXAMPLE. Take span / = 200 feet, number of panels n = 14, central height of double parabola h = 42.585 feet, uniform live load nL = 200 tons = 400,000 pounds, width of 2|~inch oak floor q 16 feet, length of transverse I floor beams ^=18 feet. (See Fig. 16.) Loads applied at all apices equally by means of struts and suspenders which sustain the floor system in the plane of the axes of girders. Assume that all cross-sections of members may be strictly adjusted to the developed strains. Weight of live load 400000 pounds, Weight of floor, article 167 . . . . 34667 pounds. Total load on longitudinal I-beams . . 434667 pounds. CALCULATION OF THE WEIGHT OF BRIDGES. 4/3 Load on i panel length of each longitudinal I-beam spaced 3.2 feet 434667 By (502), Cross-section of beam = S = o.oooisf ? X + 2000 X \ 3 = 4.65714 square inches. In order to satisfy the condition in (501), we must have, article 62, (550) ion S = bd - b^ = 4.65714; (55 1 ) from which equations we find O = (# _ bb*)d* - $b 2 Sd 2 -f tfS* + i.2-^/ - S3. (552) Take = 4.00 inches = breadth of flange. b b,_ = 0.26 inch = thickness of web. Then, from (552) and (551), d = 9.080 inches = depth of beam, d 6 ^/ 8 2 X 10000 9.08 by reason of (502) and (52) ; the load being uniformly distrib- uted on each panel length of beam, and these beams not being regarded continuous over the transverse beams. Weight of these 6 longitudinal I-beams, by (503), equals /, = 6 x fg x 12 x 200 x 4.65714 = 18628 pounds, as ^ivcn in preceding table. 474 MECHANICS OF THE GIRDER. It will be noticed that these beams are deep and compara- tively thin ; but, considering their area of cross-section, it will also be noticed that their moment of inertia is great as com- pared with ordinary beams of equal area of section. Supported by the transverse I-beams, we have Live load = 400000 pounds, Floor = 34667 pounds, Longitudinal I-beams = 18628 pounds. Total for 14 panels = 453295 pounds. Load on i beam = 32378 pounds. From (504), we have / 12 x 18 X 32378 d = 8 x 2 x 10000 = 4 by (505) ; .*. S = 21.855 square inches for vertical load. But, in order to resist the assumed wind pressure, W^ = 2,500- 7,604 pounds per panel, we must add to the cross- section due vertical load the areas found from (511), where now O 2 = - - =71500 pounds per square inch ; I + 0.93312 X 0.07 2 X 7604. 3 x .4 x = 8.149 square inches, ist and i3th beams; = 6.944 square inches, 2d and i2th beams; = 5.835 square inches, 3d and nth beams; = 4.822 square inches, 4th and loth beams; = 3.906 square inches, 5th and 9th beams; = 3.086 square inches, 6th and 8th beams; = 2.363 square inches, 7th beam. CALCULATION OF THE WEIGHT OF BRIDGES. 4/5 TOTAL SECTIONS. S + Si = 30.004 square inches, ist and i3th beams; 28.799 square inches, 2d and i2th beams; 27.690 square inches, 3d and nth beams; 26.677 square inches, 4th and roth beams; 25.761 square inches, 5th and 9th beams; 24.941 square inches, 6th and 8th beams ; 24.218 square inches, 7th beam. 351.962 = 2 x 0.78329; 2n X 15000 sin (^ 0.023114. 0.023114 x 14 X 13 = 4.206 square inches = section of ist diagonal, 0.023114 x 13 X 12 = 3.605 square inches = section of 2d diagonal, 0.023114 x 12 x IT = 3.051 square inches = section of 3d diagonal, 0.023114 x ii X 10 = 2.543 square inches = section of 4th diagonal, 0.023114 x 10 x 9 = 2.080 square inches = section of 5th diagonal, 0.023114 x 9 X 8 = 1.664 square inches = section of 6th diagonal, 0.023114 X 8 x 7 = 1.295 square inches = section of 7th diagonal. 18.444 X 4 = 3S = 73.776 square inches. = 12 x X - 1 8 s = 5652 pounds, Weight of 28 horizontal diagonals = 12 x X -7^ X 73.776 1 8 sin < x as given in the table. The cross-section of each panel length of a wind chord is shown in (514), thus : J*V_ = _ 7604 X 200 _ **4iQ 2 x 14 x 1 8 x 6400 0.471478 x 13 = 6.129 square inches, ist panel; O.47I478-X 24 = 11.316 square inches, 2d panel; 0.471478 x 33 = 15.559 square inches, 3d panel; 0.471478 x 40 = 18.859 square inches, 4th panel; 0.471478 x 45 = 21.217 square inches, 5th panel; 0.471478 x 48 = 22.631 square inches, 6th panel ; 0.471478 x 49 = 23.103 square inches, 7th panel. Total, 118.814 square inches for one-half of i girder. .47$ MECHANICS OF THE GIRDER. .-. Weight of both wind chords = 4 X -5- X I2 X 2 x 118.814 18 14 i = 22631 pounds, as by (515). We now have, upon all vertical supports and abutments, Live load ..... 400000 pounds, Floor ....... 34667 pounds, Longitudinal I-beams . 18628 pounds, Horizontal diagonals . . 5652 pounds, Wind chords . . . . 22631 pounds. % x 481578 pounds = 17199 pounds, H -- X weight of transverse I-beams = - -pounds = 812 pounds. 26 26 Load on each vertical, article 163, = s n =18011 pounds. Therefore, by (516), ,5 = %%" = 3-377 1 square inches = cross-section of a strut, due vertical forces ; and, by (518), (ToV" = 3ooi8 square inches = cross-section of a suspender, due vertical forces. From (520), 75 200 S 2 = -j^ x ~^~ 6.3025 square inches = cross-section of each vertical due bending-moment of assumed wind force ; /. S + S 2 = 9.6796 square inches for each strut, S, -}- S 2 = 9.3043 square inches for each suspender. CALCULATION OF THE WEIGHT OF BRIDGES. 479 From (473), we have length of verticals, Suspenders. Struts. y = 0.43404 x 13 = 5.6425 feet. r = i 0.43404 x 24 = 10.4170 feet. 2 0.43404 x 33 = 14-3233 feet. 3 , 0.43404 X 40 = 17.3616 feet. 4 0.43404 X 45 = T 9-53 l8 feet. 5 0.43404 X 48 = 20.8340 feet. 6 0.43404 x 49 = 21.2680 feet. 7 Sum required = 200.5268 feet. 194.4504 feet, for all. Longest strut, 20.834 feet = 250 inches; therefore Required radius of gyration = -fgj} = 2\ inches. Each vertical may be made of 4 channels, 6 inches wide, each having an area of 2.4199 square inches for struts, 2.3261 square inches for suspenders, latticed in pairs, and two pairs in one brace. Weight of all vertical struts = -fg x 12 x 194.4504 x 9.679 = 6273 pounds, Weight of all vertical suspenders = -f s x 12 x 200.5268 x 9.304 = 6218 pounds. Total, 12491 pounds. Add one-tenth for lattice braces, 1 249 pounds. 13740 pounds, which accords with (522). Equation (523) gives the cross-section of each head diagonal thus : s = O.Q2081 X 42.585 X 200 = square inch) 14 x 18 X 0.84609 which requires a round rod 1.056 inches in diameter if the ends are enlarged for cutting threads of screws. 480 MECHANICS OF THE GIRDER. Weight of the 8 head diagonals, by (524), is 8 x ^ X I2 X 2 * 20 X 0.831 = 749 pounds. 18 14 x 0.84609 Cross-section of each head strut, by (525), is yV X 4 2 -5 8 5 = 3-549 square inches. Add one-tenth for latticing, 0.355 square inch. 3.904 square inches. Weight of 5 head struts, by (526) = 25 x 42.585 = 1065 pounds. Add one-tenth for braces = 106 pounds. Total, 1171 pounds. Since for these head struts we have assumed 8000 Q = 2500 = - - pounds per square inch, 2OOOO 12 X 18 OC 2IO ^-^ p P .. p = |ij = 1.03 inches = radius of gyration. We may therefore use, for each head strut, 2 4-inch channels latticed so that the web shall be 4 inches apart. The increment of section of each top chord due to diagonal strain in head system is given by (527), thus : 42.585 x 200 S = 0.00271267 X - = 1.65 square inches. *4 The total weight thus added along the 4 panel lengths of head system is 2 X 4 X -5- X I2 X 2 X 20 x !.6 5 = 1257 pounds, 18 14 as by (528). CALCULATION OF THE WEIGHT OF BRIDGES. 48 I The strain in the top chords is given by equations (476), (481), and (482), where now W -f- L = 7.72265 -f- 14.28571 = 22.00836 tons. For the segments of each top chord, the total strains due n ( W + L) are P t = 101.15 tons > ^2 = 9 6 -55 t011S > /> = 93.10 tons, P 4 = 91.55 tons. Dividing these strains by the allowed inch strain, Q = 3.2 tons, we get cross-sections, 5, = 31.6075 square inches + ) S, = 30. 1 704 square inches } for head s y stem > S 3 = 29.0937 square inches -f- 1.65 square inches, *S 4 = 28.6078 square inches + 1.65 square inches. Now, the longest unsupported segment of top chord is 2/ '- = 29.405 feet = 352.86 inches. n cos 2 /. 3.5286 inches = radius of gyration. Therefore the top chord may be made up of 2 Q-inch chan- nels and a plate, or 2 plates 14 inches wide, and having such thickness as is required to complete the area of section. Weight of top chords due load = X X -^2^ = 44741 pounds, 10 18 n cos a 1257 pounds, due head system. Total, 45998 pounds. 482 MECHANICS OF THE GIRDER. Similarly, for the segments of each bottom chord, the total strains due n(W + L) are, from equations (477), (481), and 4483), /i = 100.128 tons, U 2 = 99.513 tons, U z = 92.133 tons, ^4 = 9 1 -3 74 tons. These strains divided by 5, the allowed inch strain in ten- sion, give the cross-sections of the successive segments of bottom chord in each girder, St = 20.0256 square inches, S 2 = 18.9025 square inches, S 5 = 18.4265 square inches, S 4 = 18.2748 square inches, from which the links can easily be made up according to speci- fied forms of body and head. No change is here made on account of longitudinal compo- nent of lateral diagonal strain, since in the present case there is no lateral system between bottom chords, by reason of gravity. Weight of all bottom chords increased by -^ = X -5- x ^2 = 28695 pounds. 10 18 11 cos/? The equations (490), (491), and (478) give cross-sections of alternate girder diagonals, thus : St = 5.115 square inches, 5 2 = 7.212 square inches, 5 3 = 8.664 square inches, 5 4 = 8. 88 1 square inches, 5 5 = 7.750 square inches, 5 6 = 5.132 square inches, CALCULATION OF THE WEIGHT OF BRIDGES. 483 for each of the two girders, the alternate set being the same inverted ; and the weight of all is, calling <2i f > and multi- plying by 1.8 X -i"J, as specified, 4 x J_ x 3 X 1.8 X ii x 12 x 200^ S _ 990^ S 1 8 8 x 10 14 cos0 7 cos# = 23184 pounds. Now, since the longest unsupported length of any girder diagonal is - X = 22.253 f e t =267 inches. 2 n cos A we have radius of gyration = 2.67 inches ; and therefore 2 8-inch channels latticed 8 inches between webs will suffice for the longest diagonals. We have thus determined the size and weight of all parts of this bridge, and find the total 216,233 pounds, as by table. STRAIN SHEET. STRAINS, TONS; CROSS-SECTIONS, SQUARE INCHES. For each of Two Girders. 20. FIG. 118. Span, 200 feet; central height, 42.585 feet (best) ; uniform live load, i ton = 2,000 pounds per linear foot, applied at all apices by vertical members. Regarding the greatest strains upon the chord pins as acting in "quadruple shear," and allowing 6,000 pounds as the inch strain in shearing, these pins will* require a diameter of 3^ inches. 484 MECHANICS OF THE GIRDER. It remains to compute the deflection of this girder under the allowed chord strain of BI = -J(3-2 -f 5) = 4.1 tons per square inch. For this purpose, use equations (318) and (319) combined thus : D = fr ^-3862950 - 2. 3 02 5 8 5 [0 + *)log(a + x) jfA t + (a - x)log(a x) - 2a log a] \, (559) where a = J/ = 100 feet = half-span, and x is measured from the centre. Also /^ = h 42.585 feet, and we will call E = 24,000,000 pounds = 12,000 tons per square inch. We have then, from (559), Deflection D l = 1.3347 inches for x = o, centre ; 100 J} 2 = 1.3150 inches for x = 200 D 3 = 1.2549 inches for x = > 300 /? 4 = 1.1521 inches for x = > 400 > 5 = i .0006 inches for x = > 500 < D 6 =. 0.7911 inch for x = > 600 D 7 = 0.4954 inch for x > A = o inch for x = ends - The proper camber may be given to the girders by equation (366), thus : X = X length of one parabolic chord. CALCULATION OF THE WEIGHT OF BRIDGES. 485 This length is given by equation (140) ; viz., = ( 2 oo 2 + 4 x 42-585 2 )* 200 2 2h -f (/ 2 + 4/; 2 )* + ' 287823 X log - -" ~ = 2 5'5 feet > .-. A = - - x 205.5 = 1-685 inches 12000 = length to be added to top chord. Or, each segment should be lengthened by -~- 0.241 = J inch nearly. SECTION 3. The Brunei Girder of Double System. 170. We now take the girder shown in Fig. 22 ; but we will apply the dead and live loads at all apices by means of verticals whose upper half will act in tension, and lower half in com- pression. These verticals must also resist bending-moment due wind. Each girder has two equal parabolic chords, and the floor system is in the plane of girders' axes ; each panel length of chord is straight, and the number of panels may be odd or even ; each system will be assumed to do one-half of the work. The height between the two parabolic arcs at the centre being /i, the height at any apex is given by (474). Equation (473) gives y. tan r = tan ft = - ^ r - = ~r( n r r -*)> (5 6 ) cos 2 a r cos 2 ft n 2 l 2 486 MECHANICS OF THE GIRDER. , n _ ( \ n I I I . A./1 r- f >.. x \ ~1 2 //T\ 171. Moments at all apices due total dead and live uniform loads are given by (65), jf -W + L, . 2Vllf f(7l / J I y 2/2 and the horizontal component of chord strain is -Z)^; (564) that is, this component is uniform throughout the girder under uniform load. TT Strain in top chords = P = , COStt TT Strain in bottom chords = U = , COSytf Cross-section of top chords = P -i- Q, ^=3.7647; Cross-section of bottom chords = U H- T t T = 5.0000. Volume of a segment of top chord is, therefore, * 2lH = UW + Z) cubic inches. (565) nQcos 2 a J Q/icos 2 a Weight of top chords, in pounds, 4g _ i\ ( 6) r 3^ 2 \ / + 0.14757^ ~ ^V | 2 x CALCULATION OF THE WEIGHT OF BRIDGES. 487 W 4- L n ' \ j-t 0.442709/2 4- 0.55338^ 4 h /* -5533 8 6/ 2 4- 0.70833/^2 5 0.664063^ 4- o.86o82/z 2 6 0.774740/2 4- I.OII9O/* 2 7 0.885418/2 4- 1.16211/^2 8 0.996095/2 4- 1.31172/^2 9 i.io6772/ 2 4- 1.46093^ 10 1.217449/2 4- i. 60984^ ii i-328i26/ 2 4- i. 75853^ 12 i.4388o4/ 2 4- 1.90705^ 13 1.549481/2 4- 2.05542/^2 14 1.660158/2 4- 2.20370/42 15 1.770835/2 4- 2.35188/^2 16 1.881512/2 4- 2.50000^ 17 1.992190/2 4- 2.64804/^2 18 2.102867/2 4- 2.79610/^2 19 2.213544/2 4- 2.94400/1* 20 2.324221/2 4- 3.09192/^2 21 2.434898/2 + 3.2398^2 22 2-545576/2 + 3.38767^ 23 2.656253/2 4- 3.53554^ 24 Similarly, we have Weight of bottom chords, in pounds, yn(W 4- Z)/ 2 X 2/a (IV -\-L}l 2 ( , 4// 2 / i\ ) , , , W 4- L A */ ) __ ** \^ L' ~h Q-333333/ 2 + 0.41667/^2 4 /* 0.416667/2 4- 0.53333/^2 5 0.500000/2 4- o. 648 1 5/; 2 6 -5 8 3333 /2 + 0.76191/^2 7 0.666667/ 2 4- 0.87500^ 8 0.750000/ 2 4- 0.98765/^2 9 - 8 33333 /2 + i.ioooo/^ 2 10 MECHANICS OF THE GIRDER. W + L i.oooooo/ 2 -f- 1.32407^ i-o83333/ a 4- i.4359 0/ * 2 i.i6666y/ 2 4- 1.54762^2 i. 250000/2 -|- 1.65926/^2 1-333333^ + I.77Q83/* 2 1.416667/2 4- I.88235// 2 I.500000/ 2 4- I.99383/* 2 i-5 8 3333 /2 + 2.10526/^2 I.66666;/ 2 -1- 2.21667/^2 I.75OOOO/ 2 4- 2.32804^ I-833333/ 2 + 2.43939/^2 i.9i6667/ 2 4- 2.55072/z 2 < 2.OOOOOO/ 2 4- 2.66204/^2 II 12 13 14 J 5 16 i7 18 J 9 20 21 22 23 24 172. For the advancing uniform live load of \L at each upper and lower apex, or of L at each vertical section through apices, we have at foremost end, by (64) and (474), z/, 2?z 2 - r) ^ n r) ' = (r + (568) and at one interval before the foremost end of live load, by (68) and (474), ~r(r + i)( - r - i) (H L )^ T = ^t- 1 - ^7 = ^r. (569) (?* 4- i) ( r i) /z 2 Therefore (570) which is the horizontal component of strain on both diagonals of a panel, on the present assumption that the two diagonals do equal work, and that the whole load is on one girder. CALCULATION OF THE WEIGHT OF BRIDGES. 489 Hence, for each of two girders, we shall have Cross-section of a girder diagonal = t cos (571) according to our specifications for members alternately in com pression and tension. Putting m = -f$, <2 X f, we find Weight of girder diagonals, pounds, = 4 X \zrnl X - onh Qi ! T f- _4 n i5/ 2 n 2 L h 0.140625/2 4- i-722667z 2 4 fi o.i6875o/ 2 4- 3.09600/52 5 0.187500/2 4" 4'7783/^ 2 6 0.200892/2 4- 6.743447^ 7 0.210937/2 4- 9-oi3i8^ 2 8 0.218750/2 4- 11.58025^2 9 0.225000/2 4- 14.445007^ 10 0.230113/2 4- 17.607817^ ii 0.234375/2 -f- 2i.o69Oi7z 2 12 0.237980/2 4- 24.828S67; 2 T 3 0.241071/2 4- 28.88759^ 14 0.243750/2 4- 33.245337^ 15 0.246093/2 4- 37-90228^ 2 16 o.248i6i/ 2 4- 42.858547^ 17 0.250000/2 4- 48.11420/^2 18 0.25 1644/ 2 4- 53-66934/J 2 X 9 o.253i25/ 2 -f 59.524037^ 20 0.254463/ 2 4- 65.678337^ 21 0.255682/ 2 4- 72.I3228/* 2 22 0.256803/2 -h 78.885927^ 2 23 0.257812/2 4- S5.93I79/* 2 24 L\ / 2\ / 30 63 30 = -r< 0.28 1 25! i I/ 2 4- o.o7s( 2/z 2 5 H - 2 ; h ( D \ ;z/ /:> V n n n z 490 MECHANICS OF THE GIRDER. All girder diagonals must be so constructed as to transmit stresses of tension or compression. 173. Collecting the weights now found for top and bottom chords and girder diagonals, we find Weight of girders due to loads, pounds, w h 2 4- 0.97005^ 0.970053/2 4- i.i64o63/ 2 4- i-5o897/2 2 1.358073/2 4- i-7738i/2 2 .1.552085/2 4- 2.03711^2 1.746095/2 4- 2.29937/^2 1.940105/2 4- 2.56093/22 2.134116/2 4~ 2.82196/22 2.328126/2 4- 3.08260/22 2.522137/2 4- 3-34295/2 2 2.716148/2 4- 3-60304/22 2.9ioi58/ 2 4- 3.86296^ 3.104168/2 4- 4.I227I/2 2 3.298179/2 4- 4-38235^ 3.492190/2 4- 4.64i87/2 2 3.686200/ 2 4- 4.90136^ 3.88021 1/ 2 4- 5.16067^ 4.O7422I/ 2 4- 5.41996/22 4.268231/2 4- 5.67920/22 4.462243/2 4- 5.93839^ 4.656253/2 4- 6.19758/22 0.916667/2 4- 2.69271^2 1.138803/2 4- 4.33766/22 I -35 I 5 6 3^ 2 + 6.27980^ 1-558965^ + 8.51725/22 1.763022/2 4- 11.05029/22 1.964845/2 4- 13.87962/22 2.165105/2 4- I7-OO593/2 2 2.364229/2 4- 20.42977/22 2.562501/2 4- 24.15161/22 2.760117/2 4- 28.17181/22 2.957219/2 4- 3 2.4 9 o63/2 2 3.I53908/ 2 + 37.io82 9 /2 2 3.35O26i/ 2 4- 42.02499^ 3-546340/ 2 4- 47.24089/22 3.742190/2 4- 52.75607/22 3.937844/2 4- 58.57070/22 4-I33336/ 2 + 64.68470/22 4.328684/2 -h 71.09829/^2 4.523913/2 + 77.81148/22 4.719046/2 4- 84.82431/22 4-9i4o65/ 2 4- 92.I2937/2 2 5 6 7 8 9 10 ii 12 13 14 15 16 17 18 20 21 22 23 24 This weight of girders is to be increased by one-tenth of itself, as explained in article 165. Also, it will be augmented to meet the strain brought upon the top chords by the head system. 174. Make the floor of 2^-inch oak planks, 52 pounds per cubic foot, and of the width of q feet. Then, if q = 16 feet, Weight of floor = X 52^ = / pounds. (573) CALCULATION OF THE WEIGHT OF BRIDGES. 491 175- Longitudinal I Floor Beams ; Conditions and Weights given in Article 157. We may further explain the assumption in (501) thus : Taking an analytical table of ordi- nary wrought-iron I-beams, we may easily see, that, for depths of 8 inches and upwards, we have approximately J = 0.15, (574) r being the radius of gyration, and d the depth of beam. Now, by equation (184), _ S' 2 I hence if we take, as we manifestly may, d = - -, and eliminate r, we shall find 3 * = S, ion as in (501), where the same notation is used, d being in inches, and / in feet. We obtain, from (503), Weight of 6 wrought-iron longitudinal I-beams, in pounds, = 6Lt n O.I3OOOOO 4 O.IO4OOOO 5 0.0866667 6 0.0742857 7 O.O65OOOO 8 0.0577778 9 0.0520000 10 0.0472727 n -433333 12 0.0400000 13 0.0371429 H n - ji v:>/:>/ n = 6Z/ + /' 0.0346667 '5 0.0325000 16 0.0305882 17 0.0288889 18 0.0273632 19 0.0260000 20 0.0247619 21 0.0236364 22 0.0226087 23 0.0216667 24 492 MECHANICS OF THE GIRDER. 176. Also, let the transverse I-beams be conditioned as in article (158) ; then (507) yields, taking 5 from (506), Weight of (n i) transverse I-beams due load, pounds, = (n i) X A X 12 x i8S (576) n 5.2650 4- I 2 0.003949 + LI 0.18225 + < 243 4 5.6160 0.003369 0.19440 324 5 5.8500 0.002925 0.20250 405 6 6.0171 0.002579 0.20828 486 7 6.1425 0.002303 0.21262 567 8 6.2400 0.002080 0.21600 648 9 6.3180 0.001895 0.21870 729 10 6.3817 0.001740 0.22091 810 ii 6.435 0.001609 0.22275 891 12 6.4800 0.001495 0.22431 972 j 13 6.5186 0.001397 0.22564 I0 53 X 4 6.5520 0.001304 0.22680 H34 15 6.5812 0.001234 0.22781 1215 16 6.6071 o. 001166 0.22871 1296 \I 7 6.6300 0.001105 0.22950 1377 18 6.6505 0.001050 0.23021 1458 19 6.6690 O.OOIOOO 0.23085 1539 20 6.6857 0.000955 0.23143 1620 21 6.7009 0.000914 0.23195 1701 22 6.7148 0.000876 0.23244 1782 23 6.7275 0.000841 0.23287 1863 24 177. Equation (512) becomes, when ;/ is odd, Weight of iron to be added to transverse I-beams on account of wind, in pounds, ~ I2 (577) CALCULATION OF THE WEIGHT OF BRIDGES. 493 25.000 -j 39.285 53.703 68.182 82.692 116.65 n 5 256.62 7 451-01 699-85 9 ii 1003.11 13 i = h 97.222 - 3 1360.80 n '5 111.765 126.315 140.873 1772.93 2239.49 2760.48 21 155435 3335-9 1 23 - terms in summing (509), for the two 178. When n is odd, we use adding 4 X -J( a i) = 2f n diagonals of the middle panel, and find, as in (509), Weight of horizontal diagonals, in pounds, n(n i) + (n i)(n 2) } + (n - 2 )( - 3) + (n - 3)O - 4) = 4 X sin 2 (/>, X m X n i terms + J( 2 i) (578) = h 28.000 + hi 2 0.0034568 n 5 40.000 0.0025195 7 51.852 0.0019758 9 63.636 O.OOI6232 ii 75.385 0.0013767 13 87.111 O.OOII949 15 98.823 O.OOIO554 17 110.526 O.OOO945O 19 122.222 0.0008554 21 I33-9I3 0.0007813 23 179. In summing (515) for odd values of * n I 1 IV P 11P terms of the series, and add \(n 2 i) for middle panel, then multiply the sum by 4, since the two wind chords are to be alike. 494 MECHANICS OF THE GIRDER. Weight of wind chords, in pounds, 18 n 2nq^Q = 0.006028164(2 + - 2 ^J n - i + 2(n - 2) -f- 3( - 3) n 2 0.0150463 0.0143414 0.0138920 0.0135871 0.0133679 0.0132030 0.0130747 0.0129721 0.0128882 0.0128183 terms + n = (579) 5 7 9 ii 13 15 17 19 21 23 180. We shall now have, at the centre of each vertical, the load, e n , as defined in article 163. Therefore Cross-section of lower half of vertical due , in square inches, 1 *O*C^ Cft ^ Q 3 1066 7 J Weight of all lower halves of vertical due , in pounds, = 2 X = 0.00021 18 10667 Cross-section of upper half of vertical due e n , in square inches, 12000 (580) (581) (582) Weight of all upper halves of verticals due e nt in pounds, = 2 x - X 18 12000 = 0.000185^2 -J/^. (583) CALCULATION OF THE WEIGHT OF BRIDGES. 495 As in the second part of article 163, so here, suiting the expression to the changed length of chord segments, we have, from the assumed wind pressure, the moment 62.5^ = fS 2 x AsyBt, B, = 5667. Therefore Cross-section of any vertical due bending-moment, in square inches, (584) Weight of verticals required to resist bending-moment due wind, in pounds, = 4 x -^ X - X i2$y= o. 9 8o3 9 22( i - -\il. (585) io oon \ n I Adding together the three expressions, (581), (583), and (585), the sum is Weight of all verticals, pounds, = /j 0.000393518^ - iv + 0.9803922^1 - ~yi (586) -h o.ooi I52Z7 4- 1.5425^ 4- 0.0000007 44/ 3 + O.OOOO2495/ 2 -|- 0.956317 4- 0.0032 o.ooi I79Z7 4- 1.9654^ 4- o.oooooo699/ 3 -j- o.oooo2oo2/ 2 -f 0.977477 -f- 0.0055 o.ooi 194^7 + 2.3885^ + o.oooooo677/ 3 -f o.ooooi725/ 2 -j- 0.990437 -(- o.ooSi o.ooi 203Z/ + 2.8077 L -f 0.000000650/ 3 + o.ooooi49o/ 2 + 0.998217 + o.oi 1 5 0.001209^7 -j- 3.2245^ + 0.0000006337 3 + o.ooooi3io7 2 -f 1.003837 + 0.0152 O.OOI2I3Z7 + 3.6396^ + o.oooooo6i67 3 + o.ooooi i687 2 + 1.007707 -f- 0.0197 o.ooi2i6Z7 -f 4.0536^ + o.oooooo6o67 3 -\- o.ooooio547 2 -f- i.oio6o7 -f 0.0245 o.ooi 2i8Z7 -}- 4466SZ + o.oooooo5937 3 + o.ooooo96o7 2 -f 1.012897 -f 0.0300 o.coi22oZ7 -f 4.8793Z -f 0.00000058 57 3 -f- o.ooooo88i7 2 -f 1.014757 -f 0.0359 o.oot22iZ7 + 5-29I4Z -f 0.0000005777 3 + o.ooooo8i47 2 + 1.016377 + 0.0425 o.ooi 222Z7 4- 5-703iZ + o.oooooo57o7 3 + 0.000007 567 2 + 1.017677 + 0.0494 O.OOI223Z7 4- 6.II45Z 4- o.oooooo5647 3 4- o.ooooo7o67 2 4- 1.018857 4- 0.0571 o.ooi 224Z7 4- 6-5257Z 4- o.oooooo5597 3 4- o.ooooo6637 2 4- 1.019927 4- 0.0651 o.ooi 224Z7 4- 6-9367Z 4- o.oooooo5557 3 4- o.ooooo6237 2 4- 1.020907 4- 0.0739 a 9 ro 496 MECHANICS OF THE GIRDER. = h o.ooi225Z7 4- 7-3474Z 4- o.oooooo549/ 3 + o.ooooo59o7 2 4- i. 021787 + 0.0829 o.ooi225Z7 4- 7.5811^ 4- O.OOOOOO546/ 3 4- o.ooooo553/ 2 4- I.O2263/ + 0.0928 O.OOI225Z/ 4- 8.i686Z -f o.oooooo543/ 3 4- o.ooooo53i/ 2 4- 1.023427 -j- 0.1029 o.ooi226Z7 4- 8-5797Z 4- o.oooooo539/ 3 4- o.ooooo505/ 2 -f i.o24i8/ -f- o.i 138 O.OOI226Z7 4- 8-9S94Z 4- o.oooooo537/ 3 4- 0.0000048 3/ 2 4- 1.024917 -f 0.1250 o.ooi226Z7 -f- 9-3997Z + o.oooooo533/ 3 4- o.ooooo462/ 2 4- 1.025607 -f 0.1370 o.ooi226Z7 4- 9.8o92Z + 0.0000005327 3 + o.ooooo4437 2 + 1.026277 4- 0.1492 18 20 24 since for the even values of n we have e n , given in article 163, and for the odd values e s = 0.62430^7 + 1040. 5^ 4- o.oio82i7 2 4- 18.03537 + o.ooi85o/;/ 2 4- 5.925^ 4- 14.58-, e 7 =z 0.44593Z7+ IO4O-5Z 4- o.oo552i7 2 4- 12.88237 4- o.ooi204>7 2 4-6.i3i// 4- 21.38-, e 9 = 0-34683Z7 + 1040. 5Z 4- O.OO334O7 2 4- 10.01967 4- 0.000881 hi 2 4- 6.237^ 4- 28.19-, e a = 0.28378Z/ 4- 1040. 5Z 4- O.0022367 2 4- 8.19787 4- o.ooo69i//7 2 4- 6.302^ 4- 34.99-, i 3 = 0.2401 2Z7 4- 1040. 5Z 4- o.ooi6oo7 2 4- 6.93667 4- o.ooo 567/^7 2 4- 6.345/6 4- 41.80-, e ls = o.2o8ioZ74- I040-5Z 4-O.OOI2027 2 4- 6.01 177 4- 0.000480/^7 2 4- 6.376^ 4-48.60-, e J7 o.i8362Z7 4- 1040. 5Z + o.ooo9357 2 4- 5.30457 4- 0.00041 67/7 2 4- 6.399/6 4- 55.40-, e X9 = o.i6429Z74- 1040. 5Z 4- o.ooo7427 2 4- 4.74617 4- 0.000366/^7 2 4- 6.417^ 4-62.21-, e 2 i = O.I4864Z7 4- 1040. 5Z 4- o.ooo6i37 2 4- 4.29407 4- o.ooo327/z7 2 4- 6.432^ 4- 69.01-, 23 = 0.1357 iZ7 4- 1040. 5Z 4- 0.00051 17 2 4- 3.92057 4- o.ooo295/// 2 4- 6.444,4 4- 75.82-, = the load applied at centre of each support. Here, as in article 163, we have put ^/for h in the last three terms of the value of t n ; a substitution introducing no practical error in the small resulting terms, but enabling us to keep our final equation down to the second degree with respect to h. 181. For the head lateral system, we proceed as in article 164, now having a pair of diagonals and a strut for each panel so far as the head system extends, say to n 6 of the central panels when head room is sufficient. CALCULATION OF THE WEIGHT OF BRIDGES. 497- ( ;/ j\th - J panel point is, since W l 2,500-, 2n \ 2/2 32 n = 78.125^ ~ T ^/ (K odd); ( ; Ath -J panel point, M = 5 /( - )- = - 2^ \ 2/2 32 = 78.125/z/ (w even); = 78.125 ( even); fi = jy x - = 156.25^^^ ^ (n odd), ;z 3 ^ x = 156.25 (even); nq l requiring each diagonal tie to resist I 5 6 -'5 X- VL x ^iTL- 1 pounds (n odd), cos cos <^ r ^ x 3 or T 5 * 2 ^ x pounds ( even), cos a cos (/>! ^ x and to have a cross-section I 5 6.2 5 ///(^ - I) 15000 cos a cos (/>i/2 3 ^/ - i)/z/ cos 0, " square mches ^ l 498 MECHANICS OF THE GIRDER. or O.OI04W S = nq t cos 0, square mches (* even >; (5 88 ) 'calling, as before, cos a = i. iLength of each head diagonal = n cos a cos (f> 1 n cos <, practically. Weight of 2 (n 6) wrought-iron head diagonals, in pounds, = a( - 6) X jg X 5^5- X ;z 2 \)hl 0.003858 ^ - o.oo 3 8 5 8O 2 - i) (n - cos , ( odd), (589) 0.00.385 8 ( 6)f 4- 324^) ( even), (590) n n - +hl* 4 = /; H.I999+/2/ 2 0.00015363 15 - - 5 12.5000 0.00015070 16 - 6 13.70*3 0.00014634 17 1.2245 0.00007713 7 15.0000 0.00014289 18 2.5000 0.00012056 8 16.2049 0.00013855 19 3-7037 0.00014113 9 17.5000 0.00013503 20 5.0000 0.00015432 10 18.7074 0.00013093 21 6.1983 0.00015811 ii 20.0000 0.00012754 22 7.5000 0.00016075 12 21.2092 0.00012375 23 8.6982 0.00015885 T 3 22.5000 0.00012056 24 IO.OOOO 0.00015747 14 > a. 1 8 feet, nnd Z T 4- ( l % n \ T i ta n 2A CALCULATION OF THE WEIGHT OF BRIDGES. 499 Multiplying the cross-section, (587), (588), of head diagonal by 2 X 1 0,000 cos c^tan < we find, after dividing by 2,500 pounds inch strain, o = 0.0104^ x 2 x ioooo( 2 i ;r i h square inches (n odd), 12 n 2 0.0104! x 2 x ioooo///tan r 2500;^ = -f^h square inches (n even), = cross-section of head strut. Weight of (n 5) head struts, in pounds, (591) (592) 5; 18 n 2 12 V593 (/* - 5)O 2 - i -h (n odd), n 2 = &(* - 5)^ X 12 x 1 8, (594) = 5( n 5)^ ( n even), n n - 4 = h .49-7777 15 o 5 55.0000 16 5.0000 6 59-7924 17 9-7959 7 65.0000 1 8 15.0000 8 69.8061 19 I 9-753 I 9 75.0000 20 25.0000 10 79.8186 21 29.7521 ii 85.0000 22 35.0000 12 89.8299 2 3 39-7633 13 95.0000 24 45.0000 14 500 MECHANICS OF THE GIRDER. Calling the compression along each segment of top chord due head diagonals equal to l or 156.25-. ^ according as n is odd or even, and taking the allowed inch strain in compression, as above, viz., 8000 2 = 75 2 9 pounds = 3.764 tons, *" 40000 we have Cross-section of iron to be added to segments of top chord in head system, square inches, 156.25 n 2 - i hi = S = 0.00115295;^, (595) 156.25 hi = 0.00115295. (596) n Weight of added iron in (n 6) panels for top chords, pounds, = 2 (_6) X ^ X ^S, (597) 18 n = 0.0076863 6 ~ T ^/ (n odd), (n even), CALCULATION OF THE WEIGHT OF BRIDGES. 501 n n = hi 2 - 4 = hi 2 0.00030609 15 - 5 0.00030024 16 6 0.00029155 J 7 0.00015366 7 0.00028468 18 0.00024020 8 0.00027603 19 0.00028116 9 0.00026902 20 0.00030745 10 0.00026085 21 0.00031427 ii 0.00025409 22 0.00032026 12 0.00024654 2 3 0.00031648 13 0.00024020 24 0.00031373 14 182. As explained in article 165, we shall here augment, by one-tenth of itself, each of the following expressions just found; viz., The girders proper, The vertical supports, and The lateral head struts. Then, adding together all the parts of the complete bridge, and putting the sum = 2,ooon W> the weight of any bridge in pounds, we derive the following values of W, in terms of L, /, and /i, for the different values of ;/. Then, by assigning values to L and /, differentiating, and putting = o, we get W a minimum^ and h best, as in dh article 140, equations (469) and (470). 502 MECHANICS OF THE GIRDER. CALCULATION OF THE WEIGHT OF BRIDGES. 503 1 "T o " * O^ N n" 5 + 4 vO DO tn 504 MECHANICS OF THE GIRDER. I ^ + + 810 [Z ;6. 1 l-A" O . o + + 00 i- ro ^ 2 + sir ^ + + "^ ^ o + + O ^ d + % s s ro 'f, d + CALCULATION OF THE WEIGHT OF BRIDGES. 505 CN d oo 9 I o + 1? j- CO 8 VL til 66 2 <&% 00 o + r-T ^ % 2 % vq" ^ 8ld + + VO > OO 506 MECHANICS OF THE GIRDER. ro 458) + 0.0 ^ W ) ON M S II ?0 oo O S 00 CALCULATION OF THE WEIGHT OF BRIDGES. 507 a r c * .. > en i V 1 ^ < f 1 "cu ' ^T" g 1 i bO i-T 4^ *T 4-T '^3 ^V, to *"* cu cu cu cu ' ON rj- ct 2 v2 ^ <^i .g to ^* ^ M rT to ro ^*- ro tT ^^ rt* 7 - ro t^. VO to 5/j II II II II J3 * 1 1 !_ M II H II II ^~* , ^ 1 1 _l_ CO + 11 II 8 + . + II H ^ 2 "^ 8 S ^ co 5 N ^ K^ en ^ CO q s- n n 1 cf II II r CT 1 c + IS 1 " + ad o ^cj cu v o cu J5 vo 1=1 00 O s + CB 4. co 4- ^'"j, ^_ ^^ to ro - **"* *^ ^ co f} en en O I-M CN.O C fl to i_ 1-1 O O O rl -5 9 + q . co & o ^ + o ^ ,00 -5- + 0.02171 1 ijh z . 0.000068 1 7 34/i 2 .978 tons, .969 tons if / = i VO CO vO ro to *. . i- O 4 M ^ to i- to I-H oS > & s II II il II II | | > >-, >-i >-, >> C G "4~ ^ vO 5 i ^ ^ + ^ J Mh Mk, ^-C 4-J ^-> -*-) ^_ ^-1 rt "^2 *rj- !> O J hH S K fi 8 2 I a g Iw B * j o 11 J w J g "S 5 O 6 ;D ^ u W _2 jy) M -5-S K 1 d M -a 6 <: 5 =5 Two DOUBLE PAR Live and Dead Loads applie co r^ N o in * in * in rx < o * rovo woo M H ro r^ o * t^ inoNt^ooNTt- ro 3 , in N tx M o n ri. ^ co oo co 0, ^ ro ^ w_ rooo^oo -oo O TO 00 O M3 f-N fx t>^ 4 ^ H H H rovo H O H in N Set ?g W -^-vO roroO OvO N O O> t^*o to w t^oo o t^ c^ t^co M OM^ONNNOvo-^-HO t^ H (N oo" T? o"vo" -f4roc>t^ ro\o HMMNMM C^OQ ^^ o oo ^ o O O" H -<*- ' ^HMH OO t- O N w o ^ O ro H 4 ? N . OO m t->vo CJ oo r^i oo 10 O t^-^- t^-^- crr**M "^H CO OH ro^o ^ 4. V _ 4iH N 00 H 4 u^r* ct M'OO cooro!N->i-ONt>. 8 A 1 )_vo w N ^-vot^. ^-r^. o^t^ H ^ O* O* co c?ro w 1 ^? ON H< vO rOO CJ H t^O>ro-^- -^-\O ON H vO VO M O oo in rooo ro in M in ON in r^oo M O K , H ro M >n oro ~ fi^8; ro -* vo_ 'O, ro H; H^ in H O ** ro i g SoS- SSvo^S^cTSN vg > t>. ro O ^O O u"> t^. COOO CO t*N.vD O W Ost^-^t-O (N^hN ^- rovo M oot^M-M q moo ^ d d S MM 2i?^ H( ^ *-TO <* roiNr-ON-. t;. (N ON H N ^ ^ I ^M^NN^roin^NHOO OOl-IHHMHONt^ -^O ^ro t^ O ** roco vOMCOO^ro t-trs-^ OvOvO WOOvDt^Hin-<-ON 01 *** M pi w* M" oo" cT 4 " Jf J g> o M CALCULATION OF THE WEIGHT OF BRIDGES. 509 CO t^CO Ot^OrOOCOCOOOWl^^O '-' t>. -* CO txvO ** ^" I M co* vo* ro coco <>in-v?(NMin'^-wrN.c i r ci) ^ M . O^O^icd o^^ iA iA CO m H vO O^O vO^Nt^iO M M O NCOt^O>CJN a i i i . feT a H I ,2 -s m s * n S -S H I la g I '. o -ai ^ || z .tj '5 D O _S 2" 1 ."la o j a & s * ll 9 1 in N O> r^^O OO t^ M T*-CO m u-> __ ^ CO O ro 000_ ro^ i CO >0 ^ oT I lO M VO 4* * *0 d roiH 0*06 ^ *o i H o w co r^ o M mco r^M ;_,,.s- ^ a vo OvO m M 5 i- MH o in Koo PIOOONO M o ONON^-M t^ inoo ^. PI pi rt* ONOO qv k ti-co q, qo5 qjco^ c^ o^ * r- in*- q, ^ t s *-'*-'<*-QOOO'**'O'OCOONrol-^ r^ vo mONONO rnt^3N*-vO O^^- t-^CO ;rrvO m N t~x ONVO O IH rON NOvO T)- mH_ CO ON CO -*vO 00 OO i-( H OO ON in R.-O" 't ON 5 co t^ m PI t^vo ON in in * in 30 * ^ J*S J" 1 R$ H H 5 IH O roco ON IH T-VO mvo ON PI MM PI ON t^ t~ ONOO CO t^ t^l". * ^ In mco_^ ^NO 5n C?'"U ON ^ O O v* i vc % . *rw . . ^* v H TMH in M rH oo O>H M S ^ 2^ "^c? o""^ ? S" 10 eT M " M" "2 ^ ? pi vo in co ^4- m rt o d cT 1 O ON* ll l III . .pj . co ON ON .&S g TJ -32-53 ,^-n^in J ..gM ' ' 1 3 ;^ ~i ' M "i" ' 2 ' * ' a l -3; I ii i /iii - if : h : i : ft M Hi! ffl ;|S :l : Mitlfl B S'C S g- SuJ^u g l||g,^||||||l|l1|||-S-|-2|-3 4 i|| s i 1 OS 2 3 5 S 3 S 53 3 5 5 5 2 5 S 5 3 > 2 | 3 3 S 8 3 q q q q c p q q q q o. q q q q q q q q o q q q o >n in in q q q o o II H ,f II O KJO w W CO "^ lOvO 1>>OO ^ O M C* r^, *- iovO tx.cO O O M w ro ^- LTJO txCO Ov O 510 MECHANICS OF THE GIRDER. PQ a * o ~ g I I 6 * I CO I'll . o 's 3 s - a I o =3 t-^oo vo oo ON CM m o oo Tt-oo vo moo rx ON inwOCMCMOOCM IOCS CO ON H ,- . t> t^ i -H ^1 t^ > i3 I'll II | -- 13 ' 4it m -- '- * - st number idge weig of panels eight weight of panels eight weight 5,-S, ;/: idg ber of tral hei ridge mber ral h ge I; % K &$& " * ^ Cc'CCc'C MtXlbJt H fi O C u" - 5 - 5 ' 5 || 4j*jcn*j*j^ _ S IIJJIJ a Lim Lim q q q q q q q q q q q q q q q q q q q q q o o in m m o o q -vO t.OO ON O * invO t^CO ON O w CM CALCULATION OF THE WEIGHT OF BRIDGES. WW r** M ro m t^oo ON ^f-vO ^ ON 1000 OO vO CNI \O ^" ** iofx ^" ON N 10 i/^OO O O -^O 10 Tj-0 o< M m txoo M 01 M c* -rf- in N 1-1 ON oo ON vo O vO rHtx roH NM VO ^% JHOO.ro f< h. M OO \O \O O VO O ioioroa-O 10 oo ovt^fl' M ^ -il- ro t-* O C* mvO I s - tf csi ro * oo^) H N M M ro ^- C* m ON O ro O OOO ^v> M N ro O OO vO W N Ot^OWu-)OOOO' 1 ^-iHmO'*' T t* rr ) O ON oo 00..-.*^-%.**^^* l ^.^ l Q*O. ^ Srooo \r, r^ rOTj- OOx ? CNl" w" ^ M'OO" t^ccco <$ o *? Tp in c? 00 mvD ro ro O O roo O M w o NinMNHHinro rooo t^ 10 t% OO H MU-)O vo (N ONO t-vO O rooo 10 io\o MOr~.r-. Mf^QNOoo M ^- ON \OvO voro vvq oo^ 10 o^vq^oo^ * ,<>* >o c^ ro ijf!^ e < oo 55 ON 10^ ro t^ * M*OO" hTt-TtCroOM't^^^ ^ ^ - ro 6 d vO o ro ON ^00 H N ON CO O ^000"^-^ H cN 10 T Tt- M roo MOO ht-troOM ^ 00 ON t^ rovo CO ro ro ON ^00 H N ON M q;O^ c^ H o H in ro ON g -"g *-g *j*j*^ 2 J 2 t 2 J aa -W j '.Pi: | : il i ^ -^2 -B & -.s .-3 -2 .-2 . *s- g s 5 be wei nels t ht anels t ht 0^ 5/1 5/1 ill fi .-&S. '^f -I - "UltHl ^ lll^L-Pifl ! likllfelt, II | ^11iiillii' s .* s ; ^iiiiii r m'Sj_. R.a a * =^.v. *~ 8 s e'n s "-a i*s 2 g^ge- ! | 4 ^^rStn^S (2 S ^^^^^^ pan pan pan Lim Limi Lim q q q q o q q q q q q q q q q q q q q q q q q 10 10 in q q q txOO O O M ft ro t n\O rxOO ON O M CM ro TJ- U1^ 512 MECHANICS OF THE GIRDER. 183. Among the deductions to be drawn from this table, for the Brunei double-bow bridge of double web system, are the following : 1st, For a given uniform live load, n oc /* nearly; (620) and generally n oc (JJ x /* nearly. (621) 2d, For spans less than 400 feet, For spans of 400 feet and upwards, nearly - (623) 3d, For different spans with same live load per running-foot, W oc Ih nearly; (624) and for the same span under different uniform live loads, nW oc ~ nearly. (625) Many other conclusions may be drawn from this table, and weights of intermediate spans may be derived by interpolation ; but the equations (598) to (619), inclusive, cover the whole case. 184. EXAMPLE. We now proceed to find the strain sheet for the 2oo-feet span of the table in article 182 in a manner similar to that employed in article 169. CALCULATION OF THE WEIGHT OF BRIDGES. 513 We now have / = 200 feet, q = 1 6 feet ; // = 40.788 feet, g l = 18 feet. n = 13. tiL = 200 tons, L = 15 A tons > n\V = 103.288 tons, W = 7.945 tons; /r -h Z = 23.330 tons. Weight of floor, by article 174, ^ X 200 = 34667 pounds ; Total live load, //Z, = 400000 pounds. Total load on longitudinal I-beams = 434667 pounds. Load on each panel length of every longitudinal I-beam spaced 3.2 feet _ x = 6687 pounds. Then, by (502), Cross-section of beam 434667 = S = 0.00015 x - = 5.0154 square inches. Take b = 4 inches = breadth of flange. b b, = 0.26 inch = thickness of web. Then, from (552), (551), and (550), d 9-774 inches = depth of beam, // tf t = 0.66 1 inch = depth of two flanges, /= 75.416 = moment of inertia of section, which is lur-vr than / for the sections given by ordinary beams of the same area of section. ht of longitudinal I-beams, 6 in number, = 6 X -j^ X 12 X 200 X 5.01538 = 20062 pounds. 5 14 MECHANICS OF THE GIRDER. Upon the transverse I-beams we have Live load, 400000 pounds, Floor, 34667 pounds, Longitudinal I-beams, 20062 pounds. Total for 13 panels, 454729 pounds. Load on i beam, 34979 pounds. From (504), / 12 x 18 X 34979 d = 8 x 2 x IQOQO = 47.2216 = 2S, by (505) ; .*. S = 23.6108 square inches for vertical load, and for the wind pressure, 40.788 Wi = 2500 x - = 7844 pounds per panel, = 7542 ' 5 = 7.6798 square inches, ist and i2th beams; = 6.4532 square inches, 2d and nth beams; = 5.3332 square inches, 3d and loth beams ; = 4.3199 square inches, 4th and gth beams; = 3.4132 square inches, 5th and 8th beams; = 2.6133 square inches, 6th and 7th beams. The total cross-sections for each half-span are S = 31.2906 square inches, = 30.0640 square inches,* = 28.9440 square inches, = 27.9307 square inches, = 27.0240 square inches, = 26.2241 square inches. CALCULATION OF THE WEIGHT OF BRIDGES. 515 Satisfying the condition (505), we may assign values to d and d v and use (557) and (558) in finding the thickness of each transverse beam. Put 2 light 12-inch beams at each panel point, the section of each being \S. Then we have d = 12 inches = depth of beam, d di = 2 inches = depth of 2 flanges ; and (558) becomes, for breadth of flanges, b = ( 24 ~ I2 -f ) X iS = 0.17803^ = 5.5707 inches, y 2 X 2 2 1 2 J = 5-35 2 3 inches, = 5.1529 inches, = 4.9725 inches, = 4.8111 inches, = 4.6687 inches; and (557) gives bl = iQ 4 x 2" 2"* X % S = ' l6 3 6 3^ = 5- I2 3 inches, = 4.9083 inches, = 4-73 6 3 inches, = 4.5705 inches, = 4.4221 inches, = 4.2912 inches. Thickness of web = 0.4504 inch = b b ly = 0.4440 inch, = 0.4166 inch, = 0.4020 inch, = 0.3890 inch, = 0.3775 inch. The weight of these 24 transverse I-bearhs is 12 x 1 8 x T 5 s^ = 205 7 7 pounds^ since 25 ( sum of all the cross-sections) is 342.9548 square inches. 516 MECHAA T ICS OF THE GIRDER. Cross-sections of horizontal diagonals are found by dividing the strains in (508) by 15,000, where we now have W, = 7844, -^- = 0.52293, 15000 sin^j = 0.76017, 0.^2203 5 = , X 13 X o. 7 6o, 7 ('3 X , i X i,, ii X 10, 10 x 9, 9 X 8, 8 x 7, 7 X 6) = 0.026458 X 156 = 4.1274 square inches, = 0.026458 X 132 = 3.4925 square inches, = 0.026458 X no = 2.9104 square inches, = 0.026458 X 90 = 2.3812 square inches, = 0.026458 X 72 = 1.9050 square inches, = 0.026458 X 56 = 1.4816 square inches, = 0.026458 x 42 = i.i 1 12 square inches, for the respective panels. .*. 2S = 67.4150 square inches, and weight of 26 horizontal diagonals is I2 . X . 1 X 5- X 67.415 = 5321 pounds. sin<, 1 8 The cross-section of each panel length of each wind chord is given by (514), thus, 7844 X 200 2 X 13 X 18 X 6400 V " 3 X = -5 2 377 X 12 = = -5 2 377 X 22 = = o-5 2 377 X 30 = = 0-5 2 377 X 36 = = -5 2 377 X 40 = = -5 2 377 X 42 = = o-5 2 377 X 4 2 = ^, ^ /\ a. i, 10, 4 x 9, 5 X 8, 6 6.2852 square inches, 11.5229 square inches, 15.7131 square inches, 18.8557 square inches, 20.9508 square inches, 21.9983 square inches, 21.9983 square inches. = 106.3253 square inches. CALCULATION OF THE WEIGHT OF BRIDGES, These sections can easily be made up of channels and plates, or of beams and plates, with the required radius of gyration given in article 162. In summing these sections for the weight formula, all are to be taken four times, except the last, which is taken twice only. 12 x 200 x 106.3253 5 Weight of wind chords = X yg = 21810 pounds. O Supported by verticals, we have Live load, 400000 pounds, Floor, ^34667 pounds, Longitudinal I-beams, 20062 pounds, Horizontal diagonals, 5321 pounds, Wind chords, 21810 pounds. 481860 26 = 18533 Transverse I-beams, 20577 -r- 24 = 857 Weight on each vertical = z n = 19390 Therefore we have the cross-sections, by (580), = 1.81776 square inches; w \j i by (582), I039O S l = - - = 1.61583 square inches; 12000 for the lower and upper halves respectively of the verticals due load; and, for the bending-moment due wind, (584) gives S 2 = 5 ^^ = ^-3Q'?67 square inches, 68 x 13 Section of compressed half = 5.21143 square inches, Section of extended half = 5.00950 square inches. And, since the upper and lower halves of the girder are symmetrical, and the sum of the lengths of the verticals = ^y = %h(n - ) by (521), = J X 40.788 X ^g 8 -, we have 518 MECHANICS OF THE GIRDER. = 2 X -5- x 5.21143 X X 40.788 X - = 6103 pounds, lS 3 J 3 Weight of lower halves = 2 X -5- x 5. lS Weight of upper halves = 2 X -^- X 5.0095 X X 40.788 X - = 5866 pounds. 18 3 1 3 _ Total weight = 11969 x 10 = 13165 pounds. after adding -fa for braces, etc. The sections may be made up of 2 channels, the one ver- tical, the other inclined at an angle whose tangent is ^. According to the principles of article 165, the bars in the bracing of these supports should have a cross-section of about -| inch ; that is, about J^ of (5 + S 2 )> Equation (587) gives the cross-section of each head diagonal thus : S = ____ j ^__ = 6 X i3 3 X 18 X 0.64972 From (589) comes the weight of 14 head diagonals in the seven central panels equal to 14 x -**- X X 2 X 0.5556 = 614 pounds. 18 13 0.64972 Cross-section of head struts, by (591), = I 68 X 4^788 = 8 incheS) 12 X 169 requiring 2 light 4-inch channels latticed not less than 4 inches apart. Weight of 8 head struts, -xSx-5-Xi2Xi8x 3.3789 = 1784 pounds, 10 18 after adding T ^ for bracing. CALCULATION OF THE WEIGHT OF BRIDGES. 519 Increment of section of top chord due to head diagonal strain is given by (595), thus : 156.25 x 168 x 40.788 x 200 S = 7529 X i 3 3 X 18 = ' 7I 9 2 Square mch > the strain being = 0.7192 X $$$ 2.707 tons. Weight added to top chords = 2 x 7 X -- X T2 X 2 X 0.7192 = 516 pounds. 18 13 For each of two girders, the horizontal component of chord strain is, by (564), and the chord strains are P = - = U = - JT = 99.317 tons, ist panel; = 97.415 tons, 2d panel; = 95.830 tons, 3d panel; = 94.580 tons, 4th panel : = 93.672 tons, 5th panel; = 93.130 tons, 6th panel; = 92.947 tons, 7th panel. Cross-section of top chord due load n = 26.381 square inches, ist panel; = 25.876 square inches, 2d panel ; = 25.455 square inches, 3d panel; = 25.123 square inches, 4th panel; = 24.882 square inches, 5th panel ; = 24.738 square inches, 6th panel ; = 24.689 square inches, 7th panel. Augment 4th, 5th, 6th, 7th, 8th, 9th, loth, by 0.7192. 520 MECHANICS OF THE GIRDER. Cross-section of bottom chord = = 19.863 square inches, ist panel; = 19.483 square inches, 2d panel ; = 19.166 square inches, 3d panel ; = 18.916 square inches, 4th panel ; '= 18.734 square inches, 5th panel ; = 18.626 square inches, 6th panel ; = 18.589 square inches, yth panel. The top chord may be composed of 2 Q-inch channels and i plate ; the bottom chord, of 3 bars and 2 bars in alternate panels. From (561), we find 2sec 2 = 27.4329; and (566) gives, adding -J^, Weight of top chords = x 10 200 2 40.788 X 27.4329 = 38206 pounds. From (567),. Weight of bottom chords = X 38206 = 28767 pounds. From (571), the strain on a girder diagonal is called ' T 5r\ X 200 x 1.8 x ^0 = 2 /? 7 = 0.583 inch for # = noo -f- 13 ; Z> 8 = o inch for # = 1300 -s- 13, ends. Equation (366) yields the excess of length required in top chord to give the proper camber, x 2 5 ' 34 x I2 = since length of polygonal top chord is equal to X 2seca = X 13.3466 = 205.34 feet; 13 !3 .-. Mean excess per panel = -- = 0.138 inch, or a little more than 4 inch. BRIDGES OF CLASS II. $2$ CHAPTER XI. BRIDGES OF CLASS II. BEST NUMBER OF PANELS AND BEST HEIGHT DETERMINED FOR A GIVEN SPAN UNDER A GIVEN UNIFORM LIVE LOAD. LEAST BRIDGE WEIGHT AND LIMITING SPAN FOUND. SECTION i. The Parabolic Bowstring Girder of Double Triangular System (Fig. with the Extreme Diagonals omitted, and a Vertical Suspender at Extreme Panel Point. 185. Let / = span, in feet. // = height of girder at centre, in feet. ;/ = number of panels. ' L := panel weight of uniform live load, in tons. W = panel weight of bridge, in tons. The height of girder at any point, ;r, is given by (472), and at all vertices by (473), if we make r = i for the first point, and put 2/1 for h throughout, thus : (626) + i)( -r- i), - > = ^(n - 2r - i), (627) tan = by -r- - = ^(n - 2 r - i), (628) n nl sec 2 = i + tan 2 ** = i + ^-(n 2r i) 2 , (629) n 2 l 2 524 MECHANICS OF THE GIRDER. tan# r = tan ., = -y, - '- = -4-( _ r), i6/r ?0 = i + tan 2 6> = i 4- -y^O ~ ? *) 2 - (630) (631) I a, #, and are defined in article 49. , The live load and a large part of the dead load are applied at the panel points of the bottom chord, and are transmitted by the diagonals to the parabolic arch, which is equilibrated by the uniform load, leaving only a tensile strain on the diagonals from full uniform load. We shall assume that the two diag- onals which support any panel weight of uniform dead load carry each one-half of the same. 186. Moments at all panel points due ;/( W + L) t the total load, are, from equation (65), W + 211 and the horizontal component of chord strain under same load is equal to If = M = i(^4-Z)^, as in (564), and is uniform throughout for maximum. H } Greatest strains in top chord = P cos Greatest strains in bottom chord = U H Cross-section of top chord, ,S = P -5- Q Cross-section of bottom chord, ,5, = U -f- T \ Take = 3.7647 tons (632) (633) (634) 1 40000 T = 5 tons as the allowed inch strains on top and bottom chords respec- tively. BRIDGES OF CLASS IL 525 I Length of segment of top chord = - inches, 72 COS Volume of segment of top chord n Q cos 2 - = \(W + L) Q/i cos 2 cubic inches, (635) by summing (629) for values of r from o to ;/ I, inclusive. Therefore, calling weight of a cubic inch of wrought-iron, as in all cases, -f$ pound, we find Weight of top chords, in pounds, = X (636) 12 x 3-7 6 47 / M 3/2 V n) \ r.^MH- 71/2/2 4- 0.59028^ n n W 4- Z 0.885417/2 4- 4.64844/^2 8 k 0.996095/2 4- 5.24688/^2 9 1.106772/2 4- 5.84372^ 10 1.217449/2 4- 6.43936^ ii 1.328126/2 4- 7.03412/^2 12 1.438804/2 4- 7.62820/^2 T 3 1.549481/2 4- 8.22i68/z 2 14 1.660158/2 + 8.81480^ 15 1.770835/2 4- 9.40752/^2 16 1.881512/2 -f IO.OOOOO// 2 i7 1.992190/2 4- 10.59216/^2 18 2.102867/2 4- 11.18440/^2 19 2.213544/2 4- 11.77600/^2 20 526 MECHANICS OF THE GIRDER. Weight of bottom chords, in pounds, = -^- X ^ X nl 2 , (637) 1 8 8 Tli W 4- Z l*n w + z n 0.666667/ 2 8 O.75OOOO/ 2 9 Q.833333/ 2 10 o.9i6667/ 2 ii .oooooo/ 2 12 OS3333/ 2 I 3 .i6666 7 / 2 14 .25OOOO/ 2 15 333333 /2 16 4i6667/ 2 17 .^ooooo/ 2 18 .66666 7/ 2 20 187. The Girder Diagonals. Separating the double sys- tem into the single web systems of Fig. 35*2 and Fig. 27, let us consider first that of Fig. 350, and find the difference, A//, of horizontal strains at the foremost end of the advancing uniform discontinuous load, nL, and for the same instant at the next two forward panel points of the double system. Putting Z for W in equation (60), and taking r 2 = J, we find the moment at any point, x, at or before the foremost end, to be M x = (r + ) 2 (/ - x), (638) 2/ where c = length of whole interval in the single system = n in the double system, r + \ number of panel points of bot- tom chord loaded, x = distance from left end to the point where moment is taken. BRIDGES OF CLASS II. 527 From (472), putting 2/1 for //, (639) Therefore the simultaneous horizontal strains due live load at the three consecutive panel points, of which the foremost end of live load is at the rear one, are H = ^ = ( r + i) 2 ' y " 8/1 ' x 1 (f _|_ IV = - -*- if x re, foremost end ; 8/1 r (640) == ^ 8/1 _Ll (r (r 4- -|) if x = (r + %)c, next panel point; if x = (r 4- i)^, next panel point; I where r takes the successive values J, f, f ,,... Then we find the greatest differences, that is, the greatest horizontal component of diagonal strain due live load, thus : 1 = _// r + I 16/2 r + i (641) for the first single system ; r taking the successive values J, f , f, J, etc. Similarly, for the second single system, we get greatest horizontal component of diagonal strain due live load, i6/i r 4- Ll r 16/1 r + , (642) where r becomes r, 2, 3, 4, etc. 528 MECHANICS OF THE GIRDER. Now, if we consider the horizontal components of the two diagonals in the same panel of the double system, Fig. 35, with its extreme tie made vertical, we see that ^H of (641) and A|// of (642) belong to the odd panels, and ^H of (641) and A!//" of (642) belong to the even panels. Also, for the odd panels, r of (641) is less by ^ than r of (642) ; and, for the even panels, r of (641) is greater by J than r of (642). Therefore, reducing so that r belongs to the second system, we find LI r Compression, A x Zf = odd panels, (643) Tension, Tension, T 1 67* r + -I - j. even panels, (044) LI T Compression, ^H = -- i6h r -f \ } which expressions are identical ; and the sum of either pair is as already given by (570) for the total horizontal component of diagonal strain due live load in any panel. Since these diagonals are to be alternately in tension and compression, the load travelling either way, and our specifica- tions would multiply the compressive strains by 1.8, we shall, for convenience, take due live load for each diagonal in a panel, instead of multiply- BRIDGES OF CLASS If, 529 ing by 1.8, and shall treat all diagonals as in compression under the inch strain, Q = 4 = ? tons. 3 2OOOO This procedure varies a little from the specifications, but on the safe side, since 27* -f- i > i.Sr. Moreover, since the dead load, with the exception of the top chords and head system and wind braces, is suspended at all times on two diagonals which transmit it to the equilibrated top chord, and since the tensile section will, for practical spans, not be greater than the compressive section due to live load as above augmented, and to be provided for in compression, we may, as appears below, leave the tensile strain which will come upon the diagonals acting as suspenders, almost entirely to the material put into them to resist maximum compression. It is to be observed, that, when the live load is fully on the bridge, there is no compression on the girder diagonals, but each one acts simply as a suspender to transmit ^( W 2 + L) to the equilibrated top chord ; W 2 being that part of the dead load at any lower apex. Now, from the results tabulated in Chap. X., we may doubtless, in the present case, for spans not over 600 feet, consider W 2 as ranging from | W to f W, while W ranges from ^L to 2L approximately. Therefore \( W 2 + L) ranges from \%L to \L nearly, in spans from 100 feet to 600 feet. Taking the greater, |Z, as the vertical component of ten- sion on each girder diagonal, we have U . n SZcot0 = ^- X ^- ^ 24 h r(n r) as the horizontal component of tension on diagonals acting as 530 MECHANICS OF THE GIRDER. suspenders ; the greatest value of which is found when r = i, or r = n i. That is, x = L&. _JL_ for tension. 24 /* A./7 = for compression. Dividing A//" max and A//" by 5 and by f, the allowed inch strains in tons respectively for tension and compression, and multiplying by sec 0, we find these resulting cross-sections for comparison : S 1 ! = X . X - nearly 17 hcQsO n - i . (646) 64 h cos d Now, S r will be greater than 5 only when r = i, or r = n i ; that is, the girder diagonals which meet at the first and second and at the (n 2)* and (n i) th lower apices, will need addi- tional section under full load to the extent of the difference between 5 r and 5. And this we shall supply in the vertical braces at these points. ^ Cross-section of a girder diagonal = S = -- -; (647) 64/7 cos v (648) Weight of 2(n 2) girder diagonals, in pounds, = 2X x-S-X-^-X - n 18 64 h i6nh\ # 2 / 2 \3O 30 _ ZjsQi- ay , (if 5 h\ i6n BRIDGES OF CLASS II. 531 z = h o-234375 /2 + 10. 1 8 ^ ^ 5^ 2 8 0.243056/ 2 + 1 3 .05899'^ 9 0.250000/2 + l6.26OQO/2 2 10 0.255682/2 + 19. 78964^ ii 0.260417/2 + 23.64873^ 12 0.264423/2 + 27.83796** 13 0.267857/ 2 + 32.35788/22 14 o.27o8 33 / 2 + 37.20889/2 2 15 0.273437/ 2 + 42.39I36/2 2 16 Q.275735/ 2 + 47-90556/2 2 17 o.277 7 78/ 2 + 53-75 I 7 2 ^ 2 18 o.2796o5/ 2 + 59.93002/22 19 0.28I250/ 2 + 66.44O62/2 2 20 Adding together (636), (637), and (648), we find Weight of girders due to loads, pounds, _ Z = h i.786459/ 2 4- M-83399/2 2 + - h ,55,084^ 4.64844^ ft 8 1.989151/2 + 18.30587^ I.746095/ 2 + 5.24688/22 9 2.I9OIO5/ 2 4- 22.IO372/2 2 I.94OIO5/2 + 5.84372^ 10 2. 389798^ 4- 26.22900/22 2.i34ii6/ 2 + 6.43936^2 ii ' 2.58S543/ 2 4- 30.68285^ 2-328I26/ 2 + 7.O34i2^ 2 12 2. 786560^ + 35.46616/22 2.522I37/ 2 + 7.62820^ 13 2.984005/2 4- 4O-57956/2 2 2.7i6i48/ 2 + 8.22168/1* 14 3.i8o99i/ 2 4- 46. 02369/1* 2.9IOI58/ 2 + 8.81480^ 15- 3.377605/ 2 4- 51.79888/^2 3.i04i68/ 2 + 9.40752/^2 16 3-5739I4/ 2 + 57.90556^ 3.298179/2 + IO.OOOOO/2 2 I 7 3.769968/2 + 64.34388^ 3.492190/2 + I 0.59216^ 18 3.965805/2+ 71.11442/^2 3.686200/ 2 + 11.18440^ 19 4-i6i46i/ 2 + 78.2i662/2 2 3.88021 1/ 2 + 11.77600^ 20 To be augmented by one-tenth. 532 MECHANICS OF THE GIRDER. 188. Floor to be the same as in article 174. Weight of floor = &p/ pounds. (649) 189. Take longitudinal I floor beams, as in articles 157, 175, equation (575). 190. The transverse I-beams supporting live load, floor, and longitudinal beams are here conditioned as in article 158, and their cross-section due vertical forces is given by equation (506). Their weight due same forces is given by (576). The cross-sections of transverse I-beams due to wind are expressed in (511). The weight of iron to be added to the transverse I-beams on account of wind, is given by (512) and (577) for n even and n odd. The whole effect of wind pressure is to be transferred to the horizontal system, in the plane of the bottom chords, by means of vertical braces connecting each transverse beam with both top chords. It may be observed, that in this and all like cases a shear- ing-stress is generated throughout the transverse beam by the wind pressure transmitted through these vertical braces ; but there will be sufficient reserve material in the web of the beam to resist this shearing-stress, as becomes evident on reflection. 191. If we divide equation (508) by 15,000, we have the cross-section of any horizontal diagonal in the floor system. And equations (509) and (578) give us the weight of the hori- zontal diagonals, in pounds, for the even and odd values of n. 192. For wind chords, let us use the bottom chords of the girders, augmenting their cross-sections by the quantity in (513) divided by the tensile inch strain, 10,000 pounds. Although this augment will only resist tension, while the compressive chord strain due to wind will sometimes be greater than the tensile chord strain due to dead load, yet, as the BRIDGES OF CLASS II. 533 excess of compression is not great, it may be left safely to the outside longitudinal I-beams, which, it will be remembered, are otherwise only half loaded. We may compare the chord strains due dead load and due wind by means of Fig. 112, thus : For^F, N= ^; 2n/i For W N = ^-J-. in^h These co-efficients of strain will be equal when and this might be made a condition determining the width, q of the bridge, so that no compressive strain would prevail in a bottom chord. But we shall not now change the uniform value of <7, = 1 8, assumed at first. Therefore the increase of section of each bottom chord due to wind, as derived from (513), is, in square inches, S= WJ \ (n - i),a(- 2), 3 (- 3 ),etc.l. (651) 0072^, ( ) 2000072^ The weight of this wind augment to bottom chords is found by putting 10,000, instead of 6,400, for Q in equations (515) and (579), thus : Weight of bottom chords due to wind, pounds, = o.oo3858o2496( 2 + - 2 )hl 2 (n even) / ^ = 0.00385802496^2 + | - I - |JA/' ( odd) (65^) 534 MECHANICS OF THE GIRDER. = hi 2 0.0090422 n = 8 = hi 2 0.0084499 = 15 0.0088909 9 0.0084093 16 0.0087963 10 0.0083678 i? 0.0086957 ii 0.0083352 18 0.0086272 12 0.0083021 19 0- o85555 13 0.0082755 20 0.0085034 14 193. Assuming that a wind pressure per panel of 125- pounds (/ being in feet) acts in a direction normal to the plane of girder at each apex in each top chord, we have the moment of each brace due wind equal to 125 X y = (653) if 5 = cross-section of flanges of a brace, B^ = inch strain allowed for bending-moment = (5,33 3 + 6000) = -^f , and if the length of brace is to its width at broadest end in the ratio of 10 to i, as in article 163, where the flanges of each brace meet at one end, and diagonal lattice work forms the web. From (653), Cross-section of a brace due wind, in square inches, nB l 34/2* o (654) Increase this section by 50 per cent for the first and second braces, to take a part of load, as explained in article 187. See value of ^y. Weight of vertical braces, pounds, = 2 X IS X 34* X = 1.9607844 (655) BRIDGES OF CLASS II. 535 = /// 2.366729 n = 8 = hi 2.091502 *= 15 2.291617 9 2.076631 16 2.235294 10 2.064152 17 2.192072 ii 2-053578 18 2.158224 12 2.044544 19 2.131248 13 2.036764 20 2.109415 14 since we now have the special value ^ _ 4/* ( (n i) + 2(n 2) + $(n 3) -f . . . (n i) terms) n 2 \ -f (n i) 4- 2( 2) for suspenders j n 2 194. The Head Lateral System. Cross-section of head diagonals is given by equations (587), (588). Weight of head diagonals found in (589) and (590). Cross-section of head strut expressed in (591) and (592). Weight of head struts is given by (593) and (594). Cross-section of iron to be added to segments of top chord shown in (595), (596). Weight of added iron in (n 6) panels of top chords is to be found in equation (597). 195. We may now collect the weights of all the parts of the bridge, and, after augmenting by one-tenth of itself the weight of the girders, the vertical braces, and the head struts, as explained in article 165, we may equate the weight so found to 2OOO;/ W, and so determine W in terms of L, /, and //, for differ- ent values of ;/, and from = o we find // rendering W a all minimum. 536 MECHANICS OF THE GIRDER. (o "^ co" 'ON 'o' to to <-O ""> \O vo vo vo^ vo vo _ ' 2 ~ ' o ^_J 6 o 1 \- " .|. ^ 1- "l i- T .]. "ON N ^2 ^ to to ^ 3. II Ii & II - n ^> II 3 l-O vo ^ ON K^ ri- >^i CO i^ to 4. ^ 4- ss 4. 4- ^ M ^ "S 10 VO Vk ON i a 4- + a ^ II 1 1X4 iC II "ro "> co KS, O\ v 1 1 v to |ii ||gn I-" ^o ^ to ^ n 2 vd ^ ^ vo *o o tj *^ o 1 2 .5,9 ~ "Xi O 2 ^5 ! 2 O to 4- o r^ Q 1 | 2 1? 4- 8 co^ " co" 4- ^ o ^J r^ O ^ " ^8 o\ ^ **>" ^ sp- |.^ ^5%- i+ o" *? 4. t> ^ 0\ ~ 1 . ^i O> "^T O ' ?^ ^ + 10 VO <- f^ t^ ro ^ ON 4- I,- i|| h-( S8 + ^I 1 + 7 ^ + ^ 4- *"!' ft "* ^* "+J to cT 4J Ol 4J fO "o to ^ ON O HH CJ "^ (U t 4- VM ^ co" to d to 4" 1-1 <-O d ro -4 n (U O to co q to O ? + ?. 1 M fO | ro 4- Tf i CO VO^ II ^ II ^ II ^ II ON II 1 1 *T s, ^ ^ CO J c C + 4- "^ 4- c * VO M 3T 1 co p t CN c O rt "^ I V) n fcfl W to w N CO w CO to VO .B vo _S vd .S vo vd c s '1 ^i 1 S '1 i 'I S 1 * o.S II i I- S rt O H C

  • - in g. r-H 0" HVO voot^t^ vOi^vO ON C >O CO rO f oom->J- vo vo r ^ - H - MO O m ON H ^^^ I " *" H- d ri-ovOrO r-- H H vo M t< ro <* *? *? OO rt-oo w 1Q M w ^ CO H O O^ ^ ^' O>I^O \o vO H H 06 H i? Tj-t>. tv COOO O M * M M J^in OO OO" ro ro ! !! ' " 1 ' s : Ml . 8 8 c *- c *- : IJIilll! !!i{ lllili J iflllll gi,g u ,--T3 a s tS lu ^oo o SuSSjasg^a -a .& n -8,9 -2 S|;S*55 - - e q q q q q q q q o q q o q q q q q o q q q q q >o to 10 q q q _0 MH - MHMMMMMM1H MM,HMMMH MMHMMMNO " ^- C* ro-^tOvOr^ 540 MECHANICS OF THE GIRDER. 196. A few simple relations may be stated here, as resulting from our investigation of this bridge having two parabolic bow- string girders with double triangular web. ist, For a given uniform live load, n oc I* nearly, (669) W oc Ih nearly. (670) 2d, For different live loads, hW oc nL nearly. (671) 197. EXAMPLE. Specifications for the bridge of 200 feet span, as tabulated in article 194. We have found / = 200 feet, n = 13, L = 15.38460 tons; h = 39.726 feet, W = 8.16590 tons; q = 16 feet, %(W -{- L) 11.77525 tons; g, = 1 8 feet. Maximum horizontal component of chord strain, each of 2 girders, M n.775 2 5 X 13 X 200 = H = -J = 8 x 39.726 - - 9 6 '333 tons, by equation (564). Therefore, by (632), Strains in top chord due loads TT = P= = 119.464 tons, /. Sections = 31.739 square inches; . cos a = 110.330 tons, = 29.312 square inches; = 107.231 tons, = 28.489 square inches; == 102.605 tons, = 27.260 square inches; = 99. 1 70 tons, = 26.347 square inches ; = 97.050 tons, = 25.783 square inches; = U = 96.333 tons, = 25.593 square inches; for each half-span of each girder. BRIDGES OF CLASS II. 541 Additional cross-section of n 6 central panels of top chords to resist lateral displacement, by (595), is S = 0.70049 square inch. Corresponding chord strain == 0.70049 X 3.764 = 2.637 tons. /. Total top chord strains = 119.464 tons, ist and i3th panels; = 110.330 tons, 2d and i2th panels; = 107.231 tons, 3d and nth panels; = 105.242 tons, 4th and loth panels ; = 101.807 tons, 5th and 9th panels; = 99.687 tons, 6th and 8th panels; = 98.970 tons, 7th panel. From (65 1), the varying sections of a bottom chord due wind are 2500 x 39.7265 x 200 S = 20000 x 13; X 18 ^ I2 > 2 X ii, 3 X 10, etc.) = 3.918 square inches, .'. Strains = 19. 590 tons; = 7.183 square inches, = 35.915 tons; = 9.794 square inches, = 48.970 tons; = 11.753 square inches, = 58.765 tons; = 13.059 square inches, = 65.295 tons; = 13.713 square inches, = 68.565 tons; Add 19.267 square inches, Add 96.333 tons, for total sections and for total strains. Putting \L for L in (647), we have The cross-section of any girder diagonal = 3 x 400QO sec fl = jg ^ 64 x 13 x 39.7265 IST SYSTEM. 2ND SYSTEM. 2d panel = 2.250 square inches = 3.041 square inches, 3d panel = 3.791 square inches = 3.041 square inches, 4th panel = 3.791 square inches = 4.387 square inches, 5th panel = 4.795 square inches = 4.387 square inches, 6th panel = 4-795 sq uar e inches = 5 ooi square inches, 7th panel = 5.001 square inches = 5.001 square inches. 542 MECHANICS OF THE GIRDER, The actual strains on these girder diagonals given in the diagram below, Fig. 120, where compressive strains are marked negative, have been derived from equations (641) and (642), using the proper value of r, and dividing by the proper value of cos 0. The floor and the longitudinal I-beams will be the same here as in article 184 ; viz., Floor of 2^-inch oak. I-beams, depth d = 9.7740 inches. d d l 0.6610 inch. 8 = o inch for x = 100 at end. Length of top chord = 222 x 5 sec a = 219.297 feet. Contraction due strain, by (336), 3.7646 + 5 = I200 Q x 2I 9- 2 97 X 12 = 1.922 inches. 1.922 Mean excess per panel = - = 0.1478 = -^ inch nearly. SECTION 2. The Post Truss with Parabolic Top Chord (Fig. 36). 198. Let our previous notation be continued as far as appli- cable ; viz., / = span, in feet. // = height of girder at centre, in feet. n = number of panels, counting on the bottom chord, and odd. L = panel weight of uniform live load, in tons, given. W = panel weight of bridge, in tons, to be determined. Symmetry here requires an odd number of panels for the bottom chord, and an even number for the top chord. As the BRIDGES OF CLASS II. 547 live load will here be applied to the bottom chord, we shall take n odd, and ranging from 9 to 21, inclusive. Each upper apex is in the middle of a panel's length. There is but a single sys- tem of counter diagonals, while there are two systems of mains. We shall here assume the difference of level between the centre and end of the top chord to be one-tenth of the whole central height, h ; and consequently the height at the end of a top chord is -&h. This, of course, is wholly arbitrary, except, possibly, in some cases where the head room at the ends would be too little. The top chord is to be polygonal (that is, straight from joint to joint), and we will take it parabolic in this case. Puttin 2 X -/ for /, and for x> in (472), we have the height of any upper apex, y = Q.^r(n r i) (n - i) 2 = /fc(say); (673) r to be counted on top chord from o to - , inclusive (that is, to the centre). VALUES OF e IN (673). n = 9 11 13 15 17 19 21 r = o 0.90000 0.900 0.900000 0.900000 0.900000 0.900000 0.900 i 0-94375 0.936 0.930556 0.926531 0.923439 0.920988 0.919 2 0.97500 0.964 0-95555 6 0.948980 0-94375 0.939506 0.936 3 Q-99375 0.984 0.975000 0.967347 0.960938 0.955556 0.951 4 1. 00000 0.996 0.988889 0.981633 0.975000 0.969136 0.964 5 I.OOO 0.997222 0.991837 0.985938 0.980247 0-975 6 I.OOOOOO 0-997959 0-993750 0.988889 0.984 7 I.OOOOOO 0.998437 0.995062 0.991 8 I.OOOOOO 0.998765 0.996 9 I.OOOOOO 0-999 10 I.OOO 548 MECHANICS OF THE GIRDER. Since the top chord for every panel length slopes uniformly, we have manifestly the height of girder, if measured in the vertical through any lower apex, equal to the mean of the two heights at the adjacent upper apices just found. If, then, we put r + i for r in (673), and add the resulting equation to (673), we have, after dividing by 2, ( o.Ar(n r 2) 4- o.2(n 2) ) y = ; '|-9 + - (n _ ,). - } = A (674) which is the height through any lower apex ; r taking the values o, i, 2, 3, n - 3 , counted on upper apices. VALUES OF e t IN (674). = 9 11 13 15 17 19 21 r = 0.921875 0.918 0.915278 0.913265 0.911718 0.910494 0.9095 i 0-959375 0.950 0.943056 0-937755 0-933595 0.930247 0.9275 2 0-984375 0.974 0.965278 0.958163 o.95 2 344 0-947 53 i 0-9435 3 0.996875 0.990 0.981944 0.974490 0.967969 0.962346 0-9575 4 0.998 0.993055 0.986735 0.980469 0.974691 0-9695 5 0.99861 1 0.994898 0.989844 0.984568 0-9795 6 0.998980 0.996094 0.991975 0.9875 7 0.999218 0.99691 3 0-9935 8 0.999382 0-9975 9 0-9995 Calling a the slope of any segment of the top chord, we have snce \ = - y,. = [ 2(r + i)]. BRIDGES OF CLASS II. 549 sec 2 = o.i6/i 2 n 2 x -,, i)] 2 , (675) = i + - a , where r is to be counted o, i, 2, 3, etc., and VALUES OF e 2 IN (675). n = 9 11 13 15 17 19 21 r = o O.I55 39 0.156816 0.157785 0.158372 0.158752 0.159014 0.159201 i 0.079102 0.094864 0.105625 0.113390 0.119241 0.123799 0.127449 2 0.028477 0.048400 0.063896 0.075906 0,085374 0.092987 0.099225 3 0.003164 0.017424 0.032600 0.045918 0.057151 0.066577 0.074529 4 0.001936 0.011736 0.023428 0.034573 0.044568 0.053361 5 0.001304 0.008434 0.017639 0.026961 0.035721 6 0.000937 0.006350 0.013755 O.O2 1 609 7 0.000706 0.004952 O.OII025 8 0.000550 0.003969 9 O.OOO44I 199. Moments due a Total Dead Load of Uniform Panel Weight, W + L. Although the total load is here uniform, the separate or single systems are in no case uni- formly loaded throughout the girder's length ; and we may find, by equations (40) and (43), the effect of each single weight, IV + Z, at all required points in each single system, or we may sum the values of a' in these two equations for the several cases, as follows : ist, When 21, etc. i is an integer; that is, when n = 9, 13, 17, 550 MECHANICS OF THE GIRDER. FIRST SYSTEM. NINE PANELS. y o o o '1234 56789 FIG. 121. Let r denote the number of intervals, each = 2c = , n between any weight in the right half-span and the left end of the girder. Then, when x 7 a', equation (43) applies, giving M= W + L a , l _ x Beginning at the left weight, and summing the values of a\ we have M = r(r (676) which is the moment due all the weights on the length, 2cr, measured from the left end, at any point not distant less than 2cr from the left end of girder. r?^ For the moments due the weights on the remaining part, / 2cr, we sum equation (40), where now M = and x \ a'. (/ a e )x, BRIDGES OF CLASS II. 551 Beginning at the point 2c(r + i), we thus sum : 20' = 2c\_(r + i) + (r + 2) + (r + 3) + . . . rj = <:(>, + r + i)(r, r), 2#' = r x r = number of terms ; .% M = ^ + Z [(r, - r) - (r, + r + i)(r x - r)]:e fX = Ifl^ ( _ 2) - _ a)( _ ar )*, (677) L 11 2 since here r, = -- . 2 Adding (676) to (677) gives n - 2T - 2)(n - ar)*, (678) which is the moment due all weights in the first system, at any point in the second half-span, when - - is an integer ; and 4 for the panel points in this system, second half-span, r becomes + i n + 5 n + 9 . , , , etc., 444 and 2?'/ x = = 2cr. n Therefore, putting this value of x in (678), M = (2r -h i)( 2r), (679) which is the moment due all the first system weights at loaded points in the second half-span. 552 MECHANICS OF THE GIRDER. If, in (678), x = 2(r + j)-, it becomes n M on 9], (680) which is the moment due all weights at the unloaded panel points in second half-span, first system. Similarly we proceed with the second system when is an integer. SECOND SYSTEM. NINE PANELS. (JO O O 1 234567 8 9 FIG. 122. Beginning at the left weight, and summing the values of a' in (43), there results n - 4- .,.,) (681) which is the moment due all the weights on the length, 2rc, measured from the left end of the girder, at any point not dis- BRIDGES OF CLASS II. 553 tant less than -- from the left end ; r being not less than /72 _ j\ , and increasing by unity for the loaded points in second half-span. For the remainder, / 2cr, we use (40), and find equation (677), which, since now r I - , becomes M = W + L (n - 2T - i) 2 x, (682) 4/2 where x cannot be greater than 2c(r -f- i), and r not less than ;/ i 4 Adding (68 1) to (682), the result is, if, as usual, we call the sum M instead of 2M y M=W- - 2T- i)^, (683) which is the moment due all weights in the second system, at any point in the second half-span, when - - is an integer ; 4 the limits of x being 2rc and 2(r + i)c> an d tne limits of r, ;/ i -, n i - and - . 4 2 If, in (683), we put x = , we have, for the loaded points n in second half-span, second system, ( W 4- M = - ~ [( - *r)(2r - i) + i] ; (684) 554 MECHANICS OF THE GIRDER. and, if x = 2(r + |)-> (683) becomes which is the moment at all upper or unloaded apices in second half-span, second system, the sign of the last moment to be changed from to +. 2d, When - is an integer; that is, n = n, 15, 19, etc. 4 FIRST SYSTEM. ELEVEN PANELS. 56 789 10 11 FIG. 123. Proceeding, as above, to sum a' in equations (40) and (43), in the present case we write snce where ^r 7 a'. (6 36) BRIDGES OF CLASS IL 555 Again, when x \ a', M = W + L ^(la' - a'}x = W L(l ' 1 ~ T)(n ~ r ' ~ r ~ X) *' (687) since 2tf' = r t r = number of terms, and 20' = 2t[(r + i) + (r + 2) + (r + 3) + . . . rj = (r, - r)(r, + r + i). Add (687) to (686), and put r, = ~-~ ; - ar- 2)*, (688) which is the moment due all weights in the first system, at any point in the second half-span, when - is an integer ; r 4 n I 11 -f- 3 ;/ -}- 7 being - , -- , - , etc., and x lying between 2rc and 2(r + iy. If x = 2rc, equation (688) becomes (639) which is the moment due all the first system weights at loaded points in the second half-span. 556 MECHANICS OF THE GIRDER. If, in (688), x 2(r + f )*, we have (690) which is the moment due all weights at the unloaded apices in the second half-span, first system. Also, for the second system, when - is an integer, we 4 write : SECOND SYSTEM. ELEVEN PANELS. U U 1234 From (43), we have x 7 a', U U 567 FIG. 124. n i i)], 11 - *) (691) n And, from (40), # \ a', (r, r) terms ; S0' = 2^r[(r + i) + (r + 2) + (r + 3) + rj, ifr,= BRIDGES OF CLASS II. 557 2tf' = [-(r, + r + i)(r. - r) + (r, - r)> = }*(# 2r i) I 2 _ 2r _ The sum of (691) and (692) is M= W + L \\n + i + *r(r + i)](/ - x) 4/1 + (n - 2r - i) 2 ^j, (693) which is the moment due all weights in the second system, at any point in the second half-span, when - is an integer ; 4 the limits of x being 2rc and 2(r + i)c, and the limits of r ;/ 3 n i being and Substituting 2rc for x in (693), we get M = - ~~[* + i + -(2 - 4^ - 2)], (694) which is the moment due all weights at the loaded apices, second half-span, second system, - being an integer. 4 And, if x = 2(r + fX in (693), we have finally M = - CKS- - 7) + *r( - ar - 4)], (695) which is the moment due all weights at all upper or unloaded apices in second half-span, second system ; the sign of the last moment to be changed from to +> as i n equation (685). Of course, the total moments for each and both systems will be equal at corresponding points in the two half-spans. 558 MECHANICS OF THE GIRDER. 200. Weights of Top and Bottom Chords due a Total Dead Load of Uniform Panel Weight, W + L. Dividing (679) by (674) gives 3 (696) which is the horizontal component of strain in top chord over loaded points in first system if (2T , = 4* and - - is an integer. 4 Also, dividing (684) by (674), calling (n 2r) (2r i) 4- i 6d = - 9 4n we get which is horizontal component of strain in top chord over loaded points in second system, for this case. Then, adding the strains due to each panel length of top chord from the two systems, we have total horizontal component of strain due n(W -f- L) in each panel of top chord, s = x if 5 = the sum of the proper values of - and - j t Similarly, in case - is an integer, making 4 _ (n - 2r)(2r - i) *'-- -^T BRIDGES OF CLASS II. 559 in equation (689), and (n 2r)(2r + i) -f i in (694). Computing in this manner, we have, for each half-span, VALUES OF s s IN (698). n = 9 11 13 15 17 19 21 Panel Pt. i 0.704320 0.679863 o-74i537 0.718544 0.762139 0.741653 0.775200 2 0.914764 1.036126 1.047237 .122328 1.121732 1.174723 1.169531 3 1.120256 1.204315 1.341205 37989S 1.468452 1.487771 1.55^306 4 1.120256 1.370786 1.480524 .629998 1.690715 1.791639 1.828413 5 1.370786 1.621512 .749982 1.909621 1.988183 2.IOOOOI 6 1.621512 .871858 2.014554 2.182537 2.275878 7 .871858 2.122266 2.276187 2.451159 8 2.122266 2.372470 2.535517 9 2.372470 2.622746 10 2.622746 2e 5 7.719192 11.323752 15.707054 20.688926 26.423490 32.775266 39.864994 Strain on top chords = H sec a Section of top chords = H sec a Q Weight of top chords, in pounds, due ( W -f Z) n _ 5 x I2/ " 18 * nO (^ + Z)/* x ^o^ sec ^ h W + L v 10 , /N (699) (700) since, by (675), sec 2 a = i + j- 2 s 2 . 5 6o MECHANICS OF THE GIRDER. Using the proper values of e 2 and 5 , we find VALUES OF s 2 s s IN (700). n = 9 11 13 15 17 19 21 Panel Pt^ i 0.109197 0.106613 0.117003 0.113797 0.120991 o."7933 0.123413 2 0.072376 0.098291 0.110614 0.127261 0.133756 0.145430 0.149056 3 0.031901 0.058289 0.085697 0.104742 0.125368 0.138343 0.153928 4 0.003545 0.023632 0.048265 0.074846 0.096626 0.119282 0.136270 5 0.002654 0.019030 0.040999 0.066021 0.088609 0.112058 6 0.002114 0.015787 0-035535 0.058843 0.081297 7 0.001754 0.013476 0.031309 0.052967 8 0.001498 0.011748 0.027954 9 0.001305 0.010410 10 0.001157 2e 2 f 6 0.434038 0.578958 0.765446 0.958372 1.186542 1.425604 1.697020 Hence, if we take, as in article 186, equation (634), Q = 3.7647 tons, we have finally Weight of top chords due n(W -f Z), in pounds, W 4- L 10 3 X 3-7647^ v7 oi) o.7594 12 ^ 2 4" 0.042 yoi/* 2 o.9o6448/ 2 4- c -h 0.0521347^ i.22i223/ 2 4- 0.056571/z 2 I.376226/ 2 + o.o6i799/j 2 I.527358/ 2 4- 0.0664347^ i. 68081 1/ 2 4- 0.07155 i/i 2 n = 9 TI 21 BRIDGES OF CLASS II. 561 Again, dividing (680) by (673) gives H = II for the bottom chord strain under an upper apex in the first n i 4r(n 2r 4) + 5// 9 system, an integer, and 6 = g Also, dividing (685) by (673), and putting g T( 1 1 _ r _ A + ** - * we have H = h ' x I 7 (7 3) as the bottom chord strain under an upper apex in the second system ; Jl -^ - being an integer. 4 Then, adding the two strains thus found for each panel length of bottom chord, we find, for this case, H = V" ^ ">' X eg, (704) which is the total strain on bottom chord due n(W + L). Proceed in like manner in case is an integer, making 4 r/n \ , n * 6 = -(- - r - i) + -* n\2 J 8n in (690), and r 7 = (" - - 4) in (695). 5 62 MECHANICS OF THE GIRDER, Computing thus, we find for each half-span, including middle panel, VALUES OF s s IN (704). = 9 11 13 15 17 19 21 Panel. i 0.246913 0.252525 0.256410 0.259259 0.261438 0.263158 0.264550 2 0.417791 0.489510 0.458860 0.506825 0.481072 0.516987 0.494988 3 0.807154 0.812868 0.894161 0.887470 0.941122 0.932223 0.973214 4 0.957264 1.117273 1.155222 1.249844 1.264131 1.330849 I-336554 5 i. linn 1.238360 1.408483 1.471186 1.578340 1.613270 1.689301 6 1.363636 1-509075 1.687720 1.770084 1.888460 1.940049 7 1.615385 1.774622 1.960186 2.058469 | 2.186634 8 1.866667 2.036256 2.227605 2-339043 9 2.117647 2.295612 2.491804 10 2.368422 2.553070 ii 2.619047 2e 8 5.969355 9.184709 12.979807 17.540519 22.702905 28.621688 33.966985 All except the middle panel taken twice for ^ 8 . Taking T 5 tons, the allowed inch strain in tension, as in (634), we find, from (704), Cross-section of bottom chords, S = ( w x g . ( 7 o 5 ) Weight of bottom chords due ( W + Z) n, in pounds, W+ L x x 18 0.442174/2 O.556649/ 2 O.66563I/ 2 -7795 79 /2 0.8903 1 o/ 2 I.OO4269/ 2 1.07831 7 / 2 n = (706) ii 13 15 17 J 9 21 BRIDGES OF CLASS II. 563 201. To find the Greatest Strains in the Girder Diag- onals, and their Weights. Equation (148) gives the strain on the counter diagonals in this case in terms of the simulta- neous moments in the vertical planes, A A >> etc., Fig. 38. These moments let us compute for the uniform panel load, L tons, advancing over the panel points O, B, D, etc., of the horizontal bottom chord. The difference of the horizontal strains due to these moments at consecutive panel points will be greatest when the foremost panel weight of load is at the foot of a Y diagonal or counter. We use, therefore, the ordi- nary formulae (64) and (68), giving simultaneous moments at O and B, B and D, etc. ; then, for the moment at A A CC lt etc., we take one-half the sum of (64) and (68, thus : = Z/ 9 (say), (707) which is the value of M l in (148), and would also be obtained by putting r 2 o, x = (r + $)c = (r + -|)-, and L for W, m (60)' for the half-intervals OA, BC, etc., Fig. 38. M r+1 in equation (68) takes the place of M 2 in (148). /*, in (148) = AA 19 Fig. 38, . = y in (673); and h 2 in (148) = BB { , Fig. 38, = y in (674). = i)V+i, (673)- With these values of M lt M 2 , h lt // 2 , a b lt we compute Y cos of (148), which, with sign changed, becomes 10 . (708) 564 MECHANICS OF THE GIRDER. VALUES OF e IO IN (708). n = 9 11 13 15 17 19 21 Panel. 2 0.024108 0.016761 0.012321 0.009436 0.007455 0.006038 0.004992 3 0.063502 0.044923 0.033488 0.025928 0.020671 0.016869 0.014025 4 0.115315 0.082009 0.061579 0.048030 0-038533 0.031615 0.026415 5 0.180545 0.127461 0.095784 0.074941 0.060364 0.049719 0.041690 6 0.182225 0.136113 0.106409 0.086241 0.070841 0-059543 7 0.183280 0.142579 0.114847 o 094826 0-079795 8 0.183999 0.147601 0.121698 0.102391 9 0.184512 0.151622 0.127379 10 0.184900 0.154894 ii 0.185203 2e 10 0.766940 0.906758 1.045130 1.182644 1.320448 1.456256 1.592654 Strain on counter = y = - X I0 sec<. (709) fi Cross-section of counter = S = Lh Th (710) Weight of (n i) counters, pounds, o - - -^ 7^ c i< 1 8 2fi 5/i nh = -[ -2 IO + 4 ^ 2 ^ A ""9 / 0.085 2 1 6/ 2 + 3-339 12 ^ 2 0.082433/2+ 4-386878/z 2 O.O78843/ 2 + 7.7942407^ o.o 77 67 3 / 2 4- 9-855875 Aa O.O76645/ 2 4- 12.140274^ 0.075841/2 4- 14.6659897^ n = 9 n J 3 15 17 19 21 BRIDGES OF CLASS II. 565 since we take T = 5 tons per square inch in tension, and Manifestly e 2 is to be taken from (673), always beginning with r = 2. Main Diagonals. To find the strains, sections, and weights of the main diagonals of the Post truss with parabolic top chord, we proceed as follows : When - - is an integer, use equation (676) for the live 4 load, nL, making W o ; and for moment in second half-span, first-system apices, At foremost end of live load, put x = , giving M ] n At point \\ panels ahead of foremost end, put 2(7* -h J)/ x = -, giving MI', At point 2 panels ahead of foremost end, put 2(r + i)/ x = , giving M l ; these three moments being simultaneous. Then M = 2 (4 .(713) In a similar manner, for the second half-span, second-system 566 MECHANICS OF THE GIRDER. apices, we find, from (68 1), simultaneous moments due live load, r+I )- !_![( -2r-f)|. (714) , = jr(r + i) - ^-^U* - 2r - 2) Dividing each of these moments, (713), (714), by the height, y t of truss at the section where the moment is taken, we find the horizontal strains^at the panel points in second half-span, M At loaded points, H = -- from (713), (714), (674); Jo At unloaded points, H % = ^ from (713), (714), (673); M, At unloaded points, H^ - from (713), (714), (674). The difference of the two simultaneous horizontal strains at vertical sections through the ends of a diagonal at and next ahead of foremost end of live uniform load is the horizontal component of maximum strain on that diagonal due live load, and is tension on the diagonal whose foot is at the foremost end, but compression on the next. LI = N - HZ = ^ x (tension), (715) = H^ - HI = -^ X I2 (compression); (716) X1 and e l2 being functions of n and r in (673), (674), (713), (714). For the moments due the dead load, ;/ W, at the same points where the simultaneous moments due live load have been found, BRIDGES OF CLASS II. 567 I being an integer, we use equations (679) and (680), and (679) with r 4- i for r, L = o, thus : First system, M Q = (2T + i)(n - 2r) Wl W7 M, = --(2T + 3 )( - 2 r - 2) 4* Second system, use (684) and (685), U/7 M = --[( - 2 r)(2r - i) + i) (717) = [( - 2T - 2)(2T 4/2 (713) Dividing each moment by the proper value of y from (673) and (674), we obtain horizontal strains at all required apices in each system, from the differences of which consecutive horizon- tal strains comes the horizontal component of maximum diag- onal strain due dead load, thus : = H HI = X e I3 (tension), (719) Wl = HI H 10 = X I4 (compression); (720) 4/? > J3 and a I4 being functions of n and r in (673), (674), (717), (718). INDEX. PAGE Advantage of steel over wrought-iron 246 Arm of resultant couple 14 Arm of the couple defined 10 Authorities cited 153 Bollman Truss classified 124 Bowstring Girder classified 101 Braced arch 93 Brunei Girder classified 89 Brunei Girder of double system 485 Brunei Girder of double system, strain sheet 521 Brunei Girder of single system 423 Brunei Girder of single system, strain sheet 483 Camber calculated for certain cases 281-288 Camber, Change of length for 278 Camber defined 277 Camber, Effective depth not altered by 286 Camber, Length of diagonals changed by 287 Cast-iron pillars 296 Chord strains 66 f All members but one inclined 89 Class I. < Method of finding linear dimensions 94 ' One or both chords parabolic 98 p. I Bottom chord horizontal. Other members inclined 101 ' I Formulae for 106 ri Til i Top chord horizontal. Other members inclined 109 ' f Formulae for no p. j, r j Both chords horizontal. Web members inclined no ' Formulae for 113 ( Both chords inclined, j Verticals in compression J II4 Class V. < ( Diagonals in tension ' Formulae for 116 (Both chords inclined j Verticals in tension j JJ? Class VI. \ ' Diagonals in compression ) v Formulae for 118 569 570 INDEX. PAGE (Bottom chord horizontal, j Verticals in compression j Class VII. s i Diagonals in tension ' Formulae for 122 Cl VIII f Top chord horizontal. Struts vertical 124 ( Formulae for 126 / T-, ,, , j , . l Verticals in compression } I Both chords horizontal. 127 Class IX. < ( Diagonals in tension 'Formulae for 129 (Bottom chord horizontal, j Verticals in tension J Class X. < ( Diagonals in compression ) * Formulae for 131 ri XT I ^P cnor d horizontal. Struts inclined. Ties vertical 132 " ' ( Formulae for 132 Class XII I ^ ot ^ cnorc ^ s horizontal. Struts inclined. Ties vertical 133 ( Formulae for 133 Classification of girders 88 Classification of girders, General formulae for 86 Col. Merrill's example for Pratt Truss 399 Combined live and dead loads, Point of greatest moment due 49 Component pressures or forces defined 2 Composition of forces defined, with example 6 Compound web systems 101 Concentrated loads, Formulae for 19 Constant first difference 37 Contrary flexure, Points of, defined 192 Correction for normal difference of moments due end moments 60 Arm of, defined 10 Arm of the resultant 14 defined, with example 10 Direction of the resultant 14 Couples, Resultant of many co-axal 13 Crescent Girder 89 Deflection at centre of uniform beam 172 Deflection at the free end of parabolic semi-bowstring 234 Deflection of a beam, Influence of fixed ends on 189 Deflection of a beam with one fixed end 192-213 Deflection of girder of variable cross-section 224-277 Deflection of semi-beam 158 Deflection of semi-beam at any interval due all weights 166 Deflection of semi-beam at its free extremity 162 Deflection of semi-girder of uniform height and strength 224 Deflection of uniform beam due any number of equal weights at equal intervals . . . 183 Deflection of uniform beam due a concentrated load 170 Deflection of uniform beam due its own weight 169 Differences, Method of first 36 Couple, INDEX. 571 PAGE Differences of simultaneous moments 36 Dimensions of beam found 249 Direction of resultant couple, with example 14 Dome Principal 116 Double-bow or Brunei Girder . . . 89 Economical proportions for girders, with examples 351 Effect of pier moment 62 Elastic curve, Equation of 157 End moments accounted for 60 Equilibrium defined 2 Experimenters cited 154 External forces. Girder supported at both ends 23-26 External forces for semi-girder, Moments of 19 /''defined 312 Fink Truss classified 124 Floor beams 314 Force defined I Forces acting at same point in equilibrium 7 Forces in one plane 4 Forces of translation defined 9 Forces, Polygon of, with example 6 Forces, Resultant of, equal to zero 6 G defined 317 General formula? for classification of girders 86 Generalized re-statement of locomotive example 58 Girders classified 86 Girder supported at both ends. Moments of external forces found 23 Gordon's formula 299 Greatest difference of simultaneous moments 39 H denned 87 h defined 87 Hodgkinson's formulae for pillars 296 Horizontal girder of one span 59 Howe Truss classified 130,133 Inertia, Moment of 154 / defined 314 Joists, Proportions and weights of 312 K defined 317 Kansas City Bridge system 105 5/2 INDEX. PAGE L defined 87 / defined 24 Lateral supporting system 316 Linville Truss classified 127 Maximum deflection of a beam with partial uniform load 178 Maximum moment at any point due any uniformly continuous load 48 Maximum moment at foremost weight 41 Maximum moment due dead and live loads, Point of 44 Method of first differences 36 Modulus of rupture 136 Moment at fixed end of semi-beam 19 Moment at foremost end a maximum for combined dead and live loads 48 Moment due both dead and live load at any point ahead of the latter 46 Moment of a force defined 9 Moment of inertia defined 154 Moment of resistance estimated 134 Moments, Differences of simultaneous 36 Moments due uniform discontinuous load 27 Moments of external forces found for semi-girder 18 Moments reversed 31 Multiple web system 101 P defined 87 Parabolic Bowstring classified 101 Parabolic Bowstring. Strain sheet .....545 Parallelogram of forces 2 Pier moment, Effect of 62 Pillars 290 Pillars, Formula for. Case 1 291 Pillars, Formula for. Case II 292 Pillars, Formula for. Case III 294 Point of greatest moment due both to dead and live loads 44 Point of greatest moment due dead and live loads lying beyond the live load .... 45 Point of greatest moment due full continuous uniform load 48 Point of greatest moment due uniform discontinuous load 40 Points of contrary flexure 192 Points of contrary flexure for concentrated load 196 Polygon of forces, with example 6 Position of foremost end of live load when moment at that end is maximum .... 47 Post Truss. Calculations for bridge weight 546-567 Post Tiuss classified 112 Pratt Truss. Best number of panels and best height found 353 Pratt Truss classified 127 Pratt Truss, Maxima strains in 324 Pratt Truss of single intersection, varying live load ; Strain sheet for 398 INDEX. 573 PAGE Pratt Truss of single system, uniform live load ; Strain sheet for 334 Pratt Truss. Strains found 318 Pratt Truss. Weight determined 324 Pressures defined I r defined . . . * 27 Radius of gyration defined 154 Rankine's modification of Gordon's formula 300 Re-actions at the piers 24 Resistance estimated. Beam of elliptical cross-section 145 Resistance estimated. Beam of equal flanges 137 Resistance estimated. Beam of hollow rectangular section 137 Resistance estimated. Beam of rectangular cross-section 136 Resistance estimated. Beam of two vertical channels and two horizontal plates . . . 140 Resistance estimated. Beam of two vertical I-beams and two horizontal plates . . . 139 Resistance estimated. Beam of two vertical plates and two horizontal channels . . . 138 Resistance estimated. Solid or hollow beam of circular cross-section 144 Resistance estimated. Solid or hollow beam of square cross-section and diagonal vertical . 142 Resistance estimated. T-shaped beam 140 Resistance, Moment of 134 Resolution of a force, with example 4 Resolution of many forces 4 Resultant of forces, equal to zero 6 Resultant pressures or forces defined 2 Rupture, Modulus of 136 Schaffhausen Truss 131 Semi-beam defined 17 Shearing-force defined 67 Shearing-strain defined 67 Specifications for iron bridges 413 St. Louis Bridge system 93 Strains deducible from moments 64 Strains determined from shearing- forces 75 Strain sheet. Brunei Girder of single system 483 Strain sheet. Parabolic Bowstring Girder 545 Strain sheet. Pratt Truss of single intersection, varying live load 398 Strain sheet. Pratt Truss of single system, uniform load 334 Strains in rectangular beams, Formulae for 147 Table I. Ultimate resistance of materials to shearing, in pounds, per square inch . 68 Table II. Ultimate resistance of materials to tension, compression, and cross-breaking, 149 Table III. Formulae for moment of inertia and square of radius of gyration .... 155 Table IV. Values of / and a in the Gordon and Rankine formulae 302 Table V. Wrought-iron pillars 304 Table VI. Solid rectangular pillars of wrought-iron 305 574 INDEX. PAGE Table VII. Rectangular tubular pillars of wrought-iron. Thin 306 Table VIII. Hollow cylindrical pillars of wrought-iron 307 Table IX. Solid cylindrical pillars of cast-iron. Ends flat 308 Table X. Solid cylindrical pillars of cast-iron. Ends rounded 309 Table XI. Solid steel pillars. Fixed ends 310 Table XII. Solid square pillars of pine 311 Table. Computation for greatest moments and differences 38 Table. Computation for greatest moments and greatest simultaneous differences . . 40 Table. Deflection of open-webbed girders of uniform strength 242 Table. Differences and maxima differences of moments (two locomotives) 56 Table giving moments at any section of beam supported at both ends 26 Table giving moments of forces applied to semi-beam. Length / 19 Table giving moments due live load (two locomotives) 55 Table giving simultaneous moments at all panel points as foremost weight of load passes each 43 Table giving simultaneous moments due each panel weight at panel point 50 Table giving sum of moments by (91) 50 Table. Horizontal strains at the joints, in tons 402 Table showing best height for least bridge weight. Brunei Girder of double system . 508 Table showing least bridge weight for two Double Parabolic Bow or Brunei Girders . 467 Table. Simultaneous moments due advancing uniform load 32 Table. Solution for dimensions and strains. Method of moments 77 Table. Solution for dimensions and strains. Method of shearing-strains 79 Table. Strains deduced from moments and shearing-forces 83 Table. Strains found from moments for dead and live loads 85 Table. Weight of two locomotives uniformly distributed 57 Thickness of beam 246 Timber pillars 298 Triangle of forces, with example . . . . 3 Trussed rib 115 Twelve classes of girders 88 Twin Fishes 122 Two locomotives as live load on Pratt Truss 392 Two locomotives, Discussion of, as moving-load 51 Typical form for W 360 U defined 87 Uniform discontinuous load. Moment at any weight 35 Uniform discontinuous load. Moment at foremost end 33 Uniform discontinuous load. Moment at the r th weight 34 Uniform discontinuous load, Moments due 27 Uniform discontinuous load, Point of greatest moment due 40 Uniformly continuous load, Maximum moment due at any point 48 Uniformly continuous load, Point of greatest moment due 48 Uniformly distributed loads, Formulae for 19 INDEX. 575 PAGE l\ defined 24 V z defined 24 v defined 87 ^defined 87 W) Typical form for 360 w defined 24 Wind pressure, Calculations for 339-35* x defined . 24 Y denned 87 y defined 24 Z defined 87 z defined 87 THIS BOOK IS DUB 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 5O CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. NOV 1L, H33 LD 21-100m-7,'33

    • since the load is applied on the windward side in the direction of the wind's motion, and subtracting the pressures then upon the end struts, since no struts or I-beams are used on the abutments, will not alter ^P. Weight of I-beams j _ynfqJV*P due to wind, from > Q 3 (426), (n odd), where (433) . 18 ad 2 / := length, in feet, between centres of end pins. f t = numerator of Gordon formula (400). n =: number of panels. u = constant. (See Table IV.) h =. height of girders, in feet, between centres of chords. q l = entire length of floor beam, in feet. d = depth of beam, in inches. = height of train or moving wind-resisting surface. w = pressure of wind per square foot, in tons. 136. In finding the diagonals of the horizontal systems, top and bottom, due to wind pressure applied on either side, we must plainly make all the diagonals mains, and the two in any one panel each equal to the original main tie in that panel. Using the strain sheet, Fig. 112, as a horizontal system now, putting W* for W, L t for Z, q for //, sin <, = n< ? V/ 2 -|- n*q* for sin , Y l for Y, the strain in any horizontal diagonal tie due 342 MECHANICS OF THE GIRDER. to wind, we have, for the horizontal system between the loaded chords, Sum of horizontal \ ,. 4 W\ (n z = m of horizontal \ ,. 4 W\ (n z / n \ ) 1 diagonal strains = ^^ {('+3+S+7 + - terms) j due to wind +2 [i +3+6+IQ+ . . . (|-i) terms! j W,n* Z, ( , j '2 sin 0i 12 sin 0! (n even), (439) + 3 + 6 + 10 . . . ^- 3 terms^ + the ( n ~ l \ term of the series (i + 3 + 6 + 10 ^ for the middle panel J (440) (n odd). Therefore, for horizontal system uniting loaded chords, Weight of horizontal 1 diagonals due to \ = SPi X wind pressure rsin 2 ^ (n even), '. + ?,) (a odd). (440 GIRDERS WITH EQUAL AND PARALLEL CHORDS. 343 137. It may be noted here, that, however complete and effi- cient the horizontal systems are made, they will be unable to maintain the stability of the bridge under the action of wind if the posts and horizontal struts at the ends of the bridge are not sufficient to resist the lateral pressure transmitted to them from these horizontal systems. That is to say, the end frame- work of the bridge must be, with regard to the wind force, incapable of lateral motion, whether of translation, rotation, or distortion. The required stability may be secured by making sufficiently large end posts fast to the abutments for light and high struc- tures, and by attaching these end posts to rigid head struts by means of diagonal braces. But as all this excess of weight over the ordinary panel weight rests directly upon the abut- ments, it does not enter into the formulae for strains due to the uniform panel pressures, W, L ; W lt L T . This excess of weight, however, has an influence on the best values of n and h ; and, calling the excess E w pounds, we here proceed to formulate its value, and find the conditions of stability. 138. To find the additional strains and weights of the end members of a bridge of two girders of Class IX. required to resist a given wind pressure, let Fig. 114 represent the elevation of the end frame of a through bridge of this class, together with its full moving-load. Then, according to our previous notation, the total hori- zontal pressure at A is (442) and at B, P, = \n(W,+ A) = \wl(h + 2 e). (443) The vertical pressure on each abutment is \n(W -f- L). 344 MECHANICS OF THE GIRDER. Now, supposing these ends of iron rest upon a plane stone surface, and calling the "co-efficient of friction" for iron upon stone | (see any good treatise on elementary mechanics), we must have, according to the received law of friction, Z), (444) which is the condition that prevents lateral translation along a plane stone surface, BE, Fig. 114. For stability against overturning, ist, Without live load. Take the moments about E\ we need, since AF = q, and FE = h y as above, (445) GIRDERS WITH EQUAL AND PARALLEL CHORDS. 345 or, if each end of the girders is tied to the abutment with a force, /, qt, (446) the condition that prevents rotation of unloaded bridge about the points of support. 2d, With live load resting upon a beam attached to the girders at the ends B and , we require the condition (447) 4 girders not tied down ; or t (448) when they are tied with the force /. But, if the end of the live load rests directly on the abut- ment, and is not connected with the girders, the condition of stability is P*h< \q\Wn +L(n- i)], (449) or P^h < Iq^Wn + L(n - i)] + q*. (45) 3d, For the stability of the load itself against turning on its own points of support, we must have P 4 ej To enable each end post to resist this additional moment, (453), it would require R = Bacd as the moment of the inter- nal stresses on the added material, if it be added to the outsides of the post, at the distance \d, in inches, from the neutral axis ; B being the ultimate bending unit strength of section, a = width of post, in inches, and c = the uniform thickness of additional iron on each side due to the greatest moment at C or G. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 347 Hence Whole thickness of added 1 i2P 2 (h />cos/3)/ . > 2C -- * - iron j aBd Cross-section of added | = ^ = i a />(A - Jcosffl/ inches< ron Volume of 4 posts due j _ 4 X i2 2 P 2 (/i bco$(3)fh , . to wind J Bd Weight to be added to ' 4 posts at end due wind bending, Bd (457) Similarly, from (456), for the two top horizontal end struts, of length q feet, and depth */, inches, Weight to be added to 2 end struts to 1 __ 2 x resist bending from wind force J ~ ' Bd l (458) pounds. If ^ 2 is the least diameter, in inches, of a brace of length b feet, with fixed ends, to resist the longitudinal pressure D, (455), then, by the Gordon formula, (400), we have Cross-section of brace, 6" = j- square inches ; /, as before, being the factor of safety, and f s the numerator of Gordon formula. /. Volume of 4 braces = 48^ = r cubic inches. 348 MECHANICS OF THE GIRDER. Weight of 4 braces = /. sin fi cofy^ = -7. 5 5 - pounds. /! sin p cos p \ h -- Bd zBdt 24/ x sm /3 cos (3 j (46o) which is the excess of weight, in pounds, of wrought-iron, on the two abutments, due to wind pressure, and not affecting the uniform panel weight of bridge, W. 139. In the preceding investigation, we have assumed that the entire effort of the wind to distort the rectangular cross- section of the bridge is to be resisted by the two end frames alone. Instead of this provision, however, we may fix firmly each horizontal strut of the unsupported lateral system throughout the bridge to the ends of the posts abutting upon it, and thus transfer the whole wind pressure to that lateral system which is between the chords resting upon the supports. The same transfer would also be accomplished should we connect the posts rigidly to the other, or supported, lateral struts. This procedure would enable us to dispense with the horizontal diagonals of the unsupported system but for the necessity of retaining them to keep the chords they connect from deflecting horizontally. In this case, of course, the horizontal diagonals and struts of the supported system will have twice as great horizontal pressure to resist as in the former case, and (438) and (441) must be multiplied by 2. The horizontal struts of the unsupported system must be GIRDERS WITH EQUAL AND PARALLEL CHORDS. 349 able to resist, in a vertical direction, the bending-moment found by (456), when for P 2 we put W iy or its value \wh- , giving If = \wh- = (461) n 127 for each strut ; d 2 being the depth of horizontal strut, in inches, a the width, and c the thickness of each of two plates of iron added at the distance \d 2 from the neutral axis of the strut. B -T- f = allowed bending unit strain, in tons, per square inch, since w is in tons. Then the weight of all these horizontal struts due to the bending-moment, (461), must be the same as in (458) if we put d 2 for d lt and regard the two extreme struts as one, since each sustains but half a panel pressure. At the same time, these horizontal struts must resist, in the direction of their least diameters, the bending-moment due to the longitudinal strain brought upon them by the attached hori- zontal diagonals in adjusting the bridge. Now these horizontal diagonals, between unsupported chords, may be of uniform size, having a cross-section S, (say) of not less than about I square inch. Then, if the allowed unit strain upon them is T 4- f, and if their inclination to the plane of the girder is < we have the longitudinal pressure of two diag- onals, from adjustment, to be provided for, equal to (46*) And if = - - - - - = the allowed pressure per square \ inch upon a strut, Q 2 , P a , / and T being of the same denom- 35O MECHANICS OF THE GIRDER. ination, and f t = numerator of Gordon formula, f = factor of safety, then , Q 2 _ 27IS'sin M = - = inch-tons (466) 2 X 4?2 / as the bending-moment allowed at the weakest part of the post, each end post having but \M instead of M. Therefore Whole thickness of added iron for i 1 _ _ yvflh* post anBd GIRDERS WITH EQUAL AND PARALLEL CHORDS. 351 Cross-section to be added to each | = 2 ^ = 3!^! e inches . post j nBd Weight to be added to 2n posts to resist distortion of rectangular cross-section of bridge pounds. (467) Finally, the weight of 2n wrought-iron braces for this case also is given by (459) ; and the cross-section of one brace is since D in (455) now becomes wh 2 ! ^nb sin ft cos ft 140. We will now exemplify the method of article 138, which provides, in the end frames alone, the means of resisting the distorting influence of the wind. EXAMPLE. To find the best number of panels, n, and the best height, 7z, for the two wrought-iron girders of a highway "through" bridge of 100 feet span = /, and 1 8 feet wide between centres of chords = q, single system of Class IX., Pratt Truss, under a uniform rolling load of I ton = 2,000 pounds per running foot, in addition to the weight of bridge. Also, to find the weight, n W, of the bridge corresponding to the best values of n and /i, using 4 as the factor of safety for iron, and 10 for wood, and taking account of wind pressure. Let us compute for n = 5, 6, 7, 8, 9, 10, ir, 12, in succes- sion, as explained in article 134, retaining h and W in all the expressions for weight. 352 MECHANICS OF THE GIRDER. ist, The floor of pine, called 50 pounds per cubic foot. Thickness t ^| foot. Width q' =17.5 feet. Length / = 100 feet. 2.5 Weight of floor F x 17.5 X 100 x 50 = 18229 pounds. 2d, The joists of pine at 50 pounds per cubic foot. g = 2 feet between centres. B =. 7,000 pounds per square inch = ultimate resistance to cross-breaking. / = 10, factor of safety for pine. /4- n = panel length of joist, in feet. d = b 2 depth of joist, in inches, by (431). Then, by (432), we have Thickness of) X 10 x 2 x 100 H .. 200000) _- ins.; a joist, *'X i 7 . 5 X and, from (433), W 1,, f - * 7 100 X 17.5 X 50/7.965443^ 153548 Weight of joists, / = I44 x 2 - V-^fy = -*an P unds - 3d, The wrought-iron I-beams, w i in number, supporting the joists, floor, and moving-load L = - = tons per panel. ;^ 72 Take j5 = 50,000 pounds, Table II. Length of beam q^ = 18.5 feet. K 1 5 3548 \i8.5 x - // -f 2I8229V = 0.4331885^- - -- J inches, from (412), using the proportions assumed in finding that equa- tion. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 353 By (434)> Weight of I-beams, P = 15. 46068 (n i) X -^ 1 8 l8 - X 50000 // + 218229X1 = i. 03 1 824 (n i)( - - - ) pounds.. 4th, The horizontal struts of the top lateral system of this "through " bridge. In this example of a highway bridge, let us assume, as actual pressure of wind per square foot, the large value 75 pounds ; also that the two open girders offer a resisting surface equiva- lent to | of the surface presented if the bridge were covered, that is, equal to f ///. Then the whole wind force to be resisted is | X 75/// = 45/*/ pounds. Wind force per running foot 45/2 pounds. Wind force per square foot = w = 2$$$ = 0.0225 ton. Although this wind force is actually applied to both girders, we shall regard it as distributed equally to the panel points of the two windward chords, no account being here taken of the action of wind on passing carriages. Suppose the top horizontal struts to be I-beams, the square of whose least radius of gyration is r 2 = 0.5 inch, which cor- responds to a six-inch beam of ordinary make. Then, using equation (385), and calling C = 40,000, E = 27,300,000, we have, in (437), 20 Q 2 = = 4. 680056 tons; 40000 x 2 io 2 47T 2 X 27300000 X 0.5 354 MECHANICS OF THE GIRDER. and (437) gives Weight of top horizontal struts due to wind 0.0225 x iQohfn 18 x 4.680056 = 28.84581^- + 2^ (n even), = 28.8458I// 7 * 2 ~ * + 2^ ( odd). 5th, The horizontal diagonals, top and bottom. From (441), where now L l o, since we take no account here of wind against live load on this highway bridge, we have, making T = 24 tons, q = 1 8 feet, m = -f$ for wrought-iron (as above), Weight of horizon- } tal diagonals, top = X= 2 X 5 X 4 X 18 X 6 ^ ( even) and bottom, J 18X24 sin'*, i ( z - i)(odd), where i ^ -L. JL = i + IOQOQ sin 2 .^ ~ n 2 q 2 ~ ~ iS 2 n 2 ' 6th, Let the residual weight, Y y be 1,000 pounds for all val- ues of n. 7th, The additional weight of iron needed in the I-beams, by reason of their acting as horizontal struts for the wind pressure on lower chord, is found by (438), after computing d as already formulated for the floor beams. Here W t = - tons ; 211 L I O] / x = 1 8 tons; q l = 18.5 feet; a = 750, since the ends ,0 are not fixed. Hence, from (438), where Q 3 = - , i J- 2222 GIRDERS WITH EQUAL AND PARALLEL CHORDS. 355 Weightto beadded | to floor beams \ due to wind 3 x 5 x 4 x l8 . 5 x O . O225 x ~ 75^ i8x i8x 2 (n even), 3.85416667/1 + \ X n (n even), x ^!nJ ( dd) Collecting the terms of K thus found, we have, in terms of //, WEIGHTS OF THE COMPONENTS OF K, IN POUNDS. 5 6 1 8 Floor . . . 18229.0000 18229.0000 18229.0000 18229.0000 Joists . . . 22258.0000 17884.0000 14864.0000 12663.0000 ( I-Beams . . 5459.0000 5969.0000 6408.0000 6796.0000 I Do. wind . . 23.39847* 30.12457* 35-37QI/& 42^30867* Hor. struts 1 26.92 1 6/1 144.22917* 156.59147* 173.07497* Hor. diags. . 120.6662/1 125.37047* 125.7336^ 13340257* Residual . . IOOO.OOOO IOOO.OOOO IOOO.OOOO IOOO.OOOO K . . \ 46946.0000 43082.0000 40501.0000 38688.0000 } +270.9862/6 + 299.7 240/fc +317.69517* +348.78607* n 9 10 11 12 Floor . . . 18229.0000 18229.0000 18229.0000 18229.0000 Joists . . . 10994.0000 9688.0000 8641.0000 7784.0000 ( I-Beams . . 7146.0000 7465.0000 7760.0000 8035.0000 Do. wind . . 48.118 i/i 55-33 12/ ' 61.62267* 69.1289/5 Hor. struts 185.89507* 201.9207/6 215.02947* 230.76657* Hor. diags. 138.10407* 147.22207* 154.03257* i63-935 8/ ' Residual . . IOOO.OOOO IOOO.OOOO IOOO.OOOO ' IOOO.OOOO * \ 37369.0000 36382.0000 35630.0000 35048.0000 \ +372.11717* +404.47397* +430.6845/5 +463.83127* 356 MECHANICS OF THE GIRDER. 8th, The top chords, of 2 channels and 2 plates of wrought- iron. In each panel let the ratio of chord's length to least diam- eter be 15. Then, in (424), 18 = 16.7442 tons, + by (400). 3000 Weight of top chords due to ver- ) _ 5 X 4 X ioo 2 / yp , ioo\ tical pressures, in pounds, j 2 x 18 x 1 6.7442^ \ n ) X 2 " 2 + 3 - a (n even), ~ 3 (n odd). And, from (435), Weight of top chords due ) _ 5 x 4 X ioo 3 X 0.0225^ to wind, in pounds, j 2 x 18 x 16.7442 x 18 - 2 - 3 (w odd). 9th, The bottom chords, of flat links or I-bars. From (425), Weight of bottom chords due to ) _ 5 x 4 X ioo 2 / ^ + ioo\ vertical forces, in pounds, j 2 x 18 x 2^\ n ) 3 2 4- - 24 21 (n even), (odd). GIRDERS WITH EQUAL AND PARALLEL CHORDS. 357 And, from (436), s being zero, Weight of bottom chords ) 5 X 4 X TOO* x 0.0225^ due wind, in pounds, ) " 2 X 18 X 24 X 18 2^3 _ 3^2 + 22n 24 3 - 3 2 4- 22H - 21 (n even), (odd). i oth, The verticals. Take ratio of length to least diameter 30; then, in (426), - 8.8 750 if the ends are not fixed, and we have Weight of verticals, in pounds, * 4- 5 X 4 X iQQ/fr hn* + 3^ - i \ n ) 18x245! 18x24 sin 4^ ( odd). 358 MECHANICS OF THE GIRDER. Computing for the different values of ;z, collecting, and arranging, we have, including the values of K above, WEIGHTS IN POUNDS, W IN TONS, h IN FEET. Live load = nL = 100 tons, I = 100 feet. 11 5 Top chords .} Bo,o m chords |-. Verticals 4140-742 2444.444 W h 9.77778 Wh 82815 48889 i k - h* 103.5185 61.1111 208.5926 88 8889 K . . 46946 270.9862 2ooonlf = 7918.519 +13.11111 +167259 +46946 +733-0973 6 Top chords .J5- Bottom chords { Load I Wind . Verticals 4866.256 2777.778 14 66667 81104 46296 - 101.3803 57,8704 1388 88q K . . 43082 299.7240 2OOO W = 9032 923 +19.66667 +157407 +43082 +861.2381 7 Top chords .{- Bottom chords j Load ' Wind . Verticals S52S-323 3174.604 78933 45351 - 98.6665 56.6894 6 66666 K . . 40501 317-6951 zooonW = 10060.471 +26.22222 +150199 +40501 +906.0064 8 Top chords .}- Bottom chords 1 Load ' ' Wind . Verticals .... 6221.067 3559-027 26 07408 77763 44488 - 97.2042 55-6098 392.1296 Diagonals K . . 1388.889 8.88889 22787 38688 I45-8333 348.7860 2ooon W 11168.983 +34.96297 +145038 +38688 +1039.5629 GIRDERS WITH EQUAL AND PARALLEL CHORDS. 359 WEIGHTS IN POUNDS, W IN TONS, h IN FEET. Live Load = nL = 100 tons, / = 100 feet. It 9 Bottom chords j Load ' , f Wind . Verticals 6881.576 3978.051 W k 32.59260 Wh 76462 44201 i - r 95-5774 55-2507 402 3777 ii. mil 20322 K . . 37369 372.1171 2OOOH W = 12231.369 +43.70371 +140985 +37369 +1089 9319 10 Top chords . I Load ' ' Wind . Bottom chords j Load ' Wind . 7564.819 4388.889 75648 43889 - 94-5602 54.8611 488 8889 Diagonals 1388.889 13.88889 18 K . . - 36382 404.4739 20oonJV = 13342.597 +54.62964 +137870 + 36382 +1226.1174 ii Top chords . j Load ' Bottom chords \ Load ' f Wind . Verticals 8226.202 4820.936 48 88888 74784 43827 - 93.4796 547834 1377.410 1 6 66667 16696 35630 430.6845 2OOO W 14424.548 +65-55555 +135307 +3563 +1279.2841 12 Top chords .jLoad. ' Wind . Bottom chords j Load ' ' Wind . Verticals 8903.037 5246.912 eR 66667 74192 437 2 4 - 92.7400 54.6553 Diagonals 1388 889 K . . 35048 463-8312 2000^ = 15538.838 + 78.66667 +133241 +35048 +1416.9878 --3959 2 59 + o-S* ~ 7-87035 -f 2.1541^ + 0.04306191^ , 0.4516461 -f 0.6/1 0.000983333^ 7-5995 + 2.02505^ + 0.04530032^ 360 MECHANICS OF THE GIRDER. Multiplying each of these eight equations by - , we find 20000 the uniform panel weight of bridge, W, in terms of // ; thus : 8.36295 4- 2.3473^ 4- 0.03665487/^2 n = 5, W = n = 6, W = n = 8, W = n = 9, W = n = 10, W = n = n, W = /Z ?*""" 1 2 t^ff . 0.7769419 4- 1.2/1 0.00393333/^2 In differentiating these and similar expressions for W, it will be convenient to have a typical form or mode of operation. Let W = a + bh + ^ (469) be a type of these equations ; then, after putting - = o, and all reducing, we have the equation 'J> 4- (ac l atc)(2h) + (be* b^c)h* t (47) 0.50302355 -f 0.7^ o. 7.2519 + 1.9344^ + 0.05197814^ 0.55844915 + 0.8/1 7.04925 + 1.86845^ 0.6115684 + 0.9^ 0.002185185^ 6-8935 + 1.8191^ 4- 0.06130587^ 0.6671298 + h O.OO273I482/Z 2 4- 1.7815^ + 0.06396421^ 0.7212274 + i.iA 0.0032777777^ 6.66205 + 1.7524^ 4- o. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 361 from which h is easily found : and there is no need, in these cases, of taking the second differential to ascertain whether the positive value of k, to be found from (470), renders W a maxi- mum or a minimum ; for the substitution of a member a little less or a little greater than the positive value of h so found will at once serve to verify the work, and show W to be a mini- mum in (469) when h takes the value given by (470). Taking the case where n = 9, and using logarithms, we may solve thus : n = 9. Logs. Log Co-efficients. Co-efficients. Equation (470). a = 7.049250000 0.8481429 logabi = 0.8023854 ab l 6.3443300 _ a-i 0.611568400 9.7864451 Iogai3 = 0.0579266 a-^b 1.1426900 7.4870200 b = 1.868450000 0.2714815 log&Ti =7.6109697 bc\ 0.0040829 - bi = 0.900000000 9.9542425 logfiic = 8.6906118 b\c 0.0490469 = 0.0531298/42 c = 0.054496590 8.7363693 log ccii = 8.5228144 ca.1 +0.0333284 - c\ = 0.002185185 7.3394882 \ogcia 8.1876311 da 0.0154039 0.0179245 (zh) 0.0531 298/5* 8.7253382 0.0179245(2/0 8.2534471 9.5281089 - 0.33737(2/5) 9.0562178 2.1493213 1.0746607 = 7.4870200 0.8743090 2.1489708 = 140.9194000 = log +0.1138000 = log 141.0332000 = log 11. 87 57400 +0.3373700 log co-efficients, log quotients, 2 log 0.33737, log (141.0332)* co-efficient of When W is a minimum, h =. 12.21311 feet. log/5, 1.0868263 a zr 7.049250 log 6/1, 1.3583078 bh rr 22.819590 log// 2 , 2.1736526 log ch z , 0.9100219 ch z = 8.128720 37.9975 6 o I -S797SS7 = log num. log c x // 2 , 9.5131408 ctf = - 0.325942 , of the girder diagonals to the horizon is about 2| degrees above 45 degrees ; and from this point increases or decreases with 11 if the best value is given to h. The best ratio of length to height of girder for this span and load is 8.1879; and, near the best simultaneous values of n and /z, we have approximately h = '-. (47i) 1 2th, Let us now find the value of E w , equation (460), the quantity of wrought-iron to be added to the end framework to resist wind force tending to produce distortion, assuming that the bridge is so fixed to the abutments that neither sliding nor overturning can take place. In equation (460), take b = 4 feet = length of brace, d^ =z 6 inches, /3 = 45 degrees = inclination of brace to post. /. sin ft = cos/3 = 0.70711, <5cos/? = 2.82844 feet. Take d = 12 inches, width of end post to resist bending. di = 12 inches, depth of end horizontal strut. f, = 1 8 tons. B = 25 tons. q 1 8 feet. m = -fg pound. / = 4- / 100. w = 0.0225 ton. Then, computing E w for the eight values of h already found, we obtain, from (460) and from the table just given, the follow- ing results : 364 MECHANICS OF THE GIRDER. W in Tons, h in Feet. No. of Panels, n. 5 6 7 8 Height, h 16.500 14.71 S 1-5 806 12 74O n times panel weight, nW . . Added iron, Ew tons .... Weight of bridge, n W + E w . Weight of wood 37.178 3-935 41.113 2O 24.4 35-927 2.898 38.825 l8.(X7 34.643 2.489 37-I32 1 6 C47 34-510 1.980 36.490 I ^.4.46 20869 20.768 "o. ;8<; 2 1 .044 Cost of iron, at $150 . . . . Cost of wood, at $15 . . . . $313 35 303 66 7474 01 $3115 20 270 86 3386 06 $3087 75 248 21 -5-J7C 06 $3156 60 231 69 -n88 20 Q8 (X so 10 ^2 77 No. of Panels,. n. 9 10 11 12 Height h 12 21 7 1 1. 70.8 II O2I 10.408 n times panel weight, n W . . Added iron, Ew tons .... Weight of bridge, n W + E w . 34.013 1773 35-786 I4.6l2 34.302 1.480 35782 I7.QCQ 34-157 1-357 35-5I4 I"V4T? 34.635 1.170 35-805 13.007 ^Veight of iron ... 21 174 21 823 22.O7Q 22.708 Cost of iron, at $150 . . . . Cost of wood, at $15 . . . . Cost of bridge $3176 10 219 18 3-JQC 28 $3273 45 209 39 3482 84 $33" 85 201 53 5 C I 'J ~l8 ^3419 70 T 95 ii 7614 81 Excess over least .... Jjyj f* CQ -?2 146 88 177 42 278 8q Here we see that n = u and/2 = 11.021 are the conditions yielding least total weight of bridge, while the whole cost is a minimum if n = 7, h = 13.896, and (/ -f- n) = 14^. Notice that both of these minima of weight and cost corre- spond to an odd number of panels, and that the excess of cost above the lowest would in all cases more than compensate the GIRDERS WITH EQUAL AND PARALLEL CHORDS. 365 manufacturer for having the best simultaneous values of n and h determined by calculation, as above. If, however, there would be sufficient head-room, we may, for this span and load, adopt either 8 or 9 panels, giving more iron, less wood, and less total weight, with a small increase of cost. In each of these 8 cases it will be seen the bridge weight is a little more than one-third the uniform moving-load, 100 tons = nL, and that the total dead load is slightly greater than one-fourth the sum of dead and live loads. 141. To exemplify the Method of Article 139, which pro- vides, at Every Post, the Means of resisting the Distorting Influence of the Wind. Taking the example of article 140, and calling the top horizontal diagonals I inch in diameter (that is, 0.7854 square inch cross-section), and weighing 2.654 pounds to the foot, we have Weight of 2n horizontal top) = 6 54 V/i8* + * diagonals ) "V n * 10000 pounds, Weight of bottom horizontal) __ y (as found in article 140, for both diagonals ) L> ( top and bottom. Strain on a top horizontal strut, from ^ = 6 tons per square inch on two top diagonals, is equal to 2 x 6 x 0.7854 sin 0! = 9.4248 sin fa tons. Now we already have the breaking inch strain on top hori- zontal struts == 4.680056 tons, and 366 MECHANICS OF THE GIRDER. Therefore, in square inches, Cross-section of a \ top strut to re- I 9.4248 . I i sist initial strain = 1.170014 sin *' = 8 ' 553 / ^ IOOOQ" S *' on diagonals J V 324722 And, from (461), Cross-section of a top 1 , /.,,, 6wf//i 2 6 h 2 . , strut to resist distort- \ = 2ac = -=* = 0.30- - square inches - r j Bd 2 n 7 n mg force of wind j if w = 0.0225 ton, f = 4, / = 100, B = 25 tons, and d 2 = 7 inches. From (465), since \2mqn 12 X y\ X i8w = Weight of top hori- J + ^^fc^t 1 unds zontal struts ) V n 2 -f- 30.8642 /& / ~~i = i85i.428-+48^.3i8^ 2 l/ pounds, ?2 V /z 2 -}- 30.8642 approximately, by reason of (471), to avoid the second power of h, for convenience. Weight to be added to floor beams due to wind = two times P f , as already given. In (467) take d == 8 inches ; then Weight to be added to all posts) i2 2 x 5 X 4 X 0.0225 x 100 to resist distortion J ~ 2 x 18 X 25 x 8 22* pounds, by (471). GIRDERS WITH EQUAL AMD PARALLEL CHORDS. 367 In the previous case the quantity of iron of uniform thick- ness to be added to each post is that due to the greatest mo- ment given by equation (453). It is plain from that equation that the added iron may vary in thickness from C, Fig. 114, where it should be greatest, to the bottom, where it may be nothing. Or, without increasing the thickness of the iron, the post may be made broader at top than at bottom, and thus resist the bending-moment whenever this broadening is not accompanied by too great reduction of the thickness of the iron composing the post. In the present case x = \h. Finally, from (459), calling d 2 = 4 inches, Weight of all) 6 X 5 X 4 X 0.0225 x ioo braces \ 18 x 18 X o.yoyn 2 = i. 746 6 7 A a = 174.667- pounds, n by (471). Computing for 8 values of n, we find, Weights of the Components of K, in Pounds. ^ V >o X 4 2 / 3000 x n. 5 6 7 8 Floor 18229 oooo 18229 oooo 18220 oooo l82''9 OOOO Joists 22258 oooo 17884 oooo 14864 oooo 12667 oooo ( I floor beams .... ( Do. wind 5459.0000 46 7968/2 5969.0000 60 2490^ 6408.0000 7O 7AO' J /i 6796.0000 84 6172^ Horizontal top struts < Horizontal diagonals < 1617.0000 370.2857/2 714.0000 120.6662/5 T.A.QT.T.'ik 2128.0000 308.5714^ 781.0000 125.3704* 29.1111^ 2650.0000 264.4898/5 854.0000 125.7336/5 24. Q?24/J 3176.0000 231.4286/2 931.0000 1334035* 21.8777^ IOOO.OOOO JOOO.OOOO I OOO.OOOO IOOO.OOOO '{ 572.6820/& + 49 2 77 523.3019/2 +45991 485.9160/5 + 44005 47I.28l6/2 + 42795 368 MECHANICS OF THE GIRDER. Weights of the Components of K, in Pounds. Concluded. n. 9 10 11 12 Floor 18229.0000 18229.0000 18229.0000 18229.0000 Joists 10994.0000 9688.0000 8641.0000 7784.0000 j I floor beams .... ( Do wind . . . 7146.0000 06.2^62^; 7465.0000 110.6624^ 7760.0000 1 2^.24.^2/1 8035.0000 n8. 2^78/2 Horizontal top struts < Horizontal diagonals < 3701.0000 205.7143,5 IOII.OOOO 138.1040/6 19.4074/2 4225.0000 185.1429/2 1093.0000 147.2220/2 17.4667/2 4746.0000 168.3117/2 1177.0000 154.0325/2 15.8788/5 5263.0000 154.2857/5 1263.0000 1 63-93 5 8/ ' 14.5555/2 Residual . . IOOO.OOOO IOOO.OOOO IOOO.OOOO IOOO.OOOO *;{ 459.4619/5 +42081 460.4940/2 +41700 461.4682/5 +41553 471.0348/2 +41574 The strain throughout each top chord due to the initial strain, 6 X 0.7854 = 4.7124 tons on each diagonal between top chords, is 4. 7 1 24 cos 0! tons, and the allowed inch strain here is 16.7442 - = 4.18605 tons. , Therefore the additional cross-section of iron for both top chords due to initial strain on top diagonals is, in square inches, 225.148 - 4.18605 Additional weight 1 for top chords, pounds, due initial strain on top diagonals = 2.25148 cos fr = 1 Y32472 2 +- 10000 _ 12 x IPO X 5 X 225.148 7549-333 ioo 324^ +- i oooo GIRDERS WITH EQUAL AND PARALLEL CHORDS. 369 The effect of wind on the bottom chords in this' case will be twice what it was in the example of article 140, and may be taken from the table therein given. Also, the weights of the girder diagonals will be the same as given in that article. We may expect a heavier bridge this time than was found in the last example, by reason of the initial strain now assumed on the top diagonals, and the smaller values of d for the top struts, the posts, and the braces, in comparison with the values used in the two end frames to resist wind. Computing weights for the different values of n, and col- lecting results, we have, / = ioo Foot- Weights in Pounds, IV and L in Tons, h in Feet, nL = 100 Tons. n 5 Top chords j L ad < Initial st., 4140.742 W k - Wh 82815 i 7i 558 h" - Bottom chords j Load ' 2444.444 - 48889 - - < Wind . 122.2222 Verticals, total . . . - 9.77778 - - 568.5926 Girder diagonals . . . 1333-333 3-33333 35555 - 88.8889 K . . - - - 49277 572.6820 2000W W 7918.519 +13.11111 +167259 +49835 + 1352.3857 6 Top chords \ Load ' ' 4866.256 - 81104 _ - ( Initial St., - - - 510 - Bottom chords { Load ' ( Wind . 2777.778 _ 46296 _ 115.7408 Verticals, total . . . - 14.66667 - - 544-2387 Girder diagonals . . . 1388.889 5.00000 30007 - 108.0247 K . . - - - 45991 523-30I9 znW = 9032.923 +19.66667 +157407 +46501 + 1291.3061 7 Top chords j Load l Initial st. 5525-323 ; 78933 467 - Bottom chords j Load ( Wind 3174.604 _ 45351 - "3.3788 Verticals, total . . - I9-55556 - - 489.6447 Girder diagonals . . 1360.544 6.66666 25915 - 126.9841 K . - - - 44005 485.9160 2aaonW = 10060.471 +26.22222 +150199 +44472 + 1215.9236 370 MECHANICS OF THE GIRDER. 1= zoo Foot-Weights in Pounds, W and L in Tons, h in Feet, nL = 100 Tons. M 8 Top chords \ Load ' 6221.067 _ Wh 77763 i _ If _ h ' Initial St., - - - " 418 - Bottom chords j Load ' 3559-027 - 44488 - - < Wind . - - - - 111.2196 Verticals, total . . . - 26.07408 - - 532.7546 Girder diagonals . . . 1388.889 8.88889 22787 - I45-8333 K . . - ~ - 42795 471.2816 2 ooon^ = 11168.983 +34.96297 +145038 +43213 +1261.0891 Top chords j Load ' ' 6881.576 - 76462 - - ' Initial St., - - - 394 - Bottom chords J Load ' 3978.051 - 44201 - - f Wind . - - - - 110.5014 Verticals, total . . . - 32.59260 - - 513.4888 Girder diagonals . 1371.742 ii. mil 20322 - 164.6090 K . . - - - 42081 459.4619 2ooon W = 12231.369 +43.70371 +140985 -t-42475 +1248.0611 * TO Top chords { Load ' ' 7564.819 - 75648 - _ ' Initial St., - - - 364 - Bottom chords I Load ' 4388.889 - 43889 - - < Wind . - - - - 109.7222 Verticals, total . . . - 40.74075 - - 578.8889 Girder diagonals . . . 1388.889 13.88889 18333 - l8 3.3333 K . . - - - 41700 460.4940 2OOO W ' = 13342.597 +54.62964 +137870 +42064 +1332.4384 II Top chords \ Load ' ' t Initial St., 8226.202 - 74784 338 - Bottom chords j ' ' Wind . 4820.936 43827 _ 109.5668 Verticals, total . . . - 48.88888 - - 572.6966 Girder diagonals . . . 1377.410 16.66667 16696 - 202.0202 K . . - - - 41553 461.4682 2000 W = 14424.548 +65.55555 +135307 +41891 +1345.7518 Top chords j Load ' 8903.037 - 74192 - - i Initial St., - - - 315 - Bottom chords j Load ' 5246.912 - 43724 - - f Wind . - - - - 109.3106 Verticals, total . . . - 58.66667 - - 647-5823 Girder diagonals . . . 1388.889 20.00000 15325 - 220.6790 K .. . - - - 41574 471.0348 2000M W = 15538.838 + 78.66667 +133241 +41889 +1448.6067 GIRDERS WITH EQUAL AND PARALLEL CHORDS. 3/1 Multiplying each of these eight equations by - , we find 20000 the uniform panel weight of bridge, W, in terms of h ; thus : 8.36295 + 2.49175^ + 0.06761929^ = 5, W n = 6, W n = 7, W = n= 8, W = 0.3959259 4- o.5/& 0.00065555 7.87035 + 2.32505^ + - -0.4516462 + o.6h - 0.0009833333^ 7.50995 4- 2.2236/2 + 0.06079618^ 0.5030235 + 0.7^ o.ooi3iiiin^ 2 7.2519 + 2.16065^ 4- 0.06305446^ 0.5584492 -f- o.8/^ 0.001748 148^" 7.04925 -|- 2.12375^ 4- 0.06240306^ 0.61156845 -f 0.9^ o.oo2i85i86/^ 2 4- 2.1032^ 4- 0.06662192^ ^2 T (*\ tfl/ "~" 0.667129854-^ 0.002731482^2" n = I]: ^ _ 6 -7 6 535 + 2.09455^ 4- 0.06728759^ 0.7212274 -f i.iA 0.003277777/^2' ft = 12 V 6.66205 4 2.09445/2 -f- 0.07243034^ 0.7769419 4- 1.2/1 0.00393333^ Differentiating these equations, and putting = o, we ah find results as here tabulated; h corresponding to the least value of n W. 372 MECHANICS OF THE GIRDER. Span / = 100 Feet, Uniform Live Load = nL, = 100 Tons. Number of Panels, n. 5 6 7 8 12.60087 I2.4OQQQ I2.^o^6 1 1 7Q2IO Weight of bridge, tons, n W '. . Panel length / * n . 43-5II95 20 ooooo 40.91832 1 6? 38.99626 IJ& 3845946 i H 1 k 7 87Q7O 8 05800 8 12640 8 480^0 Slope of diagonals, .... Ratio of dead to live load . . . Ratio of dead to total load . . Weight of bridge per lin. ft., Ibs., Weight of wood, tons .... Weight of iron, tons 32 23' 50" 0.43510 0.30320 SyO.OOOOO 2O.224OO 23 288OO 36 40' 17" 0.40920 0.29040 818.00000 18.05700 22 86lOO 40 44' 28" 0.39000 0.28050 770.00000 16.54700 22 4.4QOO 43 19' 5 1 " 0.38460 0.27780 769.00000 1 5.44600 2^01 7OO Cost of iron, at $150 . . . . Cost of wood, at $15 . . . . Cost of bridge $3493 20 303 36 ?7n6 c6 $3429 T 5 270 86 77OO OI $3367 35 248 20 76l ^ ^H #3451 95 231 69 ^68 i 6A Excess over least ... . 181 01 84. d6 J W1 J JJ o 68 09 Cost per linear foot 37 97 37 oo 36 16 3684 Number of Panels, . 9 10 11 12 Height in feet, h . II ^Q2' 7 6 1 1 067 1 4 io8-'859 10 37666 Weight of bridge, tons, nW ' . . Panel length, / n 37-83525 Ili 38.08053 IO OOOOO 38.04843 oJr 38.60209 8k / -L. h 8.62640 Q OI ^80 0.27480 Q 6l7OO Slope of diagonals, . . . . Ratio of dead to live load . . . Ratio of dead to total load . . Weight of bridge per lin. ft., Ibs., Weight of wood, tons .... Weight of iron, tons 46 I 2' 51" 0.37840 0.27450 757.00000 I4.6l^)O 2~l.22T.OO 4754'o" 0.38080 0.27580 762.00000 13.95900 24.12200 49 59' 8" 0.38050 0.27560 761.00000 13-435 24 6 1 300 5i 13' 57" 0.38600 0.27850 772.00000 13.00700 2 e co coo Cost of iron, at $150 . . . . Cost of wood, at $15 .... $3483 45 219 18 5702 67 $3618 30 209 38 3827 68 $369! 95 201 53 ogcn j.8 ^joyj uv -' $3839 25 i95 " A.O1A. T.6 Excess over least 87 08 212 17 /tiS ST Cost per linear foot 37 03 38 28 ^11 VJ 38 93 4 34 GIRDERS WITH EQUAL AND PARALLEL CHORDS. 373 Here, again, we find least weight, nW = 37.83525 tons, answering to the odd number of panels, 9, and the height, k = 11.59226 feet ; while the inclination of diagonals to hori- zon, , is about i^ degrees above 45 degrees. The least cost, at the rates here assumed, corresponds to 7 panels ; it being understood that we have once or twice employed the approximation involved in (471). 142. Again, by the method of article 139, take the same example, except that the uniform live load is now 2 tons to the linear foot, instead of i ton, as in article 141. ist, The floor, as before, weighs F = X 17.5 X ioo x 50 = 18229 pounds. 12 2d, By (432), Thickness of a joist, b = (9 X 10 X 2 x ioo (l8229 + 4OO ooo) I* ( n 2 x 17.5 X 7000 ) fc 22"i inches. t And, from (433), ioo x 17.^ X qo/g.O72i? 3 \ 2268^54 Weight of joists, / = - I44 7 X 5 2 L ( 9 -4^) = -^rf P unds - 3d, Depth of I floor beams, from (412), as in article 140, i // + 4i822o\i . = 0.4331885^ -^- -i) inches. By (434), Weight of I-beams, c // + 418229 18. P = I5 . 4 6o68( - I) X ^ X 18.5^ -i- X I J + 4l8229\| = i.o 3 i824( - i)f - - ^ - \ pounds. 3/4 MECHANICS OF THE GIRDER. 4th, Take top horizontal diagonals, each i^ inches in diam- eter. Cross-section = 0.99402 square inches; weight = 3-359 pounds per foot. Then Weight of 2n top horizontal diagonals = 2n x 3-359V/i8 2 -f - = 6.718^324^ + 10000 pounds. Weight of bottom horizontal diag- onals _ ^ _ 2X5X4X18X6 yy{ nZ ( H even )> 18 X 24 sin 2 x ' ( (n 2 i) (n odd), as in article 140, for both top and bottom. hwl i , loooo 2n' sin 2 iS 2 n 2 ' 5th, Strain on each top horizontal strut from % = 6 tons per square inch on two top diagonals = 2 X 6 X 0.99402 = 11.928245^^ tons; allowed inch strain on strut 4.680056 , = 1.170014 tons. Therefore Cross-section of a top strut ^to ^.92824 . resist initial strain on diag- onals . sm0, = 10.1040^ 1.170014 ri y ^ yo , i oooo n From (461), Gross-section of a top strut to resist distorting force of wind 2ac = "^;"~ = o. 3 of^ square inch. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 3/5 From (465), Weight of n top hori- J = 6n.6 97 n*J ? pounds zontal struts D * J V/ \ n 2 + ^0.8642 F 2 + 30.8642 pounds, by reason of (471). 6th, Weight to be added to floor-beams, due to wind, = 2 X P' in article 140, changing \ n ) i8X24sin 2 (n even), g^A 3x5x4* (odd). We therefore have, Weights in Pounds, W in Tons, h in Feet, nL = 200 Tons. n 5 Top chords j Load ' ' 4140.742 W J _ Wh 165630 T _ h* _ f Initial st., - - _ m 706 _ Bottom chords \ Load ' 2444-444 - 97777 - < Wind . - - - - 122.2222 o Girder diagonals . . . 1333-333 3-33333 71111 - 177-7778 K . . - - - 63566 569.3262 2ooonW 7918.519 +13.11111 +334518 +64272 +1646.5114 6 Top chords Load ' ' 4866.256 _ 162208 _ _ Initial st., - - - 645 - Bottom chords { Load ' 2777.777 - 92592 - - t Wind . - - - - 115-7408 Verticals - 14.66667 - - 838.4774 Girder diagonals . . . 1388.889 5.00000 60014 - 216.0494 K . . - - - 58635 518.4826 2000H W = 9032.922 +19.66667 +314814 +59280 +1688.7502 7 Top chords { Load ' 5525-323 _ 157866 - - I Initial st., - - - 59 - Bottom chords i Load ' 3174.604 - 90702 - - f Wind . - - - - 113.3788 Verticals 19.55556 795.6160 Girder diagonals . . . 1360.544 6.66666 51830 _ 253.9682 K . . - - - 55614 479-7376 2000M W = 10060.471 +26.22222 +300398 +56204 +1642.7006 GIRDERS WITH EQUAL AND PARALLEL CHORDS, 379 Weights in Pounds, W in Tons, h in Feet, nL = 200 Tons. ;/ 8 Top chords j Load ' ' ' Initial St., Bottom chords \ Load ' ( Wind . Verticals 6221.067 3559-027 IV k 26.07408 Wh 155526 88976 i ~k 542 if 111.2196 924 8842 Girder diagonals . . . K . . 1388.889 8.88889 45574 5*732 291.6666 463-333I 2OOOW W = 11168.983 +34.96297 +290076 +54274 +1791.1035 9 Top chords j Load ' ' f Initial St., Bottom chords j Load f Wind . Verticals 6881.576 3978.051 32.59260 152924 88402 499 110.5014 915.8665 Girder diagonals . . . K . . 1371.742 ii. mil 40644 52579 329.2180 449.8421 2000M W = 12231.369 +43.70371 +281970 +53078 +1805.4280 10 Top chords { Load ' ' I Initial St., Bottom chords 1 Load ' < Wind . Verticals 7564.819 4388.889 40.74075 151296 87778 461 109.7222 1067.7777 Girder diagonals . . . K . . 1388.889 13.88889 36667 5 I 9 I 5 366.6667 448.8212 2OOO IV = I3342-597 +54.62964 +27574 1 +52376 +1992.9878 ii Top chords \ Load ' ' ' Initial st., Bottom chords j Load ' ' Wind . Verticals 8226.202 4820.936 48 88888 149568 87654 428 109.5668 1071 0130 Girder diagonals . . . K . . 1377.410 16.66667 33392 51595 404.0404 447-8397 2000M W = 14424.548 +65.55555 +270614 +52023 +2032.4599 12 Top chords jL*d . . ' Initial st., Bottom chords j Load ' Wind . Verticals 8903.037 5246.912 58.66667 148384 87448 399 109.3106 1232.6646 Girder diagonals . . . K . . 1388.889 20.00000 30650 51521 441.3580 455.0907 2000H^ = 15538.838 +78.66667 +266482 +51920 +2238.4239 380 MECHANICS OF THE GIRDER. Multiplying each of these 8 equations by (Ji -4- 20000), we find the uniform panel weight, W, of bridge, in terms of k, thus : 16.7259 4- 3.2136/2 4- 0.08232557^ n = 5> W n = 6, W = = 7, = s, n= 0.395926 + 0.5/2 o.ooo65555/j 2 15.7407 + 2.964/5 + o. 0.4516461 + o.6h 0.00098333^ 15.0199 + 2.8102/2 + 0.08213503/22 0.5030235 -f 0.7^ 0.001311111^2 14.5038 + 2.7137/2 + 0.08955518^2 0.5584491 + 0.8/1 0.001748148/^2 14.0985 + 2-6539^ 4- 0.0902714/22 0.6115685 -f- 0.9^ 0.002185185^2 13.78705 4- 2.6188/2 4- 0.09964939/22 0.66712985 4- h 0.002731482/22 13.5307 4- 2.60115^ 4* 0.10162299/^2 0.7212274 4- i.i^ 0.00327777/22 = 12, 13.3241 4- 2.596/2 4- 0.11192120/^2 -0.7769419 4- 1.2^ 0.00393333/22 Differentiating, and putting - - = o, according to equation ah (470), we find, r GIRDERS WITH EQUAL AND PARALLEL CHORDS. 381 HEIGHT, h, ANSWERING TO MINIMUM VALUE OF nW. Span / = 100 Feet, Uniform Live Load nL = 200 Tons. Number of Panels, n. 5 6 7 8 I C.47O76 14 6l W^ 14 72O74 1 7 471 S4 Weight of bridge, n W. . , . Panel length, feet, / -f n . . . Ratio of length to height . . . Slope of diagonals, . . . . Ratio of dead to live load . . . Ratio of dead to total load . . Weight of bridge per lin. ft., Ibs., Weight of wood, tons .... 59-97030 2O.OOOOO 6.48060 37 39' 6" 0.29985 0.23068 1199.00000 25-55650 34.4 1 780 57.05609 16! 6.84180 41 M' 58" 0.28528 0.22196 1141.00000 22.32550 74.7 ^060 54-55331 14* 6.98290 45 4' 13" 0.27277 0.21431 1091.00000 20.99450 n.qqSSo 1 O-^-J 1 j^ 54.38682 12* 744520 47 3' 27" 0.27193 0.21379 1088.00000 18.46850 ^H.QlS^O Cost of iron, at $150 . . . . Cost of wood, at $15 . . . . Cost of bridge $5162 07 383 35 ccxr A'* $5209 59 33488 r CAA 47 $5033 82 301 42 e^qe 24 15387 75 277 03 c66i 78 Excess over least . . . DJ^J *K 2IO l8 2OQ 27 jjjj * o j^n CA Cost per linear' foot 55 46 55 45 53 35 5665 Number of Panels, n. 9 10 11 12 Height in feet h . . I -5 IO6OI 12 777OI 12 O4S7' > I I 40^76 Weight of bridge, n W . . . . Panel length, feet, / -- n . . . Ratio of length to height . . . Slope of diagonals, .... Ratio of dead to live load . . . Ratio of dead to total load . . Weight of bridge per lin. ft., Ibs., Weight of wood, tons .... "Weight of iron tons 53.61297 ii| 7.63010 49 42' 33" 0.26806 0.21140 1072.00000 17.23600 "?6 777OO 1 ' i -JjJ UI 54-435 10 IO.OOOOO 8.10830 50 57' 49" 0.27218 0.21394 1089.00000 16.27100 ?3 16410 54-39900 9rV 8.30170 52 57' 30" 0.27199 0.21384 1088.00000 1 5.49800 ?3 90100 55-64659 8* 8.76900 53 50' 33" 0.27823 0.21767 1113.00000 14.86500 40 78160 Cost of iron, at $150 . . . . Cost of wood, at $15 .... #5456 55 258 54 C7i c oQ #5724 62 244 07 cq68 60 #5835 15 232 47 6067 62 6117 24 222 98 6*340 22 Excess over least Cost per linear foot 379 85 57 15 633 45 59 69 732 38 60 68 1004 98 63 40 382 MECHANICS OF THE GIRDER. Here, for weight, the minimum minimorum is 53.61297 tons, n = 9, cf) 49 42' 3 3"; while for cost, at the assumed prices, the least is $5,335.24, answering to n = 7, and ^ = 45 4' 13". Comparing these results with the corresponding ones in article 141, we conclude : ist, For a given span and number of panels, if we increase the live load, we should increase the height. 2d, As the live load increases, the ratio of dead to both live and total loads diminishes. 143. As another example, let the span /= 200 feet; uniform live load nL = 200 tons, or i ton per linear foot ; other data as in articles 141 and 142. Compute for n = 8, 9, 10, n, 12, 13, ist, The floor weighs F = -^ X 17.5 X 200 x 50 = 36458 pounds. 2d, By (432), Thickness of a joist, b = \ 9 X I0 X 2 X 20O ( 3 6458 + 400000) i* ( n 2 X 17.5 X 7000 v * X I 10.151042 = ^ inches. # by (471). GIRDERS WITH EQUAL AND PARALLEL CHORDS. 385 Computing for 8 values of n, we find, Weights of Components of K, in Pounds. / = 200 Feet, nL =. 200 Tons. n S 9 10 * 11 Floor 36458.0000 36458.0000 36458.0000 36458.0000 Joists 58184.0000 sou t;.oooo 441516.0000 397 05. oooo ( I floor beams . . . 11294.0000 11809.0000 12282.0000 12721.0000 ( Do. from wind . . . i5-95387z 170.5809^ 194.9868^ 215.9620/5 Horizontal top struts \ 2859.0000 925.7142/5 3465.0000 822.8571^ 4092.0000 740.5714^ 4734.0000 673.2467^ Horizontal diagonals < 1656.0000 527.2219/4 1729.0000 504.8319^ 1808.0000 502.7783^ 1891.0000 495.8917/5 Braces So.sS^ 79.6296^ 71.66667* 6"v I"?!^ Residual 2OOO OOOO 2OOO.OOOO 2OOO.OOOO 20OO.OOOO K \ 1693.4732^ I577-8995^ I5I0.003I/^ I450.25IO>5 ' \ +112451 + 105976 + IOII56 +97509 n 12 13 14 15 Floor 36458.0000 36458.0000 36458.0000 36458.0000 Joists 35768.0000 724Q2.OOOO 2Q727.OOOO 27^6^.0000 j I floor beams . . . 13131.0000 I35l8.OOOO 13884.0000 14231.0000 1 Do. from wind . . . 240.9957/5 263.1340^ 288.8560/; 312.0545/5 Horizontal top struts < 5386.0000 617.1428/6 6031.0000 569.6703^ 67O8.OOOO 528.9796^ 7373.0000 493.7143/5 Horizontal diagonals < 1978.0000 501.4819^ 2O67.OOOO 503.6705^ 2l6l.OOOO 512.2314/4 2257.0000 520.3630^ Braces 597222/5 55.1282^ SI.IQOS/* 47.7777/5 Residual 2OOO OOOO 2OOO OOOO 2OOO OOOO 2OOO OOOO K\ 1419.3426/5 1391.6030^ 1381.2575^ I373-9095^ I +94721 +92566 +90938 +89684 9th, Taking Q = 16.7442 tons, as before, and L = - = n n tons, we find 386 MECHANICS OF THE GIRDER. Weight of top chords due vertical ) 5 x 4 X 20O 2 ( vy \ 2QO\ pressures, in pounds ( 2 x 18 x 16.7442/1^ n ) -f- $n 2 (n even), 4. 3 ^2 _ 2n _ (0 Odd). Strain throughout each top chord due to initial strain of - 2 ? 4 - X 0.99402 = 5.96412 tons, along each diagonal between top chords, is 5.96412 cos $! tons. Allowed pressure on top chords = 16.7442 = 4.18605 tons per square inch. .Additional cross-section of iron for 1 both top chords due to initial \ = - 777-^ - cos 4.18605 strain on top diagonals J .Additional weight for top chords due in- itial strain on top diagonals, pounds '324^ + 40000 _ 12 X 200 X 5 X 569.904 _ square inches. 4- 200 2 i oth, From (425), ight of bottom chor vertical forces, pounds Weight of bottom chords due ) _ 5 X 4 X 2OO 2 / ^ , 20Q\ ) 2 x 18 x 24/1 \ n ) 3 2 4- 24 ( even), 22H 21 (n odd). GIRDERS WITH EQUAL AND PARALLEL CHORDS. 387 From (436), t being zero, multiplying by 2, Weight of bottom chords due ) _ 5 X 4 X 2OO 3 x 0.0225^ ) = ~ wind, in pounds 18 x 24 x 18 27z 3 3 2 4- 227Z 24 (n even), 272 3 _ 3^2 4- 2272 21 X - * - (n odd). nth, From (426), Q, being 8. 18181 8 tons, Weight of verticals due load, pounds, _ 3 X 5 X ^n 2 , 5 X 4 X zoo ^n 2 + $n io 18 X 8.181818 2 X 18 X 8.i8i8i8\ n (n even), 3 X 5 X $Wh(n 2 i) , 5 x 4 X 200/1 Hn^ $n 2 jn + 3\ \ n 2 ) i8 X 8.181818 2X18x8.181818 (n odd). Weight of verticals due wind, pounds, by (467), 144 x 5 X 4 X 0.0225 * 200 72000^ 18 X 2 x 25 X 8 n 2 approximately, (471). T 7 2 1 2th, From (428), where = i +- L -, sin 2 < ;/ 2 /; 2 Weight of girder diagonals 4 X 5 X 200 X 4/i/n 2 i\ , 3X5X4^ -^ 2 18 X 24 sin 2 \ n / 18 X 24 sin 2 (n even), _ 4 X 5 X 200 X 4/i/n 2 i\ 3x5x4^ 18 X 24sin 2 c/> \ n / 18 X 24 sin 2 (^ Oodd). 388 MECHANICS OF THE GIRDER. Weights in Pounds, W in Tons, h in Feet, nL = 200 Tons. ;; 8 Top chords \ ^ ad < Initial St., Bottom chords j Load ' < Wind . 24884 14236 W k Wh 622100 355900 i 7i 1542 h" 889.757 Girder diagonals . . . K. . 5556 8.8889 182292 112451 291.667 1693.473 2000fF = 44676 +34-9630 +1160292 +"3993 +4784.155 9 Top chords I Load ' ' f Initial St., Bottom chords { Load i \V md . Verticals 27526 15912 611689 353600 1476 884.012 Girder diagonals . . . K . . 5487 II.IIII 162577 105976 329.218 1577.900 2000W W = 48925 +43-7037 +1127866 +107452 +4484.774 TO Top chords { Load ' i Initial St., Bottom chords {< Verticals 30259 17556 40.7408 605180 351120 1412 877.778 1697.778 Girder diagonals . . . K . . 5556 13.8889 146667 101156 366.667 1510.003 zooonW ' = 5337 1 +54.6297 +1102967 +102568 +4452.226 II Top chords j Load ' < Initial st., Bottom chords J Load ' ( Wind . Verticals 32905 19284 48.8889 598272 350618 1350 876.534 1591.675 Girder diagonals . . . K . . 55io 16.6666 133567 97509 404.040 1450.252 2nW = 57099 +65-5555 +1082457 +98859 +4322.501 12 Top chords \ Load ' ' ' Initial St., Bottom chords j Load ' Wind . Verticals 35612 20988 58.6667 593533 3498oo 1291 874.486 1670.165 Girder diagonals . . . K . . 5556 20.0000 122599 94721 441-357 I4I9-343 2OOO W = 62156 + 78.6667 +1065932 +96012 +4405.351 GIRDERS WITH EQUAL AND PARALLEL CHORDS. 389 Weights in Pounds, W in tons, h in Feet, nL = 200 Tons. 13 Top chords j L ad 38260 IK 1 _ Wh 588615 i _ * _ ' Initial st., - - - "35 - Bottom chords J Load ' 22801 - 350785 - - I Wind . - - - - 874.930 Verticals - 68.4444 - - 1614.025 Girder diagonals . . . 5523 23-3333 113286 - 478.633 K . . - - - 92566 1391-603 2000^ = 66584 +91.7777 +1052686 +93801 +4359.191 M Top chords ] ' ' 40952 - 585028 - - < Initial st., - 1181 Bottom chords J Load ' 24490 - 349857 - - < Wind . - - - - 874-636 Verticals - 79.8519 - - 1729.182 Girder diagonals . . . 5556 27.2222 105280 - 515-874 K . . ~ - - 90938 1381-257 zooonW = 70998 +107.0741 +1040165 +92119 +4500.949 Top chords \ Load ' ' 43602 _ 581360 _ _ ' Initial st., - - - 1131 - Bottom chords J Load ' 26272 - 350293 - - Wind . - - - - 875.720 Verticals 91.2593 1699.028 Girder diagonals . . . 5531 31.1111 98326 - 553-o86 K . . - - - 89684 1373-909 2000^ = 75405 +122.3704 +1029979 +90815 +4501.743 Multiplying each of these 8 equations by (h -4- 20000), we find the uniform panel weight, W, of bridge, in terms of h t thus: 58.0146 + 5.69965/2 4- 0.2392078^ W = = 9, n = 10, W 2.2338 4- o.8h o.ooi748i5/z 2 5 6 -3933 + 5-37 2 ^ + 0.2242387^ 2.44625 -f- 0.9^ o. 55.14835 -h 5.1284^ + O.2226II3/Z 2 2.66855 H- h 0.002 73 1485^ 390 MECHANICS OF THE GIRDER. n = ii, W = n = 12, JF = = 13, W =. n = 14, W = = 15, W = 54.12285 + 4.94295/2 + O.2l6l25/J 2 2.88495 -j- i.i/j O.OO327777/* 2 53.2966 -J- 4.8OO6/! -4- O.22O2675/J 2 3.1078 + j.2/i 0.00393333/; 2 52.6343 + 4-69005^ -f- 0.2I79595/; 2 3.3292 + 1.3/4 0.00458888^ 52.00825 + 4.60595/2 -f- 0.2250475^ -3-5499 + i -4^ - 0.005353705;^ 51.49895 -f 4-54Q75 7 * + 0.2250872^ 3.77025 + 1.5^ o.oo6u852/j 2 Differentiating these equations according to the form (470), and solving for h and W, we find as follows : HEIGHT, //, ANSWERING TO MINIMUM VALUE OF nW. Span / = 200 Feet, Uniform Live Load nL = 200 Tons. Number of Panels, n. 8 9 10 11 10.42424 IQ.4O^H IQ.0^2 D 18.89406 Weight of bridge, n W. . . . *-v7' A T*-T. ^- 163.83300 * ;7 *r OOO i55-3 8 700 * J ^O 3 151.80500 147.73400 Panel length, feet, / -f- n . . . 25.00000 22| 2o.ocxfoo ISA Ratio of length to height . . . 10.29700 10.30800 10.50800 10.58500 Slope of diagonals, ^ . . . . 37 So' 46" 4i7 / 33 // 43 34' 48" 46 5' 54" Ratio of dead to live load . . . 0.81900 0.77700 0.75900 0.73800 Ratio of dead to total load . . 0.45030 0.43720 0.43150 0.42490 Weight of bridge, Ibs. to lin. ft., 1638.00000 1554.00000 1518.00000 1477.00000 Weight of wood, tons .... 47.32100 43.48700 40.48700 38.08100 Weight of iron, tons .... 116.51200 111.90000 111.31800 109.65300 Cost of iron, at $150 . . . . $17476 80 $16785 oo $16697 70 $16447 95 Cost of wood, at $15 .... 709 82 652 3 1 607 31 5/1 22 18186 62 174-27 -Ji iT'Soi; 01 I7OIQ 17 Excess over least . . Il6? 4.S. / HO/ O 4l8 14 x /O w j w "SS 84 t y I o Cost per linear foot i * *^jj ^.^ 90 93 T*** T" 87 19 -i< TT'.X V \) 147.06400 145.26000 A / tJWi|Y \j 146.09100 1 1 oy^T-j 145.51100 Panel length, feet, / -^- n . . . I6| ISA I4f 13* Ratio of length to height . . . 10.84100 10.94900 11.23300 11.36600 Slope of diagonals, . . . . 47 o S4 ' 24 " 49 53' 42" 51 1 5' 29" 52 50' 52" Ratio of dead to live load . . . 0.73500 0.72600 0.73000 0.72800 Ratio of dead to total load . . 0.42370 0.42070 0.42210 0.42110 Weight of bridge, Ibs. to lin. ft., 1471.00000 1453.00000 1461.00000 1455.00000 Weight of wood, tons .... 36.11300 34.47500 33.09200 31.91200 Weight of iron, tons .... 110.95100 110.78500 112.99900 113.59900 Cost of iron, at $150 . . . . $16642 65 $16617 75 $16949 85 $17039 85 Cost of wood, at $15 . . ... 54i 7i Si? 13 49638 478 68 17184 ^6 17 1 74 88 17446 21 I7Hl8 $1 Excess over least / i^t 3 i6c IQ * / * Or 1 1 c 7i * 1 l V J *j Aty 06 ^/j 1 " jj 4QQ "?6 *V3 t*f ** j / * <\~ 1 w T-yy j ^ Cost per linear foot ... 8c cp 85 68 87 21 87 so "3 y~ "/ *J **/ jy Of the bridge weights in this case, the minimum minimorum is 145.260 tons, n = 13, c/> 49 53' 42"; while of the costs at the assumed prices, the least is $17,019.17, corresponding to = 11, = 46 5' 54". 144. From articles 141 and 143, exemplifying 2 bridges of different spans but under the same live load per linear foot, we may deduce, ist, That, as the length increases, the bridge weight per linear foot increases ; or, the ratio of dead to live load increases nearly as the length. 2d, That the dead load increases nearly as the square of the length. 3d, That an odd number of panels is more favorable to weight than an even number. 392 MECHANICS OF THE GIRDER. 4th, That the height of each panel should be a little greater than its length. 5th, That the ratio of length to height of girder depends upon the span, as well as upon the live load, seen by comparing articles 141, 142, 143. These principles are to be seen in this table. COMPARATIVE VIEW OF RESULTS. Span, Feet, I. Uniform Live Load, Tons, Best Number of Panels, Best Height of Girder, Feet, Least Weight of Bridge, Tons, Slope of Diagonals, *. Ratio of Dead to Live Load, Bridge Weight per Lin. Ft., Ratio of Length to Height, 7 . T. Panel Length, Feet, 7 nL. n. h. nW. W + L. Pounds. 100 100 9 11.592 37-335 46i2 / 5i" 0.378 757 8.626 rt| 100 200 9 13.106 53-6I3 49 42' 33" 0.268 1072 7.630 ii* 200 200 13 18.266 145.260 49 S3' 42" 0.726 H53 10.949 tsA These examples may suffice to illustrate a mode of deter- mining economical proportions for girders of all classes. SECTION 2. The Pratt Truss of Single System under Varying Live Load, without taking Account of Wind Pressure. 145. We shall here resume the example of article 36, the span being 100 feet of 10 panels, and the live load 2 locomotives of given weight and wheel base. Take n number of panels. W = unknown panel weight of bridge. h = 20 feet = height of girders, pin to pin. q = 14 feet = width of bridge, in clear. q^ = 1 6 feet = width of bridge, extreme. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 393 Single track, 2 rails, 56 pounds per yard each. Ties, 6 X 8 X 84 inches, spaced 8 inches in clear. 2 track stringers, 12 X d inches each. Ties and stringers, pine, 40 pounds per cubic foot. Weight of 2 rails = 2 X 100 X ^ = 3,733 pounds. x- / O Weight of 75 ties = A X 75 X 7 X 40 7,ooo pounds. 144 Panel length of stringers = 120 inches. Panel weight of rails = 373 pounds. Panel weight of ties = 700 pounds. 2 X weight on I pair of drivers = 42,000 pounds distributed. Maximum weight on 2 stringers = 43,073 pounds uniformly distributed. Then, for both stringers, b = breadth = 24 inches. d = height =15 inches. Take f = 10 = factor of safety for pine. B = 8,000 = breaking-weight for pine. From equation (52), M = \wl 2 \ X 4373 X 120, where wl = 43,073 pounds ; and, from (160), R + f = ^ Bbd * = 24 X 8000^ __ 43073 X 120 X 60 8 X 24 x 8000 ' d = 14.21 inches. Call d 1 5 inches, 2 x 12 x 15 X TOO x 40 Weight of 2 stringers = = 10000 pounds. Suppose 2 wrought-iron I-beams suspended at each panel joint, and assume the load on these beams to be concentrated at their centre. 394 MECHANICS OF THE GIRDER. Greatest load on 2 beams, From rails and ties, 1073 pounds, From stringers, 1000 pounds, From locomotive, 28612 pounds (article 36), Total, 30685 pounds, at centre has the momental effect of 61,370 pounds uniformly distributed along the double beam. Hence, for each single I-beam, D of (412) is equal to 30,685 pounds, and q l = 16 feet. Take f = 6 factor of safety. B = 50,000 pounds. From (412), 730685 x 16 X 6\j d 2 = 3.80122^- 5QQQQ 1 = 14-79 inches = required depth. From (413), 730685 x 16 X 6\f Area of cross-section of i beam = S = 1.28839! - QQ J = 19.508 inches. Now the "heavy 1 5-inch I-beam" of the Union Iron Mills, Pittsburgh, Penn., weighs 67 pounds to the foot, and its section consequently = 67 X -^ = 20.1 inches. We will, therefore, use the heavy 1 5-inch beam of 67 pounds to the foot. Weight of 9 pairs 15 -inch I-beams, 67 pounds, 16 feet ; 2 X 9 X 16 X 67 = 19296 pounds. Weight of 1 1 head struts, 14 feet, 20 pounds ; ii x 14 X 20 = 3080 pounds. Weight of 40 horizontal diagonals, \\ diameter, 3.359 pounds, 1 8 feet; 40 X 18 x 3-359 = 2419 pounds. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 395 Weight of the residue, 10 x 200 = 2000 pounds. RECAPITULATION. Rails = 3733 pounds, Ties = 7000 pounds, Stringers = 10000 pounds, Beams = 19296 pounds, Head struts = 3080 pounds, Horizontal diagonals = 2419 pounds, Residue = 2000 pounds. K 47528 pounds. For determining the girder strains in this example, we have already found, article 36, the greatest moments and greatest differences of moment due the given rolling-load. From (65), we have the moments due W, M = r(n r) = $Wr(io r), .-. M, = 45 w + 478-32, ff t = 2.25 W 4 23.916 tons; M 2 = 80 W 4 863.71, H 2 4.oolV 4- 43.186 tons; M 3 = 105 IV 4 1105.88, HI = 5.25^ 4- 55.294 tons; M 4 = i2oW 4 1236.95, H A = 6.00 W + 61.848 tons; M s = \2$W + 1276.38, HS = 6.25^ 4 63.819 tons. A r jY = 2.2$IV 4 23.916 = difference of horizontal strains. A 2 Zf = 1.75^+ 19-270 ^H = i.2$W 4 14.050 A 4 /f = 0.75 W 4- 10.451 ^H = 0.25 W 4 7-395 4.785 2.579 0.932 0.058 396 MECHANICS OF THE GIRDER. Let < = angle of elevation of any diagonal, .*. tan0 = f = 2, log tan 0.3010300, logsin< = 9.9515452, log cos < = 9.6505152. In top chord, take ratio of panel length to least diameter = 12] then, by (400), 18 P = of the Gordon formula = 17.176 tons. I I22 3000 Take ratio of length of vertical to its least diameter = 40 ; then, by (400), 18 O I = = 11.74 tons. i + ^ 3000 Let T = 50,000 pounds = 25 tons = limit of tension. f 5 factor of safety. Summing the strains on the equal panel lengths, we have Weight of top chords = ^(23.75^ + 248.063) X 10 X A? = 460.93^ + 4815 pounds. Calling the strain on end panels of the bottom chord the same as the strain on the adjacent panel, we have strains in bottom chord, HI = 2.2$W + 23.916 H 2 = 2.25^ + 23.916 H z = 4.00 W + 43.186 ^4= 5-25^+ 55- 2 94 H s = 6.00 W + 61.848 19.75^ + 208.160 GIRDERS WITH EQUAL AND PARALLEL CHORDS. 397 /. Weight of bottom chords, ^-(19.75^ + 208.16) x 10 x ^ = 263.33^ + 2776 pounds. Strain on a vertical = Z = A.// tan <, by the formulas for Class IX. ; /. Z, = 4.50 F + 47.832 Z 2 = 3.50^ + 38.540 Z 3 = 2.50 F -f- 28.100 Z 4 = i.$QW -f- 20.902 Z 5 = 0.50^ + 14-790 ^6 = _ 9-57 2Z = 12.50^ + 159.734 Weight of verticals, ^(12.5 W + 159.734) X 20 X *- Qt = ( I2 -5^ + I 59-734) X ajfl- = 709.86^+ 9071 pounds, where Z 6 is used twice to provide resistance to lateral shocks. Strain on a diagonal = Y = kH H- cos <. If we call the strain on each of the first 5 counters equal to that on the fifth, from live load alone, we have y t = y 2 = y 3 = y 4 = y 5 = 4 . 7 8 5 - C os<, ^6 = ( 7-395 + 0.25 W) -5- cos0, Y 7 = (10.541 + 0.75 W) -r- cos^, y 8 = (14.050 + 1.25 J-F) -s- cos, F 9 = (19.270 -f iJ$W) -f- cos0, Ko = (23.916 + .25^) -j- cos<. 5F = (99.007 + 6.25^) -^ cos<^>. Weight of diagonals = ^(99.007 + 6.25^) x IO X IO T 3 cos 2 $ = 416.17?^ + 6601 pounds. 393 MECHANICS OF THE GIRDER. Weight of top chords Weight of bottom chords Weight of verticals Weight of diagonals Weight of girders = G RECAPITULATION. = 460.93^ 4- 4815 pounds, = 263.33^ + 2776 pounds, = 709.86^ -f- 9071 pounds, = 416.17^ 4- 6601 pounds, = 1850.29^ 4- 23263 pounds, K = 47528 pounds, Weight of bridge = 20>o>o?iW = 1850.29^ 4- 70791 = G 4 K. :. W = 3.9004 tons = panel weight. Weight of bridge = nW = 39.004 tons. Substituting 3.9004 for W in the expressions for H, Y, and Z, we have this strain sheet : MAXIMA STRAINS, TONS. LOADS APPLIED AT BOTTOM JOINTS. CROSS-SECTIONS IN SQUARE INCHES. .27 3.27 EACH OF Two GIRDERS. FIG. 115. 146. If the dead and live loads are applied at the upper joints, instead of the lower joints, the structure becomes a deck bridge ; and the compressions here found for the verticals must be increased by the panel weight of dead load plus the greatest apex load from the locomotives ; viz., for each girder we must augment Z by 3.9004 4- 14.3063 = 9.1034 tons. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 399 _ j 147. As another example of varying load applied at the lower joints of the Pratt Truss, we will, in accordance with the practice of some engineers, assume a certain panel weight of engine, of tender, and of train, and determine the strains thence resulting, and also the weight of the bridge. Let us take the example given by Col. Merrill for this truss (see " Iron Truss Bridges for Railroads," by Col. William E. Merrill, U.S.A.); viz., Span = 200 feet = nc = I. Length of panel =12.5 feet = c. Height of truss = 18.75 f eet = h- Number of panels = 16 n. On each of 2 trusses, Panel weight of engine = 17,600 pounds. Panel weight of tender = 16,160 pounds. Panel weight of cars = 13,152 pounds. The engine is supposed to cover 2 panels, the tender 2 pan- els, and the cars follow. We therefore have, ist panel weight of moving-load = W s = 8.800 tons, engine; 2d panel weight of moving-load = W^ = 8.800 tons, engine ; 3d panel weight of moving-load = W^ = 8.080 tons, tender ; 4th panel weight of moving-load = W 2 = 8.080 tons, tender ; 5th panel weight of moving-load = W^ = 6.576 tons, cars; 6th, etc., the same. To find the strains due to this rolling-load, we employ equa- tions (91) ; and for convenience, after dividing by the height, h = 18.75 feet, we may let r, denote the number of panel weights of cars, and u = the number of panel weights of engine and tender on the girder at any time. We shall then have for the different positions of the load, by summing equa- tions (91), and putting X = ( - r x - i) W 2 + (n - r, - 2) W z + (n - r, - 3) W^ + . . . (n - r x - ) 400 MECHANICS OF THE GIRDER. = ri(?<1 2 + -^ + (r x + i) W 2 + (r, + 2) JF 3 HORIZONTAL STRAINS AT JOINTS FOR ROLLING-LOAD. h ' , + i) / - , - 2) JF a + (r, + 2>( - r x - 2) , - 3 )^ 4 + ...(r f + 2) ( - r l -u GIRDERS WITH EQUAL AND PARALLEL CHORDS. 40 1 , + i) (n - r t - 3) W 2 + (r t 4- 2) ( - r, - 3) + (r, + i) ( - r, - 4) ^ 2 + fr + 2) ( - r, - 4) ^ 3 + (. + 3)( -'ft -4) ^4 + ..-(ri+4)( ->.^> * ~ ri " ^ ' (r x H- i)( - r f - F 2 (r, 4- 2) ( - r x - ) ?F 3 + (r f + 3) ( - ^i - ) ^4 . . . (r, + u)(n - r, - u)W u g ( - r, - u) - c(n r t u 2) - : nh T7 c V J-J-n i ~ -f nh Computing these values of H from the above data, for every position of the train as it advances, a panel at a time, from left to right, preceded by the engine and tender, we find this table of horizontal strains at the joints, in tons ; viz., as also given directly by the set of equations (91), 402 MECHANICS OF THE GIRDER. CWO Mi p em -to IP " ^. -to * co 2 pi co CJ p) A jjr fi. jjf p 8 M 1 R *? CO &f cito * vo -to * oo s ^ 1 5 &f i 1 "in co vO CO CO w ; eg 0* H tj 3 ow 00 oo" o 1 1 8 f 1 1 vo CO (55 S CTi OO Hn VO S M CO N J ! $ 1 ei 1 vb CO 1 co .11 N CO * in VO ^ OO O H CO - in GIRDERS WITH EQUAL AND PARALLEL CHORDS. 403 The blanks in this table may be filled by continually adding to itself each number in the right-hand column. It follows, therefore, that this right-hand column expresses the negative differences of simultaneous horizontal strains at adjacent joints due to rolling-load. It is evident, that, in this case, these negative differences are numerically the greatest differences of horizontal strains at adjacent joints, and may therefore be employed to find the maxima vertical and diagonal strains due live load. The table shows (as was to be expected from this load, but contrary to the assumption made by Col. Merrill) that the hori- zontal strains are not maxima throughout when the foremost end of the engine is at the last joint, but that the greater part of these strains reach their greatest values when the foremost end (that is, forward panel weight) of the engine is at the four- teenth joint. Since we require only the greatest horizontal strains, we need not compute the whole table, but only enough of the higher values of // to be certain that we find the highest at each joint. In the present example, it will suffice to compute values of H for the positions of load when r, = 11, r x 10, r l =. 9, and these for only the last 8 joints, since the first 7 horizontal joint strains are smaller than the last 7 by reason of the unequal loading. We have, then, the following brief solution : i. For maxima differences of horizontal strain due live load. nh nh\ 2 + (r t + 2) W, + (r, + 3) ^4 + (r, + 4> 404 MECHANICS OF THE GIRDER. c = 12.5, h = 18.75, = 16, Wi = 6.576, W 2 = fF 3 = 8.08, JF 4 = *F 5 = 8.8. ri JPi * w* Wi ^5 *_ ii -3 o 8.8 ^4X8.8 0.3! 2 8.8 8.8 (^+A)8.8 i.i I o 8.08 8.8 8.8 &x 8.8+^x8.08 2.17 O 8.08 8.08 8.8 8.8 ^X 8.8+^X8.08 3-571 I 6.576 8.08 8.08 8.8 8.8 -2 S 4 - x 8.8+tfV x 8.08+^4 x 6.576 5-257i 2 6.576 8.08 8.08 8.8 8.8 5.257^+27(2x8.08+2x8.8+2x6.576) 7.212 3 6.576 8.08 8.08 8.8 8.8 7.212 +^(33.76+3 x 6.576) 9.440! 4 6.576 8.08 8.08 8.8 8.8 9.4401+1.401+^x6.576 n-947 5 6.576 8.08 8.08 8.8 8.8 1 1.94^+ 1.40!+ A-x 6.576 14.72 6 6.576 8.08 8.08 8.8 8.8 14.72 +1.40!+ 6X0.274 17.770! 7 6.576 8.08 8.08 8.8 8.8 17.7701+1.40!+ 7X0.274 21.0954 8 6.576 8.08 8.08 8.8 8.8 21.095^+1.40!+ 8X0.274 24.694 9 6.576 8.08 8.08 8.8 8.8 24.694 +1.40!+ 9X0.274 28.5! 10 6.576 8.08 8.08 8.8 8.8 28.51+1.401+10X0.274 32-74 ii 6.576 8.08 8.08 8.8 8.8 32.71^+1.401+11X0.274 37.134 2. For the maxima horizontal strains due live load, as already computed and tabulated above. GIRDERS WITH EQUAL AND PARALLEL CHORDS. 405 333 P P P 3 M ll ll ll ll ll ll II II X X X I I I I I P\ O\ P^ 4x 4. -P. 4*. to to to XXX w to co XXX po po po ab b co oo Co Co CO xxx XXX po po po a oo oo 4>- 4>- CO XXX 1-1 tO tO XXX po po po bo bo oo Ln 4^ CO XXX XXX oo oo oo ll ll ll ll ll ll II II II ll ll ll ll ll II $- M o vo oo ll S ON CO O l-n to \O X X X 1 1 1 X X X X 1 1 jv Oo I ON ON ON [s, [\ _p^ S X X x X ?>- M ON ON ON ON L fv ^ ***$&$ 5^^** Z 2 3 o o o o XXX X X X X to Co 4* to CO 4 Cn XXX X X X X po po po po po po oo a oo oo b b b b oo 5o So oo M M HH XXX X X X X to Co 4. tO CO 4v Lr XXX X X X X oo oo oo po oo oo oo bob oo oo oo ab b b oo oo oo CO CO to 5 M M S xxx X X X X XXX X X X X CO 00 00 bo bo 60 po oo oo oo bo bo bo bo W W M 4* Co to Co Co to xxx X X X X M to to W CO CO CO XXX X X X X po oc oo 00 00 00 00 bo _bo ^bo ^ bo bo bo II II II II II II 1 1 II II II II II II II II HH HH M M H Leo- & s^a^t j 0^3 ccH 4- c\ =4- ON ) ON Ln i-H^J^NON^J 4 < e4- wjw 04- Co OO 4*- " Co < S S -i "i .1 406 MECHANICS OF THE GIRDER. It is manifest that the labor of computing the maxima values of H would be much lessened if we could legitimately assume that the horizontal strains are greatest throughout when the head of the engine is at the last joint or at any particular joint. To find the strains due the unknown bridge weight, 2/2 W, we have the panel weight of bridge on each girder = W, and find from equation (65), after dividing by //, H = ig r( , _ r) . ^T maximum, tons. ( ISt = o, N = o,| difference> 5 * r = i, # T = 5 JF, 5 W + 37.134 r = 2, .//z = 9-5^* q\W -\- 68.401-! 3 7* ~ "2 ^^ ~~ T t \V O) 3 O u 13 W + 93.802 Maximum Diagonal Strain. ! o 5 W o or 5 2 0.3! 4* W o or 5 3 i.i 3!^ o or 5 4 2.17 - 3 fr o or 5 5 3-571 23-ff o or 5 6 5.257* i%W o or 5 7 7.212 - fF ( 7.212 Jf)sec0 8 9.440! \W 9 !( 9-440!- W} ( 9.440! * W) sec 9 11.94* + fW 10 l(ii.94* + W (11.94* + *^)sec