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No. 58. THE VENTILATION OF COAL-MINES. By W. Fairley. M.E.. F.S.S., and Geo. J. Andre, No. 59. RAILROAD ECONOMICS ; OR, NOTES WITH COMMENTS. By S. W. Robinso" O.E THEORY AND CALCULATION OF CANTILEVER BRIDGES. BY R. M. WILCOX, PH. B., Instructor in Civil Engineering in Lehigh University. NEW YOEK >. VAN NOSTRAND COMPANY $3 MURBAY AND 27 WARREN STREET 1898 Copyright, 1898, BY D. VAN NOSTBAND COMPANY. PREFACE. THIS volume replaces the original No. 25 of Van Nostrand's Science Series, bear- ing the title " Theory and Calculation of Continuous Bridges," by Prof. Mansfield Merriman, which was published in 1875. The continuous girder, though exten- sively built in Europe prior to 1875, has now gone entirely out of use, except for revolving draw-bridges, and the cantilever bridge has taken its place. Indeed, the modern cantilever bridge is simply a continuous girder with the chords cut, a form of construction which lacks most of the theoretical objections of its ances- tor, and at the same time possesses the very great advantage over simple trusses of erection without false work. This book has been written with the object of presenting as clearly as possible the theory and methods of calculating the stresses in the trusses of cantilever bridg- IV. PREFACE. es. Both highway and railroad structures are discussed. In each case a sufficient number of the stresses have been worked out to illustrate the application of the methods, and the stresses in all the mem- bers are given in tables. B.M WILCOX SOUTH BETHLEHEM, PA., March, 1898. CONTENTS. CHAPTER I. Introduction. AKT. 1 . History of Cantilever Bridges. " 2. Classification. CHAPTER II. Highway Bridges. AET. 3. Definitions. 1 ' 4. Dead Load. * l 5. Reactions Due to Dead Load. " 6. Shear and Shear Diagrams. ' * 7. Moment and Moment Diagrams. " 8. Cantilevers with Horizontal Chords; Stress in Web Members. " 9. Cantilevers with Horizontal Chords; Stress in Chord Members. * ' 10. Cantilevers with One Chord Inclined. "11. Shears and Moments Due to Concen- trated Live Load. " 12. Max. + and Shear Due to Uniform Live Load. "13. Max. + and Moment Due to Uniform Live Load. " 14. Cantilever with Horizontal Chords, Uni- form Live Load Stresses. 71. CONTENTS. ABT. 15. Snow Load and Snow Load Stresses* " 16. Stresses Due to Wind. * ' 17. False Members for Purposes of Erection. ' * 18. Final Max. and Min. Stresses. CHAPTEE III. Railroad Bridges. ABT. 19. Loads on Cantilever E. E. Bridges. " 20. Eeaction Due to Dead Load. " 21. Stresses Due to Dead Load. " 22. Live Load. " 23. Live Load Stresses. " 24. Wind Load Stresses. CHAPTER I. INTRODUCTION. ARTICLE 1. HISTORY. THE cantilever bridge is a develop- ment of the continuous girder ; in fact it is the continuous girder with the chords cut and hinged (properly) at the points of reversion of flexure. The first real practical type of canti- lever bridge consisted of two sets of logs projecting out from the two opposite shores of a stream and the space between the ends of these arms spanned by other logs or beams. Such a bridge was built in Thibet about 240 years ago. For a de- scription see R. R. Gazette, 1882, p. 2. A book entitled "A Treatise on Bridge Architecture/ 7 by Thomas Pope, published in New York in 1811, sets forth a scheme for bridging the Hudson river. It was called " Pope's Flying Pendant Lever 2 Bridge/' and contained the principles of the cantilever bridge ; but Pope's ideas were decidedly erroneous as regards the stresses. At a time when tubular bridges and continuous girders were in favor, about 1850, it was suggested, by Edwin Clark, that the chords in continuous girders be severed at the points of contrary flexure, and the central portion be hung at those points. This plan though not carried into practical operation until some twenty-five years later, was nevertheless the essential principle of the modern cantilever bridge. In 1833 M. A. Canfield built a bridge at Paterson, N. J., which is claimed to be the first cantilever bridge ever built in America. In 1876-77 C. Shailer Smith built the " Kentucky River Bridge," 300 feet above water. A suspension bridge was originally intended, and towers were built for that purpose. The bridge was built out from the shore panel by panel until the towers were reached, and then continued on until connections were made at the middle. Then, in order to avoid alternate stresses which would be pro- duced if the bridge was perfectly stiff, the chords were cut on the shore arms near the piers. This bridge is located on the Kentucky river, about 112 miles from Cincinnati. For full description see Transactions of the Am. Soc. C. E., Nov. 1878. In 1867-68 Prof. W. P. Trowbridge, of the School of Mines of Columbia College, New York, conceived of and executed a plan for the first long-span cantilever in America. It was designed to span the East river opposite 76th street, New York, and involved the construction of two im- mense masonry piers 135 feet high placed on BlackwelPs Island. On top of these masonry piers it was intended to place iron towers 150 feet higher. For full description of the proposed bridge see Eng. News, Dec. 29, 1883. " The Niagara Cantilever Bridge " was begun in April, and completed in Decem- ber, 1883. It was considered a wonderful piece of engineering, both in the rapidity of construction and obstacles overcome. It was designed by C. C. Schneider, M. Am. Soc. C. E., and built by the " Cen- tral Bridge Works" of Buffalo, N. Y. The principal dimensions are : Length over all 910 feet, each cantilever 375 feet, and central span 120 feet. It has two points of support 25 feet apart at the piers, which are simply iron towers. The structure carried two lines of railroad 299 feet above the surface of the water of Niagara river. A paper on this bridge is to be found in vol. XIV of the Transactions of Am. Soc., C. E. The next cantilever bridge of impor- tance built in America was the " St. John River Bridge." It was opened for traffic in September, 1885, and was another ex- ample of rapid construction. It had the following general dimensions : Total length 812J feet, two cantilevers of 287 and 382 feet respectively, and a central span of 143J feet. A full account of this bridge is to be found in R. R. Gazette, 1885, p. 691. " The Louisville Bridge," over the Ohio river, connecting the cities of Lou- isville, Ky., and New Albany, Ind., con- sists of two cantilever spans 480 and 483 feet long respectively, separated by a con- tinuous span of 360 feet ; two anchor spans of 260 feet each, a swing span of 370 feet, and a fixed span on the New Albany side of 240 feet making a total length of 2453 feet. The distance from the under side of the trusses to the water is 95 feet. It was built in 1886, by the Union Bridge Co. under trying difficulties, and was made of open-hearth steel. See Eng. News, Nov. 27, 1886. " The Poughkeepsie Bridge/ 7 over the Hudson river at Poughkeepsie, N. Y., has a total length (not including viaduct ap- proaches) of 3093 feet. It is 212 feet above high water and consists of five spans of continuous and cantilever trusses. It was built by the Union Bridge Co., in 1887-88. The foundations for the piers of this bridge were very deep, one being 129 feet below the surface of the river. See R. R. Gazette, July 1, 1887, and also Eng. News, Oct. 29, 1887. The " Philadelphia Cantilever Bridge," 6 over the Schuylkill river at Market street, completed in 1888, is about 409 feet long and 77 feet wide, and consists of two canti- lever spans 166 feet lOf indies, and one central span 76 feet long. The " Great Forth Bridge " was com- menced in 1881 and completed in 1890. li crosses the "Firth of Forth" in Scotland, and consists of three gigantic cantilevers connected by two central spans each 350 feet long. The middle cantilever is 1620 feet long, and rests on two supports 260 feet apart. The other two cantilevers are each 1505 feet long, and rest on two sup- ports of 145 feet apart. The total length of the bridge is, therefore, 5330 feet. This length does not include the approaches, which in themselves are immense struc- tures. The maximum distance between piers is 1700 feet, the longest span in the world. The clearance of the central spans above high water is 150 feet. A full his- tory and description of this bridge is given in London Engineering, of 1890, p. 213. Other important cantilever bridges have been built, principal among which in America may be mentioned the " Bed Rock Cantilever Bridge " in California, and the "Memphis Bridge" at Memphis, Term. The former was built in 1890, and its main span is 660 feet long. See R. R. Gazette, April 25, 1890, and Eng. News, Sept. 27 and Oct. 4, 1890. The " Memphis Bridge " was opened for traffic in 1892. Largest span 790 feet. See Eng. News, May 12, 1892. ARTICLE 2. CLASSIFICATION. A cantilever bridge, as defined in the Century dictionary, consists of bracket- shaped beam trusses extending inward from their supports and connected at the middle of the span, either directly or by an intermediate span of ordinary construc- tion. This arrangement is shown in Fig. I, and is the simplest type of the modern cantilever bridge. Various modifications of the arrangement of the trusses exist in cantilever bridges, but all contain the principle of the bracket or arm supporting a weight^ which is kept in equilibrium by a counter- weight or reaction. Cantilever bridges are arbitrarily divid- ed into two general classes, depending upon the arrangement of the supports. The first includes those which have two points of support at the pier, as is shown in Fig. xxm. The " Niagara Cantilever' 7 and " Great Forth " bridges are examples of this class. The second includes those cantilever bridges which at the pier are supported at a single point. Fig. vil represents the arrangement of the reac- tions for the second class, and the " St. John River " and " Louisville " bridges are good examples of it. The calculations of the reactions for the two cases is quite different, as will be seen by reference to the formulas in Articles 21 and 5 respect- ively. CHAPTEE II. HIGHWAY BRIDGES. ARTICLE 3. DEFINITIONS. The following definitions of shore arm, river arm, and central span apply gener- ally to both classes of cantilevers, but par- ticularly to the truss arrangement repre- sented by the Niagara cantilever bridge r in which there are two piers and two abutments. Fig. vil. shows this arrangement of trusses, and reference to it will make the definitions clearer. Shore arm is that part of the bridge in- cluded between the abutment and pier, or AF. River arm is that part included between the pier and central span, or FJ. Central span is a simple truss supported by the ends of the river arms. Only one 10 half of the central span is shown in Fig. VII. All forces acting upward are to be taken as positive, and all forces acting downward negative. Thus a positive reaction is one acting upward, while a negative reaction is downward. Maximum stress means the greatest possible stress, either positive or negative, that can occur in a member. Minimum stress means the least possible stress of the same nature as the maximum stress, or if possible the greatest stress of the opposite kind. Maximum and minimum represent, therefore, the greatest range of stress. Shear diagrams are drawn to represent the distribution of shear throughout the bridge due to the position of the load shown, while moment diagrams represent the distribution of moments for the par- ticular position of the load shown. The plus sign placed before a stress means tension, and the minus sign com- pression. 11 ARTICLE 4. DEAD LOAD. The problem of deducing a general for- mula for dead load in a cantilever bridge, in order to calculate the dead load stresses, is one very difficult to solve, either theo- retically or empirically. There are so many different forms of cantilevers, varying in so many ways, that each one seems to be a distinct problem in itself. With such conditions to contend with, it seems al- most impossible to derive a formula for dead load. No satisfactory formula for dead load in cantilever bridges has ever been found, to the author's knowledge, until very re- cently. In a little book called " De Pontibus," by J. A. L. Waddell (N. Y., John Wiley & Sons, 1898) is presented a formula or diagram for dead load. From this the dead load for each apex of shore and river arms can be found by means of what is called a " percentage curve." This curve is plotted from values taken from a number of typical cantilever bridges, and represents the ratio of the dead apex 12 load of any panel of the shore and river arms, and the dead apex load of the sus- pended or central span. It checks with remarkable precision the estimated weight of the proposed North River Bridge at New York. Although Mr. Waddell does not guarantee it to be accurate for all forms of cantilevers, nevertheless, suitable modifications of it can probably be so made, as it seems to be based on the right principle. In the absence of any formula, the only way to get the dead load is to weigh the material, or get the actual shop weights. This is laborious, and involves the calcu- lation of the stresses in certain members : say the end panel of the river arm, due to half the dead weight of the central span, live load on central span, the effect of wind, together with an assumed weight of the members themselves. If the bridge is a highway bridge, a stress due to snow load should be included. With this maxi- mum stress the members considered are designed (rather roughly at first), and their weight compared with the assumed 13 dead weight. If there is but slight differ- ence between the assumed and actual weights of the members, all well and good ; but if too great a difference exists between them, the work should be repeated to the extent necessary for close agreement in assumed and actual weights. This book is not intended to explain the method of designing bridges, but to show how to calculate the stresses in the mem- bers of a cantilever bridge : hence, it is of little importance whether the results ob- tained are the stresses caused by the act- ual weight of the bridge and the possible weights which may act upon it. The dead apex loads, therefore, will be assumed, and such values taken as to give simple numerical computation. ARTICLE 5. REACTIONS DUE TO DEAD LOAD. Let w = load per linear foot. R! = shore reaction. R 8 = river reaction. / = length of shore arm* 14 m = length of river arm. n = length of central span. W = total weight on bridge. j*/-^ 1 i i tg ^ + MI )'-' B m ........(4) I -f- w These equations are sufficient to deter- mine the reactions in any cantilever with supports arranged as shown in Fig. I. RI may have a positive, negative, or zero value, depending upon the relative length of I, m and n. The criterion that RI shall equal zero is expressed by the equation 7 2 -f- m 2 + m n - 0, from which I - V m 1 + m n. When I is greater than V m' 2 -j- m n, the value of Rj is greater than zero, or positive^ and acts upward ; also, when I is less than V m 2 + m n, the value of Ri is negative, and acts downward. ARTICLE 6. SHEAR DUE TO DEAD LOAD. Since the shear in any section of a beam is equal to the algebraic sum of the vertical forces on the left of that section, it follows that the shear in the shore arm at any section distant x from Ri (see Fig. l) may be expressed by the equation RI 16 w x. For the river arm the expression for shear in any section distant x from Ri is, RI + R 8 w x. The shear in any section of central span distant x from its left end is ^-^ w x. The distribution of shears due to dead load for the different sections through- out, for the case when I is greater than y m 2 -f- m n is represented by the dia- gram of Fig. n. ? R] being positive. . ii Fig. ill shows the distribution of shears for the case when I is less than V m 2 4- m n^ or when R t is negative. 17 ----- 1 ---- *h TO ---- n --- - ? i i V i \L/" \. r T E ' 1 lilt 1 i | IflHirmL^ Fig. Ill ARTICLE 7. DEAD LOAD BENDING MOMENT. In order to find the moment at any sec- tion of the shore and river arms, it is neces- sary first to find the values of B 2 and R t from equations (1) and (2). The moment at any section of the shore arm distant x from Ri is represented bv the equation w x 2 M = Ri x 2~- ' The point of maximum moment is where the vertical shear equals zero. Put the equation for shear R t wx equal to zero, 18 solve for #, and substituting this value of x in the above equation will give the maxi- mum moment in shore arm. When RI is positive there exists, in the shore arm, an inflection point or point at which the moment changes from positive to negative, and is sometimes called the point of reverse flexure. To find where w x 2 this point is, put R : x ^ equal to zero, and solve for x. The moment in the river arm at any section distant x from R a is expressed by This moment is always negative. The same result should be obtained if the mo- ment of the forces on the right of section is taken, or ,.. w n x w x* , . ,, ,. JM = ^ h ~ , where x is the dis- tance from section to the end of river arm. This is often a simpler equa- tion to use in computation than the former, in which RI may be positive or negative according as I is greater 19 or less than A/m 2 + in n . This should be determined and the proper sign given to Rj in the equation of moment of forces on left. In the central span the moment is found just as in the case of a simple beam. The % equation of moments at section distant x from its lift end is, .. r w n x w x* Mz ~2 2~" The moment diagram for the case when I is greater than vm 2 -f- m n, or when R, is positive, is shown in Fig. IV. K- *- I 1 1 V s \/ SJ f* J R < i ! 1 1 i ^rrrniiinillllirnTT*^ 1 1 ^rrr""^ ' ' , ; " ^^^t^ When R : is negative, Fig. V represents the distribution of moments. 20 . 1 1 n i 1 / \ \ X \! f Ri | Rz i | 1 1 1 . I | ^rmTTmTTTT^rrr^' ^^^i IIIIIIF^ Fig. "V ARTICLE 8. CANTILEVERS WITH HORI- ZONTAL CHORDS; STRESSES IN WEB MEMBERS DUE TO DEAD LOAD. The rule for finding the stress in the web members of a truss with horizontal upper and and lower chords is as follows : Pass a section cutting the member, the stress in which is to b& found, and multiply tlw shear in the section l>y the secant of the angle which the member makes ivith the vertical. Let the cantilever shown in Fig. VI have length of shore arm equal to 100 feet, river arm 80 f eet, central span 80 feet and 21 depth 16 feet. Then sec# = 1.18. Let uniform dead load equal 250 pounds per linear foot, or 5000 pounds per panel. p w m (I + m) + w (I -f w) f R ' ~~ -21- R ._ 250 X 80 x 180 + 250 x 180' 8 200 = 4-58 500 pounds. + m + R, = 250 (100 + 80 + 40) 58 500 E, = 3500 pounds. This shows that RI is a negative reac- tion ', that is, it acts downward. The stress in a A equals (R, P ) sec. B, or a A = + (3500 + 2500) 1.18 = + 7080 pounds. A b = (3500 + 2500) 1.18 = 7080. B c = (3500 + 2500 + 5000) 1.18 = 12 980 pounds. 22 F<7 = ( R, + R 2 P P, etc., ......... P 8 ) sec : 0. ( F# = ( 3500 + 58 500 2500 5 X 5000) 1.18 = + 32 450 pounds. Ij -.= ( 3500 + 58 500 2500 8 x 5000) 1.18 = +14 750 pounds. 3j = *^sec0-= 7500 x 1.18 = 8850 pounds. Enough of the members have been taken to show how the stresses are calculated for web members throughout the bridge due to dead load. ARTICLE 9. CANTILEVERS WITH HORI- ZONTAL CHORDS STRESSES IN CHORD MEMBERS. There are two methods of finding the stress in the chord members of trusses with horizontal chords : first, by the " Method of Moments " ; and, second, by the method of " Chord Increments." The method of chord increments does not hold good when one of the chords is inclined j so the method of moments will 23 be used in illustrating the calculation of chord stress. Draw a section cutting three members, take the origin of moments at the intersection of the two members cut, other than the one in which the stress is to be found ; then state the equa- tion of moments between flie stress and tlie ap- plied forces on the lift of the section, and solve for the unknown stress. It is necessary at first to calculate the reactions R 8 and RI, just as was done in Art. 8. If same example, data, etc., be taken in this case as used in the last ar- ticle, then R 2 = 58 500 and R t = 3500 pounds. Let it be required to find the stress in be, see Fig. VI. Pass a section cutting A B, B b and b c, and take the origin of moments at B. The equation of moment is then (_R i _P o )30 P 1 x 10 + bcx 16 = - 6000 X 30 5000 Xl0 + 6cxl6 = and b c = 14 375 pounds. Take the member D E. The origin of moments is at e and the equation is 6000 x 80 15 000 X 40 + D E x 16 = and D E = + 51 875 pounds. The origin of moments for the chord member fg is at F, and the equation of moments of the forces on the left and the stress in/# is ( R! 2500) 110 + R 2 x 10 25 000 X 50 +fg x 16 = 6000 x 110 + 58 500 x 10 25 000 X 50 +fg X 16 = and/# = 82 813 pounds. The moment equation may express the moment of the forces on the right instead of those on the left and it is often a sav- ing of labor to express it in that way ; as, for example, to find the stress in H I the origin is at i, and the equation of moments is H I x 16 = 12 500 x 20 H I = 250 000 -f- 16 = + 15 625 pounds. Stressing = 125QQ X 1Q = + 7313 pounds. 16 The examples given are sufficient to show how the stress in any member of a highway cantilever with horizontal chords, due to dead load, can be found. 25 ARTICLE 10. CANTILEVERS WITH ONE CHORD INCLINED. STRESS IN MEM- BERS DUE TO DEAD LOAD. ^ A favorite form of cantilever bridge is that in which one of the chords is inclined. When this arrangement of the chords exists, the principle that the stress in any web member is equal to the shear in the sections, multiplied by the secant of the angle which the member makes with the vertical no longer holds true, for the reason that the inclined chord member takes up a part of the shear. Let cantilever shown in Fig. VII have length of shore arm equal to 100 feet, E F, G H river arm 80 feet, central span 80 feet, and distance apart of trusses 16 feet ; B b = 20 feet, F / = 24 feet, I i = 21 feet and K Jc = 21 feet. Let the dead load per linear foot be, 500 pounds, air on the upper chord. Since the length of the arms is the same and the load per linear foot double that of the examples given in Art. 8, R a = + 58 500 x 2 = + 117 000 and R, = 3500 X 2 = 7000 pounds. Taking b as the center of moments, the stress in A B is given by the equation A B X 20 Ri x 20 5000 x 20 = and A B = B C = + 12 000 pounds. B & = 10 000 pounds, or the weight of the apex load that comes upon it. For the stress in A 5, take the center of moments at B. Then A b X 14.142 = 12 000 x 20 or A b = 16 960 pounds. For the stress in the member & C, pass a section through, cutting B C, & C and b c ; take the origin of moments at the inter- section of B C and It c, which is 420 feet to the left of C and the lever arm of & C is 296.98 feet. The equation of moments of the forces on the left of the section is 6C X 296.98 = 12 000 x 380 + 10 000 x 400 and b C '+ 28 740 pounds. The stress in be is found by taking the origin of moments at C. The lever arm of fee is 20.97 feet giving the equation fee X 20.97 12 000 x 40 10 000 x 20 = and I c = 32 400 pounds. The origin of moments for C c is at the intersection of C D and b c, or 420 feet to the left of C, and the equation gives Cc X 420 + 12 000 X 380 + 20 000 X 410 and C c = 30 380 pounds. A sufficient number of the members have been taken to illustrate the method of calculating the stress due to uniform load in the members of the shore and river arms. The member F/ may, however, offer some difficulty if treated according to the method shown. If the section be passed, cutting efj fg and F/, the solution becomes very simple by placing the vertical com- ponent of the stress in ef, fg and F/J equal to the reaction R 2 . This equation will contain only one unknown quantity F/ ? since ef and fg can be computed by the method of moments, and R 2 is known. The equation is : 28 orF/=R 8 -(Ve/+V/, in which Ve/and V/^r represent the vei*ti- cal components of the stress in ef and fg respectively. These stresses, taken from the table, are 133 600 and 133 500 pounds, and vertical components of them 6680 and 6675 pounds. The above equation for F/ reduces then to F/= 117 000 (6680 + 6675) = 103 600 pounds. The stress in the members of the cen- tral span are calculated just like those of a simple deck truss. Since it has its chords parallel, the stress in any web member is equal to the shear multiplied by the secant of 8. The stress in all the members of the cantilever truss shown in Fig. VII, due to dead load, are calculated in the manner shown, and the stresses given in the fol- lowing table. The object in giving the tables of stresses complete, throughout the book, is to have them serve as an- swers to any self-selected problem that the student may take. For instance, if the stu- dent wishes to test his ability in working 29 out the stress in any member not already given he may take for example F Gr and apply the same principles as used in finding the stress in A B, and verify his result from the table. If the student will pursue this course it will be found to be of very great help to him in better understanding the prob- lems. Dead Load Stresses for Cantilever shown in Fig. VII. Member. Stress in Pounds. Member. Stress in Pounds. AB=:BC + 12000 kl + 14280 CD + 32380 Eb - 10000 DE + 60000 bC + 28820 EF + 94000 Cc - 30400 FG + 91300 cD + 40120 GH + 54550 Dd - 39000 HI = IJ + 23800 dE + 50450 JK - 14280 Ee - 47300 KL 19050 eF + 6000 Ab - 16960 F/ -103650 be - 32460 F# + 64090 cd 60000 gG - 50440 de - 94040 Gh + 54700 ef -133600 m - 42280 fg -133500 Hi + 44650 gh - 91400 il 1000 hi - 54600 m - 15000 iJ - 34530 Kl + 6900 JJc + 20700 U - 10000 31 ARTICLE 11. SHEARS AND MOMENTS DUE TO CONCENTRATED LIVE LOAD. In addition to the dead and snow load, cantilever bridges are subjected to a live load stress. This live load consists, in the case of highway bridges, of foot people, horses and wagons, electric cars, etc. In order to determine the maximum and minimum stress in all the members affected by this load, involves a knowledge of the proper position of live load to pro- duce it. What, then, is the possible ar- rangement of the live load ? Unlike the dead load, the live load may occupy a part of, or different parts of, the bridge at the same time ; it may also, like the dead load, cover the entire bridge at once. It is necessary, therefore, to consider all possible arrangements of the live load, and to find that position for it which will produce the maximum and minimum shear and moment for any section. Consider first, one concentrated load, P, which is in the nature of an electric car or heavily loaded wagon. 32 (a) For a load P on the shore arm, (5) The effect is just like a load P on a simple beam, and the distribution of shears and moments is as shown in Fig. Vill. Fig. (6) For a load P on river arm, R , = - P -', and R,= I (6) and the shears and the moments are dis- tributed as shown in Fig. ix. 33 L rz --- n --- - ~\ TH I . IX (c) For load P on central span, 1= _P. , and K,= I- n I n\ I Fig. X 34 and Fig. x. shows distribution of shears and moments. An examination of the shear diagrams in the three cases (a), (b) and (c) shows that the maximum positive shear in any section, due to a load P, occurs when the load is placed just to the right of the sec- tion, and that the maximum negative shear occurs when the load is placed just to the left of the section. An examination of the moment diagrams shows that, for case (a), the moment is positive, and is a maximum for any sec- tion in lj when P is over the section ; that, for case (&), a negative moment is pro- duced in the shore arm and negative in river arm, both increasing as m' increases, and having a maximum value when m 1 = m, or when P is placed at the end of the river arm 5 and that, in case (c), a negative moment is produced in both the shore and river arms, and a posi- tive moment in the central span j also that the shore and river arm moments are a maximum when P is placed at the left end of the central span, and a maximum 35 in any section of the central span when P is over the section. The above conclusions are given, in con- densed form, in the following table : Table showing Position of Load P to give Maxi- mum Positive and Negative Moments. Case (a). P ou Shore Arm. Case (6). P on River Arm. Case (c). P on Central Span. Moment in I. f Max. , j when P is ~n over the t Section. f Max. { when P is ^ at m'=m f Max. 4 when P is ^ at n'=n. Moment in m. f Max. { when P is [at m'=m ( Max. \ when P is ( at n' =n. Moment in n. f Max. , ( when P is "*" | over L Section. Table showing Position of Load P for Maximum Positive and Negative Shear*. Max. -f Shear. Max. Shear. Shore Arm. P just to right of Section. P just to Section. left of River Arm. P at end of River None. Arm. Central Span. P just to right of Section. P just to Section. left of 36 ARTICLE 12. MAXIMUM -f- AND SHEAR DUE TO UNIFORM LIVE LOAD. The diagrams of shears and moments in the preceding article represent the effect of a single load P. The effect of any number of loads, as P P 2 , P a , etc., may be shown in the same manner, the resulting diagram being the same as the combined diagrams for each load taken separately, When these loads act sufficiently close together and are of the same intensity, the result is a uniformly distributed load. Such a load is represented in highway bridges by a mass of foot-people moving in a continuous or broken line across the bridge. It is necessary, therefore, to con- sider the effect of such load, and the posi- tion or possible arrangement of it to give the maximum + and shear and moment at any section. The live load in highway bridges may consist of both concentrated and uniform load, acting at the same time on different parts of the bridge. This combination is 37 not, however, generally made in highway bridges, as the result is not as injurious as that due to the full uniform load. If such a combination should be desired, the rules of Art. 11, 12 and 13 are to be followed. The shearing effect of uniform live load will now be considered. For any section in the shore arm, the greatest positive shear occurs when the load is so placed as to give to RI the greatest possible positive value, with no load on the left of the section to subtract from it. Any load on the river arm and central span causes a negative reaction at RI; therefore, the maximum positive shear in any section of the shore arm occurs when the shore arm is covered with the uniform load to the right of the section, which makes (8) and gives a shear diagram, as shown in Fig. XL 38 i- In the river arm, tne maximum positive shear for any section occurs when the load covers the central span and the river arm to right of the section. m 1 ivm 1 I and R 2 ==%D+_ L*l Since V R!+R 2? substituting for RI and R 2 their values as given above, and reducing, gives Ywn i / /m = -\-wrn' (9) which equation is also the algebraic sum of the vertical forces on the right of the .section. 39 The position of load for maximum posi- tive shear, for any section of river arm, is shown in Fig. XII. In order to obtain the greatest negative shear in any section of the shore arm, it is apparent, from the equation V = R ] wx, that Rj should have the greatest negative value possible. Since RI is negative due to the live load on the river arm and cen- tral span, it is evident that the proper position of live load to fulfill the condition is, to cover the river arm and central span with the uniform load, and also the shore arm to the left of the section. The value of RI is found from the equa- tion 40 wmn I 21 ~ 21 and the shear from V = Ri wx (10) The position of load and shear diagram is shown in Fig. xm. Kig. XIII In the river arm there is no negative shear due to live load. In other words, there is no possible arrangement of the load to produce a negative shear in the river arm. A load on left of section gives The maximum positive shear in central span for any section, occurs when load is 41 on the right of the section; maximum negative, when central span is loaded on the left of section, just as in the case of a simple beam. Table giving Positions of Uniform Live Load to Give Max. Positive and Negative Shears. Max. + Shear. Max. Shear. Shore Ann. Load to cover Shore Arm right of Section. Load on River Arm and Central Span, also Shore Arm, left of Section. River Arm. Load on Central Span and River Arm, right of Section. None. Central Span. Right of Section. Left of Section. ARTICLE 13. MAXIMUM POSITIVE AND NEGATIVE MOMENT DUE TO UNIFORM LIVE LOAD. The position of 'live load to produce maximum moment in any section of the shore arm is when the live load covers the entire shore arm. The equation of bending moment at any section distant x from the left end is RI x wx* wl Substituting for Rj its value in the u above equation gives nx/r WlX WX* If any live load is placed on the river arm or central span, its effect is to pro- duce a negative value in RI, thereby de- creasing the value of M. The diagram in Fig. XIV shows the dis- tribution of maximum positive moments for the shore arm. K OG- H v%%2% 'ww/twtfMww/ ^R i ! "R 2 I ! | i nHIHIU!rrtTrr. i ! J r, TTT'V 43 There is no position of the live load that will give a positive moment in the river arm ; the moment in the river arm is always negative. The central span being like a simple beam, its maximum positive moment for any section occurs when it is fully loaded, as shown in Fig. XV, in which wnx ivx 2 '2 -- WHJU 'IV JU .(12) u* 1 > *~ 1 i k%%% %%%%%%% a, - 1 R LfTTTnfll "TZhTT^ Fig. XV The maximum negative moment in the shore arm will evidently occur when the load is so placed as to give the greatest negative value for Ri, and to have no load 44 on the shore arm to cause a positive value for Ri. To produce this result, the river arm and central span should be fully loaded and no load on the shore arm. The diagram of moments is shown in Fig. xvi, and the value of the moment is found from M - -Ria (13) H--O? j ! ^ 11 1 I Wftffiw/'/- ' I" 1 i r i ! i i i l The position of load to produce maxi- mum negative moments in river arm is of course that which will produce the great- est negative value of RI. This will occur when the river arm and central span are loaded, with no load on shore arm, as in Fig. XVI. The equation of moments for 45 any section in river arm distant x from is M = Rj (I -f- x) ^-^- + R 2 x. The moment at the section due to that part of the load on the left of the section is zero ; consequently, the quantity - -i in the above equation becomes zero if the load is placed on the central span and on the right of the section only in the river arm. The effect is the same, and the above equation is simplified to M = R 1 (I + jc) + R t x (14) and the diagram of moments for this posi- tion is shown in Fig. XVII. There is no negative moment possible in the central span, since it is a simple truss. The following table gives, in condensed form, the position of live load for max. positive and negative moments in the dif- ferent parts of cantilevers. Table Showing Position of Uniform Live Load to Give Maximum + and Moments. Max. -\- Moment. Max. moment. Shore Arm (0- Load on entire Shore Arm. Load on entire River Arm and Cen- tral Span. Kiver Arm (m). No -f- moment possible. Load on Central Span and River Arm, right of Section. Central Span (ri) Load on entire Cen- tral Span. No moment possible. ARTICLE 14. CANTILEVER BRIDGE WITH HORIZONTAL CHORD; STRESSES DUE TO UNIFORM LIVE LOAD. To illustrate method of calculating stresses in a cantilever due to live load, let same example as given in Article 3 be taken. Shore arm 100 feet, river arm 80 feet, central span 80 feet, and depth of truss 16 feet. Sec# = 1.18. Let live load be taken at 70 pounds per square foot of floor surface. Assuming the bridge to be 30 feet wide, 1)he weight per linear foot for one truss is 70 X 15 = 1050 pounds. This multiplied by the panel length 20 feet gives 21 000, or say 20 000 pounds, as the live panel load which is all applied to the chord that supports the floor system. H i j K ! VvAAi To find the stress in A a caused by the maximum positive and maximum negative shear. For maximum positive shear, load covers shore arm on right of section, (see table in Art. 12). R, = 2 X 20 000 = 40 000 pounds. Stress in Aa = 40 000 X 1.18 = 47 200 pounds. For maximum negative shear, oad covers river arm, central span, and 48 shore arm left of section. K _ 60 000 x 40+ 50 000 x 80 ~100~ R l= r 64 000 pounds. Shear = V = R t = 64 000 pounds. Stress in Aa = 64 000 X 1.18= + 75 520 pounds. For stress in Qh due to maximum posi- tive shear, the live load covers central span and river arm on the right of section. The panel points loaded are h, i, j, k and I. Th n -R 4Q QQQ X 5Q + 5Q QQQ x 8Q 100 R 4 = 60 000. From (2) Ri + R f =W, therefore 60 000 + R 2 = 90 000 and R 2 = + 150 000 pounds. V in Grh = 60 000+150 000=90 000 and stress inG/&=90 000 x 1.18 = +106 200 pounds. There is no negative shear possible in river arm 5 theref ore, no compressive stress mOfc For the stresses in C D the maximum positive moment will be first considered, 49 aud this occurs by reference to table in Art. 13, when live load covers the entire shore arm. R, = 2 x 20 000 = 40 000 pounds, and center of moments is at d, then n-n _ 40 000 X 60 40000x30 IT" - _ 75 000 pounds. Maximum negative moment for section through C D occurs when river arm and central span is loaded. For this position of load, T? _ 60 000 x 40 + 50 000 X 80 T00~ and Ri = 64 000 CD= 6*X60 =+340000 ^ 16 Take the lower chord member of the river arm, ij. There is no positive mo- ment possible, so the maximum negative moment alone will be considered. This takes place when live load covers central span and river arm on right of section. T. 50 000 V 80 Ki = -~p := 40 000 pounds. R 2 rr -f 90 000 pounds. 50 With center of moments at J, stress in ij equals - 40 OOP X(1Q + 7Q ) + 90 000X70 16 ij ~ 31 250 pounds. The following is perhaps a shorter and more rapid method of calculating the reac- tions in a cantilever truss, due to live Ioad 7 than the one just used. Let P , P], P 2 , P 8 , etc., be the panel loads at the apex points, a, &, c, d, etc., of the truss shown in Fig. XVIII. Then, since 5 / 5 of P is supported by R 17 4 / 5 of P } goes to Rj and 8 / 5 of P 2 etc., goes to the same reaction, a table of coefficients can be formed giving the part of each load sup- ported by Rj and R 2 . If, then, the load is placed in the proper position for maximum, positive, and nega- tive shear or moments, the reaction R 1? due to the. loads on the required panel points is found by adding together the product obtained by multiplying each panel load by its coefficient in the column R r In the same way the value of R 2 can be found. This applies to loads on the 51 river arm and central span, as well as to loads on the shore arm. Load. Part Supported by KL Part Supported by Ha. PO + 1- XPo o. XP. Pi + 0.8 X PI + 0.2 X Pi P 8 4- 0.6 etc. + 0.4 etc. PS + 0.4 + 0.6 p, + 0.2 + 0.8 ^5 0. + 1. Pe -0.2 + 1.2 P, -0.4 + 1.4 PS -0.6 + 1.6 P 9 0.8 + 1.8 P,o -0.8 + 1.8 P,, 0.8 + 1.8 To find the stress in G/&, the load covers the river arm and central span to the right of section or -apex points h, i,j, Jc and Z, are loaded each with 20 000 pounds, except that at I, which is one half or 2Q ^ QQQ . Then by 2 the use of the table R, is found to be 52 20 000 ( .4 .6 8, -- 8,) + 10 000 X .8 = - - 60 000. and R 8 is 20 000 ( + 1.4 + 1.6 + 1.8 4- 1.8 ) + 10 000 X + 1.8 = + 150 000 then V = 60 000 4- 150000 = 4- 90000 and stress in Qh = 90 000 X 1-18 = 106 200 pounds, the same value as that found by the other method. This method has its greatest advantage when the loads are unequal, or when a uniform live load with excess loads is used, as will be shown in the discussion of live load in railroad bridges. ARTICLE 15. SNOW LOAD AND SNOW LOAD STRESSES. In addition to dead and live loads, high- way bridges are subjected to another kind of vertical load, at times in certain climates j namely, snow load. This varies according to climate from to 20 pounds per square foot of floor surface (see Roofs & Bridges, Merriman's and Jacoby, Part I, Art. 41.) Snow load is assumed to be distributed uniformly over the floor surface, and con- 53 sequently acts on the members of the truss in the same manner as the dead load. This fact makes the calculation of snow load stresses an easy matter, if the dead load stresses are known. Let iv be the dead load and w 1 the snow load per linear foot per truss ; S the stress in a member due to dead load, and S l the stress in the same member due to snow load, then = S andS' = S *.. ...(15) ' To find the stresses due to snow load, multiply the dead load stresses by the ratio between dead and snow load. It is to be borne in mind that the dead load always acts, while the snow load may or may not act ; and although the stresses are of the same nature, they are kept separate in order that the maximum and minimum stresses due to dead, live, snow and wind loads combined may be determined, as is shown in the table of maximum and mini- mum stresses at the end of the chapter. Assuming the snow load to be 15 54 pounds per square foot of floor surface, the snow load per linear foot per truss for the cantilever of Article 10 is if the dis- tance of trusses apart be 16 feet and there are two sidewalks each 5 feet wide 4. -i e^ 15(16 + 5 + 5) outside of the trusses, 2 T 2 390 - or say 200 pounds per linear foot per truss. The stress in A & due to snow load is S' = S ~ = 16960 ^ == 6780 pounds. w 500 ARTICLE 16. STRESSES DUE TO WIND. To counteract the effect of wind, which, acting horizontally, tends to deflect the truss in a horizontal plane, just as the ver- tical forces tend to deflect the truss in a vertical plane, members called struts are introduced, extending from the chord apex point of one truss to the same apex point of the other truss, and also tension members or tie -rods extending from the chord apex point of one truss to the next 55 apex point of the other truss. The ar- rangement of the members of this lateral system is like the members of a Pratt truss, as shown in Fig. xix. Upper Lateral System A' B' c f p' E' F' G' H' i' j' K' i/ IXIXLXIXJXJXIXIXDxJXCKI ABCDEFGHi J KL ABC DEFGHI J KL N 7 \/ f X / /N \ \ /\ & a d e *J~Q h LXIXIXXXJXlXXIXXlXf Abode / V h i J HI Lower Lateral, System Fig. XIX The wind blowing in one direction stresses one system, and blowing in the opposite direction stresses the other. This arrange- ment causes the diagonal member to take tension only. The actual surface exposed to wind in the cantilever bridge is an unknown quan- tity before the bridge is designed ; there- fore, some approximate value must be assumed in order that the stresses due to 56 wind may be calculated. A closely ap- proximate wind load is found for simple trusses (see Merriman's Roofs and Bridges Part I, Art. 42,) by assuming the mem- bers of the truss to be each one foot wide ; then the total area exposed to wind is twice as many square feet as there are linear feet in the skeleton outline of the truss. The pressure per square foot exerted by wind may be taken at about 30 pounds, although a pressure as high as 40 pounds per square foot is sometimes taken. Wind load on the truss is taken as act- ing uniformly over the entire length. It is. therefore similar to the dead load, ex- cept that it acts horizontally and produces tension in the leeward chords and com- pression in the windward chords. The stresses in the horizontal system effected by wind are calculated just as the stresses would be for a Pratt truss system. The distribution of shears and moments due to wind on truss is represented by the dia- grams of shears and moments due to dead load, shown in Figs. II, in, IV and v. 57 The wind on the upper chord apex points is transmitted by the upper lateral system of truss shown in Fig. XIX, directly to the abutments and piers. That on the lower chord is transmitted to the piers at one end, and to the abutment at the other, by means of the inclined end posts. The wind-load stresses given in the table have been calculated as follows : (The results are necessarily an approxima- tion, since the true area exposed to wind is not known.) The skeleton outline of the cantilever, Fig. XIX (dimensions of which are given in Art. 10), is about 1070 feet. Assuming the wind pressure per square foot at 30 pounds, the total wind pressure on one truss is 1070 x 30 = 32 120 pounds. Assuming two-thirds to be applied to the upper chord, since it carries the floor sys- tem, and one-third to the lower chord, gives 2 X 32 120 3X11 = 1946, , 1 X 32 120 07Q -, and ; stt97o pounds o X 11 58 respectively for the upper and lower chord apex wind loads. To be on the side of safety, and giving at the same time better values for computation, these may be in- creased to 2000 and 1000 pounds respect- ively. To find the stresses in the upper lateral system (see Fig. xix), proceed as follows : R 2 X 100 = 8 X 4000 x *|+10 000 X 180. A- R a = 28 800 + 18 000 = 46 800 pounds, and R ] X100^4X4000X50 10000X80 - 3 X 4000 X 40. R T = 4800. Let 6' angle which diagonals in upper lateral system make with the vertical; then secant 6', = 1.6 and tan 6' 1.25. Stress in A' B equals 4800 X 1.6 = + 7680 pounds. Stress in C D 4800 X 40 + 4000 X 20 _ for wind West, and 33 000 for wind East. For C' D the wind blows West and the shear is 4800 4000 4000 = 12 800 59 pounds. This multiplied by the secant of the angle C' D D' gives 4- 12 800 X 1.6 : + C'D = + 20 480 pounds. When the wind blows in the opposite direction, or East, the member C D' is stressed an equal amount -f 20 480 pounds. The effect of the wind on the upper chord is to turn the bridge over, as in 70200 7O2OO Fig. XX Fig. XX, which is a cross-section of truss shown in Fig. xix at F/. The total wind force acting at F is 46 800 pounds, and, if the two trusses are rigidly connected by members F/ 7 and F 7 / and struts, this force of 46 800 pounds tends to produce rotation about the point/ 7 , and is held in equilibrium by a downward force R 2 of 70 200 pounds. The equation of moments about/ 7 is - 46 800 X 24 + 70 200 X 16 = 0. The same result obtains if the center of moments . be considered as midway be- tween R a and R 7 2 , which latter act as a couple. If the dead-load reaction R 2 is less than the value R 2 , the downward force neces- sary to produce equilibrium, the bridge will overturn. R 2 due to dead load is 117 000 ; therefore there is a good factor of safety against overturning, due to the assumed wind pressure of 30.84 pounds per square foot, since in order to overturn the bridge the wind would have to exert a pressure per square foot equal to 117000X16X30.84 46 800 X 24 = 51.4 Ibs. 61 The members F'/ and F/', known as cross-bracing, are designed to take tension only. The stress in F'/is found as fol- lows: Take the center of moments at/% and state the equation of moments. - 46 800 X 24 + F 7 / Xp = p = 16 X cos/F/' = 16 X .83 = 13.28. Therefore, F'/= 46 S * 24 ^+8458Qlbs> 13.^8 In the table of final maximum and mini- mum stresses, the stresses due to overturn- ing effect of wind on truss are not given, and are omitted, because their effect is so small as not to materially change the final results. The stresses due to overturning effect of wind on truss and train are given in the table of final maximum and minimum stresses in a railroad cantilever bridge and the method of calculation given in Art. 25. In actual practice it would be well to compare the assumed apex wind loads with the actual wind apex loads as the result of multiplying the assumed pressure in pounds per square foot by the actual surface exposed in the designed structure. 62 This should be done at least to make sure that the assumed wind apex loads are on the side of safety. Stresses in Lateral Systems due to Wind. Member. Wind East. Wind West. Member. Wind East. Wind West. A A' 8800 3800 H'l 4- 22 400 BB' - 681 >0 6800 I' J 4- 16 UOO CC' 10 800 10 800 J'K + 9200 DD' 14 800 14 800 K'L 4 3200 EE' 1H 800 18 800 A' b 4 4870 FF' 44 800 - 44 }*00 Ab' 4 ' 4870 GG' 20 OuO 20 ( 00 b' c 4- 6640 HH' 10 000 16 COO be' + 6640 IF 12 000 12 OUO c' d 4-10240 J J' 501,0 5000 cd' + 10 240 XK' . 4000 4000 d' e 4 13 440 LL' 2000 de' 4 13 440 A'B 4 7680 e'f + 16 640 B'C 4 14080 ef 4 16 640 7400 7400 CD' -j- 20 480 ee' f 9400 9400 D E' 4- 26 880 22 400 22 400 EF' 4- 33 280 9 9' 10 000 10 000 F'G 4 35 200 h h' 80CO 8000 G'H -j- 28 800 i i' 6000 6000 63 ARTICLE 17. FALSE MEMBERS INTRO- DUCED FOR PURPOSES OF ERECTION. The principal advantage that the can- tilever bridge possesses over other forms of bridges, the suspension bridge excepted, consists in its economy of erection under unfavorable conditions. Comparatively little false work is required. The bridge is erected by beginning at the pier, and building out on both the shore and river arms until the abutment is reached on one side, and connection made at the middle of the central span on the other. In order to make connection in the mid- dle it is necessary that the central span or a part of it be made, temporarily, a con- tinuation of the river arm ; by means of false members introduced merely to support the arm and the necessary apparatus, etc., used in erection. It is readily seen that, in the case of the cantilever shown in Fig. VII, the compres- sion member extending from i toj &ndj to Jc with a vertical member J j, would 64 make the central span, or as much of it as is necessary, a part of the river arm of the cantilever. This change in the arrangement of the members causes a change in the nature and magnitude cf stress in some of the members of the truss. What this change is remains to be deter- mined; so that, if necessary, provision may be made in the cross-section of the mem- bers effected to safely erect the bridge. Fig. XXI represents the skeleton diagram of Fig. VII changed by the false members ij,j k and 3j being introduced for purposes of erection. I J K L The dimensions cf truss are the same as that of Fig. vn, and dead apex loads the same, 10 000 pounds ; but the live load will be assumed to consist of a single concen- trated weight to represent a traveler used 65 in erection. This will be taken at 40 000 pounds. Secant of angle which K I and J fc, etc. make with vertical is 1.38. The position farthest out on the arm that the traveler is likely to occupy is at K, since all the members L I, K I, k I etc., are erected with it in that position and connection made. This position gives greatest moment, and consequently great- est stress in all chord members, to the left. The stress in L I is 10 000, pounds due to dead load. K I = 10 000 X 1.38 = + 13 800 pounds. *l= 9520 pounds. K k = 10 000 10 000 - 40 000 = - 60 000 pounds, which is greater than the maximum stress due to dead snow and live load as given in table. J k = - 60 000 X 1.38 = + 82 800 pounds. . i . . , 50 000 X 20 + 10 000 X 40 -21- - 66 660 pounds, i J = 70 000 X 1.38 = 96 600 pounds. When the traveler is brought over the point I the stress in I i is - 40 000 + 10 000 = 50 000 pounds. No change takes place in any other 66 members throughout the truss, at least to the extent of changing the maximum and minimum stresses due to dead snow and live load. The following table shows what members are stressed during erection greater than when subjected to dead snow and live load. In addition to the stress due to wind on the truss there may be stress, due to wind on the traveler, which amounts to consider- able, depending of course upon its position and the amount of surface exposed. Possible Stresses During Erection. Member Dead Load. Traveler. Wind. Maximum LZ 10 000 10000 Kl + 13 800 ! + 13800 Kk 20 000 40 000 60000 J k -|- 27 600 + 55 200 4 82800 Jj f Assumed. \+ 2000 + 20UO J i 41 400 55 200 96 600 li 10 000 40 000 50000 KL 6400 6400 JK + 9520 12 800 + 22320 I J -j- 42 900 + 76 200 i 19 200 -f- 138 300 kl 9520 3200 12720 jk 28 600 38 100 6400 73100 ij 28 600 38 100 66660 The work of erection cannot be safely 67 carried on at times when the wind blows at a high velocity, at which time the travel- er should be run back to a point of safety. On this account no allowance has been made for stress in the members due to wind on the traveler placed in a position to effect the members given in the table. ARTICLE 18. FINAL MAXIMUM AND MIN- IMUM STRESSES. A table of stresses due to dead load, live load, snow load, and wind on truss is given for the cantilever bridge shown in Fig's vii and xix, and the final maximum and minimum stresses given in the last two columns. The overturning effect of wind on the truss has not been considered. It amounts to but little any way and would not change the final results much in this case. But in a through bridge it should be consid- ered. Impact has been omitted because of its complication. Initial tension would enter into the final results of some of the 68 members. All of these omitted forces are mentioned merely to call attention to them, so that the student may investigate the subject in works in which they are treated. Attention is called to the final results, as showing in some members the reversal of stress from tension to compression and visa versa. 69 T T T T % s s f a oo CM o 3 O ^ CM 00 I I I esioco^i ++++ I \ 0000000 in LO O * O ** * i ^ w -o t-oo-oo 00 ^ ' 1 1 1 1 1 1 1 + Snow Load. % Sssi^s* 5 5 SSI?i^l 1 1 1 1 II I++ 1+1+ 1+1+ OIOOOCOOOOOOOOOOOOU3OO CCcir-iOT5-^6c_ ^^i.^C^SSt-CsXL CSOOrH rH i-( C3 S*l r- ( T 1 i 1 + 1 + 1 1 ! 1 1 1 ++ 1 1 + 1 + 1 + 1 + ! o o88o~^-S oSSSooc-o O *H CO r-( * ft O^ 00 O O O O ?0 Wi i m i i i i+ 1 + 1 + i + 1 + 1 s ^ ^^tS^3 S25S,55 71 S Illlilil 1 i n 155 000 X 60 + 90 000 X 30 ^ ~130~ 400 000 pounds. This proves that the stress in P Q is 15 000 pounds greater when the load ex- tends only as far as the middle of the panel to the right of the center of mo- ments than when it is brought up to the section. For the stress in p r take the center of moments at P, then 155 000 X 90 + 150 000 X 45 * r -' -sr or'j? r 690 000 pounds. The calculation of stresses in the web members of the river arm involves the very same principles of loading that were used in the calculation of web stresses in river arm of highway bridge. Art. 14. Take, for example, the member R s. Its stress is equal to the shear in the section multiplied by 1.41. The maximum shear will take place when all the load possible is put on the right of the section, or when the central span and river arm right of section is loaded with uniform load, and with excess load at some point between section and end of river arm. The maximum shear in section is then 245 000 pounds, and Us = 245 000 X 1.41 == + 345450 pounds. To find the stresses in the web members of the shore arm is the most troublesome part of the whole problem, but with care in placing the loads in the proper position to produce the greatest possible positive 89 and negative shears, the stresses become readily known when Rj is known. Take, for example, the member E d. The greatest possible tensile stress in this member will occur when the river arm and central span is loaded with uniform load and with excess load at end of river arm and the shore arm covered left of sec- tion. The reaction R, due to this loading is from formula (19), R z = 155 OOP x 120+210 OOP x 6090 OOP X 150 180 = _ 98 335. The shear in section is then 98 335 90 000 = 188 355 and E d 188 335 x 1.41 = : + 265 550 pounds. This result is obtained on a rather re- diculous supposition, in that the load on left of the the section on shore arm, though isolated from the other load on river arm, is assumed to come into the desired position without any locomotive or excess load to place it there. A more reasonable suppo- sition would be to place an excess load at 90 the head of the uniform load left of the section on shore arm. Finding RI by means of formula (19) gives, Ri X 180 = 155 000 X 120 + 210 000 X 60 + 90000 X 150 20 000 X 135, and R, = 83 335 pounds. The shear in section is 83 335 110 000 = 193 335 and E d = 193 335 X 1.41 = + 272 600 pounds. The stress c d is equal to the stress in E d minus the stress in E d caused by the apex load at D or c d = 272 600 35 250 = + 237 350 pounds. Since c d supports C c the stress in C c must equal the vertical component of the stress in c d ; therefore C c equals 237 350 * 1.41. or G c = 168 330 pounds. The stress in N n = R 8 = P 2 = 395 000 pounds, and M m equals R 2 , but from formula (18), R 2 = P, R,. The greatest value for R 2 will occur when the bridge is covered with live load. That on the river arm and central span being in the position occupied for maxi- 91 mum negative moment in shore arm, while the shore arm is covered with uniform live load with excess load at M. This gives P,== (12x30000+200004-15000) = - 395 000 pounds, and Rj X 180 = 155 000 X 120 + 210 000 X 60 -- (11 X 30000) 90 -- 15000 X 180 or Rj = -|- 6666 pounds, and R 8 = M m = 395 000 6666 = 401 666 pounds. The maximum, negative live load stress in c d and d E is produced when the shore arm is loaded on the right of a section cutting D E, d E and c e with the excess load at E. RI due to this position of the load is 20 000 = 103 335 and c d = d E = 103 335 X 1.41 = 145 700 pounds. Here is a member which shows itself to be subject to alternate tension and compression for different positions of the live load, which is an objectionable con- dition, and can be avoided by the introduc- tion of a counter member d e. which will 92 prevent the members c d and d E from taking compression. The actual effective compressive stress that can occur in c d, is the algebraic sum of the stresses in c d due to dead live and wind on train loads, which, taken from the table of stress are + 42300 -- 145 700 and 11 320 respectively, the sum of which is 114 720 pounds. This is very nearly the value of the stress in the counter de. A counter is needed, therefore, in any panel in which the live load and wind overturning load negative shear exceeds numerically the dead-load positive shear. By reference to the table of final maximum and mini- mum stresses, it is readily seen that the only panels which need counter bracing are the first three at the end of the shore arm or be, d e and fg. In practice another panel might be counter braced for the sake of security. ARTICLE 25. WIND LOAD STRESSES. Wind blowing on a bridge produces a 93 double effect. First, it has the effect of stressing the members of the lateral-sys- tem, and thereby producing compression in the windward chords and tension in the leeward chords. Second, it has the effect of overturning the bridge. This latter effect produces an additional vertical load on the leeward truss, and consequently greater stress in the members of it, while at the same time decreasing the stress in the members of the windward truss. The change of stress in web members of trusses due to overturning effect of wind is caused, however, by the wind on train or live load alone, while the chord mem- bers are effected by both wind on train and wind on truss. Let the wind apex load on both the upper and lower chords due to wind on truss be 2000 pounds, except the end apex load, which is 1000 pounds. The stresses in the lateral system and chord members are now found by apply- ing the principles given in the case of highway bridge, Art. 16. The reactions Np "t* ( ',*-?> :!' 'x' O c 1^ {-?tf I: I" x: :i s x" M z\ flj _c> UJ s X x: x 1 K. / .x A' 5 \ / , \ / A x 95 are found by reference to Fig. XXIY, to be as follows : For the upper lateral system from formula (18), R 8 =:+56000, and for lower lateral system R 3 = +16 000 pounds. From formula (19), Rj for upper and lower lateral system/ equals 3333 and + 8000 pounds respectively. These results are obtained on the supposition that the wind apex loads on the central span are all transmitted by the lateral systems of the central span to the end of the river arm, and then acts through the lateral system of the upper chord. This is a rather more reasonable supposition than that in the case of wind in the high- way bridge of Art. 16, where the wind apex loads on the lower chord of the cen- tral span were assumed to be transmitted to the end of the river arm, and then into the lower chord of the river arm by means of the inclined transverse bracing J i 1 and J' i. See Fig. xix. Rj for upper system is found from formula (21) to be + 53 333 pounds, and for lower system +18 000 pounds. To 96 find the stress in any web member of the upper lateral system due to wind on truss, multiply the shear into the secant of the angle which the member makes with the vertical. For N O' shear is 52 000 pounds, and Sec<9 = ?| = 1.4, and N O' = 52 000 16 X 1.4 = + 72 800 pounds. The same stress takes effect in N' O when wind is reversed. Since these members are duplicates the stress is given for only one system. Stress in P P' is simply the shear or P P' = -46 000 pounds. The overturning effect of wind on the truss is, in the case of a cantilever bridge, a doubtful quantity, and very difficult of satisfactory determination. It is perfectly evident in the problem at hand that the wind blowing on the shore arm affects the chord stresses in connection with the lat- eral bracing, and that this effect is trans- mitted by the lateral system to the ends of the shore arm, where, by means of the cross-frame it is transmitted directly to the abutment and pier. The wind on the river arm and central span, or that part of 97 it acting on the lower chord, has, however, the effect of twisting the river arm, and thereby causing some additional stress to the chord members of the central span, as, well as additional stress in both chord and web members of the river arm. This change of stress must take place either when the train is on the bridge or when the bridge is unloaded. In the first case the overturning effect of wind on train would have the opposite effect to the wind on truss 5 or, in other words, would counter- act the overturning effect of wind on truss. In the second case, the wind blow- ing at a time when no train is on the bridge, the overturning effect of which on truss would give stresses which combined with the dead-load stresses would give re- sults very much less than the possible maximum stresses caused when the bridge is loaded. For these reasons the stresses in the members of truss due to the over- turning effect of wind on the truss will be omitted. The overturning effect of wind on the train, however, gives additional stresses in 98 members of the truss, which, acting at the time when the live load acts, should be taken into account to give the maxim/urn stresses. Assume the train to consist of box cars 10 feet high, and the wind pres- sure per square foot 30 pounds. This gives 300 pounds per linear foot, or 300 X 15 = 4500 pounds per panel. Taking the center of pressure of the wind at 9.5 feet above the center of the upper chord, the overturning moment at each panel point is then 4500 X 9.5 = 42 750 pounds. This causes an additional vertical weight to act at each apex point of the leeward girder, equal to 42 750 -4- 16 = 2675 pounds, and relieves the windward girder by the same amount. This apex load effects the chord and web members of the truss in the same manner as a live apex load, and the pro- cess of finding the stress is consequently a repetition of that for live load. Since the final maximum and minimum stresses are the result of combining those stresses caused by the different possible loading, it is necessary that care should 99 be taken to get the stress in the members due to overturning effect of wind, when the live load occupies the same position on the bridge that it occupied when the live-load stresses were calculated. The stress in the members of the cross frames of the bridge have not been calcu- lated, since the method of calculating them has been explained at the end of Article 16, in highway bridges, and differs in the railroad bridge only in the addi- tional surface exposed to the wind by the train, or 4500 pounds per panel. This force is to be considered only when the bridge is a deck structure, since the wind on the train is transmitted through the wheels and track to the chord on which it rests. 100 Section of Truss and Train with Forces and Lever Arms. Fig. XXV 101 SgogS I + 1 + 1 ssnj} no pupto. SB auras ssaj;s ninuiix'BjVL sb S ^S^^^ ni'Ba;uopatA\.o; anp raa^s^g -d 'S r-o -M-O Tt 5 k t^ Tpja , I -f I + I uo sassaa^g B ||||g||^||||||||g l + l I + I + I + I-M + I + I 1 + OQ - 1 +7+T+T+T +T+T+T+T+ 102 Wind Trus + l 4-1 1 H-l Ma S ____ T __ c ______ aaras Baw S nmniix^ ui'Bai no pniAv o, aup 8 -88^8 OK o a 103 a M . M M sb S 3. Maxi Id ' $. 6 ^ aj % ^o 2 IOO O 9 *O O O JO IQ O O O Q 1C ig ad IO MS O 16 ;oo5^;cxSc;tscoo:C"3oTp5!r::*::5 tO-*lO-*?O->i'OOTf[-'*t -^COtS'fOT^t-TC'. -*< R***s*i***M*i*3* i i i i i M i ii i i : M Mill ~ t-< ~ ~ i ~ -T- M--t-i. o o t- -.s c; to -N --0 c; -o ts VD w ^ ys * --o - -^ n o to^c^io'^t-^c^r-^^w^o^-H^ 00 ' r-l r-i ^H rH (M (M COM C^ (M ++++ 1 M I + ++++-H-++ i-f-H- O O U5 O tt O O O O O 10 O L g=g ? ' ^^- OOOOCrsO(MiOt-OCOlO CJOXOOOOOOO 7 M M M M ii nT iT M ii i ! 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