THE UNIVERSITY OF ILLINOIS « • \ INTERNATIONAL LIBRARY of TECHNOLOGY A SERIES OF TEXTBOOKS FOR PERSONS ENGAGED IN THE ENGINEERING . PROFESSIONS AND TRADES OR FOR THOSE WHO DESIRE INFORMATION CONCERNING THEM. FULLY ILLUSTRATED AND CONTAINING NUMEROUS PRACTICAL EXAMPLES AND THEIR SOLUTIONS BRIDGE SPECIFICATIONS DESIGN OF PLATE GIRDERS DESIGN OF A HIGHWAY TRUSS BRIDGE DESIGN OF A RAILROAD TRUSS BRIDGE WOODEN BRIDGES ROOF TRUSSES BRIDGE PIERS AND ABUTMENTS BRIDGE DRAWING SCRANTON: INTERNATIONAL TEXTBOOK COMPANY 97 Copyright, 1908, by International Textbook Company. Entered at Stationers’ Hall, London. Bridge Specifications: Copyright, 1907, by International Textbook Company. Entered at Stationers’ Hall, London. Design of Plate Girders: Copyright, 1907, by International Textbook Company. Entered at Stationers’ Hall, London. Design of a Highway Truss Bridge: Copyright, 1907, by International Textbook Company. Entered at Stationers’ Hall, London. Design of a Railroad Truss Bridge: Copyright, 1907, by International Textbook Company. Entered at Stationers’ Hall, London. Wooden Bridges: Copyright, 1907, by International Textbook Company. Entered at Stationers’ Hall, London. Roof Trusses: Copyright, 1907, by International Textbook Company. Entered at Stationers’ Hall, London. Bridge Piers and Abutments: Copyright, 1907, by International Textbook Com¬ pany. Entered at Stationers’ Hall, London. Bridge Drawing: Copyright, 1907, by International Textbook Company. Entered at Stationers’ Hall, London. All rights reserved. Printed in the United States BURR PRINTING HOUSE FRANKFORT AND JACOB STREETS NEW YORK x A PREFACE The International Library of Technology is the outgrowth of a large and increasing demand that has arisen for the Reference Libraries of the International Correspondence Schools on the part of those who are not students of the Schools. As the volumes composing this Library are all printed from the same plates used in printing the Reference Libraries above mentioned, a few words are necessary regarding the scope and purpose of the instruction imparted to the students of—and the class of students taught by— these Schools, in order to afford a clear understanding of their salient and unique features. The only requirement for admission to any of the courses offered by the International Correspondence Schools, is that the applicant shall be able to read the English language and to write it sufficiently well to make his written answers to the questions asked him intelligible. Each course is com¬ plete in itself, and no textbooks are required other than those prepared by the Schools for the particular course selected. The students themselves are from every class, trade, and profession and from every country; they are, almost without exception, busily engaged in some vocation, and can spare but little time for study, and that usually outside of their regular working hours. The information desired is such as can be immediately applied in practice, so that the student may be enabled to exchange his present vocation for a more congenial one, or to rise to a higher level in the one he now pursues. Furthermore, he wishes to obtain a good working knowledge of the subjects treated in the shortest time and in the most direct manner possible. iii 342995 IV PREFACE In meeting these requirements, we have produced a set of books that in many respects, and particularly in the general plan followed, are absolutely unique. In the majority of subjects treated the knowledge of mathematics required is limited to the simplest principles of arithmetic and mensu¬ ration, and in no case is any gre'ater knowledge of mathe* matics needed than the simplest elementary principles of algebra, geometry, and trigonometry, with a thorough, practical acquaintance with the use of the logarithmic table. To effect this result, derivations of rules and formulas are omitted, but thorough and complete instructions are given regarding how, when, and under what circumstances any particular rule, formula, or process should be applied; and whenever possible one or more examples, such as would be likely to arise in actual practice—together with their solu¬ tions—are given to illustrate and explain its application. In preparing these textbooks, it has been our constant endeavor to view the matter from the student’s standpoint, and to try and anticipate everything that would cause him trouble. The utmost pains have been taken to avoid and correct any and all ambiguous expressions—both those due to faulty rhetoric and those due to insufficiency of statement or explanation. As the best way to make a statement, explanation, or description clear is to give a picture or a diagram in connection with it, illustrations have been used almost without limit. The illustrations have in all cases been adapted to the requirements of the text, and projec¬ tions and sections or outline, partially shaded, or full-shaded perspectives have been used, according to which will best produce the desired results. Half-tones have been used rather sparingly, except in those cases where the general effect is desired rather than the actual details. It is obvious that books prepared along the lines men¬ tioned must not only be clear and concise beyond anything heretofore attempted, but they must also possess unequaled value for reference purposes. They not only give the maxi¬ mum of information in a minimum space, but this infor¬ mation is so ingeniously arranged and correlated, and the PREFACE v indexes are so full and complete, that it can at once be made available to the reader. The numerous examples and explanatory remarks, together with the absence of long demonstrations and abstruse mathematical calculations, are of great assistance in helping one select the proper for¬ mula, method, or process and in teaching him how and when it should be used. This volume treats of the design and construction of bridges and roof trusses. As an introduction to the subject, a full set of bridge specifications is given. These specifica¬ tions have been carefully selected and compiled from those representing the best modern practice among bridge engi¬ neers. The principles of design are fully illustrated by the complete design, including all details, of several plate-girder bridges, a highway bridge, and a railroad bridge. The part on wooden bridges is a very simple and most convenient presentation of a subject on which there is almost no easily accessible literature. The design and construction of bridge piers and abutments, which forms so important a part of bridge engineering, is treated with the thoroughness that its importance demands. Several bridge-drawing plates—com¬ plete working drawings—are given and fully described in connection with the design work dealt with in the text. The method of numbering the pages, cuts, articles, etc. is such that each subject or part, when the subject is divided into two or more parts, is complete in itself; hence, in order to make the index intelligible, it was necessary to give each subject or part a number. This number is placed at the top of each page, on the headline, opposite the page number; and to distinguish it from the page number it is preceded by the printer’s section mark (§). Consequently, a reference such as § 16 , page 26 , will be readily found by looking along the inside edges of the headlines until § 16 is found, and then through § 16 until page 26 is found. International Textbook Company 97 CONTENTS Bridge Specifications Section Introduction.74 Plans and Proposals.74 Design of Railroad Bridges.74 Design of Highway and Street-Railway Bridges.74 Construction of Steel Bridges: Materials . 74 Construction of Steel Bridges: Workman¬ ship ..74 General Remarks.74 Design of Plate Girders Beams.75 Plate Girders: Section Modulus.75 Plate Girders: Design of Flanges .... 75 Plate Girders: Design of Web .75 Plate Girders: General Design ..... 75 Plate Girders: Splices.75 Plate Girders: Bearings.75 Design of an I-Beam Highway Bridge . . 76 Design of an I-Beam Railroad Bridge . . 76 Design of a Plate-Girder Railroad Bridge 76 Design of a Highway Truss Bridge General Features.77 Design of Floor System: Stringers ... 77 Design of Floor System: Intermediate Floorbeams and Brackets.77 Page 1 3 5 22 42 46 54 1 3 5 10 17 19 30 1 12 19 1 4 13 in iv CONTENTS Design of a Highway Truss Bridge Section Page —Continued Design of Floor System: End Floorbeam and Bracket . . . '. 77 22 Stresses in Main Members. 77 26 Design of Main Members. 77 34 Design of Lateral System. 77 42 Floor Connections.78 1 Splices.78 14 Design of Pins and Pin Plates .78 16 Bearings. 78 40 Lateral Connections. 78 43 Bridge-Seat Plan . 78 45 Design of a Railroad Truss Bridge Data.79 1 Design of Floor System: Stringers ... 79 3 Design of Floor System: Intermediate Floorbeams.79 10 Design of Floor System: End Struts . . 79 18 Stresses in Main Members. 79 21 Design of Main Members. 79 27 Design of Lateral System. 79 38 • Design of Chord Splices. 79 48 Design of Truss Joints. 79 51 Bearings. 79 61 Bridge-Seat Plan . 79 64 Wooden Bridges Uses and Types of Wooden Bridges ... 80 1 Materials and Working Stresses .... 80 4 Kingpost and Kingrod Trusses.80 7 Queenpost and Queenrod Trusses .... 80 14 Howe Truss. 80 24 Towne Lattice Truss . .. 80 32 Combination Trusses . 80 35 Roof Trusses Introduction .81 1 Loads and Stresses .81 8 V CONTENTS v Roof Trusses— Continued Section Page First Illustrative Example.81 14 Second Illustrative Example.81 21 Third Illustrative Example.81 30 Design of Trusses and Lateral Systems . 81 41 Design of Connections.81 44 Bridge Piers and Abutments Definitions.82 1 Location and Estimates.82 4 Pier Design: Practical Considerations . . 82 10 Pier Design: Theoretical Considerations . 82 23 Abutment Design. 82 48 Construction . 82 62 Bridge Drawing Introduction.83 1 Drawing Plates: General Considerations . 83 8 Drawing Plate 107, Title: Highway-Bridge Details.83 11 Drawing Plate 108, Title: Highway-Bridge Details.83 19 Drawing Plate 109, Title: Highway-Bridge Details. 83 25 Drawing Plate 110, Title: Highway-Bridge Details. 83 31 i t BRIDGE SPECIFICATIONS INTRODUCTION 1. In the early days of bridge trusses, when wood, being plentiful and cheap, was used to a great extent, bridges were invariably built without plans, and in a great many cases without previously calculating any stresses or design¬ ing any members. They were built by men called bridge carpenters , and the experience of the foreman or superin¬ tendent of the gang usually enabled him to decide on the sizes of members to be used. As the loads were then light and timber was much more plentiful than it is now, the errors were generally on the side of safety; that is, the bridges were made more than strong enough for the loads. 2. When wood was replaced by wrought iron, it became necessary to manufacture in the shops most of the members, and designs and plans were made. There were no systematic scientific methods of design, however; the details, instead of being proportioned according to the forces they were to resist, were designed and arranged as seemed most conve¬ nient. Many bridges at that time were designed by shop foremen that had no knowledge of stresses; the members were laid out full size on the shop floors, and made so as to utilize the material on hand in the stock yard. Very few, if any, bridges were designed to allow an increase in the loads they were to carry in the future. 3. Later, engineers began to realize that both safety and economy required that bridges should be designed according to scientific principles and constructed in conformity with COPYRIGHTED BY INTERNATIONAL TEXTBOOK COMPANY. ALL RIGHTS RESERVED §74 135-2 2 BRIDGE SPECIFICATIONS 74 fixed rules derived from both experience and theoretical investigations. Such rules, when assembled together for the guidance of the designer, builder, or contractor, are called specifications. Specifications differed greatly at first, but after a short time they began to approach each other, and today the points in which the various specifications differ from one another are comparatively few. It is not unlikely that standard specifications for the construction of all bridges of the same type will be adopted in the near future. The introduction of such uniform system will greatly facilitate bridge design and construction. 4. When the earlier bridges were finished, the plans, if any, that had been used in the design and construction were either destroyed or lost, as the importance of saving them for future reference was apparently not fully realized. As a result, there are at the present time many bridges in use for which no plans can be found; when it is desired to know if they can support with safety heavier loads than they have been carrying, it is difficult and very expensive to calculate their strength, for it is first necessary to measure accurately the span, panel length, and depth of girders, and trusses, the cross-sections of stringers, floorbeams, girders, and truss members, and the details of all connections. For this reason, it has become the custom to keep on file detail plans of every new bridge; these plans show the location of every rivet and the size of every piece of metal in the structure, and are of great value for future reference. 5. In the following articles are given bridge specifica¬ tions agreeing with the best practice in the United States at the present time. The clauses are those that actual practice has shown to be most suitably adapted to the purposes stated. Some of these specifications will be discussed at the end of this Section; others, such as those relating to plate girders, will be given in the Sections on design. It will be sufficient for the student to read these specifications very carefully, so as to get a good idea of their contents; it is not neces¬ sary to memorize them. §74 BRIDGE SPECIFICATIONS 3 * The bridges in the following Sections will be designed according to these specifications. In case it is necessary to design bridges according to other specifications, as is usually the case when a designer works for a bridge or railroad company, it will simply be necessary to read over the other specifications and design the work accordingly. SPECIFICATIONS FOR THE DESIGN OF STEEL BRIDGES PLANS AND PROPOSALS 6. Engineer and Contractor.— The term Engineer, where used in these specifications, refers to the Chief or Con¬ sulting Engineer in charge of the design and construction of the bridge, and to his duly authorized assistants or repre¬ sentatives. The term Contractor refers to the bidder to whom the contract for the work has been awarded, and to his duly authorized representatives. The decision of the Engineer shall be authoritative in all cases of uncertainty. 7. Letter of Invitation.— A copy of these specifications will be furnished each bidder; in addition, he will be given a letter of invitation to bid, stating the general descrip¬ tion of the work for which bids are desired and any additional # facts that may be necessary. In case the requirements given in the letter of invitation conflict in any way with those in the specifications, those in the letter of invitation will rule. 8. Bids.—Bids shall state the total sum for which the work, as described in the letter of invitation, will be done, the estimated weight, and price per pound, of each class of material, and the amount of time required to complete the work. They shall be made with the understanding that the Engineer reserves the right to make such changes in the plans, before the commencement of work, as may be considered advisable by him to render the bridge a satisfac¬ tory piece of work. The increase or decrease in price due 4 BRIDGE SPECIFICATIONS §74 to such change shall be estimated from the pound price of the original bid, and shall be added to or deducted from the contract price. 9. Extension of Time. —The Contractor shall be respon¬ sible for damages on account of delay from any cause during the progress of the work. If any unforeseen delay shall arise, it will entitle him to an extension of time, to be granted in writing by the Engineer at the time of the delay. 10. Patent Devices. —The Contractor shall assume all responsibility for the use of patent devices in any part of the bridge, or in connection with the work of construction. 11. Subcontractors. —No part of the work shall be sub¬ let, nor shall the contract for the whole or any part of the work be assigned by the Contractor, without the written consent of the Engineer. No part of the work shall be done in a shop hot properly equipped with modern facilities. These specifi¬ cations shall be binding on subcontractors in every respect. 12. Plans and. Stress Sheets. —As a rule, the Engineer will provide each bidder with plans and stress sheets, show¬ ing the loads assumed, the resulting stresses, the proposed sizes and sectional areas of the members, and the style of the details and connections, as well as lengths, heights, and clearances. The bidder shall verify the plans before he submits his bid, and he alone shall be responsible for any errors, except as to general layout. He shall return the Engineer’s plans if his bid is not accepted. If the Engineer does not furnish plans and stress sheets as described, the bidder shall furnish them with his bid, if requested to do so by the Engineer. 13. Working Drawings. —After a contract has been awarded and before any material is ordered or work com¬ menced, the Contractor shall submit to the Engineer three complete sets of working drawings, including erection dia¬ grams. When satisfactory, one set of such drawings and diagrams will be approved and returned to the Contractor, and all work shall be done in accordance with them. The 74 BRIDGE SPECIFICATIONS 5 * Contractor alone shall be responsible for the correctness of these drawing's, even if they have been approved by the Engineer. No changes shall be made on the drawings after they have been approved, unless authorized in writing by the Engineer. On the completion of the work, the Con¬ tractor shall furnish the Engineer one complete set of tra¬ cings of the working drawings, which will be permanently filed in the office of the Engineer. The Contractor shall, when required, furnish also the necessary plans for design¬ ing the masonry. Drawings shall preferably be not more than 24 in. X 36 in., with details drawn to a scale of 1 or 1 inch to 1 foot. DESIGN OF RAILROAD BRIDGES GENERAL DIMENSIONS 14. Kinds of Bridges. —The following kinds of bridges shall preferably be used: For spans less than 25 feet in length, rolled beams. For spans from 25 to 100 feet in length, plate girders. For spans from 100 to 150 feet in length, riveted trusses. For spans over 150 feet in length, pin-connected trusses. If, for any reason, it is desired to depart more than 10 feet from these limits, permission in writing must be obtained from the Engineer. Deck bridges will have the preference wherever the con¬ ditions permit their use. 15. Panel Eengths and Depths. —The depth of plate girders shall preferably be one-eighth, and in no case less than one-twelfth, of the span. The depth of trusses shall preferably be not less than one-sixth of the span. Panel lengths shall preferably be from 10 to 25 feet, and in truss bridges shall be so chosen that the angle between diagonal web members and the lower chord shall be not less than 50°. 16. Spacing of Stringers and Deck Girders.—* Stringers shall be spaced 6 feet 6 inches center to center. \ 6 BRIDGE SPECIFICATIONS S74 Deck girders less than 70 feet long shall be spaced 6 feet 6 inches center to center; deck girders over 70 feet long shall be spaced 6 inches farther apart for each 10 feet increase in length. In bridges on curves, the center line between stringers, and between deck girders, shall be parallel to the chord of the | curve between abutments, and at a distance "b from it equal to two-thirds the middle '•§ ordinate. 17. Half-Through Bridges. —Half- through truss bridges shall be avoided when f ig .i possible. Where used,Ohe flanges and brackets shall not come closer to the center line of track than shown in outline in Fig. 1. 18. Th rough Bridges. — In through bridges on straight track, no part of the structure shall come closer to the center line of the nearest track than shown in outline in Fig. 2. In bridges on curves, there shall be provided 1 inch additional clearance on each side of the track for each degree of curvature, and 2| inches additional clearance on the inside of the curve for each inch of superelevation of track. 19. Spacing and Gauge of Tracks. — Tracks shall be spaced 13 feet center to center, unless other¬ wise specified. The gauge of track is 4 feet 8i inches. 20. Spacing of Trusses. —The distance center to center of trusses shall preferably be not less than one-twelfth the span, and in no case less than one- half the depth of trusses. Fig. 2 21. Double-Track Spans. — Double-track spans shall preferably have only two trusses or girders. Where, on §74 BRIDGE SPECIFICATIONS 7 * account of thin floors, three-truss bridges are advisable, the tracks shall be spread so as to provide the proper clearance between the trusses for each track. LOADING 22. Loads. —Bridges shall be designed to resist properly the stresses caused by the following forces: dead load; live , or moving, load; impaci and vibration; centrifugal force; wind pressure; and the longitudi?ial force due to suddenly stopping trains. 23. Dead Load. —The dead load shall consist of the estimated weight of the entire structure. The weight of ties, guard timbers, and rails shall be taken as 400 pounds per linear foot of track, of timber as 4i pounds per board foot, and of ballast as 120 pounds per cubic foot. In truss bridges, two-thirds of the dead load shall, in general, be assumed as applied at the loaded chord, and one-third at the unloaded chord. 24. Live Load. — The live load on each track shall consist of two engines followed by a uniform load of 5,000 pounds per linear foot, as represented in Fig. 3, or a loading o o o to .04 o o O o O O O o o o o o o o o o io in to in QQQQ o o O o in 04 04 CO CO o o o o in in 5*^- 9 - t. -8^-8 o o o o o o CD CD 5*5**-9 - -8 8 5*5*5‘ 5> (d) Fig. 11 trusses. The sections of top chord then act as beams as well as compression members, and are subjected to simultaneous compressive and bending stresses. This practice is con¬ demned by the best engineers. It is best in all cases to provide a floor system in the top chord, the ends of the floor- beams being connected to the insides of the trusses or resting on top of the trusses at the panel points. §74 BRIDGE SPECIFICATIONS 61 235. Dive Loads.—One of the most important steps in the design of a bridge is to ascertain the live or moving load the bridge is to carry. The live load on a railroad bridge consists of locomotives and cars. As explained in Stresses in Bridge Trusses , Part 4, it is customary to use typical load¬ ings that represent the heaviest loads it is expected the bridge will ever have to carry. Fig. 11 shows four typical loadings in use on leading railroads in the United States; Cooper’s loadings also have been adopted by many of the leading railroad companies. At the present time, Cooper’s E50 well represents the heaviest loads on most American railroads. In a few special cases, where the loads in use are somewhat heavier than this, each load of Cooper’s E50 may be multiplied by 1.1 or 1.2, as desired, giving what may be called E55 and E60, respectively, approximately equivalent to the actual loads, and the resultant systems substituted for E50 in the specifications. For bridges on branch lines and on lines on which the locomotives and cars are light, E40 will give sufficiently heavy loads. In case extra heavy loco¬ motives and cars are used for any purpose, the bridges on lines over which they operate should be designed for the actual loads, increased 10 per cent, in some cases to allow for future increase. 236. Up to the present time, the only type of concen¬ trated load that has been considered for railroad bridges has been the steam loco¬ es o o 10 o o o 10 o o o 10 'fr o o o in o o o 3 o o o JO 5000 ib. per Linear o c ) c i one 1 ' oof of Track T/yyyyyyyy/yyyyyyyy/// *625 *625 *- fo. 0 '-5*6255625 Fig. 12 *5.0* motive and train of cars. There are now in use electric loco¬ motives very nearly equal in weight to the steam locomotives. Fig. 12 shows the distances between axles and the weights on the axles of one of the heaviest electric locomotives that has been built. Bridges over which such heavy locomotives are to pass, or over which it is likely they will pass in the future, should be designed for Cooper’s E50, in the same way as other railroad bridges, or for the actual loads, as explained above. 62 BRIDGE SPECIFICATIONS 74 237. On electric roads designed for the multiple-unit system, which uses no locomotives, each car carrying its own motors, it is necessary to ascertain if there is any possi¬ bility of the road being used in the future by steam or electric locomotives; if so, due allowance should be made so that the bridges will be strong enough for them. If it is likely that nothing but the cars will ever run on the road, the bridges should be designed for the actual weights of the cars when fully loaded, increased 10 or 20 per cent, to provide for a possible increase in the weight of future cars. Fig. 13 rep- 1 o o o o O o 10 10 . = — (r 0 7 + rS+ r a ’+ . . .) r 0 or, denoting by Z r 1 the sum r 0 5 + r* + ?% + • • • of the squares of the distances r 0 , r lf r 2 , etc., r 0 r 0 r 0 v Mr = . ( 1 ) r* i It should be clearly understood that, in applying this formula, the distance of every rivet from the neutral axis should be squared, and the results added. When several rivets are at the same distance from the neutral axis, as is the case when rivets are arranged in horizontal rows, or when some rivets are as far above as others are below the neutral axis, the work is much simplified by squaring that distance and multiplying the result by the number of rivets to which the distance is common. Thus, in Fig. 7 (a), each distance, as r,, is common to four rivets, two above and two below the neutral axis. In this case, 2V = 4(r 0 a + r 1 2 + r2 J + . . .) In order that the splice rivets may have sufficient strength, the value of M r must be at least as great as the resisting moment of the web, and K must not exceed the value of the outside rivet. We must, therefore, have -x.v = ^ (2) s r 0 o In applying this formula, it is customary to assume first a spacing of rivets and calculate their resisting moment. If the value is not sufficient, more rivets are added, either by spacing the rivets closer together, or by adding another row outside of the first two. For this purpose, the arrange¬ ment represented in Fig. 8 is frequently employed, the §75 DESIGN OF PLATE GIRDERS 23 advantage being that most of the rivets are near the flanges, where the value of the resisting moment is greatest. The thickness of these plates may be found by means of the formula t x — t X “ (Art. 19), substi- h x tuting for //, the sum of the heights of the three portions of the splice plate; the thickness on each side, however, is usually made about 75 per cent, of that of the web. Example.— If the plate girder represented in Fig. 8 has a web 72 in. X g- in., spliced with §-inch rivets and splice plates as shown, what is: (a) the required thickness of splice plates on each side? ( b ) the resisting moment of the rivets on each side of the splice, assuming the value of a rivet to be 9,600 pounds? Solution.—( a ) As the splice plates are made up of three parts on each side of the web, the value of h x is 14 -f 3l|- + 14 = 59.5 in. Then, since t is .5 and h is 72 in., t x — .5 X rTTTa = *5 X 1.46 = . /3 in. 59.5 'i for both sides, or .73 4- 2 = .365 in. (say f in.) for each side. Ans. (b) The distances of the rivets from the neutral axis are 2, 6, 10, 14, 17^, 21, 24^, and 28 in., respectively. There are four rivets at each of the first four distances, and eight rivets at each of the last four. To apply formula 1, we have K — 9,600 lb., r = 28 in., and IV = 4 X (2 2 + 6 2 + 10 2 + 14 2 ) + 8 X(17.5 2 + 21 2 + 24.5 2 + 28 2 )= 18,396 Therefore, M r = X 18,396 = 6,307,200_in.-lb. Ans. 21. Flange-Angle Splices. —Flange angles are spliced by riveting angles to them, as represented in Fig. 9. In this figure, (a) is the elevation of a portion of the bottom flange, (b) is the cross-section at B B, and (c) is the plan of the flange and a cross-section at C C. The angles d are the flange angles, the angles e are the splice angles, and the line / is the section at which the flange angle is spliced. Practice 24 DESIGN OF PLATE GIRDERS §75 varies as to the method of designing a flange-angle splice. Some engineers use one splice angle riveted to the flange angle that is spliced; this is open to the objection that the splice angle must be very long in order to get sufficient rivets to develop the stress, and therefore interferes with other details, such as stiffeners. Others use two angles, as in Fig. 9, and assume that the stress in the flange angle is equally divided between the two. On account of the fact that the angle on the side opposite the splice gets its stress through the web and the other flange angle, it is probable that the angle in contact with the flange angle that is spliced d B O Q Q e Q Q Q f o Q O Q O O (a) B ^ d O O Q Q Q Q Q Q Q Q Q Q Q o O -—-—-1--- O Q Q r/ O Q o ' e O O Q i O Q Q i . Q (c) Fig. 9 gets somewhat more than one-half the stress. The assump¬ tion most frequently made is that the splice angle in contact with the flange angle that is spliced takes three-quarters of the stress; this angle is designed on this basis to have a cross-section three-quarters that of the flange angle, and made long enough to get sufficient rivets on each side of the splice to transmit three-quarters of the stress. The other splice angle is then made the same size and length. If F A is the area of section (net area for tension, gross area for com¬ pression) of one flange angle, then the area FJ of section of each splice angle is given by the formula F a > = ,75 F a (1) §75 DESIGN OF PLATE GIRDERS 25 If is the allowable intensity of bending- stress, the total stress in the flange angle is F A s , and in the splice angle, .75 F a s. The number n of rivets in the splice angle on each side of the splice is given by the formula n .75 F a s K " ( 2 ) in which K is the value of one rivet. Splice angles are cut from angles having legs the same width as the flange angles, so that the edges of splice and flange angles will be even, and not as shown by dotted lines in Fig. .9 ( b ). The area of cross-section of such an angle can be found by deducting from the area given in Tables IX and X for the original angle the amount that is sheared off. Example. —The flange angles in the tension flange of a plate gir¬ der are 6 in. X 6 -in. X \ in., and the allowable intensity of stress is 16,000 pounds per square inch. The diameter of the rivets is •§■ inch. (a) What size of splice angle must be used if the angle is spliced as represented in Fig. 9? (b) If the value of one rivet is 5,000 pounds, how many rivets are required on each side of the splice? Solution.—( a ) The area of a 6" X 6" X angle is 5.75 sq. in. As we are considering the tension flange, the net section is required; as there are two rivet holes in each section, and the area of a hole for a -g-inch rivet is .5 sq. in., as given in Table XXVII, the net section is 5.75 — 2 X .5 = 4.75 sq. in. (= Fa). The required net area Fa' of one splice angle is, then, by formula 1, .75 X 4.75 = 3.56 sq. in. As the flange angle is % in. thick, -g- in. must be sheared off each leg of each splice angle. The thinnest 6" X 6" angle, which is f- in. thick, will be tried first. The area of this angle, as given in Table IX, is 4.36 sq. in.; if ^ in. is cut from each leg, the area is reduced by 2 X .5 X .375 = .37 sq. in., nearly. As there are two holes in each angle, the area will be still further reduced by 2 X .375 = .75 sq. in. Then, the net area of one angle 5^- in. X 5^- in. X | in. will be 4.36 — .37 — .75 = 3.24 sq. in. As 3.56 sq. in. is required, this is not enough. The next size, 6 in. X 6 in. X tq in., the gross area of which is 5.06 sq. in., will be tried. If \ in. is sheared from each leg, the area is reduced by 2 X .5 X .44 = .44 sq. in. The two rivet holes still further reduce the section by 2 X .44 = .88 sq. in. Then, the net area of one angle in. X 5|- in. X -fe in. is 5.06 — .88 — .44 = 3.74 sq. in., which is sufficient. Ans. (b) The total stress in one flange angle is 4.75 X 16,000 = 76,000 lb., and the portion to be transmitted by the rivets is three-fourths of this, 26 DEwSIGN OF PLATE GIRDERS §75 or 57,000 lb. Then, the required number of rivets is 57,000 -4- 5,000 = 11.4, or, say, 12 rivets on each side of the splice. 22. Flange-Plate Splices. —Flange plates are spliced either by additional plates, or by continuing the outer plates beyond their theoretical ends a sufficient distance to splice the plates below them. The plates nearest the flange angles are the longest, and are the ones that require to be spliced. 23. Additional Splice Plates. —In the first method of splicing, the joint in the plate that is to be spliced is usually located somewhere near the center of the girder, as repre¬ sented at /, Fig. 10, and a splice plate g of the same thick¬ ness as the flange plate is riveted to the outside of the flange. The splice plate is made long enough to contain sufficient rivets on each side of the joint to properly transmit the stress from one part of the flange plate to the other. The number of rivets that is required to transmit the stress to or from the splice plate can be found by dividing the stress by the value of one rivet; the latter will usually be the value in single shear. When, as in Fig. 10, there are intermediate §75 DESIGN OF PLATE GIRDERS 27 plates between the splice plate and the plate that is spliced, there is some uncertainty as to how the stress is carried around the joint; to provide for this, the number of rivets in the splice plate is increased as specified in Arts. 61 and 134 of B. S. In the present case, since there are three interme¬ diate plates, the number of rivets will be increased 3 X 20 = 60 per cent. Any flange plate may be spliced in the same way. 24. Continuing Outer Plates. —In the second method of splicing, the location of the joint in the plate that is to be spliced is found by means of the curve of flange areas. In Fig. 10, the curve of flange areas and one-half of the flange are laid out to the same horizontal scale; the points d , c , and b are the theoretical ends of the three outside flange plates, and it is desired to splice the first flange plate. The ordinate O I' represents the flange area as found by the formula^ = -^-(Art. 6), and 01 represents the actual area sh e of flange. From /, IE is laid off to scale to represent the area of the plate that is to be spliced; the line EE' is drawn parallel to X'OX to its intersection E' with the curve, and the line E' e is drawn perpendicular to X'OX. Then, e' E' represents the entire area of flange, exclusive of the plate a ; that is, if all the plates are carried beyond e, the first plate a can be spliced at that section. For this purpose, the outer plate, instead of being stopped at d } is carried beyond e , as represented by the dotted lines. As there is no stress in the first flange plate a at the section e , and there is full stress in the outer plate d , it remains to find how many rivets must be contained in the plate d beyond e, in order to transmit its stress to a ; this is found by dividing the stress in the out¬ side plate by the value of one rivet in single shear. In the present case, as there are two plates between d and a , the required number of rivets will be 2 X 20 = 40 per cent, greater than the theoretical number. The plate d will be continued to a point d\ the distance e d' being made such that there will be sufficient room for the required number 28 DESIGN OF PLATE GIRDERS 75 of rivets. In a similar manner, any flange plate may be spliced. It will seldom be found necessary to splice any flange plate at more than one section, nor more than two plates in any flange. When two plates must be spliced, the first plate can be spliced at one end of the girder, as at e , Fig. 10, and the second plate at a corresponding point on the other side of the center. Example. —Let Fig. 10 represent the top flange of a plate girder in which the allowable intensity of stress is 16,000 pounds per square inch, and the sizes of the plates are as follows: a, 16 in. X \ in.; b, 16 in. X \ in.; c, 16 in. X yq in.; and d, 16 in. X -f in. (a) If it is desired to splice plate a at section j, what is the required size of the splice plate g? ( b ) If the value of one rivet in single shear is 6,600 pounds, how many rivets must be included in plate g? ( c ) If it is desired to splice plate a at section e , how many rivets must be included in plate d between e and d', assuming the value of one rivet to be 6,600 pounds? Solution. — (a) As plate g is an additional splice plate, it must be the same size as the plate that is to be spliced, that is, 16 in. X 1 in. Ans. (b) The area of cross-section of plate g is 8 sq. in., and, since the intensity of stress is 16,000 lb. per sq. in., the stress in plate g is 8 X 16,000 = 128,000 lb. As the value of one rivet in single shear is 6,600 lb., the number of rivets required to transmit this stress to plate g is 128,000 -r- 6,600 = 19.4. Since there are three intermediate plates between g and a, the number must be increased 3 X 20 = 60 per cent. Then, the total number of rivets required on each side of / is 19.4 + yyo X 19.4 = 31 As the flange rivets in the plates are driven in pairs, it is necessary to have 32 rivets. Ans. (c) The area of cross-section of plate d is 16 in. X f in. =6 sq. in.; and, since the intensity of stress is 16,000 lb. per sq. in., the stress in plate d is 6 X 16,000 = 96,000 lb. As the value of one rivet is 6,600 lb., the number of rivets required to transmit the stress to the plate d is 96,000 -7- 6,600 = 14.5 rivets. Since there are two intermediate plates between d and a , the number must be increased 2 X 20 = 40 per cent. Then, the total number of rivets required in plate d between d' and e is 14.5 + ^X 14.5 = 20.3 As this is so close to 20, 20 rivets will be sufficient, although it is better to use 22. Ans. 25. Splices in Secondary Flange Angles and Verti¬ cal Flange Plates.—Secondary flange angles are usually §75 DESIGN OF PLATE GIRDERS 29 Q O Q Q O O O Q Q Q O Q O O O O Q 0) a Q Q O Q O O O 1 Q O O Q O O O O P O 1 •-- — 1 Fig. 11 spliced, as represented in Fig. 11, by means of one splice angle riveted to the inside of the flange angle and a splice plate having a width about the same as that of the flange angle, riveted to the back of the latter. The area of the splice angle is made about 75 per cent, and that of the splice plate about 25' per cent, that of the flange angle, the area of the two together being not less than the area of the flange angle. Vertical flange plates are spliced, as represented in Fig. 12, by means of one vertical splice plate riveted to the vertical legs of the flange angles on the same side of the web as the vertical plate that is spliced. The area of the splice i plate is made not less than that of the flange plate. O O Q Q Q> 0 a O O Q Q Q Q OOOO O Q Q _ Q Q Q Q Q O Q O Q O O O O Q Q 1 —— S/?//ce Pfaf&' Fig. 12 EXAMPLES FOR PRACTICE 1. What is the resisting moment of a 36"X -f" web, if the maximum intensity of the bending stress is 16,000 pounds per square inch? Ans. 1,620,000 in.-lb. 2. If the web represented in Fig. 7 is 70 inches wide and inch thick, what is the required thickness of splice plate on each side, assu¬ ming the height to be 57.5 inches? Ans. yq hi. thick on each side 3. If the rivets in Fig. 7 are spaced as shown at the left-hand side, and the value of one rivet is 10,800 pounds, what is the resisting moment of the rivets in the splice? Ans. 3,849,600 in.-lb. 135—8 30 75 DESIGN OF PLATE GIRDERS 4. The flange angles in the compression flange of a plate girder are 8 in. X 8 in. X f in., and the allowable intensity of stress is 16,000 pounds per square inch, (a) What size of splice angle must be used? (b) If the value of one rivet is 6,600 pounds, how many rivets are required in the splice” angle? Ans {( a ) ^ angles, 8 in. X 8in. X f in., cut to 7-y in. X 7^ in. X f in. ’ 1 (b) 22 rivets on each side of the joint 5. (a) If the vertical flange plate in Fig. 12 is 15 inches wide and inch thick, and the splice plate is 13 inches wide, what is the required thickness of the latter? (b) If the allowable intensity of stress is 16,000 pounds per square inch, and the value of one rivet is 6,600 pounds, how many rivets are required in the splice plate? Ans. { M t in - thick ' \ (b) 18 or 20 rivets on each side of the joint BEARINGS 26. Size of Bedplates.—The end of a girder, where the girder rests on the masonry or other support, must be i u p- (c) d © © © 0 © © © © © © © © © © © © 0 © / © C C © s 0 © © 0 © © © © © © © © © © © © © 0 \ © © © © 0 © P (a) u l—O : ©“I jo © © ©I (b) Fig. 13 ©| © # © I©; J©Ql ' l© © © © n (b) Fig. 14 strong enough to resist the reactions. The required area of DESIGN OF PLATE GIRDERS 31 a 75 bearing, if the girder rests on masonry, is found by dividing the maximum reaction by the allowable intensity of bearing on the masonry. For example, if the reaction is 200,000 pounds, and the allowable intensity of bearing is 500 pounds per square inch, the required area of bearing is 200,000 -f- 500 = 400 square inches. Bedplates are usually made rect¬ angular and very nearly square; in the present case, each will need to be about v400 = 20 inches on each side. c \ Q O/C'Q Q P~ Q O Q O J2 Q Q 9 O o Q O Q d (a) ho p.^lVT1 (*>) Pig. 15 V 27. End Stiffeners. —The ends of plate girders are stiffened by stiffeners at each end of the bed¬ plates. The arrangements usually employed are represented in Figs. 13, 14, and 15, in which c, c are the stif¬ feners; d , the sole plates; and e , the bedplates. The arrangement rep¬ resented in Fig. 13 is most used, although that shown in Fig. 14 is somewhat better on account of the fact that in it the bearing of the stiffeners is not so close to the edge of the bedplate; the pressure is therefore more evenly distributed over the area of the bedplate. In the arrangement represented in Fig. 15, the additional stiffeners d are added. The plates g are called reinforcing plates, and are added to distribute the stress more evenly over the web. 0-0 0 O 0 o o o o o 28. D'i stribiition of Reaction. —The reaction is assumed to be equally divided among the end stiffeners, and evenly distributed over the area of the outstanding legs of the stiffeners where they bear on the upper side of the lower flange angle. The length of the portion of a stiffener in contact with the lower flange angle is usually about i inch less than the nominal width of the outstanding leg. If t' is 32 DESIGN OF PLATE GIRDERS §75 the thickness of a stiffener angle; b , the nominal width of the outstanding leg; and s 6 , the allowable intensity of bearing on the end of,a stiffener, the stress that one stiffener can resist is t' \b — If R is the reaction and n the number of end stiffeners, the amount of pressure that is transmitted by each stiffener is —; and,, in order that the stiffener may be n sufficiently strong, the following equation must be satisfied: /'(/.- \)s b = K n From this equation, we have f = _ X _, n s b (b — i) from which the required thickness of the stiffeners can be found. The nominal width of leg is controlled by Arts. 55 and 128 of B. S. 29. Rivets and Stiffeners. —The stress in each stif¬ fener is transmitted to the web by means of rivets. The required number of rivets in a stiffener can be found by dividing the stress in one stiffener by the value of a rivet in single shear, or in bearing on the angle; or, since the same rivets connect two opposite angles, by dividing the stress in two stiffeners 2 7? by the value of a rivet in double \ n / shear, in bearing on two angles, or in bearing on the web, whichever is least. In practice, it is considered advisable to transmit the greater part of the stress in the stiffeners to the web below the neutral axis. When it is impossible to get sufficient rivets below the neutral axis in four stiffeners, as in Figs. 13 and 14, more stiffeners are used, as in Fig. 15. 30. Crimped Stiffeners. —Stiffeners are sometimes placed in contact with the web, and the ends crimped; that is, bent out around the vertical legs of the flange angles, as represented in Fig. 13 (r). This arrangement is open to the objection that it is difficult to bend the angles so they will bear evenly on the flange angle, especially on the outstanding leg. §75 DESIGN OF' PLATE GIRDERS 33 31. Loose Fillers. —The best practice at the present time is to make the stiffener angles straight from top to bottom, and fill in the space between them and the web by means of bars the same width as the adjacent leg of the stiffener and the same thickness as the flange angles, as represented in Figs. 13 and 14. These bars are called loose fillers; they simply serve to fill the space between the angle and the web. The number of rivets required to connect the stiffener to the web must be increased 20 per cent, when this form of filler is used. 32. Reinforcing: Plates or Tight Fillers. —The filler under the stiffeners in Fig. 15 is continued under all the angles, and riveted to the web by rivets located outside of the stiffeners. The filler is then called a tight filler, or reinforcing plate; it distributes the stress over the area of the web. Such a plate is not to be considered an inter¬ mediate plate, but rather a part of the web, as it is firmly riveted to‘it, and in calculating the bearing value of the rivet on the web the thickness of these plates must be included. Example 1.—The maximum reaction at the end of the plate girder represented in Fig. 13 is 135,000 pounds, (a) If the stiffeners are 5 in. X 3^ in., and the allowable intensity of bearing is 18,000 pounds per square inch, -what is the required thickness of the stiffener angles? (b) If the value of one rivet in single shear is 6,600 pounds, in bearing on the web, 10,800 pounds, and in bearing on each stiffener angle, 8,400 pounds, how many rivets are required to connect the stiffeners to the web? Solution. — (a) The required thickness of stiffeners can be found by the formula in Art. 28, e _ _ n s b (b - |) In the present case', R — 135,000, n = 4, s b Substituting these values in the formula gives 135,000 18,000, and b = 5. t< = = .42 in., or in., nearly. Ans. 4 X 18,000 X 4.5 ( b ) The rivets that connect the stiffeners to the web are in single shear on each side of the web, and in bearing on the yy-inch stiffener angle. Considering first the stress in one angle, the value in single shear is evidently less than the value in bearing on the yV-inch angle. 34 DESIGN OF PLATE GIRDERS §75 The stress in each stiffener is 135,000 4- 4 = 33,750 lb., and the required number of rivets is 33,750 4- 6,600 = 5.1. Considering now the stress in two angles, the rivets are in double shear at 2 X 6,600 = 13,200 lb.; in bearing on two -pg-inch angles, at 2 X 8,400 = 16,800 lb.; and in bearing bn the web, at 10,800 lb. The latter value being the smallest, and the stress in two stiffeners being 135,000-7-2 = 67,5001b., the required number of rivets is 67,500 4-10,800 = 6.25. This number is larger than that first found, and must be used. As there are loose fillers (see Art. 31), the actual number required is 6.25 + ~ ! nPo X 6.25 = 7.5, say, 8 rivets. Ans. Example 2.—The maximum reaction at the end of the plate girder represented in Fig. 15 is 230,000 pounds; the thickness of web is inch, and the thickness of the flange angles is f- inch. Using the working stresses given in Art. 29 of B. S. , it is desired to find: (a) the required thickness of stiffeners, if the outstanding leg is 5 inches wide; ( b ) the number of rivets required to connect the stiffeners to the web, if •§--inch rivets are used. Solution. — (a) In Art. 29 of B. S. , the allowable intensity of bearing on the ends of stiffeners is given as 18,000 lb. per sq. in. Then, since R = 230,000, n = 8, and b — 5, we have, applying the formula in Art. 28, * " 8X 18,000 X 4.5 = - 355 ’ ° r ’ Say ’ 1 m ' AnS ' (b) Considering a single stiffener, the value in bearing on a -f-inch angle, as given in Table XL, is 7,220 lb.; and in single shear, 6,600 lb. The latter is the smaller. The stress in one stiffener is 230,000 -4- 8 = 28,750 lb., and the required number of rivets is 28,750 4- 6,600 = 4.4. Taking two stiffeners, it is necessary to consider the value of the rivet in double shear at 13,230 lb., in bearing on two stiffeners at 14,440 lb., and in bearing on the combined thickness of web and two reinforcing plates if- in. The last is evidently greater than either of the other two values; the value in double shear is the smallest, and must be used. Since the stress in two stiffeners is twice that in one, and the value in double shear is twice that in single shear, the number of rivets will be the same as in the first case considered, that > \ is, 4.4, or, say, 5 rivets in each stiffener. Ans. 33. Rocker Bearings.— The ends of spans over 75 feet in length are supported on rockers and supplied at one end with rollers, as explained in Bridge Members and Details. When it is necessary, on account of high water, to keep the bridge seat close to the bottom of the girders, the § 75 DESIGN OF PLATE GIRDERS 35 arrangement represented in Fig. 16 is used. The outstand¬ ing legs of the lower flange angles are cut away for a short distance at each end to allow the lower flange to enter between the vertical plates of a pedestal. The web of the girder is reinforced at the end, and a pin is passed through the pedestal and web; the pin should never be less than 6 inches in diameter. The design of the pin, pin plates, and pedestal is made by the same general principles that apply to pin-connected trusses, as explained in another Section. EXAMPLES FOR PRACTICE 1. If the maximum reaction at the end of the plate girder repre¬ sented in Fig. 13 is 157,000 pounds, stiffeners 5 in. X 3-J in., and allowable intensity of bearing 18,000 pounds per square inch, what is the required thickness of the stiffener angles? Ans. \ in. 2. If in example 1 the value of a rivet in single shear is 6,600 pounds, and in bearing on the web 15,000 pounds, Jiow many rivets must be used in each stiffener to transmit the stress to the web? Ans. 6 rivets 3. If the maximum reaction at the end of the plate girder repre¬ sented in Fig. 15 is 275,000 pounds, stiffeners 5 in. X 3-g- in., and allowable intensity of bearing 18,000 pounds per square inch, what is the required thickness of the stiffener angles? Ans. -p 6 - in. DESIGN OF PLATE GIRDERS (PART 2) DESIGN OF AN I-BEAM HIGHWAY BRIDGE 1. Introduction. —In this and in the following- Sections will be given complete designs of several classes and types of bridges. The designs will be made according to the rules given in Bridge Specifications (a title that, for convenience, will be abbreviated to B. S.). These examples will famil¬ iarize the student with the principles involved and the methods used, which he can without any difficulty extend to forms and conditions not specifically covered in the present instruction. 2. Data. —As a first example, an I-beam highway bridge will be designed from the data given in the data sheet on page 4. The words in Italics are supposed to have been written to fill out the general form, which contains only the words printed in Roman type (see B. S., Art. 226). 3. Determination of Span. —It is first necessary to determine the location of the abutments. According to B. S., Art. 18, no part of a bridge should be less than 7 feet from the center line of the nearest track, nor less than 22 feet above the base of the rail. This condition applies also to abutments and underneath clearance lines for overhead bridges. As the faces of abutments are usually rough and extend somewhat beyond the neat lines, it is well to locate the neat lines at the base of the rail 7 feet 6 inches from the COPYRIGHTED BY INTERNATIONAL TEXTBOOK COMPANY. ALL RIGHTS RESERVED §70 * 2 DESIGN OF PLATE GIRDERS §76 center line of the track, making them (as there are two tracks 13 feet center to center) 7 feet 6 jnches -f- 13 feet -f 7 feet 6 inches = 28 feet apart. The faces of abutments are sometimes made plumb (vertical), in order to shorten the span, and sometimes battered, according to the judg¬ ment of the engineer. In the present case, allowance will be made for a batter of i inch per foot in each abutment: then, the distance between the faces will increase at the rate 76 DESIGN OF PLATE GIRDERS 3 of i + 2 = 1 inch for every foot above the base of the rail, as far as the bottom of the bridge seat or coping. Bridge seat stones are usually made from 12 inches thick for short spans to 24 inches, and in some cases more, for long spans. In the present case, a thickness of 16 inches will be sufficient. Allowing 1 inch for the thickness of the sole plate and 1 inch for the bedplate makes the distance from the underneath clearance line (the bottom of the I beam) to the bottom of the bridge seat 16 + 1 + 1 = 18 inches, and the distance from the base of the rail to the bottom of the bridge seat 22 feet — 1 foot 6 inches = 20 feet 6 inches. Then, as the distance between the abutments increases 1 inch for each foot, it will increase 20i inches, or 1 foot 8i inches, in 20 feet 6 inches, and the distance between the neat lines under the bridge seat will be 28 feet + 1 foot 8i inches = 29 feet 8i inches, as represented in Fig. 1, which is the bridge-seat plan, that is,, the drawing of the bridge seat. [It will be noticed that here the word plan is used in the sense of drawing or drawings , not in the sense of a top view. In reality, the drawings in Fig. 1 show both a top view (a) and a cross-section (£).] 4. For a span of this length, the edge of the bedplate should not come closer than 3 or 4 inches to the neat line, and should not be set much farther back than this, as it lengthens the span. In the present case, it will be set 4f inches back at each abutment, making the clear distance between bedplates 29 feet 8i inches + 4f inches + 4| inches = 30 feet 6 inches. Bedplates are seldom made less than 12 inches in length; this is long enough for this span, and makes the total length of I beams 30 feet 6 inches + 1 foot + 1 foot = 32 feet 6 inches, and the distance center to center of bedplates (the span) 31 feet 6 inches^ or 378 inches. The parapets are usually set 3 inches from the ends of the beams; that makes them, in this case, 33 feet apart. 5. Deptli and Spacing of Beams.— As this span is less than 35 feet, rolled beams will be used. According to B. S ., Art. 92, they cannot be less than 378 30 — 12.6 inches in 4 DESIGN OF ELATE GIRDERS GENERAL DATA For bridge over Delaware^ Lac kawa?ina, & Western Railroad a j. _ Elmhurst , Pennsylva7iia __ Length and general dimensions_ T° s P an ^ wo tracks 3 3jeet _ ce nter to center ___ Skew or angle of abutments with center line of bridge Width of bridge and location of trusses-—- /^ c J ea T .. N o trusses ___ Floor system ® ne ^ a y er °f 3-inch oak plank-on nailing pieces and steel beams Number and location of tracks_ W<7 i rac ^ s __ Loadin'*' Art. US (-5) Bridge Specificatio?is Description of abutments Cement-concrete abutments _ Distance from floor to clearance line_ Not more than 3 feet high water_ «< u i < u (« (l <1 14 low water_ river bottom. I Character of river bottom- _ Usual season for floods _ Name of nearest railroad station Elmhurst, Pe?insylvama, DL. and W. R. R. 2 miles Distance to nearest railroad station __ Time limit_ months from d ate of award of ccmtract Ndme of Engineer International Textbook Company Address of Engineer Scranton , Pemisylvania Remarks i‘ ave a tight b oard fence at each side at least 5 feet 6 inches above the top of the floor §76 DESIGN OF PLATE GIRDERS 5 depth. Table XIV* shows that the next depth of beam is 15 inches. In finding the spacing of beams, it is first neces¬ sary to determine the distance between the outside beams. In the present case, a roadway of 20 feet between wheel- guards is required; the fence will be placed 6 inches outside of the inner edges of the wheel-guards; then, the clear dis¬ tance between the fences will be 21 feet. A tight board fence is specified to prevent injury to the traffic from cinders and sparks from the locomotives that will pass underneath. Fig. 2 shows the cross-section of a good Style of fence and connection to the beams; the posts are placed 5 or 6 feet apart. The outside beam is usually a channel, prefer¬ VT _.»T 1 -2/-5 Between out&'cte Cftanne/s Fig. 2 ably of the same depth as the I beams. To assist in keeping the fence in place and preventing it from being blown over, the channel may be connected to the next beam, at intervals of about 10 feet, by means of the diaphragm shown in Fig. 2. *A11 tables referred to in this Section are found in Bridge Tables, unless otherwise stated. 6 DESIGN OF PLATE GIRDERS §76 This may be made of the lightest material allowed—in this case, as the bridge is over a railroad, f inch thick ( B . S ., Art. 112). With the arrangement represented in Fig. 2, the face of the channel is 8 i inches outside of the edge of the wheel-guard, making the channels 20 feet + 82 inches + 82 - inches — 21 feet 5 inches from face to face. As there will be one layer of floor planks 3 inches thick, the beams cannot be more than 3 feet center to center ( B. S ., Art. 123). Five spaces at 3 feet makes 15 feet, leaving 21 feet 5 inches — 15 feet = 6 feet 5 inches, to be made up in the sides of the bridge. This requires the face of each channel to be placed one-half of 6 feet 5 inches, or 3 feet 2i inches, from the center of the first beam; if a spiking piece 4 inches wide is used on the channel, its center will be just 3 feet from the center of the next beam. This spacing of beams will satisfy all conditions. 6 . Live Load. —According to B. S ., Arts. 98 (3) and 110, the live load should be either 80 pounds per square foot or a steam road roller acting on each beam as two concentrated loads of 5,000 pounds, 11 feet apart. For the former, as the beams are 3 feet apart, the load per linear foot is 3 X 80 = 240 pounds. The maximum bending moment occurs at the center, and, as the span is 31.5 feet, is 240 X 3L5 X 3 L5 = 29,770 foot-pounds The maximum bending moment due to the two concentrated loads occurs under one of the loads when that load and the center of gravity of the two loads are equidistant from the center of the beam, and, as explained in Stresses i?i Bridge Trusses , Part 4, is equal to AQ i PPQ ^ X 1 3 _ 53 f 00 |-_ 31.5 pounds at 2.75 feet from the center of the span. As the bending moment due to the road roller is the greater, it is unnecessary to further consider that due to the uniform load. The shear will not be considered in this example, as it has been found in practice that, in all ordinary cases in bridge work, an I beam or channel that is strong enough to resist the bending moment is also strong enough to resist the shear. §76 DESIGN OF PLATE GIRDERS 7 The bending moment on the channel may be taken equal to one-half that on the I beams, or 53,650 -f 2 = 26,820 foot¬ pounds, as the load on the channel is practically one-half the load on the beam. 7. Dead Load. —As 4.5 pounds per board foot, the weight stated in B. S., Art. 97 , is rather high for the timber used in bridge floors, we shall assume it to include the weight of the spikes that hold the floor down, and the bolts that hold the spiking pieces to the beams. As the floor plank is 3 inches thick, its weight is 3 X 4.5 = 13.5 pounds per square foot, or, since the beams are 3 feet apart, 3 X 13.5 — 40.5 poiind per linear foot of beam. The floor plank will be fastened down by spiking it to wooden nailing pieces 2 to 6 inches thick that are bolted to the tops of the I beams (see B. S., Art. 122). In order to prevent water from standing on the floor, as it would do if the floor were level, it is customary to make the floor a little higher at the center of the span than at the ends; this is done by varying the depth of the nailing pieces. In the present case, the nailing pieces will be made 2 inches deep at the ends of the span and 4 inches deep at the center. As the top of the nailing piece will be curved, its average depth will be about 3i inches. Its width should be at least equal to the flange of the beam; it will be assumed that 6 inches is sufficient. The average cross-section is then 3i in. X 6 in., equivalent to If board feet; this makes the weight per linear foot 1.75 X 4.5 = 7.9 pounds. The total weight of timber sup¬ ported by each beam is, then, very nearly 40.5 -f 7.9 = 48.4 pounds per linear foot. As the maximum bending moment due to the live load occurs 2.75 feet from the center, or 13 feet from the end of the span, it is necessary to find the dead-load bending moment at that point. For the floor plank and nailing pieces, it is 43.4 X 31.5 x 13 _ ( 48A x 13 ) x jj. = 5,820 foot-pounds A for each beam, and approximately one-half of this, or 2,910 foot-pounds, for each channel. 8 DESIGN OF PLATE GIRDERS 70 8. There is still to be considered the bending moment due to the weight of the diaphragms represented in Fig. 2. The distance from the center of the I beam to the back of the channel is 3 feet 2j inches; the diaphragm will be about 3 feet If inches long, and, if 15-inch beams are used, about 12 inches deep. Then, as the weight of a 12" X t " plate is 15.3 pounds per linear foot, the weight of 3 feet If inches is 15.3 X 3.125 — 47.8 pounds. The total length of angle is 3 feet Li inches + 3 feet li inches -f 12 inches + 12 inches = 8 feet 3 inches, or 8.25 feet. As the weight of a 2i" X 2i // X i" angle is 5.9 pounds per linear foot, the total weight of angles is 5.9 X 8.25 = 48.7 pounds. In Fig. 2, it may be seen that there are thirty f-inch* rivets in one diaphragm, and the weight of their heads must be found. Table XXI gives 8.5 pounds for the weight of one hundred rivet heads for f-inch rivets; then, the weight of sixty rivet heads will be iifo X 8.5 = 5.10 pounds. The weight of one diaphragm is, therefore, 47.8 + 48.7 + 5.1 = 101.6 pounds. Using four of these placed symmetrically on the span at distances of about 10 feet, as shown in Fig. 3, the bending moment on the I beam and on the channel due to them is, since half of each goes to one beam, 101.6 X 10.75 - — - X 10 = 580 foot-pounds Zi The bending moments due to the weight of the beams and channels cannot be found until their weights are known. It is well to assume some value, however, and, if necessary, DESIGN OF PLATE GIRDERS 9 § 7(5 correct it later. An experienced designer will come very close the first time. It has been shown that the depth cannot be less than 15 inches; we shall use the weight of the lightest 15-inch beam and channel, 42 and 33 pounds per linear foot, respectively, as given in Tables XIII and XIV. For the I beams, the bending moment at the point of maxi¬ mum moment, 2.75 feet from the center, is X 13 — (42 X 13) X V — 5,050 foot-pounds And for the channel, 33 X 3 U> x 13 _ (33 x 13 ) x Jji = 3 ;9 70 foot-pounds A The guard timber and fence will be found to weigh very nearly 45 pounds per linear foot, and, as this weight is almost all carried by the channel, the moment on the channel due to it is — ^ ^ X 13 — 45 X 13 X x £~ — 5,410 foot-pounds A The total maximum bending moment on an I beam is, then, 53,650 + 5,820 + 580 + 5,050 = 65,100 foot-pounds, or 781,200 inch-pounds. The total maximum bending moment on a channel is 26,820 + 2,910 + 580 + 3,970 + 5,410 = 39,690 foot-pounds, or 476,300 inch-pounds.' 9 . Allowable Working Stress. — The allowable inten¬ sity of stress given in B. S., Art. 103 , -for the compression flange is 20,000 — 200 X —. It is not considered good prac- w tice to assume that a wooden floor gives lateral support to the top flanges of the I beams; so, if no bracing is placed between the beams, the unsupported length will be 31.5 X 12 = 378 inches. Suppose that a 15-inch beam were used, then, as the flange is about 5.5 inches wide, the ratio l 378 — would be = 68.7, and the allowable intensity of stress w 5.o would be 20,000 — 200 X 68.7 = 6,260 pounds per square inch. If, however, small struts are placed between the beams near the top flanges, in the same relative positions 135—9 •10 DESIGN OF PLATE GIRDERS 70 as the diaphragms already referred to, the unsupported lengths may be taken as 10 feet, or 120 inches, and the allowable intensity of stress for the I beam will then be 20,000 - 200 x — = 15,640 pounds 5.5 In general, when the ratio — exceeds 40, the top flanges w should be supported laterally. As the bending moment is 781,200 inch-pounds, the required value of the section modulus is 781,200 15,640 = 49.95. Consulting Table XIV, it is found that the lightest 15-inch beam—that is, a 15-inch 42-pound beam—has a sec- tion modulus of 58.9. This beam, therefore, can and will be used. « 10. Assuming that the lightest 15-inch channel is used, for which the flange is 3.4 inches wide, the allowable intensity of stress is 20,000 — 200 X --- = 12,940 pounds per square inch 3.4 Then, as the maximum bending moment is 476,300 inch- pounds, the required value of the section modulus is 476,300 - 12,940 = 36.81. Consulting Table XIII, it is found that a 15-inch 33-pound channel has a section modulus of 41.7, and so this will be used. 11. The bracing angles will add a little to the dead load on the I beams; but,* as the section modulus of the 15-inch 42-pound I beam is larger than required, it is not necessary to consider the effect of those angles. Note. —In case the required value of the section modulus had come out larger than that corresponding to the assumed size of the beam, it would have been necessary to revise the design. Suppose that it had come out 86.0. Then, if a 15-inch beam were used, it would be necessary to use a 15-inch 70-pound beam, for which the section modu¬ lus is 88.5; the same strength could be had, however, by using an 18-inch 55-pound beam, for which the section modulus is 88.4, thereby getting the same strength with a lighter beam. It would then be necessary to recompute the bending moment and allowable intensity of stress for the 18-inch beam. 12. Depth of Floor. —The distance from the clearance line and from the top of the bridge seat to the top of the 135 § 76 -/ L 6m ZPfs. 8§xg'x$$ "mo's?? \r r £0 Frames Mark F* (9) FZfxZfrfx8§' L, $' r &. R/t/e/s -g /n d/ame/er s ¥ Bo/fs-g m d/amefer (d) ' • 1 : ' ' ' ■ UU4& .W §76 DESIGN OF PLATE GIRDERS 11 floor can now be found. The vertical distances at the center of the span are 3 inches of plank, 4 inches of nailing strip, and 15 inches for the I beams and channels, making 22 inches for the depth of floor at the center, as represented in Fig. 1 ( b ). In Art. 3, 2 inches was allowed for sole plates and bedplates under the beams; the bridge seat is, therefore, 24 inches below the floor at the center of the span, and, since the nailing pieces are 2 inches deep at the ends, the bridge seat is 22 inches below the floor at the ends. If the tops of the parapets a, a, Fig. 1 (b ), are made level with the floor, they must be 22 inches high. 13. General Plan or Detail Drawings. —Fig. 4 is a detail drawing of the bridge that has just been designed, and gives all the information necessary for its manufacture. It is customary, in drawing the plan and elevation, to show a portion of the floor and fence in its finished condi¬ tion, as at ( a) and (b ), and the remainder with the floor and fence removed, as at ( c) and {d ). One end of a channel and several beams are usually shown with the top flange cut away, in order to show the detail of the connection of the beams to the sole plates. The diaphragms, sometimes called frames, are marked A,, and the cross-struts, which also are called frames, are marked F*. Frames of both kinds are shown to a larger scale at (/) and (g ). The holes for the anchor bolts at one end are made circu¬ lar and i inch larger in diameter than the bolts; at the other end, they are made li inches wide and If inches long, allowing f inch for expansion and contraction. In a bridge of this size, the diaphragms or frames are sometimes bolted instead of riveted to the beams, to avoid the expense of setting up a riveting plant for so small a number of rivets. 12 DESIGN OF PLATE GIRDERS S 76 DESIGN OF AN I-BEAM RAILROAD BRIDGE 14. Data.—An I-beam railroad bridge will now be designed from the data given on page 13. 15. Determination of Span.— The distance between neat lines under bridge seats is 20 feet. If the bearing DESIGN OF PLATE GIRDERS S 70 o o For bridge over at __ GENERAL DATA French Creek Redlands, California Length and general dimensions_ 20 feet between ncal lines under bridge seats ' __ Skew or angle of abutments with center line of bridge_J?^_ Width of bridge and location of trusses^_ No_trusses_ _ Floor system Standard tie-floor on steel s tringe rs or I beams Number and location of tracks % tracks lo feet center to center Loading Cooper's E50, as represented in Bridge Specifications, Art. 24 _ T r\(- oVMiftvmnlc CC III Cll t COllC 1C tC Distance from floor to clearance line Not greater than distance to high water Distance from floor to high water - 8 feet t» «< << (< , low water 13 feet <► << (( (C . river bottom 15 feet Character of river bottom 2 feet gravel, 3 Let shale, then solid rock 20 feet below base of rail Usual season for floods April and May Name of nearest railroad station Redlands, California Distance to nearest railroad station. 5 miles Time limit¬ ed days Name of Engineer____ Address Berkeley, California Remarks_ Henry Jones 14 DESIGN OF PLATE GIRDERS 76 plates are made 12 inches wide, and set so that their edges are 6 inches back of the neat line, the total length of the beams will be 20 feet + 1 foot 6 inches 4- 1 foot 6 inches = 23 feet; and the distance center to center of bearings (the span), 1 foot less, or 22 feet. If 3 inches is left at each end, the parapets will be 23 feet 6 inches apart. The bridge seats will be made 1 foot 6 inches thick. The bridge-seat plan is represented in Fig. 5. The distance from the base of the rail to the top of the bridge seat cannot be found until after the bridge is designed. The top of the parapet is made 7i inches below the base of the rail. 16. Depth and Spacing of Beams. —The depth and the spacing of beams depend on the maximum bending moment and required value of the section modulus. It is considered better practice to place two beams under each rail than one; where necessary, three are used. The beams under each rail are bolted together and made to act as one by means of cast-iron separators or spacers, as illustrated in Table XV; these are located about 5 or 6 feet apart along the beams, but are not to be assumed as support¬ ing the top flanges laterally. Lateral support is furnished by lateral bracing so arranged that the ratio of unsupported length to width of flange shall not exceed 12 (see B. S ., Art. 87). 17. Live Load.—The live load is represented in Fig. 3 of B. S . By applying the conditions for maximum moment, it is found that it occurs when there are four driving axles on the span, under the second (or third) driver when that driver is 1.25 feet from the center of the span, and is equal to 4 X 50,000 X 9-75 X 9 .75 _ 50j000 x 5 = 614,200 foot-pounds. A A « 18. Impact and Vibration. —The formula for impact and vibration is 1 = - X S (B. S., Art. 25). In this A T ouu case, as the moment is required, the value of the moment found in the last article should be substituted for S; and, as §76 DESIGN OF PLATE GIRDERS 15 the entire span must be loaded in order to produce the maximum moment, L — 22. Therefore, / = X 614,200 — 572,200 foot-pounds 22 + 300 19. Dead Load. —The weight w per linear foot of I-beam bridges of this class is given very closely by the formula w = 25 / ( B . S., Art. 242). In this case, / = 22, and, therefore, w — 25 X 22 — 550 pounds per linear foot. The weight of track can be taken as 400 pounds per linear foot ( B . S. y Art. 23). The maximum live-load moment occurs very near the center of the span, and it will be suffi¬ ciently accurate to find the dead-load moment at the center, which is (550 + 400) X 22 X 22 8 57,500 foot-pounds 20. Wind Load.— It is unnecessary to compute the stresses in the laterals of I-beam bridges. If the smallest- sized angle allowed is used—in this case, in. X 3| in. X f in. —it is sufficiently strong to resist any wind stresses. Accord¬ ing to B. S. y Art. 27, the wind pressure on the train is 300 pounds per linear foot, applied 7 feet above the top of the rail. Ordinarily, the lateral system will be about 2 feet below the top of the rail, or about 9 feet below the center of wind pressure. Then, as the beams will be placed 6 feet 6 inches center to center, the additional load on the leeward beams (as explained in Stresses in Bridge Trusses, Part 5) will be 300 X 9 6.5 = 415 pounds per linear foot, and the bend¬ ing moment at the center due to it will be 415_X22X_22 = 25 l00 foot . pounds 8 As this is so small, it is sometimes neglected. In the present case, it is less than 2 per cent, of the total moment, but, there being no reason why it should not be considered, it will be taken into account. 21. Total Moment.—The total moment is equal to the sum of the several moments just found, and is as follows: DESIGN OF PLATE GIRDERS 16 614,200 + 572,200 + 57,500 + 25,100 = 1,269,000 foot-pounds = 15,228,000 inch-pounds. 22. Section Modulus. —As it is required that the flanges shall be supported laterally, the full value of 16,000 pounds per square inch can be used for the allowable intensity of stress, since the ratio of width to unsupported length will be less than 20 (see B. S. } Art. 29). Then, the required value of section modulus is 15,228,000 -r- 16,000 = 951.75. If four beams are used (two under each rail), the required value of the section modulus for each will be 951.75 -r- 4 = 237.94. Consulting Table XIV, it is found that the heaviest I beam made has a section modulus of, 198.4, which is not enough. Therefore, more beams must be used. If six beams are used (three under each rail), the required value of the section modulus for each will be 951.75 — 6 = 158.62. Consulting Table XIV, it is found that a.20-inch 95-pound I beam has a section modulus of 160.7, and a 24-inch 80-pound I beam has a section modulus of 174. If it were necessary to keep the distance from the base of the rail to the underneath clearance line as small as possible, the 20-inch beams would be used. In the present case, as there are practically no restrictions, the 24-inch beams will be used, as they are lighter and therefore more economical. On account of their larger value of section modulus, the 24-inch beams are also stronger than the 20-inch beams. 23. Deptli of Bridge. —In B. S., Art. 48, it is stated that standard ties are framed to 7i inches in depth over , : . . : . . t ■i ' '■ . ■ - e-e ■ Base of f?a// 135 §76 Fic ! I Cross Secf/on on B-. ; ■ *;• . . . - . *, -4 v. : - ■ •' - ‘ - ■ •** . * J ■ '/■ ■-I: . ' . L‘ " v. , §76 DESIGN OF PLATE GIRDERS 17 stringers and girders 6 feet 6 inches center to center. Then, the distance from the base of rail to the underneath clearance line, as represented in cross-section in Fig. 6, is 7h inches + 24 inches = 2 feet 7i inches. Allowing 1 inch for the sole plate and 1 inch for the bedplate gives 2 feet 9} inches from the base of the rail to the top of the bridge seat, as represented in Fig. 5. 24. Plan.- —Fig. 7 is a detail drawing of the bridge that has just been designed, and shows the customary method of arranging the lateral system. The lateral angles are con¬ nected to plates that are riveted to angles attached to the inside beams. The lateral truss is placed as high as is pos¬ sible without interfering with the top flanges of the beams. The two sets of I beams are connected near the ends by diaphragms or frames, and at the panel points of the lateral truss by means of single angles. It is customary to show the top view of the I beams, and to consider a portion of the top flange removed at each lateral connection and at the end of one set of beams, in order to show more clearly the detail of the connection of the laterals to the beam and of the sole plates to the lower flanges. For I-beam bridges longer than about 18 feet, three panels are used in the lateral systems; for shorter bridges, two panels are employed. The panels are usually made equal. In locating the rivets in the lateral connection plates, great care must be taken that the different angles connecting to the plate do not interfere with one another. It is well to lay out each connection plate on a large sheet full or half size and draw each angle in its proper position, leaving about i inch clearance between the different angles that come close together. The process of laying out the lateral system is frequently perplexing to a beginner, but is very simple after a little experience has been had. The different steps will be briefly outlined. The end frames are first located near the ends in such a position that they will not interfere with the anchor bolts nor with the rivets that connect the sole plates to the 18 DESIGN OF PLATE GIRDERS §76 beams. This can be done by locating the frames about 1 inch from the sole plates. In Fig. 7, the backs of the angles that serve as flanges for the frames are 1 foot 1 inch from the ends of the I beams. These angles are 3i inches wide, and, according to Table XII, the gauge lines are 2 inches from the back edges. This makes the gauge lines of the end frames 1 foot 3 inches from the ends of the span and 23 feet — 1 foot 3 inches — 1 foot 3 inches — 20 feet 6 inches from each other. This distance is divided into three equal spaces of 6 feet 10 inches each, thus locating the gauge lines of the angles that act as struts between the beams at the intermediate points. The three beams that form each side of the bridge are bolted together with their centers 7f inches apart, as given in Table XV, and each set of three is placed with its center 6 feet 6 inches from that of the other. This makes the inside beams 6 feet 6 inches — 7f inches — 7| inches = 5 feet 2i inches apart. Since the webs of these beams are i inch in thickness (Table XIV), the clear distance between them is 5 feet 2 inches. The lug or hitch angles that are used to connect the plates to the webs of the inside I beams are 3i inches wide, with the gauge lines 2 inches from the webs. This makes these gauge lines 5 feet 2 inches — 2 inches — 2 inches = 4 feet 10 inches apart. The gauge lines of these hitch angles and those of the cross-frames or struts, located in the preceding paragraph, are commonly called working lines for the lateral trusses. They can be considered as the center lines of the chords and the panel points, respectively. The diagonal lines connecting the intersections of these gauge lines locate the diagonals of the lateral truss. In Fig. 7, they are taken as the gauge lines of the laterals. In some cases, the center of gravity of the angle is made to coincide with the diagonal between the working lines; in a bridge as small as that now under consideration, it matters little which method is used. The next step is to find the length of the diagonal. Each of these lines, together with the working lines, forms a right 76 DESIGN OF PLATE GIRDERS 19 triangle whose two legs are 6 feet 10 inches and 4 feet 10 inches, respectively. Then, the length of the diagonal is V(6'10")’ -M4'10") ; = 8 feet 4* inches The rivets at the ends of the laterals are next located by laying out the connection plates full or half size, as previously described. In detailing bridge work, it is frequently necessary to find the length of the hypotenuse of a right triangle when the legs, commonly called the coordinates in this work, are known. The principle for finding the hypotenuse is simple, but the arithmetical work is laborious, especially when the coordinates are given to small fractions, such as sixteenths of an inch. To shorten this work, tables are commonly employed. There are several books on the market that contain tables giving the squares of distances that occur in feet, inches, and frac¬ tions of an inch. In using such tables, the squares of the coordinates are copied and added; the length corresponding to the square root of the sum is then taken directly from the table. DESIGN OF A PLATE-GIRDER RAILROAD BRIDGE 25. Data.— As the next example of practical design will be taken a deck plate-girder railroad bridge, the data sheet for the construction of which is given on page 20. 26. Determination of Span.— It is first necessary to determine the location of the abutments. It is required that the clear distance between them at the level of the curb be made 50 feet. There is one electric-car track in the center of tfte street, and it may be assumed that the top of the rail is level with the top of the curb. According to B. S., Art. 94, the lowest line of overhead bracing in through bridges for street railways shall not be lower than 15 feet from the top of the rail. This condition applies equally well to the underneath clearance of bridges over street- railroad tracks. In the present case, the underneath clear¬ ance line of the bridge will be placed 15 feet above the top GENERAL. DATA For bridge over __ Sumne r S treet _ Cincinnatii Ohio Length and general dimensions ® ne s P an 0V€r ' street 50 feet wide at the level of the curb , with one electric-car track at the coiter Skew or angle of abutments with center line of bridge—?.— _ Width of bridge and location of trusses ho trusses. _ Sp ace__ deck girders according to Bridge Specifications , Art. 16 __ Floor system Standard-tie floor. See Bridge Specifications , Art. 48 _--_ - t : Number and location of tracks % tracks lb feet center to center Loading Cooper's E50 , as represented in Bridge Specifications , Art. 24 Description of abutments _ Granite abutments; front faces even with street lines Distance from floor to clearance line N°t more than 6 feet 6 inches Distance from floor to high water. < < a u No zv ater a it (i i i low water_ river bottom. Character of river bottom_ Usual season for floods__ _ i i i i Name of nearest railroad station Cincinna ti , Oh io Distance to nearest railroad station__ Time limit_.90 days 3 miles Name of Engineer. Address of Engineer_ Remarks Above-described bridge is to replace the bridge at present in use. New abutments will be built by the Railroad Company. Contractor will furnish bridge-seat plan within 10 days from award of contract 20 §76 DESIGN OF PLATE GIRDERS 21 of the rail, and the top of the bridge seat will be made level with that line. Bridge seats for this length of span should be not less than 18 inches thick; using this thickness, the distance from the top of the rail to the neat line under the bridge seat will be 15 feet — 1 foot 6 inches, or 13 feet 6 inches. If the face of each abutment is battered i inch per foot, the abutments will be 13i inches farther apart under the bridge seat than at the top of the curb, or, in this case, 50 feet + 1 foot li inches = 51 feet li inches. Bed¬ plates for deck plate girders are usually made about 20 inches long for spans of 50 feet, and about 24 inches long for spans 75 feet long, with the front edge from 6 to 9 inches behind the neat line. For the span under consideration, the plates will be made 22 inches in length and set with the front edges 6 i inches behind the neat line. The center of the bedplate is then 11 inches + 6i inches = 17 t inches, behind at each end of the span; the span is 51 feet li inches -f 1 foot 5| inches + 1 foot 5i inches = 54 feet; and the total length of the girder is 22 inches more than this, or 55 feet 10 inches. For a girder of this length, the parapets should not come nearer than 4 inches to the ends. In this case, they will be made 56 feet 6 inches apart. 27. Depth and Spacing of Girders.— Deck girders are usually made about one-ninth or one-tenth of the span in depth. When the distance from the base of the rail to the clearance line is specified, the depth of the girder must be chosen so as not to exceed it. The depth of the tie and the thickness of the flange plates will usually occupy about 1 foot of the depth. In the present case, as the specified distance is 6 feet 6 inches, the depth of the girder, or the width of the web, should not exceed 5 feet 6 inches. If the depth of the girder is made one-tenth of the span, it will be 54 4- 10 = 5.4 feet. It is well, when possible, to use an even foot or half foot for the width of web; in this case, 5.5 feet, or 66 inches, will be used. The backs of the flange angles are usually placed i inch farther apart, or, in this case, 66i inches. This allows for slight irregularities in the 22 DESIGN OF PLATE GIRDERS 76 edges of the web-plates, and prevents the edges from inter¬ fering with the flange plates. As the span is less than 70 feet, the girders will be 6 feet 6 inches center to center (see B. A., Art. 16). The bridge-seat plan is represented in Fig. 8. The dis¬ tance from the base of the rail to the clearance line or the B B Center L/ne of G/rcter ,, Center Line of Track Center Line of G/rcter "ck found that the greatest moment occurs under the third driver of the second engine, when that driver is .127 foot to the left of the center. The value of this moment is 2,703,200 foot-pounds. The maximum moment at the cen¬ ter, in this case, occurs when the same driver is at the center, and is 2,702,500 foot-pounds. As will be seen, these two values are very nearly equal. In general, it may be stated that for spans over 50 feet it is sufficiently accurate to use in design the greatest moment that can occiir at the center of the span , neglecting the consideration that involves the locatio?i of the center of gravity of the loads . The maximum live-load moments and shears are then as follows: At the center (under third driver, second engine), the live-load moment, in foot-pounds, is 2,702,500. At 20 feet from the end (under third driver, first engine), 2,563,000. At 15 feet from the end (under second driver, first engine), 2,247,200. At 10 feet from the end (under second driver, second engine), 1,703,700. At 5 feot from the end (under first driver, second engine), 976,900. At the end (first driver, second engine), the live-load shear, in pounds, is 228,900. At 5 feet from the end (first driver, second engine), 195,400. At 10 feet from the end (first driver, first engine), 163,100. At 15 feet from the end (first driver, first engine), 130,900. At 20 feet from the end (first driver, first engine), 101,600. At the center (first driver, first engine), 65,200. 30. Impact and Vibration. —The formula for impact and vibration is I = X A ( B . S., Art. 25). In the A -j- oUU present case, the allowance in terms of the moments and shears must be found. For the moments, it is necessary to load the entire span; then, L = 54, and / = m X M = .84746 M §76 DESIGN OF PLATE GIRDERS 25 The allowances are: Location Center 20 feet from the end 15*feet from the end 10 feet from the end 5 feet from the end Moment, in Foot-Pounds .84746 X 2,702,500 = 2,290,300 .84746 X 2,563,000 = 2,172,000 .84746 X 2,247,200 = 1,904,400 .84746 X 1,703,700 = 1,443,800 .84746 X 976,900 = 827,900 As the length of track that must be loaded to produce the greatest shears is different for different sections, the allow¬ ance for impact and vibration will be a different proportion of the maximum shear in each case. It has been found that the maximum shear at each section occurs when the first driver is at the section, for which position the first wheel is 8 feet beyond the section. The length of loaded track for sections within 8 feet of the end of the span is, therefore, 54 feet; for other sections it is equal to the distance of the section from the other end of the span plus 8 feet. The proportional allowances are as follows: At the end and 5 feet from the end, the length L of loaded portion is 54; the proportional allowance, / = .84746 6*. At 10 feet from the end, L — 52; / = Mi S = .8523 At 15 feet from the end, L = 47; / =Hf5 = .8646 5 At 20 feet from the end, L = 42; / = Iff = .8772 S At the center, L = 35; / == iHHr S = .8955 6* The allowances for shear are, therefore, as follows: Location End 5 feet from the end 10 feet from the end 15 feet from the end 20 feet from the end Center Shear, in Pounds .84746 X 228,900 = 194,000 .84746 X 195,400 = 165,600 .8523 X 163,100 = 139,000 .8646 X 130,900 = 113,200 .8772 X 101,600 = 89,100 .8955 X 65,200 = 58,400 31 . Wind Pressure. —In this article, we shall consider only the increase in moments and shears caused in the leeward girder by the overturning effect of the wind; this increase is P h calculated by the formula w — ——, in which P is the wind b 135—10 26 DESIGN OF PLATE GIRDERS §76 pressure per linear foot of train, and w is the vertical load per linear foot of girder, due to the wind pressure (see Stresses in Bridge Trusses , Part 5). The center of the wind pressure is 7 feet above the top of the rail (B. S., Art. 27 ), and the top lateral bracing is generally about 2 feet below the top of the rail; then, h , the distance from the center of wind pressure to the top lateral bracing, is 7 + 2 = 9 feet; » and b is 6.5 feet. Therefore, w = = 415 pounds per linear foot 6.5 The moments due to a uniform load of 1,330 pounds per linear foot have already been found; those due to a uniform load of 415 pounds may be found by multiplying the former moments by iWo, or .31203. The results are as follows: Location Moment, in Foot-Pounds Center 20 feet from the end 15 feet from the end ■* 10 feet from the end 5 feet from the end .31203 X 484,900 = 151,300 .31203 X 452,200 = 141,100 .31203 X 389,000 = 121,400 .31203 X 292,600 = 91,300 .31203 X 162,900 = 50,800 As the wind pressure under consideration is that on a moving train, the shears will be found as for a moving load, that is, by loading the portion of the span on one side of a section. They are as follows: Location Shear, in Pounds End 5 feet from the end 10 feet from the end 15 feet from the end 20 feet from the end 415 X 54 2 11,200 415 X 49 X 49 2 X 54 9,200 415 X 44 X 44 2 X 54 7,400 415 X 39 X 39 2 X 54 5,800 415 X 34 X 34 2 X 54 4,400 415 X 27 X 27 2 X 54 2,800 Center § 70 DESIGN OF PLATE GIRDERS 32. Total Moments and Shears.—As the dead load, live load, impact and vibration, and wind pressure may act simultaneously, the total moments and shears may be found 0 by adding- the values that have been found for the different conditions. In doing so, however, it must be remembered that the moments and shears due to dead load, live load, and impact and vibration have been found for the load on the entire width of track, that is, on two girders, while those due to the wind pressure are the effects on one girder. Therefore, to find the total moment or shear at any section of one girder, that due to the wind pressure may be added to one-half the sum of those due to dead load, live load, and impact and vibration, as just found. The total moments and shears on each girder can now be found. The total moments, in foot-pounds, at various positions on the girder are as follows: At the center, 484,900 + _ 2 ,702,50 0 +j , 290jOO + 15^300 = 2,890,200 2 At 20 feet from the end, 452,200 + 2,563,000 + 2,172,000 { lii m = 2 ,734,700 2 At 15 feet from the end, 389,0 00 -f- 2,247,200 -f~ 1,904,400 ^21 400 = 2 391 700 2 At 10 feet from the end, 292 ,600 + 1,703, 700 + 1,443,800 + 9im = lj811>400 2 At 5 feet from the end, 162,900 + 976, 900_+ 827jOO + 50g00 = ^034,700 2 The total shears, in pounds, at the different sections of. the girder are: At the end, 35,910 + 228,900 + 194,0 00- + 112 0Q = 240,600 28 DESIGN OF PLATE GIRDERS §76 At 5 feet from the end, 29,260 + _195,400 + 165,600 + 9 200 = 2 04,300 2 At 10 feet from the end, 22,610 + 163,100 + 139,000 + 7400 = 169,800 2 At 15 feet from the end, 15,960 + 130,900 + 113,200 + g 80Q 2 At 20 feet from the end, 9,310 + 101,600 + 89,100 4 40Q 2 + ’ At the center, = 135,800 104,400 0 ~t~ 6 5,200 - f~ 5 8,400 ^ ^ §qq 2 = 64,600 33. Design of Web. —A i^e-inch web will be tried first. The gross section is 66 X A = 28.875 square inches. As the total shear at the end is 240,600 pounds, the intensity of the shearing stress is 240,600 -f- 28.875 = 8,330 pounds per square inch. Consulting Table XXXVI, finding the point on the curve for the re-inch web corresponding to an inten¬ sity of stress of 8,330 pounds per square inch, and looking horizontally to the right, it is found that the stiffeners must be spaced about 16 inches apart. According to B. S., Art. 55, the spacing of stiffeners should be not less than one-third the depth of the web. In this case, therefore, the stiffeners should be not less than 22 inches apart. As the spacing given in Table XXXVI is 16 inches, it is necessary to try a thicker web. A web i inch thick will be tried next. The gross section is 66 X i = 33 square inches, and the intensity of shearing stress, 240,600 33 = 7,290 pounds per square inch. Consulting Table XXXVI, finding the point on the curve for the i-inch web corresponding to an intensity of stress of 7,290 pounds per square inch, and looking hori¬ zontally to the right, it is found that the stiffeners must be spaced 22 inches apart. As this thickness of web satis¬ fies the conditions as far as stiffener spacing at the end is 76 DESIGN OF PLATE GIRDERS 29 concerned, the required spacing of stiffeners at other sections will next be found. The intensities of shearing stress are: Location Intensity of Shearing Stress, in Pounds per Square Inch End 5 feet from the end 10 feet from the end 15 feet from the end 20 feet from the end Center of span 240,600 -p 33 = 7,290 204,300 -r 33 = 6,190 169.800 + 33 = 5,150 135.800 -- 33 = 4,120 104,400 -f- 33 = 3,160 64,600 -f- 33 = 1,960 34. Spacing of Stiffeners. —Consulting Table XXXVI, finding the points on the curve for i-inch web corresponding to the intensities just found, and looking horizontally to the right or left (whichever is nearer), the required spacings of stiffeners are found to be as follows: Spacing of Location Stiffeners, in Inches End 5 feet from the 10 feet from the 15 feet from the 20 feet from the Center of span 22 end 27 end 32 end 38 end 46 62 The stiffener spacing at other points on the girder may be found by interpolating between the values just found. These distances will not give the actual distances between the stiffeners, but simply the distances that must not be exceeded at the various sections. The actual distances between stiffener-s depend on other details, such as rivet spacing, and are usually found by the detailer or draftsman when the plans are being made. 35. Pitch of Flange Rivets.— The required spacing of the rivets that connect the flanges to the web at any section is found by the following formula, given in Desig?i oi Plate Girders , Part 1: . Kh r V 30 DESIGN OF PLATE GIRDERS §76 In the present case, the rivets are in double shear and in bearing on the i-inch web-plate. (They are also in bearing on the two flange angles, but it may be assumed that the thickness of the latter is greater than that of the web; it is found in practice that this is invariably true. For this reason, the bearing on the flange angles need not be con¬ sidered.) The rivets used in deck plate-girder railroad bridges are i inch in diameter for all spans. Those in the flanges are always shop-driven rivets and, according to B. 5., Art. 29, the values in Table XL will be used. Consulting Table XL, the value of one f-inch rivet in double shear is found to be 13,230 pounds, and in bearing on a plate i inch thick, 9,630 pounds. The latter value, being the smaller, must be used in the formula. As the flanges have not yet been designed, it is mot known what size of angles will be used, so that the value of hr cannot be calculated. In actual practice, however, the designer soon learns what sizes of angles are generally used for spans of different lengths. In the type of bridge now under consideration, 6" X 6" angles are used for all spans up to about 70 feet, and 8 in. X 8 in. for longer spans. In the present case, 6" X 6" angles will be used. Consulting Table XII, it is found that there will be two rows of rivets in each leg, the line midway between the two rows being 2i -f —- — 31 inches from the A back of the angle, as represented in Fig. 9. The distance between the backs of the angles in the flanges has already been found to be 66] inches. The distance h r is, there¬ fore, 66i — 3« •— 3f = 59|- inches. By substituting the proper values in the formula, and using the intensities of § 7(5 DESIGN OF PLATE GIRDERS 31 shearing stress found in Art. 33, the following pitches are obtained: Location Rivet Pitch, in Inches Bottom Flange Top Flange End 9,630 X 59.5 0 Q o ' 240,600 = 2 ' 38; 2.38 X .9 = 2.14 5 feet from the end 9,630 X 59.5 0 Qn . 204,300 * ’ 2.80 X .9 = 2.52 10 feet from the end 9,630 X 59.5 0 on , 169,800 ” ' ’ 3.37 X .9 = 3.03 15 feet from the end 9,630 X 59.5 _ , 00 . 135,800 4.22 X .9 = 3.80 20 feet from the end 9,630 X 59.5 . 104,400 ' ’ ’ 5.49 X .9 - 4.94 Center 9,630 X 59.5 Q Q7 . 64,600 ~ = 8 ' 8/ ’ 8.87 X .9 = 7.98 As a deck railroad bridge is under consideration, the pitch K h in the bottom flange, found from the formula p = —must be multiplied by .9 to get the required pitch in the top flange (see B. A., Art. 57). The pitch is sometimes made the same in both flanges, that found for the top flange in the manner just explained being used for the bottom as well. As in the case of stiffener spacing, the foregoing values will not represent the actual spacing of rivets at any section, but simply the values that must not be exceeded at the different sections. The spacing at other points may be found by interpolating between the given values. As the pitch at 20 feet from the end came out greater than that allowed in B. S., Art. 57, there was no necessity for com¬ puting the pitch at sections nearer the center. 36. Design of Flanges. —The required area of cross- section of the flanges at any section is found by means of the formula A =- ~ (Design of Plate Girders , Part 1). s h g 8 It is impossible to calculate the value of h s , as the areas of the flanges are not yet known; for a first trial, however, h s will be assumed equal to the depth h of wA>, the flanges 32 DESIGN OF PLATE GIRDERS 76 at the center of the span will be designed on this basis, using the bending moment at the center of the span, and the dis¬ tance h g between the centers of gravity of the trial flanges will be computed. With this corrected value for h gy the areas of the flanges will again be found, and the necessary correction made. It will seldom be found necessary to change the cross-section of the flanges as found by assuming h g = h . In the present case, h = 66 inches, ^ = 16,000 pounds, and / = i inch. To apply the formula, the bending moment already found must be multiplied by 12, to reduce it to inch-pounds. For the first trial cross-section at the center of the span, we have, therefore, = 28.72 square inches This is the value for the gross area of the top flange and the net area of the bottom flange. 37. The actual choice of the sizes of angles and plates is wholly a matter of practice. The designer usually follows certain established rules and relies to a great extent on his experience. It is considered bad practice to use very small or thin angles and a large number of plates. It is also con¬ sidered bad practice to make the entire flange section of angles; this is not economical, as the entire section of flange must be continued the whole length of the girder. In B. S., Art. 59, it is required that one-third to one-half the flange area shall be composed of angles. In the present case, that ., . , 28.72 q „ . 28.72 - . oc would require from —-—, or 9.57, to —-—, or 14.36, square 3 inches in two angles, or, as there are two angles, 4.79 to 7.18 square inches in each angle. 6" X 6" angles with thick¬ nesses from i to yi inch are commonly used in the flanges of plate girders for railroad bridges up to about 70 feet in length; the flange plates are never narrower than the total width of the two flange angles together with the thickness of the web, nor thicker than the flange angles. In the pres¬ ent case, the following sections will be used: DESIGN OF PLATE GIRDERS 33 76 Top Flange c Secti °?* IN Square Inches Two angles 6 in. X 6 in. X i in. @ 5.75 . 1 1.5 One plate 14 in. X I in. 5.2 5 One plate 14 in. X tV in. 6.1 2 5 One plate 14 in. X tV in. .. 6.1 2 5 \ _ Total gross area ..2 9.0 In B. S., Art. 60, it is required that, when plates of dif¬ ferent thicknesses are used, they shall diminish in thickness outwards from the flange angles. An exception is sometimes made to this rule, and will be made in this case, when it is required that one plate shall extend the full length of the top flange. This is done to keep water and dirt from working down between the flange angles and the web, and to give the girder a better finish. A thin plate serves the purpose just as well as a thicker | d one, and is more eco- 1 . -Iq—I I q q -- Q - — - -3 ^ nomical, as all that part beyond its theo- -©- -e- - 0 - -e- -e- -0- - q —o- -0- H0- - 0 - - 0 - 3 0— 1 - 0 - 0 - 0 - - 0 - (C) • j i -0-0-0-0-0-0-0- retical end is wasted, /--,- y~-~; r (a) so far as necessary flange section is con¬ cerned. For the bottom flange it is necessary to deduct from the gross section the areas of cross-section of the rivet holes. This is most easily done by deducting from each angle and plate the holes that are in the plate. Fig. 10 shows the method of riveting flange angles to the web and the flange plates to the angles. Each rivet d in the hori¬ zontal leg of any angle is directly opposite one e in the ver¬ tical leg; this brings two rivets <7, d directly opposite each other in each plate. There are then two holes to be deducted from each angle and two from each plate. The following sections will be used; 34 DESIGN OF PLATE GIRDERS §76 t, Section, in Bottom Flange Square It J ches Two angles 6 in. X 6 in. X h in.; 11.50 — 2 X 2 X .50 = One plate 16 in. X i in. 7 One plate 16 in. X Te in. One plate 16 in. X tq in. 8.0 — 2 X .5 = 7.0 - 2 X .4375 = 7.0 - 2 X .4375 = 9.5 0 7.0 0 6.1 2 5 6.1 2 5 Total net area, 2 8.7 5 <0 JN r, y // '6* fc '! -y n /6*% T s •*> I TT Cv Fig. 12 Oy * 2 9.0 4 4.6 3 2 §76 DESIGN OF PLATE GIRDERS 35 The distance of the center of gravity of the top flange is, therefore, 44.632 -f- 29 = 1.54 inches from the outside of the section. As the plates have a total thickness of tV + h + t = 1.25 inches, the center of gravity of the top flange is 1.54 — 1.25 = .29 inch below the back of the angles. The statical moments for the bottom flange are as follows: 6.1 2 5 X .2188 = 1.3 4 0 6.1 2 5 X .6562 = 4.0 1 9 7.0 X 1.125 = 7.8 7 5 9.5 0 X 3.05 = 2 8.9 7 5 2 8.7 5 4 2.2 0 9 The distance of the center of gravity of this flange from the outside edge of the flange is, therefore, 42.209 28.75 = 1.468 inches from the outside of the section. As the plates have a total thickness of -£■ + ts - - h-Te = 1.375 inches, the center of gravity of the bottom flange is 1.468 — 1.375 = .09 inch from the back of the angles. As the distance back to back of the flange angles is 66.25 inches, the distance h g between the centers of gravity of flanges is 66.25 — .29 — .09 = 65.87 inches. This is very nearly equal to the assumed distance of 66 inches. Substi¬ tuting this value of h e in the formula for area, the result is A = 2 ’ 890 ’ 2 —- i X 2 X 66 = 32.91 - 4.12 65.87 X 16,000 — 28.79 square inches This is slightly greater than the net area of the trial bottom flange, but the difference is so slight (.04 square inch) that it is inadvisable to make any changes. Had the difference been greater—say, .2 or .3 square inch—it might have been advisable to increase the thickness of one of the flange plates by tV inch. It will be seen that the flange plates are wider and thicker in the lower flange than in the upper. They are sometimes made the same width, in which case the lower flange plates must be still thicker, or else more plates must be used. If those in the lower flange are made about 2 inches wider than those in the top, and the same number of plates is 38 DESIGN OF PLATE GIRDERS 78 used, it will usually be found that the theoretical lengths of corresponding plates in the two flanges are the same or very nearly so; this condition is very convenient in designing, especially if the same rivet spacing is used in both flanges. Some engineers do not design the top flange, but make it the same size as the lower flange. This gives additional section, and therefore additional strength to the top flange, but is not economical. If this were done in the present case, as the gross area of the lower flange is 11.50 + 84-7+7 = 33.5 square inches, and the required gross area of the upper flange is 28.79 square inches, the difference, which is 4.71 square inches, would be wasted in the upper flange. Unless it is stated in the specifications that both flanges must have the same gross area, they should be designed separately. 39. Lengths of Flange Plates. —The flange angles in both flanges and the plate next to the flange angles in the top flange are continued the full length of the girder. The other plates are cut off where they fare no longer needed. For this purpose, the areas required at the different sections at which the moments have been computed will be deter¬ mined, and the curves of flange areas will be plotted. The distances between the centers of gravity of the flanges at sections other than at the center are not known, but they may be assumed for trial equal to that at the center, and corrected later. Using the bending moments found in Art. 32, the required flange areas, in square inches, are: At the center, 2,890,200 X 12 65.87 X 16,000 4.12 = 32.91 - 4.12 = 28.79 At 20 feet from the end, 2, 734,700 X 1 2 65.87 X 16,000 - 4.12 = 31.15 - 4.12 = 27.03 At 15 feet from the end, 2,391,700 X 12 65.87 X 16,000 - 4.12 = 27.24 - 4.12 = 23.12 §76 DESIGN OF PLATE GIRDERS 37 At 10 feet from the end, h 8 ^’ 40 L X in - 4 - 12 = 20.63 - 4 ' 12 65.87 X 16,000 At 5 feet from the end, 1,034,700 X 12 16.51 65.87 X 16,000 4.12=11.79-4.12= 7.67 Fig - . 13 shows the curves of flange areas: AB represents the half span; C, D , E, and E , the sections at 5, 10, 15, and 20 feet from A , respectively; C C', D D ', E £', EE', and B B ', the required areas of the flanges at C, D , E , F, and B, respect- th ively, B L and B L' representing —, or the portion of web 8 that is included in flange area; L L x , L X L*, A 2 A 3 , and A 3 A 4 , the areas of the angles and plates in the upper flange; and L’ L ,', A/ LJ, LJ, and LJ LJ , the areas of the angles and plates in the lower flange. Dotted curves are then drawn through A, C’, D ', E ', F', and B'. Drawing lines through L , L x , A 2 , etc., and noting where they intersect the curves, the points at which the plates are no longer required are determined. At C , no plates are required; at D, one plate on each flange is. required; at E , two plates on each flange are required; and at F and B , three plates on each flange are required. At F and B there is no need to revise the flange 38 DESIGN OF PLATE GIRDERS §76 area, as the entire section is required. In some cases, as at E, the end of a plate is just included in the section, but, as the plate at that point does not carry much stress, it should not be counted as part of the flange area in calcu¬ lating the distance between the centers of gravity of the flanges. The actual location of the center of gravity at each section can be calculated in the same way as in Art. 38 for the entire flange; it is not necessary to repeat the numerical steps. They are, for the top flange, .64 inch at E , 1.09 inches at D, 1.68 inches at C, below the backs of the flange angles; and for the bottom flange, .43 inch at E f .86 inch at D , and 1.68 inches at C, above the backs of the flange angles. The distances between the centers of gravity of the flanges are: At E, 66.25 — .64 — .43 = 65.18 inches At A 66.25 - 1.09 - ;86 = 64.30 inches At C, 66.25 - 1.68 - 1.68 = 62.89 inches The revised flange areas, in square inches, are, therefore, as follows: At 15 feet from the end, 2,391,700 X 12 65.18 X 16,000 At 10 feet from the end, 1,811,400 X 12 _ 4 12 64.30 X 16,000 At 5 feet from the end, 1,034,700 X 12 - 4.12 = 27.52 - 4.12 = 23.40 = 21.13 - 4.12 = 17.01 „ _ 4.12 = 12.34 — 4.12 = 8.22 62.89 X 16,000 The curve of flange areas may now be corrected by plotting these values at C, D, and E y drawing the curves shown in full lines, and locating the theoretical ends of the flange plates. The distance scaled from the theoretical end of a plate to the center line is one-half the length of plate; in this case, the theoretical lengths of plates in the top flange are approxi¬ mately 25, 34.5, and 40.5 feet, and in the bottom flange, 25.5, 34.75, and 43 feet. According to />. S., Art. f>0, each plate §76 DESIGN OF PLATE GIRDERS 39 shall extend 12 inches at each end beyond the theoretical end; this will increase the length of each plate by 2 feet. The first plate in the top flange and all the angles will continue the full length of the girder, that is, 55 feet 10 inches. The flanges are then made up as follows: Top Flange Two angles 6 in. X 6 in. X i in. X 55 ft. 10 in One plate 14 in. X "o in. X 55 ft. 10 in. One plate 14 in. X A in. X 36 ft. 6 in. One plate 14 in. X iV in. X 27 ft. Bottom Flange Two angles 6 in. X 6 in. X i in. X 55 ft. 10 in. One plate 16 in. X i in. X 45 ft. One plate 16 in. X tV in. X 36 ft. 9 in. One plate 16 in. X tV in. X 27 ft. 6 in. The actual lengths of the plates will probably be slightly different from the lengths given, the difference being due to rivet spacing in the flanges. 40. Splices. —As no plate or angle is longer than 70 feet, it is unnecessary to splice any flange member (B. S., Art. 61). Consulting Table V, it is found that the longest plate 66 in. X i in. that it is possible to get is 34 feet long;, it is therefore necessary to splice the web. It will be spliced at the center,' making each half 27 feet 11 inches long, nearly. The size of the splice plates will first be determined. Consulting Table XI, it is found that for a 6" X 6" X i'' angle the nominal and actual widths are equal; then, the actual size of the leg of a 6" X 6" X i" angle is 6i inches. As the distance back to back of the flange angles is 66i inches, the clear distance between the vertical legs is 66i — 6i — 6i = 54 inches. Allowing i inch clearance at top and bottom leaves 53f inches as the height of the splice plate. According to B. S., Art. 56, each splice plate shall have a sectional area equal to 75 per cent, that of the web. In the present case, that of the web is 33 square inches; then, the area of each plate must be .75 X 33 = 24.75 square inches. As the plates are 40 DESIGN OF PLATE GIRDERS 76 53f inches deep, the required thickness is 24.75 -f- 53.75 = .46 inch. The nearest standard thickness is 1 inch; then, each splice plate will be 531 in. X h in. A spacing of rivets will now be assumed; if it gives sufficient resistance, it will be used; if not, the num¬ ber of rivets will be increased. The spacing shown in Fig. 14 will be assumed. The rivets are in double shear and in bearing on a 1-inch web-plate. It has already been found (Art. 35) that the latter gives the smaller value for a l-inch rivet; this value is 9,630 pounds. As the distances of the rivets from the neutral axis are 2, 6, 10, 13, 16, 19, 22, and 25s inches, and as there are four rivets at each of these distances, the moment of re¬ sistance of the rivets is, accord¬ ing to the formula given in Design of Plate Girders, Part 1, 9,630 X 4 X (2* + 6 2 + 10 2 + 13 2 + 16 2 + 19* -f 22* + 25.125 2 ) 25.125 Pig. 14 = 3,130,000 inch-pounds The moment that the web can bear is given by the formula s t hi M = 8 (Design of Plate Girders , Part 1). In the present case, the value of s for bending stress is 16,000 pounds per square inch, t = i inch, and h — 66 inches; then, M = 16,0 00 X .5 X 66 2 8 4,356,000 inch-pounds Since this is greater than the resisting moment that has been found, it is necessary to use more rivets. As the rivets farthest from the neutral axis are the most effective, we shall begin another row outside of the first two rows, and first compute the moment of resistance of a few rivets near 76 DESIGN OF PLATE GIRDERS 41 * the flanges. Let us first try two rivets at the top and two at the bottom, at distances of 25.125 and 22 inches from the neutral axis. The moment of resistance of these four rivets is 9,630 X (22 s + 25.125 s ) 2 X 855,000 inch-pounds, o o o o o o 1° o o o o o j!° o o o o o :! ° o o o o o o O tjO o !i o o o o a o o 25.125 * which, added to that already found, gives 3,130,000 + 855,000 = 3,985,000 inch-pounds. This is still too small. Let us try one more rivet at the top and one at the bottom in the same row, each 19 inches from the neutral axis. The moment of resistance of these two is 2 X Myxlj g = 277,000 inch-pounds, 20.125 which, added to that already found, gives 3,985,000 + 277,000 = 4,262,000 inch-pounds. This is still too small. Let us try one more rivet at the top and one more at the bottom in the same row, each 16 inches from the neutral axis. The moment of resistance of these two is O .. 9,630 X 16 s X 25.125 = 196,000 inch-pounds, which, added to that already found, gives 4,262,000 + 196,000 = 4,458,000 inch-pounds. This is a little larger than the mo¬ ment of the web, and is, therefore, sufficient. Three splice plates on each side of the web will now be used instead of one, the top and bottom plates having three rows of rivets on each side of the splice and the middle plate two rows. It is neces¬ sary to rearrange the spacing of the rivets, so that there will be the proper distance from the edge of each plate to 135-11 Pig. 15 42 DESIGN OF PLATE GIRDERS §76 the nearest rivet, and i inch clearance between the plates (see B. S., Art. 41). The spacing: represented in Fig:. 15 will be used; this changes the location of the rivets, and it is well to recompute the moment of resistance of the rivets in the entire splice. That moment is as follows: M = 9,630 X [4 X (2 2 + 5.5 2 + 9 2 + 12.5 2 ) + 6 X (16.125 2 + 19.125* + 22.125 2 + 25.125 2 )] -r 25.125 = 4,433,000 inch-pounds, which is larger than the moment of resistance of the web, and therefore sufficient. It is unnecessary to calculate the moment of resistance of the splice plates; if the plates on each side have an area on a vertical section 75 per cent, that of the web, their moment of resistance will be greater than that of the web. 41. Bearings. —Since the abutments are granite, for which, according to B. S., Art. 29, the allowable intensity of pressure is 500' pounds per square inch, and the end shear, which is equal to the reaction, is 240,600 pounds, the required area of bearing is 240,,600 500 «= 481.2 square inches. In Art. 26, it was stated that the bedplates would be made 22 inches long; the required width is, therefore, 481.2 -h 22 = 21.9 inches. They will be made 22 inches long. 42. End Stiffeners. —The formula given in Design of Plate Girders , Part 1, for the required thickness of end stif- R feners is t’ In the present case, R = 240,600 ns b (b — 2 ) pounds, and, according to B. S. y Art. 29, the allowable intensity of bearing is 18,000 pounds per square inch. The outstanding legs of the flange angles are 6 inches wide; then, according to B. S., Art. 55, the stiffeners will be 5 in. X 3i in. It will first be assumed that there are four stiffeners; this gives 240,600 t’ = = .743 inch 4 X 18,000 X (5 - i) ' When the required thickness of stiffeners comes out greater than f inch, as in this case, it is generally con¬ sidered advisable to use more stiffeners. Using eight gives „ 240,600 8 X 18,000 X (5 -i) = .371 inch, say, f inch DESIGN OF PLATE GIRDERS 43 As this thickness is less than f inch, it will be adopted, and eight stiffeners 5 in. X 3i in. X f in. with reinforcing plates under them will b.e used. The stiffeners will be riveted to the girder with i-inch rivets. The value in single shear, 6,610 pounds (Table XL), is found to be the smallest value. Then, the number of rivets required to connect each stiffener is 240,600 8 X 6,601 4.55, or, say, 5 rivets 43. Lateral System. —The lateral truss represented in Fig. 16 will be used for both the upper and the lower flange; the end frames will be about 52 feet apart, which makes it possible to use eight panels at 6 feet 6 inches each. The wind load is given in B. S ., Art. 27. The pressure on the lower half of the girder, about 3 feet in depth, is resisted by the lower laterals; the pressure of 50 pounds per square foot, or, in this case, 3 X 50 — 150 pounds per linear foot, evi¬ dently causes the greatest stresses in the lower lateral truss. Then, the panel load for the lower lateral truss is 150 X 6.5 = 975 pounds. The pressure on the upper half of the girder, and on the rails and ties, say 4 feet in depth, together with the pressure of 300 pounds per linear foot on the train, are resisted by the upper lateral system. The pressure on the train—together with 30 pounds per square foot, or 4 x 30 = 120 pounds per linear .foot on the girders, ties, and rails—evidently causes greater stresses than 50 pounds per square foot on the girders, ties, and rails alone. The live wind panel load for the upper lateral system is, there¬ fore, 300 X 6.5 — 1,950 pounds, and the dead wind panel load, 120 X 6.5 = 780 pounds. 44 DESIGN OF PLATE GIRDERS §76 44. The end panel of the upper lateral truss will first be considered. The live-load shear is ^>950 X 7 _ gg25 z 780 x 7 pounds, and the clead-load shear is -——-= 2,730 pounds. A The total shear is, therefore, 6,825 + 2,730 = 9,555 pounds. The panel length is the same as the distance center to. center of girders; so the inclination of the laterals is about 45°; esc 45° = 1.414. The direct stress in the diagonal is 9,555 X 1.414 — 13,510 pounds, tension when the wind blows in one direction, and compression when in the other direction. According to B. S., Art. 34, the member must be designed for 13,510 + .8 X 13,510 = 24,320 pounds tension and com¬ pression. Dividing by 16,000 gives 24,320 -r- 16,000 = 1.52 square inches net section required to resist the tension. 45. According to B. S., Art. 86, the smallest angle that can be used for lateral bracing is 3i in. X 3i in. X I in. In Table IX the gross area of this angle is given as 2.48 square inches. The number of holes to be deducted depends on the method of riveting the angle to the con- ,// nection plate. Fig. 17 Y i l of era/ Xngfe 3^ X 3$ X g Log /\ng/e3j> X3%X § shows a connection frequently used: the short angle a riveted to the main angle at the end is called a lug angle, and serves the purpose of transmit¬ ting the stress from the leg b of the main angle to the connec¬ tion plates. In practice, the number of rivets connecting the two angles is usually made one more than half the number required to connect the lateral to the connection plate. With rivets spaced as shown, 1.5 holes must be deducted, according to B. S., Art. 33. As the angle is f inch thick, the area of cross-section of one hole is .375, and of 1.5 holes, 10 0 Fig. 17 §76 DESIGN OF PLATE GIRDERS 45 1.5 X .375 = .5625 square inch. Deducting this from 2.48 leaves 1.92 square inches net section. As only 1.52 is required, this angle is large enough so far as tension is concerned. 46. The distance center to center of girders, measured along a diagonal, is 6.5 X 1.414 — 9.19 feet = 110 inches, nearly. The ends of the laterals are riveted to the con¬ nection plates, so that the unsupported length may be taken as the distance between connections, or about 18 inches at each end shorter than the distance between girders, leaving 110 — 2 X 18 = 74 inches unsupported. Table IX gives the least radius of gyration as .68 inch; then, / 74 - = = 108.82 r .68 Table XXXV gives the allowable intensity of compressive stress as 9,650 pounds; as the gross area is 2.48 square inches, the strength of the angle is 2.48 X 9,650 = 23,930 pounds. This is very nearly equal to 24,320, the required strength, and so this angle is sufficiently large. 47. As the span under consideration is less than 75 feet long, it will be shipped riveted up complete. That is, the rivets connecting the laterals to the lateral plates will be shop-driven, and, according to B. S., Art. 29, the values given in Table XL will be used. Table XL gives the value of a |--inch rivet in single shear as 6,610 pounds; the required number of rivets is then 24,320 -f- 6,610 =, 3.7, or, say, 4 rivets. As the 3i" X 3i" X t" angle is strong enough in the end panel of the upper lateral system, it is sufficient in all other panels, and so there is no need in this case of making any computations for the other angles. 48. The amount of wind pressure that is transmitted to the abutments by the end frames (a diagram of which is shown in Fig. 18) can be found 40 DESIGN OF PLATE GIRDERS § 70 by multiplying the sum of the live and dead wind loads per linear foot on the upper lateral truss (300 -f 120 = 420 pounds) by the total length of the girder, which is 55 feet 10 inches, or 55.83 feet. As one-half of this load is transmitted by each ( f • 420 X 55.8333 11 7Q r •, A frame, the amount is--- = 11,/25 pounds. As A there are two diagonals, one will be assumed to be out of action when the other is in tension. The height of the frame is about 4.5 feet; the tension in a diagonal is, therefore, 11,725 X AA±J— = 11,725 X ~ = 14,230 pounds 6.5 6.5 It has already been found that a 34" X 34" X t" angle is more than sufficient for a stress in tension of 24,320 pounds; so that it will be used in this case. DESIGN OF A HIGHWAY TRUSS BRIDGE (PART 1) GENERAL FEATURES «* 1. General Data. —The principles of design, as applied to through pin-connected highway bridges, will be illustrated by the complete design of a bridge of this type. The bridge will be designed in conformity with the following general data, and according to the specifications given in B. S* GENERAL DATA For bridge over Lackawanna River ____ at _ Scranton , Pennsylvania _ _ Length and general dimensions ® ne s P an 160 ^ ee l center to center of bearings Skew or angle of abutments with center line of bridge Width of bridge and location of trusses ^ sidewalks, 6 feet clear width; one roadway 25 feet between wheel-guards. Trusses located between sidewalks and roadway^ __ Floor system ® ne ^ a y er oa & plank 2 inches thick; one layer 3 inches* thick. Steel stringers *The abbreviation B. S. stands for Bridge Specifications , to which Section frequent reference is made in this and in some of the fol¬ lowing Sections. All tables referred to are found in Bridge fables , unless otherwise stated. COPYRIGHTED BY INTERNATIONAL TEXTBOOK COMPANY. ALL RIGHTS RESERVED § 77 2 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 Number and location of tracks_ 0 n estr ect-rai Iway track at center of roadway _ As given in B. S., Art. 98 ( 2) and (5) Cement-concrete abutments Loading_ Description of abutments. Distance from floor to clearance line_ N ot more than 6 feet high water_ ^ _ < < (i << (< {« < < 20 feet 25 feet low water_ <( <( << << -i river bottom_ Character of river bottom * ee t mu d an d Sl ^i> then solid rock ____ Usual season for floods_ March and April _ Name of nearest railroad station Scranton , Pennsylvania Distance to nearest railroad station_ ojmzles __ Time limit ^ months from award of contract Name of Engineer. International Textbook Company Scranton , Pennsylvania Address of Engineer_:_ Remarks Bridge to have ornamental railing 3 feet 9 inches high at each side of bridge. The top of bridge seat must be kept above high-water line 2. Plans.—The plans for some parts of this bridge are shown in Bridge Drawing. The student is advised to consult those plates frequently during the study of this Section. 3. Kind of Bridge.—The first step is to determine the kind of bridge. As the distance from the floor to the clear¬ ance line is limited to 6 feet,, it is evident that a through §77 DESIGN OF A HIGHWAY TRUSS BRIDGE 3 bridge must be used, there not being sufficient room for a deck bridge. As the span is 160 feet, pin-con¬ nected trusses must be used (B. S., Art. 91 ): Ac¬ cording to B. S ., Art. 229 , panels from 20 to 30 feet in length are best for pin- connected trusses: eight panels 20 feet long will be used. If a simple Pratt truss is chosen, with a depth of 27 feet, which is about one-sixth of the span, the angle between the inclined web members and the lower chord will be about 53° 30k This angle is greater than 50°, as required in B. A., Art. 92 , and hence a depth of 27 feet will be adopted. The outline of the truss is shown in Fig. 1. 4 . Width of Bridge. The data require that the trusses shall be located between the roadway and the sidewalks, and that the roadway shall have a clear width of 25 feet between wheel-guards. According to B. S. } Art. 125 , the edges of the wheel-guards toward the roadway will be 6 inches from the clearance 4 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 lines of the trusses, making the clear distance between the latter 26 feet. The width occupied by a truss depends on the width of the widest members; for pin-connected trusses it is seldom less than 18 inches. In the present case, the width will be assumed to be 24 inches. If the design works out more or less than this, it may be slightly changed. With a width of 2 feet for each truss, the distance center to center of trusses is 28 feet, and the distance between the clearance lines of the trusses toward the sidewalks is 30 feet. As the data require 2 sidewalks 6 feet wide, the distance between the railings will be 30 + 6 + 6 = 42 feet. Fig. 2 shows the assumed width and location of trusses and the location of railings. _ DESIGN OF FLOOR SYSTEM / STRINGERS 5. Cross-Section of Floor.— The cross-section of floor usually employed for this type of bridge is shown in Fig. 3. The rails h , h for the street-railway track rest on 6" X 6" ties 8 feet long, marked^. When the rails are the same height as the thickness of the floor plank, the lower layer of plank is spiked directly to the top of the ties. When, as assumed in the figure, the height of the rail is greater than the thick¬ ness of the floor plank, longitudinal nailing pieces /, as required in B. S ., Art. 121, are spiked to the top of the ties, and the lower layer of plank is spiked to the nailing pieces. For the remainder of the roadway, the lower layer of plank d is fastened to spiking pieces that are bolted to the tops of the steel beams. The upper layer c is then spiked to the top of the lower layer. At each side of the roadway, 6" X 6" wheel-guards e are bolted to the top of the lower layer of plank; and the side¬ walk plank by for which one layer is sufficient, is continued right through the truss out over the wheel-guard. The side¬ walk plank is supported at other points by steel beams with spiking pieces on top. To prevent the ends of the sidewalk §77 DESIGN OF A HIGHWAY TRUSS BRIDGE 5 plank b and the edges of the wheel-guards e from being worn by the wheels, a guard angle a , usually 2i in. X in. X i in., is fastened by screws to the ends of the sidewalk plank, as shown in Fig. 4. It is customary, in order to facilitate drainage, to make the sides of the roadway from 3 to 6 inches lower than the center; they will be made 3 inches lower than the center in the present case. The top of the guard angle is required by B. S., Art. 125, to be 6 inches above the floor; this brings it 3 inches above the top of the floor at the center. fig. 4 The sidewalk plank rises from the wheel-guard toward the railing at the rate of i inch per foot. 6 . Spacing; of Stringers. —In B. S ., Art. 93 , it is specified that, for bridges carrying a street railway only, stringers shall be spaced 6 feet 6 inches center to center, and that for other bridges they shall be arranged to accom¬ modate the traffic. In the present case, I beams will be placed 3 feet 3 inches on each side of the center line of bridge under the ties, and at intervals of 3 feet for the remainder of the floor, as shown in Fig. 5. An I beam h will also be placed at the center to carry any loads that move along the center of the car track. The beam a, a , at the out¬ side of the sidewalk, under the railing, is frequently made deeper than the other sidewalk beams, and is composed of a web and one flange angle at top and bottom; it is then called a fascia girder. The actual location of the fascia girder depends on the type of fence railing that is used, and on the connection of the posts that support the fence. In the present case, the web of the fascia girder will be assumed to be 3 inches outside of the inner edge of the fence, making the web 3 feet from the next beam. If necessary, this distance can be changed slightly, when detailing, to conform to other details. ~0 6 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 7 ' EH M- i/5 7. In order to decrease the unsup¬ ported length of the upper flange of the stringers, so that a higher working stress may be used, cross-struts, similar to those placed between the I beams in the highway bridge designed in Design of Plate Girders , Part 2, will be placed between the stringers at the center of the panels. The unsupported length will then be 10 feet, or 120 inches. The weight of these struts is so small that it will be neglected in calculating moments and shears. 8. Design of Sidewalk String¬ ers. —In B. S., Art. 98 (2), the live load is given as either 100 pounds per square foot or a road roller weighing 15 tons. The latter need not be con¬ sidered on the sidewalk. Since the 1 beams are 3 feet apart, the live load due to the uniform load is 300 pounds per linear foot on beam b, Fig. 5. The floor plank 2 inches thick weighs 2 X 4.5 = 9 pounds per square foot (see B. S ., Art. 97), and the part sup¬ ported by beam b is 3 X 9 = 27 pounds per linear foot. The spiking piece will be assumed to weigh 8 pounds per linear foot, and the I beam 20 pounds per linear foot. The total load sup¬ ported by the beam is, therefore, 300 + 27 + 8 + 20 = 355 pounds per linear foot, and the maximum bending mo¬ ment, since the floorbeams are 20 feet , , , . 355 X 20 X 20 . 0 center to center, is---X 12 = 213,000 inch-pounds 77 DESIGN OF A HIGHWAY TRUSS BRIDGE 7 If a 9-inch 21-pound I beam is used, the width of the top flange is, by Table XIV, 4.33 inches; and, since the unsup¬ ported length is 120 inches, the allowable working stress, according to B. S ., Art. 103 , is 120 20,000 - 200 X 4.33 = 14,460 pounds per square inch © © © *5 -//'-O 0 - © © § Fig. 6 The required value of section modulus is 213,000 -r- 14,460 = 14.73. As the section modulus of a 9-inch 21-pound I beam is greater than this (18.9, Table XIV), this beam will be used. An 8-inch 20.5-pound beam could have been used, but, as there is only .5 pound per foot difference between the two, and as the 9-inch beam is a standard beam, it is better to use the latter. The live load supported by beam c is somewhat less than that supported by b, as part of the width between beams c and d is occupied by the width of the truss. It is not customary to make allowance for this in the design of stringers, however, so that c will be made the same size as b. It is unnecessary to design the fascia girder. 'The thinnest allowable web, ~rs inch (B. X., Art. 112 ), and the smallest allowable angles, 2\ in. X 2i in. X ijs in. {B. S ., Art. 113 ) , will usually be found to give sufficient strength. These sections will be used in the present case. 7.Z5' © © § 275 ‘ •/ * /O'-O' 6 . 25 ' © © © *5 'V 5 qj oon , % vl •qj ooii. qi 0091 vl L Sv ? * §t $ 5Q •qj f 69 ii » I: ^ a Qi 7 * vi £ ^ 3$ O »-( &H -® s =A *Q to the floorbeam at each roadway stringer is 20 X 107.5 = 2,150 pounds (d, e , and /, Fig. 13). The dead load on each of the stringers g under the railway track was found in Art. 13 to be 180 pounds per linear foot; there¬ fore, the load that comes to the floorbeam at each of these string¬ ers is 20 X 180 = 3,600 pounds (g, Fig. 13). The dead load on the center stringer was found in Art. 15 to be 165 pounds per linear foot; therefore,, the load that comes on the floorbeam at the center is 20 X 165 = 3,300 pounds (/z, Fig. 13). 19. No rule or formula can be given by means of which the weight of brackets and floorbeams can be calculated. In every case it is necessary to estimate first, and then correct the estimate. An experienced designer will esti¬ mate closely the first time. In the present case, the weight of each bracket will be assumed to be 75 pounds per linear foot, and the weight of the floorbeam 125 pounds per linear foot. As the loading is symmetrical about the two supports A and B,i Fig. 13, each reaction is equal to one-half the sum of all the loads on the beam and brackets, together with one-half the weight of beam and bracket, or 17,694 pounds. Fig. 13 § 77 DESIGN OF A HIGHWAY TRUSS BRIDGE 15 shows the total dead load on the floorbeam and the dead¬ load reactions. 20. Dead-Toad Shears.—As a rule, it is sufficient to find the shears on the bracket and on the floorbeam at sec¬ tions close to the trusses. Considering the left truss A, Fig. 13, the shear on the bracket just to the left of A is negative and equal to 4,240 pounds, and that on the floorbeam just to the right of A is positive and equal to 13,450 pounds. If it is found necessary in the design of the web and the computation of rivet pitch to know the shears at other sec¬ tions, they can be computed later at sections between the stringer connections. 21. Dead-Toad Moments. —As a rule, it is sufficient to find the moment on the bracket where it connects to the truss, and on the floorbeam at the center of the bridge. The moment on the bracket where it connects to the truss is negative, and equal to 226,800 inch-pounds. The moment on the floorbeam at the center is positive, and equal to 1,029,500 inch-pounds. If it is found necessary in the design of the flanges to know the moments at other sections, they can be computed later at the sections where the stringers connect with the floorbeams. 22. Tive Toad. —The uniform live load is 100 pounds per square foot [ B . S., Art. 98 (2)]. According to B. S ., Ci Oi Ci so N ' jC / /7 n Y N f ... . "i-/ 1 " —H <- fo u •—► ' r ... ... 1 , J *- < U' ■■ 00 . IS Fig. 14 Art. 98 (5), the uniform load is to be taken as covering the entire floor except a width of 10 feet for the car track. In the design of floorbeams in a bridge that carries a street-car 16 DESIGN OF A HIGHWAY TRUSS BRIDGE 77 I 1 '5 05 * ^3 1 5 "Q2 OOOOZ Y !P track, it is customary to consider only the uniform load and the weight of a street car, and to ignore the effect of the road roller, as it will probably never cross the bridge at the same time as a street car. The portion of the width occupied by the truss and wheel- guard—in this case, 2 feet 6 inches at each side—is not assumed to sup¬ port any live load. The maximum load that comes to a floorbeam from the stringers marked g, Fig. 5, occurs when the loads in the panel ahead of and in the panel behind the floorbeam are located as represented in Fig. 14. The portion of the weight on one wheel that goes to a beam g has been found to be .7,692 pounds; then, the load on the floorbeam is 15,384 pounds. In like manner, the load on the floorbeam at the center stringer is found to be 9,230 pounds. As these loads come from the car, the stresses due to them must be increased to allow for impact and vibration; in the present case, the work will be much simplified and the results will be the same if the foregoing floorbeam loads are increased by the proper amount. The load, including impact and vibra¬ tion on the floorbeam at the center stringer, is, then, 9,230 + Ao X 9,230 = 12,000 pounds, and at each of the stringers marked g, Fig. 5,‘ 15,384 + Ao X 15,384 = 20,000 'Q2 OOO Zl V T ooooz\ vl § 'Vi 1. •V 4 ^ § I «Vi vl rH C5 M, pounds, as shown in Fig. 15. The increase in load on 77 DESIGN OF A HIGHWAY TRUSS BRIDGE 17 the leeward stringer is not considered in the design of the floorbeam. 23. The clear width of each sidewalk is 6 feet. Since the panels are 20 feet in length, the uniform load on each bracket is 20 X 100 = 2,000 pounds per linear foot for a distance of 6 feet, and extends to within 1 foot of the center line of the truss, as represented in Fig. 15. The clear width of roadway between wheel-guards is 25 feet. Allowing 10 feet for the load on the car track leaves a width of 15 feet, or 7.5 feet on each side of the track that is covered by the uniform load of 2,000 pounds per linear foot. Since the edge of the wheel-guard toward the roadway is 1 foot 6 inches from the center line of the truss, the uniform load will extend to within 1 foot 6 inches of the center line, as shown in Fig. 15. 24. It is unnecessary to compute the amount of load that comes to the floorbeam at each sidewalk and roadway stringer. If the moments and shears are found for the load¬ ing shown in Fig. 15, the results will be close enough to the actual values for all practical purposes. Since the loading f-6" 7 - 6 ‘ 2000 /£>. per foot • A • •Q © © © © © © © © © © © r-f *3 L 3"-* ’ ' ’ ’ /'-9^ 7 L 6‘ 2000tk perfoot /_'H 28-0‘ Fig. 16 is symmetrical, the reactions at A and B are each equal to one-half the total load on the beam, or, in this case, 53,000 pounds. 25. Dive-Load Shears. —The live-load shear on the bracket just to the left of A is negative and equal to 12,000 pounds. The live-load shear on the floorbeam just to the right of A is positive and equal to 41,000 pounds. IS DESIGN OF A HIGHWAY TRUSS BRIDGE § 77 26. Dive-Load Moments.—The live-load moment on the bracket is negative and greatest at A ; it is equal to 576,000 inch-pounds. The live-load moment on the floor- beam is greatest at the center, when there is no live load on the overhanging or sidewalk brackets, and there is a full load on the floorbeam, as shown in Fig. 16; it is positive and equal to 4,533,000 inch-pounds. 27. Design of Floorbeam Web.—The total shear on the floorbeam at the end is 13,450 + 41,000 — 54,450 pounds. It was decided in Art. 17 that a web 42 inches wide would be employed. It is good practice in this type of floor to use a thickness of web that will require no stiffeners between the stringer connections. Then, the smallest allowable unsupported distance is the clear distance between the flange angles, probably about 34 inches in this case, or the clear distance between stringer connections, which is about the same. The connection of stringer d, Fig. 5, to the floor- beam is so close to the end that no stiffeners will be required between the end of the floorbeam and this connection; hence, the intensity of shearing stress at the end of the floorbeam need not be computed. The total shear on the floorbeam in the next space, that is, between stringers d and must now be found, so that the allowable unsupported distance can be determined. The shear is practically constant from d to = 12 g • h 16,240 As this is greater than the maximum allowable pitch [6 inches ( B . A., Art. 130)], the latter will be used through¬ out the flanges. _ END FLOORBEAM AND BRACKET 32. Length and Depth.—It is generally advisable to leave the design of the end floorbeam until after the design of the trusses, as the depth of the floorbeam must be made to conform to details that depend on the design of the truss. For convenience, however, the end floorbeam will now be designed, the depth being taken as 34f inches, as found in a subsequent article. The length and distances along the beam and bracket are the same, as for the other floorbeams, as found in Art. 6. 33. Dead Load. —The portion of the dead load that goes to the end floorbeam from the stringers is one-half that shown in Fig. 13; the weight of bracket will be assumed to be 75 pounds per linear foot, and of the floorbeam 100 pounds per linear foot. The dead loads are shown in Fig. 18. 34. Dead-Load Shears and Moments. —The shear on the floorbeam at its connection to the truss is 7,250 pounds, and the moment at the center of the floorbeam is 547,000 inch- pounds. It is not necessary to compute the shear and bending moment on the bracket, which will have sufficient strength if S 77 DESIGN OF A HIGHWAY TRUSS BRIDGE * **5 si *qi 0991 % si •qi 00 8T J v I **> •qi 9LOT * S| •qi 9 lot V N •qi 9LOT 's I 1 v •qi fT96 n 3 < ■vr - «1 is , I, K 1 > l o> vl § 00 l-H d M fe •qi 009 LT J vl 'O *97 0090 r «*> vl <0 ’<7/ 00^ f ^ vl \ » vl •qi099 J V| •<2? 0## J V? <*> L, \ . I K ^ 1 < °* ‘QI 09Z98 y Fig. 19 24 DESIGN OF A HIGHWAY TRUSS BRIDGE 77 the web is made inch thick, and each flange is composed of two X 2i" X iV' angles. 35. Dive Load.— The live load on the end floorbeam, due to the uniform load on the floor, is one-half that shown in Fig. 16, or 1,000 pounds per linear foot. The live loads at the stringers g and h are more than one-half those shown in Fig. 16, and are greatest when the wheels are in the posi¬ tion shown in Fig. 12. The loads at g and h , including the allowance for impact and vibration, are, respectively, 17,500 and 10,500 pounds. The live loads on the end floorbeam are shown in Fig. 19. 36. Dive-Doad Shears and Moments. —The live-load shear on the end floorbeam at its connection with the truss is 30,250 pounds, and the bending moment at the center of the floorbeam (when there is no live load on the brackets) is 3,612,000 inch-pounds. 37. Design of Web. —The total shear on the floorbeam is 7,250 + 30,250 = 37,500 pounds. If a iVinch web is used, the area of cross-section is 34 X tY = 10.625 square inches, and the intensity of shearing stress is 37,500 -5- 10.625 = 3,530 pounds per square inch. Consulting Table XXXVI, it is seen that the allowable unsupported distance of the web is 27 inches. If 4-inch angles are used in the . flanges, the clear distance between them will be 26 inches, and no stiffeners will be required. 38. Design of Flanges.— The required area of M th flange will be found from the formula A = 5 he 8 the In the present case, th is 10.625 square inches, M is 547,000 -h 3,612,000 = 4,159,000 inch-pounds, s is 16,000 pounds per square inch, and h s is not known. In the design of the intermediate floorbeams, it was found that the value of h g was 1.18 1.21 = 2.39 inches less than the distance back to back of flange angles. For trial, it will be assumed that h e in the end floorbeam is the same amount less than the §77 DESIGN OF A HIGHWAY TRUSS BRIDGE 25 distance back to back of flange angles, or 34.25 — 2.39 = 31.86 inches. Then, 4,159,000 10.625 A = 8.16 - 1.33 16,000 X 31.86 8 = 6.83 square inches For the top flange, two 4" X 4" X i" angles will be tried. The gross area is 2 X 3.75 = 7.50 square inches, and the center of gravity is 1.18 inches from the back of the angles. For the bottom flange, two 4" X 4" X iV / angles will be tried. The net area of each angle, if one f-inch rivet hole is deducted, is 4.18 — .49 = 3.69 square inches, and of two angles, 7.38 square inches. The center of gravity is 1.21 inches from the back of the angles. The distance h g between the centers of gravity of the flanges is, then, 34.25 — 1.18 — 1.21 = 31.86 inches, as was assumed. The trial value of the required area found is, therefore, the actual flange area required. 39. Flange Rivets. —The pitch of rivets will be found Kh r by the formula p = V In the present case, the value of Fis 7,250 + 30,250 = 37,500 pounds, and h r = 34.25 - 4.5 = 29.75 inches. The rivets are in double shear and in bear¬ ing on the web ve inch thick; the latter value is the smaller, and is 5,160 pounds. Then, . 5,160 X 29.75 . nQ . . P = ~ 37.500 = 4 '° 9 lnCh6S This is very nearly the same as the pitch at the end of the intermediate floorbeam; hence, the same rivet spacing will be used for both. 40. Sidewalk Bracket.— There is no need to design the sidewalk bracket; the web will be made 34 inches deep at the truss and ins inch thick, and two 2| // X 2i" X i V' angles will be used in each flange. 26 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 DESIGN OF MAIN MEMBERS AND LATERAL SYSTEM STRESSES IN MAIN MEMBERS 41. Live-L oad Stresses. — According to A*. S. t Art. 98 (2), the uniform load on the floor to. be used in the design of the trusses is 80 pounds per square foot for bridges 100 feet in length, and 60 pounds per square foot for bridges 200 feet in length. The uniform load for the bridge under consideration (160 feet in length) is, therefore, 68 pounds per square foot. The total load on the two side¬ walks is 2 X 6 X 68 = 816 pounds per linear foot, and on the roadway, exclusive of a width of 10 feet for the car track [ B . A., Art. 98 (5)], is 2 X 7.5 X 68 = 1,020 pounds per linear foot, making the total live uniform load on the floor 1,020 + 816 = 1,836 pounds per linear foot. The panel loads are each 1,836 X 2 0 = 18 860 pounds> 2 and the reaction for each truss when fully loaded (neglecting the half-panel loads at each end) is 18,360 x7 = 64,260 pounds z The stresses caused in the chord members by this portion of the live load are determined as explained in Stresses in Bridge Trusses , Part 2, and are as follows (see Fig. 20): Member Stress, in Pounds cib,b c 47,600 (tension) c d 81,600 (tension) de 102,000 (tension) BC 81,600 (compression) CD 102,000 (compression) DE 108,800 (compression) §77 DESIGN OF A HIGHWAY TRUSS BRIDGE 27 The shears in the panels due to this portion of the load are as follows: Panel a b be c d de Shear, in Pounds 64,260 48,195 34,425 22,950 Panel e d' d'e' c' b f b’ a! Shear, in Pounds 13,770 6,885 2,295 0 42. According to B. S ., Art. 98 (4), the load on the car track for a span of 160 feet must be taken as 1,345 pounds per linear foot for the computation of the chord stresses. This gives a panel load for each truss of 13,450 pounds, and each reaction for a full load (neglecting the half-panel loads at the ends) is 13,450 X 7 A rj A — 5 — = 47,0/5 pounds The stresses caused in the chord members by this portion of the live load are as follows: Member Stress, in Pounds a b, dc 34,870 (tension) c d 59,780 (tension) d e 74,720 (tension) BC 59,780 (compression) CD 74,720 (compression) DE 79,700 (compression) 43. These stresses must be increased to allow for the effect of impact and vibration. The increase is computed by the following formula (B. S., Art. 99): 300 - L I = 1,000 X 5 135—13 28 DESIGN OF A HIGHWAY TRUSS BRIDGE S77 As chord stresses are under consideration, the entire span is loaded when they are greatest; then, L = 160, and, there¬ fore, 300 - 160 1,000 .US The total live-load chord stresses, in pounds, including impact and vibration, are as follows: Member Tension ab, be 47,600 + 34,870 + .14 X 34,870 = 87,350 cd 81,600 + 59,780 + .14 X 59,780 = 149,750 de 102,000 + 74,720 + .14 X 74,720 = 187,180 Compression B C 81,600 + 59,780 + .14 X 59,780 = 149,750 CD 102,000 + 74,720 + .14 X 74,720 = 187,180 D E 108,800 + 79,700 + .14 X 79,700 - 199,660 44. In computing the stresses caused in the web mem- bers by the load on the car track, the load at each floor- beam is to be taken as 1,600 p at ^5 floorbeams [ B . S., P Art. 98 (4,)]. In the present case, as the panel length p is 20 feet, this gives 1,600 X 20 = 32,000 pounds at each floor- beam, one-half of which, or 16,000 pounds, is the panel load on one truss; and the number of panel points that must be loaded is 100 -i- 20 = 5.* Those panel points must be loaded that will cause the desired stresses to be greatest. For example, to find the maximum shear caused by this por¬ tion of the load in the panel a b, the five joints b, c, d, e, and d' must be loaded; for the panel be, the five joints c, d,e, d', and c', etc. must be loaded. When there are less than five panel points to the right of a panel, all the joints to the right are loaded, and the remaining loads are considered to be off the *When -— does not give a whole number, the next larger whole P number must be used. Thus, if = 5.34, this should be called 6. P §77 DESIGN OP A HIGHWAY TRUSS BRIDGE 29 bridge. The are as follows: live-load positive shears due to this loading Panel Shear, in Pounds Panel Shear, in Pounds a b 50,000 e d’ 12,000 be 40,000 d’ c> 6,000 cd 30,000 c'b ' 2,000 de 20,000 b ' a f 0 45. These shears must be increased to allow for the effect of impact and vibration. As before, the increase is computed by the formula r _ 300 — L Q. ~ 1,000“ x As web stresses are under consideration, however, the length of track that is loaded to produce the maximum stresses is different for different members. It is customary to take for the loaded length in this case the distance from the right-hand end of the span up to the first panel load, or the panel load that is farthest to the left. Then, the loaded length for the panel ab is 140 feet; for the panel be , 120 feet; for the panel cd, 100 feet; for the panel de , 80 feet; for the panel ed\ 60 feet; for the panel d' c', 40 feet; and for the panel c' b ', 20 feet. The total live-load positive shears, in the different panels, including the allowances for impact and vibration, are as follows: Panel Shear, in r Pounds a b 64,260 + 50,000 + .16 X 50,000 - 122,260 b c 48,195 + 40,000 + .18 X 40,000 = 95,395 c d 34,425 + 30,000 + .20 X 30,000 - 70,425 de 22,950 + 20,000 + .22 X 20,000 = 47,350 ed f 13,770 + 12,000 f .24 X 12,000 - 28,650 d' c’ 6,885 + 6,000 + .26 X 6,000 = 14,445 c'b' 2,295 + 2,000 + .30 X 2,000 = 4,895 b'a' 0 46. Dead-Toad Stresses.—The dead load consists of the weight of the floor, including the weight of the stringers and floorbeams, and the weight of the trusses, including the 30 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 lateral system. The weight of the floor, as found in Art. 19 and illustrated in Fig. 13, is 17,694 pounds at each end of each floorbeam, that is, at each panel point. As the panels are 20 feet On length, this corresponds to a weight of 17,694 20 = 884.7, or nearly 900 pounds, per linear foot for one truss. The latter value will be used. Each panel load is then 900 X 20 = 18,000 pounds. The approximate weight w of one truss is given by the formula w (See B. A., Art. 243) jh 1 + 2 /o5 12 L \ 100 in which l — span; W — the total load per linear foot supported by the truss, exclusive of its own weight. The dead load supported by the truss was found in the preceding paragraph to be 900 pounds per linear foot. The total live load on the floor, due to the uniform load, was found in Art. 41 to be 1,836 pounds per linear foot, or 918 pounds per linear foot per truss. The load on the car track was found in Art. 42 to be 1,345 pounds per linear foot, or 672.5 pounds per linear foot per truss. Since the stresses due to the last-named load are increased to allow for impact and vibration, it is well, in calculating the load sup¬ ported by the truss, to increase the live load by the proper amount. This gives the total load coming to one truss from the car track, including allowance for impact and vibration, as 672.5 + .14 X 672.5 — 766.65 pounds per linear foot. Total load W supported by one truss is, then, 900 -f- 918 -f- 766.65 = 2,584.65 pounds per linear foot; therefore, 1 + 2 X (- 160 ~ 90 2,584.65 ,, w = ■ ■ *— —— X 12 \ ioo — 426.5 lb. per lin. ft. The panel load due to the weight of the truss is, then, 426.5 X 20 = 8,530 pounds, of which one-half, or 4,265 pounds, is assumed to be applied at the loaded chord, and one-half at the unloaded chord (see B. S., Art. 97). The entire weight of the floor is taken at the loaded chord. The lateral system will be assumed to cause a panel load of 1,000 pounds, one- half at each chord. Then, each panel load of the unloaded §77 DESIGN OF A HIGHWAY TRUSS BRIDGE 31 chord is 4,265 -f 500 = 4,765 pounds, and each panel load of the loaded chord is 4,765 + 18,000 = 22,765 pounds. The total dead panel load is 22,765 + 4,765 = 27,530 pounds, and each reaction is 96,355 pounds. The dead-load chord stresses are as follows: Member Stress, in Pounds a b , be 71,370 (tension) c d 122,360 (tension) de 152,940 (tension) BC 122,360 (compression) CD 152,940 (compression) DE 163,140 (compression) The dead-load shears are as follows: Panel Positive Shear, in Pounds Panel Negative Shear, in Pounds a b 96,355 cd 1 13,765 b c 68,825 d'e ' 41,295 c d 41,295 b' 68,825 d e 13,765 V a' 96,355 47. Wind Pressure.—In the design of trusses, it is customary to assume that the maximum wind pressure does not occur simultaneously with the maximum live load, so that the stresses caused in the chord members and end posts of the trusses by the wind pressure should not be added to the combined dead- and live-load stresses without some additional allowance. Practice varies as to the method of allowing for these wind stresses. It is frequently specified that, when they are less than 25 per cent, of the total com¬ bined dead- and live-load stresses, the wind stresses may be ignored; when greater, the members are designed for the sum of the maximum dead, live, and wind stresses, using working stresses 25 per cent, greater than those allowed for the dead- and live-load stresses alone. This method will be used here. As a rule, the members are first designed for the combined dead- and live-load stresses; the exposed area of the trusses and floor, the wind pressure on them, and the stresses due to this wind pressure are then calculated, and 32 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 the cross-sections of the members corrected, if necessary, to allow for the wind stresses. 48. longitudinal Force.—In this type of bridge, the longitudinal thrust of a suddenly stopping car is so dis¬ tributed throughout the length and width of the bridge by the rails, stringers, and floor plank that it will not be con¬ sidered. If the trusses were supported on a steel trestle, the force would be taken at the top of the bent. 49. Combined Stresses. —The combined dead- and live-load stresses in the chord members are as follows: Member Stress, in Pounds ab,bc 87,350 + 71,370 — 158,720 (tension) cd 149,750 + 122,360 = 272,110 (tension) de 187,180 + 152,940 = 340,120 (tension) B C 149,750 + 122,360 = 272,110 (compression) CD 187,180 + 152,940 = 340,120 (compression) DE 199,660 + 163,140 = 362,800 (compression) The combined dead- and live-load positive shears in the different panels are as follows: Panel Shear, in Pounds a b 122,260 + 96,355 = 218,615 be 95,395 + 68,825 - 164,220 cd 70,425 + 41,295 = 111,720 de 47,350 + 13,765 61,115 ed' 28,650 — 13,765 = 14,885 d'e* 14,445 — 41,295 = - 26,850 As the combined shear in the panel e d' comes out positive, a counter is required in that panel. As the combined shear in the panel d' c' comes out negative, no counter is required in that panel. The stresses in the diagonals can be found by multiplying the shears in the respective panels by esc H. The length of a diagonal is V20 2 + 27* = 33.6006, and esc H = 33.6006 -r 27 = 1.2445. Then, the stresses in the diagonals are as follows: §77 DESIGN OF A HIGHWAY TRUSS BRIDGE 33 Diagonal a B Be Cd De Ed'(dE) Stress, in Pounds 218,615 X 1.2445 = 272,100 (compression) 164,220 X 1.2445 - 204,400 (tension) 111,720 X 1.2445 = 139,000 (tension) 61,115 X 1.2445 = 76,100 (tension) 14,885 X 1.2445 = 18,500 (tension) The stresses in the verticals, except the hip vertical, are each equal to the sum of the shear in the panel to the right of the vertical and the dead load at the top joint (4,765 pounds, Art. 46). The stresses are as follows: Member Stress, in Pounds Cc 111,720 + 4,765 = 116,500 (compression) Dd 61,115 + 4,765 = 65,880 (compression) Ee 14,885 + 4,765 = 19,650 (compression) According to B. S ., Art. 98 (2), the stress in the hip ver¬ tical must be found from the same loading as the stress in the floorbeam. In Art. 19, the dead load on the truss from one floorbeam was found to be 17,694 pounds; and in Art. 24 the live load, including impact and vibration, was found to be 53,000 pounds. In addition, there is a dead load of 4,765 pounds at each panel point of the loaded chord, making the total stress in the hip vertical 17,694 + 53,000 + 4,765 = 75,459, or about 75,500 pounds, tension. The total dead- and live-load stresses in the members are shown in Fig. 21. 34 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 DESIGN OF MAIN MEMBERS 50. Bottom Cliord. —The bottom chord will be com- oosed of eyebars (see B. S ., Art. 139). The working stress in tension is given in B. S., Art. 103, as 16,000 pounds per square inch. Then, the required net areas of the bottom chord members are as follows: Member Required Net Area, in Square Inches ab, be 158,700 4- 16,000 = 9.92 cd 272,100 -f- 16,000 = 17.01 de 340,100 + 16,000 = 21.26 The required areas can be made up in a number of ways by using different widths of eyebars, and no fixed rule can be given for the width that should be u§ed. In general, how¬ ever, eyebars from 4 to 8 inches in width are preferable for chord members of highway-bridge trusses from 150 to 175 feet in length. The smallest thicknesses that can be used are given in Table XXX. The actual thickness should, in general, be not greater than about twice the minimum thickness, although it is better not to exceed the minimum thickness by a very large amount. The bending moments on the pins are as a rule less when thin eyebars are used than when thick bars are used. 1. Member de .—In the present case, four bars 5 in. X ItV in. will be used for the member de. The area of one bar of this size is 5.3125 square inches, and of four bars, 21.25 square inches, which is near enough to the required area. 2. Member c d .—For the member c rf, four bars 5 in. X i in. will be used. The area of one bar of this size is 4.375 square inches, and of four bars, 17.5 square inches, which is slightly greater than required. 3. Members ab and be .—For the members ab and cd , it is specified in B. S., Art. 139, that the bars composing them must be connected to each other by latticing; this is to make these members capable of resisting a small amount of com¬ pression, if for any reason, sugh as a heavy wind storm, the stresses in them are reversed. The method of connecting §77 DESIGN OF A HIGHWAY TRUSS BRIDGE 35 I them by latticing is shown in Fig. 22: an angle is riveted to the inside of each eyebar, and the outstanding legs are con¬ nected by single latticing. The area of each bar so connected must be reduced by the area of one rivet hole; f-inch rivets will be used. If two bars 6 in. X 1 in. are used, the gross area of each bar is 6 square inches, and the net area is 6 — .875 = 5.125 square inches; then, the area of two bars is 10.25 square inches. As this is slightly greater than the required area, these bars will be used for a b and c d. 51. Inclined Web Members.—The required net areas of the inclined web members, not including the end post, which will be considered later, are as follows: Member Required Net Area,, in Square Inches Be 204,400 4- 16,000 - 12.775 Cd 139,000 4- 16,000 = 8.69 De 76,100 4- 16,000 - 4.76 dE • 18,500 4- 16,000 = 1.16 Member Be. — -For Be, two bars 6 in. X li in. will be used. The area of one bar is 6.75 square inches, and of two bars, 13.5 square inches, which is sufficient. 2. Member Cd .—For Cd, two bars 5 in. X i in. will be used. The area of one bar is 4.375 square inches, and of two bars, 8.75 square inches, which is sufficient. 36 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 3. Member D e .—For De, two bars 3 in. X it in. will be used. The area of one bar is 2.4375 square inches, and of two bars, 4.875 square inches, which is sufficient. 4. Member dE .—The member dE is a counter, and, according to B. S ., Art. 141, cannot have a sectional area less than 2 square inches, although but 1.16 square inches is required by the stress. One bar 3 in. X t in. will be used; its area of cross-section is 2.25 square inches, which is greater than 2 square inches. 52. Vertical Web Members. —The form of cross-sec¬ tion that is best adapted to the compression web members is shown in Fig. 23. It is composed of two channels with the flanges pointing toward each other; the flanges are con¬ nected by tie-plates and lattice bars, as explained in Bridge Members and Details , Part 1. The size of channel that must be used for any member depends in general on the stress in the member, but for those verticals near the center of the span, in which the stress is comparatively small, other conditions frequently govern. For example, it is specified in B. S., Art. 105, that the value of - shall not exceed 100 r for main members. Since the length of each vertical from center to center of pins is 27 feet = 324 inches, the smallest allowable value of the radius of gyration r is 324 -f- 100 = 3.24 inches. Consulting Table XIII, it is seen that the smallest channel having a value of r greater than 3.24 inches is a 9-inch 13.25-pound channel. The web of this is only .23 inch thick, and, as it is specified in B. S ., Art. 112, that webs of channels shall be not less than'! inch thick, this channel cannot be used. A 9-inch 15-pound channel is found to be the smallest channel that can be used. Its radius of 77 DESIGN OF A HIGHWAY TRUSS BRIDGE 37 gyration is 3.4 inches, and - = — 95.3. The gross area r 3.4 of two channels is (Table XIII) 2 X 4.41 = 8.82 square inches. Consulting Table XXXV, the allowable working stress corre¬ sponding to a value of - of 95.3 is found to be 10,640 pounds r per square inch. Since the gross area of two channels is 8.82 square inches, the total compressive stress that can be resisted by a member 27 feet long and composed of two 9-inch 15-pound channels is 8.82 X 10,640 = 93,800 pounds. As this is greater than the stress in either D d or Ee , these channels will be used for these two members. As the stress in Cc (+ 116,500 pounds) is greater than 93,800 pounds, the two channels just considered are not large enough for this member, and larger or heavier channels must be used. The radius of gyration of each of the heavier 9-inch channels is less than the smallest allowable value, 3.24 inches, as found in the preceding paragraph, so that the next heavier 10-inch channel, 20 pounds, will be tried. The radius of gyration of a 10-inch 20-pound channel is given in Table XIII as 3.66 inches; then, - = = 88.5, and, by r 3.66 Table XXXV, the allowable working stress is 11,150 pounds per square inch. The gross area of two channels is (Table XIII) 2 X 5.88 = 11,76 square inches, and.the total stress that can be resisted by two channels is 11.76 X 11,150 = 131,100 pounds. As this is greater than the stress in Cc , these channels will be used for this member. 53. Hip Vertical. —The stress in Bb is 75,500 pounds, ten¬ sion, and, since the allowable working stress ( B . S., Art. 103) is 16,000 pounds per square inch, the required net area of the member is 75,500 -r- 16,000 = 4.7 square inches. The same form of cross-section can be used for this member as for the other verticals, but that shown in Fig. 24 is frequently used, and will be employed here. The outstanding leg should be at least equal in width to the outstanding leg of the floorbeam connection angle. According to B. S. f Art. 126, the 38 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 connection angles cannot be less than 3 inches wide. Then, 3" X 2a 7 ' X tw" angles will be tried, as they are the smallest that can be used. To get the net area of each angle, it will be sufficient to deduct the area of one hole for a f-inch rivet from each angle, that is, .27 square inch (Table XXVII). The gross area of one angle is (Table X) 1.62 square inches, and the net area, 1.62 — .27 — 1.35 square inches. Then, the area of the member B b, if composed of four angles, is 4 X 1.35 = 5.4 square inches, which is greater than the required value, and therefore sufficient. It will be found in connection with the details, however, that it is advisable to Fig. 24 use Si-inch angles for connecting the floorbeams to the verticals. For this reason, it is well to use four angles 3^ in. X 2i in. X ~iw in. for B b. 54. Width, of Verticals.—The width of the hip vertical is usually made about the same as the compression verticals. The channels that compose the latter are placed far enough apart so that the radius of gyration about an axis parallel to the web is at least equal to that about an axis perpendicular to the webs. The distances between the webs of two chan¬ nels that form a compression member, when the flanges are turned outwards, so that the radii of gyration in the two directions will be the same, is given in column 17, Table XIII. The distance ZX, Fig. 23, between the backs of the channels when the flanges are turned inwards or toward each other can be found by means of the formula D' =■ D + 4x in which D = distance found in column 17, Table XIII; jv = distance from back of channel to center of gravity, as found in column 12, Table XIII. 77 DESIGN OF A HIGHWAY TRUSS BRIDGE 39 For two 10-inch 20-pound channels, D' — 5.97 + 4 X .61 = 8.41 inches; and for two 9-inch 15-pound channels, D' = 5.49 + 4 X .59 = 7.85 inches. They will be placed 8^ inches apart in the present case. 55. Top Chord.— 'The form of cross-section shown in Fig. 25 is frequently used for the tpp chords and end posts of trusses. The pins are located near the center of gravity of the section, which is nearer the top than the bottom; and, in order to provide room between the pin and the top cover- plate to accommodate the eyebar heads, it is sometimes necessary to use deeper chord members than would otherwise be desirable. For this reason, this form of cross-section is not as desirable for light pin-connected trusses as the symmet¬ rical form shown in Fig. 26, in which the center line is located a little below the center of gravity. This form will be used 1 r TH r t J -- c -- L M 1J — c - --*■ • • . ...■'J w LI Fig. 25 Fig. 26 in the present case. The depth or width W should be at least great enough to allow the head of the largest eyebar to go inside. The largest eyebar that connects to the top chord is 6 inches wide. In Table XXX, the sizes of heads for 6-inch eyebars are given from 13* inches to 15i inches in diameter; a depth of 15 inches will be tried first. The distance c between webs depends on the width of the vertical members and on the arrangement of the ends of the members with respect to each other. It is customary to place the heads of the eyebars that form the main diagonals between the vertical members and the inside of the chord members. As the width of the vertical members is 8|- inches (Art. 54), the clear distance c between the webs of the top chord will be taken for the present as 12 inches. If neces¬ sary, this can be changed slightly when detailing. 40 DESIGN .OP A HIGHWAY TRUSS BRIDGE §77 In Bridge Mevibers and Details, Part 1, it was explained that, when, as in this case, the distance c is greater than I W , the least radius of gyration is approximately equal to 3 W, in this case 5 inches. Since the unsupported length of the top chord members is 240 inches, the approximate value of - is 240 -s- 5 = 48, and the allowable working stress r (Table XXXV) is 14,180 pounds per square inch. Then, the trial values of the required cross-section of the top chord members are as follows (see Fig. 21): Member Trial Value, in Square Inche^ BC 272,100 -r- 14,180 = 19.19 CD 340,100 -h 14,180 = 23.99 DE 362,800 14,180 - 25.59 The top chord is usually made in sections from 30 to 60 feet in length, which are spliced in the field. In the present case, the splices will be located near D and D', Fig. 20, and the chord will be composed of three parts. 1. Member B C .—For the member B C, the following sections will be tried: Square Inches Two web-plates 15 in. X A in.9.38 Four angles 3i in. X 3iin. X t in. . . . . . 9.92 Total. ' .19.30 The flange angles will be placed with the backs of their outstanding legs 15i inches apart, as shown in Fig. 26. The radius of gyration about axis X’ X through the center of gravity of the section is computed by the method explained in Bridge Members and Details , Part 1, and is found to be 5.67 inches; the value of - is 240 5.67 = 42.3; and r the allowable working stress is 14,550 pounds per square inch. Then, the corrected value of the required area of cross-section is 272,100 —■ 14,550 = 18.70 square inches. As this is so close to the area tried, the assumed section will be used for B C. 77 DESIGN OF A HIGHWAY TRUSS BRIDGE 41 2. Member CD .—As CD is a continuation of B C> all the simple sections of which B C is composed are also parts of CD , and any additional area required for CD is provided by adding plates. The flange angles are far enough apart to allow an 8-inch vertical plate between the vertical legs of the angles, as shown at a,a in Fig. 27. The following section will be tried for CD: Square Inches Two web-plates 15 in. X in. 9.3 8 Four angles 3i in. X 3i in. X fin. 9.9 2 Two vertical plates 8 in. X it in. 5.0 0 Total .2 4.3 0 The radius of gyration with reference to the axis X'X through the center of gravity of the section is found to be 5.16 inches; the value of - is r 240 -h 5.16 = 46.5; and the al¬ lowable working stress is 14,290 pounds per square inch. Then, the corrected value of the required area of cross-section is 340,100 -j- 14,290 = 23.80 square inches. As this is so close to the area tried, the assumed section will be used for CD. 3. Member D E .—For member D E, the same form will be used as for CD. The following section will be tried: Square Inches Two web-plates 15 in. X in. 9.3 8 Four angles 3i in. X 3i in. X I in. . 9.9 2 Two vertical plates 8 in. X in. 7.0 0 Total .2 6.3 0 The radius of gyration with reference to X' Xis 5.01 inches; the value of - is 240 -r- 5.01 = 47.9; and the allowable work- r ing stress is 14,170 pounds per square inch. Then, the corrected value of the required area of cross-section is m J Cen t er of Gravity J a Fig. 27 42 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 -362,800 4- 14,170 = 25.60 square inches. As this is so close to the area tried, the assumed section will be used for DE. 5G. End Post.—The end post will be made of the same general form as the top chords. The unsupported length of the end post is 33.6 feet, or 403.2 inches; the radius of gyration will be assumed equal to 5 inches. The value of the ratio - is, then, 403.2 4- 5 — 80.6, and the allowable r working stress is 11,760 pounds per square inch. There¬ fore, the trial value of the required area of cross-section is 272,100 4- 11,760 — 23.14 square inches. The same section will be used for this member as for the top chord mem¬ ber CD, unless it is necessary to revise the section later on account of the wind stresses. DESIGN OF LATERAL SYSTEM WIND STRESSES 57. Introductory Statement.-— Before designing the details, it is advisable to compute the wind stresses, so that, if necessary, the sections of the main members may be changed. The lateral system will also be designed at this time. 58. Wind Pressure.—The wind pressure will be taken as 50 pounds per square foot on twice the exposed area of one truss together with the floor {B. S., Art. 100). In calculating the exposed area of a member, it is customary to multiply its width by the distance between the centers of its connections. In tension members composed of eyebars, the width of the eyebar is used; in built-up members, 1 inch is added to the depth of the web or channel, or the width of angles, to allow for the latticing. The exposed widths of the members at one end of the truss are as follows (see Fig. 20): for a B , B C, CD, and D E, 16 inches; for B b, 7 inches; for Cc , 11 inches; for D d and Ee , 10 inches; for §77 DESIGN OF A HIGHWAY TRUSS BRIDGE 43 Ac, 6 inches; for Cd, 5 inches; for De and Ed, 3 inches; for a b and be, 6 inches; and for cd and de, 5 inches. Multiply¬ ing each of these widths by the length of the member, the exposed area of the upper chord of one truss is found to be 160 square feet; of the lower chord, 73.33 square feet; and of the web members, 333.3 square feet. Then, the wind pressure on twice the exposed area of one truss on the top chords is 2 X 160 X 50 = 16,000 pounds; on the bottom chords, 2 X 73.3 X 50 = 7,330 pounds; and on the web members, 2 X 333.3 X 50 = 33,330 pounds The exposed area of the floor is equal to the length of the span (160 feet) multiplied by the vertical distance from the highest portion of the floor (the edge of the sidewalk) to the bottom of the lowest I beam, about 3 feet, Fig. 5. It is, therefore, 160 X 3 = 480 square feet, and the wind pressure on it is 480 X 50 = 24,000 pounds. The wind pressure on the railings will be taken as 75 pounds per linear foot, or 75 X 160 = 12,000 pounds on the entire length of railings. 59. Upper Lateral Truss.— The upper lateral truss is shown in Fig. 28. It is customary to assume that the wind pressure on the top chord and one-half that on the web members—in this case, 16,000 + 33,330 ~ 2 = 32,665 pounds —is resisted by the top lateral truss. Since the top chord is 135—14 t 44 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 120 feet Ion", this corresponds to 32,665 -r- 120 — 272.2 pounds per linear foot, and the load per panel is 272.2 X 20 = 5,444pounds, as represented in Fig - . 28. For convenience, the entire panel load is assumed to be applied at the wind¬ ward side. There are also to be considered the two hall- panel loads of 2,722 pounds each at B and B'. The wind stresses in the upper lateral truss are shown in Fig. 29. 60 . Dower Lateral Truss.— The lower lateral truss is clearly shown in Fig. 30. It is customary to assume that the wind pressure on the railings, the floor, the bottom chord, and one-half that on the web, which in this case is 12,000 + 24,000 + 7,330 + 33,330 -s- 2 - 60,000 pounds, is resisted by the bottom lateral truss. As the bottom chord is 160 feet long, this corresponds to 60,000 ■— 160 = 375 pounds per linear foot. The panel load is 375 X 20 — 7,500 pounds, as shown in Fig. 30. For convenience, all the panel 77 DESIGN OF A HIGHWAY TRUSS BRIDGE 45 load is assumed to be applied at the windward side. The half-panel loads at a and a' will be neglected; they are transmitted directly to the abutments by the pedestals. The stresses in the lower lateral truss are shown in Fig. 31. 61. Portal and Transverse Frames.—The depth of the portal and transverse frames depends on the amount of room above the overhead clearance line, and this in turn depends on the position of the floor. In the present case, as the allowable distance from the top of the floor to the underneath clearance line is 6 feet (Art. 1), the floorbeams will be connected to the vertical posts above the pins, leaving the lower chord to project a short distance below the floor- beams. To clear the main diagonals, which will be outside the vertical posts, as- will be shown in Design of a High¬ way Truss Bridge , Part 2, the bottom of the floorbeam will be placed 1 foot above the center of the pins in the lower chord. It will be seen later, in connection with the details, that the top of the floor at the center of the bridge is 9 inches above the top of the floorbeam; as the floorbeam is 3 feet 64 inches deep, the top of the floor is 9 inches 3 feet 64 inches + 1 foot = 5 feet 34 inches above the center of the lower chord, and 27 feet — 5 feet 34 inches. = 21 feet 8f inches below the center of the top chord. As this is more than 20 feet, transverse frames are required at each panel point ( B . S., Art. 159). In B. S., Art. 94, the required headroom is specified as 15 feet. This leaves 6 feet 8f inches from the overhead 46 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 clearance line to the center of the top chord. The top laterals are connected to the top of the top chord about 8 inches from * - 28 L 0" -* Top of Top CLore/ I r * N * 1 S* > 5 oy Of d 1 % 1 i > 4 5 r% \ -a, -5 5 * ■> ' T of Lower C/?ora' Pig. 32 the center of the chord, making the total depth of the trans¬ verse frames about 6 feet 8f inches + 8 inches — 7 feet 4i inches. Curved brackets will be placed at each end of each transverse frame. Since the edges of the wheel-guards are 1 foot 8 inches from the center of the trusses, the brackets §77 DESIGN OF A HIGHWAY TRUSS BRIDGE 47 will extend 4 feet 6 inches from the center of the trusses (B. S ., Art. 94). They will also be made about 4 feet 6 inches in depth, as represented in Fig. 32, which is a cross- section of the bridge showing a type of frame and bracket adapted to the present case. The holes a , a in the curved brackets are simply for ornamentation. The same general type of frame will be used for the portal, as represented in Fig. 33. The brackets will extend 4 feet 6 inches out from the center of the trusses; the other dimensions shown in Fig. 33 are found from those shown in Fig. 32 by multiplying the latter by esc H (1.2445). 62. The force that acts on the portal is one-half the total wind pressure on the upper chord, or 16,330 pounds. Making use of the formulas given in Stresses in Bridge Trusses , Part 5, assuming the point of inflection of the end post to be half way between the bottom of the bracket and the bottom of the post, the required stresses are found to be as follows: Direct stress in end posts, Phi 16,330 X 24.63 b 28 Bending moment on end posts, P 14,360 pounds {/d — d — d') — 8,165 X 9.8 X 12 = 960,200 inch-pounds Direct stress in each web diagonal, P/d l = 16,330 X 24.63 X 11 .12 nbd 6x28x9.20 The stress in the top flange is = 2,894 pounds P/d P 2d + 2 P Id x ~ bd At the section opposite D, x = 4.5, and the stress is 16,330 X 24.6 3 16,330 _ 16,330 X 24.63 X 4.5 2 X 9.20 + 2 28 X 9.20 = 23,000 pounds, compression At the section opposite D\ x — 23.5, and the stress is 16,330 X 24.63 16,330 _ 16,330 X 24.63 X 23.5 2 X 9.20 + ” 2 28 X 9.20 = — 6,640 pounds, tension 48 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 The stress in the bottom flange at D is 16,330 X 24.63 _ 16,330 X 24. 6 3 X 4 ,5 2 X 9.20 28 X 9.20 = 14,800 pounds, tension The stress in the bottom flange at D' is 16,330 X 24.6 3 _ 16,330 X 24.63 X 2 3.5 2 x 9.20 28 X 9.20 = 14,800 pounds, compression There is also a direct stress caused in the bottom chord of the main truss by the direct stress in the end post; the former is found by multiplying the latter by cos H (20 -*• 33.6 = .5952); this gives 14,360 X .5952 = 8,550 pounds. This stress is tension on the leeward side, which is the only side that will be considered, since the compression on the wind¬ ward side simply decreases the tension in the chord. DESIGN OF MEMBERS . 63. Upper Lateral Truss. —Comparing the wind stresses in the top chord members with the combined dead- and live-load stresses, it is seen that the former are less than 25 per cent, of the latter, and need not be further considered (Art. 47). The diagonals are usually composed of one angle; in the present case, the required net area in the end panel is 16,730 -- 16,000 = 1.05 square inches. The net area of a 2k" X 2V' X iV 7 angle, after deducting one f-inch rivet hole, is 1.2 square inches, which is sufficient. If this angle is used, however, the bending stress due to its own weight is greater than the allowable intensity of bending stress; so that a larger angle must be used. For this reason one 3!" X X Te" angle will be used for each diagonal in each panel, the longer leg being placed vertical. 64. The rivets connecting the laterals to the connection plates are f-inch rivets, field driven, and in single shear, the value being 3,980 pounds. Then, the number of rivets required at each end, since the greatest stress is 16,730 pounds, is 16,730 3,980 = 4.2, or, say, 5 rivets. In the next § 77 DESIGN OF A HIGHWAY TRUSS BRIDGE 40 two panels, three rivets and one rivet are sufficient, respect¬ ively, but five rivets will be used, the same as in the end panel. 65. The maximum stress in a transverse strut is shown in Fig. 29 to be 13,610 pounds. It is customary to design the upper flange of the frames shown in Fig. 32 to resist this stress, and to make the lower flange the same size as the upper. In B. S., Art. 105, it is specified that no lateral member shall have a ratio of - greater than 120. In the r present case, since / is 28 feet, it is advisable to insert one 2i" X X iV' angle connecting the center of each trans¬ verse strut with the intersections of the diagonals, as shown by the dotted lines a a in Fig. 29, thereby reducing the value of l to 14 feet, or 168 inches. Then, the smallest allowable value of r is 168 -4- 120 = 1.4 inches. The top flange of the frame will be made of two angles, with their backs spaced -fe inch apart. Consulting Table XXVI, it is seen that the smallest angles that can be used are two angles 3 in. X in. X A in., for which the radius of gyration is 1.44 inches when the short legs are placed t 5 w inch apart. These angles are found to contain sufficient area. 66. Lower Lateral Truss.—In the design of the upper lateral truss, it was found that the smallest angle that can be used for a diagonal is 3i in. X 2i in. X t& in. The net area of such an angle, after deducting one t-inch rivet hole, is 1.51 square inches, and its value in tension is 1.51 X 16*000 — 24,160 pounds. This angle can be used in all but the end panel. The number of rivets required to connect the angle in the panel be at each end, since the stress is 23,040 pounds and the value of one rivet (f-inch rivet, field driven, and in single shear) is 3,980 pounds, is 23,040 -f- 3,980 = 5.8, or, say, 6 rivets. Six rivets will be used in the panel be , and also, to make the laterals alike, in the panels cd and de. The tension in the diagonal in the panel ab is 32,260 pounds, and the required net area is 32,260 16,000 = 2.02 square inches. One angle 4 in. X 3 in. X I in. will be used; 50 DESIGN OF A HIGHWAY TRUSS BRIDGE §77 the net area, after deducting one i-inch rivet hole, is 2.15 square inches. The number of rivets required to connect the lateral at each end, since the value of one rivet is 3,980 pounds, is 32,260 -f- 3,980 = 8.1, or, say, 9 rivets. Lug angles will be placed on all the lateral^. The end of the lateral that has just been designed is shown in Fig. 34. 67. It is unnecessary in this case to consider the wind stresses in the floorbeams, such as + 26,250 pounds in bb'\ they usually increase the intensity of stress in the floorbeam by a very small amount, which can be neglected. 68. In comparing the wind stresses in the lower chord members with those due to dead and live loads, it is necessary to add the stress in the leeward chord that is caused by the direct stress in the end post (8,550 pounds, tension) to the stresses in the leeward chord members due to their positions as mem¬ bers of the lower lateral stress. In the present case, none of the wind stresses in the lower / / L ^"x3"x§ r Fig. 34 chord members is so great as 25 per cent, of those due to combined dead and live loads, and so they will not be further considered. 69. Portal. —The direct stress in a lattice bar was found in Art. 62 to be 2,894 pounds. As the stress in any one of these bars is compression when the wind comes from one direction, and tension when it comes from the other, each must be designed for 2,894 -J- .8 X 2,894 = 5,209 pounds, tension and compression ( B . Art. 107). Flat bars are sometimes used for the diagonals, but when the depth of portal is greater than about 3 feet it is better to use small angles, as they give greater stiffness to the web. In the present case, one 2" X 2" X angle will be used for each diagonal. This is smaller than the smallest allowable angle, as given in B. S., Art. 113, but, since this may be considered 77 DESIGN OF A HIGHWAY TRUSS BRIDGE 51 an unimportant detail, it will be used. Each end will be connected by two f-inch rivets. The same-sized angles will be used for the lattice web of the transverse struts. The stress in the top flange was found in Art. G2 to be 23,000 pounds compression at one end, and 6,640 pounds tension at the other end. When the wind blows in the opposite direction, these stresses are reversed. The top flange should therefore be designed for 23,000 -f .8 X 6,640 = 28,310 pounds compression, and 6,640 + .8 X 23,000 * = 25,040 pounds tension. The center of the top flange will be held by one 2^" X 2i" X iV' angle in the same way as the transverse frames, as shown at b, Fig. 29, thereby reducing the unsupported length to 14 feet, or 168 inches. The smallest allowable value of r is, then, 168 — 120 = 1.4 inches. The smallest angles that can be used are two 3" X 2|" X !%" angles, for which the radius of gyration is 1.44 inches. These angles are found to have sufficient area, and so will be adopted. The stress for which the bottom flange should be designed is 14,800 + .8 X 14,800 = 26,600 pounds tension and com¬ pression. The same-sized angles will be used here as for the top flange, that is, two 3" X 2i" X tV" angles, with the 2o-inch legs t*? inch apart and the 3-inch legs outstanding. There is no need to design the bracket. It is customary to use a web about inch thick and angles on three sides about the same size as those in the flanges of the portal. 70. Effect of Wind Stresses on End Post. —On account of the bending moment on the end post, it is not sufficient to see that the direct stress due to the wind does not exceed 25 per cent, of the combined dead- and live-load stresses, but it must be seen in every case that the maximum intensity of stress due to the combined direct and bending stresses does not exceed the allowable working stress by more than 25 per cent. The formula used in the determination of the maximum intensity of stress s is S Me A + / ^ s DESIGN OF A HIGHWAY TRUSS BRIDGE K»> t)j < < in which S' = combined dead, live, and wind direct stresses; A = gross area of member; M = bending moment due to wind; c = distance of extreme fiber from center of end post, measured at right angles to truss; / = moment of inertia of cross-section of end post about an axis parallel to webs and half way between them. In the present case, A = 272,100 (Art. 49) + 14,360 (Art. 62) = 286,500 pounds, and M = 960,200 inch- pounds (Art. 62). For the section that was decided on in Art. 56, A — 24.3 square inches, c — 6 -f- -ft- + 3i = 9if inches, and / = 1,108. Then, the actual intensity of stress s is 286,500 . 960,200 X 9.81 on , nn , . , - — -1----- = 20,300 pounds per square inch 24.3 1,108 The allowable working stress in the end post, without considering the effect of the wind, is given in Art. 56 as 11,760 pounds per square inch. When the wind is consid¬ ered, the allowable working stress is 25 per cent, greater than this, or 11,760 -f- .25 X 11,760 = 14,700 pounds per square inch. As this is much less than the actual intensity of stress, 20,300 pounds per square inch, it is necessary to increase the section of the end post. The following section will be tried: Square Inches Two web-plates 15 in. X ft - in. ..1 3.1 2 Four angles 3i in. X 3a in. X i in.1 3.0 0 Two side plates 8 in. X i in. 8.0 0 Total.3 4.12 The least radius of gyration of this section is 5.03 inches, and since the unsupported length of the end post is 403.2 inches, the value of - is 80.2. By Table’ XXXV, the allowable work- r ing stress for combined dead- and live-load stresses is 11,790 pounds per square inch. Increasing this 25 per cent, gives 14,740 pounds per square inch as the allowable §77 DESIGN OF A HIGHWAY TRUSS BRIDGE 53 working stress for combined dead, live, and wind stresses. The clear distance between webs will be made Ilf inches, or i inch less than the top chord; this will bring the vertical legs of the flange angles Ilf + = 12f inches apart, the same as in the top chord, and the extreme fiber 6.31 + 3.5 = 9.81 inches, from the center of the section. ' Then, the moment of inertia about an axis parallel to the web is 1,553. Substituting the new values in the formula gives 286,500,960,200 x 9.81 1A ,- n , . , s = - — + -— ! ——tA-= 14,470 pounds per square inch 34.12 1,553 As this is less than the allowable working stress, 14,740 pounds per square inch, the section that has been tried is sufficient, and will be adopted. It is not necessary to com¬ pute the shearing stress caused in the end post by the wind / when the bending moment is considered. 71 . The sections that have been decided on for the vari¬ ous members of the vertical trusses are, for convenience, shown on the members in Fig. 35. DESIGN OF A HIGHWAY TRUSS BRIDGE (PART 2*) DETAILS FLOOR CONNECTIONS 1. Cross-Section. —The location of the top of the floor with respect to the top of the floorbeam is shown in Fig-. 1 . It is customary to keep the top flange of the floor beam as high as it is possible to have it without interfering with the floor plank. This can be done by so placing the stringers d , e , and / under the roadway that their top flanges are very near the top of the floorbeam, and making the nail¬ ing pieces i on top of the stringers d at the edge of the road¬ way of sufficient thickness to bring the bottom of the floor plank at the lowest pointy above the top of the floorbeam. In Fig. 1, the three roadway stringers d , e , and / are placed with their tops 1 inch below the top of the floorbeam' and the top of the wheel-guard is placed 1 foot above the top of the floorbeam. This makes the nailing piece on stringer d about 2 inches thick, and brings the bottom of the 3-inch plank at j about 1 inch above the top of the floorbeam. As the sidewalk plank rises -4 inch per foot toward the railing, the distance from the top of the sidewalk bracket (level with x 'The abbreviation B . .S’, stands for Bridge Specifications , to which .Section frequent reference is made in this and in some of the following Sections. All tables referred to are found in Bridge Tables , unless otherwise stated. COPYRIGHTED BY INTERNATIONAL TEXTBOOK COMPANY. ALL RIOHTS RESERVED §78 2 DESIGN OF A HIGHWAY TRUSS BRIDGE §78 8&pug the top of the floor- beam) to the top of the sidewalk plank at stringer c is about 12f inches; this makes it possible to place the sidewalk stringers b and c, 9 inches deep, on top of the top flange of the bracket, leaving room for a nailing piece about If inches in thick¬ ness on top of stringers. 2. The location of the tops of the stringers g and h under the car track depends on the 6 crown of the roadway and on the depth of rail and tie. It has been stated (Design of a Highway Truss Bridge , Part 1) that the wheel¬ er ^ guard would be placed 6 inches above the floor ^ at the edge of the road- <§ way, and the floor given a crown of 3 inches, bringing the top of the floor at the center of the bridge (level with the top of the rail) 3 inches below the wheel-guard; since the latter is 12 inches above the top of the floorbeam, the top • of the rail is 9 inches § 78 DESIGN OF A HIGHWAY TRUSS BRIDGE above the top of the floorbeam, as shown in Fig. 1. If the rails are 7 inches high, and the ties are framed to 5i inches in depth where they bear on the stringers, the tops of the stringers will be 7 -f- 52 = 12i inches below the top of the floor, and 12i — 9 = 3-2 inches below the top of the floorbeam. 3 . Connection of Fence and Fascia Girders to End of Bracket. —The detail of the connection of the fence and fascia girders to the end of the sidewalk bracket is shown in Fig. 2: (a) is the elevation of the fence and fascia girder, and (b) is the cross-section of fence and fascia girder and elevation of end of bracket. This is a very popular and ser¬ viceable type of fence; it consists of a number of flat bars a, a crossing each other at intervals to form a lattice web and held at their intersections by small ornamental castings b. Near the top and bottom of the fence are pairs of longitudinal angles c , c> about 2 in. X 2 in. X i in. The fence posts d, d are usually composed of two angles 2i in. X 2i in. X Ar in., and enclose the web e at the end of the bracket; the fence and fascia girders are riveted to the outstanding legs of the fence angles, as shown at (a). The rivets are f inch in diameter and spaced about 2J inches apart at /, the connection of the fence-post angles to the web of the bracket; above the bracket, the angles are connected to each other by means of a rivet every 10 or 12 inches, a washer^ of the same thickness as the web e being inserted between the angles at each rivet. The rivets h in the connection of the fascia girder to the fence post are spaced about 4i inches apart; those in the top and bottom flanges of the fascia girder are spaced about 6 inches apart. In locating the rivet holes for the connection of the fence posts to the bracket, the clear width of sidewalk can be made just 6 feet, as required. 4 . Connection of Sidewalk Stringers to Brackets. The sidewalk stringers b and c, Fig. 1, are made about 1 inch shorter than the panel length, so as to leave some clearar.ce between the ends of the stringers over the bracket. The ends of the stringers are connected to each other, as shown in Fig. 3, by means of two plates d, Ar inch thick, riveted to (N e £ 4 §78 DESIGN OF A HIGHWAY TRUSS BRIDGE 5 the webs of both beams. In the figure' (a) is the cross- section of the top flange of the bracket and the elevation of the adjacent ends of the stringers; (b) is a longitudi¬ nal section on the line cc, and shows the plan of the bottom flanges of the string¬ ers and the top flange of the bracket. The lower flanges of the stringers are riveted to the upper flanges of the brackets, as shown at e. Under each sidewalk string¬ er, a pair of stiffener angles /, 2i in. X 2i in. X in., and about 15 inches long, are riveted to the web of the bracket. The reaction from the stringers is so small that it is not necessary to calcu¬ late the number of rivets required in the stiffener; it is customary to place five or six rivets in each. 5. Connection of Roadway Stringers to Floorbeam. The roadway stringers d,e, and /, Fig. 1, are riveted to the web of the floorbeam by means of connection angles d, Fig. 4; and short shelf angles e, e, and stiffeners /, / are placed under them. In Fig. 4, (a) is the cross-section of a part of the top of the floorbeam and an elevation of the ends of two stringers resting on the shelf angles; (b) is a longi¬ tudinal section on cc and a plan of the bottom flanges of the stringers. The upper parts of the- stringers z, i are cut back to clear the vertical legs of the floorbeam flange angles j, and the connection angles d are placed below the flange angles. With this kind of connection, it is customary to assume that the rivets in the connection angles transmit the entire 135—15 6 DESIGN OF A HIGHWAY TRUSS BRIDGE §78 reaction (end shear on the stringer) to the floorbeam, and to ignore the supporting effect of the shelf angles and stiff¬ eners; the shelf angles and stiffeners, however, add consider¬ ably to the strength of the connections, and should always be inserted when there is room for them under the stringers. 6. In Design of a Highway Truss Bridge , Part 1, it was found that the maxi¬ mum end shear in one of the roadway string¬ ers is 8,325 pounds. The rivets g, Fig. 4, that connect the an¬ gles to the webs of Fig. 4 the stringers are 1 inch in diameter, shop driven, in double shear, and in bearing both on the angles and on the web of the I beam, which has a thickness of .45 inch (Table XIV). The bearing value of the web is the smallest; as the rivets are t inch in diameter and the allowable bearing stress is 22,000 pounds per square inch ( B . S., Art. 103), the value is f X .45 X 22,000 = 7,425 pounds. Then, the required number of rivets is 8,325 7,425 = 1.1; three rivets will be used to connect each pair of connection angles to the end of the stringer {B. S ., Art. 155). To find the number of rivets h required to connect the connection angles to the web of the floorbeam, two cases must be considered. These rivets are field driven, and in bearing both on the connection angles and on the web of the floorbeam. When the load in one panel is"considered, the amount of stress transmitted to the floorbeam is 8,325 pounds, and the rivets are in single shear. The values of the rivets for the allowable stresses given in B. S ., 78 DESIGN OF A HIGHWAY TRUSS BRIDGE 7 Art. 103 , are given in Table XXXIX. The value in single shear is the smallest, and is 3,980 pounds. Then, the required number of rivets for this condition is 8,325 -f- 3,980 = 2.1. When the load in two adjacent panels is considered, the amount of stress transmitted to the floorbeam is 2 X 8,325 pounds = 16,650 pounds, and the rivets are in double shear. In this case, however, the bearing value of a rivet on the f-inch web of the floorbeam is the smallest value, and is 5,060 pounds. Then, the required number of rivets in this case is 16,650 -f- 5,060 = 3.3. The next larger even number will be used, in this case four. According to B. S ., Art. 126, the connection angles cannot be smaller than 3 in. X 3 in. X tV in. If a leg 3 inches wide is placed next to the web of the stringer, it will be impossible to get three rivets in it without spacing them too close together; so a leg 5 inches wide will be used, and the connection angles will be made 5 in. X 3 in. X h in. The shelf angles will be made 3 in. X 3 in. X "re in., and the stiff¬ eners will be made 2i in. X 2i in. X in. These are usually put in according to the judgment of the designer, and not by rule. 7. Connection of Stringers Under Railway Track. The stringers marked g in Fig. 1 are connected to the floor- beam in the same way as those just considered. The connec¬ tion is shown in Fig. 5, in which ( a) is the cross-section of the upper part of the floorbeam, and shows the elevation of the ends of two stringers, resting on shelf angles d, d; (b) is a cross-section of one of the stringers on the section cc , and shows the elevation of a part of the floor- beam. The stringers are cut back at e to clear the top flange angles of the floorbeam. In Design of a Highway Truss Bridge , Part 1, it was found that the maximum end shear in the stringer g is 21,610 pounds. The web of the I beam is .46 inch thick, and the value of one f-mch rivet that con¬ nects the angles to the stringer is .75 X .46 X 22,000 = 7,590 pounds. The required number of rivets is, then, 21,610 -j- 7,590 = 2.8, or, say, 3. 8 DESIGN OF A HIGHWAY TRUSS BRIDGE §78 To determine the number of rivets / required to connect the connection angles to the web of the floorbeam, two cases must be considered. These rivets are field driven, and in bearing both on the connection angle and on the web of the floorbeam. When the load in one panel is considered, the amount of stress transmitted to the floorbeam is 21,610 pounds, and the rivets are in single shear (value = 3,980 pounds). The required number of rivets for this condition is 21,610 -4- 3,980 = 5.4, or, say, 6. When the load in two adjacent panels is considered, it is necessary to compute the amount of load that is transmitted to the floor- beam. This is found in the same way as in Design of a Highway Truss Bridge , Part 1, except that, since stringer connections are under consideration, it is necessary to allow for the increase due to the overturning effect of the wind. The amount of dead load is 3,600 pounds, and of live load 15,384 pounds. The allowance for impact and vibration is 1-0“ X 15,384 = 4,615 pounds, and for the overturning effect of the wind, 2,308 pounds. Then, the total load is 25,907 pounds. The value of a rivet in bearing on the f-inch floor- beam web (5,060 pounds) is the smallest; and the required 78 DESIGN OF A HIGHWAY TRUSS BRIDGE 9 number of rivets for this condition is 25,907 -r 5,060 — 5.1, or, say, 6, as before. It is well, when possible, to add one or two rivets to the required number in stringers under railway tracks, and to use good-sized connection angles. For this reason, five and eight rivets, respectively, will be used, as shown in Fig. 5, and the connection angles will be made 32 in. X 32 in. X 2 in., which is slightly larger than the smallest allowable connection angle (A. S., Art. 126). 8. Connection of Center Stringer. —The same .con¬ nection will be used for the stringer marked h, Fig. 1, as for^, except that in the connection angles only four rivets will be driven in the legs that are in contact with the stringers, and six rivets in the legs that are in contact with the floorbeam web. 9. Connection of Bracket and Floorbeam to Truss. The connection of the bracket and that of the floorbeam to the truss are shown in Fig. 6, in which (a) is an elevation of the ends' of the bracket and floorbeam that connect to the truss; ( b ) is a cross-section on D D; and (e) is a top view of the bracket and doorbeam. The floorbeam is connected to one side of the vertical post EE, and the bracket to the other. When, as in this case, the vertical post consists of two channels, a diaphragm /, consisting of a web and two angles at each side, is riveted between the channels. The leg of the angle adjacent to the web of the diaphragm is made 2 \ inches wide; the other is made the same width as the outstanding leg of the floorbeam connection angle. 10. The end shear on the floorbeam was found in Design of a Highway Truss Bridge , Part 1, to be 54,450 pounds. The value of one of the rivets^ that connect the connection angle h to the web of the floorbeam (f-inch rivet, shop driven, in bearing on f-inch web) is 6,190 pounds. Then, the required number is 54,450 -r- 6,190 = 8.8, or, say, 9. The value of one of the rivets i that connect the angles h to the truss (f-inch rivet, field driven, in single shear) is 3,980 pounds. Then, the required number is 54,450 -f- 3,980 = 13.7, or, say 14. In the leg of the connection angle adjacent to the web, ten rivets g will be used; in the other leg, a rivet i will be 10 Fig. 78 DESIGN OF A HIGHWAY TRUSS BRIDGE 11 placed half way between each two of the former, making nine rivets in each of the outstanding legs. It is customary, as in this case, to have a few extra rivets. For the floorbeam connection angles, angles 3i in. X 3i in. X i in. will be used. 11. The shear on the bracket where it connects to the truss is 16,240 pounds ( Design of a Highway Truss Bridge , Part 1). The number of rivets required in the leg of the angle / that is adjacent to the web of the bracket is 3.1, and in the outstanding legs, 4.1. The rivets in these connection angles, which will be made 3i in. X 3 in. X ■ft in., will be spaced as far apart as allowable, that is, 6 inches apart, thereby bringing seven rivets through the web and twelve through the vertical post. , 12. Connection of Bracket to Floorbeam. —To assist in making the bracket and floorbeam act as a single beam, the top flanges should be connected to each other. A good method of connecting them, when the bracket and floorbeam are attached to different sides of a vertical post, is shown in Fig. 6. Two angles k , k , the same size as those in the top flange of the bracket, are placed close to the vertical post, one angle on each side, and their ends are connected to the ends of the bracket and floorbeam flange angles by means of plates /, /. The stress transmitted by the angles k is found by divi¬ ding the greatest negative bending moment on the bracket by the depth of the bracket at the truss. In the present case, the stress is 802,800 -f- 42 = 19,100 pounds, tension, and, as there are two angles, each transmits 9,050 pounds. The value of one of the rivets m that connect the end of an angle k to the plate /, Fig. 6 {c) (f-inch rivet, field driven, in single shear) is 3,980 pounds; then, the required number of rivets is 9,050 4- 3,980 = 2.3, or, say, 3. The same number is used to connect the other ends of the angles and to connect the plates to the top flanges at n and o. The required thickness t of the plate / is given, approxi¬ mately, by the formula 4 Sd ~ sw' \ \ 12 DESIGN OF A HIGHWAY TRUSS BRIDGE 8 78 in which A = stress transmitted by one angle k ; d = width of widest vertical member to which floorbeams are connected—that is, the clear distance between the angles k\ s = working stress in bending; w = width of plate /, measured along bracket. The width w is controlled by the rivet spacing in the plate; in the present case, it is 8 inches. Then, 4 X 9,050 X 10 ~ 16,000 X 8 2 .354, or, say, f inch The angles k and plates / are frequently omitted, and the rivets p that connect the bracket to the vertical post are depended on to transmit the stress in the bracket. This construction is not recommended, as it tends to overstrain the rivets and allow the brackets to sag at the ends. The bottom flanges of the bracket and floorbeam are not connected, but are made to bear against the outsides of the vertical posts, the stress being transmitted from one to the other by means of angles r that are riveted to the bottom of the diaphragm. 13 . Connection of End Floorbeam to Truss.—As there is no vertical member at the end joint, the end floor- beam cannot be connected to the truss in the same way as the other floorbeams. It will be supported on a chair riveted to the end post. The detail of the chair will be treated later, in connection with the design of the pin at the joint a. The portion of the end floorbeam directly over the truss is shown in Fig. 7, in which ( a ) is the elevation and ( b ) the cross-sec¬ tion of the floorbeam; the latter view shows also the relative location of floorbeam, end post, and bottom chord. In order that the bottom flange angles of the floorbeam shall not interfere with the top flange angles of the end post at e , it is necessary to place the bottom of the floorbeam 1 foot 8 inches above the center line of the bottom chord, making the floorbeam 8 inches shallower than the intermediate floor- beams, that is, 34i inches, as assumed in Design of a High¬ way Truss Bridge , Part 1. §78 DESIGN OF A HIGHWAY TRUSS BRIDGE 13 Bottom Chore/ 14 DESIGN OF A HIGHWAY TRUSS BRIDGE §78 The web of the bracket and floorbeam is in one piece extending the full length. Plates c having the same area as the flange angles of the bracket are riveted to the top and bottom flanges to splice the bracket flange angles to the floorbeam flange angles. Stiffeners d are riveted to the web where the beam rests on the chair at the truss. The stringers are connected to the end floorbeam and brackets in the same way as to the other floorbeams and brackets, except that there is a shelf angle and stiffener on but one side of the web. SPLICES 14. The only splices that are necessary are in the top chord. In Design of a Highway Truss Bridge , Part 1, it was decided that the splices would be located in the panels CD and D' C' close to the joints D and D\ respectively. The form of cross-section used for the member CD is shown in Fig. 27 of the Section just referred to. This member will be f J SI Q -—-S-■-1 ^ • • • j Q G Q G O o G — • • G G • • • ! G G Q °\ Q • • ] G G G Q • • • | G O G G VssS' vsrJ' VSSS' \ c (a) Fig. 8 spliced by means of four bars 3i inches wide riveted to the outstanding legs of the flange angles, and by vertical splice plates inside and outside of the section, as shown in Fig. 8. The following splice plates will be used: Square Inches Four plates 3^ in. X A in., gross area = 6.1 2 5 Two plates 15 in. X A in., gross area = 9.3 7 5 Two plates 14 in. X Ain., gross area = 8.7 5 Total gross area = 2 4.2 5 0 §78 DESIGN OF A HIGHWAY TRUSS BRIDGE 15 This section is slightly less than that of CD, blit, as it is so little less, and as it is greater than the required area found for CD , these splice plates will be used. The rivets in the splice will be made i inch in diameter. Each splice plate must have sufficient rivets on each side of the splice to transmit its stress to and from the ends of the members. It is customary to rivet the splice plates to the end of one member in the shop, and connect the other end in the field. The required number of field rivets in a splice plate is found by dividing the product of the area of the plate and the work¬ ing stress by the value of one rivet. In the present case, the working stress in CD, as found in Part 1, is 14,290 pounds per square inch. The smallest value of a rivet in this case is its value in single shear, or 5,410 pounds. Then,,the number of field rivets required to connect each splice plate is as follows: ~ i * oi • v/ 7 • 1.531 X 14,290 a nz ' a One plate Sf in. X tg in.,- „ J 4 362800 E in Fig. 10 {e ). The dead panel load of 4,765 pounds is dis¬ tributed over the pin by various members, but will for convenience be assumed to be applied at the centers of the eyebars. The diameter of the pin will be assumed to be 4 inches. Then, since the total stress in the vertical is 116,500 pounds, and the working stress in bearing on the pin is 22,000 pounds per square inch, the required thickness of bearing for Cc is 116,500 4 X 22,000 — 1.324 inches The web of a 10-inch 20-pound channel (member Cc) is .38 inch thick, so that the thickness to be made up by pin plates is 1.32 — 2 X .38 = .56 inch, or .28 inch on each channel. Two pin plates e, e, Fig. 10 («) and (b ), -re inch thick, will be used. Then, the thickness of bearing for each channel will be .38 4 .31 = .69 inch, the center of which is .69 -i- 2 = .345 inch from the back of the channel. The §78 DESIGN OF A HIGHWAY TRUSS BRIDGE 21 centers of bearings for the two sides of the vertical member are, then, 8.5 — .345 — .345 = 7.81 inches apart, as shown at h and h' in Fig. 10 (c ). The required thickness of bearing of the top chord is 82,700 4 X 22,000 = .94 inch or .94 -f- 2 = .47 inch on each side. If the 8" X tV' side plates that form part of the member CD are continued a short distance beyond C, as shown at g, Fig. 10 («), the thickness of bearing on each side will be' f inch, which is sufficient, and the centers of bearing for the two sides of the chord will be 12 + .3125 -f .3125 = 12.625 inches apart, as shown in Fig. 10 (d). As the eyebars are 9f inches apart and 1 inch thick, the distance between their centers is 10i inches, as shown in Fig. 10 (c) and (d). Assuming that the stress in each half of a member acts as a concentrated load at its center of bearing, the horizontal forces that act on the pin at C are shown in Fig. 10 (d) and the vertical forces in Fig. 10 (c). It will be observed that the bending moment at any point between h and h! , Fig. 10 {c), is equal to the moment of the couple formed by the two equal forces at either end of the pin. The lever arm of this couple is i X (10.25 - 7.81) = 2.44 -s- 2 = 1.22 inches The moment is, therefore, 58,250 X 1.22 = 71,065 inch-pounds (== My) It is evident that the moment at the left of h or at the right of h' is less than this. Similarly, the bending moment at any point between g and^', Fig. 10 (d) , is 41,350 X -- — = 49,100 inch-pounds (= M H ) & For the resultant moment we have, therefore, M — V71,065* + 49,100 2 = 86,400 inch-pounds Consulting Table XLI, since the working stress in bend¬ ing on the pin is 22,000 pounds per square inch, it is seen that a pin 3a inches in diameter is the smallest that has 135—16 22 DESIGN OF A HIGHWAY TRUSS BRIDGE §78 the required value of resisting- moment. The various steps will now be repeated, using 3i inches for the diameter of the pin. The required thickness of bearing for the vertical is now 116,500 3.5 X 22,000 — 1.513 inches Two |-inch pin plates will be used, making the thickness of bearing on each side .38 + .38 = .76 inch, and the distance between the centers of bearings 7.74 inches, as shown in Fig. 12. The bending moment due to the verti¬ cal forces is 'now 58,250 X 1.255 - 73,100 inch- pounds; that due to the horizontal forces re¬ mains the same as be¬ fore, since the thickness of bearing of top chord used before is sufficient. Then, the corrected value of the maximum bending moment on the pin is V^OTOO 1 + 73,100 2 — 88,100 inch-pounds A pin Sk inches in diameter has a sufficient resisting moment, and will be used. 20. The total width or thickness of bearing on each side of the vertical Cc, Fig. 10, is .76 inch. If the stress is assumed to be evenly distributed over this width, then, since the thickness of the pin plate is t inch, the amount of stress transmitted by each plate is X 58,250 = 29,125 pounds The rivets that connect the pin plates to the channel are i inch in diameter and shop driven; the value in single shear, 4,860 pounds, is the smaller. The number of rivets required to transmit the stress in the pin plate to the channel is, therefore, 29,125 -=- 4,860 = 6. In addition to the required number, which will be placed below the pin, it is customary to put a few rivets above the pin to hold the pin plates firmly to the web of the channel, as shown in Fig. 13. Since the i §78 DESIGN OP A HIGHWAY TRUSS BRIDGE 23 distance c (Table XIII) for a 10-inch 20-pound channel is 8i inches, the pin plates will be made 8 inches wide. The section of the top chord is decreased in area by the pinhole, but in this case this is more than made up by the 8-inch plates on the sides. 21 . Pins and Pin Plates at Joints D and E .— The required sizes of the pins at D and E, Fig. 9, can be found by proceeding in the same way as for joint C. In the present case, and almost in¬ variably in the kind of truss under consideration, they both work out smaller than that required at C, but for the sake of uniformity they are made the same size as at C. Pin plates a , Fig. 14, 7 in. X A in., with four rivets below and two above the pinhole will be riveted to the insides of the channels of D d and Ee\ no pin plates are required on Ee , but it is cus¬ tomary in the best practice to provide at least one plate. The upper chord is decreased in section by the pinholes at D and E\ to restore the lost section, vertical plates a , Fig. 15, will be riveted to the insides of the webs. As iio ©i io oii !! O 1 ; O i © 1 Q !Q Q |! !| © « i 11 ! Q ©!! fit© ©J ! \ © / t \_ / i i t i L of 1 1 1 1 1 1 «- - _ 1 Fig. 13 Fig. 14 ! Q i © © © © 1 © ! O © © © © !© / i © o © © "a 9 . © ! Q Q> i © © © ! © © © © © L i Fig. 15 the thickness of metal at D and E through which the pin passes is f inch on each side of the chord, the lost section is X f = 2.625 square inches/ A plate 14 in. X f in. will be added on each side; the net width, deducting the diameter of the pinhole, is '10i inches, and the area is 3.94 square 24 DESIGN OF A HIGHWAY TRUSS BRIDGE §78 inches. This is more than is necessary, but the plate is made thicker than required, on account of the fact that at D the rivets connecting the plate to the web will have to be countersunk on the inside, to make room for the eyebars composing De, and that the depth of a countersunk head of a |-inch rivet is f inch (Table XIX). Hence, thinner plate cannot be used. The working stress in D E is 14,170 pounds per square inch (see Design of a Highway Truss Bridge , Part 1); and, since the section is decreased 2.625 square inches on each side, the amount of stress that must be transmitted by each of these reinforcing plates is 2.625 X 14,170 = 37,200 pounds. The rivets that connect these plates to the web of the chord are f inch in diameter and shop driven; the value in single shear, 4,860 pounds, is the smaller. The number of rivets required in each plate on each side of the pin is, then, 37,200 -r- 4,860 = 7.7, or, say, 8. The countersunk rivets in these plates are not considered as transmitting any stress (B.S., Art. 109 ). 268000 *r 22. Pin and Pin Plates at Joint B. —The stresses in the members that meet at B are shown in Fig. 11. They are the maximum stresses, and are not simultaneous; the stress in B b is greatest when there is a full panel load at b , those in a B and B C are greatest when the truss is fully loaded, and that in Be is greatest when the joints from c to the right are loaded. It is customary, in the design of the pin at the hip joint B , to assume that the stresses in a B and B b are maximum at the same time, and then, using the stresses in these members shown in Fig. 11, to find by the method of joints the simultaneous stresses in B C and Be, using the equations 2l X — 0 and 2' Y = 0. On this assumption, the simultaneous forces that act on the pin are shown in Fig. 16; it is unnecessary to consider the dead panel load at B. Fig. 16 §78 DESIGN OF A HIGHWAY TRUSS BRIDGE 25 The customary arrangement of members at joint B is shown in Fig. 17, in which (a) is the elevation of the joint; ( b ) is the cross-section of the top chord and side elevation of the pin, showing the arrangement of members; and {c) and ( d) are half top views and half bottom views of the top chord and end post, respectively. The top chord and end posts are cut off on the line ee, leaving about i inch clear¬ ance between them; the flange angles and webs of the two members are almost the same distance from the center of the truss, and the members bear against opposite sides of the pin. Pin plates / are riveted to the inside of the end post and extend beyond the pin up into the top chord; pin plates g are riveted to the outside of the top chord, and extend beyond the pin, enclosing the end of the end post. The eyebars b, b that form the diagonal Be are placed inside the chord and the end post, as close as possible to the inside pin plates / of the end post, the rivets in this plate being countersunk on the inside of the plate to allow the eyebar to get close to it. The hip vertical i is attached to the pin by means of hanger plates /,/ that are as close as possible to the inside of the eyebars. The plates k and / are tie-plates; the adjacent ends of the top tie-plates are covered by the bent plate m that is riveted to the end of the top chord. The two angles shown in cross-section at n are the top flange angles of the portal; it is customary to place them above and to continue them across the top chord, as shown, connecting them to it by means of the bent plates o and p. 23. In the discussion of joint C, the clear distance between the webs of the top chord was found to be 12 inches; it is customary to have the distance at B the same. Since the webs are Te inch thick, the distance between the vertical legs of the flange angles is 12f inches. The flange angles of the end post will also be placed 12f inches apart. Since the webs of the end post are tV inch thick, the clear distance between them will be Ilf inches. The required diameter of the pin, found by trial, is 4f inches. This value will now be verified. The maximum 78 DESIGN OF A HIGHWAY TRUSS BRIDGE 27 stress in aB and B C being- 272,100 pounds, and the working stress in bearing 22,000 pounds per square inch, the required thickness of bearing for each of these members is 272,100 4.75 X 22,000 — 2.6 inches, or 1.3 inches on each side of the member. The webs of the end posts are tV inch thick, and the side plates 2 inch thick. Pin plates /,/, Fig. 17, | inch thick, will be riveted to the inside, making a total thickness of Tre inches bearing on each side. As the clear distance between the webs is Ilf inches, the clear distance between the plates / is 11 inches, and the distance between the cen¬ ters of the bearings is 12 ts inches. Since the pin plates are f inch thick, and the total width of bearing on each side is lxe inches, the amount of stress transmitted by one pin plate is — 1 -- -— X 136,050 = 38,900 pounds, 1.3125 and the number of rivets required to connect the plate to the end post is 38,900 -r- 4,860 = 8. The webs of the top chord are re inch thick. Pin plates y, 8 in. X t in., will be placed on the outside of the member between the vertical legs of the flange angles; pin plates r, 14 in. X i in., will be placed outside of the vertical legs of the angles and the plates q\ and pin plates g , 14 in. X I in., will be placed outside of the plates r and enclosing the end of the end post. This makes the total thickness of bearing at each side li^e inches, and the distance between the centers of bearings, 131^6 inches. The numbers of rivets required to connect these plates to the top chord are: Plate, 14 in., X t in., Plate, 14 in., X i in., .375 1.3125 X 136,050 .25 1.3125 4,860 X 136,050 4,130 = 8 rivets = 6.3, or, say, 7 rivets .375 1.3125 X 136,050 8 rivets Plate, 8 in. X f in., 4,860 28 DESIGN OF A HIGHWAY TRUSS BRIDGE §78 In Fig. 17 (a), there are ten rivets (not counting counter¬ sunk rivets) in the right-hand end of plate g, five rivets in the plate r beyond the end of g, and eight rivets in the plate q beyond the end of plate r. This is the customary way of arranging the rivets when more than one pin plate is used. 24. The clear distance between the pin plates / on the end post was found to be 11 inches; leaving t& inch clearance on each side makes the outside surfaces of the eyebars !0i inches apart, their centers being 101 — li = 9f inches apart, and their inside surfaces 10l — li — li = 8f inches apart. Leaving tV inch clearance on each side makes the outside surfaces of the hanger plates j of the hip vertical 81 inches apart, as shown at the top of Fig. 17 (b ). It is customary to make the hanger plates not less than.f inch thick, although in many cases thinner plates give sufficient bearing on the pin. A plate t inch thick will be used in the present case. No calculation will be made for the required thickness of bearing, as the actual thickness is greater than necessary; the required net section of the hanger plates, however, must be computed. In Design of a Highway Truss Bridge , Part 1, the required net section of B b was found to be 4.7 square inches. In B. S., Art. 143, it is specified that riveted tension members shall have a net section through the pinhole 25 per cent, greater than the net section of the member. Since the required net section of B b is 4.7 square inches, the required net section of the hanger plates through the pinholes is 1.25 X 4.7 = 5.88 square inches. As there are two plates, each plate must have a net section equal to 5.88 -f- 2 = 2.94 square inches. Hanger plates 10 in. X t in. will be used; the net width (deducting the diameter of the pin) is 10 — 41 = 5i inches, and the net area is 5i X i = 3.28 square inches, which is sufficient. The rivets that connect the hanger plates to Bb are f inch in diameter, shop driven, and in single shear; their value is 4,860 pounds. Then, since the stress in Bb is 75,500 pounds, the number of rivets required to connect each plate is 37,750 - 7 - 4,860 = 7.8, or. §78 DESIGN OF A HIGHWAY TRUSS BRIDGE 29 say, 8. In Fig. 17 (b) , the distance between the outer sur¬ faces of the hanger plates is shown as 8i inches; then, the distance between centers of bearings is 8i — i = 7i inches. 25. The horizontal forces acting on the pin are shown in Fig. 18 (a), and the vertical forces in Fig. 18 (b). The greatest moment due to the horizontal forces is between \ Fig. 18 c and c', and is constant; it is equal to the moment, about c , of the forces on the left of c , or 134,000 X 1.781 - 81,000 X 1.281 - 134,900 inch-pounds The greatest moment due to the vertical forces is between d and d'\ it is constant between these two points, and is equal to the moment, about d, of the forces on the left of d, or 109,300 X 2.219 — 71,600 X .9375 = 175,400 inch-pounds Then, the greatest bending moment on the pin is from d to d', and its value is Vl34,900 2 + 175,400 2 = 221,300 inch-pounds Consulting Table XLI, it is seen that a 4f-inch pin has sufficient resisting moment. 26 . Packing of Bottom Chord.— The process of arranging the members on the bottom-chord pins is spoken 30 DESIGN OF A HIGHWAY TRUSS BRIDGE §78 of as packing the bottom chord, and the arrangement is called the bottom-chord packing. The packing - for the truss under consideration is shown in Fig. 19. It is cus¬ tomary to draw first the cross-sections /, / of the verticals, and show the location of the webs h , h and side plates g,g of the end post. Next, the diagonals z, z are placed close to the verticals; in the present case, their inner surfaces will be placed tg inch from the backs of the channels, the same as at joint C, to allow f inch for the height of rivet heads and iV inch for clearance. The counter j is connected to the center of the pin. Next, the eyebars that form the bottom chord are placed in position on the pins outside of the diag¬ onals, and as close to them and to each other as possible. In calculating the location of the eyebars on the pins, it is assumed that there is -re inch space between the heads of eyebars. The eyebars of the bottom chord are always alternated on the pin; that is, there is first placed one that comes from the panel on one side, as k , Fig. 19 ( Oc" , 3375 — / v /op] IQ3(, (a) © © © (b) Fig. 22 The bending moments, in inch-pounds, caused at the different points by the horizontal forces are as follows: at c, 85,025 X 1.031 .= 87,660 at d, 85,025 X 2.062 - 68,025 X 1.031 .= 105,190 at (i on the pin are shown in Fig. 23. The horizontal forces acting on the pin are shown in Fig. 24 (a), and the vertical forces in Fig. 24 (6). The bending moment due to the vertical forces is greatest from g to g\ and its value is (a) Cs VS oc vs vs $ vs 9 © *2 ///s" * - _ rrn" i r.y --► (b) Fig. 24 55,850 X 1.175 = 65,600 inch-pounds. The bending moments, in inch-pounds, caused at different points by the horizontal forces are as follows: at c t 77,360 X 1.031 .= 79,760 at d, 77,360 X 2.062 - 56,670 X 1.031 .= 101,090 at e, 77,360 X (1.031 + 3.093) — 56,670 X 2.062 . . = 202,180 at /, 77,360 X (1.969 -f 4.031) — 56,670 X (.9375 + 3) = 241,000 The greatest bending moment is from g to g', and is V65,600 2 + 241,000 2 = 249,800 inch-pounds Consulting Table XLI, it is seen that a pin 4i inches in diameter has sufficient resisting moment; hence, it will be used for the joint d. In the present case, the required diameters, as found for the two conditions, are the same. This is simply a coincidence; if the diameters had come out different, it would have been necessary to use the larger. 78 DESIGN OF A HIGHWAY TRUSS BRIDGE 37 30. Pins at Joints' c and e. —The method of finding the diameters of the bottom-chord pins at joints c and e is entirely similar to that given in the preceding articles, and should present no difficulty. In the present case, they both work out less than that required at d> but for uniformity they will both be made the same size as at d, that is, 4i inches in diameter. It is customary to make all the bottom-chord pins the same size, so that it is necessary in each case to find the pin on which the bending moment is the greatest. This pin can only be found by trial and computation. 31. Pin at Joint b. —At the joint b there is no stress transmitted to the chord from the vertical, but the pin is passed through plates in the bottom of the vertical, similar to the hanger plates at the top. The only stresses that need be considered at this joint are the stresses in the bottom chords ab and be. The eyebars on this pin are not placed close to the vertical, but are so placed as to run as straight as possible from joint c to joint a; at the latter joint the eye- bar is usually placed outside the end post and as close to it as practicable. 32. Pin at Joint a .—‘The forces acting on the pin at the joint a are the stresses in a B and a b , the load from the end floorbeam, and the reaction. The maximum stress in a B is + 272,100 pounds. The load from the end floorbeam is given in Design of a Highway Truss Bridge , Part 1, and is 9,644 + 36,250 = 45,894 pounds. The reaction is equal to the sum of the vertical component in aB and the load at a , or 218,615 + 45,894 = 264,510 pounds. The stress in a b is given in Fig. 11; but, as the stresses in a B and a b were found for different conditions of loading, there is a slight incon¬ sistency at this joint; that is, the maximum stress in a b is not exactly equal to the horizontal component of the maxi¬ mum stress found in a B. Under such conditions, it is best to take for the stress in a b , in the design of the pin at joint a , the horizontal component of the stress in a B, or 272,100 X = 161,960 pounds, 33.6 135—17 38 DESIGN OF A HIGHWAY TRUSS BRIDGE §78 which in this case is a small amount in excess of the stress in a b. The external forces acting at the joint a are shown in Fig. 25. A 4j-inch pin will be tried first, this being the size of the remainder of the bottom-chord pins. % The load from the end floorbeam is applied to the pin by means of a riveted chair or seat shown at /, Fig. 7 {b ), which is riveted to the end post, so that this chair -and the end post act as a single member in so far as the load on the pin is con¬ cerned. To find the required combined thickness of bearing of the chair and end FlG * 25 post, it is necessary to find the resultant of the load from the floorbeam and the stress in aB , Fig. 25. This is evidently the same as the equilibrant of the reaction and the stress in a b } that is, a/ 264,510” + 161,960’ = 310,160 pounds The actual direction of this resultant is of no importance; it is shown in its approximate position by the dotted line in Fig. 25. The required thickness of bearing for the com¬ bined chair and end post is, then, 310,160 4.875 X 22,000 = 2.89 inches, or 1.44 inches, say lA, inches on each side. The side plates of the end post are i inch thick; the webs are A inch thick; the side plates / of the chair, Fig. 7 ( b) } will be made i inch thick; the total thickness on each side is, therefore, Ire inches. It is not necessary to calculate the number of rivets required to transmit the stress in the plate / to the floorbeam and end post; as a rule, if the ordinary rules of rivet spacing are followed, there will be an excess. Since the webs of the end post are Ilf inches apart, the chair plates will be lOf inches apart, and the centers of bearing 101 + 1A = 12A inches apart. The outside surfaces of the side plates of the end post are, then, 12re + lA = 13f inches apart. Leaving i inch clearance on each side makes the inside surfaces of the eyebars 13i inches apart, §78 DESIGN OF A HIGHWAY TRUSS BRIDGE 39 and the centers of the eyebars 14i inches apart, as shown in Fig. 19 {a). The load is transmitted to the abutment by means of a built-up pedestal, the vertical webs ?i , n , Fig. 19 (a ), of which are inside the end post. The required thickness of the vertical plates is 264,510 4.875 X 22,000 2.47 inches, or 1.23, say li inches, on each side. It is somewhat better to make this up with two plates t inch thick than to use one li-inch plate. The inside surfaces of the chair plates are 101 inches apart. Leaving i inch clearance on each side makes the vertical webs of the pedestal 9| inches apart center to center. The horizontal forces acting on the pin at a are shown in Fig. 26 {a); the vertical forces, in Fig. 26 {b ). The bend¬ ing moment due to the horizontal forces is 80,980 X 1.344 = 108,840 inch-pounds, and is constant from c to c'; that *1344 — Z2./875 (a) © oo Ci © 11 © 00 C5 o 00 /.344 Fig. 26 due to the vertical forces is 132,265 X 1.469 = 194,300 inch- pounds, and is constant from d to d'. The greatest bending moment on the pin is then from d to d', and is Vl08340“ + 194,300" = 222,700 inch-pounds 40 DESIGN OF A HIGHWAY TRUSS BRIDGE §78 Consulting Table XLI, it is seen that a pin 4f inches in diameter has a sufficient resisting moment; but, as x the remaining bottom-chord pins are to be made 4% inches in diameter, the latter size will be used for joint a also. The detail of joint a will be discussed presently. BEARINGS 33, Required Area. —The required area of bearing for each end of each truss is found by dividing the reaction by the working pressure on the masonry. Since the reaction is 264,510 pounds, and the masonry is cement concrete, on which the working pressure is 400 pounds per square inch ( B . S., Art. 103), the required area of bearing is 264,510 -7- 400 = 661.3 square inches. If the bedplate is made square, it should be about 26 inches square. The actual dimensions depend on other details, such as rollers, etc. 34. Pedestal. —Built-up pedestals like that shown in Fig. 27 are used for pin-connected trusses. The bottom of the pedestal is made the same length as the length of the bedplate; the width depends somewhat on the distance between the vertical webs, which are just inside the end post. The height of the pedestal depends on the allowable distance from the bottom of the truss to the top of the bridge seat; it is desirable to keep the top of the bridge seat above high water. In general, the height should be about one-half the length; hence, in this case it will be made 15 inches. In Fig. 27, (a) is the elevation of joint a, and shows the pedestal, the rollers, and the chair e that supports the end floorbeam; ( b ) is an end view of the joint; (c) is a top view of the pedestal; and (d) is a top view of the rollers. The bottom plate / of the pedestal is generally made about li inches thick, and the angles g,g that connect the vertical webs h to the base plate / are made 6 in. X 6 in. X f in. The rivets that connect the angles g to the plate / are counter¬ sunk on the under side. The diaphragms i, i are riveted to the vertical webs to stiffen the pedestal. In the chair e , a diaphragm j is inserted between the side plates, in order to (C) p m n 2 n r*» ±± (d) -o P Til i #1 a JSir 2= Q Q Q Q Q i r * i I U--V • I > -»-Fl , I pr«g h J S' mil// S A; # -# 41 Fig. 27 42 DESIGN, OF A HIGHWAY TRUSS BRIDGE §78 distribute the load from the floorbeam more evenly between the two sides. At one end of the bridge, the fixed end, the base plate / rests directly on the masonry, and is anchored to it by means of anchor bolts. At the other end, rollers are placed,under the plate /, as shown at k , in order to allow that end to move back and forth as the temperature changes. 35. Rollers.—The allowable pressure per linear inch on the rollers is given in B. A., Art. 103, as 600 D. In the present case, the smallest-sized roller (3 inches, B. S., Art. 153) gives sufficient resistance to transmit the reac¬ tion, as will be shown presently. It is customary to space the rollers about i inch apart for the full length of the pedestal. In the present case, if the end rollers /, / are each placed i inch from the end of the plate /, there is just room for nine rollers in the length of the pedestal. The rollers are usually made at least as long as the distance between the outside edges of the base angles in this case about 23 inches. Nine rollers at 23 inches gives 207 linear inches of rollers, which, at 600 X 3 = 1,800 pounds per linear inch, can support 372,600 pounds. As this is greater than the reaction, these rollers (that is, nine rollers 23 inches long) are sufficient. If the allowable load had been less than the reaction, it would have been necessary to use either longer rollers or rollers of larger diameter. The rollers are fastened together so as to form what is known as a roller nest, in the manner shown in Fig. 27 (d). Short bolts rn,m , known as studs, about f inch in diameter, are set into the ends of each roller about li inches, and the ends left to project 2 inch at each end. A flat ,bar 71 is placed on each end of the rollers, and the short bolts fit into holes in this bar. The bars n are kept in position by the bolts 0 and additional bars p at each end of the nest; the bars p are i inch longer than the rollers, and are inserted between the ends of the side bars n . A plate about 2 inches in thickness is inserted under the rollers to give them a smooth-surface to roll on, and also to distribute the pressure more evenly over the masonry. The bearing surfaces above 78 DESIGN OF A HIGHWAY TRUSS BRIDGE 43 and below the rollers are provided with projections q that fit into grooves r planed in the rollers, and are for the purpose of preventing the pedestal and rollers from moving sidewise. The base plate*of the pedestal and the bedplate are con¬ tinued out beyond the ends of the rollers a sufficient distance (generally 3i or 4 inches) to allow the insertion of the anchor bolts s that hold the pedestal and bedplate in posi¬ tion. At the expansion end, the holes in the base plate through which the anchor bolts pass are made just wide enough to allow the insertion of the bolts, and long enough, as sho\yn at t , to allow the base plate and pedestal to move backwards and forwards. LATERAL CONNECTIONS 36. Upper ^Lateral Truss. —Fig. 28 shows the connec¬ tion of the diagonals of the upper lateral truss to the top of the top chord: a and b are the diagonals, and c and d are Fig. 28 lug angles that help to connect a and b to the connection plate to one-half of Cooper’s E50 loading, as shown in Fig. 2. The maximum bending moment caused on a stringer by this load¬ ing is found as explained in Stresses in Bridge Trusses , Part 4: it occurs at the center of the panel when the loads are in the position shown in Fig. 3, and is equal to 212,500 foot-pounds. The maximum end Fig. 3 0009Z 0009Z 0002Z 9009Z G G G G ooszrQ 09Z9TQ 09Z9TQ 09Z9TQ 09Z9TQ OOOffZ ooosz 9009Z 0009Z G G G G oosziQ 24 24 " o' M ‘O G *4 VO A g *4 . "D u shear occurs when the loads are in the position shown in Fig. 4, and is equal to 58,330 pounds. The maximum live-load shear at a section 5 feet from the end, which must be computed for the determi¬ nation of the rivet pitch (see Design of Plate Girders , Part 1), is found to be 33,330 pounds. As will be seen presently, it is 25000 © C> © JO ** 25000 r s' o' * m D *■ n t * O -H J * Fig. 4 unnecessary to compute the maximum shear at any other section. 6. Impact and Vibration. —Since the stringer is a floor member, the amounts to be added to the shears and moments, to allow for impact and vibration, are as fol¬ lows ( B . S., Art. 25): to the moment, 212,500 foot-pounds; to the shear at the 79 DESIGN OF A RAILROAD TRUSS BRIDGE 5 end, 58,330 pounds; to the shear at 5 feet from the end. 33,330 pounds. 7. Wind Pressure.— The wind pressure on the train increases the amount of load that goes to the leeward stringer. The center of wind pressure is 7 feet above the top of the rail ( B . S., Art. 27) and about 8 feet above the top of the stringers. The center of moments will be taken at the top of the stringers. Then, since the wind pressure is 300 pounds per linear foot ( B . S ., Art. 27), and the stringers are 6 feet 6 inches center to center ( B . S., Art. 48), the increase in load on the leeward stringer is 300 X 8 6.5 = 369.23 pounds per linear foot. The bending moment ’due to this increase, at the section where the live-load bending moment is greatest, that is, at the center of the panel, is 14,950 foot-pounds; the shear at the end is 3,323 pounds, and at a section 5 feet from the end, 1,477 pounds. 8. Dead-Load Moments and Shears. —The weight of the track will be assumed as 400 pounds per linear foot, one-half of which, or 200 pounds per linear foot, is carried by each stringer. The weight of the stringer will be assumed to be 150 pounds per linear foot, which makes the total dead load 350 pounds per linear foot. The bending moment due to this dead load, at the section where the live-load moment is greatest, that is, at the center of the panel, is 14,180 foot¬ pounds; the shear at the end is 3,150 pounds, and at a section 5 feet from the end, 1,400 pounds. 9. Total Moments and Shears. —The total shears and moments are as follows: Total maximum moment, 212,500 4- 212,500 + 14,950 -f 14,180 = 454,130 foot-pounds. Total shear at the end, 58,330 + 58,330 + 3,320 -f 3,150 = 123,130 pounds. Total shear 5 feet from the end, 33,330 -f- 33,330 +1,480 + 1,400 = 69,540 pounds. 10. Design of Web.— It was decided in Art. 4 to have the web 30 inches deep. The thickness will be made inch. 135—18 6 DESIGN OF A RAILROAD TRUSS BRIDGE §79 Then, the intensity of shearing stress at the end is —- v- = 7,297 pounds per square inch. Consult- 30 X Te (16.8/5) ing Table XXXVI, it is seen that the least unsupported distance for the web is 25 inches, and, as the vertical legs of the flange angles will probably be closer together than this, no stiffeners will be required. It is therefore unnecessary to find the intensity of shearing stress at any other section. 11. Design of Flanges. —The required area of flange will be found by the following formula, given in Design of Plate Girders , Part 1: A = M s h 9 th T In the present case, everything is known except h s . It has been seen in the preceding Sections that h g is usually less than the width of web; in this case, it will be assumed to be 27 inches. Then, A = 454,100 X 12 30 X 1 6 12.61 - 2.11 16,000 X 27 8 = 10.50 square inches Two angles 6 in. X 6 in. X i in. will be used for both the top and the bottom flanges. The gross area of the two angles (Table IX) is 11.5 square inches, and the net area, deducting one -g-inch rivet hole from each angle, is 10.5 square inches. The distance between the centers of gravity of the flanges is, then, 30.25 — 1.68 — 1.68 = 26.89 inches, and, therefore, 454,100 X 12 A = 2.11 = 12.67 - 2.11 16,000 X 26.89 = 10.56 square inches. Two angles 6 in. X 6 in. X i in. are, therefore, sufficient for the top flange and near enough the required size for the bottom flange. 12. Lateral Bracing Between Stringers. —Since the web is inch thick and the top flange angles are 6 inches wide, the width of the top flange is 12 i 9 6 inches. Since the panel length is 18 feet, it is greater than 12 times the width §79 DESIGN OF A RAILROAD TRUSS BRIDGE 7 of the flange, and, according to B. S. y Art. 87, lateral bracing is required between the top flanges. This will be made of the same general form as used in the Section on Deng?i of Plate Girders , Part 2. 13. Flange Iiivets. —The flange rivets are f inch in diameter, shop driven, in double shear, and in bearing on the iVinch web; the latter value, 10,830 pounds, is the smaller. Since the center line between the two rivet lines of a 6-inch leg is 3f inches from the back of the angle, and the flanges are 30f inches back to back, the distance h r between the centers of the rivet lines is 30f — 3f — 3f = 23i inches. Then, the required pitch of the rivets at K h the end of the stringer, as found from the formula p — - (Design of Plate Girders , Part 1), is 10,830 X 23.5 123,130 = 2.07 inches; and at 5 feet from the end, 10,830 X 23.5 69,540 3.66 inches As the ties will rest directly on top of the top flange, the required pitch of rivets in the top flange is nine-tenths of the above, that is, 1.86 inches at the end and 3.29 inches at 5 feet from the end. For practical reasons, the rivet spacing in the top flanges of the stringers in railroad bridges should not exceed 3i inches, the principal reason being the fact that these rivets transmit the load directly to the web. The pitch is usually taken to the next lower i inch. Then, the rivets in the top flange will be spaced If inches apart at the end and 3i inches apart at 5 feet from the end. Between these sec¬ tions, the pitch is gradually changed at the rate of about 2 inch each time, making the successive pitches If, 2f, 2f, and 3f inches. It is customary to make each of these cover approximately the same distance as far as conditions will permit. For example, in the present case, the following spacing may be used: Base of Rail o s> G VA. • ® •'? • • • M' ^ Vj^ • • • Vi • • • G G G G , ,G^G G G G G $ G o >-• fe ^ £ d (a) 2; "=/B § 71) DESIGN OF A RAILROAD TRUSS BRIDGE 9 12 spaces at If inches = 1 foot 9 inches 10 spaces at 2f inches = 1 foot 10i inches 6 spaces at 2f inches = 1 foot 4i inches Total distance = 5 feet Beyond the section at 5 feet from the end, a few spaces at 3f inches can be inserted, and then a pitch of 3i inches used throughout the remainder of the flange as far as the 3i-inch pitch at the other end. 14. Connection of Stringer to Floorbeam. —In the design of the highway bridge in the preceding Sections, the design of the floor connections was left until after the design of the trusses. In the present case, the connections of the stringer to the floorbeam will be designed now, for in rail¬ road bridges it is much less difficult to decide on the type of connection. Fig. 5 shows a typical connection: A A, Fig. 5 ( a ), is the usual cross-section of the floorbeam, and B B, Fig. 5 (b) , is the cross-section of the stringer. The latter rests on a shelf angle c that is riveted to the bottom flange angles of the floorbeam; the connection angles d are placed outside of the flange angles of the stringer, and tight fillers e are placed between them and the web. Tight fillers or reinforcing plates / are also riveted to the web of the floorbeam. With this style of connection, the smallest value of the rivets that connect the angles to the stringer is 13,200 pounds (f-inch rivets, shop driven, in double shear), and that of the rivets that connect the angles to the floorbeam is 5,410 pounds (|--inch rivets, field driven, in single shear). Then, the number of rivets required to connect the angles to the stringer and to the floorbeam is, respectively, since the end shear is 123,130 pounds, = 9.3 rivets, and 193 130 — ? -= 22.8 rivets. It is customary to use the next 5,410 larger even number, 24 in this case, for the latter, and half this number, 12 in this case, for the former, as shown in Fig. 5. As it is impossible to get this number of rivets in a single row without getting the rivets too close together, 10 DESIGN OF A RAILROAD TRUSS BRIDGE §79 6-inch angles are used, giving room for two rows in each leg. The least allowable thickness of connection angle is h inch ( B . S ., Art. 51), so 6" X 6" X A" angles are used. It is well to make the thickness of the connection angles at least equal to that of the stringer web. Had the stringer web been f inch thick, 6" X 6" X i" connection angles would have been used. INTERMEDIATE FLOORBEAMS 15. Length and Depth.— The length of the floorbeam will be taken as 16 feet, the distance center to center of the trusses. When the connection represented in Fig. 5 is used, Fig. 6 the floorbeam is made 1 foot deeper than the stringer; if this does not give a depth equal to one-sixth the length ( B . S., Art. 49), the shelf angles are placed farther from the bottom flange, and stiffeners are placed under them. The top flange of the floorbeam should never be more than about 6 inches above the top of the string¬ ers, as it will inter¬ fere with the rail if it *> . § is higher than this. In the present case, a web 42 inches in width will be used for the floorbeam. 16. Live-Load Shears and Moments. —The live load that comes on the floorbeam from one line of stringers is greatest when the loads are in the position shown in Fig. 6, and is equal to 75,830 pounds. As the stringers are 6 feet 6 inches apart, and the floorbeam is 16 feet long, the loads Fig. 7 § 79 DESIGN OF A RAILROAD TRUSS BRIDGE 11 are applied to the floorbeam as shown in Fig. 7, the shear from the truss to the stringer being 75,830 pounds, and the bending moment on the floorbeam at the stringer connection being 75,830 X 4.75 = 360,200 foot-pounds. This is also the value of the bending moment at the center, as it is constant between the stringer connections. 17. Impact and Viblation. —Since the floorbeam is a floor member, the amounts that must be added to the shear and moment to provide for the effect of impact and vibration are as follows: to the shear, 75,830 pounds; and to the moment, 360,200 foot-pounds ( B . S., Art. 25). Or § © © © © 4.75 1 © § © 6.5 - / 6 ‘ © © N 0 * 4.?5- 18. Wind Pressure. —In Art. 7 it was found that the wind pressure on the train increased the load on the leeward stringer by 369.23 pounds per linear foot. Then, the increase in the load on the floorbeam at the leeward stringer connec¬ tion is 18 X 369.23 = 6,650 pounds. There will be a cor¬ responding decrease in the load on the windward stringer, as shown in Fig. 8, in which the increase and the decrease are shown as a downward and an upward force, respectively, instead of adding them to ai*d subtracting them from the loads shown in Fig. 7. Taking moments about the right end of the floorbeam, the reaction at the left end, which is also the shear from the truss to the stringer, is found to be 2,700 pounds, and the bending moment at the left-hand stringer, 2,700 X 4.75 = 12,800 foot-pounds. It is unnecessary to find the shear and moment at the other end, as they simply tend to decrease those due to live load. Fig. 8 19. Dead Doad.—The dead load on each stringer (Art. 8) is 350 pounds per linear foot; the amount that goes to the floorbeam at each stringer connection is then 350 X 18 = 6,300 pounds. The weight of the floorbeam will be assumed to be 200 pounds per linear foot. The dead-load 12 DESIGN OF A RAILROAD TRUSS BRIDGE §79 shear at the end is then 7,900 pounds, and the bending moment at the stringer connection is 36,300 foot-pounds. 20. Total Shear and Moment. —The total shear and moment are as follows: Total shear at the end, 75,830 + 75,830 + 2,700 + 7,900 = 162,260 pounds. Total bending moment at the stringer, 360,200 -f 360,200 + 12,800 -j- 36,300 = 769,500 foot-pounds. » 21. Design of Web. —In Art. 15 it was decided to make the web of the floorbeam 42 inches deep. A thickness of f inch will be used. Then, the intensity of shearing stress at the end is = 6,180 pounds per square inch; and, as the shear is nearly constant from the truss to the stringer, this will be taken as the intensity of shearing stress on that portion of the floorbeam. Consulting Table XXXVI, it is seen that the allowable unsupported distance on the web is 33 inches. The vertical legs of the flange angles will prob¬ ably be 6 inches wide, leaving the distance between them 30 inches; as this is less than the allowable distance, no stiffeners are required. 22. Design of Flanges. —The required area of flange will be found by the formula A — M _ th s h s 8 The value of h s will be assumed to be 3 inches less than the distance back to back of the flange angles, that is, 42.25 — 3 = 39.25 inches. Then, a 769,500 X 12 42 X i o oq A ~ 16,000 X 39.25 ~ ~ UJ ° ~ 328 = 11.42 square inches. For the bottom flange, two angles 6 in. X 6 in. X in. will be used. As there will be no flange plates in this case, there is no necessity for rivets in the outstanding legs of the bottom flange angles at the point where the moment is greatest, so that it will be sufficient to deduct the cross-sec¬ tion of one hole from each angle. The area of one angle is §79 DESIGN OF A RAILROAD TRUSS BRIDGE 13 6.43 square inches (Table IX); the area to be deducted for one i-inch rivet hole in material t& inch thick is .56 square inch. Then, the net area of the two angles is 2 (6.43 — .56) = 11.74 square inches, which is sufficient. The center of gravity is 1.71 inches from the back of the angles. For the top flange, two angles 6 in. X 3i in. X tw in. and one flange plate 8 in. X i in. will be used. They give a gross area of 11.92 square inches, which is sufficient. The center of gravity is 1.28 inches from the top of the angles. This gives h s = 42.25 — 1.71 — 1.28 = 39.26 inches, which is very close to the assumed distance. It is customary to make the top flange of the floorbeam as narrow as practi¬ cable, on account of the fact that it projects above the string¬ ers and therefore lies between the ties (see^, Fig. 5), which must be spread to make room for it. 23. Length of Flange Plate. —The required length of flange plate can be found by the curve of the flange areas for the portion of the floorbeam between the truss and the stringer (see Design of Plate Girders , Part 1). This curve is shown in Fig. 9: A B represents the distance from the truss to the stringer, B D the required area B B' one-eighth KJ of the web & C area t ^ ie two A an £ e angles, and CD' the area of the flange plate. As the bending moment from A to B , due to all the loads except the weight of the floorbeam, varies uniformly from A to B , it is customary to 14 DESIGN OF A RAILROAD TRUSS BRIDGE §79 assume that the curve of the flange areas AD is a straight line. It is unnecessary to consider the required flange area between the stringers, as the entire flange section is carried between them. The line C C is drawn parallel to A B , to its intersection C with A D, and C c is drawn at right angles to A B; then, c is the section at which the flange plate can be cut off. The distance A c can be found very readily by the formula A c = Oc BD XAB In the present case, C c — 3.28 + 7.92 = 11.20, BD = 14.70, and A B — 4.75; therefore,' A c = x 4.75 = 3.62 feet 24. Flange Rivets. —The flange rivets are i inch in diameter, shop driven, in double shear, and in bearing on the f-inch web; the bearing value, 12,000 pounds, is the smaller. Since the flange angles are 42i inches back to back, and the center line between the gauge lines is 3f inches from the back of the angles (Table XII), the rivet lines of the flanges are 42.25 —3.375 —3.375 = 35.5 inches center to center. Then, (see formula in Art. 13), the required pitch of the rivets at the end of the floorbeam is = 2.63 inches. 162,260 As the shear is very nearly constant from the truss to the stringer connection, this pitch will be used for that entire distance. The shear between the two stringers is practi¬ cally equal to zero, so that the smallest allowable pitch, or 4| inches, will be used in the floorbeam flanges between them ( B . X., Art. 57). As the width of the outstanding leg of the top flange angles is 3i inches, only one line of rivets can be driven in it, while in the vertical leg, 6 inches wide, two rows can be driven. The required pitch or number of rivets in the outstanding leg in this case can be found by considering the stress in the 8 " X flange plate. As the area of the plate is 4 square inches, and the working stress in compression is 16,000 §79 DESIGN OF A RAILROAD TRUSS BRIDGE 15 pounds per square inch (B. S., Art. 29), the total stress in the plate is 4 X 16,000 - 64,000 pounds. The value of one rivet (& inch in diameter, shop driven, in single shear) is 6,610 pounds. Then, the number of rivets required to trans¬ mit the stress to the plate is = 9.7, or, say, 10. That is, 6,610 there must be 10 rivets in the top flange plate, between the stringer connection and the end of the plate. If it is impos¬ sible to get this number of rivets in the distance c B, Fig. 9, without violating the rules for minimum rivet pitch, the top flange plate will be made longer than required by the curve of flange areas. In any case, it is well to make it slightly longer than required. 25. Connection of Floorbeam to Truss. —The floor- beams of railroad bridges, especially in pin-connected truss bridges, are frequently connected to the vertical posts above the lower chord in the same way as in the highway bridge designed in the two preceding Sections of this Course. A much more usual connection, however, especially for riveted trusses, is shown in Fig. 10. In this figure, (a) is the eleva¬ tion of one end of the floorbeam, showing the holes for the stringer connection and the cross-section of the bottom chord of the truss, together with the elevation of a vertical post; ( b) is an end view of the connection angles of the floorbeam; ( c ) is a cross-section on the plane C C and a plan of the lower flange; and ( d) is a top view of a portion of the top flange. The stringer connection was considered in Art. 14, and will not be further discussed. The splice in the floorbeam web and the connection to the truss will now be considered. 26. The main object in this type of connection is to dis¬ tribute the load that comes from the floorbeam over a greater length of the vertical member, thereby insuring a better dis¬ tribution of the load between the two sides of the truss. For this purpose, the depth of the floorbeam web is increased at the end; this is accomplished by splicing the web between the truss and the stringer at^, and allowing a portion / of the end of the web to project up through the top flange angles. V Fig. 10 16 §79 DESIGN OF A RAILROAD TRUSS BRIDGE 17 The deep end portion of the web, frequently called the floor- beam gusset, should not come inside the clearance line, as given in B. S., Art. 18, and shown at^ in Fig. 10. It is not customary, in splicing a floorbeam web, to apply the formulas given in Design of Plate Girders , Part 1, but rather to design the splice so as to transmit the shear from one portion of the web to the other. As a rule, if the floor- beam is over 3 feet deep, and each splice plate h is made the same thickness as the web, and three rows of rivets spaced 3 inches apart, as shown in Fig. 10, are put in on each side of the splice, it will be unnecessary to make any computation. When the floorbeam is less than 3 feet deep, the splice plates are turned into reinforcing plates and continued from the end of the floorbeam out beyond the splice and stringer connec¬ tion. In the splice shown in Fig. 10, the resisting moment of the rivets is not so great as that of the web, but at this section there is sufficient flange area even if the effect of the web in resisting the moment is not counted; so that it is not necessary to have the resisting moment of the rivets so great as that of the web at this section. The rivets connecting the connection angles i to the gusset are t inch in diameter, and shop driven. Their smallest value is that in bearing on the i-inch gusset, and is 12,000 pounds. Since the end shear on the floorbeam is 162,260 pounds, the number of rivets required is 162,260 12,000 = 13.5, or, say, 14. The rivets connecting the connection angles to the truss are field driven; their value in single shear, 5,410 pounds, is the smallest. The required number of rivets is --■■■ * — = 30. 5,410 In Fig. 10, there are more rivets than are necessary. This is a common practice, as it is considered advisable to have a few extra rivets. 27. No rule can be given as to the height of the floor- beam gusset at the end. In half-through plate-girder bridges, the gusset is continued up to the under side of the top flange; 18 DESIGN OF A RAILROAD TRUSS BRIDGE §79 in pony- truss bridges, it is made as large and as high as the clearance line will allow; in through truss bridges, as in the present case, its height depends on the depth of the floor- beam and somewhat on the required number of rivets in the connection angle. In no case should the gusset encroach on the required clearance. The gusset is cut near the bottom at j to allow for the out¬ standing leg of the top flange angle of the bottom chord, and the connection angle i on each side is put on in two pieces. The connection angles will be made 4 in. X 4 in. X I in. 28. Between the bottom of the floorbeam and the top side of the outstanding leg of the lower flange angle of the bottom chord is placed a plate k , i inch thick, which is riveted to the chord and to the bottom flange of the floor- beam, and serves the purpose of connecting the diagonals of the lateral truss to the floorbeam and bottom chord. The lateral connections will be discussed further on. 29. In Fig. 5, the distance from the base of the rail to the top of the stringer is 7} inches (B. S ., Art. 48), and the top of the shelf angle is 6f inches above the bottom of the bottom flange of the floorbeam, making the bottom of the floorbeam 3 feet 8 i inches below the base of the rail. Adding to this 2 inch for the thickness of the lateral con¬ nection plate k , and f inch for the thickness of the bottom chord angle l (to be found later), gives the distance from the base of the rail to the bottom of the bottom chord as 3 feet 9f inches, as shown in Fig. 10. This distance will be referred to again in the following articles. END STRUTS OR FRAMES 30. For the sake of variety, no end floorbeams will be provided, although in the best practice they are always put in. In the present case, the end stringers will be allowed to rest directly on the masonry, and their ends will be connected to each other and to the ends of the trusses by frames, as Base of Ra/7 I § 7<) DESIGN OF A RAILROAD TRUSS BRIDGE 1!) 20 DESIGN OF A RAILROAD TRUSS BRIDGE § 79 shown in Fig. 11. Four stiffeners are placed at the end of each stringer where it rests on the masonry, and sole plates and bedplates 1 inch thick are provided, as shown in Fig. 12. In the present case, the bedplates will be made 20 inches square, which is a customary size. It is unnecessary to calcu¬ late the required thickness of the end stiffeners on stringers; if the smallest allowable angles are used—in the present case, 5 in. X 3i in. xt in. (B. S., Art. 55)—they will be large enough. End stringers are usually made somewhat longer than the intermediate stringers; the difference is equal to one-half the length of the bedplate, which makes the distance from the center of the bedplate to the next floorbeam equal to a panel length. Otherwise, the end stringers are the same as the intermediate stringers. As shown in Fig. 11, end frames are usually made shallower than the stringers. The tops of the frames are placed below the tops of the stringers that the frames may not interfere with the ties. The bottoms of the frames are placed above the bottoms of the stringers so as to have the former above the bridge seat. The connection of the outside frames b ) b, Fig. 11, to the trusses will be considered later. § 79 DESIGN OF A RAILROAD TRUSS BRIDGE 21 DESIGN OF TRUSSES STRESSES 31. Instead of finding the actual stresses caused in the different members by the different loadings, as in Stresses in Bridge Trusses , Part 2, the shears and moments will be found, and, after properly combining them, the stresses will be cal¬ culated. 32. Eive-Eoad Shears and Moments. —The portion of the live load that goes to each truss is one-half of Cooper’s E50, as shown in Fig. 2. The portion that the load must occupy in order to cause the greatest shear in a panel or the greatest moment at a panel point is found by means of the principles explained in Stresses in Bridge Trusses , Part 4. The maximum positive live-load shears in the different panels, and the corresponding positions of the loads, are as follows (see Fig. 13): Panel Position of Load Maximum Positive Shear Pounds a b 3 at b 207,850 b c 3 at c 156,910 c d 3 at d 111,400 d e 2 at e 72,220 e d f 2 at d' 42,570 d'c' 2 at c f 20,310 c'b ' i at b f 4,170 The maximum live-load bending moments at the different panel points, and the corresponding positions of the loads, are as follows: 135—19 22 DESIGN OF A RAILROAD TRUSS BRIDGE §79 Panel Point Position of Load Maximum Bending Moment Foot-Pounds b 3 at b 3>74L300 c 5 at c 6 , 180,000 d 8 at d 7,57L9oo c 11 at e 8 , 165,600 33. Impact and Vibration.—The amount to be added to each shear and bending moment found above to allow for the effect of impact and vibration depends on the length L of the track that is loaded when the shear or moment is a max¬ imum. When load 1 is off the left end, the entire bridge is loaded, and L is 144 feet; when load 1 is on the bridge, L is the distance from the right end to load 1. For example, when the shear in the panel c d is a maximum, load 3 is at d; in this case, since load 1 is 13 feet to the left of d, and d is 90 feet to the left of the right end of the truss, load 1 is 13 + 90 = 103 feet from the right end; that is, L = 103 feet. Apply¬ ing the formula given in B. S., Art. 25 (I — — gives \ L + 300/ the amounts to be added to the shears and bending moments as follows: Panel a b I = b c / = c d / = Shear, in Pounds is ifs» * 207 - 0 ” - 142 -°“ iiSs x 15e '*“ - nim wfm xnhm - 82 - 900 § TO DESIGN OF A RAILROAD TRUSS BRIDGE 28 Panel d e e d f d'c ' c'b' Panel Point I = / = Shear, in Pounds X 72,200 = 57,000 / = / = 80 + 300 300 62 + 300 300 44 + 300 300 X 42,600 = 35,300 X 20,300 = 17,700 18 + 300 X 4 ’ 17 °= 3 ’ 93 ° Bending Moment, in Foot-Pounds * 7 - BsSoo x 8 ' 741 ' 300 - 2 ’ 556 ’™ * /= - 131 3 °° 300 X 6,180,000 = 4,301,700 ' 7 - ibSoo x 7 - 571 -” 0 - ' 7 ' mVSxi x 8 ' 165 ’ 600 - 5 ' 618 ' M0 34. Dead-Load Shears and Moments.— The dead load consists of the weight of the track and the weight of the bridge. In the general data, it is stated that the floor shall consist of standard ties, and in B. S., Art. 23, it is specified that this type of floor shall be assumed to weigh 400 pounds per linear foot. The approximate weight w x per linear foot of bridge can be found by means of the formula given in B. therefore, w x = 1,500 S., Art. 242. In the present case, / = 144; 1,937.4 lb. per linear foot The total dead load per linear foot is, therefore, 400 + 1,937.4 = 2,337.4 pounds, of which one-half, or 1,168.7 pounds per linear foot, goes to each truss. The dead panel load is equal to 1,168.7 X 18 = 21,036.6 pounds, or, practically, 21,000 pounds; of this load, one-third, or 7,000 pounds, will be treated as applied at each top joint, and the remainder, 14,000 pounds, as applied at each lower joint of the truss. (B. S. t Art. 23.) 24 DESIGN OF A RAILROAD TRUSS BRIDGE § 79 The dead-load shears in the different panels are then as follows (see Fig. 13): Panel a b be cd de ed' d'e' c'b ' Vat Shear, in Pounds 73.500 (positive) 52.500 (positive) 31.500 (positive) 10.500 (positive) 10.500 (negative) 31.500 (negative) 52.500 (negative) 73.500 (negative) The dead-load bending moments at the different panel points are as follows: Panel Point b c d e Bending Moment, in Foot-Pounds 1,323,000 2,268,000 2,835,000 3,024,000 35. Increase in Shears and Moments on Account of Wind Pressure on tlie Train. —The load on the leeward truss is increased by the overturning effect of the wind. The amount of the increase can be found by the fol¬ lowing formula, given in Stresses in Bridge Trusses , Part 5: w Ph b In this formula, P is the intensity of wind pressure, 300 pounds per linear foot ( B . A., Art. 27); 1v is the dis¬ tance from the center of wind pressure to the lower lateral system, and will be taken as 11.25 feet (see Fig. 10); b is the distance center to center of trusses, in this case 16 feet. Then, w 300 X 11.25 16 210.94 pounds per linear foot The increase in the panel loads is then 210.94 X 18 — 3,797, or, practically, 3,800 pounds. The shears due to this increase are as follows: §79 DESIGN OF A RAILROAD TRUSS BRIDGE 25 Panel Positive Shear, in Pounds a b 13,300 be 9,975 cd 7,125 de 4,750 ed' 2,850 d'e' 1,425 c' b f 475 The bending moments due to this increase are as follows: Panel Bending Moment, Point in Foot-Pounds b 239,400 c 410,400 d 513,000 e 547,200 Those stresses to which the members are subjected as members of the lateral trusses will be considered later, and treated in the same way as for highway bridges. 36. Longitudinal Force. —The longitudinal force due to suddenly stopping trains is taken care of by connecting the stringers to the diagonals of the lower lateral truss wher* ever they intersect. No calculations are needed for this connection; it is made in the most convenient way. The laterals transmit the force to the trusses, which in turn transmit it to the abutments. 37. Combined Maximum Shears and Moments.- The combined maximum shears are as follows: Panel Combined Maximum Shear, in Pounds a b 207,800 4* 142,000 + 73,500 + 13,300 = 436,600 be 156,900 + 111,800 + 52,500 + : 10,000 = 331,200 c d 111,400 + 82,900 + 31,500 + 7,100 = 232,900 de 72,200 + 57,000 + 10,500 + 4,700 = 144,400 ed' 42,600 + 35,300 — 10,500 + 2,800 70,200 d'e' 20,300 + 17,700 — 31,500 4" 1,400 = 7,900 c'b' 4,200 + 3,900 — 52,500 + 500 = - 43,900 Since the combined shears in the panels ed' and d' c f are positive, counters are required in these two panels. Since 20 DESIGN OF A RAILROAD TRUSS BRIDGE § 79 the combined shear in the panel c' b f is negative, no counter is required in this panel. The combined maximum bending moments are as follows: Panel Combined Maximum Bending Moment, Point in Foot-Pounds b 3,741,300 + 2,556,700+1,323,000+239,400 = 7,860,400 if 6,180,000+4,301,700 + 2,268,000 + 410,400 = 13,160,100 d 7,571,900+5,246,100 + 2,835,000 + 513,000 = 16,166,Q00 e 8,165,600+5,618,500+3,024,000 + 547,200 = 17,355,300 38. Stresses in the Members.—Since the depth of the truss is 26.5 feet, the maximum chord stresses, to the nearest 100 pounds, are as follows: Member Stress, in Pounds a by be = 296,600 (tension) c d 13 ’26°5 100 = 496 > 600 (tension) dc 16,166 000 = 610>000 ( tension ) Zo.b BC 496,600 (compression) CD 610,000 (compression) DE 17^355^00 _ 054 ,900 (compression) Zu .0 The length of the diagonal is V26.-5* + 18 2 = 32.035 feet, 32.035 _ 2 209. Then, the stresses in the diag- and esc H = 26.5 onals are practically as follows: Diagonal a B Be Cd De Ed' or dE D’ c' or c D Stress, in Pounds 436,600 X 1.209 = 527,800 (compression) 331,200 X 1.209 = 400,400 (tension) 232,900 X 1.209 = 281,600 (tension) 144,400 X 1.209 = 174,600 (tension) 70,200 X 1.209 = 84,900 (tension) 7,900 X 1.209 = 9,600 (tension) The stress in the hip vertical is equal to the end shear on the floorbeam, which is 162,300 pounds. Then, the stress in B b is 162,300 pounds, tension. § 79 DESIGN OF A RAILROAD TRUSS BRIDGE 27 The stresses in the other verticals are as follows: Member Cc Dd Ee Compressive Stress, in Pounds 232,900 + 7,000 = 239,900 144,400 + 7,000 = 151,400 70,200 + 7,000 = 77,200 The above combined stresses are shown on the members in Fig. 14. _ DESIGN OF MEMBERS 39. Bottom Chord.— Since the working stress in ten¬ sion is 16,000 pounds per square inch, the required net areas of the bottom chord members are as follows: Member Required Net Area, in Square Inches a b y be 296,60(1 = lg g4 16,000 cd 496,600 _ 31 04 16,000 de i 610,000 ,q jo 16,000 Fig. 15 shows the usual type of member used for the bottom chords of riveted trusses. It consists of two vertical webs c , four flange angles d , and, when necessary, two side plates e. The two sides of the section are connected by double latticing / at the top and bottom. The customary method of riveting the parts of the section together is shown in the figure. In computing the net section, allowance must 28 DESIGN OF A RAILROAD TRUSS BRIDGE §79 be made for the greatest number of rivet holes in each part of the section. The rivets are i inch in diameter. The usual depth of this type of member varies from about 12 to 24 inches; in the present case, it will be made 15 inches. (b) The clear width between the webs depends on the width of the web members and other details, and is generally from 12 to 24 inches. 40. Design of Bottom Chord Members. —1. Mem - bers ab a?id be (Fig. 14).—The bottom chord will be spliced in panels cd and d' c' close to the joints d and d\ so that the parts that form a b and b c will also form part of c d . The fol- * lowing shapes will be used for ab and be: 2 web plates 15 in. X I in. 4 angles 3i in. X 3|- in. X I in. Three holes must be deducted from each web, and one hole from each angle; then, the net area of the section is as follows: 2 plates 15 in. X I in., net section = 2 (5.625 — 3 X .375) . . . = 9.0 square inches 4 angles 3i in. X 3h in. X t in., net section — 4 (3.98 —.625) . . = 13.42 square inches Total net section . . = 22.42 square inches This is somewhat more than necessary, but it is desirable to use these sizes in the present case, because, by the addi¬ tion of two side plates of about the same thickness as the angles, the next member cd can be formed. 2. Member cd. —As the web is 15 inches deep and the flange angles are 3i inches in width, the side plates will be made 8 inches in width, and two rivet holes will be deducted §79 DESIGN OF A RAILROAD TRUSS BRIDGE 29 from each plate. The net area of two plates 8 in. X t in. is 2 (6.0 — 2 X .75) = 9.0 square inches; this, added to the net area of ab and be, gives 9.0 + 22.42 = 31.42 square inches, which is sufficient. 3. Member de. —The same general form will be used as for c d, except that the webs will be made f inch thick instead of f inch. Then, the net area is as follows: 2 plates 15 in. X 1 in., net area = 2 (11.25 — 3 X .75) . . = 18.0 square inches 4 angles 3i in. X 3| in. X t in., net area = 4(3.98 — .625) = 13.42 square inches 2 plates 8 in. X f in., net area = 2 (5.0 — 2 X .625) . . = 7.50 square inches Total net area . . = 38.92 square inches 41. Location of Center Line of Bottom Chord. The center line of the bottom chord will be placed a short distance above the center of gravity of the section, so that the bending moment caused in the chord members by the eccentricity of the stress will neutralize that due to the weight of each member considered as a beam having a span equal to a panel length. The eccentricity will be found for the heaviest member (de), and will be made the same for all the bottom chord members. The gross area of de is 48.4 square inches, and this member weighs 161 pounds per linear foot. To allow for the weight of latticing, the weight will be taken about 10 per cent, greater, or, say, 17.5 pounds per linear foot. The required eccentricity^, in inches, is computed by the fol¬ lowing formula, given in Bridge Members and Details, Part 1: w r e = 8 ^ X 12 In the present case, w = 175, l = 18, and 6 * = 610,000 (Art. 38);. therefore, _ 175 X 18 y = .14 inch, or, say, i inch 8X610,000 Then, as the center of gravity is 7f inches from the bottom of the angles, the center line will be 7f + i = 7f inches above the bottom of the bottom angles. 30 DESIGN OF A RAILROAD TRUSS BRIDGE §79 42. Main Diagonals and Counters. —In this article, only the tension diagonals will be designed; the design of the end post will be considered in connection with the design of the top chord. The required net areas are as follows: Member Required Net Area in Square Inches Be 400,400 = 25 02 16,000 cd 281,600 = 17 60 16,000 De 174,600 = 10 91 16,000 dE 84, 900 = 5gl 16,000 cD 9,600 = g0 16,000 For members requiring more than about 15 square inches net area, the form of cross-section most frequently used is /Q)(Q\ mm if \QVQ/ \QXQ/ Fig. 16 lb 1 -It- i ir Fig. 17 the same as that of the bottom chord members that have just been considered. For the others, either two or four angles latticed together as shown in Figs. 16 and 17 are used. §79 DESIGN OF A RAILROAD TRUSS BRIDGE 31 1. Member Be .—Where the diagonals are riveted to the connection plates, the web will be reduced by three i-inch rivet holes, and each angle by one rivet hole. The follow¬ ing section will be used for this member: Square Inches 2 plates 15 in. X i in., net area = 2(7.5 — 3 X .5) = 12.00 4 angles 3i in. X 3i in. xf in., net area = 4(3.98 —.625) = 13.42 i -— Total net area = 25.42 2. Member Cd. —The following section will be used for this member: Square Inches 2 plates 14 in. X i in., net area = 2(7.0 — 3 X .5) = 11.00 4 angles 3 in. X 3 in. X t in.,net area = 4(2.11 — .375) = 6.94 Total net area = 17.94 3. Member De .—As the required net area is less than 15 square inches, this member will be composed of four angles. The section of each angle will be decreased by one 1-inch rivet hole. The following section will be used for this member: 4 angles 4 in. X 3 in. X 2 in., net area = 4 (3.25 — .5) = 11 square inches 4. Member dE .—This member is a counter, and the same type of member will be used as for D O or 5| inches, below the top of the angles; then, the center line is 5f + i = 5i inches below the top of the top flange angles. The different positions of the center of gravity of the top chord members create an objection to the use of unsymmetrical sections. 47. End Post.—In Desig7i of a Highway Truss Bridge, it was found necessary to revise the design of the end post after the wind stresses on it had been computed; so, in the present case, the end post will not be designed before the wind stresses have been found. For the purpose of compu¬ ting the exposed area of the web members, the width of the end post will be taken equal to that of the top chord. 135—20 38 DESIGN OF A RAILROAD TRUSS BRIDGE § 79 DESIGN OF LATERAL SYSTEM WIND STRESSES 48. Wind Pressure. —The force to be resisted by the lateral system consists of the wind pressure on the train and that on the trusses, as specified in B. S. y Art. 27. The wind pressure on the train is given as 300 pounds per linear foot of track; that on the trusses depends on the exposed area of the members. To find the exposed area, the length of each member, center to center of joints, will be multiplied by the width as seen in elevation; the latter width will be taken 1 inch greater than the width of web, channel, or angle that forms the member. On this basis, the exposed area of the web members of one truss is 548 square feet; of the top chord, 144 square feet; and of the bottom chord, 192 square feet. The exposed height of the floor will be taken as the distance from the top of the rail to the top of the bottom chord, since the portion of the floor system lower than this is sheltered by the bottom chord. In Fig. 10 (a), it can be seen that the exposed height of the floor is about 3 feet; then, the exposed area will be 3 X 144 = 432 square feet. 49. Upper Lateral Truss. —The upper lateral truss will be designed to resist the wind pressure on the top chord and one-half that on the web members. The exposed area of the top chord is 144 square feet, and one-half that of the 548 web members is —- = 274 square feet; the combined A exposed area is, therefore, 418 square feet. In B. S., Art. 27, it is specified that the wind pressure shall be taken as 50 pounds per square foot on twice the exposed area of one truss, when, as in this case, it produces greater stresses than any other specified wind loading. Then, the total pressure 30 § 70 DESIGN OF A RAILROAD TRUSS BRIDGE to be resisted by the top lateral truss is 2 X 418 X 50 = 41,800 pounds. As the top chord is 108 feet long, this corresponds to or P oun( ^ s P er li near foot; and, since the panels are 18 feet in length, the panel load is 387 X 18 = 6,970 pounds. The upper lateral truss, with the wind panel loads, is shown loads; the latter do not affect the stresses in the truss, but will be considered later in connection with the portal. The stresses in the members, determined in the usual way, are shown in Fig. 24. 50. Lower Lateral Truss.—The lower lateral truss will be designed to resist the wind pressure on the train, and that on the floor, bottom chord, and one-half the web mem¬ bers. It is necessary in every case to ascertain which of the two alternative, loadings given in B.S., Art. 27, causes the greater stresses. The wind pressure of 300 pounds per linear foot on the train and 30 pounds per square foot on one truss and the floor will first be considered. The panel load due to the pressure on the train is 300 X 18 = 5,400 pounds, and will be taken 40 DESIGN OF A RAILROAD TRUSS BRIDGE §79 as a live panel load. The exposed area of the floor is 432 square feet; that of the bottom chord, 192 square feet, and that of one-half the web members, 274 square feet; the total exposed area is, then, 898 square feet, the pressure on which, at 30 pounds per square foot, is 30 X 898 = 26,940 pounds. Since the bottom chord is 144 feet long, this pres¬ sure corresponds to ———Q, or 187,.pounds per linear foot; 144 and, since the panels are 18 feet in length, the panel load is 187 X 18 = 3,370 pounds. The total panel load, including the wind pressure on the train, is, therefore, 5,400 + 3,370 = 8,770 pounds. The alternative wind pressure of 50 pounds per square foot on the exposed area of the floor plus twice the exposed area of one truss will now be considered. The exposed area of the bottom chord is 192 square feet; one- half that of the web members, 274 square feet; and that of the floor, 432 square feet. Then, the total wind pressure on the lower lateral truss is 50 X [2(192-f 274) + 432] = 68,200 pounds. Since the bottom chord is 144 feet long, this pressure corresponds to -, or 473.6, pounds per 144 linear foot, and, since the panels are 18 feet in length, the panel load is 18 X 473.6 = 8,525 pounds. Since this panel load is less than that found in the preceding paragraph, the latter will be used, as it will cause greater stresses in the members of the lateral system. The lower lateral truss and wind panel loads are shown in Fig. 25. There are seven full panel loads of 3,370 pounds each, dead wind pressure, and seven panel loads of 5,400 §79 DESIGN OF A RAILROAD TRUSS BRIDGE 41 pounds each, live wind pressure. In finding the stresses in the members of the lateral truss, the live panel loads are assumed to be placed so as to cause the greatest stresses in the members, in the same way that vertical live load is con¬ sidered in finding the stresses in vertical trusses. The stresses caused in the members of the bottom lateral truss by the combined dead and live wind loads are shown in Fig. 26. The half panel loads at a and a' have not been considered, because they are transmitted directly to the masonry and do not affect the stresses. 51. Transverse Frames. The depth of portal and transverse frames depends on the amount of room above the overhead clearance line. The distance between the center lines of the chords is 26 feet 6 inches; the bottom of the bottom chord is 7} inches below the center line, and the top of the cover-plate of the top chord is about 6 inches above the center line. The vertical distance from the bottom of the bottom chord to the top of the top chord is, therefore, 27 feet 7f inches, as shown in Fig. 27. Consulting Fig. 10 (a), it will be seen that the base of the rail is 3 feet 9f inches above Fig. 27 42 DESIGN OF A RAILROAD TRUSS BRIDGE § 79 the bottom of the bottom chord. The overhead clearance line is 22 feet above the base of the rail (B. S., Art. 18) and 25 feet 9f inches above the bottom of the bottom chord. This leaves 27 feet 7f inches — 25 feet 9f inches = 1 foot 10i inches from the top of the top chord to the overhead clearance line. This will be taken as the depth of the transverse frame. Curved brackets will be placed at each end of each frame. They will be made to extend out as far as the clearance diagram (B. S ., Art. 18) will allow; in the present case, 4 feet 6 inches from the center of the truss,, They can be made con¬ siderably deeper than this, but it is customary to make the depth but little greater than the width. In this case, they will be made to extend 5 feet below the bottom of the frame. The general type of frame and bracket is shown in Fig. 27. 52. Portal. — Fig. 28 is a plan of the portal, portal brackets, and end posts. The same general type of frame and bracket will be used as for the transverse frames, except that, as the portal is only about 2 feet deep, a plate-girder portal will be used. The portal brackets will extend 4.5 feet out from the center of the end post; the other dimensions shown in Fig. 28 are found from the corresponding dimen¬ sions in Fig. 27, by multiplying the latter by esc H (1.209). The force P that acts at the top of the portal is one-half the total wind pressure on the upper chord, or 20,900 Fig. 28 79 DESIGN OF A RAILROAD TRUSS BRIDGE 43 pounds. Making - use of the formulas given in Stresses in Bridge Trusses, Part 5, assuming the point of inflection of the end post to be half way between the bottom of the bracket and the bottom of the post, the stresses are found to be practically as follows: Direct stress in end posts, — = 20,900 X 20.8 4 = 27 200 pounds b 16 Bending moment on end posts, P (h f — d — d') = 10,450 X 12.58 X 12 = 1,577,500 inch-pounds Shear on portal, — = 20 ’ 900 * 20,8 4 = 27,200 pounds b 16 In the plate-girder portal, the flanges are assumed to resist the entire bending moment, and the required area at any section is found by means of the formula A = M s h in •g which M is the moment about a point in the opposite flange directly opposite the section under consideration. The formula for moments about points in the top flange is jyr _ P h' _P Jl' X ~ ~2 At F, x = 4.5, and at F ', x = 11.5; then, 20,900 X 20.84 20,900 X 20.84 X 4.5 moment at F = moment at F' = 2 16 = 95,300 foot-pounds; 20,900 X 20.84 20,900 X 20.84 X 11.5 2 16 = — 95,300 foot-pounds The formula for moments about points in the bottom flange is M= Ph! Pd Ph'x 2 ' 2 b At D, x — 4.5, and at D', x = 11.5; then, moment at D = 20,900 X 20.84 . 20,900 X 2.22 2 * 2 _ 20,900 X 20.8 4 X = 118 500 foot-pounds; 16 44 DESIGN OF A RAILROAD TRUSS BRIDGE § 79 . f n/ 20,900 X 20.84 . 20,900 X 2.22 moment at D' — —- --— —- 2 2 20,900 X 20.84 X 11.5 16 = 72,100 foot-pounds The value of h s is not known, but may be assumed as a little less than d; in this case, it will be assumed as equal to 24 inches. Since the above moments are of one kind when the wind blows in the direction shown, and of the other kind when it blows in the opposite direction, the stresses in the flanges reverse, and in finding the required area of flange, the greatest moment in each flange must be increased by .8 of the moment at the other point in the same flange ( B . S., Art. 34). The greatest moment in the lower- flange is at D. Then, the required area of the upper flange is (118,50° + .8 X 72,100)j2 = 6 . 51 square inche , The 16,000 X 24 moments at th.e two ends of the upper flange are the same. Then, the required area of the lower flange is l.f X 95,300 X 12 K da • i — --—r- = 5.36 square inches j.6,000 X 24 There is also a direct stress caused in the bottom chords of the trusses by the direct stress in the end post; it is P h' found bv multiplying-by cos H , which gives 27,200 b X .5619 = 15,300 pounds, compression on the windward, and tension on the leeward, side; only the latter stress need be considered, since the former simply decreases the tension in the bottom chord. DESIGN OF MEMBERS 53. Upper Lateral Truss.- —Comparing the wind stresses in the top chord members as given in Fig. 24 with the combined dead- and live-load stresses as given in Fig. 14, it is seen that the former are in no case as great as 25 per cent, of the latter, so that they need not be considered. For the diagonals, it is well first to find the allowable stress in the smallest member that can be used. In B. S. } Art. 86, it is specified that no member of a lateral truss § 79 DESIGN OF A RAILROAD TRUSS BRIDGE 45 shall be less than Si in. X Si in. X t in. Allowing for one i-inch rivet hole, the net area of one angle of these dimensions is found to be 2.48 — .38 = 2.10 square inches. Since the working stress in tension is 16,000 pounds per square inch, the allowable stress in one angle is 2.1 X 16,000 = 33,600 pounds. As this is greater than the stress in any diagonal of the upper lateral truss, one Si" X Si" X i" angle will be used for each member. The transverse strut will be made as shown in Fig. 29, which is a customary form. Each flange will be composed of two Si" X Si" X i" angles connected by double latticing of S" X t" bars. In the present case, as the greatest stress Fig. 29 in a transverse strut is 17,400 pounds (member CC lf Fig. 24), these angles are sufficiently large. 54. Lower Lateral Truss.- —In the'design of the upper lateral truss, it was found that the allowable stress in one Si" X Si" X a" angle is 33,600 pounds. As this is greater than the stresses in the diagonals in the panels cd and de. Fig. 26, one Si" X Si" X i" angle will be used for each of the diagonals in these panels. In the panel be , the required net area of each diagonal is = 2.13 square inches. One Si" X Si" X tV' angle will be used.(net area = 2.87 — .44 = 2.43 square inches). In the panel a b f the required net area of each diagonal is LTTTr!?! = 2*89 square inches. One Si" X Si" X tV' angle 16,000 will be used (net area = 3.62 — .56 — 3.06 square inches). 46 DESIGN OF A RAILROAD TRUSS BRIDGE The wind stresses in the lower chord members are less than 25 per cent, of the combined dead- and live-load stresses, and so need not be considered. 55. Portal.—The shear on the portal was found in Art. 52 to be 27,200 pounds. This is treated in exactly the same way as the shear on any other plate girder. A web I inch thick will be used; the depth 2.22 feet will, for convenience, be called 26 inches. Then, the intensity of shearing stress is ^,200 = 2,790 pounds per square inch. 26 X t Consulting Table XXXVI, it is seen that the allowable unsupported distance is 37 inches; so that no stiffeners are required. The required net area of the top flange is 5.51 square inches, and that of the lower flange is 5.36 square inches (Art. 52). It is customary to make the section of the top and bottom flanges the same. Two b" X 3" X iV 7 angles will be used for each [net area = 2(3.31 — .44) — 5.74 square inches]. The required pitch of flange rivets, computed by the formula used in Art. 13, comes out greater than 6 inches, so that a pitch of 6 inches will be used ( B . S., Art. 41). 56. Design of End Post. —The combined dead- and live-load stress in the end post is shown in Fig. 14 to be 527,800 pounds. The direct stress due to the wind is 27,200 pounds; the total stress is, therefore, 527,800 + 27,200 = 555,000 pounds In addition, there is a bending moment due to the wind equal to 1,577,500 inch-pounds (Art. 52). The following section will be used: Gross Area, in Square Inches 1 cover-plate 22 in. X f in. 1 3.7 5 2 top flange angles 3|- in. X 3i in. X f in. 7.9 6 2 web-plates 15 in. X I in. 1 5.00 2 side plates 8 in. X i in. 1 0.00 2 bottom flange angles 3i in. X 3i in. X i in. 7.9 6 Total gross area = 5 4.67 79 DESIGN OF A RAILROAD TRUSS BRIDGE 47 These shapes will be placed as shown in Fig. 30, the edge of the top flange angles being made flush with the edge of the cover-plate. Consulting Table XI, and applying the rule given in Bridge Members and .Details, the actual width of the leg of a 3J" X 3-i" X i" angle is found to be 3f inches. Since the width of the cover-plate is 22 inches, the distance between the vertical legs of the flange angles is 22 — 3i — 3f = 14| inches; since the webs are i inch thick, the clear distance between them is 13f inches. The length of the end post center to center of connections is 32.035 feet = 384.4 inches; the radius of gyration with respect to the axis X' X pass¬ ing through the center of gravity of the section, which is 5.63 inches from the top of the top flange angles, and 9.62 inches from the bottom of the bottom flange angles, is 5.55 inches. Then, — = r 384.4 = 69.3, J x* \ Cenfer of Gravity ITT „_ /S 3 "- I ^ k | Ik aj T 0\ 1 J . w > j 5.55 and the working stress (Table XXXV) is 12,630 pounds per square inch. The working stress for the combined dead, live, and wind stresses will be taken 25 per cent, greater than this, or 1.25 X 12,630 = 15,790 pounds per square inch, as explained in Design of a Highway Truss Bridge. Since the total combined direct stress is 555,000 pounds, and the area of cross-section is 54.67 square inches, the intensity of r Fig 9 f 30 direct stress is 555,000 = 10,150 pounds per square inch. The 54.67 intensity of stress due to bending will be found from the formula Me s — I in which M = bending moment; c = distance to the extreme fiber; / = moment of inertia of the section about the axis Y f Y, Fig. 30. 48 DESIGN OF A RAILROAD TRUSS BRIDGE § 71) The value of / is found to be 3,068. Then, s = 1*577,500 X 11 _ f^ 00 q p OUn ds per square inch 3,068 Then, the total fiber stress is 10,150 -f- 5,660 = 15,810 pounds per square inch. This is greater than the allow¬ able intensity of stress, but, as the difference is very little, there is no necessity of revising the section. DETAILS CHORD SPLICES 57. Splice in Top Chord. — The top chord will be spliced in panel CD, Fig. 14, just to the left of D. The splice will be composed of the following shapes, the total gross area of which is slightly less than the gross area of CD, but is sufficient: Gross Area, in Square Inches 1 cover-plate 22 in. X 2 in. 11.00 2 inside web-plates 15 in. X "re in. 13.125 2 outside side plates 14 in. X 2 in. 14.0 0 2 bottom flange plates 3i in. X t in. 2.625 Total area = 40.750 Each part should have sufficient rivets on each side of the splice to transmit its proportion of the total stress. The pro¬ portion transmitted by each portion is found by multiplying its gross area by the working stress for the member CD (14,740 pounds per square inch, Art. 45). The number of rivets is found by dividing this stress by the value of one rivet; l-inch rivets are used, shop driven on one side of the splice, and field driven on the other side. The value of a field rivet in single shear, 5,410 pounds, will be used and the same number of rivets put in on each side of the joint. The required number of rivets in each of the different splice plates is as follows: i r w oo- v/i- 11.00 X 14,740 onr > . 1 splice plate 22 m.X 2 in.,-- J -= 30.0 rivets 5,410 79 DESIGN OF A RAILROAD TRUSS BRIDGE 49 i r , , 1 r • w 7 . 6.56 x 14,740 1^7 0 in • , 1 web-plate 15 in. X ie in.,--P 7 - 7- 2 -= 17.9, say 18, rivets 5,410 i • j i 4 , i a ‘ w i • 7.00 X 14,740 io i on ■ 1 side plate 14 in. X £ in., - -= 19.1, say 20, rivets 5,410 1 bottom plate 3^ in. X f in., ^^J* — = 3.6, say 4, rivets 5,410 The splice is shown in Fig. 31, in which (a) is the top view, Q O O ON L O >" • iQ Q Q Q O Q JQ • • • • • • • • • • • • O Q Q jO O O O Oj O O O )O i O^ O O O \ • • • O O O O'/ O 2 P/crfeTJpAf? f Plate /4"Ai (b) the elevation, and ( c ) the section and bottom view. The correct number of rivets is shown in each splice plate; it is customary to space the rivets in splices from 2 to 3 inches apart when staggered, as shown; in the figure, thfe pitch is 50 DESIGN OF A RAILROAD TRUSS BRIDGE §79 about 2a inches The latticing is continued right along by the splice, in the same way as for the rest of the member. 58. Splice in Bottom Cliord.—The bottom chord will be spliced in panel cd (Fig. 14) just to the left of d. The method of splicing is precisely the same as for the top chord, / / / • • • • |0 O 0 0 \ \ i -^- ! 0 j • • • • ~l 0 0 0 01 of 0 ! • • • *10000 10 0 ;• • • • | 0 0 0 0 ] 0 \ \ \ • • • *10000 / / (b) except that in this case net area must be considered instead of gross area. The following splice plates will be used: Net Area, in Square Inches 4 plates 4 in. X i ^6 in., 4(2.25 — .56) = 6.7 6 2 inside plates 15 in. X us in., 2(8.44 — 3 X .56) = 1 3.5 2 2 outside plates 14 in. X iin., 2(7 — 3 X .50) = 1 1.0 0 Total area = 3 1.2 8 i 79 DESIGN OF A RAILROAD TRUSS BRIDGE 51 The number of rivets required to connect each of the splice plates on each side of the joint is as follows: h i , A . 9 • 1.69 X 16,000 c a • + 1 plate 4 in. X Te in.,- „ - = 5.0 rivets 5,410 i . i c • w 9 • 6.76 X 16,000 on . 1 plate 15 in. X Te m.,-- - = 20 rivets 5,410 1 plate 14 in. X 2 in.,--- - - = 16.3 rivets 5,410 The splice is shown in Fig. 32, in which (a) is the top view, ( b) the elevation, and ( c ) a section and bottom view. The required number of rivets is shown in each splice plate. TRUSS JOINTS 59. The general method of connecting the members of riveted trusses to each other by means of connection plates or gussets has been explained in Bridge Members a?id Details . The connection of the members to each other at the differ¬ ent joints of the truss under consideration will now be dis¬ cussed. The rivets that connect the members to the gussets are i inch in diameter, and are part shop driven and part field driven. The value of each rivet in single shear is the small¬ est and the only one that need be considered. As most of the rivets are field driven, the value for field-driven rivets, which is 5,410 pounds, will be used, and the number of rivets required to transmit the stress to and from each member will be first calculated. These numbers are as follows: Member Number of Rivets BC a B Be Cd De 4 96.600 5,410 5 27,800 5,410 400,400 5,410 281.600 5,410 174,600 5,410 = 92, or 46 on each side — 98, or 49 on each side = 74, or 37 on each side = 52, or 26 on each side = 33, or 17 on each side 52 ' DESIGN OF A RAILROAD TRUSS BRIDGE §79 Member dE a b Bb Cc Dd Ee Number of Rivets 84.900 5,410 296,600 5,410 162,300 5,410 239.900 5,410 151,400 5,410 77,200 5,410 16, or 8 on each side 55, or 28 on each side 30, or 15 on each side 45, or 23 on each side 28, or 14 on each side 14, or 7 on each side As c D was made larger than necessary, to conform to the rule that no counter shall have an area less than 3 square inches, the number of rivets will be found for the stress that the section chosen can transmit at 16,000 pounds per square inch. Since the net area of cD is 3.47 square inches (Art. 42), the required number of rivets is 3.47 X 16,0 00 = 1Q 3 6 h ; d 5,410 60. Joint B. —Fig. 33 shows joint B : (a) is the eleva¬ tion of the joint, and shows one of the gussets e that are riveted to the inside of the top chord and end post, and to the ends of the members aB, B b, Be , and B C that meet at the joint; ( b ) is the elevation of the end of the portal^ and portal bracket h where they connect to the end post; (e) is the top view of the end of the top angles of the portal, and shows the bent plate i that connects the angles to the top chord of the truss; (d) is the plan of the top chord, and shows in section at the lower right-hand side a portion of the bottom flange of the top chord. The lines aB , B b, Be , and B C are the center lines of the different members, and meet at the point B\ the end post and top chord are cut off on the line // that bisects the angle between them. In drawing a riveted joint, the center lines of the members are first drawn, and then the lines that represent the members I 79 DESIGN OF A RAILROAD TRUSS BRIDGE 53 themselves are drawn parallel to the center lines. The gauge lines of the rivets are next drawn, and the rivets are spaced on these lines according to the rules for rivet spacing. In locating the gauge lines of angles, the distances to be used are the standard distances given in Table XII; in locating gauge lines on plates, two lines are first drawn parallel to the edges of the plate and I 2 or If inches from them, as /,/ in the end post. If they are more than about 4i inches apart, one or more lines from 2} to 4 inches apart are drawn between them, as in the case of the top chord. The rivets are located on these lines and staggered in such manner that they will be not nearer to each other than 3 inches; they are in general spaced closer in compression members than in tension members. The pitch of rivets in the end post, Fig. 33 ( a ), is about l! inches; in the top chord, 2f inches; and in the main diagonal, about 2\ inches. When convenient, lug angles / are riveted to the outstanding legs of the angles of the main members, so as to spread or distribute the stress over a greater width of gusset. Plates i inch thick will be used for the gussets e , as they are found to furnish sufficient section. One-half of B c is placed outside of the gusset on each side; the two halves are connected below the gusset by tie-plates 771 and lattice bars. The two channels that com¬ pose the vertical Bb are inserted between the gussets and connected to each other below them by tie-plates n and latticing. There are more rivets than necessary in this member, on account of the fact that the rivets in the end post and main diagonal control the size of the gusset, and the rivets in the vertical are spaced about 3i inches apart to hold the gussets and channels together more tightly. The rivets in the upper end of the vertical are also counted with the end post and top chord. Some designers prefer not to count a rivet twice; but when, as in this case, the members are on opposite sides of the gusset, there is no reason why they should not be counted; in such a case, however, allow¬ ance should be made by providing more rivets than are required by the computation. 135—21 54 DESIGN OF A RAILROAD TRUSS BRIDGE §79 In order to save extra work in handling and riveting, the gussets are usually riveted in the shop to one of the members. In Fig. 33, they are shown riveted to the top chord; on this account less rivets are required, since the value of a shop- driven rivet is greater than that of a field-driven rivet. 61. The cross-section of the portal is shown at oo , Fig. 33 ( a ). The lowest point p of the lower flange of the portal is made level with the bottom of the transverse struts of the intermediate panel points. The top flange angles are placed high enough to continue right across the top chord, as shown at g, and are connected to it by means of the bent plates i and r. The web of the portal is spliced at ss, Fig. 33 (b) , and the portion toward the truss is continued down to form the web of the bracket. This portion is con¬ nected to the end post by means of the connection angles t, which are usually 3fin. X 3iin. X fin. The rivets in these connection angles are spaced about 6 inches apart, except at the upper end, where those which also transmit the stress from the gusset to the end post are spaced 3 inches apart. The diagonal in the end panel of the top lateral truss is connected to the bent plate i, as shown at u in Fig. 33 ( d ), a lug angle being riveted to the main angle. 62. Joint C. —Fig. 34 shows the connection at joint C: (a) is the elevation of the joint, showing one of the gussets d and portions of the members B C, CD, Cc, and Cd; it also shows a cross-section h h of the strut and the connection of the strut to the vertical; (b) is an elevation of an intermediate cross¬ strut and bracket; (c) is a top view of the top chord, and shows the connection of the diagonals z, i of the upper lateral truss to the chord by means of the plate /. At the top of the vertical, which lies between the gussets, a diaphragm^ is riv¬ eted between the channels, in order to distribute the stress between the sides of the truss. The same remarks with regard to spacing of rivets, etc. apply here as in the case of joint B. There is a small excess of rivets in the vertical, but this is a frequent occurrence, and is done to get even spa¬ cing in the gussets. In the remaining top chord joints, the , . ' ' 1 i. --ass- ■ ■ 135 § 79 •'iQ. 34 ' §79 DESIGN OF A RAILROAD TRUSS BRIDGE 55 cross-strut and bracket connection will not be shown, but the holes for the connection of the strut to the vertical will be D shown. In Fig. 34 (c ), a short portion of the top chord at the lower right-hand side is given in section, so as to show the tie-plate j and part of the lattice bar k. E e • Pig. 36 63. Joints Z) and E .—Figs. 35 and 36 represent the joints at D and E t respectively. The connection of the 56 DESIGN OF A RAILROAD TRUSS BRIDGE §79 laterals and of the transverse frame and bracket is the same here as at joint C. In Fig. 35, the counter c D at the left is placed inside the gussets, so as to clear the main diagonal Cd where they cross each other at the center of the panel. For the same reason, counter dE, Fig. 36, is placed inside the gusset so it will not interfere with the main diagonal D e. The counters consist of two angles each, one on each side, and it is customary to place the angles so that the center of gravity shall coincide with the center line of the member. This is frequently done in the case of laterals that consist of single angles; but in the latter case it is customary to connect them as shown in Fig. 34 (<:), making the back of the angle coincide with the center line. * Diaphragms are placed inside the ver¬ ticals D d and Ee at the ends, and the same spacing of rivets is used as for Cc, so that the connection of the transverse struts will all be the same. 64. Joint a. —Fig. 37 represents the joint a. In this figure, ( a ) is the elevation, and shows one of the gussets b that connect the end post and the bottom chord, both these members being placed outside the gussets. The center lines and gauge lines are first laid off, and then the rivets are spaced along these lines. Those in the end post are spaced about 2i inches apart and staggered as shown, until a suffi¬ cient number of rivets is obtained, when the gusset is cut off square with the end post, as shown at c. The right-hand edge of the gusset is then carried vertically downwards into the bottom chord, as shown at d , and rivets are spaced about 3 inches apart from d to the end of this member. It is sel¬ dom necessary to count the numbers of rivets in the bottom chord connection, for the above spacing invariably gives sufficient rivets. Stiffeners e are placed both outside and inside of the connection over the pedestal to transmit the load to the bearing. The stiffener/, together with T60 X 14 X 15.5 s X (5 X , 15 ' 5 + 6 X 60) = 650 lb. per sq. in., compression. Ans. 18. Example of Kingrod Bridge. —Fig. 7 shows a small highway bridge composed of one layer of floor plank 2 inches thick, the floor joists /, the floorbeam g, and the kingrod trusses. In the figure, (a) is the plan, (b) is the elevation, and (c) shows the connection of the floorbeam g to the kingrod e of the truss. The floorbeams are continued 4 feet 3 inches beyond the truss on each side, and inclined struts z, usually 4 in. X 6 in., are placed between the end of the floorbeam and the top joint of the truss to hold it in place. This is a very economical and serviceable bridge for localities to which it is adapted. 6x6 Fast 13 14 WOODEN BRIDGES 80 EXAMPLES FOR PRACTICE 1. The panel load W of the truss shown in outline in Pig. 5 is 20,000 pounds. If the span is 20 feet, and the height is 4 feet, what is the direct stress in the top chord? Ans. 25,000 lb., compression 2. The load w per linear foot on the truss shown in outline in Fig. 6 is 2,400 pounds. If the span is 25 feet, and the height is 5 feet, what is the direct stress in the member b Ans. 37,500 lb., compression 3. What is the direct stress in the member B a' of the truss referred to in example 2? Ans. 50,500 lb., tension 4. What is the direct stress in the member b a' of the truss referred to in example 2? Ans. 46,900 lb., compression 5. What is the greatest bending moment on the top chord of the truss referred to in example 2? Ans. 562,500 in. -lb. QUEENPOST AND QUEENROD TRUSSES 19. Description of the Queenpost Truss. —Fig. 8 represents a queenpost truss, which is similar to the king¬ post truss, except that the former has three panels instead of two. The inclined struts e> e are placed in the center panel to provide for the shear in that panel; the top chord a a is made continuous from end to end. The rod cc is some¬ times made in one piece, and sometimes in two pieces; in the latter case, a turnbuckle is inserted in the center panel. The castings /, / at the bottoms of the posts provide a bear¬ ing area for the posts and also for the ends of the inclined struts. The bottoms of the castings are grooved in order to afford the rods a better bearing. 20. Description of tlie Queenrod Truss. —When there is not sufficient room for the posts and rods to project below the floor, the queenrod truss shown in Fig. 9 is used. This truss is similar in every way to the kingrod truss, except that there are three panels in the former instead of two. Castings g are bolted to the upper ends of the inclined struts, to provide a bearing for the upper ends of the inclined web members J Fig. 21 be spliced, and the whole is bolted together as shown in the figure. The stress is transmitted to the cleats and by them is transmitted around the joint. Another form of splice is shown in Fig. 22; it differs from the splice shown in Fig. 21 principally in that the cleats d, d are made of iron or steel instead of wood. Fig. 22 Fig. 23 shows a combination of parts that is sometimes used to splice the bottom chords of Howe trusses. In this figure, ( a) and {b) are castings with short cylindrical pro¬ jections e , e. Holes are bored in the sides of the members 28 WOODEN BRIDGES §80 to be spliced, and the castings placed in such a position that the projections e , e fit exactly into the bored holes. The castings are bolted in this position. The cleat / shown at ( d ) is then hooked over the projections g,g on the cast¬ ings, and the wedge w, shown at (e ), is inserted between Q=r = r-. __f ) (d) Fig. 23 the end of the right-hand projection and the inside of the cleat. The wedge is driven in so as to make the cleat fit tight at both ends. Two cleats are used in this case, one on each side, in the same way as two are used in Figs. 21 and 22. 33. Design of Splice.—The splice shown in Fig. 21 may fail in any one of five ways. Referring to Fig. 24, Fig. 24 they are as follows: (1) by the breaking of the cleats between a and a' in tension; (2) by the shearing of the cleats from a to b along the grain; (3) by the crushing of the ends of the fibers where the cleats and the shoulders on the stick come together at ac in bearing along the grain; (4) by §80 WOODEN BRIDGES 29 the shearing of the end of, the stick from c to c’ along the grain; (5) by the breaking of the main stick where it is cut into the most; that is, between c and d , in direct tension. In a well-designed joint, the resistance to failure from each of these causes should be as nearly as possible the same, and, in designing, the dimensions are so proportioned that the working stresses given in Art. 8 are not exceeded. 34. Case I. —In the first case mentioned in the pre¬ ceding article, the amount of stress that can be transmitted by the cleats in tension is the product of the net area of the cleats and the working stress in tension. In finding the net area of a cleat, it is necessary to deduct from the gross section the area of the bolt holes. As a rule, it is suffi¬ ciently accurate to take the net depth of the cleat as 2 inches less than the gross depth. For example, if each cleat is made of white oak 16 inches deep and li inches thick, the net area of two cleats [since the net depth is 2 inches less than the gross depth, Fig. 21 (b)~\ is 2 X (16 — 2) X H = 42 square inches. The working stress s t in tension for white oak is given in Art. 8 as 900 pounds per square inch; then, the amount of stress that can be transmitted by the two cleats in tension is 42 X 900 = 37,800 pounds. 35. Case II. —In the second case, the amount of stress that can be transmitted by the two cleats in shearing along the grain is the product of the area in shear and the allow¬ able shearing stress. For example, if each cleat is made of white oak 16 inches deep, and the length ab (Fig. 24) sub¬ jected to shear is 12 inches, the area subjected to shearing is 2 X 12 X 16 = 384 square inches. Then, since the working stress s s in shear along the grain is, for white oak, 100 pounds per square inch (Art. 8), the amount of stress that can be transmitted is 384 X 100 = 38,400 pounds. 36. Case III. —In the third case, the amount of stress that can be transmitted by the bearing area between the cleats and the shoulders on the member is the product of the bearing area and the working stress in bearing along the grain. For example, if the shoulders ac are li inches wide, 30 WOODEN BRIDGES §80 the depth of the member is 16 inches, and the cleats are of white oak and the member of white pine, the area of bearing is 2 X 16 X 1 i = 48 square inches. The allowable intensity Sg of bearing along the grain of white oak is given in Art. 8 as 1,250 pounds, and on white pine as 800 pounds per square inch. Since the latter is the smaller, it must be used; the stress that can be transmitted is, therefore, 48 X 800 = 38,400 pounds. 37. Case IV. —In the fourth case, the amount of stress that can be transmitted by the section cc\ Fig. 24, of the member in shearing along the grain is the product of the area in shear and the allowable shearing stress. For exam¬ ple, if the member is white pine, the depth of the member 16 inches, and the distance cc' 24 inches, the area in shear is 2 X 16 X 24 = 768 square inches. Then, since the working stress s* in shear is, for white pine, 50 pounds per square inch, the amount of stress that can be transmitted is 768 X 50 = 38,400 pounds. 38. Case V.—In the fifth case, the amount of stress that can be transmitted by the member where it is cut into to admit the cleats is the product of the net area of the decreased portion of the member and the working stress in tension. In finding the net area of the stick, it is customary to deduct the area of the bolt holes from the gross section of the member. As a rule, the bolts are staggered at this connection, so that it is sufficient to take the net depth of the stick as 1 inch less than the gross depth. For example, if the stick is white pine and has a gross width of 7 inches, the gross depth is 16 inches and the notches are li inches deep, the net width is 7 — 2 X li = 4 inches, and the net depth (decreased by one bolt, Fig. 21) is 16 — 1 = 15 inches; then, the net area is 4 X 15 = 60 square inches. Since the working stress s t in tension is 650 pounds per square inch (Art. 8), the amount of stress that can be transmitted is 60 X 650 = 39,000 pounds. 39. The dimensions of the splice that has been consid¬ ered in the examples in the preceding articles are shown in 80 WOODEN BRIDGES 31 Fig. 25. This form of splice is frequently used. The bolts that hold together the various parts of the splice should not be considered as transmitting any part of the stress; they Fig. 25 serve simply to hold the parts so that they will work together. 40 . [Lateral System.— The lateral trusses are formed in the same way as the vertical trusses. The transverse members are composed of rods and the inclined members of t timber. They are connected to the sides of the chords by chord blocks, or castings, in the same way as the web mem¬ bers of the vertical trusses are connected to the chords. EXAMPLES FOR PRACTICE 1. If the cleats shown in Fig. 24 are of white oak, 14 inches deep and lj inches thick, how much stress can they transmit without exceeding the working stress in tension? Ans. 27,000 lb. 2. If, in example 1, the length of each cleat subjected to shear is 12 inches, how much stress can the cleats transmit without exceeding the working stress in shearing along the grain? Ans. 33,600 lb. 3. If the main , member is of white pine and the width of the shoulder where the cleat bears on the member is 1-j- inches, how much stress can be transmitted without exceeding the working stress in bearing along the grain?^ Ans. 28,000 lb. 4. If the length of the member from the shoulder to the end is 20 inches, how much stress can be transmitted by the member without exceeding the working stress in shearing along the grain? Ans. 28,000 lb. 5. If the thickness of the member where it is cut into for the ends of the cleats is 3g inches, how much stress can be transmitted without exceeding the working stress in tension? Ans. 29,600 lb. 32 WOODEN BRIDGES § 80 / TOWNE LATTICE TRUSS 41. General Description. —Fig. 26 shows a type of truss, called the Towne lattice truss, that is used to some extent at the present time: (a) is the elevation of the end portion of a truss, and ( b ) is a cross-section on C C. In this truss, all the members are wood, the only metal being the bolts used to hold the members together. The chords consist of the horizontal members cc, dd, ee , and //. The two portions cc and dd form the top chord; the two portions ee and // form the bottom chord. Each portion of the chords is composed of several sticks, as shown in cross-section in Fig. 26 ( b .) The web consists of a large number of flat planks, the ends of which are connected to each other and to the chords. The horizontal member gg consists of two pieces and serves the purpose of stiffening the web members near the center of the length. The trusses are stiffened at the ends by means of vertical timbers h h and i i bolted to the sides of the web members. . The ends of the trusses rest on blocking j j that is supported on the bridge seats. 42. Floorbeams. —When these trusses are used, the floorbeams are connected below the bottom chord as shown at kk in Fig. 26. The bolts /, / pass through the lower portion of the bottom chord and through the ends of the * floorbeams. The latter are usually spaced from 2 to 3 feet apart in this type of bridge. 43. Connection of Web Members to Chords. —The method of connecting the web members to the chords is shown in detail in Fig. 27, in which a a and bb are web members and cc is the chord member. At d, the intersection of the center lines of the members, an iron bolt about 1 inch in diameter is driven through a bored hole, and tightened up. Oak pins e,e, commonly called treenails, having the Fig. 26 34 WOODEN BRIDGES §80 same diameter and nearly the same length as the bolt, are also driven through bored holes, and serve to transmit the stresses to and from the members. The amount of stress that can be transmitted to any of the members by the joint shown in Fig. 27 is usually assumed in practice to be 7,500 pounds. 44. Splices in Chord Members. —The sticks that form the top chord are joined by cutting the ends square and bringing them into good contact in much the same way as in the Howe truss. The several sticks that form the bottom chord are usually spliced as shown in Fig. 28. The ends are brought together, and wrought-iron or steel pieces d y d are inserted in holes cut in the members; long U-shaped bolts e, e are placed over the ends of the pieces d , d. The nuts /, / serve to tighten up the bolts §80 WOODEN BRIDGES 35 so that they will bear firmly against the top and bottom of the pieces d, d. 45. .Lateral System.— The lateral trusses are usually made the same as for the Howe truss, and the members con¬ nected to the sides of the members by castings or chord blocks. In addition, transverse frames are put in, as shown in Fig. 26 (b), and curved knee braces or brackets m are placed between the bottom of the transverse frame and the web members of the truss. 46. Protection of Truss. —To prevent deterioration on account of the weather, it is customary to shelter this truss, and in some cases Howe trusses also, by building a roof and sides, as shown in Fig. 26 (b). The roof n n is usually composed of 1-inch boards supported by and nailed to joists o,o on top of the truss. The sides p,p also consist of 1-inch boards that are nailed to the pieces q, q bolted to the sides of the trusses. COMBINATION TRUSSES 47. General Description.— Although, strictly speak¬ ing, a combination truss is any truss in which some mem¬ bers are of wood and some of metal, the name is generally restricted to pin-connected trusses in which the compression members are wood and the tension members are steel or iron. The Pratt and the Baltimore are the two forms of truss that are most used for this construction. In all com¬ bination trusses, the details of the connections of the wooden members are arranged in such a way that the members can be removed and replaced without the necessity of taking down the truss. This arrangement is necessary because the wooden members decay and must be renewed from time to time, while the steel, if properly [taken care of, will last almost indefinitely. 48. Bottom Chord Joint. —Fig. 29 shows a bottom chord joint of a combination truss: {a) is the elevation and (b) the side view of the joint. The only difference between 36 WOODEN BRIDGES 80 this joint and that in a steel pin-connected truss is in the connection of the vertical member to the pin. This is accom- D r \ EE Fig. 29 plished by attaching- to the end of the member a casting /, having a projection^ that fits into a groove in the end of the, / Fig. 30 member. The casting has two sides or webs h that bear on the pin. The bolt i is simply for the purpose of holding the §80 WOODEN BRIDGES 37 member in place on the casting, and does not transmit any stress. The thickness of metal in the casting, which should always be a steel casting, is usually not less than 1 inch. 49. Top Chord Joint. —Fig. 30 shows a top chord joint of a combination truss: {a) is the top view, (b) is the eleva¬ tion, and (c) is the side view of the joint. The ends of the top chord members abut on the sides of a casting//, and are $ A A.rSi a Fig. 31 held in place on the casting by means of bolts g,g that pass through the ends of the members and through projec¬ tions h, h on the sides of the casting. The vertical abuts on the under side of the casting, and is held in place by a bolt i that passes through a projection j on the under side. The sides of the casting that are in contact with the ends of the chord members are connected by webs or diaphragms k> k 38 WOODEN BRIDGES §80 about opposite the centers of the sticks. These webs pro¬ vide also a bearing for the pin that passes through the cast¬ ing and holds the ends of the eyebars in place. The casting is further stiffened by inclined webs /, / between the webs k, k and the bearing surfaces. 50 . Hip Joint. —The detail at the hip joint is somewhat different from the other top chord joints. Fig. 31 shows a hip joint: (a) is the top view, and (b) the elevation of the joint. The end post a B and the top chord B C are bolted to separate castings; these are furnished with webs or diaphragms for bearing on the pin arranged in such a way that the hip vertical b B and the end diagonal Be can also connect to the pin. In the figure, the hip vertical is placed inside of both castings, and the end diagonal is placed between the two castings. 51 . Design of Pins. —The pins in a combination truss are designed in the same general manner as those in a steel pin-connected truss. The method of procedure is fully described in Design of a Highway Truss Bridge , Part 2. ROOF TRUSSES INTRODUCTION 1. Conditions Governing Use of Roof Trusses. Trusses are frequently employed to support the roofs of buildings in cases where a large area of floor clear of inter¬ mediate walls and columns* is desired, and when so used are called roof trusses. Roof trusses are usually set at right angles to the length of the building, so as to make the span as short as possible; and their ends either rest on top of the side walls or are supported by columns embedded therein. 2 . Types of Roof Trusses. —The same general types of trusses are used for.roofs as for bridges, except that the inclinations of the chords of roof trusses are made to conform with the slope of the roof and the required underneath clear¬ ance. This gives rise to special types, some of which are illustrated in Figs. 1 to 15. The simplest type of roof truss is shown in Fig. 1; it consists simply of the two inclined b d Fig. 2 Fig. 1 struts a b and be , the lower joints of which are connected by the horizontal chord or bottom tie ac. This type of truss may be used for spans up to 20 feet. In Fig. 2, the long horizontal chord or tie ac is supported at the center by the vertical tie b d. This type of truss may be used for spans COPYRIGHTED BY INTERNATIONAL TEXTBOOK COMPANY. ALL RIGHTS RESERVED §81 ROOF TRUSSES §81 up to 30 feet. For spans from 30 to 40 feet, the form shown in Fig. 3 may be used. In this truss, the inclined struts ab and be are supported at their center points by the inclined struts de and df ; the lower ends of these struts are sup¬ ported by the vertical tie b d. For spans up to 40 feet, the type shown in Fig. 4 is sometimes used. When built of timber, with the tie ac continuous, or in one piece from a to c , the diagonals in the center panel are sometimes omitted. The types shown in Figs. 5, 6, 7, 8, and 9 are used a Fig. 5 for all spans, the particular type chosen for any special case depending on local conditions, and, to a great extent, on the judgment of the designer. In Figs. 5, 6, and 7, the slopes a b and be are each divided into a number of equal spaces. In Fig. 7, the web members dd andee are at right angles to the 81 ROOF TRUSSES 3 slope. In Figs., 8 and 9, the construction shown above the trusses in dotted lines is for the purpose of giving the sur¬ face of the roof an even slope. 3 . All the trusses considered in the preceding article have horizontal lower chords. It is often desired to have greater headroom, or clearance, beneath the trusses at the center of the building than at the sides. In such a case, the walls or col¬ umns that support the ends of the roof trusses are frequently not carried as high as the required height at the center of the building, and one of the types of roof trusses shown in Figs. 10 to 13 is used. \ 4 ROOF TRUSSES § 81 4. For supporting- the roofs of station platforms, grand stands, etc., the arrangement shown in Fig. 14 is frequently employed. It con¬ sists of a truss resting on two columns; the ends of the truss over¬ hang the supports, forming cantilevers. In shop and mill buildings, in which the roof trusses are supported on col¬ umns, the lower chord of the truss is frequently connected to the columns by inclined struts, as shown at b in Fig. 15; these struts are usually pro¬ jections of one of the web members. The projection at the top of Fig. 15 is commonly called a monitor, and serves the purpose of providing vertical skylights at c , c. The mem¬ bers of the monitor do not form part of the truss, but are simply for the pur¬ pose of transmitting to the truss any loads that may come on the monitor. 5. Distance Be¬ tween Trusses. The distance between trusses depends to some extent on the architectural design; that is, the location of windows, doors, etc. It is customary, in designing a building, to arrange it in a number of sec¬ tions, or bays, and to have the same openings for windows in each bay. Between the bays, the wall is carried up solid from the foundation. The trusses usually rest on this solid Fig. 14 §81 ROOF TRUSSES 5 wall or on columns embedded in it. It is bad practice to have a truss rest on a wall directly over the opening for a door or a window. There is usually a roof truss at the junc¬ tion of each two bays and in each end wall. It has been found in practice that it is well to have the distance between trusses about one-fourth the span. They are seldom placed less than 10 feet and never more than about 50 feet center to center. 6. Span and 1=5 Fig 15 Rise. —The horizontal distance /, Fig. 16, between the sup¬ ports of a roof truss is called the span. The vertical dis¬ tance r from the top of the truss to the level of the supports is called the rise. The vertical distance h from the inter¬ sections of the inclined top members to the center of the lower chord at the center of the span is called the depth of the truss. In some cases, h is made equal to r, in others h is made less 2, 3, 5, 6, 7, are called main rafters, or simply rafters. The lower chord or member ac is called simply the chord. The trusses are connected to each other at the joints of the rafters by beams called purlins, as shown in cross-section 6 ROOF TRUSSES §81 in Fig. 17 (a). When the distance between trusses is small, channels or I beams are used for purlins; when the distance between trusses is greater than 20 or 25 feet, riveted girders or trusses are used. The ends of the purlins sometimes rest on top of the rafters, as shown in Fig. 17 ( a ), and are some¬ times connected to the web members, as shown in Fig. 17 ( b ). Fig. 17 As the surface of the roof is not horizontal, there is a tend¬ ency for the purlins to sag or deflect sidewise. To counter¬ act this tendency, the purlins are connected at one or more points intermediate between the trusses by means of tie-rods c , Fig. 17 ( a ), that run from the purlin at the top, or ridge, down both sides of the roof. 8. On top of the purlins, and at right angles to them, are smaller beams d , Fig. 17 (a ), close together. They are called common rafters, and are usually spaced about 2 feet apart. They support the roofing material and covering, and transmit any load on the roof to the purlins. The purlins transmit the load to the trusses at the joints of the main rafter. 9. Panel Length.—The panel length of a roof truss is the distance between two successive joints of the rafter, measured along the slope. It may be found by the formula , V/’ + 4 r' P =- n in which p is the panel length and n is the number of panels. §81 ROOF TRUSSES 7 10 . Slope of Roof. —The slope of the roof is frequently spoken of in terms of the ratio between the rise and the span. This ratio is called the pitch. Thus, a pitch of i means that the rise of the roof is i of the span. The slope is also spoken of as the ratio of the vertical distance, called the rise, to the corresponding- horizontal distance, called the run; thus, 1 of rise to 2 of run means that for every 2 feet measured hori¬ zontally, the roof rises 1 foot vertically. This ratio is usually spoken of as 1 in 2, 1 in 3, etc. This form will be used in the following articles. 11 . Roof Covering. —The materials commonly used for roof coverings are shingles, slate, tile, copper, tin, corru¬ gated iron, felt, asphalt, tar, and gravel. Thin slabs of rein” forced concrete have also been used in a number of cases. With all of these materials, except the reinforced-concrete slab, it is customary to put wooden sheathing about 1 inch thick directly on top of the common rafters, so as to afford $ the roofing material a flat surface on which to rest. Shingles and tile may be used on all slopes greater than 1 in 2, and slate and corrugated iron on all slopes greater than 1 in 3. Copper and tin with the joints well soldered, and also rein¬ forced concrete, may be used on all slopes. Felt, with asphalt or tar and gravel, is used on flat roofs having slopes less than 1 in 4. If used on greater slopes, the tar or asphalt will run down when it gets warm, and leave the upper part of the roof exposed. The kind of roof covering adopted depends mainly, on the type of building and on the amount of money available. 8 ROOF TRUSSES 81 LOADS AND STRESSES 12. Loads. —The loads that must be supported by roof trusses are the dead load, which consists of the weight of roof covering, rafters, purlins, and trusses; the live load, which consists of the heaviest snowfall; and the wind load, which is due to the pressure of the wind. In addition, if ceilings or balconies are attached to the chords of the trusses, the latter must be designed to support them. 13. Lead Load. —The dead load depends on the kind of roof covering used, the distance between rafters and purlins, and the distance between trusses and supports. The; approximate weights of the usual roof coverings and ceiling are as follows: Pounds Materials per Square Foot Shingles. 2 Slate. 8 Corrugated tile. 9 Tile on 3-inch fireproof blocks .35 Copper and tin . \\ Corrugated iron. 2 Felt with tar or asphalt and gravel. 9 Wooden sheathing 1 inch thick. 4 Lath-and-plaster ceiling .10 Glass skylight, including frames . 8 Concrete slab, including reinforcement (per inch of thickness) ..12 The weight of the common rafters, purlins, and trusses should first be assumed, and the weights revised after the members have been designed. If balconies are attached to the trusses, their weight must be calculated. 14. Snow Load. —The weight of snow that may fall and remain on a roof in winter depends on the slope of the roof §81 ROOF TRUSSES 0 and on the climate. It is customary to assume a weight of 30 pounds per square foot in the northern part of the United States, and 10 pounds per square foot in the southern part, on roofs the slope of which is not greater than 1 in 2. These loads are gradually decreased for steeper roofs, up to a slope of 2 in 1, for which the snow load is neglected, it being assumed that the snow will slide off by its own weight. 15. Wind Pressure. —In considering the pressure of the wind on a roof, it is customary to resolve it into two components, one parallel with the surface of the roof, arid the other normal to that surface. In calculating the stresses due to the wind pressure, the former component is neglected. The intensity of the normal component varies with the slope, being greater for steep than for flat roofs. The normal intensities of wind pressure per square foot on roofs of dif¬ ferent slopes are usually taken as follows: Normal Slope of Roof Pounds per Square Foot Flat roof. 0 , 1 in 5 (tV pitch).10 1 in 2i (i pitch).20 1 in 2 (i pitch).25 1 in lv (i pitch).30 1 in 1 (| pitch) . . . ..36 1 in I, and steeper. 40 Since wind can blow in but one direction at any one time, it is customary to consider but one side of a roof truss loaded with wind load at any one time. 16. Panel Loads. —Each truss, except those at the ends of the building, is assumed to support one-half of the load in each of the two adjacent bays. If the lengths b of the bays are all equal, and the panel lengths p of the rafter are equal, and the load per square foot of roof surface for any loading is w , the panel load W for that loading is given by the formula W = w bp 10 ROOF TRUSSES 81 For dead and snow loads, the panel loads are vertical, and the stresses in the members are greatest when there is a full panel load at each of the joints of the rafter, except at the supports, where the panel loads will W_ 2 ’ Fig. 18. For wind loads, the panel loads are inclined and con¬ sidered normal to the roof. The stresses caused in the mem¬ bers by the wind load are greatest when there is a full panel load at each joint of the rafter on one-half of the truss, except at the top and bottom joints, where the loads will be as Plo . 19 Fig. 18 shown in Fig. 19. The panel loads on the trusses at the ends of the building are, as a rule, one-half those on the other trusses. 17. The panel loads on the sloping parts of the roof and monitor shown in Fig. 20 are found in the same way as for other trusses. When the length of the sloping part of the roof of the monitor is equal to a panel length of the W' truss, the panel loads —- on the roof of the monitor are A equal to the half panel loads -—- on the truss. In addition, A W" there are horizontal wind forces of ——- at the top and bottom A of the vertical side of the monitor, W" being the total wind pressure on one bay of the vertical side, computed for a wind pressure of 40 pounds per square foot. 18. If the side walls of the building are of masonry, the wind pressure on them need not be considered in the design 81 ROOF TRUSSES 11 of the roof trusses. If the sides are exposed to the action of the wind, and are of wood, corrugated iron, or other build¬ ing material attached to the outside of the columns, the wind pressure on the side of the building causes stresses in the roof trusses, and hence must be taken into account. It will be assumed that the wind pressure on the side of the building for a length of one bay is IV"', and that one-half of this is transmitted to the w" bottom and one-half 2 “ to the top of the Fig. 20 column, giving a concentrated load at each of these points, as shown in Fig. 20. 19. Reactions. —The reactions due to the dead and snow loads are vertical, and are found in the same way as for a simple beam. The reactions due to the wind load may be vertical or inclined, according to the manner in which the ends of the trusses are connected to the supports. When the ends of the trusses rest on top of the walls, and both ends are anchored down, as is usual with spans less than about 12 ROOF TRUSSES § si 70 feet long, it is customary to assume that both reactions are parallel to the direction of the wind panel loads, as shown in Fig. 21. The magnitude of the reactions R x and R t can be found by resolving the resultant F of the wind ROOF TRUSSES 13 § 81 panel loads into two components passing through the points of support and parallel to the resultant. 20. For longer spans, it is customary to place rollers under one end to provide for changes in length due to changes in temperature. Since the truss is free to move horizontally at the expansion end, no horizontal force can be transmitted to the abutment or support at that end, except the friction of the rollers, which is usually neglected. All the horizontal components of the inclined wind panel loads are then assumed to be transmitted to the support at the fixed end. The directions of the reactions when the wind is blowing on the expansion end of the truss are shown in Fig. 22. The directions of the reactions when the wind is blowing on the fixed end of the truss are shown in Fig. 23. In each of these figures the circle at the left-hand end repre¬ sents the rollers under that end of the truss, at which end the reaction is vertical. 21. In the case represented in Fig. 24, it is customary to find the horizontal and vertical components of the reac¬ tions separately. For this purpose, the various wind forces are resolved into their vertical and horizontal components. The vertical components of the reactions at C and C are then found by taking moments about C' and C, respectively. The horizontal components of the reactions are found on 135—25 14 .ROOF TRUSSES 81 * ' ■ « ■ ■ the assumption that one-half the sum of the horizontal com¬ ponents of the wind forces goes to each support. If the columns are fixed in direction at the bottom, as is usually the case, points of inflection may be taken half way between the lower ends of the inclined braces and the bases of the columns, the same as in the case of the end post of a through truss bridge. % 22. Method of Calculation. —The stresses in the mem¬ bers of roof trusses can be found by either the analytic or the graphic method. The members of most roof trusses have so many different inclinations, however, that the work required by the analytic method is comparatively great; so this method is seldom used in practice. The graphic method is especially useful for this purpose. The stresses should be found sep¬ arately for the vertical and for the inclined loads. When one end of the truss rests on rollers, the stresses due to the wind should be found separately for the wind blowing in each direction. In the latter case, there are three conditions of loading for which the stresses in the members must be found; namely, (1) full snow load together with the dead load; (2) wind pressure on the expansion end; (3) wind pressure on the fixed end. The stresses found in (1) must be combined with those found in (2) or (3) to obtain the greatest stress in each member. In some trusses, there are members in which the stress is tension when the wind acts on one side, and compression when it acts on the other. The methods of calculation will now be illustrated by prac¬ tical examples. FIRST ILLUSTRATIVE EXAMPLE 23. Data. —The first illustrative example will be the truss shown in Fig. 16 (a), the data for which will be assumed as follows: Distance between trusses. b — 15 feet Span./ = 60 feet Rise . .. r = 15 feet Number of panels. n — 8 §81 ROOF TRUSSES 15 Roofing material . Slate on 1-inch wooden sheathing Dead load of rafters, purlins, and trusses .... 10 pounds per square foot Snow load.30 pounds per square foot Supports . . Truss anchored to the tops of the walls at both ends 24. Panel Length. —The panel length p is found by the formula in Art. 9. Since / — 60 feet, r — 15 feet, and n = 8, the formula gives y ^60* + 4 x 15* 8 = 8.385 feet 25. Panel Loads.—The panel load W for any loading is found by the formula in Art. 16. In the present case, b — 15 feet and p = 8.385 feet. 1. Dead Panel Load .—The dead panel load consists of the weight of the slate, the sheathing, and the rafters, purlins, and trusses. The weight of the slate is given in Art. 13 as 8 pounds per square foot, and the weight of the sheathing as 4 pounds per square foot. The weight of the rafters, pur¬ lins, and trusses is given in the data as 10 pounds per square foot. Then, the total dead load w is 8 + 4 + 10 = 22 pounds per square foot, and for the dead panel load W d we have W d = 22 X. 15 X 8.385 = 2,767 pounds There will be seven full panel loads on the truss, and, in addition, two half panel loads over the supports. Since the latter loads do not cause any stresses in the members of the truss, they will not be considered here. 2. Snow Panel Load .—The snow load is 30 pounds per square fool. Then, for the snow panel load W s , we have W s = 30 X 15 X 8.385 - 3,773 pounds 3. Wind Pa7iel Load. —Since the rise is 15 feet and the span is 60 feet, the roof has a pitch of i, or a slope of 1 in 2. In Art. 15, the normal pressure of the wind on a roof having a slope of 1 in 2 is given as 25 pounds per square foot. Then, denoting the wind panel load by W w , Wio — 25 X 15 X 8.385 = 3,144 pounds 16 ROOF TRUSSES §81 There will be three full wind panel loads, and, in addition, the two half panel loads at the bottom and top, respectively. Since the wind panel loads are inclined, it is well to consider the panel load over the support. 26. Stresses Due to Dead and Snow Loads. —In this example, it is best to consider the reactions and stresses due to the vertical loads (combined dead and snow loads) sepa¬ rately from those due to the inclined loads (wind pressure). The dead panel load was found in Art. 25 to be 2,767 pounds; and the snow panel load, 3,773 pounds. Then, the total ver¬ tical panel load is 2,767 + 3,773 = 6,540 pounds. There are seven full panel loads, as shown in Fig. 25 {a). Since the loading is symmetrical, the reactions are equal, and each is Z - X 6,540 __ 22,890 pounds. The stress diagram for this loading is shown in Fig. 25 (b). The panel loads are laid off on the line 0-7 , and, since the reactions are equal, the point 8 is located half way between 0 and 7. The remainder of the diagram is constructed, as usual, by drawing vectors in the stress diagram parallel to the members in the truss. It has been explained in Graphic Statics that, in construct¬ ing the stress diagram for a truss, it is necessary to consider, one after another, the joints at which but two of the stresses are unknown. In the truss under consideration, if the dia¬ gram is started for the joint a , it will be found possible to con¬ struct it also for the joints b and B. At each of the joints c and C there will then be three members in which the stresses are unknown, and the stress diagram cannot be drawn directly. In this case, it is customary to consider the next joint d on the rafter, and to assume, temporarily, the stress in one of the rafter members. This assumed stress is only an auxiliary quantity used for the purposes of calculation. It is known that the point 13 in the stress diagram will lie on a line 2-13" through 2 parallel to the member cd, and that the point 15 lies in a line through 8 parallel to C C. The stress in cd will be temporarily assumed equal to 13'-2 t §81 ROOF TRUSSES 17 and the polygon 13'-2-3-14'-13' drawn for the joint d. The polygon 13'-14'—15'—12'-13' is then drawn for the joint D. This locates the point 12 '. The stress in c C, however, will be represented by a vector through the point 11 parallel to c C, and point 12 must lie on this line. The actual location of point 12 is then found by drawing the line 12'-12, parallel to 13'-2 and 3-14', to its intersection with 11-12", the latter 18 ROOF TRUSSES §81 line being parallel to c C. This gives the stress in c C {11-12) and makes it possible to complete the stress diagram without difficulty. When the loading is symmetrical, the stress diagram is sometimes drawn for only one-half of the truss, but it is well, as in the present case, to draw the entire diagram for a check. The stresses in the members on one side of the center, as scaled from the stress diagram, are as follows, using / the plus sign for compression and the minus sign for tension: Member Vector Stress (Pounds) a b 0- 9 4- 51,200 be 1-10 + 48,300 cd 2-13 + 45,300 d e 3-14 + 42,400 a B 8- 9 - 45,800 BC 8-11 - 39,200 CO 8-15 - 26,200 bB , 9-10 + 5,800 Be 10-11 - 6,500 c C 11-12 + 11,700 cD 12-13 - 6,500 dD 13-14 + 5,800 CD 12-15 - 13,100 De 14-15 - 19,600 27. Stresses Due to Wind Toad.—The wind panel load was found in Art. 25 to be 3,144 pounds. The load¬ ing on the truss is shown in Fig. 26 {a). Since the loading is inclined and unsymmetrical, the reactions can be found by the graphic method with much less work than by the analytic method. For convenience of reference, the same notation is used as in Fig. 25; the forces 4-5, 5-5, and 6-7 may be taken as zero. The loads are laid off on the line 8'-4, Fig. 26 (6), the pole P is chosen, and the rays P-8' and P-4 are drawn. The resultant of all the wind panel loads is equal to 8'-4 , and acts through the center of the left-hand side of the rafter; that is, through the joint c, and in the direction eg. 19 20 ROOF TRUSSES §Sl The funicular gfhg is drawn, and the reactions 4-8 and 8-8' are found by drawing the ray P-8 parallel to the closing line ./ h. The remainder of the work consists simply in drawing the stress diagram shown in Fig. 26 {b ). The points 16, 17, 18, 19, 20, and 21 are not shown in the stress diagram, since they coincide with 15. This indicates that for this loading there are no stresses in the web members to the right of 15. The wind stresses in the members, as scaled from the stress diagram, are as follows: Member Vector Stress (Pounds) a b, bc,c d,de a B BC CC , C'B/B'a’ b B,d D B c,c D cC CD De e d', d f c', c' b', b' a' eD', D' C', d'D',D'c',\ c' C',c'B',b' B' ) 0-9 , 1-10, 2-13, 3-14 8-9 8-11 8-15 9- 10, 13-14 10- 11, 12-13 11-12 12-15 14-15 4-15 + 14,100 - 15,800 - 12,300 * - 5,300 + 3,100 - 3,500 + 6,300 - 7,000 - 10,500 + 7,900 0 28. Total or Combined Stresses.—The maximum total stresses in the members are found by combining the stresses due to the vertical loads with those due to the inclined loads. The total combined stresses are: Member Stress (Pounds) Member Stress (Pounds) a b + 65,300 bB + 8,900 be + 62,400 Be - 10,000 c d + 59,400 eC + 18,000 d e + 56,500 eD - 10,000 a B - 61,600 dD + 8,900 BC - 51,500 CD - 20,100 CC - 31,500 De - 30,100 It is not necessary to find the stresses in the members on the right-hand side of the center, as they are the same as those in the corresponding members on the left-hand side. 21 ROOF TRUSSES SECOND ILLUSTRATIVE EXAMPLE 29. Data. —The second illustrative example will be of the same general type as the truss shown in Fig. 5. The data will be assumed as follows: Distance between trusses . b = 22 feet Span. I = 100 feet Rise. r — 20 feet Number of panels. n — 10 Roofing material . . Corrugated iron, no sheathing Dead load of rafters, purlins, and trusses . 20 pounds per square foot Snow load.25 pounds per square foot Supports .... Rollers at left end; right end fixed The outline of the truss is shown in Fig. 27. 30. Panel Length. —Since l — 100 feet, r — 20 feet, and n = 10, the formula of Art. 9 gives VlOO’ +Vx~20* 10 10.77 feet 31. Panel Loads. —The panel load W for any loading is given by the formula in Art. 16. In the present case, b — 22 feet and p — 10.77 feet. 1. Dead Panel Load .—The dead load consists of the weight of the corrugated iron, and the rafters, purlins, and trusses. The weight of the corrugated iron is given in Art. 13 as 2 pounds per square foot. The weight of the rafters, purlins, and trusses is given in the data as 20 pounds per square foot. Then, the total dead load is 2 + 20 = 22 pounds per square foot, and for the dead panel load W d we have IV d — 22 X 22 X 10.77 = 5,213 pounds There will be nine full panel loads, and, in addition, two half panel loads over the supports. Since the latter loads 22 ROOF TRUSSES §81 do not cause any stresses in the members of the truss, they will not be considered. 2. Snow Panel Load .—The snow load is 25 pounds per square foot (Art. 29). Then, for the snow panel load W s we have Ws = 25 X 22 X 10.77 = 5,923 pounds 3. Wind Panel Load .—Since the rise is 20 feet and the span is 100 feet, the roof has a i pitch or a slope of 1 in 2^. In Art. 15, the normal pressure of the wind on a roof having § 81 ROOF TRUSSES 23 a slope of 1 in 2i is given as 20 pounds per square foot. Then, for the wind panel load W w we have W w = 20 X 22 X 10.77 = 4,739 pounds There will be four full wind panel loads, and, in addition, the two half panel loads at the bottom and top, respectively. Since the wind panel loads are inclined, the load over the support may cause stresses in the members of the truss, and it must be considered. 32. Stresses Due to Dead and Snow Loads.— -In this example, it is best to consider the reactions and stresses due to the vertical loads (combined dead and snow loads) separately from those due to the inclined loads (wind pres¬ sure). The dead panel load was found in Art. 31 to be 5,213 pounds; and the snow panel load, 5,923 pounds. Then, the total vertical panel load is 5,213 + 5,923 = 11,136 pounds. There are nine full panel loads, as shown in Fig. 28 (a). Since the loading is symmetrical, each reaction is 9 X 11,1 36 2 = 50,112 pounds. The stress diagram for this loading is shown in Fig. 28 (b) . The panel loads are laid off on the line 0-9, and, since the reactions are equal, the point 10 is located half way between 0 and 9. The remainder of the diagram is drawn as usual, presenting no special difficulty. This diagram is symmetrical about the line 10-28 , so that it is necessary to draw but one-half. For a check, however, it is always advisable to draw the complete diagram. The stresses in the members on one side of the center, as scaled from the stress diagram, are as follows: Member Vector Stress (Pounds) a b 0-11 + 134,900 be 1-12 + 134,900 cd 2-14 + 119,900 de 3-16 + 104,900 cf 4-18 + 90,000 a B 10-11 - 125,300 BC 10-13 - 111,400 CD 10-15 - 97,400 24 ROOF TRUSSES §81 Member Vector Stress (Pounds) DE 10-17 - 83,500 EE 10-19 - 69,600 bB . 11-12 + 11,100 Be 12-13 - 17,800 cC 13-14 + 16,700 Cd 14-15 - 21,700 dD 15-16 + 22,300 De 16-17 - 26,300 eE 17-18 + 27,800 Ef 18-19 - 31,100 - IF 19-20 0 33. Stresses Due to Wind Toad.— As it is stated in the data that one end of the truss rests on rollers, it is necessary to find the stresses both when the wind blows on the expansion end and when the wind blows on the fixed end. The wind panel load was given in Art. 31 as 4,739 pounds. 1. Wind on Expa?ision End .—The loading when the wind blows on the expansion end is shown in Fig. 29 («). The reaction at the left end is vertical; that at the right end is inclined, and its direction is found by resolving the result¬ ant A of the wind panel loads into two components, one ver¬ tical and passing through a , and the other inclined and passing through a'. A vertical line through a intersects the resultant of the wind panel loads at g, a'nd the line ga' gives the line of action of the reaction at a'. The wind panel loads are laid off on the line 10'-5 , Fig. 29 (b) . The magni¬ tudes of the reactions are found by drawing a vertical line through 10 f to its intersection with a line through 5 parallel to a'g. The stresses in the members are found by com¬ pleting the stress diagram in the usual way. Since the dia¬ gram is unsymmetrical, it is necessary to draw it complete. The stresses are as follows: i 25 Fig. 29 26 ROOF TRUSSES §81 Member Vector Stress (Pounds) a b 0-11 + 36,100 be 1-12 + 38,000 c d 2-14 + 33,100 de 3-16 + 28,100 e f 4-18 + 23,100 a B 10-11 - 32,700 BC 10-13 - 26,300 CD 10-15 - 19,900 DE 10-17 - 13,500 EF,FE',E'D'A D' C’, OB', B'a'J 10-19 - 7,200 bB 11-12 + 5,100 Be 12-13 - 8,200 cC 13-14 + 7,700 Cd 14-15 - 10,000 dD 15-16 + 10,200 De 16-17 - 12,000 e E 17-18 + 12,800 Ef 18-19 - 14,300 if F, fE', E'e', e' D', D' d ', d'C', C'c', \ d B', B' b' fe ', ef d, d' d, d b', b' a' 5-19 + 17,200 2. Wind oil Fixed End .—The loading when the wind blows on the fixed end is shown in Fig. 30 (a). The reaction at the left end is vertical; that at the right end is inclined, and its direction is found by resolving the resultant F of the wind panel loads into two components, one vertical and passing through a, and the other inclined and passing through a '. A vertical line through a intersects the resultant of the wind panel loads at g, and the line g a’ gives the direction of the reaction at a'. The wind panel loads are laid off on the line 4-10', Fig. 30 {b). The magnitudes of the reactions are found by drawing a vertical line through 4 to its intersection with a line through 10' parallel to ga'. The stresses in the members are found by computing the stress diagram in the usual way. They are as follows: 6380 27 Fig. 30 ROOF TRUSSES §81 Member Vector Stress (Pounds) ab, be, c d, de, e f b B, Bc, c C, Cd, dD,\ 4-20 + 17,200 0 De , eE, E /, fF ) iaB , BC, CD\ \DE, EE, EE') 10-20 - 16,000 /£' 20-21 - 14,300 E'e' 21-22 -f 12,800 e'D' 22-23 - 12,000 D'd’ 1 23-24 + 10,200 d'C 24-25 - 10,000 C'c r 25-26 + 7,700 c ' B' 26-27 - 8,200 B' b' 27-28 + 5,100 E'D' 10-22 - 22,300 D'C' 10-24 - 28,700 C'B' 10-26 - 35,100 B' a' 10-28 - 41,500 fe' 5-21 + 23,100 e'd' 6-23 + 28,100 d' c' 7-25 4- 33,100 c' b r 8-27 + 38,000 V a’ 9-28 + 36,100 34. Combined Stresses.—In order to find the total maximum stresses in the members, it is advisable to tabulate the stresses due to the different loadings, as shown in the table on the next page. Column 1 gives the members, column 2 gives the stresses in the members due to the vertical load, column 3 gives the stresses due to the wind pressure on the expansion end, and column 4 gives the stresses due to the wind pressure on the fixed end. The stresses given in column 5 are the sums of those given in columns 2 and 3; those given in column 6 are the sums of those given in columns 2 and 4. The stresses in column 7 are the maximum stresses in the members. They are taken from columns 5 and 6, the larger being taken in each case. TABLE OF STRESSES IN MEMBERS ( Pounds) 1 2 3 4 5 6 7 Mem¬ ber Stresses Due to Vertical Loads Stress Due to Wind on Expan¬ sion End Stress Due to Wind on Fixed End Total Stress Wind on Expansion ' End Total Stress Wind on Fixed End Maximum Total Stress a b + 134 , 9 °° + 36,100 + 17,200 + 171,000 + 152,100 + 171,000 b c + 134 , 9 °° + 38,000 + 17,200 + 172,900 + 152,100 + 172,900 c d + 119,900 + 33 . 100 + 17,200 + 153.000 + 137 , 100 + 153,000 d e + 104,900 + 28,100 + 17,200 + 133.000 + 122,100 + 133,000 ef + 90,000 + 23,100 + 17,200 + 113,100 + 107,200 + 113,100 a B -125,300 -32,700 — 16,000 -158,009 -141,30° — 158,000 B C — 111,400 — 26,300 — 16,000 -137,700 — 127,400 -137,700 CD - 97 > 4 °° — 19,900 — 16,000 -117,300 -113,400 -117,300 DE - 83,500 - 13 . 5 °° —16,000 — 97,000 - 99,500 - 99,500 EF — 69,600 — 7,200 — 16,000 — 76,800 — 85,600 — 85,600 bB + 11,100 + 5,100 0 + 16,200 + 11,100 + 16,200 B c — 17,800 — 8,200 O — 26,000 — 17,800 — 26,000 cC + 16,700 + O O 0 + 24,400 + 16,700 + 24,400 Cd — 21,700 —10,000 0 - 31,700 — 21,700 - 31,700 dD + 22,300 +10,200 0 + 3 2 > 5 °o + 22,300 + 32,500 De — 26,300 —12,000 0 - 38,300 — 26,300 - 38,300 eE + 27,800 +12,800 O + 40,600 + 27,800 + 40,600 Ef - 3Lioo -i 4 , 3 °° 0 - 45,400 - 31,100 — 45,400 fE 0 0 0 0 0 0 fE ' - 31,100 0 -14,300 - 31,100 - 45,400 - 45,400 E' e' + 27,800 0 + 12,800 + 27,800 + 40,600 + 40,600 e' D' — 26,300 0 — 12,000 — 26,300 - 38,300 - 38,300 D'd' + 22,300 0 + 10,200 + 22,300 + 32,500 + 32 , 5 00 d' a — 21,700 0 — 10,000 — 21,700 -' 31,700 - 30,700 C' c' + 16,700 0 + 7 . 7 °° + 16,700 + 24,400 + 24,400 c' B' 1 H 00 0 0 0 — 8,200 — 17,800 — 26,000 — 26,000 B' b' + 11,100 0 + 5,100 + 11,100 + 16,200 + 16,200 FE' — 69,600 — 7,200 —16,000 — 76,800 — 85,600 — 85,600 E' D' - 83,500 — 7,200 — 22,300 — 90,700 —105,800 —105,800 D'C' - 97,400 — 7,200 — 28,700 — 104,600 — 126,100 — 126,100 a B r — 111,400 — 7,200 -35.100 — 118,600 -146,500 -146,500 B' a' -125,300 — 7,200 —41,500 -132,500 — 166,800 —166,800 fe' + 90,000 +17,200 + 23,100 +107,200 + 113,100 + 113,100 e'd' +104,900 + 17,200 + 28,100 + 122,100 +133,000 + 133 , 00 ° d' c' + 119,900 + 17,200 + 33+00 + 137 , 100 + 153,000 + 153,000 c' b' + 134,900 + 17,200 + 38,000 + 152,100 + 172,900 + 172,900 b' a' + 134,9°° + 17,200 + 36+00 + 152,100 + 171,000 + 171,000 29 135—26 30 ROOF TRUSSES §81 It is not necessary to find the minimum stresses in the members of roof trusses, unless the stresses in some of the members' are reversed by the wind. The stress in //'is given as zero. This member simply serves to support the bottom chord at the center, and is sometimes omitted. THIRD ILLUSTR ATIVE EXAMPLE 35. Data.— The third illustrative example will be of the same general type as the truss shown in Fig. 15. The data will be assumed as follows: Distance between trusses. b = 18 feet Span. I = 80 feet Rise. r — 20 feet Height of monitor. 11 feet Number of panels. n — 8 Distance from bottom chord to base of columns. 30 feet Roofing material . . Concrete slab inches thick Vertical sides of monitor .... Glass skylights Dead load of rafters, purlins, and trusses . 18 pounds per square foot Snow load.30 pounds per square foot Supports.Columns fixed at the bottoms The outline of the truss is shown in Fig. 31. 36. Panel Length. —Since l = 80 feet, r = 20 feet, and n — 8, the formula of Art. 9 gives V80 2 + 4 X 20 2 8 11.18 feet 37. Panel Loads. —The panel load W for any loading is given by the formula in Art. 16. In the present case, b = 18 feet and p = 11.18 feet. 1. Dead Paiiel Load .—The dead load consists of the weight of the concrete slab, and that of the rafters, purlins, and trusses. In addition, the weight of the vertical sides of the monitor must be considered. The weight of concrete per square foot is given in Art. 13 as 12 pounds per inch §81 ROOF TRUSSES 31 of thickness; since the slab is 2 k inches thick, the weight is 2k X 12 = 30 pounds per square foot. The weight of rafters, purlins, and trusses is 18 pounds per square foot. Then, the total dead load on the sloping portion is 30 + 18 = 48 pounds per square foot, and for the dead panel load Wa we have: Wd = 48 X 18 X 11.18 = 9,660 pounds There are full panel loads of 9,660 pounds at b,c,g,c', and b', Fig. 32 [a), and half panel loads of 4,830 pounds, at a, d, e, e’ , d', and a \ Since the half panel loads at e and d (and the same applies to e’ and d') are in the same straight line, they will be considered together in laying out the force polygon as though they were both applied at e (e f for the other side). The half panel loads at a and a' must be considered, since they cause stresses in the columns. The weight of the sides of the monitor is 8 pounds per square foot. (See Art. 13.) Then, for the weight W m of each side per bay, we have Wm = 8x11x 18 = 1,584 pounds All this weight is applied at d and d f , but in laying out the force polygon it is better to consider it applied at e and e ’. The effect of considering the loads at d and d' as applied at e and e\ is to make the stresses in the verticals de and d' e' greater than the actual stresses by the amount of the loads at d and d' . The correction can easily be made later when the stresses are scaled from the stress diagram. 2. Snow Pa?iel Load .—The snow load is 30 pounds per square foot. Then, for the snow panel load W s we have W s = 30 X 18 X 11.18 = 6,037 pounds There are full panel loads at b , c , g , c\ and b\ and half panel loads at a, d, e , e\ d\ and a 1 ’. The half panel loads at 0981 © ^ U)8Cf9 §81 ROOF TRUSSES 33 e and d (e' and d') will be considered together as full panel loads at e ( e') in laying out the force polygon. There is no snow load on the sides of the monitor. 3. Wind Panel Load. —Since the rise is 20 feet and the span is 80 feet, the roof has a i pitch or a slope of 1 in 2. In Art. 15, the normal pressure on a roof having a slope of 1 in 2 is given as 25 pounds per square foot. Then, for the wind panel load W w on the inclined portion of the roof, we have W w = 25 X 18 X 11.18 = 5,031 pounds There are full panel loads at b and c , and half panel loads at a, d , e, and g. In addition to the above, the wind pressure on the vertical sides of the monitor and the building will be considered. It is customary to consider this pressure as horizontal and equal to 40 pounds per square foot. The total pressure on one bay of one side of the monitor is 40 X 18 X 11 = 7,920 pounds, one-half of which, 3,960 pounds, may be taken at each of the joints e and d. The total pressure on one bay of the vertical side of the building is 40 X 18 X 30 = 21,600 pounds, one- half of which, 10,800 pounds, may be taken at joint a. 38. Stresses Due to Dead and Snow Loads. —In this example, as in those preceding, the stresses due to the verti¬ cal loads will be treated separately from those due to the inclined loads. The vertical loads at the different joints of the truss are shown in Fig. 32 (a). Those at a, b, c,g, c', b', and a' are the sum of the snow panel loads and the loads due to the weight of concrete slab and roof framing. The loads at e and e' include the weight of the vertical sides. Since the loading is symmetrical, the reactions are equal. In the present case of vertical loading, it is customary to assume that the stresses in A B and B' A' are zero. The stress diagram is shown in Fig. 32 {b). It is drawn in the same way as though the truss were simply supported at a and a ', the members A B and B' A' being omitted. The i panel loads are laid off on the line 0-9, and, since the reac¬ tions are equal, the point 10 is located midway between 0 and 9. The stress diagram can be drawn for joints a, b, and B u ROOF TRUSSES §81 K without difficulty. It is then necessary to consider joints, then e and e ', and then d. At the latter joint, the stress in df is temporarily assumed equal to 19-17' , and the dia¬ gram 19-17'-16'-3-19 is drawn in dotted lines for the joints d\ joint D is then considered, and the diagram 16'—17'-18'-15'—16' drawn in dotted lines. The point 14 has already been found, and it is known that 15 lies on a line 14-15" passing through 14 and parallel to c C. The point 15 is then found by drawing 15-15' parallel to 3-16' to its intersection 15 with 14-15". The remainder of the stress diagram can now be drawn without further difficulty. The stresses in de and d'e', as indicated by the stress diagram, are each too large by 9,430 pounds, on account of the loads at d and d' having been considered as applied at e and e'. The stresses in the members on one side of the center, as scaled from the stress diagram, are as follows: Member o t Vector Stress (Pounds) EA 10- 0 + 64,400 A a 10- 0 + 64,400 AD 10-11 0 a b 1-12 + 126,400 be 2-13 + 126,400 cd 3-16 + 147,500 df 19-17 + 147,500 fe 19-20 - 18,300 dc 3-19 (34,500 - 9,400 + 25,100) eg 4-20 + 17,600 a B 11-12 - 113,100 DC 10-14 - 97,400 CF 10-18 - 56,500 bB 12-13 + 15,700 Be 13-14 - 22,200 eC 14-15 + 40,800 cD 15-16 - 34,600 CD 15-18 - 57,700 dD 16-17 + 34,600 Df 17-18 - 106,600 fF 18-23 0 • ... '» ' ' x ’ . I 135 §81 14894 §81 ROOF TRUSSES^ 35 39. Stresses Due to Wind Doad. —The diagram of the truss and the wind loads is shown in Fig. 33 (a). At joint a there is a horizontal pressure of 10,800 pounds due to the pressure against the side of the building, and also the inclined pressure of pounds, due to the normal pressure on the roof. At joipts b and c there are full inclined panel loads of 5,031 pounds, due to the normal pressure on the roof. At cL and e there are inclined panel loads of pounds due to the normal pressure on the roof, and horizontal pres¬ sures of 3,960 pounds due to the pressure against the side of the monitor. At the peak g there is an inclined pressure of --- 2 ”- pounds. The forces shown in dotted lines will be explained later. 40. In finding the reactions, it is necessary to consider the condition of the columns at E and E' . In Art. 35, it is stated that the columns are fixed at the bottoms, so that points of inflection may be assumed at i and i', half way between A and A, A' and E' , respectively. The reactions at i and i' will be inclined, and it is impossible to determine their * directions except by means of some arbitrary assumption con¬ cerning the distribution of the forces between i and i’. In practice it is frequently assumed that the horizontal com¬ ponent of each reaction is equal to one-half the sum of the horizontal components of the wind forces. This assumption probably gives the reactions as close to their actual values as any other assumption that can be made. It becomes neces¬ sary, therefore, to find the resultant of all the wind forces, and then to resolve it into two components, one passing through i and one passing through i', their horizontal com¬ ponents being equal to each other. The force polygon 0-1-2-3-4-5-6-7-8-9, Fig. 33 ( b ), is first laid out in a convenient position, the first point, 0 , being located for convenience on the line i i f connecting the points i and i'. The point P is then chosen for a pole, and is located arbitrarily, to simplify the work, on a horizontal line through the last point, 9, of the force polygon. The rays P-0, P-2, P-3, P-4, P-6, P-8, and P-9 are now drawn. 36 ROOF TRUSSES 81 The ray P-1 is left out because the dotted line 0-2 in the force polygon and ja in the space diagram represent the resultant of the forces 0-1 and 1-2, and this resultant can be considered instead of its two components. In like manner, the force 4-6 in the force polygon and md in the space diagram may be considered instead of the two components 4-5 and 5-6; the resultant en may be considered instead of 6-7 and 7-8. The line 0-9, Fig. 33 {5), gives the magnitude of the resultant of all the wind forces; the intersection q of the end strings of the funicular qjklmnp q, drawn in the usual way, gives a point in the line of action of the resultant. A line is then drawn through q parallel to 0-9 to represent the resultant. The vertical and horizontal com¬ ponents of 0-9 are given by 0-s and s-9, Fig. 33 (b). The point 9", half way between ^ and 9, divides the horizontal component of 0-9 into two equal parts, each of which is equal to the horizontal component of one of the reactions. 41. The vertical components of the reactions are found by considering the resultant F resolved into its vertical and horizontal components at t, the intersection of its line of action with it'. The vertical component 0-s is laid off on A E, vertically below i, so that iu = 0-s. The line u z 7 , the vertical t v, and horizontal v v' are then drawn. Then u v' is the vertical component of the reaction at z 7 , and v' i is the vertical component of the reaction at i. The point 10, where a vertical through 9" intersects vv f , is a point on the lines in the force polygon that represent the reactions. Drawing the lines 9-10 and 10-0, the magnitudes and directions of the reactions are found. The lines of action of the reactions are then found by drawing ir through i parallel to 10-0, and i' r through z 7 parallel to 9-10. The lines i r and z 7 r should intersect on the line of action of the resultant. This is a very good check on the work thus far. The vertical components of the reactions cause direct stresses in the columns. The horizontal components of the reactions cause bending moments and shearing stresses in the columns. §81 ROOF TRUSSES 37 42. In order to find the stresses in the members of the truss by the method of the stress diagram, it is necessary to treat the shearing stresses in the posts or columns a i and a! i’ as external forces applied at the joints a , A , a', and A'. To find the values of these shearing stresses, vertical lines are drawn through s and 9", Fig. 33 ( b ), the line s-9" repre¬ senting the horizontal component of the reaction at i. The line drawn through s intersects a horizontal line drawn through A at the point s'; the line through 9” intersects a horizontal line drawn |through a , which in this case coincides with the chord of the truss, at the point w. The line w s' is then drawn and continued to the point 10", its intersection with i i’. Then, 10"-0 , equal to 13,860 pounds, is equal to the shear in a A (and also in a' A') and may be represented as an external force acting at a (and also at a'), as shown by the dotted lines in the figure. The shear in iA, equal to 13,860 pounds, is equal to lO'-O; then, the horizontal force acting at the joint A due to the shears in a A and A i is equal to the sum of these shears, which is 27,720 pounds. This may be treated as an external force acting at A (and A ’), as shown by dotted lines. Those portions of the columns that lie below A and A' can be assumed to be removed and the direct stresses in them represented by the vertical external forces shown at A and A'. The force polygon for the external forces, including the forces shown in dotted lines at a , A , a', and A', is 10-10'-10"-0-l-2-3-4-5-6-7-8-9-9'-9"-10, Fig. 33 (b). The stress diagram can now be drawn in the usual way, ignoring the reaction at i and z 7 , and considering the parts iE and z 7 E' of the posts to be removed. It is not necessary to draw the stress diagram for wind load when the wind is blowing on the right of the truss; for it is obvious that when the wind blows from the right, the stress in any member is the same as the stress in the corresponding mem¬ ber on the other side of the center when the wind blows from the left. The stresses in the members, when the wind is blowing from the left, have the values given in the table on page 38, these values having been scaled, as usual, from the stress diagram. 38 ROOF TRUSSES §81 Member Vector Stress (Pounds) a b 2-12 4- 63,900 be 3-13 4- 66,400 cd 4-16 + 38,200 df 19-17 + 43,900 f e 19-20 + 4,000 eg 8-20 + 1,900 AB 10-11 - 39,200 bB 12-13 + 5,600 Be 13-14 - 47,200 eC 14-15 4- 25,200 cD 15-16 ' - 5,800 CD 15-18 - 35,700 Df 17-18 - 43,900 dD 16-17 + 5,800 ed 6-19 4- 1,100 a B 11-12 - 31,400 BC 10-14 - 25,700 CF 10-18 - 500 FC 10-23 - 500 C'B f 10-26 4- 11,800 B'a' 28-29 + 11,800 d'd 21- 9 4- 3,100 d'D' 22-24 + 3,100 D'f 22-23 4- 13,000 CD' 23-25 4- 17,400 d D> 24-25 - 3,100 d C 25-26 - 12,300 B'd 26-27 4- 39,200 b' B' 27-28 0 A' B' 10-29 4- 39,200 d g 20- 9 4- 3,100 fd 20-21 - 3,300 d' / 21-22 4- 5,800 d d' 9-24 4- 5,800 b'd 9-27 - 28,700 b' a' 9-28 - 28,700 E A' 10-10' 4- 3,100 A a 10"-11 4- 30,800 E' A' 9"—10 4- 14,900 A' a' 9'-29 - 12,800 fF 18-23 0 §81 ROOF TRUSSES 39 43. Combined Stresses.—The combined stresses are given in the table on page 40. In column 2 are given the stresses due to vertical loads, as found in Art. 38. In column 3 are given the stresses in all the members due to wind on the left, as found in Art. 42. The combined or total stresses in the members when the wind blows from the left are given in column 4; they are found by adding alge¬ braically for each member the stresses given for that member in columns 2 and 3. The combined or total stresses in the members when the wind blows from the right are given in column 5; they are found by taking for each member the stress given in column 4 for the corresponding member on the other side of the center. The maximum and minimum combined stresses are given in column 6; the former are printed in heavy type for the members on the left of the center; the latter are printed in ordinary type for the mem¬ bers on the right of the center. The maximum stresses, given in column 6 for the members on the left of the center, are the same for corresponding members on the right of the center; so it is necessary to give them but once. The same statement is true as regards the minimum stresses. EXAMPLES FOR PRACTICE 1. A truss having the form shown in Fig. 1 has a span of 20 feet and a rise of 5 feet. If the trusses are 8 feet apart, and the dead load is 40 lb. per square foot, what is the dead-load stress in a c? Ans. —3,600 1b. 2. A truss having the form shown in Fig. 2 has a span of 30 feet and a rise of 6 feet. If the trusses are 10 feet apart, and the ends are fixed at the tops of both walls, what is the wind stress in be when the wind is coming from the left? Ans. + 2,300 lb. 3. A truss having the form shown in Fig. 3 has a span of 36 feet and a rise of 12 feet. If the trusses are 9 feet apart, and the left end of each truss rests on rollers, what is the wind stress in a d when the wind is blowing on the left side of .the truss? Ans. —2,000 lb. 4. What is the wind stress in a d in the truss described in example 3, when the wind is blowing on the right side of the truss? Ans. — 2,600 lb. TABLE OF STRESSES ( Pounds ) 1 2 3 4 5 6 Mem¬ ber Stress Due to Vertical Loads Stress Due to Wind on Left Combined Stress Wind on Left Combined Stress Wind on Right Maximum or Minimum Combined Stress a b + 126,400 + 63,900 + 190,300 + 97,700 + 190,300 b c + 126,400 + 66,400 + 192,800 + 97,700 + 192,800 c d + 147,500 + 38,200 + 185,700 + 153,300 + 185,700 df + 147,500 + 43,900 + 191,400 + 153,300 + 191,400 fe — 18,300 + 4,000 — 14,300 — 21,600 — 21,600 eg + 17,600 + 1,900 + 19,500 + 20,700 + 20,700 A B 0 -39,200 — 39,200 + 39,200 + 39,200 b B + i 5 > 7 °° + 5,600 + 21,300 + 15,700 + 21,300 B c — 22,200 — 47,200 — 69,400 + 17,000 — 69,400 cC + 40,800 + 25,200 + 66,000 + 28,500 + 66,000 cD % 34,600 - 5,800 — 40,400 — 37,700 — 40,400 CD — 57,700 - 35 , 7 oo — 93,400 — 40,300 — 93,400 Df — 106,600 -43,900 — 150,500 — 93,600 — 150,600 dD + 34,600 + 5,800 + 40,400 + 37,700 + 40,400 e d + 25,100 + 1,100 + 26,200 + 28,200 + 28,200 a B — 113,100 -31,400 — 144,500 — 101,300 — 144,500 B C — 97,400 -25,700 — 123,100 — 85,600 — 123,100 C F — 5 6 > 5 °° - 5 °° — 57 , 00 ° — 57 ,ooo — 57,000 F C ' — 5 6 > 5 °° — 5 °° — 57 ,ooo — 57 ,ooo — 56,500 C ' B ' — 97,400 +11,800 — 85,600 — 123,100 — 85,600 B ' a ' — 113,100 +11,800 — 101,300 — 144,500 — 101,300 d ' e ' + 25,100 + 3 > IO ° + 28,200 + 26,200 + 25,100 d ' D ' + 34,600 + 3 > IO ° + 37 , 7 oo + 40,400 + 34,600 D'f — 106,600 +13,000 — 93,600 — 150,500 — 93,600 C ' D ' — 57 » 7 oo +17,400 — 40,300 — 93,400 — 40,300 c ' D ' — 34,600 - 3,100 — 37 , 7 oo — 40,400 — 34,600 c ' C ' + 40,800 -12,300 + 28,500 + 66,000 + 28,500 B ' c ' — 22,200 + 39,200 + 17,000 — 69,400 + 17,000 b ' B ' + 15.700 0 + 15,700 + 21,300 + I 5 , 7 0 0 A ' B ' 0 + 39,200 + 39,200 — 39,200 — 39,200 e ' g + 17,600 + 3 , 100 + 20,700 + 19,500 + 17,600 fe ' — 18,300 -- 3 , 3 oo - 21,600 — 14,300 — 14,300 d'f + 147,500 + 5,800 + i 53 , 3 oo + 191,400 + 147,500 c ' d ' + 147,500 + 5,800 + i 53 , 3 oo + 185,700 + 147,500 b ' c' + 126,400 — 28,700 + 97 , 7 oo + 192,800 + 97,700 b ' a ' + 126,400 — 28,700 + 97 , 7 oo + 190,300 + 97,700 EA + 64,400 + 3 , 100 + 67,500 + 79 , 3 oo + 79,300 A a " 4 ~ 64,400 + 30,800 + 95,200 + 51,600 + 95,200 E ' A ' + 64,400 + 14,900 + 79 , 3 oo + 67,500 + 64,400 A ' a ' + 64,400 — 12,800 + 51,600 + 95,200 + 51.600 0 0 0 0 0 40 81 ROOF TRUSSES 41 DESIGN TRUSSES AND LATERAL SYSTEMS 44. Kinds of Trusses. —Roof trusses are made of wood and of steel. The kind of material used depends on the particular conditions in each case. In temporary buildings, and in buildings that are not required to be fireproof, wooden trusses may be used. In all permanent work of any impor¬ tance, and in all fireproof buildings, steel trusses are used. Riveted trusses are used for short spans, less than 75 to 100 feet, and pin-connected trusses for long spans, greater than 75 to 100 feet. The dividing line between these two kinds of trusses is not so clearly defined in the case of roof trusses as in the case of bridge-trusses. In general, riveted trusses are preferable when the stresses in some of the members are reversed by the wind. 45. Working Stresses. — The allowable or working stresses for steel roof trusses are the same as those given for highway bridges in Bridge Specifications. The working stresses for wooden roof trusses are the same as those given in Wooden Bridges. 46. Design of Main Members. —The same general types of members are used for roof trusses as for bridge trusses. In addition, loop-welded rods are sometimes used for main members. The principal difference lies in the fact that, as the stresses in the members of roof trusses are, as a rule, less than the stresses in the members of bridge trusses, the members in the former are smaller than in the latter. For this reason, roof trusses are much narrower than bridge trusses, and in the shorter spans, the web members are connected to the chord and rafter by means of one web connection plate instead of two. The principles that govern the design of the main members of roof trusses are the ROOF TRUSSES 42 same as those that have been illustrated and explained in connection with bridge design. 47. [Lateral Systems.—Roof trusses are connected by lateral bracing in the planes of the rafters, by transverse bracing between the trusses in vertical planes or at right angles to the rafters, and in some cases by lateral bracing in the plane of the lower chord. The same general style of bracing is used as for bridge trusses; but, there being, as a rule, more roof trusses in a roof than bridge trusses in a bridge, the arrangement of the lateral systems is some¬ what different. It is customary to insert lateral trusses in each end bay, and also in every second or third intermediate bay through¬ out the length of the building, as shown in Fig. 34. The figure is the plan of the roof of a building: A A' and B B' are the end trusses, and C C are the intermediate trusses. The purlins A B, D D ', and A' B’ run the entire length of the building, being spliced at each truss, or at every other truss. The common rafters that run at right angles to the purlins are not shown in the figure. The end bays and every second bay throughout the building are supplied with lateral trusses. The lateral truss in each end bay is designed to resist the wind pressure on that end of the building; the other lateral trusses are then made the same as those in the end bays. Lateral and loop-welded rods are frequently used for the diagonals of lateral systems of roofs. a Lap Screw- Bloch 32*8*8$ I2 Bolt Upset, Fig. 40 44 ROOF TRUSSES 81 CONNECTIONS »• * ' * ‘ WOODEN TRUSSES 48. Rafter Joints. —Figs. 35 to 40 show typical joints for the rafters of wooden trusses. In Fig. 35, the purlin, shown in cross-section, is placed on top of the^ rafter and at right angles to it. Both the purlin and rafter are framed where they connect. The purlin is held in position by the inclined brace a. The vertical rod b passes through the rafters, and the strut c is cut at the end to fit into a shoulder cut in the bottom of the rafter. In Fig. 86, the purlin is attached to the rafters by means of the beam hanger b. The inclined strut c is connected to the castings that bears against the rafter. In Fig. 37, a wooden blocks is inserted in the bottom of the rafter, and the inclined strut c is connected to it by means of the bolt b. The end of the strut fits into a triangular hole cut in the block. In Figs. 38 and 39, the purlins are not shown; these figures show methods of providing a greater bearing area for the upper ends of the vertical rods and inclined struts by means of plates and castings. Fig. 40 shows the connection when the rod is inclined and the strut is vertical. In this figure, the purlin is shown vertical, and one side of it is in contact with the casting b that provides a bearing for the inclined rod a. 49. Peak Joint. —The joint at the top of a roof truss is usually called the peak joint. Figs. 41 to 45 show several peak joints. In Fig. 41, the rod c passes between the ends of the two rafters a and b; the plate d provides a bearing for the end of the rod. In Fig. 42, the plates a are bolted to the sides of the rafter sticks to assist in * holding them in place. In Fig. 43, the upper ends of the rafter members are cut off square and bear on a casting a. The rods (two in this case) pass through flanges in the casting, and are held in place by the nuts b. The chair or Fig. 43 45 46 ROOF TRUSSES §81 shelf c is for the purpose of supporting the ends of the' purlins. Fig. 44 shows another method of connecting the tension rods and compression mem¬ bers; in the form of connection here rep¬ resented, gussets or, connection plates a are bolted to the sides of the rafter sticks, and the inclined rods b are connected fig. 46 to them by means of pins c, which pass through the gussets. The vertical rod passes through the sticks and bears on the plate d in the same way as in Fig. 41. In the con¬ nection shown in Fig. 45, the rafter sticks bear against the casting a , and the inclined rods con- i nect to the pin b that passes through the casting. The vertical rod is connected by a pin c to a projection on the casting at the bottom. 50. Chord Joints.—Figs. 46 and 47 show the connec¬ tions of a vertical rod, two inclined struts, and the chord. The blocks c are set into the top of the chord and beveled on the ends to give the struts a square bear¬ ing. The dowels • shown in Fig. 47 are sometimes omitted. §81 ROOF TRUSSES 47 51. Heel Joints.—The end joint of a roof truss, where the rafter and the chord connect, is sometimes called the lieel. Fig. 48 shows the simplest form of heel joint. Fig. 49 shows a very good joint where the inclined member is composed of two sticks. Both the in¬ clined sticks and the horizontal chord are cut away so as to form shoulders, and the ends of the sticks are bolted together. Fig. 50 shows a form of heel joint that is used when the roof has a flat slope and it is not desired to cut too much into the end of the chord. The casting dd is fitted and bolted to the top of the chord, and provides a bearing for the inclined stick. A short piece of timbers is bolted and keyed to the bottom of the chord by the bolts a and the keys k , k. The bolt g is not assumed to 48 ROOF TRUSSES 81 transmit any stress; it is simply for the purpose of holding the lower end of the inclined member in place. The joint shown in Fig. 51 is somewhat similar to that shown in Fig. 50; but the method of providing a bearing for the rafter is different. In Fig. 51, the purlin a is shown on top of the rafter b; the common rafter c with the roof covering d is on top of the purlins. 52 . Combination Chord Joint.— Fig. 52 shows a chord joint of a truss in which the com¬ pression members are wood and the tension members steel. Pin plates a are bolted on the sides of the stick b , and the ends of all the members are connected by the pin c. § si ROOF TRUSSES 49 STEEL TRUSSES 53. Rafter Joint. —Figs. 53 to 57 show rafter joints of steel roof trusses. In Fig. 53, a wooden purlin is shown in cross-section; it is held in place by the bracket a. The rafter in this case consists of a single web b and two top flange angles c. The web members are riveted directly to the web-plate, and there is no need of a gusset. Fig. 54 shows the method of connecting a steel purlin a to the rafter by means of the Fig. 54 Fig. 55 connection angle b. This figure also shows the wooden com¬ mon rafter c on top of the purlin, held in place by the steel clip d. Fig. 55 shows another method of connecting a pur¬ lin a to the truss. In this case, a gusset b is used for the connection of the web members to the rafter. In Fig. 56, 50 ROOF TRUSSES the purlin a is shown vertical, and the common rafter b is con¬ nected to it by means of a clip. Fig. 57 is a rafter joint in a pin-connected truss. Both the rafter and the compres¬ sion . web member are com¬ posed of two channels. The rafter is spliced at this joint, being cut at the center of the pin. This splice is simply for the purpose of decreasing the length of the rafter pieces, so that they can be handled more easily. The compression web member is connected to the pin by means of pin plates. 54. Peak Joint. —Figs. 58 and 59 are peak joints of riveted trusses, and show how the members are connected at the top. In Fig. 58, the rafter members are each com¬ posed of two angles, and the vertical member is composed §81 ROOF TRUSSES 51 of two flat bars. The gusset a is riveted in the shop to one of the rafter members, and the other members are riveted to it in the field. The purlin c is composed of two channels, and is connected to the truss by means of the bent bars d. This joint also shows the way in which the common rafters b may be connected to the top of the purlin. Fig. 59 is the peak joint of a much larger truss. The vertical webs a> a that form part of the section of the rafter are spliced at b } b to the gusset c. The gusset is shown connected at the left end of the rafter and one of the web members by shop rivets, and to all the other members by means of field rivets. It is customary to rivet as great a portion of a roof truss together Fig. 59 in the shop as can be conveniently transported to the place where it is to be erected. 55. Heel Joint. —Fig. 60 shows a method of forming the heel joint of a riveted truss when the end rests on a pedestal or on the wall. The chord a is continued across 52 ROOF TRUSSES §81 the support, and the rafter b is connected to it by means of the gusset c. Fig. 61 shows a method of forming this joint when the end of the truss is supported on a column. The rafter a is continued across the top of the column, and the angles or other shapes b of which the column is composed are continued up to the under side of the rafter. The gusset c connects the end of the chord and that of the rafter to the top of the column. In the figure, the web of the column is stopped at dd , and the lower edge of the gusset is inserted between the angles at the top. The gusset and web-plate are spliced by the side plates e\ when there are no side plates, additional plates are used for splicing. §81 ROOF TRUSSES 53 56 . Chord Joints.—Fig. 62 shows a method of form¬ ing a chord joint of a pin-connected roof truss. In this joint, all the tension members are placed inside the channels that form the compression member. Fig. 63 shows a chord joint of a riveted roof truss, the chord being spliced at the joint. In this figure, every member is composed of two angles; the ends are connected by a single gusset inserted between the angles. This gusset acts also as a splice plate for the chord. BRIDGE PIERS AND ABUTMENTS PRELIMINARIES DEFINITIONS 1. Abutments and Piers.—The supports provided for the ends of a bridge are generally required to hold back the banks, thus acting both as supports for the bridge and as Fig. 1 retaining walls; supports of this character are shown at A and A, Fig. 1, and are called abutments. When, as at inter¬ mediate points, such as £, C, E>, and E , Fig. 1, there is no bank to restrain, the supports are called piers. INTERNATIONAL TEXTBOOK COMPANY. ALL RIGHTS RESERVED COPYRIGHTED BY 2 BRIDGE PIERS AND ABUTMENTS §82 2. An abutment usually has lateral extensions, a , a , Fig. 1 (a ), to hold back the earth in the sloping bank. These extensions are called wings, and the abutment is some¬ times called a wing abutment, if it is desired to indicate the existence of wings. Nearly all abutments are wing abutments. When the wings are omitted, as at A, Fig. 1 ( a ), and the filling material is allowed to run around in front of the abut¬ ment, the latter is called a pier abutment, or abutment pier. 3. Piers may be carried up to such a height as to give a direct support to the bridge, as at C and D, Fig. 1 (3), or they may be carried only to such a height as to protect the structure from the destructive influences below or near the surface of the earth, as at B and E , a trestle b , b being intro¬ duced between the top of the pier and the bottom of the bridge. These low piers are frequently divided into two separate supports, as shown at A', Fig. 1 {a ), called pedestals, or pedestal piers. 4. Superstructure and Substructure.—The bridge proper, which rests on the piers and abutments, above the level G G, Fig. 1 (b), is called the superstructure. All supporting work below the level G G, including trestles or columns, as at B and A, and all the masonry, is called the substructure. 5. Bridge Seat.—The top surface /,/, Fig. l'(£),of an abutment or pier to which the weight of the bridge is directly transmitted is called the bridge seat. The course g, g of masonry directly under the bridge seat is called the bridge- seat course, coping course, or top course; it is some¬ times called simply the bridge seat or coping. This course usually projects a few inches beyond the outer edge of the masonry directly under it. This projection is called the overhang of the bridge seat; it improves the appearance of the pier, and protects the masonry beneath it from the weather. The outer edge h,h, Fig. 1 (a), of the masonry immediately under the top course is called the neat line. §82 BRIDGE PIERS AND ABUTMENTS 3 6. Pedestal Blocks.—In cases where stringers or floor- beams rest on the masonry at elevations different from the elevations of the base of the trusses or girders, separate top courses are sometimes laid; but more frequently the top course is finished level at the proper elevation for the trusses or girders, and is called the main bridge seat. The proper elevation is then secured for the stringers or floor- beams by placing stone blocks, called pedestal blocks, on top of the main bridge seat. In many cases, the desired elevation is secured by means of steel castings, called pedestals. 7. Parapet.—In order to keep the filling from running over on the bridge seat, a back wall c, c, Fig. 1, commonly called a parapet, is usually built on top of the bridge seat. 8. Footings.—The lower part d y d } Fig. 1 (b ), of a pier or abutment is called the footing. It is usually made to project somewhat beyond the masonry of the main body above it. Footings serve the purpose of securing greater bearing area on the soil and of giving greater stability to the structure. 9. Cutwaters.—When piers are placed in running streams, as at D, Fig. 1, they are usually placed with their longest dimension parallel to the direction of the current, and the ends are made pointed instead of flat, in order to reduce the obstruction to the flow of water. The up-stream end is generally made sharper than the down¬ stream end. The sharp ends e, e, Fig. 1 (a), are called nosings or cutwaters. In cold climates, where floating ice is common, cutwaters are sometimes called ice breakers, and are frequently protected by means of a bent steel plate fastened to the masonry. This protection should always be provided where large masses of ice are likely to float down the stream. 4 BRIDGE PIERS AND ABUTMENTS §82 LOCATION AND ESTIMATES 10. Direction of Crossing.—The direction of a bridge, especially of a railroad bridge, should, if practicable, be so selected that the crossing will be at right angles, or nearly so, to the direction of the stream, road, or depression to be crossed. This is desirable both because such location reduces the length, and accordingly the cost, of the structure, and also because a skew end is objectionable from a purely structural point of view. The problem of making the crossing square or nearly so is not always considered the duty of the bridge engineer; but the locating engineer should endeavor so to arrange the alinement as to avoid the necessity of skew bridges. 11 . In the case of a railroad bridge, unless it is made unusually heavy and stiff, the rigidity of the support at the abutment and the elasticity of the unsupported part of the bridge at the opposite side cause an undesirable swing to passing trains. The cause of this swing may be understood account of the rigidity of the abutments, the deflection at such a point as A is practically negligible, while at such a point as A, directly opposite A , the truss deflects an appre¬ ciable amount. The result is that the floor, and conse¬ quently the train, is inclined sidewise from A to B. This §82 BRIDGE PIERS AND ABUTMENTS 5 objectionable condition may sometimes be avoided by squar¬ ing up the ends, as shown by dotted lines in Fig. 2, but this still further lengthens and increases the cost of the crossing; and, if more than one span is required, the intermediate piers must still be generally left on a skew, so as to present as little obstruction as possible to the current or traffic beneath. 12. Location of Piers. —With the. location of the crossing determined, the problem is not necessarily wholly solved. If the bridge is over a railroad, a highway, or a navigable stream, the position and length of one or more spans may be fixed by the requirements of the traffic beneath; for the way previously existing is entitled to the prefer¬ ence and controls to a great extent the locations and dimen¬ sions of the piers. The first thing to do after the direction and location of the crossing have been decided is to deter¬ mine the location of the abutments, and to ascertain to what extent the traffic underneath will affect the selection of the number and length of spans. The piers should be located so as to offer the least possible amount of obstruction. This condition frequently controls the location of every pier. Another element that must be taken into account in deter¬ mining the positions and number of piers is the cost. It should be borne in mind that a reduction in the number of piers requires an increase in the lengths of the spans, and that the increased cost of the spans may upset the saving resulting from a reduction in the number of piers. If the crossing is a low one, and a good foundation can be obtained at a moderate depth, short spans are often more economical than long spans. 13. Economical Arrangement. —For preliminary estimates, use may be made of the empirical rule that the economic arrangement is approximately that in which the cost of the substructure equals the cost of the superstructure (the trestles, if any, being included in the substructure). Also, in this preliminary study, the cost of the main trusses or girders may be considered as varying directly as the square of the span, and the cost of the floor directly as the 6 BRIDGE PIERS AND ABUTMENTS 82 length of the span, unless the change of length of spa?i is suffi¬ cient to change the character of the structure. The cost of the substructure for any one support depends on the length of span, the height of the structure, and the depth of the founda¬ tion. By roughly designing piers for both the longest and the shortest span to be considered, the cost of each pier can be approximately determined. If the crossing is short, the most economical number of spans and length of span may generally be determined immediately by finding a span such that the total cost of sub¬ structure divided by the length of the bridge will be approx¬ imately equal to the total cost of superstructure divided by the length of the bridge. But if the crossing is long, a more detailed estimate of cost is usually necessary. 14. Approximate Costs.—For the purpose of com¬ parison, the cost of masonry for the substructure and of the steelwork for the superstructure may be assumed. In this Section, the cost of the masonry will be taken as $10 per cubic yard, and that of the excavation for foundations as 50 cents per cubic yard, including the disposal of the material excavated. For rough estimates, the cost per linear foot of the steelwork of the bridge spans and the necessary dimen¬ sions of bridge seat for different spans may be taken approximately as Span Feet follows: Cost per Required Size - Linear Foot Dollars of Bridge Seat Feet 50 50 4 X 10 100 80 5 X 15 150 110 6 X 20 200 140 7 X 25 250 170 8 X 30 ILLUSTRATIVE EXAMPLE 15. Data. —As an illustration, let it be assumed that a bridge 1,000 feet long for a single-track railroad is to be built across a valley 80 feet deep, the masonry to be carried » §82 BRIDGE PIERS AND ABUTMENTS / to a depth of 5 feet below the bottom of the valley and up to such height as to give direct support to the bridge span; and the distance between the tops of piers and the base of rails to be one-fifth of the length of the spans. i 16. Possible Spans. —It will be readily seen that a length of 1,000 feet may be divided into two spans of 500 feet each, or three spans of 383 feet 4 inches, or four spans of 250 feet, or any greater number of spans of correspond¬ ingly shorter length. But it is also obvious that with the longer spans the cost of the superstructure will be much greater than if short spans are used. 17. Cost of Superstructure and Substructure. —For rough estimates of this kind, it is sufficiently close to assume the top of the pier to have the dimensions given for the bridge seat, and the four sides to have a batter of 1 inch per foot down to the level of the ground; whence, with an offset of 1 foot on each side and one at each end, the same dimensions are maintained down to the subfoundation, 5 feet below the ground surface. With these assumptions, the approximate quantities and cost for the substructure are found to be as given in the following table: Span Feet Masonry Excavation Total Cost per Pier Total Cost of Sub¬ structure, per Foot of Span Total Cost of Super¬ structure per Foot of Span Cubic Yards Cost Cubic Yards Cost 50 510 $ 5 UOO 78 $39 $5,139 $ 102.80 $ 50 75 533 5,330 82 4i 5,371 71.60 65 IOO 548 5,480 85 43 5,523 55-20 80 150 558 5,58o 92 46 5,626 37.50 I IO 200 537 5,370 98 49 5,419 27.10 140 250 488 4,880 103 5i 4,931 19.70 170 18. In finding the quantities given in the second column of the table, it is convenient to draw a diagram similar to that shown in Fig. 3, which is a pier for a 100-foot span. 135—28 8 BRIDGE PIERS AND ABUTMENTS 82 £feration of Bait. /■Bridge Seat 5 X/5 I I / -^— r Fig. 15 35. At the down-stream end, the two planes that come together are given the same batter as the sides. In cold regions, where much ice may be expected, the nosing at the up-stream end of the pier is given a greater batter, so that the slope of the nose will incline at the rate of 3 to 6 inches per foot. The sharp edge is frequently shod with iron, which acts as an ice breaker. The momentum of the moving ice and the pressure from the current of water force the cakes BRIDGE PIERS AND ABUTMENTS § 82 23 of ice to slide up this smooth edge; when high enough out of the water, the ice breaks of its own weight, or by the blows of other ice forced against it. The increased cost of curved ends on large piers is very small, and since the obstruction to the water is large, it is customary in practice to curve the ends, as previously shown in Fig. 12. If the ends were given the theoretically proper form shown in Fig. 13, the sharp point would soon be worn away. Semicircular ends are sometimes used; but, if there is much current, bad eddies and whirlpools result, and there is danger of the water being so agitated as to undermine the pier. When the pier does not extend more than 10 or 15 feet above the high-water line, it is customary to have the cut¬ water extend up to the bottom of the bridge-seat course, as shown in Fig. 9. In very high piers, in which this procedure would involve much needless expense, the nosings are usually carried but a few feet above high-water mark, and the remainder of the pier is made with square ends, as shown in Figs. 10 and 11. When it is desired to give high piers a better appearance, the ends of the upper part may be made semicircular, as shown in Fig. 12. 36. Belt Course.—In case the upper part of the pier is arranged differently from the lower, they are separated from each other by means of a belt course, similar to a bridge- seat course. The belt course is indicated by the letters A , A in Figs. 11 and 12. - , THEORETICAL CONDITIONS 37. After the approximate dimensions of a pier have been decided according to the foregoing practical considera¬ tions, it is necessary to investigate the stability of the pier and to make all the calculations necessary to determine whether the actual stresses will anywhere exceed the allow¬ able stresses. 135—29 24 BRIDGE PIERS AND ABUTMENTS §82 CAUSES OF FAILURE 38. The stability of a bridge pier may be destroyed by any one of the following causes: 1. Overturning at any section or on the subfoundation on account of: (a) pressure of the wind on the structure and on the loads carried by the structure; ( b ) the current in the stream, and the pressure and blows from ice and other float¬ ing objects; (c) the centrifugal force of trains; and ( d ) the longitudinal thrust due either to the friction of the wheels on the rails when brakes are set, or to the traction of the engine. 2. Sliding at any cross-section or on the subfoundation. 3. Crushing , either at the bridge seat, at any lower section, or at the base. 4. Failure of the foundation or sub foundation, either from insufficient strength, from the upheaval by frost, or from other disturbing agents. 5. Breaking apart or collapsing, on account of poor mortar, poor bond between the stones, or poor concrete. 39. Overturning. —In a direction parallel to the axis of the bridge, the forces tending to overturn the pier are the pressure of the wind, the longitudinal thrust from trains in railroad bridges, and the packing of ice or drift between the piers or between a pier and the bank of a stream. In case the earth extends to a greater height on one side of a pier than on the other, the unbalanced pressure must not be overlooked. These forces act to overturn the pier in the direction of its smallest dimension and least resistance. In a direction at right angles to the axis of bridge, there may be a pressure of the wind not only on the pier but also on the bridge and on the train or other load, and the force of the current may be added to the pressure of the ice and drift; there may also be centrifugal force if the track is on a curve. The centrifugal force can act only toward the outside of the curve, and the force of the current can act only in the direction of the flow; but the other forces may be exerted in either direction. §82 BRIDGE PIERS AND ABUTMENTS 25 • Although the overturning forces acting on a pier are dif¬ ferent in their origin from the pressure of a bank, their effects and the necessary dimensions to resist them are determined in the same way as for retaining walls, and here it will be sufficient to make a few general remarks as to their character and magnitude. 40. With a low pier, or with one of moderate height, the dimensions required for supporting the bridge are such that there is little danger of overturning; but with a very high pier, the overturning forces should be carefully looked into and provision made to resist them. A pier will never overturn about a sharp edge, unless the sharp edge crushes or sinks into the soil. This must be prevented by so designing the pier that the safe crushing strength of the masonry and the safe bearing power of the soil will not be exceeded. Since the maximum force of the wind and of the longitudinal thrust are of rare occurrence and of brief duration, the intensities of pressure usually permitted may be increased 25 per cent, when these over¬ turning forces are considered as acting at the same time. In the Section on Foundatio?is , Part 1, may be found tables giving the safe bearing power of various soils and the safe crushing stresses for masonry. 41. Sliding. —The forces tending to produce sliding of a pier on its subfoundation, or along any horizontal plane within the pier, are the same as those that tend to overturn it. The method of determining the effect of these forces and of providing for them is the same as for retaining walls. 42. Crushing. —The weight of the bridge and its load is usually transmitted to the pier at two or more points so far apart that, if ample provision is made for distributing this load over a sufficient area at the bridge seat, there need be no fear that crushing at any lower plane in the masonry may be caused by that weight. The only surfaces that heed be considered are the bridge seat, the bottom of the pier, where it rests on the footing, and the base of the footing. The overturning forces cause the center of pressure to fall near BRIDGE PIERS AND ABUTMENTS §82 26 the edge of the masonry, and the danger of crushing from this cause must be carefully investigated. 43. Crushing' at Bridge Seat. —In determining the area of bearing of the bridge on the masonry, it should be borne in mind that the load is not generally uniformly distributed over the whole area of bearing; even with a perfectly rigid bearing plate there may be a greater load on one edge than on another, on account of the deflection of the bridge under its load, or on account of the overturning forces. Since it is impossible to compute the actual distribution of the pressure, it is customary to take into account only the vertical loads and to use a much smaller working stress than would be justifiable if the uneven distribution could be properly taken into account. On this basis, a working pressure of about 300 pounds per square inch is commonly used for limestone or sandstone, 400 for concrete, and 500 for granite. This gives an apparent factor of safety of about 10 or more. 44. Crushing at Bower Sections. —In considering the crushing strength of a pier at any horizontal plane below the bridge seat, it is necessary to take into account the uneven pressure due to the overturning forces. The prin¬ ciples involved, including the rule of the middle third, are fully explained in Foundations , Part 1; how they are applied will be illustrated by an example to be given presently. Crushing of the masonry at special places may also be caused by other agents than the forces heretofore considered, as by blows from the wheels of vehicles, from vessels, ice, etc. These causes of destruction or damage will be taken up in a subsequent article. 45. Failure of Foundation. —The most frequent cause of failure of a bridge pier or abutment is a poor foundation, which may either crush entirely under the load so as to cause a settlement of the structure, or yield slightly at the point of maximum pressure, and so increase the already excessive overturning tendency by decreasing the leverage of the vertical forces and thus decreasing the resisting moment. The subfoundation, by its lack of uniformity, may cause BRIDGE BIERS AND ABUTMENTS 27 . uneven settlement and the rupture of the 'masonry. The upheaving tendency of frost is also a disturbing agent. A subfoundation that may have enough resistance to support the load may fail by being softened or washed away by the current, if in a running stream. A flood may sometimes cause the undermining of a structure that at ordinary stages of the water is far away from the water. FORCES TO BE RESISTED 46. Wind.—In the design of a bridge and its supports, it is customary to provide for a wind pressure of 50 pounds per square foot of exposed surface when there are no loads on the bridge, or an alternative pressure (for railroad bridges only) of 30 pounds per square foot on the exposed surface of the bridge and the loads combined when there are loads on the bridge. ' The pressure of 50 pounds per square foot corresponds to a velocity of over 100 miles per hour,'and is equivalent to a hurricane that would be commonly reported as “destroying everything in its path.” Such wind is of extremely rare occurrence, but since it may occur at any time, even if for a very short time only, it must be provided for. The alternative pressure of 30 pounds per square foot i corresponds to a velocity of over 80 miles per hour, and is the greatest pressure considered on moving loads, such as railroad cars, since it is sufficient to overturn empty cars; and it is assumed in practice that if the wind blows harder than this, it will be impossible to operate cars or other vehi¬ cles, so that there will probably be no loads on the bridge. In determining the total pressure on a bridge, the intensity is multiplied by the exposed area. This area is usually taken equal to the area of the longitudinal vertical projection of the bridge, and is found by multiplying the width of each mem¬ ber by its length; in a truss bridge, the sum of the areas of all the trusses should be taken. In the design of highway bridges, the exposed area of the loads is usually very small in comparison with the exposed area of the bridge, and it is customary to neglect it. In railroad bridges, the cars present 28 BRIDGE PIERS AND ABUTMENTS §82 a vertical surface about 10 feet high and for the full length of an ordinary bridge. This gives 30 X 10, or 300, pounds per linear foot wind pressure on a train of cars. For purposes of computation, this pressure is treated as a single force acting through the center of pressure, which is from 6 to 9 feet above the rail. In what follows, the center of pres¬ sure will be assumed to be 7 feet above the top of the rail. 47. Current of Stream. —The pressure of the water in a flowing stream is given by the following approximate formula: p = it®! = 2.96 V\ ( 1 ) O in which p — pressure, in pounds per square foot, of exposed surface normal to the current; v = velocity of current, in feet per second; V = velocity of current, in miles per hour. The velocity of the current varies with the depth, being greatest near the surface and much less at the bottom of the stream. Between these depths it varies approximately as the square root of the distance from the bottom. This makes the intensity of pressure due to the current vary approxi¬ mately as the distance from the bottom, and the average pres¬ sure is usually assumed as about one-half that at the surface, with the center of pressure at two-thirds the height from the bottom to the top of the stream. In a pier having a cutwater with inclined sides, the resistance to the flow of the water, and consequently the pressure at the end, are reduced. The pressure on the end of such a pier is usually assumed to be three-quarters of the pressure on a pier with a square end. For a pier with a cutwater, formula 1 may, therefore, be written p = 1.03 v 2 = 2.22 X s (2) The pressure decreases from the surface to the bed of the stream the same as for a square-ended pier. Let p<> = intensity of pressure at surface, in pounds per square foot; ^ol , ., t n . r ■ [feet per second; t = velocity of flow at surface in { . ^ ’ V 0 ] I miles per hour; BRIDGE PIERS AND ABUTMENTS 29 §82 A = area, in square feet, of section of pier made by a plane perpendicular to the current, between the surface and the bottom of the stream; P = total pressure, in pounds, on the submerged portion. Then, since the average intensity of pressure is i p 0 , P=±Ap 0 (3) The value of p 0 is found by substituting v 0 or V 0 in for¬ mula 1 or in formula 2, according to the character of the end. Making the substitution, formula 3 becomes, for square-end piers, P = .69 A Vo 2 = 1.48 A V 0 2 or, practically, P = .7 A Vo* = 1.5 A V n 3 (4) For piers with cutwaters, P = .52 A Vo 2 = 1.11 A Vo or, practically, P = .5 A Vo* = 1.1^ Vo 2 (5) As already stated, the center of pressure is taken at a distance from the bottom equal to two-thirds of the distance between the bottom and the surface. 48. Ice and Drift. —In cold climates, ice, and in all climates, drift, may pack around the pier, and, by adhering to it, present a much wider obstruction to the current than does the pier alone. In some cases, a dam may be formed by the ice or drift packing solid from pier to pier; the result is that the obstruction increases, becoming higher and higher, until the piers fall down or the obstruction is forced through the opening by the pressure of the water. Damage may also result if large masses of ice floating with high velocities strike the pier; in this case, the pier may be turned over or part of it may be torn away. There is also an element of danger in the expansion of water when freezing. If the entire surface of the water between two piers freezes to a considerable depth, there will be a pressure on each pier, which, under some conditions, may be sufficient to dislodge them. :;o BRIDGE PIERS AND ABUTMENTS 8 The dangers outlined ’in the preceding paragraph are real dangers and should never be ignored or minimized in the design of piers. The history of bridge failures and disasters shows that more failures have been due to the collapse of the piers on account of ice and drift than to any other one cause. Owing to the extreme difficulty in obtaining definite information regarding the amount of these forces, it is practi¬ cally impossible to give any values that will be of help to the designer. Provision is usually made by the use of a large factor of safety. 49. With regard to the method of providing against these dangers, it may be said that the problem is one of judgment rather than of computation. Each case must be considered by itself, and all the conditions and possibilities fully studied before locating the piers. In many cases, the danger from ice and drift will be the determining factor in the location of the piers, and may result in the selection of a single span over the waterway, in order to keep the piers safe. In any case, the problem is rather one for the experienced engineer than for the draftsman or designer. A young* engineer should not hesitate to seek advice from more experienced engineers when he has a problem of this kind. 50. Centrifugal Force.—In railroad bridges on which the track is curved, the centrifugal force A of a train causes an outward thrust, the magnitude of which is given by the formula F - .00001167 V'D W, (1) in which V = velocity of train, in miles per hour; D = degree of curve; IV — weight of train. For practical work, the following formula may be used (see Bridge Specifications)'. F = -- 4 -- 5 .T _i ? _ g) WD (2) 100 * Formula 2 provides for about 61 miles per hour for a 1° curve, 55 miles per hour for a 5° curve, and 46 miles per hour for a 10° curve, which are as high speeds as usually BRIDGE PIERS AND ABUTMENTS O - obtain. The centrifugal force F is usually assumed to act about 6 feet above the top of the rail. 51. Longitudinal Thrust.—The longitudinal thrust is due to the friction of the wheels on the rails. When a loco¬ motive is on a bridge and is exerting its maximum tractive force, the rails transmit an equal force to the bridge, and the latter transmits it to the supports. The tractive force has been found in some cases to be as high as 37 per cent, of the total weight on the drivers of the locomotive. When the brakes of a train are set, the wheels tend to slide on the rails, and this causes a frictional force that is transmitted to the supports. In practice, it is customary to assume that the greatest longitudinal thrust that need be provided for is equal to 20 per cent, of the greatest vertical load that can come on the bridge. The longitudinal thrust is usually.considered as a single horizontal force acting above the pier at the height of the bridge seat; it should never be neglected, as it acts in the direction in which the pier is weakest. The longitudinal thrust is transmitted to the piers and abutments of the fixed ends of the spans. The expansion ends, especially of long spans, are provided with rollers, so these ends cannot be relied on to transmit any of the thrust. For this reason, it is necessary to consider the arrangement of the spans and find out which will be expansion and which fixed ends. In case two fixed ends come on the same pier, that pier must be designed to provide for the longitudinal thrust coming from two spans. ILLUSTRATIVE EXAMPLE 52. Data. —Let the pier shown in Fig. 16 support the fixed ends of two 150-foot spans of a single-track railroad bridge on a tangent. Let the data be assumed as follows: Dimensions as given in figure, determined from practical considerations. Weight of granite masonry, 150 pounds per cubic foot. Weight of bridge, 2,500 pounds per linear foot. Fig. 16 32 82 BRIDGE PIERS AND ABUTMENTS 33 . Surface of each truss exposed to the wind, 8 square feet per linear foot. Surface of bridge floor exposed to the wind, 4 square feet per linear foot. Minimum weight of train, 800 pounds per linear foot. ' Maximum weight of train, 4,000 pounds per linear foot. Velocity of the current at the surface of the water, 6 miles per hour. Ice and drift collected about the up-stream end of the pier for a depth of 6 feet below high-water level, and for an average width of 18 feet. It is required to determine the stability of the pier in regard to overturning and sliding, and also to find the maximum intensity of pressure on the foundation. 53. Overturning. —In investigating the factor of safety against overturning, it is necessary to consider overturning moments in both directions of the pier; that is, at right angles and parallel to the axis of the bridge. It is also necessary to consider the two conditions of a loaded and an unloaded bridge. In discussing this subject, three cases will be considered. In Case I, the wind pressure is taken as 30 pounds per square foot (see Art. 46); in Case II, it is taken as 50 pounds per square foot. In Case I, it is customary to use the weight of an empty train, since the overturning moment is then the same as though a full train were considered, and the resisting moment will be less. The axis of moments will be taken along one edge of the bottom of the footing. Case III will deal with the overturning tendency of the forces parallel to the axis of the bridge. 54. Overturning Moments of Wind Pressure: Case I. —The wind pressure on the train is equal to 300 pounds per linear foot, applied 7 feet above the top of the rail (Art. 46). The pressure on one-half of each span may be assumed to be transmitted to the pier; this makes the total wind pressure on the train 300 X (75 + 3 + 75) = 45,900 pounds, acting at a distance (see Fig. 16, side elevation) of i BRIDGE PIERS AND ABUTMENTS O D 1 7 + 4+1 + 22 + 3 + 2 + 16 + 2 + 4 = 61 feet above the bottom of the footing. The overturning moment of this pressure is, then, 45,900 X 61 = 2,799,900 foot-pounds The wind pressure on the floor is 30 X 4 — 120 pounds per linear foot, applied 2 feet below the top of the rail. Then, the total wind pressure on the floor is 120 X (75 + 3 + 75) = 18,360 pounds (The distance of 3 feet between the centers of bearings of the trusses is. usually counted in computing wind pressure, because the floor and truss members project beyond the theoretical ends.) The distance from the center of this pressure to the bottom of the pier is 61 — 7 — i — 52 feet. Then, the overturning moment of this pressure is 18,360 X 52 — 954,700 foot-pounds The wind pressure on each truss is 30 X 8 = 240 pounds per linear foot, and where, as in this case, there is a train on the bridge, it is assumed that there is no shelter for the leeward truss; hence, the total wind pressure on the trusses transmitted to the pier is 2 X 240 X (75 + 3 + 75) = 73,440 pounds. This pressure will be assumed to be concentrated half way between the chords of the trusses; that is, 38 feet above the bottom of the pier. Then, the overturning moment of this pressure is 73,440 x 38 = 2,790,700 foot-pounds In calculating the pressure of the wind on the pier, and that of the water current, it is inadvisable to waste time in any unnecessary refinement, as the values of the wind and water pressures are but roughly approximate. The width of the pier subjected to the pressure of the wind is 6 feet 6 inches at the bridge seat, 6 feet at the bottom of the bridge-seat course, and 6 feet 8 inches at high-water level; then, for purposes of calculation, a uniform width of 6 feet 6 inches, or 6.5 feet, may be used, and the center of wind pressure taken 3 feet below the bridge seat. Then, the exposed area is 6 X 6.5 = 39 square feet, and the wind pressure is 30 X 39 = 1,170 pounds, the center of which may be taken as 4 + 2 +16 + 2 — 3 = 21 feet above the bottom BRIDGE PIERS AND ABUTMENTS S of the pier. The overturning moment of this pressure is, then, 1,170 X 21 = 24,600 foot-pounds 55. o yertnrning Moment of Pressure Due to Current: Case I. —The intensity of pressure due to the current is given by formula 1, Art. 47. In the present case, since the velocity of the water at the surface is 6 miles per hour, the pressure at the surface is 2.96 X 6 2 — 106.56 pounds per square foot. The intensity decreases uniformly with the depth, being 1 Oft —^— = 53.28 pounds per square foot at the bottom of the ice and drift, half way between the surface and the bottom. Then, since the exposed area of the ice and drift is given in the data as 6 X 18 = 108 square feet, the total pressure is 106.56 + 53.28 N 108 X = 8,631 pounds If p represents the intensity of pressure at the surface, A the intensity of pressure at the bottom of the ice, and d x the distance from the surface to the bottom of the ice, the dis¬ tance x 0 from the surface to the center of pressure is given by the formula _ (2A + p) d x ° 3 (A + p) In the present case, p = 106.56, A — 53.28, and d x — 6 feet. Substituting in the formula gives 2 X 53.28 + 106.56 X 0 = X 6 = 2 i feet 3 (53.28 + 106.56) Then, the distance from the center of pressure to the bot¬ tom of the pier is 4 -f 2 + 12 — 2f — 15i feet, and the over¬ turning moment of the pressure on the ice and drift is 8,631 X 15i = 132,300 foot-pounds The pressure of the water on the pier below the ice and drift remains to be considered. It has already been found that the intensity of this pressure at the bottom of the ice is 53.28 pounds per square foot; this value may be substituted for A i n formula 3, Art. 47, which gives P = i X 53.28 A 36 BRIDGE PIERS AND ABUTMENTS §82 The exposed area is 6 feet high and from 7 feet 8 inches to 8 feet 8 inches in width. The area is, then, 6 X 8i = 49 square feet; and, therefore, P = i X 53.28 X 49 = 1,305 pounds This pressure may be assumed to be concentrated at one- third the distance from the bottom of the ice to the bottom of the river; that is, 4 feet above the latter, or 10 feet above the bottom of the footing. Then, the overturning moment due to this part of the pressure is 1,305 X 10 = 13,100 foot-pounds 56. Total Overturning Moment: Case I. —The total overturning moment can now be found by adding the six products just found. It is as follows: Foot-Pounds Wind on train. 2,799,900 Wind on floor. 954,700 Wind on trusses. 2,790,700 Wind on pier. 24,600 Water on ice and drift . . .. 132,300 Water on pier below. 13,100 Total overturning moment . . . . . 6,715,300 If the track were on a curve, the centrifugal force of the train would be computed, and its moment about the base of the footing added to the moment already found. 57. Resistance to Overturning: Case I. —The resist¬ ance to overturning is provided by the weight of the train, bridge, and pier. The resisting moment of each is found by multiplying the weight by the distance of a vertical line through its center of gravity from the down-stream lower edge of the footing. 58. Resisting Moment Due to Weight of Bridge and Cars: Case I.—It will be assumed that, when both spans supported by the pier are fully loaded, one-half the load on each span, together with 3 feet between the spans, is supported by the pier, which makes the total weight of unloaded or empty cars 800 X (75 + 3 + 75) = 122,400 BRIDGE PIERS AND ABUTMENTS pounds. The horizontal distance from the center of the track to the lower edge of the footing is seen in Fig. 16 to be 17 feet; then, the resisting moment of the empty train is 122,400 X 17 = 2,080,800 foot-pounds In a similar manner, the resisting moment of the bridge is found to be 2,500 X (75 + 3 + 75) X 17 = 382,500 X 17 = 6,502,500 foot-pounds 59. Resisting Moment Due to Weiglit of Pier: Case I. —The weight of the pier must be computed in two parts. The part above the water level is taken at its actual weight. The part immersed in water is decreased in weight on account of the buoyant effort of the water; deduction is made by decreasing the weight per cubic foot of the masonry by 62.5 pounds, the weight of a cubic foot of water. The accurate calculation of the location of the center of gravity for each part of the pier is very tedious, and it is sufficiently accurate in practice to make use of an approxi¬ mate method that re¬ quires less work and gives close enough results. In the present case, the pier may be assumed to be composed of horizontal layers 2 feet thick, each layer having a uniform horizontal cross-section equal to that at the center of its height, and the center of gravity being taken at the center of symmetry. The horizontal cross-section of each layer with pointed ends will then have the appearance shown in Fig. 17, the area A being given by the formula: a + a! s A =lc + X b The values of c , a , and b for the different layers can be calculated from the dimensions of the pier given in Fig. 16. Commencing with the bridge-seat course, and continuing downwards, the volumes of the layers 2 feet thick are as follows: BRIDGE PIERS AND ABUTMENTS *>Q />o §82 Bridge-seat course, 2 X (20' 3" + 3 ~) x 6 ' 6 "1 First 2-foot layer, 2 X 20' 4" + 3 ' 1 " + 3'_1 " ) x e' 2" ] Second 2-foot layer, 3' 3" + 3' 3'" 2 X 2P + X 6' 6" ] = 305.5 cu. ft. 288.81 cu. ft. = 315.25 cu. ft. Third 2-foot layer, 2 X (21' 8" + 3 ' 5 " + 3 ' 5 "^ x 6' 10"! = 342.81 cu. ft. Fourth 2-foot layer, 3/ 7// + 3/ 7//> 2 X 22 ' 4 " + X T 2" ] Fifth 2-foot layer, 3' 9" + 3' 9 //N 2 X 23' + X T 6" ] = 371.47 cu. ft. = 401.25 cu. ft. Sixth 2-foot layer, 23' 8" + - H " + 3 ' n " j x T 10"j = 432.14 cu. ft. 2 X Seventh 2-foot layer, 2 X Eighth 2-foot layer, 2 X (24' 4" + £- !" . + 4' 1" ^ x2 „j = 404.14 cu. ft. 25' + 4 ' ?J " + — 3 - 'j x 8' 6"J = 497.25 cu. ft. Ninth 2-foot layer, 2' 5" + 2' 5" N 2 X 30' 8" + X 9' 10" ] 650.64 cu. ft. The base, 4 feet deep, may be considered as one layer: 4 X 37 X 11 - 1,628 cubic feet The resisting moment of the portion above the water will now be found by multiplying the volume of each of the layers by the assumed weight of the masonry, 150 pounds per cubic foot, and then multiplying each of these products by the lever arm of the respective layer. The resisting moment § 82 BRIDGE PIERS AND ABUTMENTS 39 * of the portion in the water will be found by multiplying: the volume of each layer by 150 — 62.5, or 87.5, pounds per cubic foot, and then multiplying: each product by the lever arm of that layer. The resisting* moments are as follows: Bridge-seat course, Foot-Pounds (305.5 X 150) X 17' First 2-foot layer, (288.81 X 150) X 17'1" = Second 2-foot layer, (315.25 X 150) X 17'3" = Third 2-foot layer, (342.81 X 150) X 17' 5" = Fourth 2-foot layer, (371.47 X 87.5) X 17' 7" = Fifth 2-foot layer, (401.25 X 87.5) X 17' 9" = Sixth 2-foot layer, (432.14 X 87.5) X 17' 11" = Seventh 2-foot layer, (464.14 X 87.5) X 18' 1" = Eighth 2-foot layer, (497.25 X 87.5) X 18' 3" = Ninth 2-foot layer, (650.64 X 150) X 18' 6" = Base layer, (1,628 X 150) X 18'6" = = 45,825 X 17' - 779,025 43,322 X 17'1" = 740,084 47,288 x 17'3" = 815,718 51,422 x 17'5" = 895,600 32,504 X 17' 7" = 571,530 35,109 X 17'9" = 623,185 37,812 X 17'11" = 677,465 40,612 X 18' 1" = 734,400 43,509 X 18'3" = 794,039 97,596 X 18'6" = 1,805,526 244,200 X 18'6" = Total, 4,517,700 12,954,300 The total moment is given to six significant figures, as a closer value would be an unnecessary refinement. 60 . Total Resisting Moment: Case I.—The total resisting moment is equal to the sum of those just found, and is as follows: Foot-Pounds Weight of empty train .. 2,080,800 Weight of bridge . 6,502,500 Weight of pier. 12,954,300 Total resisting moment. 21,537,600 135—30 40 BRIDGE PIERS AND ABUTMENTS §82 61. Factor of Safety Against Overturning: Case I. Dividing- the total resisting moment found in the preceding article by the total overturning moment found in Art. 56 gives for the factor of safety against overturning 21,537,600 = 3 2 6,715,300 62. Overturning Moments: Case II. —In the second case, it is assumed that there is no train on the bridge, and so there is no overturning moment due to the wind pressure on the train. The wind pressure in this case is assumed to be 50 pounds per square foot (Art. 46); therefore, the over¬ turning moments due to this pressure on the bridge and on the pier can be found by multiplying those found in Art. 54 by U. This gives the moments as follows: Wind on floor, 954,700 X to = 1,591,200 foot-pounds Wind on trusses, 2,790,700 X to = 4,651,200 foot-pounds Wind on pier, 24,600 X to = 41,000 foot-pounds The overturning moments due to the current are the same as found in Art. 55. Then, the total overturning moment is 1,591,200 + 4,651,200 + 41,000 + 132,300 + 13,100 = 6,428,800 foot-pounds 63. Resisting Moments: Case II. —The resisting moments in this case are the same as before, with the excep¬ tion of that due to the weight of the train, which is here omitted. The total resisting moment is, then: Foot-Pounds Weight of bridge. 6,502,500 Weight of pier. 12,954,300 Total resisting moment. 19,456,800 64. Factor of Safety Against Overturning: Case II. The factor of safety against overturning, in this case, is 19.456.800 _ o 6.428.800 §82 BRIDGE PIERS AND ABUTMENTS 41 65. Overturning Parallel to Axis of Bridge: Case III. —The principal force tending- to overturn the pier in the direction of the bridge is the longitudinal thrust of the train. Under some conditions, wind may blow approximately in the direction of the bridge, but the effect of this force is so small compared with the longitudinal thrust that it can be neglected with safety. In order to get the greatest force, the maximum weight of the train (4,000 pounds per linear foot, Art. 52) must be used. Since the pier supports two fixed ends (Art. 52), the weight of 306 feet of train, or 1,224,000 pounds, may be assumed to affect one pier. Then (Art. 51), the longitudinal thrust is .20 X 1,224,000 -- 244,800 pounds, acting horizontally at the bridge seat, or 24 feet above the bottom of the footing. Half of this is usually assumed to be resisted by the rails. The other half causes an overturning moment about the lower side edge of the footing, whose value is 122,400 X 24 = 2,937,600 foot-pounds 66. Resisting Moment: Case III. —The resisting moment is provided by the weight of the bridge and trains for a distance of 75 + 3 -f- 75 feet, and by the weight of the pier. The resisting moment offered by the bridge and train is (2,500 -f 4,000) X 153 X 5.5 = 5,469,750 foot-pounds The weight of the pier, making allowance for the buoyant effort of the water, can be found from Art. 59 to be 719,200 pounds. The resisting moment due to this weight is 719,200 X 5.5 = 3,955,600 foot-pounds. The total resisting moment is 5,469,750 + 3,955,600 = 9,425,400 foot-pounds 67. Factor of Safety Against Overturning: Case III. Dividing the resisting moment by the overturning moment gives the factor of safety against overturning: 9 ,425,40 0 = 3 9 2,937,600 68. Sliding.—The horizontal forces, such as wind pres¬ sure and longitudinal thrust, tend to make the upper part of 42 BRIDGE PIERS AND ABUTMENTS 82 the pier slide on the lower part. The resistance to sliding is equal to the vertical load multiplied by the coefficient of friction. The coefficient of friction for granite may be taken as .65. Three cases must be considered, in the same way as in the case of overturning. 69. Factor of Safety Against Sliding: Case I. —The horizontal forces in this case have been found in the prece¬ ding pages. They are: wind on train, 45,900 pounds; wind on floor, 18,360 pounds; wind on trusses, 73,440 pounds; wind on pier, 1,170 pounds; water on ice and drift, 8,631 pounds; water on pier, 1,305 pounds; total, 148,800 pounds. The vertical loads are as follows: weight of empty train, 122,400 pounds; weight of bridge, 382,500 pounds; weight of pier, 719,200 pounds; total, 1,224,100 pounds. Then, the resistance to sliding is 1,224,100 X .65 = 795,665 pounds, and the factor of safety against sliding is 795,665 _ r o 148,800 " ’ 70. Factor of Safety Against Sliding: Case II. —The horizontal forces in this case can be found in the same way as in the preceding pages. They are: wind on floor, M X 18,360 = 30,600 pounds; wind on trusses, f o' X 73,440 = 122,400 pounds; wind on pier, X 1,170 = 1,950 pounds; water pressure, 9,936 pounds; total, 164,900 pounds. The vertical loads are as follows: weight of bridge, 382,500 pounds; weight of pier, 719,200 pounds; total, 1,101,700 pounds. Then, the resistance to sliding is 1,101,700 X .65 = 716,100 pounds, and the factor of safety against sliding is 716,100 = . 3 164,900 71. Factor of Safety Against Sliding: Case III. The only horizontal force that need be considered in the present case is the longitudinal thrust of 122,400 pounds (Art. 65). The vertical forces are as follows: weight of train and bridge, 6,500 X 153 = 994,500 pounds; weight of pier, 719,200 pounds; total, 1,713,700 pounds. Then, the §82 BRIDGE PIERS AND ABUTMENTS 43 resistance to sliding is 1,713,700 X .65 = 1,113,900 pounds, and the factor of safety against sliding is 1,113,900 = 9 1 122,400 72. Sliding on Foundation.— In the three preceding articles, the entire weight of pier was used, so that the factor of safety relates to a surface at or near the bottom of the pier. The tendency to slide is greater at the bottom than at any other height, and it has been assumed that the pier rested on rock. Piers of the size under consideration usually extend down to rock or very hard strata, and so the method outlined will give the proper factor of safety if the foundations are first class. 73. Pressure on Subfoundation. —When the load on a pier is vertical and the center of gravity of the base is ver¬ tically under the center of the load, the pressure on the subfoundation is evenly distributed, and the intensity is found by dividing the load by the area of the base. When the center of gravity of the base is not directly under the center of the load, or when horizontal forces act on a pier, the pressure is unevenly distributed over the base. The effect of horizontal forces is shown in Figs. 18 and 19. When the only load on a pier is a vertical load, the line of action of the pressure on the subfoundation may be represented by the ver¬ tical line l m. When a horizontal force also acts on the pier, it is represented by the line In; the line lo, the resultant of /rh and In , represents the line of action of the pressure on the base. Fig. 18 shows the condition when the horizontal force is at right angles to the pier, and Fig. 19 shows the condition when the horizontal force is parallel with the pier. Under these conditions, the line of pressure does not pass through the center of the base, and the intensity of pressure varies. The formulas for finding the maximum and minimum intensities of pressure are: W 6 Wd 6M = W 6(Wd + M) A + AL + AL A + A L W _ 6 Wd _ 6M = W_ 6(Wd + M) A AL AL A AL (1) ( 2 ) 44 BRIDGE PIERS AND ABUTMENTS §82 in which p x = maximum intensity of pressure; p 3 = minimum intensity of pressure; W — total vertical load; A = area of base of pier; d = eccentricity, or distance from center of grav¬ ity of vertical load to center of base; L = length of base in direction of horizontal force; M — moment of horizontal forces about base. In order to find the maximum intensity of pressure on the foundation, two cases must be considered: Case I, when there is a full load on the bridge and the horizontal forces are greatest and acting in the direction of the pier; Case II, when there is a full load on the bridge and the horizontal forces acting at right angles to the pier are greatest. BRIDGE PIERS AND ABUTMENTS 45 S 74. Formula 1 of the preceding article is used to find the maximum intensity of pressure, in order to see if it exceeds the allowable. Formula 2 is of interest only when the maximum intensity of pressure is greater than twice the average pressure; in this case, there is a tendency for one end of the pier to rise, and under this condition water may get under it and disturb the subfoundation. In calculating the value Wd it is not necessary to get the location of the center of gravity of the vertical loads. It is invariably easier to multiply each vertical load by the distance from its line of action to the center of the base; the algebraic sum of all these products is equal to Wd. 75. Maximum Intensity of Pressure: Case I. The value of L is given in Fig. 16 as 37 feet, and, since the width is 11 feet, the value of A is 37 X 11 = 407 square feet. The value of W is found by adding the weight of the train, that of the bridge, and that of the pier, the last being decreased to allow for the buoyant effort of the water. Since the maximum weight of the train is 4,000 pounds and that of the bridge is 2,500 pounds per linear foot (Art. 52), their combined weight is (4,000 + 2,500) X (75 + 3 + 75) = 994,500 pounds The weight of the pier, decreased by the buoyant effort of the water, is found from Art. 69 to be 719,200 pounds. The value of W is, then, 994,500 + 719,200 = 1,713,700 pounds. 76. Since the vertical loads do not act through the center of the base, it is also necessary to find the value of Wd. The total weight of bridge and train that comes on the pier was found above to be 994,500 pounds. This acts 17 feet from the down-stream end of the pier (see Fig. 16); that is, 1.5 feet from the center, making the value of Wd for this load equal to 994,500 X 1.5 = 1,491,800 foot-pounds The value of Wd due to the weight of the pier can most easily be found by calculating the position of the center of gravity of the pier. Since the moment of the weight of the 46 BRIDGE PIERS AND ABUTMENTS 82 pier about the down-stream end (Art. 59) is 12,954,300 foot¬ pounds, and the weight of the pier (Art. 66) is 719,200 pounds, the distance of the center of gravity from the down¬ stream end is 12,954,3 00 = 18 012 i f ee t 719,200 This gives 18.5 — 18.0121 = .4879 foot for the distance from the center of gravity of the pier to the center of the foundation, and the value of Wd for this load is 719,200 X .4879 = 350,900 foot-pounds The total value of Wd is, therefore, 1,491,800 + 350,900 = 1,842,700 foot-pounds The moment of all the horizontal forces about the base is the same as the overturning moment found in Art. 56; namely, 6,715,300 foot-pounds. Substituting the proper values in formula 1, Art. 73, we have = 1,713,70 0 1,842,700 X 6 6,715,300 X 6 px 407 + 407 X 37 + 407 X 37 = 7,620 pounds per square foot Since this result is much less than twice the average value 4,211 pounds per square foot^, it is not necessary to consider the minimum intensity of stress 77. Maximum Intensity of Pressure: Case II.— In this case, the resultant of all the vertical loads passes through the center of the base. Then, since the eccentricity is zero, the third term in formulas 1 and 2, Art. 73, becomes zero, and need not be considered. The values of W and A are the same as in Case I; but since the horizontal forces in this case are assumed to act at right angles to the length of the pier, A =11 feet. The moment of the horizontal forces about the base is the same as that found in Art. 65, or 2,937,600 foot-pounds. Substituting the proper values in formula 1, Art. 73, gives = 1,713,70 0 2,937,600 X 6 Pl 407 + 407 X 11 = 8,148 pounds per square foot §82 BRIDGE PIERS AND ABUTMENTS 47 . Since this is less than twice the average value, it is not necessary to consider the minimum intensity. 78. Horizontal Forces Acting in Both Directions. The maximum intensities found in Cases I and II are those that occur when the horizontal forces act in but one direc¬ tion. Since it is possible for them to act in both directions at the same time, this case must also be considered. Under this condition, however, since it will probably be of such rare occurrence, the intensity of pressure is allowed to exceed the working pressure by 25 per cent. In Case I, the maxi¬ mum intensity of pressure extends for the full width of the pier at one end; in Case II, the maximum intensity of pres¬ sure extends for the full length of the pier at one side. The maximum intensity then occurs at the corner where these two edges meet, and is equal to the sum of the different terms obtained in Arts. 76 and 77, each term being used but once. For example, although the term — occurs in A each case, it is used but once in finding the sum. Then, the total intensity at one corner is 4,211 + 733 + 2,676 + 3,937 = 11,557 pounds per square foot If the intensity of pressure found in Cases I and II does not exceed the safe intensity, and if that just found does not exceed the safe intensity by more than 25 per cent., the design is assumed to be safe. Otherwise, the dimensions of the pier are increased and the foregoing computations repeated. Although the maximum intensity found by com¬ bining the two cases is greater than twice the average, no further attention need be paid to it. This high pressure will simply affect a very small corner of the pier. 48 BRIDGE BIERS AND ABUTMENTS 82 ABUTMENT DESIGN 79. The method of arriving - at the dimensions of an abutment are much the same as those used for piers. In the first place, the dimensions necessary to satisfy practical conditions are found; then, calculations are made to ascertain whether the abutment satisfies the theoretical conditions. PRACTICAL CONSIDERATIONS GENERAL FEATURES 80. Classes of Abutments. —There are three general classes, or forms, of abutments: the wing abutment; the T abutment; and the U abutment. The wing abutment con¬ sists of a mass of masonry extending the full width of the bank it restrains. It usually decreases in height from the edges of the bridge to the foot of the bank. Several forms of wing abutments are shown in the following pages. In the T abutment there is a face wall, called the head, on which the bridge rests, and a wall running back under the track or road to the top of the bank, as shown in Figs. 20 and 21. This wall is called the stem, or tail. The head usually extends beyond the stem at both sides. In the U abutment, there is a head similar to that in the T abutment; instead of one stem, there are two walls running back as far as the top of the slope, as shown in Fig. 22. These walls are even with the outer edges of the roadway. The space between the walls may be filled with earth or loose rock. 81. Bridge Seat.—As in the case of piers, the first thing to be considered is the size of the top of the abutment. An abutment usually supports one end only of a span, so the width of the bridge seat for this purpose need only be about §82 BRIDGE PIERS AND ABUTMENTS 49 /Base o f Ba// % . / r i ! | —- i i i i i i ! ! i i i_ j / .. \ /Base ofBa/7 Fig. 21 50 BRIDGE PIERS AND ABUTMENTS §82 one-half as wide as that of a pier. On account of the parapet, the bridge-seat course is usually continued back under it. The required width of bridge seat and the width of the parapet at the elevation of the bridge seat control the width of the top of the abutment. The ends of the bridge seat are carried a few feet beyond the trusses, or girders, or to the edge of the roadway. 82. Parapet.—The parapet, or back wall, of an abut¬ ment is simply a retaining wall to prevent the earth above the bridge seat from falling down. In order to keep it from ,Base o/ Ra/Y tipping over or sliding forwards on account of the pressure of the earth and of the traffic, the parapet should have a width at the bottom of from .4 to .5 the height. For a distance of about 3 feet down from the top, the back of the parapet should be given a smooth batter, usually called the frost batter, so that the action of the frost cannot move it. The face of the parapet should be set back far enough from the end of the bridge so that that end will not hit the parapet when the bridge expands. A distance of 3 inches is § 82 BRIDGE PIERS AND ABUTMENTS 51 ' I usually specified as the closest the bridge must ever come to the parapet. 83. Pedestal Blocks.—The remarks given in Art. 23 with regard to pedestal blocks for piers apply to those required for abutments. When used on abutments, it is advisable to have one surface of the block in contact with the parapet. 84. Pockets in Bridge Seats.—Great care should be taken in the arrangement of blocks and parapets on bridge seats that no pockets are formed. The word “pocket” is usually applied to portions of the work where dirt may collect and from which it is difficult to remove it. Since the earth behind an abutment usually extends up to nearly the top, winds usually blow some of it over the parapet on the bridge seat. If the steelwork is near the pocket, the dirt will pile up against it, absorb moisture from the air, and cause the steel to corrode rapidly. Pockets can be avoided by the use of end floorbeams, which do away with the pedestal blocks, and by continuing the parapet straight from end to end of the abutment so the wind can have a clean sweep across. For this purpose it is advisable to have the main trusses or girders at least 1 foot above the bridge seat. 85. Batter.—The front faces of many abutments are made plumb, because the appearance is not marred here as in the case of a pier with both faces plumb. It is somewhat better, however, to give the face a batter of about 2 inch per foot. It should be borne in mind that the batter of the front face increases the length of the span, and, therefore, that batter should not be made unnecessarily large. No standard batter is used at the back, the slope being con¬ trolled by the top width and the required width of base. 86. Width, of Base.—The width of base should be determined by means of theoretical considerations, allowing for the pressure at the back due to the earth filling increased by the weight and jar of the traffic. Generally, it is not usual in practice to make a theoretical determination of the 52 BRIDGE PIERS AND ABUTMENTS 82 required width of base, but to make the width a certain fraction of the height; in the case of high or unusual abut¬ ments, however, calculations should be made. It has been found by practical experience that the width of base should be from three-eighths to one-half the distance from the top of the fill to the level of the base. For first-class stone masonry or solid concrete, three-eighths of the height is sufficient; for second-class masonry, one-half the height should be used. 87. Footing;. —The footing course of an abutment does not have to extend beyond the back of the masonry, but it should extend beyond the front face. This is due to the fact that the overturning tendency causes higher pressure at the face of the abutment, and the footing must be extended here in order to distribute the pressure over a greater area. In some cases, several footing courses, each projecting beyond the one above, are used. T ABUTMENTS 88. Since the head of a T abutment has no bank to restrain, it may be made just wide enough at the top to accommodate the bearing plates of the bridge, and the front and back faces may be continued down to the base with a slight batter. The base of the head is at the foot of the slope, and the bank slopes up to the end of the stem. T abutments were formerly the most common, because of their simplicity and of the fact that low-grade masonry could be used. On the other hand, the T abutment requires much more masonry, on account of the stem, than abutments of other forms. For double-track railroads or wide streets, the amount of masonry is excessive. The most serious objection to this form of abutment, when used for railroad bridges, is that it gives a support to the bridge that is too rigid, making riding uncomfortable to passengers and injurious to the rolling stock and track. In some cases, 1 or 2 feet of ballast is inserted between the track and the masonry, but this does not wholly remove the bad effects of rigidity. §82 BRIDGE PIERS AND ABUTMENTS 53 U ABUTMENTS 89. The U abutment is practically a wing - abutment in which the wings are parallel to the track, although it has more the appearance of a T abutment. The slope of the bank is sometimes carried down outside the wings, but more frequently the space between the wings is filled with earth or broken stone. In the latter case, the two wings must be designed in the same way as a retaining wall. When the earth is allowed to run down outside the wings, they may be made about three-quarters the width required by the other condition; this is permissible because the earth outside the wings helps to support them. WING ABUTMENTS 90. Types of Wings.—There are three common types of wing abutments; namely, the straight-wing abutment, Fig. 23 Fig. 23, in which the face of the wings is in the same plane as the face of the abutment proper; the flaring-wing 54 §82 BRIDGE PIERS AND ABUTMENTS 55 135—31 56 BRIDGE PIERS AND ABUTMENTS §82 abutment, Fig. 24, in which the faces of the wings make an angle with the face of the abutment proper; and the curved-wing abutment, Fig. 25, in which the wings are curved. In general, the curved surfaces of the wings start almost normal with the face of the abutment and turn through about 90° until they are at right angles to the track. 91. Size of Wings. —Each wing starts at the top of the abutment and decreases in height until it reaches the foot of the bank, where the height reaches zero. The top surface of the wing follows approximately the intersection of the plane of the wing with the plane of the slope of the bank. In Fig. 26, (a) represents the form of wing that might be used with concrete, while (b) and (c) indicate the forms for stone masonry. The inclined line in (a) is smooth, because it is easier to finish concrete in this manner. The forms {b) and (r) are stepped, the height of the steps being the same as the thickness of the courses of the stone. In order to keep the earth from covering the wing, the steps should be carried beyond the slope, as shown in {b). To save masonry, the steps are sometimes made shorter and the earth allowed to cover the wing, as at (c ). In the design of wings, it is assumed that the bank has a slope of li horizontal to 1 vertical. When the earth is allowed to run over on the steps, as at (<:), the abutment presents a very untidy appearance, and the extra expense of the form {b) is warranted by the improved appearance. In the form shown in Fig. 26 (a), the top of the wing should be from 6 to 12 inches above the earth. 92. Toe of Wing.— When it is necessary to keep the earth from running around in front of the wing, the latter is BRIDGE PIERS AND ABUTMENTS 57 CO O-J continued down to the foot of the slope. When flaring wings are used, however, it is allowable for the earth to run around somewhat, unless the abutment is on the shore of the stream. A sketch of a wing in which this was done is shown in Fig. 24, the broken line at the left in (a) showing the foot of slope. Some masonry is-saved by adopting this method. The thickness of a wing is usually the same at the top as that of the abutment proper; the width or thickness gradually decreases until the toe is reached, where the width is usually about 2 feet. 93. Straight-Wing Abutments. —The straight-wing form of abutment is particularly fitted for bridges over city streets, and is probably the most economical for this purpose. The face of the abutment and wings is placed at the street line, and gives a good appearance. Since in this case it is impossible to have any earth in front of the abutment, the wings must be carried to the ends of the slopes. The principal advantage in the use of straight-wing abut¬ ments for railroad bridges is that in case an increase in the number of tracks is made, the courses can be so arranged that there will be little difficulty in extending the abutment. An outline of a straight-wing abutment is shown in Fig. 23; (a) is a plan of one end of the abutment, (b) is an elevation of the portion shown in (a ), and (c) is a cross-section through the bridge seat. In some cases, this form of abutment is slightly modified by making the back straight, and thus causing the wings to make a slight angle with the face of the abutment. This is not desirable, as it mars the appearance somewhat, and also because it involves extra expense in stone cutting in order to form the angle. * 94. Flaring-Wing Abutments. — For abutments at the edge of a river, flaring wings are usually preferable to other wings. They give a greater opening for the stream above the bridge, and help to direct the current into the proper channel, thus preventing the water from injuring the foundation by getting behind the abutment. Where the 58 BRIDGE PIERS AND ABUTMENTS 82 greatest economy is desired, the wings may be made shorter and the earth allowed to run around in front, as previously shown in Fig. 24. This reduces the amount of masonry. If the masonry is first class and well bonded, the flaring- wing abutment gives greater stability, for overturning or sliding can occur only when one part of the abutment tears away from the other part. In order to use the least amount of masonry in the wings when the earth is allowed to run around in front, they should make an angle of about 30° with the plane of the face of the abutment produced. This is the angle generally used; however, for greatest efficiency at a stream in directing the flow toward the opening, the angle should be about 10°, and the wings should be carried to the extreme foot of the bank, in order to keep the current from washing away the earth and thereby undermining the abutment. 95. Curved-Wing Abutments. —The principal reason for the use of abutments with curved wings is the improved 82 BRIDGE PIERS AND ABUTMENTS 59 appearance. With very high abutments, where filling mate¬ rial cannot be easily obtained, there is the additional advan¬ tage that they save filling. Fig. 25 shows one end of an abutment with curved wings. The front elevation, shown at (b ), indicates that the top of the curved wing follows the slope of the bank. Another form of curved wing is shown in Fig. 27. This form has the advantage that, like flaring wings,' it easily deflects the current. It also has a pleasing appearance, and is not so expensive as some other forms of curved wings. SKEW ABUTMENTS 96 . It is frequently necessary to build abutments at some other angle with the bridge than 90°. The bridge is then a skew bridge, and the abutments are called skew abut¬ ments. For such construction, straight wings may be used, as shown in outline by AD in Fig. 28. One end of each abutment can be made shorter than the other; this can be seen in the figure at A and D, both of these points being at the proper distance from the top of the slope C. In many cases, the wing is made to extend from C to A at right angles to the bridge. This really makes a flaring wing for this end of the abutment, as shown in Fig. 29. In case there is no water at the face of the abutment, masonry may be saved by allowing the earth to run around the end of the wing. /Nufura/ Surface of Ground §82 BRIDGE BIERS AND ABUTMENTS <>1 In the design of skew abutments, great care should be taken that the bridge seat is wide enough. As the bearing plates are set on a skew, they take up more width on the bridge seat than when there is no skew. PIER ABUTMENTS 97 . When a trestle is built across a valley, the greater part of the distance is spanned by girders or trusses. Near the ends, however, it is usually found more economical to fill for a short distance. It is then necessary to build an \ 7' If—1 1 1 / \ e abutment in this new ground for the end of the trestle. This is usually accomplished las shown in Fig. 30. A pier is built on the natural surface of the ground, and the fill allowed to run around in front of it. This form of structure is called a pier abutment (see Art. 2). 62 BRIDGE PIERS AND ABUTMENTS §82 THEORETICAL CONSIDERATIONS 98. Forces to be Resisted. —On account of the fact that an abutment is in contact with an earth fill, it can over¬ turn or slide in but one direction; namely, away from the bank. It is prevented from moving sidewise by the friction of the earth behind it and also by the wings. The forces that act to overturn the abutment are the pressure of the earth and the longitudinal thrust of the train. 99. Calculations. —The pressure of the earth is found in the manner explained in Retaining Walls , the weight of the train or other loads being added to the weight of the earth as the surcharge. The longitudinal thrust is found in the manner given for piers in Art. 51. The methods of determining the factors of safety against overturning and sliding, and the intensity of pressure on the subfounda¬ tion, are the same as for piers. CONSTRUCTION MATERIALS 100. The materials most used for piers and abutments are stone, concrete, and timber. In addition to these, large cylinders having steel or cast-iron outside surfaces and con¬ crete interiors are sometimes used. 101. Stone.—The most common material is stone. The very best stone available is generally selected. There is little economy in using poor masonry, as more is required than when good masonry is used. Granite, when available, is the best stone, because it is hard, strong, and durable, and can be brought to required shapes more easily than other stones of equal strength. Its only weakness is that it disintegrates easily when exposed to fire; but this danger seldom occurs in piers and abutments. 82 BRIDGE PIERS AND ABUTMENTS 63 Syenite, gneiss, quartz, trap, and porphyry are also excel¬ lent stones, but are hard to work, and so are seldom used for bridge masonry. Quartzite also is a good stone. When these stones are not available, some grades of sandstone or lime¬ stone may take their place. Certain kinds of stone are not suited for bridge masonry, on account of their tendency to disintegrate from the action of frost, acids, attrition, or fire. As a general rule, there is less danger from frost in those stones that are compact and non-absorptive than in porous stones, which permit water to enter and freeze inside. A compact stone is also less liable to absorb acids from the air or the water than a porous stone. Most sandstones are very absorptive, and the cementing material is easily dissolved if it consists of lime, iron oxide, clay, magnesia, or feldspar. Quartzite, or sandstone in which the cementipg material is of quartz, is less permeable and contains less matter that can easily dissolve. A stratified stone is also more liable to disintegration than an unstratified stone, because it offers straight paths between the strata for moisture to reach the interior, and also because it offers easy cleavage lines. In case stratified stones are used, they should be placed so that the strata will be horizontal. Limestone consists chiefly of carbonate of lime, and is easily injured by nitric and by hydrochloric acid. A mag¬ nesian limestone is much more durable than a pure limestone. Exposure to fire reduces the limestone to almost pure lime, which is soluble in water. i 102. Concrete.—Concrete, both plain and reinforced, is much used for bridge masonry. It is more durable than many limestones and sandstones, and, when properly made and laid, is sometimes as strong and durable as granite. The principal advantages of concrete are its comparatively low cost and the fact that it makes a monolithic structure. A Portland-cement concrete mixed in the proportions of 1:2:4 with hard, angular, well-graded stones for the aggregate, is a sufficiently good material for all bridge masonry. In many 64 BRIDGE PIERS AND ABUTMENTS 82 cases stone is used for the face of an abutment or pier, and the interior is made of concrete. 103. Timber.—In temporary construction, and in some cases where time does not permit the erection of masonry supports, bridges are supported by timber structures. These usually consist of several framed or pile bents set parallel, and close to one another. The abutments in this case are similar to the end bents described in Trestles , except that more than one bent may be used. In permanent structures, wood should never be used except for the foundations, and when so used the top of the timber construction should be below the lowest level the water ever reaches. 104. Joints.—The joints in stone masonry should be made as thin as possible, and rich Portland-cement mortar used. There is no economy in using a first-class stone and then binding the whole together with a poor mortar. 105. Steel Cylinders.—Sometimes, piers are made of steel or cast-iron cylinders filled with concrete. These are easy to place and are well adapted to soft soils. Short sec¬ tions are placed and bolted one on top of another, and the whole is forced into the ground. The interior is usually excavated while the cylinder is being driven, and then filled with concrete. In this form of construction, a pier usually consists of two cylinders with a set of heavy girders extend¬ ing across their tops. DETAILS 106. Specifications.—The specifications for bridge masonry may be substantially the same as for retaining walls, the principal difference being in the bridge seat. All the masonry above the footing should be first-class stone laid in Portland-cement mortar. The stones should be laid with horizontal and vertical joints, the courses being from 18 to 24 inches in height. The faces can be left rough, provided no point projects more than 3 inches beyond the edge of the joint. All joints should be perfectly straight. §82 BRIDGE PIERS AND ABUTMENTS 65 The bridge-seat course should be bush-hammered on top, and brought to a level surface at the proper elevation for the bearing plates. All joints above the footing should be pointed with neat cement with a i-inch semicircular bead, except at the top of the bridge seat, where they should be level with the stone. For bonding purposes, not less than one-quarter of the face should be composed of headers. All stones should be at least 4 feet long, but no stone should be longer than five times its height. On account of the impact of ice and drift, the masonry near the water-line of a pier in a stream may be of a better class than the remainder. The footing may be of coarse rubble, with stones about i cubic yard in size. 107. Bridge-Seat Course. —The bridge-seat course should be so designed that the stones will have a good bond. Except where the bearing is very large, there should be but one stone under the bearing plate. In general, the joints should be kept as far as possible from the edges of the bear¬ ing plates. The top of each stone under a bearing plate should be dressed smooth. Pedestal blocks should not be used for a height less than 8 inches, and the height should not be less than one-fifth the length of the block. For a small difference in elevation of the bearings, castings may be used, or the bridge-seat stones may be made of different thicknesses, as previously explained in connection with Fig. 5 (b). 108. Parapets. —The parapet of an abutment helps to bond the bridge-seat stones. The latter should extend at least 9 inches behind the face of the parapet. When prac¬ ticable, it is advisable to have the bridge-seat stones extend entirely across the top of the abutment. There is no neces¬ sity for dressing the face of a parapet. In railroad bridges, it is customary to have the top of the parapet from 8 to 12 inches below the base of rail, and to rest a tie on its top. In highway bridges, it was customary to have the top of the parapet even with the street surface. This gave rise to an uncomfortable jar in passing over the masonry. At present, 66 BRIDGE PIERS AND ABUTMENTS §82 the top of the parapet is finished off below the street sur¬ face, and the flooring is continued across. 109. Pedestals and Piers for Trestles. —The small depth of a trestle pier requires great care in properly bonding the separate stones, so that the loads shall be properly dis¬ tributed over the base. Fig. 7 shows an outline of a pier suitable for the support of a high trestle bent. When sepa¬ rate pedestals are used, the top of each should be a single stone. Even when concrete is used, the cap should be of a good hard stone. Fig. 31 shows a stepped pedestal of stone masonry. When the lower part is concrete, the sides are sloped, as shown in dotted lines. 110. Batter. — The batter of the face of a pier or of an abutment is obtained by cutting the face of the stone so that it will slope the proper amount. The batter at the back of an abutment, where it is hidden by the earth, is obtained by allowing the ends of the stones to pro¬ ject beyond those above; the back ends are usually left in the shape in which they come from the quarry. 111. Undermining. —Unless piers or abutments rest on solid rock, water is likely to wash away the material under them. This is called undermining, and is especially likely to take place in piers placed in running streams. The danger from this is not apparent, since the damage is done under water. In order to avoid undermining, the pier should be carried to a great depth, and broken stone piled up around it. This broken stone, when used for this purpose, is called riprap. BRIDGE DRAWING INTRODUCTION 1. The general principles that govern the preparation of all bridge drawings are the same as those that have been explained and illustrated in Introductioyi to Constructio?i Draw¬ ing. In the present Section, the additional special informa¬ tion required for bridge drawing will be given, the application of the principles being illustrated by four detail drawing plates. The student is required to send the tracings of the drawing plates, one at a time, to the Schools for correction. He should retain the pencil drawings for future reference; it may be necessary for him to trace them again. 2. Structural Shapes and Standards.—Almost all the parts used in structural and bridge work have been standard¬ ized to a certain extent, and their dimensions and properties have been tabulated and printed in handbooks by the manufac¬ turers of rolled steel. Bridge draftsmen should have a copy of the handbook used by the designer. All the infor¬ mation of use to bridge draftsmen that is usually contained in structural-steel handbooks is given in Bridge Tables. The use of these tables is explained in Bridge Members and Details , Parts 1 and 2. In preparing the drawing plates, however, it will not be necessary to consult tables to any great extent, as all the dimensions required for the laying out of the work are shown on the plates that he will receive, or are given in the following pages. 3. Center Bines.—In making a drawing of a part of a bridge, it is customary to draw first the center line of the COPYRIGHTED BY INTERNATIONAL TEXTBOOK COMPANY. ALL RIGHTS RESERVED §83 9 BRIDGE DRAWING 83 part, and then refer all points to that line. In addition to the first or main center line, other lines are drawn to repre¬ sent the centers of details. For example, the line on which the centers of a number of rivets in a row are located is drawn before locating .the rivets. That line is sometimes called the center line of tlie rivets; more frequently, however, it is called the gauge line. Strictly speaking, only symmetrical objects have center lines; but, in bridge work, lines that serve as bases of reference in locating details on non-symmetrical objects are spoken of as center lines. Center lines on a drawing are very important, and should be laid out very accurately, as they are used as standards of reference from which to lay out distances and locate points. 4 . Dimensions. —A frequent source of trouble in making a bridge drawing is the location of the dimension lines. The drawing is usually so filled with views of details that it is difficult to draw the dimension lines without interfering with other lines. Draftsmen should be very careful so to locate the dimension lines and write the dimensions that no ambiguity can arise. When there are several consecutive dimensions, such as a number of rivet spaces that are unequal, it is well to show them r in a line, so that they can be added readily when the drawing is being ?■ checked. In such a case, it is cus- \ * tomary to give both the separate distances between rivets and the distance between the end points, or between two easily distinguished points near the ends. This distance is the sum of a number of smaller distances, and, by giving it, the necessity is avoided of adding the smaller distances whenever the total distance is wanted. The method is illustrated in Fig. 1, which represents the connection angle at the end of a stringer. The lines a a and bb are the center lines of the flange rivets, and are 2? inches from the top and bottom, respectively, of the flange angles. The spacing of the rivets in the connection % A l7~ ( < ) ( ( > < > H ) V- <\o > -s-b )- > 7— —TO Fig. 1 83 BRIDGE DRAWING 3 angle is given by the left-hand line of dimensions. For convenience of reference, the sum of all the dimensions con¬ tained in the left-hand line is given at the right. This sum is the distance from the top of the top flange angle to the bottom of the bottom flange angle, and is usually spoken of as the vertical distance back to back of the flange angles. 5. When a dimension line is too short to permit the dimension to be written legibly between its ends, the number indicating the dimension is placed outside the line and close to it, as the dimension of 2 t inches at the top and bottom of the stringer in Fig. 1. Dimension numbers should, as a rule, be placed as close as possible to the objects or parts to which they refer, except where, when so located, they inter¬ fere with other lines or numbers. In all cases, they are located with respect to the other lines or numbers in such a way that there will be no doubt as to their meaning. 6. Shortened Views. —When a part of a bridge has the same form, dimensions, and parts for a considerable dis¬ tance, as from a a to b b } Fig. 2, it is unnecessary to show the whole of it to scale if the space on the drawing is limited. In such a case, a portion of the member may be left out, and the ends moved closer together. The parts that are then shown will be drawn to scale just as though the entire member were drawn, but the dimensions referring to the omitted or broken portion are not shown to scale. In Fig. 3 is shown the member shown in Fig. 2, but short¬ ened so as to occupy about one-half the space. An opening is usually left, as at c, Fig. 3, and lines d , e are drawn to indi¬ cate that part of the member has been omitted. The cutting lines are sometimes straight, as d and e , Fig. 3 ( a ), and some¬ times irregular, as / and g, Fig. 3 (b ). When a member is cut and shortened, as just explained, its total length (13 feet 3 inches in this case) is written on a dimension line between its ends, although this line does not represent that length to scale. The total length of a member is frequently called the over-all dimension, or the length over all of the mem¬ ber. This dimension should always be given. « O) 0 E Cb. i jt-|—©- o----e- 0 TT ? <0 •> i x 0i *> ■i •--©- *n , > to ) lie T > I 1 I 1 s •tr 4 Fig. §83 BRIDGE DRAWING 5 7 . Representation of Structural Sliapes by Dines. The usual methods of representing structural shapes in cross-section were explained and illustrated in Introdiiction to Construction Drawing. These shapes are represented in plan and elevation as shown in Fig. 4, the curved corners Fig. 4 being represented by single lines at the points where the surfaces joined by the curves would intersect. For example, the curved corners at a and b, Fig. 4 (a), are represented in the elevation by the line de , and the curved surfaces b and c by the line fg in the plan. Similarly, in Fig. 4 (b) , the curved corners at h h are represented in the elevation by the lines ij ; those at k k, by the lines l m in the elevation and the line n o in the plan. When channels and I beams are drawn to a small scale, the lines ij and l m are very close together, and it is customary to draw but one line to represent both corners, this line being about midway between the two, as shown in Fig. 5. 8. Cross-Sections of Thin Shapes. —When shapes appear very thin on drawings, they are not section-lined or cross-latched as explained in Introduction to Construction Drawing , but are filled in solid black, as shown in Figs. 4 and 5. When two surfaces are in contact, a narrow space is 135—32 BRIDGE DRAWING 8 Fig. 6 Conventional Signs for Structural Rivets Shop Field Two full heads. O e Countersunk inside and chipped. 0 <§> Countersunk outside and chipped .... Q m Countersunk both sides and chipped . . . 0 m - Inside Outside Both Sides Flatten to i inch high or counter¬ sunk and not chipped. Flatten to i inch high 0 (B Flatten to f inch high oo Of 0 # Fig. 7 83 BRIDGE DRAWING 7 left between them, as shown in Fig. 6. This method of showing cross-sections is employed on almost all drawings made to a scale of 1 inch to the foot or to a smaller scale. 9. Conventional Signs for Rivets. —The two conven¬ tional signs for rivets are explained in Bridge Membei's and Details , Part 1, and illustrated in Bridge Tables. The Osborne standard is believed to be the simpler and plainer, and will be used in the drawing a plates of this Section. For venience, the conventional are repeated in Fig. 7. In riveted members, rivets are shown only in IG ' 8 those views in which they appear in end elevation, except where but one view of the member is shown, in which case some of the rivets may be shown in side elevation, as at a, a , Fig. 8. 10. Eyebar Heads. —Usually, the sizes of heads of eye- bars are not given on bridge drawings, but they are drawn to scale. For this purpose, the diameter of the circular head is taken from a table, and the head is laid out as explained in Bitroduction to Constric¬ tion Drawing . The radius of the curves that join the outline of the head with the sides of the bar is usually equal to twice that of the circular head, as illustrated in Fig. 9. 11. Pin Nuts. The long diameter of each pin nut can be Fig. 9 taken from a table. The outlines of these nuts are drawn by inscribing hexagons in circles having diameters equal to the long diameters of the respective nuts. The thick¬ nesses and other necessary dimensions are also taken from tables. 8 BRIDGE DRAWING §83 12. Numbering Drawings. —There are various sys¬ tems in use for numbering detail and working drawings. In some offices, each drawing is given a separate number, as was done in Introduction to Constructio?i Drawing and Con¬ struction Drawing. In other offices, each “job” or contract is numbered, and all drawings relating to that contract are given the same number. When this system is followed, there is usually given a secondary set of numbers indicating the num¬ ber of each drawing and the number of drawings that relate to the contract. The form used in some offices is as follows: Contract 976 Sheet 3 of 6 sheets and in others, as follows: Contract 976: (5) of In each case, the notation indicates that the drawing is one of a set of drawings that relate to contract 976, that « it is drawing number 3 of such set, and that the complete set consists of six sheets or drawings. The former system will be used in the plates in this Section. DRAWING PLATES GENERAL CONSIDERATIONS 13. The drawing in this Section consists of four bridge plates showing all the joints on one side of the center of the pin-connected truss treated in Desig?i of a Highway Truss Bridge , Parts 1 and 2. The details of the ends of all the members that connect at the joints are also given, together with the side elevation of an intermediate floorbeam and sidewalk bracket. The student is advised not to start the drawing of these plates until after he has completed the two Sections mentioned. 14. In making a drawing of a bridge of this kind, it is the usual practice to lay out on a large sheet the skeleton drawing of one-half of the truss to a scale of 7 , i, or \ inch 83 BRIDGE DRAWING 9 to the foot. This gives the directions of the members that meet at each joint, and shows the joints and members in their relative positions. The appearance of the drawing when the skeleton outline has been drawn is shown in Fig. 10, in which fghi is the border of the sheet, and ciB Ee the outline of one-half of the truss. The details of the ends of the members are then drawn at the joints, and the members broken between the joints to indicate that por¬ tions are omitted. The details of the members are usually drawn to a scale of ! or 1 inch to the foot. Each office or drafting room has a rule as to the scale to be used. In the following drawing plates, the distances between the joints are drawn to a scale of f inch to the foot, and the details of the ends of the members are drawn to a scale of 1 inch to the foot. In explaining the layouts of the following plates, some of the dimensions given in the text refer to sizes of parts and distances on the members, which are to be laid out to the scale of 1 inch to the foot. Other dimensions, such as those referring to the location of views on the plates are to be laid out full size on the drawings. There will be no difficulty in recognizing which dimensions are to be scaled and which are to be laid out full size. 15. On account of the difficulty in handling and mailing, the Schools have found it advisable to limit the size of drawing plates to 13 in. X 17 in. inside the border lines. It is impossible to draw an entire half truss on a sheet of this size without using an inconveniently small scale; hence, to conform to the size of the plate and the usual scale, the truss is shown on four plates. The sheet shown in Fig. 10 10 BRIDGE DRAWING 83 may be considered to be divided by the lines KK and LL , and the part of the truss in each of the four corners drawn on a separate sheet. 16. Erection Diagram. —When a truss is too large to be conveniently shown on a sheet, as just described, each member is drawn separately, and in the lower right-hand corner of the sheet is placed a skeleton drawing of the truss to a very small scale. Each member is given a letter and number so as to show its position in the assembled truss. The general arrangement of the views on such a drawing is illustrated in Fig. 11. In most cases, however, more than one sheet is required in order to show all the members. 17. General Directions. —The plates will all be 13 in. X 17 in. inside the border lines, with the views arranged as explained in detail in the following articles. A space of I 2 in. X 4 in. is reserved in each plate for the title, but it is not always possible to locate the titles at the lower right- hand corners of the sheets. They are located either on the lower or on the right-hand border line. All the information necessary for the drawing of the plates is given in detail in the following articles. When the distance - / // . f o ^ ----- ---.- O'S' / // ■ / o > < ■ ..... 3 ' 5 $> 6 =/ 7 - 6 ‘ / » zo-o- II / // / " - 1 - 6 — 135 l 83 r ~' " * -- — - —• W^\\ cA viVi\\\%N. 83 BRIDGE DRAWING 11 from the edge of a piece to the last rivet is not given on the plate, it may in general be taken as I 2 inches. In case any difficulty is experienced in finding the dimensions of a section in the text or on the plate, they can be found in Bridge Tables. The heads of the shop rivets on the plates are li inches in diameter, and the holes for the field rivets are El inch in diameter. The letters in the title are i inch in height, and all the other lettering on the sheets is -32 inch in height, made as described in Introduction to Construction Drawhig. DRAWING PLATE 107, TITLE: HIGHWAY-BRIDGE DETAILS 18. This plate shows the elevation of the joint at the left end of the bottom chord and two joints to the right of it. In addition, it shows side elevations of the two verticals, cross-sections of the two verticals and end post, and a plan of the bottom chord members. The border lines enclose an area of 13 in. X 17 in. 19. Center Lines of Members. —The center lines of the members are drawn first. The center line of the lower chord in elevation is drawn parallel to and 82 inches from the top border line. The center of the end joint is then located on this center line 1 inch from the left border line, and the centers of the other joints are located from the first by laying off two distances of 20 feet each to the scale of f inch to the foot. The joints will be referred to by the letters given in Fig. 10, the three joints on this plate being a , b, and c. Next, draw vertical lines through a , b, and c, to represent the cen¬ ters of the chair at a and the verticals at b and c. Next, lay out the center line of the end post from a , and the center line of the diagonal from c. There are various methods of specifying the slope of a diagonal line on a bridge drawing; the most common is to give the coordinates of a point on the line with reference to the joint from which it starts, or to give the lengths of the sides of a right triangle of which the diagonal or part of it is the hypotenuse. See Fig. 12. The 12 BRIDGE DRAWING 83 slope of a line, when given in this way, is called the skew. In Fig. 12 ( a ), the total distance passed over vertically and horizontally by a diagonal of the truss under consideration is given; this method is convenient and well adapted to detail drawings. For shop drawings, the method shown in Fig. 12 ( b ) is preferable; this gives the distance passed Pig. 12 horizontally in 1 foot vertical. It is preferred by some to give the skew in 2 feet instead of in 1, since the workmen use steel squares 2 feet on a side. The skew in 2 feet, in this case, is M X 24 = 17fi inches. Since, in this case, the skew is given in terms of the total distances, it is best to measure along ab and bc> Fig. 13, and lay off a a' and cc\ each equal to 20 feet, to a convenient scale, say s inch to the foot. Then, erect perpendiculars at a r and c' and lay off a ' a" and c f c", equal to 27 feet, to the same scale as a a 1 and cc'. Then, a a" and cc" are the center lines of the inclined mem¬ bers. When, on the following plates, the right triangle giving the skew is drawn, the long leg of the right angle will in every case be made 1 inch in length. 83 BRIDGE DRAWING 13 . 20. Joint a .—Next, lay off the width of the eyebar that runs from a to the right, 6 inches wide, one-half above and one-half below the center line, and break it off about 4f inches from a. Now, lay off the top and bottom of the angle that shows dotted inside the eyebar; this leg is 2i inches wide, If inches of which is above the center line. The left end of the angle is 1 foot If inches and the first rivet is 1 foot 3 inches from a; the rivets are 6 inches apart. It is not necessary to make circles for the rivets on the pencil draw¬ ings, but simply to indicate their centers by short lines at right angles to the gauge lines. The head of the eyebar can now be drawn. In Bridge Tables , the radius of this head is given as 6f inches. Then, the radius of the curves joining the head to the straight portion is 132" inches. The long diameter of the nut is 7i inches; the diameter of the pin is 4f inches; and the diameter of the threaded end of the pin is 4.inches. These can now be drawn. 21. Next, lay out the outlines of the end post, breaking it off about 6f inches above the pin. First, lay off the top and bottom lines 7f and 7i inches, respectively, from the center line, and draw them parallel to it. Next, lay out the thicknesses of the outstanding legs, and draw the inside lines of the angles. Next, measure 7i inches below the center of the pin, and draw the end line of the end post to its intersection with the center line. Then, draw the other line at the end of the post, making the same angle with the center line as the first line. Measure in 3| inches from each side of the post to locate'the edges of the flange angles, and draw them parallel to the center line. Next, draw two lines very close to the edges of the angles and parallel to them, to represent the edges of the 8-inch side plate. Now, lay off the gauge lines of the rivets, and put in the rivets according to the dimensions given in the drawing. The ends of the tie-plates that appear on the sides of the end post can be located by measuring from the nearest rivets. 22. Measure up 4 inches on a vertical through a , and draw the end of the end post as shown on the drawing plate, 14 BRIDGE DRAWING 83 breaking it off 4 inches from the center of the pinhole. This is shown by itself to avoid confusion with the pin nut and eyebar head. The top of the chair is 1 foot 8 inches above the center of the pin, and 9 inches wide. The outside of the gusset at the left is straight from the outer corner of the flange angle of the end post to the outer corner of the chair. The angles of the diaphragm inside the gussets are ih inch apart, and have the 3-inch leg in view. The other dimensions can be scaled as given. Some of the rivets in the end post are countersunk on one or both sides to give room for the eyebar head on the outside and the pedestal on the inside. 23. Next, locate the center of the cross-section of the end post 8 f inches (on the inclined center line) from the center of the pin, and lay out, according to the dimensions given, first the webs 15 in. X in., then the flange angles 3l in. X 3i in. X I in., and then the side plates 8 in. X 1 in. 24. Joint b. —Now, draw the eyebars, eyebar heads, pin and pin nut, and angles inside the eyebars at the joint b, in the same way as for joint a ; the dimensions are the same. Next, draw the vertical lines representing the angles in the hip vertical, starting them 2 f inches (to scale) above the center of the pin and breaking them off 62 inches above. The adjacent backs of the angles are inch apart; the out¬ standing legs are 7ins inches apart. The gauge lines for the rivets and rivet holes are 4-re inches apart, 2 3 V inches on each side of the center. The rivet spacing is given at the left of the member. The pin plate at the lower end of this vertical extends 5 inches below the pin, and is 10 inches wide at the bottom. The top of the pin plate is lOf inches above the center of the pin. Next, put in the cross-section of this member, locating its center 7i inches above the center of the pin, and make the angles 3iin. X 2lin. X ns in. 25. Next, draw the side view of this vertical, locating its center line 2 i inches from that of the other view, and break¬ ing it off about 7 inches above the center line of the bottom chord. This view is 71 inches wide; the legs of the angles §83 BRIDGE DRAWING 15 * are 2 \ inches wide; and the gauge lines of the rivets are 4k inches apart. At the lower end of this member, the * angles are connected by a web-plate, which extends up to 4 feet 10 inches above the center of the lower chord, and is connected to the angles by means of rivets spaced as shown. Near the bottom of the member, a short horizontal angle 2 k inches wide is riveted to the inside of the member. Above the web-plate, the angles are connected by latticing. This latticing is drawn by first locating the centers of the rivets in the ends of the lattice bars from the dimensions given, then drawing the center lines of the lattice bars, then draw¬ ing a semicircle 2 \ inches in diameter at the end of each bar and connecting these semicircles by lines parallel to the center lines and Is inches from them on each side. 26. The outstanding legs of the angles are blacked in at intervals in this view to indicate that there are rivet holes in them. Those at the left of the view can be located by projecting across from the front view shown directly over the pin. In the right-hand side, the lower rivet is 1 foot 7 i inches above the center line of the lower chord, and above this there are eight spaces of 3k inches each. It is not customary to show rivet holes in this way wherever they occur, but only when it is desired to emphasize some special feature. In the present case, this view shows that the holes are not spaced the same on both sides of the truss. 27. Joint c» —The details of the members meeting at the joint c are next drawn. The eyebar that starts toward joint b is the same as the eyebars at that joint. The eyebar that extends toward the right is 5 inches in depth, and the circular portion of the head is 6i inches in radius. The eye- bar that is shown as the diagonal is 6 inches wide, and the circular part of the head has a radius of 6f inches. The pin and pin nut at this joint are the same as at joints a and b. 28. The vertical shown in this view is composed of two channels 10 inches in width, extending from 5 inches (to scale) below the center of the lower chord to about 3k inches above it. The tie-plates that show on the sides of this view 16 BRIDGE DRAWING §83 near the top are located by projecting across from the side view, which will be explained presently. The inside lines of the angles that are shown in dotted lines are A inch apart, and the outer edges are 7~iq inches apart. The rivet holes in these two angles are located on gauge lines 4 ts inches apart, their distances from the center of the bottom chord being shown on the right of the view. The top and bottom lines of these angles can also be located from the dimen¬ sions, and also the angles seen in end view below the angles just referred to. Next, put in the outline of the pin plate, 8 inches wide, at the bottom of the vertical, and then the gauge lines of the rivets, as given below the pin. The rivets connecting the pin plate to the channel are next located. These rivets are shown dotted where they are hidden by the eyebar heads; they are also shown flattened to a height of f inch on the outside of the member, to allow the eyebars to lie closer to the vertical. The cross-section of this member can now be drawn above the front elevation; the center of the cross-section is 7i inches above the center line of the lower chord. The channels are 82 inches back to back, and the flanges are 2 f inches in width. On the inside of the section is shown the outline of the dia¬ phragm that is inserted between the channels near the lower end. The web of this diaphragm is -re inch thick. The legs of the angles next to the web are 2 i inches in width, and the legs in contact with the channels are 3a inches in width. 29. The side elevation of this vertical is next drawn, the center line being located parallel to and 2 f inches to the left of the center line of the other view. It is customary to show side views on the right of front elevations; but, in this case, there being no available room on the right, the side view is shown on the left. The top of the side view is broken off 7 inches above the center line of the lower chord. The lattice bars near the top of the view are drawn in the same way as for the hip vertical. The tie-plate is 1 foot in height and located, as shown, by the dimension lines at the left of the side view. Below the tie-plate there are no rivets in the 83 BRIDGE DRAWING 17, flanges of the channels, so they are assumed .to be cut away, thus showing the webs of the channels in section. This is for the purpose of showing the detail of the diaphragm more clearly; if the flanges were hot cut away, the diaphragm would be hidden behind them to such an extent as to obscure the details. The legs of the diaphragm angles that show in this view are 2* inches in width. The gauge lines of the rivets in the diaphragm are 2i inches on each side of the center line, and the rivets are located on them as shown by the dimensions at each side. At the lower end of this view, the pin plates are shown in cross-section. 30. Plan of Bottom Cliord.— The plan of the bottom chord is shown below the elevation. The center line of the plan is parallel to and If inches above the lower border line. At joint a , only the ends of the eyebars are shown; it is assumed that the end post and pin have been removed. At joint b y the hip vertical is shown in cross-section. This view is broken off 3f inches from a and 4 inches from b. The eyebars are not parallel to the center line, but diverge from it. The ends, however, may be shown parallel as far as the breaks, and then offset, as shown on the drawing plate. This better illustrates the fact that the eyebars diverge, and shows the direction in which they diverge. The two eyebars that form each member are connected by lattice bars. When the divergence is slight, the gauge lines of the rivets that connect the lattice bars to the outstanding legs of the angles riveted to the insides of the eyebars are made parallel to each other and to the center line. This makes it easy to get the lattice bars all the same length, and is accomplished by skewing the gauge lines on the angles. In this case, the gauge line of each angle is lil inches from the back of the angle at the left end and lli inches from the back of the angle at the right end; the difference of if inch does not practically affect the strength of the latticing. The lattice bars can be put in as follows: First, draw the gauge lines of the rivets in the angles as shown; then, locate the rivets on these lines, and draw the center lines of the 18 BRIDGE DRAWING §83 lattice bars. Then, draw semicircles 2i inches in diameter at each rivet, and draw two lines 2i inches apart and parallel to each center line to represent the sides of the lattice bars. The appearance of the two eyebars latticed in this way is shown in Fig - . 14 to an exaggerated scale. This is the most satisfactory method of arranging the latticing between diver¬ ging eyebars. Other methods require the bars to be of different lengths, which involves much more computation on the part of the draftsman and more trouble in the shop. When the divergence is greater than shown on the drawing plate, the same method of connecting the lattice bars can be used, except that, in such a case, it is necessary to use wider angles on the eyebars. 31. The nuts shown on the pin at the joint b are li inches thick and are recessed i inch; the threaded ends of the pin are 1& inches in length at each end. Between the inside eye- bars and the pin plates of the hip vertical, and also between the pin plates, the pin is shown enclosed in thin rings or cyl¬ inders. These are usually made of cast iron about I or f inch in thickness, and of lengths just sufficient to fill the spaces between the members on the pin; they serve the purpose of preventing the members that connect to the pin from moving sidewise. Wherever there are spaces in pins to which members do not connect, they should be filled with fillers. The outside diameter of the fillers at the joint b is inches, and the inside diameter is 5 inches. 32. On all bridge drawings, it is customary to refer the center line of the chords to some plane of reference, such as the top of the floorbeam, the top of the floor, or some other ' “ ' 135 § 83 . 83 BRIDGE DRAWING 19 . convenient elevation. In the present case, the top of the floor at the center of the bridge is used, and is shown 5 feet 3i inches above the center line of the lower chord. This line serves to locate different parts of the bridge. DRAWING PLATE 108, TITLE: HIGHWAY-BRIDGE DETAILS 33. This plate shows the elevation of the center joint of the lower chord and the joint next to the left of it. Refer¬ ring to Fig. 10, it is seen that these joints are lettered d and