2 jf, // 
 
 /<?. 
 
 1013 12th St., Oakland, Cal. J 
 
 Not to !>< taken from the Library. ? 
 
 THE LIBRARY 
 
 OF 
 
 THE UNIVERSITY 
 OF CALIFORNIA 
 
 LOS ANGELES 
 
 GIFT OF 
 
 John S.Prell
 
 * 
 
 ii 
 
 L. M. Clement. 
 
 THE 
 
 THEORY OF STRAINS. 
 
 A COMPENDIUM FOB THE 
 
 CALCULATION AND CONSTRUCTION OF BRIDGES, 
 ROOFS AND CRANES, 
 
 WITH THE APPLICATION OF 
 
 TRIGONOMETRICAL NOTES. 
 
 CONTAINING 
 
 THE MOST COMPREHENSIVE INFORMATION IN REGARD TO THE 
 
 RESULTING STRAINS FOR A PERMANENT LOAD, AS ALSO FOR 
 
 A COMBINED (PERMANENT AND ROLLINGS LOAD. 
 
 IN TWO SECTIONS. 
 
 ADAPTED TO THE REQUIREMENTS OF THE PRESENT TIME, 
 
 <g> 
 
 BY 
 
 ?,<? 4rt 
 
 JOHN H. DTEDRFCHS, 
 
 tt> J*S a. j^- 1 
 
 &A 
 
 CIVIL AND MKCHAMCAL ENGINEER. 
 
 rOv 
 
 
 ILLUSTRATED BY NUMEROUS PLATES AXD DIAGRAMS 
 
 \- 
 
 i- 
 
 BALTIMORE: 
 
 SCHMIDT & T E O W E . 
 1871. 
 
 PRELL 
 
 Civil 6- Mechanical Engineer. 
 
 SAN FBAN CISCO, CAL.
 
 Entered according to Act of Congress, in the year 1871, by 
 
 JOHN H. DIEDR1CHS, 
 In tne Office of the Librarian of Congress, at "Washington. 
 
 WESTCOTT & THOMSON, 
 Stereotypcrs, Philada.
 
 Engineering 
 Library 
 
 INSCRIBED TO 
 
 WENDEL BOLLMAN, ESQ., 
 
 IN TESTIMONY OF HIS INTEREST IN SCIENCE AND ART. 
 
 - 
 
 l\ 
 
 k 
 
 713709 
 
 Library
 
 CONTENTS. 
 
 PREFACE .. ....... 7 
 
 EXPLANATION OF CHARACTERS USED IN THE CALCXTLATIONS 11 
 
 SECTION I. 
 
 A. INTRODUCTION 13 
 
 I. THE LEVER 15 
 
 II. SUSPENDED WEIGHTS AND THE RESULTING STRAINS 18 
 
 III. TRUSSES WITH SINGLE AND EQUALLY-DISTRIBUTED LOAD 21 
 
 Suspension Truss Bridge , 23 
 
 Suspension Bridge 27 
 
 B. ROOF CONSTRUCTION 31 
 
 C. SEMI-GIRDERS 43 
 
 I. SEMI-GIRDERS LOADED AT THE EXTREMITY... 43 
 
 Cranes . , 44 
 
 II. SEMI-GIRDERS LOADED AT EACH APEX 45 
 
 D. GIRDERS WITH PARALLEL TOP AND BOTTOM FLANGES 48 
 
 I. STRAIN IN DIAGONALS AND VERTICALS .. 48 
 
 II. STRAIN IN FLANGES , ... 51 
 
 III. TRANSFORMATIONS .. 53 
 
 IV. GENERAL REMARKS...... ..... 53 
 
 Directions for the Calculation of Complex Systems 54 
 
 E. COMPARATIVE TABLES OF RESULTING STRAINS FOR A " 
 
 PERMANENT LOAD ., 56 
 
 I. SYSTEM OF RiGicr-ANu;r> TRIAVGI.KS , 50 
 
 II. SYSTKM OF ISOSTRI.KS BHACIXO.... oS 
 
 i * t>
 
 CONTENTS. 
 
 SECTION II. 
 
 GIRDEKS CALCULATED FOR COMBINED (.PERMANENT AND 
 ROLLING) LOAD. 
 
 PAG* 
 
 A. GIRDERS WITH PARALLEL TOP AND BOTTOM FLANGES 59 
 
 I. THE RIGHT-ANGLED SYSTEM 5!) 
 
 II. ISOSCELES BRACING 66 
 
 1. Triangular Truss 66 
 
 Calculation of Strains y and u in Diagonals 67 
 
 2. Isometrical Truss 69 
 
 a. Calculation of Diagonals 70 
 
 b. Calculation of Top and Bottom Chords (Flanges) 72 
 
 B. CAMBER IN TRUSSES, WITH PARALLEL TOP AND BOT- 
 
 TOM CHORDS 73 
 
 TABLES CONTAINING THE LENGTH OF ARCHES FOR DEGREES, 
 
 MINUTES AND SECONDS, FOR A RADIUS AS UNIT 77 
 
 0. PARABOLIC GIRDER OF 48 FEET, OR 16 METRES, SPAN 80 
 
 D. THE ARCHED TRUSS 86 
 
 Calculation of Strain y in the Diagonals 88 
 
 Calculation of Strain in the Verticals V. 90 
 
 Transformations 93 
 
 E. THRUST CONSTRUCTION 94 
 
 I. GIRDER 20 FEET IN LENGTH (WITH A SINGLE WEIGHT AT THE 
 
 CENTRE). 
 
 II. GIRDER 72 FEET IN LENGTH (CALCULATED FOR PERMANENT 
 AND ROLLING LOAD). 
 
 Definition of Strain x in the Horizontal Flanges 97 
 
 Definition of Strain y in the Diagonals 98 
 
 Calculation of the Tensile Strains z in the Lower Flanges 99 
 
 Calculation of the Verticals u 101 
 
 III. CALCULATION OF A TRUSS SUSTAINING A DOME 102 
 
 Calculation of Strains x of the Outside Arch 104 
 
 Calculation of Strains z of the Inside Arch 10o 
 
 Calculation of the Diagonals y 106
 
 PREFACE. 
 
 THE want of a compact, universal and popular treatise on 
 the construction of Hoofs and Bridges especially one treating 
 of the influence of a variable load and the unsatisfactory 
 essays of different authors on the subject, induced me to pre- 
 pare the following work. 
 
 Bridge-building has been and always will be an important 
 branch of industry, not only to engineers, but also to the masses 
 for the purposes of travel and trade, and, as Colonel Merrill in 
 his recent essay on Bridge-building remarks, " important to 
 railroad companies on account of the large amount of capital 
 invested in their construction." 
 
 Bridge literature has often been used by rival parties for the 
 purpose of advancing their own private interests, their motive 
 being competition. Imposing upon the faith and credulity of 
 those whom they pretend to serve, there is no guarantee that 
 worthless structures will not be erected. 
 
 Thoroughly independent of any such motive, my aim is to 
 give, especially to bridge-builders and to engineers and archi- 
 tects, the results of my investigation on the subject of calcu- 
 lating strains, in order that capitalists and the public may be 
 benefited and protected.
 
 _J PREFACE. 
 
 These calculations will also unable those who have but a 
 limited knowledge of mathematics to acquire the necessary 
 information. For this reason special attention is paid to the 
 arrangement of the work, the whole being made as plain ami 
 simple as possible, in order to meet the wants of the common 
 mechanic as well as the experienced engineer. 
 
 Though there are many valuable treatises of this kind, there 
 has as yet been no work published serviceable to the degree 
 desired by the practical builder or mechanic most of the dis- 
 sertations being too theoretical and hard to comprehend by one 
 not versed in the higher mathematics; and some are so ar- 
 ranged that a clear understanding of the calculations is very 
 difficult.* 
 
 The most valuable work in the language is doubtless Mr. 
 Stoney's " Theory of Strains," though the Method of Moments 
 is not developed to that degree which I think necessary for the 
 practical man. 
 
 We owe to the renowned German engineers Hitter and Von 
 Kaven the universal application of this Method in the work 
 entitled "Dach und Brucken Constructionen," in which it is 
 fully explained by examples and illustrated by diagrams, these 
 being often carelessly neglected in other works. 
 
 The above-mentioned work served me very much in the 
 arrangement of this, which I hope will be kindly received. 
 
 The work being expressly prepared, as aforesaid, for the use 
 of beginners in the study of mathematics, as well as for the 
 
 * As an exception, may be named Mr. Shreve's brief but popular treatise 
 in Van Nostrand's "Engineering Magazine," No. 3 x., August, 1870; Vol. III.
 
 PliEI'ACJE, 
 
 more advanced practical engineer, it will enable them, after a 
 short perusal, to acquire all the necessary information, for which 
 even the trigonometrical notes accompanying the general results 
 are not really required. 
 
 The higher classes of colleges and other institutions of learn- 
 ing will find the work very valuable. 
 
 On account of the expense, 'an intended Appendix, contain- 
 ing a rational and concise investigation on " The Strength of 
 Materials," had to be dispensed with; yet I hope with this 
 volume to gratify not only the desire of friends, but to be able 
 with great satisfaction to assist engineers in the pursuit oi' 
 
 their high and noble calling* 
 
 THE AUTHOR, 
 OCTOBER, 1870.
 
 EXPLANATION OF CHAKACTEES USED IN THE 
 CALCULATIONS. 
 
 = Equal, or the sign of equality. 
 
 -f- Plus, or the sign of addition; also, the symbol of positive (tensile) strain. 
 
 Minus, or the sign of subtraction ; also, the symbol of negative (compres- 
 
 eive) strain. 
 
 X or . Sign of multiplication. 
 : or -5- Sign of division. 
 , Sign of decimals ; also, of thousands, 
 oo Sign of infinite. 
 < Sign of angle, signified by the Grecian cyphers oc, /?, y, 6 f e. 
 
 2 Sign of square of a number. 
 1/~~ Sign of square root of a number. 
 Sign of degrees. 
 ' Sign of minutes; also, of feet. 
 ff Sign of seconds; also, of inches. 
 ( ) or [ ] Brackets, to enclose the mathematical expression bound to the 
 
 same operation. 
 
 TT The number 3, 14, or periphery for a unit of the diameter. 
 E Eight angle, or 90. 
 J_ Vertical. 
 
 II
 
 THE THEORY OF STRAINS. 
 
 SECTION I. 
 
 A. INTRODUCTION. 
 
 To enable the student to comprehend the work, and have a 
 thorough knowledge of certain conditions and examples without 
 studying the whole, it is necessary for him to understand the 
 arrangement of the following pages. 
 
 On the first few pages and the appertaining figures at the close 
 of the chapter is found a short description of the lever in its differ- 
 ent appliances, the application being only a key to the calculations 
 of strains which follow. 
 
 The trigonometrical notes are in many cases almost superfluous. 
 Still, it may be advantageous in this way to accustom the reader to 
 their use. The " Suspended Weights and Resulting Strains" are 
 developed by the parallelogram of forces, and for a plain illustra- 
 tion the results are appended to the figures, which will also be 
 observed on figures of "Trusses with Single and Distributed Load." 
 
 In the "Suspended Weights and Resulting Strains" a more 
 elaborate calculation was thought necessary, and therefore an Intro- 
 duction to the calculation by the " Method of Moments" may be 
 found in its proper place. 
 
 This Introduction presents the beginner with a clear and com- 
 prehensive knowledge of the formation of Moments; and the equa- 
 tions for Figs. 20, 21, etc., explain the equilibrium of force and 
 leverage. 
 
 At the close of this chapter is found the explanation of maxi-v 
 mum compressive and tensile strain in the top and bottom chords 
 of a parallel-flanged truss or girder. 
 
 On "Roof Construction" (B), Plates 6 to 11, remarks are u-i- 
 
 2 13
 
 14 THE THEORY OF STRAINS. 
 
 necessary. The builder can with ease find from the figures a 
 system to suit his purpose. (See also " Arched Trusses," D, Sec- 
 tion II.) ' 
 
 On "Semi-Girders" (C) the calculation of strains is treated in 
 the way heretofore generally known (determining from the centre 
 toward the abutment), after which the "Method of Moments" is 
 applied to the same example, followed by a more elaborate expla- 
 nation of the principles of Moments on the crane skeleton (Plate 
 14), whose single members are altogether divergent. 
 
 The thorough calculation of a truss with horizontal top and 
 bottom flanges (right-angled system D, Sect. I.), with the resulting 
 strains for a system of braces, all of which are inclined ii^ the 
 same direction, shows how easily by transformation of the strains 
 a system of bracing just reversed can be formed. (Comp. D, 
 Sect, II.) 
 
 From the comparative tables, E, L, II. (Plate 18, 19) those who 
 are not mathematicians can find, by a little study (for an assumed 
 load " W" a variable load not considered), the strains in flanges, 
 braces and ties. (See D, IV., Sect. I.) 
 
 The progress of panels, and by this the increase of stress in the 
 different members, are ad libitum to be extended (are optional). 
 
 Where no composite strains appear, in the skeletons double lines 
 make the compressive ( ) strains more obvious, the tensile (+) 
 strains being always represented by single lines. The assumed 
 load in the calculations is equally divided on the apexes ; but in 
 general some attention may be paid to a peculiar load say from a 
 single locomotive at a certain apex, this being observed in exam- 
 ples on "Suspension Truss" (III., Sect. I.). 
 
 The calculations in Sect. II. with regard to the influence of a 
 variable load are more difficult to understand. Still, by the 
 results of strains in the skeletons it js easy enough to form an idea 
 about this matter, and to see the importance of counter-bracing or 
 tying at centre of railroad-bridge trusses. 
 
 What experience and observation have already taught to the 
 practical railroad man is here/w% shown by figures. 
 
 Information is given on parallel-flanged trusses for the so-called 
 " Camber in Bridges" at B, Sect. II., which to many builders has 
 heretofore been only a matter of experiment. 
 
 Yet it is to be remarked that for the calculations E, Sect. II., Plates
 
 29 to 34, a variation in 
 will be perceived, the 
 being the horizontal 
 vertex (centre), an' 
 ments. 
 
 Farther expl 
 referred to. 
 
 The calcuh 
 nectioris at 
 relieving tf~~ " 
 strain. T] 
 ture are ;_J 
 
 * 
 
 / 
 
 r i i 1 
 
 f ; 
 
 p. -f * - 
 
 ur) (*>-) W 
 
 ^ 
 
 5 
 
 * f 
 
 ^ ^
 
 ' 'STRAIN 
 
 'necessary. The builder can with east the fulcrum eq ual 
 
 system to suit his purpose, (bee also A 
 tion II.) 
 
 On "Semi-Girders" (C) the calculation of 
 the way heretofore generally known (determine 
 toward the abutment), after which the " Method 
 applied to the same example, followed by a more < 
 nation of the principles of Moments on the crane s. 
 14), whose single members are altogether divergent. is divided in 
 
 The thorough calculation of a truss with horizont 
 bottom flanges (right-angled system D, Sect. I.), with th 
 strains for a system of braces, all of which are inclin 
 same direction, shows how easily by transformation of th 
 a system of bracing just reversed can be formed. (Co) 
 Sect, II.) 
 
 From the comparative tables, E, I., II. (Plate 18, 19) thosi 
 are not mathematicians can find, by a little study (for an assu 
 load "\V" a variable load not considered), the strains in flanpad, 
 braces and ties. (See D, IV., Sect. I.) 
 
 The progress of panels, and by this the increase of stress in thi 
 different members, are ad libitum to be extended (are optional). 
 
 Where no composite strains appear, in the skeletons double lines 
 make the compressive ( ) strains more obvious, the tensile (-(-) 
 strains being always represented by single lines. The assumed 
 load in the calculations is equally divided on the apexes ; but in 
 general some attention may be paid to a peculiar load say from a 
 single locomotive at a certain apex, this being observed in exam- 
 ples on "Suspension Truss" (III., Sect. I.). 
 
 The calculations in Sect. II. with regard to the influence of a 
 variable load are more difficult to understand. Still, by the 
 results of strains in the skeletons it |s easy enough to form an idea 
 about this matter, and to see the importance of counter-bracing or 
 tying at centre of railroad-bridge trusses. 
 
 What experience and observation have already taught to the 
 practical railroad man is here/M/ shown by figures. 
 
 Information is given on parallel-flanged trusses for the so-called 
 " Camber in Bridges" at B, Sect. II., which to many builders has 
 heretofore been only a matter of experiment. 
 
 Yet it is to be remarked that for the calculations E, Sect. II., Plates
 
 p- 
 
 a, 
 *S 
 
 /r* 
 
 * 7 
 
 * > 
 
 1 * 
 
 /' 
 
 t / 
 
 
 r- f r * 
 
 ^) (?r) (S) ' ^ 
 
 5
 
 THE LEVER. 17 
 
 In every triangle the sum of enclosed angles, 
 2 R. = 2 X 90, 
 
 g-j so in the right angled triangle ABC, Fig. 6, the angles A 
 and C together = 90 = R, because angle B = 90. 
 
 In Trigonometry we say 
 
 b b c 
 
 - sine oc ; = tangent <x ; - = secant cc. 
 c a a 
 
 an c 
 
 - cosine oc; = cotangent cc; - cosecant oc; or, contracted, 
 c b b 
 
 sin., cos., tang, or tg., cotang. or cotg., sec. and cosec. 
 
 For a radius, AC, as a unit, the line AB simply is called sine; 
 the central distance, BC, cosine; and BE the versed sine of the 
 angle <x. 
 
 For certain angles oc the relation -,-,-, etc., have certain and 
 
 c a a 
 
 distinct numerical values. (See Haslett's or HaswelPs " Tables of 
 natural sines, cosines, tangents and cotangents from 1 to 90 
 degrees.") 
 
 Each triangle, AB.C, A^C (Fig. 7), consists of six members 
 i.e., three sides and three angles, from which always three are de- 
 pendent on the rest ; therefore, when three out of these six mem- 
 bers are known, we can construct, or with more exactness we 
 can calculate, the others, provided one at least of the given parts 
 is a side. 
 
 For the transformation of trigonometrical functions, short notices 
 in the form of a table, also the numerical values (natural sin, cos, 
 etc.) of the principal angles, may be serviceable, viz. : 
 
 1 tang x 
 
 sin x = - = - = v 1 cos 2 x. 
 cosec x sec x 
 
 1 cotang x /- r~ 
 
 cos x = - - V 1 sin x. 
 
 sec x cosec x 
 
 1 sin x 
 
 tang .r = = V sec'' 1. 
 
 cutang x cos x 
 
 cotang x = - - = V cosec 2 x 1. 
 
 tang .r sin x 
 
 2* R
 
 18 
 
 THE 
 1 
 
 sec x = 
 
 cosec x 
 
 cos x 
 1 
 
 .:Y OF STRAINS. 
 
 cosec x 
 
 cotang x 
 
 sec x 
 
 , -i i , 2 
 = v ! + tau S 
 
 tang a? 
 
 = V\ -4- cotang 4 x. 
 
 DEGREES. 
 
 SIN. 
 
 Cos. 
 
 TANG. 
 
 COTANG. 
 
 
 
 
 
 1 
 
 
 
 OC 
 
 15 
 
 0,258 
 
 0,965 
 
 0,267 
 
 3,732 
 
 20 
 
 0,342 
 
 0,939 
 
 0,363 
 
 2,747 
 
 25 
 30 
 
 0,422 
 1=0,8 
 
 ,- 0,906 
 
 1/3 = 0.866 
 
 ,- 0,466 
 |T/3 = 0,577 
 
 /- 2,144 
 V 3 = 1,732 
 
 40 
 45 
 
 ,_ 0,642 
 $1/2 = 0,707 
 
 ._ 0,766 
 JK 2 = 0,707 
 
 0,839 
 1,00 
 
 1,191 
 1,00 
 
 50 
 
 (JO 
 
 ,_ 0,766 
 11/8 = 0,866 
 
 0,642 
 
 i=o,5 
 
 ,-_ 1,191 
 1/3=1,732 
 
 ._ 0,839 
 11/3 = 0,377 
 
 65 
 
 0,906 
 
 0,422 
 
 2,144 
 
 0,466 
 
 75 
 
 0,965 
 
 0,258 
 
 3,732 
 
 0,267 
 
 12=-- 90 
 
 + 1 
 
 
 
 -f 
 
 
 
 2 JJ = 180 
 
 
 
 j 
 
 
 
 or: 
 
 3 11 = 270 
 
 -" 1 
 
 a: 
 
 
 
 4 /2 yen i o 
 
 + 1 
 
 
 
 + <* 
 
 sin (R ye) = + cos .r, and cos (7? #) = + sin x. 
 sin (J2 -}- .T) = + cos z, and cos (S -j- a:) = ~ sin x. 
 
 II. SUSPENDED WEIGHTS AND THE RESULTING 
 STRAINS. 
 
 In Fig. 8", when IF 5000 Ibs, 06 = 6e = 10, 
 = 8, and ac = ce 12,8 feet ; the vertical strain at 
 
 Plate 2,1 
 
 Fig. 8*, 
 
 " 8 h . 
 
 , . W 5000 
 
 c on each string = = - -. 
 
 2 2 
 
 4000 Ibs, 
 
 And, further, the actual strain in the direction of the string, 
 
 W ac 
 
 P. --=9= ir.;- 
 
 2 bo 
 
 or '= = 00 9 12 ' 8 
 
 2 8 
 
 All other information is given in Fig. 8 h . 
 
 When a heavy body, ABCD (Fig. 8), is suspended by two 
 8 ,-| oblique strings, DH and CH, in a vertical plane, a straight 
 line drawn through the intersection \vill pass through the 
 centffe.of gravity, G, of the body.
 
 SUSPENDED WEIGHTS AND RESULTING STRAINS. 
 
 For the force in the direction ad, represented by g, we find 
 
 9>] from Fig. 9, 
 
 od , " 
 
 i o 
 s& ab . ~ 
 
 tci === 0/1 + <z cos 
 
 W bd 
 
 
 mid in the same manner from similarity of triangles, 
 
 w a ^ 
 
 Iu the equation for q is TF.|> the vertical strain at d for the 
 string ad ; in the second equation is F. f ' the vertical strain at 
 d for the string ^-equal to the shearing strains 7 and 7, on the 
 supports. 
 
 Example. When, again, 
 
 W -. 5000 Ibs., L 100 feet, 
 ab '=* 10', be 90', and bd = 8', 
 ad = 12,84', and c<Z = 90,35', 
 
 so p = 5000 X -- - X -^-- 564 ^ lbs - 
 
 90 12,cS4 0( y^> >:^ 
 
 n ' 1 f7 rrrz OOUw x\ '~ i/\ 
 
 ' ioo t> 
 
 then the result* for the horisoiitel .train x and the vertical strain 
 Fat the right support are- 
 eft Trr a6 **'. 
 
 - 5000 X X = 5625 Ib, ;
 
 20 THE THEORY OF STRAINS. 
 
 V = 5000 X - = 500 Ibs. 
 
 The results for the horizontal strain Xi and the vertical strain 7, 
 
 at the left support are 
 
 be ab 
 
 *i = "-T-rj' 
 L bd 
 
 he 
 
 and \\ = W.~; 
 
 lj 
 
 thus x, = 5000 X - ': X -\ = 5625 IKs., 
 
 100 o 
 
 on 
 and }\ = 5000 . == 4500 Ibs. ; 
 
 therefore, also, x = x lt 
 
 ... ab be Trr be ab ,^. 1 - , 
 
 for W.~-. r =W. T -.- (Fig, 17.) 
 
 I, M L Id 
 
 1Q , For Fig. 10, when IF = 5000 Ibs. ; aft, ad and bd the 
 
 same as before 
 
 /if? 1 '> .H4 
 
 F = IF. = 5000 X - - = 8025 Ibs., 
 bd 8 
 
 and Y l IF. f 6 7 = 5000 X - 3 =' 6250 Ibs. 
 
 bd o 
 
 11.] In Fig. 12, 
 
 P : Q : R sin . fdb : sin . ac/6 : sin adc. 
 12.] In general, for every triangle, 
 
 y = x-\-z, (Fig. 12.) 
 
 and, as here, x -f- y = 2 -j- m = J?, 
 
 or, m = 2x. 
 
 Also, from similarity of triangles ABC, ADC and DRC, 
 
 c b b> 
 
 - =-- -, ore = - ; 
 o a a 
 
 t* 
 
 and as -4.B = c -f- the diameter = - -f- a; 
 
 a 
 
 t. e., the diameter equals the square of one-half tjie chord divided 
 by the height of the arc added to the height of the arc. 
 
 The height, of the arc CH results from the chord CD in the 
 arne way.
 
 TRUSSES WITH SINGLE AND EQUALLY-DISTRIBUTED LOAD. 21 
 
 Application in " Camber in Bridges," B, Sect. II. There, also, 
 the geometrical rule, 
 
 arc angle at centre 
 
 circumference ' 360 
 
 [Plate 2 Figs. 8 to 12.] 
 
 III. TRUSSES WITH SINGLE AND EQUALL ^-DIS- 
 TRIBUTED LOAD. 
 
 Plate 3,1 A most frequent structure is the trussed benm (Fig. 
 Fig. 13J 13). 
 
 The post at the centre is called the king-post. 
 
 The whole is a combined system, in which the horizontal 
 beam, according to its stiffness, relieves the tie-rods from an 
 aliquot amount of strain. 
 
 For the greatest exertion to which the tie-rods in the most un- 
 favorable case could be exposed, we 1 may use the result from 
 Fig. 8 b , under the supposition that the horizontal beam counteracts 
 only the horizontal forces. (For instance, when butted at the 
 centre.) 
 
 To compute the stress we have the following: 
 
 W ac W 
 
 p = n = -- sec . ac, 
 2 be 2 
 
 for a single weight at centre. 
 
 Example. The assumed weight 20000 Ibs. ; between, supports, 
 24 feet. 
 
 The length be = 5/>9', and ac 13,24 (which can be measured 
 near enough for most purposes from a skeleton) ; 
 
 W rir 1 ^ 24- 
 
 P = q = ~-r= : 10,000 X ^- = 23,700 (appror.). 
 2 be 5,59 
 
 The angle <x being in this case 65, for a calculation, using the 
 preceding tables, 
 
 = g= - sec = X -- - 10000 X -= 23700, 
 
 2 2 cos 0,422 
 
 be = ub . tang 25= 12 X 0,4663 = 5,595, 
 . - * . c . 25 = rf . = 12 X JL. = 13,24.
 
 22 THE THEORY OF STRAINS. 
 
 The vertical pressure in the king-post under the supposition be* 
 fore = 20,000 Ibs. ; at each support 10,000 Ibs. ; and the conv 
 preision in the horizontal beam, 
 
 H = K. fl l == 10000 X - *= 21467 Ibs. 
 2 be 5,59 
 
 When in Fig. 15" the strains p or q and the angle ? are 
 known, we find the resulting vertical strain, E, also by 
 means of the parallelogram of forces, viz., 
 
 -^ Vp 1 -j- (f -(- 2 p . 5 . cos f ; 
 or, because in this case at = ft, . therefore, jp = q ; 
 also, r = 2.65 = 130, or = E + 40, 
 
 and cos (R -\- 40) = sin 40 (see preceding table), 
 
 BO R = 1/2 p2_j_ 2 ^ 2 ( sin 40) = 1/2 j9 2 (1 sin 40), 
 15". J E = 1/2 x 23700 2 (1 0,642) = 20000 Ibs.* 
 
 For a structure, Fig. 15 b , reserved to the preceding one (15 a ), the 
 numerical value of strains is quite the same, but of opposite cha- 
 racter, provided the enclosed angles are the same* 
 
 If, as in Fig. 16, the load 40000 Ibs., equally distributed on the 
 beam, then each support will sustain again one-half of the load ; 
 1 -i but the reaction, D, of each support will be only one-quar- 
 ter of the load = 10000 Ibs. ; and for the same exertion a 
 truss or beam, charged with an equally-distributed load, will sustain 
 twice as much as when loaded with a single weight at the centre, 
 (Comp. Fig. 14.) 
 
 The distribution of forces on supports and at the centre is 
 explained by Fig. 16*. 
 
 For an angle, <* = 63 26', of, also, ft = 26 34', 
 the depth of the truss being always equal to one-fourth of its length, 
 for tang . 26 34' = 0,5. 
 
 Kow, e - = tang . ft, or, be = db . 0,5 = 12 X 0,5 = 6, 
 
 00 
 
 * For the angle aed = <f in Fig. 15 h , 
 
 R = yp 2 + 9* + 2 pq cos rf, and for the angle edS = f, 
 R = tf~Pi* +P- 2 p, . p. cose, (See Fig, 8 b and Fig. 9.)
 
 ^r-rs*
 
 SUSPENSION TRUSS BRIDGE, 23 
 
 and (Fig. 14), the horizontal strain, 
 
 ff= jri2 = 20g0012 *= 20000, 
 
 26 2 6 
 
 thus being in this case the same as the weight, W, at the centre. 
 
 The horizontal strain (thrust) and the strain on the oblique rods 
 increase with the angle <xj thus being GO for an angle <x. = 90. 
 
 SUSPENSION TRUSS BRIDGE. 
 
 We find a combination of trussing in the well-known "Suspen- 
 sion Truss" bridges, the principles for calculation of the strains 
 being contained in the preceding. 
 
 The Bollmann truss forms a continuous system of independent 
 trusses, in number equal to the number of vertical posts combined 
 to a common top chord (stretcher). 
 
 Plate 4,1 By Fig. 17 the dimensions of such a truss may be 
 Fig. 17.J represented. 
 
 When for a single-track railroad-bridge the assumed load, in^ 
 eluding the weight of structure 1 tons = 3360 Ibs. per lineal 
 foot, or for one rib (single truss) = 1680 Ibs. per lineal foot i. e., 
 for the given dimensions 12 X 1680 = 20160 Ibs. on each post, 
 for which may be said in round figures 20000 Ibs. = W, for calcu- 
 lation, then in Fig. 17, according to Fig. 9, the tension in the first 
 rod nearest the abutment, 
 
 Strain No. 1, rod = W- - Ak - 9* 20000 X - X ^-=27300 Ibs., 
 
 the section of which for a value of iron = 10000 Ibs. per square 
 inch (five to six times security) 2,73 square inches ; thus, when 
 two rods are applied, the size of each rod = 1 X It". 
 
 Strain No. 2, rod == W- ~ = 20000 X - X = 39000 Ibs, 
 AF d 8 10 
 
 Section 3,9 sq. in., or 2 rods, each 1 X 2". 
 
 Strain No. 3, rod = W- . = 20000 X .- X ^- = 46625 Ibs. 
 AF dm 8 10 
 
 Sect. = 4,66, or 2 rods, each 1 X 2f". 
 
 FF AN 4. 4Q 
 
 Strain No. 4, rod = W- -- = 20000 X - X = 49000 Ibs. 
 
 AF EN. 8 10 
 
 Sect, 33 4,90, or 2 rods, each 1 X 2i".
 
 24 THE THEORY OF STRAINS. 
 
 Strain No. 5, rod = W- ~' = 2000 X X ~ = 45600 Ibs. 
 
 Sect. = 4,56, or 2 rods, each 1 X 2t". 
 
 Strain No. 6, rod = W- - ff ~- ~ P = 20000 X f X "^~ = 36300 Ibs. 
 yljP gp 8 10 
 
 Sect. = 3,63, or 2 rods, each 1 X It". 
 
 Strain No. 7, rod = W- ?- = 20000 X - X = 21150 Ibs. 
 AF hq 8 10 
 
 Sect, = 2,11, or 2 rods, each i X It". 
 
 When for a partial load at a certain panel the exertion of a pair 
 of suspenders is greater than for a distributed load in calculation, 
 those rods would be strained more than to one-fifth or one-sixth of 
 their ultimate strength. So, when a locomotive of 84000 Ibs. 
 weight rests at a certain panel on a wheel-base of 12 feet, to each 
 of the four supporting posts would be transmitted one-fourth of its 
 weight = 21000 Ibs. this differing very little from the calcula- 
 tion in the example. Additional rods (panel-rods) are applied, 
 sustaining the main , suspenders and at the same time the top 
 chord, transmitting and distributing the weight on the posts, these 
 being always in a state of compression equal to the weight on the 
 post. 
 
 Without the panel-rods for an over-grade bridge (through bridge) 
 there would be in the post no further compression than that pro- 
 duced by the weight of the top chord and appendages, leaving for 
 a strong cambered truss (B, Sect, II.), in case of a partial load, the 
 possibility of raising. 
 
 The following is the strain in panel-rods according to Fig. 8 b : 
 
 Sect. = 1,52, or 1 rod = 1 X U". 
 
 The influence of temperature upon the single systems of main 
 suspenders (their length being different) is regulated by a link 
 connection. 
 
 For the compressive strain in the top chord the rule for a girder, 
 sustained at both ends and charged with an equally-distributed 
 load, may be applied (see at the close of this chapter), then,
 
 SUSPENSION TRUSS BRIDGE. 25 
 
 ?ooooxj_>< = 168000 * 
 
 10 
 
 The compression in the top chord is the same all over. 
 For the result we have as momentum one-half of the entire 
 weight on posts at one-fourth the length of truss as leverage, 
 
 Q L 
 
 or Mom. = X ' 
 
 2 4 
 
 which, when divided by the depth of truss, gives the compression = 
 168000 Ibs. as before. 
 
 For a single load, P, tit the centre would be 
 
 P L 
 
 Mom. X iy > 
 
 which, when divided by the depth of truss, gives for the compres- 
 sion twice as much, or 336000 Ibs. ; but for an addition of the re- 
 sults of each single truss with its single load, according to x and 
 #! in Fig. 9, it would be 
 
 P '} L 
 Mom. = x 
 
 2 8 
 
 this being one-half of the result for a single load, P, at the centre, 
 added to one-half of the result for an equally-distributed load, Q. 
 
 18.] The Fink truss (Fig. 18) is different in principle. 
 
 Whilst in the Bollmann system there are as many independent 
 trusses as there are posts, in the Fink all the trusses are dependent 
 on each other and transfer the load toward the centre. 
 
 The centre post (king-post) has to sustain the compression of 
 one-half of the entire load on the truss, including one-fourth of the 
 weight of the rib, the main suspenders (tie-rods) depending again, 
 as before, on the depth of the truss. 
 
 * For a simple compressive strain an area of section of stretcher = 9 square 
 inches would be sufficient (18000 Ibs. safe load fur cast iron) the actual 
 dimensions to be taken by Hodgkinson's formula ou the strength of hollow 
 cast-iron pillars: 
 
 JJ3.6 J3,6 
 
 W = breaking weight in tons = 44,3 ; 
 
 therefore, when for a pillar, (he external diameter, 7), in inches, and the 
 length, /, in feet, are known ; and for six times security, with the weight, W, 
 multiplied by 6, we can define the internal diameter, d, consequently the 
 thickness of metal. 
 3
 
 26 THE THEORY OF STRAINS. 
 
 The calculation of an example in its simplicity will give the 
 best explanation. 
 
 Taking the same dimensions and the same load as in the calcu- 
 lation for the preceding (Fig. 17), according to Fig. 8 b we have, 
 
 Strain inA.N or IN = ~ X 
 
 2 
 
 80000 ^, 49 
 i.e., X 196000 IDS., 
 
 the section of wlHch for a value of iron 10000 Ibs. per square 
 inch = 19,6 square inches. Thus, when two rods are applied, the 
 size of each = 2 X 5 inches. 
 
 Strain in kc, Ak or cm =? - X = 26000 Ibs. 
 
 2 5 
 
 Section of a single rod = 1 X 2$ in. (full). 
 Strain in Alor IE= - - ~ = 52000 Ibs. 
 
 Zi JLv/ 
 
 Section of a single rod = 2 X 2J in., or 1 X 5 in. 
 
 For a single locomotive (weight 84000 Ibs.), resting at cd on a 
 wheel-base of 12 feet, the vertical force for one post at c or d = 
 42000 
 
 2 
 
 = 21000 Ibs. = W\ 
 
 then, Strain in tie-rods -- X = 27300 Ibs. ; 
 
 2 5 
 
 so the size of rod kc- Ak or cm should be corrected to 1 X 2J in. 
 
 For the compressive strain on the top chord (stretcher), accord- 
 ing to Fig. 14, 
 
 . W AE 
 = . , 
 
 2 EN 
 
 and for a vertical strain W = 80000 Ibs. in the centre post, us 
 before mentioned, 
 
 2i LO 
 
 In this truss, as in the Bollmann, the compression in the top 
 chord is the same all over. 
 
 For this truss, when applied for an over-grade railroad-bridge, a 
 safe longitudinal connection (bracing or tying) will be essential on 
 account of the variable load.
 
 ?<., ovvfit. *ff 
 
 TT~R 
 i
 
 SUSPENSION BRIDGE. 27 
 
 To the same category of bridges belongs the 
 
 SUSPENSION BRIDGE, 
 
 though, in regard to mathematics, very different. 
 
 The curve formed by a chain or cable lies between the parabola 
 and the catenary, and is very nearly an ellipse. The curve in a 
 loaded state approaches the parabola ; in an unloaded state, the 
 catenary. (Weisbach, vol. II.) 
 
 In the following example the curve may be considered a parabola 
 or the bridge in its loaded state. 
 
 The thesis is 
 
 The vertical force at every point of the chain equals the weight 
 on the chain from the point in consideration unto the vertex. 
 Plate 4,1 So, when y in Fig. F = 25 feet and the length of 
 Fig. F.J bridge = 150 feet, 
 
 Width = 4 feet ; 
 
 load, 50 Ibs. per square foot, or 200 Ibs. per lineal foot ; 
 
 maximum load = 200 X 150 30000 Ibs. ; 
 
 the vertical force at D 200 X 25 = 5000 Ibs. 
 
 Further, the horizontal force at every point of the chain is equal, 
 and therefore equal to the horizontal strain in the vertex. 
 
 Thus, when by p is represented the weight pro unit of horizontal 
 projection = 200 Ibs., for our example, 
 
 pi 2 
 I. // = ( ' which is at the same time the horizontal force in 
 
 ~ f I 
 
 A and C, to overturn the towers, and amounting here to 
 200 X 75' 
 
 20 
 
 = 56250 Ibs. 
 
 II. 8 = j~ Y# + 4 A - ^~ X 77,6 = 58500 Ibs. 
 
 III. * = *-.*, 
 
 k 
 
 The length, L, of chain results by the formula, 
 L = 2 I + i X f = 151,75.* 
 
 o t 
 * For more specifications I ought to refer to Weisbach and other authors.
 
 28 THE THEORY OF STRAINS. 
 
 For a trussed system with two posts (queen-posts), between sup- 
 ports 36 feet, and an assumed load = 30000 Ibs., equally dis- 
 p, f. -j tributed, the distribution of forces <n tho bearings is 
 Fig 19J Denoted in Fig. 19, and for the calculation of strain, when 
 compared with Fig. 10, we fend for 
 
 r= w.f-, 
 
 bd 
 
 IQ 04 
 
 T= 10000 X - - = 23700 (approx.) ; 
 5,69 
 
 and for 1", = W. ; ^ 
 
 bd 
 
 r, = 10000 x- - = 2ur>7. 
 
 5,59 
 
 The compression on the vertical posts 10000 Ibs. ; the vortical 
 pressure on supports 15000 Ibs. ; find, reduced by lliose 5000 Ibs. 
 directly sustained (comp. Fig. 16), the reactive force of supports, 
 signified by D = 10000 Ibs. 
 
 Upon this structure, the Metfiod of Moments being applied (see 
 Preface), we suppose a section separated from the original by a 
 cut, */. 
 
 Considering the forces acting upon such a section, we form the 
 equation of equilibrium for a suitable point of rotation, by solution 
 of which we find the strain in the member in question ; and observe 
 the rule, that, for a strain, Y, the point of rotation ought to be chosen, 
 in the intersection of x and z, making their lever 0, when these are. 
 the members of the structure, separated by a cut, </, likewise a Y. 
 9ft , But when by st only one member, excepting Y, is separated, 
 we lay tho point of rotation on the next joint, as, per ex- 
 ample, for Fin b, Fig. 20. 
 
 F. 5,07 = 10000 X 12, 
 or, I., 0=-- -Y. 5,07 + 10000 X 12 (rotation round b) ; 
 
 r = Low>x_i2 = + 23700, ' '; 
 
 and because in the following the form of equation always will be 
 kept similar to I., the forces in their aim to turn to the left, like 
 Y round b, will be signified by , the same as a compressive 
 stniin ; and the forces to the right, like the hands on a watch, or D 
 round b, will be Dignified by -(-, the same as a tensile strain.
 
 SUSPENSION BRIDGE. 29 
 
 21.] According to this we have for Z, Fig. 21, 
 
 Q = Z. 5,59 + D . 12 (rot. r . d\ 
 MOOOXI2 
 
 5,59 
 
 and for F,, Fig. 22, 
 22.] = Fj.5,59 + Z>.12(rot. r.6); 
 
 Fi= 10000X12 
 
 5,59 
 
 When a diagonal, s, sustains the parallelogram, so that by a cut, 
 $1, three members, Z 1? Y l and s are separated, we have for the defi- 
 nition of s the point of rotation, as before mentioned, in the inter- 
 section of Z l and Y l ; but Z v and F, in their direction are parallel, 
 and therefore without intersection at all. In this case (the same as 
 for diagonals in girders with parallel top and bottom flanges) we 
 OQ -] suppose a point of rotation, 0, at any distance in the direc- 
 tion (axis) x, and find thus from Fig. 23, where x = oo , or in- 
 finite i.e., the lever of all forces, acting in a vertical direction 
 upon the section = cc . 
 
 = * . oo sin <p D . oc -f p . oo (rot. r . 0) ; 
 or, because oo is a factor of each part, 
 
 s . sin y 10000 + 10000. 
 
 In this equation, s . sin . <? (comp. Fig. 56 on semi-girders) is 
 9 , 1 the vertical component of a (Fig. 24), and acts right-angled 
 
 to the axis, x, like D and p. 
 The angle <f for our example 25 ; 
 
 sin <p = 0,422 (see table), 
 
 und therefore = s X 0,422 10000 + 10000, 
 or s = 0, 
 
 showing that, as in Fig. 19, the diagonal, s, is without any strain, 
 and only useful for preventing dislocation. (Compare parabolic 
 25 1 S"' c ^ ers ' to which this case is similar, because a parabola can 
 be constructed through the sustaining points a, d, e and c, and 
 therefore differs in this from Figs. 25 and (53.) 
 n/> -i For a reversed structure (Fig. 2(i) the strains will be the 
 same, but of reversed signs. 
 
 In the following calculations of rooTs and bridges it will be shown
 
 30 THE THEOEY OF STRAINS. 
 
 that the Method of Moments is thoroughly applicable, leading di- 
 rectly and in the most comprehensive manner to distinct results ; 
 but for a preliminary estimate of strain in the top and bottom 
 chords at the centre of the structure the most simple way to define 
 this strain may be stated by the following: 
 
 As the flanged girder in Fig. 27, charged with an equally-dis- 
 tributed load, O, will be exposed at its centre to the same 
 27 1 
 on'l exertion as the girder in Fig. 28, fixed at the centre, the 
 
 moments will be for both, when V= depth of girder and 
 I = length (the load being equally distributed). 
 
 Mom = -- - = V. x (rot. r . o) ; 
 
 ~ TC 
 
 ' 
 
 2 4 IQ.l 
 
 - = compression or tension in flanges ; 
 
 so Q = 30000 Ibs. ; V == 5,59'; and I = 36'. - 
 
 and - 24150 Ibs., the strain in the chords. 
 
 5,o9 
 
 It is to be observed that the result is rather too high wtien ap- 
 plied upon a truss with few panels, as in Fig. 26, on account of the 
 reactive pressure of the support, diminished by the partition of the 
 direct load on this place. 
 
 On account of its being a very convenient method we recommend 
 it; and in the following bridge skeletons we refer to it very fre- 
 quently. (See D, Example I., and Sect II., note.) 
 29.] ^ or a 8 ir<3er Charged with a single weight at the centre 
 (Fig. 29), we make a comparison with Fig. 30, and find for 
 both 
 
 .1 p i 
 30.] and -f = horizontal strain ; 
 
 t. e., compression or tension in the top or bottom chords, the for- 
 mation of moments for a point of rotation, o, being very compre- 
 hensive. 
 
 When P= 15000 Ibs.,
 
 e>..
 
 B. ROOF CONSTRUCTION. 31 
 
 but V = 5,59, and I = 36 (as before), 
 
 so the Mom = * f = 135000 > 
 
 15000 36 
 
 22 
 
 and - = 24150 Ibs., the strain in flanges ; 
 
 5,59 
 
 showing that, for the same exertion, a beam loaded with a single 
 weight, P, at the centre can bear only one-half of an equally- 
 distributed load, Q. 
 
 [Plates 3, 4 and 5 embracing Figs. 13 to 30.] 
 
 B. EOOF CONSTRUCTION. 
 
 For small and not complicated roofs, experience is a common and 
 in general, also, a sufficient guide. But experience is very limited, 
 and not every constructor has opportunity and time to acquire it. 
 
 The true and acceptable guide for a safe practice will always be 
 the calculation ; and since, especially for complicated and more 
 extensive combinations, the application of mechanical science luis 
 become unavoidable, the following compendium, leading from the 
 simplest to the most complicated structures, will give for almost 
 every purpose sufficient information. 
 
 By construction, when EG represents the weight at the centre 
 of gravity, G (Fig. 31"), the body will be at rest when 
 
 Fi^Sl* J the plane DF is ri g llt - {ln g led to tlie line DE 'i GE being 
 horizontal i. e., right-angled to the vertical line EG. 
 
 Let EG be an assumed length, then in the parallelogram of 
 forces the intensity of EK and EH is measured in proportion by 
 the same rule. 
 
 For an angle, 13 = 25 of a rafter with the horizontal line 
 op -i (Fig. 31 b similar to Fig. 16 reversed) leaning with the top 
 end against a wall, the heel at A being morticed ; we then 
 have 
 
 - = 0,422; 
 
 sin 2 13 = 0,17; 
 sin 2/3 = bin 50 =0,766;
 
 32 THE THEORY OF STRAINS. 
 
 ^ = cos t = -i|- = 0,906; 
 ac 13,24 
 
 cos 2 ;? = 0,82. 
 When W is an equally-distributed load of 20000 Ibs., then 
 
 *= ffl = ]r x d goooo x =21467 lbs . 
 
 2 A 2 o,o9 
 
 The vertical force, F, at the top of the rafter = 0, and the ver- 
 tical pressure F 2 of the heel = 20000 lbs., or equal to the entire 
 load. 
 
 Also the vertical force, FI at the centre = 20000 lbs. 
 
 The pressure in the rafter itself (compression) = 
 
 W 1 1 3 24 
 
 - X - = 10000 X - = 23685 (23700 lbs.). 
 2 h 5,59 
 
 The entire pressure, ^R, df rafter toward the support, 
 
 R = Vvf+H? = 29300 lbs. 
 
 Its direction can be constructed in making K 2h 11,18 feet, 
 HI and F 2 forming the sides of the parallelogram. 
 
 When P 10000 lbs., a single weight at centre of rafter, 
 
 TT TT * ** -I n7Q . 
 
 Jtl = Xli . _ lyjidfj 
 
 2 h 
 K=o, 
 
 and V, F, = 10000 lbs. 
 
 When in Fig. 31", by FM the weight of the body AB CD is rep- 
 o-jo -i resented, then FN, the force toward the wall, results in the 
 horizontal and vertical forces CH and (7F. 
 
 FL, the force acting perpendicular to the plane SI, in the direc- 
 tion BF. 
 
 G, the centre of gravity. 
 
 GF, vertical. 
 
 GFl CB, or < FCB = 90. 
 
 BI l FB, then BI is the direction of the plane required to 
 make the body at rest. 
 
 Also FN = P = *-Q CM * 
 
 H P sin oc = i G sin 2 .
 
 M=JVilZs*<Z^
 
 \ 
 
 B. ROOF CONSTRUCTION. 33 
 
 oid -, Let the sloping body ABCD (Fig. 31 d ) be supported by a 
 wall at its lower end, D, which coincides with the surface of 
 the body ; 
 
 Let G be again the centre of gravity ; 
 
 It is required to cut a notch out of the body at the upper end, 
 C, so that it may rest upon the top of a wall which is made to fit 
 the notch. 
 
 Make GE vertical ; 
 
 From D drzwDE l to CD ; 
 
 Join EC, and draw CF at right angles to it ; then the notch ufc 
 C being cut, the body ABCD will be at rest. 
 q-, e -i A body, ABCD, resting on supports (Fig. 31"), will only 
 
 DE EL 
 
 produce the vertical strains Q and Q at the sup- 
 
 ports. DL DL 
 
 Plate 7,1 For Figs. 32, 33 and 34 we have, again, 
 Fig. 32.J 
 
 be . ab , be ab 
 
 = sm p ; = cos p ; = tang p ; == cotg p 
 
 ac ac ab be 
 
 sin 2 p = sin 50 = 0,766; 
 
 and when upon each rafter W = 20000 Ibs., equally distributed, we 
 have for Fig. 32, 
 
 W rl W 
 H= H 1 = -j- = cotg /9 = 21467 Ibs. 
 
 2 A 
 
 W I 
 
 The compression in the rafter = -== 23685 Ibs., 
 
 2 fi 
 
 and, again, R = T/JJ^ _j_ y* = 29300 Ibs. 
 
 For an angle /9 = 26 34'; the horizontal thrust will 20000 
 Ibs. i. e., the same as the entire weight. (Comp. Fig. 16.) 
 go -i For Fig. 33 is, as the rafter in a vertical direction, sus- 
 tained on the top, 
 
 JBT= J3i = ^-sin 2-/J = 5000 X 0,766 X 3830; 
 
 4 
 
 V=^- cos 3 = 10000 X 0,82 8200 ; 
 
 Z 
 
 W 
 V,= Z- (1 + 8 in'/3) = 10000 (1 + 0,17) + 11700;
 
 34 THE THEORY OF STRAINS, 
 
 and the compression in the rafter, 
 
 sin = 10000 X 0,422 = 4220 Ibs. 
 2 
 
 For Fig. 34, when the rafter is sustained at the top by a vertical 
 post, 
 
 H = H, = sin 2 /3 = 3830 ; 
 4 
 
 34.] V = W cos 2 /3 == 20000 X 0,82 = 16400 ; 
 
 W 
 
 y^lL(lJ r s i n * ft 10000 (1 -f 0,17) = 11700 ; 
 
 z 
 and the compression in the rafter = 
 
 W 
 
 - sin /9 =.- 4220 Ibs. 
 
 In the cases in Figs. 33 aud 34 the post relieves the tie-rod or 
 (as here, in the absence of a tie) the wall from a part of the thrust 
 of the rafters, and compared with the truss in the preceding we 
 see that the king-post acts in a different way. 
 
 The different combinations of this single hanging-and-thrust 
 construction may be disregarded, and by the following calculations 
 the method of moments thoroughly applied. 
 
 A rafter being constructed in this way (Fig. 35), without con- 
 nection between the point e and the horizontal tie-rod aa, there are 
 two different systems, that of a trussed beam, aec, similar to Fig. 16, 
 or -| and that of a single triangle like Fig. 32,|and it is as there for 
 the same angle, /?, and a load = 20000 Ibs. upon each rafter ; 
 the horizontal thrust, H, at the top, as also the tension in the hori- 
 zontal tie-rod = 21467 Ibs. The trussed rafter is understood to be 
 calculated like Fig. 16 in combination with Fig. 31 b . 
 
 When the weight of structure, snow and wind-pressure upon a 
 gg -, roof (Fig. 36), for one square foot of horizontal projection, 
 in all = 50 Ibs., and the width between the supports or AC 
 40 feet, the distance of rafters = 10 feet, then 20000 Ibs. is the 
 entire load, or 10000 Ibs. upon each rafter, supported at A, B, C, 
 D and E. 
 
 The pressure of one-half of the weight, or 10000 Ibs., on the 
 supports is counteracted by the direct load of 2500 Ibs. It is there- 
 fore the reacting force, D, only 7500 Ibs. (See Howe Truss, 
 Sect. I., D.) 
 
 For a section separated by st we can define at once the strains in
 
 B. ROOF CONSTRUCTION. 35 
 
 z, y and x 1} so for x lf considering the forces acting upon this section 
 and the point of rotation in the intersection of y and z or F. 
 (Fig. 36 a .) 
 36".] = x, . a + D . 20 5000 X 10, 
 
 (Comp. Fig. 20, Equat. I.) 
 
 The arm or lever, a, can be measured near enough from a skele- 
 ton =: 9,6'. 
 
 Besides, for the calculation of a we have for the angle ABG 
 (Fig. 36), 
 
 AG= GB . tang < ABG, 
 
 t. e., <ABG = 53 7', and sin 53 T= 0,8 ; 
 
 a = BF sin < ABG = 12 X 0,8 = 9,6 ; 
 
 therefore = x, . 9,6 -f- 7500 X 20 5000 X 10 ; 
 
 100000 
 x l = -- - = 10412 11)3. 
 
 9,6 
 For z (rot. r . E, Fig. 36 a ), 
 
 = z . 6 + 7500 X 10, 
 * = + 7 -f = + 12500; 
 
 and for y (rot. r . A, Fig. 36 a ), 
 
 Q = y .c+ 5000.10; 
 Q y. 10,75 + 50000, 
 50000 
 
 y = -7o^ = - 4650 - 
 
 Plate 8, 1 For the strain in x (rot. r . F, Fig. 36 b ) is, 
 Fig. 36 b .J = x 9)6 + 750Q x 20> 
 
 or x = - - = 15625, 
 
 J,o 
 
 In regard to the vertical V, we use for its definition the strain of 
 the joining brace, ^ 10412, and make st a curved line ; then 
 we have, for a rot. r . D (Fig. 36 b ), 
 
 = V. 10 ( 10412) . 10,9, 
 or = F. 10 + 113490;
 
 36 THE THEORY OF STRAINS. 
 
 _ 113490 
 t.., V= -\ -- = + 11349. 
 
 37.] The results are combined in Fig. 37. 
 
 For Fig. 38, the entire load (equally distributed) again being 
 20000 Ibs., the depth 15', and between the supports 40'. 
 When here the cut st separates the line %$&, we have for #j 
 (rot. in the intersection of t/jZi, or F, Fig. 39), 
 
 = Xl . 9,1 5000 X 5* + 7500 X 15* ; 
 
 39.] Il= _ 9752 . 
 
 y,i 
 
 For yi (rot. r . A\ = y l . 15 -f 5000 X 10, 
 
 y ,= + 522p = +3,333; 
 
 and for z, (rot. r . C), 
 
 = -^. 15 5000 X 10-^7500 X 20; 
 
 2 , = -f- 6,666. 
 
 40.] For a section, st, through x and z, we have for x (Fig. 40), 
 = x. 9,1 + 7500 X 151 (rot. r.F}; 
 116250 
 
 TOT " 12774; 
 
 and for z (rot. r . D\ 
 
 = 2 . 7,4 -f 7500 X 10 ; 
 
 8 
 
 For y (rot. r . A) is, = y . 12,5 + 5000 X 10 ; 
 
 , = -^ = -4000. ' 
 
 41.] The results combined in Fig. 41. 
 
 42.] When the figure before is changed in the depth, like Fig. 
 
 42, we have the following equation : 
 = Xl . 5,8 5000 X 3,25 + 7500 X 13,25 (rot. r . F, Fig. 43) 
 
 43.] For tlic definition of y\, the intersection of a* and \ will be 
 in G, and it is for G as rotation, 
 
 = y, . 8,25 4- 5000 X 6 -f 7500 X 4 ; (Fig. 43.)
 
 -l^rT'i
 
 B. ROOF CONSTRUCTION. 37 
 
 H yi =+?2 00 5 
 
 8,25 
 
 For Zi (rot. r . (7) we have 
 
 = z, . 12 5000 X 10 + 7500 X 20; 
 
 ooooo 
 
 J. ^J 
 
 44.] For a section, s, through # and z (Fig. 44), it i 
 = x . 5,8 + 7500 X 13,25 (rot. r . F) ; 
 993750 
 
 5,8 
 and = z . 5,2 + 7500 X 10 (rot. r . D) ; 
 
 75000 
 
 ~5j~ ' ; 
 
 and for y, = y . 12,5 + 5000 X 10 (rot. r.A); 
 
 50000 
 
 y = 4000. 
 
 12,5 
 
 45.] The results combined in Fig. 45. 
 
 Plate 9,1 For the definition of X in Fig. 46, the point of rotation 
 
 Fig. 46. J i n E, or in the intersection of Y and Z, will be from 
 
 Fig 47. 
 
 47.] = X. x P.CE+ D.AE; 
 
 P.CED.AE 
 
 or X= 
 
 x 
 
 For Y we choose A, or the intersection of X and Z, as the point 
 of rotation, and the equation will be 
 
 = Y. y + P. AC + Q . AE, 
 or Y= 
 
 y 
 
 and in the same way for Z, rot. r . H. 
 
 z -^ 
 
 It will tiot be necessary to show, by repetition of the foregoing, 
 the equations i'ur the other parts of the structure.
 
 38 THE THEORY OF STRAINS. 
 
 4S -, In more complicated systems (Fig. 48), it may happen that 
 by a cut, st (which can be made curved as \vell as straight), 
 different braces or rods are spared, like FG, D G and DE. 
 
 In this case it is possible to come to a direct result when st 
 only can be laid so that all the braces or rods cut by st meet at 
 one point, except that one whose strain is in question. 
 49.] So for the strain V in FG (rot. r . H, Fig. 49), 
 
 E.r; 
 
 V- 
 
 FH 
 
 50.] In the same manner the strain U in DG (rot. r . H, Fig. 50), 
 Q= U.u E.r; 
 
 u 
 
 thus we find also the strain in KT and LT. . 
 r-i -i Being by the foregoing in possession of a value for U in 
 1 DG, we find for the strain X in DF, Y in DE, and Z in CE 
 the following equations from Fig. 51 : 
 
 = X.DE + U.v Q. NO P. MO + W. AO (rot. r . E} ; 
 = Y. AD + U. I + Q . AN+ P. AM (rot. r . A) ; 
 = Z . z + W. AN P. MN (rot. r . D) ; 
 
 each 6ne enabling us to obtain a direct result for the strain in 
 
 question. 
 
 Plate 10,1 For a roof (Fig. 52), the weight of which, 11,3 Ibs. 
 
 Fig. 52. J p er square foot of its horizontal plan, may be calcu- 
 
 lated 20 Ibs. for wind pressure and snow, making together 31,3 Ibs. 
 
 per square foot. 
 
 The distance of rafters being 15J feet, the width, 100 feet, makes 
 for each rafter 15 J X 100 X 31,3 = 48000 Ibs. (approx.). 
 
 The load at each apex, therefore, will be = 6000 Ibs., the 
 
 8 
 
 distribution of which is shown by the skeleton. 
 For the reactive force on the supports is again 
 
 D = 24000 3000, or D = 21000 Ibs. 
 
 There are, in all, seven times 6000 Ibs. acting downward, and 
 twice 21000 Ibs. acting vertically upward upon the system.
 
 B. ROOF CONSTRUCTION. 39 
 
 go -j The section, A, s, t (Fig. 53), kept in equilibrium by the 
 replaced forces, x, y and z, may be regarded first as a lever 
 with the fulcrum at D ; then the strain in x for the middle sec- 
 tion is 
 0= x . 18,6 + 21000 X 50 6000 X 12,5 6000 X 25 6000 X 
 
 37,5, 
 or x = 32300 Ibs. ; 
 
 and in y, when A is the point of rotation, 
 
 = y . 38,4 + 6000 X 12,5 + 6000 X 25 + 6000 X 37,5 (rot. A) ; 
 
 y = 1 1700 Ibs., 
 and 
 
 0^ 2.15+21000x37,5 6000X12,5 6000X25 (rot. r.E.), 
 
 Z = + 37500 Ibs. 
 54.] For V in Fig. 54 the rotation also round A is 
 
 = V. 37,5 + 6000 X 12,5 a 6000 X 25 ; 
 
 V= + 6000 Ibs. 
 For the other members in Fig. 52, 
 
 == a?, . 13,9 + 21000 X 37,5 6000 X 12,5 6000 X 25 
 (rot. r.F*); 
 
 x l = 40400 ; 
 
 = y l . 23,5 + 6000 X 12,5 + 6000 X 25 (rot. r.A); 
 
 yi = 9570; 
 = z, . 10 + 21000 X 25 6000 X 12,5 (rot. r.G); 
 
 Zl = + 45000 ; 
 = Fi . 25 + 6000 X 12,5 (rot. r.A); 
 
 F, = + 3000 ; 
 = x 2 . 9,3 + 21000 X 25 6000 X 12,5 (rot. r . H ) ; 
 
 x 2 = 48400 ; 
 = y, . 9,3 -f 6000 X 12,5 (rot. r.A)', 
 
 y* -=-8100; 
 
 = z, . 5 + 21000 X 12,5 (rot. r . /) ; 
 z, = + 52500. 
 
 For the strain in x 3 we choose a convenient point for rotation in 
 the line z, per Example D, Fig. 55.
 
 40 THE THEORY OF STRAINS. 
 
 55.] The equation in this case will be 
 
 Q = x 3 .18,Q + 21000 X 50; 
 
 jT 8 56500. 
 
 The only strain not directly deducible is U in the vertical line 
 CD at the centre. 
 As in Fig. 36, we use the strain of the joining brace, 
 
 x = 32300 Ibs. 
 
 56.] For B as the point of rotation (Fig. 56), our equation is 
 = U. 50 6000 X 50 ( 32300) . 37,2 ; 
 
 U= 18000 Ibs. 
 67.] The results combined in Fig. 57. 
 
 The weight and load of a roof (Fig. 58) being estimated, in- 
 
 f-ft -. k eluding wind-pressure and snow, to 50 Ibs. per square foot 
 
 of its horizontal plan, the distance of rafters being 12 feet, 
 
 and the space between the walls 99 feet, which gives 50 . 12 . 99 
 
 59400 Ibs. for one rafter, or, in round figures, 60000 Ibs. 
 
 The calculation of the top structure can be made as in the pre- 
 ceding example (Fig. 36). 
 
 In the main construction are six supporting points, charged as 
 P-o-i in Fig. 58. The top structure transmits one-third of the 
 
 entire load, or on each post 10000 Ibs. to the apexes, ff. 
 Each wall has to bear 30000 Ibs. ; and after subtraction of the 
 direct load the reactive force is 26666 Ibs., or, by calculation, 
 
 _ 6666 (11 + 22) 13333 (33 + 66) 6666 (77 + 88) t 
 99 99 99 ; 
 
 D = 26666 Ibs. 
 59.] For the strain x 3 we have in Fig. 59, 
 
 = *, . 21 13333 X 161 6666 (27* + 381) -f- D. 491 
 
 (rot. r . /i) ; 
 or, also, 
 
 = x, . 21 6666 (11 + 22) + 26666 X 33 (rot. r ./) ; 
 659967 
 
 Further, 
 
 = Z t . 21 13333 X 161 6666 (271 + 381) + D . 491 
 (rot. r . g).
 
 %.jr */ I %* .ir| 
 
 .^-^Q^ V;/':
 
 B. ROOF CONSTRUCTION, 
 
 659967 
 
 and 0=y 3 . 39,5 -f 13333 X 33 -f 6666 (11 + 22) (rot. r . o) ; 
 y ? = _ 16708. 
 
 The tie-rod, gh, transmits the strain to the top flange, and is herg 
 sustained by the counter-brace, eh. 
 60.] From Fig. 60 is 
 
 == x 3 . 14 6666 X 11 + D . 22 (rot. r , ef) ; 
 
 513304 
 x t = - = ooboo ; 
 
 = z 3 . 171 6666 (11 + 22) -f 26666 X 33 (rot. r . ) j 
 
 II s ;:"; :; : 
 
 = 2/ 2 . 26,7 + 6666 X 11 + 6666 X 22 (rot. r . o) ; 
 
 7/ 2 = 8223. 
 61.] Fig, 61 gives the equations, 
 
 = Xl , 7 + 26666 X 11 (rot r , 6) ; 
 
 ^ = + 41902 ; 
 
 = 2 2 ,13 6666 X 11 + 26666 X 22 (rot. r.c); 
 2 2 = 39485 ; 
 
 = y, , 13 + 6666 X 11 C rot - f ^ a ) 5 
 j/,^ 5640; 
 
 nd for i we find from the same figure, 
 
 = .! . 13 + 26666 X 22 ; 
 
 z l = 45125. 
 62.] For the strain in tie-rods we find from Fig, 62. 
 
 = F, , 33 + 6666 . (11 + 22) (rot. r , a) ; 
 
 V 3 = + 6666 ; (Comp. Fig. 68.) 
 
 = V. . 22 + 6666 X 11 (rot. r . a) ; 
 
 F 2 = + 3333 ; 
 Q = Fi.11 + (rot. r.a); 
 
 Fi = (and is therefore not egsential)^
 
 42 THE THEORY OF STRAINS. 
 
 ^ 
 
 go -i The strain in V t at the centre rod, according to 8 b , can be 
 
 defined thus : 
 
 01 
 V< = 2 X -=?- X 16708 =.26200 Ibs. 
 
 The results are combined in Fig. 63. 
 
 When in Fig. 64 the rafters are trussed i. e., stiffened by a 
 A -i king-post at b there will be only four supporting points in 
 the main construction, because the load in this case is trans- 
 ferred to the wall. 
 
 9999X22 13333(33 + 66). 9999X77 9qqq9 
 ~99~ ~~99~ ~99~~ 
 
 Plate 12,1 Further in Fig. 65, 
 Fig. 65.J 
 
 == x z . 21 13333 X 16J 9999 X 27 + 23332 X 49* 
 (rot. r . Ji) ; 
 
 x, 31427 ; 
 
 = z 3 . 21 13333 X 16J 9999 X 27* + 23332 X 49* 
 (rot. r . g) ; 
 
 s 3 31427 ; 
 
 = y, . 39,5 -f 9999 X 22 + 13333 X 33 (rot. r . a) ; 
 
 y , = 16708. 
 66.] And from Fig. 66, 
 
 = *! . 14 + D. 22 = s, . 14 + 23332 X 22 (rot. r . d) ; 
 
 X! = + 36665 ; 
 = 2 . 171 + 23332 X 33 9999 X 11 (rot. r.e}-, 
 
 z t = - 37180; 
 = y, . 26,75 + 9999 X 22 (rot, r . a) ; 
 
 y,= -8223. 
 67.] For Z we have from Fig. 67, 
 
 = ^ . 13 + 23332 X 22 (rot. r.e); 
 
 Z,= -39485. 
 68.] See the results in Fig. 68 combined. 
 
 The strain F 4 = 26200 Ibs. can be defined independently of the 
 Method of Moments by the parallelogram of forces, as in Fig. 63, 
 already shown, 
 
 F 4 =2( 16708). cos oc;
 
 x., 
 
 >' \ ^SV 9
 
 SEMI-GIRDERS LOADED AT THE EXTREMITY. 43 
 
 and when by means of counter-braces, e, h, the top chord is re- 
 lieved from the strain, so that one-half to each side is transported 
 to the tie-rods, e,f, then here the strain will increase to 13000 -f- 
 6666 = 19666 Ibs. 
 
 In a combination of rafters (Figs. 69, 70), the pressure of 
 Y^' I the end rafters upon the wall results in an outward horizontal 
 
 and vertical force. 
 
 Different from this is the action of the intermediate rafters, 
 being similar to an oblique bridge-truss, sustained at the top chord. 
 The horizontal force at the heels of the intermediate rafters is 
 opposed to the horizontal force of the end rafters. 
 
 [Plates 6, 7, 8, 9, 10, 11 and 12 embracing Figs. 31 to 70.] 
 
 C. SEMI-GIRDERS. 
 
 I. SEMI-GIRDERS LOADED AT THE EXTREMITY. 
 
 pi , i o -| As the most simple presentation for a weight, W, the 
 Fijr. 7lJ stress in struts and tie- rods is inscribed in Figs. 71 to 
 
 74, and the parallelogram of forces connected. 
 73.] To compute in Fig. 73 the stress- in the lower flange, we 
 have Z 
 
 de 2 
 
 = sec QC, and = sm oc, 
 
 df W. sec cc 
 
 & 
 
 or - = W. sec oc. sin cc ; 
 
 Z= W. sec a. since W. sec cc sin QC, 
 or Z = 2 W ' . sec cc . sin QC ; 
 
 , . tang oc 
 
 and since sin cc a , 
 
 sec a 
 
 Z^ -2TF.seca^^ =-2W. tang oc 
 sec oc 
 
 (much easier determined in Fig. 77 by the Method of Moments). 
 
 From 'Fig. 73 and the following we see that, for a load at the 
 extremity, the diagonals are strained equally and alternately with 
 tensile (-J-) and compressive ( ) strains. (Comp. Fig. 23.)
 
 44 
 
 THEORY OF STRAINS; 
 
 But the strain in the flanges increases toward the support ift 
 
 each, 
 
 2 W. tang : , 
 
 where : is the angle of diagonals with a vertical line* 
 75.] For a better presentation of this, see Fig. 7o, and for the 
 calculation apply the Method of Moments. 
 
 g -i When by a cut, st, a section of the structure is separated; 
 77* J we nave f r the flanges as equation of equilibrium (Figst 
 76 and 77), 
 
 = - Xi . cb -f- W. ca (rot. r . i), 
 or for cb A v and ac = I ; 
 
 x= + W. l - = + l 
 h 
 
 78] and by Figs. 76 and 79 : 
 79] 
 
 = x 2 . h + W. 3 f (rot. r . e) ; 
 a- 2 = -f 3TF- - = 3TT. tang oc. 
 
 In the same tnanner is 
 ,0 = a?, . h + W. 5 1 (rot. r. $0 ; 
 , i 
 
 = + 2, . h + IT". 2 / (rot. r . d) 
 
 = 
 
 = 
 
 4-F;<n(rot.r./)! 
 
 ^_ 
 
 A~ 
 
 PI t 14 1 ^ wrou gh t '' ron crane (Fig. A), constructed of braces 
 Fig A. J wittl ^ n ^-J omts ma y be loaded at the extremity with 
 30000 Ibs. = P; so is (for the dimensions noted in the 
 skeleton) the horizontal strain, 8, in a and 6. 
 
 0= --5. 6 + P. 12 (rot. r.d); 
 
 fif = 60000 Ibs ; 
 
 B.] and for the other members we have 
 = *i 0,75 -f (rot. in the intersection of ft and 4* or g, Fig. B) | 
 
 o = ft.i,l P. 2,1 (rot. r.O; 
 
 , "0000X2,1 
 ^ ~ ' T; 7272 5
 
 If 
 
 u
 
 y^fcjhs*. ^ 
 
 ^\ *^s^ 
 
 if 3t% |t 
 
 fi^. 
 
 k^T t ,^/t. 
 
 -J "1 ,-' ./ 
 
 -;/_ J ^, --:-- ;."'"' 
 
 % ;>k\ 
 
 '^. *v^* 
 
 TV 
 
 / 
 
 > 
 
 ?: 
 
 X ^
 
 LOADED AT EACH Affix. 45 
 
 = ^. 1,8 + P. 4 (rot./); 
 30000.4 
 
 1,8 
 = z, . 1,99 + P. 4 (rot. r . e, Fig. C) ; 
 
 120000 
 2, = - + oOdOi ; 
 
 1,99 
 = 2/ 2 . 4,4 P% 1,2 (rot. r . /O J 
 
 >.8 (rotr.d); 
 
 :r 2 = 80000 ; 
 ft.] = e s . 3,2 + P. 8 (rot. r . c, Fig. D) { 
 
 2 3 = 75000 ; 
 
 = fc, . 4,2 + P. 12 (rot. r. 6) J 
 a- 3 =r 85700. ' 
 
 For $, the intersection of rc 3 and z 3 is to the left of the sus- 
 pended weight, and the symbql reversed. 
 
 = y, . 7,25 + P. 2,6 (rot. r . I) } 
 y, = 10758, 
 
 which would be = when the intersection is in the vertical line of 
 the suspended weight, as the lines oe and pc in Fig, A indicate. 
 For the verticals, V, we have from Fig. D, 
 
 = Fi -. 10,1 P. 6,1 (rot. r . m) ; 
 
 F, = 18118 ; 
 = F 2 . 26,5 P. 18,5 (rot. r.n)\ 
 
 V, = 20943; 
 E.] The results combined in Fig. E.* 
 
 II. SEMI-GIRDERS LOADED AT EACH APEX. 
 
 In Fig. 25 is occasionally explained how to compute the stress irt 
 diagonals, as there is no intersection of joining flanges, x and z, and 
 fcs in the case here considered the diagonals receive at each loaded 
 
 * For most purpose? the above will be sufficient. In Giynn's rudimentary 
 'treatise on the Construction of Crnncs we find valuable and complete drawings
 
 46 THE THEORY OF STRAINS. 
 
 PI 1^1 a P ex an increment of strain, prior to the calculation 
 pj gQj may be given the general thesis that the strain in two 
 diagonals whose intersection is at an unloaded point is 
 the same in numerical, value, but of opposite character. (Fig. 80.) 
 
 (See IV. General Remarks.) 
 
 The strain in diagonals, meeting at a loaded point, is in numeri- 
 cal value different. 
 
 The strain in flanges increases from apex to apex in geometrical 
 progression. 
 
 * -. In Fig. 81 suppose the angle <p of diagonals with a hori- 
 zontal line = 45, so, also, angle oc = 45; and by the table, 
 sec oc = 1,414. 
 
 When, again, in the axis x = GO a point of rotation, o, is sup- 
 posed, we have per example for diagonal, i/ 5 . 
 
 / W\ 
 
 = -f ?/ 5 . x sin <p -f I W+ W + - . x (rot. r . o), 
 
 where y- 3 . sin <p is the vertical component of y 5 , or a.b ?.& in 
 the parallelogram of forces (Fig. 81), presenting by y 5 the resulting 
 strain or diagonal. Divided by x, it follows : 
 
 \ 
 and as <p = 45, and sin 45 = 0,707, 
 
 = + y 6 . 0,707 + I-.TF, 
 or y, = 3,535 W. 
 
 The same results from f W. sec oc , or f . W . 1,414, which 13 
 also = 3,535 W. 
 
 82.] * n ^ ie sarue manner * n Fig. 82 for y 2 , the direction of 
 which, when separated by a cut, st, is reversed to the weight. 
 (See Figs. 22 and 23.) 
 
 W 
 
 - y, . sin 45 + (rot. o) ; 
 
 0,707 ' 
 and for the other diagonals, 
 
 0^ + 2/3.0,707+ TF + , 
 
 ib---, 
 
 0,707 ^ r 0.707 '
 
 ' * 
 
 * I
 
 SEMI-GIRDERS LOADED AT EACH APEX. 47 
 
 TT fir 
 
 
 f-TF 
 or yr = r 
 
 0,707 0,707 
 
 83.] For the strain in flanges we have from Fig. 83, 
 
 W 
 0== Xi . I . cot. QC -\ I (rot. r . 1) ; 
 
 %W.l 1 
 
 KI = * or as = tang oc , 
 
 I. cotg QC cotg oc 
 
 x 1 = W. tang oc ; 
 
 and when W= 10 tons, < y = 60; therefore < oc = 30, 
 and tang 30 = 0,577 (" Example Stoney") ; 
 
 Xl = 5 . X 0,577 = + 2,9 tons ; 
 or when, for an easier understanding, in Fig. 83, 
 I . cotg oc = h, 
 
 84.] we have for x 2 in Fig. 84, 
 
 W 
 
 Q = x . 1 .h+W.l+ -Zl (rot. r. 3) ; 
 
 JO 
 
 IW.l 
 
 n- A . 
 
 k 
 
 and as = tang oc = 0,577, 
 
 h 
 for our example, 
 
 x, = f . W. 0,577 = | X 10 X 0,577 = 14,5 tons. 
 85.] So for x s in Fig. 85, 
 
 = a,, h + W. I + TF. 3 I + 5 1 (rot. r . 5) ; 
 
 Zt 
 
 x. = LS. |p. o,577 = 37,3 tons, 
 and so further. 
 
 86.] For the strain, 2, in the lower flanges, 
 
 W 
 
 = + 2l . h + 2 / (rot. r . 2 in Fig. 86) ; 
 <u 
 
 i = Tr4 = 10 X 0,577 = 5,7 tons. 
 h
 
 48 THE THEORY OP STRAINS 
 
 87.] Fig. 87 gives 
 
 r.4); 
 
 Jh== _4jp. == _ 4 X 10 X 0,577 = 23,2 tons, 
 h 
 
 In the same way for z q , 
 
 2 3 9 W- - = 52 tons. 
 /i. 
 
 In case the load should be connected to the lower apexes 
 (Fig. 88), the equation of equilibrium, per example for s 3l 
 
 would be 
 
 W 
 
 -51 (rot, r .m) ; 
 
 Z3 =- if. W- r = 37,3 tons ; 
 
 /I/ 
 
 L e., the strain is the same as in the flange of the reversed figure, 
 but of opposite character. So also is the strain the same for the 
 other flanges, (See Fig. 88.) 
 
 Remark. In the given example the strains are determined with- 
 out a certain length for h or L This is easily explained by the 
 relation which the angle <p or <x. bears to h and I, as by the exten- 
 sion of one, the other will increase in the same ratio. 
 
 [Plates 13, 14 and 15 embracing Figs. 71 to 88.] 
 
 D. GIRDERS WITH PARALLEL TOP AND BOTTOM 
 FLANGES. 
 
 ( Calculated for a Permanent Load.) 
 I. STRAIN IN DIAGONALS AND VERTICALS. 
 
 The calculation is very similar to the preceding. Provided, 
 again, the load to be connected to the upper or lower apexes for 
 the application of the Method of Moments, we now consider as a 
 special force the reaction of the supports toward the system. 
 
 D, may represent one-half the weight of loaded truss or the
 
 STRAIN IN DIAGONALS ASfD VERTICALS. 49 
 
 pressure upon each support (prop), and diminished by the par- 
 tition of load on this place directly sustained. (Comp. Fig. 21.) 
 The reactive force of support wanted for our calculation will be 
 signified by D. 
 
 To compute D, we refer to Fig. 4 in Sect. I., and define first for 
 the following example its numerical value : 
 
 Plate 16,1 Through bridge (over-grade bridge), between supports, 
 Fig. 89.J 48 feet; 
 
 8 panels, each 6 feet = I ; 
 
 depth of truss = 6 feet = h, from centre to centre of top and 
 
 bottom chords ; 
 
 the weight of structure = 3000 Ibs., and the load = 16000 Ibs. ; 
 gives a permanent load = 18000 Ibs. per panel ; 
 rolling load = 0. (See Sect. II.) 
 
 For the distribution of load, see Fig. 89. 
 
 Remark. The strange impression which the arrangement of 
 diagonals, unsymmetrical toward the centre, may first produce, will 
 soon disappear after observation of the advantages for transforma- 
 tion upon succeeding systems. 
 
 Whole pressure of truss upon supports = 8 X 18000 = 144000 Ibs. ; 
 A = 18000(i + f + f + t + * + i + l) + 9000XI-72000 Ibs.; 
 D = 18000 (^+1 + 1 + 1+1 + 1 + 1) = 63000 Ibs. 
 
 The strain in the post, V , is 0, because x^ = 0, 
 and F 8 = 63000.* 
 
 q^ -, Excepting the vertical component of y t (i. e., yi sin $0), for 
 a section (Fig. 90), only D = 63000 Ibs. is a second vertical 
 force. 
 
 Both turn to the left around o in the axis x ; therefore, their 
 symbol, . (Comp. Fig. 22.) 
 
 * For a deck-bridge (under-grade bridge) i. e., when the upper apexes ar 
 loaded would be 
 
 V = 9000, and F 8 = 72000. 
 
 The angle of diagonals with a horizontal line will be 45, 
 5 D