H-l" 
 
 # 
 
 LIBRARY 
 
 OF THK 
 
 UNIVERSITY OF CALIFORNIA. 
 
 Received 
 Accessions No:s$& ^ ^\f Shelf No. 
 
SIMPLE PRACTICAL METHODS 
 
 OF CALCULATING 
 
 STEAINS ON GIRDEES, AECHES, 
 AND TEUSSES; 
 
 WITH A SUPPLEMENTARY ESSAT ON 
 ECONOMY IN SUSPENSION BRIDGES. 
 
 BY 
 
 E. W. YOUNG, 
 
 \ 
 
 ASSOCIATE OP KINO'S COLLEGE, LONDON, 
 AND MEMBER OP THE INSTITUTION OP CIVIL ENGINEERS. 
 
 UNIVERSITY 1 
 
 ILonfcon : 
 MACMILLAN AND CO. 
 
 1873 
 
 [All Rights reserved.] 
 
\J 
 
 'V 
 
 PBIMTED BY C. J. CLAY. M.A. 
 AT TUB UNIVEESITY PEESa 
 
PREFACE. 
 
 THE bulk of the following work was written several years ago, 
 and was designed to supply a want which was then much felt, 
 that of a treatise giving simple methods of calculating strains 
 on girders, trusses, arches, roofs, and other kindred structures. 
 The treatises existing at that time were either too meagre or 
 too mathematical for the average engineering student. One class 
 gave him results without explaining to him how they were ob- 
 tained ; the other, in making the required explanation, bewildered 
 him with mathematical formulae beyond his comprehension. 
 
 Several excellent works on this subject having appeared 
 during the last few years, the issue of another from the press 
 might be thought superfluous. I have judged otherwise, because 
 the methods adopted in this little work are mostly original, 
 and, as I consider, simpler than usual. A knowledge of simple 
 equations, excepting on the subject of deflection, where a little 
 elementary trigonometry is required, is the extent of the 
 mathematical attainment required for working out the problems 
 presented in this work to the reader. 
 
 The article on " Economy in Suspension Bridges" from " Engi- 
 neering" is added as a supplement to the chapter on Suspen- 
 sion Bridges. 
 
 THE AUTHOR. 
 
CONTENTS. 
 
 PAOB 
 
 PREFACE iii 
 
 CHAPTER I. 
 
 OF LEVERS AND THE RESOLUTION OF FORCES. 
 
 The Lever. Fulcrum. Arm or leverage. Moment. Levers of three 
 kinds. Bent Lever. Examples. Parallelogram of Forces. Resul- 
 tant. Triangle of Forces. Corollaries. Polygon of Forces. 
 Examples of the Polygon of Forces. Resolution or Decomposition 
 of Forces. Example. Alternative methods, the Lever or Parallel- 
 ogram of Forces. Lever method. Method of Parallelogram of 
 Forces. Forces acting in different planes 
 
 CHAPTER II. 
 GIRDERS WITH PARALLEL FLANGES. 
 
 Load at centre of span. Load evenly distributed, strain at centre. Strain 
 at any point of the Flange. Example. Irregularly loaded Girders. 
 Strain on the web. Shearing force. Cantilevers. Load at the ex- 
 tremity. Distributed load. Wooden beams. Neutral axis. The 
 web in Iron Girders. Examples. Strains on the web. Result 
 checked by method described, p. 14. Example. Formula for 
 Shearing Force. Irregular loading, First case, Bays equal. Second 
 case, Bays of an unequal size, Load symmetrical. Third case, Bays 
 and Loading irregular. Trusses and Lattice Girders. Without 
 verticals. Warren Truss. Strains on Diagonals . . . .12 
 
Tl CONTENTS. 
 
 .CHAPTER III. 
 
 HOGBACKED GlBDERS. 
 
 PAGF. 
 
 Strains on the Flanges. Strains on the Diagonals. Comparison with a 
 parallel Flanged Girder. Diagonals as struts. Diagonals of both 
 kinds used . . . . . V . . . . . . 41 
 
 CHAPTER IV. 
 
 THE BOWSTRING GIRDER. 
 
 Strains on the Flanges. Curve of Equilibrium. Method of drawing the 
 Curve of Equilibrium. Regular Loading. To draw the Curve of 
 Equilibrium with a given versine. Irregular Loading, Bays equal. 
 Irregular Loading, Bays unequal. Line of greatest Resistance. 
 Methods of resisting Distortion. First Method. Second Method. 
 Third Method. Fourth Method. Fifth Method. Effect of elasticity 
 in distributing the Load among the Diagonals. Load resting on 
 the Top . . , , . .. . , :,. r . . 45 
 
 CHAPTER V. 
 THE ARCH. 
 
 Method of finding the Thrust at Crown. Curve of Equilibrium for an 
 Arch of Masonry. Other methods of resisting Distortion of the 
 Arch. Strains on the Spandrils. Other forms of Spandril filling. 
 Arched bridges of multiple span. Stability of Pier. Effect of 
 Horizontal Girder. Transverse strain on Pier. Stability of a 
 Wrought-iron Arched Viaduct .61 
 
 CHAPTER VI. 
 
 SUSPENSION BRIDGES. 
 
 Strain at centre of Span. Strain at any portion of the Chain. Strain 
 
 on the Back-tie. Methods of stiffening ..... 72 
 
CONTENTS. Vll 
 
 CHAPTER VII. 
 
 DEFLECTION. 
 
 PAGE 
 
 Definition. Laws of Deflection. Radius of Curvature. To find the 
 Deflection. Example. Irregularly strained Girder. Deflection of 
 an Arched Rib. Example. Deflection of a Bowstring Girder. 
 Deflection of a Suspension Bridge. Strength of a Girder, how far 
 deducible from its deflection under load , . . . . .76 
 
 CHAPTER VIII. 
 CONTINUOUS GIRDERS. 
 
 Definition. Continuous Girder of two spans. Both spans loaded. One 
 span only loaded. Method of finding position of Point of Contrary 
 Flexure. Example. Two unequal spans. Both spans loaded. 
 Longer span loaded. Extremely disproportioned spans. Shorter 
 span loaded. Anchoring down the shorter Girder. Multiple spans. 
 Three spans equally loaded. Spans unequally loaded. Position of 
 Point of Contrary Flexure in three-span Girder. Girders of more 
 than three spans. Continuous Girders of varying depth. Practical 
 remarks 86 
 
 CHAPTER IX. 
 
 ROOFING TRUSSES. 
 
 Elementary form of Principal. Method of finding the strains. More 
 complicated forms of Principal. Arched Roofs. First System. 
 Second System. Example, Diagonals, both Ties. Third System. 
 Equally loaded. Unequally loaded. Force of Wind. Pressure of 
 Wind on Inclined Planes. Methods of finding strains produced by 
 force of Wind 109 
 
Vlll CONTENTS. 
 
 CHAPTER X. 
 
 ECONOMY IN SUSPENSION BRIDGES. 
 
 PAGE 
 
 The question hitherto not sufficiently investigated. Subject considered 
 under two heads. First head : The most economical height for the 
 towers. For load at centre of span. For distributed load. Method 
 of estimating for towers of different kinds of material. Effect of 
 weight of chains not taken into account. Second head : The most 
 economical arrangements of parts in the superstructure. Suspension 
 and cantilever systems compared. For a dead load. For a live 
 load. Proper position for the point of junction of Cantilever and 
 Suspension Systems. Practical part of the question. Bending of 
 platform caused by rolling load. Caused by changes of tempera- 
 ture. Arrangements thereby necessitated. Great economy of the 
 Compound System . . . . . ..... .119 
 
 ERRATA. 
 
 Page 33, line 3, for F have two read F has two. 
 
 Page 93, line 8, for loading strains read loading shewn. 
 
 Page 119, line 15, for proportion of depth of space read proportion of depth to space. 
 
STRAINS ON GIRDERS, ARCHES, 
 AND TRUSSES. 
 
 CHAPTER I. 
 
 OF LEVERS AND THE RESOLUTION OF FORCES. 
 
 The Lever. By " a lever" is commonly understood a bar of 
 wood or iron such as a handspike or crowbar employed for moving 
 great weights. As we shall have under our consideration in the 
 course of this work structures consisting of a combination of many 
 levers, the whole of the structure itself being sometimes considered 
 as one lever, it will be advisable to commence with a few remarks 
 on levers and leverage, a correct knowledge of the principles of 
 which is indispensable for the determination of the strains to 
 which such complex structures are subjected by the pressures 
 which they have to bear. 
 
 Fulcrum. The bar abc (fig. 1) 
 is a representation of a simple 
 lever resting on a support at b 
 called the fulcrum, or fixed point 
 about which the lever would turn 
 if the forces at each end did not 
 balance each other. A weight of 100 Ibs. is suspended from the 
 end c. 
 
 Ann or Leverage. Suppose the lever to have no appreciable 
 weight, to be 6 feet long, to lie horizontally, and the fulcrum 7> 
 to be placed 2 feet from the end c, and therefore 4 feet from a. 
 
 Y. 1 
 
2 OF LEVERS AND FORCES. 
 
 The distance be, being the perpendicular distance of the line of 
 action of the load from the fulcrum, is termed the Arm or 
 Leverage at which the 100 Ib. weight acts. The tendency of the 
 load at c is to depress that extremity of the lever and raise the 
 other. The fulcrum b may be considered as the centre of two 
 concentric circles of which be and ba are ra,dii ; and, assuming 
 the lever to oscillate about the point 6, the arcs described by 
 the points a and c and their velocities will be directly proportional 
 to their respective distances from the point b. 
 
 Moment. Since it is a first principle that the moments of two 
 opposing forces must be equal in order that they may balance one 
 another, and the moment of any force about a fulcrum is the force 
 multiplied by the leverage, if we divide the moment of the 100 Ib. 
 weight by the leverage of the counterbalancing force at a, we shall 
 obtain the value of the latter. 
 
 Now the moment of the 100 Ib. weight = 100 x 2 = 200 ; 
 
 200 
 
 .*. 7- = 50 Ibs. = counterbalancing force or power acting down- 
 wards at a. 
 
 The pressure on the fulcrum at b will therefore be 150 Ibs. 
 
 Levers of three kinds. Figs. 1, 2 and 3 represent what are 
 sometimes called the three kinds of levers, because the relative 
 
 positions of the power, the weight, and the fulcrum are different 
 in each case. 
 
 Such a distinction is however un- 
 necessary, and may be confusing to 
 some. We will therefore consider the 
 pressures exerted by the power, the ^ 
 
 weight, and the fulcrum, simply as r * "~j * 
 
 forces, and represent them by the O O 
 
 pull of weights on strings (see fig. 4). 
 
OF LEVERS AND FORCES. 3 
 
 We can then assume at will any of the three points as a fulcrum, 
 and thus avoid confusion of ideas. 
 
 Bent Lever. In estimating the 
 moment of the force acting upon 
 the extremity d of the bent lever 
 shewn in fig. 5, the distance bd must 
 not be assumed as the arm or lever- 
 age but be, since be is the perpen- O 
 dicular distance of the line of action 
 of this force from the fulcrum at b. 
 
 Examples. In the example il- 
 lustrated in fig. 6, the moments of 
 the three forces on the left side of 
 the fulcrum at b are 
 
 10 x 2 = 20 
 10 x 4 = 40 
 10 x 6 = 60 
 
 120 
 
 120 
 .'. -g- = 60, the force required at c to maintain equilibrium. 
 
 The pressure on the fulcrum at b = 30 + 60 = 90. The mo- 
 ment of the three forces of 10, each acting at leverages of 2, 4 
 and 6 respectively, is equal to that of a force of 30 acting at their 
 mean leverage of 4 thus 30 x 4 = 120 their moment, and gene- 
 rally The moment of any number of parallel forces about a 
 fulcrum is equal to the sum of the forces acting at a point which 
 may be called their centre of gravity. 
 
 For instance, in fig. 6, the point b is the centre of gravity of 
 the downward forces acting on the lever abc, and their resultant 
 is a pressure of 90, their sum on the fulcrum at that point. 
 
 It must be carefully remembered that the moments of forces 
 acting in contrary directions must be added together when they 
 are situated on opposite sides of the fulcrum, because each tends 
 to turn the lever in the same direction. 
 
 Fig. 7 is a representation of a lever acted upon by 7 forces, 
 acting either vertically upwards or downwards as indicated by 
 
4 OF LEVERS AND FORCES. 
 
 the arrows. Their points of application are equidistant from 
 
 -" . 
 
 I I I 
 
 3 27 17 
 
 each other, and their intensities, being represented by numbers 
 written against the arrows, are so arranged as to maintain the lever 
 in equilibrium. 
 
 The sum of the upward forces is equal to the sum of the 
 downward forces, and any point in the lever being assumed as a 
 fulcrum, the moments of opposing forces about that point balance 
 each other, e.g. 
 
 Let the force 21 represent the reaction of a fulcrum, and let 
 the forces tending to depress the right-hand extremity of the 
 lever have the sign +, and those tending to turn the lever in the 
 opposite direction the sign , then 
 
 Moments of forces on left I 
 
 ., -- , \+ 10x2 = + 20 
 
 side of fulcrum are 
 
 I + 4 x 1 = + 4 
 
 Total ....................................... + 9 
 
 J' _ 22 x 1 = 22 
 
 side of fulcrum are ] 
 
 I 
 
 7x3 = 21 
 
 Total ....................................... -9 
 
 Again, take the left-hand extremity of the lever as the 
 fulcrum. 
 
 The moments are as follows : 
 
 - 7x6 = - 42 
 
 + 17 x 5 = +85 
 
 -22x4 = - 88 
 
 + 21x3= +63 
 
 - 4x2=- 8 
 -10xl=- 10 
 
 Totals ................ -148 +148 
 
OF LEVEES AND FORCES. 
 
 Once more, take a point half way between the points of appli- 
 cation of the forces 10 and 4 as the fulcrum. 
 
 The moments are 
 
 -lOx } = - 5 
 
 + 4x } = + 2 
 -21xlJ= -31} 
 
 + 22 x 2} = -f 55 
 -17x3}= -59} 
 
 Totals ................. +96 -96 
 
 The lever is thus shewn to be in equilibrium. 
 
 Fig. 8 is a representation of an irre- 
 gularly shaped body acted upon by three 
 forces f l , f\ and f*, in the same plane, 
 whose lines of action are shewn by dot- 
 ted lines, and their directions by arrows. 
 If these forces are in equilibrium, what- 
 ever point in the body may be assumed as the fulcrum, the op- 
 posing moments of the forces about that point will be found to 
 balance. 
 
 Any point a in the body being assumed as the fulcrum the 
 leverages of the forces/ 1 , / 2 , and / 3 may severally be represented 
 by the lines ab, ac, and ad drawn perpendicularly to their respec- 
 tive lines of action from the fulcrum a. 
 
 If equilibrium be maintained 
 
 Although the strains on engineering structures of a very com- 
 plicated kind can be found by considering them as combinations 
 of levers and applying the principles which have been laid down 
 in the preceding pages, nevertheless the strains in certain cases 
 can be determined more rapidly, and with less liability of error, 
 by the application of the principle known as The parallelogram 
 or triangle of forces, which is as follows. 
 
6 OF LEVEES AND FORCES. 
 
 Parallelogram of Forces. If two adjacent sides of a parallelo- 
 gram represent both in magnitude and direction two forces meeting 
 in a point, their resultant may be represented by the diagonal of 
 the parallelogram drawn to the point where they meet. 
 
 For example, let a (fig. 9) be a body 
 acted upon by two forces represented by 
 the lines ac and ab, and their directions 
 by the arrows. If cd* be drawn parallel 
 to ab, and bd to ac, the diagonal ad repre- 
 sents both in magnitude and direction a single force equal to the 
 combined effect of the forces ab and ac upon the body a. 
 
 Resultant. The line ae, which is the equal of the diagonal ad, 
 called the resultant of the farces ab and ac, shews the magni- 
 tude and direction of the force necessary to balance the forces ab 
 and ac, 
 
 Triangle of Forces. The forces which keep the body a in 
 equilibrium are represented by the sides of the 
 triangle abd, both as to direction and amount. 
 The arrows indicating direction arrange them- 
 selves as in fig 10, in consecutive order. 
 
 Corollary 1. When three forces which meet in a point, and 
 whose directions lie in the same plane, maintain a body in equi- 
 librium ; their magnitudes are proportional to the sides of a tri- 
 angle, each of which is drawn parallel to one of the forces. 
 
 Corollary 2. One of the three forces being known, the magni- 
 tude of the other two may be found by constructing the triangle. 
 
 Corollary 3. Any one force in the triangle is the resultant 
 of the other two. 
 
 Polygon of Forces. In fig. 10 substitute for the force ad the 
 two forces represented by the dotted lines ac and ad, of which it 
 may be regarded as the resultant. Also for the force ab substi- 
 tute the forces be and ea. The condition of equilibrium remains 
 unimpaired. 
 
 Thus we have a body maintained in equilibrium by the action 
 of five forces represented by the sides of the polygon acdbe. 
 
 * This operation is generally called completing the parallelogram. 
 
OF LEVEES AND FOKCES. 
 
 Hence we may gather that if a body in equilibrium is acted 
 upon by more than three forces meeting in a point, they are 
 proportional to the sides of a polygon to which their directions 
 are severally parallel. 
 
 But since many polygons may be drawn whose homologous 
 sides shall be parallel, it is obvious that we cannot by this means 
 discover the value of unknown forces, except in certain cases. 
 
 A body o, fig. 11, is acted upon by 6 forces p, q, r, s, t, u, which 
 may be respectively represented by the sides ab, be, cd, de, ef, 
 and fa of the polygon abode/, of which ab is drawn parallel 
 to the direction of p, be to q, and so on, or the forces may be 
 represented by any other polygon as abcghik whose sides are 
 respectively parallel to the forces. 
 
 Examples of the Polygon of Forces. To find the values of 
 unknown forces by the polygon of forces. 
 
 Of the forces p, q, r, s, t, u, let the values of p, q, r and s be 
 known, and of t and u unknown. 
 
 To find t and u. 
 
 Draw the line ab parallel to the direction of force p, and make 
 the length of ab to represent the intensity of p. From the point 
 b draw the line be in a similar manner parallel and equal to 
 force q, cd equal to r, and de equal to s. From the point e draw 
 a line of indefinite length parallel to t, and from point a a line 
 parallel to u intersecting the line drawn from e parallel to t in /; 
 the lines ef and fa represent the values of the forces t and u 
 respectively. 
 
 Again, let the forces q and t represented by the sides be and ef 
 in the polygon be unknown, required their values. 
 
8 OF LEVEES AND FORCES. 
 
 Now as the sides fa, ab, cd and de 
 represent known forces, their lengths 
 and inclination are fixed, consequently 
 the relative position of the points /and b 
 is fixed, also that of c and e. Join ec (fig. 
 11). Through b (fig. 12) draw a line of 
 indefinite length parallel to the direction 
 of q, also a line through point/ parallel 
 to t ; in this line take any point #, draw xy 
 parallel and equal to ec (fig. 11), and 
 through the point y draw a line parallel 
 to fx intersecting the line drawn through b parallel to,^ in the 
 point c. From c draw a line parallel to the line ce (fig. 11), and 
 meeting the line fas in e ; the lines be and fe represent the values 
 of the two unknown forces. 
 
 If the directions of the two unknown forces be parallel, their 
 values cannot be discovered. 
 
 This method is frequently of use in determining the strains on 
 roofs or other structures where a number of struts and ties meet 
 in a point ; the strains on some of them being known, this fur- 
 nishes a means of arriving at the strains on the others. 
 
 Corollary. In the polygon of forces any one force, its direction 
 being reversed, is the resultant of all the other forces. 
 
 Thus if the body o be acted upon by 5 forces represented by 
 sides ab, be, cd, de and ef of the polygon abcdef (fig. 11), their 
 combined effect tends to move the body in the direction of a to / 
 with a force represented by the line of, since the addition of the 
 force fa will keep the body at rest. 
 
 Resolution or Decomposition of Forces. In constructing the 
 polygon acdbe (fig. 10), the force ad was resolved or decomposed 
 into two forces ac and cd. Forces inclined at an angle with the 
 horizontal may be resolved into a horizontal and a vertical force. 
 For example: the force ac (fig. 13) may be j*ia.ia 
 
 resolved into the horizontal force ab, and the 
 vertical force bc- } these forces being termed 
 respectively the horizontal and vertical ele- 
 ments or components of the force ac. 
 
OF LEVERS AND FORCES. 
 
 We shall find it very convenient, in tracing the strains througf 
 complicated structures, not to attempt to find the direct strain on 
 every bar in the first instance, but the horizontal or vertical strain 
 merely ; from either of these the direct strain can afterwards be 
 calculated. 
 
 Example. By the resolution of forces 
 we obtain the following proof that the 
 sides of a polygon represent a set of 
 forces in equilibrium. 
 
 Let the sides of the polygon abcdef t 
 represent six forces acting upon a point, N b 
 the arrow-heads shewing the directions 
 of the forces which are consecutive. 
 
 Describe the rectangle ABCD about the polygon abcdef. 
 From c draw eg parallel to AB, and ch parallel to AD, and from 
 e draw ei parallel to BC, and ek parallel to AB. 
 
 Let the effect of any force towards the upper part of the 
 parallelogram be called northing, represented by the letter N, that 
 towards the right hand easting, represented by E, and similarly 
 southing and westing by S. and W. 
 
 Now the force ab exerts a pressure represented by the line aD 
 in a westerly direction, and in a northerly a pressure equal to Db. 
 
 Kesolving each force into its N. S. E. or W. directions, and 
 adding all the N.s, S.s, E.s and W.s together, we get the following 
 result. 
 
 
 N. S. 
 
 E. 
 
 W. 
 
 Force ab = 
 
 Db 
 
 
 aD 
 
 be = 
 
 bg | 
 
 (gc = )Ah 
 
 
 cd = ( ( 
 
 ;/& = ) gA 
 
 hd 
 
 
 de = 
 
 (ie = ) Bk 
 
 di 
 
 
 . rf = 
 
 ( 
 
 (ek = ) iB 
 
 
 ,, fa- 
 
 fo 
 
 
 Ca 
 
 Sums are equal to DA BC AB CD 
 
 But the forces DA and BC are equal and opposite. 
 Also the forces AB and CD are equal and opposite. 
 
10 OF LEVERS AND FORCES. 
 
 Therefore they are in equilibrium, and as the combined effect of 
 the forces ab, be, cd, de, ef, fa has been shewn to be equal to that 
 of the forces DA,BC, AB and CD, therefore the forces represented 
 by the sides of the polygon abcdef are in equilibrium. 
 
 Alternative methods, the Lever or Parallelogram of Forces. 
 
 As it has been already stated, the 
 strains upon the parts of a struc- 
 ture may be found by means of 
 the parallelogram of forces as well 
 as by the principles of the lever. ,^_ 
 
 abed (fig. 15) is a simple truss, " 
 
 in which ac is horizontal and bd vertical, supporting a load of 8 at 
 
 n (* a (* 
 
 the point b. The depth Id = -r , and the distance be = -j- . Then 
 
 the load on the point of support at a will be 2 and that on c 
 will be 6. 
 
 Lever method. Taking the distance be as our unit of length, 
 we have 6 as the moment of each of the reactionary pressures of 
 the abutments. Assuming the point b as the fulcrum, 6 being the 
 moment of the force acting at a, the balancing strain on the bar ad 
 
 must =-7, the distance eb being its leverage. Similarly the 
 
 n 
 
 strain on the bar cd must = T> . 
 
 
 
 Again, taking point d as the fulcrum, we find the strain on the 
 
 s* 
 
 bars ab and be to be ^= = 6, since bd bc = l. 
 oa 
 
 Since eb : bd :: ab : ad, we perceive that the strains on the 
 bars ab and ad are to each other as their lengths, and the same 
 holds good of the bars be and cd. 
 
 Method of Parallelogram of Forces. To find the strains by 
 the parallelogram of forces, produce the line bd to h making 
 dh = the load of 8 to any convenient scale. Produce the lines ad 
 and cd indefinitely. From the point h draw hg parallel to cd, and 
 hi parallel to ad. From the points i and g draw horizontal lines 
 
 11 and glc to meet the line dh in the points / and k. 
 
 The line dh being the diagonal of the parallelogram dghi 
 represents a force which is the resultant of two forces represented 
 
OF LEVERS AND FORCES. 11 
 
 by the lines di and dg, wherefore by applying the same scale to 
 these lines as we have used for dh, we ascertain the strains on the 
 bars ad and cd, which they severally represent. 
 
 The distance dk is the vertical element of the strain dg on the 
 bar ad = 2, and dl is the vertical element of the strain on dc = 6. 
 
 Forces acting in different planes. When three forces whose 
 directions are not in the same plane meet in a point, their resultant 
 is the diagonal of a parallelopiped of which three consecutive edges 
 joining the opposite extremities of the diagonal represent the 
 forces. 
 
 Thus the diagonal ad of 
 the parallelopiped (rig. 16) 
 is the resultant of three 
 forces represented by the 
 sides abj be and cd. These are 
 the equivalent of the three forces shewn by the lines AB, BC 
 and CD, of which the first two only are assumed to be in the same 
 plane, that of the paper. These two may be resolved into one 
 force AC, and the forces AC and CD may be resolved into AD, a 
 force lying in a plane common to both, and the equivalent of the 
 diagonal ad. 
 
 In a similar manner the resultant of any number of forces 
 acting in different planes and meeting in a point may be found by 
 taking them in pairs and finding their resultants, again pairing 
 the resultants and so on, until only two forces remain, whose 
 resultant is the required force. 
 
CHAPTER IL 
 
 GIRDERS WITH PARALLEL FLANGES, 
 
 Load at Centre of Span. Fig. 17 is a representation of a 
 parallel flanged girder of a length I v ^^ ^ 
 
 and depth d, loaded at its middle 
 with a weight W. Required the 
 strains on the flanges at the points ^ 
 a and b at the centre of the span. 
 
 W being half way between the points of support, the upward 
 
 TIT -I 
 
 reactionary force is acting at an arm - . The balancing mo- 
 ment of the strains at a and b is Sd, S being the strain, and d t 
 the depth of the girder, being the leverage. 
 
 Equilibrium requires that - . - = Sd, whence S= -r-j . 
 
 The strain at a is compression, that at b tension. 
 
 The strain on the flanges at any distance x from the nearest 
 
 W x 
 
 abutment (see fig. 17) will be . -^ ; this shews that the strain on 
 
 the flanges at any point is inversely proportional to its distance 
 from the centre of the span. 
 
 To find the strain on the flanges at the point of application of 
 the load when that point is not at the centre of the span. 
 
 Let x be the distance of the load from one bearing, and I x 
 the distance from the other bearing, I being the span. 
 
 Let W and W" be the two reactionary pressures on the 
 bearings, and let 
 
 W : W" :: x : l-x, 
 then W (l-x) = W"x- 
 
GIRDERS WITH PARALLEL FLANGES. 13 
 
 which expressions represent the moments of the pressures on the 
 bearings about the point where the load is applied, where the 
 strain on the flanges is 
 
 W'l-x W"x 
 
 Load evenly distributed, strain at centre. To find the strain 
 on the flanges at the centre of the girder when the load is evenly 
 distributed over it. 
 
 ' f * 
 
 Let the point a (fig. 18) be the ' 
 
 fulcrum. Let TTbe the distributed 
 load, I the length, and d the depth 
 of the girder as before. 
 
 Now since the moment of any number of parallel forces about 
 a fulcrum is equal to the sum of the forces acting at their centre 
 of gravity (see p. 3), we may consider the whole load on the 
 part Aa of the girder to be collected a distance half way between 
 
 I W 
 
 A and a, or - from either point, its value will- of course be . 
 
 TD 2i 
 
 Now the reactionary force of the abutment acts at an arm of ^ 
 
 2t 
 
 about the fulcrum a, its moment therefore is 
 
 W l_Wl 
 2 X 2~~4~* 
 
 It is balanced by S, the strain on the flange at b acting at a 
 
 W I 
 
 leverage of d, and the force of -=- acting at a leverage of -j ; 
 
 Wl W I Wl 
 
 .-. =^ + T . 5 , whence 8=-^. 
 
 This is a very useful formula for finding the strain on the 
 flanges at the centre of a uniformly loaded girder. 
 
 Corollary. The strain produced at the centre of the span by 
 an evenly distributed load is one half that produced by the same 
 load collected at the centre of the span. 
 
14 GIRDERS WITH PARALLEL FLANGES. 
 
 Strain at any point of the flange. To find the strain on the 
 flanges at any point distant x from the nearest bearing. 
 
 Let I be the length of the girder, vg iff./9. of 
 
 d depth 
 
 w unit of loading, | 
 and wl the total load. 
 
 Let a (fig. 19) be the point in the flange where the strain is 
 required, and x its distance from the nearest bearing. 
 
 Assuming the weight of each of the two portions into which 
 the girder is divided by the point a to be collected at its centre 
 of gravity, the moments about the fulcrum a giving differing signs 
 to the opposing forces will be as follows : 
 
 On the left-hand side of the fulcrum, 
 force leverage sign 
 
 WX X I : 
 
 y X * + -T-J-. 
 
 , On the right-hand side of the fulcrum, 
 force leverage sign 
 
 Since these moments balance one another, 
 
 wlx wx* _ wl (I x) w (I xf I x 
 
 S being the strain on the flanges at a, 
 
 ~,_ I x A Q t x 
 
 an expression for the strain on the flanges of a uniformly loaded 
 parallel flanged girder at any distance x from a bearing point. 
 
GIRDERS WITH PARALLEL FLANGES. 15 
 
 Example. A girder 12 feet long and 1 foot in depth is loaded 
 with two tons per foot of its length, required the strain on the 
 flanges at a point 3 feet from the bearing. 
 
 Here Z = 12, d = 1, cc=3, w = 2. 
 
 I x 
 
 Giving these values to I, d, and x in the expression wx , , we get 
 
 {12 3) 
 [ = 27, the strain required. 
 Zxl) 
 
 If the strain on the flanges be 
 calculated at a few points in the 
 girder, and a perpendicular propor- 
 tional to the strain thereat erected 
 from each point, a curved line drawn 
 through the extremities of these lines 
 (see fig 20) will afford a ready means 
 
 of arriving at the strain on intermediate points by measuring the 
 distance of the flange from the curve in a direction perpendicular 
 to the former. 
 
 When a girder is evenly loaded all over, this curve is a para- 
 bola of which the apex is situated at the extremity of the perpen- 
 dicular raised from the centre of the girder, which is the axis of 
 the curve. 
 
 In the parabola if from points in any straight line ab 
 (fig. 20) drawn perpendicular to the axis cutting the curve in a 
 and b ordinates be drawn to the curve, their lengths vary as the 
 areas of the rectangles included by the parts into which they 
 severally divide the line ab. 
 
 Thus the length of the perpendicular at point x in line ab 
 (fig. 20), is in the same proportion to that at y as ax x bx is to 
 ay x by. 
 
 Now in the equation S=wx 5-7- , oj is a constant quantity ; 
 
 therefore the value of 8 varies as x (I x), or the strain on the 
 flanges at any point varies as the rectangle included by the parts 
 into which the length I is divided by the point x, and therefore if 
 the strain on the flange at several points in the girder be expressed 
 
16 GIRDERS WITH PARALLEL FLANGES. 
 
 by ordinates, the curve which passes through their extremities 
 will be a parabola. 
 
 Knowing the strain on the flanges of a girder at one point, we 
 can easily find the strain at any other by the following proportion : 
 
 S':S::(l-y)y.(l-x}x, 
 
 in which S is the known strain at a point distant x from one 
 extremity of the girder, and S' the required strain at a distance y 
 from the extremity. 
 
 Irregularly loaded Girders. To find the strain at any point 
 a in the flanges of a girder at a distance x from one of the 
 bearings when the load is not evenly distributed. 
 
 First find the load on the bearing (see p. 33) and then the 
 centre of gravity of the portion x, which we will assume to be 
 at a distance y from the point a. 
 
 Let W be the load on the bearing, 
 and W portion x. 
 Then the moment about the point a will be 
 Wx-W'y=Sd; 
 
 Wx- W'y .. ., a 
 
 .'. S= -j =the strain on the flanges at a. 
 
 a 
 
 Strain on the Web. Shearing Force. We have now to 
 consider the strain to which the web of a girder is subjected: this 
 will depend upon the amount of vertical or shearing force, as it 
 has been called, to which it is subjected. 
 
 The vertical or shearing force at any vertical as ab (fig. 19) 
 in a girder is the total vertical downward effect of all external 
 forces on the one side of the plane, which of course equals the total 
 vertical upward effect on the other side of it. 
 
 Required the shearing force at the plane ab (fig. 19). 
 
 The upward vertical force on the left-hand side of ab is 
 
 wl 
 ^--wx. 
 
 The downward vertical force on the right-hand side of ab is 
 
GIRDERS WITH PARALLEL FLANGES. 17 
 
 The shearing force or F 
 
 wl n N ,- 
 
 = -x- wx = w (I x) g" 
 
 From this it appears that in a uniformly loaded girder the 
 shearing force on the web at any point equals the total load 
 between that and the centre of the girder. At the centre of the 
 span all shearing stress disappears. 
 
 To find the value of F at any plane ab distant x from one 
 bearing when the girder is not uniformly loaded. 
 
 Find the load on the bearing (which may conveniently be done 
 by finding the centre of gravity of the load) and the weight of the 
 portion between it and the plane ab ; the difference between these 
 loads is the shearing force. 
 
 Cantilevers. When a beam or girder supports a load which 
 does not lie between the points of support, it is called a cantilever. 
 
 Load at the extremity. Fig. 21 is a representation of a 
 cantilever in which the load W is situated at one extremity a of 
 the girder, the other extremity c and an intermediate point b 
 being the bearings. 
 
 Here the upward pressure on the bearing f . f/ 
 
 at c = Wj~, and the load on the bearing 
 
 I 
 
 ~. 
 be 
 
 The strain on the top flange at b = W~ . The strain on the 
 flange at any distance x from the extremity of the cantilever 
 
 -< 
 
 The vertical or shearing force at any point in the cantilever 
 
 between a and b = W. between b and c = W j- . 
 
 be 
 
 Distributed Load. When the load is 
 evenly distributed > the strain at b (fig. 22) is 
 equal to that which would result if the total 
 load were concentrated half-way between a 
 and b. 
 Y. 
 
18 GIRDERS WITH PARALLEL FLANGES. 
 
 If w = unit of loading, 7 the length of the cantilever, d the 
 depth, the strain on the flange at b or 8 
 
 The strain on the flange at any distance x from the extremity 
 
 X 
 WX X 
 
 a is 
 
 wx x - o 
 
 2 wx* 
 
 The shearing force at x = wx. 
 
 In the above two cases the strain on the top flange is tension, 
 that on the bottom compression. 
 
 The preceding investigations are chiefly applicable to the cases 
 of girders or beams supporting evenly distributed dead weights, 
 as the walls of houses or floors of warehouses. 
 
 Wooden Beams. In applying the formulae to wooden beams 
 izi which the whole of the section of the beam is considered to 
 resist the bending moment, the upper half by compression, the 
 lower by tension, we must find a new value for d. 
 
 Neutral Axis. If, as is usual, we assume that the strain on 
 the fibres increases in direct proportion to their distance from the 
 neutral axis or neutral plane, which is the part of the beam where 
 the longitudinal strains of tension and compression merge into one 
 another, and that this is at the centre of the beam ; the forces in 
 compression and tension on each side of it may be 
 represented by shaded wedges (fig. 23) the centres 
 
 of gravity of which are at the distance - d from the ^^ 
 
 2 
 neutral axis, and therefore at a distance = d apart, 
 
 o 
 
 if d be the whole depth of the beam. 
 
 The moment of resistance to bending will therefore be the 
 
 2 
 sum of the strains on the fibres on one side of the neutral axis x -= 
 
 o 
 
 the depth of the beam. . 
 
GIKDEES WITH PARALLEL FLANGES. 19 
 
 Thus, let a = the breadth of the beam in inches, 
 d = depth 
 
 . - = sectional area of half the beam in inches, 
 
 Zi 
 
 s = strain per sq. in. on the outermost fibre, 
 Then <8f-Y.| = ^, 
 
 and substituting - d for d in the expression -^-j- (see p. 13), 
 
 Wl ZWl 
 2 
 
 e ads 3WI 
 
 therefore : 
 
 whence s 
 
 ^ 4ad?s 
 and W = jrr- 
 
 = the load which the beam will carry. 
 
 Actual experiment proves that timber beams will carry a 
 much greater load than theory would lead us to suppose. The 
 reasons for this are still involved in mystery. One of them, there 
 can be little doubt, is the fact that the fibres in the neighbourhood 
 of the neutral axis are strained much more than is generally 
 supposed, and therefore that, in fig. 23, the shaded part represent- 
 ing the strain on the fibres would more correctly have been 
 bounded by the curved dotted lines, than by the straight lines as 
 in the figure. 
 
 The Web in Iron Girders. In iron girders the resistance to 
 bending offered by the web is not usually taken into account, 
 and d is taken to be the distance between the respect- 
 ive centres of gravity of the top and bottom flanges, 
 as shewn in fig. 24. The web however does assist the 
 flanges in resisting transverse strain, and in some cases 
 it may be necessary to determine the amount of this 
 assistance. 
 
20 GIRDERS WITH PARALLEL FLANGES. 
 
 When the girder is of wrought iron, a material which contracts 
 under compression and extends under tension in about an equal 
 ratio, the top and bottom flanges being of equal area, the neutral 
 axis is situated half-way between the flanges, and the formula 
 
 A .rJ2 
 
 W= 7- (see p. 19) will enable us to ascertain the proportion of 
 
 of/ 
 
 the load which may be considered to be supported by the web. 
 
 Should the material of the girder be one which compresses 
 more easily than it extends, or the reverse, the position of the 
 neutral axis will not be equidistant between the flanges if they 
 are of equal area. 
 
 Fig. 25 represents a portion of beam com- JF^.SS 
 
 posed of a material which offers 3 times as much 
 resistance to compression as to extension. Let 
 the line ab represent an imaginary vertical 
 section through the beam. Let de be the posi- 
 tion assumed by the line ab when the beam is loaded, the distance 
 ad representing the amount of compression in the upper layer of 
 the fibres, and eb the amount of extension in the lower layer. The 
 point c where these lines intersect lies in the neutral axis. Since 
 the triangles adc and bee are similar to each other, 
 
 ac : be :: ad : eb :: 1 : 3, 
 
 therefore the distance of the neutral axis from the upper or under 
 surfaces of a rectangular beam is inversely proportional to the 
 resistance per unit of section of the fibres at those surfaces to the 
 strain which they experience. 
 
 Examples. No. 1. What amount of evenly distributed load 
 will a beam 12 inches deep and 6 inches wide laid across an 
 opening 10 feet wide support, the strain on the outermost fibres to 
 be at the rate of 1 ton per square in-ch ? 
 
 Here a = 6in., ^=12 in., s = 1 ton, I 120 in., 
 
 .a, f rrr 4 X 6 X 144 X 1 n , 
 
 therefore W Tiifi = 9'6 tons. 
 
 o x \2i\j 
 
 No. 2. A wrought-iron girder is 1 foot in depth and 12 feet 
 span. The web at the centre of the span is J" thick ; what dis- 
 tributed load is the web capable of sustaining, taking it as 10'' 
 
GIEDERS WITH PARALLEL FLANGES. 
 
 21 
 
 deep over all, and the strain on the outermost fibre as 5 tons per 
 square inch? 
 
 Here a ='5 in., cZ = 10in., s = 5tons, =144 in. 
 4 x -5 x 100 x 5 
 
 3 x 144 
 
 = 2-314 tons. 
 
 Beams and Girders of Irregular Section. To find the load 
 which a beam or girder of irregular section will sustain, the 
 location of the neutral axis must be determined, which will be 
 such that the total moments about the neutral axis of all the 
 forces on the one side of it balance the moments of all the 
 opposing forces on the other side ; d the working depth of the 
 beam or girder will be the distance between the centres of gravity 
 of the two sums of opposing forces. 
 
 When the section of the beam or girder cannot be equally and 
 symmetrically divided by a line parallel to the line of action of 
 
 fiff.se 
 
 
 
 the load, generally a vertical line, allowance must be made for the 
 tendency to buckle unless the beam or girder be prevented from 
 so doing by stays. 
 
 Sections 1 arid 2, fig. 26, are examples of this want of symmetry. 
 No. 3 is symmetrical. 
 
 Practical method of determining the strain on an ordinary 
 wrought-iron girder. 
 
 Ordinary wrought-iron Girders. In the commonest type of 
 wrought-iron girder bridge, the roadway is carried by cross girders 
 which are supported by main girders. The points of attachment 
 of the cross girders, whether they are suspended from the bottom 
 or rest upon the top flange, are stiffened in the main girder by 
 upright gusset plates or T irons. The load is considered to be 
 concentrated at these points, and the strain on the flange between 
 two adjacent gussets to be uniform. 
 
22 GIRDERS WITH PARALLEL FLANGES. 
 
 Diagram 1, Plate I., is a diagram for a girder in which the 
 cross girders attached to the bottom flange are equally loaded, 
 and placed at equal distances apart. The web of the girder is 
 assumed to be so thin as to be incapable of resisting compression, 
 and therefore to do duty in tension only. 
 
 Let the load at each gusset be 8 tons. 
 
 Since the girder is uniformly loaded, there will be an equal 
 load of 44 tons on each pier, consequently of the load of 8 at the 
 centre of the span 4 will be borne by the right-hand, and 4 by the 
 left-hand abutment. The load 4 is supported by the web repre- 
 sented by the diagonal line, and is conveyed up to the point where 
 the adjoining gusset meets the top flange. Here it puts a hori- 
 zontal compressive strain on the top flange, and a vertical strain 
 of 4 on the gusset. This force of 4 being added, the load of 8 
 makes a total load of 12 to be carried by the web in the 2nd bay 
 from the centre. In this manner the load on the web in each bay 
 increases towards the pier. 
 
 Strains on the Flanges. The strains on the flanges in each 
 bay are found by commencing with the bay next the abutment 
 and working towards the centre of the span. 
 
 For the sake of convenience, the depth of the girder has been 
 taken equal to the width of a bay and therefore the horizontal 
 strain on the top flange on the last bay is 44, which is also the 
 vertical load on the pier, for the point a is kept in equilibrium 
 by three forces represented by the sides of a right-angled triangle, 
 in which the two sides enclosing the right angle are equal to one 
 another. 
 
 In like manner the point b in the bottom flange is kept in 
 equilibrium by three forces, of which the vertical is 44, conse- 
 quently the horizontal strain on the bottom flange of the second 
 bay from the abutment is 44, being equal to the vertical. 
 
 These horizontal strains on the flanges are carried through till 
 they meet with reacting strains coming from the other end of the 
 girder. 
 
 The vertical load of 36 carried by the diagonal of the second 
 bay adds 36 to the strain on the flanges, making it 80 + in the 
 top flange, and 80 in the bottom. 
 
GIRDERS WITH PARALLEL FLANGES. 23 
 
 Adding in the horizontal effect of each diagonal as we proceed 
 towards the centre we thus get the strain on the flanges in each 
 bay, and a maximum strain of 144 tons at the centre. 
 
 Diagram 2, Plate I., shews the strains on the flanges, the web 
 being assumed to act only in compression, while Diagram 3, 
 Plate I., gives the strains which result when the web resists equally 
 extension and compression. 
 
 It may be observed that the sum of the strains on the top and 
 bottom flange of any bay is the same for each of the three kinds 
 of web, while in the last case the strains on the top and bottom 
 flanges are equal, being the mean of the strains on the two flanges 
 by either of the other methods. 
 
 To prove the truth of the third diagram as in Diagrams 1 and 2, 
 Plate I., take the load per bay as 4 instead of 8 ; find the strains 
 on the flanges, which will be just half those given in the figure, 
 and consider Diagram 1 to be placed upon Diagram 2, so that the 
 flanges and verticals of each coincide, and the strains of each to be 
 added together : the result will be Diagram 3, Plate I. 
 
 In practice, find the strains by the method of either Diagram 1, 
 or Diagram 2, and take the mean as the correct strain for the 
 flanges, the web being strained equally in tension and compres- 
 sion. 
 
 Since As depth of girder : breadth of bay 
 
 :: vertical load on a diagonal : horizontal strain 
 caused thereby, 
 
 , , , breadth of bay b 
 
 /. horizontal strain = vertical load x 1 -r- <> . , or -j , 
 
 depth ol girder a 
 
 if 6 = breadth of bay, 
 
 and d = depth of girder. 
 
 In the example we have just considered -= = 1. 
 
 The strains on the flanges in Diagrams 1, 2, and 3, multiplied 
 
 by the fraction -^ , will give the correct strains for any similarly 
 d 
 
 loaded girders in which the proportion of the depth to the breadth 
 of a bay is as d to b. 
 
24 GIRDERS WITH PARALLEL FLANGES. 
 
 Strains on the Web. The web in plate girders should be 
 considered to act as in Diagram 3, in this case it is made to do 
 double duty; the same fibres which are resisting extension in one 
 direction, resist a force of compression in a direction at right angles 
 to the other, without impairing the efficiency of the plate in 
 offering resistance to either force. 
 
 On the system of Diagram No. 3 the webs need only be one- 
 half the thickness of those required for Diagrams Nos. 1 and 2. 
 
 It is a further advantage attending the system of Diagram 3 
 over Diagrams 1 and 2, that it allows the top and bottom flanges 
 to be made exactly alike without waste of metal. 
 
 Diagram No. 4, Plate L, is similar to Diagram No. 1, but the 
 letter w is substituted for the load of 8 on each bay. It shews a 
 more expeditious way of finding the strains on the flanges, saving 
 figures. 
 
 Adding the top and bottom flanges together in any bay, and 
 dividing by 2, we get an expression for the strain on the system 
 of Diagram No. 3. 
 
 Thus the strains on the third bay from the abutment on this 
 
 ^A I, 13 J W+IV W 113 TT7 OA 'f 
 
 system would be - -= = llf W = 94, if w = 8. 
 
 Result checked by method described p. 14. The strains on 
 the 'flanges in any bay may also be found by the method indi- 
 cated at the commencement of this Chapter, for finding the strain 
 on the flanges at any point of a uniformly loaded girder. We 
 
 cannot make use of the formula 8=wx , , as our load is not 
 evenly distributed, but concentrated at intervals. 
 
GIRDERS WITH PARALLEL FLANGES. 25 
 
 Taking a girder of which Diagram No. 3, Plate I., is a diagram, 
 required the strain on the flanges of the 'third bay from the abut- 
 ment. Draw the vertical line ab (fig. 27) through the centre of 
 the bay, and therefore passing through the point of intersection 
 of the diagonals. 
 
 The fulcrum being supposed to be either the point a or 6, we 
 have an upward force representing the reaction of the pier = 5 \ W, 
 which acts at a leverage of 2, if the length of a bay be taken 
 as 1. In opposition to this we have two forces, W acting at a 
 leverage of J, and again IF at a leverage of 1 J. 
 
 The difference between the moments of these forces is the 
 actual bending moment about point a or b. 
 
 Since the length of each bay is 1, and the depth 1, 
 The moment of the reaction 
 of the abutment is 5J TFx2J=13j 
 
 The moment of the load be- 
 tween the fulcrum and the - * 
 abutment is . . W x -^ 
 
 Of? 
 17 ' 
 
 Bending moment or $=111 W 
 
 If TF = 8, =94. 
 
 It is proper to consider the flanges throughout the whole bay 
 as subject to a strain of 11JTF, though if the plane ab in which 
 the fulcrum lies were taken at a point in the bay nearer the centre 
 of the girder, the value of 8 would be greater, while on the other 
 hand it would be less if it cut the bay at a point nearer to the 
 abutment than the centre of the bay. In such case though the 
 strain on the flange would be the same, since the line ab does not 
 pass through the point of intersection of the diagonals, their effects 
 would have to be reckoned in determining the total bending 
 moment. The moments of the diagonals depend upon the vertical 
 distance apart of the points where they are intersected by the 
 imaginary plane. 
 
 Example. In the girder just given, required the total bending 
 moment on S at a distance of 2f bays from the abutment, being 
 at the section a'b' } fig. 27. 
 
26 GIRDERS WITH PARALLEL FLANGES. 
 
 Proceeding as in the previous example, 
 Moment of reaction of 
 
 abutment = 5J- W x 2 f = 15J W + 
 
 Moment of loads at) 'J * I _iw_ 
 
 each bay j~1 Wx ^J 
 
 =12|TF 
 when width of each bay and depth of girder, or d, = 1. 
 
 The value of 8 for the flanges of bay No. 3 we have seen to be 
 11JJF, therefore 
 
 which must be the bending moment of the diagonals acting at a 
 leverage of d. 
 
 Now the line a'b' intersects the diagonals in the points p and q 
 and the* distance apart of these points is half the depth of the 
 
 d 
 
 girder, or ^. 
 
 The horizontal effect of each diagonal in bay No. 3 is equal to 
 its vertical effect, since the diagonals are inclined at an angle 
 of 45. 
 
 By referring to Diagram No. 4, Plate I., it will be seen the 
 total vertical or shearing force on bay No. 3 is 3 J W. The 
 vertical force on each diagonal when they are crossed is therefore 
 1| W, which is also the amount of horizontal force exerted by each, 
 the diagonal in which the point p lies giving a compressive hori- 
 zontal force and the other a tensive force. 
 
 Now 1 J W acting at a leverage of - 
 
 therefore considering all the horizontal forces as acting at a 
 leverage of d, 
 
 HfTF+f JF=12fTF, 
 
 the value of S at a distance of 2| bays from the abutment. 
 
 In finding the value of 8 at a point distant 2J bays from the 
 abutment, the horizontal effect of the diagonal has to be deducted 
 
GIRDERS WITH PARALLEL FLANGES. 27 
 
 from the flange, as the strain on the diagonal is of a different sign 
 to that on the adjacent flange. 
 
 General Formula for the Strain on any Bay. By means of 
 the foregoing diagrams formulae may be constructed for finding 
 the strains on the flanges of any bay of a girder. 
 
 When loaded as per Diagram No. 1, Plate I., the web being in 
 tension only, 
 
 Let N = whole number of bays in the girder, 
 
 n = number of any bay reckoned from the abutment, 
 
 b = breadth of a bay, 
 
 W = load per bay, 
 
 d = depth of girder, 
 
 S = strain on flange of any bay n, 
 
 Nn^ T b f 
 
 o =n - yy.-j,tor top tiange (1), 
 
 S = 0-1) ~ ^ " V 'W. j for bottom flange (2). 
 
 When the web is under compression only as in Diagram No. 2, 
 Plate L, equation No. 1 will give the strain on the bottom flange, 
 and equation No. 2 on the top flange. 
 
 When the web plate acts equally in compression and extension, 
 the mean of equations 1 and 2 will give the strain on either 
 flange ; wherefore 
 
 S = 
 
 
 2 v } 2 
 
 b 
 
 2 
 
 'd' 
 
 Wb ( N+I\ 
 
 
 2 d} n ^ l ' 21 
 
 
 Shearing strain on each Bay. The method described on 
 p. 16 would enable us to determine the shearing force on any bay 
 of a girder loaded at intervals, but as for each alteration in the 
 position of the moving load the proportion of load borne by each 
 abutment has to be calculated afresh, a more expeditious method 
 of ascertaining the maximum shearing strain on any bay of the 
 girder is to construct a series of diagrams as in Plate II., in which 
 
28 
 
 GIRDERS WITH PARALLEL FLANGES. 
 
 the shearing effects produced by the different positions of the load 
 are shewn in separate diagrams. 
 
 The diagrams 1 to 14, Plate II., re'present a girder of 12 bays, 
 having a dead load of 2 tons, and a live load of G tons per bay. 
 
 Diagram No. 1 shews the shearing strains produced by the 
 dead load which are constant, the diagonal line across each bay 
 serves to shew by which abutment the load on that bay is carried, 
 the upper end of the line lying towards the particular abutment. 
 
 The diagonal lines may be regarded as tension rods and the 
 figures written against them as the vertical elements of the strains 
 to which they are subjected. 
 
 Diagram No. 2 shews the strains caused by loading the 1st 
 bay, counting from the left hand, with 6 tons. 
 
 Diagram No. 3, the strains caused by the loading of the 2nd 
 bay, and so on. 
 
 The strain on bay No. 2, Diagram No. 3, is 2 -2 5 only ; this is 
 the difference between 2-5, the proportion of the load of 3 resting 
 at the junction of bays 2 and 3, which goes to the left hand abut- 
 ment ; and '25, the proportion of the load of 3 situated between 
 bays 1 and 2, which goes to the right hand abutment. For this 
 bay diagonals crossing one another having the numbers 2'5 and 
 *2o written against them respectively would represent the shearing 
 forces due to each of the loads of 3 in turn. In all cases where 
 diagonals representing direction of shearing stress cross, the only 
 vertical force upon the bay is the excess of the preponderating 
 diagonal over the other, and the only diagonal to be shewn is the 
 one representing the intenser force of the two. 
 
 If the vertical element of two diagonals crossing one another 
 is the same, the shearing strain on the bay is nil : such is the case 
 in the central bay of a symmetrically loaded girder with an odd 
 number of bays. 
 
 The propriety of subtracting opposing diagonal effects may 
 further be seen if we remember that shearing force produces strain 
 
GIRDERS WITH PARALLEL FLANGES. 29 
 
 on the diagonal, consequently distortion of the normal form of the 
 bay. When the diagonal ab (fig. 28) is subjected to tension it 
 stretches ; in consequence of this the originally square form of the 
 bay abed (fig. 28) becomes a rhombus af^d^ the points a and 
 b being forced further apart and the points c and d brought closer 
 together. 
 
 When the diagonal cd (fig. 29) is in tension, a reverse action 
 takes place, so that were the respective hypothetical strains on the 
 diagonals ab and cd equal, the rectangular form of the bay would 
 be preserved, in which case there can be no actual strain on the 
 diagonals. If the hypothetical strain on the one exceed that on 
 the other, the actual strain will be the difference of the two. 
 
 Diagram No. 14 exhibits the greatest shearing force for every 
 bay of the girder. The 4 centre bays have two sets of figures and 
 diagonals crossing one another : this signifies that under one con- 
 dition of loading the direction of the diagonal differs from that 
 under another possible condition of loading. The figures written 
 against each diagonal shew the vertical component of the greatest 
 strain to which the diagonal can be exposed. 
 
 In Diagram 14, the vertical strain of 44 in bay No. 12 is 
 obtained by taking the sum of the strains on bay No. 12 in all the 
 diagrams 1 to 13 inclusive. In bay No. 11 the strain of 36'25 is 
 the sum of the strains on bay No, 2, in Diagrams 1 to 12 inclusive, 
 that is to say the live load is supposed to be removed from bay 
 No. 12, the girder being loaded throughout with that exception. 
 When the girder is loaded all over the strain on bay No. 11 is 36, 
 since "25 the strain on the opposing diagonal of bay No. 11, 
 Diagram 13, has to be deducted from 36'25. 
 
 Similarly in bay No. 10 the shearing strain is 29, being the 
 sum of the strains in Diagrams 1 to 11 inclusive. 
 
 In bay No. 7 the strain is 5 '25, bays Nos. 8 to 12 being loaded. 
 5*25 is the sum of the strains on bay No. 7 in Diagrams 9 to 13 
 inclusive, 1 the strain on bay No. 7 in Diagram No. 1. Since the 
 dead load is always constant, this neutralizing strain must not be 
 neglected. 
 
 The shearing strain on bay No. 7 when bays 1 to 7 are loaded 
 will be 10'25, being the sum of the strains in Diagrams 1 to 8 
 inclusive, the direction of the diagonal in this case being reversed. 
 
30 GIRDERS WITH PARALLEL FLANGES. 
 
 The shearing strain on bay No. 5 caused by loading bays 
 No. 1 to No. 4 is 1 only. It is the sum of strains on bay No. 5 
 in Diagrams 2 to 5 less 3, the opposing strain on bay No. 5 in 
 Diagram 1. 
 
 There is no shearing force to left abutment on bay No. 9, for 
 if we add together the strains on bays No. 9 in Diagrams 11 to 13 
 we get a shearing force to left hand of 2*25, but the shearing force 
 to the right-hand abutment on bay No. 9 caused by the dead load 
 is 5 as given in Diagram 1. Subtracting these forces from one 
 another we have a remaining shearing force of 2*75 to the right 
 hand. 
 
 Hence it is evident that if Diagram 14 represents a lattice 
 girder with vertical struts and diagonal tension bars, bars crossing 
 one another should be inserted in the 4 central bays. For a plate 
 girder the web in each of the 4 centre bays must be made strong 
 enough to sustain the greater of the two shearing strains. Since 
 the web does duty equally in compression as in tension, the 
 vertical shearing force will be halved for each diagonal ; thus in bay 
 No. 1 the web will have to sustain a vertical force of 22 by tension 
 and a like amount by compression. 
 
 The bays in this example being squares, the actual strains on 
 the diagonals will .be the vertical load upon them x V2. 
 
 Formula for Shearing 1 Force. Formula for the maximum 
 shearing force upon any bay. 
 
 If we examine Diagram No. 4, Plate I., we shall see that the 
 shearing force on the bay next the pier may be expressed by the 
 formula 
 
 ~ 2 ' 
 
 W being the load per bay and.-ZV the total number of bays in the 
 girder. 
 
 N 3 
 In the 2nd bay from the abutment F= W ^ 
 
 Srd F= W ~^ 
 
 UlU. 5, ,, 2 ' 
 
 A 1 1 Tjl T/T7" ^ 
 
 n ^^n ,, -t 7 r ) ' 
 

 GIRDERS WITH PARALLEL FLANGES. 31 
 
 and so on. Up to the 6th bay the value of F is positive, but 
 if after reaching the centre of the girder we continue to count the 
 number of the bays from the same abutment, we shall get F with 
 a negative sign ; this indicates the change in the direction of the 
 diagonal. The value of F will however be correct, as, for instance, 
 it may be written thus for the bay next the abutment, 
 
 either F= W^^ for bay No. 1 = W^^- = 5 JIT, 
 
 or F = W N ~^ for bay No. 12 = W ~ = - 5 J W, 
 
 according as we count from the right or left-hand abutment. 
 A negative value to F shews that the shearing force is carried by 
 the abutment opposite to that from which the counting has com- 
 menced. 
 
 Putting n for the number of the bay, the formula for the 
 vertical or shearing force on any bay of a symmetrically loaded 
 girder becomes 
 
 By deducting the values of F as given in Diagram 1, from its 
 values in Diagram 14, Plate IL, it will be seen that the greatest 
 shearing force caused by a live load of W per bay is 
 
 On bay No. 1, W ( " 
 
 >> 4 W 
 
 and so on. 
 
 N-l N-3 JV-5 
 
 For the expressions -r, ^ , - , &c. substitute the 
 
 2 22 2 
 
32 GIRDERS WITH PARALLEL FLANGES. 
 
 _ 
 general formula - -~ -- - , n being the number of the bay, 
 
 14 9 , , (n-l)* 
 and express , , , &c. by . 
 
 Thus the general formula for the shearing stress on any bay n 
 due to the live load becomes 
 
 2 ~2^ 
 
 Therefore the maximum shearing force on any bay w. caused 
 by the live and dead loads combined will be expressed by the 
 formula 
 
 when the number of the bays in the girder is even. When the 
 number of bays is uneven, the formula is true when n is not 
 
 N 
 greater than -= ; but when this is the case, a quantity represented 
 
 W 
 by the expression ^-^ (N 2% 1) must be added to the right- 
 
 zIlV 
 
 hand side of the above equation to obtain the true value of F. 
 Therefore when the bays are uneven, the value of F in any bay 
 whose number is more than half the total number of bays in the 
 girder may be expressed by the formula 
 
 W 
 
 ~ ( N ~ ( Zn ~ !)1 + K - 1) 2 
 
 In the foregoing formulas, 
 
 F maximum shearing force on the bay n t 
 
 n = the number of the bay counting from one abutment, 
 
 N = the total number of bays in the girder, 
 
 W = the live load per bay, 
 
 W = dead 
 
GIRDERS WITH PARALLEL FLANGES. S3 
 
 N.B. Negative values of F obtained from the above formulas 
 should be neglected. For the bays near the centre of the girder F 
 have two positive values, varying accordingly as the number (ri) 
 of the bay is counted from one or the other abutment. Of these 
 the greater will be that obtained by counting the number n of the 
 bay from the nearer abutment. This value of F will be the only 
 one required for a plate-girder. For a lattice-girder both values 
 of F must be found, and diagonals crossing one another intro- 
 duced, as shewn in Diagram 14, Plate II. 
 
 Irregular loading, First case, Bays equal. In Diagram 5, 
 Plate I. are given the strains on the flanges of one girder of a 
 skew bridge, the load on the left-hand end of the girder being 
 less than that on the right. For simplicity's sake, the web is 
 assumed as acting in tension only. 
 
 There are two convenient methods of finding the load borne 
 by each abutment, the first by dealing with each load singly, and 
 adding the separate effects of all the loads as follows : 
 
 Proportion to left-hand Proportion to right-hand 
 
 abutment abutment 
 
 of bad i ; g | v ^ 
 
 3 3x 10 - 30 Sx 2 - 6 
 
 X 12~12 X I2~I2 
 
 A_i5 * JL-is 
 
 X 12~12 X 12~12 
 
 7 Tx^- 5 ? 7xA-?? 
 
 12 12 12 12 
 
 _7 _ 56 5 __ 40 
 
 X 12~12 X l2~T2 
 
 8 8x-i- 48 8x^-M 
 
 X 12~l2 X 12~12 
 
 j>_40 1_-!M 
 
 X 12~l2 X 12~12 
 
 8 8x 4 - 32 8x 8 - 6 * 
 
 X 12~l2 K 12~12 
 
 Y. 3 
 
34 GIRDERS WITH PARALLEL FLANGES. 
 
 Proportion to left-hand Proportion to right-hand 
 
 abutment abutment 
 
 OfloadS 8x^-g 8x = g ' 
 
 j^_16 10_80 
 
 " " X 12"12 * X l2~l2 
 
 18 11 _ 88 
 
 X 12~12 X 12~12 
 
 Totals = 30-5 
 
 Since the total load on the left-hand abutment is 30*5 
 and right-hand ,. 41'5, 
 
 we must so divide the loads that the sums of those which go to 
 the left shall be 30*5, and the sum of those which go to the right 
 41'5. To do this, assume that 6*5 of the load of 8 at the centre 
 of the girder goes to the left, and 1/5 of it to the right-hand 
 abutment, 
 
 then 1 + 3 + 5 + 7 + 8 + 6*5 = 3O5 load on left-hand abutment 
 and 8 +8 + 8 + 8 + 8 + 1-5 = 41-5 right-hand 
 
 The moments of these loads balance one another. 
 
 Moments of Loads 
 
 On left-hand abutment On right-hand abutment 
 
 load leverage moment load leverage moment 
 
 1x1=1 8x1=8 
 
 3x2=6 8x2= 16 
 
 5x3= 15 8x3= 24 
 
 7x4= 28 8x4= 32 
 
 8 x 5 = 40 8 x 5 = 40 
 
 6-5 x 6 = 39 1-5x6=9 
 
 Total moment 129 Total moment 129 
 
 The second method is to find the moments of all the loads 
 about one abutment which are equal to the moment of the 
 pressure on the other abutment about the first abutment. Thus 
 the moments about the left-hand abutment are 
 
GIRDERS WITH PARALLEL FLANGES. 35 
 
 1x1=1 
 3x2=6 
 5x 3 = 15 
 
 7 x 4 = 28 
 
 8 x 5 = 40 
 8x 6 = 48 
 8x 7 = 56 
 8x 8 = 64 
 8x 9 = 72 
 8 x 10 = 80 
 8x11 = 88 
 
 498 
 
 Let P be the pressure on the right-hand abutment, P acts at 
 an arm of 12, 
 
 .-. Pxl2=498, 
 
 z> 498 ^ - 
 P== __ = 41-o, 
 
 which agrees with result obtained by the first process. 
 
 Second Case, Bays of an unequal size, Load symmetrical. 
 
 Diagram 6, Plate I. is the diagram for a cross girder 25 feet 
 long and 2 feet deep. The load being symmetrically disposed on 
 each side of the centre line, there will be no shearing strain on 
 the centre bay under full load, but an equal load of 16 on each 
 abutment. Assuming the web to act only in tension, the strains 
 on the flanges would be those given on the left-hand side of the 
 centre line of the diagram. On the right-hand side are given 
 the strains which arise when the web acts equally in tension 
 and compression. 
 
 The strain of 36 on the flange of the 4'. 6" bay to the left is 
 obtained by multiplying 16, the vertical load carried by the 
 diagonal, by 4'. 6" the length of the bay, and dividing by 2 the 
 depth of the bay, 4*6 being the leverage at which the load acts, 
 and 2 feet the leverage of the counterbalancing strain on the 
 flange. The method by the parallelogram of forces also shews us 
 that the forces which keep the upper left-hand corner of the 
 girder in equilibrium are proportional to the sides of a triangle 
 composed of the depth of the girder 2', the length of the bay 
 
 32 
 
36 GIEDEKS WITH PARALLEL FLANGES. 
 
 4'. 6" and the diagonal of the bay, since the depth of the girder 
 2 represents the vertical force 16. 
 
 /. As 2' : 16 :: 4'. 6" : 36, the horizontal strain on the flange. 
 
 It may be shewn in a similar manner that 8 x = 20 is the 
 
 ft 
 
 additional strain on the flange caused by the load of 8 at the 
 junction between the second and third bays, and 36 + 20 = 56 is 
 the strain on the top flanges of second and third bays. 
 
 On page 23 is given the following equation for the strain on 
 the flanges caused by a vertical load on any bay. Horizontal 
 
 strain = vertical load x -3 . 
 a 
 
 When the bays of a girder are unequal in length, the fraction 
 
 -j will vary for each bay. This is the only point of difference be- 
 cL 
 
 tween the case we have just considered and that of the uniformly 
 loaded girder treated of on page 22. 
 
 Third Case, Bays and Loading irregular. In Diagram 7, 
 Plate I. the bays and loads are both irregular. 
 
 Proceeding as in the case of Diagram 5, Plate I. 
 
 Proportion to left-hand Proportion to right-hand 
 
 abutment abutment 
 
 0,,-S 
 
 6 
 
 " 
 
 - 25 25~ 25 
 
 nc> 15 180 _. 10 120 
 
 2 12X 25 = 25 12X 25=25 
 
 JL_ 81 16 _ 144 
 
 X 25~ 25 y X 25~~25~ 
 
 4 _ 16 21 _ 84 
 
 X 25 " 25 X 25 ~ 25 
 
 Totals = 2 3, 
 
 To get a load of 23 on the left-hand abutment, and of 16 on 
 the right-hand abutment, we must split the load of 12, taking 
 9 to the left and 3 to the right-hand side. 
 
GIRDERS WITH PARALLEL FLANGES. 37 
 
 The moments of loads are as follows : 
 On left-hand abutment On right-hand abutment 
 
 load leverage moment load leverage moment 
 
 8x2 = 16 4x4 = 16 
 
 6x6 = 36 9x9 = 81 
 
 9 x 10 = 90 3 x 15 = 45 
 
 Total moment 142 Total moment 142 
 
 This moment of 142 divided by 2, the depth of the girder 
 which represents the arm or leverage at which the strain on the 
 flange acts, gives 71 as the accumulated or maximum strain on the 
 flanges of the girder. 
 
 In order to find the greatest shearing stress on any bay of 
 girder with irregular bays, caused by a moving load, the method 
 illustrated by Plate II. should be adopted. 
 
 Trusses and Lattice Girders. If for a solid plate web we 
 substitute open work of bars, whether acting as ties or struts, 
 inclined at an angle with the vertical, we have what is called a 
 truss or a lattice girder. 
 
 Diagram 1, Plate II. may be taken to represent a simple truss 
 in which the diagonals are tension-bars, and the verticals struts, 
 and Diagram 2 a truss in which the diagonals are struts, and the 
 verticals tension- bars. When several sets of diagonal bars cross- 
 ing one another are used, the girder is styled a lattice, and some- 
 times a trellis girder, from its resemblance to trellis-work. 
 
 Without Verticals. Diagram 1, Plate III. is a diagram of a 
 truss in which there are no verticals; the diagonals are inclined at 
 an angle of 45 with the vertical. The left-hand half of the 
 diagram gives the strains on the flanges, and the vertical stress 
 on the diagonals caused by resting the load on the top flange. 
 The right-hand portion gives the strains when the load is sus- 
 pended from the bottom flange. 
 
 It will be observed that the strains may be made to agree in 
 the two parts by turning either half upside down and changing 
 the signs of the strains. 
 
38 GIRDERS WITH PARALLEL FLANGES. 
 
 The formula 
 
 8= T \ {" (N ~ n + 1} ~ ^} (see page 27)> 
 
 will give us the mean strain, i. e. half the sum of the strains on the 
 top and bottom flanges of any bay n. 
 
 When the load rests upon the top flange, the strain on the 
 bottom flange will exceed that on the top flange of the same bay 
 
 W 
 
 by the amount -$ . But when the load is suspended from the 
 2i 
 
 bottom flange, the strain on the bottom flange will be less than 
 
 W 
 that on the top flange by the amount -5- ; accordingly the true 
 
 strain on the flanges will be obtained, as the case may be, by 
 
 W W 
 
 adding -r- to, or subtracting ~j- from, the value of S obtained from 
 4 T 
 
 the formula above. 
 
 Plate IV. shews by a series of diagrams the shearing effect of 
 the line load considered at each of its points of application in 
 succession. 
 
 Diagram 14 gives the extreme vertical strains. From this 
 diagram it will appear that the diagonals at the centre of the 
 truss have to be made capable of resisting both tension and com- 
 pression. 
 
 The sums of the extreme strains on the diagonals of any 
 bay n may be found by the formula 
 
 WA- W (r) 1 V 2 
 
 F=-^f- {N-(2n-\}}+W' ( ^f- (see page 32). 
 
 In each bay the vertical strain on the one diagonal exceeds 
 
 W+ W 
 
 that on the other by 5 . 
 
 z 
 
 When the loapl is suspended from the bottom flange the dia- 
 gonal in tension is subject to the greater strain, but if the load 
 rest on the top flange the reverse is the case; see Diagram 1, 
 Plate III. 
 
 Warren Truss. The peculiarity of the Warren truss is that 
 the struts and ties make angles of 60 or thereabouts with the 
 flanges. 
 
GIRDERS WITH PARALLEL FLANGES. 39 
 
 In Diagram 2, Plate III. are given the strains on a Warren 
 truss in which the load rests upon the top flange. Diagram 3, 
 Plate III. shews the strains caused by suspending the load from 
 the bottom flange. For the sake of comparison with the diagrams 
 of girders and trusses already given, the depth of these Warren 
 
 trusses has been taken as 1 ~ of the span, consequently the angle 
 
 made by the diagonals with the flanges is more than 60, being 
 about 63. 
 
 Referring to Diagram 3, Plate I. it will be seen that the strains 
 on the bottom flange of that diagram agree with those on the top 
 flange of Diagram 2, Plate III. and with those on the bottom 
 flange of Diagram 3, Plate III. These strains may consequently 
 be found by equation 3, page 27. 
 
 The strains on the eleven bays in the bottom flange of Dia- 
 grams 2 and 4, and in the top flange of Diagrams 3 and 5, Plate 
 III. may be found by the equation 
 
 a N (n 1) TTT b 
 
 S = n- -^ -Wj', seep. 27; 
 
 N being 11 in this case. 
 
 The strains on the top flange of Diagram 4, and bottom flange 
 of Diagram 5, Plate III. are given by the formula 
 
 and on the bottom flange of Diagram 3, Plate III. by equation 3, 
 p. 27. 
 
 In the example Diagram 6, Plate III. the load is supported by 
 both top and bottom flanges. It may be regarded as a combina- 
 nation of the two systems illustrated by Diagrams 3 and 5, 
 Plate III. 
 
 Since the load at each point of intersection of diagonals with 
 the flanges is 4 in Diagram 6 ; we may, in order to find the strains, 
 take 4 as the load per bay in each of the Diagrams 3 and 5, cal- 
 culate the strains, and consider the one to be superimposed upon 
 the other, adding the strains. 
 
 The strains on the top flange of Diagram 6, Plate III. may be 
 
40 GIRDERS WITH PARALLEL FLANGES. 
 
 found by the equation above given for the top flanges of Diagrams 
 3 and 5. For the bottom flange the strains may be found by 
 the formula 
 
 It must be remembered that N is the whole number of bays 
 (12 in the case of Diagram 6, Plate III.) and TFthe load per bay 
 ( = 8 in this case) . 
 
 Strains on the Diagonals. The strains on the diagonals of a 
 Warren truss may be found by constructing a series of diagrams 
 as in Plate II., or by means of the formula given on page 32. 
 
 Diagram 7, Plate III. shews the strains on the diagonals of a 
 Warren truss with a live load of 6, and a dead load of 2 per bay. 
 
CHAPTER in. {(UNIVERSITY 
 
 HOGBACKED GIRDERS. 
 
 WE now come to the consideration of girders in which the flanges 
 are not parallel. Girders which have a slightly curved top flange 
 are very common, these have been called Hoghacked girders. The 
 effect of thus diminishing the depth towards the abutments is to 
 increase the strain on the flanges and relieve the web. 
 
 Equations 1, 2 and 3, p. 27, will give the strains on the flanges 
 of any bay of a hogbacked girder according as the diagonals do 
 duty in tension only, compression only, or in tension and compres- 
 
 sion alike. The fraction -j in those equations will vary for each 
 
 bay. 
 
 Diagrams 1 to 6, Plate V. represent a hogbacked truss with 
 diagonal tie-bars under various conditions of loading. Diagram 7 
 gives the extreme horizontal strains on the diagonals, and the 
 greatest strains on the verticals ; Diagram 1 the extreme hori- 
 zontal strains on the flanges. 
 
 Strains on the Flanges. The horizontal strains on the flanges 
 may be found conveniently by the method described on pp. 24 and 
 25. Thus at the points a and , Diagram 1, 
 
 S= 45x^=58-06; 
 ' 1 7o 
 
 at points c and d, 
 
 = 45 x ^-18 x ^ = 7579; 
 95 *9o 
 
 at points e and/, 
 
 Strains on the Diagonals- To find the greatest strain on the 
 diagonals, it will be necessary to calculate the strains on the 
 flanges of the girder under the various conditions of loading. 
 
42 HOGBACKED GIRDERS. 
 
 The extreme strains on the diagonals of bays Nos. 1 and 6 
 occurs when the girder is fully loaded, as in Diagram 1, in which 
 case the horizontal element of the strain on the diagonal is obviously 
 58*06, as it forms the medium of communicating the horizontal 
 strain of 58'06 in the top flange to the bottom flange at b. 
 
 The greatest strain occurs on bay No. 2, when the bays Nos. 2 
 to 6 of the girder are fully loaded, as shewn in Diagram 2. 
 
 The horizontal strain on the diagonal of this bay is equal to 
 the difference between the horizontal strains on the top flange 
 of bays 1 and 2, or between the strains on the bottom flange of 
 bays 2 and 3. 
 
 This is obvious, for to maintain point g or point k in equi- 
 librium, the opposing horizontal as well as vertical forces upon g 
 must be equal ; for instance, the horizontal thrust upon g by the 
 top flange in bay 2 being 71 '58, it is opposed by 51*61 in bay 
 No. 1, and 71*58 -51*61, or 19*97 must be the horizontal pull of 
 the diagonal to preserve equilibrium. 
 
 Thus, the horizontal strain on a diagonal is equal to the dif- 
 ference between the horizontal strains on the flanges on each side 
 of the point of its junction therewith. 
 
 Having found the horizontal element of the strain on the 
 diagonal, it is easy to calculate therefrom the vertical element; 
 this determines the strain on the vertical : e.g. take point 6, 
 Diagram 1, which is subjected to an upward vertical pull of 29*03 
 from the diagonal, while a load of 18 tends to pull it vertically 
 downwards, to maintain equilibrium therefore we require a force 
 of 29*03 18, or 11*03 compressive strain on the vertical. On 
 the other hand, to keep point d in equilibrium an upward force 
 of 18 13*735 or 4*265 is required in tension on the vertical. 
 
 Diagram 7 contains the maximum strains on diagonals and 
 verticals collated from Diagrams 1 to 6. Diagrams 1 to 6 give the 
 strains on every part of the truss under each condition of loading. 
 The correctness of the figures given may be checked by examining, 
 in turn, each point where the diagonals and verticals meet the 
 flange, and observing whether the conditions of equilibrium are 
 maintained. 
 
 For instance, point a Diagram 1 is acted upon by a horizontal 
 force of 75*79 which is balanced by two horizontal forces 58*06 
 
HOGBACKED GIRDERS. 
 
 43 
 
 arid 17'73, whose sum equals 75*79 : thus equilibrium as regards 
 horizontal motion is proved. Again, the vertical forces tending 
 to force point a downwards are 13'265 and 13*735, their sum = 27. 
 These are balanced by the forces 15'97 and 11'03, sum = 27, 
 tending to force point a upwards : thus equilibrium as regards 
 vertical motion is proved, and the correctness of the figures ascer- 
 tained. 
 
 A further check upon the figures may be obtained by adding 
 together the total vertical effects of flange and diagonal in any 
 bay, the sum of which, be it remembered, should equal the total 
 shearing force on the bay (see p. 16). For example, the vertical 
 force taken by the top flange in bay No. 2, Diagram 3, is 10*3, 
 which being added to 147 the vertical force taken by the diagonal 
 gives a total of 25 ; an examination of the diagram will shew that 
 31 6 or 25, is the shearing force on the bay. 
 
 In bay No. 4, of Diagrams 4 and 5, the vertical effect of the 
 flange must be subtracted from that of the diagonal to get the 
 true shearing force on the bay, since the direction of flange and 
 diagonal with reference to the horizontal is similar, and their signs 
 are opposite. Thus the flange of bay No. 4, Diagram 4, tends to 
 push point I upwards with a force of 2'84, and the diagonal to 
 pull it downwards with the same force, therefore the shearing force 
 on the bay is 0. 
 
 In practice it would not be necessary to find the strains on 
 every part of every bay as is done in Diagrams 1 to 6, but only in 
 those bays which immediately adjoin the point where the live 
 load changes from the dead load only, to dead and live load com- 
 bined. Reference to Diagrams 1 to 6 will shew that it is in 
 these bays that the maximum strains are to be looked for. 
 
 Comparison with a parallel Flanged Girder. For the sake 
 of comparison, the strains on a parallel flanged girder whose depth 
 is the same as that of the hogbacked girder at the centre, and 
 whose loading is similar, are given in fig. 30. 
 
44 HOGBACKED GIRDERS. 
 
 It will be observed that the strains on the flanges of the hog- 
 backed girder are in general the greater, and the strains on the 
 diagonals the lesser of the two. 
 
 All the verticals in the parallel girder are in compression, 
 whereas on the hogbacked some are in tension, and others have 
 to resist both tension and compression. 
 
 Diagonals as Struts. Fig 31 represents a truss of the same 
 dimensions and loading as that illustrated in Plate V., but having 
 the position of the diagonals reversed, they being struts in this 
 case. The same process may be adopted in this case as when 
 tension-bars are used for calculating the strains on the diagonals. 
 
 Diagonals of both kinds used. When it is desirable to make 
 use of diagonals both as struts and ties, or when a web of plate 
 iron is employed, the most convenient plan will be to make the 
 horizontal strains on the top and bottom flanges of each bay alike 
 by taking a mean. 
 
 Fig. 32 represents a truss with equal horizonal strains on the 
 top and bottom flanges of each bay. Examination of the diagram 
 will shew that this necessitates equal horizontal strains on the 
 diagonals in each bay, and that the verticals are in tension. 
 
CHAPTER IV. 
 
 THE BOWSTRING GIRDER 
 
 Is so named from its resemblance to a strung bow. In all the 
 various forms of girder that we have hitherto considered, the 
 thrust of the top flange was conveyed to the bottom flange by 
 means of a web or by struts and ties. The peculiar feature of the 
 bowstring girder is that the thrust of the top flange bears directly 
 against the end of the tie. In fact, the top member may be 
 regarded as an arch, the thrust of which is taken by a horizontal 
 tie instead of by abutments. 
 
 On p. 16, it is shewn that the strains on the flange of a hori- 
 zontal parallel girder at any points are in the same proportion to 
 each other as are the vertical lines drawn from these points to a 
 parabolic arc, the apex of which is over the centre of the girder 
 and whose extremities coincide with those of the flange. Since 
 the strain on the flange of a girder is inversely proportional to the 
 depth of the girder, it follows that by making the depth at every 
 point along the girder proportionate to the vertical ordinate to the 
 parabolic arc, we should obtain a girder in which the horizontal 
 strain on the flanges was equal throughout. In fact, we should 
 obtain a bowstring girder, the form of the bow being that of a 
 parabola, the axis of which bisects the girder at right angles. 
 When such a girder was uniformly loaded at the top the web 
 would have (practically) no duty to perform and might be dis- 
 pensed with, but it is required to resist the effects of unequal 
 loading. 
 
 Curve of Equilibrium. A parabola is the curve of equili- 
 brium for a uniformly distributed load, that is to say, the line of 
 stress passes entirely along the curved flange so that it has no 
 tendency to bend. 
 
46 THE BOWSTRING GIRDER. 
 
 A curve of equilibrium is simply a line of pressure whose direc- 
 tion is curved ; the direction of the pressure at any point in the 
 curve being tangential to the curve at that point. A chain 
 hangs in a curve of equilibrium, forming what is called a catenary 
 (from catena, a chain). If it were so loaded as to have the same 
 weight per horizontal unit of length it would assume a parabolic 
 curve ; by loading it irregularly it might be made to assume an 
 irregular curve. 
 
 Method of drawing the Curve of Equilibrium. As we shall 
 have to make use of the curve of equilibrium presently we must 
 know how to draw it. 
 
 In most cases which come before us we shall find the load con- 
 centrated at intervals, so that our line of pressure will be poly- 
 gonal instead of curved. We may however call the curved line 
 which would pass through the angles of the polygon our curve of 
 equilibrium. 
 
 Regular Loading. The simplest case will be that of a number 
 of equal weights, horizontally equidistant, which have to be sup- 
 ported by a polygonal arched framing hinged at the joints. Let 
 it be required to draw such a framing. 
 
 Let the span of the arch be divided up into six equal hori- 
 
 zontal bays, and at the junction of each bay let a load W act 
 vertically. (See fig. 33.) 
 
 Then the load will be symmetrically distributed over the arch, 
 one of the loads W being at the apex. Of this load half will go 
 to each pier. Let ab and ab' be two beams meeting at a, making 
 the same angle with the horizontal and supporting the load W. 
 
 W 
 
 Each of these beams will support a vertical load of -- , which may 
 
 be represented by the vertical line be or b'c drawn from the point 
 
THE BOWSTRING GIRDER. 47 
 
 b or H to meet the horizontal line drawn through point a in c or c'. 
 The lines ac and ac will represent the horizontal thrusts on the 
 beams ab and ab', which are equal by construction, so that point a, 
 is in equilibrium. 
 
 At point b (dealing with one side of the polygon only, for the 
 two sides must obviously be alike) an additional vertical force 
 of W comes into play; thus the beam Id will have to carry a 
 
 W 
 
 vertical load of W+ -~- , or three times that on ab, and to re- 
 sist the horizontal thrust of the beam ab. Therefore to find the 
 position for the beam bd, draw from point b the horizontal line be, 
 equal to ac ; from point e let fall a perpendicular ed three times 
 the length of the line be ; join bd: bd is the correct position for the 
 beam. 
 
 Similarly, as the beam df has to carry a vertical load of 
 
 W 
 W+ W+-=- , the vertical gf must be made five times the length 
 
 of cb, gd being made equal to ac. The points d' and f on the 
 other side of the polygon being found in the same manner, a 
 curved line passing through the points fdbab'd'f would be 
 called the curve of equilibrium. It is a parabola in this case, and 
 would represent the line of pressure in an arch whose versed sine 
 was ah caused by an evenly distributed load. 
 
 To draw the Curve of Equilibrium with a given versine. 
 
 It is obvious that the curve fdbab'd'f' is not the only form of 
 arched polygonal framing that would support the weights W. 
 
 The angle at which the beams ab, ab' are inclined with the 
 horizontal is arbitrary. The lines cb, ed and gf are in the proportions 
 of the numbers 1, 3 and 5 respectively, and provided these pro- 
 portions are preserved the lines may be of any length. That is to 
 say, the versine of the curve may be made of any required dimen- 
 sion. The line ah representing the versine of the arch is equal to 
 the sum of the lines cb, ed and gf, or = 1 + 3 + 5 = 9. We have 
 only therefore to divide the versine in the proportion of 1, 3 and 5, 
 as shewn in fig. 33, and to draw horizontal lines from the points so 
 obtained to the corresponding vertical lines of action of the load 
 and trace the curve through these points. 
 
48 THE BOWSTRING GIRDER. 
 
 It will be observed that when one of the loads comes at the 
 centre of the span, that is, when the number of bays is even, the 
 versine or rise of the arch, as we may in fact call it, will have to 
 be divided up into lengths which are terms of an arithmetical 
 series 1 + 3 + 5 + 7 + &c. according to the number of bays. Also 
 
 Fiy.34: 
 
 that when the number of bays is odd the versine will be repre- 
 sented by the sum of a series 2 + 4 + 6 + &c. (see fig. 34), and in 
 each case the last term of the series is one less than the number of 
 ~bays in the span. 
 
 If n = the number of bays in the span, 
 
 2 
 = the sum of the series for a girder with an even number of bays, 
 
 and 
 
 uneven 
 
 The most convenient practical way to set out the curve is as 
 follows. Take with the compasses from any convenient scale a 
 length representing the sum of the series answering to the versine 
 of the arch, and with the crown of the arch as a centre and this 
 length as radius, describe an arc cutting the chord of the arch 
 (see fig. 34) : join this point of intersection with the centre from 
 which the arc was struck, and mark off along this line divisions cor- 
 responding to the several terms of the series, commencing with the 
 first term at the crown of the arch. Through the points so 
 obtained draw horizontal lines to meet their corresponding verti- 
 cals ; the points of meeting will be in the curve of equilibrium. 
 
 Irregular loading, bays equal. It is obvious that the curve 
 will be of the same form whether W the load per bay be great 
 or small, but if the value of W be not the same for every bay the 
 curve will be altered accordingly. 
 
THE BOWSTRING GIRDER. 
 
 49 
 
 To draw the curve of equilibrium when the loads on the 
 bays are irregular, the first step is to ascertain the total load which 
 
 goes to each abutment. Fig. 35 represents the irregularly loaded 
 framing: in this case the load on the left-hand abutment is 19 and 
 on the right-hand abutment 20, therefore we must take 4 of the 
 load of 6 at the crown to the left and 2 to the right hand. 
 
 The vertical loads supported by the beams 
 on the fe/fc-hand side will be respectively 4, 4 + 3 and 4 + 3 + 12, 
 
 right 2, 2 + 6 and 2 + 6 + 12, 
 
 that is 4, 7 and 19, sum = 30, 
 
 and 2, 8 and 20, =; 30. 
 
 Taking the point at the crown where the load 6 divides as 
 centre, describe an arc with a radius = 30 cutting the chord of the 
 arch in two places ; join each of these points of intersection with 
 the centre, mark off divisions equal to 4, 7 and 19 on the left-hand 
 line and divisions equal to 2, 8 and 20 on the right-hand line. 
 Draw horizontal lines through the points of division to meet their 
 respective verticals on either hand, as shewn in the figure; the 
 points so obtained are the hinges of the framing, the beams of 
 which coincide with the lines of thrust. 
 
 Irregular Loading, Bays unequal. It will be observed that 
 the sums of the two irregular series are equal, being 30 for both 
 sides. This corresponds to the equality of moment at the centre 
 of the forces which go to each abutment, as previously pointed out 
 in treating of irregularly loaded parallel girders (see p. 34). This 
 correspondence will afford a convenient means of setting out the 
 polygon of equilibrium when the bays are of unequal lengths. 
 Y. 4 
 
 
50 
 
 THE BOWSTRING GIRDER. 
 
 Fig. 36 represents a polygonal framing in which the bays and 
 loads are both unequal. 
 
 The moments of the loads which go to the abutments are 
 as follows. 
 
 Left-hand abutment, 
 loads sums leverage moment 
 84-18 = 31 x 1 = 31 
 5 + 8 =13 x 3 = 39 
 5 x 2 = 10 
 
 80 
 
 Right-hand abutment, 
 loads sums leverage moment 
 
 7+6 + 10 = 23 x 2 = 46 
 
 7 + 6=13x1 = 13 
 
 7 x 3 = 21 
 
 80 
 
 That the above table of moments may be seen at a glance to 
 be correct, we give in fig. 37 a diagram of a parallel girder loaded 
 
 similarly to the framing, the additional horizontal moment on each 
 several bay agreeing with that given in the tables. Now the 
 Jwrizontal strain throughout our polygonal framing must be the 
 same for every beam ; consequently the depth of the arch or 
 vertical distance from any hinge to the chord must vary directly 
 as the strain on the flange of the parallel girder at the point 
 corresponding to the hinge. Therefore taking the versine of the 
 arch as equal to 80, the heights of the first and second hinges to 
 the left above the chord will be 31 and 70 respectively, on the 
 right they will be 46 and 59. 
 
 By describing an arc with a radius = 80, having its centre at 
 the hinge where the load divides, and cutting the chord in two 
 
THE BOWSTRING GIRDER. 51 
 
 places, and proceeding in a manner similar to that described on 
 p. 49 for an irregularly loaded framing with equal bays, the 
 positions of the hinges may be found. 
 
 Line of greatest resistance. Though in the preceding exam- 
 ples the quasi curve obtained was the true curve of equilibrium for 
 the conditions given, since it must of necessity pass through the 
 centre of each hinge-joint to preserve equilibrium, yet it does not 
 follow that a curve of equilibrium drawn within an arch or curved 
 rib will be the curve of equilibrium for that case. The line of 
 pressure must always be the line of greatest resistance, its direction 
 will therefore greatly depend upon the material as well as the form 
 of the arch or curved rib. The true curve of equilibrium in an 
 arch is the line of pressure which strains the arch least, so that if 
 we draw the curve of equilibrium for an arch, and can shew that, 
 assuming the pressure to pass along this curve, the arch is not 
 overstrained, we may consider the arch as safe. 
 
 Methods of resisting Distortion. When a bowstring girder 
 with a parabolic top flange is loaded evenly all over, the curve of 
 equilibrium passes along the centre of the flange, and there is no 
 tendency to distortion ; but when the girder is loaded on one side 
 only, the curve passes outside the bow on the one side of the span 
 and inside on the other, if the flange be narrow, or close to the 
 outer edge on one side and inner on the other if the bow be 
 of considerable depth, as when it forms a curved box beam. 
 
 The distorting action on the bow may be resisted, 
 
 1. By making the bow sufficiently deep to include within 
 its boundaries the curve of equilibrium under every condition of 
 loading, it being wholly in compression. 
 
 2; By the transverse strength of the bow. 
 
 3. By the transverse strength of the tie. 
 
 4. By the transverse strength of the bow and tie com- 
 bined. 
 
 5. By means of a web or diagonals. 
 
 First Method. Fig. 38 illustrates the first method. The 
 dotted line touching the boundaries of the top flange or bow at 
 a, b and c is the curve of equilibrium under one condition of 
 
 42 
 
 
52 
 
 THE BOWSTRING GIRDER. 
 
 unequal loading, say one half span fully loaded. (In describing 
 the curve of equilibrium for this and the following cases we must 
 
 Fiy.38. 
 
 loaded fytlf ^ 
 
 endeavour to draw it so that its maximum or minimum distance 
 from the outer and inner edges of the bow is the same, as this is 
 very near the truth, and gives the most economical results. It 
 will be almost impossible in drawing the curve to hit off on the 
 first occasion its true position ; but by lengthening or shortening 
 the ordinates to the curve in the same proportion, we can raise or 
 depress the curve until its position is approximately correct.) 
 In Plate VI. the strain on the top flange of a certain truss fully 
 loaded is given as 81 at the centre ; when the same truss is only 
 half loaded, the strain at the centre is given as 6075. Supposing 
 our bowstring girder to have the same span, depth at centre, and 
 conditions of loading as the truss in Plate VI., there would be a 
 horizontal strain of 6075 at points a, b and c. By changing the 
 position of the load we can cause the curve of equilibrium to pass 
 very near to almost every point in the outer and inner edges of the 
 bow, and at all of these points the strain would have a horizontal 
 element of 6075, or thereabouts. Thus it will be seen that metal 
 will have to be provided to take a horizontal thrust of 6075 
 throughout the outer arid inner edges of the bow, or 121*5 in all, 
 whereas had the curve of equilibrium passed along the centre of 
 the flange, a strain of but 81 would have to be provided for. The 
 curve of equilibrium may however fall considerably within the 
 
 edge of the top flange ; in such case the compressive force on either 
 edge will be inversely proportional to its distance from the line of 
 
THE BOWSTRING GIRDER. 
 
 53 
 
 pressure, the sum of the horizontal strains on the two edges being 
 equal to the thrust on the tie ; for example, 
 
 S being the total force acting along the dotted line at I, fig. 39, 
 
 be/ 
 
 Strain at a = S, 
 ac 
 
 ab 
 
 Second method. In fig. 40, the curve of equilibrium shewn 
 by the dotted line passes outside the limits of the top flange 
 
 Fig. 40. 
 
 in two places, the extreme distances being ab in the one case 
 and de in the other. With the curve in this position, the tendency 
 of the bow is to flatten at b and bulge outwards at e. 
 
 S being the pressure acting along the dotted line, 
 Strain on inside edge at c = j~ S tension, b being the fulcrum, 
 
 outside 
 
 inside 
 
 outside 
 
 ac 
 
 b = 7- 8 compression, c 
 
 7/> 
 
 e = --j S tension, / 
 e J 
 
 fj,S compression, e 
 
 It must be remembered that in this and in the last example, 
 S is not a uniform force, it will vary with the angle of inclination 
 of the curve of equilibrium with the horizontal. The horizontal 
 element of 8 is uniform. 
 
 Third method. Fig. 41 represents a bowstring girder in 
 which the bow is quite flexible, that is hinged at the points where 
 
THE BOWSTRING GIRDER. 
 
 it joins the verticals. The tie affords the means of resisting 
 distortion. When the left-hand side of the span is loaded, the 
 
 F iff. 41 
 
 """ 
 
 arch tends to flatten at a and bulge outwards at b, causing the 
 horizontal tie to bend downwards between d and c, and upwards 
 between c and e. It assumes in fact the form of an S curve as 
 shewn by the dotted line d, c, e, the point of contrary flexure being 
 at or very near to the centre of the span. 
 
 To prevent the top flange from buckling it is evident that the 
 bottom flange must have sufficient transverse strength to resist 
 the bending action to which it is subject. Let us endeavour to 
 ascertain what this should be. 
 
 Let us suppose the left-hand half of the span to be loaded, 
 and the load to be suspended from the bottom tie, this will 
 throw all the verticals into tension. Now if the bottom tie were 
 flexible and the bow stiff as in the immediately preceding cases, 
 the strain on the verticals on the left would be P= dead + live 
 load, while on those to the right it would be p = dead load. But 
 as our parabolic arc is perfectly flexible and is the curve of equili- 
 brium for an evenly distributed load, it is evident that in order to 
 preserve its form the strain on each vertical must be the 'same 
 
 throughout the span, it will therefore be 
 
 But since 
 
 a load of P is attached to the bottom flange underneath each 
 vertical on the left-hand half of the span of which an amount 
 
 P + v 
 ^ 
 
 is carried by the vertical, the remainder 
 
 must be supported by the transverse strength of the tie offering 
 resistance to a vertical downward force. Again, the upward pull 
 
THE BOWSTRING GIRDER. 
 
 of ^ on the verticals of the left-hand side of the span is 
 
 met by a load of p only, therefore that part of the tie is subject to 
 a vertical upward force of 
 
 P 
 
 Now P- p the live load on each bay, therefore the loaded 
 half dc of the tie is subject to a load = half the live load per bay 
 acting downwards, and the other half of the tie ce to the same 
 force acting upwards. When the position of the load is reversed 
 the directions of the transverse strains on these half-lengths of 
 horizontal tie will also be reversed, so that each half of the tie 
 must be made capable of resisting an upward and downward 
 transverse force equal to one half the live Iqad on half the span, 
 or one quarter the whole live load on the span 1 . The tie may be 
 hinged at c the centre, if necessary, without impairing its effect in 
 resisting distortion. Each half of the tie must be treated as a 
 distinct girder of a length equal to half the whole span of the 
 bridge. Since the tie is in tension throughout all its cross-section, 
 the transverse strain, unless very great indeed, will not put any 
 part of the tie into compression. Thus if the horizontal strain on 
 the tie, when the bridge is loaded over one half of its length only, 
 be 100 or 50 on each flange, neg- 
 lecting web (see fig. 42), and the 
 strains due to transverse downward , ..... _ 
 force be 10 + in the top flange and \P 
 10 in the bottom flange, the addi- 
 
 tion of these strains will give 40 for the top flange and 60 for 
 the bottom: when the transverse force acts upwards the strains will 
 be 60 in the top flange and 40 in the bottom flange. It would 
 therefore be necessary to insure rigidity to increase the sectional 
 area of both^ flanges at the centre of each half span. The extra 
 metal may be diminished to nothing at the centre and ends of the 
 span of the arch. 
 
 1 The "live load" here mentioned means that which comes upon each girder, 
 
 which will of course be - the whole load on the span if there are only two trusses. 
 
 X 
 
56 THE BOWSTRING GIRDER. 
 
 Fourth Method. Bowstring girders are often made with a 
 deep trough-like bow and tie ; by utilizing the transverse stiffness 
 of both these, resistance to distortion can be obtained. 
 
 Let us assume that it is desired to divide the work equally 
 between the top and bottom members of the girder. This as 
 regards the tie is equivalent to reducing the live load by one half 
 in the preceding case, and proceeding as before. For the bow, we 
 must draw the curve of equilibrium for one side only loaded, but 
 the total load at each vertical of the loaded side must be taken as 
 
 = dead load per bay -\ ~ . Having correctly drawn 
 
 L 
 
 the curve the strain on the flanges of the bow can easily be deter- 
 mined, as explained on p. 53. 
 
 Fifth Method. The general method of providing against the 
 distorting effect of unequal loading is by means of a web or cross- 
 bracing between the bow and the tie. 
 
 Plate VI. contains a series of Diagrams of the strains on every 
 part of a bowstring girder under unequal loading. It is neces- 
 sary to find the strains on the flanges first and thence to deduce 
 the strains on the diagonals, as in the case of the hogbacked girder. 
 The mode of procedure is precisely the same as that adopted for 
 the hogbacked girder ; see p, 41. The depth at centre, number 
 and width of bays and loading are the same as those taken for the 
 hogbacked girder, Plate V., so that the strains on the two may be 
 compared. 
 
 Effect of elasticity in distributing the Load among the 
 Diagonals. As we have the bow of our girder in the form of a 
 parabolic curve, when the load is uniformly distributed the curve 
 of equilibrium will pass along the centre line of the top flange, the 
 strain on the tie will be the same throughout its whole length, and 
 there is no need of diagonals to prevent distortion. Diagram, 
 No. 1, Plate VI. represents the girder under this condition of 
 things. In this Diagram the diagonals are wholly omitted, and the 
 load is assumed to be carried by the vertical rods only. In reality 
 the strains are not such as are here given, because in consequence 
 of the stretching of the verticals a strain comes upon the diagonals. 
 
THE BOWSTRING GIRDER. 57 
 
 The accompanying figure represents a pair of diagonals inclined at 
 
 an angle of 45 with the vertical and a ver- 
 
 tical at whose point of meeting a a load is 
 
 suddenly applied. The effect is to bring 
 
 the point a down to b, stretching the ver- 
 
 tical to the extent ab and bringing the 
 
 diagonals into the position shewn by the 
 
 dotted lines. Taking the upper extremity 
 
 of one of the diagonals as a centre, and describing an arc through 
 
 the point a cutting the dotted line in c, it becomes evident that be 
 
 represents the amount of stretching of the diagonal. Now since 
 
 the distance ab is very small, the triangle acb is practically a right- 
 
 angled triangle in which ab is the hypothenuse and ac and cb are 
 
 equal sides. And the line ab is to the line be as V2 (or 1*41 
 
 nearly) is to 1. 
 
 Let a& = -3-rrrth of the length of the vertical before it was 
 loaded, the proportion being expressed by the fraction 
 
 100 _ length of vertical 
 1 amount of its extension ' 
 
 Now, diagonal : vertical :: V2 : 1, and the amount of the 
 diagonal's extension expressed in terms of ab 
 
 h l 
 
 ~ ab ~ V2 ' 
 
 length of diagonal _ 100 x V2 
 amount of its extension 1 
 
 Vl 
 
 V2 x V2 x 100 200 
 
 From this it appears that if each of the diagonals and the 
 vertical be of the same sectional area, the sum of the strains on 
 the two diagonals will be as nearly as possible equal to that on the 
 vertical. 
 
 In this case the strain on the diagonals is 1'41 times the load 
 supported by them, therefore if 
 
58 THE BOWSTRING GIRDER. 
 
 x proportion of load taken by vertical, 
 y = diagonals, 
 x -f y = W total load carried ; 
 but x = 1'41 y ; 
 
 /. 2-41 y = W, 
 W 
 
 If W=I, 2/ = -4149 and x = '5851. 
 
 These values of x and y are of course only true when the 
 diagonals are inclined at an angle of 45 with the vertical and the 
 sectional area of each diagonal is equal to that of the vertical. 
 To find the proportions of load taken by diagonals and vertical or 
 any system of bars, the general rule is : Find the proportion of 
 extension to its length which each bar experiences, assuming the 
 load to descend vertically through a given space. The strain per 
 square inch thus placed upon the bar multiplied by its area in 
 square inches gives the actual stra : n upon that bar. Finally, ascer- 
 tain the vertical element of the strain so obtained. This being 
 done for each bar the share of the load carried by each will be in 
 proportion to the vertical strain upon it. 
 
 For example. The three bars ab, ac, ad (fig. 44) support 
 a vertical load of 100 between them : 
 parts 
 
 4 x 100 
 ab supporting 4 = ^ = 20 actual load carried. -tfgM*. 
 
 * 
 
 7 - 
 
 9 x 100 
 
 e\r\ 
 
 7x100 
 
 Total 20 100 
 
 Tn dealing with the diagonals of a truss it must be remem- 
 bered that the extension of the bars may be effected by the yield- 
 ing of their supports at the upper extremities. Thus if b, c, d, 
 fig. 44, represent the top flange of a truss, its compression under 
 
THE BOWSTRING GIRDER. 59 
 
 load would cause the points b, cand dto approach one another, and 
 thus affect the lengths of the bars ab, ac, ad. Further, if the 
 points b, c and d were supported as to vertical motion by struts, 
 their relative depressions would depend upon the strength of the 
 struts which supported them, and again upon the strength of the 
 ties which supported these struts, and so on ad infinitum. 
 
 Thus it is found practically impossible to tell the exact strain 
 upon a number of bars in a truss inclined at different angles and 
 meeting at a point where a load is applied. 
 
 The usual practice is to calculate the strain on every bar under 
 the most unfavourable condition and to neglect the easing effect 
 upon it of other bars, the strain upon which may be much or little, 
 but is uncertain. 
 
 The strains on Diagrams 1 to 6, Plate VI., are strictly correct for 
 trusses of the form represented in those Diagrams, but they are 
 not true as regards the verticals for one similar to Diagram 7, in 
 which the diagonals cross one another. However, by taking the 
 strains as given in Diagrams 1 to 6 as true for a truss of the form 
 of Diagram No. 7, we shall be safe. 
 
 The reader will observe this peculiarity in the Diagrams for 
 the bowstring girder, that the live load being removed from one 
 end of the girder the diagonals which come into play slope all in 
 the same direction. As a natural consequence the horizontal strain 
 on the top members increases in each bay till it reaches its maxi- 
 mum in the last bay at the loaded end of the girder. If the 
 diagonals were struts their slope would of course be reversed. The 
 diagonals throughout may however be made capable of acting 
 either as struts or ties. In such case the extreme strain on each 
 single diagonal given in Diagram 7 would have to be divided 
 between the two diagonals of the bay. This is analogous to the 
 effect of a solid web capable of resisting compression and extension ; 
 see page 24. 
 
 Figs. 45 and 46 shew two other alternative methods of arrang- 
 
60 
 
 THE BOWSTRING GIRDER. 
 
 ing the diagonals which should be made capable of resisting both 
 extension and compression. For the purpose of shewing this the 
 
 strains are given for each system, the girders being taken as half- 
 loaded. 
 
 Although not to be commended for appearance, trusses with 
 single double-acting diagonals are cheap, and have this advantage, 
 that the strains can be predicted with certainty. 
 
 Load resting on the top. The effect of resting the load 
 upon the top flange instead of suspending it from the bottom is to 
 
 alter the strains upon the verticals only, they also becoming 
 struts. Fig. 47 corresponds to Diagram No. 2, Plate VI, but gives 
 the strains when the load rests upon the top of the girder. With 
 this example before him, the student should be able to construct 
 without difficulty the Diagrams for the other positions of the load. 
 The curve of the bow is often made an arc of a. circle for 
 appearance sake and convenience of manufacture. 
 
CHAPTER V. 
 
 THE ARCH. 
 
 THE preceding Chapter contains much that we shall have to 
 refer to in treating of the Arch in consequence of its affinity to the 
 bowstring girder as indicated in the opening paragraph of that 
 chapter. 
 
 Method of finding the Thrust at Crown. The horizontal 
 thrust on the crown of an arch may be found in the following 
 manner. 
 
 Consider the arch with its spandrils, fiy.4.8. 
 
 roadway, and live load, as forming two 
 rigid masses divided vertically at c the 
 crown of the arch, and hinged at a the 
 the centre of the skew-back, (see fig. 
 48). Find the centre of gravity of the mass , b, c, being one-half 
 of the whole arch. Draw a vertical line through this point to 
 
 W 
 
 represent the line of action of the weight which we will call . 
 
 This acts at an arm of z, about the fulcrum a. Draw a hori- 
 zontal line cb through the centre c of the arch to represent the 
 line of action of the thrust. Through a, draw a\ perpendicular 
 to cb] the line ab represents the leverage at which the thrust acts 
 
 W 
 
 about point a, and as this thrust is the force that balances _ ; 
 
 W 
 
 .*. S x ab = -- x z, if 8 = thrust, 
 
 Curve of equilibrium for an Arch of Masonry. An arch of 
 masonry is stable when the curve of equilibrium lies well within 
 the arch under all conditions of loading. The mode of drawing 
 
62 THE ARCH. 
 
 the curve of equilibrium has been fully described in the preceding 
 chapter. An example of the method of procedure in the case of 
 an arch of masonry is given in fig. 49. The arch should be 
 
 divided up into equal horizontal segments ; the weight of each 
 portion should be calculated, and the mean weight of two ad- 
 joining segments considered as acting in the direction of the 
 vertical line which divides them. To the dead load must be 
 added the live load when necessary. For the loads so obtained 
 a curve of equilibrium must be drawn through those points in the 
 crown and in the haunch of the arch which appear likely to give 
 the best results, which are obtained when the distance of the 
 curve from the outer or inner face of the arch is a maximum. 
 The available area of masonry for resisting thrust is equal in width 
 to twice the minimum distance of the line of pressure from the 
 extrados or intrados of the arch. If the points through which the 
 curve has been drawn have been injudiciously chosen, the position 
 of the curve can be changed by altering all the ordinates in the 
 same proportion. 
 
 Other methods of resisting distortion of the Arch* In the 
 
 case of metal ribs, the resistance to distortion may be effected as 
 in the cases of the bow-string girder by the five methods men- 
 tioned on p. 51, substituting the word rib for bow, horizontal girder 
 for tie, and adding in the spandrils after the word diagonals to 
 the description of the 5th method. The first four methods are 
 exactly analogous for the bowstring girder and the arched rib, so 
 that it is unnecessary to say more about them, but the method of 
 finding the strains on the diagonals in the spandril of an arch 
 differs from that used for the bowstring girder. 
 
 Strains on the Spandrils. Diagrams 1 to 6, Plate VII., shew 
 the strains caused upon the various parts of a parabolic arched rib, 
 
THE ARCH. 63 
 
 whose spandrils are braced with a single diagonal, under various 
 conditions of loading. Diagram 7 gives the greatest strains col- 
 lected from Diagrams 1 6. The form of the arch, and conditions 
 of loading, are the same as those of the bowstring truss illustrated 
 in Plate VI. In Diagram No. 1 are given the strains caused by 
 the full load. The arch being parabolic, there is no strain on the 
 diagonals under this condition of loading. To find the strains 
 when the load is unequally distributed, and the diagonals are 
 placed as in Plate VII., the most convenient method is to com- 
 mence at the springing of the arch, and work towards the centre. 
 
 We will now deal with Diagram, No. 2, Plate VII., com- 
 mencing at the left-hand corner. 
 
 In the first place, we find the vertical load on the abutments 
 which amount to 40 for the left-hand, and 44 for the right-hand 
 abutment (see p. 33). In the next place, we find the thrust of 
 the arch on the principle mentioned on p. 61. Thus 
 
 load leverage moment 
 
 12 x 1 = 12 
 
 18 x 2 = 36 
 10 x 3 = 30 
 
 78 horizontal thrust. 
 
 Of course the thrust on the right-hand abutment is the same. 
 
 Thus 
 
 load leverage moment 
 
 18 x 1 = 18 
 18 x 2 = 36 
 
 8 x 3 = 24 
 
 78 horizontal thrust. 
 
 Now the segment of the arch forming the lower member of 
 bay No. 1 being inclined at the rate of 5 vertical to 9 horizontal, 
 the vertical pressure which it exerts is to its horizontal as 5 is to 9, 
 
 .-. 78 x I =43-33, 
 the vertical pressure. 
 
64 THE AECH. 
 
 But since there can only be a vertical pressure of 40 on the 
 abutment, the vertical bar which meets the rib at the springing 
 must exert an upward pull of 43'33 - 40 = 3*33 tension. 
 
 From this we derive a vertical thrust of 3'33 on the diagonal 
 of bay No. 1, which answers to a horizontal of 7*5, which again 
 requires a strain of 7 '5 in tension on the top member of bay 
 No. 1. To maintain equilibrium, there must be a horizontal 
 
 thrust of 78 + 7'5 = 85'5 by the rib in bay No. 2. The vertical 
 2 
 
 element being - of this = 28*5. 
 y 
 
 The rib in bay No. 1 is pressing point a upwards with a force 
 of 43'33, while the rib in bay No. 2 and the diagonal in bay No. 1 
 are pressing it downwards with a force of 28'5 + 3'33 = 31'83, 
 
 therefore 43'33 - 31'83 = 11'5 
 
 is the pressure which the vertical bar must exert to maintain 
 equilibrium. Therefore of the load of 12 which is supported at 
 the upper extremity of this vertical *5 will be carried by the 
 diagonal of bay No. 2. This bar will exert a horizontal thrust 
 of 4 '5, which will cause the total horizontal strain on the top 
 member in bay No. 2 to be 12 ; this strain will be carried 
 through to bay No. 3. 
 
 Now 85'5 + 4'5 = 90, the combined horizontal thrust of rib 
 and diagonal in bay No. 2, which must be met by a thrust of 90 
 on the rib in bay No. 3. The vertical element of the thrust on 
 
 this part of the rib will be 90 x - = 10, and 10 is the proportion of 
 
 v/ 
 
 the load of 18 at the centre which goes to the left. 
 
 Again, the point b is pressed upwards with a force of 28'5 by 
 the rib in bay No. 2 and downwards with a force of '5 by the 
 diagonal of bay No. 2, of 10 by the rib in bay No. 3, and of 18 
 by the vertical bar dividing the bays ; the total downward force 
 being 28'5. 
 
 Further, 90 12 = 78 the thrust at the crown, which agrees 
 with that at the haunches. We find therefore, that the conditions 
 of equilibrium are satisfied throughout, and we may conclude that 
 our strains are correct. 
 
THE ARCH. 65 
 
 The strains on the right-hand side of the arch may be obtained 
 by a similar mode of proceeding. In this part of the arch the 
 diagonals of the spandril are in tension. 
 
 For the arrangement of diagonals given in Plate VIII., the 
 most convenient way of calculating the strains is to commence 
 from the centre of the arch. 
 
 For the arrangement of loading given in Diagram No. 2, 
 Plate VIII., we know that of the load 18 resting at the crown 10 
 goes to the left, consequently there must be a vertical load of 10 
 on the rib in bay No. 3 ; this will produce a horizontal thrust 
 of 90 upon it, which will produce a horizontal thrust also of 90 
 on the rib in bay No. 2 of which the vertical thrust will be 30, 
 necessitating a downward pressure of 20 by the vertical at point b. 
 To produce this, the diagonal in bay No. 2 must exert a down- 
 ward pull of 2, and therefore a horizontal pull of 4*5. Thence we 
 determine the total horizontal thrust on point a to be 90 4*5 
 = 85*5, which must be the thrust on the rib in bay No. 1. The 
 vertical pressure exerted by this part of the rib will be 47'5, 
 necessitating an upward pull of 7*5 by the diagonal in bay No. 1, 
 to reduce the total load on the abutment to 40. This diagonal 
 will exert a horizontal pull of 7*5 also on the top member of bay 
 No. 2, and cause the total load on the vertical at a to be 19 '5. 
 
 The diagonal in bay No. 2 causes the downward pressure of 
 the vertical at b to amount to 20, and adds 4*5 of horizontal 
 strain to the top member, making a total of 12 tension there- 
 upon. 
 
 Comparing the corresponding diagrams of Plates VII. and VIII., 
 we find that the horizontal strains on the diagonals are the same 
 for corresponding bays under similar conditions of loading. 
 
 In Diagram No. 8, Plate VIII., are given the greatest hori- 
 zontal strains on diagonals, rib, and top member, when the dia- 
 gonals forming the spandrils are crossed. Diagram No. 8 is the 
 sum of Diagrams 7 in Plates VII. and VIII. It also gives the 
 greatest strain on the verticals. The load is 36 per bay for this 
 diagram; for a load of 18 it would be necessary to halve the 
 amounts. 
 
 We observe, therefore, that in the spandrils of arches con- 
 structed on the system of Diagram 8, Plate VIII., the horizontal 
 Y. 5 
 
66 THE ARCH. 
 
 strain on each diagonal of ,a bay is the same. Wherefore it is 
 sufficient, so far as the diagonals are concerned, to calculate the 
 strains for one system only. 
 
 The two following facts are to be noted with regard to the 
 Diagrams on Plates VII. and VIII. 
 
 1st. If in each bay of a Diagram the horizontal strains on 
 the diagonal, top member, and rib, be added together, the 
 cancelling the +, the remaining + will be the same for each bay, 
 and will equal the thrust of the arch at the crown. 
 
 2nd. The sum of the vertical strains on diagonal and rib in 
 any bay, the cancelling the +, is equal to the shearing force on 
 that bay. 
 
 In illustration of the first statement, add together the hori- 
 zontal strains on the severals bay of Diagram No. 2, Plate VII., 
 and the resulting strain will be found to be + 78h. 
 
 Again, adding the vertical effect on each bay, we get 
 
 for bay No. 1 40 
 
 ,, 2 28 
 
 3 10 
 
 4 8 
 
 ,, 5 26 
 
 ,, 6 44. 
 
 These amounts are the correct shearing forces for their several 
 bays. 
 
 The extreme strain which these diagrams shew on the rib in 
 bays 3 and 4, would not arise in actual practice, for the horizontal 
 girder and rib are united together for some distance on each side 
 of the centre of the arch in most arched bridges ; this being the 
 case, the tension on the horizontal girder neutralizes some of the 
 compression on the rib. If, however, the horizontal girder be at 
 a higher level than the crown of the arch, the variation of strains 
 would not be so great as it appears in the case of which we have 
 just treated. 
 
 Other forms of Spandril filling. Figs. 50, 51, and 52, repre- 
 sent three other regular forms of spandril filling ; for the first of 
 these systems the strains may be obtained by proceeding as in the 
 
THE ARCH. 
 
 67 
 
 last example, making a separate series of diagrams for each system 
 of diagonals. In figs. 51 and 52 the resistance to distortion is 
 effected by the transverse stiffness of the rib and horizontal girder, 
 (see p. 62). 
 
 Fiy.51. 
 
 . so 
 
 Fig. ss. 
 
 The inclined position of the spandril bars in fig. 52 will cause 
 the curve of equilibrium for the strains on the rib to be an ap- 
 proximation to an arc of a circle when the bridge is fully loaded ; 
 indeed the angle of the bars might be so adjusted that the curve 
 of equilibrium should be exactly an arc of a circle under the full 
 load, and correspond with the centre of the rib. The effect of 
 unequal loading in a bridge of this description on the horizontal 
 girder would be to introduce an additional horizontal stretching 
 or compressing besides transverse strain. 
 
 Arched bridges of multiple span. The preceding pages treat 
 of the strains which come upon an arch of a single span. We have 
 now to consider the strains which arise from unequal loading 
 on an arched bridge of many spans. 
 
 When the piers of an arched viaduct are short, it may so 
 happen that the system of arching may be stable without the 
 aid of the spandrils. 
 
 In fig. 53 are shewn two arches of a viaduct, one of which 
 only is loaded. The curved dotted lines drawn within the 
 
 52 
 
68 THE ARCH. 
 
 limits of the arch represent the curves of equilibrium for each 
 arch. From the point of their intersection a, lines ab, ac are drawn 
 tangential to the curves, and representing to scale the strain on 
 the arches, ab being that on the loaded arch, and ac that on the 
 unloaded arch. The parallelogram being completed, the diagonal 
 ad represents the resulting line of pressure. In this example the 
 line of thrust falls well within the base of the pier, so that the 
 arching may be pronounced stable. The dotted prolongation of 
 the pier and line of thrust below the ground-line, is intended to 
 shew that if the pier were of the height shewn by the dotted 
 lines, the line of thrust would fall outside the base and the arching 
 be unstable. 
 
 Stability of Pier. The dead weight of the pier contributes 
 to the stability of a system of arches, as fig. 54 will serve to shew. 
 Let ab be the resultant thrust of the two arches, o the 
 centre of gravity of the pier; through o draw a perpen- 
 dicular to intersect the line of thrust in a, take ac = the 
 weight of the pier, complete the parallelogram abed, 
 join ad ; ad is the resulting line of thrust which now falls 
 within the base. 
 
 fzg.se. 
 
 II J 11 
 
 Effect of horizontal Girder. Figs. 55 and 56 are intended 
 to shew the action of a horizontal girder in preserving the stability 
 of an arch. In fig. 55, there being no horizontal girder, the weight 
 of the pier is the sole source of stability ; this being overcome, the 
 upper ends of the piers a and b are forced apart. If these are 
 tied together by means of a horizontal girder the arch can only 
 yield by the pier breaking as at c, fig. 56. 
 
 Transverse strain on Pier. In estimating the transverse 
 strain upon the pier at c, find the surplus horizontal thrust at c 
 arising from the loaded span, consider the whole pier as a lever 
 fastened at its two extremities and acted upon by a transverse 
 
ffiy.57. 
 
 THE ARCH. 
 
 force at c, the breaking tendency can then easily be 
 
 find the value of the surplus horizontal force, resolve 
 
 the thrusts of the two adjacent arches into their 
 
 horizontal and vertical elements, consider the vertical 
 
 effect of the thrusts as acting downwards through the 
 
 point c (see fig. 57), the neutral axis of the pier at the 
 
 springing, add to this the weight of that part of the 
 
 pier above c, the point of rupture, and call the whole 
 
 weight W. Again, call the weight of the lower part 
 
 of the pier acting through its centre of gravity o, P. 
 
 Choose a point in the base of the pier as a fulcrum or turning 
 
 point. 
 
 Let x be the leverage of P, y the leverage of TF, and z the 
 distance from base of pier to point of rupture. 
 
 Then 
 
 is the resistance which the dead weight offers 
 
 to rupture. 
 
 Add this force to the horizontal thrust of the unloaded arch, 
 and subtract the whole from the horizontal thrust of the loaded 
 arch, the remainder is the surplus horizontal thrust. 
 
 Tt is manifest that the resistance of the horizontal girder in 
 the unloaded span to compression, helps to prevent the pier from 
 overturning. 
 
 In viaducts of masonry the spandril filling acts partly as a 
 horizontal girder in preventing the overturning of the pier, and 
 partly as bracing preventing alteration of form in the arch. 
 
 In dealing with a viaduct of masonry, and describing the 
 curve of equilibrium for loaded and unloaded spans for the pur- 
 pose of finding the resultant thrust, the curve in the loaded 
 span should be carried as near to the extrados of the arch at 
 the crown, and as close to the intrados at the springing as is 
 consistent with safety; while for the unloaded span, the curve 
 should be brought as close to the intrados at the crown as safety 
 permits, and be made to intersect the curve of the loaded arch 
 as high up as is possible without causing the resultant pressure 
 to pass so near to the face of the masonry at the springing as 
 to crush it. Fig. 53 illustrates this -method. 
 
70 
 
 THE ARCH. 
 
 Stability of a Wrought-iron Arched Viaduct. In a wrought- 
 iron arched viaduct the question of stability is somewhat compli- 
 cated. 
 
 Fig. 58 represents a wrought iron viaduct, perfectly continuous 
 throughout. We might calculate its stability in the manner 
 adopted for the case illustrated in fig. 56, and supposing it were 
 found that the piers were too weak to ensure the stability of 
 the system, this might be secured by making the rib stiff enough 
 at the crown to resist the upward force which comes upon it. 
 The arching must in that case be regarded as a continuous girder 
 of varying depth in which the effect of the load is to flatten 
 
 the loaded arch and to raise the crown of the unloaded arch. The 
 piers must be assumed to be hinged at the springing and at 
 the ground-level, so that no assistance is obtained from them, 
 because we are now considering the sole effect of the metal at 
 the crown in affording stability. 
 
 The first step in the calculation is to find the surplus hori- 
 zontal thrust at the springing. (In this case there is no resistance 
 
 p 
 afforded by the pier.) Call this force P, take half this force or as 
 
 acting at a leverage of ab (the versine of the arch), fig. 58, tending 
 to crush the bottom flange of the rib at b, and to tear it open at c. 
 
 p 
 Let S be the strain in tension at c caused by the force - , then 
 
 P.ab 
 
 or 
 
 The compressive strain at b will be -^ + S, 
 
 P 
 2 
 
THE AECH. 71 
 
 We take the horizontal force tending to force up the crown of 
 
 p 
 the unloaded arch as ^ only, because half the force P is absorbed 
 
 Zi 
 
 in flattening the loaded arch. 
 
 p 
 
 On the loaded arch the force ^ acts at an arm of al>' also ; 
 
 therefore if $' be the strain in compression on the top flange caused 
 
 by the force -= , 
 
 ~ . 
 
 P.a'V 
 
 The tensive strain at b' will be ~- + $', 
 
 or ^ 
 
 These strains must be added to the strains which result from 
 the position of the curve of equilibrium to obtain the extreme 
 strain upon the flanges. 
 
 In practice the stability of the pier must be taken into ac- 
 count. Indeed in the majority of cases it will be on the pier that 
 we must rely : for before any considerable strain can be produced 
 on the rib in the unloaded span, the span of the loaded arch must 
 have increased through the forcing apart of the piers, as shewn in 
 fig. 58. The amount of this alteration in length of the span will 
 vary according to the height, thickness, and material of the pier. 
 By measuring the actual increase in length of the loaded span, it 
 is possible to determine how much of the resistance to overturning 
 is offered by the stability of the piers, and how much by the stiff- 
 ness of the ribs. This cannot however be done without a know- 
 ledge of the laws of deflection, 
 
CHAPTER VI. 
 
 SUSPENSION BKIDGES. 
 
 A SUSPENSION bridge is merely an arched bridge inverted, the 
 chain of the suspension bridge being subject to a strain in tension 
 corresponding to the strain in compression on the rib of the arch. 
 
 There is one important practical difference between the two : 
 this, that in the suspension bridge the chain being flexible is able 
 to adapt itself to the curve of equilibrium, whereas no such pro- 
 perty belongs to the arched rib. 
 
 This peculiarity of the suspension bridge causes the platform of 
 the bridge to alter its form under the passage of a rolling load, 
 unless the bridge be rendered rigid by some kind of bracing. 
 
 Strain of centre of Span. The strain on the chains at the 
 
 WL 
 
 centre of the span may ordinarily be found by the formula --y- , 
 
 since the weight of the chains is small compared with that of the 
 platform and live load, and consequently the load may be taken as 
 evenly distributed. In this formula 
 
 W the total load on the bridge. 
 L the span between towers. 
 
 d = the depression of the chain at centre below 
 the points of support on the towers. 
 
 In very large bridges, however, where the weight of the chains 
 is relatively large, this formula will not give correct results, and it 
 will be necessary to describe the curve of equilibrium by the 
 method described on p. 46, in order to obtain the strains on the 
 chain at the centre of the span. 
 
 In ordinary cases the chains will hang in a curve which is 
 a close approximation to a parabola. In cases where the chains 
 
SUSPENSION BRIDGES. 73 
 
 are relatively heavy, the curve will approach more nearly the form 
 of a catenary. 
 
 Strain at any portion of the Chain. The strain at the 
 centre of the span being known, the strain at any point a (fig. 60) 
 in the chain may be found by drawing a tangent to the curve at 
 that point, cutting at b the horizontal line which is the tangent to 
 the lowest point in the curve. If be 
 the angle that the tangent makes with 
 the horizontal, and 8 the strain on 
 the chain where it is horizontal ; then 
 
 a, or S.SQC 0, is the strain at a. 
 
 COS0' 
 
 Or the strain may be found by geometrical construction, thus : 
 let fall from point a a perpendicular to meet the horizontal 
 tangent in c. 
 
 Then, strain at a : S :: ab : be ; 
 
 ~ ab 
 
 .*. strain at a = S 7- . 
 be 
 
 In the triangle abc, ab represents the actual strain at a. 
 be horizontal element = S. 
 and ac vertical element of the strain 
 = weight of superstructure between c and centre of span. 
 
 Strain on the back tie. As in the arch, the thrust of the 
 arch is taken by abutments, so in the suspension bridge the pull 
 of the chains is resisted by back ties, which are in general anchored 
 to masses of masonry at a considerable depth below the surface of 
 the ground in the abutments of the bridge. The strain on the 
 back tie is found by means of a parallelogram of which the adja- 
 cent sides are tangents to the chain at its junction with the saddle 
 on the pier and the back tie. The strain on the chain being 
 known, that on the back tie is obtained by completing the paral- 
 lelogram, the vertical diagonal of which represents the load on the 
 tower. 
 
 In those suspension bridges in which no means are provided 
 for resisting the distorting effect of unequal loading, the curve of 
 the chain alters its form during the passage of the moving load. 
 
74 SUSPENSION BRIDGES. 
 
 Usually the moving load is small compared with the dead load, 
 and consequently the alteration in the form of the chain is not of 
 serious moment. Such bridges are, however, liable to accident 
 from the undulations produced by high winds. 
 
 Rigidity can be obtained by means similar to those which are 
 applicable to the bowstring girder as given on p. 51. 
 
 The adoption of the 1st, 2nd or 4th of these systems neces- 
 sitates the substitution of a curved rib for a flexible chain, thus 
 producing what has been called " an inverted arch bridge." 
 
 Whatever system of stiffening may be adopted, the whole 
 structure should be hinged at the centre of the span, to allow 
 for the rise and fall of the chain caused by variations of tem- 
 perature. 
 
 The methods of finding the strains on the various parts of the 
 suspension bridge are precisely similar to those already given for 
 finding the strains on the corresponding parts of the bowstring 
 girder, to which the reader is referred. 
 
 The usual practice is to stiffen the suspension bridge by means 
 of a horizontal girder. The method of calculating the necessary 
 strength of this girder is given on pp. 54, 55, and 56. The hori- 
 zontal girder answers to the tie in the bowstring girder, but is not 
 like it subject to longitudinal strain (see p. 55). It acts only as a 
 stiffener. The ends on the abutments should be anchored, but 
 should be free to move horizontally to allow for change of tem- 
 perature. 
 
 To restate the rule, in a form applicable to the suspension 
 bridge. If W be the whole live load on the bridge, then the 
 horizontal girder, which reaches from the abutment to the centre 
 of the span, will, under the most favourable conditions of loading, 
 be subject to an upward and downward transverse force equal 
 
 W 
 
 distributed. 
 
 o 
 
 If I be the span of the bridge, 
 
 d depth of the horizontal girder, 
 S the strain on the flange at centre, 
 
SUSPENSION BRIDGES. 75 
 
 Plates VII. and VIII. turned upside down with the signs of 
 the strains changed, give the strains on a suspension bridge 
 stiffened by means of diagonal bracing. As under the condition 
 of loading, given in Diagram 5, Plate VII. or VIII., the rib of 
 the arch becomes subject to tension near the crown, so, under the 
 same condition of loading, the suspension chains are subject to 
 compression. 
 
 For further remarks on suspension bridges see supplementary 
 Chapter on Economy in Suspension bridges. 
 
CHAPTER VII. 
 
 DEFLECTION. 
 
 BEFORE investigating the properties of continuous girders, it is 
 necessary to become acquainted with the laws of Deflection, which, 
 in the continuous girder, very importantly determine the magni- 
 tudes and directions of the strains. 
 Deflection is due to elasticity. 
 
 Definition. The deflection of a beam or girder is its vertical 
 displacement by a load from the position that it occupied when 
 unloaded, its bearing points being fixed. 
 
 So that were the bottom flange of the girder horizontal in its 
 unloaded state, when loaded it would be curved downwards, the 
 distance between the lowest point in this curve and the original 
 horizontal line being the measure of the deflection, generally 
 called the deflection of the girder. 
 
 If the flanges of the girder be parallel, straight, and of uni- 
 formly proportioned strength, the curve which it would assume 
 when loaded would be an arc of a circle whose versine is the 
 deflection. 
 
 If the girder be constructed with a camber, the curve of the 
 camber will be flattened when the girder is 
 loaded. The amount of deflection will not 
 be affected by cambering the girder. 
 
 Laws of Deflection. Let ABCD, fig. 61, 
 represent a girder, originally horizontal, un- 
 der deflection, having its flanges so propor- 
 tioned to the strain, that the strain per 
 square inch is uniform throughout the whole 
 of each flange. Then will the flange AB be 
 uniformly contracted throughout its whole 
 length, and CD uniformly extended. 
 
 I if/. 61. 
 
DEFLECTION. 77 
 
 Suppose that on the girder, when undeflected, two parallel 
 vertical lines pr, and qs, be drawn perpendicular to the flanges 
 at a distance x apart, then, when the girder is deflected, 
 their upper extremities p and q will be drawn together to a 
 small extent, which we will call e, so that the distance of p 
 from q will now be x e. In like manner, their lower extremi- 
 ties r and s will be forced apart to an amount e', the distance 
 between them being now x + e'. The lines pr and qs being no 
 longer parallel, will, if produced, meet in a point 0. 
 
 Since the flanges are uniformly contracted or extended 
 throughout their whole length, it is obvious that lines drawn 
 perpendicular to the flanges upon the face of the girder, would, 
 when the girder is deflected, all meet in the same point ; and 
 therefore AB and CD would be arcs of circles, of which was the 
 centre. 
 
 Radius of Curvature. To find the distance Op. 
 
 The distance x being assumed very small, pq and rs may be 
 
 regarded as parallel straight lines. 
 i 
 
 From point p, draw pr parallel to qs. Then r's = pq, and 
 
 rr = e -f e. 
 
 Since the angle pOq is equal to the angle rpr\ the triangle 
 rpr' is similar to the triangle pOq (very nearly). 
 
 /. rr or e 4- e' : pr :: pq : Op-, 
 
 pr.pq_d.pq 
 u P--= e + e > '- e + e 
 
 since pr = d, the depth of the girder. Whence Op the radius of 
 curvature for the top flange may be found. The neutral axis of 
 the girder, represented by the dotted line MN t does not alter in. 
 length. 
 
 The radius of curvature of the line MN, or 
 
 R : x :: d : e + e, 
 by similar triangles, 
 
 -r> xd 
 
 therefore M= 7. 
 
 e + e 
 
78 
 
 DEFLECTION. 
 
 When e = e, substitute 2E for e + e' in the foregoing ex- 
 pression, and taking x = 1, 
 
 we have 
 
 E = v 
 
 ^ represents the actual alterations of a unit of length, say 
 1 foot lineal, of either flange under load in terms of the unit of 
 length, in feet therefore in this case ; d represents the depth of the 
 girder, and R the radius of curvature in the same terms. 
 
 To find the Deflection. In fig. 62, let the arc CAD be the 
 neutral axis of the girder when deflected. Let 
 be the centre, and OA the radius of curvature. 
 
 Complete the circle AGED. 
 
 Produce A to E. 
 
 By Euclid, Book III. Prop. 35, AB . BE= BD\ 
 
 Since AE=2B, 
 
 CD = I the span of the girder, 
 and AB = 8 the deflection, 
 
 AB.BE= (25 - 8) 8 
 
 But & being extremely small in comparison with R, (25 S) S 
 may be assumed equal to 25 . S, without sensible error. 
 
 ' 
 
 
 
 Substituting for R the expression -=r^ , we obtain 
 
 -3 
 
 < 
 
 * The method here given of finding the radius of curvature is not absolutely 
 correct, for lines drawn on the side of the girder at right angles to the flanges in its 
 unloaded state will not make the same angle when the girder is deflected, but will 
 make angles with the flanges increasing in acuteness as the deflection and their 
 distance from the centre of the girder increases. Thus the angles ABC, SAD 
 (fig. 61), will be acute angles when the girder is deflected, though right angles when 
 it is undeflected. The consequence is that the true radius of curvature is less, and 
 therefore the deflection greater than our theory, which however is true enough for 
 practical purposes, gives. 
 
DEFLECTION. 79 
 
 Example. A wrought-iron girder 60 feet long, and 10 feet 
 deep, receives a strain of 4 tons per square inch throughout its 
 
 flanges when loaded ; required the deflection. 
 
 .QA 
 Since a wrought-iron bar contracts or extends TrTr of its 
 
 length for every ton of strain per square inch of its section, in 
 this case 
 
 I = 60 feet, 
 d = 10 feet, 
 
 E=CS, where 0= a constant, varying for different 
 materials, and S= the strain in tons per square inch, caused by 
 the load*. 
 
 To put equation 1 into a more convenient form, substitute 
 CS for E y then 
 
 Irregularly strained girder. When the flanges are not sub- 
 ject to a uniform strain, the foregoing equation is not available, 
 since the value of S is not the same throughout the girder, and 
 consequently the radius of curvature will vary for different parts. 
 
 We may, however, assume the value of S to be the same 
 throughout the whole of the flange in each bay, consequently 
 there will be a distinct radius of curvature for each bay. 
 
 Let us take the case of a girder in which the value of s is 
 largest for the bays at the centre of the span. Then it is 
 obvious that as S diminishes, the value of E, or e + e, diminishes 
 also (see fig. 61, p. 76), and consequently the radius of curva- 
 ture R increases in proportion as we approach the abutments. 
 
 Let us assume that the girder, when unloaded, lies hori- 
 zontally, and is constructed and loaded symmetrically on each 
 
 * If the girder be already partially deflected by a dead load, S will be the addi- 
 tional strain caused by the live load. 
 
80 DEFLECTION. 
 
 side of the centre of the span p (fig. 63), then p will be the 
 lowest point in the girder when loaded, 
 and the centre of curvature for the bay qp 
 will lie in the line po drawn vertically 
 through p. Let o be this point. Join 
 q, o, and produce the line qo indefinitely. 
 Make qo = R for the bay rq. Join ro and 
 make ro" = r for the bay sr. Join s, o". 
 Through the points q, r and s draw hori- 
 zontal dotted lines to meet perpendiculars 
 erected from points p, q and r respectively 
 in the points p, q and /. 
 
 The extremity 5 of the girder resting 
 on the abutment, it is obvious that the sum 
 of the distances pp, qq' and rr equals the 
 total deflection of the girder. 
 
 If pq, qr and rs be drawn as straight lines, the sine of the 
 angle pqp' is 2L; 
 
 /. pp'=pqsmpqp'', 
 
 similarly qq = qr sin qrq, 
 
 and rr rs sin rsr '. 
 
 Now the angle pqp / *-- ; 
 
 2 
 
 , qor 
 
 and qrq = tpoq + ^ - , 
 
 rsr = tpoq + ^ qo r + 
 
 .*. the deflection or S pp 1 + qq -f rr' 
 pq sin ^ + qr . sin (poq + 2~J 
 
 + rs.sin ^o^ + qor + -^J ......... (3). 
 
 In ordinary girders the lengths of the bays are all equal, con- 
 sequently pq, qr and rs may in such cases be represented by the 
 
DEFLECTION. 81 
 
 expression -~ , in which I is the length of the girder and N is the 
 number of bays. 
 
 Equation 3 may then be simplified to this, 
 
 I ( . poq / qo'r\ f ro"s\) 
 
 8 = -j- {sin" + sin (poq + -$-} + sm poq + qor + -~- } Y . 
 IV ( ^ V 25 / \ * / J 
 
 The general expression for the deflection of a symmetrical 
 girder of N number of bays is when N is even, 
 
 N_ * 
 
 ? rl -i-i)J --w- 
 
 When N is uneven, 
 
 ?-i ^. 
 
 (5), 
 
 6 being the angle of the centre bay. 
 
 To obtain the value of the angles poq, qo'r and rd's, we have 
 only to find the value of E for each bay, which will be given by 
 
 the equation R = ^ (see p. 78). 
 
 Now the value of any angle 6 in degrees may be found from 
 the equation 
 
 _. 57'29578; 
 
 radius 
 
 t 
 or 6 = 2 . 57*29578 = 
 
 Substituting -= for pg and OS for E (see p. 79), we obtain 
 
 finally for any bay 
 
 . 
 
 In calculating the value of E in plate girders the effect of the 
 web must be taken into account. 
 
 Y. 6 
 
82 DEFLECTION. 
 
 For accuracy, therefore, ^th of the area of the cross section of 
 
 the whole web should be added to the area of the flange (see 
 p. 19). 
 
 Deflection of an arched rib. To find the deflection of an 
 arched rib, it will be necessary to know . 
 
 whether the abutments yield sensibly to 
 
 the thrust when the arch is loaded. 
 
 Assuming the abutment to be unyield- 
 
 ing, let a, c, b (fig. 64) represent half the rib in its normal posi- 
 tion, b being the vertex and a the springing of the arch, of which 
 Id is the rise. 
 
 Let the dotted line aeb' represent the position of the rib 
 when loaded, then W is the amount of deflection. 
 
 Draw the chords ab, ab'. 
 
 Since the distance W is very small, 
 
 the arc acb : arc acb' :: ab : ab', very approximately. 
 
 Knowing the strain in compression which the load gives to 
 the rib, the length of the arc acb' can be easily found, for 
 arc acb' = arc acb (arc acb x CS). 
 
 The half-span ad and the rise db of the arc are given 
 quantities ; 
 
 is known. 
 
 . , acV xab 
 
 .out ab = 
 
 * 7 
 
 acb acb 
 
 Now db f * = ab' 2 -ad*-, 
 
 and W or 8 = db 
 
 (acb - acb. Cfyjad* + db 
 
 Example. Required the deflection of a cast-iron arched rib of 
 20 feet span and 5 feet rise, under load. (See Fig. 65.) 
 
DEFLECTION. 83 
 
 Let the strain on the cross section of the rib average 1 ton per 
 square inch when the bridge is unloaded, and 
 3 tons when loaded, then the strain caused by 
 the live load or S = 2. 
 
 For cast iron the value of the constant 
 0= -00018. 
 
 In this case db = 5 feet, 
 
 acb = 1I'6 feet; 
 ad =10 feet; 
 
 8 = 5 - 4-99086 = '00914 feet = 10968 inches. 
 
 Deflection of Bowstring Girder. If the abutments yield, the 
 case becomes analogous to that of the bowstring girder, where 
 the span is lengthened by the stretching of the tie under load. 
 
 Let the strong line (Fig. 66) shew 
 the position of the flanges of a bowstring Fl ?' & e$ - 
 
 girder when unloaded, and the dotted 
 lines their position when loaded. 
 
 Here, as in the previous case, 
 
 arc acb : arc ac'b' :: ab : a'b', approximately; 
 
 , ,,, ac'b' x ab 
 
 whence a b = 7 . 
 
 acb 
 
 But acb' = acb -acb. CS. See ante, p. 82. 
 
 And ab 
 
 ,, , (acb - acb . CS) *Jad*+db* 
 
 .*. a o = T , as before. 
 
 acb 
 
 Assuming ad to be a horizontal, and db' to be a vertical line, 
 
 a f d*. 
 
 62 
 
84 DEFLECTION. 
 
 Now ad=ad+ad.CS; 
 
 /. db 
 
 /((acb 
 V IV 
 
 by substitution ......... (9) . 
 
 And W or S = db-db f ', 
 
 
 . . , (IQ). 
 
 Deflection of Suspension Bridge. The case of the suspension 
 bridge, though analogous to that of the arch, presents certain 
 peculiarities. 
 
 The greatest amount of deflection in a bridge of three spans 
 occurs when the centre span only is loaded. 
 
 In consequence of the sliding of the chain on the top of the 
 tower, the chain in the centre opening will become lengthened 
 under load by the amount of extension, not only of itself but also 
 of the back tie. 
 
 Wherefore in equation No. 8, p. 82, it will be necessary to 
 substitute for the expression acb acb . (7$, 
 
 acb + acb . CS + amount of extension of back tie caused by live load. 
 
 Strength of a girder how far deducible from its deflection 
 under load. From the deflection of a bridge as actually observed 
 under test load, an idea can be formed of its strength. With the 
 span as the chord of an arc, and the deflection as its versine for 
 data, we can obtain R the radius of curvature, while from the 
 equation 
 
 the value of E, and the average strain per square inch on the 
 section of the flanges is deducible. 
 
 By comparing the deflection of a girder under a moving load 
 with its deflection under a dead load an approximate estimate 
 may be made of the strain to which it is subjected by the moving 
 load. It should, however, be borne in mind that a load moving 
 
DEFLECTION. 85 
 
 rapidly on to a bridge deflects it to a greater extent than the 
 same load does when remaining at rest upon the bridge. 
 
 The mere stiffness of a girder is no criterion of its strength, for 
 the former depends upon the average sectional area of the flanges, 
 while the latter depends upon the net sectional area at any point. 
 For example, a girder whose flanges were made of the same 
 sectional area from end to end would be stiffer than a girder 
 which had this identical sectional area of flange at the centre of 
 the span and a sectional area diminishing towards the abutments, 
 but it would not be stronger, as the strength would be dependent 
 upon the sectional area at the weakest point the centre of the, 
 span. Again, if one of the plates in the flange of a girder of two 
 or three thicknesses of plate were cut completely through, the 
 deflection of the girder after the injury would not be perceptibly 
 greater than before, though the strength would be seriously- 
 diminished. 
 
 Mere stiffness alone does not prove the strength of a badly 
 constructed girder. 
 
CHAPTER VIII. 
 
 CONTINUOUS GIKDEKS. 
 
 THE term " continuous " is applied to a girder when it is carried 
 without break over two or more spans. The peculiarity of the 
 continuous girder is, that a certain portion of it acts as an ordinary 
 girder, while another part is acting as a cantilever. In the one 
 part the top flange is in compression, and bottom flange in tension, 
 in the other part these conditions are reversed. 
 
 Fig. 67 shews the form assumed by two plain girders over 
 adjacent spans when deflected. Assuming that in their unloaded 
 state the flanges of the two girders formed two unbroken parallel 
 
 straight lines, and that the ends of the girders at a over the pier 
 were in contact for the whole of their depth ; upon deflection it is 
 obvious that the extremities of the top flanges of the two girders 
 would part at a, and a wedge-shaped gap be formed. 
 
 Now let us suppose that before causing the girders to deflect, 
 we had connected together the extremities of their top flanges 
 
 Jfcff.08. 
 
 at a. The effect of this would be to produce tension in the top 
 
CONTINUOUS GIRDERS. 87 
 
 flanges of the girders for a certain distance on each side of a, and 
 compression in the bottom flanges. The girders would assume 
 the form shewn in fig. 68, in which the centre of curvature for 
 the flanges in the neighbourhood of the pier is below the girder, 
 while for the other parts the centre of curvature is above the 
 girders. The points where the curve changes from convex to 
 concave, are called points of contrary flexure; at these points 
 there is no strain on the flanges. The parts of the girders in which 
 the top flange is in tension are in fact cantilevers, and the girders 
 of fig. 67, united as in fig. 68, form one " continuous girder." 
 
 It is evident that the strains on the flanges will greatly depend 
 upon the position of the point of contrary flexure. This fluctuates 
 with every change in the position of the load, and depends upon 
 the sectional area of the flanges at each point in the girder's 
 length. It is impossible to calculate the exact position of this 
 point in practice, we can, however, define its position within 
 certain limits. 
 
 Continuous Girder of two spans. Assuming the girder to be 
 of uniformly proportioned strength, both spans to be alike and 
 similarly loaded, the point of contrary flexure will be at a distance 
 
 o 
 
 of about Z^TV ths of the span from the pier, provided the relative 
 
 level of the three supporting points is similar to that of the three 
 bearing points in the girder in its unloaded state. For instance, 
 if the bearing points of the girder at the extremities and centre 
 are in the same straight line, the supporting points on the pier 
 and abutments must be so also. Or, if one bearing point in the 
 girder is, say 1 inch, higher than the other, its supporting point 
 must be made 1 inch higher than the other points of support. 
 
 ' A corollary may hence be deduced, that the position of the 
 point of contrary flexure in a continuous girder may be shifted at 
 will to a certain extent by raising or depressing any one of the 
 
 points of support. 
 
 g 
 
 The reader may satisfy himself that T7\ths of the span is ap- 
 proximately the true position of the point of contrary flexure in 
 the following manner. 
 
88 CONTINUOUS GIRDERS. 
 
 Draw a horizontal line ab (fig. 69) to represent the girder in 
 its unloaded state. Let a be the end 
 of the girder resting on the abutment, 
 and b the bearing point over the pier, 
 the distance ab representing the span. 
 
 From point b drop a perpendicular 
 bo, making bo of convenient length. 
 With o as a centre, describe the arc be. 
 With a as a centre, and a distance 
 equal to ob as a radius, describe an 
 arc on the upper side of the line ab. 
 With o as a centre, and a distance equal to twice ob as a radius, 
 describe an arc cutting the last-mentioned arc in p. With p as 
 a centre, and ap as a radius, describe the arc ac touching the arc 
 be in c. 
 
 A line joining points o and p will cut the arcs be and ac at 
 their point of contact in c, which is the point of contrary flexure. 
 
 If ab = l, it will be found by actual measurement that the 
 horizontal distance of the point c from the line ob will be about 
 & 
 
 The curved line acb represents the position assumed by an 
 originally horizontal girder under load, for, supposing the girder, 
 as previously stated, to be symmetrically constructed and loaded, 
 it is obvious that the curvature of the girder in each span will be 
 the same, therefore the centre of curvature of the cantilever 
 portion must lie in the centre line ob of the pier, and as it is 
 assumed to be equally strained throughout, the rate of curvature 
 must be the same everywhere, (see Chapter on Deflection). This 
 assumption, however, can never be realized in practice ; to do so, 
 it would be necessary to diminish the sectional area of the flanges 
 at c to nothing. 
 
 The extra proportional strength of the flanges in the neigh- 
 bourhood of the point of contrary flexure flattens the curves ac 
 and cb for a short distance on each side of the point c, and would 
 throw c rather further out from the pier. On the other hand, 
 in girders which are constructed to support a moving load, the 
 curvature of the girder portion ac is flatter than that of the canti- 
 
CONTINUOUS GIKDERS. 
 
 89 
 
 lever portion, in consequence of the extra metal in the flanges 
 of the girder portion, which is required to enable it to bear the 
 strain which arises when the adjoining span is unloaded and the 
 length of the girder portion increased, as will be seen subse- 
 quently : this would throw point c nearer to the pier. For girders 
 that have to bear a dead load only, the points of contrary flexure 
 should be taken as one-third of the span from the pier for safety. 
 This condition of loading gives the greatest possible strain to the 
 flanges at b over the pier, and for girders which have to carry 
 a dead load, such as the wall of a house, no variation in the 
 position of the point of contrary flexure can occur, but when a 
 moving load has to be carried, we must provide for the change 
 caused by the loading of one span only. 
 
 Fig. 70 represents a continuous girder of two spans, one only 
 
 Fiy. 70. 
 
 JM6-* 
 
 of which, that on the left, is loaded, c and c are assumed po- 
 sitions of the point of contrary flexure ; be and be the respective 
 cantilever portions of the girder in the spans I and I'. 
 
 The conditions of equilibrium require that 
 
 weight of ac ,. , . , r , distance be 
 
 2_ - x distance bo + weight of be x -- ^ 
 
 weight of dc ,. , , . , , r 7 , distance be' 
 
 x distance le -\- weight of be x -- -= -- . 
 
 _ ^ 
 
 Let w be the unit of the dead load, 
 
 w live 
 Then 
 
 ac (w + w) 7 , , . ,,bc do . w , , , , be 
 i-| - '- be + be (w + w ) -^ = g oo + be . w , 
 
 or 
 
 (ac + 6c) (w + w'} j _ (dc' + bc^w , , 
 
 ~~ J> " " 
 
90 CONTINUOUS GIRDERS. 
 
 but ac + be = I, and dc + be' = Z', 
 
 .-. l(w+w^ bc=l' .w. be (1). 
 
 If Z=f, (w + w')bc=wbc (2), 
 
 whence be : be :: w : (w + w'). 
 
 That is to say, when the spans are equal, the lengths of the canti- 
 lever portions of each span are inversely proportional to the loads 
 on the spans. 
 
 be, the cantilever portion of the unloaded span, may be any 
 length between '3Z' and I'. It will be the former when w = 0, but 
 what value of w will give be' = I can only be determined approxi- 
 mately by very careful calculation, w, however, is generally a 
 known quantity. 
 
 Referring to fig. 70, it will be seen that the load on the abut- 
 ment d is equal to half the weight of the girder portion c'd. Now 
 if the point of contrary flexure c be assumed to coincide with d, 
 there will be no load upon the abutment d. This condition gives 
 the greatest possible length for the cantilever be, when the spans 
 are unequally loaded. 
 
 In calculating the length of the girder portion ac, be' should 
 always be assumed as considerably under the maximum length I'. 
 
 Let e and e (rig. 70) be the positions of the points of contrary 
 flexure, when the girder abd is fully loaded, then we see that the 
 effect of removing the live load from one span is to cause a con- 
 siderable strain upon parts of the girder which are unstrained 
 when both spans are fully loaded. 
 
 The positions of the points of contrary flexure c and c will 
 plainly be dependent upon the curvature of the girder, which is 
 regulated by the transverse stiffness. The transverse stiffness 
 again varies directly as the area of metal in the flanges. 
 
 In attempting to find approximately the true positions of the 
 points of contrary flexure it will therefore be necessary to assume 
 a certain transverse stiffness of girder, i.e. a certain area of metal 
 in the flanges of each bay. 
 
 Method of finding position of Point of Contrary Flexure. 
 
 The method of procedure should be as follows. Assuming both 
 spans to be fully loaded, calculate the strains on the cantilever 
 portions on each side of the pier 
 
CONTINUOUS GIRDERS. 
 
 91 
 
 Now assume each span in turn to be loaded, the adjoining 
 span being unloaded. 
 
 Take c, the point of contrary flexure for the unloaded span 
 (see fig. 70), as two-thirds of the span I' from b, the pier, and 
 calculate the position of c in the loaded span from equation 2, 
 and the strains on the girder portion ac, and cantilever por- 
 tion be. 
 
 Construct a diagram of the girder, shewing the maximum 
 strains on the flanges of each bay, as deduced from the foregoing 
 calculations. 
 
 Arrange the plates so as to give at least the required area 
 in each bay. 
 
 Take out the actual area of metal in the flanges of each bay, 
 and add thereto one-sixth the area of the web. 
 
 By equation 6, p. 81, find the value of 6 for each bay, (see 
 pp. 80, 81). 
 
 Proceed to calculate the slope of the flange at l } the pier, with 
 the horizontal, in the following manner. 
 
 / / 1 I 
 
 ', \ 3*9 
 \ \ 
 
 .7A 
 
 
 
 
 i 
 
 /III 
 
 \ \ \ 
 
 
 
 
 
 1 i 
 
 
 \ \ \ e 
 
 3 
 
 
 
 
 ! i ! 
 
 ^=*^Zt .!" " 
 
 \=---\.---z\ _--ir _^=d 
 \ J ! V 
 
 ? 
 
 
 
 
 
 _^ 
 
 ^=rTTL \ 
 
 1 
 
 
 
 
 ' 
 
 
 \ 
 
 1 
 i 
 
 
 
 
 1 I 
 
 
 \ 
 
 j 
 
 
 
 
 i i 
 
 
 \ 
 
 t 
 I 
 
 I 
 i 
 
 i 
 
 
 
 I [ 
 
 Consider the flange at the centre point of the assumed girder 
 portion of the loaded span to be horizontal. 
 
 Let^> (fig. 71) be this point. 
 
 Calculate by equation 4, or equation 5, as the case may be, 
 p. 81, the vertical distance of point p below a, the extremity of 
 the girder resting on the abutment, which = 8 in this equation, 
 and also the vertical height of point ~b above p. 
 
 Let ob' be a line drawn parallel to the flange at p, and there- 
 fore horizontally, and intersecting a perpendicular line drawn 
 through b in V. 
 
92 CONTINUOUS GIRDERS. 
 
 If the height of a be less than that of I above pointy, as in 
 this case, b f will fall below 6, but if greater it will be above it. 
 
 The value of 6 for each bay being known, the angle of incli- 
 nation of the flange with the horizontal line ab' is also known for 
 each bay. 
 
 But the truly horizontal line passes through points a, b and d, 
 consequently the true angle of inclination of the flange at b will 
 be obtained by subtracting the angle bob' from the angle of incli- 
 nation which it makes with the line ab', when, as in this case, 
 the point b' is below 5, or by adding it thereto when it is situated 
 above. 
 
 mi, ili> W 180 j 
 
 The angle bab = y in degrees. 
 
 Knowing the angle of inclination of the flange at b with the 
 true horizontal line abd, proceed to calculate step by step the 
 angle of inclination of the flange in each bay up to the abutment 
 at d, and thence ascertain the vertical height of the d extremity 
 of the girder above or below the horizontal line. 
 
 If it fall above, the girder is safe, but if below, the girder is 
 weak ; the remedy for which is, to take the point of contrary 
 flexure c rather nearer the pier. On the other hand, if the 
 extremity of the girder at d fall much above the horizontal, 
 the point c' may be moved a little further from the pier, and 
 economy effected. 
 
 Example. A girder, in two spans of 60 ft., see fig. 1, Plate 
 IX., has to sustain a rolling load three-fourths of a ton per 
 foot run, its depth is 5 feet, and it is divided up into bays of 
 5 feet, at which distance apart the cross girders are affixed. 
 
 Taking the dead load at one quarter ton per foot run, we 
 
 3 1 
 
 have -r + -r 1 ton per foot run of girder, or 5 tons per bay. 
 
 Following the method laid down on page 90, we commence by 
 ascertaining the strains on the cantilever portions of the girder 
 when both spans are fully loaded. 
 
 In fig. 2, Plate IX., are given the strains obtained by taking 
 the point of contrary flexure one-third the span from the pier, that 
 is, at the line dividing bays 8 and 9. 
 
CONTINUOUS GIRDERS. 93 
 
 In fig. 1 are given the strains when the. left-hand span only is 
 loaded; c' being taken two-thirds of the span from the pier, c falls 
 in the loaded span at the line dividing bays 10 and 11. 
 
 In fig. 3 -are given the maximum strains collated from figs. 1 
 and 2. 
 
 In fig. 4 are given the arrangement of plates, areas of metal, 
 values of 8 and 6 for each bay of the girder under the condition 
 of loading strains in fig. 1. 
 
 Now e = 57'29578, see page 81, 
 
 S = strain per sq. inch in tons, 
 
 Here 
 
 ioootr 
 
 = -009626^, 
 
 whence the value of 6 for each bay is quickly obtained. 
 
 Assume the lowest point in the girder portion (coinciding with 
 point p, fig. 71) to be in the line dividing bays 5 and 6, which is 
 the centre of the girder portion, and consider this line to be 
 vertical when the girder is deflected. 
 
 Let 1 , fl 2 , ... 12 be the respective angles for bays 1 to 12 of 
 left-hand span, and 1? 2 , ... 12 be the respective angles for bays 
 (1) to (12) of right-hand span. Then the vertical height of the 
 lower left-hand corner of the girder portion above p, or 
 
 = 5 jsin |+ sin (>+ |) + si 
 
 sn -f + 
 
 = 5 (sin -0151 4- sin "04405 + sin -07146 + sin -09633 + sin -11148) 
 =5(-0002635ft.+'0007689ft.-f0012472ft.+-0016812ft.-H-0019457ft.) 
 
 = -0295325 ft 
 
94 CONTINUOUS GIRDERS. 
 
 On the right-hand side of p, the vertical height of the flange 
 at the extremity of the girder portion, being in the line dividing 
 bays 10 and 11, above p, or 
 
 B = 5 {sin ~ + sin ( 6+ ~] + sin ( s + 6 7 
 
 ( A \ A / \ 
 
 = 5 (sin -0151 + sin -04367 + sin'07021+ sin '09412+ sin -10792) 
 
 =5(-0002635ft.+-0007622ft.4-'0012254ft.+-0016426ft.+-0018835ft.) 
 
 =- -028886 ft. 
 
 8 for left side of p = -0295325 ft. 
 S right ='0288860 ft. 
 
 Difference . -0006465ft. 
 
 Value of difference at pier = '0006465 ft. x = '0007758 feet. 
 
 For bays 11 and 12 the centre of curvature is below the 
 horizontal line, and consequently the angle of slope of the flange 
 with the horizontal begins to decrease as we go to the right. 
 
 The angle made with the vertical by the line dividing bays 
 10 and 11 = the sum of the angles of the bays between this and 
 the vertical line at the centre of the girder portion 
 
 = ff> + O 7 + (9 8 + 9 + 6 = -11090 
 and 0* + ff* ='01853 
 
 Difference = '09237 
 
 = angle with the vertical made by the line dividing bays 12 and 
 (12) at the pier, and angle of flange with the horizontal subject to 
 correction as below. 
 
 Now the height of point b (fig. 71) above point e, the lower 
 right-hand extremity of the girder portion, 
 
 5 jsin (-1109 - 6 ^}+ sin ('1109 - 11 - ^ 
 5 (sin -10826 + sin -09899) = '018165 ft. 
 
CON11NUOUS GIRDERS. 95 
 
 But -018165ft. - -0007758 ft. (see ante) = '0173892 ft 
 the distance bb', 
 
 , ,, '0173892 ft..x 180 
 
 and / bob = ^TT^ 7,^, . -. = '0166 , 
 
 60 ft. x 3'1416 
 
 .-. -09237 -'0166 = -07577, 
 the true angle of the flange with the horizontal at the pier. 
 
 Turning our attention to the unloaded span, we observe that 
 the highest part of the flange will be in the fifth bay from the 
 pier, bay (8), where it will be nearly horizontal. It will make an 
 angle therewith 
 
 = -07577 - (e, 9 + tt + 6> + 6 g + ^} = -00304. 
 
 The flange of bay (7) will slope with an angle of 
 
 12 + 0n + M + 0. + 0. + | = '004085 
 in the contrary direction. 
 
 Therefore the highest point of the flange will be in the line 
 dividing bays (7) and (8). 
 
 The rise or 8 = 5 jsin (-07577 - |- 2 ) + sin (-07577 - 12 - 
 + sin (-07577 - 12 - U - *p) + sin (-07577 - 12 - tt - 10 - 1 
 
 + sin (-07577 - - 6 n - 10 - 9, - )} 
 
 = 5 {sin -06759+ sin '05061 + sin -03311 + sin -01606 + sin -00304} 
 = 5 (-0011797ft. + -0008831 ft. + -0005779 ft. + -0005779 ft. 
 
 + -0002803 ft. + -0000530 ft.) = '01487 ft. 
 or about ^ ths of an inch. 
 
 The drop from the summit towards d the right-hand abut- 
 ment or S 
 
 (since "004085 = slope of flange in bay (7) and '004085 +- r 
 
 = -00654, the slope with the vertical of the line dividing bays 
 (7) and (8)) 
 
96 CONTINUOUS GIRDERS. 
 
 = 5 {sin -004085 + sin (-00654 + %} + sin ( -00654 + 0. + 
 (. V */ V * 
 
 + sin (-00654 + 0^ + 6 - \ + sin (-00654 + 0, + <9 5 - 6, - -*} 
 + sin (-00654 + ft + e -0 4 -0 s -) 
 
 + sin f -00654 + 6 + B - 4 - 3 - 2 - | 
 \ 2 
 
 = 5 (sin -004085 + sin '00779 + sin '00933 + sin '00938 
 
 + sin -008465 + sin '006055* + sin '00244) 
 
 = 5 (-0000713 ft. + -0001359 ft, + '0001628 ft. + '0001637 ft. 
 
 + -0001477ft. + '0001057 ft. + '0000426 ft.) 
 = -00415 feet. 
 
 01487 ft. - '00415 ft. = '01072 ft. = J th of an inch, 
 
 the height of the extremity d of the girder above the abutment. 
 But as by hypothesis the end of the girder rests upon the abut- 
 ment at d, it is obvious' that in order to fulfil this condition of 
 things, the angle of the bays between points a and e, fig. 71, 
 should be smaller, or those between e and the point of contrary 
 flexure in the unloaded span, that is to say the angles of the part 
 acting as cantilever, should be larger. Both these effects will 
 result if the points of contrary flexure be taken further from the 
 pier than they are in the present example ; thus proving that 
 the length of the girder portion in the loaded span cannot in 
 practice be as long as that we have assumed, and therefore our 
 girder is so far safe, since the maximum strain on the portion 
 over the pier is obtained when both spans are fully loaded, and 
 the maximum strain on the girder portion is less than that we 
 have allowed for. 
 
 . " As the point of contrary flexure in the unloaded span will fall 
 further from the pier than we have supposed, for safety's sake we 
 will assume the unloaded end of the girder as being lifted off the 
 pier. By this means we get the whole of the unloaded span in the 
 condition of a cantilever. By working out the strains produced by 
 this condition of things we find that on the flanges of the third and 
 
CONTINUOUS GIRDERS. 97 
 
 fourth bays, counting from the pier each way, there will be strains 
 of 52f and 41| tons respectively, instead of 36J and 31 J tons, as 
 given in fig. 3, Plate IX. To meet this it will be necessary to 
 have two thicknesses of J inch plate in bays 9 and (9), and to con- 
 tinue the -fg plate through in bays 10 and (10). 
 
 The question now arises whether we might not save metal. 
 
 By the proportion be : be :: w : (w+w'} (p. 90), we get bc= 15ft. 
 when be = 60 ft., that is, when c' coincides with the extremity of 
 the girder at d (fig. 70). It would, however, be manifestly unsafe 
 in fixing the position of the point of contrary flexure in the left- 
 hand span to assume the girder to be lifted off its bearing at d, 
 when the left-hand span only is loaded, for an almost inappreciable 
 error in the fixing or construction of the girder would prevent this 
 from taking place, and the result would be excessive strain on the 
 girder portion of the loaded span arising from increased span. 
 Again, the end of the girder might be lifted off the abutment to an 
 extent that would produce a dangerous jar when the load came on 
 to the unloaded span: for a species of blow would then be given 
 to the abutment by the underside of the girder, destructive of both. 
 If we assume be' = 50', the utmost we can with safety, then be = 
 12' 6". 
 
 That is to say, the point c would fall in the centre of bay 10, 
 and onr girder would be 2' 6" shorter than what we have assumed. 
 
 Very little metal would be saved hereby, and that at increased 
 risk. An error of but of an inch in the relative height of the 
 pier and abutments would only be just covered by our arrangement 
 in Plate IX. As it is, we know c' must fall further away from the 
 pier than we have assumed. The extra metal in the girder portion, 
 by lessening the value of 6 for the bays of the girder portion, helps 
 to keep c' nearer the pier than would be the case were the metal 
 in the girder portion reduced to the utmost limit of safety. 
 
 We conclude then, that considering all circumstances, it would 
 not be safe to take c further from the pier than we have done. 
 
 From the foregoing example the reader will perceive how much 
 of uncertainty and guess-work belongs to the calculations for con- 
 tinuous girders. In actual practice engineers do not think of thus 
 elaborately treating every case of continuous girder that comes 
 before them. They know by experience where to locate the points 
 Y. 7 
 
98 CONTINUOUS GIRDERS. 
 
 of contrary flexure so as to keep within safe bounds. In small 
 bridges the labour of calculation would not be repaid by the metal 
 saved in the bridge. In large and important bridges the chances 
 of error are very much diminished, because the proportion of dead 
 weight is much increased, and the amount of variation in the 
 position of the point of contrary flexure is proportional to the 
 difference between w and w + w', as shewn on p. 90. 
 
 In our example the variation of the point c amounts to 10 ft. or J 
 of the span, and c is taken as distant f of the span from the pier. 
 In calculating for a large bridge where the dead load reaches a high 
 figure, we should be careful riot to take c' too far from the pier. 
 We might perhaps take it so far as to get the cantilever portion of 
 the loaded girder as long, or even longer than, one third of the span ; 
 and as we know it must be less than this when the adjoining span 
 is unloaded, we know that c must be assumed at such a distance 
 from the pier as will give it a reduced length. 
 
 No universal rule can be given. The safest plan for the 
 student will be to work out carefully a few cases for himself, after 
 the manner of the preceding example, taking different proportions 
 of live and dead load, and thus satisfy himself as to what should be 
 the proportions of cantilever and girder in all cases likely to occur 
 in practice. 
 
 Two unequal spans. When the spans are unequal the deter- 
 mination of the position of the points of contrary flexure is still 
 more laborious. 
 
 Let fig. 72 represent a continuous girder of two unequal spans 
 I and I', having a uniformly proportioned strength. 
 
 When both spans are fully loaded, in which case we may 
 assume the Unit of load to be alike for both, we find by equation 
 No. 1 (p. 90) 
 
 I . be = I' . be'; 
 whence the proportion be ib'c :: I' : I. 
 
CONTINUOUS GIRDERS. 99 
 
 In this case the centre of curvature for the cantilever portion 
 will not, as in that of the girder of two equal spans (p. 88), fall in 
 the centre line of the pier. For since the rate of curvature for 
 all parts of the girder is assumed to be alike, it is obvious that the 
 flange at c will make a greater angle of slope with the horizontal 
 than the flange at c, I being the greater span, and as moreover 
 the pier is nearer to c than to c', it is plain that the flange at 6 
 must make an angle with the horizontal, and the perpendicular 
 to the flange in which lies the centre of curvature, must fall to the 
 right of the pier, as shewn by the dotted line in fig. 72. 
 
 To find point of c. flexure when both spans are loaded. The 
 
 following is a rough practical method of determining the position 
 of the points of contrary flexure when both spans are fully loaded. 
 
 Draw with the same radius three curves touching one another, 
 and forming an undulating line similar to acbc'd, fig. 73. 
 
 Join the centres of adjoining curves by straight lines cutting 
 the curves at their points of contact in c and c. 
 
 Divide the distance cc' into two parts be and 6c', 
 
 such that be : be' :: I' : I. 
 
 Through the point b draw a straight line cutting the two side 
 arcs in a and d, so that 
 
 ab : bd :: I : I'. 
 
 This may be done by fixing a needle in the paper at b, and 
 keeping against it the edge of a straight edge, which should be 
 moved about until its edge cuts the arcs at a and d, so as to give 
 the proper proportion between the spans. 
 
 By the diagram so obtained the proportion between the canti- 
 lever and girder portions of the structure is made visible at a glance. 
 
 We have now to determine the positions of the points of con- 
 trary flexure when one span only is loaded. 
 
 Adopting the notation made use of in the case of fig. 70, we call 
 the point of contrary flexure in the loaded span c, and that in the 
 unloaded span c. 
 
 72 
 
100 CONTINUOUS GIRDERS. 
 
 Longer span loaded. When the longer span only is loaded 
 the position of c in the shorter span should be taken nearer to 
 the pier than in the case of equal spans, and the greater the dis- 
 proportion in the spans the more should c' approach the pier. 
 The object of this precaution is to prevent the end of the unloaded 
 span d being lifted off the abutment, a contingency very likely 
 to occur if there is a deficiency of metal in the girder portion of the 
 longer span, in consequence of the large angle of slope which the 
 flanges would then make with the horizontal at the point c. 
 
 The actual position of c in practice will, however, be nearer to 
 the abutment d than in the case of the equal spans. From this it 
 will be understood that the girder portion of the longer span will 
 .be subject to less strain than the other parts of the structure. 
 
 Extremely disproportioned spans. If the disproportion be- 
 tween the spans be very great it may be impossible to prevent the 
 unloaded end of the girder from lifting off the abutment. 
 
 Ft'ff- 74. 
 
 Fig. 74 explains this pictorial ly. Here the structure over the 
 longer span is treated as entirely girder, which occurs when be 
 has a minimum value = 6. Notwithstanding this the slope of the 
 flange at the pier is so great as to lift up the end of the shorter 
 girder from the abutment. By putting a great deal of extra metal 
 in the longer span it might be possible so to flatten the curvature 
 of the girder as to avoid this defect, but as even now we have suffi- 
 cient metal in the structure to enable the longer span to support 
 itself alone, it is obvious that the proper thing is to span the two 
 openings by unconnected girders. 
 
 Shorter span loaded. To find what portion of the shorter 
 span when it alone is loaded should be taken as girder, determine 
 the position which the point of contrary flexure would occupy if 
 the adjoining span were equal in length to the shorter span and 
 unloaded. 
 
 The true position will be somewhere between this and its posi- 
 tion when both spans are loaded. The greater the disproportion 
 
CONTINUOUS GIRDERS. 101 
 
 of the two spans the nearer it will approach to this latter point, 
 and of course the longer will be the cantilever portion. 
 
 We cannot do more than indicate the law which regulates this 
 variation in position of the point of contrary flexure. The engineer 
 must exercise his own judgment in fixing its position. The limits, 
 as we see, are pretty well known. The only precaution to be ob- 
 served is not to take the point of contrary flexure too far from the 
 pier, when estimating the strains on the girder portion. Of course 
 it would be possible to find approximately the positions of the 
 points of contrary flexure by calculation in the manner of the 
 last example, but the labour would be doubled in cases where the 
 spans are unequal, and the result would be rendered inaccurate by 
 such slight irregularities in the methods of constructing or fixing 
 the girder, that it would be a waste of time to make the attempt. 
 
 Anchoring down the shorter girder. It is sometimes advisable, 
 when the spans are very disproportionate, to obtain the effect of 
 continuity by weighting or fastening down the abutment extremity 
 of the shorter girder. In such cases the shorter span acts entirely 
 as cantilever when the longer span only is loaded. 
 
 As the most usual cases of anchoring down occur in bridges 
 having two short side spans and one centre long span, we will 
 indicate the method of determining the amount of keying up neces- 
 sary to give the required position of the points of contrary flexure 
 in such three-span structures ; a precisely analogous method will 
 enable the engineer to do the same for a two-span bridge. 
 
 In consequence of the facility afforded by a keying apparatus, 
 it is possible to locate the points of contrary flexure in the longer 
 span at convenient places within limits. 
 
 Suppose we desire that they should occur at a distance from 
 
 the pier = - of the longer span. 
 
 Let abcdefg (fig. 75) be a representation of the curve assumed 
 by the lower flange of the girder. Then, as the structure will 
 receive nearly its greatest strain everywhere when the centre 
 
102 CONTINUOUS GIRDERS. 
 
 span only is loaded, we may take for granted that the metal in the 
 flanges has been arranged so as to give an approximately uniform 
 working strain throughout, and consequently that the rate of 
 curvature is the same everywhere. 
 
 That being so r it is plain that since the arcs cd, ef are each of 
 them by construction half the length of the arc de, the radii of 
 curvature of all three arcs 'must be parallel vertical lines; and if 
 distances cb and fg equal to cd and ef be marked off on the curve 
 of the flange of the shorter spans, the four points 5, d } e and g will 
 be in the same straight line. 
 
 This line is a certain vertical distance below the bearings on 
 the piers at c and f, which distance can be ascertained either by 
 
 z 2 // 7 
 calculation or experiment. The formula S = j-r (page 78) will 
 
 answer for the former method^ for the latter, it will be necessary 
 to key up the 'bridge by guess approximately to its true position, 
 then to load the centre span, and observe by measurement the 
 depression of points d and e. If the depression given to the points 
 b and g by the keying up be incorrect, the amount of keying up 
 should be altered until the lines 6, d, e and g are in one straight 
 line. 
 
 As most bridges are built with camber, to simplify matters 
 it is well to set out on the web or upright gussets of the girder 
 when unstrained a horizontal line to be the sole guide for observa- 
 tion. When a bridge of this kind is loaded all over, the curves 
 ac, fh become slightly flatter as the strain on these parts is some- 
 what relieved. The effect of this is to shift the points d and e a 
 little further from the pier. 
 
 . Multiple spans. We come now to continuous girders of three 
 or more spans. 
 
 f iff. 76* 
 
 Three spans equally loaded. Let the curved line abcdefyh 
 (fig. 76) represent a continuous girder of 3 spans and of uniformly 
 proportioned strength, under deflection from a uniformly distributed 
 load. 
 
CONTINUOUS GIRDEKS. 103 
 
 Let b, d, e and g be points of contrary flexure. Then, if the 
 points of support a, c,/and h are to be in one horizontal line, the 
 span of the centre opening must be to the span of the side open- 
 ings as 12 is to 10. 
 
 For when the centre of curvature of the cantilever portions 
 Id and eg lies in the vertical line passing through the points of 
 
 support c and f t the length of the cantilevers be and fg will be 
 g 
 of the spans ac and fh (see p. 88). And since the girder is by 
 
 assumption of uniformly proportioned strength, the radius of curva- 
 ture of the girder portion de is equal to that of the cantilever 
 portions bd and eg, and consequently the length of de is equal to 
 that of bd or eg\ wherefore the whole distance cf=4< times cd. 
 
 Now, if ac = 10, 6c = 3, cc = 3 and de = 6, 
 
 .'. spanac = 10, span cf= 12 and span fh = 10. 
 
 If therefore we wish to bridge an opening with a continuous 
 girder of three spans, these proportions of spans will be the most 
 convenient for purposes of calculation and erection. And if we use 
 more than three spans, the intermediate spans should be all alike, 
 and bear the same proportion to the two side spans as does the 
 centre span in fig. 76, viz. 12 to 10. 
 
 Spans unequally loaded. It is seldom, however, that multiple- 
 span continuous girders of any size are required to carry only a 
 uniform dead load. In the great majority of cases they are used 
 for railway purposes, where the- live load is large in proportion to 
 the dead load. In such cases the most trying position of the live 
 load for the girder will generally be when the two side spans only 
 are loaded, if there are but three spans. 
 
 The girder abcdefg (fig. 77) is a representation of a continuous 
 girder of three equal spans, of which the two side spans only are 
 loaded. In this case the proportion of live to dead load is assumed 
 to be so great as to throw the point of contrary flexure in the 
 
104 CONTINUOUS GIRDEES. 
 
 unloaded span so far from the pier as to pass the centre of the span 
 at d. Under these circumstances, both side spans being loaded, 
 there is no contrary flexure in the centre span, but the whole of 
 the top flange is in tension, and the bottom flange in compression. 
 
 Fig. 78 represents the same girder under similar conditions of 
 loading, except that the proportion of live to dead load is such 
 
 
 that the points of contrary flexure d, e in the unloaded centre 
 span are at a distance of less than half the span from the piers 
 c and/. 
 
 As in the case of the two-span girder, so here it is evident that 
 the position of the points of contrary flexure in the loaded span 
 will depend upon the curvature of the unloaded span. 
 
 In order to obtain the position of the point of contrary flexure 
 with approximate correctness, it would be necessary in the first 
 place to assume its position under different conditions of loading, 
 to arrange the metal in the flanges so as to give sufficient area of 
 metal to take the strain in each bay, to calculate the amount of 
 bending in each bay, and thence deduce the slope of the flange at 
 the point of contrary flexure in a manner similar to that employed 
 in dealing with the two-span girder, p. 91, and following pages. 
 
 If the slope at b (fig. 77), as derived from a calculation of the 
 bending of the portion of the girder bd, agree with that obtained 
 by calculating the bending of the portion ab, the position of the 
 point of contrary flexure which has been assumed is the correct 
 one ; but if not, it must be shifted in a direction which will tend to 
 make the slopes so obtained by calculation agree. 
 
 The case of fig. 78 is very unusual, and as the variation in the 
 position of the points of contrary flexure in the side spans in such 
 a case is very slight, it is easy to locate them so as to ensure safety 
 without wasting metal. The ordinary case is that of fig. 77. 
 
 Diag. 1, Plate X. gives the strains on a continuous girder of 
 three equal spans, when the two side spans only are fully loaded. 
 In this case it will be seen that supposing the point of contrary 
 
CONTINUOUS 
 
 flexure in the side spans to be at a distance of 
 span from the piers, the lifting force at the middle of the centre 
 span will be equal to 12J, the strain on the flanges being that 
 which would be caused by a load of 12J, supported at the centre 
 of a girder of the same length as that of the centre span. The 
 span, loading of the girder, and position of the point of contrary 
 flexure in the side span, are taken the same as in Diag. ], Plate 
 IX. for comparison's sake. 
 
 Diag. 2, Plate X. shews the strains on the flanges when the 
 centre span only is loaded, assuming the points of contrary flexure 
 to be one-sixth of the span from the piers, and Diag. 3 gives the 
 strains for the centre span when the whole bridge is loaded, and 
 the points of contrary flexure for the centre span are taken at a 
 distance of one-fourth the span from the piers*. This position 
 gives a strain of only 67J over the piers, and whereas there is a 
 strain of 60 at these points when the side spans only are loaded, 
 it is clear that the difference between the curvature of the centre 
 span for the condition of a strain on the top flange at the centre 
 of 37 J in tension, and for the condition of 27 J in compression at 
 the same point, would alter the curvature of the side spans more 
 than is implied by the difference of only 7J in tension over the 
 piers. We therefore judge that the positions of the points of 
 contrary flexure in the centre span should be nearer the centre of 
 the span, or more in accordance with that shewn in fig. 78. 
 
 If equality of span be indispensable, economy may be effected 
 by lowering the bearing of the girder on the abutments, so as to 
 throw the points of contrary flexure, b and f (fig. 77), further 
 from the piers. 
 
 Reference to Diag. 1, Plate X., will shew that under this con- 
 dition of loading the centre girder is strained to its maximum 
 
 * It happens that in the examples given the points of contrary flexure have been 
 assumed to fall at the line dividing two bays, they may however fall at any inter- 
 mediate point in a bay. When it falls in the centre of a bay there is no strain on 
 the flanges, if the web be doing equal duty in tension and compression. There will 
 be strains in tension and compression on the flanges of the bay according as the 
 point falls nearer the girder side or cantilever side of the bay. Diagrams 4 and 5, 
 
 Plate X. are examples. In Diagram 5 the point falls j of the bay from the girder 
 side of it. 
 
106 CONTINUOUS GIRDERS. 
 
 almost throughout its whole length. Now Diag. I, Plate IX. shews 
 that in the case of the two-span girder the unloaded span is, for 
 the greater part of its length, only slightly strained, and conse- 
 quently its bending is much less than that of the former girder, 
 whence it results that the angle of slope with the horizontal of the 
 flange at the piers is greater in the three-span than in the two- 
 span girder, and therefore the position of c (fig. 70, p. 89), the 
 point of contrary flexure in the unloaded span, falls nearer to the 
 pier in the three-span girder than in the two-span. 
 
 Practical rule for determining the position of the point of 
 contrary flexure in three-span girder. In fixing the position of 
 the point of contrary flexure, the course we recommend is to de- 
 termine in the first place the position of the point c (fig. 69),, on 
 the assumption that we are dealing with a two-span girder, the 
 unloaded span of which corresponds with the centre span of our 
 three-span girder, in the manner pointed out on pages 89 and 90. 
 
 Having done so, to take for the three-span girder the position 
 of point c as somewhat nearer the pier. How much nearer is a 
 matter upon which the engineer must exercise his discretion, as it 
 will be greatly dependent upon the difference between the live 
 and dead loads. If he determines to lower the bearing on the 
 abutment, c need not be taken nearer to the pier. 
 
 Girders of more than three spans. When a continuous girder 
 spans four openings, the two centre spans are not subject to so 
 much strain as the centre span in a girder of three spans, for the 
 upward transverse strain upon them is less. 
 
 Continuous girders of varying depth. Economy may, in gene- 
 ral, be effected by increasing the depth over the piers whereby the 
 area of the flanges at this point is reduced, at the expense, how- 
 ever, of an increase in the amount of web. The greater local 
 stiffness thus produced will affect the position of the points of 
 contrary flexure; its tendency is to throw them further from the 
 piers. 
 
 When the load to be carried is entirely dead, or when the pro- 
 portion of live to dead load is small, the system is undoubtedly 
 attended with considerable economy; but when the proportion of 
 live to dead load is very large, the strains to which the unloaded 
 
CONTINUOUS GIRDERS. 107 
 
 spans are subjected, in consequence of the greater length of the 
 cantilever portions in the loaded spans, are increased, and much of 
 the saving in the side spans counterbalanced. 
 
 Figs. 79 and 80 represent girders of this kind of three spans; 
 in fig. 79 the girder portions have parallel flanges ; in fig. 80 they 
 are hogbacked. 
 
 In these examples the position of the points of contrary flexure 
 will fluctuate more or less on either side of the points b, c, d and e t 
 according to the manner in which the girder is loaded. 
 
 In the example, fig. 81, the depth of the girder is assumed to 
 be nothing at the points b and e in the side spans, the structure 
 must therefore be hinged at these points. The points of contrary 
 flexure in the side spans are thus fixed under all conditions of 
 loading to b and e. The advantages attending this peculiarity of 
 construction are, that the strains on all parts of the bridge can be 
 calculated with certainty and ease, while the metal, which would 
 be required in the examples of figs. 79 and 80 in the flanges in 
 the neighbourhood of the points b and e to provide for the fluctua- 
 tion in the positions of the points of contrary flexure, is saved. 
 
 The economy and convenience of this system is most conspicu- 
 ously shewn when circumstances require a large centre span and 
 two short side spans (see fig. 82). As in case of fig. 75 already 
 mentioned, the ends of must be anchored down to the abutments. 
 The points of contrary flexure being fixed, all uncertainty about 
 
108 . CONTINUOUS GIRDERS. 
 
 the strains disappears, and no necessity exists for that extreme 
 nicety in adjusting the bearings of the girder upon which stress 
 
 r 
 
 was laid in the case of fig. 75, and which is required in all other 
 kinds of continuous girders. 
 
 The structure shewn in fig. 82, properly speaking, does not 
 come under the category of continuous girders; continuity is broken 
 at the points c and d. It must be considered as a compound 
 structure formed of a girder supported by two cantilevers. 
 
 Practical remarks. Continuous girders of more than three 
 spans are rarely to be met with. The difficulties of erection and 
 the niceties of construction requisite to ensure the full advantages 
 of continuity are, to many engineers, serious objections to their use. 
 The large amount of contraction and expansion in long girders is 
 another objection. 
 
 The attention of the reader is once more called to the import- 
 ance of careful superintendence in the construction and erection of 
 continuous girders. The chief precaution to be observed is, that 
 the girders be so manufactured that when they are placed in situ 
 they shall not be subjected to initial strains except such as the 
 engineer may desire. A simple means of securing this result is to 
 bone a horizontal line on the side of the girder, before it leaves the 
 manufacturer's yard, making a notch with a file in the gussets at 
 the bearings. When the girder rests on its bearings, if it be 
 found that these notches are not in the same straight line, the 
 girder must be packed up, or the bearings lowered so as to make 
 them so, unless it be desired to put an initial strain upon some 
 parts of the girder. The amount of this strain will depend on the 
 deviation of the girder from the original straight line. 
 
 As any settlement of the piers or abutments would seriously affect 
 the strains in a continuous girder, it is obvious that very great care 
 must be 'taken that they are well and substantially built. By 
 means of the horizontal line boned along the face of the girder, 
 settlement of bearings could be observed, and remedied by packing 
 up the girder. 
 
CHAPTER IX. 
 
 EOOFING TRUSSES. 
 
 THE great variety of forms adopted for the principals of roofs 
 precludes a notice of any but types of the most common. 
 
 The most elementary form of principal is that shewn in fig. 83. 
 It consists of two inclined rafters 
 ab and be, the upper extremities 
 of which abut against each other, 
 while the lower extremities are 
 connected by a tie rod adc, gene- 
 rally cambered at the centre of 
 the span by means of a vertical suspending bar. 
 
 The strains on a principal of this character may be expe- 
 ditiously found in the following manner. 
 
 Draw through the point of intersection of tie rod and rafter 
 the horizontal line ae. 
 
 Erect a perpendicular to the line ae. 
 
 With any convenient scale mark off a distance on this perpen- 
 dicular representing the total load on the abutment, i. e. half the 
 whole load on the principal*. 
 
 Through the point so obtained draw a line parallel to ae, 
 intersecting the centre line of the rafter ab in f. 
 
 Through /drop the perpendicular^^ to the line ae. 
 
 The scale applied to af will give the strain on the rafter : to 
 
 * By. "the whole load on the principal" is meant the load which may be 
 assumed to be concentrated at the apex b in this case. The reaction of the abutment 
 must be assumed to be the share of the load which comes upon it minus the pres- 
 sure produced upon it by the load carried by the transverse strength of the rafter 
 itself. Thus, supposing the distributed load on the principal abc, Fig. 83, to be 
 4, the rafter abc (Fig. 83) carries a load of 2 by its own transverse strength, and 
 the point a is kept in equilibrium by three forces, the thrust of the rafter, the pull 
 of the tie-rod, and the vertical reaction of the abutment minus half the load carried 
 by the rafter or 2-1=1. 
 
110 ROOFING TRUSSES. 
 
 ag, the strain on the tie, and to gh, half the strain on the sus- 
 pending bar. 
 
 Or the strains may be found by calculation thus. 
 As be : ab : ae : 2de :: load on abutment : strain on rafter 
 
 : strain on tie : strain on suspending bar. 
 
 Fig. 84 is a diagram of a principal in which the rafters ac 
 and ce are supported at intermediate points, b and d, by struts 
 If and df. If W be the total load on 
 
 W 
 the principal, there will be a load of 
 
 concentrated at each of the points b, c, 
 and d. 
 
 The strains on a truss such as that 
 shewn in fig. 84 might be determined by considering it as one 
 primary, and two secondary trusses, finding the strains on the 
 secondary trusses abf and fdc, and afterwards on the primary 
 truss ace. A similar method may be adopted with the more 
 complicated trusses illustrated in figs. 86, 87, and 88, but we 
 think the simplest method is that indicated on pp. 24 and 25, 
 viz. to find the bending moment at different parts of the truss, 
 and thence deduce the strains on the subsidiary parts. 
 
 We propose therefore to adhere to this method as much as 
 possible throughout these investigations. 
 
 In the case before us, let us first find the bending moment on 
 the vertical line passing through the point b, and intersecting the 
 tie in the point g. It will be obviously 
 
 SW 
 
 - x horizontal distance between a and g 
 
 o 
 
 bg 
 
 = horizontal element of the strain on the tie ag 
 
 rafter ab. 
 
 Again, horizontal strain at/ 
 
 3 W /horizontal distance \ W /horizontal distance \ 
 8 V between a and / / 4 \ between g and / / 
 
 "7T" 
 = horizontal strain on rafter be. 
 
EOOFINQ TRUSSES. 
 
 Ill 
 
 Now vertical strain on be 4- vertical strain on cd 
 
 W 
 
 = vertical strain on cf, 
 
 and horizontal strain on af horizontal strain at/ 
 
 = horizontal strain on bf. 
 Also for checking purposes vertical strain on bf 
 
 strain on cf ,. , . 
 
 = - ~ vertical strain on af. 
 
 Fig. 85 shews another common arrangement of parts for roofs 
 of small span. The bars are marked 
 with the 4- and signs, to shew which 
 are in compression, and which are in 
 tension. 
 
 To find the strains on a truss of this shape, it will be neces- 
 sary to find the horizontal element of the bending moment on the 
 parts at three vertical planes passing through the points b, f, 
 and c, in the manner indicated in the foregoing example. 
 
 Figs. 86, 87, and 83 are examples of roofs of larger span. The 
 strains may be found by taking the bending moment at vertical 
 sections in fig. 86, through the points b, h, c, i, and d ; in fig. 87, 
 through the points b, c, and d\ and in fig. 88, through the points 
 b, k, c, and e ; for fig. 88, however, the strains on the subsidiary 
 
112 HOOFING TRUSSES. 
 
 trusses cdeo and efgp must be obtained separately and added to 
 the strains obtained by the first process. 
 
 The strains on the top and bottom members in figs. 86, 87, 
 and 88, being thus ascertained, the strains on the diagonal struts 
 and ties are easily found by methods similar to those adopted for 
 finding the strains on the diagonals of the hogbacked and bow- 
 string girders. 
 
 Arched Roofs. Roofs of large span are generally of an arched 
 form. 
 
 First System. Fig. 89 represents a plain arched principal 
 intended to act mainly by compression. 
 
 If the only load that could come upon the principal were that 
 arising from its own weight, by making 
 the arch of the form of an inverted cate- 
 nary, a very shallow rib might be used, 
 as the line of pressure would always fall 
 within the flanges ; but since all roofs are 
 subject to unequal loading, arising either 
 from the pressure of the wind or snow, it does not follow that 
 a catenary will be the best form for our rib. It will also be subject 
 to transverse strain at times, and should therefore have a con- 
 siderable depth. 
 
 To determine the form and scantling of the various parts of 
 the rib. 
 
 Having settled upon the amount of rise that it is desirable to 
 have for the crown of the arch, draw the curve of equilibrium for 
 the states of equal and unequal loading through the springing 
 points a and e, and the crown c (see p. 46, and following pages). 
 Draw the dotted line abode, representing the position of the curve 
 of equilibrium when the roof is acted upon by a high wind coming 
 from the left in the direction of the arrow. Suppose the line to 
 fall outside the centre of the rib by a maximum distance bf in 
 the one case, and inside by a distance dg. The most economical 
 form of rib will be that which makes the distances bf and dg 
 equal. Although the rib so obtained may not be graceful, any 
 alteration of form will be at the expense of economy. 
 
 The method of finding the strain on the rib at f and g is 
 
ROOFING TRUSSES. 113 
 
 given' on p. 53. The thrust of the rib is taken by the abut- 
 ments a and e. 
 
 Second System. The resistance to the distortion of the rib 
 may be effected by bracing, as in the ordinary bowstring truss. 
 In roofing trusses the tie is generally very mwh cambered. 
 
 The strains on the top and bottom members of the truss shewn 
 in fig. 90, may be obtained by treating it as a girder of varying 
 
 Fiff. 90. 
 
 depth, the mode we have already adopted when dealing with 
 hogbacked and bowstring girders. 
 
 The most convenient way of finding the strain on' the diagonals 
 will be to proceed as follows. Ascertain the vertical or shearing 
 force on each bay (see pp. 16 and 43) and the strains on the top 
 and bottom members of each particular bay in that condition of 
 loading which gives this shearing force upon it. Resolve the 
 strain of the top and bottom members of the bay into their hori- 
 zontal and vertical components, the difference (in this case) be- 
 tween these vertical components of the strains on the flanges will 
 tie the total shearing force taken ly them; this amount must be sub- 
 tracted from the total shearing force on the bay, the remainder 
 will be the shearing or vertical forte borne by the diagonals. 
 
 If both the diagonals in the bay are struts only or ties only, 
 the whole of this force will be borne' by one of them' alone. If 
 they are capable of acting both as strut's and ties, it will be shared 
 between them. The amount taken by the one or the other will 
 depend mainly upon the sectional areas of metal in the top and 
 bottom members of the bay. 
 
 Example. Diagonals, both Ties. Let us by way of example 
 investigate a case in which the diagonals are both ties. 
 
 We will assume that when the truss is unequally loaded, it 
 is loaded in ; a-" manner similar to that shewn in Diagram 4,, 
 Plate VI. 
 
 Y. 8 
 
114 ROOFING TRUSSES. 
 
 Now the shearing force on bay No. 2 is 24 6 = 18, and the 
 strains on the flanges db and cd may be found by the method 
 described on pages 25 and 41. Thus : 
 
 If I be the span, and ^ = the length of one bay, 
 
 I 
 
 horizontal strain on cd = 24 x 
 
 ac 
 
 24x4- 
 
 bd 
 
 The horizontal elements of the strains on the flanges being 
 known, the vertical are easily calculated. 
 
 If m = the vertical thrust of the top flange, 
 
 and n = the vertical pull of the bottom flange, 
 
 m n their total shearing effect, since they counteract one 
 another ; 
 
 and 18 (m n) is the vertical element of the strain on the 
 diagonal ad. 
 
 In this manner the strain on the acting diagonal of each bay 
 may easily be found. 
 
 When the left side only of the truss is heavily loaded, the 
 other set of diagonals come into play. Thus, cb and gf will be 
 subject to strain while ad and eh will be unstrained. 
 
 Third System. Fig. 91 represents an arched rib with parallel 
 flanges resting on horizontal bearings having one end free to 
 move. 
 
 Equally Loaded. When the load acts in a vertical direction 
 only, the strain on the flanges at any point may be easily found 
 by the method described on p. 25 for finding the strain at any part 
 of a girder. 
 
 Thus, required the strain on the top and bottom flanges of the 
 rib at point c. 
 
HOOFING TRUSSES. 
 
 115 
 
 We have a force equal to one half the whole load on the rib, 
 
 W 
 or -- , pressing upwards in the direction of the line aa, and acting 
 
 at a leverage of x, tending to break off the part ac from the re- 
 mainder of the rib cb. This force is partially counteracted by 
 loads at points e and / resisting the tendency to break at c ; 
 therefore the moments of the load at e and /, acting respectively 
 at leverages represented by their horizontal distances from a 
 vertical line passing through point c, must be deducted from the 
 
 W 
 
 force -5- x x ; the remainder divided by the depth of the rib at c 
 
 will give the strain on the flanges at that point. 
 
 The method of ascertaining the strain on the diagonals has 
 just been described (p. 114), and consists in ascertaining the total 
 shearing force to be borne by any particular bay, deducting from 
 that the amount taken by the flanges, and assuming the re- 
 mainder as borne by one or both diagonals according as they are 
 ties only, or struts also. 
 
 ' Unequally Loaded. The action of the wind upon trusses of 
 this description is very powerful, on account of their great com- 
 parative height : it considerably complicates the process of deter- 
 mining the strains upon the various parts. Perhaps the most 
 satisfactory method will be to resolve the force of the wind into 
 its vertical and horizontal components; to treat the rib as subjected 
 to a system of unequal vertical loads, and obtain the strains so 
 arising ; then, making another diagram of the rib, to consider it as 
 acted upon by the horizontal forces only. The sum of the two 
 
 82 
 
116 ROOFING TRUSSES. 
 
 systems of strains so obtained will give the actual strain upon 
 each part. 
 
 Force of Wind. The force of the wind will act perpendicularly 
 to the curve of the rib ; but for the purpose of ascertaining this 
 force it will be sufficient to assume the wind as acting perpendicu- 
 larly to a tangent to the rib at its junction with the diagonals. 
 
 Thus, to find the force of the wind at the point b of the rib 
 (see fig. 92), draw abc tangential to the curve 
 of the rib at b, making the distances ab and 
 be respectively equal to half the lengths of the 
 bays on each side of the point b. The distance 
 ac x the distance between the principals is the 
 surface exposed to the wind, and we may take 
 the angle made by abc with the horizontal as that at which the 
 wind acts. 
 
 Pressure of Wind on Inclined Planes. The pressure of wind 
 on inclined surfaces is a subject on which until quite lately we 
 were without any information whatever. Under the auspices of 
 the Aeronautical Society of Great Britain, a few rough experiments 
 have been tried, from which the following table has been compiled 
 as a sufficiently approximate guide to the engineer in calculating 
 the load on a roof from the pressure of wind. 
 
 Pressure of wind on plane 1 foot square being taken as 1, 
 
 Pressure of wind vertically, not perpendicularly to the plane, 
 on same plane inclined at an angle with the horizontal of 
 
 15 vertical pressure = '4, horizontal pressure = *11 ; 
 20 ='5, =16; 
 
 4,XO -7 _ I 7. 
 
 I J> " ) 
 
 60 ='47, ='9. 
 
 These experiments were made with smooth planes and low 
 wind pressures, for high pressures and rough surfaces the hori- 
 zontal pressure should be taken higher than is given in the 
 table. 
 
 The maximum pressure of the wind in this country is about 
 40 Ibs. per square foot against an exposed vertical surface. Usually, 
 however, roofs are very much sheltered, and it is only in the case 
 
ROOFING TRUSSES. 117 
 
 of lofty and exposed buildings that we need assume the pressure 
 so high. 
 
 Let p lt p 2 , p 3 , Pi and p 5 (fig. 91) represent the horizontal 
 element of the pressures of the wind at the junctions of the bays 
 of the truss, and let l lt 2 , 1 3 , Z 4 and 1 5 be the leverages at which 
 these forces respectively act about the point 6, that being taken as 
 the fixed end of the truss and a as the free end. 
 
 Then M+M+M+M + PA = g = th e vertical pressure on 
 distance ab 
 
 the abutment a, and the lifting force at b. 
 
 We have therefore a horizontal force of p l + p a + p 9 + P 4 + P 5 
 on the bearing at 6, and a lifting force of - on each of the 
 
 Lt 
 
 flanges of the arch at that point. With these data we can, by 
 means of the polygon of forces, find the strains caused by the 
 horizontal force of the wind on the various parts of the structure. 
 When b is the free end and a .the fixed end, it will be necessary 
 to commence from the a end of the span, taking as our data 
 the same horizontal force as before, but a vertical pressure of 
 
 -^ instead of a lifting force of that amount on the extrados and 
 
 2t 
 
 intrados of the arch at a. 
 
 It must be remembered to take one system of diagonals only 
 at a time as doing duty, otherwise the problem is insoluble, and 
 to take into account the forces p lf p 2 , p s , p 4 and p 5 when resolving 
 the forces at the points where they impinge. 
 
 The method of estimating the effects of the vertical loads upon 
 the truss, being similar to that adopted in cases of the bowstring 
 and other roofing trusses, will, we trust, need no further explanation. 
 
 On Diagram 6 are given the strains on one half of a circular 
 rib due to the horizontal force of the wind calculated by means of 
 the polygon of forces. Being determined by a graphic method, 
 and not by mathematical calculation, they are only approximately 
 correct, but are sufficiently near for practical purposes. The bar 
 aa (Diagram 6) of the truss resting on the abutment is assumed 
 "to be held down by its centre to the abutment with a force 
 = 1@10 x 2 = 3220, the total lifting force of the wind; it is there- 
 
118 ROOFING TRUSSES. 
 
 fore assumed to be capable of standing a transverse strain of that 
 amount with safety. 
 
 The strains caused by the horizontal force of the wind may also 
 be found in the same way as those caused by its vertical effect 
 (see p. 115); for instance, the strain on the bottom flange about 
 the fulcrum 6 (Diagram 6, Plate X.) may be obtained by multiply- 
 ing the reactionary upward pressure of the abutment on the 
 right-hand side of the truss by its leverage about point b, and 
 subtracting therefrom the moments about the same point of the 
 forces, 2100 and 900, which tend to turn the truss in a contrary 
 direction (see pp. 4 and 5). The remainder, divided by the depth 
 of the truss, should give the strain on the bottom flange in 
 tension. 
 
CHAPTER X. 
 
 ECONOMY IN SUSPENSION BRIDGES. 
 
 (Reprinted from " Engineering.") 
 
 WHILE mathematicians and- engineers of the highest ability have 
 devoted themselves to elucidating the principles which govern the 
 construction of girder bridges of almost every variety of type, but 
 comparatively little of their attention has been bestowed upon 
 the question of economy in bridges on the suspension principle. 
 
 This neglect may perhaps be owing to the fact, that, in this 
 country at least, suspension bridges are but rarely constructed, 
 and consequently the element of wonder which attends the con- 
 struction of bridges of very large span has not yet generally 
 subsided. It is still regarded as so much a feat to throw an iron 
 structure over a clear span of from 600 to 1000 ft., that to criti- 
 cise its cost would appear ungracious. It is enough that the work 
 has been successfully carried out. 
 
 Every one who has given attention to the subject is aware of 
 the important part which the proportion of depth of span plays in 
 the economy of girders of various types, and, as might be expected, 
 it has received due attention at the hands of investigators. But 
 while we are not left in ignorance as to the most economical forms 
 of truss or girder, no one, so far as the writer is aware, has 
 attempted to point out the most economical proportions for the 
 suspension bridge, or the most advantageous method of disposing 
 the materials in its construction. 
 
 In view of these facts, it has appeared to the author worth 
 while to make some attempt to supply the deficiency by the ele- 
 mentary investigations presented in this communication, which, 
 
120 ECONOMY IN SUSPENSION BRIDGES? 
 
 wanting as they are in mathematical exactness, are yet, it is 
 believed, sufficiently approximate to be of real value to the prac- 
 tical engineer, and may, it is hoped, lead to further and more 
 elaborate treatment at the hands of the experienced mathema- 
 tician. 
 
 The main questions which it is now proposed to consider are 
 the two following : 
 
 1. What proportion should the height of the tower of the 
 suspension bridge .above the roadway bear to the span ? and 
 
 2. What should be the arrangement of parts in the super- 
 structure to obtain the greatest economy ? 
 
 In endeavouring to answer the first of these questions, let us 
 consider an elementary case. 
 
 Let us suppose that we have to support a given load W (fig. 1, 
 Plate XI.), halfway between the sides of a chasm, and are desirous 
 of knowing what is the most economical position for W. All that 
 is necessary is to construct a diagram such as that shewn in fig. 1, 
 Plate XL, in which AB and AC are the suspending bars, whose 
 .upper extremities B and G lie at the same level, and whose lower 
 ends meet in A, the point of application of the load. 
 
 Let the vertical dotted line AE represent the load W, then 
 the dotted lines ED and EF represent the strains on the bars 
 AB and A G, to which respectively they are drawn parallel ; and 
 since the sectional area of each bar is propior.tional to the strain 
 upon it, by multiplying each bar by its sectional area, a quantity 
 is obtained which represents the amount of material required to 
 carry the load in the given position. 
 
 W being equidistant between the points of support B and (7, 
 it will be found that the position of greatest economy is attained 
 when the bars AB and A G make an angle of 45 with the hori- 
 zontal, that is to say, when the load is depressed below the points 
 of support to an amount equal to half the horizontal distance be- 
 tween them. 
 
 If the point A be raised, the bars AB and AC are shortened, 
 it is true, but the strains ED and EF are more than propor- 
 tionally increased. On the other hand, if point A be lowered, the 
 strains are diminished, but the bars are more than proportionally 
 lengthened. 
 
ECONOMY IN SUSPENSION BRIDGES. 121 
 
 'The above statement is only true on the assumption that the 
 bars themselves have no weight. If this be taken into account, 
 the most economical position for the point A will be higher than 
 4)hat shewn in fig. 1, Plate XL, and the greater the weight of the 
 bars in proportion to the load W, the less will be the depression 
 of the point A below the points of support requisite to obtain 
 the greatest economy. 
 
 But few cases, however, occur in practice when it is possible 
 to make use of perpendicular cliffs as points of attachment for 
 t/he chains of a suspension bridge. It is therefore necessary to 
 take into account the cost of the towers which we have to raise 
 to carry our chains, and the back-ties, in order to arrive at a just 
 estimate of the cost of the whole structure. 
 
 For this purpose the diagram, fig. 2, Plate XL should be 
 constructed, in which (dealing with one side only for simplicity's 
 sake) AS is the suspending bar, BD the back-tie, and BE the 
 tower. 
 
 We have now to find that position of the point A which will 
 give (AB x its sectional area) + (BD x its sectional area) + (BE 
 x its sectional area) a minimum. 
 
 Assuming the sectional area of the bars AB and BD, and of 
 the tower BE, to be in the ratio of the strains upon them, this 
 will be attained when the depression of the point A below the top 
 of the tower ^ = '354 of the distance BG (nearly), that is to say, 
 when the height of the tower is rather more than one-third of the 
 span. 
 
 But since in practice, in order to obtain the necessary stability, 
 we require to make the sectional area of the tower much greater 
 in proportion to the strain upon it than is the sectional area of the 
 main chains, it will be obvious that the height of tower just given 
 will be too great to give the most economical practical result, and 
 that the larger the proportion which the sectional area of the tower 
 bears to the strain upon it, the less should be its height in pro- 
 portion to the span. 
 
 Fig. 3, Plate XI. is a diagram from which can be seen at a glance 
 that proportionate height of tower to span which gives the greatest 
 economy for any given sectional area of tower up to 24 times the 
 ratio of section to strain existing in the chains. The vertical 
 
122 ECONOMY IN SUSPENSION BRIDGES. 
 
 distance between the horizontal lines represents the half span. 
 The heights of the tower are represented by the vertical ordinates 
 from the lower horizontal line to the curves, the upper of which 
 terminates the ordinates which give the height of the tower when 
 the load is at the centre of the span, and the lower, the ordinates 
 which shew the most economical height of tower when the load is 
 evenly distributed. 
 
 When the load is concentrated at the centre of the span, and 
 there are no towers, that is to say, when the sectional area of the 
 tower = 0, the depression of the platform below the points of 
 support should be, as we have already pointed out, equal to half 
 the span, therefore the ordinate at the extreme left of the diagram, 
 with the figure underwritten, is made equal to '5 of the span. 
 Calling the ratio of section to strain which exists in the chains 
 1, the height of the tower, when its sectional area bears the same 
 ratio to the strain, is represented by ordinate No. 1. When the 
 ratio of the sectional area of the tower to the strain is twice that 
 in the chains, the height is represented by ordinate No. 2, and so 
 on up to a ratio of 24 times. 
 
 From the lower curve we obtain in a similar manner the most 
 economical heights for towers of various characters, when an 
 evenly distributed load has to be carried. These heights are, as 
 might be expected, less than those for towers of similar section, 
 when the load is to be carried at the centre of the span. 
 
 To estimate the amount of material in the superstructure of a 
 suspension bridge, such as that shewn in fig. 4, Plate XI., carrying 
 an evenly distributed load and with a varying height of tower, 
 would involve a great amount of labour; it therefore became an 
 object to discover a less laborious method of arriving at a suffi- 
 ciently approximate estimate of the required quantities. The first 
 step towards the accomplishment of this result was made when 
 it was found that the quantities required to carry a distributed 
 load by the system of fig. 4, Plate XL were exactly equalled by 
 the quantities required by the system of fig. 5, Plate XI. Finally, 
 it was ascertained by experiment that an identical result was 
 obtained when the whole distributed load was taken as concen- 
 trated at two points, each situated at a distance equal to one- 
 fourth of the span from the pier, as shewn in fig. 6, Plate XI. 
 
ECONOMY IN SUSPENSION BRIDGES. 123 
 
 It thus became an easy matter to calculate the quantities re- 
 quired for the superstructure with the different heights of tower. 
 
 If AB, fig. 7, Plate XII., be the span, and CD the most econo- 
 mical height of tower of a certain section when the load is con- 
 centrated at the centre of the span, the vertical EF will represent 
 the most economical height of tower when the load is concen- 
 trated, after the manner shewn in fig. 6, Plate XL, at two points 
 situated at a distance AE from the pier. If a series of vertical 
 ordinates representing the most economical height of tower at 
 various positions of the load be raised from the line AB, the point 
 from which each ordinate is raised representing the position in the 
 span of the load to which it corresponds, it will be found that 
 a line drawn through the upper extremities of the ordinates will 
 give a curve similar to that shewn in fig. 7, Plate XII. This 
 curve resembles somewhat the hyperbola. When the point E is 
 situated at a distance from A equal to one-fourth of AB, the 
 ordinate EF will be about five-sevenths of CD, that is to say, for 
 a distributed load the height of the tower should be about five- 
 sevenths of the height when a load is to be carried at the centre 
 of the span only. 
 
 The ordinates in fig. 3, Plate XL are calculated on the as- 
 sumption that the tower is composed of the same material as 
 the chains, but this is rarely the case in practice ; cast iron or 
 masonry being most commonly employed in the tower. In order, 
 therefore, to utilise the diagram, it will be necessary to find the 
 equivalent of the tower in the material of which the chains are 
 composed. 
 
 For example : It is required to determine the most economical 
 height of tower for a suspension bridge of which the chains are to 
 be of wrought iron, subject to a maximum strain of six tons per 
 square inch. 
 
 Let us estimate for a cast-iron tower. Assuming that the 
 engineer desires that the working-strain on the cast iron shall not 
 exceed three tons per square inch section, and that he estimates 
 the requisite bracing and architectural ornament to absorb twice 
 as much metal as that employed in carrying the load, he will then 
 have one square inch section of metal in his tower for every ton 
 upon it. 
 
124 ECONOMY IN SUSPENSION BRIDGES. 
 
 Let us assume the cost of cast iron to be in this case one-half 
 that of wrought iron ; then we may take the tower as formed of 
 wrought iron, and containing one square inch of metal for every 
 two tons upon it. Now since there is a strain of six tons per 
 square inch on the chains, the ratio of section to strain in the 
 tower is three times that in the chains, therefore ordinate No. 3, 
 fig. 3, Plate XL, will give us the proper height for the tower, which 
 is about eighteen, or rather more than one-sixth of the span. 
 
 In a similar manner by obtaining the equivalent of a tower 
 of masonry in wrought iron, the most economical height for a 
 tower of this kind may readily be determined. 
 
 In deciding upon the height of the tower, it must not, how- 
 ever, be forgotten, that the greater the height, the greater the 
 expense of erecting the bridge. In order, therefore, to obtain the 
 'greatest economy, it will be advisable to find this extra cost in 
 terms of quantity of material. For instance : supposing it is esti- 
 mated that beyond a certain height from the ground the cost of 
 erection will increase by one shilling a ton for every additional 
 foot of elevation, then if at this height the cost of the material 
 in place be, say 25 per ton, an extra elevation of ten feet would 
 add one-fiftieth to the cost of the bridge, which is the equivalent 
 of an addition of one-fiftieth to the quantities in the structure. 
 
 The raising of the height of the tower in this case to the 
 .amount of ten feet beyond the given elevation, would only be 
 justified, so far as economy is concerned, when more than one- 
 fiftieth of the quantities was thereby saved. 
 
 In fig. 8, Plate XII. is given a series of curves, from the 
 ordinates to which may readily be seen the comparative economy 
 of various proportions of height of tower to span from one-fourth 
 up to one-fourteenth. The figures at the extremity of each curve 
 shew the ratios of the sectional area of the tower to chains, and 
 correspond with the numbers of the ordinates in fig. 3, Plate XL 
 From this diagram it will be seen that with a tower whose ratio 
 is 3, nearly one-third more material would be required for a 
 bridge with a height of tower equal to one-fourteenth of the span 
 4han for a similar bridge whose tower was one- sixth of the span. 
 But when a very expensive character -of tower is adopted, the 
 ^extra cost incurred by keeping the tower low is comparatively 
 
ECONOMY IN SUSPENSION BRIDGES. 125 
 
 small. From this diagram the engineer can tell at a glance what 
 extra expense he will incur by adopting ornamental instead of 
 plain towers. 
 
 Fig. 9, Plate XII. shews by means of ordinates to a curve the 
 great economy which can be attained by adopting an inexpensive 
 class of tower. The ordinates represent the minimum total quanti- 
 ties in tower and chains .necessitated by the adoption of any ratio 
 of sectional area of tower to sectional area of chains from to 24. 
 
 The whole of the foregoing diagrams are based on the assump- 
 tion that the chains have no weight, they do not, therefore, afford 
 strictly correct information as to the most economical heights of 
 tower under varying circumstances. For when the weight of the 
 structure is taken into account, the load to be carried varies with 
 the height of the tower. The higher the tower, within certain 
 limits, the less the quantities in the- superstructure, and conse- 
 quently the less the load to be carried. 
 
 For bridges of very large span, where the proportion of dead 
 load is very large, this variation will be so serious as to affect 
 very materially the question of height of tower. In all cases the 
 most economical height of tower is really greater than that given 
 by the diagrams. In small bridges the error will be inconsider- 
 able, but in large bridges it is quite worthy of being taken into 
 account. 
 
 To construct a diagram in which this variation of the load is 
 allowed for, would be a somewhat laborious operation. It has not 
 therefore been thought advisable to go minutely into this ques- 
 tion on the present occasion, more especially as a variety of con- 
 siderations other than that of mere economy often determine the 
 decision of the engineer as to the height of the tower. And 
 whereas the object of the preceding remarks is rather to shew that 
 the proportions of height of tower to span hitherto adopted are 
 much too small, if economy be an object, than to insist upon any 
 particular proportion, it will be enough to point out that if the- 
 weight of the superstructure be taken into account, a diagram 
 would result still more condemnatory of the proportions of height 
 of tower to span hitherto generally adopted, than are the diagrams, 
 accompanying this article. 
 
 Let us now turn our attention to the second part of our sub- 
 
126 ECONOMY IN SUSPENSION BRIDGES. 
 
 ject: "What should be the arrangement of parts in the super- 
 structure to obtain the greatest economy?" 
 
 From the point A, fig. 6, Plate XI. erect a perpendicular AB 
 equal to CD, the height of the tower above the platform. Join 
 ^(7 and BE. 
 
 Since BC=AE by construction, BE is parallel to AC. So 
 then, if the line AB represent the magnitude of the load con- 
 centrated at A, AE or GB will represent the strain on the bar 
 AE, and A or EB the strain on the bar AG. 
 
 Let us now suppose the bar AE, which is subject to tension, 
 to be removed, and its place to be supplied by a strut occupying 
 the position AD. The strain upon this strut and its length are 
 both equal to that of the bar AE, and as the bar AE is common 
 to both arrangements of carrying the load, it follows that if the 
 ratio of strain to sectional area upon the strut be identical with 
 that in the bar AE, the same amount of material will be required 
 whether we make use of the tension bar AE or the strut AD, to 
 carry the load at A. 
 
 But if the load be shifted to a point F nearer the pier, the 
 strut FD being shorter than the bar FE while the strain upon 
 both is the same, the strut arrangement will be the more econo- 
 mical of the two. On the other hand, if the load be removed to 
 G further from the pier, the tension bar arrangement will be pre- 
 ferable. It is therefore evident, assuming bars acting in compres- 
 sion to be as effective as bars acting in tension, that the most 
 economical method of carrying a dead load is by means of sus- 
 pension for that half of the span which lies equally on each side 
 of the centre, and by means of cantilevers for the remaining parts 
 of the span which adjoin the piers. 
 
 For a moving load this will not hold good, because, whereas 
 the cantilever parts are not liable to distortion, the same cannot 
 be said of the suspension portion of the structure, to which certain 
 additions are requisite in order to preserve its form under varying 
 conditions of load. It is thus apparent, that for ordinary bridges 
 which have to carry moving loads, it would be necessary to have 
 a preponderance of cantilever, in order to obtain the most econo- 
 mical results. 
 
 To determine exactly the proper relative proportions of these 
 
ECONOMY IN SUSPENSION BRIDGES. 127 
 
 two forms of construction in any bridge would be a matter of con- 
 siderable difficulty. We can, however, conveniently do so under 
 two extreme conditions the first, when the value of the live 
 load = 0; the second, when the value of the dead load = 0. 
 
 The first of these conditions we have just considered; to elu- 
 cidate the second, let us inquire how the distorting tendency of 
 a moving load can be most economically counteracted. 
 
 If we compare the quantities required to carry a given load in 
 the manner shewn in fig. 10, Plate XII., with the systems of 
 figs. 4 and 5, Plate XI., we shall find that the method of fig. 10, 
 Plate XII. consumes about 50 per cent, more material than its 
 rivals. It has, however, this advantage over them, that it is a 
 perfectly rigid system of construction. 
 
 If, now, we take the system of fig. 4, Plate XI., and stiffen it 
 by means of two girders of an economical form, in the manner 
 shewn in fig. 11, Plate XIII. , we shall find, on taking out the 
 quantities, that they exceed somewhat those of system fig. 10, 
 Plate XII., which may, in fact, be regarded as absolutely the 
 most economical method of carrying by suspension a load entirely 
 moving. We will, therefore, now compare suspension by this 
 system with the cantilever method. 
 
 The quantities required to carry the load of 100 by the bars 
 AC and AB, fig. 12, Plate XIII., will be found to be exactly 
 equal to the quantities required to carry the same load by the 
 bar A G and the strut A G, when the distance AG is one-third 
 of the span. The suspension system is the more economical 
 when the distance of the load from the pier exceeds one-third 
 of the span, but the cantilever system when this distance is less, 
 than one-third of the span. 
 
 We have thus defined the distances of one-fourth and one-third 
 of the span from the pier to be the limits between which it will 
 be necessary to fix the point where the character of the structure 
 should change from that of suspension to cantilever, if economy is 
 to be a supreme consideration. 
 
 Now, if we compare together the systems of fig. 10, Plate XIL 
 and fig. 11, Plate XIII., when the load to be carried consists 
 partly of dead and partly of live load, we shall find the latter 
 system (fig. 11, Plate XIII.) to be more economical than the 
 
128 ECONOMY IN SUSPENSION BRIDGES. 
 
 former in proportion to the quantity of dead load which has to be 
 carried. For whereas by the system of fig. 10, Plate XII. the 
 quantities required will be the same, whether the load be live or 
 dead, in the system of fig. 11, Plate XIIL, when the live load = 0, 
 the quantities in the stiffening trusses become = 0, and the bridge 
 takes the character of fig. 4, Plate XI. Hence it is evident that, 
 in ordinary suspension bridges, where the dead load forms a serious 
 item in the load, the system of fig. 11, Plate XIIL, will be prefer- 
 able to that of fig. 10, Plate XII. 
 
 From the foregoing facts we deduce the principle, that when 
 the proportion of dead load is excessive, the point where the sus- 
 pension part of the structure ends and the cantilever portion 
 begins, should approach the limit nearest the pier. On the other 
 hand, when the live load preponderates, it should lie nearer the 
 limit furthest from the pier. 
 
 In these calculations the effect of the weight of the structure 
 itself has been completely left out of account. It will, however, 
 modify our conclusions to a certain extent, and in the direction of 
 further increasing the proportions of cantilever. 
 
 If we turn to fig. 6, Plate XI., and once more compare the 
 suspension with the cantilever system, we shall see that since the 
 tension bar AE occupies a position nearer the centre of the 
 spans than the strut AD, the weight? of the structure is more 
 advantageously situated for economy in the cantilever than in the 
 suspension part. Hence the truly economic position of the point 
 of change from suspension to cantilever method would be at a 
 distance, more or less, exceeding one-fourth of the span from the 
 pier, as the weight of the bars bore a large or small proportion to 
 the load carried. 
 
 Thus much for the economical part of the subject. Let us 
 now devote a short time to the consideration of the practical 
 part. 
 
 However economical any arrangement of parts may be, its 
 adoption must be barred if it involve danger to the safety of the 
 bridge. It probably has been with the idea chiefly of keeping down 
 the amount of oscillation to which a suspension bridge is liable, that 
 the curve of the chains in many existing suspension bridges has 
 been made so excessively flat. And if no means of stiffening the 
 
ECONOMY IN SUSPENSION BRIDGES. 129 
 
 platform is to be adopted, the arrangement is so far calculated to 
 attain the desired object. But this is a very expensive way of 
 accomplishing in a very imperfect manner a result which can be 
 completely effected by the use of stiffening girders. The trans- ' 
 verse strain to which these are subjected is always the same with 
 certain proportions of live to dead load, and is entirely unaffected 
 by the rate of curvature of the chain. There can, therefore, be no 
 objection on this score in a bridge provided with stiffening girders 
 to proportionally high towers. 
 
 The higher the towers the less the deflection of the bridge 
 under load, and the less the alteration of form caused by changes 
 of temperature. This, to some engineers, will appear an impor- 
 tant consideration. 
 
 Let us now consider the behaviour of structure, such as that 
 shewn in fig. 13, Plate XIII., compounded of a suspension portion 
 AB between two cantilever portions AC and BD, under a rolling 
 load. 
 
 The load travelling on to the bridge from left to right, a 
 bending will take place at each of the points E } F, and G, in the 
 platform. At these points it should therefore be hinged. The 
 alteration in the form of the platform thus caused will, however, 
 be slight. In a bridge of 1600 feet span, where the proportion 
 of dead load would be large, it was found by calculation that the 
 effect of an ordinary train running on to the bridge as far as point 
 E t would be to depress the platform at that point to an amount of 
 about one inch. We may therefore take it that the distortion of 
 the platform produced by unequal loading would be so insig- 
 nificant as to be unworthy of consideration. 
 
 The alteration in the form of the platform caused by changes 
 of temperature will, however, be considerable. The expansion of 
 the metal will cause the extremities E and G of the two canti- 
 levers to approach one another, and consequently the platform at 
 F to drop. In order to prevent cross strains, it will be advisable 
 to make the hinge joints at E and G such as to admit of a certain 
 amount of end motion in the platform. This can be done in a 
 variety of ways. 
 
 In order to avoid the unsightly appearance which would be 
 produced by a reverse camber in the platform, it is recommended 
 that the platform should be laid so as to have a slight gradient 
 Y. 9 
 
130 ECONOMY IN SUSPENSION BRIDGES. 
 
 each way towards the centre of the span, as shewn by the straight 
 lines H, E } F, F, G, /, fig. 14, Plate XIIL, when the temperature 
 is at its lowest. When the temperature is at its highest, the 
 point F would drop to F 1 , a point not below the level of the 
 hinges E and G. 
 
 By this arrangement, all unsightliness can be avoided. The 
 consequence of these alterations in form of platform would amount 
 simply to this, that the gradients on the bridge would vary with 
 the season of the year, a peculiarity in no way affecting the safety 
 of the structure. Nor would the fact that the platform is capable 
 of end motion at the points E and G affect its lateral stiffness, 
 provided it be treated as a continuous girder, of which HE and 
 GI are the cantilever portions and EG the girder portion, E 
 and G being the points of contrary flexure. 
 
 The erection of such a bridge as that shewn in fig. 13, Plate 
 XIIL, could be conveniently effected by building out from the 
 piers as far as the extremities of the cantilevers, and erecting the 
 suspension portion in the usual manner. 
 
 We have not here entered into the question of the most econo- 
 mical height of tower for a bridge of single span : to do so would 
 unnecessarily lengthen this article; we may, however, consider 
 that the proper height of tower for a bridge of this kind should be 
 much the same as for a bridge of several spans, inasmuch as the 
 quantities in the back tie of the former would be not very different 
 from those in the half span of the latter description of bridge. 
 When the towers of a bridge of single span are backed by high 
 ground, of course it would be advisable to adopt a loftier tower 
 than would be proper if no such natural feature characterised the 
 site. 
 
 The result of the investigations briefly summarised in the fore- 
 going remarks has been to convince the author that considerable 
 unnecessary expense has been incurred in bridging large spans, 
 through the adoption of towers deficient in height, and an arrange- 
 ment of parts in the superstructure not of the most economical 
 description. In confirmation of these opinions he would state that 
 he has made numerous careful and detailed estimates for bridges 
 of various spans on the system of fig. 13, Plate XIIL, and the result 
 has fully justified his anticipations. 
 
INDEX. 
 
 ABUTMENTS of an arch, yielding of, 83 
 
 Alternative methods of finding strains, 
 10 
 
 Anchoring down continuous girder, 101 
 
 Arch, 6 1 ; thrust at crown of, ib.; me- 
 thods of resisting distortion, 62 ; strain 
 on spandrils of, ib. ; stability of pier of, 
 68; effect of horizontal girder in, 16.; 
 transverse strain on piers of, ib. 
 
 of masonry, curve of equilibrium 
 
 for, 6 1 
 
 Arched bridges of multiple span, 67 
 viaduct of wrought iron, 70 
 
 Arm, i 
 
 Axis, neutral, 18 
 
 Back tie of suspension bridge, 73 
 Beams, wooden, 18 
 
 of irregular section, 2 1 
 
 Bowstring girder, 45; curve of equili- 
 brium, ib.; method of drawing the 
 curve, 46 
 
 regularly loaded, 46; irregularly, 
 
 48; method of resisting distortion of, 
 51 ; load resting on the top, 60 
 
 Cantilevers, 17 
 
 Continuous girders, 86; of two spans, 87 ; 
 of two unequal spans, 98 ; example, 
 92 ; method of finding position of point 
 of contrary flexure when both spans 
 are loaded, 99 ; when longer span load- 
 ed, 100; extremely disproportion ed 
 span, ib.; shorter span loaded, ib.; an- 
 choring down the shorter girder, 101 ; 
 multiple spans, 102; three-span girder, 
 1 06 ; girders of more than three spans, 
 
 ib.; of varying depth, ib.; practical re- 
 marks on, 1 08 
 
 Curve of equilibrium, 45 ; regular load- 
 ing, 46; with a given versine, 47; irre- 
 gular loading, bays equal, 48; with 
 bays unequal, 49 
 
 Decomposition of forces, 8 
 
 Definition of deflection, 76 
 
 Deflection, laws of, 76; of arched rib, 
 82; of bowstring girder, 83; of con- 
 tinuous girders, 92; of suspension 
 bridge, 84; strength of a girder, how 
 far deducible from, ib.; of an irregularly 
 strained girder, 79.; to find the, 76; 
 definition, ib. 
 
 Diagonals, effect of elasticity upon, 56; 
 resisting extension and compression, 60 
 
 Economy in suspension bridges, 119; 
 proper proportion of height of tower to 
 span, ib. ; method of estimating proper 
 height for towers of various kinds, 123 ; 
 economical arrangement of parts in 
 superstructure of suspension bridge, 
 126; suspension and cantilever system 
 compared, for dead loads, Oh.; for live 
 loads, 127; behaviour of a compound 
 cantilever and suspension bridge under 
 loads and changes of temperature, 128; 
 summary of results, 130 
 
 Effect of horizontal girder on arch, 68 
 
 Elasticity, effect of, in distributing load 
 among diagonals, 56 
 
 Equilibrium, curve of, 45 
 
 Evenly distributed load, 17; on canti- 
 lever, ib.; on parallel flanged girder, 5. 
 
132 
 
 INDEX. 
 
 Forces, resolution or decomposition of, 
 
 8; acting in different planes, II 
 Fulcrum, i 
 
 Girders with parallel flanges, 12 ; load at 
 centre of span, ib.; load evenly distri- 
 buted, 13; irregularly loaded, i6;bays 
 equal, 33; bays unequal, 36; load 
 symmetrical, bays unequal, 35 
 
 lattice, 37 
 
 warren, 38 
 
 hogbacked, 41 
 
 of irregular section, 21 
 
 ordinary wrought-iron, ib. 
 
 bowstring, 45 
 
 continuous : see Continuous Girders 
 
 Hogbacked girders, 41 ; compared with 
 
 parallel flanged girder, 43 
 Horizontal girder to arching, 62; effect 
 
 of on piers in arched viaducts, 68 
 
 Irregular loading of arches, 63; of bow- 
 string girders, 48 ; of parallel flanged 
 girders, 16; of hogbacked girders, 41; 
 of suspension bridge, 73; of girders, 
 89 
 
 Lattice girders, 37 
 
 Lever, i 
 
 Lever, bent, 3 
 
 Leverage, ib. 
 
 Line of greatest resistance, 51 
 
 Load at centre of span, 12 
 
 Masonry, arch of, 61 
 
 Method of finding position of point of 
 
 contrary flexure in continuous girders, 
 
 90 
 Methods of resisting distortion in the 
 
 arch, 62 ; in the bowstring girder, 51 
 Moment, 2 
 Multiple span, arched bridges of, 67 
 
 Multiple span, continuous girders of, 102 
 Neutral axis, 18 
 
 Ordinary wrought-iron girders, 2 1 ; strains 
 on the flanges, 22; strains on the web, 
 24; shearing strain on each bay, 27 
 
 Parallelogram of forces, 6 
 Polygon of forces, ib. 
 Practical remarks on continuous girders, 
 108 
 
 Radius of curvature of deflected beams, 77 
 Resolution of forces, 8 
 Resultant, 6 
 
 Shearing force, 16; formulas for, 30 
 
 Spandrils of an arch, 62 
 
 filling, sundry forms of, 66 
 
 Stability of piers, 68 
 
 Strain on flanges at centre of span, 13; 
 at any point of the flange, 14; on the 
 web, 16, 24; on the flanges of hog- 
 backed girders, 41 ; on diagonals, ib. 
 
 Suspension bridges, 72; strain at centre 
 of, 73; strain at any portion of chain, 
 $>.-, methods of stiffening, 74 
 
 Transverse strain on piers of arched via- 
 ducts, 68 
 
 Triangle of forces, 6 
 Truss, warren, 38 
 Trusses, 37 
 
 Vertical, or shearing force, 16 
 Viaduct, arched, of masonry, 67 ; of 
 wrought-iron, 70 
 
 Warren truss, 38; strains on diagonals 
 
 of, 40 
 
 Web in iron girdera, 19 
 Wooden beams, 18 
 
 CAMBRIDGE : PRINTED BY C. 3. CLAY, M.A. AT THE UNIVERSITY PRESS. 
 
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