UC-NRLF 
 
ON THE 
 
 THEORY AND CALCULATION 
 
 CP 
 
 BY 
 
 MANSFIELD MERRIMAN, C.E., Ph.D., 
 
 SHEFFIELD 
 
 INSTRUCTOR IN 
 
 SCIEN 
 
 THE 
 
 NEW YORK: 
 
 D. VAN NOSTRAND, PUBLISHER, 
 23 MURRAY AND 27 WARREN STREET, 
 
 1876. 
 
\ 
 
 If 
 
 o 
 
PKEFACE, 
 
 The following pages are divided into three 
 chapters. The first presents by way of intro 
 duction some of the elementary principles of 
 continuous girders, and the fundamental ideas 
 relating to the calculation of strains. The sec 
 ond gives the theory of flexure as applied to 
 the continuous trjiss^ of constant cross section, 
 and exhibits it in formulae (I) to (VI), ready 
 for application to any particular case ; and the 
 third gives an example of the computation of 
 strains in a continuous truss of five unequal 
 spans, with some useful hints concerning the 
 practical building of such bridges. 
 
 The theory of flexure indicates that, by the 
 use of continuous instead of single span 
 bridges, a saving in material of from twenty to 
 forty per cent, may be effected. f It is easy in 
 deed to say that this advantage will be entirely 
 swallowed up by the effect of changes of tem 
 perature, increased labor of erection, or addi 
 tional cost of workmanship, but by no amount 
 of reasoning can such disadvantages be esti 
 mated. Theory indicates a large saving, 
 
whether or not it can be realized, may only be 
 determined by trial. Other nations have built 
 and are building continuous bridges, and their 
 experience has not yet shown that the system 
 is inferior to that of single spans. The inter 
 est now prevailing among American engineers 
 in the subject, and the fact that at some recent 
 bridge let tings plans have been offered for a 
 continuous structure, seem to indicate that the 
 system will also be tried here. 
 
 This little book may then perhaps be of val 
 ue to bridge engineers, as well as to students 
 in general. 
 
 M. M. 
 
 New Haven, Conn. , July 10, 1876. 
 
THEORY AND CALCULATION 
 
 OP 
 
 CONTINUOUS BRIDGES. 
 
 WHEN a straight bridge consists of 
 several spans, each entirely independent 
 of the others, it is said to be composed 
 of simple girders. If, on the other hand, 
 it consists of a single truss extending 
 from one abutment to the other without 
 any disconnection of parts over the piers 
 it is called a continuous girder. A load 
 placed upon any span of a continuous 
 beam influences, to some extent, each of 
 the other spans, and hence its complete 
 theory is much more complex than that 
 of the simple one. This very complexity 
 however has rendered the subject an at 
 tractive one to mathematicians, who, 
 pursuing science for science s sake, have 
 
6 
 
 investigated the laws of equilibrium 
 which govern it. These laws with the 
 many beautiful consequences attending 
 them form one of the most interesting 
 chapters of mathematical analysis, and 
 as such have interest and value inde 
 pendent of their application in engineer 
 ing art. 
 
 It is the object of the present paper to 
 present in as simple a form as possible 
 some of the main principles and laws 
 most needed by the engineer, and to il 
 lustrate their application as fully as 
 space will permit to the practical de 
 signing of continuous bridges. 
 
 CHAPTER 1. 
 
 The first point to be observed in con 
 sidering either a simple or continuous 
 girder is that all the exterior forces 
 which act upon it are in equilibrium. 
 The exterior forces embrace the weight 
 of the girder and the loads upon it which 
 act downward, and the pressures or re 
 actions of the supports which act up- 
 
ward. In order that these may be in 
 equilibrium, it is necessary that the sum 
 of the reactions of all the supports must 
 be equal to the total weight of the girder 
 and its load. 
 
 Thus, if a simple girder of uniform 
 section and weight rest at its ends upon 
 two supports, the reaction of each sup 
 port will be one-half the weight. Exactly 
 in the center between the two supports 
 or abutments, let us suppose a pier to 
 be placed just touching, but not pressing 
 against the beam, which, at that point, 
 has a deflection below a straight line 
 joining the two abutments. Then the 
 condition of things is in no way altered, 
 for the weight being W, each abutment 
 reacts with a force \ W, while the pier 
 bears no load. Raise now the pier so as 
 to lift the girder above the line of de 
 flection and it receives a part of the 
 weight W, while Jie reactions of the 
 abutments become less than \ W. If 
 the pier be raised higher and higher, it 
 will at length lift the girder entirely 
 from the abutments and bear itself the 
 
8 
 
 whole load W. In every position, how 
 ever, the sum of the reactions of the 
 three supports is equal to the total load. 
 For example, when the three are on the 
 same level it may be shown that the re 
 action of each abutment is -fs W, and 
 that of the pier ii W. 
 
 This illustration shows also that small 
 differences of level in the supports occur 
 ring after the erection of a bridge cause 
 large variations in the reactions of its 
 supports and in the strains in its several 
 parts. A simple girder having a deflec 
 tion of one inch, would, if raised one 
 and three-fifth inches at the center, be 
 entirely lifted from the abutments. In 
 the first case the upper fiber would be in 
 compression, the lower in tension; in the 
 second case, the upper would be in ten 
 sion, the lower in compression. If the 
 center were raised only one inch, the re 
 versal would be only partial, the upper 
 fiber becoming subject to tension for a 
 short distance on each side of the mid 
 dle. This fact often used as an argu 
 ment against continuous bridges, is really 
 
9 
 
 an objection only when the piers are 
 liable to settle after erection. Difference 8 
 of level, previously existing, do not act 
 prejudicial when the bridge is built upon 
 the piers, and with a profile correspond 
 ing to them. 
 
 The mathematical theory of the con 
 tinuous girder enables its reactions and 
 internal strains to be found for any as 
 sumed levels of the supports, provided 
 only that the differences of level are very 
 small compared w T ith the length of the 
 spans. However interesting such inves 
 tigations may be in themselves, they are 
 of little importance in practice, since it 
 has been shown that when all the points 
 of support are on the same level, the 
 greatest economy of material results.* 
 In all that follows, then, we shall regard 
 the girder as resting on level supports, 
 or, what is the same thing, that it was 
 built with a profile corresponding to that 
 of the piers. 
 
 The loads upon a bridge and the reac- 
 
 * Weyrauch ; Theorie der continuir lichen Trdger, p. 129. 
 Winkler ; Lehre der Elasticitdt, p. 155. 
 
10 
 
 tions of the supports are external forces. 
 The equilibrium between them is main 
 tained by means of internal forces, which, 
 in a framed truss, are transmitted longi 
 tudinally along the pieces as strains of 
 tension and compression. When all the 
 external forces are known, these internal 
 forces or strains can be readily found. 
 This very important point we shall now 
 proceed to illustrate. 
 
 Fig. 1 represents a portion of a con- 
 Fig, i, 
 
 A 
 
 R l R 
 
 tinuous girder; the first span on the left 
 is called Z l5 which also represents its 
 length, the second J 2 , the third / 3 , etc. ; 
 in like manner,- the supports beginning 
 on the left are designated by the indices 
 1, 2, 3, etc., and their reactions by R,, 
 R 2 , R 3 , etc. Let the load per linear unit 
 be iv, supposed uniformly distributed, 
 then the weight of the first span will be 
 wl of the second wl of the third w l 
 
11 
 
 etc. If there are four spans the total 
 
 weight will be 
 
 and from the fundamental idea of equili 
 brium, we must have the equation 
 
 Each of the reactions is then a fraction 
 al part of the total load, and by methods 
 hereafter to be explained, their values 
 may be readily computed, whatever be 
 the number of spans. Granting for the 
 present that they may be found, let us 
 inquire how we may obtain the internal 
 forces or strains in any part of the gir 
 der. 
 
 In the span l z let a vertical plane be 
 passed, cutting the beam at a point 
 whose distance from the support 3 is x. 
 All the internal forces acting in this sec 
 tion may be considered as resolved into 
 two components, one vertical and the 
 other horizontal. The sum of all the 
 vertical components is a force which pre 
 vents the two parts of the beam from 
 
12 
 
 shearing asunder, and is called the shear 
 for that section ; the horizontal compo 
 nents acting in parallel planes are the 
 resisting strains of tension and compres 
 sion in the horizontal fibers, and the sum 
 of their moments with reference to any 
 point in the section is called the moment 
 of resistance, or simply the moment for 
 that section. The internal strains in any 
 section are thus completely represented 
 by the shear and moment. For example, 
 if the girder in Fig. 1 be a framed truss 
 of which Fig. 2 represents the span 1 3 
 
 Fig. 2, 
 A 
 
 enlarged, and the section be passed cut 
 ting the three pieces E F, F e, and ef, 
 the vertical components of the chord 
 strains will be zero, and that of the di 
 agonal strain e F will be the shear. 
 Hence, if the shear be known, and the 
 angle included between the vertical and 
 a diagonal be 6, we have only to multi- 
 
13 
 
 ply the shear by sec. 6 to find the strain 
 in the diagonal. Again, let the section 
 be moved to the left so as to pass through 
 the point e, and let that point be taken as 
 the center of moments. Then the mo 
 ment of resistance will be the moment 
 of the chord strain E F. Hence, if that 
 moment be known, we have only to di 
 vide it by the depth of the truss to get 
 the strain in E F. 
 
 The internal shear and moment at 
 any section are easily found from the 
 fundamental conditions of statical equili 
 brium. The shear being an internal 
 vertical force is the resultant of 
 the exterior vertical forces on either 
 side of the section. The exterior forces 
 on the left of the section, for instance, 
 have for a resultant their algebraic sum ; 
 considering the upward forces as posi 
 tive, and the downward ones as negative 
 we have from Fig. 1, their sum 
 
 wx 
 
 as the expression for the shear in the 
 section x. To get the internal moment 
 
14 
 
 for the same section, we have only to 
 consider in like manner that it is equal 
 to the sum of the moments of all the ex 
 terior forces on either side of the section, 
 for if otherwise, there would be a ten 
 dency to rotation. The moment of the 
 force Ri with reference to x is R : 
 (I, + Z a + a) , of R 2 is R 2 (I, + x) , of the load 
 t0i l9 iatoi, (i^ + ^+aJ), etc. Thus from 
 a mere inspection of Fig. 1 we write the 
 value of the moment M, regarding those 
 moments as positive which cause a ten 
 sile strain in the upper fiber at x, and 
 those as negative which cause a compres- 
 sive one. The expression is 
 
 + x ) + w I* (i k + x ) 1^3 x + wx \ x 
 Now in these expressions for the in 
 ternal shear S and the moment M at any 
 point x, the lengths , 2 , x are given by 
 the conditions of the case in hand, and 
 the same is true of the load per linear 
 unit w. Hence the shear and moment, 
 and consequently the internal strains are 
 easily obtained as soon as the reactions 
 
15 
 
 of the supports are known. We shall 
 hereafter give methods by which the re 
 actions may be readily determined. 
 
 By the same reasoning if we pass a 
 section in the span Z 2 , (Fig. 1) at a point 
 whose distance from the support 2 is x, 
 the shear S and the moment M for that 
 section will be 
 
 W X* 
 
 ^^ w 2 w x 
 
 Mc= -R x (Z, + x) + w 1 1 (\ 1, + x) 
 
 In a simple girder whose length is 
 each of whose reactions is R or ^ w I the 
 shear and moment for any section x will 
 be 
 
 I x) 
 
 WX* W 
 
 2 
 
 When the number of spans is large, 
 the expressions for the shear and moment 
 as above deduced become long and in 
 volve much arithmetical computation. 
 We are, however, fortunately able to 
 
16 
 
 place them under a much simpler form. 
 First they may be written thus 
 
 wx 
 M= [-K, ( + 
 
 wl^ + R 3 ) is the 
 shear in the span 1 3 at a point infinitely 
 near to the support 3 ; let this be called 
 S 3 . Also the quantity enclosed in [ ] in 
 the second equation is the moment of 
 the exterior forces with reference to the 
 point 3 ; let this be called M 3 . Then the 
 equations become 
 
 Therefore the internal shear and mo 
 ment at any section can immediately be 
 found) without the necessity of determin 
 ing the reactions, provided ice know the 
 shear and the moment for the preceding 
 support. This method, due to Clapeyron, 
 of using the moment at the supports in 
 stead of the reactions greatly simplifies 
 
17 
 
 the numerical computations of a continu 
 ous truss. We designate the moment at 
 3 by M 3 , the reaction being R 3 , or the 
 sum of the shear S 3 , in the span 1 3 and 
 of the shear S 2 in the span 1 2 both infi 
 nitely near to the support 3. In like 
 manner the moments at the supports 2 
 and 4, will be designated by M 2 and M , 
 the shears just to the right of those 
 points by S 2 and S 4 , and those to the left 
 by S j, and S 3 . In general for any sup 
 port whose index is n, w r e have (Fig. 3) 
 
 Fig. 3, 
 
 < 7 * 7 *- 7. _ _ 
 
 -u-i n I ^ n+i 
 
 n-\ ^ n-t-i 
 
 ^^^^^"^^^ji -X 
 
 - 
 
 on the left, the span n _i, on the right the 
 span / n ; the shear infinitely near to n 
 on the left is S / n _i, on the right S n ; the 
 sum of S n _i, and S n makes the reaction 
 R n ; and the moment over the support 
 is M G . If the load be uniform and 
 equal to w per linear unit, the internal 
 
18 
 
 
 
 shear S and moment M for any section x 
 are given by 
 
 S = S n wx 
 
 M=M n S n x + 1 w x* 
 
 To find then the internal strains in 
 the diagonals of a continuous truss due 
 to dead load only, we have to pass a 
 section cutting each diagonal and find 
 the value of S, this is the shear which 
 the strain in the diagonal must resist 
 
 o 
 
 and multiplied by sec. 8 (8 being the in 
 clination of the diagonal to the vertical,) 
 it gives the required strain. To find the 
 s trains in the upper chord we have to 
 take the lower chord apices as centers 
 of moments and compute the values of 
 M ; these divided by the depth of the 
 truss give the strains, which are tensile 
 if M is positive, compressive if M is neg 
 ative. To find the lower chord strains, 
 we choose the upper apices from which 
 to measure the values of a?, and divide 
 the resulting values of M by the depth 
 of the truss ; if M is positive these give 
 compressive strains ; if negative, tensile 
 ones. 
 
Everything is thus known, except the 
 shears and moments at the supports, and 
 for these formulae and methods will be 
 presented in Chapter II, by which they 
 may be found for all cases. We give 
 here, however, two tables from which 
 they may be found for the common 
 case when all the spans are equal, and 
 which, by a simple law, may be extend 
 ed to include any number of such spans. 
 
 As before, let w be the unifornily dis 
 tributed load per linear unit; let I be the 
 length of each span, then will w I be the 
 weight of one span. The shear at any 
 support is a fractional part of w I, or 
 
 s A w I 
 
 A being a fraction given in the follow 
 ing triangle : 
 
20 
 
21 
 
 Each of the squares composing this 
 triangle represents one of the supports, 
 and its left hand division gives the left 
 hand shear S n _i, and the right hand one 
 the shear S n (Fig. 3). Thus, in the gir 
 der of three spans, the triangle shows 
 that the first support, beginning at the 
 left, has on the left no shear, and on the 
 right YQ 10 /, that the second support has 
 on the left a shear of ^ w /, and on the 
 right one of -f w I. The sum of the two 
 shears for any supports is of course its 
 reaction. For example, a girder of six 
 equal spans has at its middle support a 
 reaction of I8t w I. 
 
 The moment at any support will be a 
 fractional part of ivl^or 
 
 7. 
 
 Moment = B w l^ 
 
 B being a fraction given in the following 
 triangle ; in which like the preceding 
 one the spaces indicate the supports of 
 the girder. Thus, the fourth horizontal 
 line refers to a girder of four spans, the 
 moments at the first and last supports 
 being 0, at the second and fourth - 2 3 8 - w I* 
 and at the middle one w I 2 . 
 
22 
 
23 
 
 The triangles can be extended to any 
 required length by the application of the 
 following law which obtains in &\\ oblique 
 columns. Any fraction belonging to an 
 even number of spans, may be obtained 
 by multiplying both numerator and de 
 nominator of the preceding fraction by 
 two and adding the numerator and de 
 nominator of the fraction preceding that. 
 
 33 
 Thus for eight spans, the fraction 
 
 2X11 + 11 
 
 is equal to - - or to 
 
 2X142 + 104 
 
 <388 
 
 ,. 
 2X142 + 104 accordm S as we use one 
 
 oblique column or the other. For an odd 
 number of spans, any fraction is found 
 by adding the two preceding fractions, 
 numerator to numerator and denomina 
 tor to denominator. Thus for seven 
 
 spans, 
 
 12 8+4 12 9 + 3 
 
 or 
 
 142 104 + 38 142 104 + 38 
 
 These tables* furnish the data for solv- 
 
 * These triangles were first given by the author in the 
 Journal of the Franklin Institute for March, 1875. A de 
 monstration of the laws governing them may be seen in 
 the same Journal for April, 1875. 
 
24 
 
 ing all questions concerning continuous 
 girders whose supports are on the same 
 level, whose spans are all equal, and 
 which are loaded uniformly throughout 
 their entire length. The reader should 
 first acquire facility in the use of the 
 tables. We give, therefore, a few ex 
 amples for practice : 
 
 1. In a girder of six spans, what is the 
 reaction at the second support ? 
 
 118 
 
 Ans. K. = - - w I. 
 104 
 
 2. In one of eight spans, what is the 
 reaction at the middle support? 
 
 386 
 
 Ans.K=-wl 
 
 3. In one of ten spans, what is the mo 
 ment over the fourth support from the 
 left? 
 
 4. In one of seven spans, what are the 
 shears S 2 and S ? (see Fig. 3.) 
 
25 
 
 5. In one of six spans, what is the mo 
 ment M and the shear S 6 ? 
 
 Ans. M= 
 
 Having thus found from the triangles, 
 .the moment M n and the shear S n for the 
 n th support, the shear S and the moment 
 M for any section in the n ih span are 
 readily found from the formulae 
 
 O - On W X 
 
 Tf C I 1 nn rf 
 
 -LT_Ln c?n th ~r 2^ w **j 
 
 which we have demonstrated above, and 
 
 Fig. 3. 
 
 n-i n n+i 
 
 
 in which x is the distance from the sup 
 port n to the assumed section. If in 
 these x be made equal to I, they will, of 
 course, give the shear S n at the left of 
 the w + l th support, and the moment 
 M n+ i over that support. We will illus 
 trate their use by a few examples : 
 
26 
 
 6. In a continuous girder of three 
 spans, what is the shear and the moment 
 
 at the center of the middle span ? 
 
 5 
 
 We have from the table S a =--- wl and 
 1 10 
 
 M,=-r. 
 
 Hence 
 
 S== w I iv x 
 
 1 5 
 
 M= ivF-w l 
 
 and placing x equal to /, we have 
 
 S=0 M=-irP 
 
 40 
 
 7. In a girder of six spans, find the 
 shear and moment at the center of the 
 second span ? 
 
 Ans. S= A wl M = -- - w ? 
 104 208 
 
 8. In one of four spans, what is the 
 shear and moment in the third span for 
 x-=.\ I and x=%l? 
 
 Ans. S= wl and S= -- wl 
 14 14 
 
 
27 
 
 In computing a framed truss we need 
 to find the value of S for a section cut 
 ting every diagonal and that of M for 
 one passing through each panel apex, 
 from these we readily derive the strains 
 in the webbing and chords by the rules 
 explained above. A single example will 
 render the whole process clear. (We 
 here treat of the dead load only ; com 
 putations involving the live or rolling 
 load will be presented hereafter.) 
 
 Let the truss represented in Fig. 2 
 consist of seven continuous spans, each 
 sixty feet in length. Let the uniformly 
 distributed load per linear foot be two 
 hundred pounds, one half of which rests 
 upon the upper chord and the other half 
 upon the lower. The lower chord is di 
 vided into six bays, each of ten feet, and 
 is connected with the upper one by a 
 Warren system of diagonals. The depth 
 of the truss is seven feet. Let it be re 
 quired to compute the strains in all the 
 pieces of the third span, due to this dead 
 load. 
 
 We first take from the triangles for a 
 
28 
 
 girder of seven spans. S q = wl and 
 
 M 3 =- wl*. We have for our case, 
 
 w=200 Ibs. and 1=60 feet. Hence 
 S 3 5916 Ibs. andM 3 =55775 Ibs. ft. 
 
 Inserting these and the value of w in the 
 above general formulae, we have 
 
 S = 5916 wx 
 
 M = 55775 5916 ^ + 100 x* 
 
 as the value of the shear and moment for 
 anv section x. 
 
 v 
 
 Now, since this is a framed truss, and 
 the several pieces are to be subjected 
 
 only to longitudinal strains, the load 
 should not be strictly uniformly distrib 
 uted but concentrated on the upper 
 chord at the panel points B, C, etc. and 
 on the lower chord at a, b, c, etc. Allow 
 ing that each of these points receives an 
 
29 
 
 equal weight and that a and g, count as 
 but one point, we have at a 500 Ibs., at 
 g 500 Ibs. and at each of the others 1000 
 Ibs. In finding the shear for the diag 
 onal a B, we pass the section anywhere 
 between a and B and takewx as 500, 
 for B b 10 x is 1500, for b C 2500 and so 
 on ; these subtracted from S 3 give the 
 required shears. This is in fact nothing 
 but taking the algebraic sum of all the 
 exterior forces between the left hand of 
 the truss and the diagonal under consid 
 eration, for S 3 is the sum of those forces 
 from the left end to the beginning of the 
 span. For the diagonal Fe we have, for 
 example, 
 
 S = 5916 - - 8500 = - - 2584 Ibs. 
 
 Thus by successive subtraction we 
 find the shears for all diagonals. Multi 
 plying them by the secant of the angle 
 between a diagonal and the vertical or 
 
30 
 
 and we have the required strains. To 
 determine their character we have simply 
 to consider that a positive shear causes 
 
 ( tensile ) 
 
 a -{ I strain in a diagonal 
 
 ( compressive j 
 
 which slopes \ u P ward \ toward the 
 ( downward ) 
 
 left hand support, while a negative shear 
 produces the reverse. In the following 
 table, the results thus determined are 
 recapitulated. The first column shows 
 the name of the diagonal corresponding 
 to Fig. 2, the second gives the shears, 
 the third gives the slopes, + indicating 
 an upward inclination toward the left, 
 and - a downward one, and the last 
 column gives the strains, -f indicating 
 tension, and compression. In forming 
 the last column from the two preceding 
 ones, it will be noticed that the rule of 
 signs is observed : 
 
31 
 
 DIAGONALS. (See Fig. 2.) 
 
 Piece. 
 
 Shear. 
 
 Slope. Strain. 
 
 Cb 
 
 J 
 
 -5416 
 -3416 
 
 + 
 
 -66561bs. 
 
 +5427 
 -4198 
 
 Cc 
 DC 
 
 
 h2416 
 -1416 
 
 
 
 -U2969 
 -1740 
 
 Ed 
 
 + 416 
 - 584 
 
 -1584 
 
 - 
 
 + 511 
 + 718 
 -1947 
 
 Ye 
 
 -2584 
 -3584 
 
 ^ 
 
 +3176 
 -4405 
 
 S{ 
 
 -4584 
 
 
 +5634 
 -6863 
 
 We will now pass to the computation 
 of the chord strains. In the above ex 
 pression for M the quantity ^ 10 cc 2 or 
 100 x* is the moment of the load between 
 the point 3 and the assumed section and 
 its value is the same whether the load be 
 considered as uniformly distributed or 
 concentrated at the apices as above. 
 Hence to find the moments for the upper 
 chord we have in the expression 
 
 M 55775 -- 5916^+lOOar 2 
 simply to give to x the successive values 
 
32 
 
 0, 10, 20, etc., since for the bays A B, 
 B C, C D, etc., the opposite vertices #, b, c 
 etc., must be taken as centers of moments. 
 Thus if the bay C D be cut rotation will 
 at once begin around the point c ; we 
 take then x =20 and find for the mo 
 ment of the strain in C D, 
 
 M= - - 22545 Ibs. ft. 
 
 and dividing this by its lever arm or 
 seven feet we have 3221 Ibs for the 
 strain. The character of the strain is 
 found by recollecting that a positive mo- 
 
 ( tensile } 
 
 ment causes a \ . > strain in 
 
 ( compress^ve \ 
 
 the ** [ chord, while a negative 
 ( lower } 
 
 moment produces the reverse. If we 
 designate then tension by -f and com 
 pression by , the signs of the strains 
 in the upper chord will be the same as 
 those of the moments. In this way it is 
 easy to compute the following results : 
 
33 
 
 UPPER CHORD. (See Fig. 2). 
 
 Bay. 
 
 Moment. 
 
 Strain. 
 
 AB 
 BC 
 CD 
 DE 
 EF 
 FG 
 GH 
 
 +55775 Ibs. ft. 
 + 6615 
 -22545 
 -31705 
 -20865 
 + 9985 
 +60845 
 
 +7968 Ibs. 
 + 945 
 -3221 
 -4529 
 -2981 
 +1426 
 +8692 
 
 For the lower chord the calculation is 
 very similar. The centres of moments 
 are taken at the points B, C, etc., the 
 successive values of x are 5, 15, 25, etc., 
 the strains are numerically one-seventh 
 of the moments, and their signs are op 
 posite to that of the moments. Thus for 
 the bay e/,a?=45, M= - 7945 Ibs. ft. 
 and the strain in ef is 1135 Ibs. tension. 
 The results are in the following table : 
 
34 
 
 LOWER CHORD. (See Fig. 2.) 
 
 Bay. 
 
 Moment. 
 
 Strain. 
 
 ab 
 
 +28695 Ibs. ft. 
 
 -4099 Ibs. 
 
 be 
 
 -10465 
 
 
 -1495 
 
 cd 
 
 -29625 
 
 
 -4218 
 
 de 
 
 -28785 
 
 _ 
 
 -4112 
 
 f 
 
 - 7945 
 
 
 -1135 
 
 fg 
 
 +32895 
 
 -4699 
 
 and the strain sheet for the span is now 
 complete. 
 
 In the same way the strains for each of 
 the other spans may be readily found. 
 From the symmetry of the truss it is 
 evident that the fifth span will be exact 
 ly the same as the third, the sixth the 
 same as the second, and the seventh the 
 same as the first. For the fourth span 
 the value of S and M for any section x 
 
 are 
 
 w x 
 
 M=60848 6000& + 100 x 3 
 and the strains will be the same on each 
 side of its center. For the first span 
 Sj is the same as the reaction R 1 the mo 
 ment M, is zero and we have 
 
35 
 
 8 = 4732 wx 
 M= 4732 x+100 
 
 It will be seen then that the compu 
 tation of the strains in a continuous 
 girder is exactly the same as in a simple 
 one, except only in the preliminary de 
 termination of the shears and moments 
 at the supports. In a simple girder the 
 end shears are the same as the reactions 
 which are known from the law of the 
 lever, and the moments at the supports 
 are zero. In a continuous one these 
 quantities must be determined by formu 
 lae, or, for the case of equal spans uni 
 formly loaded,* taken from the triangles 
 which we have given above. They may 
 also be found by a graphical process. 
 
 If the truss above discussed were built 
 with seven simple girders, the strains in 
 each would be the same. It may prove 
 interesting then to compare the results 
 above found with those for a simple gir- 
 
 * Other convenient tables for concentrated loads and 
 for uniform loads over one span only are given in the ar 
 ticle above referred to in the Journal of the Franklin In. 
 stitute. 
 
36 
 
 der. The mode of computation is essen 
 tially the same ; the end shears are each 
 one half the total load, the pieces A B 
 and G PI in Fig. 2 disappear, and for any 
 section x we have 
 
 8=6000 wx 
 
 Considering as before that the load is 
 concentrated at the panel points we have 
 500 Ibs. at a and y and 1000 Ibs. at B, C, 
 D, , c, d, etc., respectively. We then 
 find the shears and moments and from 
 them deduce the strains as above de 
 scribed. The results are given below 
 compared with those for the third span 
 of the continuous truss. 
 
 (See Table on following pages.} 
 
 Adding these strains regardless of 
 sign we find the two sums to be the 
 same. It can be easily demonstrated 
 that for the dead load such should be 
 
DIAGONALS. (See Fig. 2.) 
 
 Piece. 
 
 Continuous 
 truss. 
 
 Simple truss. 
 
 Ea 
 
 -6656 Ibs. 
 
 6760 Ibs. 
 
 Cb 
 
 +5427 
 -4198 
 
 +5531 
 4301 
 
 Cc 
 DC 
 
 +2969 
 1740 
 
 +3072 
 1844 
 
 Dd 
 
 Erf 
 
 + 511 
 + 718 
 -1947 
 
 + 614 
 + 614 
 1844 
 
 e 
 
 +3176 
 -4405 
 
 +3072 
 4301 
 
 G 
 
 +5634 
 6863 
 
 +5531 
 6760 
 
 Sams. . . 44244 Ibs. 
 
 44244 Ibs. 
 
 the case. As far as the diagonals are 
 concerned, the two structures require an 
 equal amount of material. 
 
 UPPER CHORD. (See Fig. 2.) 
 
 Bay. 
 
 Continuous 
 truss. 
 
 Simple truss. 
 
 AB 
 BC 
 CD 
 DE 
 EF 
 FG 
 GH 
 
 +7968 Ibs, 
 + 945 
 3221 
 4529 
 
 2981 
 +1426 
 +8692 
 
 7143 Ibs. 
 11429 
 
 12857 
 11429 
 7143 
 
 Sums. . . 29762 Ibs. 
 
 50001 Ibs. 
 
38 
 
 Adding the strains in the upper chord 
 we observe that the sum for the simple 
 truss is about 1.7 times that of the other. 
 If the amount of material is to be pro 
 portional to the strain, a considerable 
 saving will here be expected. 
 
 LOWER CHORD. (See Fig. 2.) 
 
 Bay. 
 
 Continuous 
 truss. 
 
 Simple truss. 
 
 ab 
 
 -4099 Ibs. 
 
 3929 Ibs. 
 
 be 
 
 
 -1495 
 
 9643 
 
 cd 
 
 
 -4218 
 
 12500 
 
 de 
 
 
 -4112 
 
 12500 
 
 ef 
 
 
 -1135 
 
 9643 
 
 fff 
 
 -4699 
 
 3929 
 
 Sum.... 19758 Ibs. 
 
 52144 Ibs. 
 
 The lower chord in the simple truss 
 would then be subjected to about 2.6 
 times as much strain as in the continuous 
 
 one. 
 
 If we suppose that the same working 
 strength may be allowed for compression 
 as for tension, we may obtain an estimate 
 of the saving in material by employing a 
 
39 
 
 continuous truss instead of a simple one. 
 The amount of material will be propor 
 tional to the strain and to the length of 
 the piece strained. Regarding the bays of 
 the chord as unity, the diagonals will be 
 represented in length by 0.86. The pro 
 portionate amounts of iron will then be 
 found by multiplying the above sums by 
 unity for the chords and by 0.86 for the 
 diagonals. Thus we have a 
 
 COMPARISON. 
 
 
 Continuous 
 truss. 
 
 Simple truss. 
 
 Diagonals .... 
 Upper Chord. . 
 Lower Chord. . 
 
 38050 
 29762 
 19758 
 
 38050 
 50001 
 52144 
 
 Total 
 
 87570 
 
 140195 
 
 
 
 
 from which we see that the amounts of 
 material in the two cases are in the ratio 
 of the numbers 87570 and 140195 or as 
 1 to 1.6. For this particular span then 
 a saving in material of thirty-seven and 
 a half per cent, is effected by using a con 
 tinuous truss instead of a common one. 
 
40 
 
 It is capable of demonstration that for 
 girders subjected only to dead load, the 
 total amount of strain in the webbing 
 will be the same for simple as for con 
 tinuous trusses, and also that under the 
 most favorable circumstances, the total 
 strain in the chords of the first is to that 
 in the chords of the second as /\/2^ is to 
 2 or nearly as 2.6 to 1. 
 
 In studying the theory of girders many 
 interesting questions arise which are of 
 little importance in practice. One of 
 these is the determination of the inflec 
 tion points. At these points the curva 
 ture of the beam changes, the strain 
 passes from tension to compression and 
 the moment is zero. At any point in the 
 nth span the moment is 
 
 M = M n - - S n x +-J- w x* 
 
 Making in thisM equal to zero and solv 
 ing the equation with reference to x we 
 find 
 
 ; 2 " w 
 
41 
 
 For a girder of equal spans and uniform 
 ly loaded we may hence write for the two 
 inflection points, 
 
 in which A and B are to be taken from 
 the above triangles, A always being taken 
 for the right hand side of the support 
 under consideration, for example, in a 
 girder of eight spans the inflection points 
 for the fourth span are at the points 
 
 195 
 
 V388/ 388 
 or for 33=0.22 I and ^=0.79 L 
 
 The point of maximum moment, or 
 the point near the center of the beam, 
 where the chord strain is the greatest is 
 more important and readily determined 
 from the above general value for M. 
 Differentiating it with reference to x we 
 have 
 
 = S n + W 23=0 
 
 a x 
 that is, the maximum moment obtains at 
 
42 
 
 the point where the shear is zero. Its 
 
 S 
 value is found by replacing x by 
 
 which gives 
 
 ft 2 
 Max. M=M n 
 
 2 w 
 
 as the greatest negative moment. 
 
 The following examples will enable 
 the reader to test his knowledge of the 
 preceding principles : 
 
 9. In a girder of two spans uniformly 
 loaded what are the maximum positive 
 and negative moments ? 
 
 Ans. 0.125 wT and 0.071 wl\ 
 
 10. In one of three spans what is the 
 maximum negative moment in the mid 
 dle span ? 
 
 Ans. 0.025 wF. 
 
 11. In one of eight spans where are 
 the inflection points in the fifth span ? 
 
 Ans. x= 0.2lZandcc=( 
 
 12. A continuous girder of three spans, 
 each equal to fifty feet, is divided into 
 
43 
 
 five panels on the lower chord, and has 
 bracing similar to that shown in Fig. 2. 
 Supposing a load of five tons applied at 
 each of the lower panel points, what are 
 the strains in each of the pieces of the 
 middle span ? the height of the truss 
 being six feet. 
 
 Ans. In a &, --11.7 tons; in b c, +1.7; 
 in B&, +13 ; etc. 
 
 In this chapter we have treated of the 
 continuous girder when affected only by 
 dead load or its own weight. In the fol 
 lowing chapters we shall take up the ac 
 tion of the live or variable load. 
 
45 
 
 CHAPTER II. 
 
 A continuous girder, loaded in any 
 manner, is held in equilibrium by the up 
 ward pressures or reactions of the sup 
 ports, and, as we have seen, these reac 
 tions are alone sufficient for the complete 
 determination of the strains in every 
 part of the girder. But if, regarding the 
 question as one of pure statics, we con 
 sider the beam as rigid, we find it impos 
 sible to determine the reactions when the 
 number of supports is greater than two. 
 This does not arise from the fact that in 
 an actual case the question is indeterm 
 inate, but simply because in considering 
 the girder as rigid we have restricted the 
 data to the mere weight, neglecting en 
 tirely the physical properties of the ma 
 terial. By taking into account the elas- 
 
46 
 
 ticity of the girder, the problem becomes 
 determinate ; we find the reactions, or 
 what is equivalent, the shears and mo 
 ments at the supports, and from these 
 the investigation of the internal forces 
 or strains is easy. 
 
 THE ELASTIC LINE. 
 
 When a girder is acted upon by verti 
 cal forces, a change of shape arises, 
 which causes the originally parallel fibers, 
 to be on one side lengthened, and on the 
 other shortened. Between the lengthen 
 ed and shortened fibers, there is a plane 
 which undergoes no change of length ; 
 the central line of this plane is called the 
 neutral axis or the elastic line. Thus, in 
 the bent beam represented in Fig. 4, m o 
 is the neutral axis, the fibers above it be 
 ing shortened "or compressed, and those 
 below it lengthened or tensioned. 
 
 We derive the equation of the elastic 
 line upon three hypotheses : 1st, that all 
 planes perpendicular to the neutral axis 
 before the bending or flexure, preserve 
 
4 
 
 - 
 
 during the bending their perpendiculari 
 ty and their form as planes ; 2d, that the 
 change of length of a body subjected to 
 a force is, within certain limits called the 
 elastic limits, proportional to the intensi 
 ty of the force ; and 3rf, that the change 
 of shape is so little that the length of the 
 neutral axis is sensibly the same as its 
 horizontal projection. 
 
 In Fig. 4 we have a longitudinal sec 
 tion of a portion of a bent beam ; the two 
 planes a b and d 6, originally parallel, re 
 maining perpendicular to the neutral 
 axis m o, and intersecting in c the center 
 of curvature. Hence, drawing f g paral 
 lel to ab through o, the lines /V7, ge, etc., 
 denote the elongations and compressions 
 of the respective fibers, and we see from 
 the figure that 
 
 od:od r \\df: d f 
 
 or the change of length in the fibers is 
 proportional to their distances from the 
 neutral axis. This is the consequence of 
 the first hypothesis. 
 
 Designating by H and H the force 
 
48 
 
 acting in the fibers df and d /", the sec 
 ond hypothesis says that 
 
 H:H :;<7/: d f 
 FIG. 4. 
 
 Hence, by combining these two propor 
 tions, 
 
 H : H ; I o d : o d r 
 or, the horizontal forces are directly pro- 
 
portional to their distances from the 
 neutral axis. Therefore, if we denote 
 the distance of any fiber from the axis 
 by 2 the strain upon it by H , the dis 
 tance of the remotest fiber by e and the 
 strain upon it by H, we have 
 
 H : H ; ; z : e or H = 
 
 e 
 
 We have thus far considered the cross- 
 section of the fibers as unity. If the 
 actual section be a, the force in each is 
 
 evidently - - . Each of these forces, 
 
 as for instance H in the figure, tends to 
 turn the beam around the point o with a 
 lever arm o d or z, and its moment or the 
 measure of that tendency to rotation is 
 
 the product of the force * by the dis- 
 
 TT y 8 
 
 tance z, or - . The sum of all these 
 
 moments is 
 
 M=- 
 
 e 
 
50 
 
 or, since 2 a is the moment of inertia 
 of the section a &, we have 
 
 6 
 
 as the expression for the sum of the mo 
 ments of the internal forces, H being the 
 strain in the remotest fiber, e its distance 
 from the neutral axis, and I the moment 
 of inertia of the cross-section. 
 
 The line df denotes the change of 
 length of the fiber ad due to the force 
 H. Hence if E be the coefficient of 
 elasticity,* 
 
 ad: df: ; E: H 
 
 Designating the radius c o by r we have 
 from the similar figures o df and cad 
 (since mo=ad), 
 
 a d : df. \r : e 
 
 * The Coefficient of Elasticity is the ratio of the force 
 of displacement to the amount of displacement taken 
 upon a cube whose edge is unity ; Hence for the above 
 
 J 4* 
 
 case E=H-$ - . The term modulus of elasticity pro- 
 
 wtt 
 
 perly relates to the impact of bodies, and is a measure of 
 elasticity in the common sense of the word, unity indi 
 cating perfect elasticity or restitution of form. These 
 terms are often confounded by writers. 
 
51 
 
 Combining these proportions we find 
 
 H_E 
 
 e r 
 
 and hence, for the value of the internal 
 moment, we have 
 
 The radius of curvature of any plane 
 curve, whose length is u, and co-ordinates 
 x and y is* 
 
 r= 
 
 d x d* y 
 
 And as by our third hypothesis we may 
 place du=dx, this becomes 
 
 r _dx* 
 
 /*y ^ i 
 
 \At */ 
 
 which, inserted in the above value of M, 
 gives 
 
 (1) 
 
 d* y M 
 
 dx* El 
 as the differential equation of the elastic 
 
 * See any work on the Differential Calculus. 
 
52 
 
 curve, applicable to all bodies subject to 
 flexure, which fulfill the condition im 
 posed by the third hypothesis. The co 
 efficient of elasticity and the moment of 
 inertia may be different in every section. 
 
 CONDITIONS OF EQUILIBRIUM. 
 
 Let us consider the r thv span of a con 
 tinuous girder whose length is l Ty and 
 let a single concentrated load P r be 
 placed on this span at a distance M r from 
 the left hand support r. This load, the 
 loads on the other spans, and the weight 
 of the girder itself, are held in equili 
 brium by the vertical reactions R r i, 
 R Tj etc., of the several supports. (See 
 Fig. 6.) 
 
 Let us pass a section between the load 
 P r and the support r-j-1 at a distance x 
 from the r th support. As shown in the 
 last chapter, all the internal forces in this 
 section are represented by a shear S and 
 a moment M. The shear S is equal to 
 the algebraic sum of all the external 
 forces upon the left hand side of the sec- 
 
53 
 
 tion, and the moment M is equal to the 
 sum of the moments of those forces with 
 respect to the section as a center. Hence, 
 regarding upward forces as positive, and 
 a moment as positive when it tends to 
 cause tension in the upper fiber of the 
 section, w r e have 
 
 (2) S = S, - P r 
 
 in which S r is the shear at the right of 
 the r th support, and M r the moment at 
 that support. In like manner for a sec 
 tion between r and P r> we have 
 
 S=S r 
 
 M=M r - - S r x 
 
 The internal forces at any section can 
 then be found as soon as the shear and 
 moment at the preceding support are 
 known. 
 
 If, in the above expression, we make x 
 equal to / r , M becomes M r _j_ i, and we de 
 duce 
 
54 
 
 The shear and moment at any section 
 can then be determined as soon as M r and 
 M r 4_ i, the moments at the preceding and 
 following supports, are known. 
 
 These conditions of equilibrium are 
 entirely independent of variations in the 
 dimensions or material of the beam, or 
 in the relative heights of the supports of 
 tlie girder. 
 
 THE EQUATION OF THE ELASTIC LINE. 
 
 In order to apply equation (l) to the 
 case of continuous girders, we have to 
 insert for M, E and I their values as func 
 tions of x and integrate the equation 
 twice. E, however, cannot under any 
 ordinary circumstances be a function of 
 x, it being dependent upon the elasticity 
 of the material alone, which is nearly the 
 same in one and the same beam, and we 
 hence regard it as constant. In a beam 
 of uniform section I is constant, and 
 although it varies in common bridge 
 
 o o 
 
 trusses, we shall be obliged in order to 
 bring the investigation within the limits 
 
55 
 
 of this paper to consider it always as con 
 stant, taking care to point out after 
 wards the slight error thus introduced. 
 Inserting then in (1) the value of M 
 from (2), we have 
 
 as the differential equation, applicable 
 to girders of constant elasticity and 
 uniform cross section. Integrating this, 
 the constant is t r , the tangent of the in 
 clination of the elastic line at the sup 
 port r and 
 
 dy 2M r x-S r a 
 
 * + -- 
 
 we have thus far taken no account of 
 the relative heights of the supports. 
 For the reasons mentioned in the last 
 chapter, we shall consider them as all 
 upon the same level. The constant for 
 the second integration is then 0, the 
 origin being at r, and we have 
 
 3 M x 2 -S x* + P r x 
 
 
56 
 
 as the equation of the elastic curve be 
 tween the load P r and r-fl th support 
 (Fig 5). If there be several loads we 
 have only to affix the sign of summation 
 2 to the term involving P r> and if that 
 term be omitted we shall have the equa 
 tion between the load and the r ih sup 
 port, since for any section between those 
 points M=M r S r #. 
 
 If in (5) we make # / r , y becomes 0, 
 and inserting for S r its value from (3), 
 we find 
 
 (6) 6EH=-2M r t 
 
 Thus the equation of the curve is 
 completely determined, when we know 
 M r and M r + 1 the moments at the sup 
 ports r and r-f-1. These may be found 
 by the remarkable theorem of three mo 
 ments. 
 
 THE THEOREM OF THREE MOMENTS. 
 
 In Fig. 5 is represented a portion of a 
 continuous truss. Beginning at the left 
 hand end, the lengths of the spans are 
 
57 
 
 denoted by l^ / 2 , ..... l r , etc , and the sup 
 ports are designated as 1, 2, ..... r, etc. 
 Upon the spans l r \ and / r are loads 
 P r i and P r> whose distances from the 
 nearest left hand supports are // r i an d 
 Jc l^ k being any fraction less than unity, 
 and not necessarily the same in the two 
 cases. The equation of the elastic line 
 between P r and the support r + 1 is 
 given by (5), and the tangent of the 
 angle which the curve at the section x 
 makes with the axis of abscissae is iven 
 
 (4). 
 
 Fio. 5. 
 
 ^ ---- if-i ~" x ----- it- 
 
 
 If in (4) we substitute for S r its value 
 from (3), and for t r its value from (6), 
 
 and make cc=/ r , -~ becomes t^\\ the 
 ax 
 
 tangent of the inclination of the curve 
 at r-f-1, and we find 
 
58 
 
 Now, if we consider the origin moved 
 from the support r back to r 1, we may 
 derive a value for t r by simply diminish 
 ing each of the indices in the above ex 
 pression by unity, hence 
 
 (7) 6 EH=M r _i? r _i + 2 M r / r _i 
 
 -P r _!Z 2 r _!(&-& ) 
 
 Equating the values of 6 E I t r given by 
 (6) and (7), we have 
 
 (8) M r _ 1 / r _ 
 
 which is the most general form of the 
 theorem of Three Moments for girders 
 of constant cross-section. By prefixing 
 the sign 2 to the terms in the second 
 member, it becomes applicable to any 
 number of single loads. For uniform 
 
 c_> 
 
 loads ^ r i and w r per linear unit, we 
 have only to place 
 
 P r _ i = w r id(kl r i) = w r il r \dk 
 P r = w r d (k l r )=io r l r d k 
 
 And to replace the sign of summation 
 2 by that of integration /. If these 
 
59 
 
 loads extend over the entire spans l r and 
 / r 1> we take the integrals between the 
 limits k=Q and &=1 and have 
 
 r (/r + ?r - l) + M r + l 1 T 
 W _ 1 3 _ W T I s T 
 
 
 which is the theorem as first announced 
 by Clapeyron. 
 
 For each support of a continuous 
 girder an equation may be therefore 
 written involving the moment at that 
 support, and those at the preceding and 
 following support. In a girder of s 
 spans there are 5+1 supports, and since 
 the moments at the first and last sup 
 ports are zero we have s 1 moments, 
 whose values may be found by the so 
 lution of the 51 equations. The mo 
 ments give the shear at any support, and 
 by (2) the internal forces or strains may 
 be determined for every section of the 
 girder. 
 
 REMARKS ON THE PRECEDING THEORY. 
 
 The laws of the theory of continuity 
 above deduced can be regarded as onlyap- 
 
60 
 
 proximate. Of the three hypotheses upon 
 which the differential equation of the 
 elastic line is deduced, the first, although 
 a most reasonable assumption, has not 
 been definitely verified by experiment, 
 and the second is rendered somewhat 
 doubtful by the extreme difficulty in 
 delicate experiments of assigning the 
 elastic limits. Nevertheless they are 
 universally regarded by all writers as 
 sufficiently accurate to form the basis of 
 a working theory, and must continue to 
 be thus used until we attain to a more 
 thorough knowledge of matter and force. 
 The third hypothesis, however, is a limi 
 tation of the data, which we are at per 
 fect liberty to make, since we know that 
 the increase in length of the girder by 
 deflection is too small to be practically 
 measured. We may conclude then that 
 the equation (1) is an extremely close 
 approximation to the actual law govern 
 ing straight elastic beams. From the 
 time of Navier to the present it has 
 been so accepted and used. 
 
 The next hypothesis or limitation of 
 
61 
 
 data, which we make, is that E, the co 
 efficient of elasticity, is constant through 
 out the girder. In a solid beam of 
 ordinary homogeneous material, there 
 can be no reason for supposing it other 
 wise. 
 
 In the Journal of the American So 
 ciety of Civil Engineers for May, 1876, 
 appeared an article by Charles Bender, 
 C. E., in which the use of continuous 
 bridges is strongly opposed. One of his 
 main objections is that the theory upon 
 which such bridges are computed is un 
 reliable in consequence of the assump 
 tion of a constant coefficient of elasticity, 
 and he quotes the records of experi 
 menters to show that values for the 
 coefficient of iron and steel have been 
 observed ranging from 17,000,000 to 
 40,000,000 pounds per square inch. 
 These limiting values are, however, 
 decidedly exceptional, but even grant 
 ing that such variations may exist 
 in materials and forms like soft iron 
 wire, steel rails and wrought iron eye- 
 bars, it cannot be supposed that they 
 
62 
 
 will occur in one and the same structure, 
 where the material is of one kind, of 
 similar cross sections and which has 
 been subjected in the same mill to the 
 same process of manufacture. The mere 
 statement that Morin has observed val 
 ues of the coefficient of elasticity as low 
 as 17,000,000 has very little weight when 
 unaccompanied by any reference to the 
 kind of iron experimented upon. Let 
 us see what Morin himself actually says 
 in recapitulating the results of experi 
 ments upon wrought iron.* 
 
 " Iron of superior quality, which comes 
 from standard ores, and which has been 
 manufactured exclusively with charcoal, 
 or iron from sheet metal, many times 
 refined, may give for the coefficient of 
 elasticity values as high as E=28,- 
 400,000, or even E= 31, 200,000 Ibs. per 
 square inch, equal to those furnished by 
 ordinary steel. Iron of ordinary manu 
 facture reduced with common coal, and 
 drawn into forms like rails, T irons, 
 
 *Morin ; Resistance des Materiaux, Paris, 1862. Vol. 
 I, p. 443. 
 
63 
 
 flanges, etc., give such values as 
 E = 24,100,000 and E = 25,600,000. 
 Finally the most soft and ductile iron 
 furnishes values as low as E=21, 300,000, 
 or E= 19,800,000, or even E=: 17,000,000. 
 It is well, then, in calculations upon the 
 strength of iron, to ascertain the quality 
 of the material and the process of manu 
 facture." 
 
 Mr. Bender likewise alludes to ex 
 periments upon wrought iron bars in 
 which the coefficient of elasticity was 
 found to be 40,000,000, and it seems to 
 be implied by his language that such 
 values are of common occurrence. The 
 fact, however, that a standard writer on 
 the strength of materials, like Morin, 
 regards 34,000,000 as an exceptionally 
 high figure for iron, may justify us in 
 demanding that when a value like 
 40,000,000 is quoted, we should be fur 
 nished with some details concerning the 
 quality of such iron, the process of 
 manufacture, as well as a description of 
 the testing machine and the manner of 
 measuring the small extensions or com- 
 
V 64 
 
 pressions, from which the coefficient is. 
 calculated, or at least that we should be 
 referred to the book or journal where 
 such experiments are described. And 
 further as it is well known that by strain 
 ing a bar beyond the elastic limits, very 
 low values of E can be deduced, are we 
 not justified in asking similar informa 
 tion concerning experiments which fur 
 nish such values ? 
 
 Undoubtedly there is some variation 
 in the elasticity of different pieces of 
 iron, even when great care has been 
 taken to ensure uniformity of material 
 and manufacture, and it is greatly to be 
 desired that experiments should be made 
 to determine how it varies with the 
 cross section and length of the piece. 
 As soon as such a law of variation is 
 discovered (if any exist), we shall be 
 obliged to consider E as variable in in 
 vestigating a continuous truss. But if 
 no law exists and we know only the fact 
 that there are slight variations in the 
 elasticity of different pieces in the same 
 truss, what is to be done ? Nothing but 
 
65 
 
 to regard E as constant, 
 assured that the distribution of the 
 variable pieces throughout the structure 
 will be governed by the law of proba 
 bility, and that hence the girder as a 
 whole will conform closely to the 
 theoretic form of the elastic line. 
 
 The next argument which it is our 
 duty to criticise in that article is, that 
 the theory of continuous girders is un 
 reliable, because the calculated deflection 
 does not generally agree with the 
 actually measured deflection. The ac 
 curacy of the computed strains must 
 depend upon the accuracy of the 
 theorem of three moments, and this it is 
 asserted depends upon the calculated 
 deflection. And because the deflection 
 as actually measured is sometimes no 
 more than one-half of the calculated 
 one, hence, it is said, the same differences 
 may occur in the strains, and the whole 
 theory is unworthy of consideration. 
 
 This we can only regard as a striking 
 instance of the jncompetency of prac 
 tical men to draw conclusions from even 
 
66 
 
 simple experiments. The reader who 
 has followed onr presentation of the 
 theory of the elastic line, will see at 
 once that the value of the deflection 
 given by (5) only enters the discussion 
 as an auxiliary for finding (6), the 
 tangent of the inclination angle at the 
 support. The process supposes, indeed, 
 that E is constant, but it supposes noth 
 ing whatever concerning the value of 
 the deflection at any point. When we 
 pass to the next span and find again in 
 (7) a second value of the tangent, the 
 actual value of the deflection there is 
 likewise not considered. And when by 
 the combination of (6) and (7), we de 
 duce (8) in which E does not appear, its 
 very absence is a proof that the moments 
 and hence the strains are entirely in 
 dependent of its value or of the actual 
 deflection. If two trusses of the same 
 spans, height and form are continuous 
 over several supports, one of steel hav 
 ing a coefficient of elasticity of 31,000,000 
 Ibs. per square inch, and the other of 
 wood having a coefficient of only one- 
 
67 
 
 twentieth as much, the reactions, shears, 
 moments and strains would be the same 
 in each. The measurement of the actual 
 deflections of these bridges under given 
 loads is only useful for determining and 
 comparing their stiffness or elasticity, or 
 in connection with theory for finding the 
 values of E and I. The theory of con 
 tinuity rests not upon absolute deflec 
 tions, but on relative ones on the form 
 of the elastic curve, and this again upon 
 the three universally accepted hypotheses 
 included in our equation (1), with the 
 additional assumption that the coefficient 
 of elasticity is practically constant. 
 
 One more remark and we close for to 
 day a discussion which shall be resumed 
 in our next chapter. Mr. Bender advises 
 us to abandon the theory of flexure, to 
 make no further advance in bridge 
 building, to remain content with the 
 simple lever, or at the utmost with the 
 continuous (sic) patent hinged truss. 
 But until such advice is enforced by 
 more logical arguments than we have 
 yet seen, we must continue our work 
 
68 
 
 in support of a theory and system 
 which is universally accepted as only 
 slightly deviating from the exact ex 
 isting conditions, which is applied in 
 the erection of continuous bridges by 
 every nation except our own, and which 
 perhaps if carried out by us might lead 
 to more perfect structures than the 
 world has yet seen. The great majority 
 of coefficients of elasticity quoted in his 
 paper, made by such men as Staudinger, 
 Baker, Morin and Wohler, were in fact 
 found by measuring the deflections of 
 beams, and then from the theory of 
 flexure computing the value of E. He 
 accepts those values, and on their evi 
 dence condemns the theory by which 
 they were deduced ! Is not Morin s 
 conclusion, which we have quoted above, 
 by far the most reasonable? 
 
 MOMENTS AT THE SUPPORTS. 
 
 The theorem of three moments given 
 by (8) furnishes the means of finding the 
 moments at the supports due to any as 
 signed system of loading. The actual 
 
69 
 
 solution of those equations is, however, 
 quite tedious when the number of spans 
 is large, and we proceed therefore to de 
 velop a general solution, by which the 
 values of the moments may be formulated 
 and placed in a convenient form for nu 
 merical computation. 
 
 In the designing of continuous bridges 
 it is only necessary to consider single 
 loads concentrated at the panel apices or 
 uniform loads extending over an entire 
 span. Let us, then, consider a continu 
 ous girder of constant cross-section and 
 homogeneous material whose supports 
 are on the same level. Let, as in Fig. 
 6, the supports beginning at the left 
 hand end be designated by the indices 1, 
 2, 3, .... r, etc, and the lengths of the 
 spans Z t , l# l l n etc. Call the num 
 ber of spans s; then the last span will 
 be / and the last support s+l. The 
 ends of the girders rest upon abutments 
 in the usual manner, the lengths of the 
 spans may be all different and their 
 number may range from one to infinity. 
 In the span 1 T let a single load P be 
 
70 
 
 placed at a distance k 1 T from the sup 
 port r, (k being any fraction less than 
 
 FIG. 6. 
 
 
 unity), or let this span be loaded uni 
 formly from r to r+ 1 with a weight W, 
 (W being equal to w / r , if w is the load 
 per linear unit), all the other spans be 
 ing unloaded. By reference to (8) we 
 notice that there are two functions of 
 such loads which enter the equations of 
 moments. If the single load P is alone 
 
 V--- 
 
 considered these functions are 
 
 the first entering into the equation for 
 the preceding support r, and the second 
 into the one for the following support 
 r 4- 1. If the uniform load over the whole 
 span is alone considered, these become 
 each equal to J W C, as we have shown 
 above in discussing the theorem of three 
 
71 
 
 moments. In the following investiga 
 tion, we place therefore for abbreviation 
 
 A=P 5 (2* 3 ** + * ) ) for a single 
 B=P II (* i*) j load in span I,. 
 
 (I) 
 
 A \ -WZJ ) for a uniform load over 
 B=J WZ* f whole span Z r . 
 
 By introducing the letters A and B to 
 represent these functions, our discussion 
 will apply equally well to a single load 
 P, or to a weight W uniformly distribut 
 ed over the whole of a single span. 
 
 Since the girder is not fastened at the 
 abutments 1 and s-fl, the moments at 
 those points will be zero. The moments 
 at 2, 3, .... r, etc., we designate by M 2 , 
 M 3 , .... M r , etc., and from (8) we may 
 write an equation for each of those sup 
 ports. As there is no load considered 
 except on the span Z r , the right hand 
 member of the equation for the support 
 r will be A, of that for r + 1 will be B 
 and of all the others will be zero. Thus 
 we have the following equations : 
 
>7f 
 
 2 
 
 t 
 The number of these equations is s 1, 
 
 the same as the number of unknown mo 
 ments. Their solution is best effected 
 by the method of indeterminate multi 
 pliers. Let then the first equation be 
 multiplied by a number c 2 , the second 
 by c a , etc., the index of the as yet inde 
 terminate numbers corresponding with 
 that of the M in the middle term. Then 
 let all the equations, thus multiplied, be 
 added, and the resulting equation be ar 
 ranged according to the coefficients of 
 the unknown moments M 2 , M 3 , etc. Now, 
 if we require that such relations exist 
 between the multipliers, that all the 
 terms in the first member shall reduce to 
 
zero, except the last containing M 8> the 
 value of M 8 is 
 
 M= 
 
 ^I/S^JL+SA (/8-i K) 
 
 And the values of the multipliers will 
 be given by the equations 
 
 etc., etc 
 
 After deducing the values of c from 
 these equations, the value M 8 is at once 
 known. 
 
 Again, if we multiply the equations of 
 moments, beginning with the last, by the 
 indeterminate numbers d d 3 , etc., all the 
 moments except M 2 may be eliminated, 
 and we have 
 
 and the multipliers will bo given by the 
 equations 
 
 etc., etc., etc. 
 
74 
 
 The values of the numbers in the series 
 c and d need only satisfy the equations 
 as given above. Assuming then c,=l, 
 and d, 1, we get the following : 
 
 c, = 
 <V=l 
 
 2 A+A 
 
 C 3 _ ^ 
 
 (II) Ci =-2c 3 
 
 3 
 
 c.= -2c,-(2c 4 + c.) J 
 
 c 6 =+2c 6 -(2c B + c 4 )-^ 
 
 6 
 
 etc., etc. 
 
 etc., etc. 
 
75 
 
 which reduce to numerical form as soon 
 as the lengths of the spans for any par 
 ticular case are substituted. 
 
 Since the 51 equations of moments 
 are of the same form as the equations of 
 the multipliers c and d, we must have 
 
 or, if n indicate the index of any sup 
 port, 
 
 M n = c n JVI 2 when n <r -f 1 
 M n = ^s_ n -f2 MS, when n>r 
 
 Inserting in these the values of M 2 and 
 M 8 as found above, we have 
 
 M n = 
 (III) 
 
 M n = 
 
 when 
 
 which give the values of the moments at 
 all supports in terms of the quantities A 
 and B, depending only upon the charac 
 ter of the load and its position in the 
 
76 
 
 span /r, and the numbers c and d depend 
 ing only upon the lengths and number 
 of the spans of the girder.* To find 
 their numerical value for any given case 
 is hence a simple arithmetical exercise. 
 
 Example 1. A continuous girder has 
 four unequal spans, ^ = 80 ft.,/ 2 =100ft., 
 1 3 = 50 ft., and / 4 =40 ft. (Let the reader 
 draw the figure). On the span Z 2 is a 
 single load P=10 tons, whose distance 
 from the support 2 is &Z 2 =40ft. To 
 find the moments at the supports. 
 
 Since // 2 = 40, and 2 =100, we have 
 &=0.4. Inserting then in (I), the values 
 of &, l z and P, we find 
 
 A=38400 tons ft. B- -33600 tons ft. 
 
 Inserting next the lengths of the spans 
 in (II), we have 
 
 ^=0 ^=0 
 
 c,= l 4, = ! 
 
 c 3.6 c?=: 3.6 
 
 * In the London Philosophical Magazine for September, 
 1875, where the above demonstration was first give ", the 
 author has- extended the method to girders witn fastened 
 or walled-in ends. 
 
77 
 
 Since the load is in the second span r 2, 
 also 5=4; hence r/ s _ r _f_ 2 =f/ 4 =10.3, c r = 
 2 =1, etc., and / 8 _ ^^^SO, etc. By- 
 inserting these values of c, d, I, A and B 
 in (III), we obtain 
 
 M n = 82.01 c n , when n<3 
 M n = --24.05 / 6 -n, when n>2 
 
 For the abutment or left hand support, 
 we have n=l r c,=0, and hence IV^ 0, 
 for the second support, n = 2, ^ 2 =1, and 
 M 2 =S2.01 tons ft. For the third sup 
 port, w=3, f/ 6 _ n r= 3.6, and M 3 = 88.56 
 tons ft. For the next, ?i=4, r/e n : =l> 
 and M 3 = 24.65 tons ft. Lastly, for the 
 right hand abutment, w = 5, and M 6 0. 
 A positive moment, it will be remember 
 ed, causes tension in the upper chord of 
 a truss, while a negative moment causes 
 the reverse. 
 
 2. A girder of four equal spans has a 
 load P at any point on the tirst span. 
 Find the moment at each pier due to P. 
 
 Ans. M 4 =if PJ (k & 3 ), etc. 
 
78 
 
 SHEARS AND REACTIONS AT THE SUPPORTS. 
 
 It is thus easy from (I), (II) and (III) 
 to find the moments at all supports due 
 either to single concentrated loads or to 
 a uniform load over an entire span, and 
 these are the only two kinds of loading 
 which we need to consider in designing 
 
 O O 
 
 a continuous bridge. We next need the 
 shear S u at the right of any support due 
 to these same loads. 
 
 In computing strains in a continuous 
 truss we take up each span separately. 
 The index n refers always to the particu 
 lar span under consideration, while the 
 index r referring to the span in which 
 the load is for the moment considered, 
 may be less than, equal to or greater 
 than n. For single loads the shear S r is 
 given directly by (3), fora uniform load 
 P (1 k) in that expression becomes ^ wl^ 
 while for an unloaded span those terms 
 disappear. Thus we have 
 
 -ft), for Fig. 6. 
 
 (IV) S r =* for Fig. 3. 
 
 
 
79 
 M n -M n 
 
 , whenn>r, orn<r, 
 
 &n 
 
 for the shear in the span / n infinitely 
 near to the support n. 
 
 The shear in the span / n at a point in 
 finitely near to the n+l ih support is 
 called S n (see Fig. 3). For its values 
 we deduce 
 
 - 
 
 S f = I- + P k, for Fig. 6. 
 
 
 
 S r = ll + $ wfti for Fig. 3. 
 
 r 
 
 Q/ M n + l M n 
 
 S n = - -,for n>r, or n<r. 
 
 n 
 
 The reaction I? n at any support n is 
 evidently the sum of the shear S n in the 
 span l n ; and of the shear S n _i in the 
 span / n i O1 " f r all cases 
 
 Rn = n + n ! 
 
 Example. A girder of four equal 
 spans has a single load P at the center of 
 the third span from the left end. Find 
 the reaction at the third support. 
 
 Ans. E = 4JP. 
 
80 
 
 SHEAR AND MOMENT AT ANY SECTION. 
 
 These are given directly by the sim 
 ple conditions of static equilibrium. 
 For a single load we have from (2), (see 
 Figs. 5 and 6), 
 
 S = S r P, for a section between 
 (Y) Pandr-fl 
 
 S=S n> for any other section. 
 
 as expressing the internal shear for any 
 section x (see Figs. 5 and 6). 
 Also, we have 
 
 M=M r - - S r x + P (xkl r \ between 
 
 (VI) Pandr-fl 
 
 M=M n -- S n ce, for any other section, 
 
 for the internal moment for any section 
 x in a span either unloaded or containing 
 the weight P. Similar expressions may 
 be also written as in Chapter I, if the 
 load be taken as uniformly distributed. 
 JZxample. A girder of three equal 
 spans has a load P at the center of the 
 first spans. What is the shear and mo 
 ment at the center of the middle span ? 
 
 Ans. S=*P, M=ftP/. 
 
81 
 
 MAXIMUM SHEARS AND MOMENTS. 
 
 The formulae (I) to (VI) inclusive are 
 sufficient in connection with an arithmeti 
 cal process of tabulation to determine 
 the maximum strains in all the pieces of 
 a properly designed continuous truss. 
 Having for instance to calculate tbe span 
 /n, we may take at every panel apex 
 throughout the bridge a single load P, 
 and compute the shear and moment at 
 any section due to every possible posi 
 tion of P. These arranged in a table, 
 afford a clear view of the distribution of 
 loading giving the maxima ; the great 
 est positive shear, for example, occurring 
 when the live load covers those portions 
 of the bridge which furnish plus values 
 of S, while at the same tim.3, it is absent 
 from those portions giving minus values 
 of S. Adding then all the plus values 
 thus found, the maximum is determined 
 by combination with that due to the al 
 ways existing dead load. 
 
 It is therefore not absolutely neces 
 sary that the engineer should be ac- 
 
82 
 
 quainted with the theory of the distri 
 bution of loading giving rise to the maxi- 
 murn strains in the various pieces of the 
 truss. As such a knowledge, however, 
 is of great assistance in checking the ac 
 curacy of the calculation, we shall here 
 state without demonstration the cases 
 
 under which such maxima and minima 
 i 
 
 arise. 
 
 First the shear ; from this we obtain 
 the strains in the webbing by the simple 
 multiplication by a constant, a positive 
 shear producing tension in a diagonal 
 which slopes upward toward the left 
 hand support. The maximum positive 
 shear in the span l u at the section whose 
 distance from the support n is cc, occurs 
 under a distribution of loading such as 
 is represented in Fig. 7, in which the 
 shaded portions denote the live or rolling 
 load. From this we see that the nearest 
 span on the left and each alternate one 
 
 FJG. 7. 
 
 
 u-/ n 
 
83 
 
 are covered with the live load ; that 
 from the section x to the support n-f-1 
 the live load extends ; and that the sec 
 ond span on the right and each alternate 
 one are also covered with the live load ; 
 all other portions being subjected only 
 to the dead or actual load of the truss. 
 The minimum positive, or, what is the 
 same thing, the maximum negative shear 
 obtains under exactly reverse conditions, 
 the loaded portions in Fig. 7 being un 
 loaded, while the empty ones receive the 
 live load. Let the reader draw a figure 
 for this case, and imagine the section x 
 to move from n to n-h 1. 
 
 Next the moment ; from this we ob 
 tain the chord strains by dividing by the 
 constant depth of the truss, a positive 
 moment producing tension in the upper 
 chord. Here the maximum positive mo 
 ment in the span / n> occurs near the sup 
 port n under a distribution of loading 
 like that represented in the first illustra 
 tion of Fig. 8, near the middle of the 
 span as in the second and near the sup 
 port tt-fl as in the last. Fig. 8 repre- 
 
84 
 
 sents one and the same beam with the 
 cases of loading causing maximum posi 
 tive moments at three different sections 
 in the span l Uf the first a section between 
 n and a point i, the second between i 
 and i and the third between i and w-f 1. 
 These points i and i are called fixed in 
 flection points, and they enjoy the pro 
 perty that all loads on the spans to the 
 
 FIG. 8. 
 
 -Ln 
 
 71H 71 
 
 71*, 
 
 atb. 
 
 Mb. 
 
 right of l Ut produce no moment at i, 
 while all loads on the spans to the left 
 of l n produce no moment at i . The 
 position of these points depend only up 
 on the lengths and number of the spans 
 
85 
 
 of the girder, or upon the numbers c and 
 d given by (II). If the distance from n 
 to i be denoted by i, and that from n to 
 i r by i 9 the following simple formulae 
 
 <?n ni 
 
 .,__ $s n-f-2 ^n 
 
 n-j-2 < n-f 1 
 
 will give the position of the fixed^inflec- 
 tion points in any span n . 
 
 In order to render these distributions 
 of load clear, let us imagine the sec 
 tion x to move from the support n to 
 Ti+1. When the section is at n the live 
 load covers the whole span n to render 
 the moment at a; a maximum, as x 
 passes toward i the load recedes rapidly 
 toward n + 1, until when x reaches i the 
 span l n becomes empty, and the loads 
 on the following spans shift as shown in 
 Fig. 8. As x passes from i to i r the 
 span l u remains empty as in the second 
 sketch and when it reaches i the loads 
 on the preceding spans shift. As soon 
 as x passes i the load begins to come on 
 
86 
 
 at n, which rapidly increases as x moves, 
 until it covers the whole span when x 
 coincides with n+l. 
 
 The arrangements of loading for caus 
 ing the maximum negative moment in 
 any section depend likewise upon the 
 position of that section with reference to 
 the fixed inflection points, and are in all 
 cases exactly the reverse of those for the 
 positive moment. 
 
 It will be seen, then, that the maxi 
 mum moments between the supports and 
 the fixed inflection points cannot be de 
 termined by cases of loading, for such 
 cases are different for every section. In 
 a giider of two equal spans for example, 
 one of these points in each span coin 
 cides with the abutments, the others are 
 at one-fifth the length of the span from 
 the pier. Between those points the 
 maximum strains are not to be found by 
 parabolic curves of moments drawn from 
 a few assumed arrangements of loading. 
 Here have some late writers fallen into 
 grave error. 
 
 The above completes, what seems to 
 
87 
 
 us a simple presentation of the theory of 
 the continuous girder of constant cross- 
 section. We have disconnected it en 
 tirely from the properties of the simple 
 girder, have avoided the use of artificial 
 angles and couples, parabolic moment 
 and shear curves, static and elastic reac 
 tions and other paraphernalia which are 
 too often introduced to complicate the 
 subject. The formulae (I) to (VI) which 
 may be written on a page of the engi 
 neer s note book include indeed the 
 whole theory, and are sufficient for the 
 determination of the maximum strains in 
 a continuous truss of any number or 
 lengths of spans. 
 
89 
 
 CHAPTER III. 
 
 We will now apply the above theory 
 to the practical calculation of the strains 
 in a continuous trus, and to show the 
 perfect generality of our method we will 
 take one of t ftve unequal spans. 
 
 Fig. 9 shows the relative lengths of 
 the several spans, each support and span 
 receiving an index according to the no 
 tation previously adopted. The first 
 span Z, has a length of 70 feet, the 
 
 FIG. 9. 
 It 72 3 Iff- 15 
 
 y & """nr" ^ & 
 
 / * 3 * 56 
 
 second l^ of 100 feet, the third l % of 
 80 feet, the fourth 1 4 of 120 feet and 
 the last J t of 90 feet. This girder is 
 
90 
 
 to be subject to a live load of 0.8 tons 
 per linear foot per truss ; the dead 
 load we estimate at 0.6 tons per 
 linear foot per truss. It is divided 
 into panels of ten feet each and its 
 height is also ten feet, the webbing being 
 a simple series of isosceles triangles as 
 shown in Fig. 10, which represents the 
 
 FIG. 10. 
 
 <u I, = SO- , 
 
 A BC D E 3 r is H tc I 
 
 span 1 3 enlarged. The live load is ap 
 plied at the panel points on the lower 
 chord. It is required to calculate the 
 maximum strains in all the pieces of the 
 span 1 9 due to the above live and dead 
 loads. 
 
 We take up first the live load of 0.8 
 tons per linear foot, or eight tons per 
 panel. Since every load in the span ^ 
 affects every section in 1 3 in a similar 
 manner, we may, instead of considering 
 
91 
 
 the separate panel loads on / take them 
 as uniformly distributed in the prelimi 
 nary determination of the shear and 
 moment at 3. (The load of eight tons at 
 the points 1, 2, 3, etc., give reactions 
 only at those points, and cannot affect 
 the span Z 3 ). On the span /, there are 
 seven panels and six apices, hence the 
 live load in that span is W, = 6XS = 48 
 tons. In the same way we have on the 
 spans J a , / 4 and Z 6 , to consider the live 
 loads applied at the panel apices as uni 
 formly distributed over the spans; but in 
 the span 1 3 we must consider each panel 
 load separately, since different arrange 
 ments of those loads give maxima for 
 different sections. Thus we have 
 
 On 1 19 the load W 1 = 48 tons, 
 On ^, the load W a =72 tons, 
 On * 3 , the loads P,, P u , P 3 , etc., 
 
 (see Fig. 10) each equal to 8 tons, 
 
 On Z 4 , the load W 4 =88 tons, 
 On Z 6 , the load W 5 =64 tons. 
 
 We now turn to formulae (I) of the 
 
92 
 
 preceding chapter, and determine the 
 quantities A and B due to each of these 
 loads. For that on /, we have, for ex 
 ample, W t =48, /\=:4900, hence A = B 
 =58800. For P t we have P=8, l\ 
 6400, k%, hence A= 10500, and B= 
 6300 ; in like manner, for P 2 , P 3 , etc., we 
 place &=t, ^^f, etc., and find for each 
 a value of A and B. Thus we have for 
 the several loads : 
 
 For W t , A=B= 58800 
 
 For W 2 , A=B= 180000 
 
 ForPj, A= 10500 B= 6300 
 
 ForP,, A=16800 B=12000 
 
 ForP 3 , A=19500 B=16500 
 
 ForP 4 , A=19200 B = 19200 
 
 ForP 6 , A 16500 B = 19500 
 
 ForP 6 , A-12000 B = 16800 
 
 ForP 7 , A= 6300 B= 10500 
 
 For W 4 , A=B = 316800 
 
 ForW 6 , A=B= 129600 
 
 We next turn to formulae (II), and sub 
 stitute the lengths ^ = 70, /,= 100, etc., 
 and thus obtain 
 
93 
 
 c,=0 d, 
 
 * "* 
 
 04. /7 Q K 
 
 3 ^ ? "3 *** 9 
 
 c= 44.567 < 54.8 
 
 o 5 
 
 as the values of the multipliers c and d 
 for the case under consideration. 
 
 We are now able to find from (III) 
 the moments M s and M 4 at the supports 
 3 and 4 for each of the above loads. 
 Since there are five spans, 55, d n = 
 54.8, c7 8 _i 16, etc. ; substituting these 
 in (III), we have 
 
 r 
 -,whenn<r-fl 
 
 *r c r -fr r 4_i 
 
 M n = -^ T -n- -, when n>r 
 
 1 t U O A 
 
 n being any index (in our case either 3 
 or 4), and r the index of that span, 
 which, for the moment, we regard as 
 loaded. Taking the load on the first 
 span, we have r=l, and since n>r, we 
 use only the second of the above formu 
 lae, which becomes 
 
94 
 
 A c, 4- B <r 
 M n - -rf T _ n _-i__ - 3.452 <7 T _ n 
 
 Substituting in this TI 3 and w 4, and 
 we have 
 
 M 3 = 3.452 (16) = 55.23 tons ft. 
 M 4 = 3.452 (3.5) = 12.08 tons ft. 
 
 Taking the load in the second span, we 
 have r=2, and 
 
 -,.- . 
 
 M n = -^ 7 _ n -- 3 - 
 
 in which, by making n=3 and n=4, we 
 get the values of M 3 and M 4 . 
 
 For the single loads on the span 1 31 we 
 must use the first of formulae (III) to 
 obtain M 3 , and the second to obtain M 4 . 
 Making then r 3, we have, 
 
 3.4 
 
 3 17032 17032 
 
 _ A ^ + B C 4 3.5 
 
 M ^2 // 
 
 4 
 
 (16A-3.5B) 
 
 17032 17032 
 
 (-3.4A + 14.05 B) 
 
 and inserting in these the values of A 
 
and B, as given above, we find the mo 
 ments at 3 and 4 from each of the loads 
 from P x to P 7 . For the load on the 
 fourth span, we make r=4; for that on 
 the fifth span, r=5; and the first of for 
 mulae (III) give the moments. Thus, by 
 very simple arithmetical work, we obtain 
 the moments M 3 and M 4 due to each of 
 the single loads in Z 3 , and each of the 
 uniform loads in the exterior spans, and 
 arrange them in the second and third 
 columns of the following table : 
 
 Load 
 
 M 3 
 
 M 4 
 
 83 
 
 
 tons ft. 
 
 tons ft. 
 
 tons. 
 
 w, 
 
 - 55.23 
 
 + 12 08 
 
 0.841 
 
 W 2 
 
 +40"). 82 
 
 - t^.77 
 
 
 - 6.182 
 
 p, 
 
 + 81.19 
 
 
 h 10.83 
 
 
 - 7.254 
 
 P 2 
 
 4- 45.36 
 
 
 - 22.85 
 
 
 - 6.281 
 
 
 
 - 50.85 
 
 
 - 83.93 
 
 
 - 6.211 
 
 pi 
 
 
 r 48.00 
 
 
 - 41.92 
 
 
 - 4 078 
 
 p.. 
 
 
 -3D. 15 
 
 
 - 44.65 
 
 
 - 2.9U 
 
 PC 
 
 
 ~ 2(5.64 
 
 
 - 39.92 
 
 + 1.834 
 
 P 7 
 
 
 - 12.81 
 
 
 - 25.85 
 
 + 0.837 
 
 W 4 
 
 - 158.10 
 
 
 -653.34 
 
 -10.143 
 
 W 8 
 
 + 25.87 
 
 -106.91 
 
 + 1.660 
 
 The last column of the table, which 
 gives the shear S 3 in the span 1 3 at a 
 
06 
 
 point infinitely near to the support 3, is 
 found from the quantities M 3 and M 4 by 
 means of formulae (IV). The load W s , 
 for example, gives a positive moment of 
 405.82 tons ft. at 3, and a negative one 
 of 88.77 tons ft. at 4. From the last 
 formula of (IV), we have then 
 
 
 80 
 
 Also for the load P 3 on the span / 3 , we 
 have P=8, &=i, and from the first for- 
 
 / o / 
 
 mulae of (IV) 
 
 p _ 50.85-33.93 (1 ax fl 91 , tnns , 
 80 
 
 and in the same way the other shears in 
 the last column are computed. All of 
 these refer, of course, only to the live 
 load of eight tons per panel. 
 
 We are now ready to proceed with the 
 computation of the maximum strains in 
 the span 7 S . And first we take up the 
 webbing. 
 
 The maximum strain in any diagonal 
 in Fig. 10, is equal to the maximum 
 
shear for that section multiplied by the 
 secant of the angle between the diagonal 
 and a vertical. We proceed first to find 
 the maximum shears. 
 
 The shear at any section due to the 
 dead load is constant, increases or de 
 creases as the live load comes upon the 
 bridge, and becomes a maximum or 
 minimum under certain particular distri 
 butions of loading. To determine these 
 
 o 
 
 it is only necessary to tabulate the shear 
 due to each separate load. This is easily 
 done from the values of S 3 and the 
 formulae (V). In the following table the 
 vertical column headed aB/> includes the 
 shears which may act upon the diagonals 
 a B and B b, b C c those for b C and C c, 
 etc. The horizontal column numbered 1 
 gives then the shears at every section 
 due to the live loads ; the load W, for 
 example producing a negative shear of 
 0.84 tons in every panel or S=S 3 , the 
 load P 3 giving in the three panels on its 
 left S S. = +5.21 tons and in the five 
 
 5 
 
 on its right S = S 3 P= + 5.21 8 2.79 
 tons. A mere inspection of this table 
 

 
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 i -H-H--H-++7 
 
 + 
 
 CO 
 7C T-I 
 
 + 1 
 
 
 *V--v=.v sV 
 
 Ss 
 
 CO 
 
 5 
 
 CQ 
 
 T; 
 oJ 
 
 O 
 
 ei 
 o 
 
 Q 
 
 + 1 
 
 
 t-t 
 
 1 
 
 CO 
 
 
 
 
 
99 
 
 shows the distributiorr of live load caus 
 ing the maximum or minimum shear in 
 
 c_? 
 
 any section. Thus for the panel c?Ee 
 the maximum occurs when those loads 
 giving positive shears are present, 
 namely, W,, P 4 , P., P., P,. and W. and 
 when all the others are absent, and the 
 minimum occurs when only those giving 
 negative shears are on the bridge. If 
 then we add all the positive quantities 
 in 1 and likewise all the negative ones 
 and place the results in the horizontal 
 column 2, we have for the panel dl&e, 
 4-17.52 tons and -16.24 tons as the 
 greatest and least shears which can occur 
 in that panel due to the live load, and 
 these need only to be combined with the 
 shear due to the dead load to obtain the 
 absolute maximum and minimum. 
 
 If the dead load be regarded like the 
 live load as concentrated at the panel 
 points on the lower chord, its effect will 
 be a fractional part of that of the live 
 considered as uniformly distributed. 
 Adding algebraically the quantities in 2 
 we have in 3 the shears produced by a 
 
100 
 
 uniformly distributed live load of eight 
 tons per panel, since this is the same as 
 taking the algebraic sum of all the quan 
 tities in 1. The live load if extending 
 over the whole bridge will then produce 
 in dTSe a shear of 4-1.28 tons, and since 
 the actual dead load is three-fourths of 
 the live, the dead load must produce in 
 ttuit panel a shear equal to fxl.28 
 = 0.96 tons. Taking then three-fourths 
 of the quantities in the horizontal col 
 umn 3 we have in 4 the shears due to 
 the dead load of six tons per panel. 
 
 The shears in 4 always must exist, 
 while those in 2 may exist under certain 
 positions of the live load. The absolute 
 maxima arc therefore found by adding 
 algebraically the quantities in those two 
 horizontal rows. Thus for<#Ee, -f-0.96 
 always obtains, and if +17.52 also oc 
 curs their sum 4-18.48 is the positive 
 maximum shear ; and if -16.24 occurs, 
 4-0.96 16.24 = 15.28 is the minimum 
 or negative maximum. Placing these 
 results in column 5 we have the required 
 maximum shears in every section of the 
 
101 
 
 span under consideration. If the dead 
 load shear in 4 is greater than the live 
 load shear of opposite sign in 2 only one 
 kind of shear can prevail ; thus in bCc 
 the greatest possible value is +12.96 
 + 29.01 =+41.97 tons, the least possible 
 is +12.96 --11.73 = + 1.23 tons and the 
 diagonals b C and Cc will be subject to 
 only one kind of strain. In the case be 
 fore us three panels cDr7, c/Ee and 
 e~Ff have both a positive and negative 
 maximum and the diagonals in those 
 panels may be subject to either tension 
 or compression. 
 
 The maximum shears in 5 multiplied 
 by sec. 9, or the secant of the inclination 
 of diagonal to vertical give the maximum 
 strains, tension of the diagonal slopes up 
 ward toward the left hand support, com 
 pression if it slopes downward. The 
 depth of the truss being ten feet and the 
 half panel length 5 feet, sec. 6 is 1.118. 
 We have then the following table of 
 maximum strains in the 
 
102 
 
 DIAGONALS. (See Fisr. 10.) 
 
 Ba 
 B6 
 Od 
 Cc 
 DC 
 T>d 
 Ed 
 
 HA 
 KA 
 JLk 
 
 61.7 tons 
 
 +61.7 
 
 46.9 
 
 +46.9 
 
 2 or +7.2 tons 
 2 or 7.2 
 
 20.6 or +17.1 
 
 +20 6 or 17 1 
 
 +28.2 or 9.4 
 
 28.2 or +9.4 
 
 +40.5 
 
 40.5 
 
 +.",4.0 
 
 54.0 
 
 +688 
 
 68.3 
 
 in which + denotes tension and com 
 pression. 
 
 In the same manner we may tabulate 
 the moments at every section due to 
 
 P 
 
 each load and deduce the maximum 
 chord strains. For the upper chord 
 panels A B, B C, C D, etc., the centers 
 of moments are at the opposite vertices 
 a, b, c, etc., and hence in formulae (VI) 
 we must take 0, 10, 20, etc., as the suc 
 cessive values of x. To find the moment 
 for CD due to the load W, on the 
 
103 
 
 span ,, we have only to insert # 20, 
 and the values of M 3 and S 3 as found 
 above for that load (see Fig. 10), giving 
 
 M = 55.23 - (0.841 X 20) 
 
 = -- 38.41 tons ft. 
 
 the negative sign denoting that W, 
 causes compression in the upper chord. 
 Also to find the moment in the same panel 
 due to the load P , we have as before 
 z=:20 and 
 
 M = 31.197.254X20 + 8X10 
 
 = 33.89 tons ft. 
 
 In this way we readily compute the mo 
 ments for every panel due to every live 
 load and arrange them as in the follow 
 ing table of 
 
MOMENTS FOR UPPER CHORD. (See Fig. 10.) 
 
 W 
 
 r 
 
 H GO 
 
 oc 
 
 v c 
 
 ^ C5i > G 
 
 Ci C 
 
 OC 
 
 ft 
 
 H^ > 
 
 o 
 
 GO 
 
 
 O5 
 t* 
 
 TH 
 
 O 
 
 T 1 
 
 O7GOC 
 
 r- GO r 
 
 + 1- 
 
 ^O7CO -rCSiCC 
 - O7 CO rf ^ CO O7 
 
 L.I <. w < 
 
 C C, 
 
 T- 
 
 9 
 
 9 
 - 
 
 1C H 
 
 coc 
 
 GO r 
 
 t 
 
 r. 
 
 
 l - 
 
 
 
 
 M 
 
 w 
 
 O Oi -: 
 
 fr- r-l t 
 
 GO O CO 
 
 or *> 
 
 
 
 GO 
 CO 
 
 GO 
 
 CO 
 GO 
 07 
 
 T1 
 
 
 GO 
 
 CO CO CO 1C Ci 
 
 2 07 CO 
 
 C 
 
 - lO C 
 <? rr 1C C 
 lO 
 
 
 
 co c 
 
 t^_ ** 
 
 
 1.^^ * 
 
 
 
 - 
 % 
 
 
 
 HH 
 HH 
 
 O 
 
 3 
 
 - 
 
 D C5 T- 
 
 K^T 
 
 CO 
 
 4 CO GO O * Tf -^ JC Z> 
 
 ^ O 
 
 < 
 
 ^ 
 
 GO 
 
 - T- 
 
 r 
 
 - r- 
 
 H CO CO C 
 7 CO 1C 
 
 - O Cf 
 O 7C O t- 
 
 d 
 
 
 
 I*** 1 ^ 
 
 GO C 
 
 I! 
 
 5 
 
 2 
 
 1C 
 f ( 
 
 1 
 
 r- CO 
 
 O T-I 
 C0_07 
 
 d 
 
 
 
 V 
 
 P 
 
 7 J> tf 
 
 
 
 -CO^r- 
 
 H O r 
 
 H r- 
 
 -i 
 
 CO 1C 
 
 CO 
 
 T* 
 
 O7 TH 
 
 : co - 
 
 - c: T- 
 
 - -: 
 
 
 
 ; - 
 
 i *C X 1- 
 
 r C-- O C, 
 
 T 
 
 r 3: Ci t- 
 
 507^^ 
 
 CO 
 
 > 
 
 ~. a 
 ^ ^ 
 
 
 \ 
 
 3 
 
 H 
 
 + 
 
 + 
 
 o co 
 
 *C CO 
 
 "Vji ^^ 
 
 + 1 
 
 H 
 
 V 
 
 f- 
 - 
 * 
 
 * 1C O O CO rH i t- 
 
 - f>- CO iO 
 
 rH O7 
 
 1 1 
 
 o 
 
 rH O7 
 
 -t GO c: ^ 
 
 7 1C I-* T; 
 
 | 
 
 : * 
 
 r * 
 
 - tC C5 
 
 - r- t- T 
 
 rn 
 
 D cr t c 
 
 pi O7 ^ "^ 
 
 + 
 
 > 
 
 4 
 
 O 2 
 
 J, 
 
 
 
 
 
 - 
 
 m 
 
 
 ^- 
 
 > 
 
 ? 
 
 If 
 
 
 
 : 
 
 CO "* 
 
 P 
 
 - 
 
 c 
 
 
 " ^"fr- ^ 
 ^ ^^ ^ 
 
 H rH Tf CO GO ^ 
 
 H CO C 
 
 i Ci 
 
 CO 
 
 CO 
 
 ~f 
 
 ^ 
 
 5 O 
 :> 07 O 
 07 
 
 - - 
 > - - 
 
 : ic 
 
 : _ 
 
 
 
 ^ -f c: o 
 
 . t> Tt 
 
 - 
 
 D O7 co r 
 
 i T^ ^^ O 
 
 rH 
 
 j 
 
 ! 
 
 ?> 
 
 .CO r 
 
 ^ 
 < 
 
 Q 
 
 * 
 
 \( 
 
 h 
 
 J 
 
 CO.TJ1 
 
 P 
 
 
 
 
 
 01 
 
 7 
 
 H 07 3 
 
 _ ? 
 
 
 -^coicc: 
 
 ?GfeO 
 
 D Ci* 
 
 > 
 
 O CO 
 
 t- 
 
 07 
 
 07 rH 
 
 5 o > r 
 
 2 GO C 
 07 
 1 
 
 r 7 
 
 : ? 
 
 ; u 
 
 : cc o c 
 
 5 CO O7 - 
 
 2 CO ^ ?> 
 
 07 a 
 
 CO ^ 
 
 H 
 
 3 
 
 ! 
 
 + 
 
 CO 
 
 CO O7 
 
 + 
 
 o* 
 n 
 
 y 
 
 r 
 
 }O^ 
 
 I 1C CO O7 GO 00 ^ t 
 
 - CO 
 
 t> 
 
 + 
 
 + 
 
 
 
 r -^ ^ 
 CO 
 
 H t< 
 
 - -- 
 
 I 
 
 ( t> C5 GO ** C 
 
 5 C5 
 
 66 c: 
 
 CC r- 
 
 : 
 
 - 
 
 PQ 
 
 O7 GO O7 ^ 3" 
 
 5 O 07 ^ 
 
 5 Xr- 
 
 < d 
 
 
 CO CO 
 
 07 
 
 i- 
 
 CO 
 
 CO 
 
 T-t 
 
 
 
 o 
 
 TH 
 
 O O rH JO O GO Si O O7 y 
 
 C O CO ^n" 1C ^r CO O7 i 
 
 zjL-H-t-H-H-- 
 
 J l!t 
 
 ft 
 
 
 1C C 
 y^ 
 
 CO O 
 
 
 
 
 > 
 
 
 S V---v> V 
 
 -~ A 
 
 > ^ 
 
 
 
 
 
 l~> 
 
 CC 
 
 a 
 
 rt 
 O 
 
 + 1 
 
 
 T-I 
 
 05 
 
 CO 
 
 !"* 
 
 
 
105 
 
 The horizontal column 1 of this table 
 shows at a glance the distribution of live 
 load giving the maximum strain in any 
 bay : in the bay B C for instance the 
 loads W 2 W 5 and P 4 to P 7 inclusive 
 produce positive moments, and hence 
 the greatest tensile strain obtains when 
 those loads are on the bridge and all 
 others are absent, while the least tensile 
 strain in B C occurs when W 1? W 4 , P,, P, 
 and P 3 are present and the others ab 
 sent. Adding separately therefore these 
 positive and negative moments, we have 
 in 2, the greatest and least moments for 
 every bay due to the live load. Adding 
 those algebraically and we have in 3 the 
 moments when the live load covers the 
 entire bridge ; taking three-fourths of 
 the quantities in 3 we have in 4 the mo 
 ments due to the actual dead load. 
 Lastly combining the moments in 4 which 
 always must exist with those in 2 which 
 may exist, we get in 5 the absolute 
 maxima and minima. For example, in 
 CD we have due to the live load the 
 moments +327.0 and -281.3, the sum 
 
106 
 
 of these or +45.7 is the moment when 
 the live load extends over the whole 
 girder, three-fourths of this or -f 34.2 is 
 the value due to the actual dead load, 
 and finally 
 
 + 34.2 -f 327.0= -f- 361.2 tons ft. 
 + 34.2 281.3:^247.1 tons ft. 
 
 are the maximum positive and negative 
 moments, C D may then be subject to 
 two kinds of strain. 
 
 Dividing these results by the depth of 
 the truss and remembering that a posi 
 tive moment causes tension in the upper 
 chord, we have the maximum strains for 
 the 
 
 UPPER CHORD. (See Fig. 10.) 
 
 AB 
 
 
 -104.0 tons 
 
 
 BG 
 
 
 - 547 
 
 
 CD 
 
 
 - 36 1 or 
 
 24.7 tons 
 
 DE 
 
 
 - 33.2 or 
 
 44 9 
 
 EP 
 
 
 - 30.1 or 
 
 41 I 
 
 FG 
 
 
 - 4~>.0 or 
 
 43.3 
 
 GH 
 
 
 - (>0.2 or 
 
 21.4 
 
 Hit 
 
 
 - 860 
 
 
 KL 
 
 
 -140.3 
 
 
107 
 
 From this we see that the whole upper 
 chord may under certain positions of the 
 rolling load be subject to tension. This 
 is due to the short length of the span 
 compared with the adjacent ones. 
 
 The calculation of the maximum 
 strains in the lower chord is entirely 
 similar, the centers of moments being at 
 the opposite vertices B, C, D, etc., and 
 the corresponding values of x being 5, 
 15, 25, etc. We leave therefore as an 
 exercise for the student the formation of 
 the tabulation, merely giving the results 
 to which it will lead, viz. 
 
 LOWER CHORD. (See Fig. 10.) 
 
 ab 
 
 76.9 tons 
 
 be 
 
 41.4 gry "+" ^"l 
 
 cd 
 
 34.7 or - 
 
 -34. 6 tons 
 
 de 
 
 84.7 or - 
 
 -478 
 
 ef 
 
 40.7 or - 
 
 -47.1 
 
 fg 
 
 51.9 or - 
 
 -34.0 
 
 gh 
 
 70.7 
 
 lik 
 
 127.9 
 
 The strain sheet for the span l z (Fig.9) 
 is now finished, and in a similar way 
 
each of the other spans may be comput 
 ed. The method we have presented is 
 entirely general and applicable to any 
 number of continuous spans, whether 
 equal or unequal. In each case we take 
 the load in the exterior spans as uniform 
 and that in the span under consideration 
 as applied at the panel apices, and find 
 for each the quantities A and B from (I) 
 and from the lengths of the spans we 
 find by (II) the multipliers c and d. 
 These enable us to deduce from (III) 
 the moments at the supports due to each 
 load, from which (IV) give us the shear. 
 It is only in this preliminary computation 
 of moments and shears that the calcula 
 tion of continuous girders differs from 
 that of ordinary simple trusses. In the 
 latter the moments at the ends are known 
 to be zero, and the shears coincide with 
 the reactions which are found from the 
 law of the lever. The simple truss is 
 thus but a particular case of the continu 
 ous one as may be readily seen by placing 
 5=1 in our formulae (I) to (IV). In an 
 end span of a continuous truss the mo- 
 
109 
 
 ment at the abutment is also zero and 
 the shear is the same as the reaction.* 
 
 GIRDERS WITH SPANS ALL EQUAL. 
 
 This is one of the most common cases. 
 Making in (II) all the Ps equal, we have 
 
 c^ d^~ 
 c,=d t = 1 
 
 C 3 " i = 4 
 C 4 =<= 15 
 c s =rf.= -56 
 
 c 6 =d 9 = 209, etc., 
 
 which are the well-known Clapeyronian 
 numbers first deduced by the discoverer 
 of the theorem of three moments. f Each 
 of these numbers is equal to four times 
 the preceding one less the one next pre 
 ceding, and their signs are alternately 
 positive and negative. They are here 
 seen to be a particular case of our gener 
 al formulae (II). 
 
 * For an example of the computation of a continuous 
 truss of two spans, see Van Nostrand s Engineering 
 Magazine for July, 1875. 
 
 t See Comptes JKendus, 1857, p. 1076. 
 
110 
 
 GIRDERS WITH SYMMETRICAL SPANS. 
 
 If the two end spans of the bridge are 
 each equal to ft I and the others each 
 equal to I, the multipliers c and d become 
 also equal. Their values are 
 
 c b d^ 26 -- 30 /3 
 c G d^ 97 + 112 /?, etc., 
 
 each being equal to four times the pre 
 ceding one less the one preceding that. 
 Having established the value of ft (usual 
 ly taken at about 0.8) these reduce at 
 once to numerical form. If ft be unity 
 the spans become all equal and the mul 
 tipliers reduce to the Clapeyronian num 
 bers. 
 
 Example. A girder of four spans has 
 a single load P in the second^ span at a 
 distance k I from the second support ; 
 the two end spans being equal to 0.8/ 
 and the central ones to I. Find the mo 
 ment at the second support. 
 
 Ans. M = P/0.52* 0.901 7 
 
Ill 
 
 CONSTANT AND VARIABLE CKOSS SECTION. 
 
 Having computed the maximum strains 
 in a continuous truss, we choose for the 
 various pieces cross sections of an area 
 and form sufficient to resist those strains, 
 thus making the girder one of uniform 
 strength. The theory by which we have 
 computed the strains supposes however 
 that the cross section of the girder is 
 constant. The question now arises what 
 error is introduced by this hypothesis. 
 
 As we have been unable to present in 
 the short limits of this paper the theory 
 of the continuous girder with variable 
 cross section, we cannot place before the 
 reader a mathematical comparison of the 
 two cases, and are hence obliged to re 
 ly on the computations of others and on 
 general considerations. 
 
 Computations of strains in continuous 
 girders have been made by Bresse, Mohr, 
 Winkler, Weyrauch and others, consid 
 ering the cross section both constant and 
 variable. The general conclusion to be 
 derived from their investigations is, that 
 
112 
 
 the maximum moments over the supports 
 are greater, when the variable cross sec 
 tion is taken into account, but rarely 
 more than six per cent, that the maximum 
 moments near the centers of the spans 
 are generally slightly less, and that the 
 shearing forces do not sensibly differ.* 
 For example, in a truss of two equal 
 spans, the maximum moment at the pier 
 is 0.125 wl* for constant cross section 
 and 0.133 wl* for variable ; the maximum 
 negative moment is 0.070 wl* for constant 
 and 0.067 w I* for variable, and the reac 
 tions of the pier are 1.250 iv I and 1.266 
 w I respectively. 
 
 If then we compute continuous trusses 
 as if they were of constant cross section, 
 we are liable to slight errors in the chord 
 strains. These strains are however com 
 puted on the assumption of a distribution 
 of live load which can never occur in 
 practice, and in proportioning the sec- 
 
 * Mohr. in Zeitschrift des Arch. u. Ing. Ver. zu Hanno 
 ver, 1860, 1862, Winkler ; Die Lehre von der Elasticitdt, 
 p. 150. Weyrauch ; Theorie der continuirlichen Trdger, 
 p. 22., p. 143. 
 
113 
 
 tions to those strains a factor of safety 
 involving five or six-fold security is in 
 troduced. Considering then that it 
 must be almost impossible for the live 
 load to be arranged on the bridge as 
 Fig. 8 represents, we may be well as 
 sured that our computations on the hy 
 pothesis of constant moment of inertia 
 give greater strains than can ever obtain. 
 After having computed on both hypo 
 theses a girder with four spans, two of 
 sixty-five meters in length and two of 
 fifty-two meters, Weyrauch says : " We 
 are now able to answer the question, 
 whether it is allowable to calculate con 
 tinuous girders with variable cross sec 
 tion by the formulae for constant cross 
 section, in the affirmative. The maxi 
 mum moments arising from the two cal 
 culations differ but slightly. In our ex 
 ample, where the cross sections vary be- 
 tw T een 1 and 24, the greatest difference 
 was 6 per cent, the next following only 
 2.7 per cent. The shears change scarcely 
 at all. Only for bridges with extremely 
 long spans, is it desirable to make a sec- 
 
114 
 
 ond computation on account of variable 
 cross section." 
 
 LEVELS OF SUPPORTS. 
 
 In Chapter I we alluded to the fact 
 that small changes in the relative levels 
 of the piers produce great variations in 
 the strains of the pieces of the truss. 
 Continuous girders should not for this 
 reason be used when the piers are liable 
 to settle. Whether the supporting points 
 are exactly upon level when the bridge 
 is built is a matter of no importance, al 
 though, of course, no great differences 
 can be allowed. 
 
 If in finding the equation of the elas 
 tic line we had considered the supports 
 on different levels, a term containing 
 those differences of level and the term 
 El would have .entered the theorem of 
 three moments. Now if a straight beam 
 be laid across two points, a downward 
 force is necessary in order that it should 
 touch a third point at a lower level. If 
 this downward force be furnished by the 
 
115 
 
 weight of the beam, a certain part of 
 this weight will be effective in producing 
 the deflection or satisfying the term E I, 
 while the remaining part will act exactly 
 as if the three supports were on the same 
 level. But if the beam, instead of being 
 originally straight, were of such a shape 
 that it exactly fitted the three points it 
 is in the same condition as the horizontal 
 one after it has undergone the deflection. 
 Hence its action is independent of the 
 variations in level, for no exterior force 
 is required to compel it to correspond with 
 the points of support. 
 
 We therefore conclude, that if a bridge 
 be built, by suitably adjusting its false 
 works, corresponding to the profile of the 
 piers, all the strains obtain exactly as if 
 those piers were on one and the same 
 level. 
 
 Except the well-known rule that the 
 lengths of the spans should be so adjust 
 ed that the cost of the piers and super 
 structure may be equal, there is little to 
 
116 
 
 be said upon this subject. Although 
 most writers give a mathematical discus 
 sion of the most economical relations of 
 spans, the fact that no two of them 
 agree except in the simplest case, only 
 indicates that the theory contains no 
 principle which will lead to general con 
 clusions. 
 
 Viewing the matter from a practical 
 point of view, but not neglecting the in 
 vestigations of mathematicians, we may 
 give the following rules. For two spans 
 the lengths should be equal. For others 
 the span should be symmetrical, the in 
 terior ones being equal and. the end ones 
 shorter by about one-fifth or one- sixth, 
 or making the central ones each equal to 
 /, the end ones should be about f I or I L 
 Such an arrangement equalizes the mo 
 ments due to the dead load and being 
 pleasing to the eye, it is advisable to re 
 gard it in designing continuous bridges. 
 
 PRACTICAL CONSTRUCTION. 
 
 In the construction of continuous 
 bridges, the following points should be 
 carefully observed : 
 
117 
 
 1. The iron should be of a uniform 
 qnality, and have undergone as near as 
 possible the same process of manufac 
 ture. 
 
 2. The truss should be built with par 
 allel chords, each capable of resisting 
 either tension or compression ; the web 
 bing should be simple. With double 
 and triple systems of webbing the strains 
 cannot be accurately determined. 
 
 3. Joints in the chords should never 
 be made over the piers. 
 
 4. The false works should be so adjust 
 ed that the bridge may be built with its 
 points of support on the same relative 
 level with the actual bed plates. 
 
 5. The bearing surface of the bridge 
 upon the piers should be as .small as pos 
 sible consistent with considerations of 
 strength and safety, and arrangements 
 for longitudinal variation, due to changes 
 in temperature, should be provided. 
 
 6. Continuous girders cannot be used 
 if the piers are liable to variations in 
 level after erection. 
 
118 
 
 ADVANTAGES OF THE CONTINUOUS SYS 
 TEM. 
 
 In favor of the continuous girder versus 
 the simple one, we may mention : 
 
 1. Greater stiffness, since the deflec 
 tion under a rolling load is much less 
 than that of independent simple spans. 
 
 2. Ease of erection in cases where 
 false works are difficult and expensive ; 
 the girder may then be built on shore 
 and pushed out over the piers. 
 
 3. Saving in material for the piers, 
 since a less bearing surface is required 
 than for two ends of single span bridges. 
 
 4. Saving in iron, amounting to from 
 twenty to forty per cent, over the ordinary 
 construction of single spans. 
 
 5. Simplicity of construction, when an 
 angle of skew exists in the piers: in such 
 cases the cross girders may be placed at 
 right angles in the continuous structure, 
 and the difficulties of oblique connec 
 tions entirely avoided. 
 
 In a simple girder, whose length is /, 
 and live load per unit of length w, the 
 
119 
 
 maximum deflection due to the live load 
 
 is ^r-f- In two continuous spans each 
 
 384 El 
 
 equal to /, the maximum deflection, which 
 occurs when one span is covered with 
 
 3.7 iv r 
 
 the live load, is - -^TTJ or only three- 
 384 xL 1 
 
 fourths as much. With many continu 
 ous spans, the deflection is much less, its 
 greatest value being in the end spans. 
 In the case of a girder with horizontally 
 fastened ends, the deflection is only one- 
 fifth of that of ordinary simple spans. 
 
 The saving in iron is large, and alone 
 sufficient to recommend the continuous 
 system, particularly for long spans. This 
 saving occurs wholly in the chords where 
 material can best be spared. In the web 
 bing the quantity of material is slightly 
 increased. The exact percentage of sav 
 ing depends upon the number and lengths 
 of spans, the proportion of live to dead 
 load, the arrangement of webbing, and will 
 be the same in no two cases. In the ex 
 ample of Fig. 9, the center span which 
 we computed does not afford scarcely 
 
120 
 
 any saving in material owing to the in 
 fluence of the larger adjacent spans. 
 For girders of two hundred feet in 
 length with spans nearly equal, calcula 
 tion indicates a saving of about thirty 
 per cent. For the extreme case a sin 
 gle span with horizontal fixed ends the 
 saving is fifty per cent. 
 
 DISADVANTAGES OF THE CONTINUOUS SYS 
 TEM. 
 
 In our last chapter we referred to an 
 article by Charles Bender, C. E., which 
 contains many ingenious arguments 
 against the use of continuous bridges. 
 There are, in fact, fifteen " conclusions 
 to which he is led and which may be 
 seen by the reader on p. 109 of the cur 
 rent volume of The Journal of the 
 American Society of Civil Engineers. 
 Of these we will give a short abstract 
 and append a running commentary. 
 
 1. The theoretical calculation of curves 
 of moments, without consideration of 
 proportions and details, is exceedingly 
 
121 
 
 fallacious. Other things being equal, 
 the calculation of strains does furnish an 
 estimate of amounts of material. The 
 method of curves moreover is neither so 
 accurate nor so quick as our process of 
 tabulation. 
 
 2. This fallacy will be greater, if the 
 theory stands upon false premises. Un 
 doubtedly. 
 
 3. The theory of continuity is fallaci 
 ous and unreliable, because it supposes 
 the coefficient of elasticity constant, 
 whereas it has been shown that it may 
 vary for wrought iron from 17,000,000 to 
 40,000,000 Ibs. per square inch. In our 
 last chapter we have shown that the 
 proper interpretation of the variability 
 of E is, that different qualities of iron 
 have different degrees of elasticity. The 
 values of E, from which it is concluded 
 that the laws of flexure are fallacious, 
 were in fact found by the theory itself 
 from the measured deflections of beams, 
 and it is hence more than fallacious to 
 regard them as condemning that theory. 
 
 4. With several diagonal systems the 
 
122 
 
 strains in a continuous girder cannot be 
 calculated, but only guessed. A good 
 objection and it applies also to simple 
 girders and draw spans. 
 
 5. The theory needs a correction if the 
 chords are made variable in cross section. 
 The amount of this correction we have 
 indicated above. 
 
 6. The calculation of continuous 
 trusses is excedingly tedious, and if gen 
 erally introduced would greatly impede 
 the business of bridge building in this 
 country. A conclusion to which those 
 who know how to calculate decidedly ob 
 ject. For a construction costing half a 
 million dollars, it matters little whether 
 one day or one week be spent in compu 
 tation, and if the one week saves ten per 
 cent or more on the cost, it is certainly 
 well employed.* 
 
 * In the discussion of this subject at the late Engineers 1 
 Convention, one of the speakers mentioned an instance 
 where the strain sheet for a continuous revolving draw 
 bridge was furnished for $40. As that strain sheet was 
 made by the author of this paper it is not improper that 
 he should remark that its price would have been consid 
 erably less bad the computations been made by tabula 
 tion of apex loads, instead of by the consideration of 
 cases of loading. 
 
123 
 
 7. Pieces which resist two kinds of 
 strain must be proportioned to resist the 
 maximum tension plus the maximum 
 compression. Further experiments are 
 much needed in this connection ; if 
 Woehler s conclusions are confirmed, 
 pieces must be so proportioned and hence 
 the percentage of saving lowered. 
 
 8. Continuous girders require very 
 accurate workmanship. Are we to infer 
 that ordinary trusses do not ? 
 
 9. The foundations and masonry of 
 the piers must be of excellent quality. 
 Does not the same hold for the simple 
 girder? See No. 11. 
 
 10. If the girders are built on shore 
 and pushed out over the piers, additional 
 computations must be made and extra 
 pieces introduced. The additional com 
 putations are of the simplest character 
 and in many cases no extra pieces would 
 be needed. 
 
 11. If improperly placed on their bed 
 plates, greater strains arise than are con 
 templated, an inch difference in level 
 producing great variations in strains. 
 
124 
 
 We have shown above that unless piers 
 are liable to settle after the erection of 
 the bridge, such differences in level pro 
 duce no effect. 
 
 12. If one chord be protected from 
 the heat of the sun the strains are much 
 disturbed. The camber of the bridge 
 is altered, as also occurs in single spans. 
 It should be remembered however that 
 although certain strains cause a curva 
 ture, a certain curvature does not neces 
 sarily cause corresponding strains. 
 
 13. Continuous bridges have proved 
 to be more economical in Europe than 
 simple spans, because the latter have 
 been improperly proportioned. Trial is 
 necessary to prove that they would or 
 would not be more economical in this 
 country. 
 
 14. By designing two bridges of 200 
 feet each, a two span continuous truss 
 twenty-five feet high, and a simple truss 
 27 feet high, with different details, Mr. 
 Bender finds that the latter is more 
 economical Perhaps for other propor 
 tions this might be reversed. In a com- 
 
125 
 
 parison of amount of material, ought not 
 other things to be taken equal ? 
 
 15. Continuous bridges deflect as much 
 as single spans. Not if they have the 
 same height and span, and are subject to 
 the same loads ; a fact which every 
 schoolboy knows. 
 
 In regard to objections 2, 6, 7, 8, 9, 10, 
 12 and 13 one remark further is neces 
 sary. They are objections which may 
 be made to every new proposed con 
 struction in engineering art. In the 
 days of wooden bridges, they were ad 
 vanced against the use of iron ; they 
 have been made against the suspension 
 system and against the braced arch. 
 But their value can be estimated in only 
 one way, by trial. On the other hand 
 theory can estimate one at least of the 
 advantages claimed for the continuous 
 systems, and that estimate is a saving of 
 twenty to forty per cent, in material; 
 how much of this must be deducted for 
 extra care in workmanship, labor of erec 
 tion, effects of temperature, or varia 
 tions in the elasticity of the material can 
 
126 
 
 only be determined by the actual erec 
 tion of continuous bridges, by experi 
 ments extending through a long series 
 of years. 
 
 mr 
 
 Having thus stated briefly, but fairly, 
 the arguments for and against the use of 
 continuous bridges, we leave it to prac 
 tical builders to decide whether or not 
 the system is worth a trial. Other na 
 tions have long been using it ; profiting 
 by their experience and by our own im 
 proved methods of manufacture and 
 modes of erection, it may, perhaps, turn 
 out that we shall find it better and more 
 economical than the present system of 
 single independent spans. 
 
 HfSTORY AND LITERATURE. 
 
 The literature on the theory of con 
 tinuous girders is very extensive in the 
 German and French languages, and very 
 limited in English. We can only give 
 here a few hints concerning its develop 
 ment and history. 
 
 About the year 1825, Navier founded 
 the present theory of flexure by intro- 
 
127 
 
 ducing the hypothesis that the exten 
 sions and compressions of the fibers on 
 each side of the neutral axis were pro 
 portional to their distances from that 
 axis. From this he deduced the equa 
 tion of the elastic line, and applied it to 
 the discussion of continuous girders. 
 His method consisted in determining 
 first the reactions of the supports, and 
 from these the internal forces or strains, 
 which, although the most logical, was 
 exceedingly tedious in practice. In the 
 following works the reader may find de 
 tailed information concerning his meth 
 od : 
 
 Kayser : Handbuch der Statik, Carls- 
 ruhe, 1836, chap. X. 
 
 Molinos et Pronnier : Traite de la 
 construction des ponts metaliques, I^aris, 
 1857. 
 
 From the time of Navier to the pub 
 lication in 1857 by Clapeyron of the 
 method of using the moments over the 
 piers as auxiliaries in the computation 
 instead of the reactions, many continu 
 ous girders were built in France, Ger- 
 
128 
 
 many and England. Among these may 
 be mentioned the Britannia tubular 
 bridge, of four spans, two of 231 feet, 
 and two of 460 feet ; the Boyne Via 
 duct with three spans of 141, 267 and 
 141 feet ; and the bridge over the 
 Weichsel at Dirschau with six continu 
 ous spans, each 397 feet in length. By 
 Clapeyron s happy discovery of the the 
 orem of three moments, a great impetus 
 seemed to be given toward the erection 
 of such bridges, for in the twenty years 
 following 1857, we find them extensively 
 built in Germany and France, although 
 in England the unfortunate example of 
 the Britannia tube caused a tendency to 
 other forms of construction. A mere 
 list of such bridges would occupy pages. 
 They are generally of shorter span than 
 those mentioned above, rarely exceeding 
 300 feet, while the number of spans 
 varies from two to seven. 
 
 To give here a list of the books which 
 has appeared since Clapeyron s time will 
 also be impossible. We can only indi 
 cate two or three which are at the same 
 time valuable and easily accessible : 
 
129 
 
 Bresse : La mecanique appliqu&e, 
 Paris, 1865. (Vol. Ill contains a tolera 
 bly complete mathematical discussion, 
 with tables for facilitating the calcula 
 tion of moments.) 
 
 Winkler: Theorie der J5rM6c&6?i, Vienna, 
 1872. (Contains the graphical method 
 of Culmann and Mohr, with also analyt 
 ical investigations.) 
 
 Weyrauch : Theorie der continuir- 
 lichen Traeger^ Leipzig, 1873. (A com 
 plete analytical discussion of the whole 
 subject. The best work which has yet 
 appeared.) 
 
 In our own country have been issued 
 during the past year the works of 
 Greene, Herschel and DuEois, each of 
 which contains valuable contributions to 
 the literature of the subject, and which 
 should be in the library of every student 
 of engineering. On the other hand, 
 works by English authors, which treat of 
 the subject at all, do it in such an imper 
 fect and unsatisfactory manner, that we 
 are forced to consider them as twenty 
 years behind the age. 
 
130 
 
 As the above mentioned works treat 
 only of the theory and calculation of 
 continuous girders, we ought perhaps to 
 say that a book giving an account of the 
 most important continuous and simple, 
 together with suspension and arch 
 bridges, is the admirable descriptive 
 work by Heinzerling, Die BruecJcen in 
 Eisen: Leipzig, 1870, which is illustrat 
 ed by over a thousand engravings, and 
 presents a complete history of iron 
 bridge construction. 
 
 ERRATA. 
 
 Page 15, line 13 ; for ^ read I. 
 11 21, "17 and 18 ; for wl 9 read wl*. 
 
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