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THE THEORY 
 
 OF 
 
 THE CONTINUOUS GIRDER 
 
 Its Applicatioq to Girders with and without Variable 
 Cross-sections, 
 
 BY 
 
 A. HOWE, C. 
 
 Professor of Civil Engineering, Rose Polytechnic Institute. 
 
 UNIVERSITY 
 
 NE\V YORK: 
 
 Eqgiqeering News Publishing Co, 
 
 1889. 
 
Entered according to an Act of Congress, 1888, by MALVERD A. HOWE, in the office of the Librarian of Congres 
 
 at Washington. 
 
 MOOKK & IjANGfEN, 
 
 Printers, 
 Terre Haute, Ind, 
 
PREFACE. 
 
 Continuous bridges with the exception of draw bridges- 
 are not considered economical and are not designed by. 
 American engineers. This probably accounts for the brief 
 treatment the theory of the continuous girder receives in 
 text books and engineering literature. 
 
 With one or two exceptions, all American treatises con- 
 sider the moment of inertia as constant and deduce two 
 equations for the moment over any support, one to be ap- 
 plied when the loads are on the left of the support and the 
 other when the loads are on the right. These two equa- 
 tions combined would, of course, give the moment over any 
 support for any load, but only when the moment of inertia 
 could be assumed as constant, as in girders with parallel 
 flanges, where it might be undue refinement to consider the 
 cross-section as variable. 
 
 American engineers have, until recently, assumed the 
 moment of inertia as constant even when the flanges of the 
 girder were considerably inclined, as in the Sabula draw 
 bridge, relying upon the factor of ignorance to cover all dis- 
 crepancies which might arise from the assumption. A few 
 years ago the modulous of elasticity was quite variable in 
 large bridges though all engineers considered it as constant 
 in computations but now, since experience shows that iron 
 and steel can be manufactured with, practically, a constant 
 modulus of elasticity, it may safely be considered as con- 
 stant, without leading to any appreciable error. 
 
 Although no material may be saved by considering the 
 
IV PRP:FACE. 
 
 moment of inertia as variable yet, if it is so considered, 
 the material will be placed where it will do the best ser- 
 vice. 
 
 Girders with inclined chords, computed and designed as 
 if their cross-sections were constant, have more material 
 than is necessary in some of the compression pieces, and 
 not enough in some of the tension members, as is clearly 
 shown by the results on page 77 of the text. 
 
 The object of the following pages is to present to com- 
 puters, engineers and students a complete mathematical 
 treatment of the theory of the continuous girder, and show how 
 it can be applied to any girder especially to fixed girders 
 and girders of two and three spans under any conditions. 
 
 With the single assumption that the modulous of elas- 
 ticity is constant a general equation has been deduced for 
 the moment over any support of any girder under any 
 conditions of loading, of any length of spans, of variable 
 or constant cross-section and for supports at any level. By 
 difference of level of the supports is meant any change of 
 level which may take place when the girder is in position. 
 It is evident that such a change alone would affect the mo- 
 ments. From this equation special equations are readily 
 deduced for any particular case. To illustrate the sim- 
 plicity of the transformations necessary for any special case 
 and also for the convenience of engineers, equations have 
 been given for all the special cases usually discussed in text 
 books, and also equations for these cases when the moment 
 of inertia is considered as variable. 
 
 The usual general equations for reactions, deflections and 
 intermediate moments and their transformed equations for 
 special cases are also presented. 
 
 Several examples are solved to illustrate the application 
 of the formulas and to show that the processes are almost 
 mechanical when the formulas are thoroughly understood. 
 From those examples which are solved considering the mo- 
 ment of inertia as constant and then variable, a good idea 
 of the manner in which the moment of inertia affects the 
 
PKEFACE. V 
 
 results can be obtained. In the case of the Sabula draw 
 there is a difference of about twenty per cent. 
 
 In all pin connected bridges the loads are considered to 
 be concentrated at the apices or panel points. In com- 
 puting the moments for such girders Table I. will be found 
 to be very convenient, as in it are found expressions for 
 
 rfk-tW+k" and A- k' 1 for all values of k= / from 0.001 to 
 0.999 inclusive. 
 
 The works enumerated under References, page 107, have 
 been consulted, and some of their parts used without any 
 material change, for which credit is given in foot notes. 
 
 The author is indebted to R. H. Brown, C. E., and Geo. 
 H. Hutchinson, C. E., for valuable assistance and sugges- 
 tions. 
 
 M. A. H. 
 
 TERRE HAUTE, IND., September, 1888. 
 
GENERAL CONTENTS. 
 
 PACiE. 
 
 Nomenclature 1-2 
 
 I. 
 
 GENERAL RELATIONS. 
 
 Conditions of equilibrium In any girder 4 
 
 General relations betweenthe moments and reactions and the loads . . 4-6 
 Equation for the moment over any support when the moment of inertia 
 
 is variable 7 
 
 Equation for the moment over any support when the momen 1 of inertia 
 
 is constant 12 
 
 General equations for Shear 15-16 
 
 General equations for Deflection . ... 17 
 
 II. 
 
 SUPPORTED GIRDERS. 
 
 I. A simple girder resting upon two supports 19-20 
 
 II. A beam continuous over three supports .... 21-26 
 
 III. A beam continuous over four supports . . . 27-28 
 
 IV. "The Tipper" 29-83 
 
 III. 
 
 BEAMS WITH FIXED ENDS. 
 
 I. A beam fixed at one end, and supported at the othf 1 ' 34-38 
 
 II. A beam fixed at both ends i(9-41 
 
 III. A beam fixed at one end, and unsupported at the othei 42-43 
 
 IV. A beam 011 two supports, and one end unsupported 44 
 
 V. A beam on one*support, having one end fixed, and the other unsup- 
 
 ported 45 
 
 VI. A beam on two supports, having neither end supported 6 
 
 IV. 
 THE POINT OF ZERO MOMENT. 
 
 General equations with graphical determination of the point of zero 
 
 moment . . 47-51 
 
vni GENERAL CONTENTS. 
 
 V. 
 
 APPLICATIONS. 
 
 PAGE. 
 Examples illustrating the application of the formulas o2-XM 
 
 APPENDIX. 
 
 Determination of the equation of the elastic line 8<>-87 
 
 Demonstration of Equation (A), p. 7 89-100 
 
 Demonstration of the equation of the elastic line 80-88 
 
 General expression for the Theorem of Three Moments &7 
 
 Demonstration of Equation (A) ' 8W-10O 
 
 TABLE I. 
 
 Values for k 2 k*+k 3 and k k* for all ratios ~ = k from .mil to .9W*, in- 
 clusive 101-UHi 
 
 REFERENCES. 
 
 Some references to monographs, periodicals, Ac., which consider the 
 
 theory of the continuous girder 1(>7-108 
 
 INDEX. 
 
 Index of Equations . XKMlo 
 
 si MMARV. 
 
 Summary of contents 111-11(1 
 
NOMENCLATURE. 
 
 number of the support just at the left of the /" span. 
 r=The horizontal length of the span r. 
 -P r =Any concentrated load in the i** span. 
 tc r =Any uniform load, per lineal foot, in the j** span. 
 a f =The distance from the left support r to any concentrated 
 
 load P r . a=J: r l r , 
 t r '=The distance from the support to the point where the 
 
 uniform load ends in the /" span. 
 O/' The distance from the left support r to the point where 
 
 the uniform load begins in the /** span. 
 
 T fl r 
 
 Kj= -T-, or, a r =k r l r , 
 
 x r =The distance from the left support r to any point in the 
 
 J w * span. 
 
 JIr==rhe bending moment over the support r. 
 bending moment over any support m. 
 
 bending moment at any section, x r from the sup- 
 port r. 
 
 JJT=The bending moment at the center of any span of a 
 girder. 
 
 shear just at the right of the support /'. 
 
 shear just at the left of the support r. 
 
 reaction at the support r, and equals S r ^-S r '. 
 
 distance the support )' is below some specified hori- 
 zontal line of reference. 
 n of summation. 
 
2 THE CONTINUOUS GIRDER. 
 
 ,.=Tangent of the angle that the elastic line makes with the 
 horizontal over the support r. 
 
 y r =The deflection of 1he girder at any section or,. ; y/,.is meas- 
 ured from the horizonal line of reference. 
 
 S The number of spans. 
 
 e, The distance from the left support r to the point where 
 the moment of inertia of the section of the girder 
 changes for the fi^t time in the span i*. 
 
 e 2 =The distance to the point where it changes the second 
 time. 
 
 e v =The distance to any point where the moment of inertia 
 changes, always measured from the left support / and 
 in the span r. 
 
 Ii=The moment of inertia between the last value of e e and 
 the end of the /*"* span. See Fig. 
 
 _t>=The moment of inertia between the support r and the 
 pointy. 
 
 j, The moment of inertia between e t and e 2 in the r ( " span. 
 
 ,7,. The moment of inertia at any section in the r m span. 
 12=The modulus of elasticity. 
 
GENERAL RELATIONS. 
 
 * In the '/** span of a continuous girder, whose length is 
 r , Fig. 1, take a point o vertically above the r'" support, 
 as the origin of co-ordinates, and the horizontal through o 
 as the axis of abscissas. Suppose any section of the girder 
 at a distance w r from the left support r, and between this 
 section and the support r a weight _P, distant from r, 
 Or=k r /,.. 
 
 Now, if the girder is continuous over any number of sup- 
 ports there will be a bending moment over each of them (if 
 the ends are not fixed the bending moment over the first 
 and last support will equal zero): JC-/ over r 1, lML r over r, 
 Hfr+i over r-\-lj etc., also, there will be a shear S r just at the 
 right and $,. just at the left of /*, $.+,, just at the right and 
 Sr+i just at the left of r-\-l, etc. If there is equilibrum the 
 following conditions must be fulfilled : 
 
 * See "Continuirlichen Und Einfachen Trager," page 4, by Prof. Weyraucb. 
 "Theory and Calculation of Continuous Bridges," page 52, by Prof. Mer- 
 riman. 
 
 "Strains in Framed Structures," page 244, by Prof. DuBois. 
 
4 THE CONTINUOUS GIRDER. 
 
 /. The algebraic sum of all the horizontal forces must be zero. 
 II. The algebraic sum of all the vertical forces must be zero. 
 III. The algebraic sum of the moments of all the forces must 
 be zero. 
 
 From the third condition we have for any section in the 
 r'" span, 
 
 M,. M r -4-S r x P,. (x r a,'.) .... xa r . . ... . . .(1) 
 
 If in (1) we make w,.~l n .M r ~M r+l , and it becomes 
 M r+i =-M,A -S f I,. P, (l r a,.), or, since a, &,. /,., we have 
 
 M r+i = M,A -S r lrPr I (1 ^ ...... . . (2) 
 
 From (2), 
 
 And if there is no load in the span r this becomes 
 
 r= M. L -M r 
 t 
 
 ^,P,S n therefore from (3), 
 ^ -K-ML.:p f t , ,,. better, 
 
 And if there is no load in the span '/* /, (o) l)ecomes 
 
 The reaction at any support equals the sum of the shears 
 at that support, or 
 
GENERAL RELATIONS. 5 
 
 The above formulas were deduced under the supposition 
 that there was but a single concentrated load JP, in the span 
 r. If there be more than one concentration, the formulas 
 become : 
 
 (1) M r ^=M f -\ $,. x,.l' P r (x,.--a,.) . . . xa, ...... (8) 
 
 (2) M^^lC+flt lr- P, /, (/-*) ......... W 
 
 (3)- N,- -^ Mf ^P, (/- A- ( .) .......... (10) 
 
 partial uniform load itt span r. If to, 
 
 represents the uniform load per lineal foot in the span /, 
 we can write 
 
 ,., fc,. /,. 
 
 I ft= ,, da,, or, since a,., fc 
 
 ,. /, d k,.=lw f I,. k f ) (12) 
 
 ar=k f l f fc '-V 
 
 The last expression in (12) indicates the difference of the 
 values of the parenthesis when A', equals - and - re- 
 spectively. 
 
 Substituting (12) in the above equations, we have, 
 From (8), 
 
 M r =M r +S r x r --Cw r /, d k r (xk f I,} . . . xa, (13) 
 
 <fv 
 
6 THE CONTINUOUS GIRDER. 
 
 From (9), 
 
 a,. 
 M r+l ^M r -S,. l-w, l* f fd k,. (7-ik,) (14) 
 
 a, 
 
 Or 
 
 M r+i ^M r -\ -8, lw r l* r (k 4- tf) :. (15) 
 
 From (10), 
 
 From (11), 
 
 M _ ]f j ^ 
 
 Uniform load over entire apan r.lf the entire 
 span is uniformly loaded, we have a,.-~l n d' f --o and a, f \ I,.. 
 
 From (13), (In this case a,---x r , a,. =o and a f $ x, since 
 the load considered cannot extend beyond jr,.). 
 
 M r =M,-{ -Sf r ai r -^- w, r; (18) 
 
 From (15), 
 
GENERAL RELATIONS. 7 
 
 From (16), 
 
 4^Vf*< ...... ....... ..(20) 
 
 From (17), 
 
 These twenty- one general equations apply to all spans 
 and conditions of loading, but before they can be solved, it 
 will be necessary to find expressions for Jf^or Sin terms of 
 the loads, lengths of the spans, etc. 
 
 This can be done by introducing the equation of the 
 elastic line and the theorem of three moments. The following 
 expression is obtained for the bending moment over any 
 support m', E alone, being constant. 
 
 A 
 
 In which the respective expressions have the following 
 values : 
 
 * A complete demonstration of this equation is given in the Appendix 
 and should be thoroughly understood before any attempt is made to apply 
 equation A practically. 
 
8 THE CONTINUOUS GIRDER. 
 
 A, 
 
 Con cent rated loads 
 
 A,= 2 P, /; V Av- -> A 1 ; #) *U See (57) ....</) 
 
 Partial uniform load 
 
 From (12), 
 
 - P r --w r I,, (k,.) therefore (i) becomes 
 
 a',=k f /, 
 
 ,-> ............ (74) 
 
 Uniform load orer all 
 
 Uniform load over all 
 
 A,= - w,. Ill <>,._, 
 
 4 
 
 B, 
 
 Concentra ted loads 
 
 B,.= -S P, I- (Av-A:;) o r . ....... See (58) . . , (./) 
 
 Partial uniform toads 
 
GENERAL RELATIONS. 9 
 
 Concentrated loads- 
 Partial uniform load 
 
 a r =k r ! r 
 " r = -F r *, /; o r _, | fe,- -L A-f | +H r o r _ t . , (78) 
 
 Uniform load over all 
 
 Concentrated loads 
 
 '^= ~2F;_, IP r _ s !,_, (lk,.^ O-ILL, O r ... See (59) . . . (//.) 
 Partial uniform load 
 
 ;L,= -2 F;_, o.w^i^, k r _ t - ~v_, [ -H r '_<o r . . (so) 
 
 Uniform load over all 
 
 ;L<= - F;L, o r w^ I^-H^ o r . (si) 
 
 H t 
 
 Concentrated loads 
 
 V=/ r 
 
10 THE CONTINUOUS GIRDER. 
 
 Partial -uniform load 
 
 Substitute the following expressions in (/) : 
 
 o r 
 I P r (r v a,y = Cw f d a,, (eajf .... a' r <e, .... (82) 
 
 y 
 
 ,- 
 
 ? P r (e v a r ) s = Cw r d a,. (c a r ) s . . - .. d r <e v ..... (83) 
 a'f 
 
 Probably the easiest way to handle this case is to consider 
 the uniform load as extending from the left support to the 
 farther end of the load and integrate between the limits 
 a r =e v and d' r =o; then consider the load as extending from 
 the left support to the nearer end of the load and integrate 
 between the limits a',=d^e v and o, and take the difference 
 of the results. 
 
 Uniform load over all 
 
 da r (e v -a r y=~w r . . . .(84) 
 
 t. y 
 
 IP r (e v a r y= fw, d a r (e a r ^=4rw r el (85) 
 
 -/ W 
 
 , 
 
 Concentrated loads 
 
 - r ** r , v .., 
 
 A ] r- -r- - P, (e c a r )- ... (r/) 
 
 '' V. <'> Vf ) 
 
GENERAL RELATIONS. 11 
 
 Partial uniform load 
 
 OT 
 
 2 P f (e r a r y=Cw r d a r (e,,-o^ . . . d,<e v (82) 
 
 a,. 
 
 a r 
 1' P r (c v a r y= Cw, d a r faOrl* a'e v .... (83) 
 
 Uniform load over all 
 e, 
 
 Z P r (r r a v y=Civ r d a r (e r ci r Y^--^r-w r ft . .(84) 
 
 J 4- 
 
 o 
 
 c, 
 
 1' P r (<>a^= Cw r d a f (e v ~^Y=-i w r c . (85) 
 
 ./ ?> 
 
 For all loads 
 
 v=l 
 
 r "=2: A ^~ .......... See (50) ' 
 
12 THE CONTINUOUS GIRDER. 
 
 / S,.=:(7 r +F;) #,. + , See (60) ..... (?) 
 
 fr r =(l r _i -f /Vl/) 0r-k(Jr+#) #r-, - - . SeC (61) . . ... . () 
 
 ^,._ ; See (62) . . . (0) 
 
 /,_; /,. ) 
 
 O r =6 E f, See (51) (e) 
 
 c,=o. c a =l. 
 
 c m - 2c m _ j -^ L r m _ s '/"~~ . . . See (66) (p) 
 
 d.= -2 rf._^=L. ._ rfM |= 
 
 / J x-i+2 P-,>i+2 
 
 . .See (68) ..... (r) 
 
 By means of the above equations we can deduce the bend- 
 ing moment over any support of any continuous girder under 
 the single assumption that the modulus of elasticity, E, shall 
 be constant. 
 
 E AND I CONSTANT. 
 
 If the moment of inertia as well as the modulus of elas- 
 ticity is constant our equations will be much more simple. 
 
 By inspection, we see that (a) equals zero, and hence jP r , 
 Fr and F,,'.' equal zero and also H r and H t ; hence X,' and 
 X^L t equal zero. By reduction we obtain 
 
 . (A,) 
 
GENERAL RELATIONS. 13 
 
 r,=o. r g =--l. 
 
 J , ? . /,-*+/,-/ In,-, 
 
 L ltt ; 'ill I 
 
 (lo. d a =l. 
 
 l S - m+ 3 + l g - m+3 l 
 
 -m+3 fc .< m+S 
 
 Concentrated loads 
 
 A = - I P r II (2 k r 3 # + fcj) (v; ; ) 
 
 Partial uniform load 
 
 A r = - w r /; ] ^-4? h-j- r ( 86 ) 
 
 4 J ;: ==ib r /,. 
 
 Uniform load over all 
 
 A r = ~ w t . /; (87) 
 
 "^ 
 
 Concen t rated lott ds 
 Partial uniform load 
 
 {O L2 7./, -j (t r =K r l r 
 %r (88) 
 
 j o;=k r i r 
 
 Uniform load over 
 
 4rW* (89) 
 
14 THE CONTINUOUS GIRDER. 
 
 E AND I CONSTANT Span* equal. 
 
 If the spans arc equal, the equations of the last case 
 reduce to, 
 
 -+* * ' 1 H ) 
 
 r/,-0. <k=l 
 
 d m = - 4 d.,,-* (I*-, (r s ) 
 
 A r 
 
 Concentrated loads 
 
 A r = - I P r l* (2 k r 3 A^-j-Ar;) ;.<,) 
 
 Partial uniform load 
 
 7U -j r- /,'.. / 
 
 /* J z^ z. i '' I /qn\ 
 
 ~7~ ( yu ) 
 
 Uniform load orer all . 
 
 ^ r = -v> (91) 
 
 B, 
 
 Concentrated loads 
 
 B,.= - 1' P,. I 3 (kty (j g ) 
 
 Partial uniform load 
 t'etr' &.* o' r =k r t 
 
 *= -^'l^Tr). (92) 
 
GENERAL RELATIONS. 15 
 
 Uniform load over all 
 
 B r = - w r F . (93) 
 
 -y 
 
 h 
 
 V I' T? f ^ " " r " Jr ' "(+/ 
 
 In case the supports are at the same level, Y r =o in all 
 equations. 
 
 From (p.}) and (/*.,) we obtain the following values for c 
 and (I: 
 
 Ci = = d, c 7 = 780 = d, 
 
 c 3 = - 4 = d s c 9 = 10864 = d 
 
 c 4 = + 15 = d, CF= + . 40545 = d lo 
 
 c 5 = - 56 = d, Cll = - 151316 = d tl 
 
 c 6 = + 209 = cl G r, 3 = -f 564719 d 
 
 Explanation of Table I In the co-efficients 
 (2 k r 3 # + #) and (k r k?), k, is a fraction and equals 
 
 -y 1 , henc^, it is immaterial about the actual values of a 
 
 '; 
 
 and I as long as the ratio is known. 
 
 The ratio k r has been assumed to have all possible values 
 from .001 to .000 inclusive, and the respective values of 
 the above co efficients, carefully computed, are given in 
 Table I. A few trials will prove the great utility of this 
 table, and convince the computer that much time and labor 
 can be saved by its use. 
 
 SHEAR. 
 
 The moments over the supports being determined, the 
 shears can be found from the folio wing formulas, which apply 
 to all cases. 
 
16 THE CONTINUOUS GIRDER, 
 
 + <2U ....... See (11) . . . . (C) 
 
 ,._; 
 
 In which Q and Q' have the following values: 
 Concet it MI ted loads 
 
 Q r =? p r (i-k,) .................. 0)4) 
 
 Q f =SP r k ..................... 0)5) 
 
 Partial uniform load 
 
 ' / / 
 
 Q r=Wr I, ]^=^j- _ ' " . . . 
 
 ( vi , d,=k, I, 
 ,.=w, l r \-f . . (97) 
 
 V ' Al /' / 
 
 a" r =k r I,. 
 Uniform load over all 
 
 INTERMEDIATE BENDING MOMENT 8. 
 
 General equations for all possible cases. 
 
 M^Mr+SrTrLr .......... See (8 ) . . - (D) 
 
 In which L has the following values: 
 
 Concentrated loads 
 L r =IP r (xa r ) .......... ax, ...... (100) 
 
 Partial uniform loads 
 
 L r =w r a r (x r -^-} ..... a' r <z, ...... (101) 
 
 ' a r =a" r 
 
 ( n ) fl r a' r x r 1 
 
 L r =w f a r (y r -) \ ..... - r n- r ,-l ..... (102) 
 
 ' a' ==<=> 
 
GENERAL. RELATIONS. 17 
 
 DEFLECTION. 
 
 General formula. E alone constant. 
 
 ? =h+t + :-*$SM T* s. tf- v P(x aY\ 
 6EI t * 
 
 + - "^ (-**- - A) \ 3Mr e *fc ,- H^V ^ 
 
 tf E/I ,/_, / *' ' 
 
 -rp r ( ey a> y 3 (xe v ) 2P r (ea r y\ . See (39) . . (E) 
 
 t r+l =. Ad_p^!L _}_ .__!_ | Mr ^_|_ ^ j/ r+/ l^^p^l (k r ls* r ) | 
 
 1 r f 
 
 rf 7 -' 1-^ -T-|| Mr i$ -T 1 )-}- Mr+1^~ 
 
 6EIJ, v=l r ^j v _ t j ' ( + * 
 
 + -f ? P r l r (l-k r )^ P r (e v -a f y-3e v S P r (e-a,^ . .(45) 
 
 For uniform, loads 
 
 a r =a r a,.=k r l r 
 Z p r = C Wr d a r = Cw r l r d k, ........... (12) 
 
 a r =a'r a'r=k r l r 
 
 Remember that in the terms containing (x a) and (e a), 
 x and e must never be less than a, or rather a^x or e 
 
 E AND I CONSTANT. 
 
 If the moment of inertia is constant the preceding equa- 
 tions reduce to, 
 
 yr=h r +t r x,+ ~jjnT r T r -\S r xt~^P r (x r -~a 1 ) :i . . (E,) 
 
 tr+^ 'f * ^J { Mr 1+2 MT+, l f +S P r l> r (k r ~K) } - - (45 a ) 
 
18 THE CONTINUOUS GIRDER. 
 
 For uniform loads 
 
 a r =a r a r =k, I,. 
 
 Z P r = Cw r d Ur= Cw r l r d k r (12) 
 
 a r =a'r a'r=k r l r 
 
 If the supports are at the same level, the terms containing 
 h in the above equations become zero; the equations re- 
 main the same in every other particular. 
 
 We have now deduced general equations by which we can 
 determine the bending moments, shears and deflections for 
 any continuous girder. In the next chapters we will give 
 numerous examples or special cases illustrating the applica- 
 tion of the formulas. 
 
II. 
 
 SUPPORTED GIRDERS. 
 
 CASE I. 
 A simple girder resting upon ttvo supports 
 
 'Jr. sfe 
 
 ft & W 
 
 %. s. y 
 
 ( a ) E alone constant. 
 From (A] we have for M, and MI 2 
 t=M 3 =0 .................... (103) 
 
 From (B) and (C) 
 ,-=Q t ...... . .............. . (104) 
 
 =e; ...................... (105) 
 
 Or 
 
 ,= P, (1 k t ) . . for concentrated loads ...... (106) 
 
 i=2 PI k, . . for concentrated loads ......... (107) 
 
 S f =Wt I, -,&/ . . .for any uniform load . . . (108) 
 
 * ! a'^k, I, 
 
 ( k 2 , ) a i~k t l t 
 
 S 2 =w i I, \-jrr -for any uniform load .... (109) 
 ' d;=k,l t 
 
 $,= w t l t , . .for uniform load overall . . . . . . (110) 
 
20 THE CONTINUOUS GIRDER. 
 
 Sg= -r- w, I, . . . for uniform load over all . . . . . .(Ill) 
 
 From ( J>) the bending moment at any section x, is 
 Jt'<=$ &, L, .................. (112) 
 
 Substituting the values of &, and L, we have, 
 
 For concentrated loads 
 M X =Z P< (l-k t ) x-Z P, (x-a,) . . .. :/<.c. . . .(113) 
 
 For any uniform load 
 
 a', a,=a'i 
 
 M*= *< w, I, { k~ f-| - w, j, (z, - ^ ) j- a,<z, (114) 
 a/ a, a',' 
 
 For uniform load over all- 
 
 1 1 1 1 
 
 M x =-g-Wi I, x w t x 2 ,= w, x t (l< x,) .... (115) 
 
 For a uniform load over all, the moment at the center of 
 the girder becomes, 
 
 The well known formula for this case. We see, therefore, 
 that our formulas are perfectly exact for the simple girder, 
 and also that a variable moment of inertia or difference of 
 level of the supports does not effect the values of the bend- 
 ing moments or the shears. 
 
 From ( JB7), we have, 
 
 6 E 7 X 
 
 v=l 
 
 ? 
 
 -r P, (^-a,r-^ (fv-^) ^P, (c -a,) ~ (117) 
 
SUPPORTED BEAMS. 
 
 h h, 
 
 21 
 . (118) 
 
 (b) 12 and I constant. 
 The moment and shear equations are the same as in (a). 
 
 y,=h,+ t l r, + -s i x' l -SP l (z,-o,)* . . . . (119) 
 
 IF the origin is taken at one of the supports, the corre- 
 sponding value of faj will, of course, equal zero. 
 
 CASE II. 
 A beam continuous over three supports 
 
 ' 
 
 (a) IE alone, constant. 
 
 From (yl 
 M,=M 3 -=0 
 
 ^=o. c 2 =L c 3 =- 
 
 - P2 
 
 (120) 
 
 (121) 
 
 (122) 
 
 ,=o. d,=l. 4= - ........... (123) 
 
 PI 
 
 Substituting (122) and (123) in (//), it reduces to, 
 
 "Pi 
 
 (124) 
 
22 THE CONTINUOUS GIRDEU. 
 
 x,;=n, 0,F,; s p, i, (ikj o, ........... (125) 
 
 x;= -H;o,-2F;' zp t i, (/-/,-, K 
 
 2 
 
 ^(<^:^! /> (,,-^. ..(127) 
 
 v=l 
 ;= Z A I ^ Pt (e ~ a ' )3 - 3 - v p, (e-a,Y \ . . ( 128) 
 
 V { I, i, ) 
 
 v=l, 
 
 -T- .......... ........ (129) 
 
 A=v- -T- (180) 
 
 ........... (183) 
 
 v=! 2 
 
 =6 E /(/,+ F;')-\-6EI t 
 
 . (134) 
 
SUPPORTED BEAMS. 23 
 
 A% and BI 
 Concentrated loads 
 
 A,= - ? P 2 II (2 k,3 kl-\-kt) 6 E I t (135) 
 
 B t = - S P, If (kkf) 6EI t ' (136) 
 
 Partial uniform loads 
 
 . d,=k, I, 
 
 /?,= -wJ^-^^^EI, . . . . (138) 
 
 a'i =k, I, 
 
 Uniform load over all 
 A 2 = -J,w,li#EJt (13 ( ,)) 
 
 n,= - -4- Wl i? 6 E i t . .... (iio) 
 
 4 
 
 a,.=a' r a' r =k r l r 
 Z P r = Cw r d a,= Cw r l r d k, (12) 
 
 " " 7 7 
 
 Shears and intermediate bending moments 
 
 For shears and the intermediate bending moments, apply 
 the general equations (IB) (c) and (I>), and for deflection 
 use equations (E) and (45). 
 
 (6) JE and I constant. 
 
 In this case, we have from the equations under (a), 
 M,=M y =o . (141) 
 
24 THE CONTINUOUS GIRDER. 
 
 Y,+A,+B, 
 
 : 
 
 In which, 
 
 Concentrated loads 
 
 = - i' P, // (Jg fc-5 */+Jk/) (14*) 
 
 ,= - S P, If (k,k?) ... (145) 
 
 Uniform load over all 
 
 <*'a=k s L 
 (146) 
 
 1 *;=*, /, 
 
 Uniform load over all 
 
 A 2 = 4-^ V ........ (148) 
 
 B t = - W7 / ^ ................... ( 149 ) 
 
 4- 
 
 As this is a case much used in practice, we will give ex- 
 pressions for the bending moments in terms of the spans 
 and the loads. 
 
 Concentrated loads 
 
 v _ V P / 2 K _ v P 1 ~ 1C 
 
 _ J * - ' r * '2 *^2 - 1 r > l 'i A / 
 
 ^r\\ 
 
 *(/,+ 
 
 In which, 
 2 =2 k,3 k/+k, ....',...., ...... (151) 
 
 And 
 
 ;=k,k; s ................ ..... (152) 
 
SUPPORTED BEAMS. 25 
 
 From(J?) and ( C), 
 
 y ___ \' p l 2 L^ _ v p / " 
 
 [+ - 
 
 _ -,,, 
 
 ~in(- ' " (164) 
 
 Or, 
 
 Uniform loads over each span 
 
 Y g WV // W7; If 
 
 - F 2 + w 2 Z/-(- w, If 
 
 Or, 
 
 The above equations at once reduce to 
 
 (158) 
 
 w, II w f If 
 
 F 2 + -y- w, 4*+ -y W, f 
 
 ' + - w< '' ..... (160) 
 
26 TIIK CONTINUOUS (J 
 
 ,- - w,i*- -w. i 
 
 (c) E, I and It 
 
 In this case, the formulas are the same as in case (ft) with 
 the term Y g =0. 
 
 (d) E, J, h and I constant. 
 Making l,~l g , the equations of case (6) reduce to 
 Conceit t rut i*<1 font Is 
 
 v /' / l\ v /' / A" 
 
 - - /y h Ay- 1 t i, A, 
 
 Or, 
 
 V /> /.' N' /> /.' 
 
 - 
 
 .' 
 
 V 
 
 , 
 
 v /' A' v P V 
 
 3,= ^- '-+P,(1 k.)=R,. .(166) 
 
 (166) 
 
 /, 
 
 i >' /> A' v /> #" 
 
 , SP.(1 /.,) ' (167) 
 
 v /' / v /' ( I 1- \ l> i ir 
 
 _/,/,, _ / ( / /, . i /, . . ( I (>, 
 
 I it i form loffd owr it/ltr, // // 
 
 '.= -"< < 170 > 
 
 , / 1.5 
 
 -f w; ^ -^- 
 
 / i / / ^ 1 7'^ \ 
 
 l ~U~ ' fc ~T ~^~ W I - -Q- V t ^ I / ) ^1 
 
 Q A? O 
 
H 
 
 
 x f _ p 
 
 > W f r f - fV 
 
 n 
 
 t!74) 
 
 .-I 
 
 CASE III 
 
 or<>r four 
 
 , J 
 
 ( (1 ) / 
 
 > : v V \ B,rf.) 
 
 .1 ) \ \ i .; \ 
 
 ',.1 ) .V \ ../?.<,} . 
 
 T * I** ( ' 
 
 Which minors to tho following: 
 f ( A- } \ .V 
 
 ,: . - ( ; , , , , 
 
 - - 
 
 f - 
 
 flTS) 
 
 (176 
 
 " X 
 
 7>> I 
 
 : ' \ \ 
 
 I'so general e<|iuitions tor tho various torms in the 
 and also tor shear, intormoiliato moments anil 
 
28 THE CONTINUOUS GIRDER. 
 
 (b) E and I constant. 
 Our equations now reduce to 
 
 M (Y,+A,+B,) 2 (k+g-^, 
 4 (ft ' 
 
 ,. ,-, 
 
 4 (/,+ (4+4)-4* 
 
 (c) H, I and h constant. 
 
 2 (4+4) (^+B,)-^ (A S +B,) 
 - 
 
 TUT v \"/^ *g/ V-"an ^~V frg ^x^-p^y /ioo\ 
 
 *s / // \i\fi i 7~\ 71 (^"^/ 
 
 (d) JEJ, jT, /i and constant. 
 
 (184) 
 
 Uniform loads 
 
 If each span is covered with a uniform load, we have 
 
 = -* "'^'- *-+* tf . . . ....... (185) 
 
 ^-^tf-tf+u,,// ..... ..... (186) 
 
 If w,=Wg=w s =w, and / / =L=/ V = ) we obtain, at once, the 
 well known forms, 
 
 M 3 = *? ...... ........... (187) 
 
 -wP .................. (188) 
 
SUPPORTED BEAMS. 29 
 
 And for the shears we have 
 = -^-"M - - w l=;^. w i=R f . , .(189) 
 
 wl ......... (190) 
 
 Jo wl 
 
 (192) 
 
 wi + --wi = j-wi ...... (194) 
 
 -ivi-\--wi = -v>i*=R 4 . ...... (196) 
 
 CASE IV. 
 
 * We will now consider the case of the "Tipper," or a 
 beam continuous over four supports, the second and third 
 supports, resting on a rigid beam, supported, generally, by a 
 single support at its center. 
 
 It is evident that if supports 2 and 3 are supported by an 
 unyielding bar, which is supported, in turn, at its center, that 
 the reaction J?., must equal the reaction It, 3 , and also that 
 -\-fi s = Ti s . If the unyielding bar is not supported at its 
 center, H,, and _B. V will be inversely proportional to their 
 lever arms around the point of support, and -\-h a and h 3 
 will be directly proportional to the lever arms. 
 
 See " Strains in Framed Structures," page 2o4, by Prof. DuBois. 
 
30 THE CONTINUOUS GIRDER. 
 
 The unyielding bar supported at its center 
 
 a_^ i 
 
 M, M, ^Tf/z> I ^ 
 
 '*^ x>fts* v ? < 
 
 HP * ^ 
 
 ^r *** * 
 
 ~2 I ~ \**z 
 
 7> 
 
 (a) E alone, constant. 
 From our general equations, by reduction, we obtain, 
 
 ''/ ^8 
 
 p - ~ 3 -L * 3 i r>' -i_ D n QK^ 
 
 i i V^9 "| ^C3 V * vOJ 
 
 Since R 3 ~R 3) we have 
 
 ^ -?- -f - ^1 i- + Q' i J r Q 2 Q' 2 Q 3== . . (199) 
 
 Q Q' -V p^ /j ^ fc \ (200) 
 
 Let 
 
 Q't+Q a Q'sQs=Q - (201) 
 
 l,=n I, and l s =m I, . . (202) 
 
 Then (199) becomes 
 n M 3 m M s -\-2 m n (M :i M 2 )-}-m n Q 1 2 =0 ..... (203) 
 
 Or, 
 n M 3 m M 2 -\-% m n (M 3 M a )= - m n Q L (204) 
 
 By inspecting the equations under Case III, we see that 
 the moment equations may be placed under the following 
 forms : 
 
 M S =C; K+C; Y,- 
 
SUPPORTED BEAMS. 31 
 
 From (205) and (206), we have 
 
 ;c t ) .... (20?) 
 
 Substituting (205), (206) and (207) in (204), it becomes 
 f +n C' 2 Y 2 +n C' 3 Y a +n C' t } 
 
 I _ TO a F f -ro C, Y 3 -m C, [=- mn Ql 2 . (208) 
 
 +<mn'<<5-G) F 2 +mn(0;-C v ) F s 
 
 -f^mn(C;-COJ 
 
 Or, by placing F and F' in place of the co-efficients of 
 Yj etc., we have 
 VY 2 +V'Y 3 = -mnQlV" ........ . . .(209) 
 
 (210) 
 
 Hence, 
 
 But, h 2 = h ;f , or, h 3 = h; therefore, 
 
 v Y *= ~ "' V ~ ~ v "' "* h * ' ' (212) 
 
 o,k,.. (214) 
 
 Or, 
 V Y 3 +VYf=U]0 a [2-] o a h a ......... . . (215) 
 
 Therefore, 
 [l-\0 2 -[2]0 2 h 2 = -mn QlV" .......... (216) 
 
 And 
 
 +m nQ l,+ V" a 
 
 ffl 2 ~ ............ (217 
 
32 THE CONTINUOUS GIRDER. 
 
 From (217), we can find the value of //.,. //. and then 
 the values of Y 2 and Y 3 from (210) and (211), whence we 
 can determine Jfi and j&f,. from (177) and (178). 
 
 The easiest method of finding the values of the terms in 
 the equation giving the value of h ai is to substitute, for any 
 particular case, the values of all the known quantities in the 
 immediately preceding equations, or find the values of the 
 constants C 1} CX, etc., and in turn, substitute them in the 
 equations containing them. The operations are long and 
 exceedingly tedious, but simple, as an examination of the 
 equations shows. 
 
 (b) E and I constant. 
 
 If the moment of inertia is constant, the deduction of the 
 constants C,, C 2 , etc., becomes much more simple. After 
 Jt> is determined from (217), and Y 2 and Y 3 from (210) and 
 (211), the bending moments are readily obtained from (179) 
 and (180). 
 
 If h i =h^=o i as is usually the case, the process becomes 
 still more simple. 
 
 (c) E and I constant. h,=h 4 =o. 1,-l^L L=n I. h= -h :i . 
 We have, as in (), by a little reduction, 
 
 n (M 3 M,)-\-2 (M S M 3 )= -nlQ . . . . ..... (218) 
 
 Or, 
 
 (n+2)(M s -M,)= -nlQ. ........ . .(219) 
 
 Letting . (/,H-4) <^+4) l,*=D I ......... (220) 
 
 We have, from (179) and (180), 
 
 *, _ t tt,, n 
 
 
 
 s 
 
 Then, 
 D (M s ~M,)=(Y a +A s +B s ) 2 (n-^l)-(A^B t +Y s ) n 
 
 (Y 3 +A 3 +B,)n -(At+Bi+Y,) 2 (n+1) . . (223) 
 
SUPPORTED BP;AMS. ;>-> 
 
 Hence, 
 
 Y a (Sn-\ 2) Y a (Sn 2} (A, ; 1L 
 
 -A S -B) (3 n+2)=(M 3 -M,) D . . (224) 
 
 Substituting (224) in (219), it reduces to 
 (YsYa+Aa+B^A^BdtSn^^fy+fy^ -DnlQ . (225) 
 
 Therefore, 
 
 Y,-Y,= (3 ~^l n % -A : ,-B 3 +A s + Bl . . . .(226) 
 Since hjh 4 =o and h s = h :ij 
 
 n= 6 E I = + :f , : + 6 / (228) 
 
 ( I ('>) (. M L ) 
 
 Therefore, 
 1V= -r, .................... (229) 
 
 And 
 Y, Y 2 = -2 K. ...... ......... (230) 
 
 Hence, 
 
 r J^ /( l_ . ^.-F-ft-^-^i 
 
 - 
 
 But, D 1=4 (/,+ (^+y-// /; 44-4 4 I, 
 
 ^^ I l t + ll_ll = ^ /*._)_ n ^_|_ C 5 ^ ^-'^p (3 U 3 
 
 l\($n+2)fy+2) ..... (232j 
 
 And 
 Q=^^~({<~Q, .............. (201) 
 
 Substituting (201) and (232) in (231), we have 
 Y _ llM+Qi-Q-Q^A^Bi-At-B, 
 
 * 2 ,> *3 I -'Jd ) 
 
 (233) completely determines 1^ and 3^ y , and now the 
 bending moments can be deduced from (179) and (180). 
 
III. 
 
 BEAMS WITH FIXED ENDS. 
 
 In this chapter we shall consider beams with one or both 
 ends fixed. 
 
 CASE I. 
 
 A beam flxed at one entl <m<l stijtjtortcd at the 
 other 
 
 For convenience, the left end will be considered as fixed. 
 
 This case is, in reality, the same as Case II, in Chapter II, 
 
 / 
 considering l,=o, h,=h.> and 7i o, hence, we can use the 
 
 equations of that case by making the proper changes. 
 
 (a) E alone constant. 
 (124) becomes 
 
 M>-(A+K4A7) (234) 
 
BEAMS WITH FIXED ENDS. 35 
 
 (134) becomes 
 * K= 3 n FJI t 7, j^r^j ........... (235) 
 
 By inspecting equations (124) to (134), inclusive, it is seen 
 
 i 
 
 that we can cancel out 6* JEJ I t Irom all the terms in (234), 
 hence, we can write 
 
 *F,= -0j/,^r^ .............. (236) 
 
 fi=l s +F s .................... (237) 
 
 2 2 
 
 I,-, I 
 
 *i 3 
 
 L 
 
 < 129 > 
 
 v=l 
 
 =^ (133) 
 
 ..(127) 
 
 f ) 
 
 v=L 
 
 ;=H 3 F:P a l s (l-k s ) .............. (238) 
 
 Concentrated loads 
 
 ,= I P, I* (2 k3 kf+kf) ........... (140 
 
 * In reality, 7 { ] ^-, Ii \ -jr \ which is indeterminate, 
 
 (. h } (. U~' ) 
 
 but the above form is the only logical one which the ex- 
 pression for Y s can take. 
 
36 THE CONTINUOUS GIRDKH. 
 
 Partial nit i form loads 
 
 f T, A \ GLo^Ko I.) 
 
 A 3 = - w, // -*/--*/+ --f- ........... (146) 
 
 -4 >:-L/, 
 
 Uniform load over nil 
 
 A.= 4- m, I* . . (148) 
 
 4 
 
 Shears and intermediate betid itt-f/ moments 
 
 For shears and the intermediate bending moments, apply 
 the general equations (.B), ( C) and (_D), and for deflection, 
 use equations (JE7) and (4o). 
 
 (6) JS7 and JT constant. 
 
 For this case, (129), (131), (133), (127) and (238) be- 
 come zero, and (237) reduces to 
 
 &=J, ..................... (239) 
 
 we have from (234) or (142) 
 
 In which the values of Y 2 and A,, are given by (236), (144), 
 (146) or (148). 
 
 As this is a case quite likely to occur in practice, we will 
 give expressions for the bending moments and shears in 
 terms of the loads and span. 
 
 Concentrated loads 
 
 Substituting in ( 144) in (240), or from (150), we have 
 
 ^= r '~^ //g ' ............ .... (241) 
 
 In which K 2 =2 k,3 H+H. 
 From (B) and ( C), or (155), 
 
 S,= r f* ^ K * 4- I p^ (1 ]<)=& ...... (242) 
 
BEAMS WITH FIXED ENDS. 37 
 
 Also, (see 157). 
 fifr=? -^P^ + - A **=#, . (243) 
 
 Uniform load over all 
 
 The above equations at once reduce to 
 
 K- 4- v*K. 
 
 ;!/=-" 
 
 !> I, 
 
 y, ^w, u 
 
 s ?= ~^rp - + -J- * ^=ft 
 +;T* .- 4- tr. I* 
 
 s:,= __i_ i 
 
 (c) .fJ, I and h constant. 
 
 In this case the formulas are the same as in case (b) with 
 the term Y 3 =0. 
 
 The formulas are : 
 
 Concentrated loads 
 
 M - Z 1\ I, K., 
 
 M <= ~J^ ........... (247) 
 
 V*=^-j^ -f R (1-L)=R, ........... (248) 
 
 S= ~r~^ +?RL=R :i ...... (249) 
 
 Uniform load over all 
 
 (244) becomes 
 
 M*= -- ?'V// - ............... (250) 
 
 (251) 
 
38 THE CONTINUOUS GIRDER. 
 
 S= - 4" Ws k+ 4- w s 4= -|- w>.o 4=#, ...... (-J52) 
 
 From (18), we have 
 JM^ : -|-'b(^t>--tf-4aS) ...... - ..(253) 
 
 Mix. Af= - 4- '" f *=^ ....' (254) 
 
 x <J 
 
 From (12,) and (4 o a), we obtain, if 7*.=^, 0, 
 y, = - 48 2 EI w s 4 (3 tt5 I y^+2 xr!) . . x*=a* . . (255) 
 
 Special case 
 
 A single concentrated load at the centre of the beam. 
 By applying Table I, (250) becomes at once 
 M*= -P, 1,0.1875= -jfiP,l* (256) 
 
 = - jfr P, + P, == W P, = - (258) 
 
 l&=- TU + ^P^-P, k-^4 *&*r t 
 
 Or, 
 
 M,== ~ P, (11 x,3 L) . . . x, <-i- 4 (.260) 
 
 The maximum moment occurs when a;,= -j- L, or, 
 
 MaX M * =: ^ PJi (2G1) 
 
BEAMS WITH FIXED ENDS. 
 
 From (J5J,), if h a =h 3 =0, 
 
 -12 U x.,-^2 II) . . z,> - 
 
 And 
 
 CASE II. 
 A beam fixed at both ends 
 
 39 
 
 . . (262) 
 . (263) 
 
 This is the same as Case II, Chapter II, with h t =h 3 , h 4 =h s , 
 
 i 3 
 
 l t =o, 1 3=0, and 7i=o,-and I t =o. 
 
 (a) ^J alone constant. 
 From (177), we have 
 
 (264) 
 
 From (178), 
 
 (265) 
 
40 THE CONTINUOUS GIRDER. 
 
 (264) reduces to 
 
 And (178) becomes 
 
 . (267) 
 
 TY-y r , 
 
 4 ft ft ft ft (. 
 In which, after dividing by 0,^=0 o s , 
 
 (268) 
 
 ........ (269) 
 
 (270) 
 
 ...... (271) 
 
 ..... (272) 
 
 -. .............. (278) 
 
 The values of the remaining terms are easily found from 
 the general equations, remembering that H, or 0,, has been 
 cancelled out of each term. 
 
 ( b ) JE and I constant. 
 From (266) or (179), we have 
 
 
 From (267) or (180), we obtain 
 
 (1- + ^)^L-(A + K) /,_^(}>^)-(A + K) 
 
 5// "5T" ' ' (2lo) 
 
 In which the terms have the values given under (<f). 
 
 See note under Equation (235). 
 
BEAMS WITH FIXED ENDS. 41 
 
 (c) JE, I and h constant. 
 
 (274) becomes 
 
 M 'L^?JL (276) 
 
 (275) becomes 
 
 *?>~^ir- (277) 
 
 For concentrated loads we can write 
 
 MS = - 3 ' ~ (2" 8) 
 
 And 
 
 2 V P I K'4-^P I K 
 
 if ~ " r * l - A 3 i ('27Q} 
 
 aJL& \ i u I 
 
 Uniform load over all 
 
 M 3 = - ^ w, U=M :1 (280) 
 
 SM= 4" ^^V^S C 281 ) 
 
 2 
 If x=0 
 
 - 12 \ 
 
 If h^=h ,=- 0. 
 
 1 
 Max. M x = - ^ w, %=M 9 =M 3 (283) 
 
 =?-g-j * x ** W2 '^v+i'/) - - .*,=: . -(284) 
 
 A single load In the centre of the beam 
 
 By using Table I, we have at once from (278) and (279), 
 
 ~ P * L - . (285) 
 
 A 
 
 
 ( 286 ) 
 
42 
 
 THE CONTINUOUS GIRDER. 
 
 Also, 
 
 *-w **< - 
 
 . - (287) 
 . (288) 
 
 If jr a =0, 
 
 And 
 Sfb= 
 
 r-*.a5r) - - iKtfrF /, - - (291) 
 
 CASE III. 
 
 A beam fixed at one end, and unsupported <it 
 
 the other 
 
 As the right end of the beam is unsupported, there can 
 be no reaction S 3 =R 3 , therefore, by (11), 
 
 Concentrated loads 
 
 S 3 = ~~ - + - P> L=0, and since M 3 =0, we have 
 
BEAMS WITH FIXED ENDS. 43 
 
 M,= -IP 2 Lk, . . .S a =SP a (292) 
 
 Showing that the moment of inertia does not enter into 
 the expression for the bending moment. 
 
 Any uniform, load 
 
 M,= ~ w, (a-al} (ai+ai) (293) 
 
 S 2 =w, (ai CQ (294) 
 
 From (8), 
 
 M x =0. xa 2 
 And 
 
 M x = I P 2 ( x a 2 ) . . . x g <a g (295) 
 
 Also, for any uniform load, 
 
 M x =w 2 (aa 2 ) x a *+ a * 1 . . . X a; (296) 
 
 For deflection, use the general formulas, if the moment 
 of inertia is variable. 
 
 If the moment of inertia is constant, we have, from (JEJ,), 
 If h,=0, 
 
 1 c ~\ 
 
 . . . '297) 
 
 Or, for a single concentrated load, 
 
 ........ (298) 
 
 And if the load is at the end of the beam, y- 3 =a 2 =l 2 , and 
 (298) becomes 
 
 P I 3 
 y a = -- ' * . .at end of beam ......... (299) 
 
 Also, for a uniform load over all, (xs=a g =l a ) t 
 
 ......... (300) 
 
44 THE CONTINUOUS GIRDER. 
 
 CASE IV. 
 
 A beam on ttro sH/tports, find one etui nnxttp- 
 ported 
 
 In this case, JLT,, Jf and 8 y equal zero, hence, from (11), 
 or (C), we have 
 
 S=4^ +<M==^,w,JC= -ft/, ......... (301) 
 
 Which is precisely the same equation we obtained in Case 
 III. The values of Q', are given in (95), (97) and (99). 
 
 From the general equations (B) and ( C ), we obtain 
 S,= . + Q I =R, ................. (302) 
 
 S= -y^ + ft 1 - (303) 
 
 V H 
 
 5,= -|P -fft J (304) 
 
 From (D), we obtain 
 ^=8, x, L,= -y-^ -f Q, x ; L ; (305) 
 
 ^^-^ 4- ft sc 4 .... (306) 
 
 Note that the moment of inertia does not appear in any 
 of the above equations. 
 
 For deflection, use the general equation (JE). 
 
BEAMS WITH FIXED ENDS. 45 
 
 CASE V. 
 
 A beam on one support, having one end 
 fixed, and the other unsupported 
 
 M, M 
 
 (a) E -alone, constant. 
 
 In this case, as before, M 4 =0, and S t =0, hence, 
 
 y = - Q 3 /, ..... .............. (307) 
 
 From (63), we have 
 
 2 ft 2 +2M;ft^=A^B^Y^X^X: ........ (308) 
 
 Substituting (307) in (308), it becomes, after reduction, 
 
 ... (309) 
 
 (6) JE and I constant. 
 (309) becomes 
 
 >*j (sio) 
 
 M s = - Q:i :i (307) 
 
 (c) J5J, I and h constant. 
 
 1 c ^ 
 
 (311) 
 
 M,= -.<g 4 (307) 
 
 The shears, intermediate bending moments and deflec- 
 tions are readily obtained from the general equations. 
 
46 
 
 THE CONTINUOUS GIRDER. 
 
 CASE VI. 
 
 A beam on two supports, hftriut/ tteitlter 
 supported 
 
 We have, at once, M, and M A equal zero, and hence, from 
 
 S, = r^- -f Q,=0, or M a = - Q, I, 
 And, from (C), 
 
 (312) 
 
 (313) 
 
 The shears, intermediate bending moments and deflec- 
 tions can now be readily obtained from the general equations. 
 
IV. 
 * THE POINT OF ZERO MOMENT. 
 
 Let us take (-D). 
 
 In which L is dependent upon the kind of loading in the 
 span r. If there is no load in the span r, then 
 
 Hf x =M r +S r x, (314) 
 
 Now, if there is a point of zero moment anywhere in the 
 span /, we can find its distance from the left support by 
 making (314) equal zero, and solving for x r ', doing this, we 
 obtain 
 
 (315) 
 
 S r 
 
 From (JB), 
 
 S r = Mf+l ~ Mr 4- ( q r in this case = 0) (B) 
 
 Now, 
 
 M m =c m '?: /? :! ' ' ' ' 3 T' M s . , . . wO-f.7 (70) 
 
 -\-l indicating that only those loads upon the right of 
 the span are considered. 
 
 Substituting (B) and (70) in (315), it reduces to 
 
 C r Cr+t ~a 
 
 Pt 
 
 (316) 
 
 See "Annalesdes Fonts etChaussees," 188(5, PaperNo. 40, by M. Collignon. 
 
48 THE CONTINUOUS GIRDER. 
 
 We also have, for loads on the left, 
 
 . And, substituting (71) and (J5) in (315), we obtain 
 
 X r = - ^=^ - ^~ I ..... LOA,, ,* THK I.KKT ..... (317) 
 
 s r+/ ~5 
 
 Pr+f 
 
 (316) and (317) are general equations. Knowing the 
 points of zero moment, we can tell .at once what loads must 
 be considered to have a certain effect. 
 
 The values of x,. are very easily computed, but they can 
 be constructed graphically as soon as c n f, +/ , etc., are known. 
 Thus: 
 
 -jx,i v *x'H 
 
 >-/'" 
 
 If 
 
 Let IB C represent any unloaded span. Then, IB H=x r 
 for m<,r+l or a load on the right, and T> K=.r, for 
 
 m>r or a load on the left, if E B=c n C F '=-- - c r +i -^ 
 
 Pr 
 
 F C=d^r+ t ^ and B E'= - </,_,+,. 
 
 /";+/ 
 
 For, from Fig. 9, 
 
 I^*H!KH -V- 
 
THE POINT OF ZERO MOMENT. 
 
 49 
 
 Also, 
 
 R E' + CF 
 
 l r =x f . . m>r 
 
 H-/ 
 
 If the span considered is loaded, then the value of x r must 
 be obtained from ( T)). 
 
 If the supports are level, and the span considered is uni- 
 formly loaded, .T r ,and also the bending moments, can be 
 found graphically. 
 
 Considering the span r as discontinuous and uniformly 
 
 1 1 
 
 loaded, the bending moment at the centre equals - - w r l' r . 
 
 o 
 
 Now, in Fig. 10, we wish to draw the moment parabola, so 
 that A M=M r and B N=M r+l . We know H K must 
 
 equal -~- w r I 2 ,.. Let D and C be the points of zero moment 
 o 
 
 for loads on the right and left of the span r. On the ver- 
 tical H K take any point Jt and draw h D in and 7t C n, 
 
 and connect in and n. Make n p = - - w r ?;' and draw 
 
 o 
 
 p r parallel to in n until it is intersected by in h, pro- 
 duced in the point r. Draw r H and s jKparallel to li, and 
 
 also jM~ ^parallel to in n. Then is H K= -~ w r I 2 n A H>f= 
 
 o 
 
 M n B y=M r+l and A E and A F=x r . 
 
50 
 
 THE CONTINUOUS GIRDER. 
 The co-efficient* c (tit<? d. 
 
 3: *, A 
 
 ^*l. 
 
 iuA -tT C ^-|. ft 
 
 z.H 
 t 
 
 *# 
 
 The co-efficients r? and r? can be found graphically, if 1 
 is considered as constant. 
 
 In Fig. 11, let b m and in k represent the first and sec- 
 ond spans of a continuous girder. Bisect b in at A, m / 
 at B, and b k at C. Make C I -2 r, - 2 and draw 
 a A ? e f, & right line passing through ^1 and , then will 
 J5 e equal, numerically, r... Thus: 
 
 but b c h k=f A k a b A=f n m k, 
 
 since den =f e g,f n m k=g d m k=L (B e), 
 
 Therefore, B e -2 ', '* -.= r :i . 
 
 tl ;i is found in a similar manner from Fig. 12. 
 
 k B= ^,bA= lr,kO= 
 
BEAMS WITH FIXED ENDS. 
 
 51 
 
 b c h k= -2 d 3 (X-fC,)= - 2 (/ S -K_,), 
 k g d m=(B e) / s _,, but b c h k=k g d m, 
 Therefore, 
 
 Then, in general, from Fig. 13, we have, if 
 
 ^L if 
 
 p b = c >ll _ 3 and 1C- 2 c m _ h 
 c b k h=d g r d+p r k b=d g k m+p d' m 6 
 but, g d m k=(B e) l m _,, p d m b=c m _.> l m _ s 
 and cbkh= -2 c m _ t (l m _,^l m _ 2 ) . 
 Therefore, 
 
 L-/ -4- l m - l m - 
 
 B e = 2 c m _, r - c m . a T - == c m . 
 
 And for d m we can write 
 
 r> rt J "8 -f2 " n 
 
 n e= -- % d m _, . -. 
 
 L-, 
 
 f'sm+3 
 
 s w-j-2 
 
V. 
 
 APPLICATIONS. 
 
 We will now give a complete analysis of a continuous 
 girder, to illustrate, more particularly, the use of Table I. 
 
 Ex. 1. Let Fig. 14 represent a continuous girder of three 
 spans on level supports, and having a constant moment of 
 inertia. 
 
 jr -jf -ft if * 
 
 I j. Z = 300 
 
 i ' 
 
 Let each panel be 30 in length and the loads repre- 
 sented as in the figure. Then 8=3, r=l, 2, 3 and 4, m=l, 
 2, 3 and 4, l l =24D'=l a and l,=300'. 
 
 First, look up in Table 1^ the values of the co-efficients 
 2 k 3 2 -}-F and k k 3l for the different values of - = k. 
 
 fc,-*. 
 
 **,-*+*! 
 
 k-V 
 
 0.123,046,875 
 0.234,375,000 
 0.322,265,625 
 0.375,000,000 
 0.380,859,375 
 0.328,125,000 
 0.205.078,125 
 
 k ? 
 
 2k^3li 
 + g 
 
 k y -kl 
 
 0.100 
 0.200 
 0.300 
 0.400 
 0.500 
 0.600 
 0.700 
 0.800 
 0.900 
 
 0.171 
 0288 
 0.357 
 0.384 
 0.375 
 0.336 
 0.273 
 0.192 
 0.099 
 
 0.099 
 0.192 
 0.273 
 0.336 
 0.375 
 0384 
 0.857 
 0.288 
 0.171 
 
 0.125 
 0.250 
 0.375 
 0.500 
 0.625 
 0.750 
 0.875 
 
 0.205,078,125 
 0.328,125,000 
 0.380,859,375 
 0.375,000,000 
 0.322,265,625 
 0,234,375,000 
 0.123,046,875 
 
APPLICATIONS. 
 From (A,), we have M,=0=M,. 
 
 53 
 
 M a = 
 
 A > 
 
 From (p,) and (l*/), we find that, since c,=0 and c a =lj 
 c a = 3.6 and C A = -j- ^-#. 
 
 Also, 
 d ; # and c? 2 =-/, d s = 3.6 and <i 4 -4- 14.93. 
 
 Finding the values of A f , A 2 , IB,, _R>, etc., from (/,) and 
 (,//), and substituting them, with the values of c and d 
 above, in the moment equations, we obtain, 
 
 TABLE VALUES OF Mo. 
 
 
 
 Pi 
 
 (0.123,046,875) ] 
 
 
 f-- 7.111,198 
 
 P] 
 
 
 
 
 (0.234,375,000) 
 
 
 
 - 13545,140 
 
 Pf 
 
 
 
 Pf 
 
 (0.322,265,625) 
 
 
 
 - 18.624,568 
 
 P! 
 
 
 
 (0.375,0 
 
 no ooo^ l ft? 
 
 79 9 6 . . 
 
 91 fi79 99S 
 
 pt 
 
 
 
 Pf 
 
 ((X380J859|375) 
 
 
 ) 
 
 - 23 010,853 
 
 j i 
 
 
 
 Pf 
 
 (0.328,125,000) 
 
 
 
 - 18.963,196 
 
 PS 
 
 
 
 PI 
 
 (0.205,078,125) J 
 
 1-11.851,998 
 
 PI 
 
 
 
 Pi 
 
 (0.171) 
 
 
 f + P^ 
 
 (0.099) 
 
 - 12.958,194 
 
 p.i 
 
 
 
 PI 
 
 (0.288) 
 
 
 + PI 
 
 (0.192) 
 
 
 - 21.190,636 
 
 p.^ 
 
 
 
 
 (0.357) 
 
 
 + P| 
 
 (0.273) 
 
 
 - 25.389,634 
 
 \>;i 
 
 1 2 
 
 
 
 P| 
 
 (0.384) 
 
 
 4- PI 
 
 (0.336) 
 
 
 - 26.247,494 
 
 ]> 1 
 
 
 
 Pi 
 
 (0.375) 
 
 h 90.3010+25.0836 { + P\ 
 
 (0.375) }- = -j 24.456,525 
 
 
 
 
 Pi 
 
 (0.336) 
 
 
 I p i 
 
 (0.384) 
 
 
 - 20.709,033 
 
 l ;f 
 
 
 
 PI 
 
 (0.273) 
 
 
 
 (0.357) 
 
 
 - 15.697,327 
 
 
 
 
 PI 
 
 (0.192) 
 
 
 + PI 
 
 (0.288) 
 
 
 - 10.113,715 
 
 ii 
 
 
 
 (0.099) 
 
 
 I + PI 
 
 (0.171) 
 
 
 4.650,503 
 
 
 4- 
 
 P^ 
 
 (0.205,078,125) ] 
 
 + 3.292,221 
 
 pj 
 
 4- 
 
 p 2 
 
 (0.328,125,000) 1 
 
 4- 5.267,554 
 
 PI 
 
 -4- 
 
 PI 
 
 (0.380,859,375) | 
 
 4- 6.114,126 
 
 PI 
 
 + 
 
 Pf 
 
 (0.375.000,000) } 16.0535- - . 
 
 ...... { -f 6.020,061 
 
 p| 
 
 -f 
 
 Pi 
 
 (0.322,265,625) 
 
 
 
 -4- 5.173,491 
 
 PI 
 
 -f 
 
 P^ 
 
 (0.234,3 
 
 75,000) 
 
 
 
 4- 3.762,365 
 
 pi; 
 
 . 
 
 Pa 7 
 
 (0.123,046,875) 
 
 
 1 -f- 1.975,333 
 
 PS 
 
THE CONTINUOUS GIRDER. 
 TABLE a Continued. 
 
 VALUES OF M,. 
 
 + Pi 
 
 (0.123,046,875)1 
 
 f -t- 1.975,333 P} 
 
 + Pf 
 
 (0.234,375,000) 
 
 
 4- 3.762,365 Pf 
 
 -f Pf 
 
 (0.322,265,625) 
 
 
 4 5.173,491 P? 
 
 
 (0.375,000,000) 1- 16.0535= \ -f- 6.020,061 Pf 
 
 f PI 
 
 (0.380,859,375) 
 
 
 4- 6.114,126 Pi 
 
 + Pf 
 
 (0.328,125,000) 
 
 
 5.267,554 P? 
 
 -rP{ 
 
 (0.205,078,125) J 
 
 L+ 3.292,221 Pi 
 
 -f Pi 
 
 (0.171)1 
 
 f P (0.099)1 
 
 - 4.650,503 PJ 
 
 4- P| 
 
 (0.288) 
 
 - P| (0.192) 
 
 - 10.113,715 PI 
 
 f PI 
 
 (0.357) 
 
 - P| (0.273) 
 
 - 15.697,327 Pi! 
 
 + P| 
 
 (0.384) 
 
 - Pi (0.336) 
 
 - 20.709,0:53 P 4 
 
 + Pi 
 
 (0.375) \ 25.0836+90.3010 \ P 5 2 (0.375) j - 24.456,525 P5 
 
 f P* 
 
 (0.336) 
 
 P (0.384) 
 
 - 26.247,494 P!) 
 
 4- P| 
 
 (0.273) 
 
 - PI (0.357) 
 
 - 25.389,634 Pi 
 
 -f P! 
 
 (0.192) 
 
 - PI (0.288) 
 
 - 21.190,6:50 V : , 
 
 -f Pf 
 
 (0.099) J 
 
 [ P (0.171) J 1 15.958,194 P:; 
 
 _ P 1 
 
 (0.205,078,125) 1 
 
 f 11.851,998 PJ 
 
 PI 
 
 (0.328,125,000) 
 
 
 - 18.963,196 P^ 
 
 -PI 
 
 (0.380,859,375) 
 
 
 22.010,853 P^ 
 
 
 (0.375,000,000) l 57.7 
 
 926 -....-{- 21.672,225 P| 
 
 pi 
 
 *- 3 
 
 (0.322,265,625) 
 
 
 - 18.624,568 PJ 
 
 pe 
 
 * 3 
 
 (0.234,375,000) 
 
 
 - 13.545,140 Pj; 
 
 
 (0.123,046,875) J 
 
 1- 7.111,198 P, 7 
 
 Compare the values of M 2 and BI ;1 , and notice that the 
 co-efficients of JPJ, JP, 6 , etc., of M it , are the same as the 
 co efficients of J /, J?/, etc., of JfC, as they should be, owing 
 to the symmetry of the girder. 
 
APPLICATIONS. 55 
 
 TABLE b 
 
 The next step is the deduction of S from (J) and (C). 
 
 
 FIRST SPAN. 
 
 SECOND SPAN. 
 
 THIRD SPAN. 
 
 s< 
 
 K 
 
 8, 
 
 & 
 
 ft 
 
 8* 
 
 Pi 
 
 +0 845,870 
 
 -i-0. 154,630 
 
 +0.030,288 
 
 
 -0.008,230 
 
 
 -p? 
 
 +0.693.5<i2 
 
 +0.306,438 
 
 +0.057,691 
 
 
 0.015,676 
 
 
 Pf 
 PI 
 Pi 
 
 +0.547,397 
 
 +0.409.WH 
 +0.283,288 
 
 +0.452,603 
 +0.590,301 
 +0.716,712 
 
 +0.079,327 
 +0.092.307 
 +0.093,749 
 
 Same as 82 
 with sign 
 
 -0.021,556 
 -0.025,083 
 -0.025,475 
 
 Same as S 3 
 with+sign 
 
 Pf 
 
 +0.170,98<> 
 
 +0.829,014 
 
 +0.080,769 
 
 
 -0.021,948 
 
 
 PI 
 
 +0.075,610 
 
 +0.924,384 
 
 +01,050,480 
 
 
 -0.013,717 
 
 
 Pi 
 
 -0.053,992 
 
 
 +0.927,692 
 
 +0.072,307 
 
 +0.019,377 
 
 
 ?! 
 
 -: 1.088,294 
 
 
 +0.836,923 
 
 +0.163,076 
 
 +0.042,140 
 
 
 pi 
 
 -0.105,790 
 
 
 +0.732,307 
 
 +0.267,692 
 
 +0.065,405 
 
 PI ' 
 
 | 
 
 -0.109,364 
 -0.101,902 
 -0.086,287 
 
 Same as Si 
 with+sign 
 
 +0.618,461 
 +0.800,000 
 
 +0.381,538 
 
 +0.381,538 
 +0.500,000 
 +0.618,461 
 
 +0.086,287 
 +0.101,902 
 +0.109,364 
 
 Same as S 3 
 with sign 
 
 p| 
 
 -U.065,405 
 
 
 +0.267,692 
 
 +0.732,307 
 
 +0.105,790 
 
 
 PI 
 
 -0.042,140 
 
 
 +0.163,076 
 
 +0.836,923 
 
 +0.088.294 
 
 
 TJ9 
 
 -0.019,377 
 
 
 +0.072,307 
 
 +0.927,692 
 
 +0.053,992 
 
 
 PJ 
 
 +0.013,717 
 
 
 -0.050,480 
 
 
 +0.924,384 
 
 +0.075,616 
 
 P 
 
 +0.021,948 
 
 
 -0.086,769 
 
 
 +0.829,014 
 
 +0.170,986 
 
 8 
 
 Pi 
 
 +0.025,475 
 +0.025,083 
 +0.021,556 
 
 Same as Si 
 with sign 
 
 -0.093,749 
 -0.092,307 
 -0.079,327 
 
 Same as S 2 
 with+sign 
 
 +0.716,712 
 +0.590,301 
 +0.452,603 
 
 +0.283,288 
 +0.409,699 
 +0.547,397 
 
 PS 
 
 +0.015,676 
 
 
 -0.057,691 
 
 
 +0.306,438 
 
 +0.693,562 
 
 p 
 
 +0.008,230 
 
 
 -0.030,288 
 
 
 +0.154,630 
 
 +0.845,370 
 
 The computation of the above table is very simple and 
 easy. Thus : 
 
 $, -y^ -f- PI (.1 &/) an d numerically becomes, for loads 
 i f 
 
 in the first span, 
 
 7.111 + 
 
 210 
 
 40 
 
 P<=. -f- 0.845 P/. 
 
 240 
 
 f - P?= +Q.698P*, etc, etc. 
 
 For loads in the second and third spans, we have merely 
 * , or the moment over the second support divided by 
 the length of the first span. 
 
56 THE CONTINUOUS GIRDER, 
 
 S g = , -f- P 2 (1 L), for loads in the second span, and 
 
 S g = -,- for loads in the other spans. 
 
 '2 
 
 -M 
 
 Sv *- -f P a (1 k s ) y for loads in the third span, and 
 
 83= 2 j for loads in the other spans. 
 
 It is necessary to compute only $ S y and $, to fill out 
 the table of shears, since S,-\-Sg==0 or jP,, S a +S 3 =0 or _P 2 , 
 etc., but it is better to compute S',, S', and $, as a check. 
 
 TABLE r VALUES OF J 
 
 FIRST SPAN. 
 
 
 
 1 
 
 * 
 
 3 
 
 4 
 
 o 
 
 6 
 
 7 
 
 i 
 
 9 
 
 
 
 M< 
 
 i 
 
 Mi 
 
 Mi 
 
 i 
 
 MI 
 
 M; 
 
 
 1 
 
 PI 
 
 + 25.361 
 
 + 20.722 
 
 + 16.083 
 
 + 11.444 
 
 + 6.805 
 
 + 2.166 
 
 - 2.472 
 
 P7, 
 
 2 
 
 P 
 
 + 20.806 
 
 + 41.613 
 
 + 32.420 
 
 + 23.237 
 
 + 14.044 
 
 + 4.851 
 
 - 4.341 
 
 PS 
 
 8 
 
 Pf 
 
 + 16.421 
 
 + 32.843 
 
 + 49.265 
 
 + a5.687 
 
 -22.109 
 
 + 8. -531 
 
 - 5.046 
 
 
 4 
 
 P| 
 
 + 12.290 
 
 + 24.581 
 
 + 36.872 
 
 + 49.163 
 
 -T 31.554 
 
 + 13.745 
 
 - 3.963 
 
 P 4 
 
 5 
 
 P 
 
 + 8.498 
 
 + 16997 
 
 + 25.495 
 
 + 33.994 
 
 + 42.493 
 
 + 20.991 
 
 - 0509 
 
 PT 
 
 6 
 
 Pf 
 
 + 5.129 
 
 + 10.259 
 
 + 15.388 
 
 + 20.518 
 
 + 25,647 
 
 + 30.777 
 
 + 5.907 
 
 p-j 
 
 7 
 
 PI 
 
 + 2.268 
 
 + 4,536 
 
 + 6.805 
 
 + 9.073 
 
 + 11.342 
 
 -r 13.610 
 
 + 15.679 
 
 Pi 
 
 8 
 
 VI 
 
 - 1.619 
 
 - 3.239 
 
 - 4.859 
 
 - 6479 
 
 - 8/9S 
 
 - 9.71* 
 
 - 11.338 
 
 p* 
 
 9 
 
 PS 
 
 - 2.648 
 
 - 5.297 
 
 - 7.916 
 
 - 10.595 
 
 - 13.244 
 
 - 15 892 
 
 - 18.541 
 
 Pg 
 
 10 
 
 
 - 3.173 
 
 - 6.347 
 
 - ' 9.521 
 
 - 12.694 - 15.868 - 19.042 
 
 - 22.215 
 
 PS 
 
 11 
 
 P| 
 
 - 3.280 
 
 - 6.561 
 
 - 9.842 
 
 - 13.123 
 
 - 16.404 
 
 - 19.685 
 
 - 22.966 
 
 
 12 
 
 PI ! - 3.057 
 
 - 6.114 
 
 - 9.171 
 
 - 12 228 
 
 - 15.285 
 
 - 18.342 
 
 - 21.399 
 
 PS 
 
 13 
 
 Pi : - 2.588 
 
 - 5.177 
 
 - 7.76-5 
 
 - 10.354 
 
 - 12.943 
 
 - 15.531 
 
 - 18.120 
 
 PA 
 
 14 
 
 PA - 1.962 
 
 - 3.924 
 
 - 5. 886 
 
 - 7.848 ; - 9.810 
 
 - 11.772 
 
 13.735 
 
 P 3 
 
 15 
 
 PI | - 1.264 
 
 - 2.528 
 
 - 3.792 
 
 - 5.056 
 
 9.321 
 
 - 7.58-5 
 
 - 8.849 
 
 PR 
 
 16 
 
 Pi 
 
 - 0.581 
 
 - 1.162 
 
 - 1.743 
 
 - 2.325 
 
 - 2.906 
 
 - 3.487 
 
 - 4.0 9 
 
 pi 
 
 17 
 
 PI 
 
 + 0.411 
 
 + 0.823 
 
 + 1.234 
 
 + 1.646 
 
 + 2.057 
 
 + 2469 
 
 + 2880 
 
 Pi 
 
 18 P^ 
 
 + 0.658 
 
 + 1.316 
 
 + 1.975 
 
 + 2.633 
 
 + 3.292 
 
 + 3.950 
 
 -r 4.609 
 
 P? 
 
 19 1 P 3 
 
 + 0.764 
 
 + 1.528 
 
 + 2.292 
 
 -r 3.057 
 
 + 3.82L 
 
 + 4.585 
 
 + 5.349 
 
 Pf 
 
 20 P| 
 
 + 0.752 
 
 + 1.504 
 
 + 2.257 
 
 + 3.009 
 
 + 3.762 
 
 + 4.514 
 
 + 5.267 
 
 Pf 
 
 21 Pi> 
 
 + 0.646 
 
 + 1.293 
 
 + 1.940 
 
 + 2.586 
 
 + 3.233 
 
 + 3.880 
 
 + 4.527 
 
 
 22 Ptj 
 
 + 0.470 
 
 + 0.940 
 
 + 1.410 
 
 + 1.881 
 
 + 2.351 
 
 + 2.821 
 
 + 3.291 
 
 P| 
 
 23 
 
 1 
 
 + 0.246 
 
 + 0.493 
 
 + 0.740 
 
 + 0.987 
 
 + 1.234 
 
 -r 1.481 
 
 + 1.728 
 
 
 
 
 3 
 
 3 
 
 3 
 
 3 
 
 ., 
 
 3 
 
 3 
 
 
 
 m 
 
 m 
 
 M* 
 
 M 
 
 Mi 
 
 m 
 
 AC 
 
 
 THIRD SPAX. 
 
APPLICATIONS. 
 
 57 
 
 TABLE c Continued, 
 
 SK( OND SPAN. 
 
 / 
 
 2 
 
 r> 
 
 4 
 
 ~) 
 
 6 
 
 7 
 
 8 
 
 9 
 
 10 
 
 11 
 
 
 m 
 
 Ml 
 
 m 
 
 Mi 
 
 M* 
 
 'MI 
 
 m 
 
 Ml 
 
 m 
 
 
 1 >_' 
 
 P? 
 
 ft 
 
 !! 
 
 - 6.202 
 - 11.824 
 Hi. 241 
 - 18.903 
 - 19.198 
 - 16.540 
 - 10.337 
 
 - 5.293 
 - 10 103 
 - 13.804 
 - 16.133 
 - 16.385 
 - 14.117 
 - 8.823 
 
 - 4.385 
 - 8.382 
 - 11.485 
 - 13.364 
 - 13.573 
 - 11.693 
 - 7.308 
 
 - 3.476 
 - 6.662 
 - 9.105 
 - 10.595 
 - 10.760 
 - 9.270 
 - 5.794 
 
 - 2.567 
 - 4.941 
 - 6.725 
 
 - 7826 
 - 7.948 
 - 6.847 
 - 4.279 
 
 - 1.659 
 - 3.220 
 - 4.345 
 - 5.05(5 
 - 5.136 
 - 4.424 
 - 2.765 
 
 - 0.750 
 - 1.500 
 - 1.965 
 - 2.287 
 - 2.323 
 - 2.001 
 - 1.251 
 
 + 0.157 
 + 0.220 
 + 0.413 
 + 0.481 
 + 0.498 
 + 0.421 
 + 0.263 
 
 + 1.006 
 1 1.041 
 + 2.793 
 + 3.250 
 + 3.301 
 + 2.844 
 + 1.777 
 
 il 
 
 ii 
 
 P 
 
 i:f 
 P! 
 
 + 14.872 
 
 + 3.917 
 - 3.420 
 - 7.693 
 - 9.456 
 
 + 12.703 
 + 29.024 
 + 18.548 
 + 10.860 
 + 5.543 
 
 + 10.534 
 + 24.132 
 + 40.517 
 + 29.413 
 + 20.543 
 
 + 8.364 
 + 19.240 
 + 32 487 
 H 47DU7 
 + 35.543 
 
 + 6.195 
 + 14.347 
 + 24.456 
 + 36.521 
 + 50.543 
 
 + 4.026+ 1.857 
 + 9.155'+ 4.563 
 + 16.425|+ 8.394 
 + 25.075 + 13.629 
 + 35.543'+ 20 543 
 
 - 0.312 
 - 0.329 
 + 0.361 
 + 2.183 
 + 5.543 
 
 - 2.481 
 - 5.221 
 - 7.666 
 - 9.263 
 - (1.450 
 
 ii 
 n 
 
 M 9 
 
 ifc 
 
 m 
 
 m 
 
 j\b 
 
 m 
 
 m 
 
 M; 
 
 m 
 
 SECOND SPAN. 
 
 M X =S, w, !*, (Wr-di) for loads in the first span, and 
 
 / 
 MxS, Xi for loads in the other spans. M x = 8 3 W 3 IP 3 
 
 (w 3 a 3 ) for loads in the third span, and JJ=& X 3 for loads 
 in the other spans. 
 
 If there were no equal spans, then Table c would have 
 the number of apices squared computations, making -its 
 formation tedious, if many spans were considered, but the 
 great worth of the table, when once computed, overbalances 
 the hard labor in its formation. 
 
 Having Table c before us, we can tell at once which of the 
 apices must be loaded, to produce a certain result, in any par- 
 ticular chord member. 
 
58 THE CONTINUOUS GIRDKK. 
 
 MAXIMUM MOMENTS. 
 
 Table (c) enables one to find the maximum bending 
 moments for any chord piece, with comparatively little labor. 
 
 r 
 
 Dead load 
 
 For the dead, or static loads, of the structure, the maximum 
 bending moment for any chord member is found by taking the 
 algebraic sum of the quantities in the proper column of Table ( c) 
 (each co-efficicient is, of course, multiplied by its IP). 
 
 For example, suppose the maximum bending moment at 
 the second panel of the first span, or, better, the second 
 panel point of the first span, is desired : Multiply each 
 co-efficient in column 3 of Table (#), by its proper JP, and 
 take the algebraic sum of the products. 
 
 Live load 
 
 For the live or moving load, the maximum negative mom< nt 
 at any panel point is found by taking the sum of the negative 
 co-efficients (multiplied by their proper IP's) in the proper column. 
 of Table (c), and vice versa for the maximum positive bending 
 moment. 
 
 For example, suppose the maximum negative bending 
 moment at the second panel point is desired : Take the 
 sum of the negative products in column 3 of Table (c). 
 For positive moment, take the sum of the positive products. 
 
 If the IP's are equal, the work is somewhat easier, as the 
 co-efficients can be at once summed, and then multiplied 
 by the common value of IP. 
 
 The maximum bending moments over the supports are 
 obtained in the same manner from Table (a). 
 
APPLICATIONS. 59 
 
 MAXIMUM SHEAR. 
 
 The maximum shear is obtained from Table (6). 
 
 .J)ead load 
 
 The maximum shear at any panel point is found by: 
 First. Taking the algebraic sum of the co-efficients (multiplied 
 
 by their JP*s) of all the loads outside of the span in which the 
 
 apex considered lies. 
 
 Second. Considering the span in which the apex lies alone, 
 and using for the left reactions the co-efficients as found in the 
 column giving the values of 8. 
 
 For example, take the second apex of the first span. The 
 maximum shear is found : First, by taking the algebraic 
 sum of the co-efficients in column $, for the second and 
 third spans. Second, by taking the sum of the co-efficients 
 opposite JP/, ff to JP 1 /, inclusive, and decreasing it by the 
 co-efficient opposite JP/ in column S' s . 
 
 Live load 
 
 For positive or negative shear : 
 
 First. Take the sum of the positive or negative co-efficients 
 (multiplied by their JP^s) for spans in which the apex does not lie. 
 
 Second. Consider the span in which the apex lies alone, and 
 use for left reactions the positive or negative co-efficients (multi- 
 plied by their J?'s) as given in the proper column of Table (b). 
 
 For example, find the maximum positive shear at the 
 second apex of the first span : 
 
 First. Take sum of co efficients (multiplied by their jP V) 
 in column $,, opposite third span. 
 
 Second. Take sum of coefficients (multiplied by their 
 jP's), opposite J?,* to JP/, inclusive. 
 
 Knowing the maximum shears and bending moments, 
 the maximum stresses are then easily obtained. 
 
60 
 
 THE CONTINUOUS GIRDER. 
 
 1=2.40' 
 
 The following is a graphical solution of the same example. 
 Tables (&) and (<") can be filled by means of graphics, as 
 can also Table (), by using Pro/. Greene's Method of Area 
 Moments, which is fully explained in Part II of Greene's 
 Trusses and Arches. 
 
 Let it be supposed that Table (ft) has been filled, either 
 by computation or by Greenes Method of Area Moments; 
 we have, then, the bending moments over the supports, or 
 the "pier ordinates." Consider only a single concentration 
 at the second apex of the first span, and call it J?*; then, if 
 the first span were discontinuous, the moment polygon 
 A c< IB would enable us to determine the bending moments 
 at other apices of the span; bat the girder is not discon- 
 tinuous, and the load Pf induces a negative moment over 
 the second support, which may be designated by IB J?,,, 
 then the other negative moments are given by the ordi- 
 nates in the triangle or polygon A B ]$>. The differ- 
 ence of the ordinates of the two polygons determines the 
 magnitude and kind of bending moment at each apex of 
 the first span. The load jP/ induces a positive moment 
 over the second support, which may be designated by C C g , 
 then the ordinates between the lines 1$ C and J2.> C 3 will 
 give us the moments at the apices of the second span. The 
 
APPLICATIONS. 61 
 
 moments in the third span are given by the ordinates be- 
 tween the lines > I) and C ID. 
 
 Having given an outline of the method, we will now pro- 
 ceed to give it in detail. 
 
 First. Assume some value for JP/, as unity, ten or one 
 hundred. 
 
 Second. Form the stress diagram, Fig. 16, assuming for 
 the pole distance some value as unity, ten or one hundred. 
 Assume the pole in such a position, that the closing line 
 A 1$ to the equilibrium polygon A c, IB shall be horizontal, 
 and construct the equilibrium polygon A c, IB. 
 
 Third. From Table (), we find the bending moment 
 over the second support to be 13.54 JPf; now, if IPf=l, 
 and the pole distance == J, then the ordinate IB J^ 2 : =13.54j 
 laid off to the tcale of A IB. Draw A IB a . Scale the ordinates 
 b, b,, c, c*, etc., and place the results in Table (c), column 3, 
 opposite _P/, fgj etc. The results will be found to agree 
 with those computed. 
 
 Fourth. From Table (a), we find that C C 2 = + 3.762 
 Pf] hence, lay off C C s =3.76 to scale of A IB, and draw 
 1$ 2 C 2 and C s ID and fill out the remainder of column 3, 
 Table (c), by scaling the ordinates between these lines and 
 the horizontal. 
 
 In like manner, each column of Table (c) can be filled 
 in a comparatively short time. 
 
 To fill Table (ft), proceed as follows: 
 Fiist. Draw G H, Fig. 16, parallel to A _R, then 
 72 H $, and F H=S;. 
 Second. Draw G If, parallel to B 3 C 2 , then K H t =8 2 
 
 '-&'. 
 Third. Draw G H 3 parallel to C, 1), then K H 2 = -& 
 
 = + ,'. 
 
 Proceed in like manner with each load. 
 
62 THE CONTINUOUS GIRDER. 
 
 * EXAMPLE 2 VARIABLE L 
 
 K-- 
 
 l t =52, lf=6fi, lf=65, 1 4 =52. 8=4. r=l, 2, 3 and 4. 
 
 m=l, 2, '>', 4 and . 
 i t / - 
 
 L=240=I, L=300=I ;i 
 
 1^=420=1, 
 
 I t =r>2o=i 
 
 Wr=6.7 
 
 I 5 =300=I S l tn =360=I 3 
 
 I t =420=1 IS I 6 = 
 
 I 2 =300=1,, / 7 =300=1, 
 
 I 3 =300='i io } 8 =240=I r> 
 
 I A =240=1, 1=300=1 
 
 Determine the value of 
 From (A), 
 
 M 2 = 
 
 
 In which the several terms have the following values : 
 
 * See page 131, Alhjem,eine Theoricund Berechnung tier continuirlic7i.cn untl 
 einfachen Trager, by J. Jacob Weyraueh, Ph. D. 
 
APPLICATIONS. 
 
 First Span 
 
 L=52. e, =e, 
 
 , 
 
 A, 
 
 *: 4 
 
 e, = 10,50 
 
 4- 0.300 
 
 + 6 
 
 e, = 13 75 
 
 /\, = -|- -337 
 
 37 
 
 -e,= 3125 
 
 - 337 
 
 198 
 
 e 4 = 31,50 
 
 - 0.300 
 
 237 
 
 c, = 43.50 
 
 + 0.433 
 
 + 685 
 
 f ' G = 46 00 
 
 + 289 
 
 541 
 
 e 7 = 48.00 
 
 + 0.201 
 
 427 
 
 e a = 50.00 
 
 A, - + 0.243 
 
 + 584 
 
 
 
 f 1825 
 
 : + -si" : + 35 - omi 
 
 Second Span, 
 
 
 A, 
 
 A'-'/ \ 
 
 A. <- 
 
 ef 
 
 c v 
 
 A ^ ^"' 
 
 (3-2 +) 
 
 A " 
 
 A. z 
 
 A, t 
 
 c, = 1.50 
 
 A/ = 
 
 -0.256 
 
 73 
 
 2 
 
 
 
 <V'=5: K.OO 
 
 Af- 
 
 - 222 
 
 124 
 
 () 
 
 
 
 e* = 4.75 
 
 A, = 
 
 -0.311 
 
 267 
 
 20 
 
 1 
 
 l 4 '=: 6.75 
 
 
 - 0.467 
 
 553 
 
 59 
 
 2 
 
 c s = 19.25 
 
 A, - 
 
 + 0.467 
 
 + 1285 
 
 + 417 
 
 + 51 
 
 ^ =r 23.50 
 
 
 [0.311 
 
 + 972 
 
 391 
 
 + 62 
 
 <? 7 5== 42.75 
 
 A? == 
 
 0.311 
 
 1261 
 
 957 
 
 374 
 
 e, == 47.00 
 
 A,- = 
 
 - 0.467 
 
 1931 
 
 1603 
 
 746 
 
 r v 5625 
 
 AP 
 
 + 0467 
 
 + 1968 
 
 + 1876 
 
 + 1278 
 
 r/0 ^: 58.75 
 
 
 | 0.311 
 
 1313 
 
 + 1279 
 
 970 
 
 P W == 60.75 
 
 A== 
 
 -I 0.222 
 
 938 
 
 927 
 
 + 765 
 
 r ; ,^ 62.25 
 
 A/* 
 
 | 0167 
 
 + 706 
 
 701 
 
 + 620 
 
 Ctt = 63.50 
 
 A 
 
 + 0.166 
 
 + 701 
 
 + 700 
 
 + 657 
 
 
 
 
 + 3674 
 
 + 3641 
 
 + 3280 
 
 + ^6f7^ 
 
 -f^ 
 
 U F + 3644 I tei 
 
 -)dl=Fa. 
 
 65 
 
 ^~ f - 65 
 
 | 50 461 -=!<;. 
 
64 THE CONTINUOUS GIRDER. 
 Third S/Hfit. 
 
 e ,, 
 
 A. 
 
 i :i -d-e v y 
 
 , 
 
 > ' e " } 
 
 i 
 11 
 
 32 
 90 
 
 -f 343 
 331 
 858 
 1443 
 + 1775 
 1202 
 865 
 + 1004 
 
 ^~30~82~ 
 
 ,; 
 
 I 
 
 A, l 
 
 ~0 
 
 
 
 1 
 
 4 
 
 H- 39 
 49 
 318 
 638 
 + 1316 
 + 972 
 + 755 
 + 937 
 
 ~Tl07 
 
 e,= 1.50 
 e= 2.75 
 e, = 4.25 
 e 4 = -. 6.25 
 r, = : 8.75 
 e,= 18.00 
 e 7 = 22.25 
 e s = 41.50 
 <> 9 = 45.75 
 e lo = 58.25 
 e,,= 60.25 
 = 62.00 
 e, 3 = 63.50 
 
 -0.154 
 -0.155 
 -0.206 
 -0.289 
 - 0.433 
 : + 0.433 
 f 0.289 
 -0.289 
 p = -0.433 
 AF= + 0.433 
 A//=+ 0.289 
 A= + 0.206 
 /\ /3= -f 0.238 
 
 44 
 
 80 
 160 
 
 319 
 644 
 + 1138 
 874 
 1163 
 ] 782 
 + 1827 
 1220 
 -870 
 1005 
 
 ~T742~ 
 
 + *W_ , ,,.,. v S M _ , 
 
 
 /r 
 
 65 
 
 65 
 
 + M)7 - 4-soo 
 
 - T 
 
 
 65 
 Fourth Spun. 
 
 < A; 
 
 i :i -(l~e,,r 
 
 f?(3-2 A_) 
 / 
 
 A, L 
 
 *, 2.00 0.110 33 
 e.., = 4.00 A, ~- -- 0.095 55 
 f,= 600 0134 111 
 C A - 8.50 A4 : 0-200 224 
 (V= 17.50 0.143 + 274 
 0; - 20.75 A = 0.151 320 
 e, ^ 38.25 0.151 101 
 e^= 41.50 A S = 0.143 384 
 
 1 
 4 
 13 
 39 
 + 102 
 143 
 337 
 346 
 
 
 -614 
 
 495 
 
 * 22807 F d** 
 
 - 
 
 E\o) * >-*& 
 
 o& o& 
 
 ' 4 
 
APPLICATIONS. 65 
 
 Collecting the values of F, we have 
 
 .F,"= -f 35.096 
 - F a = + 56.528 FS=-- -f- 56.061 F* =-- + 50 Jfil 
 
 F s = + 4%315 F,'=: + -47.^5 F/= -f 47.00 
 
 #= - ^.07 JFV = 9.519 
 
 From (e), 
 
 0, = 6 E I t = 520 E ' (by making L '=6 E) 
 O g = 6 E I t = 560 E' (by making E'=6 E) 
 
 3 = 6 E I t = 520 E (by making E'=6 E) 
 
 4 = 6 E I = 240 E ' (by making E '=6 E) 
 
 From (mj, 
 p a =0 s (l,^F.;)=520 (121.061) E'= 62,951.7 E' 
 
 ^=0 4 (l 8 -^ r F;)=240 (112.415) E'= 26,979.6 E' 
 
 From (n), 
 ^(t.+A^+ACt+R) 
 
 = 560 (87.096) E' +520 (121.523)E'= . . 111,965.7 E' 
 
 &=M4+^;) + M^+^) 
 
 = 520 (115.461) E'+560 (107.215) E "= 120,080.1 E' 
 
 /^M^ + ^')-M<(^ + ^) 
 
 = 240 (112.800) E'+520 (40.193} E'= 47,972.3 E' 
 
 From (o), 
 &=V* (l s -\-Fs)=520 (121.061} E = 62,951.7 E' 
 
 ft=0, (1 3 +F;)=560 (112.415) E = 62,952.4 E' 
 
 fc=0 3 (1 4 + F^)=520 (42.481) E= 22,090.1 E' 
 
 Since k r = ~, I P r l r (lk r )=l J P r (l r a r ). Now, we 
 i r 
 
 lr 
 
 have, from (&), for a uniform load, - P r = lw r da r , 
 
 
 
 lr 
 
 :. ^' P r l r (lk r )- = w f d a,. (l r ~a r )== - w r //, and we 
 
66 THE CONTINUOUS GIRDER. 
 
 can write 
 
 r P, i, (i-k,)= 4- . v= 4- *M^= 
 
 v P 2 I, (l-k)= 
 
 Also, 
 
 e v 
 
 IP r (e r a r y= Cw r da r (e,:a r ) 3 = -j- w r 
 
 
 ^ P f (e-a r y=-- w r d a r (e^a r ) 9 =-- -- w r 
 
 
 
 Substituting these values in (/) and (r/), they become 
 
 v=l r 
 
 v=l 
 
 4k lr 3 
 
 v=l r 
 
 v=l 
 
 ' 3 w r ( et__ \ 
 
 ~T~1 lr I 
 
 9=4 
 
APPLICATIONS. 
 VALUES OF H AND H'. 
 
 67 
 
 
 4 
 
 4 
 
 Is 
 
 i. 
 
 
 
 ^P. 
 
 
 QS 
 
 
 l~ 
 
 V 
 
 , 
 
 4 
 
 -, 
 
 i 
 
 *.u 
 
 \ 
 
 \ ^ 
 
 ' ' 
 
 ** ~ 
 
 V 
 
 ^ I 
 
 V 
 
 
 <\ 
 
 <r 
 
 <f 
 
 <f 
 
 < 
 
 <f 
 
 1 
 
 -f- 70 
 
 3 
 
 
 
 2 
 
 
 
 3 
 
 2 
 
 -f- 232 
 
 23 
 
 
 
 12 
 
 
 
 23 
 
 3 
 
 - 6180 
 
 126 
 
 2 
 
 60 
 
 1 
 
 106 
 
 4 
 
 - 8173 
 
 530 
 
 15 
 
 262 
 
 7 
 
 431 
 
 5 
 
 -f- 29815 
 
 -f- 10365 
 
 + 986 
 
 1043 
 
 39 
 
 4 2292 
 
 6 
 
 + 24884 
 
 + H767 
 
 -f- 1459 
 
 4- 8003 
 
 -|- 699 
 
 3781 
 
 7 
 
 -f 20514 
 
 49249 
 
 - 15980 
 
 4 9464 
 
 1090 
 
 15153 
 
 8 
 
 -f 29207 
 
 88767 
 
 35059 
 
 43059 
 
 -13188 
 
 16412 
 
 9 
 
 
 4 117518 
 
 + 71927 
 
 78303 
 
 - 29183 
 
 
 10 
 
 
 + 81258 
 
 + 57000 
 
 4- 112239 
 
 4 76693 
 
 
 11 
 
 
 H- 59538 
 
 4-46518 
 
 4 77063 
 
 + 58589 
 
 
 12 
 
 
 + 45396 
 
 -f 38580 
 
 4 55895 
 
 -f 46830 
 
 
 13 
 
 
 -f 45445 
 
 4- 41523 
 
 4- 65156 
 
 4 59533 
 
 
 
 //; 
 
 5 
 
 H; 
 
 H, 
 
 q 
 
 k 
 
 -67777 w, +58147- W M -W 
 
 .-. H 2 = 4 389,584.9. 
 H 3 = 4- 112,794.0. 
 JZ,^ .- 43,643.8. 
 
 g +51270 w a -150762 w 3 -6514 > 
 
 HJ= 454,105.9. 
 
 H 2 = - 1,039,860.1. 
 H 3 = 331,676.4. 
 
 From (h), 
 
 ^^ p; vp a i s (1k,) o t = 
 
 -56.06 (14,153.7) 520E= - 412,597,339 E' 
 
 <j) XT'" \'Z57 // I- \ ft 
 
 ** 1 ""* * / "1 V * ~"'i ) ''2 
 
 - 2-35,09 (9,058.4) 560 E'= - 356,002,336 E' 
 4- H, fli==+ 389,584.9 (520) '= 4 202,584,148 E' 
 
 I 
 
 -H; o,= +454,105.9 (560) E'= -f 254,299,304 E' 
 
 - 311,716,^23 E' 
 
68 THE CONTINUOUS GIRDER. 
 
 + V v P 7 / 7 I- \ 
 -* 3 ' * J kg V * "'S ) U 2 
 
 -47.41 (4647.5) 560 E ^ - 123,389,266 E' 
 
 | 2 F" - P 9 I (1 k. ) Q.= 
 
 X *= - 2 (50.46) (1 1153.7) 520 E '- - 742,763,530 E' 
 
 X;+X; +#, 0,= + 112,794.0 (560) E,= + 63,164,640 ' 
 
 -Hi #.,=+1,039,860.1 (520) E'-^- + 540,727,252 J5' 
 
 - 262,260^904 ^ r 
 
 - F 4 IP 4 l 4 (l-k 4 ) o,= 
 
 +9.52 (9055.4) 520 E'= - + 44,842,703 E' 
 
 - 2 F; ? P 3 I, (lk s ) 0,= 
 
 -47.80 (4647.5) 240 E'= 106,632,240 E' 
 
 + H t a = -- 43,643.8 (520) E^ 22,694,776 E' 
 ~ H; o,= + 331,676.4 (240) E'=-. + 79,602,336 E' 
 
 ' 4,881,977 A' 
 
 From (75), J r = w r I? O r _ t . 
 
 4 
 
 From (77), B r = - w r I? <> r+) . 
 
 Therefore, A^= 4~ w 2 1 0,= -239,198,375 E' 
 4 
 
 A,= 4- w, II o.= - 84,584,500 E 
 4 
 
 A 4 = - 4- w * V f= - 122,469,568 E 
 
 4 
 
 B<= - w, If 2 = - 131,890,304 E 
 
 B.= 4- w t U o,= 239,198,375 E 
 
 4 
 
 B,= w 3 If A = 36,250,500 E 
 
APPLICATIONS. 69 
 
 From (p), since c t =0. and c a =l, 
 
 C 3 - 
 
 2c 4 fcc s K _ ~ 2,590,360 E' 
 
 ~ ~ 
 
 From (>), since <i,~tf, and d s =l, 
 
 2 d A fad s ft, 1,109,880 E' 
 
 d 5 = -y> - , 
 
 Therefore, 
 
 ( ( 239,198,375 E' 210,013,191 E' ~\ 
 
 101,703,032 E' -oo- 
 ' ' 5.38o 
 
 -311,716,223^' J 
 f 84,584,500 E' 60,224,626 S' 239,198,375*.' ^ (1.524) 
 
 ^ __ ~ 1 j 
 
 =Tl09.HHOJP' 
 
 202,036,278 ' 
 
 262,260,904 E' 
 
 122,469.568 E + 22,147,927 E ' 36,250,500 E ' 
 
 27,029,904 E' 
 
 4,881,977 E' 
 
 5,385 (2.33) 
 
 2,590,360 E 
 
 ( 131,890,304 E') 
 
 M.= 
 
 (- 2,966,675,: 
 !_ ^i 12.565 
 
 ! + 893 ' 13 ' 719 + (-131,890,304) 
 
 I- 163,602,045 J 
 
 M 2 = -2015.66 639.68= -2656. 
 
 The moments, produced by the loads in the respective 
 spans, can be easily deduced, as follows : 
 
 First span loaded 
 
 M ~ u '~* 3 ' 2 R _L ~ V" 1 
 
 ~ (c. ft.) ft, ft, ' TdJfT) * 
 
 > (-101,703,032) (5,385) 
 
 | , -,, , _ 4 , )4 
 
 M,= - 640 
 
70 THE CONTINUOUS GIRDER. 
 
 Second span loaded 
 
 M,= 
 
 ,.- ! (( 239,198,375 210,013,191) (5,385) )_ _ 1 574 
 
 * ** 1,109,880 \ ( _ 202,036,278 - 239,198.375) (- 1,524) ) 
 
 Third span loaded 
 
 >.\ 
 
 - 1 C (- 84,584,500 - 60,224,626) ( 1 ,524) i I 1 4.O 
 
 2 = 1,109,880 j (_ 27,029,904 - 36,250,500) J ....... 
 
 Fourth span loaded 
 
 - __ 
 
 1,109,880 j _j_ 22,147,927 \ Sum- -2656 
 
 THE SAME EXAMPLE WITH I CONSTANT. 
 
 Fr-* 
 
 c 5 ^ 
 
 ' 
 
 ,=0, ^,=1, d,= 3.6, d t = -f 
 
 2901 () 
 
 ,= -- r w s I, A,= -- r v., // ^1,= - 
 
 444 
 
 <= 4- w/ ^* Br=^- ^- w, // ^.,= 
 
 -y 4 4 
 
 Then, 
 
 4 
 
 - 4- < l ? ( 13 -4) 
 
 4 
 
APPLICATIONS. 
 f 919,993.7 w -f 247,162.5 w 
 
 71 
 
 --J -h 247,162.5 v - 68,656.2 v - 2587 
 
 - 35,152,0 w 471,036.8 w 
 The partial moments are easily obtained, as follows : 
 
 First span loaded 
 
 M- 
 
 ~ 
 
 2901.6 
 
 (471,036.8 w) =-. 1088 
 
 Second span loaded 
 
 1 
 
 w + 247,162.5 w)= ... 1554 
 
 Third span loaded 
 t=5IKfi- (247,162.5 v 68,656.2 v)= . . 136 
 
 Fourth span loaded 
 (- $6,126.0 )= 
 
 Sum --= 2587 
 
 Spans loaded. 
 
 Variable I. 
 
 Constant I. 
 
 Difference. 
 
 Per cent, of 
 Variable 
 I Results. 
 
 FIRST. 
 
 1134 
 
 -1088 
 
 -46 
 
 .041 
 
 SECOND. 
 
 - 1574 
 
 -1554 
 
 -20 
 
 .012 
 
 THIRD. 
 
 + 142 
 
 -f 136 
 
 + 6 
 
 .043 
 
 FOURTH. 
 
 90 
 
 81 
 
 9 
 
 .100 
 
 ALL. 
 
 -2656 
 
 -2587 
 
 - 69 
 
 .026 
 
72 THE CONTINUOUS GIRDER. 
 
 EXAMPLE 3. THE SABULA DRAW. 
 
 We now propose to compare the bending moments in the 
 Sabula Draw, considering the moment of inertia as constant 
 and variable, respectively. 
 
 Variable moment of Inertia 
 
 From Fig. 20, is obtained the cross section of each mem- 
 
APPLICATIONS. 73 
 
 ber, and the length of each vertical; the moments of inertia 
 for each section are readily deduced as follows: neglecting 
 the moment of inertia of the section of the member about 
 its own axis, it being very small, in comparison with the 
 moment of inertia of the truss section. 
 
 =(17.7+17.6) (12.5 y- 144= 40= 
 
 =(29.1+22.5) (13.4 ) 2 -144= 60= 
 
 =(29.1+28.5) (14.25) 2 --144= 80= 
 
 =(34.5+31.5) (15.15)^144=110= 
 
 =(34.5+27.5) (16.05) 2 144=110= 
 
 =(33.8+275) (16.9 /-144=120= 
 
 =(31.5+28.5) (17.8 ) 2 -:-144=130= 
 
 =(47.5+36.0) (187 ) 2 --114=200= 
 
 =(47.5+48.0) (200 ) 2 --144+270= 
 
 FIRST SPAN. 
 
 =104.5 
 
 c 4 = 75.5 
 ^,=113.5 
 ^=132.5 
 V=151.5 
 
 From (124), we have, 
 M a -- z4-=-(- 
 
 In which, 
 
 X:=H 2 o, F 2 I P 2 1, (1 h) (> s 
 X,"= - H, 2 2 F t " P, 1, (1k,) 0, 
 
 v-=l 
 
 ( 4 1 3 
 
 v=L 
 
THK CovriNrors UIKDKK. 
 
 v=l t 
 ; /,(/,-/: )+6 Kl(L-F s ) 
 
 9=1 
 
 * f 3 
 
 IT" v A r 
 
 '* - A -IT 
 
 r I, 
 
 
 v=l 
 F s = Z 
 
 I 4 
 
 /_, 1 Ir-, I 
 
 We will first compute those co-efficients that do not depend 
 upon the loading. For a uniform load over all, 
 
 t -=/ 
 
 H != I 4-3 4 
 
 r 4 ( S ) 
 
 H,= r 
 
 9=1 
 
 ' >'~ 
 
APPLICATIONS. 
 
 75 
 
 Ft fat san 
 
 <?/ == 
 
 28.5 
 
 A, = -f 
 
 2.250 
 
 - - 1,484,437 
 
 e s = 
 
 47.5 
 
 A* ~ + 
 
 1.125 
 
 5,726,996 
 
 e 3 = 
 
 66.5 
 
 A* = + 
 
 0.921 
 
 18,011,347 
 
 e. = 
 
 104.5 
 
 A* = 
 
 0.204 
 
 24,327,379 
 
 e, = 
 
 123.5 
 
 A* = -r 
 
 0.173 
 
 40,245,185 
 
 e fi = 
 
 142.5 
 
 A = -f 
 
 0.727 
 
 - 299.773,934 
 
 e 7 = 
 
 161.5 
 
 Ar= + 
 
 0.350 
 
 - 238,099,317 
 
 Multiplying this sum by 
 
 H;. 
 
 Second span 
 
 3w t 
 
 627,668,595 
 obtain 2,615,286 
 
 
 
 , 
 
 
 L\ <> ^ 
 
 e< = 
 
 18.5 
 
 /\* = 
 
 0.052 
 
 1,215 
 
 e s = 
 
 37.5 
 
 A? = 
 
 0.107 
 
 19,043 
 
 e s = 
 
 56.5 
 
 As 
 
 0026 
 
 14,342 
 
 e 4 = 
 
 75.5 
 
 Ax 
 
 0.030 
 
 35,397 
 
 e, = 
 
 113.5 
 
 A,-- 
 
 0.137 
 
 422,318 
 
 C 6 = 
 
 132.5 
 
 A-= 
 
 0.166 
 
 691,825 
 
 e 7 = 
 
 151.5 
 
 A;-- 
 
 0.333 
 
 - 1,709,655 
 
 - 2,893,795 
 
 
 Multiplying this sum by ^-, we obtain 7 
 
 4 
 
 First span l a =180- 
 
 e<= 28.5 A^ + 2.250 A/ 
 
 ,= 47.5 A='+ 1-125 A 
 
 e.^ 66.5 A.? + 0.921 A. 
 
 c.= 104.5 A, -f 0.204 Ax 
 
 *,=123.5 A= -f ai 73 A, 
 -f 0.727 
 
 u-.,~H* 
 
 = -h 
 
 52,085 
 120,568 
 270,847 
 232,797 
 325,871 
 
 6^=142.5 
 
 7 =-161.5 A r = 4- 0.350 
 
 AX- 4- 
 
 A, ^ + 
 
 A* <S= -f 2,103,676 
 
 A; J= + 1,474,299 
 
 /f=:3*2,400 ) 4,580,143 ( 141.3(5 /' . 
 
70 THE CONTINUOUS GIRDER. 
 
 Second span L= 180 . 
 
 a , i' 
 
 { '* &> 's "" ('!) I \ 'l '2 V'2 '17 I, / \ 
 
 <?,= 18.5 -0052- 8,951 84,225 
 
 e, 37.5 -0.107- 69,967 314,404 
 
 e~_ 56.5 0.026- 35,440 102,657 
 
 r/ 75,5 4= 0.030- 66,521 140,725 
 
 r, 113.5 0.137 552,403 - 758,695 
 
 r,, 132.5 A- -0.166 801,442 - 950,321 
 
 c, 151.5 0.3331,813,227 -1,936.284 
 
 =32,400 ) -3,347,951 /!^-32,400 ) 4,287,311 
 
 103.33=*V 132.32=*; 
 
 Next find the values of X* and X,', for a uniform load 
 over all. 
 
 o,=.27(t E', (t y 40 E', in which E' ~n K. 
 + H. o,= - 723,449 (270) E' w. = - 195,331,230 w, K' 
 
 F, . 103.33 
 
 -- j- w* 1: <>,= H -- - - 
 
 (32,400) (270) " MJ, = + 451,965,420 w s E' 
 
 + 256,634,190 IP, A 1 ' 
 
 _;/ VA, // ?/ /*= -141.36 (32,400)(40)w, '= - 183,202,560 w t E ' 
 -H, o s = -f 2,615,286 (40) w t E'= + 104,611,440 w t E' 
 
 78,591,120^ E' 
 &=40 (180+141.36) '+270(180 132.32) E'= + 25,728 E'. 
 
 A= ~ "'" ^' 2 70 "= 393,660,000 E' w,. 
 
 4 
 
 B,= -- 4- ^ // 40 E'= 58,320,000 E' w,. 
 4 
 
 Then, 
 
 ( 393,660,000 w,E'} 
 1 \ 58,320,000 w, E ' \ 
 " - 5T4M7? 78,59i;i20 ,; ' "< lf ^ 
 
 H 256,634,190 t 3 B{ 
 
APPLICATIONS. 
 
 77 
 
 Jfirst spun alone loaded 
 
 f 58,320,000 w/i 
 - 78^91,120 W| 
 
 I 137,000,000 w, } 
 
 Second span alone loaded 
 
 ( 393,660,000 w,} 
 M ^ + 256,834,190 w, I 
 
 - 137,000,000 w s J 
 
 EXAMPLE 3o. 
 
 This is the same as Example 3, with the moment of inertia 
 considered as constant. 
 
 From (142), 
 
 720 
 
 A,= w. 11= 145,800 w,. 
 
 4 
 
 B,= 7- w< lf--= 145,800 w,. 
 
 4 
 
 Therefore, 
 M a - ~ r and if w,=w, F =w 1 M s = - 4050 w. 
 
 No. of 
 
 Loaded 
 Span. 
 
 FIRST. 
 
 Bending Moment 
 Variable 1. 
 
 - 2661 w, 
 
 SECOND. 2661 w. 
 
 BOTH. 5322 w 
 
 M 2 
 
 Bending Moment 
 Constant I. 
 
 - 2025 w, 
 
 2025 w 
 
 4050 w 
 
 Difference. 
 
 636 w, 
 
 636 w 
 
 1272 w 
 
7-S THE CONTINUOUS GIKDEK. 
 
 EXAMPLE 4. CONCENTRATED LOADS. 
 
 Let us consider the Sabula Draw again, but with single 
 concentrated loads, instead of uniform loads. Supports 
 level. a,=7. o. ? /?>. 
 
 See pages 73 and 74 for the value of M,. 
 
 From Example 3 : 
 
 #=40 (180+141.36) #'+270 (180132.32) "= + 25,728 E'. 
 F t "=-- + 141.36. #=103.33. F,= -132.32. 
 ^=270 E'. ^=40 E'. 
 
 The values of H 3 and Hi depend upon the position of 
 the loads. Let us take a load in the first span 7' from the 
 left support, and one in the second span 123' from the 
 left support; then we have for Hi and H,, the following: 
 
 First span- l,=180'. a,==7'. 
 
 V=l 
 
 *= 28.5 ea<. 
 
 e,= 47.5 
 
 e s = 66.5 + 9.5 A*= + 0.921 88 P, 
 
 e 4 = 104.5 + 47.5 , = + 0.204 680 P t 
 
 + 66.5 A 5= + 0.173 1,292 P, 
 
 + 85.5 A<r= + 0.727 - 10,097 P, 
 
 + 104.5 t\!= + 0.350 8,068 P, 
 
 - 20,225 P,= 
 NOTE. e r must always be greater than or equal to a,. 
 
 Second span- 1=180'. a x =123'. 
 
 v=l 
 H,= I A | JjPg (^ a >I ^ t -ii ^ J P, (c,a 2 Y | 
 
APPLICATIONS. 79 
 
 e, - 18.5 
 
 e, = 37.5 ea, 
 
 e 3 - 56.5 
 
 ej, - 75.5 
 
 e s == 113.5 
 
 e, == 132.5 4- 9.5 A 5 = -0.166 13 P 
 
 e 7 = = 151.5 + 28.5 = 0.333 172 P 2 
 
 185 P,=H, 
 NOTE. e, must always be greater than or equal to a s . 
 
 r -f #, /= - 185 (270) E' P 2 49,950 E' P g 
 
 * - - F: P, (l s -n,) (),= 
 
 ( H- 103.33 (57) (270) E' P,= + 1,590,248 E' P 3 
 
 98 E' P 2 
 ( H; 3 = + 20,225 P ; (40) E'= + 809,000 E' P, 
 
 ( - 141.36 (123) (80) E' P<= - 1,390,928 E' P, 
 
 ~581~,928 Y' P; 
 
 From (i) and (J), 
 
 / 
 
 A p 12 ( <r> i, & Z-2 i .3\ ft T? r T 
 
 ./lo JL ? t"> \ & A/ d A/ /V I C/ Xl/ _//. 
 
 Or, by Table I, 
 
 A,= - P // (0.285,144) 270 E' - - 2,494,269 E f P,. 
 ^/= - P^/ (0.285,144) 40"= 369,520 E' P,. 
 
 Therefore, 
 
 f 2,494,269 ^' P/l 
 
 M _ I j + 1,540/298 E ' P, { __ j -18 5 P 2 \ 
 
 ~ 51456 E'} - 369,520 E' P, f ~~ 1 18.5 P, } 
 581,928 E' P; 
 
 Or, if P ; =P, ? 
 3f 3 ^ - 37.0 P. 
 
 First span alone loaded 
 
(SO THE CONTINUOUS GIRDED. 
 
 Or, 
 
 51,448/>=18.5/>, 
 
 ' 556 , 
 
 Second span alone loaded 
 
 Or, 
 
 - 953,961 J>, = 18.6 P, 
 
 Since IP, and IP, are symmetrical about the center of the 
 draw, the moments over the center pier should be equal, 
 as shown above. If our work was absolutely correct, and 
 decimals had been used to several places, the quantities 
 in the parentheses above would have been the same ; as it 
 is, the quotients are practically equal. 
 
 * EXAMPLE 4 BY GRAPHICS. 
 
 For any load in the first span of a two span girder, we 
 have, from Greene's Trusses and Arches, Part //, 
 
 EI El El 
 
 i 
 In which: 
 
 distance from the left support to the ordinate y, 
 of the equilibrium polygon A C IB. Fig. 21, page 72. 
 
 ordinate of the polygon A C IB. 
 
 distance from the left support to the ordinate y' f of 
 the polygon A IB IB,. 
 
 ordinate of the polygon A IB IB,. 
 
 distance from the right support to any ordinate 
 a of a polygon in the second span, similar to AS _Z?,. 
 
 * For this excellent graphical method, the author is indebted to R. H. 
 Brown, C. E., First Assistant Engineer of Boston Bridge Works. 
 
APPLICATIONS. 81 
 
 ordinate of the above polygon. 
 
 jT The moment of inertia of the cross section of the girder 
 at any ordinate y',. 
 
 2 
 
 J The moment of inertia of the cross section of the girder 
 at any ordinate y' s . 
 
 _Z? The modulus of elasticity. 
 
 / 2 
 
 Since l,=L=l, and 1=1=1, we have, considering JE as 
 constant, 
 
 Now, - ' ^ equals an area multiplied by the distance 
 
 of its center of gravity from the left support, and hence we 
 
 or ?/ 
 may write 2 '-7^ = A a, in which A represents the area 
 
 and a the c. g. distance. 
 
 In like manner, # - ''' . ' =2 B b. Therefore, we may 
 write, ^1 a 2 B b. 
 
 With any scale, lay off the horizontal line A 12, Fig. 21, 
 and divide it into panel points at ?>, r, d, etc. 
 
 With any scale, preferably a large scale, lay off' a load 
 line A ID, equal unity, and assume If also equal unity, 
 then, assuming the pole in such a position that its closing 
 line will be horizontal, construct the equilibrium polygon 
 A C IB and scale its ordinates (the lengths are given in 
 Fig. 21.) 
 
 Not knowing the proper position of the closing line, let 
 us assume a position as A 12, making B B 2 =12.8 9 and 
 scale the ordinates b b 3 , c C 2J etc. 
 
 Divide each ordinate of the polygon A C B by its proper 
 I (given in per cent, of the center I above Fig. 21 ), and lay 
 
82 THE CONTINUOUS GIRDER. 
 
 off the results downward from A It, forming the polygon 
 Ab 5 c 5 . . . . B. The ordinates are 80.00, 118.18, 
 
 etc., as shown in Figure 21, page 72. 
 
 Now, find the area of this figure, either by computation 
 or the planimeter, and also its center of gravity. The center 
 of gravity is readily found by cutting the polygon out of 
 stiff card board and balancing it upon a needle point. 
 
 The area = 109 15.1= A, and a=nr>f>. 
 
 Proceed in like manner with the polygon A B B,, de- 
 ducing the figure A b- ; c ;; .... />,, which has an area of 
 ;>.' f (>!t.(>l=-B', and V =100.8. 
 
 Let B B, represent the true magnitude of the pier ordi- 
 nate //, then from the triangles A. B B, and A J> B n we 
 have, 
 
 y,',:y lt '.'.c c 2 :e c,, or, c e,= r e e 2 . Then, 
 
 x ,(,) ?/ v x 2 (ec.^ y^ , , 
 
 f r - j , UI ,/)!/ r 1J v . 
 
 I Vo I 'Un 
 
 -, v,> , , , j , (Ad) '/,', 
 
 Hence, A a=2 -~- II' b', and y a = ^ ^/^ , 
 
 Which becomes 
 
 10915.1X65.5X12.8 
 2X2469.61X100.8 
 
 Since H=l and Pf=l, the bending moment over the 
 center pier is 18.38 JP*. We obtained, by computation, 
 - 18.3 JP; V , which shows the graphical method to be accu- 
 rate enough for all practical purposes. 
 
 It would have been more correct to have taken the ordi- 
 nates y at the points where the values of J change, but the 
 result would have been but little different. 
 
 The above computations can be very readily made by 
 means of Thatcher's Calculating Instrument. 
 
APPLICATIONS. 
 
 83 
 
 A. rf 
 
 Fig. 17, page 72, and Fig. 18 are diagrams for the Sabula 
 Draw, with a concentration at each apex in both spans. 
 
 A aB b. or A a--= -^4- B' 6. Therefore, 
 V* 
 
 _(Aa)y 65568 X 148x94.6 
 B' b' 36865X2045 
 
 Since Hfif^ the bending moment over the 
 
 pier is 121,800x50-= 6,090,000 
 
 Considering / as constant, the bending 
 
 moment is 4,732,000 
 
 7^58~000:= Diff. 
 
APPENDIX. 
 
 * The general equation of the elastic line can be deduced 
 as follows : 
 
 Vertical forces acting upon a girder cause a change of 
 shape, lengthening the originally parallel fibres on one side, 
 and shortening or compressing them on the other. Between 
 the lengthened and shortened fibres there is a plane which 
 undergoes no change in length ; the centre line of this plane 
 is called the neutral axis or the elastic line. 
 
 Thus, in Fig. (<r), m o is the neutral axis, the fibres 
 above being compressed, and those below lengthened. Upon 
 the three following hypotheses we shall deduce the equation 
 of the elastic line. 
 
 I. All planes perpendicular to the axis before the bend- 
 ing or flexure, preserve, during the bending, their perpen- 
 dicularity and their forms as planes. 
 
 II. The change in length of a body subjected to a force, 
 is, within certain limits, called the elastic limits, propor- 
 tional to the intensity of the force. 
 
 See Merrimaii's Theory and Calculation of Continuous Beams. 
 
86 APPENDIX. 
 
 III. The change of shape is so small that the length of 
 the neutral axis is sensibly the same as its horizontal pro- 
 jection. 
 
 In Fig. (a), we have a longitudinal section of a portion 
 of a bent beam ; the two planes (i 1) and (I e origininally 
 parallel, remaining perpendicular to the neutral axis or 
 elastic line in 0, and intersecting in c the centre of curva- 
 ture. Hence, drawing / y parallel to a b through o, the 
 lines /f/, (/ e, etc., denote the elongation of the fibres, and 
 we see, from the figure, that 
 
 od:od'::df:d' f (22a) 
 
 or the change of length in the fibres is proportional to their 
 distance from the neutral axis. This is a consequence of the 
 first hypothesis. 
 
 Designating by If and If the force acting in the fibres 
 (If and <l' f , the second hypothesis gives 
 
 H:H'::df:d'f (226) 
 
 Combining (22a) and (226), we have 
 H:H'::od:od' (22r) 
 
 or, the horizontal forces are directly proportional to their 
 distances from the neutral axis. 
 
 Denote the distance of any fibre from the neutral axis by 
 2, the stress in it by II', the distance of the remotest fibre 
 by e, and its stress by H ; then, from (22r), we obtain 
 
 H f :H::z:e r QT,H f =-~- (22rZ) 
 
 Thus far the cross-section of the fibres has been considered 
 as unity. If the actual area is ft-, the force is L . Each 
 
 of these forces If tend to turn the beam around o with a 
 lever arm o d' or 2, hence the moment of the force is 
 
 H a z II a z 3 ~ ,, ., 
 
 ' X* = , and the sum of all the moments is 
 
APPENDIX. 87 
 
 M t = Zazr .................. (22c) 
 
 c 
 
 M x meaning the bending moment at any section x r of the 
 beam in the span l r . 
 
 Since - ft & is the expression for the moment of inertia 
 of the section ft ?>, (22r ; ) becomes 
 
 M x = ..................... (22/) 
 
 (5 
 
 or the moment of the internal forces equals the stress in the 
 remotest fibre times the moment of inertia of the section 
 divided by the distance of the remotest fibre from the neutral 
 axis. 
 
 The line d f denotes the change of length in the fire 
 a d, due to the force JZ, hence, if J be the co-efficient of 
 elasticity, 
 ad:df::E:H .................. (220) 
 
 Designating the radius c o by ?-,., we have, from similar 
 figures, o rl/'and c ft d. (-in o=a d). 
 a d:df::r,.:e ................... (22 h) 
 
 H E H r , 
 
 - : or ' c = - 
 
 Substituting this value of e in (22/), we have 
 
 r.- 
 
 The radius of curvature of any plane curve, whose length 
 is tt, and co ordinates x r and y n is 
 
 d K 
 7r d x,. d* y f ' ' 
 
 According to the third hypothesis, d u,.=-d x n and (2 
 becomes 
 
 !-,.= 'i^- (22m) 
 
88 APPENDIX. 
 
 Substituting (22/n) in (22A;), it becomes 
 
 ,.,- 
 
 E I x 
 
 Which is the differential equation of the elastic line, appli- 
 cable to all bodies subjected to flexure which fulfill the con- 
 ditions imposed by the third hypothesis. The values of J 
 and I may be different for each and every section. 
 
 If (22) be integrated, it becomes 
 
 ''/ / 
 ;//., 1 CM* < l V . i 
 
 o 
 
 If x r =0, then C = = t n and we have 
 
 o A * 
 Integrating again, (24) becomes 
 
 If x r =6>, then C'^=h n and we have 
 
 X r r X r 
 
 1 rM r d x f c 
 
 y r =h r +t r x r + -g-Jr- - J 
 
 o T * o 
 
 ( 23) 
 
 ^- = t f +- r I ~^ - (24) 
 
APPENDIX. 
 
 89 
 
 By examining Fig. 2, which represents a continuous gir- 
 der having a variable cross-section, and consequently a 
 variable moment of inertia, the following integration will 
 be clearly understood : 
 
 X r r 
 
 * T 4 4-- c C M x d x r 
 
 * Integration 01 I 
 
 i. 
 
 r r ., e 2 
 
 /M x d x r 1 / , r , 1 c\Jr j 
 
 - = \ M x d x r + I M x d x r . . . 
 T T J T J 
 
 o lx lo o Ji e t 
 
 x 
 
 1 / r 
 
 . . . + -\M X d 
 
 iA 
 
 x r 
 
 e x 
 
 But, 
 
 1 f* r 1 r* r 1 f* T 
 I M x d x r = I M x d x r I M x d x r . . . . 
 
 7 T T 
 
 Therefore, we can write for (27), 
 
 * See Weyrauch's Continuirlichen und Einfachen Trager, p. 168. 
 
 (27) 
 
 (28) 
 
90 APPENDIX. 
 
 x r e, e, 
 
 CM X d x,.-- 4- (*M x dx r -\- 4- faL d x,.... 
 
 I i* 'i-i *<o 
 
 e* 
 
 ..+ -4- CjLf x d'x r 
 I " 
 
 - l xl n 
 
 ( M x d x r CM X d x,. . . . - (M,d x. 
 
 j J J <J T J 
 
 2i o o * o 
 
 
 
 Which reduces to 
 
 />* or, 
 
 & r 
 
 x r 
 -f CM x d x (29) 
 
 i A 
 
 . (30) 
 
 * T '-> T > 
 
 v=x r o 
 
 Substituting (30) in (26), it reduces to 
 
 x r x r 
 y r = b + t r x r + -jr; I d x r M X d x r 
 
 T ^o o 
 
 v=l x r e r 
 
 <3i) 
 
 r T 
 
 v=x r v 
 
 From 8, we have M x ^M r ^S r x r -l' P r (x r a r ) . . a<x . . (8) 
 
APPENDIX, 91 
 
 Hence, 
 
 ''', 
 
 | V, fl x r =M f x,. + ~ $,. 4 - ^ l\ ( x~a,y . . . a<x . . . (32) 
 o 
 
 And, 
 x r x f 
 
 fd- x r p d x r = ~M r 4 + -^ S, x: 
 o o 
 
 Z p r < r,.a r ) . . a<x . , (33) 
 
 Now, 
 
 x r e,. e r c, 
 
 j'd x, j*M x dx,.= j*d x ^M x d x, 
 
 x r 
 
 Cdx r M x dx, (34) 
 
 J */ 
 
 00 00 
 
 x r t 
 
 e,. o 
 But, 
 
 -^ Z P r (e v -a,Y . . . < . . . (35) 
 
 Hence, 
 
 e r e,. 
 
 Cd x r (*M r d x r = 4- M r ef. + S r e], 
 J J 2 6* 
 
 o o 
 
 -|- JP P (e, a r ) a . . . a<e . . . (36) 
 
 x,. 
 ldx r =x r e v , 
 
92 APPENDIX. 
 
 Therefore, 
 
 x r c,. 
 
 Cd x r CM, d x r -- 4- (x r -e.) 1 2 M c, s, <: 
 
 . / ./ ~ ^ 
 
 <?,, o 
 
 -ZP,(e v -a>)'} ..... (37) 
 
 Substituting (33), (34), (36) and (37) in (31), it becomes 
 
 y r =h r + t r x r + \ -I .I/,. **+S r x*-l' P f (x r -0r?\ 
 
 6 El, ( 
 v=l 
 
 4- --7- ? 
 
 6EI >V=*r J ~ T * 
 
 +S r el-l'P r (e r -a,.Y-]t . . .(38) 
 
 Which reduces to 
 
 i r =h r -\-t r x r + \3 M r aj-f- S, x*. I P r (x r a r ] a 1 
 6EI n ( 
 
 + ?' [i - i- 
 
 6EJ X v=x r ^/ r _ ; /,. 
 
 - I P,. (e-a r )*-i (x-e,.} 1' P, (r-n,.y [-'...; (39) 
 
 If we make x r -l r , then y r =h r+1 , e f e t < and I x =Ii. 
 Let r . then, from (39), we have 
 
 r+l = h, + t r l r + - \8 M r tf-f S, lf-*S P,. (/,-",.)< 
 
 6' E I, 
 v=l 
 
 M, e,. (~> lr-e v + $ r r r (3 /,.-^ e,.) 
 
 - ^'P, (/,.-,)' v -^ (/,-r,,) ^ />, (c.-a,)^ . . . (40) 
 From (10), 
 
APPENDIX. 93 
 
 But, since AY 1 , this becomes 
 
 (10a) 
 
 Substituting (10a) in (40), and solving for f we obtain 
 h f+ . -h,. 1 ( , 
 
 -f I P r a r ft. a r ) (# /,. a r ) 
 't?=jf 
 
 - ^' A, | 2 M;. * (5^,-^^- -^-) 
 
 i 1 yf" 9 / fo ^ *'V \ 
 
 | 7lf,. + / ^ (o ) 
 
 + 4 (> -^ L ) " Pr (Ir-a,.)? P r (e~d r Y 
 
 -3 (/, e r ) I P r (e a,.) 2 1 (41) 
 
 Since O,.=AY /,, a, (la r ) (2 l r a r )=l? (2 k r 3 ^+K), 
 / r a,. 7 ( . (1 AY), and (41) reduces to 
 
 tr 7 ... I - r ,--N ,-H r 
 
 - ->' (/, e,.)l' P, (e a*)* I . . . . (42) 
 Returning to (24), 
 
 '',": =^ 4rf-^4^-: .020 
 
94 APPENDIX. 
 
 Substituting (32) and (35) in (30), and the result in (24), 
 it becomes 
 
 = t r + ^~-r \ 2 M r x~ r +S r *t-SP r (*,-a,)> 
 2 E I, ( 
 
 v=l 
 
 3/ r v-f-S r <-!' P r (<v_ a,)* . . . .(43) 
 
 Making x,.=l ri then -^|- /,. ,,, c^c h and/,. ),, and 
 
 substituting for S,. its value from (10), t, from (42), and 
 A 1 , I, for rr, in (43), it becomes 
 
 I - } 2 M r 1,+M,. , /, 
 
 J^j _/ . 
 
 - S L r \;>M f e, (3 13 e r 
 '; E I, /, , l 
 
 ,.l; (2 fc r -5 
 
 
 v=l 
 
 +9 J P, (lk r ) 1..3 l t . 2 P ; . (r r a,)* \ . . . . (44) 
 
 ) 
 
 Which reduces to 
 
 hr+l'hr + _!_ | M ^^ M ^ s p ^ ^ft) 1 
 
APPENDIX. 95 
 
 If we were to suppose loads in the v 1 th span at distances 
 <7,._ ; --Av_/ /,._/ from the left support r-l, we would find in 
 a similar manner, or by decreasing the subscripts of (45). by 
 unity, 
 
 k-h,._. 1 
 
 -(46) 
 
 Equating (42) and (46), we obtain 
 
 ^~ hf + -2 M r l r M r+l lZ P, g (^ fc r 
 
 -- ^ A. ^ f ' - ^r + 4- } 2 M - ~ - A. 4 
 
 ' 
 
 - 
 
 ' ; y 
 
 v=l 
 
 ?P r l r (i-k r ) 4 i' A, 
 
APPENDIX. 
 
 r / 
 r l^+S P,_, /;_, (fc^-#_;) -f i- "7 
 
 >S Pr-, U (1kr-,) - - ^ A /-/-'' ? I*,-, (c,-a r _,Y 
 
 V=lr-< 
 
 V-=l 
 
 v=l r _, 
 Let 
 v=l 
 
 ? A, f*- ^r- + 4] =F ' ^ 48) 
 
 v ', ',- y 
 
 t;=l 
 
 - (50) 
 (51) 
 SPf (er - ai Y-H r . .(52) 
 
 v P ( P _ /^ ^ 3 c ? ^ ^ 
 
 f ^j~ ? P,. (ecty t H r > . . (53) 
 
 Substituting (48), (49), (50), (51), (52) and (53) in (47), 
 transposing and reducing, we obtain 
 
APPENDIX. 97 
 
 o r Or-* ^=|-^- f 2 M r /, ",_,-{- 1/,^ /, C,-hS P r 1; (2 k r 3 ^ 
 
 ,'!' P r l r (lk r ) O r _, 
 
 ;_ t M,_, Br+FZ-t 2 M r 9 r +F^ 
 4 0=0 , . . (54) 
 
 Which reduces to 
 
 r+1 () o r _,= - o r o r _ t 
 
 I (,._/ t,. 
 
 -r p, // (^ ^ r -^ +) O,._F; i p r i r (ik r } o r _ s 
 
 (55) 
 Let 
 
 F) ...... ... (56) 
 
 -?P r l;(2k,.-< 3 ](*+](?) o r _ t =A, .......... (57) 
 
 - I P r l~ (kty (> r+! =B, ........ ....... (58) 
 
 -F; i 1 p r i,. (iic r ) 0,-,+Hr o r _, =x; . . | . . . . (59) 
 
 -2 F;_, Z P,_ ; U (i tU) ^-^--/ ^r=-X^/ ) ~ ' (59a) 
 A ................... (60) 
 
 =^ f . . . ................ (62) 
 
 Then we can write 
 
 M l ._ l ^_ l +2M r ft^M r+ . l ^ f =A r ^ r B r _ i ~}-Y r ^X 1 ..... (63) 
 
 Which is the general form of the Theorem of three moments. 
 It expresses the relation between the bending moments over 
 three consecutive supports in terms of the loads, spans, E t 
 Tand /i. An equation of the form of (63) can be written 
 for every support, and, as, if the girder rests on the supports, 
 the moment at the firgt and last support is zero, we shall 
 
98 APPENDIX. 
 
 have as many equations as there are unknown quantities or 
 bending moments, and hence we can determine their values. 
 
 Let there be s spans in the continuous girder represented 
 in Fig. 1, and let the /'"' span alone be loaded, then by the 
 three moment theorem, the equation for each support is as 
 follows: . 
 
 .! M, ^ 4 ^f & i 
 
 M, & \ 2 ^r ! & 4 M 4 & =o 
 
 XX XX XX 
 
 M,.- t '&_ 4- * M f K 4 M*, K K 43 A; \ ... 
 
 M r r 4- 2 M f+l ^ 4- M r+2 ^=B r 
 
 XX XX XXX 
 
 Multiplying the first equation of (64) by #,, the second 
 by c ;i} the i^* by r ,, etc , we obtain, after reduction, 
 
 ; c s _;H-^ r s /SJJ^r, <?rH X O--^, O B, r,.^ . . (65 ) 
 
 Now, supposing it is desired to determine the value of 
 J!C, it is only necessary to impose such conditions upon the 
 multiplier c that all terms shall reduce to zero, excepting 
 those containing JfcT,. Evidently, the co-efficients must 
 separately equal zero, or 
 
 -<y''s 4 c ;! fa=o *\ 
 c, p" 3 4 c, fa 4 c 4 ,=0 
 r s ^ 4 ^ c. fc - rj.ft=M9 - ........ (66) 
 
 r m _, fi' in _, 4- 2 C M [i' m 4 f l)l+/ j5 IH =0 J 
 And, for Jjf,, we have, at once, 
 
 r ) f ,^ rCf -:B, c ^ 67 
 
 ' ' ' ' 
 
APPENDIX. 99 
 
 In a similar manner, multiplying the last equation of 
 (64) by d s , the last but one by d ;i , and so on, we obtain 
 
 ; d., fc (I. &L<=0 "| 
 
 .? - 2 = 
 
 ..... (68) 
 
 ,,,_, _,,, r , 
 And 
 
 , _ (r+d^+Ard^+Brd^, 
 
 (2&rf.+ cU.,y = -d. +l fi 
 
 From (65), we see that M :i = "-^- M SJ from (66), c 3 = 
 
 2 } ft' 
 
 ; hence, assuming c f =0 and c,=l, we have M 3 =c s 
 
 Pa 
 
 ^- Mgj and in a similar manner we find that M 4 =c 4 '^'^ 
 
 Pa Pa PS 
 
 3/ 2 , M B =c B '* ( r j, ' r j, M g , or in general for any support on the 
 
 Pz Ps PA 
 
 left of the loaded span or spans. 
 
 Jm-t jir /rn\ 
 
 -7 --- M s ............. (/Uj 
 
 Pa t j :; i-i 
 
 In a like manner, we obtain for any support on the right 
 of the loaded span or spans, 
 
 m>r v 
 
 IT f l . P-J t*--' Pm I T /ni\ 
 
 M m =d s _ m+2 - - ^ - - M fl ............ (71) 
 
 /-_/ Ps 2 ' l j ,i, 
 
 Substituting (69) in (70), and (67) in (71), we obtain 
 
 /7O\ 
 
 /; 7?rtt \ ~+> ' ~+> (72) 
 
 < 73 > 
 
 In (72), as in must always be less than -/*-r i, r can have 
 values from .s to in. 
 
100 APPENDIX. 
 
 In (73"), as m must always be greater than f, /' can have 
 values from 1 to m 1. 
 
 Hence, adding (72) and (73), and substituting X' -f X,_, 
 for X n we have 
 
 
 From (J), we can obtain the bending moment over any 
 support m of a continuous girder of any number of spans s, 
 of any lengths as l t , I, . . . 1 8 , supports at any levels, the 
 moment of inertia I constant or variable, the modulus of 
 elasticity JE being constant, and the loads being placed at 
 pleasure. 
 
 Note that '</?_ &<*-/, /C/^*, and / 5 fj _ / >/? w . 
 
TABLE I. 
 
 101 
 
 *4 
 
 t-f 
 
 
 a 
 
 /l "7~ 
 
 A- -A" 
 
 
 Hr 
 
 A- -A;* 
 
 
 
 
 
 oe-o 
 
 059 784 COO 
 
 .940 
 
 .120 
 
 118 272 000 
 
 .880 
 
 .001 
 
 COO 999 999 
 
 .999 
 
 ' 61 
 
 060 773 019 
 
 39 
 
 21 
 
 119 228 439 
 
 79 
 
 2 
 
 001 999 992 
 
 8 
 
 62 
 
 061 761 672 
 
 38 
 
 22 
 
 120 184 152 
 
 78 
 
 3 
 
 002 999 973 
 
 7 
 
 63 
 
 062 749 953 
 
 37 
 
 23 
 
 121 139 133 
 
 77 
 
 4 
 
 003 999 936 
 
 6 
 
 64 
 
 063 737 856 
 
 36 
 
 24 
 
 122 093 376 
 
 76 
 
 5 
 
 004 999 875 
 
 5 
 
 65 
 
 064 725 375 
 
 35 
 
 25 
 
 123 046 875 
 
 75 
 
 6 
 
 005 999 784 
 
 4 
 
 66 
 
 065 712 504 
 
 34 
 
 26 
 
 123 999 624 
 
 74 
 
 
 006 999 657 
 
 3 
 
 67 
 
 066 699 237 
 
 33 
 
 27 
 
 124 951 617 
 
 73 
 
 8 
 
 007 999 488 
 
 2 
 
 68 
 
 067 685 568 
 
 32 
 
 28 
 
 125 902 848 
 
 72 
 
 9 
 
 COS 999 271 
 
 1 
 
 69 
 
 068 671 491 
 
 31 
 
 29 
 
 126 853 311 
 
 71 
 
 .010 
 
 009 9S9 000 
 
 .990 
 
 .070 
 
 069 657 000 
 
 .930 
 
 .130 
 
 127 803 000 
 
 .870 
 
 11 
 
 010 998 669 
 
 89 
 
 71 
 
 070 642 089 
 
 29 
 
 31 
 
 128 751 909 
 
 69 
 
 12 
 
 Oil 998 272 
 
 88 
 
 72 
 
 071 626 752 
 
 28 
 
 32 
 
 129 700 032 
 
 68 
 
 13 
 
 012 997 803 
 
 87 
 
 73 
 
 072 610 983 
 
 27 
 
 33 
 
 130 647 363 
 
 67 
 
 14 
 
 013 997 256 
 
 86 
 
 74 
 
 073 594 776 
 
 26 
 
 34 
 
 131 593 896 
 
 66 
 
 15 
 
 014 996 625 
 
 85 
 
 75 
 
 074 578 125 
 
 25 
 
 35 
 
 132 539 625 
 
 65 
 
 16 
 
 015 995 904 
 
 84 
 
 76 
 
 075 561 024 
 
 24 
 
 36 
 
 133 484 544 
 
 64 
 
 17 
 
 016 995 087 
 
 83 
 
 77 
 
 076 543 467 
 
 23 
 
 37 
 
 134 428 647 
 
 63 
 
 18 
 
 017 994 168 
 
 82 
 
 78 
 
 077 525 448 
 
 22 
 
 38 
 
 135 371 928 
 
 62 
 
 19 
 
 018 993 141 
 
 81 
 
 79 
 
 078 506 961 
 
 21 
 
 39 
 
 136 314 381 
 
 61 
 
 .020 
 
 019 992 000 
 
 .980 
 
 .080 
 
 079 488 000 
 
 .920 
 
 .140 
 
 187 256 000 
 
 .860 
 
 21 
 
 020 990 739 
 
 79 
 
 81 
 
 080 468 559 
 
 19 
 
 41 
 
 138 196 779 
 
 59 
 
 22 
 
 021 989 352 
 
 78 
 
 82 
 
 081 448 632 
 
 18 
 
 42 
 
 139 136 712 
 
 58 
 
 23 
 
 022 987 833 
 
 77 
 
 83 
 
 082 428 213 
 
 17 
 
 43 
 
 140 075 793 
 
 57 
 
 24 
 
 023 986 176 
 
 76 
 
 84 
 
 083 407 296 
 
 16 
 
 44 
 
 141 014 016 
 
 56 
 
 25 
 
 024 984 375 
 
 75 
 
 85 
 
 084 385 875 
 
 15 
 
 45 
 
 141 951 375 
 
 55 
 
 26 
 
 025 982 424 
 
 74 
 
 86 
 
 085 363 944 
 
 14 
 
 46 
 
 142 887 864 
 
 54 
 
 27 
 
 026 980 317 
 
 73 
 
 87 
 
 086 341 497 
 
 13 
 
 47 
 
 143 823 477 
 
 53 
 
 28 
 
 027 978 048 
 
 72 
 
 88 
 
 087 318 528 
 
 12 
 
 48 
 
 144 758 208 
 
 52 
 
 29 
 
 028 975 611 
 
 71 
 
 89 
 
 088 295 031 
 
 11 
 
 49 
 
 145 692 051 
 
 51 
 
 .030 
 
 029 973 000 
 
 .970 
 
 .C90 
 
 089 271 000 
 
 .910 
 
 .150 
 
 146 625 000 
 
 .850 
 
 31 
 
 030 970 209 
 
 69 
 
 91 
 
 090 246 429 
 
 9 
 
 51 
 
 147 557 049 
 
 49 
 
 32 
 
 031 967 232 
 
 68 
 
 92 
 
 091 221 312 
 
 8 
 
 52 
 
 148 488 192 
 
 48 
 
 33 
 
 032 964 063 
 
 67 
 
 93 
 
 092 195 643 
 
 7 
 
 53 
 
 149 418 423 
 
 47 
 
 34 
 
 033 960 696 
 
 66 
 
 94 
 
 093 169 416 
 
 6 
 
 54 
 
 150 347 736 
 
 46 
 
 35 
 
 034 957 125 
 
 65 
 
 95 
 
 094 142 625 
 
 5 
 
 55 
 
 151 276 125 
 
 45 
 
 36 
 
 035 953 344 
 
 64 
 
 96 
 
 095 115 264 
 
 4 
 
 56 
 
 152 208 584 
 
 44 
 
 37 
 
 0-)6 949 347 
 
 63 
 
 97 
 
 096 087 327 
 
 3 
 
 57 
 
 153 130 107 
 
 43 
 
 38 
 
 037 945 128 
 
 62 
 
 98 
 
 097 058 808 
 
 2 
 
 58 
 
 154 055 688 
 
 42 
 
 39 
 
 038 940 681 
 
 61 
 
 99 
 
 098 029 701 
 
 1 
 
 59 
 
 154 9KO 321 
 
 41 
 
 .040 
 
 039 936 000 
 
 .960 
 
 .100 
 
 099 000 000 
 
 900 
 
 .160 
 
 155 904 COO 
 
 .810 
 
 41 
 
 040 931 079 
 
 59 
 
 1 
 
 099 969 699 
 
 899 
 
 61 
 
 156 826 719 
 
 39 
 
 42 
 
 041 925 912 
 
 58 
 
 2 
 
 100 938 792 
 
 98 
 
 62 
 
 157 748 472 
 
 38 
 
 43 
 
 042 920 493 
 
 57 
 
 3 
 
 101 907 273 
 
 97 
 
 63 
 
 158 669 253 
 
 37 
 
 44 
 
 043 914 816 
 
 56 
 
 4 
 
 102 875 136 
 
 96 
 
 64 
 
 159 589 056 
 
 36 
 
 45 
 
 044 908 875 
 
 55 
 
 5 
 
 103 842 375 
 
 95 
 
 65 
 
 160 507 875 
 
 35 
 
 46 
 
 045 902 664 
 
 54 
 
 6 
 
 104 808 984 
 
 94 
 
 66 
 
 161 425 704 
 
 34 
 
 47 
 
 046 896 177 
 
 53 
 
 7 
 
 105 774 957 
 
 93 
 
 67 
 
 162 342 537 
 
 33 
 
 48 
 
 047 889 408 
 
 52 
 
 8 
 
 106 740 288 
 
 92 
 
 68 
 
 163 258 368 
 
 32 
 
 49 
 
 048 882 351 
 
 51 
 
 9 
 
 107 704 971 
 
 91 
 
 69 
 
 164 173 191 
 
 31 
 
 .050 
 
 049 875 000 
 
 .950 
 
 .110 
 
 108 669 000 
 
 .890 
 
 .170 
 
 165 087 000 
 
 .830 
 
 51 
 
 050 867 349 
 
 49 
 
 11 
 
 109 632 369 
 
 89 
 
 71 
 
 165 999 789 
 
 29 
 
 52 
 
 051 859 392 
 
 48 
 
 12 
 
 110 595 072 
 
 88 
 
 72 
 
 166 911 552 
 
 28 
 
 53 
 
 052 851 123 
 
 47 
 
 13 
 
 111 557 103 
 
 87 
 
 73 
 
 167 822 283 
 
 27 
 
 54 
 
 053 842 536 
 
 46 
 
 14 
 
 112 518 456 
 
 
 74 
 
 168 731 976 
 
 26 
 
 55 
 
 054 8 {3 625 
 
 45 
 
 15 
 
 113 479 125 
 
 85 
 
 75 
 
 169 640 625 
 
 25 
 
 56 
 
 055 824 384 
 
 44 
 
 16 
 
 114 439 104 
 
 84 
 
 76 
 
 170 548 224 
 
 24 
 
 57 
 
 056 814 807 
 
 43 
 
 17 
 
 115 398 387 
 
 83 
 
 77 
 
 171 454 767 
 
 23 
 
 58 
 
 057 804 888 
 
 42 
 
 18 
 
 116 356 968 
 
 82 
 
 78 
 
 172 360 248 
 
 22 
 
 59 
 
 058 794 621 
 
 41 
 
 19 
 
 117 314 841 
 
 81 
 
 79 
 
 173 264 661 
 
 21 
 
 
 > k _ w +l* 
 
 a 
 
 
 *i*-lM* 
 
 i. 
 
 
 ^ k-'i A"+A" V 
 
 , a 
 
 /-= 
 
 
 
 I 
 
 
 
 / 
 
 
 " i 
 
102 
 
 TABLE I. Continued. 
 
 a 
 ' 
 
 k-k' 
 
 1 
 
 1 n ! 
 
 fc=-7- I: k :: 
 
 
 i 
 
 fcfc 8 
 
 
 
 
 
 \ 
 
 
 
 
 
 .180 
 
 174 108 000 
 
 .82) 
 
 .210 
 
 220 176 000 
 
 .760 
 
 .300 
 
 273 000 000 
 
 .700 
 
 81 
 
 175 070 259 
 
 19 
 
 41 
 
 227 002 479 
 
 59 
 
 1 
 
 273 729 099 
 
 699 
 
 82 
 
 175 971 432 
 
 18 
 
 42 
 
 227 827 512 
 
 58 
 
 2 
 
 274 456 392 
 
 MS 
 
 83 
 
 176 871 513 
 
 17 
 
 43 
 
 228 651 093 
 
 57 
 
 3 
 
 275 181 873 
 
 97 
 
 84 
 
 177 770 496 
 
 16 
 
 44 
 
 2i9 473 216 
 
 56 
 
 4 
 
 275 1)05 536 
 
 06 
 
 85 
 
 178 668 375 
 
 15 
 
 45 
 
 230 293 875 
 
 55 
 
 5 
 
 276 627 375 
 
 
 86 
 
 179 565 144 
 
 14 
 
 46 
 
 231 113 061 
 
 54 
 
 6 
 
 277 347 384 
 
 01 
 
 87 
 
 180 460 797 
 
 13 
 
 47 
 
 231 930 777 
 
 68 
 
 7 
 
 278 065 557 
 
 93 
 
 88 
 
 181 355 328 
 
 12 
 
 48 
 
 J.T2 747 008 
 
 52 
 
 8 
 
 278 781 888 
 
 92 
 
 89 
 
 182 248 731 
 
 11 
 
 49 
 
 233 561 751 
 
 51 
 
 9 
 
 279 496 371 
 
 91 
 
 .190 
 
 183 141 000 
 
 .81CU 
 
 .250 
 
 234 375 000 
 
 .750 
 
 .310 
 
 280 209 000 
 
 .600 
 
 91 
 
 184 032 129 
 
 9 
 
 51 
 
 235 186 749 
 
 18 
 
 11 
 
 280 919 769 
 
 89 
 
 92 
 
 184 922 112 
 
 8 
 
 52 
 
 235 9!) !)!I2 
 
 48 
 
 12 
 
 281 628 672 
 
 88 
 
 93 
 
 185 810 943 
 
 7 
 
 J-3 
 
 236 805 723 
 
 47 
 
 13 
 
 282 335 703 
 
 87 
 
 94 
 
 186 698 616 
 
 6 
 
 54 
 
 237 612 936 
 
 46 
 
 14 
 
 283 010 856 
 
 86 
 
 95 
 
 187 585 125 
 
 5 
 
 55 
 
 238 418 6.5 
 
 45 
 
 15 
 
 283 744 12.') 
 
 85 
 
 96 
 
 18* 470 464 
 
 4 
 
 56 
 
 239 222 784 
 
 44 
 
 16 
 
 2<si 44 ') 504 
 
 84 
 
 97 
 
 189 354 627 
 
 3 
 
 57 
 
 240 025 407 
 
 43 
 
 17 
 
 285 144 987 
 
 83 
 
 98 
 
 190 237 608 
 
 2 
 
 58 
 
 240 826 488 
 
 42 
 
 18 
 
 235 842 :>(iS 
 
 82 
 
 9} 
 
 191 119 401 
 
 1 
 
 59 
 
 241 626 021 
 
 41 
 
 19 
 
 286 538 241 
 
 81 
 
 .200 
 
 192 000 000 
 
 .8 r O 
 
 .WO 
 
 $42 424 O'X) 
 
 .740 
 
 .320 
 
 287 232 000 
 
 .680 
 
 1 
 
 192 879 399 
 
 .799 
 
 61 
 
 243 220 419 
 
 39 
 
 21 
 
 287 923 88!) 
 
 79 
 
 2 
 
 193 757 592 
 
 98 
 
 62 
 
 244 015 272 
 
 38 
 
 22 
 
 288 613 752 
 
 78 
 
 3 
 
 191 634 573 
 
 97 
 
 63 
 
 244 808 553 
 
 37 
 
 23 
 
 289 301 733 
 
 77 
 
 4 
 
 195 510 336 
 
 96 
 
 64 
 
 245 600 256 
 
 88 
 
 24 
 
 289 987 776 
 
 76 
 
 5 
 
 196 384 875 
 
 95 
 
 65 
 
 246 390 375 
 
 35 
 
 25 
 
 290 671 875 
 
 75 
 
 6 
 
 197 258 184 
 
 94 
 
 66 
 
 247 178 904 
 
 34 
 
 26 
 
 291 354 024 
 
 74 
 
 7 
 
 198 130 257 
 
 93 
 
 67 
 
 247 9a5 837 
 
 33 
 
 27 
 
 292 034 217 
 
 73 
 
 8 
 
 199 001 088 
 
 92 
 
 68 
 
 248 751 168 
 
 32 
 
 28 
 
 292 712 448 
 
 72 
 
 9 
 
 199 870 671 
 
 91 
 
 69 
 
 249 534 891 
 
 31 
 
 29 
 
 293 388 711 
 
 71 
 
 .210 
 
 200 739 000 
 
 .790 
 
 .270 
 
 250 317 000 
 
 .7450 
 
 .aso 
 
 291 063 OfO 
 
 .670 
 
 11 
 
 201 606 069 
 
 89 
 
 71 
 
 251 097 489 
 
 21) 
 
 31 
 
 294 735 309 
 
 69 
 
 12 
 
 202 471 872 
 
 88 
 
 72 
 
 251 876 352 
 
 28 
 
 32 
 
 295 405 632 
 
 68 
 
 13 
 
 203 830 403 
 
 87 
 
 73 
 
 252 653 583 
 
 27 
 
 33 
 
 296 3 963 
 
 67 
 
 14 
 
 204 199 6515 
 
 86 
 
 74 
 
 253 429 17(i 
 
 26 
 
 34 
 
 2!>(i 740 296 
 
 66 
 
 15 
 
 205 061 625 
 
 85 
 
 75 
 
 254 203 125 
 
 25 
 
 35 
 
 2JJ7 404 25 
 
 66 
 
 16 
 
 205 922 304 
 
 84 
 
 76 
 
 254 975 424 
 
 24 
 
 36 
 
 298 066 944 
 
 
 17 
 
 206 781 687 
 
 83 
 
 77 
 
 255 746 067 
 
 23 
 
 37 
 
 298 727 247 
 
 88 
 
 18 
 
 207 639 76$ 
 
 82 
 
 78 
 
 256 515 048 
 
 22 
 
 38 
 
 299 385 528 
 
 62 
 
 19 
 
 208 496 541 
 
 81 
 
 79 
 
 257 282 361 
 
 21 
 
 39 
 
 300 041 781 
 
 61 
 
 .220 
 
 209 a52 000 
 
 .780 
 
 .280 
 
 258 048 000 
 
 .720 
 
 .340 
 
 30) 696 000 
 
 .660 
 
 21 
 
 210 206 139 
 
 79 
 
 81 
 
 258 811 959 
 
 19 
 
 41 
 
 301 348 171) 
 
 59 
 
 22 
 
 211 058 952 
 
 78 
 
 82 
 
 259 574 232 
 
 18 
 
 42 
 
 301 998 312 
 
 58 
 
 23 
 
 211 910 433 
 
 77 
 
 83 
 
 260 334 813 
 
 17 
 
 43 
 
 302 646 393 
 
 57 
 
 24 
 
 212 760 576 
 
 76 
 
 84 
 
 261 093 696 
 
 16 
 
 44 
 
 303 292 416 
 
 56 
 
 25 
 
 213 609 375 
 
 75 
 
 85 
 
 261 850 875 
 
 15 
 
 45 
 
 303 936 375 
 
 55 
 
 26 
 
 214 456 824 
 
 74 
 
 86 
 
 262 606 344 
 
 14 
 
 46 
 
 304 578 264 
 
 54 
 
 27 
 
 215 302 917 
 
 73 
 
 87 
 
 263 360 097 
 
 13 
 
 47 
 
 SOS 218 077 
 
 53 
 
 28 
 
 216 147 648 
 
 72 
 
 ' 88 
 
 264 112 128 
 
 12 
 
 48 
 
 305 855 808 
 
 52 
 
 29 
 
 216 991 Oil 
 
 71 
 
 89 
 
 264 862 431 
 
 11 
 
 49 
 
 306 491 451 
 
 51 
 
 .230 
 
 217 833 000 
 
 .770 
 
 .290 
 
 265 611 (XO 
 
 .710 
 
 .350 
 
 307 125 POO 
 
 .650~ 
 
 31 
 
 218 673 609 
 
 69 
 
 91 
 
 266 357 829 
 
 9 
 
 51 
 
 307 756 449 
 
 49 
 
 32 
 
 219 512 832 
 
 68 
 
 92 
 
 267 102 912 
 
 8 
 
 52 
 
 308 385 792 
 
 48 
 
 33 
 
 220 350 663 
 
 67 
 
 93 
 
 267 846 243 
 
 7 
 
 53 
 
 309 013 023 
 
 47 
 
 34 
 
 221 187 096 
 
 66 
 
 94 
 
 268 587 816 
 
 6 
 
 54 
 
 309 638 136 
 
 46 
 
 35 
 
 222 022 125 
 
 65 
 
 95 
 
 269 327 625 
 
 5 
 
 55 
 
 310 261 125 
 
 45 
 
 36 
 
 222 855 744 
 
 64 
 
 96 
 
 270 065 664 
 
 4 
 
 56 
 
 310 881 9*4 
 
 44 
 
 37 
 
 223 6S7 947 
 
 63 
 
 97 
 
 270 801 927 
 
 3 
 
 57 
 
 311 500 707 
 
 43 
 
 38 
 
 224 518 728 
 
 62 
 
 98 
 
 271 536 108 
 
 2 
 
 58 
 
 312 117 288 
 
 42 
 
 39 
 
 225 348 081 
 
 61 
 
 99 
 
 272 269 101 
 
 1 
 
 59 
 
 312 731 721 
 
 41 
 
 
 2k-3k 2 +k 3 
 
 a 
 
 
 ,'>k-i1r -I- 
 
 k- a 
 
 
 2 k3 k s -^k 
 
 /. a 
 
 
 
 / 
 
 1 
 
 I 
 
 
 
 I 
 
TABLE I. Continued. 
 
 103 
 
 7. Cl 
 
 ~T 
 
 k- k :i 
 
 
 fc= 
 
 I 
 
 k F 
 
 1 
 
 *=T 
 
 k~k ! 
 
 
 .3i,0 
 
 :{l;J 344 UW) 
 
 .641) 
 
 .420 
 
 345 912 ( (-0 
 
 .580 
 
 .4 
 
 408 CO ) 
 
 .520 
 
 61 
 
 313 954 119 
 
 39 
 
 21 
 
 346 381 539 
 
 79 
 
 81 
 
 715 359 
 
 19 
 
 m 
 
 314 562 072 
 
 88 
 
 22 
 
 346 848 552 
 
 78 
 
 82 
 
 370 019 832 
 
 18 
 
 6;} 
 
 315 167 853 
 
 37 
 
 25 
 
 347 313 033 
 
 77 
 
 83 
 
 32 L 413 
 
 17 
 
 64 
 
 315 771 456 
 
 36 
 
 24 
 
 347 774 976 
 
 76 
 
 84 
 
 620 096 
 
 16 
 
 65 
 
 316 372 875 
 
 85 
 
 25 
 
 348 234 375 
 
 7i 
 
 85 
 
 915 875 
 
 15 
 
 66 
 
 316 972 104 
 
 34 
 
 26 
 
 348 691 224 
 
 74 
 
 86 
 
 371 i08 741 
 
 14 
 
 67 
 
 317 569 137 
 
 33 
 
 27 
 
 349 145 517 
 
 73 
 
 87 
 
 498 697 
 
 13 
 
 68 
 
 318 163 968 
 
 32 
 
 28 
 
 349 597 248 
 
 72 
 
 fc8 
 
 785 728 
 
 12 
 
 09 
 
 318 756 591 
 
 31 
 
 29 
 
 350 046 411 
 
 71 
 
 89 
 
 372 069 831 
 
 11 
 
 .370 . 
 
 319 317 000 
 
 .630 
 
 .430 
 
 50 4!)3 000 
 
 .r.70 
 
 ,4U) 
 
 351 COO 
 
 .510 
 
 71 
 
 319 935 189 
 
 29 
 
 31 
 
 350 937 009 
 
 69 
 
 91 
 
 629 229 
 
 9 
 
 72 
 
 320 521 152 
 
 28 
 
 32 
 
 a51 378 432 
 
 68 
 
 92 
 
 904 512 
 
 8 
 
 73 
 
 321 104 883 
 
 27 
 
 33 
 
 851 817 263 
 
 67 
 
 93 
 
 373 176 843 
 
 7 
 
 74 
 
 321 686 376 
 
 26 
 
 34 
 
 a52 253 496 
 
 66 
 
 94 
 
 446 216 
 
 6 
 
 75 
 
 322 265 625 
 
 25 
 
 35 
 
 352 687 125 
 
 65 
 
 95 
 
 712 625 
 
 5 
 
 76 
 
 322 842 624 
 
 -24 
 
 36 
 
 353 118 144 
 
 64 
 
 96 
 
 976 064 
 
 4 
 
 77 
 
 323 417 367 
 
 23 
 
 37 
 
 353 546 547 
 
 63. 
 
 97 
 
 374 236 527 
 
 3 
 
 78 
 
 323 989 848 
 
 22 
 
 38 
 
 353 972 328 
 
 62 
 
 98 
 
 494 C08 
 
 2 
 
 79 
 
 324 560 061 
 
 21 
 
 39 
 
 354 395 481 
 
 61 
 
 99 
 
 748 501 
 
 1 
 
 .380 
 
 325 128 OCO 
 
 .620 
 
 .440 
 
 354 816 000 
 
 .560 
 
 .500 
 
 375 000 000 
 
 .500 
 
 81 
 
 325 693 659 
 
 19 
 
 41 
 
 355 233 879 
 
 59 
 
 1 
 
 375 248 499 
 
 .499 
 
 82 
 
 326 257 032 
 
 18 
 
 42 
 
 355 649 112 
 
 5S 
 
 2 
 
 493 992 
 
 98 
 
 83 
 
 326 818 113 
 
 17 
 
 43 
 
 356 061 693 
 
 57 
 
 3 
 
 736 473 
 
 97 
 
 84 
 
 i?27 376 896 
 
 16 
 
 44 
 
 356 471 616 
 
 56 
 
 4 
 
 975 936 
 
 96 
 
 85 
 
 327 933 375 
 
 15 
 
 45 
 
 356 878 875 
 
 5 
 
 5 
 
 376 212 375 
 
 95 
 
 86 
 
 323 487 544 
 
 14 
 
 46 
 
 357 283 464 
 
 54 
 
 6 
 
 445 784 
 
 94 
 
 87 
 
 329 039 397 
 
 18 
 
 47 
 
 357^685 377 
 
 53 
 
 7 
 
 676 157 
 
 93 
 
 88 
 
 329 588 928 
 
 12 
 
 48 
 
 358 084 608 
 
 52 
 
 8 
 
 903 488 
 
 92 
 
 89 
 
 33-J 136 131 
 
 11 
 
 49 
 
 358 481 151 
 
 51 
 
 9 
 
 377 127 771 
 
 91 
 
 .390 
 
 3 '0 681 000 
 
 .610 
 
 .450 
 
 358 875 000 
 
 .550 
 
 .510 
 
 349 OCO 
 
 ;490 
 
 91 
 
 331 223 529 
 
 9 
 
 51 
 
 859 266 149 
 
 49 
 
 11 
 
 567 169 
 
 89 
 
 92 
 
 331 763 712 
 
 8 
 
 52 
 
 359 654 592 
 
 48 
 
 12 
 
 782 272 
 
 88 
 
 93 
 
 332 301 543 
 
 7 
 
 53 
 
 36) 040 323 
 
 47 
 
 13 
 
 994 3,3 
 
 87 
 
 94 
 
 332 837 016 
 
 6 
 
 54 
 
 360 423 336 
 
 46 
 
 14 
 
 378 203 256 
 
 86 
 
 95 
 
 333 370 125 
 
 5 
 
 55 
 
 360 81.3 625 
 
 45 
 
 15 
 
 409 125 
 
 85 
 
 96 
 
 333 900 864 
 
 4 
 
 56 
 
 361 181 184 
 
 44 
 
 16 
 
 611 904 
 
 84 
 
 97 
 
 334 429 227 
 
 3 
 
 57 
 
 556 007 
 
 43 
 
 17 
 
 811 587 
 
 83 
 
 98 
 
 334 955 208 
 
 2 
 
 58 
 
 928 088 
 
 42 
 
 18 
 
 379 008 168 
 
 82 
 
 99 
 
 335 478 801 
 
 1 
 
 59 
 
 362 297 421 
 
 41 
 
 19 
 
 201 641 
 
 81 
 
 .400 
 
 336 000 000 
 
 .600 
 
 .460 
 
 661 000 
 
 .540 
 
 .520 
 
 392 (KO 
 
 .480 
 
 1 
 
 336 518 799 
 
 .599 
 
 61 
 
 S63 027 819 
 
 39 
 
 21 
 
 579 239 
 
 79 
 
 2 
 
 337 035 192 
 
 98 
 
 62 
 
 363 388 872 
 
 38 
 
 24 
 
 763 352 
 
 78 
 
 3 
 
 337 549 173 
 
 97 
 
 63 
 
 747 153 
 
 37 
 
 23 
 
 379 944 333 
 
 77 
 
 4 
 
 338 060 736 
 
 96 
 
 64 
 
 364 102 656 
 
 36 
 
 24 
 
 380 122 176 
 
 76 
 
 5 
 
 338 569 875 
 
 95 
 
 65 
 
 455 375 
 
 35 
 
 25 
 
 296 875 
 
 75 
 
 6 
 
 339 076 584 
 
 94 
 
 66 
 
 805 304 
 
 34 
 
 26 
 
 468 424 
 
 74 
 
 7 
 
 339 580 857 
 
 93 
 
 67 
 
 365 152 437 
 
 33 
 
 27 
 
 636 817 
 
 73 
 
 8 
 
 340 082 688 
 
 92 
 
 68 
 
 496 768 
 
 32 
 
 28 
 
 802 048 
 
 72 
 
 9 
 
 340 582 071 
 
 91 
 
 69 
 
 838 291 
 
 31 
 
 k9 
 
 964 111 
 
 71 
 
 .410 
 
 311 079 000 
 
 .590 
 
 .470 
 
 366 177 COn 
 
 .530 
 
 80 
 
 381 123 000 
 
 .470 
 
 11 
 
 341 573 469 
 
 89 
 
 71 
 
 512 8^9 
 
 29 
 
 31 
 
 278 709 
 
 69 
 
 12 
 
 342 065 472 
 
 88 
 
 72 
 
 845 952 
 
 28 
 
 32 
 
 431 232 
 
 6* 
 
 13 
 
 342 555 003 
 
 87 
 
 73 
 
 367 176 183 
 
 27 
 
 33 
 
 580 563 
 
 67 
 
 14 
 
 343 042 056 
 
 86 
 
 74 
 
 503 576 
 
 26 
 
 34 
 
 726 696 
 
 66 
 
 15 
 
 343 526 625 
 
 85 
 
 75 
 
 *28 125 
 
 25 
 
 35 
 
 869 625 
 
 65 
 
 16 
 
 344 008 704 
 
 84 
 
 76 
 
 . 368 149 824 
 
 24 
 
 36 
 
 382 (H*9 344 
 
 64 
 
 17 
 
 344 48$ 287 
 
 83 
 
 77 
 
 468 667 
 
 23 
 
 37 
 
 145 847 
 
 63 
 
 18 
 
 344 965 368 
 
 82 
 
 78 
 
 784 648 
 
 22 
 
 38 
 
 279 128 
 
 62 
 
 19 
 
 345 439 941 
 
 . 81 
 
 79 
 
 369 097 761 
 
 21 
 
 39 
 
 409 181 
 
 61 
 
 
 2 A-->' Jb*+fr' 
 
 k- a 
 
 
 **-** fc 8 
 
 A-=A 
 
 
 Sk-XV+l? 
 
 *=-- 
 
 
 
 I 
 
 
 
 / 
 
 
 
 I 
 
104 
 
 TABLE I. Continued. 
 
 *=f 
 
 fe- /," 
 
 
 /l> I 
 
 /,' I" 
 
 
 1. <( 
 n . 
 
 A- -A" 
 
 
 .540 
 
 536 Oi).' 
 
 .460 
 
 .6-JO 
 
 384 000 000 
 
 .400 
 
 .660 
 
 .304 i 00 
 
 .310 
 
 41 
 
 659 579 
 
 59 
 
 1 
 
 383 918 19!) 
 
 .399 
 
 61 
 
 195 219 
 
 90 
 
 42 
 
 779 912 
 
 58 
 
 2 
 
 832 792 
 
 98 
 
 62 
 
 371 882 472 
 
 38 
 
 43 
 
 896 993 
 
 57 
 
 3 
 
 743 773 
 
 97 
 
 63 
 
 565 753 
 
 37 
 
 44 
 
 383 010 816 
 
 56 
 
 4 
 
 651 136 
 
 96 
 
 64 
 
 245 056 
 
 86 
 
 45 
 
 121 375 
 
 55 
 
 5 
 
 554 875 
 
 95 
 
 65 
 
 370 920 375 
 
 35 
 
 46 
 
 228 664 
 
 54 
 
 6 
 
 454 984 
 
 94 
 
 60 
 
 o!)l 704 
 
 34 
 
 47 
 
 332 677 
 
 53 
 
 7 
 
 351 457 
 
 93 
 
 67 
 
 259 037 
 
 88 
 
 48 
 
 433 408 
 
 52 
 
 8 
 
 244 288 
 
 92 
 
 68 
 
 369 922 368 
 
 32 
 
 49 
 
 530 851 
 
 51 
 
 9 
 
 133 471 
 
 91 
 
 69 
 
 .581 691 
 
 M 
 
 .550 
 
 383 625 000 
 
 .450 
 
 .610 
 
 019 000 
 
 .393 
 
 .670 
 
 237 000 
 
 .330 
 
 51 
 
 715 849 
 
 49 
 
 11 
 
 382 900 869 
 
 89 
 
 71 
 
 368 888 289 
 
 29 
 
 52 
 
 803 392 
 
 48 
 
 12 
 
 77!) 072 
 
 88 
 
 72 
 
 .535552 
 
 28 
 
 53 
 
 887 623 
 
 47 
 
 13 
 
 653 603 
 
 87 
 
 73 
 
 178 783 
 
 27 
 
 54 
 
 968 536 
 
 46 
 
 14 
 
 524 456 
 
 86 
 
 74 
 
 367 817 976 
 
 26 
 
 55 
 
 384 046 125 
 
 45 
 
 15 
 
 391 625 
 
 8i 
 
 75 
 
 453 125 
 
 2.3 
 
 56 
 
 120 884 
 
 44 
 
 16 
 
 255 104 
 
 84 
 
 76- 
 
 084 224 
 
 24 
 
 57 
 
 191 307 
 
 43 
 
 . 17 
 
 114 887 
 
 83 
 
 77 
 
 366 711 2(57 
 
 23 
 
 58 
 
 2.50 888 
 
 42 
 
 18 
 
 381 970 968 
 
 82 
 
 78 
 
 334 248 
 
 22 
 
 59 
 
 323 121 
 
 41 
 
 19 
 
 823 341 
 
 81 
 
 79 
 
 365 953 161 
 
 21 
 
 .560 
 
 384 000 
 
 .440 
 
 .620 
 
 672 000 
 
 .380 
 
 .680 
 
 568 000 
 
 .320 
 
 61 
 
 441 519 
 
 39 
 
 21 
 
 516 939 
 
 79 
 
 81 
 
 ITS 75! 
 
 19 
 
 62 
 
 4^5 672 
 
 38 
 
 22 
 
 358 152 
 
 78 
 
 82 
 
 361 785 432 
 
 18 
 
 63 
 
 546 453 
 
 37 
 
 23 
 
 195 633 
 
 77 
 
 83 
 
 888 013 
 
 17 
 
 64 
 
 593 856 
 
 36 
 
 24 
 
 Oi9 376 
 
 76 
 
 84 
 
 363 986 496 
 
 16 
 
 65 
 
 637 875 
 
 35 
 
 25 
 
 380 859 375 
 
 75 
 
 85 
 
 fi80 875 
 
 15 
 
 66 
 
 678 504 
 
 34 
 
 26 
 
 685 624 
 
 74 
 
 86 
 
 171 144 
 
 14 
 
 67 
 
 715 737 
 
 33 
 
 27 
 
 508 117 
 
 73 
 
 87 
 
 362 757 297 
 
 i:; 
 
 68 
 
 749 568 
 
 32 
 
 28 
 
 326 818 
 
 72 
 
 88 
 
 339 328 
 
 12 
 
 69 
 
 779 991 
 
 31 
 
 29 
 
 141 811 
 
 71 
 
 89 
 
 361 917 231 
 
 11 
 
 570 
 
 807 000 
 
 .430 
 
 630 
 
 379 9.53 000 
 
 .370 
 
 .690 
 
 491 000 
 
 .310 
 
 71 
 
 830 589 
 
 29 
 
 31 
 
 760 409 
 
 (ij 
 
 91 
 
 060 629 
 
 9 
 
 72 
 
 850 752 
 
 28 
 
 32 
 
 564 032 
 
 68 
 
 92 
 
 360 626 112 
 
 8 
 
 73 
 
 867 483 
 
 27 
 
 33 
 
 363 863 
 
 67 
 
 93 
 
 187 443 
 
 7 
 
 74 
 
 880 776 
 
 26 
 
 34 
 
 159 896 
 
 66 
 
 94 
 
 359 744 616 
 
 6 
 
 75 
 
 890 625 
 
 25 
 
 35 
 
 378 952 125 
 
 65 
 
 
 297 625 
 
 5 
 
 76 
 
 897 024 
 
 24 
 
 36 
 
 710 44 
 
 64 
 
 96 
 
 358 846 464 
 
 4 
 
 77 
 
 899 967 
 
 23 
 
 37 
 
 525 147 
 
 s 
 
 97 
 
 391 127 
 
 3 
 
 78 
 
 899 448 
 
 22 
 
 38 
 
 305 928 
 
 62 
 
 98 
 
 357 931 608 
 
 2 
 
 79 
 
 895 461 
 
 21 
 
 39 
 
 082 881 
 
 61 
 
 99 
 
 467 901 
 
 1 
 
 .580 
 
 888 000 
 
 .420 
 
 .640 
 
 377 856 000 
 
 .3(30 
 
 .700 
 
 357 000 000 
 
 .801 
 
 81 
 
 877 059 
 
 19 
 
 41 
 
 625 279 
 
 59 
 
 1 
 
 356 527 899 
 
 .299 
 
 82 
 
 862 632 
 
 18 
 
 42 
 
 390 712 
 
 58 
 
 2 
 
 051 592 
 
 98 
 
 83 
 
 844 713 
 
 17 
 
 43 
 
 152 293 
 
 57 
 
 3 
 
 355 571 073 
 
 97 
 
 84 
 
 823 296 
 
 1H 
 
 44 
 
 376 910 016 
 
 56 
 
 4 
 
 086 336 
 
 9o 
 
 85 
 
 798 375 
 
 15 
 
 45 
 
 663 875 
 
 55 
 
 5 
 
 354 597 375 
 
 95 
 
 86 
 
 769 944 
 
 14 
 
 46 
 
 413 864 
 
 54 
 
 6 
 
 354 104 184 
 
 94 
 
 87 
 
 737 997 
 
 13 
 
 47 
 
 159 977 
 
 53 
 
 7 
 
 353 606 757 
 
 93 
 
 88 
 
 702 528 
 
 12 
 
 48 
 
 375 902 208 
 
 52 
 
 8 
 
 105 088 
 
 92 
 
 89 
 
 663 531 
 
 11 
 
 49 
 
 640 551 
 
 51 
 
 9 
 
 352 599 171 
 
 91 
 
 .590 
 
 621 000 
 
 .410 
 
 .6 
 
 375 375 000 
 
 .350 
 
 .710 
 
 089 000 
 
 .290 
 
 91 
 
 574 929 
 
 9 
 
 51 
 
 105 549 
 
 49 
 
 11 
 
 351 574 569 
 
 89 
 
 92 
 
 525 312 
 
 8 
 
 52 
 
 374 832 192 
 
 48 
 
 12 
 
 055 872 
 
 88 
 
 93 
 
 472 143 
 
 7 
 
 53 
 
 554 923 
 
 47 
 
 13 
 
 350 532 903 
 
 87 
 
 94 
 
 415 416 
 
 6 
 
 54 
 
 273 736 
 
 46 
 
 14 
 
 005 656 
 
 86 
 
 95 
 
 355 125 
 
 5 
 
 55 
 
 373 988 625 
 
 45 
 
 15 
 
 349 474 125 
 
 85 
 
 96 
 
 291 264 
 
 4 
 
 56 
 
 699 .584 
 
 44 
 
 16 
 
 348 938 304 
 
 84 
 
 97 
 
 223 827 
 
 3 
 
 57 
 
 406 607 
 
 43 
 
 17 
 
 398 187 
 
 83 
 
 98 
 
 152 808 
 
 2 
 
 58 
 
 109 688 
 
 42 
 
 18 
 
 347 853 768 
 
 82 
 
 99 
 
 078 201 
 
 1 
 
 59 
 
 372 808 821 
 
 41 
 
 19 
 
 305 041 
 
 81 
 
 
 2 k-3 A-'+fc 3 
 
 , a 
 
 
 <2k-31^1f 
 
 *. 
 
 
 21-3 V+tf 
 
 fc=-i 
 
 
 
 I 
 
 
 
 I 
 
 
 
 [ 
 
TABLE I. Continued. 
 
 105 
 
 k ~ I 
 
 
 
 k=-r- 
 
 Mf 
 
 
 *=f 
 
 k-lf 
 
 
 .720 
 
 346 752 000 
 
 .280 
 
 .780 
 
 305 448 0>0 
 
 .220 
 
 .840 
 
 247 296 000 
 
 .160 
 
 21 
 
 194 639 
 
 79 
 
 81 
 
 304 620 459 
 
 19 
 
 41 
 
 246 176 679 
 
 59 
 
 22 
 
 345 632 952 
 
 78 
 
 82 
 
 303 788 232 
 
 18 
 
 42 
 
 245 052 312 
 
 58 
 
 23 
 
 f66 933 
 
 77 
 
 83 
 
 302 951 313 
 
 17 
 
 43 
 
 243 922 893 
 
 57 
 
 24 
 
 344 496 576 
 
 76 
 
 84 
 
 302 109 696 
 
 16 
 
 44 
 
 242 788 416 
 
 56 
 
 25 
 
 343 921 875 
 
 75 
 
 85 
 
 301 263 375 
 
 15 
 
 45 
 
 241 648 875 
 
 55 
 
 26 
 
 343 342 824 
 
 74 
 
 86 
 
 301 412 344 
 
 14 
 
 46 
 
 240 504 264 
 
 54 
 
 27 
 
 342 759 417 
 
 73 
 
 87 
 
 299 566 597 
 
 13 
 
 47 
 
 239 354 577 
 
 53 
 
 28 
 
 171 648 
 
 72 
 
 88 
 
 293 696 128 
 
 12 
 
 48 
 
 238 199 8U8 
 
 52 
 
 29 
 
 341 579 511 
 
 71 
 
 89 
 
 297 830 931 
 
 11 
 
 49 
 
 237 039 951 
 
 51 
 
 .730 
 
 310 983 000 
 
 .270 
 
 .790 
 
 296 961 000 
 
 .210 
 
 .850 
 
 235 875 000 
 
 .150 
 
 31 
 
 382 109 
 
 69 
 
 91 
 
 296 086 329 
 
 9 
 
 51 
 
 234 704 919 
 
 49 
 
 32 
 
 339 776 832 
 
 68 
 
 92 
 
 295 206 912 
 
 8 
 
 52 
 
 233 529 792 
 
 48 
 
 as 
 
 167 163 
 
 67 
 
 93 
 
 294 322 743 
 
 7 
 
 53 
 
 232 349 523 
 
 47 
 
 34 
 
 338 553 096 
 
 66 
 
 94 
 
 293 433 816 
 
 6 
 
 54 
 
 231 164 136 
 
 46 
 
 35 
 
 337 934 625 
 
 65 
 
 95 
 
 292 540 125 
 
 5 
 
 55 
 
 229 973 625 
 
 45 
 
 36 
 
 311 744 
 
 64 
 
 96 
 
 291 641 664 
 
 4 
 
 56 
 
 228 777 984 
 
 44 
 
 37 
 
 336 684 447 
 
 63 
 
 97 
 
 290 738 427 
 
 3 
 
 57 
 
 227 577 207 
 
 43 
 
 38 
 
 (J52 728 
 
 62 
 
 98 
 
 289 830 408 
 
 2 
 
 58 
 
 226 371 288 
 
 42 
 
 39 
 
 335 416 581 
 
 61 
 
 99 
 
 288 917 601 
 
 1 
 
 59 
 
 225 169 221 
 
 41 
 
 .740 
 
 834 776 000 
 
 .260 
 
 .800 
 
 288 000 000 
 
 .200 
 
 .860 
 
 223 944 000 
 
 .140 
 
 41 
 
 130 979 
 
 59 
 
 1 
 
 287 077 599 
 
 .199 
 
 61 
 
 222 722 619 
 
 39 
 
 42 
 
 333 481 512 
 
 58 
 
 2 
 
 286 150 392 
 
 98 
 
 62 
 
 221 496 072 
 
 38 
 
 43 
 
 aS2 827 593 
 
 57 
 
 3 
 
 285 218 373 
 
 97 
 
 63 
 
 220 264 353 
 
 37 
 
 44 
 
 169 216 
 
 56 
 
 4 
 
 284 281 536 
 
 96 
 
 64 
 
 219 027 456 
 
 36 
 
 45 
 
 331 506 375 
 
 55 
 
 5 
 
 283 339 875 
 
 95 
 
 65 
 
 217 785 375 
 
 35 
 
 46 
 
 330 839 064 
 
 54 
 
 6 
 
 282 393 384 
 
 94 
 
 66 
 
 216 538 104 
 
 34 
 
 47 
 
 167 277 
 
 53 
 
 7 
 
 281 442 057 
 
 93 
 
 67 
 
 215 285 637 
 
 33 
 
 48 
 
 329 491 008 
 
 52 
 
 8 
 
 280 485 888 
 
 92 
 
 68 
 
 214 027 968 
 
 32 
 
 49 
 
 328 810 251 
 
 51 
 
 9 
 
 279 524 871 
 
 91 
 
 69 
 
 212 765 091 
 
 31 
 
 .750 
 
 328 125 000 
 
 .250 
 
 .810 
 
 278 559 000 
 
 .190 
 
 .870 
 
 211 497 000 
 
 .130 
 
 51 
 
 327 4a5 249 
 
 49 
 
 11 
 
 277 588 269 
 
 89 
 
 71 
 
 210 223 689 
 
 29 
 
 52 
 
 326 740 992 
 
 48 
 
 12 
 
 276 612 672 
 
 88 
 
 72 
 
 208 945 152 
 
 28 
 
 53 
 
 326 042 223 
 
 47 
 
 13 
 
 275 632 203 
 
 87 
 
 73 
 
 207 661 383 
 
 27 
 
 54 
 
 325 338 936 
 
 46 
 
 14 
 
 274 646 856 
 
 86 
 
 74 
 
 206 372 376 
 
 26 
 
 55 
 
 324 631 12> 
 
 45 
 
 15 
 
 273 656 625 
 
 85 
 
 75 
 
 205 078 125 
 
 25 
 
 56 
 
 323 918 784 
 
 44 
 
 16 
 
 272 661 504 
 
 84 
 
 76 
 
 203 778 624 
 
 24 
 
 57 
 
 323 201 907 
 
 43 
 
 17 
 
 271 661 487 
 
 83 
 
 77 
 
 202 473 867 
 
 23 
 
 58 
 
 322 480 488 
 
 42 
 
 18 
 
 270 656 568 
 
 82 
 
 78 
 
 201 163 848 
 
 22 
 
 59 
 
 321 7.54 521 
 
 41 
 
 ' 19 
 
 269 646 741 
 
 81 
 
 79 
 
 199 848 561 
 
 21 
 
 .760 
 
 321 024 000 
 
 .240 
 
 .820 
 
 268 632 000 
 
 .180 
 
 .880 
 
 198 528 000 
 
 .120 
 
 61 
 
 320 288 919 
 
 39 
 
 21 
 
 267 612 339 
 
 79 
 
 81 
 
 197 202 159 
 
 19 
 
 62 
 
 319 549 272 
 
 38 
 
 22 
 
 266 587 752 
 
 78 
 
 82 
 
 195 871 032 
 
 18 
 
 63 
 
 318 805 053 
 
 37 
 
 23 
 
 265 558 233 
 
 77 
 
 83 
 
 194 534 613 
 
 17 
 
 64 
 
 318 056 256 
 
 36 
 
 24 
 
 264 523 776 
 
 76 
 
 84 
 
 193 192 896 
 
 16 
 
 65 
 
 317 302 875 
 
 35 
 
 25 
 
 263 484 375 
 
 75 
 
 85 
 
 191 815 875 
 
 15 
 
 66 
 
 316 544 904 
 
 34 
 
 26 
 
 262 440 024 
 
 74 
 
 86 
 
 190 493 544 
 
 14 
 
 67 
 
 315 782 3i7 
 
 33 
 
 27 
 
 261 390 717 
 
 73 
 
 87 
 
 189 135 897 
 
 13 
 
 68 
 
 315 015 168 
 
 32 
 
 28 
 
 260 336 448 
 
 72 
 
 88 
 
 187 772 928 
 
 12 
 
 69 
 
 314 243 391 
 
 31 
 
 29 
 
 259 277 211 
 
 71 
 
 89 
 
 186 404*631 
 
 11 
 
 .770 
 
 313 467 000 
 
 .230 
 
 .830 
 
 258 213 000 
 
 .170 
 
 .890 
 
 185 031 000 
 
 .110 
 
 71 
 
 312 685 989 
 
 29 
 
 31 
 
 257 143 809 
 
 69 
 
 91 
 
 183' 652 029 
 
 9 
 
 72 
 
 311 900 a52 
 
 28 
 
 32 
 
 256 069 632 
 
 68 
 
 92 
 
 182 267 712 
 
 8 
 
 73 
 
 311 110 083 
 
 27 
 
 33 
 
 254 990 463 
 
 67 
 
 93 
 
 180 878 043 
 
 7 
 
 74 
 
 310 315 176 
 
 26 
 
 34 
 
 253 906 296 
 
 66 
 
 94 
 
 179 483 016 
 
 6 
 
 75 
 
 309 515 625 
 
 25 
 
 35 
 
 252 817 125 
 
 65 
 
 95 
 
 178 082 625 
 
 5 
 
 76 
 
 308 711 424 
 
 24 
 
 36 
 
 251 722 944 
 
 64 
 
 96 
 
 176 676 864 
 
 4 
 
 77 
 
 307 902 567 
 
 23 
 
 37 
 
 250 623 747 
 
 63 
 
 97 
 
 175 265 727 
 
 3 
 
 78 
 
 1307 089 048 
 
 22 
 
 38 
 
 249 519 528 
 
 62 
 
 98 
 
 173 849 208 
 
 2 
 
 79 
 
 30S 270 861 
 
 21 
 
 39 
 
 218 410 281 
 
 61 
 
 99 
 
 172 427 301 
 
 1 
 
 a 
 
 
 g / ; i>* + l>3 7. a 
 
 
 
 a 
 
 -. -* k ~~~r 
 
 
 I 
 
 
 
 t 
 
106 
 
 TABLE I. Continued. 
 
 I. a 
 
 " / 
 
 A ~ A 
 
 
 -T 
 
 ikjb 8 
 
 
 Hr 
 
 A-- -/r 
 
 
 .900 
 
 171 000 000 
 
 .100 
 
 .940 
 
 109 416 000 
 
 60 
 
 .980 
 
 03* 808 000 
 
 20 
 
 1 
 
 169 5t7 299 
 
 99 
 
 41 
 
 107 762 379 
 
 59 
 
 81 
 
 036 923 859 
 
 19 
 
 2 
 
 168 129 192 
 
 98 
 
 42 
 
 106 103 112 
 
 58 
 
 82 
 
 035 033 H32 
 
 18 
 
 3 
 
 166 685 673 
 
 97 
 
 43 
 
 104 43S 193 
 
 57 
 
 83 
 
 033 137 913 
 
 17 
 
 4 
 
 165 236 736 
 
 96 
 
 41 
 
 102 7(57 till! 
 
 66 
 
 84 
 
 031 236 098 
 
 16 
 
 5 
 
 1(53 782 375 
 
 95 
 
 45 
 
 101 091 :!7"> 
 
 55 
 
 86 
 
 029 328 375 
 
 15 
 
 6 
 
 162 322 584 
 
 94 
 
 46 
 
 099 409 464 
 
 54 
 
 86 
 
 027 414 744 
 
 14 
 
 
 160 857 357 
 
 93 
 
 47 
 
 097 721 877 
 
 53 
 
 87 
 
 025 495 197 
 
 13 
 
 8 
 
 159 386 688 
 
 92 
 
 48 
 
 096 028 608 
 
 52 
 
 88 
 
 023 5(59 728 
 
 12 
 
 9 
 
 157 910 571 
 
 91 
 
 49 
 
 094 329 651 
 
 51 
 
 89 
 
 (2L 038 331 
 
 11 
 
 .910 
 
 156 429 000 
 
 90 
 
 .950 
 
 092 (525 COO 
 
 50 
 
 .990 
 
 019 701 000 
 
 10 
 
 11 
 
 154 941 969 
 
 89 
 
 51 
 
 090 914 649 
 
 49 
 
 91 
 
 017 757 72!) 
 
 9 
 
 12 
 
 153 449 472 
 
 88 
 
 52 
 
 089 198 592 
 
 48 
 
 92 
 
 015 808 512 
 
 8 
 
 18 
 
 151 951 -503 
 
 87 
 
 53 
 
 087 47U 823 
 
 47 
 
 93 
 
 013 8.-,:! ;M3 
 
 7 
 
 14 
 
 150 448 056 
 
 86 
 
 54 
 
 085 749 :m 
 
 16 
 
 94 
 
 Oil 892 21 (i 
 
 (i 
 
 15 
 
 148 939 125 
 
 85 
 
 55 
 
 084 016 125 
 
 45 
 
 95 
 
 009 925 125 
 
 6 
 
 16 
 
 147 424 7(14 
 
 84 
 
 56 
 
 082 277 184 
 
 44 
 
 96 
 
 007 952 064 
 
 4 
 
 17 
 
 145 904 787 
 
 83 
 
 57 
 
 ( s i r,32 507 
 
 18 
 
 97 
 
 005 973 027 
 
 3 
 
 18 
 
 144 379 368 
 
 82 
 
 58 
 
 078 782 088 
 
 42 
 
 98 
 
 003 998 (X 8 
 
 2 
 
 19 
 
 142 848 441 
 
 81 
 
 59 
 
 077 025 921 
 
 41 
 
 99 
 
 TOl 997 001 
 
 .001 
 
 .920 
 
 141 312 000 
 
 80 
 
 .96) 
 
 075 264 000 
 
 40 
 
 
 
 n 
 
 21 
 
 139 770 039 
 
 79 
 
 61 
 
 073 49(5 819 
 
 39 
 
 
 2 k-3 Ir /,' 
 
 fc= 
 
 22 
 
 138 222 552 
 
 78 
 
 62 
 
 071 722 872 
 
 38 
 
 
 
 / 
 
 23 
 
 136 (569 533 
 
 77 
 
 63 
 
 069 943 653 
 
 :!7 
 
 
 
 
 24 
 
 13T> 110 976 
 
 7(5 
 
 64 
 
 068 158 (-..Hi 
 
 36 
 
 
 26 
 
 133 S 46 875 
 
 :r> 
 
 65 
 
 (KK> 3(57 875 
 
 35 
 
 
 26 
 
 131 977 224 
 
 74 
 
 66 
 
 0(51 571 304 
 
 34 
 
 
 27 
 
 130 402 017 
 
 73 
 
 67 
 
 062 7(58 937 
 
 33 
 
 
 28 
 
 128 821 248 
 
 72 
 
 (18 
 
 060 960 7(58 
 
 32 
 
 
 29 
 
 127 234 911 
 
 71 
 
 69 
 
 OE9 146 791 
 
 81 
 
 
 .930 
 
 125 643 000 
 
 70~ 
 
 ~970 
 
 057 327 000 
 
 30 
 
 
 31 
 
 124 045 f>()9 
 
 69 
 
 71 
 
 055 501 38! 
 
 29 
 
 
 32 
 
 122 442 432 
 
 68 
 
 72 
 
 053 669 !52 
 
 28 
 
 
 33 
 
 120 833 763 
 
 67 
 
 73 
 
 051 S32 liS3 
 
 -'7 
 
 
 34 
 
 119 219 496 
 
 66 
 
 74 
 
 049 !IH!t :>7(i 
 
 26 
 
 
 35 
 
 117 599 625 
 
 65 
 
 75 
 
 048 140 625 
 
 25 
 
 
 36 
 
 115 974 144 
 
 64 
 
 76 
 
 046 285 824 
 
 24 
 
 
 37 
 
 114 343 047 
 
 63 
 
 77 
 
 044 425 Ki7 
 
 23 
 
 
 38 
 
 112 706 328 
 
 62 
 
 78 
 
 C42 558 648 
 
 22 
 
 
 39 
 
 111 (63 981 
 
 61 
 
 79 
 
 040 686 261 
 
 21 
 
 
 
 2 k-3 k^k 3 
 
 Hr 
 
 
 2 k-i k^+k 1 
 
 k~ a 
 k ~ I 
 
 
REFERENCES. 
 
 The Britannia and Conway Tubular Bridge, with general inquiries on 
 Beams and on the properties of Materials used in Construction. Edwin Clark, 
 London, 1850. 
 
 Calcul d'une poutr elastique reposant librement sur des appuis inegalement 
 especes. Clapeyron. 1857. 
 
 Beitrag zur Theorie der Holz und Eisen constructionen. Mohr, 1860. 
 
 Beitrage zur Theorie continuirlicher Bruckeutrager. Winkler, Civil Iug<5- 
 nieur, 1862. 
 
 Die Lehre von der Elasticetaet und Festigkeit, Winkler, 18(i7. 
 
 -Vortrage viber Briickenbau. Winkler, 1875. 
 
 -Theorie der Briicken Aeussere Kraf tegerader Trager. Winkler, Wien, 1875. 
 
 Course de Mechanique Appliquee. Ill Partie. Bresse, Paris, 1865. 
 
 -Die graphische Statik. Culmann, 1866. 
 
 A Handy Book for the Calculation of Strains, &c.Humber, New York, 1860. 
 
 Calcul des Fonts mgtalliques a poutres droites et continus. DeMoudesir, 
 Pans. 1873. 
 
 = : Allgemeine Theorie und Berechnung der continuirlichen und einfachen 
 Trager. Weyrauch, Leipzig, 1873. 
 
 The Elements of Graphical Statics. DuBois, New York, 1875. 
 
 Graphical Method for the Analysis of Bridge Trusses. Greene, New York, 
 1875. 
 Continuous Revolving Draw Bridges. Herschel, Boston, 1875. 
 
 Der bau der Briickentrager mit Besouderer Biicksicht auf eisen-construc- 
 tiouen, pp. 161-231. Laissle-Schiibler, Stuttgart, 1876. 
 
 The Theor/ and Calculation of Continuous Bridges. Merriman, New York, 
 1876. 
 
 Practical Treatise on the Properties of Continuous Bridges. Bender, New 
 York, 1876. 
 
 Elementary Theory and Calculation of Iron Bridges and Roofs. Bitter 
 (Hankey)- London, 1879. 
 
 The Strains in Framed Structures DuBois, New York, 1888. 
 
108 
 
 REFERENCES. 
 
 Annales des Fonts et Chaussces. 
 
 1864 Paper 
 
 75 pp. 141-213 
 
 Collignon. 
 
 1866 " 
 
 126 pp. 310-408 
 
 Renaiidot. 
 
 1866 " 
 
 132 pp. 53-175 
 
 Albaret. 
 
 1867 " 
 
 158-^pp. 27-115 
 
 Regnauld. 
 
 1871 " 
 
 3 pp. 44-51 
 
 Pierre. 
 
 -1871- " 
 
 20 pp. 170-274 
 
 Des Orgeries. 
 
 1872 " 
 
 25 pp. 189-220 
 
 Poulet. 
 
 1S74 " 
 
 15 pp. 327-391 
 
 Choron. 
 
 1882 " 
 
 9 pp. 141-218 
 
 Huleu'icz. 
 
 1884 " 
 
 44- pp. 101-197 
 
 HulHvicz. 
 
 188-5 " 
 
 72 pp. 267-351 
 
 Guillaume. 
 
 1885 
 
 86 pp. 613-726 
 
 Hausserct. 
 
 1886- " 
 
 40 pp. 5-39 
 
 Collignon. 
 
 1886- " 
 
 11 pp. 304-357 
 
 Leygue. 
 
 Van Nostrand's Engineering Magazine. 
 
 *1874 pp. 55i_.V)7 . 
 
 1876 pp. 65- 73 . . 
 
 *1877 pp. 481-490 . . 
 
 1878 pp. 55:{-5f)0 . . 
 
 1881-pp. 49- 59 . . 
 
 1885 pp. 137-144 . . 
 
 Eddy. 
 
 Merriman. 
 
 Eddy. 
 
 Hiidyins. 
 
 Kidder.\ 
 
 Berfsford. 
 
 ' Variable Moment of Inertia. 
 
INDEX OF EQUATIONS. 
 
 TT"" . 
 
 -> 1 ; 
 
 A. . . . 
 B . . 
 C . . . 
 D . . . 
 E . . . 
 E, . . . 
 
 PAGK. 
 
 7 
 12 
 . ,14 
 15 
 16 
 
 10a 
 
 PAGE. 
 
 5 
 
 14-19 
 
 6 
 
 20-21 
 
 . . 7 
 . . 86 
 87 
 
 22a-22d 
 22c-22m . . 
 
 16 
 17 
 17 
 11 
 
 22-26 
 27-28 
 
 . . 88 
 89 
 
 29-31 
 32-36 
 
 . . 90 
 . . 81 
 . . 92 
 93 
 
 b . . . 
 c . . . 
 
 d . . . 
 
 e . . . 
 f 
 
 11 
 11 
 11 
 12 
 9 
 
 37-40 
 41-42 ..... 
 
 43-44 
 45a 
 
 . . 94 
 . . 17 
 95 
 
 45-46 
 
 o 
 
 10 
 
 47-53 
 54-63 . . 
 
 . . 96 
 97 
 
 h . . . 
 
 i . . . 
 i., . . . 
 L . - . 
 i 
 
 9 
 8 
 12 
 ....... ]4 
 
 8 
 
 59a 
 
 97 
 
 64-67 
 68-73 
 
 74-77 
 
 . . 98 
 . . 99 
 . . 8 
 . . 9 
 10 
 
 j> 
 
 L 
 
 12 
 14 
 . 12 
 12 
 15 
 
 78-81 
 82-85 
 
 86-89 
 
 . . 13 
 14 
 
 90-92 
 
 93 
 
 15 
 
 n . . . . 
 
 . . . . 
 
 12 
 
 94-102 
 103-110 ...... 
 111-117 
 118-124 
 
 . . 16 
 . . 19 
 . . 20 
 21 
 
 12 
 12 
 12 
 
 
 12 
 
 125-134 
 135-141 
 142-152 
 
 . . 22 
 . . 23 
 24 
 
 'D 
 
 ... 14 
 
 r . . . . 
 
 T , . . . . 
 
 1-7 
 
 12 
 12 
 14 
 4 
 
 153-162 ." 
 
 25 
 
 163-173 .... 
 
 26 
 
 174-178 
 
 27 
 
 8-13 . . . 
 
 5 
 
 179-188 
 
 . . 28 
 
110 
 
 INDEX OF EQUATIONS. 
 
 189-196 
 
 . 21) 262-2(8 . 39 
 
 197-206 .'.......:'() 
 
 207-217 31 
 
 218-223 :\-2 
 
 224-233 33 
 
 234 34 
 
 235-238 35 
 
 239-2-12 :-',(> 
 
 243-251 37 
 
 252-261 . 38 
 
 266-275 40 
 
 276-286 41 
 
 28 -291 . 42 
 
 292-300 
 301-30(5 
 307-311 
 312-313 
 314-316 
 
 43 
 44 
 
 45 
 46 
 47 
 48 
 
SUMMARY. 
 
 PAGE. 
 
 Nomenclature 1-2 
 
 I. 
 GENERAL RELATIONS. 
 
 Conditions of equilibrium (I), (II) and (III) ... 4 
 Moment equations for single concentrated load (1) 
 
 and (2) 4 
 
 Moment equations for concentrated loads (8) and (9) 5 
 Moment equations for partial uniform loads (13), (14) 
 
 and (15) 5-6 
 
 Moment equations for uniform load over all (18) and 
 
 (19) 6 
 
 MOMENTS. 
 
 General equations for the moment over any support ; the modulus 
 
 of elasticity, E, alone, being considered constant. 
 Moment equation ( A ) 7 
 
 Concentrated loads- 
 
 Values of A, and B r (i) and (j) 
 
 Values of X;, X^ t and H r (A), (h) and (/) .... 9 
 Value of H;(g) .... 10 
 
 Partial uniform loads 
 
 Values of A,, and B r (74) and (76) 
 
 Values of A7 and AZ/ (78) and (80) 9 
 
 Value of #, (f), (82) and (83) 9-10 
 
 Value of H/ (g), (82) and (83) 10-11 
 
 Uniform load over all 
 
 Values of A r and B,. (75) and (77) 
 
 Values of A7 and X;_, (79) and (81) 9 
 
 Value of Jf y . (/), (84) and (85) 9-10 
 
 Value of If/ (g), (84) and (85) 10-11 
 
112 SUMMARY. 
 
 For all loads 
 
 Values of^., F/, F," and ,(6), (c), (rf) and () . . 11 
 Values of ft r , ,V, ', }',",, e* and d,, (m), (n), (o), (/), 
 
 (e), (p) and (r) 12 
 
 The modulus of elasticity, E, and tlie moment of inertia, I, being 
 
 constant. 
 
 Moment equation (A,) 12 
 
 Concentrated loads, values of A f and B,. ( i, ) and ( j, ) . 13 
 Partial uniform loads, values of ^1, and B, (86) and (88) 13 
 Uniform load over all, values of A,, and B, (87) and (89) 13 
 
 For all loads- 
 Values of c,,,, d, H and Y,(p,), (r t ) and (/,) 13 
 
 The modulus of elasticity, E, the moment of inertia, I, and the 
 length of the spans being constant. 
 
 Moment equation (A x ) 14 
 
 Concentrated loads, values oi A, and B, (L) and (,/,) . 14 
 Partial uniform loads, values of A,, and B,. (90) and (92) 14 
 Uniform load over all, values of A,, and B, (91) and (93) 14-15 
 
 For all loads- 
 Values of c w ,rf w and Y r (p,),(r s ) and (/,) 14-15 
 
 GENERAL EQUATIONS FOR SHEAR. 
 
 General equation (B), value of S, 15 
 
 General equation (C), value of &,! 16 
 
 Concentrated loads, values of Q ; .and Q/ (94) and (95) 16 
 
 Partial uniform loads, values of Q,. and QJ (96) and (97) 16 
 
 Uniform load overall, values of Q,and Q/ (98) and (99) 1(5 
 
 GENERAL EQUATIONS FOR INTERMEDIATE BENDING MOMENTS. 
 
 General equation (D) 16 
 
 Concentrated loads, value of L r (100) 5 and 16 
 
 Partial uniform loads, value of L, (101) 6 and 16 
 
 Uniform load over all, value of L,. (102) 7 and 16 
 
 GENERAL EQUATIONS FOR DEFLECTION. 
 
 E, alone, constant. 
 
 General equation ("), value of y, 17 
 
 Value of t r+i (45) 17 
 
 E and I constant. 
 
 General equation (";), value of y, ..-.-. 17 
 
 Value of t r +, (45a) 17 
 
SUMMARY. 113 
 
 II. 
 SUPPORTED GIRDERS. 
 
 GIRDER RESTING UPON T\VO SUPPORTS. 
 
 (a) JE, alone, constant. 
 Values of M, and M (103) . 19 
 
 Concentrated loads 
 
 Values or >S and ' (106) and (107) 19 
 
 Value of J/,' (113) 20 
 
 Value of y, (117)' 20 
 
 Partial uniform loads- 
 Values of S, and ' (108) and (109) 19 
 
 Value of M; (114) 20 
 
 Value of y, (117) ... See (12), p. 18 20 
 
 Uniform load over all 
 
 Values of S, and &' (110) and (111) 19-20 
 
 Value of MJ (115) 20 
 
 Value of M (116) 20 
 
 Value of ?/, (117) - - See (12), p. 18 . 20 
 
 (6) IE and I constant. 
 Value of y, (119) . . (concentrated loads) 21 
 
 A BEAM CONTINUOUS OVER THREE SUPPORTS. 
 
 (a) jEJ, alone, constant. 
 
 Value of J/ ; , M., and M (120) and (121) 21 
 
 Value of M (124) 21 
 
 Concentrated loads, values of A.> and B t (135) and (136) 23 
 Partial uniform load, values of A., and B, (137) 
 
 and (138) 23 
 
 Uniform load over all, values of A.> and B, (139) 
 
 and (140) " 23 
 
 For all loads 
 
 Values of c,, c., and ,(122) 21 
 
 Values of </ d* and d (123) 21 
 
 Values of A7, A7, H, and' H,' (125), (126), (127) 
 
 and (128) 22 
 
 Values of J,, z\ r , FJ, F" and F w (129), (130), (131), 
 
 (132) and (133) 22 
 
 Values of /V and K (133a) and (134) 22 
 
114 SUMMAHV. 
 
 Shears, equations (B) and (C) 15-16 
 
 Moments, intermediate (D) 16 
 
 Deflection (E) and (45) . 17 
 
 (6) E and I constant. 
 
 Values of J/ ; and AT--(141) 23 
 
 Value of M (142) ... 24 
 
 Concentrated loads 
 
 Value of M (150) 24 
 
 Values of A and B (144) and (145) 24 
 
 Values of S,, ', S, and S/ (153) to (157) .... 25 
 Partial uniform loads, values of A., and B, (146) 
 
 and (147) 24 
 
 Uniform loa'l over all 
 
 Value of M,( 158) 25 
 
 Values of S,, X/, ,% and ^'(159) to ( 163) 25-26 
 
 For all loads- 
 Value of Y, (143) 24 
 
 (c) JEj I and h constant. 
 
 Same as case (6), with Y., 26 
 
 (d) E, I, h and I constant. 
 Concentrated loads 
 
 Value of M (164) 2ii 
 
 Values of S,, ', S, and S' (165) to (169) . . W 
 
 Uniform load over all 
 
 Value of M (170) 26 
 
 NaluesofX,, X/', tf/ancLS,' (171) to (174) 26-27 
 
 A BEAM CONTINUOUS OVER FOUR SUPPORTS. 
 
 (a) JJEJ, alone, constant. 
 
 Values of J/, and 3/,-(175; 27 
 
 Values of M, and M( 177) and (178) 
 
 See General Relations for other equations .... 27 
 
 (6) E and I constant. 
 
 Values of M, and , I/ (179) and (180) 28 
 
 See General Relations for other equations . 
 
 (c] JEJ, I and h constant. 
 
 Values of M, and M (181) and (182) 28 
 
 See General Relations for other equations 
 
SCMMAKY. 115 
 
 ( d } H, /, h <in<l / cniixtniit. 
 \ allies of .17, and .17 ( . 1 S3) and (184) . 2S 
 
 rniform load over all- 
 Values of .17, and .U (185) and (18(3) . . . 2S 
 
 rniform load over all and spans equal 
 Values of .17, and J/:, (187) and (1S8) . . . . 2S 
 
 Values of X ; , X/, N,, >',', X, and X," (189) to (196) . . k 2l) 
 
 TIIK TIIM'ER. 
 
 ( d ) K, (done, ronx((tnt. 
 
 Value of //.,( 217) 
 
 Values of V., and )> (210) and (211) ....... 
 
 Values of .17, and .I/, (177) and (ITS) . . . . . 
 
 See General Relations for other equations 
 
 ( I) } E and I convlant. 
 
 Values of T. and Y (233) 33 
 
 Values of .17, and .17 (179) and (180) . . 28 
 
 See General Relations for other equation^ 
 
 III. 
 
 BEAMS WITH FIXED ENDS. 
 
 A MKAM KIXKI) AT ONE KM) A XI) SI l'l'( >RTKI > AT TIIK OT 
 
 Value of Mr- (234) 34 
 
 Concentrated loads, value of A,, ( 144) 
 
 Partial uniform loads, value of A., (146; 
 
 Uniform load over all, value of A., (148) . . ... 
 For all loads 
 
 Values of )'.,, B' and A, (236) to ( 129) 35 
 
 Values of 7'V, 7-; 11, and AY (131) to (238) 35 
 
 See ( General Relations for other equations 
 
 ( ft ) K ml I roiiNffi lit. 
 Value of .17, (240) . . . 36 
 
 Concentrated loads- 
 Value of .17, (241) 36 
 
 Value of )', (236) 35 
 
 Values of X, and X' (212) and (243) . . 36-37 
 
116 SUMMARY. 
 
 Uniform load over all 
 
 Value of .)/., (244) . . . :'>7 
 
 Value of K, (236) . . . 85 
 
 Values of X and X' (245) and (24(5) . 87 
 
 (c) .H, 1 "/id It constant. 
 
 Concentrated loads 
 
 Value of Mr- (247) . 37 
 
 Values of X and X,.' ( 24s ) and ( 249 , 8 , 
 
 ("niform load over all- 
 Value of M; (250) 37 
 
 Values of X, and S,' (251 ) and (252) . 8,-8S 
 
 Value of ,!/,( 253) . 
 Value of M.r. .I/, (254) . 
 
 Load at center of beam 
 
 Value of .JA (256) 
 
 Value of X, and X,.' (257) and (25X> . 8,s 
 
 Value of M (251)) or (260) . :'.S 
 
 \ r alue of Mas. :]/, (261) . 38 
 
 N'alue of//,, if/;, />, <> (22) or (2r,:V) . 3<) 
 
 A 1JEA.M FIXKD AT I'.OTII ENDS. 
 
 (a ) /, nlnin . a m slant. 
 
 N'alues J/, and .)/, (266) and (267) . 40 
 
 Values of"^'. A', A, &", ^ and );-(268) to (273) . . 40 
 See Generai Relations for other equations. 
 
 (/;) E and I constant. 
 Values of .!/, and M : (274) and (275) 40 
 
 ( c ) .E, I an d h con *ta-nt. 
 Values of .17, and M (276) and (277) 41 
 
 Concentrated loads 
 Values of ,]/, and M : . (27) and (27?h 
 
 Uniform load over all- 
 Values of .17, and : I/. (280) . 
 \'alues of X, and X,' (281) 
 
 Value of M (282) 41 
 
 N'alue of Mas. J/ (283) . . 41 
 
 Value of?/, (284) 
 
SrM.MAKY. 11, 
 
 A single load in the center of the beam 
 
 Values of M, and M (285) 
 
 Values of & and &' (286) . . 
 
 Value of J^ (287) or (288) ... 42 
 
 Value of Mas. #(289) . 42 
 
 Valueof^ (290) or(291) . . , 42 
 
 A I5EAM FIXED AT ONE END, AND I XST !'!'( ) KTET) AT TIIK OTIIEIl. 
 
 Concentrated loads 
 
 Value of M (21)2) . . . 43 
 
 X'alue of X, (292) .... 43 
 
 Partial uniform loads 
 
 Value of M,( 293) 43 
 
 Value of X, ( 294) 43 
 
 Value of M' (295) , . 43 
 
 Value of ;?/,, I constant (297) 43 
 
 Single concentrated load, value of y, (298) 43 
 
 Load at end of beam, value of y, (299) 43 
 
 Uniform load over all, value of ;//, (300). . . 43 
 
 A I5EAM OX TWO Sl'I'I'OUTS, AND ONE KXD 1 XSI I'I'oKTED. 
 
 Value of M (301) . 44 
 
 Value of X, ,SV and X, (302) to (304) . . 44 
 
 Values of .!/, and .I/, (305) and (30(i) 44 
 
 \ P.EA.M ON ONE Sl'J'I'ORT, HAVIN(i OXE END FIXED, AXD THE OTIIEI! 
 i NSIIM'OKTED. 
 
 ( a ) 12, afoiWj constant. 
 X'alues of .)/, and M :I (309) and (307) . . . 4") 
 
 ( b ) K an d 1 constant. 
 
 N'alues of M, and .]/,- (310) and ( 307 ) . 45 
 
 (c) H, I and h cr instant. 
 
 \' ulues of .I/,, and M (31 1 ) and (307) ... 45 
 
 A I'.KA.M ON T\\ () sri'l'ORTS, IfAXINt. XKITIIEK EN D SI I' I'OUTEI). 
 
 Values of M, and .17^ (312) and (313) 40 
 
1 18 SUMMARY. 
 
 IV. 
 THE mi NT OK XKUo MOMENT. 
 
 Load on the right, value of .-,. ('>\\<>) 47 
 
 Load on the left, value of .r r (317) 4s 
 
 Values of .r, by (Jraphics L8-49 
 
 T1IK CO-KFKICIKNTS <- AND </. 
 (Iraphical determination of the values of < and <l . . -">0-5i 
 
 APPLICATIONS. 
 
 Kxainple I. A continuous girder of three spans upon 
 level supports, the two end spans being equal. 
 Determination of bending moments and shears o2 
 Values of .17, and .17.. for each load Table (H) .... 08-54 
 
 Values of N, >./, >', >',', >', and X/ Table (/>) -V> 
 
 Values of .17,. Table' (c) -^ 
 
 Maximum mom^n'ts ">* 
 
 Maximum shear -V.t 
 
 Example 1. By graphics 60-ttl 
 
 Kxample '2. A. continuous girder of four spans with 
 uniform loads and a variable cross section. De- 
 termination of moments (;_!-<;!> 
 
 Kxample 2. With a constant cross-section 7<-71 
 
 A comparison of the results obtained, considering the 
 
 moment of inertia as variable and then constant . 71 
 Kxample o. A continuous girder of two spans. The 
 Sabula Draw uniformly loaded, and having a 
 variable cross- section. Determination of the mo- 
 ment over the second support 72-7(> 
 
 Example o. The same as Example 3, but with a 
 
 constant cross-section 77 
 
 Comparison of the results of Example 3 and Example 3</ 77 
 Example 4. The Sabula Draw, with concentrated 
 loads and a variable cross-section. Determination 
 
 of the moment over the second support 7< v -*0 
 
 Kxample 4. By graphics X0-x:l 
 
SUMMARY. 119 
 
 APPENDIX. 
 
 Demonstration of the equation of the elastic line . . . 85-88 
 General expression for the Theorem of Three Moments . 97 
 Demonstration of Equation (A) 89-100 
 
 TABLE I. 
 Values for k 2 F+F and fc 1? for all ratios -j- = k 
 
 from .001 to .999, inclusive 101-106 
 
 REFERENCES. 
 
 Some references to monographs, periodicals, &c., which 
 
 consider the theory of the continuous girder . . 107-108 
 
 INDEX. 
 Index of Equations . 109-110 
 
UNIVE1 
 
 OP CALIFORNIA 
 
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