F. 



458 Royal Society : — 



let the lower limits be #=0. 



jFSf-*. ff*ff- ■ 



When the integrations extend over the whole span /, that will be 

 denoted by suffixing 1 ; for example, m lt n x , &c. 



Let — P be the upward shearing-force exerted close to the point of 

 support (#=0), <1> the bending-moment, and T the tangent of the in- 

 clination, positive downwards, at the same point. Then, by the general 

 theory of deflection, we have, at any point x of the span I, the follow- 

 ing equations : — 



moment, $=$ — ¥oc-\-m\ (2) 



deflection, y=Ta?— Pg + $o w + F (3) 



Let O x be the moment at the further end of the span I, and sup- 

 pose it given. This gives the following values for the shearing-force 

 P and slope T at the point (#=0) : — 



p. »»-*i + "> , (4) 



and because 2^=0, 



1- | ~*o^ Z 2 t J f T- p I' ' • W 



Consider, now, an adjacent span extending from the point of 

 support (<z=0) to a distance (— oe=V) in the opposite direction, and 

 let the definite integrals expressed by the formulae (1), with their 

 lower limits still at the same point (#=0), be taken for this new 

 span, being distinguished by the suffix —1 instead of 1. Let — T' 

 be the slope at the point of support (#= 0) Then we have for the 

 value of that slope, 



\l> 2 V ) I' 2 + V 2 V ' ' • l ; 

 Add together the equations (5) and (5A), and let t=T— T' denote 

 the tangent of the small angle made by the neutral layers of the 

 two spans with each other in order to give imperfect continuity. 

 Then, after clearing fractions, we have the following equation, which 

 expresses the theorem of the three moments in Mr. Heppel's theory : 



+ m 1 q 1 l' 2 + m_ 1 q_ l l 2 -FJl' 2 -F_ l l'l 2 -tl 2 P. J ' (6) 



That equation gives a linear relation between the bending mo- 

 ments &__ v *o» $1 at an y three consecutive points of support, and 

 certain known functions of known quantities. In a continuous girder 

 of N spans there are N — 1 such equations and N — 1 unknown 

 moments ; for the moments at the end of most supports are each =0. 

 The moments at the intermediate points of support are to be found 

 by elimination ; which having been done, the remaining quantities 



