EQUATIONS FOR CONTINUOUS BEAMS. 



1G1 



Multiplying (c) byJZ r and substituting in (h) 

 - El = 



For the span rs, by analogy, 



At the middle support (t) and (fe) must be equal. 



(h) 

 (i) 



(t) 



(43) 



This equation expresses what is called the theorem of three 

 moments for uniform load. In a similar manner, the theorem 

 of three moments for concentrated loads is derived. Its deriva- 

 tion is simplest when the locations of the loads are given as shown 

 in Fig. 113, in which /,-,. and fc r+1 are fractions whose values may 

 vary from to 1. The distances kl are sometimes measured 

 tm\ard the riirht in both spans, in which case, the form of the 

 equation of three moments is different in one term. The equa- 

 tion is (Fig. 113). 



kl )l; - S/' r+1 (k r+1 - kl +1 )ll +1 (44) 



Applying equation (43) to the case explained in Art. 96, 



tt l = ^wP and 3/ B = 

 Taking moments about B, 



RI y*wl y^wl = %wl as before. 

 Applying equation (43) to the case of Fig. 109, 

 Uf B l = - \wP and M B J s wP. 

 Rd =M B = - ^wP and R 3 = - &wl 

 R\l \wP = M R = -fswP and Ri == -f- ^ 



= I 



These shears act in the same direction, of course; the oppo- 

 site signs are gotten from opposite ends of the beam. 

 S = RI wx = -f$wl ivx 



= when x = {^l or x = I 

 M= R\x iwo; 2 



