162 



EQUATIONS FOR CONTINUOUS BEAMS. Art. 97. 



= when x %l 



^ n ,ax == J&wP iX-^Z 2 = J&wP for x = T y 

 = -ifWP when x = I 



Applying equation (44) twice to the case explained in Art. 

 95, the same results are gotten. It is necessary to consider the 

 beam as one span of a continuous beam of many equal spans, 

 similarly loaded, as shown in Fig. 116, in order that the elastic 

 line may be horizontal at the supports. There will evidently be 

 but two different moments at supports. 



'^L-^-ft.L b-f( t i^\ 



I <- +-T-4 



M I( L = - 



Pit* 



2M T = PLk 2 (lkl ) (a) 



This is for the left-hand and central spans Fig. 116. In like 

 manner, for the right-hand and central spans 



Subtracting (a) from (6) 



Since hL = h fci = 4~ . Also hL = /, h = -^- 



+ 2^ + H - 5) 



(/? + 2 M, + 5 



3f q ^ - P 2 as given under Fig. 99. 

 Substituting this value in equation (a) or (5) 

 M T = P as by the method of Art. 95. 



Applying equation (44) to the case of Fig. 112, 

 c nl= - P(k P)* 2 and 



M = 



n 

 2n+2 



which in the first equation gives 



