236 CONTINUOUS GIRDER. [CHAP. ML 



For the right support, 



where the expressions for 0, 0', A, 7, 7', /3, /3', a, a', are to be 

 taken from the tables and inserted. 



Thus, for five spans load in second span from right, or third 

 from left, we have at once A = 45 + 104 n + 60 n z . For the 

 moment at the left support, to find 0, we must take horizontal 

 line for III., and thus find (2 + 2 n). For 7', /3' and a' we 

 must take II'., and find, therefore, 3+4 n, 2 + 2 ft. and 3 + 2 n. 

 Hence, moment at left support is 



For moment at right support, we must take line II'. for 0' and 

 line 111. for 7, /S and a, and hence 



M ' =M = 45+104n+60n* [(8+4)*+8(2+8n)*- (9+10 )*']. 

 Since the span in question is not an end span, 1^ = 1 and 



*-- 



B 



For a load in an end span, use the formulae of Art. 150. 

 For a load in middle span of five spans, i.e., third span from 

 each end, we have 



M=M >= 



When k = -, both these moments become, as they should, equal 



for any assumed value of n, as the reader may readily prove by 

 insertion. 



For the other moments not adjacent to the loaded span, the 

 rule of Art. 140 holds good. 



Thus, Mk = 0, M^lM,, M 8 = ^M r , Mi = jM^ etc., 



V 



