So 



STRESSES IN BRIDGE TRUSSES 



Now move the loads a short distance to the left, the distance being 

 assumed so small that the distribution of the loads will not be changed, 

 and 



-P l (y l -dy 1 )+P t (y t dy t )+P t (y t + d yt ) (e) 



Subtracting (d) from (e) and solving for a maximum 



But 



i = d x tan a x = d x - 

 n / 



d y 2 = d x tan a 2 = d x 



n I 

 nl 



d y 3 = d x tan a 3 = d x - 



nl 





FIG. 5ob. INFLUENCE DIAGRAM FOR SHEAR. 



and substituting we have 



dx 

 1 ~nl 



n i d x 



- -- |-P 8 = 

 nl nl 



t P 2 (n 



and 

 and 



