224: CONTINUOUS GIRDEB. [CHAP. XIII. 



of a girder over any number of supports, att spans equal. For 

 any desired span, we can find the maximum positive and nega 

 tive moments by the cases of Fig. 80, as also the moments due 

 to various positions of the weight P. We can also find the re- 

 actions at all the supports due to these cases. From the reac- 

 tions and known forces, we can then easily find the algebraic 

 sum, or shear, at any support. The moment and shear at any 

 support due to any case of loading are, as we have seen, the 

 quantities required for calculation. 



Now it is not necessary to find all the reactions in order to 

 obtain the shear. The moments at the supports being known } 

 we can find the shear directly. 



Thus, for concentrated load in a span l t (Fig. 89) we have for 

 any point a? 



M,. - S r x + P (4 - a) - m = 0, 



where S r is the shear at the left of the loaded span, and m is 

 the moment at any point. We see at once that, to determine 

 this moment, it is the shear that we wish, and not the reaction. 

 For a uniform load we have similarly, 



ivr . W V? 



M,. S r x + ^- m = 0. 



If in both these equations we make x = 1 I} m becomes Mj+i, 

 and we have 



s. = 



where q = P (1 Jc) for a concentrated load, and q = -^ for 



uniform load. 



In an unloaded span at the left support, or when m < r t 

 q disappears, and we have 



i_ o 



when m < r, S m = 



4- 1 



For the shear at the right support of the loaded span we have 

 simply S r P or S r w I, and hence 



