206 CONTINUOUS GIRDER. [CHAP. XIIL 



load over whole length of girder from end to end, and for uni- 

 form load over the whole of any single span. It may be writ- 

 ten as follows : 



m+ + . 



If we suppose only one of the two adjacent spang as l m to 

 contain the full live load w, while all the spans are of course 

 covered with the dead load u, the above equation becomes 



M m 



+2 m+1 = 



If both spans bear the same uniform load u alone, 

 M m l m + 2 M^ [Z m + k+d + M m+2 l m+1 = [4 4- 



If the spans are equal^ the above two equations become re- 

 spectively 



M,, I + 



and 



Now in every continuous beam, whose extreme ends are not 

 fixed, two moments are always known, viz., those at the ex- 

 treme supports, which are always zero. Hence, by the applica- 

 tion of this theorem, we can form in any given case as many 

 equations as there are unknown moments, and then, by solving 

 these equations, can determine the moments themselves. 



132. Es&mple Total uniform Load all Spans equal. 

 Thus let it be required to find the moments at the supports for 

 a beam of seven equal spans, uniformly loaded over its whole 

 length. The moments at the end supports M t and Ma are zero. 

 We have then, by the application of the last equation above, the 

 following equations : 



For the first three supports 1, 2 and 3, m = 1, and 



A TUT 7 , r J U>P A *.* UP 



4 M2 Z + MS Z = , or 4 M 2 + M 8 = . 



2i A 



For supports 2, 3 and 4, m = 2, and 



11 7 s 

 M z + M 3 + M 4 = -. 



