CHAP. Xm.] ANALYTICAL FORMULAE. 239 



But GI = 0, <% = 1, 



2 ft + 4) J ,7 



=- ~ and =- 



hence, since B = P I? (k $*), 



_2P4 2 ( 

 ~ 



If in this we make 4 = 4> we have the extreme spans equal, 

 and then 



2py(*-ff)ft + 4) 

 4 (4 + 4) 2 - 4 2 



If we make in this, again, 4 = 4, we have for K spans equal 



4PZ(^-^) 

 "16 -- ' 



just what we should have from Art. 149. 

 For the reaction at the end support, we have 



or, since M x = 0, 



For all spans equal, or 4 = 4 = 4 = ^j this reduces to 



as we should have found from Art. 149. 



Ex. 1. A beam of one span is fixed horizontally at the 

 right end ; what are the reactions and the moments for concen- 

 trated load f 



Here s 1=1 or = 2, r = 1, 4 == 0, an d from the for- 

 mulae of Art. 155, GI = 0, <% = 1, and d\ = 0, d^ = 1, d$ = 2, 



M -0 M --<7 A fr 



"M W J "* ^2 ~7 



^ = = 



= - 2 + P (1 - k) = (2 - 3 k + % 



