of Continuous Girders. 187 



Making n = 2 and ti = 3, we get 



Now from (4) and (5) 



M P 



R 1 =-^ + P(l-A)=^[A-(A + 2a + 2a 3 )*+(2« + 2a 8 )* s ], 



where A denotes the quantity 3 + 8«-f-4a 2 . 



Knowing a, we can find the reactions for every value of k. 

 For instance, let a = 2; then A = 35, and 



R 2 = P (1 - 1 -3428 k 4- 0*3428 k 3 ), 



B 4 =0'0571 (*-*"). 



R 1= 2'972P and R 4 = O0214P. 



3. In a girder of two unequal spans, ends free, to find the 

 reaction at the pier due to a uniform load over a part of one of 

 the spans. 



ul I 



//////////////// 



A A— #— — R— A 



1 2 3 



Let one span equal /, the other ul, and let the uniform load be 

 placed between the limits k l -=^ ) A 2 = j; then we have 



In this case s = 2 and r = 2 ; hence from (2), 



., Ad 2 + Bd, A UivP 



2 2 2/(« + iy 2 ~2/(a + l)~ 128(1 + *) 



Then, from (6) and (7), 



Ms Ms _ 11m?/ wl _ (U + S2u)wl 

 *»"" ul + I +a ~128« + 4 ~ 128«~~* 

 Let u — \ y then 



27 

 R s = 64^ 



Suppose this same load to be concentrated at its middle point. 

 Here we have P= \ wl and k = \ ; hence A = -^ wl 6 and a = \ wl. 



