AET. 7.] SUPPLEMENT TO CHAP. XEH. 249 



supports for load between the first two. Let r 1 = 1, or r 2. Then, 

 since M, = M = 0, we have 



2M a (Z, + Z a )+M, ? a = -T a + PZ 1 (fc-yfc 3 ) = R . . (10) 

 where R stands for convenience equal to the expression on right. 

 Let r 1 = 2, or r = 3. Then the weight disappears, and since Zj = l^ 



Prom (11) we have 



But since R a must always equal R 8 , we have from (9) 

 M, - M, 2 M a - 2 M, 



Substituting (12) in (10), we have 



-Rfa 



3Z 3 a + 



Substituting (12) in (13), 



_ .- a . + , 



3 k a + 8J,*i + 4Ji l ' 



From (14) and (15), we have then 



Y, R = li Z a P L 

 Insert in this the value of R from (10), and 



Y a - Y, = P I, 3 (k- k*) + P I, Z a k = P (IS k-W+lilt *). 

 Now in the present case Ai = 0, h t = 0, and h a = h, and since also 



and Y a = 6Bi + . That is, T a = - T,. 



Hence, from our equation above, 



Y, = -|- 



Substituting these values of y a and y in (11) and (13), we can obtain at 

 once M a and M 3 , which finally substituted in eq. (9), will give us the re- 

 actions as already given in Art. 120, when we put n I in place of Z a . 



7. In similar manner we can solve other problems. Thus what are the 

 reactions for a girder continuous over three supports, the two right-hand one 

 resting upon an inflexible body which is pivoted at the centre f 



This is the case of the tipper when raised at the centre so that the ends 

 just touch, and then subjected to a load at any point of first span the 

 other end not being latched down, so that it rises freely, as though without 

 weight of its own. 



