THE INVERTED ARCH. [dlAP. XVI. 



The distance of P from the left is then half the loaded por- 

 tion, or kL Its distance from P', which acts at the centre of 

 the span, is I (1 k). 



1 1 nice we have for the left reaction A 



Axl = Pl(l-fy or A=P(l-K) = 2klm(Ify. 

 Also P'l = P x kl or P' = 2tflm. 



The chain reaction per unit of length is then 



P' 



= k 9 m. 



21 



Now let x be the distance to the point of maximum moment. 

 Now since at this point the shear must be zero, the weight 

 of the portion x must be equal to A (Art. 38). 

 We have then 



Ax- ^ = P(kl-x) = 

 whence, by substituting the value of A, 



x = 



.A. 35 .A. *7J 



But the maximum moment is A a; - = , and there- 

 fore, substituting the value of x above, 



_, k 3 (1 - k) . 



M max. = ? r-*- . 2 ? m. 



1.4-4 



This becomes a maximum for 1 & #" = 0, or for 

 k = $ VE - == 0.618034. 



Therefore, the greatest moment occurs when 0.62 of the span 

 is covered with the load. 



"We have then the 



Length of the loaded portion, = 2 k I = 0.61803 x 2 .1. 

 Reaction, A = 2klm(l-fy = ( 4/5 - 2)2Zm = 0.23607. 2/m 

 Chain reaction, P' = 2& a Zm = ( 4/5)2^ = 0.38196.2^. 

 Load per unit in loaded portion, or the difference between 

 the load m and the chain reaction m k* per unit of length 

 = m(l-F) = $(V5-l)m= 0.61803 m. 



The distance of the point of maximum moment is 



= 0.38196.81 = . 



