CHAP. X.] SPECIAL CASES OF LOADING. 149 



These values are repeated upon the PL for convenience of 

 reference. Also, p I = 48 tons = 2.4 in. = n, p Z 2 = 4608 ft. 

 tons = 4.608 in. in. For the first pole distance [Art. 98 (3)] 



we have a X. = . ' ^ I j^ I 1 in. Second pole dis- 



tance [Art. 98] I = g X = 0.32 in. 



According to Art. 98, we have now, for the cross lines, 



OP = QR = ^\ 2 \ and O' P' = Q' P' = 



Laying off these distances under the supports, we have thus the 

 cross-lines. 



We have next to construct the second equilibrium polygon. 

 This, by the aid of the cross-lines and fixed points already con- 

 structed, we can easily do, as detailed in Art. 97 (3). Then 

 the moments at the supports are given directly to moment 

 scale in the interior spans, or we can find them from the end 



spans by laying off ^ j I [Art. 89]. 

 n 



Finally, the moments thus found and laid off at the sup- 



ports, we can construct the 'moment curve by making D' E' = 

 /? \ 2 / 1 \ 2 



r and DE i^X 2 - [Art. 98 (2)], and thus draw- 



ing the end tangents and corresponding parabolas. 



According to Art. 98 (3), we can then find the shearing 



n\ 



forces by laying off a = \ and drawing parallels to the end 



tangents to intersection with verticals through supports, as 

 shown in Fig. 



We thus have both moments and shearing forces for uniform 

 load. By careful attention to the above, the reader will have 

 no difficulty in solving any case. We recommend him earn- 

 estly to perform the entire construction for himself, referring 

 to the, proper Arts, at every step. [For convenience of size, 

 we have not observed our scales strictly in the Figs. T.he 

 reader should therefore not attempt to check results with the 

 dividers.] 



1OO. Partial uniform Load. 1. When the girder is only 

 partially loaded, as, for instance, a certain portion of the span 

 I from B, the simple moment area consists of a triangle ABO 



