CHAP. X.] SPECIAL CASES OF LOADING. 151 



The intersection R of the line Q I with A B lies, then, in the 

 vertical through the centre of gravity of the simple moment 

 area. 



Thus the construction of the cross-lines is now easy. 



4. It is most convenient in the application of the above to 

 construct or calculate the distances of the cross-lines under the 

 supports once for all, for load over various parts of the span. 

 The necessary formulae can be directly deduced. Thus the 



Trianle BAT=BT x -l = - x I. 





Triangle BCT = BT x s^ls^-X H 



*j *j Cb i 



Triangle BLT = BT X :/3Z = - x I. 



O 2l Cb O 



The entire area is equal to the triangle ALB. 



But A L B = B A T-B L T = <? (1 l-\ ft l\ 



& u \& 3 / 



The triangle ACB = BAT BCT; hence 



The parabolic area is equal to the entire area minus A C B, 



D /3 s 1 



or parabolic area = ^~~ X- 8 I. 

 2 a 6 



F is distant from B T by a distance 61, N by a distance = 



- 1, NO is - of N F ; hence O is distant from B T 

 o 



Therefore, the moment of the triangular area, with reference 

 to the right support B, is 



The moment of the parabolic area is 



The total moment, with reference to the right support, is, 

 therefore, 



