CHAP. X.] SPECIAL CASES OF LOADING. 153 



We can thus find the cross-lines for any position of the load, 

 and for each position can, if we wish, draw the equilibrium 

 polygon and determine the moments according to the general 

 method of Art. 97. 



1O1. Concentrated Load. The simple moment area is in 



this case a triangle [PI. 16, Fig. 65] whose area is - I h, h be- 



2i 



ing the height C D. Therefore 



If from the centre of the span E we lay off E F = - E D, D 



o 



being the point of application, a vertical through F passes 

 through the centre of gravity. 



As the height C D is proportional to 9JZ, we may take C D as 

 second force polygon. Since h = 2 3ft, the distance of the pole 



N must be 2 x \ I = \ I, when X = I (Art. 89). Draw N P 

 o o 



parallel to A B, then is N P = - A B. 



o 



Parallel to N C and N D we may draw the cross-lines. A 

 simple construction may be given for them when they are made 

 to pass through A and B. Let A M and B L be the cross-lines. 

 Then the triangle 3 B M is similar to N C D, and 



B M : C D : : B F : N P. But 



and N P = A B ; therefore, 



B M : C D : : (2 A B-A D) : A B. 

 Make D G = A B, then B G = 2 A B-A D. 



The point M is accordingly found by drawing a straight line 

 through G and C, D G being equal to A B. In the same way 

 make D H = A B, and draw a line through H and C. We 

 thus obtain L. 



The prolongations ofJMLG andl* C, therefore, intersect A B 

 prolonged in the points G and H, distant from D by A B. 



If, then, we have to investigate a concentrated load in various 

 positions, we draw the first equilibrium polygon C X and C Y 

 [PI. 16, Fig. 66], and lay off in the same the closing lines (for 



