90 SIMPLE GIRDERS. [ciIAl'. VII 



We have, therefore, the simple construction given in PI. 13, 

 Fig. 42 (I). The positive and negative values of S equally 

 distant from the right and left supports are equal, so that it is 

 only necessary to construct S for one value. [See also Art. 

 78.] 



If the second load is to be at the cross-section, and if e is the 

 distance between the first and second, we draw first a line 



P ff* 



whose equation is y = P t - , and construct, as above, a 



I/ 



polygon, for which the second load lies on the right support B, 

 and whose second side (between second and third loads) coin- 

 cides with the above line. The ordinates to this line above the 

 axis of abscissas will give maximum of + S. 



73. maximum moments. Since, according to Art. 70, a 

 concentrated load causes a negative moment at any point, 

 wherever it may lie, we must have evidently loads upon both 

 sides of any point, in order that the moment may be a maxi- 

 mum. Since a single load causes a greater moment at any 

 point the nearer it lies to that point, the greatest load must lie 

 nearest the cross-section in question. The method of loading, 

 causing maximum moments, can be best determined for a dis- 

 tributed load (not necessarily uniform). In this case the equi- 

 librium polygon becomes a curve [PI. 13, Fig. 43]. If in this 

 curve we draw A B, and take C so that A. G : C B '.'.x : I x, 

 then C D = M for x. Suppose A B to take the position A' B', 

 the horizontal protection of C C' being indefinitely small, 

 then C' D' = M + d M. In order now that M may be a 

 maximum, C' D' must be equal to C D or C C' parallel to D D'. 

 If in the force polygon O A t is parallel to A A', O B t to B B', 

 and O D t to D D', then A t D x and D x Ej are the loads upon 

 A C and B C. 



Draw through C a vertical, and through A, A', B, B', paral- 

 lels to C C' or D D' intersecting this vertical in E, E 7 , F, F'. 



Then CE:CF::AC:BC::a;:Z a?, 



E' : C F' : ; A' C' : B' C' : : os : I - X-, 

 therefore 



C E : C F : : C E' : C F', or C E' : C E : : C F' : C F ; 

 also CE'-CE:CF'-CF::CE:CF, 



that is, E E' : F F' : : x : I - x. 



If now we draw through A' and B' parallels to C C', or D D' to 



