SIMPLE TRUSSES 5 



If the diagonal inclines in the opposite direction, as in 

 a Howe truss, the construction of the influence diagram 

 remains unchanged but the character of the stress is reversed. 



Load-position for Maximum Moment. For uniform 

 loads, the influence diagram DEGFB at once indicates the 

 portions of the span which are loaded to produce like 

 moments and hence maximum stresses. 



The criterion for giving the position of wheel loads 

 producing the maximum moment is readily found from the 

 influence diagram. For cases which usually occur in 

 practice, the portion of the span on the left of the cut 

 stringer may be considered as unloaded. For convenience W 2 

 is assumed to represent all of the loads between the verticals 

 through E and G concentrated at their center of gravity, 

 W 2 all of the loads between the verticals through G and F, 

 and Ws all of the loads between the verticals through F 

 and B. Let 2 2 , 22' and 23, be the ordinates of the influence 

 diagram directly below the wheel loads W 2 , W 2 and TF 3 . 

 The moment is 



M, = W 2 (Z2)-W2'(Z2')-W3(ZB). . . . (10) 



If the loads move toward the left a distance dx, and no 

 additional load comes on the span from the right and no 

 load moves off the left end of the cut stringer, the moment 

 becomes: 



MJ = W 2 (z 2 - dx tan 3 + to tan 2 ) 



-W 2 (z2 + dx tan 6 3 - dx tan 2 ) - W 3 (z 3 + 5 * tan to). 



The difference between these moments is, 



M,' -M,= 8M 8 = W 2 ( -tan 3 +tan 2 ) 8x 



3 -tan 2 ) 8x -W* (tan 3 ) 8x. (12) 



