GIRDERS AND TRUSSES 



127 



M,atb = (30x25,000) -(10x10,000-1-20x10,000) =450,000 ft. Ibs. 

 M,atd = (20 x 25,000) - (10 x 10,000) = 400,000 ft. Ibs. 



Dividing the moments by the depth, 



250 000 

 tension in fh = compression in eg = - ~ = 25,000 Ibs., 



400 000 

 tension in cf = compression in de = - - = 40,000 Ibs., 



450,000 

 tension in oc = ^ = 45,000 Ibs., 



Fig. 81 shows a truss having an odd number of panels. There 

 is no stress in the dotted cross diagonals in the middle panel except 



in case of wind or rolling loads, or otherwise unbalanced loading. 

 Coefficients may readily be written for uniform and symmetrical 

 loadings for this case, or the loads may be followed from the point 

 of zero shear in cases of unsymmetrical loading, or the shear method 

 may be followed. 



In Fig. 82 (c) is shown a truss with a subvertical and sub- 

 diagonal at each end. Such an arrangement involves the con- 

 sideration of an additional triangle in which half the weight is 

 added to the load at 6 and is then carried to a, the other half being 

 added to the load at c. This arrangement offers no difficulty when 

 figured by the shear method, but sometimes causes trouble and 



