126 



PRACTICAL STRUCTURAL DESIGN 



Wl 



The compression and tension per panel in the chords = 



UP 



therefore compression in eg = tension in fh = 25,000 Ibs., for the 

 ratio "3 = TK = ! The compression in de = tension in cf = 



L J.V/ 



25,000 + 15,000 = 40,000 Ibs. The tension in&c = 25,000 + 15,000 

 + 5000 = 45,000 Ibs. 

 The object of the computations being to obtain the stresses so 



Fig. 80. 



the members may be proportioned, the method above given of 

 following the loads from joint to joint and obtaining the coefficients 

 for uniformly and symmetrically loaded beams, or of obtaining 

 the shear at panel joints for unsymmetrically loaded beams, is 

 adequate and simple. 



It can be proven that a truss is merely a skeleton beam by 

 finding the shear and bending moment at each joint and then divid- 

 ing the bending moment by the depth obtain the stresses in the 

 chords, the web members carrying the shear. In Fig. 80 the end 

 reactions each equal 25,000 Ibs. Then 



M, at/ = 10 x 25,000 = 250,000 ft. Ibs. 



M , at g = 0, for the top chord rests on gh. 



M, at c = (20 X 25,000) - (10 x 10,000) = 40,000 ft. Ibs. 



M, at e = 10 x 25,000 = 250,000 ft. Ibs. 



