80 STRESSES IN A TRUSS WITH PARALLEL CHORDS. Art. 5S 



30000X4015000X20 



18 



M 8 =-0 or (/^-i/o 

 30000X20 



= 50000 Ibs. (+) 

 20 = 18L 2 = 30000X20. 



18 



= 3330011*. ( ) 



Section vw. 



2 vert, comps. = or shear in panel 9-10 = vert. 



comp. D 3 = 0. Hence Z> 3 = 0. 

 M 4 = Q, or 30000X40 15000X20 = 1817,. 

 Us = 30000X40-30000X10 _ r )()m ^ (+) 



Jtf 10 =0 or 30000X60 15000(40+20) =18 L v 

 30000X 60 30000X 30 



18 



= 50000 Ibs. ( ) 



Since the loads are symmetrical about the center line, and 

 Z) 3 = 0, the stresses will be symmetrical. It will be noted that 

 the stresses U^ and L 2 are equal and opposite as they should be 

 in order that the sum of the horizontal components, for section 

 rs, shall be zero. Likewise must U 2 =- L 3 and Z7 3 = L 3 . It is also 

 apparet at joint 9 that U 2 = U 3 and V s = 15000 Ibs. 



The signs of the stresses are apparent upon inspection. The 

 top chord is in compression, the bottom chord in tension, the di- 

 agonals in tension, and the verticals in compression. 



If the loads were applied at the lower joints in place of the 

 upper ones, the signs of the stresses would be the same, but V 3 

 would be zero (joint 9), V 2 would be decreased by P, and V 1 by 

 %P. The stress in each post is decreased by the amount of load 

 transferred from its upper to its lower end. 



/"ores 



Fig. 59. 



60. Graphical Solution by Sections. Culmann's 

 Method. Fig. 59 shows a part of a truss cut off by a section 

 for the purpose of finding 'the stresses U, D, and L. R is the 

 resultant of the external forces R v P v and P 2 ; thus the 

 system is reduced to four forces in equilibrium R t U, D, 



