72 



STRESSES IN AN UNSYMMETRICAL TRUSS. Art. 52. 



lengthens D 2J and the shear in panel 5-7 lengthens D 3 . V l is 

 In tension because it must be directly opposite to P lt 



20' 



L 3 o.o 



-20- 



L, 



J SI 



u 



f?, * 19000 



Fig. 50. 



The algebraic solution is as follows : 

 0=45. SinO=cos 0=0.707. Sec = 1.414. Tan = 1.000. 



At joint 4, from 2 vert, comps. =0, we have V 2 =Q, and 

 at joint 7, from 2 hor. comps. =0, we have L 3 =0. These two 

 stresses v/ill, therefore, not appear in the stress diagram. 



Joint 1. >!=#! sec = 17000X1.414 =24000 Ibs. (+). 



Joints. V !=P t = 15000 Ibs. (-). 



Lj =L 2 = 17000 Ibs. (-). 

 Joint 2. D l cos 0-V l =D 2 cos 0. D 2 =D l - V l sec 0. 



D 2 =24000- 15000X1.414 =2800 Ibs. (-). 



U l =D l sinO+D 2 sinO =26800 X0.707 = 19000 Ibs. ( +) 

 Joint 4. U l = U 2 = 19000 Ibs. (+). 

 Joint 7. F 3 =# 2 = 19000 Ibs. (+). 

 Joint 6. D 3 cos = F 3 = 19000. 



D 3 = 19000X1.414 = 26800. (-). 

 Joint 5. D 2 cos +D 3 cos ,0 =P 2 =21000 Ibs. 



(2800 +26800)0.707 =21000 Ibs. Check. 



Only resolution equations have been used above, but in some 

 cases the moment equations are even simpler. Thus, for joint 1, 

 JW 2 =20# 1 -20L 1 =0; hence L^=R V 



53. Bow's Notation. The method of lettering most fre- 

 quently used for graphic solutions is one devised by Bow. By 



