238 



STRESSES IN SIMPLE TRUSSES. 



Art. 127. 



and D 3 must be in compression and D 2 in tension. On account 

 of the symmetry, D^D^ Z> 2 =D 5 , Z) 3 =Z> 4 . 



Live load stresses. If P 1? P 2 , and P 3 are considered separ- 

 ately, the stresses due to P 4 and P 5 may be gotten by inspection 

 because each load is equal to 16,000 Ibs., and P 5 , for example, will 

 produce the same stress in D 2 as P produces in D 6 . 

 Stresses for P 1 =16,000 Ibs. 



1?,=% of 16,000=2,670 Ibs. 2 =of 16,000 = 13,330 Ibs. 

 Sect. 1-1. D 1 =R 1 sec. 45=2670X 1.414=3780 Ibs. 



+Df= 7) 2 =+Z> 3 = Z) 4 =+Z> 5 since the shear that 



each must carry is the same and since the shear is up in 



each panel, the signs alternate as the directions of the 



diagonals alternate. 

 Sect. 6-6. D 6 =E 2 sec. 45=13,330X 1.414=18,850 Ibs. 



Stresses for P 2 =16,000 Ibs. 

 Sect. 1-1. .B 1= f of 16,000=5,330 Ibs. 



R 2 =$. of 16,000=10,670 Ibs. 



D l =R l sec. 45=5,330X 1-414=7,550 Ibs. 



+Z) 1 =-Z) 2 =+Z) 3 =-7) 4 

 Sect. 6-6. Z> 6 =# 2 sec. 45=10,670X 1.414=15,100 Ibs. 



Stresses for P 3 =16,000 Ibs. 



R l= y 2 of 16,000=8,000=7?.. 



D!=R! sec. 45=8,OOOX1.414=11,310 Ibs. 



For P 3 the stresses are symmetrical. 



