Art. 137. 



STRESSES IN A CAMEL-BACK TIM SS. 



273 



Counters are required in the three middle panels only. 



Another method of getting the stresses in the diagonals is 

 to consider the horizontal components at the lower joints. This 

 will require the calculation of the live load stresses in the lower 

 chords for the position of the live load giving a maxiuji?m stress 

 in each diagonal; for instance, at joint 5, 



D 2 sin QI = L 3 L 2 . 



For a maximum stress in D%, loads 1 to 5 inclusive must 

 be on the truss, giving a reaction R 1 =^-P. 



ScS^OOlbs. (-) 

 57,140 Ibs. (-) 

 D 2 sin #! = 57, 140 -38, 100 = 19,040 Ibs. 



OA Q 



D 2 = 19,040X^ = 25,620 Ibs. L.L., 



which corresponds with the value found by the moment equation. 



As stated in the beginning of this article, the vertical posts 

 are subject to reversals of stress and each kind of stress must 

 be calculated. 



Vi is a suspender and gets whatever load is applied at 

 joint 6. 



Vi = (l-g)P =0.7X10,000- 7,000 Ibs. (-) D. L. 

 = IP =16,000 Ibs. (-) L.L. 



23,000 Ibs. (-) Total. 



For maximum compression in the posts, the position of the 

 loads follows the regular rule, as was proven in Art. 129. 



Center of moments at b, 



807 2 =407? 1 -60P-SO(l-g)P-4P for dead load. 



=40^i =40 X XP for live load. 



7 



F 2 = -X10,000 = 500 Ibs. ( + ) D.L. 



oU 



=-X 16,000 = 11, 430 Ibs. (+) L.L 



11,930 Ibs. (+) Total compression. 



