Art. 137. STRESSES IN A CAMEL-BACK TRUSS. 



Center of moments at 4, 



Arm of C/ 3 =?? =27.45 ft. 



271 



27.45/ 3 =60tf i - (20 +40)P = 120P, 



- 43,710 Ibs. (40 D.L. 



27.45 



120X16,000 



27.45 



= 69,940 Ibs. (+) L. L. 



113,650 Ibs. (+) Total. 



Center of moments at 3, 



Arm of U 4 =28 ft. 

 28C/ 4 =80Ri - (20 +40 + 60)P = 120P. 



_ 120X10, OOP 

 4 ~ 28 

 _ 120X16, 000 

 28 



= 68,570 Ibs. (+) L.L. 

 11 1,430 Ibs. (+) Total. 



If the main diagonals are to be tension members, they 

 ill have the same direction of inclination as in a Pratt truss, 

 as shown in Pig. 194. The position of the loads for maximum 

 stresses in the diagonals will follow the regular rule as shown 

 w Art. 129. 



For the end post the maximum stress will occur under full 

 load. Taking vertical resolutions at joint 7, 



D l 



ofi q 



L>i =3 X 10,000 X-- = 44,830 Ibs. (+) D.L. 

 18 



9fi Q 



=3Xl6,OOOX-- = 71,730 Ibs. (+) L.L. 

 18 



116,560 Ibs. (+) Total. 



For D 2 , with a section through U 2 , D 2 , and L 2 , and a center 

 of moments at b, the lever arm may be found by similar triangles. 



Arm of 



18 



80 



Length of D 2 



Arm of D 2 =53.5ft. 



53.5D 2 =40/2i -60P=60P, for dead load. 

 P, for live load. 



