Art. 137. STRESSES IN A CAMEL-BACK TRUSS. 275 



If this value of R t comes out less than the value of Ti^ for 

 loads 5 and 6 alone, use the greater value. This would show 

 that Z>3 could not reverse. 



With this value of R 1 the live load stress in V 2 may be 

 obtained without actually finding the exact position of the live 

 load which produces the stress. We know that loads 5 and 6 

 are on, plus some more which are beyond the section, and do 

 not enter into our moment equation. 



Center of moments at 6, 



80 F 2 =40#i - (60 +80)P =40 X41,500 - 140 X 16,000. 



7 2 =7,250 Ibs. (-) L.L. 

 D.L.= 500 Ibs. (+) From former calculation. 



6,750 Ibs. (-) Total tension. 



The case of V% is much simpler. It is evident that the 

 maximum tension occurs under full load as D 4 =0 for a full 

 load. Therefore, with a center of moments at a, 



140 F 3 = SOtf ! - (100 + 120 + 140)P = - 120P for live load. 



140 

 Dead load = 5,570 Ibs. (-) 



19,280 Ibs. ( -) Total tension. 



A graphic method for finding all the maximum stresses, 

 except the tension in the posts, which requires but two stress 

 diagrams, is often used for trusses with inclined chords. This 

 is illustrated in Fig. 195. 



Fjg. 195 (6) is drawn for panel loads of unity, distributed 

 between the upper and lower joints in the same proportion as 

 the dead load. If the stresses scaled from this diagram be mul- 

 tiplied by the dead load panel load, the dead load stresses in 

 all the members will be obtained, and if the scaled stresses for 

 the chords and end post be multiplied by the live load panel 

 load, the live load stresses in these members will be obtained, as 

 these get their maxima under full load. For this purpose only 

 the stresses in the members on the left of the center of the truss 

 should be scaled, as they are the ones in action for full load. 



Fig. 195 (c) is drawn for a load at joint 1 only, such that 



