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



Center of moments at a, 

 140 V 3 = 80#i - (100 + 120)P- 140(1 -q)P = -78P for dead load. 



= SO#i =80 X-XP for live load. 



78 



V 3 =- -X 10,000= -5,570 Ibs. (-) D. L. 

 140 



ox 



-X 16,000= +7,830 Ibs. (+) L. L. 

 49 



2,260 Ibs. (+) Total compression. 



To find the maximum tension in V 2 , the rule for position 

 of the live load cannot be applied because it is possible to have 

 three different trusses with different combinations of counters 

 and main ties in action, depending upon the position of the 

 live load. If the main ties were capable of resisting compression 

 as well as tension and counters were not used, the regular rule 

 would apply. The tension in the post is produced by the upward 

 component of the stresses in the top chords. This is lessened 

 by the downward component of the stress in any diagonal 

 acting at the top of the post. Therefore, a maximum tension 

 in a post will occur with as much load as possible on the bridge 

 (to produce compression in the top chords) with no stress, or 

 as little stress as possible in the diagonals connecting at the top of 

 the post. 1 



Considering V 2 , we know that with loads 6 and 5 on, diagonal 

 D 3 is not in action. The problem is to find how much more 

 load may be added without bringing this diagonal into action. 

 We can write an equation for the total stress in D 3 with loads 

 6 and 5 on the bridge and enough more loads to make D 3 =0. 

 This will give us the position of the loads for maximum tension 

 in the post. 



Center of moments at a, 



107.5Z) 3 = 80#! - (100 + 120) (D. L. +L. L.) =0. 

 Solving this 



X(D. L.H-L. L.) =71,500 Ibs. total when Z) 3 =0. 



ovl 



ZP =30,000 11 )s. 



/ti=41,500 Ibs. L.L. when =D 3 =0. 



J For an extensive discussion of this problem see Mueller-Breslau's 

 Graphischc Siatik, Vol. I. 



