Art. 130. STRESSES IN PRATT TRUSS BY COEFFICIENTS. 249 



Diagonal Stresses. 



D t = 3Psec6= 3X 7, 120 = +21,360 D. L. 

 = 3Psecd= 3 X 19,000= +57,000 L. L. 



+ 78,360 Total. 



D 2 



2Psecd= 2X 7,120 = 

 se c # =if X 19,000 = 



-14,2407). L. 

 -40,700 L. L. 



-54,940 Total. 



19,000 = -2,720 L. L. No reversal. 



Z> 3 = Psec6= 

 = P sec 0= 



7,120 = 

 ,000 = 



- 7,120 D. L. 

 -27,140 L. L. 



-34,260 Total. 



Dead load same as D 



= _ 8,140L. L. 

 =+ 7,120 D. L. 



1,020 Total. Reverses. 



D 4 = PsecO= fX 19,000 = -16,300 L. L. and total. 



The counter stress D is the same as the resultant shear in 

 its panel multiplied by sec 6. If there were no counter tie, D 3 

 would be a counter-brace and V 2 would also be in tension. 



The dead load is divided between the upper and lower panel 

 points. It is usually considered that one-half the weight of the 

 trusses and bracing is delivered at each chord and that all of the 

 weight of the floor system is delivered at the loaded chord. If q 

 equals the proportion of the total dead load panel load which is 

 delivered at the unloaded chord, the dead load coefficients for 

 the verticals will be as follows : 



D. L. Coefficient for V^l-q. 

 D. L. Coefficient for 7 2 = l-fg. 

 D. L. Coeffici3nt for V 3 =q. 



Assuming q in this case to be J, the following are the stresses 

 in the verticals : 



= - 3,600 D. L. 

 = 1 X 14,400 = - 14,400 L. L. 



-18,000 Total. 



