Art. 153. STRESSES IN A PRATT TRUSS FROM WHEEL LOADS. 301 



Wheel 3 governs from G = 7x30 =210 until G = 7X50 =350, 

 or from wheel 13 at the end of the bridge until 33 ft. of uniform 

 load is on. 



Wheel 4 governs from (7-7X50-350 until G =7x70 =490, 

 or from 33 ft. of uniform load on until 103 ft. of uniform load 

 is on. This more than covers the 154 ft. of our span. 



These fields of shear are shown plotted in Fig. 207. The 

 maximum shear in any panel will occur when the wheel in whose 

 field of shear the right-hand panel point of the panel falls, is 

 at that point. Occasionally a -panel point will fall in the part 

 where two fields of shear overlap. In that case both wheels 

 must be used to find which will give the greater shear. 



The maximum stress in Z) t occurs with wheel 4 at joint 6. 

 This is the same position of the load that was found to give a 

 maximum stress in L l and L 2 , as it should be, because the stress 

 in L! is the horizontal component of the stress in D v 



With wheel 4 a t joint 6, 7^ = 192. 8 from the calculation for 

 Lj. The portion of the load in panel 6-7 which is carried by 

 the stringers to joint 7 is obtained by moments about 6. 



10X18+20X10+20X5 480 



it 7 = = - =Z1.O. 



22 22 



Shear in panel 6-7 = 192.8 -21 .8 = 171.0. 



D! =171.0 X =217.4 (-f)L.L. 

 28 



-3x29.15X^- 6 = H1.2 ( + ) />. L, 



328.6 (+) Total 

 Wheel 3 at joint 5, 14 ft. of uniform load on: 



= 133.4 



154 

 U= ?= 10.4 



Shear = 123.0 

 D 2 = 123.0 X^- =156.4 (-)L.L. 



28 

 =2X29.15X 



35.6 



28 



= 74.1 (-) D.L O 



130.5 ( -) Total. 



