114 



RESISTANCE OF MATERIALS 



Solution. The first step is to find the tensions in the belts. Since power is the 

 rate of doing work, and 1 H.P. = 550 ft.-lb./sec., the formula for power may be 

 written 



Fv 



(164) Horse power = 



FIG. 



where F denotes the effective force, or difference in tension in the two sides of the 

 belt, expressed in pounds, and v is the belt speed in ft. /sec. Hence, for the pulley 

 of 8 in. radius transmitting 30 H.P., we have 



60 550 



whence F = 4730 Ib. Since by assumption the tension on the tight side of the belt 

 is twice that on the slack side, their values are 



Tension on tight side = 9460 Ib. 

 Tension on slack side = 4730 Ib. 



The belt tensions for the other 

 pulleys are calculated in a simi- 

 lar manner, the results being 

 indicated on Fig. 89. 



Considering the shaft as a 

 beam, the load at each pulley 

 is equal to the sum of the belt 

 tensions for that pulley, as 

 shown in Fig. 90. The reac- 

 tions of the bearings and the 

 bending-moment diagram are 

 next obtained, the results being 

 given in Fig. 90. 



The maximum bending and twisting moments thus occur at pulley No. 1, their 



numerical values being 



M b = 364,230 in.-lb., 



M t = 37,840 in.-lb. 



Moment diagram 

 FIG. 90 



