108 RESISTANCE OF MATERIALS 



and for a hollow circular shaft, from equation (139), 



32 M t l 



If Jf, is known and 6 can be measured, equations (140) and (141) 

 can be used for determining G. If G is known and 6 measured, these 

 equations can be used for finding M t ; in this way the horse power 

 can be determined from the angle of twist. 



63. Power transmitted by circular shafts. Let H denote the 

 number of horse power being transmitted by a circular shaft, n its 

 speed in revolutions per minute (R.P.M.), and M t the torque, or 

 twisting moment, acting on it, expressed in inch-pounds. Then, since 

 the angular displacement of M t in one minute is 2 Tra, the work 

 done by the torque in one minute is 2 7rnM t . Also, since one horse 

 power = 33,000 ft.-lb./min. = 396,000 in.-lb./min., the total work 

 done by the shaft in one minute is 396,000 H. Therefore 



2 7m Jf, = 396,00077; 

 whence 



(142) M t = 396 ' OOOJ * = 6 3,030 in.-lb. 



2nn n 



Therefore, if it is required to find the diameter D of a solid circular 

 shaft which shall transmit a given horse power H with safety, then, 

 from equation (138), 



16 M t 321,000 H 



whence 



(143) ^, = 68.5^. 



As safe values for the maximum unit shear q', Ewing recom- 

 mends 9000 lb./in. 2 for wrought iron, 13,500 lb./in. 2 for steel, and 

 4500 lb./in. 2 for cast iron. Inserting these values of q' in formula 

 (143), it becomes 



(144) D = 



where for steel /*= 2.88, for wrought iron p= 3.29, and for cast 

 iron i = 4.15. 



