STRENGTH OF MATERIALS 



and for a hollow circular shaft, from equation (56), 

 (58) 



If M is known and 6 can be measured, equations (57) and (58) can 

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

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

 which a rotating circular shaft is transmitting can be determint'<l. 



Problem 113. A steel wire 20 in. long and .182 in. in diameter is twisted by 

 a moment of 20 in. Ib. The angle of twist is then measured and found to be 

 6 = 18 31'. What is the value of G determined from this experiment .' 



Problem 114. If the angle of twist for the wire in Problem 118 is 6 = 40, how 

 great is the torsional moment acting on the wire ? 



95. Power transmitted by circular shafts. Let H be tin* number 

 of horse power transmitted by the shaft, and n the number of revolu- 

 tions it makes per minute. Then, if q is the force acting on a particle 

 at a distance r from the center, the moment of this force is qr, ami 



consequently the total moment transmitted by the shaft is M = I 7 /. 



Also, the 'distance traveled by q in one minute is 2 urn, and there- 

 fore the total work transmitted by the shaft is 



W 



I 27rmq. 





Since 1 horse power = 33,000 ft. Ib./min. = 396,000 in. lb./min., tin- 

 total work done by the shaft is 



W = 396,000 H in. Ib./min. 

 Therefore 



2irn I rq = W= 396,000 H t 



2 7rnM= 396,000 H-, 



Jf= 3_9|000^ =63)030? . iLlb 



z 77 n 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 (55), 



, = 2JM; = 16 M = 321,000 #. 



" -^,8 _^78 78 J 



