80 



111 which F = load due to friction ; 



V ^ velocity of surface of shaft; 

 9 = coefficient of friction for factory shafting ; 

 Wz= weight of shaft. 

 While o varies from 0.08 to 0.08 under different conditions, we have assumed 

 it to equal 0.06 for ordinary factory shafting, with more or less imperfect luhrica- 

 tion and alignment. 



If there are no pulleys on the shaft, W will equal 



"Y d- L X -'-SB pounds, where 



L = length of shaft in feet, and 

 d = diameter of shaft in inches. 



The horse-power exerted to overcome friction will then be : 



The horse-power transmitted by the shaft will be : 



- dM y 2 TT N 

 H P = — ^ — —. ( ^ ^ 



16 12 X 33000 ^ ' 



If we assume the angle of torsion not to exceed y^ degree per foot length 

 of shaft, there is obtained 



360 LX12 _360f 12L. 



2- >^* G r - -G ^ d ' ^^> 



hence : 



f = '-'?-^^^^^ = 800d 



360 X 12 L 



when G — 11,000,000; that is, when the modulus of torsion = J modulus of 

 elasticity. 



Substituting this value of f in (3), and noting that v = — r-^ — , we have: 

 H. P. = 0.0095 d^ V. ( 5 ) 



From ecjuations ( 2 ) and ( '^ ) there is obtained 



H. P„. _ 0.000006 d^ L v _ 

 H. P. ~ 0.0095 d« v ' 



that is, H. P,. = 0.000063 H. P. ^ = ^ X ?' ^' very closely. 

 ' d d 1600 •' ■' 



(6) 



We see from this that the horse-power required to overcome the friction of a 

 one-inch shaft 1600 feet long is equal to the total allowable transmitting capacity 

 of the shaft under ordinary working conditions. 



