Question

Calculate the torque developed by a motor that delivers 200 hp rotation at 3600 rpm. Use the Torque value to determine the diameter of a solid steel shaft with an allowable shear stress of 20,520 psi.

Answer 1

Given:

Power, P = 200 hp

= 110000 lbf ft/s

Speed, N = 3600 rpm

Power ,
`$P=(2 \pi NT)/(60)$`

∴ Torque,
`$T = (60 P)/(2 \pi N)$`

`$T = (60 * 110000)/(2 \pi * 3600)$`

T = 291.93 lb -ft

`$T_(allow) = 20,250 \ psi$` ---- allowable shear stress

Therefore, torque , T = 391.93 lb-ft

T = 291.93 x 12 lb-in

∴ T 3503.16 lb-in

Now from the Torsion formula,

`$(T)/(J)=(T_(allow))/( r)$`

Here,
`$r=(d)/(2) ; \ \ \ J=(\pi d^4)/(32)$` ---- polar moment of inertia.

∴
`$(T)/((\pi d^4)/(32))=(T_(allow))/( d/2)$`

`$T_(allow)=(16 T)/(\pi d^3)$`

`$\pi d^3=(16 * 3503.16)/(20520)$`

`$\pi d^3 = 2.7315$`

`$d^3 = 0.8699$`

or
`$d= (0.8699)^(1/3)$`

d = 0.9546 in

Therefore the diameter of the shaft is 0.9546 in.