Question 3.7: A motor driving a solid circular steel shaft transmits 40 hp...

(a) Motor operating at 500 rpm. Knowing the horsepower and the speed of rotation, we can find the torque T acting on the shaft by using Eq. (3-43). Solving that equation for T, we get

H=\frac{2\pi n T}{60\left(550\right) } =\frac{2\pi nT}{33,000}                  (n = rpm, T = lb-ft, H = hp)               (3-43)

T=\frac{33,000H}{2\pi n}=\frac{33,000\left(40  hp\right) }{2\pi \left(500  rpm\right) } = 420.2  lb-ft = 5042 lb-in.

This torque is transmitted by the shaft from the motor to the gear.
The maximum shear stress in the shaft can be obtained from the modified torsion formula (Eq. 3-12):

Solving that equation for the diameter d, and also substituting \tau _{allow} for \tau _{max}, we get

from which

d = 1.62 in.

The diameter of the shaft must be at least this large if the allowable shear stress is not to be exceeded.

(b) Motor operating at 3000 rpm. Following the same procedure as in part (a), we obtain

T=\frac{33,000  H}{2\pi n} =\frac{33,000\left(40  hp\right) }{2\pi \left(3000  rpm\right) } = 70.03  lb-ft = 840.3 lb-in.

d = 0.89 in.

which is less than the diameter found in part (a).
This example illustrates that the higher the speed of rotation, the smaller the required size of the shaft (for the same power and the same allowable stress).