The shaft shown below rotates at a uniform angular velocity

The shaft shown below rotates at a uniform angular velocity of 2000 RPM, and is driven by a motor attached to the middle pulley. The belt tensions are indicated, and the three pulleys are rigidly keyed to the shaft. If the allowable shear stress is 9 ksi, and the allowable twist angle is 10_, determine the motor horsepower and the diameter of the shaft. Neglect shaft bending, and take G = 80 GPa.

Solution

I assume the allowable twist is 10 degrees?

Without any safety factor: formulae are.

Maximum torque will occcur between the drive pulley and one or other driven pulleys, L=24in =24x2.54 cm

The formulae are

Angle of twist = TL/(JG)

Tr/J = max shear stress

Equating T from the two expression allows J to cancel out along with powers of r.

Entering the numbers

r = 9*g*24*2.54*10^5/( 2.54^2*8)

= .6 *.81cm

dia of 11.77 cm

Power of motor = Torque * angular speed = T* 2000*2*pi/60= 2094T NM/s

Where T = taumax* (pi/2 r^3)/2

Calculation shows Torque 4.47 Nm

Power =Torque*angular speed (r/s)=936 J/s

IHP=746 J/s ==> 1.25 HP

Sounds reasonable. But I must say that the units are mixed up and would surely confuse any student.