If the torque of 535 lbf-ft is applied on a handwheel of diameter 16 inches, Then stress in shaft connected?

Question

What will be the stress in the Shaft which is connected to Handwheel? This is for some Actuators...

Answer 1

The torsional shear stress of a shaft is given by the formula:

s =Tr/Ip

Where:

s = stress i psi

T = torque in in-lbs

Ip = polar moment of inertia of the shaft in in^4

You have to give the diameter of the shaft to get the polar moment of inertia.

For a solid circular shaft, the polar moment of inertia is;

Ip = (pi(D^4))/32;  where D is the diameter of the shaft in inches

For a hollow circular shaft, the polar moment of inertia is;

Ip = pi(D^4 - d^4)/32; where  D is the outside diameter in inches and d is the inside diameter in inches

The maximum torsional shear stress of the shaft therefore is:

s = 535x12 x r/Ip = 6420r/Ip

where r = D/2