WORLD FOR CIVIL

Topic 6.1a: Shear Stress - Example 1

Part I. In Diagram 1a we have shown a solid shaft with what we will call a driving external torque of 1000 ft-lb. at end A, and a load torque of 1000 ft-lb. at end B. The shaft is in equilibrium. We would like to determine the maximum transverse shear stress in the shaft due to the applied torque.

To solve, we first need to determine the internal torque in the shaft. We cut the shaft a distance x from end A and draw a Free Body Diagram of the left end section of the shaft as shown in Diagram 1b. Where we cut the shaft there is an internal torque, which in this case must be equal and opposite to the torque at end A for equilibrium. So for this shaft the value of the internal torque is equal to the value of the externally applied torque.

We next simply apply the torsion formula for the shear stress: = T r / J; where:
T is the internal torque in that section of the shaft = 1000 ft-lb = 12,000 in-lb.
r = the radial distance from the center of the shaft to the point where we wish to find the shear stress. In this problem the outer edge of the shaft since that is where the transverse shear stress is a maximum; r = 1 in.
J = polar moment of inertia = (/32) d⁴ for a solid shaft = (3.1416/32) (2⁴in⁴) = 1.57 in⁴.
So, = T r / J = 12,000 in-lb. * 1 in./1.57 in⁴. = 7,640 lb/in².
This is the Maximum Transverse (and longitudinal) Shear Stress in the shaft.

Part II
We now would like to consider the case where the shaft is not solid, but a hollow shaft with an outer diameter of 2" and an inner diameter of 1", as shown in Diagram 2a. We still apply the same driving and load torque, and still have the same value of the internal torque, as is shown in Diagram 2b.

We next apply the torsion formula for the shear stress for the hollow shaft:
= T r / J; where we observe that all the values are the same as in part one, except for the value of J, the polar moment of inertia.
T is the internal torque in that section of the shaft = 1000 ft-lb = 12,000 in-lb.
r = the radial distance from the center of the shaft to the point where we wish to find the shear stress. In this problem the outer edge of the shaft since that is where the transverse shear stress is a maximum; r = 1 in.
J = polar moment of inertia = (3.1416/32) [d_o⁴ - d_i⁴] for a hollow shaft = (3.1416/32) [(2⁴in⁴) - (1.0⁴in⁴)] = 1.47 in⁴.
So, = T r / J = 12,000 in-lb. * 1 in./1.47 in⁴. = 8,150 lb/in².

This then is the Maximum Transverse (and longitudinal) Shear Stress in the hollow shaft.