WORLD FOR CIVIL

Topic 6.1b: Shear Stress - Example 2

In Diagram 1 we have shown a solid compound shaft with what we will call the driving external torque of 1600 ft-lb. acting at point B, and load torque of 400 ft-lb. at end A, 900 ft-lb. at point C, and 300 ft-lb. at end D. Notice that the shaft is in rotational equilibrium. We would like to determine the maximum transverse shear stress in each section of the shaft due to the applied torque.

To solve, we first need to determine the internal torque in each section of the shaft. We cut the shaft a distance 0' <>

We next apply the torsion formula for the shear stress: = T r / J; where:
T is the internal torque in that section of the shaft = 400 ft-lb. = 4,800 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 r is to the outer edge of the shaft since that is where the transverse shear stress is a maximum; r = .5 in.
J = polar moment of inertia = (/32) d⁴ for a solid shaft = (3.1416/32) (1⁴in⁴) = .098 in⁴.
So, = T r / J = 4,800 in-lb. * .5 in./.098 in⁴. = 24,500 lb./in².

This value is the Maximum Transverse Shear Stress in the shaft section AB, and it falls in a reasonable range for allowable shear stresses for metals.

We now determine the internal torque in the next section of the shaft. We cut the shaft a distance 1' <>

We apply the torsion formula for the shear stress once again: = T r / J; where:
T is the internal torque in that section of the shaft = 1,200 ft-lb. = 14,400 in-lb.
r = the radial distance to 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 = 14,400 in-lb. * 1 in./ 1.57 in⁴. = 9,170 lb./in².

This then is the Maximum Shear Stress in shaft section BC. We note that even though the internal torque is much larger in section BC as compared to section AB, because of the size of the shaft in section BC, the shear stress is much lower in BC.

We now determine the internal torque in the next section of the shaft. We cut the shaft a distance 3' <>

We apply the torsion formula for the shear stress: = T r / J; where:
T is the internal torque in that section of the shaft = 300 ft-lb. = 3,600 in-lb.
r = the radial distance from the center of the shaft to the outer edge of the shaft since that is where the transverse shear stress is a maximum; r = .25 in.
J = polar moment of inertia = (/32) d⁴ for a solid shaft = (3.1416/32) (.5⁴in⁴) = .0061 in⁴.
So, = T r / J = 3,600 in-lb. * .25 in./ .0061 in⁴. = 147,500 lb./in².

This is the Maximum Shear Stress in shaft section CD. We note that this is much larger than the ultimate shear stress most metals, thus this section of the shaft would fail - a larger diameter is needed to carry the torque.