WORLD FOR CIVIL

Topic 8.1b: Combined Stress - Example 2

A second example of a structure where stress may be added using the principle of superposition is shown in Diagram 1.

In this example, a solid 1 foot long shaft is attached to a wall at point A, and has a disk attached at end B. A force of 1000 lb. is applied to the outer edge of the disk, as shown. We would like to determine the maximum shear stress in the shaft.
We first consider the static equilibrium conditions. The 1000 lb. force acting on the outside edge of the disk produces a torque of 2000 in-lb. with respect to the center (centroid) of the shaft. Additionally it also effectively exerts a 1000 lb. force perpendicular to the shaft. It can be considered to produce a torque and a force acting at the centroid, as shown in the box in Diagram 1.
In response to this, a torque and perpendicular force develops in the supporting wall, such that the shaft is in translational and rotational equilibrium. (Notice also that an external moment develops at the wall, since the shaft is also acting as a cantilevered beam. An axial bending stress will also develop in the beam, but we will not consider this at this point, since we are concerned with the total shear stress in the beam.)

The internal torque in the shaft produces transverse shearing stress. The perpendicular forces produce an internal shearing force in the shaft, which also produces a shear stress. The two stresses may be summed to find the maximum shear stress in the shaft.
Maximum Transverse Shear Stress = Tr/J = (2000 in-lb.) * (1") / (pi * d⁴/32) = (2000 in-lb.)*(1") / (3.1416 * 2"⁴/32) = 1,270 lb/in².

Maximum Horizontal Shear Stress = Vay'/Ib = (4/3) V/A This last is the expression for the maximum horizontal (and vertical) shear stress in a circular cross section beam. The derivation for this is a bit messy, so the result will simply be stated here. (The student is referred to a complete Strength of Materials Text for the derivation.)
Maximum Horizontal Shear Stress = (4/3)V/A = (4/3) (1000 lb.)/3.1416 * 2"²/4) = 424 lb/in².

We can now sum the two shear stresses to determine the maximum shear stress in the shaft. In Diagram 2 and Diagram 3, we have shown a section of the shaft, and cross sectional views showing the directions of the two shear stresses.

As we can see from the diagram, at the top the cross sections the two stresses add giving the result of 1,270 lb/in² + 424 lb/in² = 1,694 lb/in².
This is the maximum shear stress in the shaft.