WORLD FOR CIVIL

Topic 3.4b: Shear Stress & Strain - Example 2

Example 2
In our second example we have a inner shaft (which perhaps drives a piece of machinery) and an outer driving wheel connected by spokes to an inner ring which is connected by means of a shear key to the inner shaft. That is, the wheel, spokes, and inner ring are one structure which is connected to the inner shaft by a shear key. (See Diagram 3.) When force is applied to the wheel (through a driving belt perhaps), the wheel begins to rotate, and through the shear key causes the inner shaft to rotate. We are interested in determining the shear stress on the shear key. We will also determine the compressive stress (or bearing stress) acting on the shear key. [The purpose of a shear key is to protect machinery connected to a shaft or gear. If the driving forces and/or torque become too large the shear key will "shear off " (fail in shear) and thus disconnect the driving force/torque before it can damage the connected machinery.]
To determine the shear stress we first need to determine the force trying to shear the key. We do this by realizing, after a little thought, that the driving force is really producing a torque about the center of the shaft, and that the torque produced by the driving force(s) must equal the torque produced by the force (of the inner ring) acting on the shear key. In our problem the two 500 lb. driving forces are acting a distance of 2 ft from the center of the 1 ft diameter shaft. Calculating torque about the center of the shaft we have:

500 lb.(2 ft) + 500 lb.(2 ft) = 2000 ft-lb. This must also be the torque produced by the force acting on the upper half of the shear key (shown in Diagram 4). So we may write: 500 lb.(2 ft) + 500 lb.(2 ft) = 2000 ft-lb. = F (.5 ft), Solving for F = 4000 lb. This is the force acing on the top half of the shear key. There is a equal force in the opposite direction acting on the bottom half of the shear key. These two forces place the horizontal cross section of the key in shear. The key is 1/2 inch wide, by 3/4 inch high, by 1 inch deep as shown in Diagram 4. We calculate the shear stress by:
Shear Stress = Force parallel to area / area = 4000 lb./ (1/2" * 1") = 8000 lb/in².

In addition to the shear stress on the horizontal cross sectional area, the forces acting on the key also place the key itself into compression. The compressive stress (also called the bearing stress) on the top half of the shear key will be given by:
Compression (Bearing) Stress = Force normal to the area / area = 4000 lb. / (3/8" * 1") = 10, 700 lb/in². There is an equal compressive stress on the bottom of the shear key.