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.
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Topic 3.4: Shear Stress & Strain
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Topic 3: Stress, Strain
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