this example as a
A solid circular shaft has a diameter of 3 in. and a length of 24 in. It
is subjected to a torque T=2kp in. Determine the maximum torsional stress
in the shaft.
To determine the state of stress, we select an element of the bar d
and radius rho. Due to the torque, a straight line parallel to
will be deflected or twisted by p ,
and so the shear strain is
This equation clearly shows that the shear stress is directly proportional
to the radius of the element and hence is largest at the surface of the
Next we have to relate the shear stress to the torque which produced
it. Any element of the cross section is subjected to a tangential force
due to the external torque .
The torque is the integral over all these moments.
We know that .
Using equation (a), we have
Thus the twist per unit length is
Thus, for a bar of length
where theta; is the total angle of twist in radians.
To obtain the shear stress, we note that
Then, since ,
And, because of equation (a),
The cross section element dA is rhodrho for circular cross sections,
Rewriting this result in Mathematica syntax, we get a function
for the stress.
For the above example, we get
Converted by Mathematica
September 1, 1999