Torque Torsion
For a certain process, a circular metal rod with a 1 m diameter has its grain parallel to the neutral axis. If the allowable shearing stress parallel to the alloy's grain is 150 Pa, what is the torque transmitted by the rod?
Expand Hint
$$$\tau =\frac{Tr}{J}$$$
where
$$\tau$$
is torsion,
$$T$$
is the torque at the section of interest,
$$r$$
is the radius to the point of interest, and
$$J$$
is the section's polar moment of inertia.
Hint 2
Polar moment of inertia of a solid cylinder shaft:
$$$J_{hollow}=\frac{\pi D^{4}}{32}$$$
where
$$D$$
is the diameter.
Torsional stress in a circular solid or thick-walled shafts:
$$$\tau =\frac{Tr}{J}$$$
where
$$\tau$$
is torsion,
$$T$$
is the torque at the section of interest,
$$r$$
is the radius to the point of interest, and
$$J$$
is the section's polar moment of inertia.
Polar moment of inertia of a solid cylinder shaft:
$$$J_{hollow}=\frac{\pi D^{4}}{32}$$$
where
$$D$$
is the rod’s diameter.
$$$J=\frac{\pi (1m)^4}{32}=0.098125\:m^4$$$
Thus,
$$$T=\frac{\tau\cdot J}{\frac{D}{2}}=\frac{(2)(150N)(0.098125m^4)}{(1m)(m^2)}=29.4\:N\cdot m$$$
$$$29.4\:N\cdot m$$$
Time Analysis
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