## 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?

##
__
__**Hint**

**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**

**Hint 2**

Because we are interested in the polar moment of inertia of a solid cylinder shaft:

$$$J_{hollow}=\frac{\pi D^{4}}{32}$$$

where
$$D$$
is the diameter.

$$$\tau =\frac{Tr}{J}$$$

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.

Because we are interested in the polar moment of inertia of a solid cylinder shaft:

$$$J_{hollow}=\frac{\pi D^{4}}{32}$$$

where
$$D$$
is the rodâ€™s diameter.

$$$J=\frac{\pi (1m)^4}{32}=0.098125\:m^4$$$

Thus,

$$$T=\frac{\tau\cdot J}{\frac{D}{2}}=\frac{(2)(150N)(0.098125m^4)}{(1m)(m^2)}=29.4\:N\cdot m$$$

$$$29.4\:N\cdot m$$$