## 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
$$\tau =\frac{Tr}{J}$$$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 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 rod’s diameter.
$$J=\frac{\pi (1m)^4}{32}=0.098125\:m^4$$$Thus, $$T=\frac{\tau\cdot J}{\frac{D}{2}}=\frac{(2)(150N)(0.098125m^4)}{(1m)(m^2)}=29.4\:N\cdot m$$$
$$29.4\:N\cdot m$$\$