Solid Circular Rod
Consider a solid circular rod with a polar moment of inertia of 0.1 m^4 experiences a 100 Pa torsional stress. If the applied torque was 20 N∙m, what is the rod’s diameter in meters?
Expand Hint
Torsional stress in circular solid or thick-walled (t<0.1r) 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.
Hint 2
Solve for
$$r$$
, then multiply by 2 to get the diameter.
Torsional stress in circular solid or thick-walled (t<0.1r) 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.
Solving for radius:
$$$100 N/m^2 =\frac{(20N\cdot m)r}{0.1m^4}$$$
$$$r=\frac{100 N(0.1m^4)}{20N\cdot m^3}=0.5\:m$$$
To solve for the rod’s diameter:
$$$D=(2)r=(2)0.5m=1\:m$$$
1 m
Time Analysis
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