If the torque on a shaft doubles, the required diameter increases by approximately what factor according to the diameter formula?

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Multiple Choice

If the torque on a shaft doubles, the required diameter increases by approximately what factor according to the diameter formula?

Explanation:
In torsion of a solid circular shaft, the torque is related to the diameter by T ∝ d^3 when the allowable shear stress is fixed. The exact relation is T = (τ π d^3)/16, so solving for diameter gives d = [16 T / (π τ)]^(1/3). This shows diameter depends on the cube root of torque. If the torque doubles, the new diameter becomes d' = [16 (2T) / (π τ)]^(1/3) = 2^(1/3) d. Therefore the required diameter increases by a factor of 2^(1/3) ≈ 1.26. The other factors (2, 1, or 1.414) correspond to different power relationships and don’t apply here.

In torsion of a solid circular shaft, the torque is related to the diameter by T ∝ d^3 when the allowable shear stress is fixed. The exact relation is T = (τ π d^3)/16, so solving for diameter gives d = [16 T / (π τ)]^(1/3). This shows diameter depends on the cube root of torque. If the torque doubles, the new diameter becomes d' = [16 (2T) / (π τ)]^(1/3) = 2^(1/3) d. Therefore the required diameter increases by a factor of 2^(1/3) ≈ 1.26. The other factors (2, 1, or 1.414) correspond to different power relationships and don’t apply here.

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