From the shaft diameter formula d = [16 T /(π τ_allow)]^(1/3), the shaft diameter scales with torque to the power of what exponent?

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Multiple Choice

From the shaft diameter formula d = [16 T /(π τ_allow)]^(1/3), the shaft diameter scales with torque to the power of what exponent?

Explanation:
The relationship comes from how torsion relates torque to the shaft’s geometry. For a solid circular shaft, T = τ J / R. With J = π d^4 / 32 and R = d/2, this becomes T = τ (π d^4 / 32) / (d/2) = τ π d^3 / 16. Solving for d gives d^3 = 16 T /(π τ_allow), so d = [16 T /(π τ_allow)]^(1/3). This shows diameter is proportional to T^(1/3), i.e., the exponent is 1/3. If torque doubles, diameter grows by 2^(1/3) ≈ 1.26; eightfold torque doubles the diameter, illustrating the cube-root relationship.

The relationship comes from how torsion relates torque to the shaft’s geometry. For a solid circular shaft, T = τ J / R. With J = π d^4 / 32 and R = d/2, this becomes T = τ (π d^4 / 32) / (d/2) = τ π d^3 / 16. Solving for d gives d^3 = 16 T /(π τ_allow), so d = [16 T /(π τ_allow)]^(1/3). This shows diameter is proportional to T^(1/3), i.e., the exponent is 1/3. If torque doubles, diameter grows by 2^(1/3) ≈ 1.26; eightfold torque doubles the diameter, illustrating the cube-root relationship.

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