Torsional strength comparison: two solid circular shafts of the same material have diameters D (shaft A) and D/2 (shaft B). The torsional strength (torque capacity at the same allowable shear) of shaft B is what fraction of shaft A?
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Aone-eighth
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Bone-fourth
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Cone-half
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Dfour times
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Ethree-eighths
Answer
Correct Answer: one-eighth
Explanation
Introduction / Context:Torsional strength of a solid round shaft is proportional to its polar section modulus, which scales as the cube of the diameter. Halving the diameter therefore reduces torque capacity dramatically.
Given Data / Assumptions:
- Shafts are solid, circular, and made of the same material.
- Allowable maximum shear stress is the same for both.
- Elastic torsion theory applies.
Concept / Approach:Polar section modulus for a solid round: Z_p = J / R = (π d^4 / 32) / (d/2) = π d^3 / 16. Therefore, torque capacity T_allow ∝ Z_p ∝ d^3 for the same allowable shear stress.
Step-by-Step Solution:
Let d_A = D and d_B = D/2.Strength ratio = (d_B / d_A)^3 = (D/2D)^3 = (1/2)^3 = 1/8.Hence shaft B is one-eighth as strong as shaft A in torsion.Verification / Alternative check:Compute Z_p explicitly for each and take the ratio to confirm 1/8.
Why Other Options Are Wrong:One-fourth, one-half: These reflect area or linear scaling, not the correct cubic scaling.Four times: Opposite of the actual reduction; reducing diameter cannot increase strength.Three-eighths: Arbitrary fraction not supported by d^3 scaling.
Common Pitfalls:Assuming strength scales with d^2 (area) rather than d^3 (polar section modulus).
Final Answer:
one-eighth