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?

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:

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