For a hollow circular shaft with internal-to-external diameter ratio n = d_i / d_o, what is the ratio of the weight of the hollow shaft to that of a solid shaft of the same torsional strength (same allowable shear stress)?

Introduction / Context:
Weight efficiency comparisons between solid and hollow shafts are common in machine design. For equal torsional strength (same maximum shear stress), we compare the required sizes and then the cross-sectional areas (and hence weights) of the two shafts.

Given Data / Assumptions:

• Hollow shaft: outside diameter d_o, inside diameter d_i, with n = d_i/d_o.
• Solid shaft: diameter d_s provides the same torsional strength.
• Same allowable shear stress and material density for both shafts.

Concept / Approach:
Torsional strength ∝ polar section modulus Z_p. For a circle:Z_p(solid) = (π/16) * d_s^3Z_p(hollow) = (π/16) * (d_o^4 - d_i^4) / d_oEquate strengths to relate d_s and d_o, then form the weight ratio using areas.

Step-by-Step Solution:
Equal strength → (d_o^4 - d_i^4)/d_o = d_s^3.Put d_i = n d_o → d_s = d_o * (1 - n^4)^(1/3).Areas (∝ weight): A_h = (π/4)(d_o^2 - d_i^2) = (π/4)d_o^2(1 - n^2).A_s = (π/4)d_s^2 = (π/4)d_o^2 (1 - n^4)^{2/3}.Weight ratio = A_h / A_s = (1 - n^2) / (1 - n^4)^(2/3).

Verification / Alternative check:
Boundary checks: n = 0 (solid) → ratio 1; n → 1 (very thin tube) → ratio → 0, both physically consistent.

Why Other Options Are Wrong:

• Other expressions do not follow from equating Z_p and simplifying areas; they fail dimensional and limit checks.

Common Pitfalls:

• Using J instead of Z_p or forgetting the d in the denominator for hollow Z_p.

Final Answer:
(1 - n^2) / (1 - n^4)^(2/3)