Comparing torsional strength: two solid circular shafts A and B (same material) have diameters 50 mm and 100 mm respectively. The torsional strength of shaft B is how many times that of shaft A?

Introduction / Context:
In torsion, the strength of a circular shaft—its capacity to resist a given maximum shear stress—is proportional to its polar section modulus. For solid round shafts, this scales with the cube of diameter. Doubling the diameter therefore has a dramatic effect on strength.

Given Data / Assumptions:

• Shaft A: d_A = 50 mm; Shaft B: d_B = 100 mm.
• Same material and allowable shear stress.
• Solid circular sections; elastic behavior.

Concept / Approach:
Polar section modulus Z_p for a solid round is Z_p = J / R = (π d^4 / 32) / (d/2) = π d^3 / 16. For given allowable shear stress τ_allow, torque capacity T_allow ∝ Z_p ∝ d^3. Thus the strength ratio equals the cube of the diameter ratio.

Step-by-Step Solution:

Verification / Alternative check:
Compute Z_p explicitly: Z_pA = π (50)^3 / 16; Z_pB = π (100)^3 / 16; take the ratio to confirm 8.

Why Other Options Are Wrong:
One-half, double, or four times underestimate the cubic scaling.Sixteen times would correspond to a diameter ratio of 2.52^(something), not to a simple doubling.

Common Pitfalls:
Assuming strength scales with area (d^2) instead of with section modulus (d^3); ignoring the role of maximum radius R in Z_p.

Final Answer:

eight times