Assessing shaft capacity — is the strength of a circular shaft judged by the maximum torque it can safely transmit?

Introduction:
Shaft design involves two criteria: strength (stress) and rigidity (twist). This item asks whether strength assessment is fundamentally linked to the torque-carrying capacity.

Given Data / Assumptions:

• Homogeneous, isotropic shaft; elastic behavior.
• Failure governed by allowable shear stress.

Concept / Approach:
Strength is about stress not exceeding an allowable limit. In torsion, τ_max = TR/J; thus, the largest safe torque determines whether the shaft is strong enough. Rigidity uses θ = TL/(JG) and is a separate serviceability check.

Step-by-Step Solution:

Verification / Alternative check:
Design codes specify allowable shear stress or failure theories; solving for T gives torque capacity, directly reflecting strength.

Why Other Options Are Wrong:

• No: ignores the fundamental torsional stress relation.
• Only for hollow shafts: applies to both solid and hollow (with respective J).
• Only when angle of twist is zero: unrealistic and not a strength criterion.
• Depends only on length: length affects rigidity (θ), not the stress capacity directly.

Common Pitfalls:
Confusing strength (stress) with rigidity (twist); both must be checked, but strength is evaluated via torque capacity.

Final Answer: