Assessing shaft capacity — is the strength of a circular shaft judged by the maximum torque it can safely transmit?
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AYes
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BNo
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COnly for hollow shafts
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DOnly when angle of twist is zero
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EDepends only on length
Answer
Correct Answer: Yes
Explanation
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:
Relate torque to stress: τ_max = TR/JSet τ_max ≤ τ_allow to find T_safeTherefore, strength is indeed judged by the maximum torque safely transmittedVerification / 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:
Yes