Elastic versus plastic behaviour under axial tension A 20 mm diameter steel bar is subjected to an axial tensile stress of 2300 kg/cm² and exhibits a total strain of 0.004. If Young’s modulus E = 2.1 × 10^6 kg/cm², the bar is operating in which range?
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AIn the elastic (linear) range
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BIn the plastic (nonlinear) range
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CAt the yield point exactly
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DNone of these
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EIn the perfectly plastic plateau
Answer
Correct Answer: In the plastic (nonlinear) range
Explanation
Introduction / Context:Hooke’s law states that in the elastic range, strain is proportional to stress with the constant of proportionality being Young’s modulus. Deviations from this linear relation indicate yielding or plastic deformation.
Given Data / Assumptions:
- Applied stress σ = 2300 kg/cm².
- Measured strain ε = 0.004 (dimensionless).
- Young’s modulus E = 2.1 × 10^6 kg/cm².
Concept / Approach:In the elastic range, ε_elastic = σ / E. If the observed strain greatly exceeds σ / E, the response is not purely elastic and plastic strain is present.
Step-by-Step Solution:
Compute elastic strain prediction: ε_elastic = 2300 / (2.1 × 10^6) = 0.001095 (approx.).Compare with observed: ε_observed = 0.004.Since 0.004 » 0.001095, significant plastic strain exists → behaviour is plastic/nonlinear.Verification / Alternative check:Apparent modulus from the data: E_app = σ / ε = 2300 / 0.004 = 5.75 × 10^5 kg/cm², far below E, confirming yielding.
Why Other Options Are Wrong:Elastic range would require ε ≈ 0.0011; “at yield point exactly” cannot be inferred from one data pair; “perfectly plastic plateau” implies near-constant stress at large strain, not demonstrated by the single reading.
Common Pitfalls:Using percentage strain units inconsistently; forgetting strain is dimensionless; assuming any deviation is measurement error rather than plasticity.
Final Answer:In the plastic (nonlinear) range