A material obeying Hooke's law carries a direct stress σ₀. At its elastic limit, which statement is true?

Asked in SSC JE 23/01/2018 (Morning)

Detailed Explanation

For a bar in uniaxial tension or compression, three standard results follow from Hooke's law and the stress transformation.

Strain: \( \varepsilon = \sigma_0/E \)

Maximum shear stress (on planes at 45°): \( \tau_{max} = \sigma_0/2 \)

Strain energy: \( U = \tfrac12\sigma_0\varepsilon \times V = \dfrac{\sigma_0^2}{2E}V \)

Why “All options are correct” Is Correct

All three statements are correct, so the answer is (d).

Stress–strain curve of mild steel in tension. O–A: straight line, Hooke's law holds up to the limit of proportionality A. B: elastic limit. C and D: upper and lower yield points. D–E: yielding at nearly constant stress. E–F: strain hardening up to the ultimate stress F. G: fracture.

Why the Other Options Are Wrong

A. Strain is equal to σ₀/E: Correct, but (b) and (c) are also correct.
B. Maximum shear stress = σ₀/2: Correct, but (a) and (c) are also correct.
C. Strain energy = (σ₀²/2E) × volume: Correct, but (a) and (b) are also correct.

Key Point

Uniaxial σ: ε = σ/E, τ_max = σ/2 at 45°, U = σ²V/2E.

Key Concepts for Students

Proof resilience = σ_e²V/2E at the elastic limit.
Ductile materials in tension often fail on 45° shear planes.
Strain energy per unit volume = area under the elastic σ–ε line.

Practice more: this is question 37 in Stress and Strain MCQs – Part 3. Try the whole set and check every answer instantly.