Strength of Materials – Proof resilience of mild steel (maximum strain energy density without permanent set) E = 2 × 10^5 MPa, yield stress σ_y = 250 MPa. Compute the maximum strain energy per unit volume (Nmm/mm3).

Introduction / Context:
Proof resilience is the maximum strain energy stored per unit volume in a material without causing permanent deformation. For linearly elastic behavior up to yield, it depends only on the yield stress and Young’s modulus.

Given Data / Assumptions:

• Material: mild steel.
• Young’s modulus E = 2 × 10^5 MPa.
• Yield stress σ_y = 250 MPa.
• Elastic behavior up to σ_y.

Concept / Approach:

Strain energy density u for linear elasticity under uniaxial stress is u = ∫0^ε σ dε = 0.5 * σ * ε. At yield, ε_y = σ_y / E, so u_max = 0.5 * σ_y * (σ_y / E) = σ_y^2 / (2E). Units of MPa convert directly to N/mm^2, which equal Nmm/mm^3 for energy density.

Step-by-Step Solution:

Verification / Alternative check:

Direct formula check: u = σ_y^2 / (2E) = 62,500 / 400,000 = 0.15625 Nmm/mm3, consistent with the step calculation.

Why Other Options Are Wrong:

1.56, 15.6, and 156 are 10×, 100×, and 1000× larger, stemming from unit mistakes (e.g., mixing MPa with N/mm2 or omitting the 1/2 factor).

Common Pitfalls:

Confusing energy per unit volume units, or using ultimate stress instead of yield for proof resilience.

Final Answer:

0.156 Nmm/mm3