Strain energy equivalences For a linearly elastic member under load, the strain energy stored can be equated to which of the following expressions (most generally correct)?

Introduction / Context:Strain energy is the elastic potential energy stored in a body due to deformation. For linearly elastic loading from zero to a final value, the load–displacement curve is a straight line; the strain energy equals the triangular area under this curve. Recognizing equivalent forms helps in solving a broad class of problems quickly.

Given Data / Assumptions:

• Material follows Hooke’s law within the loading range.
• Loading is monotonic from zero to final value (no unloading).
• “Average resistance” means average load over the displacement path.

Concept / Approach:

For a generalized single-degree description, strain energy U = ∫₀^Δ P dδ. With linear behaviour, P varies from 0 to P_final, so U = (1/2) P_final * Δ, which is “average resistance × displacement”. In distributed form, U = ∫_V (1/2) σ ε dV; for uniform stress/strain this reduces to (1/2) σ ε × volume. Any correct global statement must capture the factor 1/2 from linearity and the appropriate extent (volume), not just area.

Step-by-Step Solution:

Verification / Alternative check:

For an axially loaded bar of stiffness k: P = kΔ ⇒ U = (1/2) kΔ² = (1/2) PΔ, aligning with the average resistance interpretation.

Why Other Options Are Wrong:

• “Stress × strain × area” lacks the length factor and the essential 1/2 factor.
• “Stress × strain × volume” omits the factor 1/2; correct form is (1/2) σ ε V.
• “(Stress)^2 × volume + E” is dimensionally inconsistent and incorrect.

Common Pitfalls:

• Forgetting the factor 1/2 for linear loading.
• Confusing force–displacement area (global) with distributed energy density without integrating over the volume.

Final Answer:

Average resistance × total displacement.