Thermal stress in a restrained metallic bar: which factor below does the stress not depend on, assuming uniform temperature change and linear elasticity?

Introduction / Context:Thermal stress arises when a component is prevented from freely expanding or contracting with temperature. Recognizing what governs this stress helps avoid cracking, distortion, and joint failure in process equipment and piping.

Given Data / Assumptions:

• Uniform, fully restrained bar; no yielding.
• Linear elastic behavior; small strains.
• Uniform temperature change ΔT.

Concept / Approach:For a fully restrained bar, free thermal strain would be α * ΔT, but restraint enforces zero net strain, creating a compressive or tensile stress σ such that σ/E counters α * ΔT. Therefore σ = E * α * ΔT, which depends on material properties and temperature change, not on the bar’s cross-sectional area A.

Step-by-Step Solution:Write free thermal strain: ε_free = α * ΔT.Restraint condition: ε_mech + ε_free = 0 → ε_mech = −α * ΔT.Hooke’s law: σ = E * ε_mech = −E * α * ΔT (sign by convention).Area A cancels because stress is force/area; only if we asked for force (or reaction) would A matter.

Verification / Alternative check:Dimensional analysis confirms σ depends on E (Pa), α (1/K), and ΔT (K). Reaction force would be F = σ * A, which does include area, but stress itself does not.

Why Other Options Are Wrong:(a) clearly affects σ; (d) contradicts the derivation; (e) lists parameters that determine σ; only (b) is irrelevant to σ magnitude.

Common Pitfalls:Confusing stress with total force; neglecting partial restraint or creep/relaxation at high temperature that can reduce σ over time.

Final Answer:Cross-sectional area of the bar