Composite bar of copper and steel heated to 40°C: What is the nature of stress induced in the copper bar (assuming the two are rigidly bonded and free expansion is restrained)?

Introduction / Context:
Temperature changes in composite members cause self-equilibrating thermal stresses when free expansion is restrained. Understanding the sign of thermal stress in each material is essential for safe composite design.

Given Data / Assumptions:

• Copper bar is rigidly bonded to a steel bar (composite system).
• Uniform heating to 40°C above the reference temperature.
• Free expansion is prevented by compatibility (bonding) and overall restraint.
• Typical properties: α_cu > α_steel, and E_cu, E_steel finite and positive.

Concept / Approach:
Free thermal strain for each material is epsilon_free = α * ΔT. Because α_cu > α_steel, copper wants to expand more than steel. Bonding enforces equal total strain, so copper is forced to expand less than it would freely → copper experiences compressive mechanical strain; steel is forced to expand more than freely → tensile mechanical strain.

Step-by-Step Solution:

Verification / Alternative check:
Force equilibrium requires internal compressive force in copper to balance tensile force in steel. Analytical solution yields sigma_cu = − (E_cu * E_st / (E_cu A_cu + E_st A_st)) * (α_cu − α_st) * ΔT * A_st in compression (sign negative).

Why Other Options Are Wrong:
Tensile/Zero/Shear: contradict thermal compatibility and equilibrium for α_cu > α_steel.“Depends only on cross-section” is false; coefficients of expansion are decisive.

Common Pitfalls:
Assuming no stress appears under uniform heating; ignoring restraint due to bonding; mixing sign convention for tension/compression.

Final Answer:

Compressive