Thermal stresses in composite members during cooling: If a composite body experiences a fall in temperature, the member having the greater coefficient of linear expansion will be subjected to which type of stress?

Introduction / Context:
When dissimilar materials are bonded and subjected to temperature changes, incompatible free thermal strains create internal forces. Recognizing the sign of thermal stress is important in bimetallic strips, composite bars, and reinforced concrete.

Given Data / Assumptions:

• Two members are perfectly bonded (no slip), forming a composite bar.
• Temperature decreases by ΔT.
• α1 ≠ α2 (coefficients of linear expansion differ).

Concept / Approach:
Free thermal strain is ε_th = α * ΔT. On cooling (ΔT < 0), a member with larger α desires a larger contraction in magnitude. Bonding prevents differential contraction, so the larger-α member is restrained from contracting as much as it would freely, inducing tensile stress in it; the smaller-α member experiences compressive stress.

Step-by-Step Solution:

Verification / Alternative check:
Energy or force-balance formulations lead to equal and opposite internal forces; signs match the qualitative reasoning above.

Why Other Options Are Wrong:

• Compressive stress in higher-α member is opposite of actual behavior under cooling.
• Pure shear/zero/bending-only are not consistent with axial composite action.

Common Pitfalls:
Reversing signs for heating versus cooling; under heating the higher-α member is in compression if constrained.

Final Answer:
Tensile stress (it wants to contract more but is restrained)