Composite bar under uniform heating – identify internal force sense A composite member (two different materials 'a' and 'b') was made at 25°C and then heated uniformly to 45°C. Material 'a' has a higher coefficient of thermal expansion than material 'b'. If the two are perfectly bonded so they must undergo the same overall extension, which statement about the internal forces is correct?

Introduction / Context:
Composite bars made of different materials experience internal self-stresses when the temperature changes because each material wants to change length by a different free thermal strain. Compatibility (same final extension) forces the materials to “negotiate” through internal forces, which are equal and opposite, producing compression in one and tension in the other.

Given Data / Assumptions:

• Two materials a and b, perfectly bonded, same original length at 25°C.
• Temperature is increased uniformly from 25°C to 45°C (ΔT = 20°C).
• Coefficient of thermal expansion: alpha_a > alpha_b.
• Ends are free of external load but compatibility (no slip) is enforced.

Concept / Approach:
If the two bars were free, their free thermal strains would be: epsilon_free,a = alpha_a * ΔTepsilon_free,b = alpha_b * ΔT Because alpha_a > alpha_b, material a wants to expand more. Bonding imposes a common final strain epsilon that lies between the two free strains. Hence, the higher-expanding material a must be compressed back from its free expansion, while b must be pulled ahead to catch up.

Step-by-Step Solution:
Let epsilon be the common strain. Then for a: epsilon = alpha_a * ΔT + (σ_a / E_a)For b: epsilon = alpha_b * ΔT + (σ_b / E_b)Force equilibrium for the composite (no external axial load): A_a * σ_a + A_b * σ_b = 0.Solving reveals σ_a is compressive and σ_b is tensile because alpha_a > alpha_b.

Verification / Alternative check:
Think physically: material a tries to expand more and is restrained by b, so a is squeezed (compression); b is dragged along (tension).

Why Other Options Are Wrong:

• (a) reverses the correct sense of stress.
• (b) and (c) cannot both be true due to action–reaction balance in an unloaded composite.
• (e) ignores compatibility; even with free ends, bonding creates internal forces.

Common Pitfalls:
Confusing free thermal expansion (no stress) with constrained thermal expansion (self-stress). The key is the bond imposing equal net strain.

Final Answer:
Material a will be in compression and material b in tension