Thermal stress in a restrained steel tie: A 5 m long steel rod of cross-sectional area 250 mm² is fixed rigidly between two parallel walls. The end nuts were tightened at 100 °C. Given α_steel = 0.000012 per °C and E_steel = 0.2 MN/mm², find the tensile stress developed in the rod when the temperature drops to 50 °C (walls prevent contraction).

Introduction / Context:
When a member is fully restrained against free thermal expansion or contraction, temperature change induces internal stress. For linear elastic behavior, the thermal stress equals E * α * ΔT in magnitude, tension when cooling and compression when heating (in a restrained bar).

Given Data / Assumptions:

• Length L = 5 m (length cancels in stress calculation for full restraint).
• Area A = 250 mm² (not required for stress magnitude).
• Coefficient of thermal expansion α = 0.000012 / °C.
• Elastic modulus E = 0.2 MN/mm² = 200,000 N/mm².
• Temperature drop ΔT = 50 − 100 = −50 °C (cooling).
• Ends fully restrained, no slip; linear elasticity.

Concept / Approach:

For a fully restrained bar, thermal strain α ΔT is prevented, producing mechanical strain ε = −α ΔT. Stress σ = E ε = −E α ΔT. Sign indicates tension for cooling (since free contraction is prevented).

Step-by-Step Solution:

Verification / Alternative check:

If needed, the corresponding force would be F = σ A = 120 * 250 = 30,000 N tension, consistent with restraint forces.

Why Other Options Are Wrong:

80, 100, 150 N/mm² correspond to incorrect ΔT or arithmetic; 60 N/mm² would follow if α or ΔT were halved.

Common Pitfalls:

Using negative sign to select a “compression” answer; here, cooling in a fixed tie produces tension. Also, mixing MPa and N/mm² units.

Final Answer:

120 N/mm²