A steel rod of area 25 mm² connects two parallel walls 5 m apart. The nuts were tightened when the rod was at 100 °C. If α = 0.000012 /°C and E = 0.2 MN/mm², the tensile stress developed when the temperature falls to 50 °C is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 02/03/2017 (Evening)
Detailed Explanation
As the rod cools it tries to shorten, but the walls stop it. The prevented shortening becomes a tensile strain.
\( \Delta T = 100 - 50 = 50^\circ\text{C},\ E = 0.2\ \text{MN/mm}^2 = 2\times10^5\ \text{N/mm}^2 \)
\[ \sigma = E\alpha\Delta T = 2\times10^5 \times 12\times10^{-6} \times 50 = 120\ \text{N/mm}^2 \]
Force = σA = 120 × 25 = 3000 N.
Why “120 N/mm²” Is Correct
The tensile stress is 120 N/mm². (The book's question says ‘tensile force’, but the options are stresses; the force would be 3000 N.) Note that the length (5 m) does not affect the stress.
| Situation | Stress |
|---|---|
| Bar fully restrained, temperature change ΔT | σ = EαΔT (compressive on heating, tensile on cooling) |
| Composite bar heated (ends held together) | Material with higher α → compression; lower α → tension |
| Composite bar cooled | Higher α → tension; lower α → compression |
Why the Other Options Are Wrong
- A. 80 N/mm²: 80 N/mm² does not match EαΔT.
- B. 100 N/mm²: 100 N/mm² does not match EαΔT.
- D. 150 N/mm²: 150 N/mm² does not match EαΔT.
Key Point
Fully restrained bar: σ = EαΔT, independent of length.
Key Concepts for Students
- Heating a restrained bar gives compression.
- If the walls yield by δ, σ = E(αLΔT − δ)/L.
- Railway rails need expansion gaps for this reason.
Practice more: this is question 42 in Stress and Strain MCQs – Part 3. Try the whole set and check every answer instantly.