---
title: "Thermal stress in a restrained bar"
subject: "Strength of Materials"
chapter: 2
key: "semester_4/strength_of_materials_mechanical:bar-between-rigid-walls-temperature"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/bar-between-rigid-walls-temperature
site: BCT Engineering
---

# Thermal stress in a restrained bar

Strength of Materials · Chapter 2: Stress and strain – axial load

- Asked in: year not given, 3+3 marks
- Video: [Thermal strain and stress in restrained bar](https://www.youtube.com/watch?v=QuZcTsW-sNY) — Dremph
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/bar-between-rigid-walls-temperature
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-2

## Question

A steel bar 2.5 m long and 25 mm in diameter is fixed between two rigid walls at 20 degrees C with no initial stress. The temperature rises to 70 degrees C. Take E = 200 GPa and coefficient of thermal expansion 12 x 10^-6 per degree C. Find (a) the stress and force developed in the bar, (b) the stress if the walls yield (move apart) by a total of 0.6 mm.

## Short answer

Free thermal expansion 1.5 mm. Rigid walls: compressive stress σ = EαΔT = 120 MPa, force P = 58.9 kN. If walls yield 0.6 mm, restrained expansion = 0.9 mm, σ = 72 MPa, P = 35.3 kN.

## Answer

When a bar is prevented from expanding freely, a compressive thermal stress develops.

#### Data
$L = 2500\ \text{mm}$, $A = \frac{\pi}{4}(25)^2 = 490.9\ \text{mm}^2$, $\Delta T = 70 - 20 = 50\,^\circ\text{C}$.

#### (a) Rigid walls
Free thermal expansion:

$$
\delta_T = \alpha L \Delta T = 12\times10^{-6}\times 2500\times 50 = 1.5\ \text{mm}
$$

The walls allow no expansion, so the wall reactions must push the bar back by the full 1.5 mm. The thermal strain $\alpha\Delta T$ is cancelled by an equal elastic strain:

$$
\sigma = E\alpha\Delta T = 200000\times 12\times10^{-6}\times 50 = 120\ \text{MPa (compressive)}
$$

$$
P = \sigma A = 120\times 490.9 = 58.9\times10^3\ \text{N}
$$

(The stress does not depend on the length.)

#### (b) Walls yield by 0.6 mm
Only the difference between the free expansion and the wall movement has to be suppressed:

$$
\delta_{\text{restrained}} = 1.5 - 0.6 = 0.9\ \text{mm}
$$

$$
\sigma = E\,\frac{\delta_{\text{restrained}}}{L} = 200000\times\frac{0.9}{2500} = 72\ \text{MPa (compressive)}
$$

$$
P = 72\times 490.9 = 35.3\ \text{kN}
$$

**Answer: (a) $\sigma = 120$ MPa (compression), $P = 58.9$ kN; (b) $\sigma = 72$ MPa (compression), $P = 35.3$ kN.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the nature and formula of the stress developed when the bar is fully restrained and heated?
   A. Tensile, sigma = E alpha dT
   B. Compressive, sigma = E alpha dT
   C. Compressive, sigma = alpha dT / E
   D. Tensile, sigma = E alpha dT L
   Answer: B. The wall prevents expansion, so the thermal strain is cancelled by elastic compressive strain: sigma = E alpha dT = 120 MPa. (Key point: Thermal stress between rigid walls)

2. How does the thermal stress change if the bar were longer?
   A. It increases in proportion to length
   B. It decreases in proportion to length
   C. It does not change
   D. It doubles when length doubles and area halves
   Answer: C. sigma = E alpha dT contains no length, so the stress is independent of length. (Key point: Length independence)

3. When the walls move apart by 0.6 mm, how much shortening must the load produce?
   A. 1.5 mm
   B. 0.6 mm
   C. 2.1 mm
   D. 0.9 mm
   Answer: D. Only the difference between free expansion (1.5 mm) and wall movement (0.6 mm), i.e. 0.9 mm, is suppressed, giving sigma = 72 MPa. (Key point: Yielding supports)

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