How to use this calculator
Enter the original length and the temperature change (final minus initial; negative for cooling). Choose a material to fill in the expansion coefficient and Young’s modulus, or type your own values. The first result is the free expansion. The last two show what happens if the member cannot move at all.
Formulas
A change of 1 °C equals a change of 1 K, and 1 °F equals 5/9 °C. Restrained stress does not depend on length: a short bar and a long bar of the same material reach the same stress for the same temperature change.
| Material | α (µm/m·°C) | α (µin/in·°F) | E (GPa) |
|---|---|---|---|
| Structural steel | 12 | 6.5 | 200 |
| Stainless steel 304 | 17.3 | 9.6 | 193 |
| Aluminum 6061-T6 | 23.6 | 13.1 | 68.9 |
| Copper | 17 | 9.4 | 117 |
| Titanium Ti-6Al-4V | 8.6 | 4.8 | 113.8 |
| Concrete (varies with aggregate) | 10 | 5.5 | 25 |
Worked example
10 m steel member heated by 40 °C (the default)
- ε = 12 × 10⁻⁶ × 40 = 4.8 × 10⁻⁴, or 480 µε.
- Free expansion: ΔL = 4.8 × 10⁻⁴ × 10 m = 4.8 mm.
- If the ends are fixed: σ = −200,000 MPa × 4.8 × 10⁻⁴ = −96 MPa (compression).
- With A = 5,000 mm², the restrained force is 96 × 5,000 = 480 kN.
That is why bridges, pipelines and rails need expansion joints or loops.
Assumptions and limits
- The temperature change is uniform through the member, with no gradient through its thickness.
- α and E are constant over the temperature range. Both change at high temperatures, so use values for your range.
- Restrained results assume perfectly rigid supports. Real supports give a little, which reduces the stress.
- A long, slender member heated while restrained may buckle before it reaches the calculated compressive stress.
Frequently asked questions
Why is the restrained stress negative?
Negative means compression. A heated bar wants to grow, so walls that hold it push back. A cooled, restrained bar is in tension instead.
How do I handle partial restraint?
If a support moves a distance δ, only ΔL − δ of the expansion is prevented, and the stress is E(ΔL − δ)/L.
Does this work for area or volume change?
For small strains, area changes by about 2αΔT and volume by about 3αΔT for isotropic solids.
For education and preliminary checks only. Have a qualified engineer verify any design before it is built.