Thermal expansion
Compute how much a member grows when it heats up, and the stress it develops if that growth is restrained. The second number is usually the one that cracks welds and shears anchors.
| Material | α (µm/m·K) | E (GPa) | ΔL (mm) | Restrained stress |
|---|---|---|---|---|
| Carbon steel | 12.0 | 200 | 8.640 | 144.0 MPa |
| Stainless 304 | 17.3 | 193 | 12.456 | 200.3 MPa |
| Aluminium | 23.1 | 69 | 16.632 | 95.6 MPa |
| Copper | 16.5 | 117 | 11.880 | 115.8 MPa |
| Concrete | 10.0 | 30 | 7.200 | 18.0 MPa |
| Soda glass | 9.0 | 70 | 6.480 | 37.8 MPa |
| Titanium | 8.6 | 114 | 6.192 | 58.8 MPa |
| Invar 36 | 1.2 | 141 | 0.864 | 10.2 MPa |
Restrained stress depends only on α, E and ΔT — not on length. A stiffer or more expansive material fails first.
ΔL = α·L·ΔT · α in 1/K, L in metres
ε = α·ΔT · independent of length
σ = E·α·ΔT · full restraint, elastic range
F = σ·A · what the anchors must resist
A clear path from inputs to results.
A heated member grows by αLΔT. If it is free to grow, that is the end of the story. If it is restrained, the same growth becomes a stress of EαΔT — a number that reaches yield surprisingly quickly and is the reason expansion joints exist.
How to use
- Enter the member length.
- Set the temperature change in kelvin (the same number as °C difference).
- Enter the expansion coefficient α, or read it from the material table.
- Enter Young’s modulus for the restrained-stress calculation.
- Set the cross-section area to convert stress into anchor force.
- Compare the free expansion with the restrained stress before deciding on a joint.
The formula
Reference: Standard isotropic thermoelasticity; α and E values from ASM Handbook and MatWeb material tables.
Worked example
12 m steel member, 60 K rise (default inputs)
- Length 12 m.
- Temperature change 60 K.
- α = 12 µm/m·K, E = 200 GPa.
- Section area 2,500 mm².
- 1
Expansion: 12×10⁻⁶ × 12 × 60 = 8.64 mm.
- 2
Strain: 12×10⁻⁶ × 60 = 720 microstrain.
- 3
Restrained stress: 200×10⁹ × 720×10⁻⁶ = 144 MPa.
- 4
Anchor force: 144 MPa × 2,500 mm² = 360 kN.
8.64 mm growth · 720 µε · 144 MPa if fully restrained · 360 kN of anchor force.
Interpreting the result
The 8.64 mm of growth is modest, but the 144 MPa of stress it would generate is not — that is over half the yield strength of ordinary structural steel, produced by a warm day. Notice that the stress does not depend on the length: a 1 m member and a 100 m member develop the same stress if both are fully restrained, because the strain is the same. Only the force changes, through the area.
Assumptions & limitations
- The temperature change is uniform through the member. A gradient causes bending as well as extension.
- The material is isotropic, so α is the same in all directions.
- Behaviour is linear elastic; once yield is reached, the stress relaxes rather than growing further.
- Restraint is total. Partial restraint produces a stress somewhere between zero and EαΔT.
- α and E are constant over the temperature range, which fails at high temperature and through phase changes.
Frequently asked questions
Why does length not affect the restrained stress?
Because stress is proportional to strain, and strain is αΔT — no length term. Length affects the free expansion and, through the cross-section, the force, but not the stress.
What is an expansion joint for?
It removes the restraint, letting the member grow freely so the stress never develops. The joint must accommodate the full calculated ΔL plus a margin for temperature extremes.
Does this apply to plastics and composites?
The formula holds, but α for polymers is 5–10 times higher and often anisotropic in composites. Thermal stress in bonded dissimilar materials is a common failure mode.
How much stress does steel develop per kelvin?
About 2.4 MPa per kelvin (200 GPa × 12×10⁻⁶). Fifty kelvin of restraint is 120 MPa, which is why unjointed steel structures need careful detailing.
Why does Invar behave differently?
Its α is about 1.2 µm/m·K, a tenth of steel’s, because of its nickel-iron composition. That makes it valuable for precision instruments, at a significant cost.
What about cooling?
Enter a negative ΔT. The member contracts, and if restrained it develops tensile stress instead of compressive. Both directions matter in practice.
Methodology last reviewed 2026-10-18. Results are informational and not financial or engineering advice — see our disclaimer.