Change in length ΔL
-Final length
-Strain
-ΔL / L0Restrained stress
-Restrained force
-Yield safety factor
-Sy / restrained stressAll results
| Temperature change ΔT | - |
| Coefficient used | - |
| Coefficient used, other units | - |
| Length change per 100 degrees of ΔT | - |
| Area change of a surface (exact) | - |
| Volume change (exact) | - |
| Elastic modulus used | - |
| Temperature change that would just yield a fully restrained part | - |
Method and formulas
For a straight part of uniform temperature, free to move, the length change is proportional to the length and the temperature change. These are textbook relations, so you can check every result by hand.
| Quantity | Formula |
|---|---|
| Linear expansion | ΔL = α L0 ΔT, L = L0 (1 + α ΔT) |
| Thermal strain | ε = α ΔT |
| Area change (surface) | A / A0 = (1 + α ΔT)² ≈ 1 + 2 α ΔT |
| Volume change | V / V0 = (1 + α ΔT)³ ≈ 1 + 3 α ΔT |
| Fully restrained bar, stress | σ = −E α ΔT (compression when heated) |
| Fully restrained bar, force | F = σ A |
| Temperature change to reach yield | ΔTy = Sy / (E α) |
| Unit conversion | 1 µin/in·°F = 1.8 µm/m·K; ΔT in °F = 1.8 × ΔT in K |
The coefficient is a mean value over a temperature range and rises slowly with temperature. The material list holds typical handbook values near room temperature; choose Custom to enter the figure from your material data sheet. Elastic modulus and yield strength are likewise typical values and are only used for the restrained-part result. The restrained result assumes a perfectly rigid support and a stocky part; a slender part buckles at a much lower load than it yields.
Worked example: 10 ft steel bar heated by 100 °F
Carbon steel, α = 11.7 µm/m·K = 6.5 µin/in·°F, L0 = 120 in, ΔT = 100 °F (68 to 168 °F). ΔL = 6.5×10−6 × 120 × 100 = 0.078 in (1.98 mm). If the bar were fixed at both ends, with E = 205 GPa (29.7×106 psi), the stress would be 29.7×106 × 6.5×10−6 × 100 = 19 300 psi in compression, and with A = 1 in² the force is 19 300 lbf.
Questions
How do you calculate thermal expansion?
Linear thermal expansion is ΔL = α · L0 · ΔT, where α is the coefficient of thermal expansion, L0 the original length and ΔT the temperature change. Example: a 10 ft (120 in) carbon steel bar (α ≈ 6.5 µin/in·°F) heated by 100 °F grows 6.5×10⁻⁶ × 120 × 100 = 0.078 in. In metric, a 1 m steel bar (α ≈ 11.7 µm/m·K) heated by 50 K grows 0.585 mm.
What is the thermal expansion coefficient of steel, aluminum and stainless steel?
Typical mean values near room temperature are about 11.7 µm/m·K (6.5 µin/in·°F) for carbon steel, 17.3 µm/m·K (9.6 µin/in·°F) for 304 stainless, 23.6 µm/m·K (13.1 µin/in·°F) for 6061 aluminum, 8.6 µm/m·K for Ti-6Al-4V titanium and about 1.3 µm/m·K for Invar 36. Real values depend on alloy, heat treatment and temperature range, so use the figure from your material certificate or data sheet for critical work.
How much force does a restrained part produce when heated?
If a bar is held rigidly at both ends, the thermal strain cannot occur and a stress σ = E · α · ΔT develops (compression when heated, tension when cooled). The force is that stress times the cross-sectional area. Example: steel with E = 205 GPa heated by 50 K gives 205 000 × 11.7×10⁻⁶ × 50 = 120 MPa, which is why long rigid pipe runs and rails need expansion joints. This simple result ignores buckling, which usually occurs first in slender parts.
Does area and volume expansion follow the same coefficient?
For an isotropic material the area coefficient is about 2α and the volume coefficient about 3α, because each dimension grows by the same fraction. This calculator evaluates (1 + αΔT)² and (1 + αΔT)³ exactly, so the result stays correct for large temperature changes.
Is the coefficient of thermal expansion constant?
No. α rises slowly with temperature for most metals, so a single value is a mean over a range. For temperature swings of a few hundred degrees around room temperature the error of a mean value is small. For furnace parts, cryogenic work or tight precision fits, use a mean coefficient published for the actual temperature range.
How do I leave room for thermal expansion in a design?
Compute ΔL for the highest and lowest temperatures the part will see, from the temperature at which it is assembled, and provide that clearance, a floating bearing, a slotted hole or an expansion joint. Remember that two different materials joined together expand by different amounts, and use the calculator once for each material and subtract.
Does the calculator include bending, bimetal effects or temperature gradients?
No. It assumes a uniform temperature and a free or fully restrained straight part. Bimetallic bending, temperature gradients, partial restraint, creep and phase changes are not covered. It is an engineering aid; verify critical designs.
Engineering aid for guidance. It assumes uniform temperature and a free or fully restrained straight part, and does not cover temperature gradients, partial restraint, bimetal bending, buckling, creep or phase changes. Coefficients, moduli and strengths are typical handbook values: use certified data for your material. Verify critical designs; you remain responsible for the design. Nothing you enter leaves your browser except through a share link you copy yourself.