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Thermal Expansion Calculator

Estimate linear, area or volume expansion from a temperature change and linear expansion coefficient, with metric and US dimensional units.

Use a coefficient appropriate to your material and temperature interval. Negative values are allowed.

Estimate the dimensional change of an unconstrained solid over a temperature interval. Choose length, area or volume and supply a linear expansion coefficient appropriate to the material. The same calculation supports metres, centimetres, millimetres, feet and inches, while temperature can be entered in Celsius, Fahrenheit or kelvin.

How to use this tool

  1. Choose whether the starting measure is a length, area or volume.
  2. Enter that measure in the selected dimensional unit.
  3. Enter the linear coefficient in millionths per kelvin.
  4. Enter initial and final temperatures in one scale, then read the change and final measure.

Formula

ΔL = L0αΔT; ΔA ≈ A0(2α)ΔT; ΔV ≈ V0(3α)ΔT

For an isotropic solid, the first-order area and volume coefficients are approximately two and three times the linear coefficient. The input 12 means 12 × 10⁻⁶ per kelvin. A Fahrenheit temperature difference is multiplied by 5/9; Celsius and kelvin differences have the same size.

Example

A 2 m member with α = 12 × 10⁻⁶/K warming from 20°C to 120°C changes length by 2 × 12 × 10⁻⁶ × 100 = 0.0024 m, or 2.4 mm. Its predicted final length is 2.0024 m. These numbers use the supplied coefficient rather than identifying a particular alloy.

Frequently asked questions

Can I enter a negative coefficient?

Yes. Some materials contract over particular heating intervals. The sign of the supplied coefficient and temperature change determines the sign of the result.

Can I use a liquid’s volume coefficient?

Not directly in this tool. Its volume mode starts from a solid’s linear coefficient and uses 3α; a measured liquid volume coefficient belongs in a direct volumetric model.

Does this calculate stress in a restrained member?

No. It predicts free dimensional change. Restraint, material stiffness and boundary conditions are needed to calculate thermal stress.

What limits the accuracy?

The coefficient may vary with temperature or direction. Large strain, phase changes and non-isotropic behaviour need a more suitable material model.

Sources

Last updated: 2026-10-30

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