Define the length and temperature change
L₀ is the length at the initial temperature. ΔT is final temperature minus initial temperature, and α is the linear expansion coefficient over that interval. The result is an approximate change in length; final length is L₁ = L₀ + ΔL.
The relation and its constant-coefficient approximation are explained in OpenStax University Physics, Thermal Expansion. For this calculator, assume a homogeneous, isotropic solid, uniform temperature change and freedom to expand. Supply a coefficient appropriate to the material and temperature range.
Worked example: a one-metre part warms by 50 K
Take L₀ = 1,000 mm and an illustrative α = 12 × 10⁻⁶/K. A rise from 20°C to 70°C gives ΔT = 50°C, equal to a 50 K interval. In the calculator, enter the coefficient as 12 because its input unit is µm/m/K.
ΔL = 1,000 × 0.0006 = 0.6 mm
L₁ = 1,000 + 0.6 = 1,000.6 mm
The strain is 600 microstrain, or 0.06%. The example coefficient is an assumption, not a certified value for a particular alloy. Doubling the starting length doubles the predicted movement if α and ΔT stay the same.
Convert intervals without an offset
A Celsius degree and a kelvin have the same interval size. A Fahrenheit interval is multiplied by 5/9 to obtain a kelvin interval. The +32 offset used for absolute Celsius-to-Fahrenheit temperatures does not apply to differences.
| Temperature interval | Equivalent interval | Length change |
|---|---|---|
| +50°C | +50 K | +0.6 mm |
| +90°F | +50 K | +0.6 mm |
| −50°C | −50 K | −0.6 mm |
Each row uses the same 1,000 mm reference length and coefficient. The cooling row starts from its own reference temperature; it is not a second step applied after the warming row.
Common questions
Can I use this to calculate thermal stress?
The tool calculates free movement. If anchors, adjoining parts or contact prevent that movement, compatibility and stiffness determine the forces and stresses. A temperature gradient can also bend a part. Those effects need a different model.
Which expansion coefficient should I enter?
Use a traceable value for the material, direction and operating interval. A mean coefficient over a stated range can differ from an instantaneous value at one temperature. Composites, anisotropic materials and phase changes need particular care; this tool contains no material-property database.
Does the result specify an expansion-joint gap?
No. Gap selection also depends on installation temperature, tolerances, restraint, joint capacity and the full movement envelope. The 0.6 mm example is one predicted free length change, not a recommended clearance.
Related: engineering unit converter or all engineering guides.