Length change ΔL
Positive for expansion, negative for contraction.
Thermal engineering · Free expansion
Linear thermal expansion calculator
Calculate free expansion or contraction from initial length, linear expansion coefficient and temperature difference.
Created v1.0
By Thibaut Grzelak, Mechanical Analysis Engineer
02
ExpansionResults
Final length L
m
L = L₀ + ΔL.
Thermal strain
µε
εth = α · ΔT.
Relative change
%
ΔL / L₀ as a percentage.
Calculation detailsRelations and numerical substitution
εth = α · ΔTΔL = α · L₀ · ΔT = εth · L₀L = L₀ + ΔLL₀initial lengthmm, cm, m, in, ft
Lfinal lengthlength unit
ΔLlength changeµm, mm, cm, m, in
αlinear expansion coefficientK⁻¹, °C⁻¹, °F⁻¹
ΔTdifference T𝒇 − TᵢK, °C, °F
εthfree thermal straindimensionless, %, µε
Numerical substitution
ΔT = 40 °C = 40 K
εth = 12 × 10⁻⁶ × 40 = 480 µεΔL = 12 × 10⁻⁶ × 2 × 40 = 0.96 mmL = 2 + 0.00096 = 2.00096 mFree-expansion calculation with constant α. Material values are indicative and must be checked for the actual grade and temperature range. Thermal stresses may develop when expansion is restrained.
01Indicative material coefficients
Typical α values
Values are expressed in µm/(m·K), numerically equivalent to 10⁻⁶ K⁻¹.
| Material | α [µm/(m·K)] | Note |
|---|---|---|
| Steel | 12 | Typical indicative value. Check the grade and project temperature range. |
| Stainless steel | 14 | Indicative value: about 14 × 10⁻⁶ K⁻¹, with a possible variation of roughly ±4 depending on the family. |
| Aluminium | 23 | Typical indicative value. Individual alloys may differ. |
| Concrete | 12 | Indicative value depending on mix design, aggregates, and moisture. |
| Bronze | 17.5 | Indicative value. The exact bronze composition affects the coefficient. |
| Constantan | 15.2 | Indicative value for a constantan-type alloy. |
| Copper | 17 | Typical indicative value. Check metallurgical condition and reference temperature. |
02Understanding the units
Why 12 means 12 × 10⁻⁶
With µm/(m·K), entering 12 is equivalent to 12 × 10⁻⁶ K⁻¹, or 0.000012 K⁻¹. For a temperature interval, 40 °C and 40 K represent the same change.
03Assumptions and limitations
Scope of validity
- Homogeneous part with free expansion in the studied direction.
- Coefficient α is assumed constant over the temperature interval.
- Small relative length changes.
- A restrained part requires an additional thermal-stress analysis.
Free material expansion
Thermal-expansion formula, coefficient and units
Length change depends on L₀, α and ΔT. For a temperature interval, 1 K and 1 °C have the same magnitude.
Linear-expansion equation
A positive ΔT produces expansion; a negative ΔT produces contraction.
ΔL = α · L₀ · ΔTSteel
With α ≈ 12 × 10⁻⁶ K⁻¹, a 2 m bar heated by 40 K expands by about 0.96 mm.
Aluminum
With α ≈ 23 × 10⁻⁶ K⁻¹, the same case produces about 1.84 mm.
Restrained expansion
Full restraint can create thermal stress that also depends on modulus E.
σth ≈ E · α · ΔT