ΔL = α · L · ΔT
Length change grows linearly with original length and temperature change; the expansion coefficient is a material property.
Length change grows linearly with original length and temperature change; the expansion coefficient is a material property.
Select a target and calculate.
Length change grows linearly with original length and temperature change; the expansion coefficient is a material property.
A 10 m structural-steel beam (α ≈ 12 µm/(m·K)) expands by about 6 mm on a 50 K temperature rise.
Isotropic material with an expansion coefficient constant over the range considered.
This calculator finds the missing quantity among length change ΔL, expansion coefficient α, original length L and temperature change ΔT from the other three. Using the Golden Gate Bridge as an example, it shows why large steel structures need measurable expansion joints even though individual components only expand by a few millimetres per metre.
A component's linear thermal expansion is ΔL = α·L·ΔT. Expansion coefficient α is a material constant stating what fraction of its length a material gains per kelvin of temperature change.
San Francisco's Golden Gate Bridge is built mostly from steel with an expansion coefficient of about 12 µm/(m·K). Across the roughly 55 K temperature range typical there between cold and hot extremes, the bridge calculates out to expand by several centimetres per 100 metres of structure — over a total length well over a kilometre, that adds up to a length change that dedicated expansion joints must absorb.
ΔL = α · L · ΔT
ΔL = α · L · ΔTα = ΔL / (L·ΔT)L = ΔL / (α·ΔT)ΔT = ΔL / (α·L)| Symbol / input | Meaning |
|---|---|
| Length change ΔL | Change in component length caused by the temperature change. |
| Linear expansion coefficient α | Material constant; structural steel about 12 µm/(m·K), aluminum about 23 µm/(m·K). |
| Original length L | Component length at the reference temperature. |
| Temperature change ΔT | Difference between final and initial temperature. |
Enter three of the four quantities to find the fourth. Use a table value for α matching the actual material; structural steel is about 12 µm/(m·K), aluminum about 23 µm/(m·K).
Select the target quantity and enter the three known values with units. Make sure ΔT is entered as a temperature difference in kelvin, not an absolute temperature.
Given a 10 m structural-steel beam (α ≈ 12 µm/(m·K)) and a temperature rise of ΔT = 50 K, substitution gives ΔL = 12×10⁻⁶ · 10 m · 50 K = 0.006 m = 6 mm.
A length change of 6 mm at 10 m original length and a 50 K temperature rise shows how much movement clearance a structure must provide at that point, e.g. via an expansion joint or a sliding bearing, to avoid excessive restraint stresses.
Length change and original length in the same or convertible length unit, expansion coefficient in µm/(m·K) (equal to 10⁻⁶/K) or directly in 1/K, temperature change in kelvin.
The formula is used to design expansion joints in bridges and buildings, plan pipework with expansion compensators, in precision measurement with temperature-sensitive scales, and to size fits for components operating at different temperatures.
The formula applies to isotropic materials with an expansion coefficient constant over the temperature range considered. Over very large temperature spans, α itself changes slightly with temperature; for anisotropic materials (e.g. many composites or wood), expansion must be considered separately per direction.
Common mistake: Do not confuse temperature change ΔT with an absolute temperature — only the difference between final and initial temperature matters. Also, don't substitute the expansion coefficient in the wrong unit: 12 µm/(m·K) equals 12×10⁻⁶/K, not 12/K.
Because structural steel measurably expands and contracts with temperature swings over the full length of the structure. Without expansion joints or sliding bearings, this would create excessive restraint stresses.
Structural steel is about 12 µm/(m·K), while aluminum at about 23 µm/(m·K) is nearly double — aluminum components expand correspondingly more for the same temperature change.
Yes, on cooling (negative ΔT), ΔL is also negative and describes a shortening of the component.