Materials / Related reference
Calculate a free length change.
Enter a mean linear coefficient appropriate to the material and the complete temperature interval. No material value is assumed.
Your result
Enter values and calculate. Negative length changes indicate contraction.
Formula and reproducible example
ΔL = L₀ × α × 10⁻⁶ × (T₂ − T₁); L₂ = L₀ + ΔL. α uses parts per million per degree on the selected temperature scale. Length changes use the same unit as L₀.
Assumed arithmetic inputs: L₀ = 1000 mm, α = 12 µm/m/K, 20 → 70 °C. The result is +0.6 mm and a final length of 1000.6 mm. This is not a coefficient recommendation for any grade.
What this model leaves out
It assumes free, uniform expansion and small strain, with a valid mean coefficient for the interval. Gradients, restraint, phase changes, moisture, creep, joint stiffness and material damage are outside this model. A result is not a permitted heating temperature or a shrink-fit pressure.
Sources and scope
Reviewed 23 October 2026. Process explanations and public standard summaries do not establish a treatment recipe or certify a supplied part.
Kaiser Aluminum 6061 Rod and Bar — page 2, typical physical properties
KA-RBH-6061-8.10, page 2, typical physical properties: 23.6 µm/m/K over 20–100 °C. The coefficient row does not separately specify temper or test method. This is a typical source example, not a batch guarantee or a value for every 6061 product.
Open source ↗NIST — Engineering Metrology Toolbox FAQ
Temperature effects and the need for material-appropriate expansion data in dimensional measurement.
Open source ↗NIST — Aluminum 6061-T6 cryogenic data
Historical dataset, no longer updated. The linear-expansion model covers 4–300 K and is referenced to length at 293 K; it is not a constant coefficient for elevated temperatures.
Open source ↗