Linear, area & volume thermal expansion — 4-way linear solver, material α table, and a thermal stress estimate. 100% client-side.
Formula: ΔL = α·L₀·ΔT. Only the temperature change matters, so °C and K deltas are identical; °F deltas are ×5/9.
Isotropic area expansion: ΔA = 2α·A₀·ΔT (β = 2α).
Isotropic volume expansion: ΔV = 3α·V₀·ΔT (γ = 3α). Not valid for water near 4 °C (anomalous expansion).
Material
α (×10⁻⁶/K)
Typical use
Aluminium
23
Structures, heat sinks
Copper
17
Electrical, piping
Iron / steel
12
Bridges, rails (expansion joints)
Glass (soda-lime)
9
Windows, containers
Invar
1.2
Precision instruments, clocks
Values are representative room-temperature linear coefficients, spot-checked against published reference tables. Real alloys vary by a few ×10⁻⁶/K; use the "Custom α" option for exact grades. α itself drifts slightly with temperature — this tool uses the standard constant-α engineering approximation.
Fully constrained bar: σ = E·α·ΔT. Example: steel (E = 200 GPa, α = 12×10⁻⁶/K) cooled by 100 K develops 200×10⁹ × 12×10⁻⁶ × 100 = 240,000,000 Pa = 240 MPa tensile stress. A dedicated thermal stress calculator is planned as a separate tool; boundary: this page covers expansion, stress gets its own deep-dive tool.
How thermal expansion works
Heating a solid makes its atoms vibrate more, and the asymmetric atomic bond pushes the average spacing apart — the object grows. For engineering purposes the growth is linear in temperature: ΔL = α·L₀·ΔT. Area and volume follow from geometry for isotropic materials: ΔA = 2α·A₀·ΔT, ΔV = 3α·V₀·ΔT. Engineers leave expansion joints in bridges and rails precisely because of this — a 1 km steel rail (α = 12×10⁻⁶/K) grows 12×10⁻⁶ × 1000 × 40 = 0.48 m across a 40 K summer day.
Frequently asked questions
What is the formula for linear thermal expansion?
Linear thermal expansion follows ΔL = α·L₀·ΔT, where α is the coefficient of linear expansion (per K), L₀ is the original length, and ΔT is the temperature change in kelvin (or °C degrees). For example, a 1 m aluminium rod (α = 23×10⁻⁶/K) heated from 20 °C to 100 °C grows by 23×10⁻⁶ × 1 × 80 = 0.00184 m, i.e. 1.84 mm.
How do you calculate area and volume thermal expansion?
For isotropic materials, area expansion uses β = 2α so ΔA = 2α·A₀·ΔT, and volume expansion uses γ = 3α so ΔV = 3α·V₀·ΔT. A 1 m² steel plate (α = 12×10⁻⁶/K) heated from 20 °C to 120 °C gains 2 × 12×10⁻⁶ × 1 × 100 = 0.0024 m². A 0.1 m³ copper block (α = 17×10⁻⁶/K) heated from 20 °C to 100 °C gains 3 × 17×10⁻⁶ × 0.1 × 80 = 0.000408 m³.
What are typical α values for common materials?
Typical coefficients of linear thermal expansion (×10⁻⁶ per K): aluminium 23, copper 17, iron 12, glass 9, invar 1.2. Invar (α = 1.2×10⁻⁶/K) is the standout: a 5 m invar bar heated by 100 K grows only 1.2×10⁻⁶ × 5 × 100 = 0.0006 m (0.6 mm), about 19 times less than aluminium.
How does thermal stress relate to thermal expansion?
If expansion is fully prevented, the material develops thermal stress σ = E·α·ΔT, where E is Young's modulus. A steel bar (E = 200 GPa, α = 12×10⁻⁶/K) cooled by 100 K while clamped experiences 200×10⁹ × 12×10⁻⁶ × 100 = 240,000,000 Pa = 240 MPa of tensile stress. A dedicated thermal stress calculator is planned as a separate tool; this page covers the expansion side and the basic formula only.
Can this calculator solve for α, L₀, or ΔT instead of ΔL?
Yes. The linear mode is a 4-way solver: given the other three quantities it can solve for ΔL, for the coefficient α, for the original length L₀, or for the temperature change ΔT. For instance, if a 1 m bar grows by 1.84 mm over an 80 K rise, α = 0.00184 / (1 × 80) = 23×10⁻⁶/K, identifying it as aluminium.
How do °F temperature changes convert to kelvin for expansion math?
Only the temperature CHANGE matters, so convert the delta: ΔT in kelvin = Δ°F × 5/9. Heating from 32 °F to 212 °F is a 180 °F change, equal to 180 × 5/9 = 100 K, exactly the same as heating from 0 °C to 100 °C. A 10 m iron bar (α = 12×10⁻⁶/K) through that change grows 12×10⁻⁶ × 10 × 100 = 0.012 m (12 mm).
Why is invar used for precision instruments?
Invar's coefficient (1.2×10⁻⁶/K) is roughly one-twentieth of aluminium's (23×10⁻⁶/K), so its length barely changes with temperature. That makes it ideal for pendulum clocks, precision gauges, and telescope structures where thermal drift would ruin accuracy.
Does thermal expansion apply to cooling as well?
Yes — cooling contracts materials by the same formula with a negative ΔT. A 2 m copper bar (α = 17×10⁻⁶/K) cooled from 100 °C to 20 °C shrinks by 17×10⁻⁶ × 2 × (−80) = −0.00272 m, i.e. it shortens by 2.72 mm.
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