Linear Thermal Expansion Calculator
Calculate how much a material expands or contracts when its temperature changes. Enter the initial length, expansion coefficient and temperature change to get ΔL and the final length.
Find How Much a Material Grows or Shrinks with Temperature
Linear Thermal Expansion Formula
Where:
- \(\Delta L\) — change in length (same unit as \(L_0\))
- \(\alpha\) — linear expansion coefficient of the material (1/°C, 1/K or 1/°F)
- \(L_0\) — initial length at the starting temperature
- \(\Delta T\) — temperature change (final minus initial)
- \(L\) — final length after the temperature change
The formula expresses the fact that, over ordinary temperature ranges, the length change of a solid is proportional to both its original length and the temperature change. A negative \(\Delta T\) (cooling) simply gives a negative \(\Delta L\), meaning the object shrinks.
Linear Thermal Expansion – Calculation Example
Example: a 2 m steel bar (\(\alpha = 0.000012\) 1/°C) is heated by 50 °C. How much does it lengthen, and how long is it afterwards?
\[ \Delta L = \alpha \cdot L_0 \cdot \Delta T = 0.000012 \times 2 \times 50 = 0.0012 \ \text{m} = 1.2 \ \text{mm} \\ \[4pt] L = L_0 + \Delta L = 2 + 0.0012 = 2.0012 \ \text{m}
\]
Result: the bar expands by 1.2 mm and its final length is 2.0012 m — a relative change of only +0.06 %.
How to Use the Calculator
- Enter the Initial length (L₀) of the object and pick its unit (mm, cm, m, in or ft).
- Enter the Linear expansion coefficient (α) for your material and choose the matching unit (1/°C, 1/K or 1/°F). Typical values are listed in the field hint: steel ≈ 0.000012, aluminum ≈ 0.000023, glass ≈ 0.000009 per °C.
- Enter the Temperature change (ΔT) with its unit. Use a positive value for heating and a negative value for cooling.
- Read the results: the Change in length (ΔL) and the Final length (L), shown in your chosen unit and in a table of all other common units.
What Is Linear Thermal Expansion?
When a solid is heated, its atoms vibrate more strongly and, on average, sit slightly farther apart. The object therefore grows in every dimension. Linear thermal expansion describes the change along a single dimension — the length of a bar, rail, pipe or beam. Because most materials expand by only a fraction of a percent over everyday temperature swings, the relationship between length and temperature is linear to a very good approximation: doubling the temperature change doubles the length change.
The constant of proportionality, \(\alpha\), is a property of the material. It tells you how much a 1-unit-long piece stretches per degree of warming. Metals expand relatively strongly (aluminum, brass, copper), while materials like glass, concrete and especially invar are engineered or naturally chosen for low expansion. Note that \(\alpha\) itself varies slightly with temperature; the values in the hint box are typical room-temperature figures.
The absolute change may look tiny — a few millimetres on a 2 m steel bar — but it matters enormously in engineering. Railway tracks are gapped, bridges sit on expansion joints, and pipelines include loops precisely because these small movements, if restrained, generate enormous forces. Conversely, a shrink fit (heating a ring so it slides over a shaft) exploits the same effect. When interpreting your result, remember that a negative \(\Delta L\) is just as physically real as a positive one: cooling contracts the object.
Assumptions, Limitations and Practical Tips
- Small temperature changes. The linear formula assumes \(\alpha \, \Delta T \ll 1\) and a constant \(\alpha\). For temperature swings of hundreds of degrees, or near phase changes, results become approximate.
- Isotropic material. The calculator assumes the material expands equally in all directions. Wood and many composites expand at different rates along and across the grain.
- Free expansion only. The result is the length change of an unrestrained object. If the object is clamped, it cannot expand, and stresses build up instead — no length change occurs but large forces appear.
- Unit consistency. The calculator converts everything internally, so you may freely mix units (e.g. inches with 1/°C), but the coefficient must correspond to the actual material.
- From 1/°F to 1/°C. A coefficient in 1/°F is 1.8 times the value in 1/°C, because one degree Fahrenheit is only 5/9 of a degree Celsius. The calculator handles this conversion automatically.
- Volume expansion. For liquids or full 3D changes, the volumetric coefficient is roughly \(3\alpha\). You can also estimate mass-related properties with the Density Calculator.
Typical Linear Expansion Coefficients
| Material | α (10⁻⁶ 1/°C) |
|---|---|
| Aluminum | 23 |
| Brass | 19 |
| Copper | 17 |
| Steel (typical) | 12 |
| Glass (common) | 9 |
| Invar | 1.2 |
| Concrete | ≈ 12 |
Thermal expansion is just one of the ways temperature shapes physical behaviour — it also changes air density and even the speed of sound in air. For mechanical deformation caused by forces rather than temperature, see the Hooke's Law Calculator.
Frequently Asked Questions
How do I calculate linear thermal expansion by hand?
Multiply three numbers: the linear expansion coefficient, the original length, and the temperature change. The formula is \( \Delta L = \alpha \cdot L_0 \cdot \Delta T \). For example, a 5 m copper pipe (\(\alpha \approx 0.000017\) 1/°C) heated by 30 °C expands by \(0.000017 \times 5 \times 30 = 0.00255\) m, or about 2.6 mm. Add \(\Delta L\) to \(L_0\) to get the final length.
Are 1/K, 1/°C and 1/°F coefficients interchangeable?
Values per kelvin and per degree Celsius are identical, because a change of 1 K equals a change of 1 °C — you can use them interchangeably. A coefficient in 1/°F is different: one degree Fahrenheit is only 5/9 of a degree Celsius, so \( \alpha_{(1/\text{°F})} = \tfrac{5}{9} \alpha_{(1/\text{°C})} \). A steel value of 12×10⁻⁶ 1/°C equals 6.67×10⁻⁶ 1/°F. The calculator converts this automatically.
What is a typical or normal expansion coefficient?
Most metals fall between 10×10⁻⁶ and 25×10⁻⁶ per °C: steel ≈ 12, copper ≈ 17, brass ≈ 19, aluminum ≈ 23 (all ×10⁻⁶ 1/°C). Glass and concrete are lower (≈ 9 and ≈ 12), while invar is exceptionally stable at ≈ 1.2. Plastics can be much higher, from 50 to over 200×10⁻⁶ 1/°C. Always check a materials table for your specific alloy, as composition changes the value.
Does the object shrink when it cools?
Yes. Enter a negative temperature change and the result will be negative. For example, a 3 m aluminum bar (\(\alpha = 0.000023\) 1/°C) cooled by 40 °C contracts by \(0.000023 \times 3 \times (-40) = -0.00276\) m, ending at 2.99724 m. Contraction from cooling is just as physically real as expansion, and restrained cooling can generate large tensile stresses and cracking.
Can I use this calculator for area or volume expansion?
For solids, yes approximately: the area coefficient is about \(2\alpha\) and the volumetric coefficient about \(3\alpha\), where \(\alpha\) is the linear coefficient. So you can multiply your \(\alpha\) by 2 or 3 and treat the result as the fractional change in area or volume. This approximation breaks down for large temperature swings. For liquids, use a tabulated volumetric coefficient directly, since liquids have no meaningful linear dimension.
What happens if the object is fixed at both ends and cannot expand?
Then its length stays constant and the thermal expansion turns into stress instead. The compressive thermal stress is \( \sigma = E \cdot \alpha \cdot \Delta T \), where \(E\) is Young's modulus. For steel heated by just 50 °C this already exceeds 100 MPa — enough to buckle railway tracks or crack rigid pipework. This calculator shows the free, unrestrained expansion only, so do not use it directly for clamped structures.
What are common mistakes when calculating thermal expansion?
The most frequent errors are: mixing up 1/°F and 1/°C coefficients (they differ by a factor of 1.8); entering an absolute temperature instead of a temperature change (ΔT, not the final temperature); using a length in one unit while expecting the answer in another; and forgetting the sign — cooling needs a negative ΔT. Also, wood and composites expand differently along and across the grain, so a single α may not describe them accurately.