Audited 27 Jul 2026·Last updated 28 Oct 2026·5 citations·Tier 1·0 uses

Thermal Expansion Calculator

How much a bar, pipe, rail or slab grows or shrinks with temperature: ΔL = α × L₀ × ΔT for steel, aluminium, copper, glass, concrete, Invar or your own α.

Thermal Expansion Calculator

Entered in the length unit selected below. The results are always reported in metres.
Length unit
In °C or K. Only the difference between the initial and final temperature matters, and a difference is the same in both
In °C or K. Only the difference between the initial and final temperature matters, and a difference is the same in both
Material
Change in length (ΔL)
0.0043
Change in length in metres: positive when the part grows, negative when it shrinks. Multiply by 1,000 for millimetres.
Final length (L)
12.0043
Temperature change (ΔT)
30

Background.

Warm a 12-metre steel beam from 15 °C to 45 °C and it grows by 4.32 millimetres. That sounds trivial until something stops it. Held rigidly at both ends, the same beam would press on its supports with a stress of about 72 megapascals, which is why bridges sit on sliding bearings, concrete slabs are cut with joints, long pipe runs are bent into expansion loops, and railway engineers care about the temperature at which continuous rail is welded.

The calculation behind all of these is linear thermal expansion: the change in length equals the material's expansion coefficient, times the original length, times the temperature change. Choose a material — aluminium, structural steel, copper, soda-lime glass, concrete or the low-expansion alloy Invar — or type in your own coefficient. Give the starting length in metres, millimetres, inches or feet, then the starting and finishing temperatures. Those can be in °C or kelvin, since only their difference enters the formula and a step of one degree Celsius is a step of one kelvin. The change in length and the final length come back in metres, with the temperature change beside them.

The coefficients differ more than intuition suggests. Aluminium grows almost twice as much as steel for the same warming, 23.1 against 12.0 millionths of its length per degree, so aluminium frames and cladding need wider movement gaps. Concrete and steel share the same figure, 12.0, and reinforced concrete depends on that match: the bars and the concrete around them expand together instead of working loose. Invar, at 1.2, barely moves at all.

The model treats the coefficient as constant across the temperature range, which suits solids within a few tens of degrees of room temperature but drifts over hundreds of degrees, and it describes one dimension of a part that is free to move.

What is thermal expansion calculator?

Linear thermal expansion is the change in a solid's length caused by a change in its temperature. Its size is set by the coefficient of linear expansion, α, the fractional change in length per degree: steel's 12.0 × 10⁻⁶ /°C means every metre of steel grows by 12 micrometres for each degree of warming, and shrinks by the same amount for each degree of cooling. Because a degree Celsius and a kelvin are the same size, α has the same value per °C and per K. The cause lies in the bonds between atoms, which resist being squeezed more strongly than they resist being stretched; as atoms vibrate harder with heat, their average spacing creeps outward. For a material that behaves the same in every direction, area grows at about 2α and volume at about 3α. Differences in α are put to work as often as they are designed around. A bimetallic strip, two metals with different coefficients bonded together, bends as it warms because one side grows more than the other, and that bending has long switched thermostats and tripped circuit breakers. Borosilicate laboratory glass, with a coefficient around 3.3 × 10⁻⁶ /°C against 9.0 for ordinary soda-lime glass, survives sudden heating far better because its hot and cold regions pull against each other less.

How to use this calculator.

  1. Enter the initial length and choose its unit: metres, millimetres, inches or feet. The length is converted to metres before the calculation, and both length results are given in metres; multiply by 1,000 for millimetres, divide by 0.0254 for inches or by 0.3048 for feet.
  2. Enter the initial and final temperatures, both in °C or both in kelvin. Fahrenheit is not accepted, so convert each reading to Celsius first. A final temperature below the initial one is allowed and gives a negative change: the part contracts.
  3. Choose the material. For anything not on the list, choose Custom and enter α in millionths per degree as printed on most datasheets: 3.3 for borosilicate glass means 3.3 × 10⁻⁶ /°C. The custom box is used only when Custom is selected, and it must be greater than zero.
  4. For a movement joint or clearance, start from the temperature at which the part is installed or fixed, and run the calculation twice: once up to the hottest temperature it will reach and once down to the coldest. Surfaces in direct sun can run well above the air temperature, so allow for that in the hot case.

The formula.

ΔL = α × L₀ × ΔT

The change in length is ΔL = α × L₀ × ΔT, and the final length is L = L₀ + ΔL. Here α is the coefficient for the chosen material, L₀ the initial length in metres and ΔT the final temperature minus the initial one. The length conversions are exact by definition: one inch is 0.0254 m and one foot 0.3048 m.

The equation is the first term of a more general relationship, and it works because the strains involved are small. In the example the beam stretches by 0.036% of its length, so treating α as fixed and ignoring the tiny compounding of growth on growth changes nothing that could be measured on site. Over wide ranges the coefficient itself changes, usually rising with temperature in metals, and precise work integrates it: ΔL = L₀ × ∫α dT, using tabulated values such as those in NIST's compilation for metals and alloys. The listed coefficients are room-temperature values. The same equation applies to any dimension of a part, including the diameter of a hole: a hole in a heated plate gets larger, exactly as if it were filled with the plate's material, which is how a shrink fit is made by heating a ring until it slips over its shaft.

The formula also assumes the part is free to move. If it is restrained, the expansion that cannot happen turns into stress, σ = E × α × ΔT, where E is the material's elastic modulus. With E of about 200 GPa for steel, a fully restrained steel member warmed by 30 °C carries 200 × 10⁹ × 12 × 10⁻⁶ × 30 = 72 MPa, and by 50 °C, 120 MPa, regardless of its length. A slender member can buckle sideways under such loads well before the steel itself yields.

Two cases are outside the model altogether: materials that contract when heated, which need a negative coefficient that is not accepted here, and any temperature range that crosses a change of phase, such as melting or the rearrangement of iron's crystal structure at 912 °C, where the length jumps rather than drifting.

A worked example.

Example

A 12-metre steel footbridge deck is installed at 15 °C, and the designer wants its growth on a day when the deck reaches 45 °C. Enter 12 with the unit set to metres, choose structural steel (α = 12.0 × 10⁻⁶ /°C), and enter 15 and 45 as the temperatures. The temperature change is 45 − 15 = 30 °C. The growth is ΔL = 12.0 × 10⁻⁶ × 12 × 30 = 0.0043 m, which is 4.32 mm, and the final length is 12.0043 m. As a fraction of the deck's length that is 0.036%, too small to see but large enough to jam a bearing that has no room to slide. The cold side matters as much. Running the same deck from its 15 °C installation temperature down to −10 °C gives ΔT = −25 °C and ΔL = −0.0036 m, a shortening of 3.6 mm. Between its coldest and hottest days the deck's length therefore varies by 4.32 + 3.6 = 7.92 mm, and that full range, plus a margin, is the movement its bearings and the gap at its end need to accommodate.

initial Length12
T_initial15
T_final45

Frequently asked questions.

Can I enter temperatures in Fahrenheit?
Not directly; convert each reading to Celsius first, or convert just the difference by multiplying it by 5/9. A swing of 54 °F is a swing of 30 °C, so a 12-metre steel beam warming by 54 °F grows by the same 4.32 mm as in the example. Do not enter the Fahrenheit readings themselves, which would overstate the change by 80%.
How do I get the answer in millimetres or inches?
Results are always in metres, whatever unit the length was entered in. Multiply by 1,000 for millimetres or divide by 0.0254 for inches. A 120-inch aluminium frame warming from −5 °C to 35 °C grows by 0.0028 m, which is 2.82 mm or about 0.11 inch.
What happens if the part cannot expand?
The growth it is denied becomes stress, σ = E × α × ΔT, independent of length. A fully restrained steel member warmed by 30 °C carries about 72 MPa of compression, and in a long slender piece that can be enough to make it bow sideways. Expansion joints, sliding bearings, pipe loops and flexible couplings exist to let the movement happen instead.
How do railways cope with continuous welded rail?
They restrain it rather than letting it move. A kilometre of steel rail that was free to expand would grow by 0.84 m between −20 °C and 50 °C. Instead the rail is anchored firmly to the sleepers and fastened at a chosen stress-free temperature, so it sits in tension on cold days and in compression on hot ones. If the hot-weather compression wins against the track's lateral resistance, the rail buckles, which is the familiar summer track kink.
Why does Invar expand so little?
Invar is an iron-nickel alloy with about 36% nickel. Below its magnetic transition, the alignment of its magnetic moments tends to expand the lattice in a way that falls away as it warms, almost cancelling ordinary thermal growth. Charles-Édouard Guillaume discovered it in 1896 and received the 1920 Nobel Prize in Physics for the work. The 12-metre example in Invar grows by only 4.320 × 10⁻⁴ m, under half a millimetre.
Is volume expansion just three times the linear figure?
For a material that expands equally in all directions, yes, to a very good approximation: volume grows at about 3α and area at about 2α, because each of the three dimensions grows by the same small fraction. Crystals such as quartz and graphite expand differently along different axes, and liquids are described by a volume coefficient directly, so neither suits a linear calculation.
What value should I use for plastics or other materials?
Use the manufacturer's datasheet and enter it with Custom. Plastics typically expand several times more than steel, and their coefficient can change sharply near the glass-transition temperature, so check that your temperature range sits inside the datasheet's. With a datasheet figure of 52 × 10⁻⁶ /°C, a 6-metre pipe warming by 30 °C grows 0.0094 m, more than 9 mm.
Why can't I enter a negative coefficient?
The custom coefficient must be greater than zero, so materials that shrink when heated are outside what this tool calculates. Such materials exist: zirconium tungstate contracts over a wide temperature range and has been studied as a filler to cancel expansion in composites, and liquid water shrinks as it warms from 0 °C to about 4 °C. For those, consult measured data directly.
Why are precise measurements corrected to 20 °C?
Because international standard ISO 1 fixes 20 °C as the reference temperature for dimensional specifications, and small temperature differences produce measurable errors. A 500 mm steel gauge block that is 0.1 °C too warm is longer by 6.000 × 10⁻⁷ m, 0.6 micrometres, which matters when tolerances are in microns. Metrology rooms are held close to 20 °C for this reason.
Does the coefficient stay the same at high temperatures?
No. The listed values apply near room temperature, and for most metals the coefficient rises as they get hotter, so a calculation over several hundred degrees will underestimate the growth. For furnace parts, exhaust systems or anything near a phase change, use temperature-dependent data and integrate it over the range, or take an average coefficient quoted for that specific range.

References& sources.

  1. [1]Halliday, D., Resnick, R., and Walker, J. (2013). Fundamentals of Physics, 10th ed. John Wiley & Sons. ISBN 978-1-118-23072-5
  2. [2]NIST (2004). Thermal Expansion of Pure Metals and Alloys. NIST Interagency/Internal Report (IR) 84.
  3. [3]ASTM E228-17 (2017). Standard Test Method for Linear Thermal Expansion of Solid Materials with a Push-Rod Dilatometer. ASTM International.
  4. [4]Guillaume, C.E. (1896). Recherches sur les aciers au nickel. Comptes Rendus Hebdomadaires des Séances de l'Académie des Sciences 122:1074–1076.
  5. [5]Wallace, D.C. (1972). Thermoelasticity of stressed materials and comparison of various elastic constants. Phys. Rev. B 6(4):1624–1630. doi:10.1103/PhysRevB.6.1624

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ΔL = α × L₀ × ΔT
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