Thermal expansion calculator
Calculate how much longer a part becomes when heated and how much shorter when cooled. The expansion coefficient comes from a maintained material table or from free input - the calculator shows the length change, the new length and the area and volume change, live with every input.
Thermal expansion
The calculation uses linear thermal expansion with an α held constant over the temperature range. The part expands freely and is at the same temperature across its section; residual stress, creep and phase changes are not considered.
Results
Calculating …
Calculation in your browser, inputs go to our server only when you export or save.
Formulas and fundamentals
Length change
A part that can expand freely becomes longer by the following amount for a temperature change ΔT = T₁ − T₀:
α is the linear coefficient of thermal expansion in 1/K, usually quoted in 10⁻⁶/K. On cooling ΔT is negative and the part gets shorter. No stress arises: the bar simply moves. Only when the expansion is restrained does the thermal stress σ_th = −E·α·ΔT appear, and the axial bar calculator is the right tool for that case.
Area and volume
If every edge grows by the factor (1 + α·ΔT), the area grows with its square and the volume with its cube:
The calculator uses this exact form. The common approximations 2·α·ΔT and 3·α·ΔT are its leading terms; for steel at 100 K they are off by 0.6 ‰ of the area change, for aluminium at 300 K by 7.0 ‰ of the volume change. The difference is small, but being exact costs nothing.
The expansion coefficient
α is not a constant but rises with temperature. The tabulated values are therefore mean values over a temperature range, and the sources state them for different ranges: mostly 0 or 20 to 100 °C, for some materials a wider one. For plain carbon steel the value is 12·10⁻⁶/K, for aluminium 23.5·10⁻⁶/K, for plastics a multiple of that. The calculator says so as soon as one of the two temperatures lies outside −50 to 200 °C, for plastics already above 100 °C. The calculation stays correct, only its input becomes uncertain - for a different range enter a suitable mean value as a free input.
Worked example
Calculator default: a steel beam of S235JR with L = 1000 mm is heated from 20 to 120 °C. With α = 12·10⁻⁶/K and ΔT = 100 K this gives ΔL = 12·10⁻⁶ · 1000 · 100 = 1.2 mm, the new length is 1001.2 mm. The area grows by (1.0012)² − 1 = 0.2401 %, the volume by (1.0012)³ − 1 = 0.3604 %.
The same beam in aluminium AW-6060 expands almost twice as far with α = 23.5·10⁻⁶/K: ΔL = 23.5·10⁻⁶ · 1000 · 100 = 2.35 mm. Cooling the steel beam from 20 to −30 °C instead gives ΔT = −50 K and therefore ΔL = −0.6 mm; it is then 999.4 mm long.
The 1.2 mm are harmless only as long as the beam can move. Held at both ends it develops the thermal stress σ_th = −E·α·ΔT = −210000 · 12·10⁻⁶ · 100 = −252 N/mm² instead - more than the yield strength of S235. That case belongs to the axial bar calculator, not to this one.
Frequently asked questions
Why does free thermal expansion produce no stress?
Because nothing opposes the deformation. The bar gets longer, that is all - there is no force and therefore no verification to carry out. Stress only arises when the expansion is fully or partly restrained, for instance between two fixed supports or in an assembly of two materials. The axial bar calculator covers that case; it gives σ_th = −E·α·ΔT and the safety check against the yield strength.
Why does this calculator not assess anything?
Because there is nothing here to assess. A traffic light claims that a value has been checked against a limit - with a freely possible expansion there is none. A length change is neither OK nor borderline, it is simply the length that results. Whether it is acceptable is decided by the surrounding design: a fixed bearing with a floating one, an expansion joint, clearance in a guide. As soon as the expansion is restrained there is a limit again, and then the axial bar calculator is the right tool.
How do I find out whether a fit seizes at temperature?
Calculate both parts one after the other, each with its own material and nominal size. The difference of the two length changes is the change of clearance. What matters is the product α·L of each part, not the coefficient alone: a POM-C shaft of 79.95 mm expands by 0.528 mm from 20 to 80 °C, the steel bore of 80.00 mm by only 0.058 mm - nothing is left of 50 µm initial clearance, the pair seizes. With identical materials and sizes everything grows proportionally and the clearance stays practically unchanged.
Why does the calculator use the exact area formula instead of 2·α·ΔT?
Because the exact form is no more effort. Expanded, (1 + α·ΔT)² − 1 equals 2·α·ΔT + (α·ΔT)²; the common approximation simply drops the quadratic term. For steel at 100 K that is a difference of 0.6 ‰ of the area change, for aluminium at 300 K 3.5 ‰; for the volume it is roughly twice that, so 7.0 ‰ there. Irrelevant for design, but there is no reason to be less accurate than necessary.
Does the expansion coefficient hold over the whole temperature range?
No. The tabulated values are mean values over a temperature range, mostly 0 or 20 to 100 °C, for some materials a wider one - for S235JR this calculator uses 12·10⁻⁶/K. Over a wide range that mean rises noticeably: for plain carbon steel it reaches roughly 14·10⁻⁶/K over 0 to 600 °C, about a sixth higher. The calculator therefore warns as soon as a temperature lies outside −50 to 200 °C, for plastics already above 100 °C. The calculation stays correct, only its input becomes uncertain - for higher temperatures enter a coefficient for the relevant range as a free value.
Is this the right calculator for an interference fit?
No. This calculator tells you how much longer a single part becomes. An interference fit brings joint pressure, equivalent stress and transmissible torque, and the interference fit calculator to DIN 7190 covers that. The temperature needed to expand a hub for joining, or to cool a shaft, is given by the joining temperature calculator; it uses separate coefficients for heating and cooling.
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