Thermal expansion calculator
Calculate how much longer a part becomes when heated, how the clearance of a pair made from two materials changes with temperature, and which temperature change produces a wanted length change. The expansion coefficient comes from a maintained material table or from free input - the calculator shows length, area and volume change as well as the clearance curve with the seizing temperature, 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.
Clearance of a pair
When two parts sit inside one another - shaft in bore, carriage in guide - their clearance s changes only insofar as the two parts expand by different amounts. What matters is not the difference of the coefficients alone but the product α·L of each part:
Index o denotes the outer part (the bore), i the inner part (the shaft). If the bracket is negative the clearance shrinks; it becomes zero when both parts are of the same material and the same size. The temperature at which the clearance is used up follows directly:
If that result lies below absolute zero there is no seizing temperature - the pair then never seizes.
Inverse: required temperature change
The same basic equation answers the question which temperature reaches a wanted length change - for aligning a long shaft, say, or for thermal joining:
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 pair is more instructive: a POM-C shaft of 79.95 mm runs in a steel bore of 80.00 mm, so the clearance at 20 °C is 50 µm. At 120 °C the bore expands by 12·10⁻⁶ · 80 · 100 = 0.096 mm, the shaft by 110·10⁻⁶ · 79.95 · 100 = 0.87945 mm. Arithmetically a clearance of 50 + 96 − 879.45 = −733.45 µm remains, that is an interference. The clearance is already used up at 26.4 °C - the bearing seizes before the machine is warm. With a steel shaft of the same size the clearance would stay practically unchanged.
Inverse: to make the same steel beam 100 µm longer requires ΔT = 0.1 / (12·10⁻⁶ · 1000) = 8.33 K, that is a final temperature of 28.3 °C.
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.
When does a fit seize, and when does it open up?
What decides is the product α·L of both parts, not the coefficient alone. If the inner part expands more (α_i·L_i larger than α_o·L_o), the clearance shrinks on heating and the pair eventually seizes. If the outer part expands more, the clearance grows - then cooling is the critical case. With identical materials everything grows proportionally and the clearance stays practically unchanged. The calculator also gives the seizing temperature at which the clearance becomes zero.
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 for the range around room temperature. For steel α rises from about 11.1·10⁻⁶/K over 0 to 100 °C to roughly 13.9·10⁻⁶/K over 0 to 600 °C, which is 25 %. 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 assumes a positive clearance and works out how it changes with temperature. An interference is not a clearance: there you get 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 is given by the joining temperature calculator.
Why is there no traffic light next to a plain length change?
Because there is nothing to assess. A traffic light claims that a value has been checked against a limit; with a freely possible expansion there is none. The calculator therefore shows an explicit "not rated" in blue. Only the pair is assessed, because there a criterion exists: the clearance must stay positive over the whole temperature range.
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