Audited 31 Jul 2026·Last updated 28 Oct 2026·3 citations·Tier 2·0 uses

Thermal Stress Calculator

Find the stress in a fully restrained bar from Young's modulus, the expansion coefficient and the temperature change using σ = EαΔT, in pascals.

Thermal Stress Calculator

Pa
K⁻¹
K
Fully restrained stress magnitude
84,000,000
The magnitude of the stress in a fully restrained member, in pascals (1 MPa is 1,000,000 Pa); compressive when heated, tensile when cooled.
Model scope
Magnitude for complete uniaxial restraint in a homogeneous linear-elastic member; partial restraint, gradients, yielding, creep, buckling, multiaxial states, and joints are outside it.

Background.

Heat a steel bar and it grows by about 12 millionths of its length for every kelvin. Let it grow and nothing much happens. Clamp both ends so it cannot, and the expansion it was denied turns into force: the bar pushes on whatever holds it, and the metal inside carries a compressive stress. Cool a restrained bar and the same thing happens in reverse, as tension.

For a fully restrained member the size of that stress is σ = E × α × ΔT. The expansion coefficient α times the temperature change gives the strain the bar would have taken up if it were free, and Young's modulus E converts that blocked strain into stress. For steel, with E around 200 GPa and α around 12 × 10⁻⁶ per kelvin, every kelvin of restrained heating adds 2.4 MPa.

Length appears nowhere in that formula. A one-metre bar and a kilometre of welded rail, each held rigidly at both ends, develop the same stress for the same temperature change. That is the problem railways face with continuous welded rail: the track is fastened down at a chosen stress-free temperature, and on a hot afternoon well above it the rail is in heavy compression and can buckle sideways, while on a cold night it is in tension and a weak weld can pull apart.

The number here is the most a rigid restraint can produce, an upper bound. Real structures rarely hold a member perfectly still; bearings slide, joints open and supports flex, which is exactly what expansion joints are for. The formula also assumes a uniform temperature through the member, elastic behaviour and stress along one axis only. Temperature gradients, yielding, creep and buckling itself all fall outside it.

What is thermal stress calculator?

Thermal stress is the stress that builds up in a material when a temperature change tries to make it expand or contract and something stops it. For a straight, uniform member restrained completely along one axis, its magnitude is σ = EαΔT, the stress needed to cancel the free thermal strain αΔT. Restrained heating produces compression and restrained cooling produces tension; the page reports the magnitude in pascals.

How to use this calculator.

  1. Enter Young's modulus in pascals. Steel is about 200 GPa, entered as 200000000000 or 2e11, and aluminium alloys about 70 GPa (7e10); use the datasheet value for your grade if you have it.
  2. Enter the linear expansion coefficient per kelvin, about 0.000012 for carbon steel and about 0.000023 for aluminium. A coefficient quoted per °C is the same number.
  3. Enter the temperature change as a positive number of kelvin: the difference between the temperature at which the member was fixed without stress and the temperature you want to check.
  4. Compare the result with the material's yield strength, remembering that it assumes rigid restraint, so a real support that gives a little will see less.

The formula.

|σ| = Eα|ΔT|

|σ| = E × α × |ΔT|. The product α × ΔT is the free thermal strain, the fractional change in length the member would show with nothing in its way; for steel heated by 35 K it is 0.00042, or 0.042%. Full restraint holds the actual strain at zero, so the material must carry a mechanical strain equal and opposite to the thermal one, and Hooke's law turns that into stress by multiplying by E. The three inputs enter as a straight product, so doubling any one doubles the stress, and stiff materials that also expand a lot are the most sensitive. When restraint acts in two directions, as in a plate held along all its edges, the stress rises to EαΔT ÷ (1 − ν), where ν is Poisson's ratio; that case is not covered here.

A worked example.

Example

A steel rod with E = 200 GPa and α = 12 × 10⁻⁶ per kelvin is fixed between two rigid walls and then warms by 35 K. Its free thermal strain would be 12 × 10⁻⁶ × 35 = 0.00042. Multiplying by the modulus, 200,000,000,000 × 0.00042 = 84,000,000 Pa, which is 84 MPa, in compression. If the rod's cross-section were 1,000 mm², the walls would have to push back with 84,000,000 × 0.001 = 84,000 N, about 84 kN, to stop it growing. The same 35 K rise in an aluminium rod with E = 70 GPa and α = 23 × 10⁻⁶ per kelvin gives only about 56 MPa: its expansion is nearly double steel's, but its modulus is only about a third.

temperature Change K35
youngs Modulus Pa200,000,000,000
linear Expansion Per K0

Frequently asked questions.

Does the length of a restrained bar change the thermal stress?
No. Stress depends only on E, α and ΔT. Length decides how far the bar would move if it were free, about 42 mm for 100 m of steel heated by 35 K, and the cross-sectional area decides how much force the restraint must supply, but neither changes the stress.
How big a temperature change can restrained steel take before it yields?
Divide the yield strength by E × α. For steel at 200 GPa and 12 × 10⁻⁶ per kelvin, each kelvin adds 2.4 MPa, so ASTM A36 with a 250 MPa minimum yield reaches it after about 104 K, and S355 after about 148 K, assuming nothing else loads the member.
Why is the stress higher in a plate held on all edges?
A plate restrained in two directions cannot relieve stress by bulging sideways, so the stress becomes EαΔT ÷ (1 − ν). With Poisson's ratio ν = 0.3 for steel, the 84 MPa of the worked example becomes 120 MPa, about 43% more than the bar formula gives.
Why does borosilicate glass cope with sudden temperature changes better than ordinary glass?
Its expansion coefficient is about 3.3 × 10⁻⁶ per kelvin, against roughly 9 × 10⁻⁶ for soda-lime glass, while the two have similar stiffness, so the same temperature difference produces only about a third of the stress. Real thermal shock comes from gradients through the glass, which this page does not model.
What temperature change should I enter for a rail or a pipeline?
Use the gap between the temperature at which it was fixed without stress and the most extreme temperature it will see. A steel rail anchored stress-free at 30 °C and later chilled to −10 °C has a ΔT of 40 K, giving 96 MPa of tension under full restraint.

How this page was produced

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Quanta Calculator
Primary sources
3 cited below
Method
|σ| = Eα|ΔT|
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