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  1. Home
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  3. /Materials Science & Solid Mechanics
  4. /Thermal Stress Calculator

Thermal Stress Calculator

Last updated: March 18, 2026

Calculator

Results

Enter values to see results

Free Thermal Strain (ε_th)

—

Thermal Stress (σ)

—

MPa

Thermal Stress (σ)

—

ksi

Free Expansion per meter (δ)

—

mm/m

Results

Enter values to see results

Free Thermal Strain (ε_th)

—

Thermal Stress (σ)

—

MPa

Thermal Stress (σ)

—

ksi

Free Expansion per meter (δ)

—

mm/m

The Thermal Stress Calculator determines the stress that develops in a material when it is heated or cooled while restrained from freely expanding or contracting. When a constrained structure undergoes a temperature change ΔT, thermal stress arises according to $$\sigma = E \, \alpha \, \Delta T$$ where E is Young's modulus, α is the coefficient of linear thermal expansion, and ΔT is the temperature change. The corresponding free thermal strain is $$\varepsilon_{th} = \alpha \, \Delta T$$

Thermal stress is a critical concern in engineering design. Bridges require expansion joints to prevent buckling, pipelines use expansion loops, and composite materials must account for differential expansion between layers. When thermal stresses exceed the yield strength, permanent deformation or fracture can occur. This calculator handles both fully and partially constrained conditions, enabling engineers to assess realistic loading scenarios where some degree of movement is permitted.

How It Works

When a material is heated, its atoms vibrate more energetically and occupy more space, causing the material to expand. If this expansion is completely prevented by rigid supports, the material develops internal compressive stress (for heating) or tensile stress (for cooling). The fundamental relationships are:

$$\varepsilon_{th} = \alpha \, \Delta T$$

$$\sigma = E \, \varepsilon_{th} = E \, \alpha \, \Delta T$$

Key principles governing thermal stress:

  • Sign convention: A positive ΔT (heating) with full constraint produces compressive stress (positive value here represents magnitude). Cooling produces tensile stress. The sign of the result indicates whether the stress is compressive or tensile.
  • Partial constraint: In practice, most structures are neither perfectly free nor perfectly fixed. A constraint fraction of 0.5 means half the thermal strain is accommodated by expansion and half produces stress: $$\sigma_{partial} = E \, \alpha \, \Delta T \times f_c$$
  • Material dependence: Steel (E ≈ 200 GPa, α ≈ 12 × 10⁻⁶/°C) develops much higher thermal stress than aluminum (E ≈ 70 GPa, α ≈ 23 × 10⁻⁶/°C) for the same ΔT, despite aluminum expanding nearly twice as much, because steel's stiffness dominates.
  • Free expansion: If unconstrained, the material simply changes length by δ = α × ΔT × L per unit length. No stress develops in the absence of constraint.

Thermal stress analysis is essential in pressure vessel design, railroad track engineering, electronic packaging (solder joint fatigue from thermal cycling), dental restorations, and aerospace structures exposed to extreme temperature gradients. Combined thermal and mechanical loading requires superposition of stresses to check against material limits.

For anisotropic or composite materials, directional expansion coefficients must be used, and the stress state becomes multiaxial. In such cases, the biaxial thermal stress for a thin plate constrained in two directions is $$\sigma = \frac{E \, \alpha \, \Delta T}{1 - \nu}$$ where ν is Poisson's ratio.

Understanding Your Results

The free thermal strain shows how much the material would deform if unconstrained — this is the strain that must be suppressed to generate stress. The thermal stress is the resulting internal stress when that strain is prevented. Compare this value to the material's yield strength: if σ exceeds σ_y, plastic deformation or cracking will occur. The free expansion per meter gives a practical measure of how much a one-meter bar would lengthen or shorten if free to move.

Worked Examples

Steel Bridge Girder Heated by 50°C

Inputs

E200
alpha12
delta T50
constraintfully

Results

thermal strain0.0006
thermal stress120
thermal stress ksi17.405
free expansion0.6

A fully constrained steel girder heated by 50°C develops 120 MPa of thermal stress — about half the yield strength of structural steel (250 MPa), showing why expansion joints are essential.

Aluminum Rail, Partially Constrained, ΔT = 80°C

Inputs

E70
alpha23
delta T80
constraintpartial
constraint fraction0.6

Results

thermal strain0.00184
thermal stress77.28
thermal stress ksi11.208
free expansion1.84

With 60% constraint, the aluminum develops 77.3 MPa — still significant compared to aluminum's yield strength of ~275 MPa for 6061-T6 alloy.

Frequently Asked Questions

Thermal stress is the internal stress that develops in a material when it is heated or cooled while restrained from freely expanding or contracting. It is calculated as $$\sigma = E \alpha \Delta T$$ where E is the elastic modulus, α is the thermal expansion coefficient, and ΔT is the temperature change. Without constraint, no thermal stress develops.

Bridges experience daily and seasonal temperature changes of 50°C or more. Without expansion joints, a fully constrained steel bridge would develop thermal stresses exceeding 100 MPa, potentially causing buckling in summer or cracking in winter. Expansion joints allow controlled movement to relieve these stresses.

The coefficient of linear thermal expansion (α) measures how much a material's length changes per degree of temperature change. Typical values include steel at 12 × 10⁻⁶/°C, aluminum at 23 × 10⁻⁶/°C, and concrete at 10 × 10⁻⁶/°C. Higher α means more expansion per degree.

Yes. If thermal stress exceeds the material's yield strength, plastic deformation occurs. If it exceeds the ultimate tensile strength (especially in brittle materials like ceramics or glass), fracture results. Thermal fatigue from repeated cycling can also cause failure at stresses below the static yield strength.

Partial constraint means only a fraction of the thermal expansion is prevented. The effective thermal stress becomes $$\sigma = E \alpha \Delta T \times f_c$$ where f_c is the constraint fraction (0 = free, 1 = fully fixed). Real structures typically have constraint fractions between 0.3 and 0.8 depending on support conditions.

For a uniformly heated, homogeneous body, thermal stress depends only on E, α, ΔT, and the degree of constraint — not on the object's size. However, thermal gradients within large objects create differential expansion that generates internal stresses even without external constraints.

Sources & Methodology

Gere, J.M. & Goodno, B.J. (2012). Mechanics of Materials, 8th Ed. Cengage Learning. Timoshenko, S.P. & Goodier, J.N. (1970). Theory of Elasticity, 3rd Ed. McGraw-Hill. ASM International. (2002). Atlas of Stress-Strain Curves, 2nd Ed. ASM International.
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Roboculator Team

The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.

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