When the temperature of a material changes, it expands or contracts in proportion to its coefficient of linear thermal expansion α. If it is free to do so, no stress arises, but if it is constrained, a thermal stress develops. The defaults are a steel bar 1000 mm long (α = 12×10⁻⁶ K⁻¹, E = 200 GPa) heated from 20 °C to 70 °C: free, it lengthens by 0.60 mm; held between immovable walls, it develops a compressive stress of about 120 MPa.
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Results
Temperature change ΔT
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Free elongation ΔL = αLΔT
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Thermal strain αΔT
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Thermal stress when constrained
Thermal stress σ (tension positive)
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Sign of stress
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Temperature change at which |σ| reaches the strength, ΔTc
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αs is the expansion coefficient of the frame holding the bar or of the film's substrate. The constraining part is taken as rigid (it does not deform), and the thermal stress is calculated as −E(α − αs)ΔT (divided by (1 − ν) for equibiaxial). Temperature dependence of the properties and stress relaxation by plastic deformation or creep are not considered.
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The thermal expansion equation
A solid gets longer as its temperature rises because the average distance between atoms increases. Over a not-too-wide temperature range, the elongation is proportional to the temperature change and can be written ΔL = αLΔT. α is the coefficient of linear thermal expansion: about 12×10⁻⁶ K⁻¹ for steel, about 23×10⁻⁶ K⁻¹ for aluminum, about 2.6×10⁻⁶ K⁻¹ for silicon and about 0.5×10⁻⁶ K⁻¹ for fused silica. For an isotropic material the volumetric expansion coefficient is roughly 3α. Materials with stronger interatomic bonds and higher melting points tend to have smaller α.
Thermal stress under constraint
When a bar held between immovable walls is heated, an elastic strain of equal size cancels the thermal strain αΔT that it would otherwise undergo. The thermal stress is therefore σ = −EαΔT: compressive on heating and tensile on cooling. The stress does not depend on the bar's length and is set by the product of E and α. For the default steel, 200×10⁹ × 12×10⁻⁶ × 50 = 1.2×10⁸ Pa, about 120 MPa in compression. If the frame also expands, use the difference Δα = α − αs in place of α.
Thermal stress in thin films (equibiaxial)
A thin film on a thick substrate is constrained in both in-plane directions at once. In this equibiaxial state the biaxial modulus E/(1 − ν) applies, giving σ = −EΔαΔT/(1 − ν). For example, an aluminum film on a silicon substrate (E = 70 GPa, ν = 0.35, α = 23×10⁻⁶ K⁻¹) cooled from a stress-free 125 °C to 25 °C develops a tensile stress of about 220 MPa. If the film's expansion coefficient is larger than the substrate's, the film is in tension after cooling; if smaller, in compression.
A rough guide to thermal shock
When a hot ceramic is quenched, the surface tries to shrink first but is held back by the interior, so an equibiaxial tensile stress arises at the surface. The temperature difference at which this stress reaches the strength σf can be estimated as ΔTc = σf(1 − ν)/(Eα). For a flexural strength of 300 MPa, E = 380 GPa, ν = 0.22 and α = 8.0×10⁻⁶ K⁻¹, it is about 77 K. Choosing equibiaxial in this tool with αs = 0 and σf as the strength gives the same value. In practice heat conduction and part size also matter, so use it as a guide for comparing materials.
Δα = 23×10⁻⁶ − 2.6×10⁻⁶ = 20.4×10⁻⁶ K⁻¹ and ΔT = 100 K, so σ = 70×10⁹ × 20.4×10⁻⁶ × 100/(1 − 0.35) ≈ 2.2×10⁸ Pa, about 220 MPa. On cooling, aluminum tries to shrink more than silicon, but constrained by the substrate it cannot, so the film is pulled into tension. Calling it compressive gets the sign wrong. About 143 MPa forgets to divide by (1 − ν). The film is in a biaxial stress state, constrained simultaneously in both in-plane directions, so the biaxial modulus E/(1 − ν) must be used. Thermal stresses like this cause stress migration and voids in interconnects.
On cooling, silicon shrinks more than SiO₂. The SiO₂ film is made to shrink more than it would on its own, so it is squeezed in plane and develops compressive stress. The sign of thermal stress can be judged this way: if the film's thermal expansion coefficient is larger than the substrate's, the film is in tension after cooling; if smaller, in compression. The relative expansion coefficients are the reverse of those for an aluminum film on silicon, so the stress is of the opposite sign. As long as the expansion coefficients differ, stress develops whenever the temperature changes from the deposition temperature, so the choice that no stress develops is wrong. The compressive stress in thermal oxide reaches a few hundred MPa and can be determined by measuring wafer bow.
Eα = 380×10⁹ × 8.0×10⁻⁶ = 3.04×10⁶ Pa/K and σf(1 − ν) = 300×10⁶ × 0.78 = 2.34×10⁸ Pa, so ΔTc = 2.34×10⁸/(3.04×10⁶) ≈ 77 K. About 99 K forgets to multiply by (1 − ν), and about 127 K divides by (1 − ν) instead. Thermal shock resistance is greater the higher the strength and the lower the Young's modulus and thermal expansion coefficient. Fused silica, with a very small thermal expansion coefficient of about 0.5×10⁻⁶ K⁻¹, is extremely resistant to thermal shock. A high thermal conductivity also helps, because it keeps the temperature difference between the surface and the interior small.
FAQ
Does the thermal stress depend on the length of the bar?
Not when it is fully constrained. The thermal strain αΔT does not depend on length, so neither does the stress EαΔT. What matters for long members is the amount of elongation ΔL, which is why bridges and rails have expansion gaps.
Why divide by (1 − ν) for a thin film?
A thin film is prevented from shrinking in both in-plane directions at once. A stress in one direction partly cancels the strain in the perpendicular direction through the Poisson effect, so a larger stress is needed to cancel the same thermal strain. The result is as if the modulus increased from E to E/(1 − ν).
What happens if the calculated stress exceeds the yield stress?
Beyond that point the elastic equation no longer holds; in metals plastic deformation occurs and the stress levels off at roughly the yield stress. When the temperature returns, a residual stress of the opposite sign may remain. In brittle ceramics and glass, cracking may occur on the tensile side.