Thermal Stress Index⚠ unverified
Mechanical / Materials · Compute the thermal stress index
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| alpha | α | 1/K | 1.0 | Coefficient of linear thermal expansion |
| E | E | Pa | 1.0 | Young's modulus of the material |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | σth | Pa/K | Thermal stress index, in pascals per kelvin (Pa/K) |
The science & history
Understanding the Parameters
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Thermal expansion $\alpha$ — the free strain per kelvin, $\varepsilon_{th} = \alpha\,\Delta T$. Under constraint this strain is prevented, converting directly into stress.
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Young's modulus $E$ — converts the suppressed strain into stress. A stiff material resists the expansion strongly, so it builds up more stress for the same $\alpha$ and $\Delta T$.
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Index $\alpha E$ — the per-degree thermal stress. Multiply by the actual temperature swing $\Delta T$ to get the stress; compare against yield or fatigue strength to judge whether thermal cycling will damage the part.
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Design implication — minimising $\alpha E$ (low expansion or low stiffness) reduces thermal stress; this is why low-$\alpha$ Invar and compliant (low-$E$) interfaces are used to manage thermal mismatch. Contrast with the thermal-shock-resistance index $\sigma_f/(\alpha E)$, which you instead maximise.
Derivation (Approaching a Proof)
Take a bar rigidly constrained at both ends (no axial movement). A temperature change $\Delta T$ would freely produce a thermal strain
$$\varepsilon_{th} = \alpha\,\Delta T.$$
Because the constraint forces the total strain to zero, a mechanical (elastic) strain equal and opposite to the thermal strain must develop: $\varepsilon_{mech} = -\alpha\,\Delta T$. By Hooke's law the corresponding stress is
$$\sigma_{th} = E\,\varepsilon_{mech} = -E\,\alpha\,\Delta T,$$
compressive on heating, tensile on cooling. The magnitude per degree is the index
$$\frac{|\sigma_{th}|}{\Delta T} = \alpha\,E.$$
(Full constraint is the worst case; partial constraint or free expansion reduces the stress proportionally.)
Dimensional check. $\alpha\,E = \dfrac{1}{\text{K}}\cdot\text{Pa} = \dfrac{\text{Pa}}{\text{K}}$ — stress per kelvin, as the output label states.
History and Development
Thermal stress from constrained expansion is a classical result of thermoelasticity, central wherever parts see temperature change under constraint: engine components, rail tracks (which buckle without expansion gaps), turbine blades, bimetallic strips, and solder joints in electronics. The product $\alpha E$ (and the related thermal-shock group $\sigma_f/(\alpha E)$) is a standard Ashby material index for thermal design, and the reason low-expansion alloys and controlled-expansion ceramics exist.
Related Concepts: Thermal Stress, Material Selection Index Stiffness Thermal, Material Performance Index Thermal Conductivity, Ashby Charts, Residual Stress Superposition
Notes: $\sigma_{th} = \alpha E\,\Delta T$ for full constraint (worst case); compressive on heating, tensile on cooling. Multiply the index by $\Delta T$ for the stress. Minimise $\alpha E$ to cut thermal stress; thermal-shock resistance instead maximises $\sigma_f/(\alpha E)$.