Shaft Thermal Stress⚠ unverified
Mechanical / Shafts · Compute the thermal stress in an axially constrained shaft
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| alpha | α | 1/K | 1.0 | Coefficient of thermal expansion |
| E | E | Pa | 1.0 | Young's modulus of the shaft material |
| dT | dT | K | 1.0 | Temperature change |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | σ | Pa | Thermal stress, in pascals (Pa) |
The science & history
Understanding the Parameters
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Thermal expansion coefficient $\alpha$ — how much the material strains per degree (steel ≈ 12×10⁻⁶ /K, aluminium ≈ 23×10⁻⁶ /K). It sets the free thermal strain $\alpha\,\Delta T$ that the constraint must cancel.
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Young's modulus $E$ — converts the suppressed strain into stress. Because $\sigma = E \times$ strain, a stiff material develops large thermal stress even for modest $\Delta T$ — steel's high $E$ and moderate $\alpha$ still give ~2.5 MPa per °C when fully constrained.
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Temperature change $\Delta T$ — measured from the stress-free (assembly) temperature. Heating a constrained shaft puts it in compression (it wants to grow but can't); cooling puts it in tension. Cyclic $\Delta T$ drives thermal fatigue.
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Independence from geometry — length and area cancel out: a long fat shaft and a short thin one develop the same thermal stress for the same $\Delta T$ and material. Geometry affects the force ($\sigma A$) and displacement if partially constrained, but not the stress under full constraint.
Derivation (Approaching a Proof)
Consider a shaft held rigidly between two immovable supports so its total length cannot change. Heating by $\Delta T$ would, if free, produce a thermal strain
$$\varepsilon_{\text{thermal}} = \alpha\, \Delta T.$$
Because the ends are fixed, the net strain must be zero — the constraint imposes an equal and opposite mechanical (elastic) strain that exactly cancels the thermal one:
$$\varepsilon_{\text{total}} = \varepsilon_{\text{thermal}} + \varepsilon_{\text{mechanical}} = 0 \;\Longrightarrow\; \varepsilon_{\text{mechanical}} = -\alpha\,\Delta T.$$
By Hooke's law, that mechanical strain corresponds to a stress:
$$\sigma = E\,\varepsilon_{\text{mechanical}} = -\alpha E\,\Delta T.$$
The magnitude is $\sigma = \alpha E\,\Delta T$; the sign shows heating ($\Delta T > 0$) gives compression. Nothing about the length or area entered — the constraint is on strain, which is intensive, so the resulting stress is geometry-independent. Partial constraint (a support with finite stiffness) scales the result by a compliance factor between 0 (free, no stress) and 1 (rigid, full stress).
Dimensional check. $[\sigma] = (\text{1/K}) \cdot \text{Pa} \cdot \text{K} = \text{Pa}$. ✓
History and Development
Thermal stress under constraint is classical thermoelasticity (Duhamel, 1830s) and a standard result in Timoshenko and Roark. It governs the design of expansion joints in bridges, railway track (sun kink/ buckling), steam and process piping, and precision machinery, and it drives thermal fatigue in components that cycle in temperature (engine parts, turbine shafts). Allowing free expansion, or providing expansion joints, is the usual mitigation.
Related Concepts: Hooke's Law strain, Von Mises Stress, Beam Bending Stress, Combined Stress Shaft, Factor of Safety
Notes: Full axial constraint assumed (geometry-independent stress). Partial constraint scales the result down by a compliance factor. Heating → compression, cooling → tension. Superpose on mechanical stresses; cyclic $\Delta T$ causes thermal fatigue.