Thermal Stress⚠ unverified

Understanding the Parameters

• Young's modulus $E$ — the elastic stiffness converting the suppressed strain into stress. Stiffer materials develop higher thermal stress for the same temperature change: steel and titanium build up more thermal stress than aluminium at equal $\alpha\,\Delta T$ because their $E$ is larger.

• Thermal expansion coefficient $\alpha$ — how much the material wants to grow per kelvin ($\sim 23\times10^{-6}/\text{K}$ for aluminium, $\sim 9\times10^{-6}$ for titanium, near zero for some carbon composites and Invar). Low-$\alpha$ materials develop less thermal stress and hold dimensions under temperature swings — prized for optical benches, structures, and precision spacecraft.

• Temperature change $\Delta T$ — the driver. Heating a restrained member ($\Delta T > 0$) puts it in compression (it is prevented from growing); cooling puts it in tension. The sign convention here gives compressive stress for positive $\Delta T$.

• What "fully constrained" means. The formula assumes the member is held rigidly at both ends so the net mechanical strain is exactly $-\alpha\Delta T$. A partially restrained member develops proportionally less; a free member develops none (it just expands). This is the key subtlety: thermal stress requires constraint — free thermal expansion is stress-free.

Derivation (Approaching a Proof)

The total strain of a heated elastic member is the sum of a mechanical (stress-producing) part and a thermal (free-expansion) part:

$$\varepsilon_{total} = \varepsilon_{mech} + \varepsilon_{thermal} = \frac{\sigma}{E} + \alpha\,\Delta T.$$

Full constraint means the ends cannot move, so the total strain is zero:

$$\varepsilon_{total} = 0 \quad\Longrightarrow\quad \frac{\sigma}{E} + \alpha\,\Delta T = 0.$$

Solving for the stress,

$$\sigma = -E\,\alpha\,\Delta T. \qquad\blacksquare$$

The magnitude is $E\alpha\Delta T$; the minus sign says heating ($\Delta T>0$) produces compression. Length and area cancel out entirely — a $1$-mm strut and a $10$-m spar develop the same thermal stress for the same $\Delta T$, because both the free expansion and the elastic recovery scale with length. (A member free to expand, by contrast, has $\sigma = 0$ and $\varepsilon_{total} = \alpha\Delta T$.)

Dimensional check. $$[E\,\alpha\,\Delta T] = (\text{Pa})\left(\frac{1}{\text{K}}\right)(\text{K}) = \text{Pa}.\ \checkmark$$

History and Development

• Thermoelasticity. The additive split of strain into mechanical and thermal parts is the foundation of thermoelasticity (Duhamel, 1830s; Neumann). The fully-constrained bar is its simplest and most-quoted result.

• Thermal management in flight. Thermal stress is a first-order concern wherever temperatures swing: gas-turbine blades and combustor liners, rocket nozzles (Combustion Chamber Temperature), and the skins of high-speed vehicles. The Concorde fuselage grew several inches at Mach 2 from aerodynamic heating, and was designed to accommodate that expansion rather than restrain it — the practical way to avoid thermal stress.

• Low-expansion materials. The drive to eliminate thermal distortion produced near-zero-$\alpha$ materials — Invar, ultra-low-expansion glass, and tailored carbon-fibre layups — used where dimensional stability under temperature is paramount (satellite optical benches, telescope structures). See the related Shaft Thermal Stress and Thermal Moment pages.

Related Concepts: Thermal Moment, Shaft Thermal Stress, Combustion Chamber Temperature, Von Mises Stress 2D, Wing Bending Stress, Margin Of Safety

Notes: Registry calculator thermal-stress (unverified). Fully-constrained thermoelastic stress $\sigma = E\alpha\Delta T$ — correct as shipped; independent of length and area. Heating ⇒ compression; requires constraint (free expansion is stress-free). $\alpha$ correctly labelled 1/K. All defaults $1.0$ ⇒ $\sigma = 1$ Pa.