A temperature change produces a free thermal strain in a solid. For a small temperature interval with approximately constant linear expansion coefficient $\alpha$,

$$\boxed{\varepsilon_{th}=\alpha\Delta T}.$$

If the body is free to expand, this strain changes its dimensions without requiring mechanical stress. If supports, neighboring materials, or geometric compatibility prevent some or all of that expansion, mechanical stresses develop.

Fully constrained uniaxial bar

Consider a uniform bar whose axial length is completely fixed while its temperature changes uniformly by $\Delta T$.

If it were free, the bar would develop thermal strain

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

Because its total axial strain is constrained to zero, the mechanical elastic strain must cancel the thermal strain:

$$\varepsilon_{mech}+\varepsilon_{th}=0.$$

Thus

$$\varepsilon_{mech}=-\alpha\Delta T.$$

For linear elasticity,

$$\sigma=E\varepsilon_{mech},$$

so

$$\boxed{\sigma=-E\alpha\Delta T}.$$

For positive $\alpha$, prevented heating produces compressive stress and prevented cooling produces tensile stress.

Worked example

A steel bar is fully restrained axially and heated by

$$\Delta T=50,\mathrm K.$$

Take

$$E=200,\mathrm{GPa},\qquad \alpha=12\times10^{-6},\mathrm{K^{-1}}.$$

The free thermal strain would be

$$\varepsilon_{th} =(12\times10^{-6})(50) =6.0\times10^{-4}.$$

Because this expansion is completely prevented,

$$\sigma =-(200\times10^9)(6.0\times10^{-4}) =-1.20\times10^8,\mathrm{Pa}.$$

Thus

$$\boxed{\sigma=-120,\mathrm{MPa}},$$

where the negative sign denotes compression under the chosen sign convention.

Partial restraint and compatibility

Real components are rarely either perfectly free or perfectly fixed. The actual stress must satisfy both:

• the constitutive response of the material;
• the geometric compatibility imposed by supports and connected parts.

The thermal strain $\alpha\Delta T$ acts like an imposed strain that would occur in the absence of constraint. Structural analysis then determines what mechanical strain and stress are needed to satisfy the actual boundary conditions.

Mismatched expansion

Bonded materials with different expansion coefficients would undergo different free strains:

$$\varepsilon_{th,1}=\alpha_1\Delta T,$$

$$\varepsilon_{th,2}=\alpha_2\Delta T.$$

If bonding forces them to deform compatibly, internal stresses arise. This is important in bimetallic strips, coatings, composites, electronic packages, glass-metal seals, and components subjected to thermal cycling.

A bimetallic strip bends when heated because the two bonded layers prefer different free expansions but must remain joined. The deformation relieves part of the incompatibility by curvature rather than forcing both layers to have identical unconstrained lengths.

Microscopic origin of positive thermal expansion

In a bonded solid, atoms vibrate around equilibrium separations. If the interatomic potential were perfectly symmetric about its minimum, increasing vibration amplitude would not change the average spacing.

Real interatomic potentials are generally asymmetric: strong short-range repulsion makes compression energetically steeper than comparable extension. As thermal motion increases, this asymmetry commonly shifts the average atomic separation outward, producing macroscopic positive thermal expansion.

This microscopic explanation also shows why thermal expansion is material-dependent and why unusual bonding structures can produce small or even negative expansion coefficients over some temperature ranges.

Thermal stress is therefore not caused merely by temperature itself. It arises when a temperature-induced free strain is made incompatible with the mechanical constraints of the system.