11.3 Thermal Properties of Materials

11.3 Thermal Properties of Materials

NCEES lists "Thermal properties of materials" as a Materials sub-topic. It is distinct from the Thermodynamics and Heat Transfer area: there you compute heat flow through a wall given $k$, while here you must know what $k$ is for a given material class and why, and you must handle the mechanical consequences of temperature change.

Coefficient of Thermal Expansion

$\varepsilon_{\text{thermal}} = \alpha\,\Delta T, \qquad \delta = \alpha L\,\Delta T$

The steel-and-concrete coincidence: both are near $11 \times 10^{-6}$/°C, which is precisely why reinforced concrete works. If the two expanded at appreciably different rates, every temperature swing would debond the reinforcement. Note also that aluminum expands about twice as much as steel — the source of countless mixed-assembly problems.

Restrained Thermal Stress

The key concept, and the one most often confused:

$\boxed{\sigma_{\text{thermal}} = E\alpha\,\Delta T}$

Note what is absent: length. A fully restrained bar develops the same thermal stress whether it is 1 m or 100 m long, because both the free expansion and the compression needed to cancel it scale with $L$. Candidates who look for a length in the problem statement are looking for information that does not matter.

Worked Example: Restrained Rail

A steel rail ($E = 200$ GPa, $\alpha = 11.7\times10^{-6}$/°C) is installed at 10 °C with no expansion gap and fully restrained. Find the stress at 45 °C.

$\Delta T = 45 - 10 = 35\ °\text{C}$ $\sigma = E\alpha\Delta T = (200\times10^9)(11.7\times10^{-6})(35) = 81.9\times10^6\ \text{Pa} = \boxed{81.9\ \text{MPa compression}}$

For A36 steel with $\sigma_y = 250$ MPa, that is 33% of yield from a 35 °C swing alone — before any traffic load. A 100 °C swing would reach 234 MPa, essentially at yield, which is why continuous welded rail must be laid at a controlled neutral temperature and why bridges have expansion joints.

Thermal Conductivity

$q = -kA\frac{dT}{dx}$

The physics behind the ranking: metals conduct heat through free electrons, the same carriers that conduct electricity — which is why good electrical conductors are good thermal conductors (the Wiedemann-Franz relationship). Non-metals must rely on phonons (lattice vibrations), which are far less effective. Insulators go further and exploit trapped still air, whose conductivity is only $0.026$ W/m·K.

Alloying penalty: carbon steel at 50 W/m·K conducts eight times better than 304 stainless at 16 W/m·K, because chromium and nickel atoms scatter the conduction electrons. The same mechanism raises stainless steel's electrical resistivity — one microstructural cause, two property consequences.

Specific Heat and Thermal Diffusivity

$Q = mc\,\Delta T$

Thermal diffusivity combines conduction and storage to describe how fast a temperature front moves:

$\alpha_{\text{diff}} = \frac{k}{\rho c} \qquad [\text{m}^2/\text{s}]$

Do not confuse the two $\alpha$'s. The expansion coefficient $\alpha$ has units of 1/°C; thermal diffusivity has units of m²/s. Both are conventionally written $\alpha$, and the units are the only reliable discriminator.

High diffusivity means a material reaches steady state quickly. Copper's $\alpha_{\text{diff}} \approx 1.1\times10^{-4}$ m²/s versus stainless steel's $4\times10^{-6}$ m²/s — a factor of 28, which is why a copper heat sink responds almost immediately while a thick stainless vessel wall lags.

Thermal Shock and Expansion Mismatch

Thermal shock resistance improves with high strength and conductivity and with low expansion and modulus:

$TSR \propto \frac{\sigma_f\,k}{E\,\alpha}$

This explains a familiar comparison: borosilicate glass ($\alpha = 3.3$) survives thermal shock that shatters soda-lime glass ($\alpha = 9$) — the ratio of expansion coefficients is nearly three to one, and the induced stress scales directly with $\alpha$.

Expansion mismatch in a bonded assembly generates interfacial strain on every cycle:

$\varepsilon_{\text{mismatch}} = (\alpha_1 - \alpha_2)\,\Delta T$

Worked Example: Mismatch Stress in a Bonded Joint

An aluminum plate ($\alpha = 23.6\times10^{-6}$) is bonded to a steel plate ($\alpha = 11.7\times10^{-6}$). The assembly is cured at 150 °C and cooled to 20 °C. Estimate the mismatch strain and the stress if the aluminum ($E = 69$ GPa) were fully restrained by the steel.

$\Delta T = 20 - 150 = -130\ °\text{C}$ $\varepsilon_{\text{mismatch}} = (23.6 - 11.7)\times10^{-6}(-130) = -1{,}547\times10^{-6} = -0.155\%$

$\sigma = E\,|\varepsilon| = (69\times10^9)(1.547\times10^{-3}) = \boxed{107\ \text{MPa}}$

The aluminum wants to shrink more than the steel allows, so it ends up in tension while the steel goes into compression, and the assembly bows toward the aluminum side. At 107 MPa this approaches the yield strength of many aluminum alloys — from cooling alone, with no external load. This is the dominant failure mechanism in solder joints, thermal barrier coatings, and bonded dissimilar-metal assemblies, and the standard remedies are a compliant interlayer, a graded transition, or matching the coefficients.