Duhamel–Neumann law

Remarks

The law says that the strain splits into an elastic part, related to stress by Hooke's law, and a thermal part $\alpha_{kl}\theta$ that produces no stress. It follows from a free energy that is quadratic in $\varepsilon$ and $\theta$, $\psi = \tfrac12 C_{ijkl} \varepsilon_{ij} \varepsilon_{kl} - C_{ijkl} \alpha_{kl} \theta \varepsilon_{ij} + (\text{terms in } \theta \text{ only})$, through $\sigma_{ij} = \partial \psi / \partial \varepsilon_{ij}$. Example: a bar that is free to expand has $\sigma = 0$ and strain $\alpha\theta$. A bar of Young's modulus $E$ held at fixed length has axial strain $0$ and axial stress $-E \alpha \theta$, which is compressive when it is heated. Since $3\lambda + 2\mu = 3K$ with $K$ the bulk modulus, the isotropic thermal term is $-3K\alpha\theta\delta_{ij}$.

Going deeper

For a student

This law is the constitutive equation of linear thermoelasticity. Inserted into the equilibrium equations, it shows that a temperature field acts like a body force $-\nabla (3K\alpha\theta)$. It also adds a normal boundary traction $3K\alpha\theta n$, and this is how thermal stress problems are reduced to ordinary elastic ones. A temperature field that is linear in the coordinates produces no stress in a free, simply connected body of homogeneous isotropic material, since the thermal strain is then compatible.