Generalised Hooke's law for isotropic materials

Generalised Hooke's law

For a linear elastic body each stress component is a linear function of all strain components: $\sigma_{ij} = C_{ijkl}\,\varepsilon_{kl}$, where $C_{ijkl}$ is the elasticity (stiffness) tensor. Because $\sigma_{ij}$ and $\varepsilon_{ij}$ are symmetric and a strain energy function exists, the 81 constants reduce to 21 for the most general anisotropic body.

Isotropic material

$$\begin{aligned} \varepsilon_x &= \frac{1}{E}\left[\sigma_x - \nu(\sigma_y+\sigma_z)\right] &\qquad \gamma_{xy} &= \frac{\tau_{xy}}{G} \\ \varepsilon_y &= \frac{1}{E}\left[\sigma_y - \nu(\sigma_z+\sigma_x)\right] & \gamma_{yz} &= \frac{\tau_{yz}}{G} \\ \varepsilon_z &= \frac{1}{E}\left[\sigma_z - \nu(\sigma_x+\sigma_y)\right] & \gamma_{zx} &= \frac{\tau_{zx}}{G} \end{aligned}$$

Inverse (Lamé form): $\sigma_{ij} = \lambda\,\varepsilon_{kk}\delta_{ij} + 2G\varepsilon_{ij}$, with $\lambda = \dfrac{\nu E}{(1+\nu)(1-2\nu)}$.

Elastic constants

Relations

Adding the three normal strain equations gives the volumetric strain $\varepsilon_v = \dfrac{1-2\nu}{E}(\sigma_x+\sigma_y+\sigma_z)$. For a hydrostatic pressure $p$ (taking the magnitude, with compression positive) the sum of the three stresses is $3p$ in magnitude, so

$$K = \frac{p}{\varepsilon_v} = \frac{E}{3(1-2\nu)}$$

Pure shear: the principal stresses are $\pm\tau$, so $\varepsilon_1 = \dfrac{\tau}{E}(1+\nu)$. Since $\gamma = 2\varepsilon_1$ (the principal axes are at $45^\circ$ to the shear planes) and $\gamma = \tau/G$,

$$G = \frac{E}{2(1+\nu)}$$

Hence $E = 9KG/(3K+G)$ and $\nu = (3K-2G)/(2(3K+G))$. Only two of the four constants are independent.

Number of independent constants