Constitutive relations for 3D elastic solid

The constitutive relation links stress and strain. For a linear elastic isotropic material the stresses are related to strains by Hooke's law with two constants, $E$ and $\nu$.

Strains from stresses

By superposition of the effect of each normal stress (a stress $\sigma_x$ gives $\varepsilon_x=\sigma_x/E$ and lateral strains $-\nu\sigma_x/E$):

$$\begin{aligned} \varepsilon_x&=\frac{1}{E}[\sigma_x-\nu(\sigma_y+\sigma_z)]\\ \varepsilon_y&=\frac{1}{E}[\sigma_y-\nu(\sigma_x+\sigma_z)]\\ \varepsilon_z&=\frac{1}{E}[\sigma_z-\nu(\sigma_x+\sigma_y)] \end{aligned}$$

and shear strains $\gamma_{xy}=\dfrac{\tau_{xy}}{G},\ \gamma_{yz}=\dfrac{\tau_{yz}}{G},\ \gamma_{zx}=\dfrac{\tau_{zx}}{G}$ with $G=\dfrac{E}{2(1+\nu)}$.

Stresses from strains

Add the three normal strain equations. With $e=\varepsilon_x+\varepsilon_y+\varepsilon_z$ and $\Theta=\sigma_x+\sigma_y+\sigma_z$:

$$e=\frac{1-2\nu}{E}\Theta\ \Rightarrow\ \Theta=\frac{E}{1-2\nu}e$$

Write the first equation as $E\varepsilon_x=(1+\nu)\sigma_x-\nu\Theta$, so

$$\sigma_x=\frac{E}{1+\nu}\varepsilon_x+\frac{\nu}{1+\nu}\Theta=\frac{E}{1+\nu}\varepsilon_x+\frac{\nu E}{(1+\nu)(1-2\nu)}e$$

Substituting $e$ and collecting terms gives $\sigma_x=\dfrac{E}{(1+\nu)(1-2\nu)}[(1-\nu)\varepsilon_x+\nu\varepsilon_y+\nu\varepsilon_z]$, and similarly for $\sigma_y,\sigma_z$.

Matrix form $\{\sigma\}=[D]\{\varepsilon\}$

$$\begin{Bmatrix}\sigma_x\\\sigma_y\\\sigma_z\\\tau_{xy}\\\tau_{yz}\\\tau_{zx}\end{Bmatrix}=\frac{E}{(1+\nu)(1-2\nu)}\begin{bmatrix}1-\nu&\nu&\nu&0&0&0\\\nu&1-\nu&\nu&0&0&0\\\nu&\nu&1-\nu&0&0&0\\0&0&0&\frac{1-2\nu}{2}&0&0\\0&0&0&0&\frac{1-2\nu}{2}&0\\0&0&0&0&0&\frac{1-2\nu}{2}\end{bmatrix}\begin{Bmatrix}\varepsilon_x\\\varepsilon_y\\\varepsilon_z\\\gamma_{xy}\\\gamma_{yz}\\\gamma_{zx}\end{Bmatrix}$$

The matrix $[D]$ is symmetric. In terms of Lame's constants, $\sigma_x=\lambda e+2\mu\varepsilon_x$ and $\tau_{xy}=\mu\gamma_{xy}$ with $\lambda=\dfrac{\nu E}{(1+\nu)(1-2\nu)}$, $\mu=G$.