Constitutive relation for 2D isotropic material

For a two-dimensional problem the constitutive law depends on whether it is plane stress (thin body) or plane strain (long body). Start from the 3D Hooke's law of an isotropic material:

$$\varepsilon_x=\frac{1}{E}[\sigma_x-\nu(\sigma_y+\sigma_z)],\quad \varepsilon_y=\frac{1}{E}[\sigma_y-\nu(\sigma_x+\sigma_z)],\quad \gamma_{xy}=\frac{2(1+\nu)}{E}\tau_{xy}$$

Plane stress ($\sigma_z=0$)

For plane stress, put $\sigma_z=0$:

$$\varepsilon_x=\frac{\sigma_x-\nu\sigma_y}{E},\quad \varepsilon_y=\frac{\sigma_y-\nu\sigma_x}{E},\quad \gamma_{xy}=\frac{2(1+\nu)}{E}\tau_{xy}$$

and $\varepsilon_z=-\nu(\sigma_x+\sigma_y)/E$. In matrix form $\{\varepsilon\}=[C]\{\sigma\}$:

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

Invert the first two equations: from $E\varepsilon_x=\sigma_x-\nu\sigma_y$ and $E\varepsilon_y=\sigma_y-\nu\sigma_x$, we get $\sigma_x=\dfrac{E}{1-\nu^2}(\varepsilon_x+\nu\varepsilon_y)$ and $\sigma_y=\dfrac{E}{1-\nu^2}(\varepsilon_y+\nu\varepsilon_x)$. Also $\tau_{xy}=G\gamma_{xy}=\dfrac{E}{2(1+\nu)}\gamma_{xy}$. Hence

$$\begin{Bmatrix}\sigma_x\\\sigma_y\\\tau_{xy}\end{Bmatrix}=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\\nu&1&0\\0&0&\dfrac{1-\nu}{2}\end{bmatrix}\begin{Bmatrix}\varepsilon_x\\\varepsilon_y\\\gamma_{xy}\end{Bmatrix}$$

i.e. $\{\sigma\}=[D]\{\varepsilon\}$ with the plane stress matrix $[D]$ above.

Plane strain ($\varepsilon_z=0$)

From $\varepsilon_z=\frac{1}{E}[\sigma_z-\nu(\sigma_x+\sigma_y)]=0$ we get $\sigma_z=\nu(\sigma_x+\sigma_y)$. Substituting this into the first two equations,

$$\varepsilon_x=\frac{1+\nu}{E}[(1-\nu)\sigma_x-\nu\sigma_y],\qquad \varepsilon_y=\frac{1+\nu}{E}[(1-\nu)\sigma_y-\nu\sigma_x]$$

Solving these two equations for the stresses, and using $\tau_{xy}=G\gamma_{xy}$, gives

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

The same relation follows from the plane stress one by replacing $E$ with $E/(1-\nu^2)$ and $\nu$ with $\nu/(1-\nu)$.