Plane strain, axisymmetric; plane stress constitutive

Plane strain

Plane strain: a long prism (length much larger than cross-section) with a uniform cross-section and a load that does not vary along its length and acts perpendicular to it. Movement along the axis is prevented, so $\varepsilon_z=\gamma_{xz}=\gamma_{yz}=0$, but $\sigma_z=\nu(\sigma_x+\sigma_y)\neq0$.

Examples: gravity dam, retaining wall, long tunnel, strip footing, embankment, culvert.

Using $\varepsilon_z=0$, i.e. $\sigma_z=\nu(\sigma_x+\sigma_y)$, the constitutive matrix is

$$[D]=\frac{E}{(1+\nu)(1-2\nu)}\begin{bmatrix}1-\nu&\nu&0\\\nu&1-\nu&0\\0&0&\dfrac{1-2\nu}{2}\end{bmatrix}$$

Axisymmetric problem

Axisymmetric problem: a body of revolution about an axis (the $z$-axis) whose loading and supports are also symmetric about that axis. All quantities are independent of the circumferential angle $\theta$, so the 3D problem reduces to a 2D problem in the $r$-$z$ plane. The circumferential displacement is zero ($v_\theta=0$), the shear strains $\gamma_{r\theta}=\gamma_{z\theta}=0$, and the non-zero components are $\varepsilon_r,\varepsilon_z,\varepsilon_\theta,\gamma_{rz}$.

Examples: circular footing, pile or well under axial load, pressure vessel and cylindrical tank, water tank, shaft, tunnel/borehole under uniform pressure, thick-walled cylinder under internal pressure, soil sample in a triaxial test.

The strains are $\{\varepsilon_r,\varepsilon_z,\varepsilon_\theta,\gamma_{rz}\}=\{\partial u/\partial r,\ \partial w/\partial z,\ u/r,\ \partial u/\partial z+\partial w/\partial r\}$, and the stresses $\{\sigma_r,\sigma_z,\sigma_\theta,\tau_{rz}\}$ are related to them by a $4\times4$ matrix $[D]$, which is the 3D isotropic matrix with the $r\theta$ and $z\theta$ rows and columns removed.

Constitutive law for plane stress

Start from Hooke's law for an isotropic material in 3D:

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

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.