Axisymmetric; plane stress strain-displacement

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.

Plane stress: strain-displacement relations

Let a point move by $u$ in $x$ and $v$ in $y$. For small displacements

$$\varepsilon_x=\frac{\partial u}{\partial x},\quad \varepsilon_y=\frac{\partial v}{\partial y},\quad \gamma_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}$$

In matrix form

$$\{\varepsilon\}=\begin{bmatrix}\partial/\partial x&0\\0&\partial/\partial y\\\partial/\partial y&\partial/\partial x\end{bmatrix}\begin{Bmatrix}u\\v\end{Bmatrix}$$

(derived from the change in length of $dx$ and $dy$ and the change in the right angle of a small rectangle). The out-of-plane strain $\varepsilon_z=-\nu(\sigma_x+\sigma_y)/E$ is not needed in the analysis.

Plane stress: constitutive relations

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 $\{\varepsilon\}=[C]\{\sigma\}$:

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

Inverting the first two equations: from $E\varepsilon_x=\sigma_x-\nu\sigma_y$ and $E\varepsilon_y=\sigma_y-\nu\sigma_x$, solve for $\sigma_x=\dfrac{E}{1-\nu^2}(\varepsilon_x+\nu\varepsilon_y)$ 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.