Elasticity in solids

Definitions

Plane stress: a thin plate loaded only by forces in its own plane, so the stresses through the thickness are zero: $\sigma_z=\tau_{xz}=\tau_{yz}=0$. The stress state is two-dimensional, $\{\sigma\}=\{\sigma_x,\sigma_y,\tau_{xy}\}^T$, although the strain $\varepsilon_z=-\dfrac{\nu}{E}(\sigma_x+\sigma_y)\neq0$.

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

Comparison

The plane strain matrix is obtained from the plane stress one by replacing $E\to E/(1-\nu^2)$ and $\nu\to\nu/(1-\nu)$.

Figures

 Plane stress           Plane strain
   thin plate           long dam / tunnel
  ->+------+<-          ||||||||||||| load
  ->|      |<-          +-----------+   z (long)
  ->+------+<-         ==============>  no strain
  t small, sz=0         ez = 0

In plane stress the plate is free to change thickness; in plane strain each slice of unit thickness of the long body is prevented from changing length by neighbouring slices.

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 $\lambda=\dfrac{\nu E}{(1+\nu)(1-2\nu)}$, $\mu=G$.

Lame's constants $\lambda$ and $\mu$ express the isotropic stress-strain law in the compact form

$$\sigma_{ij}=\lambda\,e\,\delta_{ij}+2\mu\,\varepsilon_{ij},\qquad e=\varepsilon_x+\varepsilon_y+\varepsilon_z$$

that is, $\sigma_x=\lambda e+2\mu\varepsilon_x$, $\sigma_y=\lambda e+2\mu\varepsilon_y$, $\sigma_z=\lambda e+2\mu\varepsilon_z$, $\tau_{xy}=\mu\gamma_{xy}$ etc.

Derivation

Hooke's law: $\varepsilon_x=\dfrac{1}{E}[(1+\nu)\sigma_x-\nu\Theta]$, with $\Theta=\sigma_x+\sigma_y+\sigma_z$.

Adding the three normal equations: $e=\dfrac{1-2\nu}{E}\Theta$, so $\Theta=\dfrac{E}{1-2\nu}e$.

Solve for $\sigma_x$:

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

Compare with $\sigma_x=\lambda e+2\mu\varepsilon_x$:

$$\boxed{\lambda=\frac{\nu E}{(1+\nu)(1-2\nu)}},\qquad \boxed{\mu=\frac{E}{2(1+\nu)}=G}$$

For shear, $\tau_{xy}=G\gamma_{xy}$ with $G=\dfrac{E}{2(1+\nu)}$, which confirms $\mu=G$ (the shear modulus).

Inverse relations: $E=\dfrac{\mu(3\lambda+2\mu)}{\lambda+\mu}$, $\nu=\dfrac{\lambda}{2(\lambda+\mu)}$. The bulk modulus is $K=\lambda+\tfrac23\mu$.

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.

Strain-displacement relations (displacements $u$ along $r$ and $w$ along $z$):

$$\{\varepsilon\}=\begin{Bmatrix}\varepsilon_r\\\varepsilon_z\\\varepsilon_\theta\\\gamma_{rz}\end{Bmatrix}=\begin{Bmatrix}\dfrac{\partial u}{\partial r}\\[2mm]\dfrac{\partial w}{\partial z}\\[2mm]\dfrac{u}{r}\\[2mm]\dfrac{\partial u}{\partial z}+\dfrac{\partial w}{\partial r}\end{Bmatrix}$$

The hoop strain $\varepsilon_\theta=u/r$ arises because a ring at radius $r$ stretches from circumference $2\pi r$ to $2\pi(r+u)$.

Constitutive relation $\{\sigma\}=[D]\{\varepsilon\}$ with $\{\sigma\}=\{\sigma_r,\sigma_z,\sigma_\theta,\tau_{rz}\}^T$:

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

        z
        ^   |p|
        |   v v v      axis of symmetry
   -----+---------
   |    |   r ->  |   Only the r-z half-plane
   |    |         |   is analysed.
   -----+---------

Plane strain

Plane strain: a long prism (length much larger than cross-section) with uniform cross-section and load that does not vary along its length and is 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. Constitutive matrix: For plane strain, $\varepsilon_z=0$, which gives $\sigma_z=\nu(\sigma_x+\sigma_y)$, and

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

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 stress-strain relation is the 4x4 matrix $[D]$ with $\{\sigma\}=\{\sigma_r,\sigma_z,\sigma_\theta,\tau_{rz}\}^T$ and $\varepsilon_\theta=u/r$ (given in the axisymmetric matrix of the standard textbooks).

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

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

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.

For a two-dimensional problem the constitutive law depends on whether it is plane stress (thin body) or plane strain (long body). Starting 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 $\{\varepsilon\}=[C]\{\sigma\}$:

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

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.

Plane strain ($\varepsilon_z=0$)

From $\varepsilon_z=0$: $\sigma_z=\nu(\sigma_x+\sigma_y)$. Substituting 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 for the stresses gives For plane strain, $\varepsilon_z=0$, which gives $\sigma_z=\nu(\sigma_x+\sigma_y)$, and

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

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

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

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

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.

Axisymmetric problems

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.

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 3D problem is analysed on a 2D mesh in the $r$-$z$ plane (ring elements).

Isotropic and anisotropic bodies

Lame's constants

For a linearly elastic isotropic body, Hooke's law is $\varepsilon_x=\dfrac{1}{E}[(1+\nu)\sigma_x-\nu\Theta]$ with $\Theta=\sigma_x+\sigma_y+\sigma_z$. Adding the normal equations gives $e=\varepsilon_x+\varepsilon_y+\varepsilon_z=\dfrac{1-2\nu}{E}\Theta$. Therefore

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

so that

$$\lambda=\frac{\nu E}{(1+\nu)(1-2\nu)},\qquad \mu=\frac{E}{2(1+\nu)}=G$$

and $\tau_{xy}=\mu\gamma_{xy}$. These are Lame's constants; $\mu$ is the shear modulus.

Plane stress

Plane stress: a thin plate loaded only by forces in its own plane, so the stresses through the thickness are zero: $\sigma_z=\tau_{xz}=\tau_{yz}=0$. The stress state is two-dimensional, $\{\sigma\}=\{\sigma_x,\sigma_y,\tau_{xy}\}^T$, although the strain $\varepsilon_z=-\dfrac{\nu}{E}(\sigma_x+\sigma_y)\neq0$. Examples: thin plate with a hole, gusset plate, deep beam, shear wall. Constitutive relation:

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

Plane strain

Plane strain: a long prism (length much larger than cross-section) with uniform cross-section and load that does not vary along its length and is 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. Constitutive relation:

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

Axisymmetric

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.

Strain-displacement relations (displacements $u$ along $r$ and $w$ along $z$):

$$\{\varepsilon\}=\begin{Bmatrix}\varepsilon_r\\\varepsilon_z\\\varepsilon_\theta\\\gamma_{rz}\end{Bmatrix}=\begin{Bmatrix}\dfrac{\partial u}{\partial r}\\[2mm]\dfrac{\partial w}{\partial z}\\[2mm]\dfrac{u}{r}\\[2mm]\dfrac{\partial u}{\partial z}+\dfrac{\partial w}{\partial r}\end{Bmatrix}$$

The hoop strain $\varepsilon_\theta=u/r$ arises because a ring at radius $r$ stretches from circumference $2\pi r$ to $2\pi(r+u)$.

Constitutive relation $\{\sigma\}=[D]\{\varepsilon\}$ with $\{\sigma\}=\{\sigma_r,\sigma_z,\sigma_\theta,\tau_{rz}\}^T$:

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

Plane stress and plane strain

Plane stress: a thin plate loaded only by forces in its own plane, so the stresses through the thickness are zero: $\sigma_z=\tau_{xz}=\tau_{yz}=0$. The stress state is two-dimensional, $\{\sigma\}=\{\sigma_x,\sigma_y,\tau_{xy}\}^T$, although the strain $\varepsilon_z=-\dfrac{\nu}{E}(\sigma_x+\sigma_y)\neq0$.

Plane strain: a long prism (length much larger than cross-section) with uniform cross-section and load that does not vary along its length and is 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: thin plate with a hole under in-plane load (plane stress); long gravity dam or retaining wall of uniform section (plane strain).

Differential equations of equilibrium in 3D

Consider an infinitesimal element $dx\,dy\,dz$ in equilibrium with body forces per unit volume $X, Y, Z$ (e.g. self-weight). Stresses vary from face to face, e.g. the normal stress on the face at $x+dx$ is $\sigma_x+\dfrac{\partial\sigma_x}{\partial x}dx$.

        y
        ^     sy + dsy
        |   +----------+
  txy   |   |          | sx + dsx
  <-----|   |  dx dy   |---->
   sx   +---+----------+---> x
            sy

Equilibrium of forces in the $x$ direction

Forces = stress $\times$ area. Faces normal to $x$ have area $dy\,dz$, those normal to $y$ have area $dx\,dz$ and those normal to $z$ have area $dx\,dy$:

$$\Big(\frac{\partial\sigma_x}{\partial x}dx\Big)dy\,dz+\Big(\frac{\partial\tau_{yx}}{\partial y}dy\Big)dx\,dz+\Big(\frac{\partial\tau_{zx}}{\partial z}dz\Big)dx\,dy+X\,dx\,dy\,dz=0$$

(the terms $\sigma_x$, $\tau_{yx}$, $\tau_{zx}$ on opposite faces cancel). Dividing by $dx\,dy\,dz$:

$$\frac{\partial\sigma_x}{\partial x}+\frac{\partial\tau_{xy}}{\partial y}+\frac{\partial\tau_{xz}}{\partial z}+X=0$$

using $\tau_{yx}=\tau_{xy}$ and $\tau_{zx}=\tau_{xz}$ (from moment equilibrium).

Similarly for $y$ and $z$

$$\begin{aligned} \frac{\partial\tau_{xy}}{\partial x}+\frac{\partial\sigma_y}{\partial y}+\frac{\partial\tau_{yz}}{\partial z}+Y&=0\\ \frac{\partial\tau_{xz}}{\partial x}+\frac{\partial\tau_{yz}}{\partial y}+\frac{\partial\sigma_z}{\partial z}+Z&=0 \end{aligned}$$

Moment equilibrium about the three axes gives $\tau_{xy}=\tau_{yx}$, $\tau_{yz}=\tau_{zy}$, $\tau_{zx}=\tau_{xz}$ (symmetry of the stress tensor).

In compact form: $\ \sigma_{ij,j}+b_i=0$.

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.

Conditions for axisymmetric stress

Axisymmetric stress exists when all of the following hold:

1. The body is a solid of revolution about an axis (the $z$-axis).
2. The loads (surface and body forces) are symmetric about the axis, so they do not depend on the angle $\theta$.
3. The supports/boundary conditions are symmetric about the axis.
4. The material is also axisymmetric (isotropic, or the same in every radial plane).

As a result displacement, strain and stress are functions of $r$ and $z$ only, the tangential displacement is zero, and $\gamma_{r\theta}=\gamma_{z\theta}=0$, $\tau_{r\theta}=\tau_{z\theta}=0$. The non-zero stresses are $\sigma_r$, $\sigma_z$, $\sigma_\theta$ and $\tau_{rz}$. Examples: circular footing, pile, water tank, pressure vessel.

Stress-strain relations

$$\begin{Bmatrix}\sigma_r\\\sigma_z\\\sigma_\theta\\\tau_{rz}\end{Bmatrix}=\frac{E(1-\nu)}{(1+\nu)(1-2\nu)}\begin{bmatrix}1&\frac{\nu}{1-\nu}&\frac{\nu}{1-\nu}&0\\\frac{\nu}{1-\nu}&1&\frac{\nu}{1-\nu}&0\\\frac{\nu}{1-\nu}&\frac{\nu}{1-\nu}&1&0\\0&0&0&\frac{1-2\nu}{2(1-\nu)}\end{bmatrix}\begin{Bmatrix}\varepsilon_r\\\varepsilon_z\\\varepsilon_\theta\\\gamma_{rz}\end{Bmatrix}$$

with $\varepsilon_r=\partial u/\partial r$, $\varepsilon_z=\partial w/\partial z$, $\varepsilon_\theta=u/r$, $\gamma_{rz}=\partial u/\partial z+\partial w/\partial r$.

Consider an infinitesimal element $dx\,dy\,dz$ in equilibrium with body forces per unit volume $X, Y, Z$ (e.g. self-weight). Stresses vary from face to face, e.g. the normal stress on the face at $x+dx$ is $\sigma_x+\dfrac{\partial\sigma_x}{\partial x}dx$.

        y
        ^     sy + dsy
        |   +----------+
  txy   |   |          | sx + dsx
  <-----|   |  dx dy   |---->
   sx   +---+----------+---> x
            sy

Equilibrium of forces in the $x$ direction

Forces = stress $\times$ area. Faces normal to $x$ have area $dy\,dz$, those normal to $y$ have area $dx\,dz$ and those normal to $z$ have area $dx\,dy$:

$$\Big(\frac{\partial\sigma_x}{\partial x}dx\Big)dy\,dz+\Big(\frac{\partial\tau_{yx}}{\partial y}dy\Big)dx\,dz+\Big(\frac{\partial\tau_{zx}}{\partial z}dz\Big)dx\,dy+X\,dx\,dy\,dz=0$$

(the terms $\sigma_x$, $\tau_{yx}$, $\tau_{zx}$ on opposite faces cancel). Dividing by $dx\,dy\,dz$:

$$\frac{\partial\sigma_x}{\partial x}+\frac{\partial\tau_{xy}}{\partial y}+\frac{\partial\tau_{xz}}{\partial z}+X=0$$

using $\tau_{yx}=\tau_{xy}$ and $\tau_{zx}=\tau_{xz}$ (from moment equilibrium).

Similarly for $y$ and $z$

$$\begin{aligned} \frac{\partial\tau_{xy}}{\partial x}+\frac{\partial\sigma_y}{\partial y}+\frac{\partial\tau_{yz}}{\partial z}+Y&=0\\ \frac{\partial\tau_{xz}}{\partial x}+\frac{\partial\tau_{yz}}{\partial y}+\frac{\partial\sigma_z}{\partial z}+Z&=0 \end{aligned}$$

Moment equilibrium about the three axes gives $\tau_{xy}=\tau_{yx}$, $\tau_{yz}=\tau_{zy}$, $\tau_{zx}=\tau_{xz}$ (symmetry of the stress tensor).

In compact form: $\ \sigma_{ij,j}+b_i=0$.