Posted in Academic Issues, Geotechnical Engineering

# Constitutive Elasticity Equations: Plane Strain and Axisymmetric Cases

In the last post we considered the three-dimensional elastic constitutive equations. Three-dimensional analyses are becoming more common in geotechnical engineering but many problems are considered and analysed using a two-dimensional paradigm. In this post we’ll look at plane strain/axisymmetric problems. The references are the same as the original post.

Let’s begin by defining what we mean by the terms “plane stress,” “plane strain” and “axisymmetric” by using and illustration.

Let’s concentrate on the last two here. With plane strain we assume that

$\epsilon_z = 0$ (1a)
$\tau_{xz} = 0$ (1b)
$\tau_{yz} = 0$ (1c)

This reduces the stress and strain vectors to

\sigma=\left[\begin{array}{c} \sigma_{{x}}\\ \noalign{\medskip}\sigma_{{y}}\\ \noalign{\medskip}\sigma_{{z}}\\ \noalign{\medskip}\tau_{{\it xy}} \end{array}\right] (2)

\epsilon=\left[\begin{array}{c} \epsilon_{{x}}\\ \noalign{\medskip}\epsilon_{{y}}\\ \noalign{\medskip}0\\ \noalign{\medskip}{\it \gamma}_{{\it xy}} \end{array}\right] (3)

The constitutive matrix is

D_{e}=\left[\begin{array}{cccc} {\frac{{\it E}\,\left(1-\nu\right)}{\left(1+\nu\right)\left(1-2\,\nu\right)}} & {\frac{{\it E}\,\nu}{\left(1+\nu\right)\left(1-2\,\nu\right)}} & {\frac{{\it E}\,\nu}{\left(1+\nu\right)\left(1-2\,\nu\right)}} & 0\\ \noalign{\medskip}{\frac{{\it E}\,\nu}{\left(1+\nu\right)\left(1-2\,\nu\right)}} & {\frac{{\it E}\,\left(1-\nu\right)}{\left(1+\nu\right)\left(1-2\,\nu\right)}} & {\frac{{\it E}\,\nu}{\left(1+\nu\right)\left(1-2\,\nu\right)}} & 0\\ \noalign{\medskip}{\frac{{\it E}\,\nu}{\left(1+\nu\right)\left(1-2\,\nu\right)}} & {\frac{{\it E}\,\nu}{\left(1+\nu\right)\left(1-2\,\nu\right)}} & {\frac{{\it E}\,\left(1-\nu\right)}{\left(1+\nu\right)\left(1-2\,\nu\right)}} & 0\\ \noalign{\medskip}0 & 0 & 0 & {\frac{{\it E}\,\left(1-\nu\right)}{\left(1+\nu\right)\left(2-2\,\nu\right)}} \end{array}\right] (4)

Applying the Lamé constants the way we did the last time,

D_{e}=\left[\begin{array}{cccc} {\frac{\left(1-\nu\right)\lambda}{\nu}} & \lambda & \lambda & 0\\ \noalign{\medskip}\lambda & {\frac{\left(1-\nu\right)\lambda}{\nu}} & \lambda & 0\\ \noalign{\medskip}\lambda & \lambda & {\frac{\left(1-\nu\right)\lambda}{\nu}} & 0\\ \noalign{\medskip}0 & 0 & 0 & G \end{array}\right] (5)

Inverting this matrix,

D_{e}^{-1}=\left[\begin{array}{cccc} -{\frac{\nu}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}} & {\frac{{\nu}^{2}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}} & {\frac{{\nu}^{2}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}} & 0\\ \noalign{\medskip}{\frac{{\nu}^{2}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}} & -{\frac{\nu}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}} & {\frac{{\nu}^{2}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}} & 0\\ \noalign{\medskip}{\frac{{\nu}^{2}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}} & {\frac{{\nu}^{2}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}} & -{\frac{\nu}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}} & 0\\ \noalign{\medskip}0 & 0 & 0 & {G}^{-1} \end{array}\right] (6)

Again using the same method as before, we determine the stress and strain vectors thus:

\sigma = \left[\begin{array}{c} \sigma_{{x}}\\ \noalign{\medskip}\sigma_{{y}}\\ \noalign{\medskip}\sigma_{{z}}\\ \noalign{\medskip}\tau_{{\it xy}} \end{array}\right]=\left[\begin{array}{c} {\frac{\left(1-\nu\right)\lambda\,\epsilon_{{x}}}{\nu}}+\lambda\,\epsilon_{{y}}\\ \noalign{\medskip}\lambda\,\epsilon_{{x}}+{\frac{\left(1-\nu\right)\lambda\,\epsilon_{{y}}}{\nu}}\\ \noalign{\medskip}\lambda\,\epsilon_{{x}}+\lambda\,\epsilon_{{y}}\\ \noalign{\medskip}G{\it \gamma}_{{\it xy}} \end{array}\right] (7)

\epsilon=\left[\begin{array}{c} \epsilon_{{x}}\\ \noalign{\medskip}\epsilon_{{y}}\\ \noalign{\medskip}0\\ \noalign{\medskip}{\it \gamma}_{{\it xy}} \end{array}\right]=\left[\begin{array}{c} -{\frac{\nu\,\sigma_{{x}}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}}+{\frac{{\nu}^{2}\sigma_{{y}}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}}+{\frac{{\nu}^{2}\sigma_{{z}}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}}\\ \noalign{\medskip}{\frac{{\nu}^{2}\sigma_{{x}}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}}-{\frac{\nu\,\sigma_{{y}}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}}+{\frac{{\nu}^{2}\sigma_{{z}}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}}\\ \noalign{\medskip}{\frac{{\nu}^{2}\sigma_{{x}}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}}+{\frac{{\nu}^{2}\sigma_{{y}}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}}-{\frac{\nu\,\sigma_{{z}}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}}\\ \noalign{\medskip}{\frac{\tau_{{\it xy}}}{G}} \end{array}\right] (8)

We can write the third row of Equation (8) as follows:

${\frac{{\nu}^{2}\sigma_{{x}}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}}+{\frac{{\nu}^{2}\sigma_{{y}}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}}-{\frac{\nu\,\sigma_{{z}}}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\right)}}=0$ (9)

From which we can determine $\sigma_z$:

$\sigma_{z}=\nu\,\left(\sigma_{{x}}+\sigma_{{y}}\right)$ (10)

This suggests that we can reduce the matrix from 4 x 4 to 3 x 3 and the vectors to 3-row. The elasticity equation $D \epsilon = \sigma$ would look thus:

\left[\begin{array}{ccc} {\frac{\left(1-\nu\right)\lambda}{\nu}} & \lambda & 0\\ \noalign{\medskip}\lambda & {\frac{\left(1-\nu\right)\lambda}{\nu}} & 0\\ \noalign{\medskip}0 & 0 & G \end{array}\right] \left[\begin{array}{c} \epsilon_{{x}}\\ \noalign{\medskip}\epsilon_{{y}}\\ \noalign{\medskip}{\it \gamma}_{{\it xy}} \end{array}\right] = \left[\begin{array}{c} \sigma_{{x}}\\ \noalign{\medskip}\sigma_{{y}}\\ \noalign{\medskip}\tau_{{\it xy}} \end{array}\right] (11)

and inverting the matrix, the equation $D^{-1}_e \sigma = \epsilon$ is

\left[\begin{array}{ccc} {\frac{\left(-1+\nu\right)\nu}{\lambda\,\left(-1+2\,\nu\right)}} & {\frac{{\nu}^{2}}{\lambda\,\left(-1+2\,\nu\right)}} & 0\\ \noalign{\medskip}{\frac{{\nu}^{2}}{\lambda\,\left(-1+2\,\nu\right)}} & {\frac{\left(-1+\nu\right)\nu}{\lambda\,\left(-1+2\,\nu\right)}} & 0\\ \noalign{\medskip}0 & 0 & {G}^{-1} \end{array}\right] \left[\begin{array}{c} \sigma_{{x}}\\ \noalign{\medskip}\sigma_{{y}}\\ \noalign{\medskip}\tau_{{\it xy}} \end{array}\right] = \left[\begin{array}{c} \epsilon_{{x}}\\ \noalign{\medskip}\epsilon_{{y}}\\ \noalign{\medskip}{\it \gamma}_{{\it xy}} \end{array}\right] (12)

Multiplying Equations (11) and (12) through,

\sigma=\left[\begin{array}{c} {\frac{\left(1-\nu\right)\lambda\,\epsilon_{{x}}}{\nu}}+\lambda\,\epsilon_{{y}}\\ \noalign{\medskip}\lambda\,\epsilon_{{x}}+{\frac{\left(1-\nu\right)\lambda\,\epsilon_{{y}}}{\nu}}\\ \noalign{\medskip}G{\it \gamma}_{{\it xy}} \end{array}\right] (13)

\epsilon= \left[\begin{array}{c} {\frac{\left(-1+\nu\right)\nu\,\sigma_{{x}}}{\lambda\,\left(-1+2\,\nu\right)}}+{\frac{{\nu}^{2}\sigma_{{y}}}{\lambda\,\left(-1+2\,\nu\right)}}\\ \noalign{\medskip}{\frac{{\nu}^{2}\sigma_{{x}}}{\lambda\,\left(-1+2\,\nu\right)}}+{\frac{\left(-1+\nu\right)\nu\,\sigma_{{y}}}{\lambda\,\left(-1+2\,\nu\right)}}\\ \noalign{\medskip}{\frac{\tau_{{\it xy}}}{G}} \end{array}\right] (14)

with Equation (10) rounding out for $\sigma_z$.

## Axisymmetric Case

The axisymmetric case is achieved if we replace Equation (2) with

\sigma=\left[\begin{array}{c} \sigma_{{r}}\\ \noalign{\medskip}\sigma_{{z}}\\ \noalign{\medskip}\sigma_{{\theta}}\\ \noalign{\medskip}\tau_{{\it rz}} \end{array}\right] (15)

and Equation (3) with

\epsilon=\left[\begin{array}{c} \epsilon_{{r}}\\ \noalign{\medskip}\epsilon_{{z}}\\ \noalign{\medskip}\epsilon_{{\theta}}\\ \noalign{\medskip}{\it \gamma}_{{\it rz}} \end{array}\right] (16)

The subscript swap for polar coordinates is as follows:

• x becomes r
• y becomes z
• z becomes θ
• xy becomes rz

The math after that is identical, including the reduction for $\sigma_{\theta}$.

Note: after this was first posted, some errors were discovered. These were corrected with the assistance of Jaeger, J.C. and Cook, N.G.W. (1979) Fundamentals of Rock Mechanics. London: Chapman and Hall. This book has an excellent treatment of basic theoretical solid mechanics in general and elasticity in particular. Although they do not put the equations in matrix form, they do use the Lamé constants in their formulation, albeit in a different way than is done here. We apologise for any inconvenience or confusion.