Posted in Academic Issues, Geotechnical Engineering

Constitutive Elasticity Equations: Plane Stress

In our previous posts we discussed three-dimensional and two-dimensional plane strain/axisymmetric constitutive elasticity equations. In this post we will consider plane stress. This case doesn’t appear very much in geotechnical engineering; it’s more of interest to mechanical engineers and deals with problems such as this.

It’s probably a good idea to look again at the graphic for this:

Plane stress, plane strain and axial symmetry cases, from Owen and Hinton (1980)

The conditions of plane stress are thus:

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

If we compare these to plane strain, we find out that we switch the z-strain for the z-stress. This means that deriving the equations–especially the 3 x 3 case–is the mirror image of the plane strain case, and we will see this clearly below.

Let’s start by noting that, as before, the starting 4 x 4 constitutive matrix is the same as plane strain, thus

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] (2)

and its inverse

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] (3)

The stress and strain vectors, however, are

\sigma=\left[\begin{array}{c} \sigma_{{x}}\\ \noalign{\medskip}\sigma_{{y}}\\ \noalign{\medskip}0\\ \noalign{\medskip}\tau_{{\it xy}} \end{array}\right] (4)

\epsilon=\left[\begin{array}{c} \epsilon_{{x}}\\ \noalign{\medskip}\epsilon_{{y}}\\ \noalign{\medskip}\epsilon_{z}\\ \noalign{\medskip}{\it \gamma}_{{\it xy}} \end{array}\right] (5)

We see Equation (1a) applied to the stress vector to yield Equation (4). Multiplying through as we did before,

\sigma=\left[\begin{array}{c} \sigma_{{x}}\\ \noalign{\medskip}\sigma_{{y}}\\ \noalign{\medskip}0\\ \noalign{\medskip}\tau_{{\it xy}} \end{array}\right]=\left[\begin{array}{c} {\frac{\left(1-\nu\right)\lambda\,\epsilon_{{x}}}{\nu}}+\lambda\,\epsilon_{{y}}+\lambda\,\epsilon_{{z}}\\ \noalign{\medskip}\lambda\,\epsilon_{{x}}+{\frac{\left(1-\nu\right)\lambda\,\epsilon_{{y}}}{\nu}}+\lambda\,\epsilon_{{z}}\\ \noalign{\medskip}\lambda\,\epsilon_{{x}}+\lambda\,\epsilon_{{y}}+{\frac{\left(1-\nu\right)\lambda\,\epsilon_{{z}}}{\nu}}\\ \noalign{\medskip}G{\it \gamma}_{xy} \end{array}\right] (6)

\epsilon=\left[\begin{array}{c} \epsilon_{{x}}\\ \noalign{\medskip}\epsilon_{{y}}\\ \noalign{\medskip}\epsilon_{z}\\ \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)}}\\ \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)}}\\ \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)}}\\ \noalign{\medskip}{\frac{\tau_{{\it xy}}}{G}} \end{array}\right] (7)

If we equation the two third row elements of the vectors in Equation (6), we have

\lambda\,\epsilon_{{x}}+\lambda\,\epsilon_{{y}}+{\frac{\left(1-\nu\right)\lambda\,\epsilon_{{z}}}{\nu}}=0 (8)

Solving for the z-axis strain,

\epsilon_{z}={\frac{\left(\epsilon_{{x}}+\epsilon_{{y}}\right)\nu}{-1+\nu}} (9)

Since the z-axis strain is a function of the x-axis and y-axis strains, the obvious question is this: can we go to a 3 x 3 matrix formulation? The answer is yes if we literally invert the process. We need to do the following:

  1. Consider the basic equation in this form: D_{e}^{-1}\sigma=\epsilon .
  2. Since \sigma_z = 0 , we can remove the third column and row from the constitutive matrix and the third row from the stress (and strain) matrix.
  3. We multiply through to yield

\left[\begin{array}{ccc} -{\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}{\lambda\,\left(-1+\nu+2\,{\nu}^{2}\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]=\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)}}\\ \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)}}\\ \noalign{\medskip}{\frac{\tau_{{\it xy}}}{G}} \end{array}\right] (10)

We then invert the inverse to obtain the 3 x 3 constitutive matrix. We use the forward formulation D\epsilon=\sigma and multiply through to obtain

\left[\begin{array}{ccc} {\frac{\lambda\,\left(-1+2\,\nu\right)}{\nu\,\left(-1+\nu\right)}} & {\frac{\lambda\,\left(-1+2\,\nu\right)}{-1+\nu}} & 0\\ \noalign{\medskip}{\frac{\lambda\,\left(-1+2\,\nu\right)}{-1+\nu}} & {\frac{\lambda\,\left(-1+2\,\nu\right)}{\nu\,\left(-1+\nu\right)}} & 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]=\left[\begin{array}{c} {\frac{\lambda\,\left(-1+2\,\nu\right)\epsilon_{{x}}}{\nu\,\left(-1+\nu\right)}}+{\frac{\lambda\,\left(-1+2\,\nu\right)\epsilon_{{y}}}{-1+\nu}}\\ \noalign{\medskip}{\frac{\lambda\,\left(-1+2\,\nu\right)\epsilon_{{x}}}{-1+\nu}}+{\frac{\lambda\,\left(-1+2\,\nu\right)\epsilon_{{y}}}{\nu\,\left(-1+\nu\right)}}\\ \noalign{\medskip}G{\it \gamma}_{{\it xy}} \end{array}\right] (11)

After these things we use Equation (9) to obtain the z-axis strain.

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