Chapter 8
Plane problems
Plane problems are problems where the system has a symmetry in a certain direction. We will here use the \(y\)-direction as the direction in which the plane conditions hold. This symmetry implies that the relevant quantities do not vary in \(y\)-direction. Note that throughout this text, we will switch this direction.
8.1 Plane strain
For a plane strain situation, the system cannot elongate or shrink in that direction and hence \(\varepsilon _{yy} = 0\). From Hooke’s law for isotropic elasticity, \begin {equation} \varepsilon _{ij} = \frac {1}{E} [(1+\nu )\sigma _{ij} - \nu \delta _{ij} \sigma _{kk}], \end {equation} we see that \begin {equation} \label {eqn:planestrainzz} \varepsilon _{yy} = \frac {1+\nu }{E} \sigma _{yy} - \frac {\nu }{E} (\sigma _{xx} + \sigma _{yy} + \sigma _{zz}) = 0 \end {equation} and hence \begin {equation} \label {eq:planestrainstressyy} \sigma _{yy} = \nu (\sigma _{xx} + \sigma _{zz}). \end {equation} With the two relations for \(\varepsilon _{yy}\) and \(\sigma _{yy}\) we can express Hooke’s law as \begin {align} \label {eq:planestrainHooksxx} \varepsilon _{xx} &= \frac {1-\nu ^2}{E} \sigma _{xx} - \frac {\nu (1+\nu )}{E} \sigma _{zz} \\ \label {eqn:planestrain_Hooksyy} \varepsilon _{zz} &= -\frac {\nu (1+\nu )}{E} \sigma _{xx} + \frac {1-\nu ^2}{E} \sigma _{zz} \end {align}
and its inverse \begin {align} \label {eq:planestrainHooksinversexx} \sigma _{xx} &= (\lambda + 2\nu ) \varepsilon _{xx} + \lambda \varepsilon _{zz} \\ \label {eqn:planestrain_Hooksinverseyy} \sigma _{zz} &= \lambda \varepsilon _{xx} + (\lambda + 2\mu ) \varepsilon _{zz} \end {align}
Note that the condition for elastic equilibrium becomes (in Cartesian coordinates) \begin {align} \label {eqn:planestrainelasticequixx} \frac {\partial \sigma _{xx}}{\partial x} + \frac {\partial \sigma _{xz}}{\partial z} &= 0 \\ \label {eqn:planestrainelasticequizz} \frac {\partial \sigma _{zz}}{\partial z} + \frac {\partial \sigma _{xz}}{\partial x} &= 0. \end {align}
8.2 Plane stress conditions
For plane stress we consider a situation with \(\sigma _{yy} = 0\), i.e. there is no stress in the \(y\)-direction. This is a good approximation for example for a thin plate. In this limit, Hooke’s law becomes \begin {align} \label {eqn:planestresse1} \varepsilon _{xx} &= \frac {1}{E} (\sigma _{xx} - \nu \sigma _{zz}) \\ \label {eqn:planestresse2} \varepsilon _{zz} &= \frac {}{E} (-\nu \sigma _{xx} + \sigma _{zz}) \end {align}
and its inverse \begin {align} \label {eqn:planestresss1} \sigma _{xx} &= \frac {E}{1-\nu ^2} (\varepsilon _{xx} + \nu \varepsilon _{zz}) \\ \label {eqn:planestresss2} \sigma _{zz} &= \frac {E}{1-\nu ^2} (\nu \varepsilon _{xx} + \varepsilon _{zz}). \end {align}
Note that plane strain and plane stress are described by the same set of differential equations but with different elastic moduli. We can convert the plane stress Eqs. \eqref{eqn:planestresse1}-\eqref{eqn:planestresss2} to the corresponding plane strain equation by substituting the elastic constants, \begin {equation} E \to \frac {E}{1-\nu ^2}\quad \text {and}\quad \nu \to \frac {\nu }{1-\nu }. \end {equation} Hence any plane stress solution can be converted into a plane strain solution using this substitution. In the following, we will continue to work with the plane stress expression (because they are simpler), but all results carry over to plane strain with this substitution.
8.3 Compatibility condition
Another condition to be fulfilled is the compatibility condition. For plane problems, the compatibility conditions is the single equation: \begin {equation} \label {eqn:Compatibility_eqn} \frac {\partial ^2 \varepsilon _{xx}}{\partial z^2} + \frac {\partial ^2 \varepsilon _{zz}}{\partial x^2} = 2\ \frac {\partial ^2 \varepsilon _{xz}}{\partial x \partial z} \end {equation} That it must hold is easily seen by expressing the strain in terms of the displacements \(\v {u}\). It is therefore a consequence of the fact that the strain field is the gradient of the displacement field. It is similar to the well-known condition that the curl of a gradient has to disappear.
The compatibility condition has a simple geometric explanation. Imaging a jigsaw puzzle that you deform in its assembled state. Even in the deformed state, all pieces still have to fit together. They deformation of neighboring pieces can therefore not be independent of each other.