I'm looking into plane strain and I'm a little confused about the concept. If we assume the material is much longer in the $z$-direction, the typical assumptions I see are \begin{equation*} |\varepsilon_{12}|, |\varepsilon_{13}|, |\varepsilon_{33}| \ll 1 \end{equation*} where $\boldsymbol{\varepsilon}$ is the strain tensor. Even if we suppose those terms are zero, it seems to yield the following stresses are nonzero: \begin{equation*} \begin{split} \sigma_{33} =& C_{1133} \varepsilon_{11} + C_{2233} \varepsilon_{22} + C_{3312} \varepsilon_{12} \\ \sigma_{23} =& C_{1123} \varepsilon_{11} + C_{2223} \varepsilon_{22} + C_{2312} \varepsilon_{12} \\ \sigma_{13} =& C_{1113} \varepsilon_{11} + C_{2213} \varepsilon_{22} + C_{1312} \varepsilon_{12} \\ \end{split} \end{equation*} In practice are the $\sigma_{i3}$ typically ignored or assumed to be small? This formulation seems to imply we have $\sigma_{33}$ is not necessarily small even in the case of isotropy, for instance $C_{1133}$ or $C_{2233}$ $\gg 1$. As we already have three (hopefully independent) equations for $\sigma_{11},\sigma_{12},\sigma_{22}$ in terms of $\varepsilon_{11}, \varepsilon_{12},\varepsilon_{22}$ we could I suppose rewrite $\sigma_{33}, \sigma_{23},\sigma_{13}$ in terms of $\sigma_{11},\sigma_{12},\sigma_{22}$, but I wonder if I am perhaps missing a core assumption of plane strain. For instance we assume $\frac{\partial u_1}{\partial z} \approx -\frac{\partial u_3}{\partial x}$ since $|\varepsilon_{13}| \ll 1$, but would it be wrong to strengthen this and assume $\frac{\partial u_1}{\partial z} = \frac{\partial u_3}{\partial x} = 0$, where $u$ is the displacement?

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    $\begingroup$ The plane strain assumption does not in any way imply the out of plane stress is negligible. $\endgroup$
    – agentp
    Oct 24 '17 at 15:38
  • $\begingroup$ Plane strain has nothing to do with the dimension of the material in the $z$ direction. What matters is the constraints on the structure, which force the three strain components you mention in your OP to be zero. For a general anisotropic material, assuming three strain components are zero might not make any sense, but it does make sense in some special cases, e.g. an orthotropic material where the material anisotropy is in the same direction as the $z$ axis. $\endgroup$
    – alephzero
    Oct 24 '17 at 20:03
  • $\begingroup$ You might get some help if you look up LEFM in Wikipedia. Of the many calculations I have seen , I never saw any for anisotropic material. $\endgroup$ Nov 24 '17 at 21:34

There are two cases typically considered for stress on a 2d material, plane stress and plane strain.

In plane stress, the stress is considered negligible in the 3 direction. This is the case for a material that has no forces in these directions. Note the strain can change - the material can get thinner or thicker, but ultimately no new stress is being applied in the 3 direction.

In plane strain, the strain is considered negligible in the 3 direction. This is the case for a material that is constrained to not get thicker or thinner in this direction. Note that the stress can change - there is some force holding the material to be a constant thickness, to ensure the strain is zero.

Many pressure loading cases are combinations of plane strain and plane stress. An analysis of both scenarios is taken, and a few FEAs of the 3D scenario can lead to an empirical equation. For example, one of my analysis of a knuckle on a flat top head for FRP composites (with a 1 1/2" radius knuckle) shows that the highest stress (at the knuckle portion), regardless of diameter, was 0.7 x the plain strain stress and 0.3 x the plain stress stress.


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