Plane stress and plane strain

Many problems in continuum mechanics can be reduced to two dimensions, when the geometry and the loading of a body allow one of the three directions to be treated separately. The two most common idealizations are plane stress, for bodies that are very thin in one direction, and plane strain, for bodies that are very long in one direction. In both cases the analysis is carried out in the plane $x_{1}x_{2}$, but the out-of-plane behaviour is different, as shown below.

In the following, $\varepsilon_{ij}$ is the (small) strain tensor and, for an isotropic linear elastic material with Young’s modulus $E$ and Poisson’s ratio $\nu$, the stresses and strains are related by Hooke’s law:

$$\varepsilon_{ij}=\frac{1+\nu}{E}\sigma_{ij}-\frac{\nu}{E}\sigma_{kk}\delta_{ij} \tag{1}$$

Plane stress

Plane stress is a special state of stress that occurs in bodies in which one dimension is very small compared to the other two. Consider a very thin plate, as shown in Fig. 1. The two faces normal to $x_{3}$ are free of tractions, so the normal and shear stresses acting on them, $\sigma_{33}$, $\sigma_{31}$ and $\sigma_{32}$, are zero there. Because the plate is very thin, we may assume that these components remain approximately zero throughout the whole body:

$$\sigma_{33}=\sigma_{31}=\sigma_{32}=0 \tag{2}$$

For this assumption to hold, the external loads must act in the plane of the plate (they have no component in the direction of $x_{3}$) and be distributed uniformly over its thickness.

The stress tensor then takes the form:

$$\sigma_{ij}=\left[\begin{array}{ccc}\sigma_{11}&\sigma_{12}&0\\ \sigma_{21}&\sigma_{22}&0\\ 0&0&0\end{array}\right] \tag{3}$$

and it can be reduced to the two-dimensional tensor:

$$\sigma_{\alpha\beta}=\left[\begin{array}{cc}\sigma_{11}&\sigma_{12}\\ \sigma_{21}&\sigma_{22}\end{array}\right],\qquad \alpha,\beta=1,2 \tag{4}$$

Since the stress tensor is symmetric, $\sigma_{12}=\sigma_{21}$, only three independent stress components remain: $\sigma_{11}$, $\sigma_{22}$ and $\sigma_{12}$. Thin bodies under plane stress can therefore be analyzed as two-dimensional. Note that the plane stress assumption is valid only if the plate does not buckle or bend out of its plane.

The axis $x_{3}$ is always a principal direction, with a principal stress equal to zero, so the third invariant of the stress tensor vanishes, $I_{3}=0$ (see Principal stresses and stress invariants). The two in-plane principal stresses and their directions follow from the two-dimensional analysis presented in the article Mohr’s circle.

Although the out-of-plane stresses are zero, the out-of-plane normal strain is not: the plate becomes thinner or thicker as it is loaded. Setting $\sigma_{33}=\sigma_{31}=\sigma_{32}=0$ in Hooke’s law, Eq. (1), gives:

$$\varepsilon_{33}=-\frac{\nu}{E}\left(\sigma_{11}+\sigma_{22}\right),\qquad \varepsilon_{31}=\varepsilon_{32}=0 \tag{5}$$

while the in-plane strains are:

$$\begin{array}{l}\varepsilon_{11}=\dfrac{1}{E}\left(\sigma_{11}-\nu\sigma_{22}\right)\\ \varepsilon_{22}=\dfrac{1}{E}\left(\sigma_{22}-\nu\sigma_{11}\right)\\ \varepsilon_{12}=\dfrac{1+\nu}{E}\sigma_{12}\end{array} \tag{6}$$

Plane strain

Plane strain is the opposite case: it occurs in bodies in which one dimension is very large compared to the other two, e.g. a long tunnel, a dam or a retaining wall (Fig. 2). For plane strain conditions:

• the geometry of the cross section does not change along the long axis $x_{3}$;
• the loads act in the plane of the cross section ($x_{1}x_{2}$) and do not vary along $x_{3}$;
• the body cannot deform in the direction of $x_{3}$, either because its ends are restrained or because the cross section of interest is far from the ends.

Under these conditions every cross section deforms in the same way, and the displacement field takes the form:

$$u_{1}=u_{1}(x_{1},x_{2}),\qquad u_{2}=u_{2}(x_{1},x_{2}),\qquad u_{3}=0 \tag{7}$$

From the definition of the small strain tensor, $\varepsilon_{ij}=\frac{1}{2}\left(u_{i,j}+u_{j,i}\right)$ (see Index notation for the comma notation), the out-of-plane strains vanish, since $u_{3}=0$ and $u_{1}$, $u_{2}$ do not depend on $x_{3}$:

$$\varepsilon_{33}=\varepsilon_{31}=\varepsilon_{32}=0 \tag{8}$$

so the strain tensor takes the form:

$$\varepsilon_{ij}=\left[\begin{array}{ccc}\varepsilon_{11}&\varepsilon_{12}&0\\ \varepsilon_{21}&\varepsilon_{22}&0\\ 0&0&0\end{array}\right] \tag{9}$$

In this case it is the out-of-plane normal stress that does not vanish. From Hooke’s law, Eq. (1), the condition $\varepsilon_{33}=0$ gives:

while $\varepsilon_{31}=\varepsilon_{32}=0$ gives $\sigma_{31}=\sigma_{32}=0$. The stress tensor therefore takes the form:

$$\sigma_{ij}=\left[\begin{array}{ccc}\sigma_{11}&\sigma_{12}&0\\ \sigma_{21}&\sigma_{22}&0\\ 0&0&\nu\left(\sigma_{11}+\sigma_{22}\right)\end{array}\right] \tag{11}$$

As in plane stress, the axis $x_{3}$ is a principal direction, but now the corresponding principal stress is $\sigma_{33}$, which is in general not zero, and so $I_{3}\neq 0$. Since $\sigma_{33}$ depends only on $\sigma_{11}$ and $\sigma_{22}$, the problem is still two-dimensional: the unknowns are the three in-plane stresses $\sigma_{11}$, $\sigma_{22}$ and $\sigma_{12}$. Substituting $\sigma_{33}$ into Hooke’s law, the in-plane strains are:

$$\begin{array}{l}\varepsilon_{11}=\dfrac{1+\nu}{E}\left[(1-\nu)\sigma_{11}-\nu\sigma_{22}\right]\\ \varepsilon_{22}=\dfrac{1+\nu}{E}\left[(1-\nu)\sigma_{22}-\nu\sigma_{11}\right]\\ \varepsilon_{12}=\dfrac{1+\nu}{E}\sigma_{12}\end{array} \tag{12}$$

Comparing Eqs. (6) and (12), the plane strain relations are obtained from the plane stress relations if $E$ and $\nu$ are replaced by:

$$E'=\frac{E}{1-\nu^{2}},\qquad \nu'=\frac{\nu}{1-\nu} \tag{13}$$

This is useful in practice: a solution derived for plane stress can be converted to plane strain (and vice versa) simply by changing the elastic constants.

In rock mechanics, plane strain is the more common assumption, since many structures in rock, such as tunnels and long excavations, are long compared to their cross section. In laboratory testing, on the other hand, thin specimens such as the discs of the Brazilian (indirect tensile) test are usually analyzed under plane stress conditions.

Comparison

	Plane stress	Plane strain
Typical geometry	Thin plate	Long body
Out-of-plane stresses	$\sigma_{33}=\sigma_{31}=\sigma_{32}=0$	$\sigma_{31}=\sigma_{32}=0$ $\sigma_{33}=\nu(\sigma_{11}+\sigma_{22})$
Out-of-plane strains	$\varepsilon_{31}=\varepsilon_{32}=0$ $\varepsilon_{33}=-\frac{\nu}{E}(\sigma_{11}+\sigma_{22})$	$\varepsilon_{33}=\varepsilon_{31}=\varepsilon_{32}=0$
Third stress invariant	$I_{3}=0$	$I_{3}\neq 0$ (in general)
Examples	Thin plates, Brazilian test discs	Tunnels, dams, retaining walls