Introduction
The concepts of
plane stress and
plane strain mechanics are integral
to the analysis of 2‑D linear elastic problems, particularly those involving fracture
mechanics. The terms relate to the stress and strain states within a 2‑D plane
and indirectly specify which quantities are zero perpendicular to it.
The plane stress condition applies when all loads and stresses are contained within
a 2‑D plane. This occurs in the 'real world' when objects are relatively thin.
The plane strain condition applies when all displacements and strains take place
in a 2‑D plane. This occurs when objects are thick.
Plane stress and plane strain conditions are limit-case idealizations
that are rarely met in a rigorous way by actual engineering problems.
Nevertheless, many scenarios do come sufficiently close.
As always, the final decision as to which case, if either, applies
to a given problem is dependent on the engineering judgment of
individuals performing the analysis.
Thin Wall Pressure Vessels
A prime example of a plane stress analysis condition
is a thin wall pressure vessel. Pressurized tanks comprise
loading scenarios in which plane stress assumptions
can be accurately applied even though the loading conditions
are neither rigorously plane stress, nor even 2-D planar. This demonstrates
the robust applicability of the analysis assumptions.
Indeed, they can be applied to a greater number of engineering
problems than one might initially think possible.
The same goes for plane strain analyses as well.
The figure shows a cross-section of a pressurized tank of
radius, \(R\), and thickness, \(t\). The hoop stress in the
tank is given by
\[
\sigma_\text{hoop} = P \left( R \over t \right)
\]
The radial stress through the tank's thickness is
\(\sigma_\text{radial} = -P\) on the tank's interior surface and
its magnitude decreases through the wall
to zero at the exterior surface. (The radial stress value
increases algebraically from \(-P\) up to zero. But its
magnitude decreases from \(|P|\) down to zero.)
In this case, the radial stress is the out-of-plane stress and the
hoop stress is the in-plane stress.
One might think that plane stress conditions apply only to
the outer surface of the tank because that is the only place
where the radial out-of-plane stress is truly zero.
Although the radial stress is zero only on the outer surface,
the value is always negligible compared to the hoop
stress, because \(R/t >> 1\). And it is for this reason
that plane stress conditions remain applicable.
Hooke's Law
The key impact of plane stress and strain assumptions is on the interaction of
the three normal stresses and strains. In order to see this, it is necessary to
be familiar with Hooke's Law, though only for normal stresses and strains, not
shears. Recall Hooke's Law can be summarized in a single tensor equation.
\[
\epsilon_{ij} = {1 \over E} \left[ (1 + \nu) \sigma_{ij} - \nu \; \delta_{ij} \; \sigma_{kk} \right]
\]
And inverted to obtain
\[
\sigma_{ij} = {E \over (1 + \nu)} \left[ \epsilon_{ij} + { \nu \over (1 - 2 \nu)} \delta_{ij} \epsilon_{kk} \right]
\]
Much info on Hooke's Law can be found at:
www.continuummechanics.org/hookeslaw.html.
For those not familiar with tensor notation, the equations can be written out brute force as
shown below. The left column shows strain as a function of stress. The right column shows
stress as a function of strain. Again, shears can still be present. But they are not the focus here
and are therefore being ignored for the moment.
\[
\epsilon_{xx} = {1 \over E} \left[ \sigma_{xx} - \nu \; (\sigma_{yy} + \sigma_{zz}) \right]
\qquad \qquad
\sigma_{xx} = {E \over (1 + \nu)} \left[ \epsilon_{xx} + { \nu \over (1 - 2 \nu)} (\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}) \right]
\]
\[
\epsilon_{yy} = {1 \over E} \left[ \sigma_{yy} - \nu \; (\sigma_{xx} + \sigma_{zz}) \right]
\qquad \qquad
\sigma_{yy} = {E \over (1 + \nu)} \left[ \epsilon_{yy} + { \nu \over (1 - 2 \nu)} (\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}) \right]
\]
\[
\epsilon_{zz} = {1 \over E} \left[ \sigma_{zz} - \nu \; (\sigma_{xx} + \sigma_{yy}) \right]
\qquad \qquad
\sigma_{zz} = {E \over (1 + \nu)} \left[ \epsilon_{zz} + { \nu \over (1 - 2 \nu)} (\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}) \right]
\]
Uniaxial Tension
For reference, simple uniaxial tension involves setting \(\sigma_{yy} = \sigma_{zz} = 0\). This leaves
\[
\begin{eqnarray}
\epsilon_{xx} & = & {\sigma_{xx} \over E}
\qquad \qquad \qquad &
\sigma_{xx} & = & E \epsilon_{xx}
\\
\\
\epsilon_{yy} & = & - \nu \left( {\sigma_{xx} \over E} \right)
\qquad \qquad \qquad &
\sigma_{yy} & = & 0
\\
\\
\epsilon_{zz} & = & - \nu \left( {\sigma_{xx} \over E} \right)
\qquad \qquad \qquad &
\sigma_{zz} & = & 0
\end{eqnarray}
\]
Note that \(\epsilon_{yy}\) and \(\epsilon_{zz}\) are not zero even though
both \(\sigma_{yy}\) and \(\sigma_{zz}\) are.
Plane Stress
Implementation of plane stress and strain analyses involves setting certain terms in the above
equations to zero. For the case of plane stress, \(\sigma_{zz} = 0\), and the equations
for strain in terms of stress reduce to
\[
\begin{eqnarray}
\epsilon_{xx} & = & {1 \over E} \left( \sigma_{xx} - \nu \sigma_{yy} \right)
\\
\\
\epsilon_{yy} & = & {1 \over E} \left( \sigma_{yy} - \nu \sigma_{xx} \right)
\\
\\
\epsilon_{zz} & = & {- \nu \over E} \left( \sigma_{xx} + \sigma_{yy} \right)
\end{eqnarray}
\]
The first two equations give \(\epsilon_{xx}\) and \(\epsilon_{yy}\) in terms
of \(\sigma_{xx}\) and \(\sigma_{yy}\). They can be thought of as
two
equations with two unknowns that need to be inverted to obtain \(\sigma_{xx}\)
and \(\sigma_{yy}\) in terms of \(\epsilon_{xx}\) and \(\epsilon_{yy}\).
This can be visualized as
\[
\left[
\matrix{
\;\;1 & -\nu \\
-\nu & \;\;1}
\right]
\left\{
\matrix{
\sigma_{xx} \\
\sigma_{yy} }
\right\}
=
\left\{
\matrix{
E \, \epsilon_{xx} \\
E \, \epsilon_{yy} }
\right\}
\]
and inverted to obtain \(\sigma_{xx}\) and \(\sigma_{yy}\) in terms of
\(\epsilon_{xx}\) and \(\epsilon_{yy}\).
The final equations for plane stress loading are
\[
\begin{eqnarray}
\epsilon_{xx} & = & {1 \over E} \left( \sigma_{xx} - \nu \sigma_{yy} \right)
\qquad \qquad &
\sigma_{xx} = {E \over 1 - \nu^2} \left( \epsilon_{xx} + \nu \epsilon_{yy} \right)
\\
\\
\epsilon_{yy} & = & {1 \over E} \left( \sigma_{yy} - \nu \sigma_{xx} \right)
\qquad \qquad &
\sigma_{yy} = {E \over 1 - \nu^2} \left( \epsilon_{yy} + \nu \epsilon_{xx} \right)
\\
\\
\epsilon_{zz} & = & {- \nu \over E} \left( \sigma_{xx} + \sigma_{yy} \right)
\qquad \qquad &
\sigma_{zz} = 0
\end{eqnarray}
\]
Note that \(\epsilon_{zz} \ne 0\) even though \(\sigma_{zz} = 0\). This is the classic
Poisson effect that is the source of much confusion when working with Hooke's Law.
Also, the only important shear term in this case is \(\tau_{xy} = G \gamma_{xy}\).
Laterally Constrained Sheet
A laterally constrained sheet is an excellent candidate for plane stress analysis.
The sheet may be constrained because its sides are attached to underlying
structure that does not move laterally as the sheet is pulled.
It is plane stress because \(\sigma_{zz} = 0\).
For this case, it is clear that \(\epsilon_{yy} = 0\) in addition to the fact that
\(\sigma_{zz} = 0\). This is easily inserted into the above equations to obtain
\[
\begin{eqnarray}
\epsilon_{xx} & = & {1 \over E} \left( \sigma_{xx} - \nu \sigma_{yy} \right)
\qquad \qquad &
\sigma_{xx} = {E \over 1 - \nu^2} \left( \epsilon_{xx} \right)
\\
\\
\epsilon_{yy} & = & 0
\qquad \qquad &
\sigma_{yy} = {E \over 1 - \nu^2} \left( \nu \epsilon_{xx} \right) = \nu \sigma_{xx}
\\
\\
\epsilon_{zz} & = & {- \nu \over E} \left( \sigma_{xx} + \sigma_{yy} \right)
\qquad \qquad &
\sigma_{zz} = 0
\end{eqnarray}
\]
Of the six equations here, the most interesting are
\[
\sigma_{xx} = {E \over 1 - \nu^2} \left( \epsilon_{xx} \right)
\qquad \quad
\text{and}
\qquad \qquad
\sigma_{yy} = \nu \sigma_{xx}
\]
And \((1-\nu^2)\) is unique in the equation relating
\(\sigma_{xx}\) to \(\epsilon_{xx}\). It is the key difference
between the laterally constrained plate in tension and an object in
conventional uniaxial tension. For many metals, \(\nu \approx 1/3\), and
the plate will effectively be 12.5% stiffer in tension/compression
due to the lateral constraint than if it were in uniaxial tension.
It is important to understand that one could have just as easily started from the
original Hooke's Law equations and still arrived at this same point. The
Plane Stress equations make the process easier by providing a starting point
that is further down stream.
The following chart compares stress-strain curves of a laterally constrained
sheet of aluminum to a bar loaded in uniaxial tension. The separation
between the blue and red curves is the 12.5% difference due to
the \((1-\nu^2)\) term in the denominator of the constrained sheet
\(\sigma_{xx}\) equation.
Plane Strain
For the case of plane strain, \(\epsilon_{zz} = 0\), and the equations for Hooke's Law
can be simplified to
\[
\begin{eqnarray}
\epsilon_{xx} & = & {1 \over E} \left[ (1 - \nu^2) \sigma_{xx} - \nu (1 + \nu) \sigma_{yy} \right]
\qquad \qquad &
\sigma_{xx} = {E \over (1 + \nu)(1 - 2\nu)} \left[ (1 - \nu) \epsilon_{xx} + \nu \epsilon_{yy} \right]
\\
\\
\epsilon_{yy} & = & {1 \over E} \left[ (1 - \nu^2) \sigma_{yy} - \nu (1 + \nu) \sigma_{xx} \right]
\qquad \qquad &
\sigma_{yy} = {E \over (1 + \nu)(1 - 2\nu)} \left[ (1 - \nu) \epsilon_{yy} + \nu \epsilon_{xx} \right]
\\
\\
\epsilon_{zz} & = & 0
\qquad \qquad &
\sigma_{zz} = \nu ( \sigma_{xx} + \sigma_{yy} )
\end{eqnarray}
\]
This time, \(\epsilon_{zz} = 0\) even though \(\sigma_{zz} \ne 0\); the exact opposite case
of plane stress in which \(\epsilon_{zz} \ne 0\) even though \(\sigma_{zz} = 0\).
And as always, \(\tau_{xy} = G \gamma_{xy}\).
Comparisons for the Case of \(\nu = 1/3\)
Here's a comparison of the stress-strain equations with \(\nu = 1/3\), a common
value of Poisson's Ratio for metals. Note how the coefficients change values
depending on the situation.
Hooke's Law
\[
\epsilon_{xx} = {1 \over E} \left[ \sigma_{xx} - {1 \over 3} ( \sigma_{yy} + \sigma_{zz} ) \right]
\qquad \qquad
\sigma_{xx} = E \left[ {3 \over 2} \epsilon_{xx} + {3 \over 4} (\epsilon_{yy}+\epsilon_{zz}) \right]
\]
\[
\epsilon_{yy} = {1 \over E} \left[ \sigma_{yy} - {1 \over 3} ( \sigma_{xx} + \sigma_{zz} ) \right]
\qquad \qquad
\sigma_{yy} = E \left[ {3 \over 2} \epsilon_{yy} + {3 \over 4} (\epsilon_{xx}+\epsilon_{zz}) \right]
\]
\[
\epsilon_{zz} = {1 \over E} \left[ \sigma_{zz} - {1 \over 3} ( \sigma_{xx} + \sigma_{yy} ) \right]
\qquad \qquad
\sigma_{zz} = E \left[ {3 \over 2} \epsilon_{zz} + {3 \over 4} (\epsilon_{xx}+\epsilon_{yy}) \right]
\]
Plane Stress ( \(\sigma_{zz} = 0\) )
\[
\begin{eqnarray}
\epsilon_{xx} & = & {1 \over E} \left( \sigma_{xx} - {1 \over 3} \sigma_{yy} \right)
\qquad \qquad &
\sigma_{xx} = E \left( {9 \over 8} \epsilon_{xx} + {3 \over 8} \epsilon_{yy} \right)
\\
\\
\epsilon_{yy} & = & {1 \over E} \left( \sigma_{yy} - {1 \over 3} \sigma_{xx} \right)
\qquad \qquad &
\sigma_{yy} = E \left( {9 \over 8} \epsilon_{yy} + {3 \over 8} \epsilon_{xx} \right)
\\
\\
\epsilon_{zz} & = & {-1 \over \;\; 3 E} \left( \sigma_{xx} + \sigma_{yy} \right)
\qquad \qquad &
\sigma_{zz} = 0
\end{eqnarray}
\]
Plane Strain ( \(\epsilon_{zz} = 0\) )
\[
\begin{eqnarray}
\epsilon_{xx} & = & {1 \over E} \left[ {8 \over 9} \sigma_{xx} - {4 \over 9} \sigma_{yy} \right]
\qquad \qquad &
\sigma_{xx} = E \left[ {3 \over 2} \epsilon_{xx} + {3 \over 4} \epsilon_{yy} \right]
\\
\\
\epsilon_{yy} & = & {1 \over E} \left[ {8 \over 9} \sigma_{yy} - {4 \over 9} \sigma_{xx} \right]
\qquad \qquad &
\sigma_{yy} = E \left[ {3 \over 2} \epsilon_{yy} + {3 \over 4} \epsilon_{xx} \right]
\\
\\
\epsilon_{zz} & = & 0
\qquad \qquad &
\sigma_{zz} = {1 \over 3} ( \sigma_{xx} + \sigma_{yy} )
\end{eqnarray}
\]
Uniaxial Tension ( \(\sigma_{yy} = \sigma_{zz} = 0\) )
\[
\sigma_{xx} = E \epsilon_{xx}
\]
