# Elasticity/Plane strain example 1
## Example 1

Given:
The plane strain solution for the stresses in a rectangular block with $0<x_{1}<a$, $-b<x_{2}<b$, and $-c<x_{3}<c$ with a given loading is
    $\sigma _{11}={\frac {3Fx_{1}x_{2}}{2b^{3}}}~;~~\sigma _{12}={\frac {3F(b^{2}-x_{2}^{2})}{4b^{3}}}~;~~\sigma _{22}=0~;~~\sigma _{33}=-{\frac {3\nu Fx_{1}x_{2}}{2b^{3}}}.$
Find:
* Find the tractions on the surfaces of the block and illustrate the results on a sketch of the block.
* We wish to use this solution to solve the corresponding problem in which the surfaces $x_{3}=\pm c$ are traction-free. Determine an approximate corrective solution for this problem by offloading the unwanted force and moment results using elementary bending theory.
* Find the maximum error in the stress $\sigma _{33}$ in the corrected solution and compare it with the maximum tensile stress in the plane strain solution.

## Solution

The tractions acting on the block are:
    ${\begin{aligned}{\text{at}}~x_{1}&=0~,~~{\widehat {\mathbf {n} }}\equiv (-1,0,0)~,~~t_{i}=n_{1}\sigma _{1i}\equiv (-\sigma _{11},-\sigma _{12},-\sigma _{13})\\{\text{at}}~x_{1}&=a~,~~{\widehat {\mathbf {n} }}\equiv (1,0,0)~,~~t_{i}=n_{1}\sigma _{1i}\equiv (\sigma _{11},\sigma _{12},\sigma _{13})\\{\text{at}}~x_{2}&=-b~,~~{\widehat {\mathbf {n} }}\equiv (0,-1,0)~,~~t_{i}=n_{2}\sigma _{2i}\equiv (-\sigma _{21},-\sigma _{22},-\sigma _{23})\\{\text{at}}~x_{2}&=b~,~~{\widehat {\mathbf {n} }}\equiv (0,1,0)~,~~t_{i}=n_{2}\sigma _{2i}\equiv (\sigma _{21},\sigma _{22},\sigma _{23})\\{\text{at}}~x_{3}&=-c~,~~{\widehat {\mathbf {n} }}\equiv (0,0,-1)~,~~t_{i}=n_{3}\sigma _{3i}\equiv (-\sigma _{31},-\sigma _{32},-\sigma _{33})\\{\text{at}}~x_{3}&=c~,~~{\widehat {\mathbf {n} }}\equiv (0,0,1)~,~~t_{i}=n_{3}\sigma _{3i}\equiv (\sigma _{31},\sigma _{32},\sigma _{33})\end{aligned}}$
Plugging in the expressions for stress,
    ${\begin{aligned}{\text{at}}~x_{1}&=0~,~~t_{i}\equiv (0,-{\frac {3F(b^{2}-x_{2}^{2})}{4b^{3}}},0)\\{\text{at}}~x_{1}&=a~,~~t_{i}\equiv ({\frac {3Fax_{2}}{2b^{3}}},{\frac {3F(b^{2}-x_{2}^{2})}{4b^{3}}},0)\\{\text{at}}~x_{2}&=-b~,~~t_{i}\equiv (0,0,0)\\{\text{at}}~x_{2}&=b~,~~t_{i}\equiv (0,0,0)\\{\text{at}}~x_{3}&=-c~,~~t_{i}\equiv (0,0,{\frac {3\nu Fx_{1}x_{2}}{2b^{3}}})\\{\text{at}}~x_{3}&=c~,~~t_{i}\equiv (0,0,-{\frac {3\nu Fx_{1}x_{2}}{2b^{3}}})\end{aligned}}$
These tractions are illustrated in the following figure
* Tractions on the beam
To unload the tractions on the faces $x_{3}=\pm c$, we have to superpose the solution to a problem with equal and opposite tractions and moments. In order to use elementary bending theory, we have set up a problem with simple boundary conditions.
Let us first consider the force distribution required for the superposed problem. Since the loading is antisymmetric at $x_{3}=\pm c$, there is no net force is the $x_{3}$ direction. Similarly, there is no net moment about the $x_{2}$ axis.
However, there is a net moment about the $x_{1}$ axis. Hence, the problem to be superposed should have a bending stress distribution $\sigma _{33}=Cx_{2}$, where $C$ is a constant that is chosen so as to make the total bending moment (original problem + superposed problem) equal to zero. (Note: Think of a beam in the $x_{3}-x_{2}$ plane subjected to bending moments at the ends.)
The total stress for the corrected problem is
    $\sigma _{33}=-{\frac {3\nu Fx_{1}x_{2}}{2b^{3}}}+Cx_{2}$
The bending moment for a cross-section of the beam in the $x_{3}-x_{2}$ plane about the $x_{1}$ axis is $M=-\sigma _{33}I/x_{2}\,$, where $I=\int _{0}^{a}\int _{-b}^{b}x_{2}^{2}dx_{2}dx_{1}$.
Since $\sigma _{33}$ varies with $x_{1}$, the total bending moment for the beam is given by
    ${\begin{aligned}M&=\int _{0}^{a}\int _{-b}^{b}\left({\frac {-\sigma _{33}}{x_{2}}}\right)x_{2}^{2}dx_{2}dx_{1}\\&=\int _{0}^{a}\int _{-b}^{b}\left({\frac {3\nu Fx_{1}x_{2}}{2b^{3}}}-Cx_{2}\right)x_{2}dx_{2}dx_{1}\\&=\int _{0}^{a}\left[{\frac {3\nu Fx_{1}x_{2}^{3}}{6b^{3}}}-{\frac {Cx_{2}^{3}}{3}}\right]_{-b}^{b}dx_{1}\\&=\int _{0}^{a}\left[{\frac {\nu Fx_{1}b^{3}}{b^{3}}}-{\frac {2Cb^{3}}{3}}\right]dx_{1}\\&=\left[{\frac {\nu Fx_{1}^{2}b^{3}}{2b^{3}}}-{\frac {2Cx_{1}b^{3}}{3}}\right]_{0}^{a}\\&={\frac {\nu Fa^{2}b^{3}}{2b^{3}}}-{\frac {2Cab^{3}}{3}}\end{aligned}}$
Setting the bending moment to zero, we have
    $C={\frac {3\nu Fa}{4b^{3}}}$
Therefore, the corrected solution is
    ${\sigma _{33}=-{\frac {3\nu Fx_{1}x_{2}}{2b^{3}}}+{\frac {3\nu Fax_{2}}{4b^{3}}}}$
Ideally, for a problem with zero tractions on $x_{3}=\pm c$, we should have $\sigma _{33}=0$. Therefore, the error in our solution is
    $\sigma _{33}^{\text{err}}={\text{abs}}\left({\frac {3\nu Fx_{1}x_{2}}{2b^{3}}}-{\frac {3\nu Fax_{2}}{4b^{3}}}\right)$
The error is maximum at $(0,-b)$, $(0,b)$, $(a,-b)$, and $(a,b)$. Thus,
    ${\begin{aligned}\left.\sigma _{33}^{\text{err}}\right|_{(0,-b)}&={\text{abs}}\left({\frac {3\nu Fab}{4b^{3}}}\right)={\text{abs}}\left({\frac {3\nu Fa}{4b^{2}}}\right)\\\left.\sigma _{33}^{\text{err}}\right|_{(0,b)}&={\text{abs}}\left({\frac {3\nu Fab}{4b^{3}}}\right)={\text{abs}}\left({\frac {3\nu Fa}{4b^{2}}}\right)\\\left.\sigma _{33}^{\text{err}}\right|_{(a,-b)}&={\text{abs}}\left({\frac {3\nu Fab}{4b^{3}}}\right)={\text{abs}}\left({\frac {3\nu Fa}{4b^{2}}}\right)\\\left.\sigma _{33}^{\text{err}}\right|_{(a,b)}&={\text{abs}}\left({\frac {3\nu Fab}{4b^{3}}}\right)={\text{abs}}\left({\frac {3\nu Fa}{4b^{2}}}\right)\end{aligned}}$
The maximum error is
    ${\frac {3\nu Fa}{4b^{2}}}$
The maximum tensile stress is
    $\sigma _{11}(a,b)={\frac {3Fab}{2b^{3}}}={\frac {3Fa}{2b^{2}}}$
Therefore, the ratio of the maximum error in $\sigma _{33}$ to the maximum tensile stress is
    ${{\text{Ratio}}={\frac {\nu }{2}}}$