# Elasticity/Equilibrium example 3
## Example 3

Given:
If a material is incompressible ($\nu$ = 0.5), a state of hydrostatic stress ($\sigma _{11}=\sigma _{22}=\sigma _{33}$) produces no strain. The corresponding stress-strain relation can be written as
    $\sigma _{ij}=2\mu \varepsilon _{ij}-p\delta _{ij}$
where $p$ is an unknown hydrostatic pressure which will generally vary with position. Also, the condition of incompressibility requires that the dilatation
    $e=\varepsilon _{kk}=0~.$
Show:
Show that the stress components and the hydrostatic pressure $p$ must satisfy the equations
    $\nabla ^{2}{p}={\boldsymbol {\nabla }}\bullet {\mathbf {b} }~;~~\sigma _{11}+\sigma _{22}=-2p$
where $\mathbf {b}$ is the body force.

### Solution

We have, $e=\varepsilon _{kk}=\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}=0~.\,$ Also,
    $\sigma _{11}=2\mu \varepsilon _{11}-p~;~~\sigma _{22}=2\mu \varepsilon _{22}-p~;~~\sigma _{33}=2\mu \varepsilon _{33}-p~.\,$
Therefore,
    ${\begin{aligned}\sigma _{11}+\sigma _{22}+\sigma _{33}&=2\mu \left(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}\right)-3p\\&=-3p\end{aligned}}$
Since $\sigma _{11}=\sigma _{22}=\sigma _{33}\,$, the above relation gives $\sigma _{11}=\sigma _{22}=\sigma _{33}=-p\,$. Therefore,
    $\sigma _{11}+\sigma _{22}=-2p\,$
The strain-stress relations are
    $2\mu \varepsilon _{11}=\sigma _{11}+p~;~~2\mu \varepsilon _{22}=\sigma _{22}+p~;~~2\mu \varepsilon _{12}=\sigma _{12}~.$
Differentiating the strains so that they correspond to the compatibilityrelation is two-dimensions, we have
    $\varepsilon _{11,22}={\frac {1}{2\mu }}\left(\sigma _{11,22}+p_{,22}\right)~;~~\varepsilon _{22,11}={\frac {1}{2\mu }}\left(\sigma _{22,11}+p_{,11}\right)~;~~\varepsilon _{12,12}={\frac {1}{2\mu }}\left(\sigma _{12,12}\right)~.$
In terms of the compatibility equation,
    ${\begin{aligned}&\varepsilon _{11,22}+\varepsilon _{22,11}-2\varepsilon _{12,12}={\frac {1}{2\mu }}\left(\sigma _{11,22}+\sigma _{22,11}-2\sigma _{12,12}+p_{,11}+p_{,22}\right)\\{\text{or,}}~&0=\sigma _{11,22}+\sigma _{22,11}-2\sigma _{12,12}+\nabla ^{2}{p}\end{aligned}}$
From the two-dimensional equilibrium equations,
    $\sigma _{11,1}+\sigma _{12,2}+b_{1}=0~;~~\sigma _{12,1}+\sigma _{22,2}+b_{2}=0$
Therefore, differentiating w.r.t $x_{1}$ and $x_{2}$ respectively,
    $\sigma _{11,11}+\sigma _{12,21}+b_{1,1}=0~;~~\sigma _{12,12}+\sigma _{22,22}+b_{2,2}=0$
Adding,
    $2\sigma _{12,12}+\sigma _{11,11}+\sigma _{22,22}+b_{1,1}+b_{2,2}=0$
Hence,
    $\sigma _{11,11}+\sigma _{22,22}+b_{1,1}+b_{2,2}=-2\sigma _{12,12}$
Substituting back into the compatibility equation,
    ${\begin{aligned}&\sigma _{11,22}+\sigma _{22,11}+\sigma _{11,11}+\sigma _{22,22}+b_{1,1}+b_{2,2}+\nabla ^{2}{p}=0\\{\text{or,}}~&\nabla ^{2}{\sigma _{11}}+\nabla ^{2}{\sigma _{22}}+\nabla ^{2}{p}+{\boldsymbol {\nabla }}\bullet {\mathbf {b} }=0\\{\text{or,}}~&\nabla ^{2}{(\sigma _{11}+\sigma _{22}+p)}+{\boldsymbol {\nabla }}\bullet {\mathbf {b} }=0\\{\text{or,}}~&\nabla ^{2}{(-2p+p)}+{\boldsymbol {\nabla }}\bullet {\mathbf {b} }=0\\{\text{or,}}~&-\nabla ^{2}{p}+{\boldsymbol {\nabla }}\bullet {\mathbf {b} }=0\end{aligned}}$
Hence,
    ${\nabla ^{2}{p}={\boldsymbol {\nabla }}\bullet {\mathbf {b} }}$