# Introduction to Elasticity/Equilibrium example 3

## Example 3

Given:

If a material is incompressible (${\displaystyle \nu }$ = 0.5), a state of hydrostatic stress (${\displaystyle \sigma _{11}=\sigma _{22}=\sigma _{33}}$) produces no strain. The corresponding stress-strain relation can be written as

${\displaystyle \sigma _{ij}=2\mu \varepsilon _{ij}-p\delta _{ij}}$

where ${\displaystyle p}$ is an unknown hydrostatic pressure which will generally vary with position. Also, the condition of incompressibility requires that the dilatation

${\displaystyle e=\varepsilon _{kk}=0~.}$

Show:

Show that the stress components and the hydrostatic pressure ${\displaystyle p}$ must satisfy the equations

${\displaystyle \nabla ^{2}{p}={\boldsymbol {\nabla }}\bullet {\mathbf {b} }~;~~\sigma _{11}+\sigma _{22}=-2p}$

where ${\displaystyle \mathbf {b} }$ is the body force.

### Solution

We have, ${\displaystyle e=\varepsilon _{kk}=\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}=0~.\,}$ Also,

${\displaystyle \sigma _{11}=2\mu \varepsilon _{11}-p~;~~\sigma _{22}=2\mu \varepsilon _{22}-p~;~~\sigma _{33}=2\mu \varepsilon _{33}-p~.\,}$

Therefore,

{\displaystyle {\begin{aligned}\sigma _{11}+\sigma _{22}+\sigma _{33}&=2\mu \left(\varepsilon _{11}+\varepsilon _{22}+\varepsilon _{33}\right)-3p\\&=-3p\end{aligned}}}

Since ${\displaystyle \sigma _{11}=\sigma _{22}=\sigma _{33}\,}$, the above relation gives ${\displaystyle \sigma _{11}=\sigma _{22}=\sigma _{33}=-p\,}$. Therefore,

${\displaystyle \sigma _{11}+\sigma _{22}=-2p\,}$

The strain-stress relations are

${\displaystyle 2\mu \varepsilon _{11}=\sigma _{11}+p~;~~2\mu \varepsilon _{22}=\sigma _{22}+p~;~~2\mu \varepsilon _{12}=\sigma _{12}~.}$

Differentiating the strains so that they correspond to the compatibilityrelation is two-dimensions, we have

${\displaystyle \varepsilon _{11,22}={\frac {1}{2\mu }}\left(\sigma _{11,22}+p_{,22}\right)~;~~\varepsilon _{22,11}={\frac {1}{2\mu }}\left(\sigma _{22,11}+p_{,11}\right)~;~~\varepsilon _{12,12}={\frac {1}{2\mu }}\left(\sigma _{12,12}\right)~.}$

In terms of the compatibility equation,

{\displaystyle {\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,

${\displaystyle \sigma _{11,1}+\sigma _{12,2}+b_{1}=0~;~~\sigma _{12,1}+\sigma _{22,2}+b_{2}=0}$

Therefore, differentiating w.r.t ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ respectively,

${\displaystyle \sigma _{11,11}+\sigma _{12,21}+b_{1,1}=0~;~~\sigma _{12,12}+\sigma _{22,22}+b_{2,2}=0}$

${\displaystyle 2\sigma _{12,12}+\sigma _{11,11}+\sigma _{22,22}+b_{1,1}+b_{2,2}=0}$

Hence,

${\displaystyle \sigma _{11,11}+\sigma _{22,22}+b_{1,1}+b_{2,2}=-2\sigma _{12,12}}$

Substituting back into the compatibility equation,

{\displaystyle {\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,

${\displaystyle {\nabla ^{2}{p}={\boldsymbol {\nabla }}\bullet {\mathbf {b} }}}$