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

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

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

$e={\epsilon }_{kk}=0.$

Show:

Show that the stress components and the hydrostatic pressure $p$ must satisfy the equations

${\mathrm{\nabla }}^{2}p=\mathrm{\nabla }\bullet 𝐛;{\sigma }_{11}+{\sigma }_{22}=-2p$

where $𝐛$ is the body force.

Solution

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

${\sigma }_{11}=2\mu {\epsilon }_{11}-p;{\sigma }_{22}=2\mu {\epsilon }_{22}-p;{\sigma }_{33}=2\mu {\epsilon }_{33}-p.$

Therefore,

$\begin{array}{cc}{\sigma }_{11}+{\sigma }_{22}+{\sigma }_{33}& =2\mu ({\epsilon }_{11}+{\epsilon }_{22}+{\epsilon }_{33})-3p\\ & =-3p\end{array}$

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

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

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

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

In terms of the compatibility equation,

$\begin{array}{cc}& {\epsilon }_{11,22}+{\epsilon }_{22,11}-2{\epsilon }_{12,12}=\frac{1}{2\mu }({\sigma }_{11,22}+{\sigma }_{22,11}-2{\sigma }_{12,12}+p_{,11}+p_{,22})\\ \text{or,}& 0={\sigma }_{11,22}+{\sigma }_{22,11}-2{\sigma }_{12,12}+{\mathrm{\nabla }}^{2}p\end{array}$

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,

$\begin{array}{cc}& {\sigma }_{11,22}+{\sigma }_{22,11}+{\sigma }_{11,11}+{\sigma }_{22,22}+b_{1,1}+b_{2,2}+{\mathrm{\nabla }}^{2}p=0\\ \text{or,}& {\mathrm{\nabla }}^2{\sigma }_{11}+{\mathrm{\nabla }}^2{\sigma }_{22}+{\mathrm{\nabla }}^{2}p+\mathrm{\nabla }\bullet 𝐛=0\\ \text{or,}& {\mathrm{\nabla }}^2({\sigma }_{11}+{\sigma }_{22}+p)+\mathrm{\nabla }\bullet 𝐛=0\\ \text{or,}& {\mathrm{\nabla }}^2(-2p+p)+\mathrm{\nabla }\bullet 𝐛=0\\ \text{or,}& -{\mathrm{\nabla }}^{2}p+\mathrm{\nabla }\bullet 𝐛=0\end{array}$

Hence,

${\mathrm{\nabla }}^{2}p=\mathrm{\nabla }\bullet 𝐛$