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{align}

\sigma_{11} + \sigma_{22} + \sigma_{33} & = 2\mu\left(\varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33}\right) - 3p\\ & = -3p \end{align}$$ Since $$\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{align}

& \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{align}$$ 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{align}

& \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{align}$$ Hence,

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