Introduction to 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}}}

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