Elasticity/Equilibrium example 2

Example 2

Given: The displacement equation of equilibrium for an isotropic inhomogeneous linear elastic material can be written as

$\mathrm{\nabla }\bullet (𝐂:\mathrm{\nabla }𝐮)+𝐛=0$

where

$𝐂=\lambda {\mathrm{𝟏}}^{(2)}\otimes {\mathrm{𝟏}}^{(2)}+2\mu {\mathrm{𝟏}}^{(4s)}$

and $\lambda (𝐱)$ and $\mu (𝐱)$ are the Lamé moduli.

Show:

Show that the displacement equation of equilibrium can be expressed as

$\mu \mathrm{\nabla }\bullet (\mathrm{\nabla }𝐮)+(\lambda +\mu )\mathrm{\nabla }(\mathrm{\nabla }\bullet 𝐮)+(\mathrm{\nabla }𝐮+\mathrm{\nabla }{𝐮}^T)\mathrm{\nabla }\mu +(\mathrm{\nabla }\bullet 𝐮)\mathrm{\nabla }\lambda +𝐛=0$

Solution

The skew part of the tensor $\mathrm{\nabla }𝐮$ does not affect the stress because it leads to a rigid displacement field. Therefore, the displacement equation of equilibrium may be written as

$\mathrm{\nabla }\bullet [𝐂:\text{symm}(\mathrm{\nabla }𝐮)]+𝐛=0$

where

$\text{symm}(\mathrm{\nabla }𝐮)=\frac{1}{2}(\mathrm{\nabla }𝐮+\mathrm{\nabla }{𝐮}^T)$

In index notataion,

$\text{symm}(\mathrm{\nabla }𝐮)=\mathit{\epsilon }\equiv {\epsilon }_{kl}=\frac{1}{2}(u_{k,l}+u_{l,k})$

and

$𝐂\equiv C_{ijkl}=\lambda {\delta }_{ij}{\delta }_{kl}+\mu ({\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk})$

Therefore,

$\begin{array}{cc}𝐂:\text{symm}(\mathrm{\nabla }𝐮)\equiv C_{ijkl}{\epsilon }_{kl}& =\lambda {\delta }_{ij}{\delta }_{kl}{\epsilon }_{kl}+\mu {\delta }_{ik}{\delta }_{jl}{\epsilon }_{kl}+\mu {\delta }_{il}{\delta }_{jk}{\epsilon }_{kl}\\ & =\lambda {\epsilon }_{mm}{\delta }_{ij}+\mu {\epsilon }_{ij}+\mu {\epsilon }_{ij}\\ & =\lambda {\epsilon }_{mm}{\delta }_{ij}+2\mu {\epsilon }_{ij}\\ & \equiv \lambda (\text{tr}\mathit{\epsilon })\mathrm{𝟏}+2\mu \mathit{\epsilon }\end{array}$

Now,

$\text{tr}\mathit{\epsilon }\equiv {\epsilon }_{mm}=\frac{1}{2}(u_{m,m}+u_{m,m})=u_{m,m}\equiv \mathrm{\nabla }\bullet 𝐮$

Hence,

$𝐂:\text{symm}(\mathrm{\nabla }𝐮)=\lambda (\mathrm{\nabla }\bullet 𝐮)\mathrm{𝟏}+\mu (\mathrm{\nabla }𝐮+\mathrm{\nabla }{𝐮}^T)$

Taking the divergence,

$\begin{array}{cc}\mathrm{\nabla }\bullet [𝐂:\text{symm}(\mathrm{\nabla }𝐮)]& =\mathrm{\nabla }\bullet [\lambda (\mathrm{\nabla }\bullet 𝐮)\mathrm{𝟏}+\mu (\mathrm{\nabla }𝐮+\mathrm{\nabla }{𝐮}^T)]\\ & =\mathrm{\nabla }\bullet [\lambda (\mathrm{\nabla }\bullet 𝐮)\mathrm{𝟏}]+\mathrm{\nabla }\bullet \left(\mu \mathrm{\nabla }𝐮\right)+\mathrm{\nabla }\bullet \left(\mu \mathrm{\nabla }{𝐮}^T\right)\end{array}$

Recall that

$\begin{array}{cc}\mathrm{\nabla }\varphi & ={\varphi }_{,j}\\ \mathrm{\nabla }𝐯& =v_{i,j}\\ \mathrm{\nabla }\bullet 𝐯& =v_{j,j}\\ \mathrm{\nabla }\bullet 𝐓& =T_{ij,j}\end{array}$

Therefore,

$\begin{array}{cc}\mathrm{\nabla }\bullet [\lambda (\mathrm{\nabla }\bullet 𝐮)\mathrm{𝟏}]& \equiv {\left(\lambda u_{k,k}{\delta }_{ij}\right)}_{,j}\\ & ={\lambda }_{,i}{u}_{k,k}+\lambda u_{k,ki}\\ & \equiv \mathrm{\nabla }\lambda (\mathrm{\nabla }\bullet 𝐮)+\lambda \mathrm{\nabla }(\mathrm{\nabla }\bullet 𝐮)\end{array}$

$\begin{array}{cc}\mathrm{\nabla }\bullet \left(\mu \mathrm{\nabla }𝐮\right)& \equiv {\left(\mu u_{i,j}\right)}_{,j}\\ & ={\mu }_{,j}{u}_{i,j}+\mu u_{i,jj}\\ & \equiv \mathrm{\nabla }\mu \mathrm{\nabla }𝐮+\mu \mathrm{\nabla }\bullet (\mathrm{\nabla }𝐮)\end{array}$

$\begin{array}{cc}\mathrm{\nabla }\bullet \left(\mu \mathrm{\nabla }{𝐮}^T\right)& \equiv {\left(\mu u_{j,i}\right)}_{,j}\\ & ={\mu }_{,j}{u}_{j,i}+\mu u_{j,ij}\\ & \equiv \mathrm{\nabla }\mu \mathrm{\nabla }{𝐮}^T+\mu \mathrm{\nabla }(\mathrm{\nabla }\bullet 𝐮)\end{array}$

Hence,

$\begin{array}{cc}\mathrm{\nabla }\bullet [𝐂:\text{symm}(\mathrm{\nabla }𝐮)]& =\mathrm{\nabla }\lambda (\mathrm{\nabla }\bullet 𝐮)+\lambda \mathrm{\nabla }(\mathrm{\nabla }\bullet 𝐮)+\mathrm{\nabla }\mu \mathrm{\nabla }𝐮+\mu \mathrm{\nabla }\bullet (\mathrm{\nabla }𝐮)+\mathrm{\nabla }\mu \mathrm{\nabla }{𝐮}^T+\mu \mathrm{\nabla }(\mathrm{\nabla }\bullet 𝐮)\\ & =\mu \mathrm{\nabla }\bullet (\mathrm{\nabla }𝐮)+(\lambda +\mu )\mathrm{\nabla }(\mathrm{\nabla }\bullet 𝐮)+\mathrm{\nabla }\mu (\mathrm{\nabla }𝐮+\mathrm{\nabla }{𝐮}^T)+\mathrm{\nabla }\lambda (\mathrm{\nabla }\bullet 𝐮)\end{array}$

Therefore, the displacement equation of equilibrium can be expressed as required, i.e,

$\mu \mathrm{\nabla }\bullet (\mathrm{\nabla }𝐮)+(\lambda +\mu )\mathrm{\nabla }(\mathrm{\nabla }\bullet 𝐮)+(\mathrm{\nabla }𝐮+\mathrm{\nabla }{𝐮}^T)\mathrm{\nabla }\mu +(\mathrm{\nabla }\bullet 𝐮)\mathrm{\nabla }\lambda +𝐛=0$