Example 2

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

${\displaystyle {\boldsymbol {\nabla }}\bullet (\mathbf {C} :{\boldsymbol {\nabla }}\mathbf {u} )+\mathbf {b} =0}$

where

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

and ${\displaystyle \lambda (\mathbf {x} )}$ and ${\displaystyle \mu (\mathbf {x} )}$ are the Lamé moduli.

Show:

Show that the displacement equation of equilibrium can be expressed as

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

Solution

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

${\displaystyle {\boldsymbol {\nabla }}\bullet \left[\mathbf {C} :{\text{symm}}({\boldsymbol {\nabla }}\mathbf {u} )\right]+\mathbf {b} =0}$

where

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

In index notataion,

${\displaystyle {\text{symm}}({\boldsymbol {\nabla }}\mathbf {u} )={\boldsymbol {\varepsilon }}\equiv \varepsilon _{kl}={\frac {1}{2}}(u_{k,l}+u_{l,k})}$

and

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

Therefore,

${\displaystyle {\begin{aligned}\mathbf {C} :{\text{symm}}({\boldsymbol {\nabla }}\mathbf {u} )\equiv C_{ijkl}~\varepsilon _{kl}&=\lambda \delta _{ij}\delta _{kl}~\varepsilon _{kl}+\mu \delta _{ik}\delta _{jl}~\varepsilon _{kl}+\mu \delta _{il}\delta _{jk}~\varepsilon _{kl}\\&=\lambda ~\varepsilon _{mm}\delta _{ij}+\mu ~\varepsilon _{ij}+\mu ~\varepsilon _{ij}\\&=\lambda ~\varepsilon _{mm}\delta _{ij}+2\mu ~\varepsilon _{ij}\\&\equiv \lambda ~({\text{tr}}~{\boldsymbol {\varepsilon }})\mathbf {1} +2\mu ~{\boldsymbol {\varepsilon }}\end{aligned}}}$

Now,

${\displaystyle {\text{tr}}~{\boldsymbol {\varepsilon }}\equiv \varepsilon _{mm}={\frac {1}{2}}(u_{m,m}+u_{m,m})=u_{m,m}\equiv {\boldsymbol {\nabla }}\bullet \mathbf {u} }$

Hence,

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

Taking the divergence,

${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\bullet {\left[\mathbf {C} :{\text{symm}}({\boldsymbol {\nabla }}\mathbf {u} )\right]}&={\boldsymbol {\nabla }}\bullet {\left[\lambda ~({\boldsymbol {\nabla }}\bullet \mathbf {u} )\mathbf {1} +\mu ~({\boldsymbol {\nabla }}\mathbf {u} +{\boldsymbol {\nabla }}\mathbf {u} ^{T})\right]}\\&={\boldsymbol {\nabla }}\bullet {\left[\lambda ~({\boldsymbol {\nabla }}\bullet \mathbf {u} )\mathbf {1} \right]}+{\boldsymbol {\nabla }}\bullet {\left(\mu ~{\boldsymbol {\nabla }}\mathbf {u} \right)}+{\boldsymbol {\nabla }}\bullet {\left(\mu ~{\boldsymbol {\nabla }}\mathbf {u} ^{T}\right)}\end{aligned}}}$

Recall that

${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}{\phi }&=\phi _{,j}\\{\boldsymbol {\nabla }}{\mathbf {v} }&=v_{i,j}\\{\boldsymbol {\nabla }}\bullet {\mathbf {v} }&=v_{j,j}\\{\boldsymbol {\nabla }}\bullet {\mathbf {T} }&=T_{ij,j}\end{aligned}}}$

Therefore,

${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\bullet {\left[\lambda ~({\boldsymbol {\nabla }}\bullet \mathbf {u} )\mathbf {1} \right]}&\equiv \left(\lambda ~u_{k,k}\delta _{ij}\right)_{,j}\\&=\lambda _{,i}~u_{k,k}+\lambda ~u_{k,ki}\\&\equiv {\boldsymbol {\nabla }}{\lambda }({\boldsymbol {\nabla }}\bullet \mathbf {u} )+\lambda {\boldsymbol {\nabla }}{({\boldsymbol {\nabla }}\bullet \mathbf {u} )}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\bullet {\left(\mu ~{\boldsymbol {\nabla }}\mathbf {u} \right)}&\equiv \left(\mu ~u_{i,j}\right)_{,j}\\&=\mu _{,j}~u_{i,j}+\mu ~u_{i,jj}\\&\equiv {\boldsymbol {\nabla }}{\mu }{\boldsymbol {\nabla }}\mathbf {u} +\mu {\boldsymbol {\nabla }}\bullet {({\boldsymbol {\nabla }}\mathbf {u} )}\end{aligned}}}$

${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\bullet {\left(\mu ~{\boldsymbol {\nabla }}\mathbf {u} ^{T}\right)}&\equiv \left(\mu ~u_{j,i}\right)_{,j}\\&=\mu _{,j}~u_{j,i}+\mu ~u_{j,ij}\\&\equiv {\boldsymbol {\nabla }}{\mu }{\boldsymbol {\nabla }}\mathbf {u} ^{T}+\mu {\boldsymbol {\nabla }}{({\boldsymbol {\nabla }}\bullet \mathbf {u} )}\end{aligned}}}$

Hence,

${\displaystyle {\begin{aligned}{\boldsymbol {\nabla }}\bullet {\left[\mathbf {C} :{\text{symm}}({\boldsymbol {\nabla }}\mathbf {u} )\right]}&={\boldsymbol {\nabla }}{\lambda }({\boldsymbol {\nabla }}\bullet \mathbf {u} )+\lambda {\boldsymbol {\nabla }}{({\boldsymbol {\nabla }}\bullet \mathbf {u} )}+{\boldsymbol {\nabla }}{\mu }{\boldsymbol {\nabla }}\mathbf {u} +\mu {\boldsymbol {\nabla }}\bullet {({\boldsymbol {\nabla }}\mathbf {u} )}+{\boldsymbol {\nabla }}{\mu }{\boldsymbol {\nabla }}\mathbf {u} ^{T}+\mu {\boldsymbol {\nabla }}{({\boldsymbol {\nabla }}\bullet \mathbf {u} )}\\&=\mu {\boldsymbol {\nabla }}\bullet {({\boldsymbol {\nabla }}\mathbf {u} )}+(\lambda +\mu ){\boldsymbol {\nabla }}{({\boldsymbol {\nabla }}\bullet \mathbf {u} )}+{\boldsymbol {\nabla }}{\mu }\left({\boldsymbol {\nabla }}\mathbf {u} +{\boldsymbol {\nabla }}\mathbf {u} ^{T}\right)+{\boldsymbol {\nabla }}{\lambda }({\boldsymbol {\nabla }}\bullet \mathbf {u} )\end{aligned}}}$

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

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