# Introduction to Elasticity/Compatibility example 1

## Example 1

Given:

The compatibility equations in terms of the strains are (in index notation)

${\displaystyle e_{ikr}~e_{jls}~\varepsilon _{ij,kl}=0}$

The stress-strain relations for a linear elastic material are

${\displaystyle \varepsilon _{ij}={\frac {1}{E}}\left[(1+\nu )~\sigma _{ij}-\nu ~\sigma _{mm}~\delta _{ij}\right]}$

Show:

Substituting the stress-strain relations into the compatibility equations, show that the compatibility equation of stress can be expressed as

${\displaystyle \sigma _{ii,jj}~\delta _{rs}-\sigma _{ii,rs}-(1+\nu )(\sigma _{ij,ij}~\delta _{rs}+\sigma _{rs,ii}-\sigma _{is,ir}-\sigma _{ir,is})=0}$

### Solution

Substituting the stress-strain relations into the left hand side of the compatibility equations and multiplying both sides by ${\displaystyle E}$, we get,

{\displaystyle {\begin{aligned}E~e_{ikr}~e_{jls}~\varepsilon _{ij,kl}&=e_{ikr}~e_{jls}~\left[(1+\nu )~\sigma _{ij,kl}-\nu ~\sigma _{mm,kl}~\delta _{ij}\right]\\&=(1+\nu )~e_{ikr}~e_{jls}~\sigma _{ij,kl}-\nu ~e_{ikr}~e_{jls}~\delta _{ij}~\sigma _{mm,kl}\\&=(1+\nu )~e_{ikr}~e_{jls}~\sigma _{ij,kl}-\nu ~e_{nkr}~e_{nls}~\sigma _{mm,kl}\\&=(1+\nu )~e_{ikr}~e_{jls}~\sigma _{ij,kl}-\nu ~e_{krn}~e_{lsn}~\sigma _{mm,kl}\end{aligned}}}

Now, the ${\displaystyle \delta -e}$ rule states that

${\displaystyle e_{ijk}~e_{pqk}=\delta _{ip}\delta _{jq}-\delta _{iq}\delta _{jp}}$

Therefore,

{\displaystyle {\begin{aligned}E~e_{ikr}~e_{jls}~\varepsilon _{ij,kl}&=(1+\nu )~e_{ikr}~e_{jls}~\sigma _{ij,kl}-\nu ~\left(\delta _{kl}\delta _{rs}-\delta _{ks}\delta _{rl}\right)~\sigma _{mm,kl}\\&=(1+\nu )~e_{ikr}~e_{jls}~\sigma _{ij,kl}-\nu ~\left(\sigma _{mm,nn}\delta _{rs}-\sigma _{mm,sr}\right)\end{aligned}}}

Recall that,

{\displaystyle {\begin{aligned}e_{ikr}~e_{jls}&={\text{det}}{\begin{bmatrix}\delta _{ij}&\delta _{il}&\delta _{is}\\\delta _{kj}&\delta _{kl}&\delta _{ks}\\\delta _{rj}&\delta _{rl}&\delta _{rs}\end{bmatrix}}\\&=\delta _{ij}\delta _{kl}\delta _{rs}-\delta _{ij}\delta _{ks}\delta _{rl}-\delta _{il}\delta _{kj}\delta _{rs}+\delta _{il}\delta _{ks}\delta _{rj}+\delta _{is}\delta _{kj}\delta _{rl}-\delta _{is}\delta _{kl}\delta _{rj}\end{aligned}}}

Therefore,

{\displaystyle {\begin{aligned}\delta _{ij}\delta _{kl}\delta _{rs}\sigma _{ij,kl}&=\sigma _{ii,jj}\delta _{rs}\\\delta _{ij}\delta _{ks}\delta _{rl}\sigma _{ij,kl}&=\sigma _{ii,rs}\\\delta _{il}\delta _{kj}\delta _{rs}\sigma _{ij,kl}&=\sigma _{ij,ij}\delta _{rs}\\\delta _{il}\delta _{ks}\delta _{rj}\sigma _{ij,kl}&=\sigma _{ir,is}\\\delta _{is}\delta _{kj}\delta _{rl}\sigma _{ij,kl}&=\sigma _{is,ir}\\\delta _{is}\delta _{kl}\delta _{rj}\sigma _{ij,kl}&=\sigma _{sr,jj}\end{aligned}}}

Hence,

{\displaystyle {\begin{aligned}E~e_{ikr}~e_{jls}~\varepsilon _{ij,kl}&=(1+\nu )\left(\sigma _{ii,jj}\delta _{rs}-\sigma _{ii,rs}-\sigma _{ij,ij}\delta _{rs}+\sigma _{ir,is}+\sigma _{is,ir}-\sigma _{sr,jj}\right)-\nu ~\left(\sigma _{ii,jj}\delta _{rs}-\sigma _{ii,rs}\right)\\&=\sigma _{ii,jj}\delta _{rs}-\sigma _{ii,rs}-(1+\nu )\left(\sigma _{ij,ij}\delta _{rs}+\sigma _{rs,ii}-\sigma _{is,ir}-\sigma _{ir,is}\right)=0\end{aligned}}}

Hence shown.