Introduction to Elasticity/Compatibility example 1

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[edit] Example 1

Given:

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

e_{ikr}~e_{jls}~\varepsilon_{ij,kl} = 0

The stress-strain relations for a linear elastic material are

\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

\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

[edit] Solution

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

\begin{align} 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{align}

Now, the δ − e rule states that

e_{ijk}~e_{pqk} = \delta_{ip}\delta_{jq} - \delta_{iq}\delta_{jp}

Therefore,

\begin{align} 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{align}

Recall that,

\begin{align} 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{align}

Therefore,

\begin{align} \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{align}

Hence,

\begin{align} 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{align}

Hence shown.

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