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\section{Elastic Deformation} 
\section{Elastic Deformation} 
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\label{ELASTIC CHAP} 
\label{ELASTIC CHAP} 
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In this section we want to discuss the deformation of a linear elastic body caused by expansion through a heat distribution. We want 
In this section we want to examine the deformation of a linear elastic body caused by expansion through a heat distribution. We want 
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to displacement field $u\hackscore{i}$ which solves the momentum equation 
a displacement field $u\hackscore{i}$ which solves the momentum equation 
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\index{momentum equation}: 
\index{momentum equation}: 
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\begin{eqnarray}\label{HEATEDBLOCK general problem} 
\begin{eqnarray}\label{HEATEDBLOCK general problem} 
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 \sigma\hackscore{ij,j}=0 
 \sigma\hackscore{ij,j}=0 
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T(x)= T\hackscore{0} e^{\beta \xx^{c}\}; 
T(x)= T\hackscore{0} e^{\beta \xx^{c}\}; 
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\end{eqnarray} 
\end{eqnarray} 
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with a given positive constant $\beta$ and location $x^{c}$ in the domain. Later in \Sec{MODELFRAME} we will use 
with a given positive constant $\beta$ and location $x^{c}$ in the domain. Later in \Sec{MODELFRAME} we will use 
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$T$ from an timedependent temperature diffusion problem as discussed in \Sec{DIFFUSION CHAP}. 
$T$ from a timedependent temperature diffusion problem as discussed in \Sec{DIFFUSION CHAP}. 
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When we insert~\eqn{HEATEDBLOCK linear elastic} we get a second oder system of linear PDEs for the displacements $u$ which is called 
When we insert~\eqn{HEATEDBLOCK linear elastic} we get a second oder system of linear PDEs for the displacements $u$ which is called 
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the Lame equation\index{Lame equation}. We want to solve 
the Lame equation\index{Lame equation}. We want to solve 
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this using the \LinearPDE class to this. For a system of PDEs and a solution with several components the \LinearPDE class 
this using the \LinearPDE class to this. For a system of PDEs and a solution with several components the \LinearPDE class 
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A\hackscore{ijkl} u\hackscore{k,l}){,j}=X\hackscore{ij,j} \; . 
A\hackscore{ijkl} u\hackscore{k,l}){,j}=X\hackscore{ij,j} \; . 
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\end{equation} 
\end{equation} 
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$A$ is of \RankFour and $X$ is of \RankTwo. We show here the coefficients relevant 
$A$ is of \RankFour and $X$ is of \RankTwo. We show here the coefficients relevant 
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for the we traying to solve. The full form is given in~\eqn{LINEARPDE.SYSTEM.1}. 
for the we trying to solve. The full form is given in~\eqn{LINEARPDE.SYSTEM.1}. 
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The natural boundary conditions \index{boundary condition!natural} take the form: 
The natural boundary conditions \index{boundary condition!natural} take the form: 
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\begin{equation}\label{LINEARPDE.SYSTEM.2 TUTORIAL} 
\begin{equation}\label{LINEARPDE.SYSTEM.2 TUTORIAL} 
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n\hackscore{j} A\hackscore{ijkl} u\hackscore{k,l}=n\hackscore{j}X\hackscore{ij} \;. 
n\hackscore{j} A\hackscore{ijkl} u\hackscore{k,l}=n\hackscore{j}X\hackscore{ij} \;. 
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\right. 
\right. 
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\end{equation} 
\end{equation} 
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Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$ 
Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$ 
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are constant setting $Y\hackscore{i}=\lambda+\frac{2}{3} \mu) \; \alpha \; T\hackscore{i}$ seems to be also possible. However, 
are constant we may use $Y\hackscore{i}=\lambda+\frac{2}{3} \mu) \; \alpha \; T\hackscore{i}$. However, 
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this choice would lead to a different natural boundary condition which does not set the normal stress component as defined 
this choice would lead to a different natural boundary condition which does not set the normal stress component as defined 
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in~\eqn{HEATEDBLOCK linear elastic} to zero. 
in~\eqn{HEATEDBLOCK linear elastic} to zero. 
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\begin{eqnarray}\label{LINEARPDE.SYSTEM.SYMMETRY TUTORIAL} 
\begin{eqnarray}\label{LINEARPDE.SYSTEM.SYMMETRY TUTORIAL} 
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A\hackscore{ijkl} =A\hackscore{klij} \\ 
A\hackscore{ijkl} =A\hackscore{klij} \\ 
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\end{eqnarray} 
\end{eqnarray} 
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Note that different from the scalar case now the coefficients $D$ and $d$ have to be inspected. It is easy to see that 
This Lame equation is in fact symmetric, given the difference in $D$ and $d$ as compared to the scalar case. 
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the Lame equation in fact is symmetric. The \LinearPDE class is notified by this fact by calling its \method{setSymmetryOn} method. 
The \LinearPDE class is notified of this fact by calling its \method{setSymmetryOn} method. 
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After we have solved the Lame equation we want to analyse the actual stress distribution. Typically the vonMises stress\index{vonMises stress} defined by 
After we have solved the Lame equation we want to analyse the actual stress distribution. Typically the vonMises stress\index{vonMises stress} defined by 
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\begin{equation} 
\begin{equation} 
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(\sigma\hackscore{22}\sigma\hackscore{00})^2) 
(\sigma\hackscore{22}\sigma\hackscore{00})^2) 
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+ \sigma\hackscore{01}^2+\sigma\hackscore{12}^2+\sigma\hackscore{20}^2} 
+ \sigma\hackscore{01}^2+\sigma\hackscore{12}^2+\sigma\hackscore{20}^2} 
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\end{equation} 
\end{equation} 
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is used to detect material damages. Here we want to calculate the vonMises and write the stress to a file for visualization. 
is used to detect material damage. Here we want to calculate the vonMises and write the stress to a file for visualization. 
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\index{scripts!\file{diffusion.py}}: 
\index{scripts!\file{diffusion.py}} 
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The following script which is available in \file{heatedbox.py} in the \ExampleDirectory solves the Lame equation 
The following script, which is available in \file{heatedbox.py} in the \ExampleDirectory, solves the Lame equation 
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and writes the displacements and the vonMises stress\index{vonMises stress} into a file \file{deform.xml} in the \VTK file format: 
and writes the displacements and the vonMises stress\index{vonMises stress} into a file \file{deform.xml} in the \VTK file format: 
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\begin{python} 
\begin{python} 
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from esys.escript import * 
from esys.escript import * 