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revision 624 by gross, Fri Mar 17 05:48:59 2006 UTC revision 625 by gross, Thu Mar 23 00:41:25 2006 UTC
# Line 1  Line 1 
1  % $Id$  % $Id$
2    %
3    %           Copyright © 2006 by ACcESS MNRF
4    %               \url{http://www.access.edu.au
5    %         Primary Business: Queensland, Australia.
6    %   Licensed under the Open Software License version 3.0
7    %      http://www.opensource.org/licenses/osl-3.0.php
8    %
9  \section{Elastic Deformation}  \section{Elastic Deformation}
10  \label{ELASTIC CHAP}  \label{ELASTIC CHAP}
11  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 discuss the deformation of a linear elastic body caused by expansion through a heat distribution. We want
# Line 43  When we insert~\eqn{HEATEDBLOCK linear e Line 50  When we insert~\eqn{HEATEDBLOCK linear e
50  the Lame equation\index{Lame equation}. We want to solve  the Lame equation\index{Lame equation}. We want to solve
51  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
52  takes PDEs of the form  takes PDEs of the form
53  \begin{equation}\label{LINEARPDE.SYSTEM.1}  \begin{equation}\label{LINEARPDE.SYSTEM.1 TUTORIAL}
54  -(A\hackscore{ijkl} u\hackscore{k,l}){,j}+(B\hackscore{ijk} u\hackscore{k})\hackscore{,j}+C\hackscore{ikl} u\hackscore{k,l}+D\hackscore{ik} u\hackscore{k} =-X\hackscore{ij,j}+Y\hackscore{i} \; .  -A\hackscore{ijkl} u\hackscore{k,l}){,j}=-X\hackscore{ij,j} \; .
55  \end{equation}  \end{equation}
56  $A$ is a \RankFour, $B$ and $C$ are each a \RankThree, $D$ and $X$ are each a \RankTwo and $Y$ is a \RankOne.  $A$ is of \RankFour and $X$ is of \RankTwo. We show here the coefficients relevant
57    for the we traying to solve. The full form is given in~\eqn{LINEARPDE.SYSTEM.1}.
58  The natural boundary conditions \index{boundary condition!natural} take the form:  The natural boundary conditions \index{boundary condition!natural} take the form:
59  \begin{equation}\label{LINEARPDE.SYSTEM.2}  \begin{equation}\label{LINEARPDE.SYSTEM.2 TUTORIAL}
60  n\hackscore{j}\left(A\hackscore{ijkl} u\hackscore{k,l}+B\hackscore{ijk} u\hackscore{k}\right)+d\hackscore{ik} u\hackscore{k}=n\hackscore{j}X\hackscore{ij}+y\hackscore{i}  \;.  n\hackscore{j} A\hackscore{ijkl} u\hackscore{k,l}=n\hackscore{j}X\hackscore{ij} \;.
61  \end{equation}  \end{equation}
62  The coefficient $d$ is a \RankTwo and $y$ is a    Constraints \index{constraint} take the form
63  \RankOne both in the \FunctionOnBoundary. Constraints \index{constraint} take the form  \begin{equation}\label{LINEARPDE.SYSTEM.3 TUTORIAL}
 \begin{equation}\label{LINEARPDE.SYSTEM.3}  
64  u\hackscore{i}=r\hackscore{i} \mbox{ where } q\hackscore{i}>0  u\hackscore{i}=r\hackscore{i} \mbox{ where } q\hackscore{i}>0
65  \end{equation}  \end{equation}
66  $r$ and $q$ are each \RankOne.  $r$ and $q$ are each \RankOne.
67  We can easily identify the coefficients in~\eqn{LINEARPDE.SYSTEM.1}:  We can easily identify the coefficients in~\eqn{LINEARPDE.SYSTEM.1 TUTORIAL}:
68  \begin{eqnarray}\label{LINEARPDE ELASTIC COEFFICIENTS}  \begin{eqnarray}\label{LINEARPDE ELASTIC COEFFICIENTS}
69  A\hackscore{ijkl}=\lambda \delta\hackscore{ij} \delta\hackscore{kl} + \mu (  A\hackscore{ijkl}=\lambda \delta\hackscore{ij} \delta\hackscore{kl} + \mu (
70  +\delta\hackscore{ik} \delta\hackscore{jl}  +\delta\hackscore{ik} \delta\hackscore{jl}
# Line 78  are constant setting $Y\hackscore{i}=\la Line 85  are constant setting $Y\hackscore{i}=\la
85  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
86  in~\eqn{HEATEDBLOCK linear elastic} to zero.  in~\eqn{HEATEDBLOCK linear elastic} to zero.
87    
88  Analogously to concept of symmetry for a single PDE, we call the PDE defined by~\eqn{LINEARPDE.SYSTEM.1} symmetric if  Analogously to concept of symmetry for a single PDE, we call the PDE defined by~\eqn{LINEARPDE.SYSTEM.1 TUTORIAL} symmetric if
89  \index{symmetric PDE}  \index{symmetric PDE}
90  \begin{eqnarray}\label{LINEARPDE.SYSTEM.SYMMETRY}  \begin{eqnarray}\label{LINEARPDE.SYSTEM.SYMMETRY TUTORIAL}
91  A\hackscore{ijkl} =A\hackscore{klij} \\  A\hackscore{ijkl} =A\hackscore{klij} \\
 B\hackscore{ijk}=C\hackscore{kij} \\  
 D\hackscore{ik}=D\hackscore{ki} \\  
 d\hackscore{ik}=d\hackscore{ki} \  
92  \end{eqnarray}  \end{eqnarray}
93  Note that different from the scalar case now the coefficients $D$ and $d$ have to be inspected. It is easy to see that  Note that different from the scalar case now the coefficients $D$ and $d$ have to be inspected. It is easy to see that
94  the Lame equation in fact is symmetric. The \LinearPDE class is notified by this fact by calling its \method{setSymmetryOn} method.  the Lame equation in fact is symmetric. The \LinearPDE class is notified by this fact by calling its \method{setSymmetryOn} method.

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