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revision 579 by gross, Tue Mar 7 01:28:23 2006 UTC revision 757 by woo409, Mon Jun 26 13:12:56 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 21  definition of stress $\sigma$ in \eqn{HE Line 28  definition of stress $\sigma$ in \eqn{HE
28  to bring in stress due to volume changes trough temperature dependent expansion.    to bring in stress due to volume changes trough temperature dependent expansion.  
29    
30  Our domain is the unit cube  Our domain is the unit cube
31  \begin{eqnarray} \label{HEATEDBLOCK natural}  \begin{eqnarray} \label{HEATEDBLOCK natural location}
32  \Omega=\{(x\hackscore{i} | 0 \le x\hackscore{i} \le 1 \}  \Omega=\{(x\hackscore{i} | 0 \le x\hackscore{i} \le 1 \}
33  \end{eqnarray}  \end{eqnarray}
34  On the boundary the normal stress component is set to zero  On the boundary the normal stress component is set to zero
# 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}(A\hackscore{ijkl} u\hackscore{k,l})\hackscore{,j}+(B\hackscore{ijk} u\hackscore{k})+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.
# Line 147  saveVTK("deform.xml",disp=u,stress=sigma Line 151  saveVTK("deform.xml",disp=u,stress=sigma
151  \end{python}  \end{python}
152    
153  \begin{figure}  \begin{figure}
154  \centerline{\includegraphics[width=\figwidth]{HeatedBlock}}  \centerline{\includegraphics[width=\figwidth]{figures/HeatedBlock.eps}}
155  \caption{von--Mises Stress and Displacement Vectors.}  \caption{von--Mises Stress and Displacement Vectors.}
156  \label{HEATEDBLOCK FIG 2}  \label{HEATEDBLOCK FIG 2}
157  \end{figure}  \end{figure}
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