# Diff of /trunk/doc/user/heatedblock.tex

revision 757 by woo409, Mon Jun 26 13:12:56 2006 UTC revision 1388 by trankine, Fri Jan 11 07:45:58 2008 UTC
# Line 1  Line 1
1    %
2  % $Id$  % $Id$
3  %  %
5  %               \url{http://www.access.edu.au  %
6  %         Primary Business: Queensland, Australia.  %           Copyright 2003-2007 by ACceSS MNRF
7  %   Licensed under the Open Software License version 3.0  %       Copyright 2007 by University of Queensland
9    %                http://esscc.uq.edu.au
10    %        Primary Business: Queensland, Australia
13  %  %
14    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
15    %
16
17  \section{Elastic Deformation}  \section{Elastic Deformation}
18  \label{ELASTIC CHAP}  \label{ELASTIC CHAP}
19  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
20  to displacement field $u\hackscore{i}$ which solves the momentum equation  a displacement field $u\hackscore{i}$ which solves the momentum equation
21  \index{momentum equation}:  \index{momentum equation}:
22  \begin{eqnarray}\label{HEATEDBLOCK general problem}  \begin{eqnarray}\label{HEATEDBLOCK general problem}
23   - \sigma\hackscore{ij,j}=0   - \sigma\hackscore{ij,j}=0
# Line 44  For the temperature distribution we use Line 52  For the temperature distribution we use
52  T(x)= T\hackscore{0} e^{-\beta \|x-x^{c}\|};  T(x)= T\hackscore{0} e^{-\beta \|x-x^{c}\|};
53  \end{eqnarray}  \end{eqnarray}
54  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
55  $T$ from an time-dependent temperature diffusion problem as discussed in \Sec{DIFFUSION CHAP}.  $T$ from a time-dependent temperature diffusion problem as discussed in \Sec{DIFFUSION CHAP}.
56
57  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
58  the Lame equation\index{Lame equation}. We want to solve  the Lame equation\index{Lame equation}. We want to solve
59  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
# Line 54  takes PDEs of the form Line 62  takes PDEs of the form
62  -A\hackscore{ijkl} u\hackscore{k,l}){,j}=-X\hackscore{ij,j} \; .  -A\hackscore{ijkl} u\hackscore{k,l}){,j}=-X\hackscore{ij,j} \; .
63
64  $A$ 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
65  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}.
66  The natural boundary conditions \index{boundary condition!natural} take the form:  The natural boundary conditions \index{boundary condition!natural} take the form:
67  \label{LINEARPDE.SYSTEM.2 TUTORIAL}  \label{LINEARPDE.SYSTEM.2 TUTORIAL}
68  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} \;.
# Line 81  q\hackscore{i}(x)=\left\{ Line 89  q\hackscore{i}(x)=\left\{
89  \right.  \right.
90
91  Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$  Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$
92  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,
93  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
94  in~\eqn{HEATEDBLOCK linear elastic} to zero.  in~\eqn{HEATEDBLOCK linear elastic} to zero.
95
# Line 90  Analogously to concept of symmetry for a Line 98  Analogously to concept of symmetry for a
98  \begin{eqnarray}\label{LINEARPDE.SYSTEM.SYMMETRY TUTORIAL}  \begin{eqnarray}\label{LINEARPDE.SYSTEM.SYMMETRY TUTORIAL}
99  A\hackscore{ijkl} =A\hackscore{klij} \\  A\hackscore{ijkl} =A\hackscore{klij} \\
100  \end{eqnarray}  \end{eqnarray}
101  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.
102  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.
103
104  After we have solved the Lame equation we want to analyse the actual stress distribution. Typically the von--Mises stress\index{von--Mises stress} defined by  After we have solved the Lame equation we want to analyse the actual stress distribution. Typically the von--Mises stress\index{von--Mises stress} defined by
105
# Line 101  After we have solved the Lame equation w Line 109  After we have solved the Lame equation w
109                 (\sigma\hackscore{22}-\sigma\hackscore{00})^2)                 (\sigma\hackscore{22}-\sigma\hackscore{00})^2)
110  +  \sigma\hackscore{01}^2+\sigma\hackscore{12}^2+\sigma\hackscore{20}^2}  +  \sigma\hackscore{01}^2+\sigma\hackscore{12}^2+\sigma\hackscore{20}^2}
111
112  is used to detect material damages. Here we want to calculate the von--Mises and write the stress to a file for visualization.  is used to detect material damage. Here we want to calculate the von--Mises and write the stress to a file for visualization.
113
114  \index{scripts!\file{diffusion.py}}:  \index{scripts!\file{diffusion.py}}
115  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
116  and writes the displacements and the von--Mises stress\index{von--Mises stress} into a file \file{deform.xml} in the \VTK file format:  and writes the displacements and the von--Mises stress\index{von--Mises stress} into a file \file{deform.xml} in the \VTK file format:
117  \begin{python}  \begin{python}
118  from esys.escript import *  from esys.escript import *

Legend:
 Removed from v.757 changed lines Added in v.1388