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+ Merge of intelc_win32 branch (revision 741:755) with trunk. Tested on iVEC altix (run_tests and py_tests all pass)

1 jgs 110 % $Id$
2 gross 625 %
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 gross 578 \section{Elastic Deformation}
10     \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
12     to displacement field $u\hackscore{i}$ which solves the momentum equation
13     \index{momentum equation}:
14 gross 579 \begin{eqnarray}\label{HEATEDBLOCK general problem}
15 gross 578 - \sigma\hackscore{ij,j}=0
16     \end{eqnarray}
17     where the stress $\sigma$ is given by
18 gross 579 \begin{eqnarray}\label{HEATEDBLOCK linear elastic}
19 gross 578 \sigma\hackscore{ij}= \lambda u\hackscore{k,k} \delta\hackscore{ij} + \mu ( u\hackscore{i,j} + u\hackscore{j,i})
20     - (\lambda+\frac{2}{3} \mu) \; \alpha \; (T-T\hackscore{ref})\delta\hackscore{ij} \;.
21     \end{eqnarray}
22     In this formula $\lambda$ and $\mu$ are the Lame coefficients, $\alpha$ is the
23     temperature expansion coefficient, $T$ is the temperature distribution and $T_{ref}$ a reference temperature. Note that
24 gross 579 \eqn{HEATEDBLOCK general problem} is similar to eqn{WAVE general problem} introduced in section~\Sec{WAVE CHAP} but the
25 gross 578 inertia term $\rho u\hackscore{i,tt}$ has been dropped as we assume a static scenario here. Moreover, in
26     comparison to the \eqn{WAVE stress}
27 gross 579 definition of stress $\sigma$ in \eqn{HEATEDBLOCK linear elastic} an extra term is introduced
28 woo409 757 to bring in stress due to volume changes trough temperature dependent expansion.
29 gross 568
30 gross 578 Our domain is the unit cube
31 gross 593 \begin{eqnarray} \label{HEATEDBLOCK natural location}
32 gross 578 \Omega=\{(x\hackscore{i} | 0 \le x\hackscore{i} \le 1 \}
33     \end{eqnarray}
34     On the boundary the normal stress component is set to zero
35 gross 579 \begin{eqnarray} \label{HEATEDBLOCK natural}
36 gross 578 \sigma\hackscore{ij}n\hackscore{j}=0
37     \end{eqnarray}
38     and on the face with $x\hackscore{i}=0$ we set the $i$-th component of the displacement to $0$
39 gross 579 \begin{eqnarray} \label{HEATEDBLOCK constraint}
40 gross 578 u\hackscore{i}(x)=0 & \mbox{ where } & x\hackscore{i}=0 \;
41     \end{eqnarray}
42     For the temperature distribution we use
43 gross 579 \begin{eqnarray} \label{HEATEDBLOCK temperature}
44     T(x)= T\hackscore{0} e^{-\beta \|x-x^{c}\|};
45 gross 578 \end{eqnarray}
46 gross 579 with a given positive constant $\beta$ and location $x^{c}$ in the domain. Later in \Sec{MODELFRAME} we will use
47     $T$ from an time-dependent temperature diffusion problem as discussed in \Sec{DIFFUSION CHAP}.
48 gross 578
49 gross 579 When we insert~\eqn{HEATEDBLOCK linear elastic} we get a second oder system of linear PDEs for the displacements $u$ which is called
50 gross 578 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
52     takes PDEs of the form
53 gross 625 \begin{equation}\label{LINEARPDE.SYSTEM.1 TUTORIAL}
54     -A\hackscore{ijkl} u\hackscore{k,l}){,j}=-X\hackscore{ij,j} \; .
55 gross 568 \end{equation}
56 gross 625 $A$ 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 gross 568 The natural boundary conditions \index{boundary condition!natural} take the form:
59 gross 625 \begin{equation}\label{LINEARPDE.SYSTEM.2 TUTORIAL}
60     n\hackscore{j} A\hackscore{ijkl} u\hackscore{k,l}=n\hackscore{j}X\hackscore{ij} \;.
61 gross 568 \end{equation}
62 gross 625 Constraints \index{constraint} take the form
63     \begin{equation}\label{LINEARPDE.SYSTEM.3 TUTORIAL}
64 gross 568 u\hackscore{i}=r\hackscore{i} \mbox{ where } q\hackscore{i}>0
65     \end{equation}
66 gross 578 $r$ and $q$ are each \RankOne.
67 gross 625 We can easily identify the coefficients in~\eqn{LINEARPDE.SYSTEM.1 TUTORIAL}:
68 gross 578 \begin{eqnarray}\label{LINEARPDE ELASTIC COEFFICIENTS}
69     A\hackscore{ijkl}=\lambda \delta\hackscore{ij} \delta\hackscore{kl} + \mu (
70     +\delta\hackscore{ik} \delta\hackscore{jl}
71     \delta\hackscore{il} \delta\hackscore{jk}) \\
72     X\hackscore{ij}=(\lambda+\frac{2}{3} \mu) \; \alpha \; (T-T\hackscore{ref})\delta\hackscore{ij} \\
73     \end{eqnarray}
74     The characteristic function $q$ defining the locations and components where constraints are set is given by:
75 gross 579 \begin{equation}\label{HEATEDBLOCK MASK}
76 gross 578 q\hackscore{i}(x)=\left\{
77     \begin{array}{cl}
78     1 & x\hackscore{i}=0 \\
79     0 & \mbox{otherwise} \\
80     \end{array}
81     \right.
82     \end{equation}
83     Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$
84     are constant setting $Y\hackscore{i}=\lambda+\frac{2}{3} \mu) \; \alpha \; T\hackscore{i}$ seems to be also possible. However,
85     this choice would lead to a different natural boundary condition which does not set the normal stress component as defined
86 gross 579 in~\eqn{HEATEDBLOCK linear elastic} to zero.
87 gross 578
88 gross 625 Analogously to concept of symmetry for a single PDE, we call the PDE defined by~\eqn{LINEARPDE.SYSTEM.1 TUTORIAL} symmetric if
89 gross 578 \index{symmetric PDE}
90 gross 625 \begin{eqnarray}\label{LINEARPDE.SYSTEM.SYMMETRY TUTORIAL}
91 gross 578 A\hackscore{ijkl} =A\hackscore{klij} \\
92 gross 568 \end{eqnarray}
93 gross 578 Note that different from the scalar case now the coefficients $D$ and $d$ have to be inspected. It is easy to see that
94 gross 579 the Lame equation in fact is symmetric. The \LinearPDE class is notified by this fact by calling its \method{setSymmetryOn} method.
95 gross 568
96 gross 579 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
97     \begin{equation}
98     \sigma\hackscore{mises} = \sqrt{
99     \frac{1}{6} ((\sigma\hackscore{00}-\sigma\hackscore{11})^2+
100     (\sigma\hackscore{11}-\sigma\hackscore{22})^2
101     (\sigma\hackscore{22}-\sigma\hackscore{00})^2)
102     + \sigma\hackscore{01}^2+\sigma\hackscore{12}^2+\sigma\hackscore{20}^2}
103     \end{equation}
104     is used to detect material damages. Here we want to calculate the von--Mises and write the stress to a file for visualization.
105 gross 568
106 gross 579 \index{scripts!\file{diffusion.py}}:
107     The following script which is available in \file{heatedbox.py} in the \ExampleDirectory solves the Lame equation
108     and writes the displacements and the von--Mises stress\index{von--Mises stress} into a file \file{deform.xml} in the \VTK file format:
109     \begin{python}
110     from esys.escript import *
111     from esys.escript.linearPDEs import LinearPDE
112     from esys.finley import Brick
113     #... set some parameters ...
114     lam=1.
115     mu=0.1
116     alpha=1.e-6
117     xc=[0.3,0.3,1.]
118     beta=8.
119     T_ref=0.
120     T_0=1.
121     #... generate domain ...
122     mydomain = Brick(l0=1.,l1=1., l2=1.,n0=10, n1=10, n2=10)
123     x=mydomain.getX()
124     #... set temperature ...
125     T=T_0*exp(-beta*length(x-xc))
126     #... open symmetric PDE ...
127     mypde=LinearPDE(mydomain)
128     mypde.setSymmetryOn()
129     #... set coefficients ...
130     C=Tensor4(0.,Function(mydomain))
131     for i in range(mydomain.getDim()):
132     for j in range(mydomain.getDim()):
133     C[i,i,j,j]+=lam
134     C[j,i,j,i]+=mu
135     C[j,i,i,j]+=mu
136     msk=whereZero(x[0])*[1.,0.,0.] \
137     +whereZero(x[1])*[0.,1.,0.] \
138     +whereZero(x[2])*[0.,0.,1.]
139     sigma0=(lam+2./3.*mu)*alpha*(T-T_ref)*kronecker(mydomain)
140     mypde.setValue(A=C,X=sigma0,q=msk)
141     #... solve pde ...
142     u=mypde.getSolution()
143     #... calculate von-Misses stress
144     g=grad(u)
145     sigma=mu*(g+transpose(g))+lam*trace(g)*kronecker(mydomain)-sigma0
146     sigma_mises=sqrt(((sigma[0,0]-sigma[1,1])**2+(sigma[1,1]-sigma[2,2])**2+ \
147     (sigma[2,2]-sigma[0,0])**2)/6. \
148     +sigma[0,1]**2 + sigma[1,2]**2 + sigma[2,0]**2)
149     #... output ...
150     saveVTK("deform.xml",disp=u,stress=sigma_mises)
151     \end{python}
152    
153     \begin{figure}
154 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/HeatedBlock.eps}}
155 gross 579 \caption{von--Mises Stress and Displacement Vectors.}
156     \label{HEATEDBLOCK FIG 2}
157     \end{figure}
158    
159     Finally the the results can be visualize by calling
160     \begin{python}
161     mayavi -d deform.xml -f CellToPointData -m VelocityVector -m SurfaceMap &
162     \end{python}
163     Note that the filter \text{CellToPointData} is applied to create smooth representation of the
164     von--Mises stress. \fig{HEATEDBLOCK FIG 2} shows the results where the vertical planes showing the
165     von--Mises stress and the horizontal plane shows the vector representing displacements.
166    

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