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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 % $Id$
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}
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 \begin{eqnarray}\label{HEATEDBLOCK general problem}
15 - \sigma\hackscore{ij,j}=0
16 \end{eqnarray}
17 where the stress $\sigma$ is given by
18 \begin{eqnarray}\label{HEATEDBLOCK linear elastic}
19 \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 \eqn{HEATEDBLOCK general problem} is similar to eqn{WAVE general problem} introduced in section~\Sec{WAVE CHAP} but the
25 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 definition of stress $\sigma$ in \eqn{HEATEDBLOCK linear elastic} an extra term is introduced
28 to bring in stress due to volume changes trough temperature dependent expansion.
29
30 Our domain is the unit cube
31 \begin{eqnarray} \label{HEATEDBLOCK natural location}
32 \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 \begin{eqnarray} \label{HEATEDBLOCK natural}
36 \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 \begin{eqnarray} \label{HEATEDBLOCK constraint}
40 u\hackscore{i}(x)=0 & \mbox{ where } & x\hackscore{i}=0 \;
41 \end{eqnarray}
42 For the temperature distribution we use
43 \begin{eqnarray} \label{HEATEDBLOCK temperature}
44 T(x)= T\hackscore{0} e^{-\beta \|x-x^{c}\|};
45 \end{eqnarray}
46 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
49 When we insert~\eqn{HEATEDBLOCK linear elastic} we get a second oder system of linear PDEs for the displacements $u$ which is called
50 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 \begin{equation}\label{LINEARPDE.SYSTEM.1 TUTORIAL}
54 -A\hackscore{ijkl} u\hackscore{k,l}){,j}=-X\hackscore{ij,j} \; .
55 \end{equation}
56 $A$ 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:
59 \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 \end{equation}
62 Constraints \index{constraint} take the form
63 \begin{equation}\label{LINEARPDE.SYSTEM.3 TUTORIAL}
64 u\hackscore{i}=r\hackscore{i} \mbox{ where } q\hackscore{i}>0
65 \end{equation}
66 $r$ and $q$ are each \RankOne.
67 We can easily identify the coefficients in~\eqn{LINEARPDE.SYSTEM.1 TUTORIAL}:
68 \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 \begin{equation}\label{HEATEDBLOCK MASK}
76 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 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 TUTORIAL} symmetric if
89 \index{symmetric PDE}
90 \begin{eqnarray}\label{LINEARPDE.SYSTEM.SYMMETRY TUTORIAL}
91 A\hackscore{ijkl} =A\hackscore{klij} \\
92 \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
94 the Lame equation in fact is symmetric. The \LinearPDE class is notified by this fact by calling its \method{setSymmetryOn} method.
95
96 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
106 \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 \centerline{\includegraphics[width=\figwidth]{figures/HeatedBlock.eps}}
155 \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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