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1
2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 %
4 % Copyright (c) 2003-2008 by University of Queensland
5 % Earth Systems Science Computational Center (ESSCC)
6 % http://www.uq.edu.au/esscc
7 %
8 % Primary Business: Queensland, Australia
9 % Licensed under the Open Software License version 3.0
10 % http://www.opensource.org/licenses/osl-3.0.php
11 %
12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13
14
15 \section{Elastic Deformation}
16 \label{ELASTIC CHAP}
17 In this section we want to examine the deformation of a linear elastic body caused by expansion through a heat distribution. We want
18 a displacement field $u\hackscore{i}$ which solves the momentum equation
19 \index{momentum equation}:
20 \begin{eqnarray}\label{HEATEDBLOCK general problem}
21 - \sigma\hackscore{ij,j}=0
22 \end{eqnarray}
23 where the stress $\sigma$ is given by
24 \begin{eqnarray}\label{HEATEDBLOCK linear elastic}
25 \sigma\hackscore{ij}= \lambda u\hackscore{k,k} \delta\hackscore{ij} + \mu ( u\hackscore{i,j} + u\hackscore{j,i})
26 - (\lambda+\frac{2}{3} \mu) \; \alpha \; (T-T\hackscore{ref})\delta\hackscore{ij} \;.
27 \end{eqnarray}
28 In this formula $\lambda$ and $\mu$ are the Lame coefficients, $\alpha$ is the
29 temperature expansion coefficient, $T$ is the temperature distribution and $T_{ref}$ a reference temperature. Note that
30 \eqn{HEATEDBLOCK general problem} is similar to eqn{WAVE general problem} introduced in section~\Sec{WAVE CHAP} but the
31 inertia term $\rho u\hackscore{i,tt}$ has been dropped as we assume a static scenario here. Moreover, in
32 comparison to the \eqn{WAVE stress}
33 definition of stress $\sigma$ in \eqn{HEATEDBLOCK linear elastic} an extra term is introduced
34 to bring in stress due to volume changes trough temperature dependent expansion.
35
36 Our domain is the unit cube
37 \begin{eqnarray} \label{HEATEDBLOCK natural location}
38 \Omega=\{(x\hackscore{i} | 0 \le x\hackscore{i} \le 1 \}
39 \end{eqnarray}
40 On the boundary the normal stress component is set to zero
41 \begin{eqnarray} \label{HEATEDBLOCK natural}
42 \sigma\hackscore{ij}n\hackscore{j}=0
43 \end{eqnarray}
44 and on the face with $x\hackscore{i}=0$ we set the $i$-th component of the displacement to $0$
45 \begin{eqnarray} \label{HEATEDBLOCK constraint}
46 u\hackscore{i}(x)=0 & \mbox{ where } & x\hackscore{i}=0 \;
47 \end{eqnarray}
48 For the temperature distribution we use
49 \begin{eqnarray} \label{HEATEDBLOCK temperature}
50 T(x)= T\hackscore{0} e^{-\beta \|x-x^{c}\|};
51 \end{eqnarray}
52 with a given positive constant $\beta$ and location $x^{c}$ in the domain. Later in \Sec{MODELFRAME} we will use
53 $T$ from a time-dependent temperature diffusion problem as discussed in \Sec{DIFFUSION CHAP}.
54
55 When we insert~\eqn{HEATEDBLOCK linear elastic} we get a second order system of linear PDEs for the displacements $u$ which is called
56 the Lame equation\index{Lame equation}. We want to solve
57 this using the \LinearPDE class to this. For a system of PDEs and a solution with several components the \LinearPDE class
58 takes PDEs of the form
59 \begin{equation}\label{LINEARPDE.SYSTEM.1 TUTORIAL}
60 -A\hackscore{ijkl} u\hackscore{k,l}){,j}=-X\hackscore{ij,j} \; .
61 \end{equation}
62 $A$ is of \RankFour and $X$ is of \RankTwo. We show here the coefficients relevant
63 for the we trying to solve. The full form is given in~\eqn{LINEARPDE.SYSTEM.1}.
64 The natural boundary conditions \index{boundary condition!natural} take the form:
65 \begin{equation}\label{LINEARPDE.SYSTEM.2 TUTORIAL}
66 n\hackscore{j} A\hackscore{ijkl} u\hackscore{k,l}=n\hackscore{j}X\hackscore{ij} \;.
67 \end{equation}
68 Constraints \index{constraint} take the form
69 \begin{equation}\label{LINEARPDE.SYSTEM.3 TUTORIAL}
70 u\hackscore{i}=r\hackscore{i} \mbox{ where } q\hackscore{i}>0
71 \end{equation}
72 $r$ and $q$ are each \RankOne.
73 We can easily identify the coefficients in~\eqn{LINEARPDE.SYSTEM.1 TUTORIAL}:
74 \begin{eqnarray}\label{LINEARPDE ELASTIC COEFFICIENTS}
75 A\hackscore{ijkl}=\lambda \delta\hackscore{ij} \delta\hackscore{kl} + \mu (
76 \delta\hackscore{ik} \delta\hackscore{jl}
77 + \delta\hackscore{il} \delta\hackscore{jk}) \\
78 X\hackscore{ij}=(\lambda+\frac{2}{3} \mu) \; \alpha \; (T-T\hackscore{ref})\delta\hackscore{ij} \\
79 \end{eqnarray}
80 The characteristic function $q$ defining the locations and components where constraints are set is given by:
81 \begin{equation}\label{HEATEDBLOCK MASK}
82 q\hackscore{i}(x)=\left\{
83 \begin{array}{cl}
84 1 & x\hackscore{i}=0 \\
85 0 & \mbox{otherwise} \\
86 \end{array}
87 \right.
88 \end{equation}
89 Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$
90 are constant we may use $Y\hackscore{i}=\lambda+\frac{2}{3} \mu) \; \alpha \; T\hackscore{i}$. However,
91 this choice would lead to a different natural boundary condition which does not set the normal stress component as defined
92 in~\eqn{HEATEDBLOCK linear elastic} to zero.
93
94 Analogously to concept of symmetry for a single PDE, we call the PDE defined by~\eqn{LINEARPDE.SYSTEM.1 TUTORIAL} symmetric if
95 \index{symmetric PDE}
96 \begin{eqnarray}\label{LINEARPDE.SYSTEM.SYMMETRY TUTORIAL}
97 A\hackscore{ijkl} =A\hackscore{klij} \\
98 \end{eqnarray}
99 This Lame equation is in fact symmetric, given the difference in $D$ and $d$ as compared to the scalar case.
100 The \LinearPDE class is notified of this fact by calling its \method{setSymmetryOn} method.
101
102 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
103 \begin{equation}
104 \sigma\hackscore{mises} = \sqrt{
105 \frac{1}{6} ((\sigma\hackscore{00}-\sigma\hackscore{11})^2
106 + (\sigma\hackscore{11}-\sigma\hackscore{22})^2
107 + (\sigma\hackscore{22}-\sigma\hackscore{00})^2)
108 + \sigma\hackscore{01}^2+\sigma\hackscore{12}^2+\sigma\hackscore{20}^2}
109 \end{equation}
110 is used to detect material damage. Here we want to calculate the von--Mises and write the stress to a file for visualization.
111
112 \index{scripts!\file{diffusion.py}}
113 The following script, which is available in \file{heatedbox.py} in the \ExampleDirectory, solves the Lame equation
114 and writes the displacements and the von--Mises stress\index{von--Mises stress} into a file \file{deform.xml} in the \VTK file format:
115 \begin{python}
116 from esys.escript import *
117 from esys.escript.linearPDEs import LinearPDE
118 from esys.finley import Brick
119 #... set some parameters ...
120 lam=1.
121 mu=0.1
122 alpha=1.e-6
123 xc=[0.3,0.3,1.]
124 beta=8.
125 T_ref=0.
126 T_0=1.
127 #... generate domain ...
128 mydomain = Brick(l0=1.,l1=1., l2=1.,n0=10, n1=10, n2=10)
129 x=mydomain.getX()
130 #... set temperature ...
131 T=T_0*exp(-beta*length(x-xc))
132 #... open symmetric PDE ...
133 mypde=LinearPDE(mydomain)
134 mypde.setSymmetryOn()
135 #... set coefficients ...
136 C=Tensor4(0.,Function(mydomain))
137 for i in range(mydomain.getDim()):
138 for j in range(mydomain.getDim()):
139 C[i,i,j,j]+=lam
140 C[j,i,j,i]+=mu
141 C[j,i,i,j]+=mu
142 msk=whereZero(x[0])*[1.,0.,0.] \
143 +whereZero(x[1])*[0.,1.,0.] \
144 +whereZero(x[2])*[0.,0.,1.]
145 sigma0=(lam+2./3.*mu)*alpha*(T-T_ref)*kronecker(mydomain)
146 mypde.setValue(A=C,X=sigma0,q=msk)
147 #... solve pde ...
148 u=mypde.getSolution()
149 #... calculate von-Misses stress
150 g=grad(u)
151 sigma=mu*(g+transpose(g))+lam*trace(g)*kronecker(mydomain)-sigma0
152 sigma_mises=sqrt(((sigma[0,0]-sigma[1,1])**2+(sigma[1,1]-sigma[2,2])**2+ \
153 (sigma[2,2]-sigma[0,0])**2)/6. \
154 +sigma[0,1]**2 + sigma[1,2]**2 + sigma[2,0]**2)
155 #... output ...
156 saveVTK("deform.xml",disp=u,stress=sigma_mises)
157 \end{python}
158
159 \begin{figure}
160 \centerline{\includegraphics[width=\figwidth]{figures/HeatedBlock.eps}}
161 \caption{von--Mises Stress and Displacement Vectors.}
162 \label{HEATEDBLOCK FIG 2}
163 \end{figure}
164
165 Finally the the results can be visualize by calling
166 \begin{python}
167 mayavi -d deform.xml -f CellToPointData -m VelocityVector -m SurfaceMap &
168 \end{python}
169 Note that the filter \text{CellToPointData} is applied to create smooth representation of the
170 von--Mises stress. \fig{HEATEDBLOCK FIG 2} shows the results where the vertical planes showing the
171 von--Mises stress and the horizontal plane shows the vector representing displacements.
172

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