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Tue Mar 7 01:28:23 2006 UTC (14 years, 6 months ago) by gross
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new section on lame equation completed
1 % $Id$
2 \section{Elastic Deformation}
3 \label{ELASTIC CHAP}
4 In this section we want to discuss the deformation of a linear elastic body caused by expansion through a heat distribution. We want
5 to displacement field $u\hackscore{i}$ which solves the momentum equation
6 \index{momentum equation}:
7 \begin{eqnarray}\label{HEATEDBLOCK general problem}
8 - \sigma\hackscore{ij,j}=0
9 \end{eqnarray}
10 where the stress $\sigma$ is given by
11 \begin{eqnarray}\label{HEATEDBLOCK linear elastic}
12 \sigma\hackscore{ij}= \lambda u\hackscore{k,k} \delta\hackscore{ij} + \mu ( u\hackscore{i,j} + u\hackscore{j,i})
13 - (\lambda+\frac{2}{3} \mu) \; \alpha \; (T-T\hackscore{ref})\delta\hackscore{ij} \;.
14 \end{eqnarray}
15 In this formula $\lambda$ and $\mu$ are the Lame coefficients, $\alpha$ is the
16 temperature expansion coefficient, $T$ is the temperature distribution and $T_{ref}$ a reference temperature. Note that
17 \eqn{HEATEDBLOCK general problem} is similar to eqn{WAVE general problem} introduced in section~\Sec{WAVE CHAP} but the
18 inertia term $\rho u\hackscore{i,tt}$ has been dropped as we assume a static scenario here. Moreover, in
19 comparison to the \eqn{WAVE stress}
20 definition of stress $\sigma$ in \eqn{HEATEDBLOCK linear elastic} an extra term is introduced
21 to bring in stress due to volume changes trough temperature dependent expansion.
22
23 Our domain is the unit cube
24 \begin{eqnarray} \label{HEATEDBLOCK natural}
25 \Omega=\{(x\hackscore{i} | 0 \le x\hackscore{i} \le 1 \}
26 \end{eqnarray}
27 On the boundary the normal stress component is set to zero
28 \begin{eqnarray} \label{HEATEDBLOCK natural}
29 \sigma\hackscore{ij}n\hackscore{j}=0
30 \end{eqnarray}
31 and on the face with $x\hackscore{i}=0$ we set the $i$-th component of the displacement to $0$
32 \begin{eqnarray} \label{HEATEDBLOCK constraint}
33 u\hackscore{i}(x)=0 & \mbox{ where } & x\hackscore{i}=0 \;
34 \end{eqnarray}
35 For the temperature distribution we use
36 \begin{eqnarray} \label{HEATEDBLOCK temperature}
37 T(x)= T\hackscore{0} e^{-\beta \|x-x^{c}\|};
38 \end{eqnarray}
39 with a given positive constant $\beta$ and location $x^{c}$ in the domain. Later in \Sec{MODELFRAME} we will use
40 $T$ from an time-dependent temperature diffusion problem as discussed in \Sec{DIFFUSION CHAP}.
41
42 When we insert~\eqn{HEATEDBLOCK linear elastic} we get a second oder system of linear PDEs for the displacements $u$ which is called
43 the Lame equation\index{Lame equation}. We want to solve
44 this using the \LinearPDE class to this. For a system of PDEs and a solution with several components the \LinearPDE class
45 takes PDEs of the form
46 \begin{equation}\label{LINEARPDE.SYSTEM.1}
47 -(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} \; .
48 \end{equation}
49 $A$ is a \RankFour, $B$ and $C$ are each a \RankThree, $D$ and $X$ are each a \RankTwo and $Y$ is a \RankOne.
50 The natural boundary conditions \index{boundary condition!natural} take the form:
51 \begin{equation}\label{LINEARPDE.SYSTEM.2}
52 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} \;.
53 \end{equation}
54 The coefficient $d$ is a \RankTwo and $y$ is a
55 \RankOne both in the \FunctionOnBoundary. Constraints \index{constraint} take the form
56 \begin{equation}\label{LINEARPDE.SYSTEM.3}
57 u\hackscore{i}=r\hackscore{i} \mbox{ where } q\hackscore{i}>0
58 \end{equation}
59 $r$ and $q$ are each \RankOne.
60 We can easily identify the coefficients in~\eqn{LINEARPDE.SYSTEM.1}:
61 \begin{eqnarray}\label{LINEARPDE ELASTIC COEFFICIENTS}
62 A\hackscore{ijkl}=\lambda \delta\hackscore{ij} \delta\hackscore{kl} + \mu (
63 +\delta\hackscore{ik} \delta\hackscore{jl}
64 \delta\hackscore{il} \delta\hackscore{jk}) \\
65 X\hackscore{ij}=(\lambda+\frac{2}{3} \mu) \; \alpha \; (T-T\hackscore{ref})\delta\hackscore{ij} \\
66 \end{eqnarray}
67 The characteristic function $q$ defining the locations and components where constraints are set is given by:
68 \begin{equation}\label{HEATEDBLOCK MASK}
69 q\hackscore{i}(x)=\left\{
70 \begin{array}{cl}
71 1 & x\hackscore{i}=0 \\
72 0 & \mbox{otherwise} \\
73 \end{array}
74 \right.
75 \end{equation}
76 Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$
77 are constant setting $Y\hackscore{i}=\lambda+\frac{2}{3} \mu) \; \alpha \; T\hackscore{i}$ seems to be also possible. However,
78 this choice would lead to a different natural boundary condition which does not set the normal stress component as defined
79 in~\eqn{HEATEDBLOCK linear elastic} to zero.
80
81 Analogously to concept of symmetry for a single PDE, we call the PDE defined by~\eqn{LINEARPDE.SYSTEM.1} symmetric if
82 \index{symmetric PDE}
83 \begin{eqnarray}\label{LINEARPDE.SYSTEM.SYMMETRY}
84 A\hackscore{ijkl} =A\hackscore{klij} \\
85 B\hackscore{ijk}=C\hackscore{kij} \\
86 D\hackscore{ik}=D\hackscore{ki} \\
87 d\hackscore{ik}=d\hackscore{ki} \
88 \end{eqnarray}
89 Note that different from the scalar case now the coefficients $D$ and $d$ have to be inspected. It is easy to see that
90 the Lame equation in fact is symmetric. The \LinearPDE class is notified by this fact by calling its \method{setSymmetryOn} method.
91
92 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
93 \begin{equation}
94 \sigma\hackscore{mises} = \sqrt{
95 \frac{1}{6} ((\sigma\hackscore{00}-\sigma\hackscore{11})^2+
96 (\sigma\hackscore{11}-\sigma\hackscore{22})^2
97 (\sigma\hackscore{22}-\sigma\hackscore{00})^2)
98 + \sigma\hackscore{01}^2+\sigma\hackscore{12}^2+\sigma\hackscore{20}^2}
99 \end{equation}
100 is used to detect material damages. Here we want to calculate the von--Mises and write the stress to a file for visualization.
101
102 \index{scripts!\file{diffusion.py}}:
103 The following script which is available in \file{heatedbox.py} in the \ExampleDirectory solves the Lame equation
104 and writes the displacements and the von--Mises stress\index{von--Mises stress} into a file \file{deform.xml} in the \VTK file format:
105 \begin{python}
106 from esys.escript import *
107 from esys.escript.linearPDEs import LinearPDE
108 from esys.finley import Brick
109 #... set some parameters ...
110 lam=1.
111 mu=0.1
112 alpha=1.e-6
113 xc=[0.3,0.3,1.]
114 beta=8.
115 T_ref=0.
116 T_0=1.
117 #... generate domain ...
118 mydomain = Brick(l0=1.,l1=1., l2=1.,n0=10, n1=10, n2=10)
119 x=mydomain.getX()
120 #... set temperature ...
121 T=T_0*exp(-beta*length(x-xc))
122 #... open symmetric PDE ...
123 mypde=LinearPDE(mydomain)
124 mypde.setSymmetryOn()
125 #... set coefficients ...
126 C=Tensor4(0.,Function(mydomain))
127 for i in range(mydomain.getDim()):
128 for j in range(mydomain.getDim()):
129 C[i,i,j,j]+=lam
130 C[j,i,j,i]+=mu
131 C[j,i,i,j]+=mu
132 msk=whereZero(x[0])*[1.,0.,0.] \
133 +whereZero(x[1])*[0.,1.,0.] \
134 +whereZero(x[2])*[0.,0.,1.]
135 sigma0=(lam+2./3.*mu)*alpha*(T-T_ref)*kronecker(mydomain)
136 mypde.setValue(A=C,X=sigma0,q=msk)
137 #... solve pde ...
138 u=mypde.getSolution()
139 #... calculate von-Misses stress
140 g=grad(u)
141 sigma=mu*(g+transpose(g))+lam*trace(g)*kronecker(mydomain)-sigma0
142 sigma_mises=sqrt(((sigma[0,0]-sigma[1,1])**2+(sigma[1,1]-sigma[2,2])**2+ \
143 (sigma[2,2]-sigma[0,0])**2)/6. \
144 +sigma[0,1]**2 + sigma[1,2]**2 + sigma[2,0]**2)
145 #... output ...
146 saveVTK("deform.xml",disp=u,stress=sigma_mises)
147 \end{python}
148
149 \begin{figure}
150 \centerline{\includegraphics[width=\figwidth]{HeatedBlock}}
151 \caption{von--Mises Stress and Displacement Vectors.}
152 \label{HEATEDBLOCK FIG 2}
153 \end{figure}
154
155 Finally the the results can be visualize by calling
156 \begin{python}
157 mayavi -d deform.xml -f CellToPointData -m VelocityVector -m SurfaceMap &
158 \end{python}
159 Note that the filter \text{CellToPointData} is applied to create smooth representation of the
160 von--Mises stress. \fig{HEATEDBLOCK FIG 2} shows the results where the vertical planes showing the
161 von--Mises stress and the horizontal plane shows the vector representing displacements.
162

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