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Mon Mar 6 06:12:04 2006 UTC (13 years, 11 months ago) by gross
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new section in the tutorial
1 jgs 110 % $Id$
2 gross 578 \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{BEAM general problem}
8     - \sigma\hackscore{ij,j}=0
9     \end{eqnarray}
10     where the stress $\sigma$ is given by
11     \begin{eqnarray}\label{BEAM 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{BEAM 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{BEAM linear elastic} an extra term is introduced
21     to bring in stress due to volume changes trough temperature dependent expansion.
22 gross 568
23 gross 578 Our domain is the unit cube
24     \begin{eqnarray} \label{BEAM 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{BEAM 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{BEAM 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{BEAM temperature}
37     T(x)= \frac{\beta}{\|x-x^{c}\|};
38     \end{eqnarray}
39     with a given positive constant $\beta$ and location $x^{c}$ in the domain\footnote{This selection of $T$ corresponds to
40     a temperature distribution in an indefinite domain created by a nodal heat source at $x^{c}$. Later in \Sec{X} we will calculate
41     the $T$ from an time-dependent temperature diffusion problem as discussed in \Sec{DIFFUSION CHAP}.}
42    
43     When we insert~\eqn{BEAM linear elastic} we get a second oder system of linear PDEs for the displacements $u$ which is called
44     the Lame equation\index{Lame equation}. We want to solve
45     this using the \LinearPDE class to this. For a system of PDEs and a solution with several components the \LinearPDE class
46     takes PDEs of the form
47 gross 568 \begin{equation}\label{LINEARPDE.SYSTEM.1}
48 gross 578 -(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} \; .
49 gross 568 \end{equation}
50     $A$ is a \RankFour, $B$ and $C$ are each a \RankThree, $D$ and $X$ are each a \RankTwo and $Y$ is a \RankOne.
51     The natural boundary conditions \index{boundary condition!natural} take the form:
52     \begin{equation}\label{LINEARPDE.SYSTEM.2}
53 gross 578 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} \;.
54 gross 568 \end{equation}
55     The coefficient $d$ is a \RankTwo and $y$ is a
56     \RankOne both in the \FunctionOnBoundary. Constraints \index{constraint} take the form
57     \begin{equation}\label{LINEARPDE.SYSTEM.3}
58     u\hackscore{i}=r\hackscore{i} \mbox{ where } q\hackscore{i}>0
59     \end{equation}
60 gross 578 $r$ and $q$ are each \RankOne.
61     We can easily identify the coefficients in~\eqn{LINEARPDE.SYSTEM.1}:
62     \begin{eqnarray}\label{LINEARPDE ELASTIC COEFFICIENTS}
63     A\hackscore{ijkl}=\lambda \delta\hackscore{ij} \delta\hackscore{kl} + \mu (
64     +\delta\hackscore{ik} \delta\hackscore{jl}
65     \delta\hackscore{il} \delta\hackscore{jk}) \\
66     X\hackscore{ij}=(\lambda+\frac{2}{3} \mu) \; \alpha \; (T-T\hackscore{ref})\delta\hackscore{ij} \\
67     \end{eqnarray}
68     The characteristic function $q$ defining the locations and components where constraints are set is given by:
69     \begin{equation}\label{BEAM MASK}
70     q\hackscore{i}(x)=\left\{
71     \begin{array}{cl}
72     1 & x\hackscore{i}=0 \\
73     0 & \mbox{otherwise} \\
74     \end{array}
75     \right.
76     \end{equation}
77     Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$
78     are constant setting $Y\hackscore{i}=\lambda+\frac{2}{3} \mu) \; \alpha \; T\hackscore{i}$ seems to be also possible. However,
79     this choice would lead to a different natural boundary condition which does not set the normal stress component as defined
80     in~\eqn{BEAM linear elastic} to zero.
81    
82     Analogously to concept of symmetry for a single PDE, we call the PDE defined by~\eqn{LINEARPDE.SYSTEM.1} symmetric if
83     \index{symmetric PDE}
84     \begin{eqnarray}\label{LINEARPDE.SYSTEM.SYMMETRY}
85     A\hackscore{ijkl} =A\hackscore{klij} \\
86 gross 568 B\hackscore{ijk}=C\hackscore{kij} \\
87     D\hackscore{ik}=D\hackscore{ki} \\
88     d\hackscore{ik}=d\hackscore{ki} \
89     \end{eqnarray}
90 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
91     the Lame equation in fact is symmetric.
92 gross 568
93    

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