ViewVC logotype

Contents of /trunk/doc/user/heatedblock.tex

Parent Directory Parent Directory | Revision Log Revision Log

Revision 578 - (show annotations)
Mon Mar 6 06:12:04 2006 UTC (14 years, 4 months ago) by gross
File MIME type: application/x-tex
File size: 5158 byte(s)
new section in the tutorial
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{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.
23 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}.}
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 \begin{equation}\label{LINEARPDE.SYSTEM.1}
48 -(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 \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 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 \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 $r$ 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.
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 B\hackscore{ijk}=C\hackscore{kij} \\
87 D\hackscore{ik}=D\hackscore{ki} \\
88 d\hackscore{ik}=d\hackscore{ki} \
89 \end{eqnarray}
90 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.


Name Value
svn:eol-style native
svn:keywords Author Date Id Revision

  ViewVC Help
Powered by ViewVC 1.1.26