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

Revision 578 - (show annotations)
Mon Mar 6 06:12:04 2006 UTC (14 years, 5 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. 22 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}.} 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 \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. 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 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. 92 93

## Properties

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