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% $Id$ |
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\section{Elastic Deformation} |
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\label{ELASTIC CHAP} |
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In this section we want to discuss the deformation of a linear elastic body caused by expansion through a heat distribution. We want |
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to displacement field $u\hackscore{i}$ which solves the momentum equation |
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\index{momentum equation}: |
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\begin{eqnarray}\label{HEATEDBLOCK general problem} |
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- \sigma\hackscore{ij,j}=0 |
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\end{eqnarray} |
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where the stress $\sigma$ is given by |
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\begin{eqnarray}\label{HEATEDBLOCK linear elastic} |
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\sigma\hackscore{ij}= \lambda u\hackscore{k,k} \delta\hackscore{ij} + \mu ( u\hackscore{i,j} + u\hackscore{j,i}) |
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- (\lambda+\frac{2}{3} \mu) \; \alpha \; (T-T\hackscore{ref})\delta\hackscore{ij} \;. |
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\end{eqnarray} |
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In this formula $\lambda$ and $\mu$ are the Lame coefficients, $\alpha$ is the |
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temperature expansion coefficient, $T$ is the temperature distribution and $T_{ref}$ a reference temperature. Note that |
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\eqn{HEATEDBLOCK general problem} is similar to eqn{WAVE general problem} introduced in section~\Sec{WAVE CHAP} but the |
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inertia term $\rho u\hackscore{i,tt}$ has been dropped as we assume a static scenario here. Moreover, in |
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comparison to the \eqn{WAVE stress} |
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definition of stress $\sigma$ in \eqn{HEATEDBLOCK linear elastic} an extra term is introduced |
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to bring in stress due to volume changes trough temperature dependent expansion. |
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|
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Our domain is the unit cube |
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\begin{eqnarray} \label{HEATEDBLOCK natural} |
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\Omega=\{(x\hackscore{i} | 0 \le x\hackscore{i} \le 1 \} |
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\end{eqnarray} |
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On the boundary the normal stress component is set to zero |
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\begin{eqnarray} \label{HEATEDBLOCK natural} |
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\sigma\hackscore{ij}n\hackscore{j}=0 |
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\end{eqnarray} |
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and on the face with $x\hackscore{i}=0$ we set the $i$-th component of the displacement to $0$ |
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\begin{eqnarray} \label{HEATEDBLOCK constraint} |
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u\hackscore{i}(x)=0 & \mbox{ where } & x\hackscore{i}=0 \; |
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\end{eqnarray} |
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For the temperature distribution we use |
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\begin{eqnarray} \label{HEATEDBLOCK temperature} |
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T(x)= T\hackscore{0} e^{-\beta \|x-x^{c}\|}; |
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\end{eqnarray} |
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with a given positive constant $\beta$ and location $x^{c}$ in the domain. Later in \Sec{MODELFRAME} we will use |
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$T$ from an time-dependent temperature diffusion problem as discussed in \Sec{DIFFUSION CHAP}. |
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|
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When we insert~\eqn{HEATEDBLOCK linear elastic} we get a second oder system of linear PDEs for the displacements $u$ which is called |
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the Lame equation\index{Lame equation}. We want to solve |
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this using the \LinearPDE class to this. For a system of PDEs and a solution with several components the \LinearPDE class |
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takes PDEs of the form |
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\begin{equation}\label{LINEARPDE.SYSTEM.1} |
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-(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} \; . |
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\end{equation} |
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$A$ is a \RankFour, $B$ and $C$ are each a \RankThree, $D$ and $X$ are each a \RankTwo and $Y$ is a \RankOne. |
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The natural boundary conditions \index{boundary condition!natural} take the form: |
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\begin{equation}\label{LINEARPDE.SYSTEM.2} |
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n\hackscore{j}\left(A\hackscore{ijkl} u\hackscore{k,l}+B\hackscore{ijk} u\hackscore{k}\right)+d\hackscore{ik} u\hackscore{k}=n\hackscore{j}X\hackscore{ij}+y\hackscore{i} \;. |
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\end{equation} |
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The coefficient $d$ is a \RankTwo and $y$ is a |
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\RankOne both in the \FunctionOnBoundary. Constraints \index{constraint} take the form |
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\begin{equation}\label{LINEARPDE.SYSTEM.3} |
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u\hackscore{i}=r\hackscore{i} \mbox{ where } q\hackscore{i}>0 |
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\end{equation} |
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$r$ and $q$ are each \RankOne. |
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We can easily identify the coefficients in~\eqn{LINEARPDE.SYSTEM.1}: |
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\begin{eqnarray}\label{LINEARPDE ELASTIC COEFFICIENTS} |
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A\hackscore{ijkl}=\lambda \delta\hackscore{ij} \delta\hackscore{kl} + \mu ( |
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+\delta\hackscore{ik} \delta\hackscore{jl} |
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\delta\hackscore{il} \delta\hackscore{jk}) \\ |
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X\hackscore{ij}=(\lambda+\frac{2}{3} \mu) \; \alpha \; (T-T\hackscore{ref})\delta\hackscore{ij} \\ |
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\end{eqnarray} |
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The characteristic function $q$ defining the locations and components where constraints are set is given by: |
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\begin{equation}\label{HEATEDBLOCK MASK} |
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q\hackscore{i}(x)=\left\{ |
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\begin{array}{cl} |
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1 & x\hackscore{i}=0 \\ |
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0 & \mbox{otherwise} \\ |
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\end{array} |
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\right. |
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\end{equation} |
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Under the assumption that $\lambda$, $\mu$, $\beta$ and $T\hackscore{ref}$ |
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are constant setting $Y\hackscore{i}=\lambda+\frac{2}{3} \mu) \; \alpha \; T\hackscore{i}$ seems to be also possible. However, |
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this choice would lead to a different natural boundary condition which does not set the normal stress component as defined |
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in~\eqn{HEATEDBLOCK linear elastic} to zero. |
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|
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Analogously to concept of symmetry for a single PDE, we call the PDE defined by~\eqn{LINEARPDE.SYSTEM.1} symmetric if |
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\index{symmetric PDE} |
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\begin{eqnarray}\label{LINEARPDE.SYSTEM.SYMMETRY} |
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A\hackscore{ijkl} =A\hackscore{klij} \\ |
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B\hackscore{ijk}=C\hackscore{kij} \\ |
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D\hackscore{ik}=D\hackscore{ki} \\ |
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d\hackscore{ik}=d\hackscore{ki} \ |
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\end{eqnarray} |
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Note that different from the scalar case now the coefficients $D$ and $d$ have to be inspected. It is easy to see that |
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the Lame equation in fact is symmetric. The \LinearPDE class is notified by this fact by calling its \method{setSymmetryOn} method. |
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|
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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 |
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\begin{equation} |
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\sigma\hackscore{mises} = \sqrt{ |
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\frac{1}{6} ((\sigma\hackscore{00}-\sigma\hackscore{11})^2+ |
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(\sigma\hackscore{11}-\sigma\hackscore{22})^2 |
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(\sigma\hackscore{22}-\sigma\hackscore{00})^2) |
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+ \sigma\hackscore{01}^2+\sigma\hackscore{12}^2+\sigma\hackscore{20}^2} |
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\end{equation} |
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is used to detect material damages. Here we want to calculate the von--Mises and write the stress to a file for visualization. |
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|
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\index{scripts!\file{diffusion.py}}: |
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The following script which is available in \file{heatedbox.py} in the \ExampleDirectory solves the Lame equation |
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and writes the displacements and the von--Mises stress\index{von--Mises stress} into a file \file{deform.xml} in the \VTK file format: |
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\begin{python} |
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from esys.escript import * |
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from esys.escript.linearPDEs import LinearPDE |
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from esys.finley import Brick |
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#... set some parameters ... |
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lam=1. |
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mu=0.1 |
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alpha=1.e-6 |
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xc=[0.3,0.3,1.] |
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beta=8. |
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T_ref=0. |
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T_0=1. |
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#... generate domain ... |
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mydomain = Brick(l0=1.,l1=1., l2=1.,n0=10, n1=10, n2=10) |
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x=mydomain.getX() |
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#... set temperature ... |
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T=T_0*exp(-beta*length(x-xc)) |
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#... open symmetric PDE ... |
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mypde=LinearPDE(mydomain) |
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mypde.setSymmetryOn() |
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#... set coefficients ... |
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C=Tensor4(0.,Function(mydomain)) |
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for i in range(mydomain.getDim()): |
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for j in range(mydomain.getDim()): |
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C[i,i,j,j]+=lam |
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C[j,i,j,i]+=mu |
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C[j,i,i,j]+=mu |
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msk=whereZero(x[0])*[1.,0.,0.] \ |
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+whereZero(x[1])*[0.,1.,0.] \ |
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+whereZero(x[2])*[0.,0.,1.] |
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sigma0=(lam+2./3.*mu)*alpha*(T-T_ref)*kronecker(mydomain) |
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mypde.setValue(A=C,X=sigma0,q=msk) |
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#... solve pde ... |
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u=mypde.getSolution() |
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#... calculate von-Misses stress |
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g=grad(u) |
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sigma=mu*(g+transpose(g))+lam*trace(g)*kronecker(mydomain)-sigma0 |
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sigma_mises=sqrt(((sigma[0,0]-sigma[1,1])**2+(sigma[1,1]-sigma[2,2])**2+ \ |
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(sigma[2,2]-sigma[0,0])**2)/6. \ |
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+sigma[0,1]**2 + sigma[1,2]**2 + sigma[2,0]**2) |
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#... output ... |
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saveVTK("deform.xml",disp=u,stress=sigma_mises) |
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\end{python} |
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|
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\begin{figure} |
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\centerline{\includegraphics[width=\figwidth]{HeatedBlock}} |
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\caption{von--Mises Stress and Displacement Vectors.} |
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\label{HEATEDBLOCK FIG 2} |
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\end{figure} |
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|
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Finally the the results can be visualize by calling |
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\begin{python} |
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mayavi -d deform.xml -f CellToPointData -m VelocityVector -m SurfaceMap & |
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\end{python} |
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Note that the filter \text{CellToPointData} is applied to create smooth representation of the |
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von--Mises stress. \fig{HEATEDBLOCK FIG 2} shows the results where the vertical planes showing the |
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von--Mises stress and the horizontal plane shows the vector representing displacements. |
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