Skip to main content

finite element method - Swinging Beam example from Help. Not Understanding/Not Working


To help me understand finite element analysis with Mathematica I have been reading the Finite Element Method User Guide in Help and am stuck on the example in Coupled PDEs. Here there is a static beam problem and a swinging beam problem.


I have taken the code from the static beam problem and this works nicely


 Needs["NDSolve`FEM`"]

\[CapitalOmega] = ImplicitRegion[True, {x, y}];
mesh = ToElementMesh[\[CapitalOmega], {{0, 5}, {0, 1}}, 

   "MaxCellMeasure" -> 0.1];
vd = NDSolve`VariableData[{"DependentVariable", 
     "Space"} -> {{u, v}, {x, y}}];
sd = NDSolve`SolutionData["Space" -> ToNumericalRegion[mesh]];

bcDu0 = DirichletCondition[u[x, y] == 0, x == 0];
bcDv0 = DirichletCondition[v[x, y] == 0, x == 0];
bcNL = NeumannValue[-1, x == 5];
initBCs = 
  InitializeBoundaryConditions[vd, sd, {{bcDu0}, {bcDv0, bcNL}}];


diffusionCoefficients = 
  "DiffusionCoefficients" -> {{{{-(Y/(1 - \[Nu]^2)), 
        0}, {0, -((Y (1 - \[Nu]))/(2 (1 - \[Nu]^2)))}}, {{0, -((
         Y \[Nu])/(1 - \[Nu]^2))}, {-((Y (1 - \[Nu]))/(
         2 (1 - \[Nu]^2))), 
        0}}}, {{{0, -((Y (1 - \[Nu]))/(2 (1 - \[Nu]^2)))}, {-((
         Y \[Nu])/(1 - \[Nu]^2)), 
        0}}, {{-((Y (1 - \[Nu]))/(2 (1 - \[Nu]^2))), 
        0}, {0, -(Y/(1 - \[Nu]^2))}}}} /. {Y -> 10^3, \[Nu] -> 33/100};

initCoeffs = 
  InitializePDECoefficients[vd, sd, {diffusionCoefficients}];


methodData = InitializePDEMethodData[vd, sd];
discretePDE = DiscretizePDE[initCoeffs, methodData, sd];
split = {#[[1]] + 1 ;; #[[2]], #[[2]] + 1 ;; #[[3]]} &[
   methodData["IncidentOffsets"]];
discreteBCs = DiscretizeBoundaryConditions[initBCs, methodData, sd];
{load, stiffness, damping, mass} = discretePDE["SystemMatrices"];


DeployBoundaryConditions[{load, stiffness}, discreteBCs];

solution = LinearSolve[stiffness, load];
uif = ElementMeshInterpolation[{mesh}, solution[[split[[1]]]]];
vif = ElementMeshInterpolation[{mesh}, solution[[split[[2]]]]];
dmesh = ElementMeshDeformation[mesh, {uif, vif}];

I can plot results as follows using


 mesh["Wireframe"]

ContourPlot[uif[x, y], {x, y} \[Element] mesh, 
 ColorFunction -> "Temperature", AspectRatio -> Automatic]
ContourPlot[vif[x, y], {x, y} \[Element] mesh, 
 ColorFunction -> "Temperature", AspectRatio -> Automatic]
Show[{
  mesh["Wireframe"],
  dmesh["Wireframe"[
    "ElementMeshDirective" -> Directive[EdgeForm[Red], FaceForm[]]]]}]

to give



Mathematica graphics


All this is good but it goes wrong for me when I move onto the swinging beam which modifies the above code. I can't get this to run even if I copy directly from Help. Here is the relevant set-up code from Help


    massCoefficients = "MassCoefficients" -> {{1, 0}, {0, 1}};
initCoeffs = 
  InitializePDECoefficients[vd, 
   sd, {diffusionCoefficients, massCoefficients}];
discretePDE = DiscretizePDE[initCoeffs, methodData, sd];
{load, stiffness, damping, mass} = discretePDE["SystemMatrices"];
rayleighDamping = 0.1*mass + 0.04*stiffness;
DeployBoundaryConditions[{load, stiffness, rayleighDamping, mass}, 

  discreteBCs];
dof = methodData["DegreesOfFreedom"];
init = dinit = ConstantArray[0, {dof, 1}];
sparsity = 
  ArrayFlatten[{{mass["PatternArray"], 
     mass["PatternArray"]}, {rayleighDamping["PatternArray"], 
     rayleighDamping["PatternArray"]}}];

It is the next bit of the solution code that gets stuck. The time monitor stops around 0.00158 so I have tried running the code to 0.0015 rather than 45 as in Help but it still gets stuck.


 Dynamic["time: " <> ToString[CForm[currentTime]]]

AbsoluteTiming[
 tif = NDSolveValue[{
    mass.u''[ t] + rayleighDamping.u'[ t] + stiffness.u[ t] == load
    , u[ 0] == init, u'[ 0] == dinit}, u, {t, 0, 0.0015}
   , Method -> {"EquationSimplification" -> "Residual"}
   , Jacobian -> {Automatic, Sparse -> sparsity}
   , EvaluationMonitor :> (currentTime = t;)
   ]]

Are there any suggestions for how to make this work?




Answer



It seems that my version of Mathematica was corrupted. I did a Shift+Control while starting Mathematica (see here for good instructions) and the code now works perfectly. Thanks to user21 and MarcoB for confirming that the code works. Wolfram Support also confirmed that the code works and suggested a restart.


Out of interest I compared a benchmark from before the Shift+Control to after the restart. The benchmark is found by running


    Needs["Benchmarking`"]
    BenchmarkReport[]

Before restart: Mathematica graphics


After restart: Mathematica graphics


As you can see I have gone from anti-penultimate to top-of-the-heap!


Thanks for the help



Comments

Post a Comment

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....
Post a Comment
Read more