Skip to main content

finite element method - NDEigensystem for structural vibration


Having installed version 11 I thought I would check an old Stack Exchange vibration problem using NDEigensytem. The old problem was Test a wooden board's vibration mode and I think this was before NDEigensystem and thus very complicated. However I get different answers to the old question and a warning I don't understand.


User21 had done all the hard work in the previous question part of which I copy now to make the mesh. The code is as follows. First define the stress operator for 3D structures



Needs["NDSolve`FEM`"]

ClearAll[stressOperator, u, v, w, x, y, z];
stressOperator[
  Y_, \[Nu]_] := {Inactive[
     Div][{{0, 0, -((Y*\[Nu])/((1 - 2*\[Nu])*(1 + \[Nu])))}, {0, 0, 
       0}, {-Y/(2*(1 + \[Nu])), 0, 0}}.Inactive[Grad][
      w[x, y, z], {x, y, z}], {x, y, z}] + 
   Inactive[
     Div][{{0, -((Y*\[Nu])/((1 - 2*\[Nu])*(1 + \[Nu]))), 

       0}, {-Y/(2*(1 + \[Nu])), 0, 0}, {0, 0, 0}}.Inactive[Grad][
      v[x, y, z], {x, y, z}], {x, y, z}] + 
   Inactive[
     Div][{{-((Y*(1 - \[Nu]))/((1 - 2*\[Nu])*(1 + \[Nu]))), 0, 
       0}, {0, -Y/(2*(1 + \[Nu])), 0}, {0, 
       0, -Y/(2*(1 + \[Nu]))}}.Inactive[Grad][
      u[x, y, z], {x, y, z}], {x, y, z}], 
  Inactive[Div][{{0, 0, 0}, {0, 
       0, -((Y*\[Nu])/((1 - 
              2*\[Nu])*(1 + \[Nu])))}, {0, -Y/(2*(1 + \[Nu])), 

       0}}.Inactive[Grad][w[x, y, z], {x, y, z}], {x, y, z}] + 
   Inactive[
     Div][{{0, -Y/(2*(1 + \[Nu])), 
       0}, {-((Y*\[Nu])/((1 - 2*\[Nu])*(1 + \[Nu]))), 0, 0}, {0, 0, 
       0}}.Inactive[Grad][u[x, y, z], {x, y, z}], {x, y, z}] + 
   Inactive[
     Div][{{-Y/(2*(1 + \[Nu])), 0, 
       0}, {0, -((Y*(1 - \[Nu]))/((1 - 2*\[Nu])*(1 + \[Nu]))), 0}, {0,
        0, -Y/(2*(1 + \[Nu]))}}.Inactive[Grad][
      v[x, y, z], {x, y, z}], {x, y, z}], 

  Inactive[Div][{{0, 0, 0}, {0, 
       0, -Y/(2*(1 + \[Nu]))}, {0, -((Y*\[Nu])/((1 - 
              2*\[Nu])*(1 + \[Nu]))), 0}}.Inactive[Grad][
      v[x, y, z], {x, y, z}], {x, y, z}] + 
   Inactive[
     Div][{{0, 0, -Y/(2*(1 + \[Nu]))}, {0, 0, 
       0}, {-((Y*\[Nu])/((1 - 2*\[Nu])*(1 + \[Nu]))), 0, 0}}.Inactive[
       Grad][u[x, y, z], {x, y, z}], {x, y, z}] + 
   Inactive[
     Div][{{-Y/(2*(1 + \[Nu])), 0, 0}, {0, -Y/(2*(1 + \[Nu])), 0}, {0,

        0, -((Y*(1 - \[Nu]))/((1 - 2*\[Nu])*(1 + \[Nu])))}}.Inactive[
       Grad][w[x, y, z], {x, y, z}], {x, y, z}]}

Next set up the geometry.


base = {0, 0, 0};
h1 = 5;
h2 = 5;
w1 = 40;
l1 = 76;
cw1 = 5;

cl1 = 68;
cw2 = 36;
cl2 = 5;
offset1 = base + {(w1 - cw1)/2, (l1 - cl1)/2, 0};
offset2 = base + {(w1 - cw2)/2, (l1 - cl2)/2, 0};
offset3 = base + {(w1 - cw1)/2, (l1 - cl2)/2, 0};
ClearAll[rect]
rect[base_, w_, l_, h_] := {base + {0, 0, h}, base + {w, 0, h}, 
  base + {w, l, h}, base + {0, l, h}}
coords = ConstantArray[{0., 0., 0.}, 4 + 4 + 12 + 12];

coords[[{1, 2, 3, 4}]] = rect[base, w1, l1, 0];
coords[[{5, 6, 7, 8}]] = rect[base, w1, l1, h1];
coords[[{9, 10, 15, 16}]] = rect[offset1, cw1, cl1, h1];
coords[[{19, 12, 13, 18}]] = rect[offset2, cw2, cl2, h1];
coords[[{20, 11, 14, 17}]] = rect[offset3, cw1, cl2, h1];
coords[[20 + Range[12]]] = ({0, 0, h2} + #) & /@ 
   coords[[8 + Range[12]]];
bmesh = ToBoundaryMesh["Coordinates" -> coords, 
   "BoundaryElements" -> {QuadElement[{{1, 2, 3, 4}, {1, 2, 6, 5}, {2,
         3, 7, 6}, {3, 4, 8, 7}, {4, 1, 5, 8}, {5, 6, 10, 9}, {6, 12, 

        11, 10}, {6, 7, 13, 12}, {7, 15, 14, 13}, {7, 8, 16, 15}, {8, 
        18, 17, 16}, {8, 5, 19, 18}, {5, 9, 20, 19}, 
       Sequence @@ ({{9, 10, 11, 20}, {11, 12, 13, 14}, {14, 15, 16, 
            17}, {17, 18, 19, 20}, {20, 11, 14, 17}} + 12), 
       Sequence @@ (Partition[Join[Range[9, 20]], 2, 1, 
           1] /. {i1_, i2_} :> {i1, i2, i2 + 12, i1 + 12})}]}];

Look at the geometry and then the mesh


Show[bmesh["Wireframe"], 
 bmesh["Wireframe"["MeshElement" -> "PointElements", 

   "MeshElementIDStyle" -> Red]]]

Mathematica graphics


mesh = ToElementMesh[bmesh, "MeshOrder" -> 1, "MaxCellMeasure" -> 10];
mesh["Wireframe"]

Mathematica graphics


Now we use NDEigensystem to find the eigenvalues and vectors


{vals, vecs} = 
  NDEigensystem[

   stressOperator[100, 1/3], {u, v, w}, {x, y, z} \[Element] mesh, 
   14];

This gives the warning


(* Eigensystem::chnpdef: Warning: there is a possibility that the second matrix SparseArray[<<1>>] in the first argument is not positive definite, which is necessary for the Arnoldi method to give accurate results. *)

So something is suspect.


The eigenvalues are as follows


TableForm[vals, TableHeadings -> {Automatic, None}]


Mathematica graphics


The first six are zero which is correct since these are the rigid body modes. The seventh is negative which is very suspect. The next values are possible but are different to the values found in the original question which were


{0.`, 0.`, 0.`, 0.`, 0.`, 0.`, 0.011403583383327644`, \
0.01526089137692353`, 0.05661022352859022`, 0.07266104128273859`}

Although they may just about agree to a couple of figures. The eigenvectors are


Column@(MeshRegion[
     ElementMeshDeformation[mesh, vecs[[#]], 
      "ScalingFactor" -> -300]] & /@ Range[7, 14])


Mathematica graphics


I also tried to run the code from the previous question directly and I could not get it to work.


In summary what does the warning mean from NDEigensystem and how can this be avoided? Also, it looks like the output from either the previous answer or this one is wrong. Finally, should the previous code work in version 11? Thanks




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