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3D Elastic waves in a glass

Take an empty glass, hit the side, the glass will make a sound that can be recorded using

s0=AudioCapture["C:\\Users\\...\\Desktop\\\\glass0.wav", MaxDuration -> 2]

Find the sound spectrum

Spectrogram[s0]

The photo shows a glass and a spectrum of sound

Now we measure the dimensions of the glass, take the density, Young's modulus, glass Poisson's ratio from the reference book, compose the equations and find the eigenvalues

<< NDSolve`FEM`;
L = .14; L1 = .01; r1 = .085/2; r2 = .055/
  2; del = .006;(*cg=3962 m/s, 3980, 5100, 5640*);
reg = RegionUnion[
   ImplicitRegion[(r2 + (r1 - r2) (z - L1)/(L - L1))^2 <= 
      x^2 + y^2 <= (r2 + (r1 - r2) (z - L1)/(L - L1) + del)^2 && 
     L1 <= z <= L, {x, y, z}], 
   ImplicitRegion[
    0 <= x^2 + y^2 <= (r2 + del)^2 && 0 <= z <= L1, {x, y, z}]];
param = {Y -> 56*10^9, ν -> 25/100}; rho = 2500;
ClearAll[stressOperator];
stressOperator[
   Y_, ν_] := {Inactive[
      Div][{{0, 0, -((Y*ν)/((1 - 2*ν)*(1 + ν)))}, {0, 0, 
        0}, {-Y/(2*(1 + ν)), 0, 0}}.Inactive[Grad][
       w[t, x, y, z], {x, y, z}], {x, y, z}] + 
    Inactive[
      Div][{{0, -((Y*ν)/((1 - 2*ν)*(1 + ν))), 
        0}, {-Y/(2*(1 + ν)), 0, 0}, {0, 0, 0}}.Inactive[Grad][
       v[t, x, y, z], {x, y, z}], {x, y, z}] + 
    Inactive[
      Div][{{-((Y*(1 - ν))/((1 - 2*ν)*(1 + ν))), 0, 
        0}, {0, -Y/(2*(1 + ν)), 0}, {0, 
        0, -Y/(2*(1 + ν))}}.Inactive[Grad][
       u[t, x, y, z], {x, y, z}], {x, y, z}], 
   Inactive[
      Div][{{0, 0, 0}, {0, 
        0, -((Y*ν)/((1 - 
               2*ν)*(1 + ν)))}, {0, -Y/(2*(1 + ν)), 
        0}}.Inactive[Grad][w[t, x, y, z], {x, y, z}], {x, y, z}] + 
    Inactive[
      Div][{{0, -Y/(2*(1 + ν)), 
        0}, {-((Y*ν)/((1 - 2*ν)*(1 + ν))), 0, 0}, {0, 0, 
        0}}.Inactive[Grad][u[t, x, y, z], {x, y, z}], {x, y, z}] + 
    Inactive[
      Div][{{-Y/(2*(1 + ν)), 0, 
        0}, {0, -((Y*(1 - ν))/((1 - 2*ν)*(1 + ν))), 
        0}, {0, 0, -Y/(2*(1 + ν))}}.Inactive[Grad][
       v[t, x, y, z], {x, y, z}], {x, y, z}], 
   Inactive[
      Div][{{0, 0, 0}, {0, 
        0, -Y/(2*(1 + ν))}, {0, -((Y*ν)/((1 - 
               2*ν)*(1 + ν))), 0}}.Inactive[Grad][
       v[t, x, y, z], {x, y, z}], {x, y, z}] + 
    Inactive[
      Div][{{0, 0, -Y/(2*(1 + ν))}, {0, 0, 
        0}, {-((Y*ν)/((1 - 2*ν)*(1 + ν))), 0, 
        0}}.Inactive[Grad][u[t, x, y, z], {x, y, z}], {x, y, z}] + 
    Inactive[
      Div][{{-Y/(2*(1 + ν)), 0, 0}, {0, -Y/(2*(1 + ν)), 
        0}, {0, 0, -((Y*(1 - ν))/((1 - 
               2*ν)*(1 + ν)))}}.Inactive[Grad][
       w[t, x, y, z], {x, y, z}], {x, y, z}]};

{vals, funs} = 
 NDEigensystem[
  stressOperator[56*10^9, 1/4] + 
    rho {D[u[t, x, y, z], {t, 2}], D[v[t, x, y, z], {t, 2}], 
      D[w[t, x, y, z], {t, 2}]} == {0, 0, 0}, {u, v, w}, 
  t, {x, y, z} ∈ reg, 15];

Frequencies in Hertz

Abs[vals ]/(2 Pi)

Out[9]= {0.000389602, 0.000865814, 0.000865814, 0.000921462, \
0.000921462, 0.00136215, 0.00136215, 0.00152256, 0.00152256, \
0.0015598, 0.0015598, 2140.67, 2140.67, 2144.36, 2144.36}

And so we see that frequencies 2140-2144 explain the result of our experiment (in the spectrogram, the peak is about 2000 H). Build 3D functions u,v,w for frequency 2144.36

DensityPlot3D[Re[funs[[15, 1]][x, y, z]], {x, y, z} ∈ reg, 
 ColorFunction -> "Rainbow", OpacityFunction -> None, Boxed -> False, 
 PlotLabel -> Row[{"f = ", Abs[vals [[15]]]/2/Pi}], 
 BoxRatios -> Automatic, PlotPoints -> 50]

DensityPlot3D[Re[funs[[15, 2]][x, y, z]], {x, y, z} ∈ reg, 
 ColorFunction -> "Rainbow", OpacityFunction -> None, Boxed -> False, 
 PlotLabel -> Row[{"f = ", Abs[vals [[15]]]/2/Pi}], 
 BoxRatios -> Automatic, PlotPoints -> 50]
DensityPlot3D[Re[funs[[15, 3]][x, y, z]], {x, y, z} ∈ reg, 
 ColorFunction -> "Rainbow", Boxed -> False, 
 PlotLabel -> Row[{"f = ", Abs[vals [[15]]]/2/Pi}], 
 BoxRatios -> Automatic, PlotPoints -> 50]

OK! Problems arise if we put del=0.003 (real glass wall thickness). First, the desired frequencies 2140-2144H disappear. Secondly, the 3D functions u,v,w look as if there are holes in the glass

Is it possible to get the desired result for del=.003?

Update 1. We use the algorithm proposed by user21 with a small modification and with the boundary condition DirichletCondition[{u[t, x, y, z] == 0, v[t, x, y, z] == 0, w[t, x, y, z] == 0}, z == 0]. Then the first 5 modes are consistent with the experiment (15 modes can be calculated with an error):

<< NDSolve`FEM`;
L = 0.14; L1 = 0.01; r1 = 0.085/2; r2 = 0.055/2; del = 0.003;


reg = RegionUnion[
   ImplicitRegion[(r2 + (r1 - r2) (z - L1)/(L - L1))^2 <= 
      x^2 + y^2 <= (r2 + (r1 - r2) (z - L1)/(L - L1) + del)^2 && 
     L1 <= z <= L, {x, y, z}], 
   ImplicitRegion[
    0 <= x^2 + y^2 <= (r2 + del)^2 && 0 <= z <= L1, {x, y, z}]];
(mesh = ToElementMesh[reg, 
    "BoundaryMeshGenerator" -> {"BoundaryDiscretizeRegion", 
      Method -> {"MarchingCubes", PlotPoints -> 31}}, 
    "MeshOrder" -> 1])["Wireframe"]

Modes

{vals, funs} = 
 NDEigensystem[{stressOperator[56*10^9, 1/4] + 
     rho {D[u[t, x, y, z], {t, 2}], D[v[t, x, y, z], {t, 2}], 
       D[w[t, x, y, z], {t, 2}]} == {0, 0, 0}, 
   DirichletCondition[{u[t, x, y, z] == 0, v[t, x, y, z] == 0, 
     w[t, x, y, z] == 0}, z == 0]}, {u, v, w}, 
  t, {x, y, z} \[Element] mesh, 5];

Modes in Hz

Abs[vals]/(2 Pi)

Out[]= {2047.63, 2048.03, 2048.03, 2336.35, 2336.35}

There are radial and azimuthal modes

Update 2. We use the algorithm proposed by Pinti with a modification and with the boundary condition DirichletCondition[{u[t, x, y, z] == 0, v[t, x, y, z] == 0, w[t, x, y, z] == 0}, y == 0]. Then the first 9 modes are consistent with the experiment (modes can be calculated without an error):

Get["MeshTools`"]

L = 0.14; L1 = 0.01; r1 = 0.085/2; r2 = 0.055/2; del = 0.003;

n1 = 5;
n2 = 31;
n3 = 5;
n4 = 12;
mesh2D = StructuredMesh[{{{r2, 0}, {r1, L}}, {{r2 - del, 
     0}, {r1 - del, L}}}, {n2, n1}]

mesh2D["Wireframe"[Axes -> True, AxesOrigin -> {0, 0}]]

Modes

{vals, funs} = 
 NDEigensystem[{stressOperator[56*10^9, 1/4] + 
     rho {D[u[t, x, y, z], {t, 2}], D[v[t, x, y, z], {t, 2}], 
       D[w[t, x, y, z], {t, 2}]} == {0, 0, 0}, 
   DirichletCondition[{u[t, x, y, z] == 0, v[t, x, y, z] == 0, 
     w[t, x, y, z] == 0}, y == 0]}, {u, v, w}, 
  t, {x, y, z} \[Element] mesh, 9];

vals in Hz

     Abs[vals]/(2 Pi)

Out[]= {23.1411, 1806.36, 1806.36, 1806.36, 1806.36, 1970.47, \
1970.47, 1970.58, 1970.58}

There are radial and azimuthal modes too

Update 3. We use the algorithm proposed by user21 for version 12.1 with a small modification

<< NDSolve`FEM`;
L = 0.14; L1 = 0.01; del = 0.003; r1 = 0.085/2; r2 = 0.055/2;

polygon = 
  Polygon[{{0, 0, 0}, {r2 + del, 0, 0}, {r2 + del, 0, L1}, {r1 + del, 
     0, L}, {r1, 0, L}, {r2, 0, L1}, {0, 0, L1}}];

Needs["OpenCascadeLink`"]
shape = OpenCascadeShape[polygon];
axis = {{0, 0, 0}, {0, 0, 3/2 L}}; sweep = 
 OpenCascadeShapeRotationalSweep[shape, axis, 2 Pi];
bmesh = OpenCascadeShapeSurfaceMeshToBoundaryMesh[sweep, 
   "ShapeSurfaceMeshOptions" -> {"LinearDeflection" -> 0.0003}];


mesh = ToElementMesh[bmesh, AccuracyGoal -> 5, PrecisionGoal -> 5, 
  "MeshOrder" -> 1];


param = {Y -> 56*10^9, \[Nu] -> 25/100}; rho = 2500; cg = 
 Sqrt[56.*10^9/rho]; 


ClearAll[stressOperator];
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[t, 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[t, 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[t, 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[t, 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[t, 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[t, 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[t, 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[t, 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[t, x, y, z], {x, y, z}], {x, y, z}]};

{vals, funs} = 
 NDEigensystem[{stressOperator[56*10^9, 1/4] + 
     rho {D[u[t, x, y, z], {t, 2}], D[v[t, x, y, z], {t, 2}], 
       D[w[t, x, y, z], {t, 2}]} == {0, 0, 0}, 
   DirichletCondition[{u[t, x, y, z] == 0, v[t, x, y, z] == 0, 
     w[t, x, y, z] == 0}, z == 0]}, {u, v, w}, 
  t, {x, y, z} \[Element] mesh, 12];

vals in Hz

     Abs[vals]/(2 Pi)

{1973.97, 1973.97, 1974.86, 1974.86, 2169.47, 2169.47, 2250.23, 2250.23, 4183.69, 4183.69, 5532.12, 5532.12} Visualisation of 3 modes

DensityPlot3D[Re[funs[[1, 1]][x, y, z]], {x, y, z} \[Element] mesh, 
 ColorFunction -> "Rainbow", OpacityFunction -> None, Boxed -> False, 
 PlotLabel -> Row[{"f = ", Abs[vals [[1]]]/2/Pi}], 
 BoxRatios -> Automatic, PlotPoints -> 50]
DensityPlot3D[Re[funs[[5, 1]][x, y, z]], {x, y, z} \[Element] mesh, 
 ColorFunction -> "Rainbow", OpacityFunction -> None, Boxed -> False, 
 PlotLabel -> Row[{"f = ", Abs[vals [[5]]]/2/Pi}], 
 BoxRatios -> Automatic, PlotPoints -> 50]
DensityPlot3D[Re[funs[[7, 1]][x, y, z]], {x, y, z} \[Element] mesh, 
 ColorFunction -> "Rainbow", OpacityFunction -> None, Boxed -> False, 
 PlotLabel -> Row[{"f = ", Abs[vals [[7]]]/2/Pi}], 
 BoxRatios -> Automatic, PlotPoints -> 50]

Answer*

You get a better mesh with a different boundary mesh generator:

    (mesh = ToElementMesh[reg, 
        "BoundaryMeshGenerator" -> \
    {"BoundaryDiscretizeRegion",
          Method -> {"MarchingCubes", PlotPoints -> 33}}, 
        "MeshOrder" -> 1,
        "MaxCellMeasure"\[Rule]0.000000005])["Wireframe"]


[![enter image description here][1]][1]

For that mesh I get

    Abs[vals]/(2 Pi)
    (*{0.000502385, 0.000502385, 0.00072869, 0.00072869, \
    0.000733392, 0.000733392, 0.0010404, 0.0010404, 0.00150767, \
    0.00150767, 0.00151325, 0.00151325, 0.308656, 2238.88, 2238.88}*)


And the 14th mode looks like:

    MeshRegion[
     ElementMeshDeformation[mesh, Re[Through[funs[[14]]["ValuesOnGrid"]]],
       "ScalingFactor" -> 10^9]]

[![enter image description here][2]][2]

Two other comments: the fact that NDEigensystem gives messages suggests to me that this mesh is still not good enough; as you see I also used `MeshOrder->1` as I did not want to wait for a second order mesh to finish. But you might want to try that and a finer mesh. Probably by using more plot points. Perhaps generate the boundary mesh manually?

A second thing that come to mind is that I think you should have some rigid body modes because the glass stands on the table. Maybe experiment with

    DirichletCondition[{u[t, x, y, z] == 0, v[t, x, y, z] == 0, 
      w[t, x, y, z] == 0}, x == 0]

Also, there is a nice Bell Acoustics customer example in the [FEMAddOns](https://github.com/WolframResearch/FEMAddOns). You can install that with 

    ResourceFunction["FEMAddOnsInstall"][]

and find it on the Applications guide page

    FEMAddOns/guide/FEMApplications

or have a look at the cloud version of that [notebook](https://www.wolframcloud.com/obj/github-cloud/blobs/e55af7e9216711be5a312d3ed6dd8e448775435a).

Hope this helps.

**Update: 12.1**

Another way to generate the mesh is to make use of the [OpenCascadeLink](https://reference.wolfram.com/language/OpenCascadeLink/tutorial/UsingOpenCascadeLink.html). For this we generate a flat cross section of the glass in 3D.

    polygon = 
      Polygon[{{0, 0, 0}, {r2 + del, 0, 0}, {r2 + del, 0, L1}, {r1 + del, 
         0, L}, {r1, 0, L}, {r2, 0, L1}, {0, 0, L1}}];
    Graphics3D[{FaceForm[], EdgeForm[Black], polygon}, Boxed -> False]

[![enter image description here][3]][3]

We load the link

    Needs["OpenCascadeLink`"]

and convert the polygon into an OCCT shape:

    shape = OpenCascadeShape[polygon];

We set up an axis of revolution and sweep the polygon.

    axis = {{0, 0, 0}, {0, 0, 3/2 L}};
    sweep = OpenCascadeShapeRotationalSweep[shape, axis, 2 \[Pi]];

Here is a visual of the result:

    bmesh = OpenCascadeShapeSurfaceMeshToBoundaryMesh[sweep, 
       "ShapeSurfaceMeshOptions" -> {"LinearDeflection" -> 0.00125}];
    Show[Graphics3D[{{Red, polygon}, {Blue, Thick, Arrow[axis]}}], 
     bmesh["Wireframe"], Boxed -> False]

[![enter image description here][4]][4]

You see the original polygon in red and the blue arrow is the rotational axis. From here we can generate the mesh in the same way:

    mesh = ToElementMesh[bmesh, "MeshOrder" -> 1(*,
      "MaxCellMeasure"\[Rule]0.000000005*)]

    mesh["Wireframe"[
      "MeshElementStyle" -> 
       Directive[Opacity[0.2], Specularity[White, 17], FaceForm[White], 
        EdgeForm[]]]]

[![enter image description here][5]][5]

This is a much better approximation of the geometry. Nevertheless finding the eigenvalues remains challenging as there is a strong dependency of the eigenvalues on the mesh.

  [1]: https://i.sstatic.net/F993n.png
  [2]: https://i.sstatic.net/Ru9Lt.jpg
  [3]: https://i.sstatic.net/q6GCU.png
  [4]: https://i.sstatic.net/kQJA4.png
  [5]: https://i.sstatic.net/l4e6d.png

Edit Summary*

Cancel

$\begingroup$ Thank you for the mesh (+1) and links. They also use a thick cylinder. I will try to create a thin glass manually. $\endgroup$

Alex Trounev
– Alex Trounev

2020-02-07 10:01:01 +00:00
Commented Feb 7, 2020 at 10:01
$\begingroup$ Finite element codes often have shell elements. It looks like they would be useful here. Is there any chance of getting such elements available in Mathematica? $\endgroup$

Hugh
– Hugh

2020-02-07 16:21:25 +00:00
Commented Feb 7, 2020 at 16:21
$\begingroup$ @Hugh, sorry I saw the comment but then forgot to reply. Shell elements are currently not planed. There is still too much other stuff I can improve before going down this alley. $\endgroup$

user21
– user21

2020-03-24 09:20:36 +00:00
Commented Mar 24, 2020 at 9:20
$\begingroup$ @user21 thank you. This is a very interesting update in version 12.1. I'll check how it works. $\endgroup$

Alex Trounev
– Alex Trounev

2020-03-24 10:48:09 +00:00
Commented Mar 24, 2020 at 10:48
$\begingroup$ @user21 See update to my post. $\endgroup$

Alex Trounev
– Alex Trounev

2020-03-25 18:28:24 +00:00
Commented Mar 25, 2020 at 18:28

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