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

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...