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 - 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[ ...
Post a Comment
Read more

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...
Post a Comment
Read more

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...
Post a Comment
Read more