Skip to main content

differential equations - Detecting collisions in FEM


Say I want to study the deformation of a pitchfork when you have it fixed on the bottom and push one side.


<< NDSolve`FEM`
Ω =
RegionDifference[Cuboid[{0, -5}, {1, 0}],
Cuboid[{0.45, -4.5}, {.55, 0}]];

Ω // DiscretizeRegion
bcs = DirichletCondition[#, {z <= -4.99}] & /@ {u[y, z] == 0.,
v[y, z] == 0.};
mesh = ToElementMesh[Ω, "MaxCellMeasure" -> 0.005]
planeStress = {Inactive[
Div][{{0, -((Y*ν)/(1 - ν^2))}, {-(Y*(1 - ν))/(2*(1 \
- ν^2)), 0}}.Inactive[Grad][v[y, z], {y, z}], {y, z}] +
Inactive[
Div][{{-(Y/(1 - ν^2)),
0}, {0, -(Y*(1 - ν))/(2*(1 - ν^2))}}.Inactive[Grad][

u[y, z], {y, z}], {y, z}],
Inactive[
Div][{{0, -(Y*(1 - ν))/(2*(1 - ν^2))}, {-((Y*ν)/(1 \
- ν^2)), 0}}.Inactive[Grad][u[y, z], {y, z}], {y, z}] +
Inactive[
Div][{{-(Y*(1 - ν))/(2*(1 - ν^2)),
0}, {0, -(Y/(1 - ν^2))}}.Inactive[Grad][
v[y, z], {y, z}], {y, z}]} /. {Y -> 10^3, ν -> 33/100};

That's my domain



enter image description here


and this my results:


{uif, vif} = 
NDSolveValue[{planeStress == {NeumannValue[1, y == 0 && z > -.1],
0}, DirichletCondition[u[y, z] == 0, z == -5],
DirichletCondition[v[y, z] == 0, z == -5]}, {u,
v}, {y, z} ∈ mesh];
dmesh = ElementMeshDeformation[mesh, {uif, vif}, "ScalingFactor" -> 1];
Show[{mesh["Wireframe"],
dmesh["Wireframe"[

"ElementMeshDirective" -> Directive[EdgeForm[Red], FaceForm[]]]]}]

enter image description here


I was expecting that the right arm of the pitchfork would somehow know when the other one was touching it, but this is not the case at all.


I don't have much experience with FEM so I don't know if what I'm asking is impossible due to the nonlocality of the problem or what not. Is there a way to obtain the correct behaviour in mma, ie how to make the left arm collide with the right one and that they both deform due to the force applied only on the left one?



Answer



Here it is necessary to determine not the impact, but the additional stress arising at the point of contact of two elastic elements. The simplest solution method is to determine the force that acts on each element in contact. I will show the simplest code that can be complicated to infinity.


<< NDSolve`FEM`
Ω = RegionDifference[Cuboid[{0, -5}, {1, 0}], Cuboid[{0.45, -4.5}, {.55, 0}]];


Ω // DiscretizeRegion

bcs = DirichletCondition[#, {z <= -4.99}] & /@ {u[y, z] == 0., v[y, z] == 0.};
mesh = ToElementMesh[Ω, "MaxCellMeasure" -> 0.005];
planeStress = {Inactive[
Div][{{0, -((Y*ν)/(1 - ν^2))}, {-(Y*(1 - ν))/(2*(1 \
- ν^2)), 0}}.Inactive[Grad][v[y, z], {y, z}], {y, z}] +
Inactive[
Div][{{-(Y/(1 - ν^2)),
0}, {0, -(Y*(1 - ν))/(2*(1 - ν^2))}}.Inactive[Grad][

u[y, z], {y, z}], {y, z}],
Inactive[
Div][{{0, -(Y*(1 - ν))/(2*(1 - ν^2))}, {-((Y*ν)/(1 \
- ν^2)), 0}}.Inactive[Grad][u[y, z], {y, z}], {y, z}] +
Inactive[
Div][{{-(Y*(1 - ν))/(2*(1 - ν^2)),
0}, {0, -(Y/(1 - ν^2))}}.Inactive[Grad][
v[y, z], {y, z}], {y, z}]} /. {Y -> 10^3, ν -> 33/100};
{uif, vif} =
NDSolveValue[{planeStress == {NeumannValue[1, y == 0 && z > -.1] +

NeumannValue[-1/3, y == 0.45 && z > -.1] +
NeumannValue[1/3, y == 0.55 && z > -.1], 0},
DirichletCondition[u[y, z] == 0, z == -5],
DirichletCondition[v[y, z] == 0, z == -5]}, {u,
v}, {y, z} ∈ mesh];
dmesh = ElementMeshDeformation[mesh, {uif, vif}, "ScalingFactor" -> 1];
{mesh["Wireframe"],
dmesh["Wireframe"[
"ElementMeshDirective" -> Directive[EdgeForm[Red], FaceForm[]]]]}


fig1


Note that in this solution, the elements do not crawl together, but are deformed together. Consider the general case of an arbitrary combination of parameters of force and elasticity.


<< NDSolve`FEM`
\[CapitalOmega] =
RegionDifference[Cuboid[{0, -5}, {1, 0}],
Cuboid[{0.45, -4.5}, {.55, 0}]];
\[CapitalOmega] // DiscretizeRegion;
bcs = DirichletCondition[#, {z <= -4.99}] & /@ {u[y, z] == 0.,
v[y, z] == 0.};
mesh = ToElementMesh[\[CapitalOmega],

"MaxCellMeasure" -> 0.005]; f = 2;
planeStress = {Inactive[
Div][{{0, -((Y*\[Nu])/(1 - \[Nu]^2))}, {-(Y*(1 - \[Nu]))/(2*(1 \
- \[Nu]^2)), 0}}.Inactive[Grad][v[y, z], {y, z}], {y, z}] +
Inactive[
Div][{{-(Y/(1 - \[Nu]^2)),
0}, {0, -(Y*(1 - \[Nu]))/(2*(1 - \[Nu]^2))}}.Inactive[Grad][
u[y, z], {y, z}], {y, z}],
Inactive[
Div][{{0, -(Y*(1 - \[Nu]))/(2*(1 - \[Nu]^2))}, {-((Y*\[Nu])/(1 \

- \[Nu]^2)), 0}}.Inactive[Grad][u[y, z], {y, z}], {y, z}] +
Inactive[
Div][{{-(Y*(1 - \[Nu]))/(2*(1 - \[Nu]^2)),
0}, {0, -(Y/(1 - \[Nu]^2))}}.Inactive[Grad][
v[y, z], {y, z}], {y, z}]} /. {Y -> 10^3, \[Nu] -> 33/100};
sol = ParametricNDSolveValue[{planeStress == {NeumannValue[f,
y == 0 && z > -.1] + NeumannValue[-g, y == 0.45 && z > -.1] +
NeumannValue[g, y == 0.55 && z > -.1], 0},
DirichletCondition[u[y, z] == 0, z == -5],
DirichletCondition[v[y, z] == 0, z == -5]}, {u,

v}, {y, z} \[Element] mesh, {g}];


sol1 =
ParametricNDSolveValue[{planeStress == {NeumannValue[f,
y == 0 && z > -.1] + NeumannValue[-g, y == 0.45 && z > -.1] +
NeumannValue[g, y == 0.55 && z > -.1], 0},
DirichletCondition[u[y, z] == 0, z == -5],
DirichletCondition[v[y, z] == 0, z == -5]},
u, {y, z} \[Element] mesh, {g}];


g0=g/.FindRoot[
sol1[g][.45, 0] - sol1[g][.55, 0] == .1, {g, f/2}] // Quiet

Out[]= 0.834936
dmesh = ElementMeshDeformation[mesh, sol[g0],
"ScalingFactor" -> 1];
{mesh["Wireframe"],
dmesh["Wireframe"[
"ElementMeshDirective" -> Directive[EdgeForm[Red], FaceForm[]]]]}


fig2 For f=1, we find g0=0.335075, which is close to 1/3 found by another method.


Comments

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}]

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}]

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....