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enter image description here

I want to drawing with Vector Plot a deflection and slope of a linearly varying cantilever. And with Contour Plot it works but with Vector Plot it didn't. Here is the code:

M = EI D[y[x], {x, 2}];
q = -1;
M1 = M /. x -> 0;
M2 = M /. x -> L;
y1 = y[x] /. x -> 0;
y2 = y[x] /. x -> L;
s = DSolve[{EI y''''[x] == q, M1 == 0, y1 == 0, M2 == 0, y2 == 0}, y,  x];
displacement = y[x] /. s[[1]];
Eb = 20000000;
b = 0.25;
t0 = 0.5;
L = 5;
t[x_] := (t0 (L + x))/L;
Ib = (b t[x]^3)/12
EI = Eb Ib
u = {-x2 D[displacement, x], displacement}
VectorPlot[u, {x, 0, L}, {x2, -t[x]/2, t[x]/2}, 
PlotLabel -> "vektoros", AspectRatio -> Automatic]

The result with constant cross section and the same code for constant cross section:

M = EI D[y[x], {x, 2}];
q = -10;
M1 = M /. x -> 0;
M2 = M /. x -> L;
y1 = y[x] /. x -> 0;
y2 = y[x] /. x -> L;
s = DSolve[{EI y''''[x] == q, M1 == 0, y1 == 0, M2 == 0, y2 == 0}, y, 
x]
displacement = y[x] /. s[[1]]
Eb = 20000000;
b = 0.25;
t = 0.5;
L = 5;
Ib = (b t^3)/12;
EI = Eb Ib;
u = {-x2 D[displacement, x], displacement}
VectorPlot[u, {x, 0, L}, {x2, -t/2, t/2}, PlotLabel -> "vektoros", 
AspectRatio -> Automatic, VectorStyle -> Black]

enter image description here

searched a similar result

enter image description here

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It's not exactly what was asked for but it's another way to display the sought-after deformation. I thought I would share it because it's probably not well known how it can be done, and it seems appropriate to the problem at hand.

First the OP's code computed this displacement:

displacement
(*  (0.0001 (-125 x + 10 x^3 - x^4))/(5 + x)^3  *)

Here is an image of the beam. I did not see where the parameters (other than the length L = 5) were specified. So I made some of it up.

Needs["NDSolve`FEM`"];
mesh = ToElementMesh[FullRegion[2], {{0, L}, {-1, 1}/2}];
u = Function[{x, y}, 0];
v = Function[{x, y}, y (x - L)/20];

uif = ElementMeshInterpolation[{mesh}, u @@@ mesh["Coordinates"]];
vif = ElementMeshInterpolation[{mesh}, v @@@ mesh["Coordinates"]];

mesh = ElementMeshDeformation[mesh, {uif, vif}];
mesh["Wireframe"]

Mathematica graphics

Then we can deform the mesh according to displacement using ElementMeshDeformation. I magnified the displacement by 1000 to make the deformation perceptible. The beam can be colored by the magnitude of the displacement at each point.

u = Function[{x, y}, 0];
v = Function @@ {{x, y}, 1000 displacement};

uif = ElementMeshInterpolation[{mesh}, u @@@ mesh["Coordinates"]];
vif = ElementMeshInterpolation[{mesh}, v @@@ mesh["Coordinates"]];

dmesh = ElementMeshDeformation[mesh, {uif, vif}];

deform = (Norm[{0, v @@ #}] & /@ mesh["Coordinates"])
Legended[
 Show[
  Graphics@ElementMeshToGraphicsComplex[dmesh, All, 
    VertexColors -> ColorData["Rainbow"] /@ Rescale[deform]],
  dmesh["Wireframe"]],
 Placed[BarLegend[{"Rainbow", Through[{Min, Max}[deform]]}, 
   LegendLayout -> "Row"], Below]
 ]

Mathematica graphics


If the arrows are standard in the industry/field, then perhaps something like this:

Show[
 BoundaryMeshRegion[mesh],
 Graphics[Table[Arrow@Thread[{x, 5000 displacement {-1, 1}}], {x, 0.5, 4.5, 0.5}]]
 ]

Mathematica graphics

A combination of arrows and coloring. The legend is scaled by 10^6.

u = Function[{x, y}, 0];
v = Function @@ {{x, y}, displacement};
deform = (Norm[{u @@ #, v @@ #}] & /@ mesh["Coordinates"]);
Legended[
 Show[
  Graphics[
   ElementMeshToGraphicsComplex[mesh, All, 
    VertexColors -> (ColorData["Rainbow"] /@ Rescale[deform])]], 
  Graphics[Table[Arrow@Thread[{x, 5000 displacement {-1, 1}}], {x, 0.5, 4.5, 0.5}]]
  ],
 Placed[BarLegend[{"Rainbow", 10^6 Through[{Min, Max}[deform]]}, 
   LegendLayout -> "Row"], Below]
 ]

Mathematica graphics


VectorPlot issue

VectorPlot with a variable domain for x2 just hangs on me for reasons I don't understand (bug, maybe?). You can use RegionFunction instead:

VectorPlot[u, {x, 0, L}, {x2, -t[L]/2, t[L]/2}, 
 PlotLabel -> "vektoros", AspectRatio -> Automatic, 
 RegionFunction -> Function[{x, x2}, -t[x]/2 < x2 < t[x]/2]]

Mathematica graphics

| improve this answer | |
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  • $\begingroup$ Thanks for the answer but I want to solve the problem with finite difference method $\endgroup$ – wlkyr Apr 14 '15 at 13:31
  • $\begingroup$ @wikyr I used your solution from your quesiton!! I don't know what you mean. $\endgroup$ – Michael E2 Apr 14 '15 at 13:42
  • $\begingroup$ Oops, yes it is, but I would be interested how can I show with vector plot? for example why my code dont work?Because for constant cross section it works fine $\endgroup$ – wlkyr Apr 14 '15 at 14:02
  • 1
    $\begingroup$ @wlkyr Some problem in VectorPlot, I guess. See my workaround. $\endgroup$ – Michael E2 Apr 14 '15 at 14:09
  • 1
    $\begingroup$ @wlkyr You're welcome. $\endgroup$ – Michael E2 Apr 14 '15 at 14:21
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Eb = 2 10^7;
b = 1/4;
t0 = 1/2;
L = 5;
t[x_] := (t0 (L + x))/L;

s = DSolve[{Eb   (b t[x]^3)/12 y''''[x] == -1, 
            y''[0] == 0,  y''[L] == 0, y[0] == 0, y[L] == 0}, y, x][[1]];
displacement = y[x] /. s[[1]];
u = {-x2 D[displacement, x], displacement};
VectorPlot[u, {x, 0, L}, {x2, -4, 4}]

Mathematica graphics

| improve this answer | |
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  • $\begingroup$ Thanks for the reply, it was very helpful. But still there is a question, that how could be detected/express the deflection with vectorplot on the support with variable cross section? I attach the picture about the support with constant cross section. $\endgroup$ – wlkyr Apr 13 '15 at 11:12

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