2
$\begingroup$

I dont know why dont work but I want to animate torsion prismatic beam.

a = 1; b = 2; L = 10; G = 1; dx = dy = dz = 0.25; J = 2.1849801331564294`; 
Clear[x, y, z, T]; k = T/( G J);
phi = (32 G k a)/Pi^3 Sum[1/n^3 (-1)^((n - 1)/2) (1 - Cosh[n Pi y/(2 a)]/
Cosh[n Pi b/(2a)]) Cos[(n Pi x)/(2 a)], {n, 1, 10, 2}];
tauyz = -D[phi, x]; tauxz = D[phi, y];
gammayz = tauyz/G;
gammaxz = tauxz/G;
Rotate[{x_, y_, z_}, Theta_] := {x Cos[Theta] - y Sin[Theta],
x Sin[Theta] + y Cos[Theta], z};
Off[Graphics3D::"gprim"];
Do[Show[Graphics3D[Table[alpha = 0.5 k z;
p1 = p2; 
p2 = Map[({x, y} = #;
Rotate[{x + 0 gammaxz/2, 
y + 0 gammayz/2, 
z + (x gammaxz + y gammayz)/
  2}, alpha]) &,
Flatten[{Table[{x, -b/2}, {x, -a/2, a/2 - dx, dx}], 
Table[{a/2, y}, {y, -b/2, b/2 - dy, dy}], 
Table[{x, b/2}, {x, a/2, dx - a/2, -dx}], 
Table[{-a/2, y}, {y, b/2, dy - b/2, -dy}]}, 1]];
n = Length[p2];
If[z == 0, Polygon[p2], 
  Table[Polygon[{p1[[i]], p2[[i]], p2[[Mod[i, n] + 1]], 
     p1[[Mod[i, n] + 1]]}], {i, 1, n}]], {z, 0, L, dz}], 
ViewPoint -> {-1, 1, 1}, ViewVertical -> {0, 1, 0}, Axes -> True, 
AxesLabel -> {"x", "y", "z"}, 
PlotRange -> {{-a, a}, {-b, b}, {0, L}}]], {T, 0, 1, 0.1}];
SelectionMove[
EvaluationNotebook[], All, GeneratedCell];
FrontEndTokenExecute["CellGroup"];
FrontEndTokenExecute["OpenCloseGroup"];
FrontEndTokenExecute["SelectionAnimate"];  
$\endgroup$
3
  • 3
    $\begingroup$ What behaviour were you expecting, that is not occurring? It's hard to help you without more information. "It doesn't work" is not very informative, unfortunately. $\endgroup$
    – Verbeia
    Oct 3, 2014 at 7:31
  • $\begingroup$ Rotate is a built-in function. You cannot redefine it just like that. Furthermore, lots of undefine variables and what's 0 gammaxz supposed to do? And as to animation, why the rather strange use of frontend tokens? Are you aware of the Animate function? $\endgroup$ Oct 3, 2014 at 10:42
  • 3
    $\begingroup$ Since we have a good answer from Junbo Lee, I think this question should be reopened, as there was clearly enough info for him to answer it. $\endgroup$
    – m_goldberg
    Oct 3, 2014 at 11:46

1 Answer 1

11
$\begingroup$

I think this code might be old version. You made some mistake so I changed your code like this.

  1. Rotate -> rotate

  2. Do -> Table

  3. SelectionAnimate -> Manipulate or Animate

We can give change variable T in Manipulate but that computation time is so long. Thus I set the Graphics3D-s to frames and apply Manipulate or Animate.

Have try like this.

a = 1; b = 1; L = 10; G = 1; dx = 
 dy = dz = 0.25; J = 2.1849801331564294`;
Clear[x, y, z, T];
k = T/(G J);
phi = ((32*G*k*a)*Sum[((-1)^((n - 1)/2)*(1 - Cosh[(n*Pi*y)/(2*a)]/
                      Cosh[(n*Pi*b)/(2*a)])*Cos[(n*Pi*x)/(2*a)])/
       n^3, {n, 1, 10, 2}])/Pi^3; 
tauyz = -D[phi, x]; tauxz = D[phi, y]; 
gammayz = tauyz/G;
gammaxz = tauxz/G;
rotate[{x_, y_, z_}, Theta_] := {x Cos[Theta] - y Sin[Theta], 
   x Sin[Theta] + y Cos[Theta], z};
Off[Graphics3D::gprim];

frames = Table[
   Graphics3D[
    Table[alpha = 0.5 k z; p1 = p2; 
     p2 = (({x, y} = #1; 
          rotate[{x + (0 gammaxz)/2, y + (0 gammayz)/2, 
            z + 1/2 (x gammaxz + y gammayz)}, alpha]) &) /@ Flatten[{
         Table[{x, -(b/2)}, {x, -(a/2), a/2 - dx, dx}], 
         Table[{a/2, y}, {y, -(b/2), b/2 - dy, dy}],
         Table[{x, b/2}, {x, a/2, dx - a/2, -dx}], 
         Table[{-(a/2), y}, {y, b/2, dy - b/2, -dy}]}, 1];
     n = Length[p2];
     If[z == 0, Polygon[p2], 
      Table[Polygon[{p1[[i]], p2[[i]], p2[[Mod[i, n] + 1]], 
         p1[[Mod[i, n] + 1]]}], {i, 1, n}]], {z, 0, L, dz}],
    ViewPoint -> {-1, 1, 1}, ViewVertical -> {0, 1, 0}, Axes -> True, 
    AxesLabel -> {"x", "y", "z"},
    PlotRange -> {{-a, a}, {-b, b}, {0, L}}],
   {T, 0, 1, 0.1}];

You can manipulate motion with this code.

Manipulate[frames[[i]], {i, 1, Length[frames], 1}]

or

Animate[frames[[i]], {i, 1, Length[frames], 1}, 
 AnimationRunning -> False]

Mathematica graphics

$\endgroup$
2
  • $\begingroup$ There is a glitch with a = 1 and b = 2 which isn't present when both are equal to 1 so I edited your answer and the plot (with a better quality :)). Feel free to rollback :) $\endgroup$
    – Öskå
    Oct 3, 2014 at 15:26
  • $\begingroup$ @Öskå I got it. thanks $\endgroup$
    – Junho Lee
    Oct 3, 2014 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.