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Solving a time dependant circular excavation

Posted 10 years ago

Hello,

I want to simulate a the deformation of a circular excavation. So I defined the displacement vector (u=ur(r)), and then the deformation and the stress tensor, and I wrote the equation I want to solve : div [Sigma] = D[u,{t,2}]

Then I wrote the initial conditions : at t=0, the field is charged normally, and at t=1, we excavate a part. Here is the code :

Excavation circulaire, dynamique

lambda = 1; mu = 1; ro = 1; ur =.;

u[r_, teta_, t_] = {ur[r, t], 0}; 
gradientu[r_, teta_, t_] = Grad[u[r, teta, t], {r, teta}, "Polar"];
epsilon[r_, teta_, t_] = 
  0.5*(gradientu[r, teta, t] + Transpose[gradientu[r, teta, t]]);
\[Sigma][r_, teta_, t_] = 
  2*mu*epsilon[r, teta, t] + lambda*Tr[epsilon[r, teta, t]]*IdentityMatrix[2];

eq1 = Div[\[Sigma][r, teta, t], {r, teta}, "Polar"][[1]];
eq2 = ro*D[ur[r, t], {t, 2}];

ursol = NDSolveValue[{eq1 == -eq2,
    \[Sigma][r, teta, 0][[1, 1]] == 1, \[Sigma][r, teta, 0][[2, 2]] == 1,
    DirichletCondition[\[Sigma][r, teta, t][[1, 1]] == 1, r == 5], 
    DirichletCondition[\[Sigma][r, teta, t][[1, 1]] == 0, r == 1 and t >= 1]},
    ur, {r, 1, 5}, {t, 0, 2}];

But when I run the code, Mathematica stops, and return nothing. Can you help me please ?

Thank you

POSTED BY: alexis rosuel
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