I tried adding a new node to the mesh.
If I call:
MeshTools::Generation::build_cube(mesh, 2, 2, 0, -1.0, 1.0, -1.0, 1.0, -1.0,
1.0,TRI3);
I simply add a new (copied) node to the mesh and change it in one element. That
gives me the desired mesh with 10 vertices. Defining a system of linear
Lagrangian elements on this mesh, the solution vector has 10 entries. All good.
On the other hand, if before adding the node, I call
MeshTools::Generation::build_cube(mesh, 1, 1, 0, -1.0, 1.0, -1.0, 1.0, -1.0,
1.0,TRI3);
MeshRefinement refinement(mesh);
refinement.uniformly_refine();
then the mesh object sees 10 vertices, but the solution vector has only 9
entries. Also the exporter does not recognize the additional node in the mesh.
This is how I’m adding the node in both scenarios:
Node * node = mesh.node_ptr(0);
auto nodeID = mesh.node_ptr(0)->id();
Node * new_node = new Node(*node);
new_node -> set_id(100);
mesh.add_node(new_node);
bool stop = false;
for (int elc = 0; elc < mesh.n_elem() ; ++elc)
{
Elem * elem = mesh.elem_ptr(elc);
for(int nn = 0; nn < elem->n_nodes(); ++nn)
{
int ID = elem->node_ptr(nn)->id();
if(ID == nodeID)
{
elem->set_node(nn) = new_node;
stop = true;
break;
}
}
if(stop) break;
}
Do you have any suggestion?
Thanks for the help,
Simone
> On Sep 11, 2018, at 12:58 PM, Roy Stogner <[email protected]> wrote:
>
>
> On Tue, 11 Sep 2018, Rossi, Simone wrote:
>
>>> If you start with the 10 node mesh
>>> on top and refine, then you will always have continuity of any C0 (or
>>> C1) solution variables at the two shared domain corner nodes.
>> This is what I would like to have. The idea is to introduce some random
>> “cuts” in the mesh, where I can apply some boundary conditions.
>> I thought the easiest way to achieve this would have been adding new
>> duplicated nodes on the side using AMR.
>
> Yeah, if that's the effect you want, then a single refinement followed
> by duplicating the non-corner internal boundary nodes is exactly what
> you want to do.
>
>> Could I “break” the mesh, use AMR, and then stitch back the nodes together?
>
> You could but it would be more of a pain. Just do one refinement and
> then break it afterward. Or, actually, you could also use TRI6
> instead of TRI3 and break the middle node between the two triangles.
> ---
> Roy
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