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Problem with 3D Fem sparse matrix ( plz help me)

Submitted by sameer2008 on
FEM

 hi,

 i have calculated the local stiffness matrix and form global stiffness matrix. but when i am solving the equation  it is saying the matrix (stiffness) is badly scaled and near to singular matrix. i think there is one problem in assembling the local matrix. anybody can give  the idea  what will be the probabal solution. i have done the 2d and its working fine. i have some doubt on  assembling. what i have done is that i have use the same algorithm of 2d( linear basis is triangle) assemble but different degrees of freedom and ofcourse node number will be differents for 3D ( basia is linear tetrahedron).

 

 

plz help me  i am on my deadline 

 

Alejandro Orti…

What I would do in your case is to try with a simple example done by hand. Thus, you know what numbers the stiffness matrix should have. There are some reasons to get a singular matrix. One of them might be that the global stiffness matrix is not variationally correct meaning that the essential boundary conditions are not properly included in your stiffness matrix. After you form the global stiffness matrix you need to modify it along with the force vector to include the essential boundary conditions. Otherwise the matrix will be singular. If this is not your problem then a wrong assembly procedure might be the cause.

Best wishes in your project,

Alejandro A. Ortiz

Wed, 10/15/2008 - 06:34 Permalink
sameer2008

 

thank you very much. i will do the things that u told. 

but if i dont include the boundary condition and simpliy find the stiffness matrix for each local tetrahedron and to get the global stiffness matrix, i just add the  local matrices in there corresponding  positions with respect to the global matrix, even after that it will show any problem? 

 for assembling the local matrices i just  did these things

 kk=zeros(total_nod,total_nod); global stiffnes matrix

kk(t(iel,:),t(iel,:))=kk(t(iel,:),t(iel,:))+k; where k is local stiffness matrix for t(iel,:) element

i am using matlab software

once again thank you ( if u have any suggestion plz suggest ) 

Wed, 10/15/2008 - 15:27 Permalink
sameer2008

 Dear Sir,

I have one question regarding the quadratic elment in 2D. if i take triangular basis in quadratic form ( six nodes) the stiffness matrix for local triangle is increases 3x3 to 6x6 matrix because of six basis function for six nodes. For linear basis it is 3x3 matrix. If my total number of triangles and node points are constant  for mesh, how i will take care of the size of global stiffness matrix? if initial node number is 1500 and triangle 3000, after considering quadratic  6 points triangle, will the global stiffness matrix incease to 4500x4500 ( according to node number)? can i keep it constant 1500x1500 matrix?

 

please give some idea 

Wed, 02/18/2009 - 19:47 Permalink
Alejandro Orti…

Not sure if I have understood well your question. Let's assume you have a mesh of N nodes with DOF number of degree of freedoms each, then the total size of the global stiffness matrix before enforcing essential boundary conditions will be (N*DOF) by (N*DOF). In your case it seems that DOF = 1, so you simply will get N by N. Therefore, if you switch from linear triangle to quadratic triangle the size of the global stiffness matrix will be affected.

Wed, 02/18/2009 - 20:16 Permalink
Temesgen Markos

Hi Sameer,

With out the BC's, the stiffness matrices are simply singular.
This isn't a numerical issue. Even when you solve a DE analytically your
solution is not uniquely defined until you supply appropriate boundary
conditions. So include the BC before you go ahead.

 

I am Temesgen and I approve this message!

Wed, 10/15/2008 - 16:29 Permalink