6 Nodes, 4 Elements, Size of Stiffness Matrix?

[ThurmanMurman]

Howdy,

If I have the following configuration of nodes:

2---4--6
| \ | /|
| \ | / |
| \| / |
| \/ |
1---3--5

What should the dimensions of my FEM stiffness be?

Thanks

 

[ThurmanMurman]

Alright my attempt at formatting got messed up once I posted. Nodes 2-4-6 are equally spaced across the top, and nodes 1-3-5 sit below 2-4-6 on the bottom. 2-4-6 are respectively connected to 1-3-5 vertically, and 2 is connected to 3 diagonally, and 3 is connected to 6 diagonally.

 

[AlephZero]

The matrix size is (number of nodes) times (number of degrees of freedom per node).

The number of elements and how they connect is irrelevant.

 

[ThurmanMurman]

So is the (number of nodes) = (number of rows) and the (DOFs per node) = (number of columns)?

 

[ThurmanMurman]

Just to check my understanding, I think that my arrangement of nodes (4 total) with 2 DOFs each (8 total) leaves me with a 4X8 stiffness matrix. Is that correct?

 

[SteamKing]

Stiffness matrices are square and symmetric. So your stiffness matrix will be 8x8.

 

[ThurmanMurman]

So is there a (nodes,DOFs) equation that states the size of a stiffness matrix for a system?

 

[AlephZero]

So is the (number of nodes) = (number of rows) and the (DOFs per node) = (number of columns)?

No, the number of rows and columns are both equal to (number of nodes) x (DOFs per node) .

Sorry if that wasn't clear - I took it as "obvious" that stiffness matrices are square.

 

[ThurmanMurman]

So for my 6 node example, the stiffness would be sized (number of nodes = 6) x (DOFs per node = 2) = 12 x 12?

 

[SteamKing]

Yes.

 

[ThurmanMurman]

So that's the most difficult part of FEM, right?