6 Nodes, 4 Elements, Size of Stiffness Matrix?

Click For Summary

Discussion Overview

The discussion revolves around determining the dimensions of the stiffness matrix in a finite element method (FEM) context, specifically for a configuration of 6 nodes and 4 elements. Participants explore the relationship between the number of nodes, degrees of freedom (DOFs) per node, and the resulting stiffness matrix size.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asks about the dimensions of the stiffness matrix based on a specific node configuration.
  • Another participant clarifies that the matrix size is determined by the number of nodes multiplied by the number of degrees of freedom per node.
  • A participant questions whether the number of nodes corresponds to the number of rows and the DOFs per node to the number of columns in the stiffness matrix.
  • One participant asserts that stiffness matrices are square and symmetric, leading to a conclusion that the stiffness matrix for their example would be 8x8.
  • Another participant inquires if there is a general equation relating nodes and DOFs to the stiffness matrix size.
  • There is a confirmation that for 6 nodes with 2 DOFs each, the stiffness matrix would be sized 12x12.
  • One participant reflects on the complexity of FEM, suggesting that understanding the stiffness matrix size is a challenging aspect of the method.

Areas of Agreement / Disagreement

Participants generally agree on the relationship between nodes, DOFs, and the stiffness matrix size, although there are clarifications and corrections regarding the interpretation of these relationships. The discussion does not reach a consensus on the complexity of FEM as a whole.

Contextual Notes

Some participants express uncertainty about the implications of the number of elements and their connections on the stiffness matrix size, indicating that further clarification may be needed.

ThurmanMurman
Messages
12
Reaction score
0
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
 
Engineering news on Phys.org
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.
 
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.
 
So is the (number of nodes) = (number of rows) and the (DOFs per node) = (number of columns)?
 
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?
 
Stiffness matrices are square and symmetric. So your stiffness matrix will be 8x8.
 
So is there a (nodes,DOFs) equation that states the size of a stiffness matrix for a system?
 
ThurmanMurman said:
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.
 
So for my 6 node example, the stiffness would be sized (number of nodes = 6) x (DOFs per node = 2) = 12 x 12?
 
  • #10
Yes.
 
  • #11
So that's the most difficult part of FEM, right?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 3 ·
Replies
3
Views
10K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
7K
Replies
5
Views
2K