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6 Nodes, 4 Elements, Size of Stiffness Matrix?

  1. Apr 19, 2013 #1
    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
     
  2. jcsd
  3. Apr 19, 2013 #2
    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.
     
  4. Apr 19, 2013 #3

    AlephZero

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    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.
     
  5. Apr 19, 2013 #4
    So is the (number of nodes) = (number of rows) and the (DOFs per node) = (number of columns)?
     
  6. Apr 19, 2013 #5
    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?
     
  7. Apr 19, 2013 #6

    SteamKing

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    Stiffness matrices are square and symmetric. So your stiffness matrix will be 8x8.
     
  8. Apr 19, 2013 #7
    So is there a (nodes,DOFs) equation that states the size of a stiffness matrix for a system?
     
  9. Apr 19, 2013 #8

    AlephZero

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    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.
     
  10. Apr 19, 2013 #9
    So for my 6 node example, the stiffness would be sized (number of nodes = 6) x (DOFs per node = 2) = 12 x 12?
     
  11. Apr 19, 2013 #10

    SteamKing

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  12. Apr 19, 2013 #11
    So that's the most difficult part of FEM, right?
     
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