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Combining sub-matrices into global matrix

  1. Sep 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Combine five element matrices into single matrix forming a global stiffness matrix for simple truss analysis (five elements, four nodes).

    2. Relevant equations

    -See attached image-

    3. The attempt at a solution

    Where the four sub-matrices overlap in the global matrix two or three values/numbers occupy the same position in the global matrix (two separate values for AA, three separate values for BB, and so on). Should I simply add these values to form a single value for the global matrix? For example: if AA1 = -10, and AA2 = +5 should I simply combine them as AA = -10+5 = -5?

    Is it really as simple as that?

    Thanks.
     

    Attached Files:

  2. jcsd
  3. Sep 1, 2012 #2
    Yes, that is the way you build a global matrix from individual elemental matricies.
     
  4. Sep 1, 2012 #3
    Thank you very much. I'm assuming your post is based on my AA = AA1 + AA2 and not just the matrix image I posted! I very much appreciate your confirmation.
     
  5. Sep 1, 2012 #4
    Suppose you had a triangular element with nodes 2,6,8 at the corners. You would add it into the global matrix as follows where A could be a 100X100 matrix. a is the 3X3 contribution from the element. Below would be the coding in Fortran.
    The first indice is the row, the second is the column.

    A(2,2)=A(2,2) + a(1,1)
    A(2,6)=A(2,6) + a(1,2)
    A(2,8)=A(2,8) + a(1,3)

    A(6,2)=A(6,2) + a(2,1)
    A(6,6)=A(6,6) + a(2,2)
    A(6,8)=A(6,8) + a(2,3)

    A(8,2)=A(8,2) + a(3,1)
    A(8,6)=A(8,6) + a(3,2)
    A(8,8)=A(8,8) + a(8,3)
     
  6. Sep 3, 2012 #5
    In my last equation there is a typo.

    a(8,3) should be a(3,3).
     
  7. Sep 4, 2012 #6
    Thank you. It was kind of you to post that, and I appreciate it. Apologies for not replying more promptly but I had a couple of deadlines to meet and my mind was away from the matrix problem for a day or two.

    Now my mind is back on this issue I am much more confident but stuck at the final stage. I’ve attached an example someone else produced which shows three stages. I can get from stage one to stage two but not two to three. I don’t know where the 160083 came from, or the three bracketed numbers. Any help with this would be very welcome!
     

    Attached Files:

  8. Sep 5, 2012 #7
    The problem is solved by using determinants where for a1 you divide the determinant of the stiffness matrix with the load vector substituted for the first column by the determinant of the original stiffness matrix. All numbers are evenly divisible by 3 so it is factored out.
     
  9. Sep 5, 2012 #8
    Thank you once again. Your reply to me is just the sort of thing I want to read in the text books, yet it is nearly always missing and so I often learn little from them. Perhaps I've just been reading the wrong books!
     
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