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Plane Trusses Finite Elements 2 - Assembled Matrix

  1. Oct 19, 2012 #1
    Folks,

    I am having difficulty understanding how this global matrix is assembled with the naming convention used as shown in attached.

    The numbers in the corners such as 1(1,2) etc in figure 4.6.3 (b) denote the global and element numbers respectively.

    Can anyone shed light on how this assembled...? Thanks
     

    Attached Files:

  2. jcsd
  3. Oct 23, 2012 #2
    Maybe if I write out the matrix attached for easier read....the blanks indicate symmetry.

    ##\begin{bmatrix}
    k^1_{11}+k^3_{11} &k^1_{12}+k^3_{12} &k^1_{13} &k^1_{14} &k^3_{13} &k^3_{14} \\
    &k^1_{22}+k^3_{22} &k^1_{23} &k^1_{24} &k^3_{23} &k^3_{24} \\
    & & k^1_{33}+k^2_{11} &k^1_{34}+k^2_{12} &k^2_{13} &k^2_{14} \\
    & & &k^1_{44}+k^2_{22} &k^2_{23} &k^2_{24} \\
    & & & & k^2_{33}+k^3_{33} &k^2_{24}+k^3_{34} \\
    & & & & & k^2_{44}+k^3_{44}
    \end{bmatrix}##

    The above is the matrix I am trying to understand how it was assembled based on the attached picture..

    The numbers in the corners such as 1(1,2) etc in figure 4.6.3 (b) denote the global and element numbers respectively.

    There are 2 displacement degrees of freedom (horizontal and vertical) at each node of the element...thanks
     

    Attached Files:

  4. Oct 25, 2012 #3
    After some searching online I have a found an easy way of assembling the global matrix for this problem.

    If we focus on element 3 which has global nodes 1 and 3. we can create the element freedom table 'EFT' for this element by the following

    2 dof's times global number 1 minus 1=1
    2 dof's times the global number 1 =2

    2 dof's times global number 3 minus 1=5
    2 dof's times the global number 3 =6

    ( I am interested to know what the above technique is based on)

    Thus the EFT is {1,2,5,6}. Similarly for the other 2 elements.

    Then one combines the EFT for each element into the global matrx (2 dof's times number elements 3= 6 gives a 6 matrix.)
     
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