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Plane Trusses Finite Elements

  1. Oct 15, 2012 #1
    Folks,

    The element equations for a uniform bar element with constant EA according to the attachment is given as

    ##\displaystyle \frac{E_a A_e}{h_e}\begin{bmatrix}
    1 &0 &-1 &0 \\0
    &0 &0 &0 \\-1
    &0 &1 &0 \\
    0 &0 &0 &0
    \end{bmatrix}\begin{Bmatrix}
    u^e_1\\v^e_1
    \\u^e_2
    \\ v^e_2

    \end{Bmatrix}=\begin{Bmatrix}
    F^e_1\\0\\F^e_2
    \\0

    \end{Bmatrix}##

    I am just wondering, can this not be also written as

    ##\displaystyle \frac{E_a A_e}{h_e}\begin{bmatrix}
    1 &-1 &0 &0 \\0
    &0 &0 &0 \\-1
    &1 &0 &0 \\
    0 &0 &0 &0
    \end{bmatrix}\begin{Bmatrix}
    u^e_1\\u^e_2
    \\v^e_1
    \\ v^e_2

    \end{Bmatrix}=\begin{Bmatrix}
    F^e_1\\0\\F^e_2
    \\0

    \end{Bmatrix}##...?
     

    Attached Files:

  2. jcsd
  3. Oct 15, 2012 #2

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    You could do that, but things will get very confusing in the long run if the vector of displacements is in a different order to the vector of forces. For example the work done by the external forces will not be just ##F^T x## and the strain energy in the structure will not be ##(1/2)x^TKx##. You would need to keep track of the different order of the two vectors.

    You could use a different consistent ordering of the variables, like

    ##\displaystyle \frac{E_a A_e}{h_e}\begin{bmatrix}
    1 &-1 &0 &0 \\
    -1 &1 &0 &0 \\
    0 &0 &0 &0 \\
    0 &0 &0 &0
    \end{bmatrix}\begin{Bmatrix}
    u^e_1\\u^e_2
    \\v^e_1
    \\ v^e_2

    \end{Bmatrix}=\begin{Bmatrix}
    F^e_1\\F^e_2
    \\0\\0
    \end{Bmatrix}##
     
  4. Oct 16, 2012 #3
    OK thanks. Further on I am just wondering how he determined the transformations between the 2 sets of coordinates systems global ##(x,y)## and local ##(x_e,y_e)## based on the attachment.

    He writes

    ##x_e=x \cos \theta_e+y \sin \theta_e##, ##y_e=-x \sin \theta_e+y \cos \theta_e##

    ##x=x_e \cos \theta_e- y_e \sin \theta_e##, ##y=x_e \sin \theta_e+y_e \cos \theta_e##

    I would have written ##x_e=x \cos \theta_e-y \sin \theta_e##, ##y_e=x \sin \theta_e+y \cos \theta_e## for ##(x_e,y_e)##

    Then based on my above i write it in matrix form and get the inverse to arrive at

    ##x=x_e \cos \theta_e+y_e \sin \theta_e##, ##y=-x_e \sin \theta_e+y_e \cos \theta_e##
    (On the side, I would like to know how I would obtain ##(x,y)## graphically)
     
    Last edited: Oct 16, 2012
  5. Oct 16, 2012 #4
    It is ok, I have it sorted. The book is correct. Cheers
     
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