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Finite element method

  1. Nov 5, 2017 #1
    1. The problem statement, all variables and given/known data

    Linear and quadratic elements differ because of the extra mid-nodes on quadratic elements. Quadratic elements can "bend", linear can't. How do you define the DOF of a element? (see at solution attempt)


    2. Relevant equations
    -

    3. The attempt at a solution
    For example: a linear triangle has 3 nodes. Each node has 2 DOFs (x and y). The total field variable will have 3 x 2 = 6 DOFs? Why does this element don't have 3 DOFs per node? (2 translations + rotation)

    Wat is the problem with following elements, why is this not possible? What with a quadratic and linear element adjacent to each other. This is not compatible, how can this be visualized?

    Thanks in advance!
     
  2. jcsd
  3. Nov 9, 2017 #2
    the hex elements along with the 10-node tetras get close to the solution and provide conservative results. The 4-node tetras, however, which are actually degenerate 8-node hex elements because ANSYS removed their 4-node tetra elements along time ago, show only half the deflection under the same load. The extra stiffness also causes the modal frequencies to be higher. In this case, the frequencies of the low-order tetras were about 40% higher than the high-order tetras. When you’re looking at operating ranges, this could also lead to unconservative conclusions.
     
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