General Mechanical

General Mechanical

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Definition of an ellipsoidal section with ASEC

    • User196534

      Hello to all,

      I am trying to define an ellipsoidal section of a beam (BEAM 188) via the ASEC function.

      Ansys documentation shows the following data to be entered in SECDATA:

      A, Iyy, Iyz, Izz, Iw, J, CGy, CGz, SHy, SHz, TKz, TKy
      A = Area of the section
      Iyy = Moment of inertia about the y axis
      Iyz = Product of inertia
      Izz = Moment of inertia about the z axis
      Iw = Deformation constant
      J = Torsion constant
      CGy = y coordinate of the centroid
      CGz = z coordinate of the centroid
      SHy = y coordinate of the shear center
      SHz = z coordinate of the shear center
      TKz = Thickness along the Z axis (maximum height)
      TKy = Thickness along the Y axis (maximum width)

      My ellipse has a dimension of 60 cm along the y axis and 30 cm along the x axis. Starting from the centroid, I have TKy = 30 cm and TKz = 15 cm.

      Here is the data I get by entering (A, Iyy, Izz, J, TKz, TKy):
      Area = 1360.5
      Iyy = 77939.
      Iyz = 0.0000
      Izz = 0.29275E+06
      Warping Constant = 0.0000
      Torsion Constant = 0.58550E+06
      Centroid Y = 0.0000
      Centroid Z = 0.0000
      Shear Center Y = 0.0000
      Shear Center Z = 0.0000
      Shear Correction-xy = 1.0000
      Shear Correction-yz = 0.0000
      Shear Correction-xz = 1.0000
      Maximum height = 15
      Maximum width = 30

      My question is the following.
      No matter what value I enter for TKz and TKy, the displacement of my beam is always the same. I tried with zero values (TKz = 0 cm, TKy = 0 cm) and extreme values (TKz = 100,000 cm, TKy = 100,000 cm).

      Do you know what is the impact of TKz and TKy and if they correspond to the values I entered?

      Here is a piece of my Apdl code if it helps:

      SECTYPE,1, BEAM, ASEC,di, 0
      SECDATA, A_di, Iy_di, , Iz_di, , J_di, , , ,TKz ,TKy, ,

      Thanks to those who will take the time to answer, I hope I have been clear enough in my explanations.

      Do not hesitate to ask me questions if this is not clear enough.

      Have a good day !

    • Ashish Khemka
      Ansys Employee


      How are you applying the load? Are you applying a force load or a displacement load?


      Ashish Khemka

      • User196534


        I am applying a force load. 

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