3D Frame Update

Following a comment here I have updated the code for the 3DFrame spreadsheet to calculate shear deflections correctly for sections with a different shear area for each transverse principal axis. The updated files can be downloaded from:

3DFrame.zip

A note has been added to the 3DFrame sheet with the definition of Section Properties I1 and I2 and SA1 and SA2:

Section Properties
I1 and I2 are the second moments of area about Axis 1 and Axis 2 respectively.
SA1 and SA2 are shear areas for forces in the directions of Axis 1 and Axis 2 respectively.

It follows that SA1 and I2 control deflections due to forces in the plane of Axis 1, and SA2 and I1 for forces in the plane of Axis 2. This is handled correctly in the main frame analysis, but not in the calculation of intermediate deflections. Note that in the workbook comparing results with a Strand7 analysis all the results are unchanged because the beam sections are rectangular, and have the same shear area in each direction.

This entry was posted in Excel, Frame Analysis, Newton, Strand7, UDFs, VBA and tagged , , , , , . Bookmark the permalink.

2 Responses to 3D Frame Update

  1. angelgalvanc says:

    Congratulations on this amazing work!
    I have one doubt regarding the distribute torsion, I am trying to calculate one cantilever using a distribute torsion along the beam but I get the wrong results.
    Maybe I am doing something wrong. Could you say me if the distribute force along the X-axis is a distribute torsion? In that case, how is possible to get 0 in the internal forces when a distribute force along the X direction?

    Like

    • dougaj4 says:

      The distributed loads are forces in the specified global direction, so if your cantilever is parallel to the X axis a distributed load in the X direction will be a distributed axial force, not a torsion.
      The only way to apply a distributed torsion is as a series of point moments about the X axis (for a beam parallel to the X axis).

      Like

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