ConBeamU 4.13

The continuous beam analysis spreadsheet, ConBeamU, is now up to version 4.13 with some minor bug-fixes, and the addition of a non-linear option for the BeamAct functions.

See ConBeamU for more details of the functions available in the spreadsheet.

Download the latest version from

The BeamAct2D and BeamAct3D user defined functions (UDFs) return beam actions and deflections for any beam subject to specified loads and end deflections.  The beam may have any number of segments with different section properties, and may be specified in either global or local coordinate systems.  The new non-linear option allows non-linear geometric effects to be taken into account.

Most of the function input is unchanged from previous versions:


The optional non-linear arguments specify if a non-linear analysis is required, the target relative error in maximum deflection, and the maximum number of iterations.  The support stiffness (translational and rotational) may also be supplied.  The default is fixed at both ends for both translation and rotation:


The output is before, with output at any number of specified locations:


Results are shown below for a 3D analysis of a 10 m long beam under combined axial and transverse load.  The blue and green lines are the results ignoring geometric non-linear effects, and the grey and red are the output from both the spreadsheet and a non-linear analysis in Strand7, with the beam divided into 16 segments:


Due to differing approximations in the two programs there are small differences in the shear results:


Bending moments are in good agreement.


The 2D function gives similar results:

conbeamu6-8Note that these functions are still under development.  All results must be independently verified using other software.

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

7 Responses to ConBeamU 4.13

  1. ivanStructEng says:

    Hi, Could you please tell me how to create more output stations for calculation at “ConbeamU1”.?

    Thank you very much.


    • dougaj4 says:

      The example is currently set up to return a specified number of points per span, as listed in Column I (Sections/Span). To change the number of spans or points and the output display:
      – Enter the details of support locations and number of sections/span
      – Go to the top-left cell of the output range (J22)
      – Press Ctrl-Shift-S
      -Reformat the extended output range as required
      If your output range is very long you might want to move or delete the results starting at cell K122

      There are other options, but I will do a new post in a few days with more details.


  2. Pingback: Setting up UDF Applications | Newton Excel Bach, not (just) an Excel Blog

  3. metrox says:

    Is it possible to output result for only one interested point and interested value.
    Lets say output only Shear force at distance 5.3m etc. ?


    • dougaj4 says:

      Sure, assuming you are using ConbeamU (or another “unit aware” function):
      Enter m (or distance unit) and 5.3 in two cells in a column
      Select those two cells as the OutPoints argument
      Set the ListOutPoints argument to True
      Press ctrl-shift-enter

      That will display all the results at X = 5.3
      To display just the shear result, wrap it in the Index function:

      Finally to display just the one result (rather than multiple copies of it), press Ctrl-Shift-S

      Note that the OutUnits range should not be changed, it still needs units for all the available output values, even though you are only displaying one.

      Using the example in the download file it becomes:


  4. says:

    I am not able to validate the calculation for the continuous beam. Either I have not set the correct support types or the excel is unable to calculate it correctly.

    Conditions –

    Continuous beam of 3 m.
    4 simple supports at x = 0,1,2 & 3 i.e. 3 equal spans of 1m.
    UDL throughout the length of 15kN/m.
    EI= 1.6667×10^6.
    GA ignored.
    I am using Conbeam Check sheet.

    According to the text book formula, for this loading condition, the shear stress at x=0 should be 0.4*15*10^3*3= 18000. This never occurs in the excel. This formula is from the text book and is for a continuous beam – 3 equal span – all span loaded with udl.

    If this not clear, could you tell what would be the support conditions for a simple supported beam? Both transitional and rotational stiffness if I am using Conbeam Check sheet?

    Many thanks,


  5. dougaj4 says:

    In your formula the length should be the span length, not the total length. Also the factor of 10^3 converts kN to N. For a result in kN the formula should be:
    0.4*15*1 = 6 kN
    which is what the spreadsheet gives.
    If you want to display the result in N use the ConbeamU1 sheet and just enter N at the top of the Shear results. The ConbeamU function adjusts for units, the Conbeam function assumes consistent units.

    The default support conditions are rigid for translation and free for rotation, so you don’t need to enter anything for support stiffness.
    Also note that the shear result is a force, not a stress.


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