Critical bending moment: How to calculate it with numerical analysis (in RFEM)

Lateral torsion buckling (LTB) is a very dangerous phenomenon, that can easily cause the collapse of a poorly designed beam. In civil engineering codes, the critical bending moment is crucial in the proper design of bent beams susceptible to LTB, as it allows for slenderness calculation.  In “typical” cases everything is ok since code equations allow engineers to obtain the value of the critical moment. Those equations, however, require a lot of conditions to be met in order to work, and if at least one is not fulfilled… problems start. Today I will show you how to calculate critical moment in any situation you may encounter in your engineering work

Critical bending moment equation and required conditions

There are many equations for the critical moment which vary slightly in terms of parameters (some are more complicated/accurate than others). If you are interested in hand calculations of the critical moment I believe this is a nice guide. Note that most available equations follow the same set of rules that need to be followed in order to use the equation.

Required conditions for calculation critical bending moment according to equation:

  • Beam must be symmetric in at least 2 planes – this is a huge drawback, forget L-sections, C-sections (even threw old code in my country stated that for C-sections you can calculate slenderness as for I-section and then reduce it by 25%), and many others including custom welded cross-sections.
  • Beam must have a constant cross-section on its length – so no “optimized” beams with thinner flanges near end hinged supports.
  • Beam must be straight (linear) – equations do not allow for curved beams like hopper rings in silos made of corrugated sheets.
  • Beam must be bended in plane of it’s symmetry – this actually is important, as when you have bending in 2 directions you do not fulfill this requirement. You would be surprised how many beams are bended in both directions.
  • Beam must be restraint in transverse movement and in cross-section plane rotations at its end – so many purlins do not fulfill this condition – it is not enough to screw the beam by it’s bottom flange – top flange have to be supported in transverse direction as well.
  • It would be nice if the beam would have a relatively simple moment distribution along it’s length – this is often the case, but from time to time it might get problematic.

As you can see there are many limitations and many engineers are not aware of them. Each time your software make a design for you, you actually assume that all of the above is correct, and unfortunately some of those assumptions, when unfulfilled may have a drastic influence on capacity reduction due to lateral torsional buckling. Of course it is impossible to verify each beam in the design, but for the most important elements or those obviously not fulfilling the requirements given above this should be verified. If you cannot find an equation to calculate critical moment in your case doesn’t worry – there is a numerical way to solve this problem.

Numerical method for critical bending moment calculation

Most of the finite element programs have the possibility to calculate the critical moment. As long as your software uses plates and can do linear buckling you should be fine 🙂

Actions to take:

  • Model the beam using plate elements: define surfaces and apply corresponding thickness to each one
  • Support your beam in a realistic way: remember that software now “sees” your beam as a 3D object, you can for instance support only one edge of the beam
  • Load your beam in a realistic way: same as above, since model “sees” your beam in 3D you can actually choose at which part of the beam load is applied
  • Perform a linear analysis: check for the maximal moment in the analyzed beam (sometimes I use a secondary simplified beam model so I do not have to integrate stresses from plates to derive bending moment in the beam).
  • Perform linear buckling analysis: outcome would be a stability failure shape and critical load multiplier.
  • Bending moment from linear analysis multiplied by critical load multiplier is the critical bending moment

Below I have recorded how to do this in RFEM software. I use it in my engineering office for beam static and simpler designs, while Femap and NX Nastran is used in mode demanding cases.

If you are interested in FEA analysis to be sure to see my free FEA course – you can get it below!

This article was created as Mathias asked me on how to calculate a critical moment in one of the elements he was designing. If you have a problem with design feel free to write me about it – I will try to help 🙂

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If you have any questions feel free to leave them below in the comments section.

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10 Comments

  1. Mathias June 2, 2016 at 1:28 pm - Reply

    Super stuff, thanks for this really good tutorial!

    After have run the analysis and got the critical moment for an optimized beam (*), I still have 2 more questions:
    1)if I wish to calculate Mb,Rd can I continue use the general method of the Eurocode?
    2)if yes, can the section modulus of the beam be integrated along the length for the formula of Mb,Rd?

    (*) optimized beam : asymmetric beam with different sections along its length

    • Łukasz Skotny June 2, 2016 at 7:15 pm - Reply

      Hi Mathias,

      Thank you for kind words and good questions 🙂

      1. Yes – this method provides the critical moment, you can then calculate slenderness as usual and then proceed with “normal” Eurocode procedure
      2. This is tricky – basically Eurocode in most places refers to members with constant cross section (hence need for numerical calculation of critical moment if the cross-section is not constant). I would say at this point we should divide capacity of a beam into 2 categories:

      A. Plastic capacity – this is the simpler part – in each cross section you have a certain moment and a certain moment of inertia so design for plastic capacity is easy and Eurocode provides guidelines for such calculations.

      B. Stability capacity – whether or not the beam will loose stability. In Eurocode procedure you are using “buckling curves” that are assign to your cross-section (the is the parameter alpha in the procedure). Those refer to beams with constant cross section, and as such are not the best fit. Also the way capacity is calculated you use the reduction factor to plastic capacity (in order to take stability into account), and that would lead to situation in which beam would have “different” stability capacity in different cross sections… while there is only one stability capacity – a certain multiplier for the loads that when applied leads to stability failure (theoretically there are more since there is unlimited amount of eigenvalues but this is beyond point – in engineering interest is in the first one since it will cause collapse).

      I would say that when you have a constant cross section (even a weird one) using critical moment and Eurocode procedure make sense, however if the cross section is not constant I would rather perform a “numerical” design which is also described in Eurocode. I will try to make a post about how to do that in near future 🙂

      Hope that helps 🙂

      • Mathias June 3, 2016 at 11:04 am - Reply

        Thank you it helps a lot, I’m waiting for the next post to have the full answer to my question then 🙂

        • Łukasz Skotny June 4, 2016 at 4:43 pm - Reply

          Hey Mathias,

          The way I organize things here it will take few weeks to make an article. For now I would suggest calculating stability capacity using minimal cross-section this is on the safe side of course 🙂

          Have a good one 🙂

  2. Michael November 22, 2016 at 1:32 pm - Reply

    Thank you Lukasz
    I from your video, LTB and stability issues is now clear to me.
    Michael

    • Łukasz Skotny November 24, 2016 at 8:08 am - Reply

      Hi Michael,

      I’m happy that you enjoyed the video 🙂

  3. Jakub Beszczyński May 30, 2018 at 11:10 am - Reply

    Hi Łukasz! Thank You for this video and the whole blog. I have succesfully tested the LBA approach to find Mcr value for an I-beam. But as I was researching the realm of LBA-to-Mcr approach, I’ve stumbled upon a problem. For some examples, in my case a RHS beam, no eigenvalue could be found. Lowering the demanded convergence factor helped find eigenvalues, but the results were trash (obviously). Making the section more slender resulted in nice results again, and then stabilizing the slender section with a lateral support in the middle made it impossible to find eigenvalues yet again.
    My take is that RHS’s are really inherently stable, so this means just that torsional instability would not occur at reasonable load level.
    Am I interpreting this correctly?

    Thanks and have o good day!

    • Łukasz Skotny May 30, 2018 at 11:20 am - Reply

      Hey Jakub!

      I’m really happy that you like the blog : )

      I guess that by RHS you mean Rectangular Hollow Section (and not “Round Hollow Section”). If yes, then it would depend on the cross-section dimensions. If it is a square in cross-section there will be no lateral torsional buckling. The effect happens when you are bent in the “strong axis” and cross section deforms in the “weak axis”. If both axes are the same, there is no effect. The same would go for a Circular pipe (it doesn’t have a strong axis so there is no lateral torsional buckling).

      However, if you have a rectangle and you apply bending in the strong axis it should show lateral torsional buckling as normal. Even though critical capacity is higher than say for an I-Section, there is still a “capacity”. Meaning that you should be able to calculate critical moment. Maybe if you are “a bit rectangular” but very close to a square, the multiplier is so high that your LBA algorithm goes crazy – hard to guess really…

      I hope this will help you out at least a bit!

      All the best
      Ł

      • Jakub Beszczyński May 30, 2018 at 1:49 pm - Reply

        Thanks for such a swift reply!
        I mean a rectangular section and it is 120x220mm bent along stronger axis. As You say, It’s probably the algorithm going crazy, like the infamous Robot Stiffness difference problem. I’ll try it on a different FEM software.

        • Łukasz Skotny May 31, 2018 at 4:01 am - Reply

          Well – this is a rather slender beam I would say. almost 2:1 in height/width. I don’t have my license with me so I can’t do a quick check, but at a gut feeling, I think you should get “something” from the analysis.

          Let us know how it goes!

          All the best
          Ł

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