Rational Buckling Analyses to AS4100 or NZS3404 (Part 6)

In this post we’ll try demystify CL 5.6.4 part b, what the hell does it really mean?

Well this is my take on it anyway… I can’t say for certain I’m correct as I’ve never seen anyone else use it, never seen an example, never used it in anger, etc. But the answers make sense.

CL 5.6.4 cryptically states the following:-

So a couple of things here:-

  • Regarding M_{os}, basically it’s saying isolate the segment for which you are trying to calculate \alpha_m for, provide ‘F’ restraints at each end of the new member irrespective of what they are in reality, allow for minor axis rotation at the ends. Do a buckling analysis on this one segment.
  • Regarding M_{oo}, basically work out the reference buckling moment using the actual member length, substituting in equation 5.6.1.1(4) L in place of L_e.
  • Divide M_{os} by M_{oo}…. boom \alpha_m

Easy right…

Now I will warn you that it’s a lot of work for almost no gain in my experience. In so far as you’ll calculate a very similar \alpha_m value for a whole lot of extra work, which may or may not matter depending on your actual moment demand.

So let’s work out \alpha_m using this approach using the same scenario we used in the previous post in this series.

Previously we had this beam with ‘F’ restraints centrally and at ends, and for the left hand critical segment \alpha_m=2.483:-

So basically we delete everything but the span of interest, and add some equivalent end moments to produce the same moment diagram.

Moment added to new right hand support
New buckling mode
Buckling results

So basically M_{os}=1380.3kNm

Then all that’s left to do is to work out M_{oo} as follows using L=8001mm in place of L_e in equation 5.6.1.1(4). So for our 610UB125:-

M_{oo}=M_o= \displaystyle{ \sqrt{\bigg\{\bigg(\frac{\pi^2\times205000\times 39.3\times10^6 }{8001^2}\bigg)\bigg[78846.15\times1560\times10^3 +\bigg(\frac{\pi^2\times205000\times 3450\times10^9 }{8001^2}\bigg) \bigg]  \bigg\} }}

M_{oo} = 535.0kNm

Therefore:-

\displaystyle{\alpha_m=\frac{M_{os}}{M_{oo}}}

\displaystyle{\alpha_m=\frac{1380.3}{535.0}}

\alpha_m=2.580

As you’ll note, a minor improvement over the previous value of \alpha_m which was 2.483

That’s all there is to it really, was it worth the extra effort, you can be the judge of that. I personally wouldn’t bother in practice. But at least now you know how to interpret CL 5.6.4 part b to amaze your friends.

4 Comments

  1. Excellent post – thankyou! This nicely ties in with part 3 of your series of posts on the topic.

  2. Perhaps an idea for a future blog post is taking buckling to a truss with compression chord restraints in various arrangements? To make it more interesting the top/btm chords are continuous and it’s loaded with uniformly distributed loads (ie combined actions case).

    • Hi Trenno

      Happy to consider if you could provide a bit more info on what type of restraint arrangements you were meaning. I was thinking you could demonstrate by using the code alignment curves vs a buckling analysis and see how close the two answers were (as never covered the use of alignment curves/charts previously). Generally, the question always seems to come up around how long should I consider the effective length for buckling of a chord for example, especially when there is a load reversal, is this kind of what you meant, or something else?

  3. Sure, I was just mucking around with a test case:

    8 bay, propped cantilever truss (2 bays cantilevering).
    3m x 3m bays.
    Transverse joint restraints at the top and bottom chords on the even bays (ie start/end and every 2nd bay).
    This means that the bottom chord over the internal support has both tension/compression between lateral restraints.
    UC254x89 sections. Verticals and diagonals pinned minor/major. Top/btm chords continuous.
    20 kN/m SDL and 40 kN/m LL applied to top chord.
    ULS Load combo: 1.35G + 1.5Q.
    Max compression = ~978 kN
    BLF = 3.14 with the top chord at Bay 1 displacing.

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