19 Feb

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 , basically it’s saying isolate the segment for which you are trying to calculate 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 , basically work out the reference buckling moment using the actual member length, substituting in equation 5.6.1.1(4) in place of .
• Divide by …. boom

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 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 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 :-

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

So basically

Then all that’s left to do is to work out as follows using in place of in equation 5.6.1.1(4). So for our 610UB125:-

Therefore:-

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

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.

## One thought on “Rational Buckling Analyses to AS4100 or NZS3404 (Part 6)”

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