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LTB restrains and effective length factors in AS4100?

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s4lmonsushi

Structural
Apr 1, 2021
4
Hi all,

A newbie engineer here and a newbie to the forum.

I'm having trouble understanding the lateral-torsional buckling restrains for beams in bending.

AS4100 cl. 5.4.2.1 states that a restrain is considered to be "Fully Restrained" if the restraint "effectively prevents lateral deflection of the critical flange and partially prevents twist rotation of the section". This contradicts with my understanding that if a restraint has partial twist restrain it should be "Partially Restrained".
The Steel Designers' Handbook seems to agree with the latter definition, whereas the Australian Guidebook For Structural Engineers seems to follow AS4100.

Another item I cannot get my head around is the twist restraint factor kt for calculating the effective length. Table 5.6.3 in AS4100 shows for restrain types containing the "U" restraint will always yield a better result (smaller kt and thus shorter effective length) than those for "P" restraint. What is the logic behind this? I am thinking along the lines where a "U" restrained end (effectively cantilever in most cases) is free to displace laterally and does not really twist which is why it's better than "P" restrain, where the member is held down in place and there is more risk for it to twist.

Thanks for reading my rather long question and I appreciate all your inputs!
 
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Lateral restraint PLUS partial twist restraint = Full restraint.

kt penalises you if your web is very flexible and you are relying on partial restraint. The web flexibility adds to the partial restraint flexibility meaning the critical flange is even less well restrained.
 
You can't compare the kt for P and U directly. These are adjustment factors compared to a base case (FF). The other factors take account of the U end, especially the alpha_m factor.
 
Thanks Tomfh. How about the unrestrained condition for kt? If an end is unrestrained wouldn't it result in a worse case than partially restrained?

Hi Steveh49, yes I believe they cannot be compared directly. But like what I wrote above, logically wouldn't an unrestrained end be worse as one that is partially restrained? This is what is troubling me to understand.
 
If there is any significant bending moment at the unrestrained end, the alpha_m accounts for that. For a constant BM, the PU capacity is a quarter of the PP.
 
OP said:
How about the unrestrained condition for kt?

Sorry I misunderstood your question. As noted above you get penalised via alpha M instead. I don’t know why they do it via alphaM as opposed to increasing effective length.

As for why PP is worse than FU in cantilever case, who knows. I’ve always found the concept of P restraints very murky, ie when does something transition from “stiff” to “flexible”, and at what point does “flexible” become ineffective?
 
It's a good question, S4lmonsushi. As Tom said, the kt factor penalises *only* for partial stiffness compared to the kt=1.0 base case for F restraint. For an unrestrained end, the lack of restraint is accounted for elsewhere. It can't be worse than completely unrestrained, so no further kt penalty. This does lead to an apparent oddity that P looks worse than U.

I would have written the code with a different equation for M_o for FU restraint rather than still use the FF case as the basis. I believe this is simply M_o(FF)/4.

EDIT: I'm wrong. This is the equation.
Screenshot_20210402-193013_Box_nolhms.jpg
 
Thanks Steve and Tom, viewing it from this perspective makes much more sense now :)
 
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