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flexural-torsional buckling of single angles

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NS4U

Structural
Apr 2, 2007
320
The code says that FTB of singly symmetrical angles does not need to be checked- my question is why?
 
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Do you mean bent about their principal axis (Z)?

If so, it's akin to bending about the weak axis of a wide flange, it simply cannot control.
 
frv, why do you say that? the commentary doesn't elude to this.
 
That's "alude"... Just commenting.

Mike McCann
MMC Engineering
 
The Z axis is the weak axis. It is physically impossible for the angle to undergo LTB when bent about this axis. The angle will fail either by some localized failure or flexural yielding before it could buckle.
 
Actually Mike,

That's "allude"...

al?lude
? ?/??lud/ [uh-lood]
–verb (used without object), -lud?ed, -lud?ing.
1. to refer casually or indirectly; make an allusion (usually fol. by to): He often alluded to his poverty.
2. to contain a casual or indirect reference (usually fol. by to): The letter alludes to something now forgotten.
Origin:
1525–35; < L all?dere to play beside, make a playful allusion to, equiv. to al- al- + l?dere to play

Synonyms:
hint, intimate, suggest.

tg
 
... frv, are you sure? I ask because I am reinforcing a single angle in a joist and I want to understand the rational for the code and if FTB is a realistic buckling mode here.

Others, if you wish to critique my grammar please do so in another thread. thanks.
 
NS4U-

Working from memory without the code in front of me was not a good idea. It appears I pulled information from my rear.

I know you were talking about FTB, not LTB but I wanted to use the analogy; However, it appears that LTB does, in fact, need to be checked for angles in flexure about the Z axis. Even so, there are some limits placed on when it does and does not control, aluding a bit to what I was talking about.

 
FTB is a realistic column buckling mode for a single angle. I think the Spec.'s point is in providing E5 is that FTB won't control in lots of cases.

You can certainly do the following if you have a case that seems outside of E5:

Compute Qs per E7.1(c).
Compute Fe per both E3-4 (FB) and E4-5 or E4-6 (FTB)
Compute Fcr per Eq. E7-2 or E7-3 (slender) or E3-2 or E3-3 (nonslender)
Compute Pn = Fcr Ag

Don't forget that your load might be attached with an eccentricity, so you might have a moment about one or both axes. Compute phiMn and then use interaction equations in Ch. H.

Compare what you get with results from E5.
 
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