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Stability bracing for beams

Stability bracing for beams

Stability bracing for beams

(OP)
Hello All,

I'm wondering if the beam bracing requirements of App. 6.3.1 of AISC, 14th ed, for nodal bracing can be divided (equally or partially) among the total number of braces provided along the span. It seems like the braces would all help resist the beam as it started to laterally buckle, but the code doesn't address it in the required strength equation A-6-7.

Thanks!

RE: Stability bracing for beams

The code equation can be difficult to understand - it first appears that the more braces you add the higher total lateral brace load occurs.

However, the way to understand this is to see it as a brace force and stiffness required based on a set Lb with a beam size consistent with that Lb.

So if you have a 30 ft. beam with a certain vertical load, and you "try" Lb = 10 ft. you'd have two brace points.
The beam size that RESULTS FROM THAT Lb will require a resulting brace strength and stiffness at those points.

Now if you add braces at 5 ft. on center, that doesn't mean you now have higher brace forces.
The brace force and stiffness should always be consistent with the maximum Lb that the beam can tolerate.

Hope this is clear.

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RE: Stability bracing for beams

(OP)
Right, but is the required brace load per brace the same if I have bracing every 10 feet versus every 5 feet? I can see the stiffness required actually increases as the braced length reduces; which intuitively doesn't make any sense.. but does the required brace strength changed at all?

RE: Stability bracing for beams

The required stiffness doesn't increase if you reduce the Lb. The formula appears to show it does but I think that would be totally counter-intuitive.
Take it to a ridiculous extreme.... Add a brace every foot. It makes no sense that the stiffness required of every brace at every foot would be sky-high from the formula.

I think the assumption embedded in this stability system is that the designer will use the least-weight beam that works based on the chosen Lb distance.
Then, with that Lb, you have a set brace force and stiffness.

So for brace strength - which doesn't include Lb in the formula, the required strength is based on a set applied moment Mr. Again, I think the assumption here is that you have a decided-upon Lb value, a resultant beam size, and then at every brace point (per the Lb chosen) you have a brace force required.

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RE: Stability bracing for beams

There is a note just kinda slapped on the end in the beam section (page 16.1-194 before torsional bracing) 'When Lb is less than Lq, the maximum unbraced length for Mr, then Lb in equation A-6-8 shall be permitted to be taken equal to Lq' which basically maxes out the required stiffness as Lb decreases in certain situations.

Attached is a paper that may give a little more insight into beam bracing.

RE: Stability bracing for beams

Of course the stiffness increases the more points you brace. You are trying to get the beam/column to bend in smaller and smaller sine waves - think of it this way - if you only have a brace every 10 feet, it will be a lot easier to get the beam to buckle between those braces than it would be to get the beam to buckle between braces every 1 foot. The real question is - what strength do you need out of the member being braced? If you NEED the strength that comes from the beam having to buckle between braces every foot, then yes, the higher bracing force is requires and is real.

RE: Stability bracing for beams

Yes - I'd agree with that - but for practical ranges of beam sizes vs. spans/loads you wouldn't ever (and shouldn't ever) have braces extremely close.

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