Unbraced length of a beam
Unbraced length of a beam
(OP)
As most of you probably know, the strength of a beam in bending depends on the unbraced length of the compression flange. My question is, what defines braced? If I have a channel spanning 20 feet with another channel alongside of it 2 feet away, and they are connected to each other with a plate welded continuously along the top flange of each, are these beams continuously braced? I have another pair of these channels 6 feet away and they could be connected to the first pair on, say, 5' centers. If the first pair aren't considered braced, would this give them an unbraced length of 5' or is it still too wobbly to be called braced?






RE: Unbraced length of a beam
To rely on elements that are parallel to the same beam is not a good idea, unless they are really stiff elements. That's a judgment call.
Mike McCann
McCann Engineering
RE: Unbraced length of a beam
RE: Unbraced length of a beam
RE: Unbraced length of a beam
DaveAtkins
RE: Unbraced length of a beam
RE: Unbraced length of a beam
I would not provide a continuous weld though, a staggered weld would be less likely to warp the channels.
If you are treating it as a built up section then you need to check the shear flow in the welds using v = VQ/It.
Bracing 2 unbraced members together does not help as they could then both buckle in the same direction. There needs to be a load path back to a rigid point of support(like your plate does).
hope this helps.
csd
RE: Unbraced length of a beam
RE: Unbraced length of a beam
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which has a very similar situation but with discrete stair treads instead of a continuous top plate. I suggested the possibility of unbraced in this situation and everyone else seemed to think it was braced.
Can someone please explain the difference to me?
RE: Unbraced length of a beam
RE: Unbraced length of a beam
This is a common misperception. Bracing two unbraced members together does help provided the bracing is designed properly. Common Example: A bridge - two beams with cross frames between.
See Appendix 6 of the 13th Edition Manual for all the explicit requirements, but essentially if cross frames are present they are considered beam nodal bracing and may either be classified as lateral or torsional bracing depending on their placement and must meet both the strength and stiffness requirements of the Code. If the cross frames are designed to meet these requirements, then yes the unbraced length of the beams is reduced to the spacing between cross frames.
A continuous plate or slab would be classified as continuous torsional bracing and need to meet the requirements of Appendix 6.3.2a and 2b.
RE: Unbraced length of a beam
It is hard for me to realize that two seperate members which are continuously connected at the top with a rigid plate will be able to torsionally twist to create a lateral torsional buckling situation!!
RE: Unbraced length of a beam
You guys would make really good politicians!
You take a quote out of context and pick it to pieces.
Maybe you should have read the rest of the paragraph:
"There needs to be a load path back to a rigid point of support(like your plate does)."
I agree that crossed bracing would to the trick to provide torsional bracing, but I dont think that was what the OP was suggesting.
Structural EIT,
A stair is not 2' wide by 20' long so the lateral capacity is much more obvious. Also I am not saying that it is unbraced only that the OP needs to do some numbers to confirm it meets the code.
csd
RE: Unbraced length of a beam
RE: Unbraced length of a beam
You'll really need the bracing seminar notes to dig into the subject, though. The entire subject is very convoluted.
RE: Unbraced length of a beam
Torsional brace may be connected anywhere on the section height, but they don't do much good without adequate web distortional stiffness.
RE: Unbraced length of a beam
RE: Unbraced length of a beam
I tend to agree, although it would be hard (for me anyway) to prove. The plate doesn't brace the channels by tying them together because both could buckle in the same direction if they're equally loaded. If only one was loaded (or loaded more), the other one could provide some bracing for the loaded one.
However, for the channel top flange at midspan to laterally displace more than that same channel flange at the 1/4 point, the plate has to bend. I think this makes it fall under the category of a relative brace, but I'd have to study the notes to be sure. I think the seminar notes show at least one case that's similar--seem to remember it being two members connected together by a horizontal truss.
RE: Unbraced length of a beam
Like I said, I've faced this situation before, hunted for a solution, and always came up empty. Always ended up treating the beams unbraced, even though I believe they are braced. Never went to the seminar and so I don't have the notes. I would be VERY interested in a solution, if it's there.
RE: Unbraced length of a beam
RE: Unbraced length of a beam
I didn't participate in the first thread. I gather what you meant to say is that, in the first thread, the stringer is braced, while here, the beam is not. The difference I see, is that the stair treads, attached to the stringer web, offer torsional restraint to the stringers at many points. Here, the plate does not offer torsional restraint, unless, of course, it can be proven that it does. Never has, in my experience.
RE: Unbraced length of a beam
That is the only real difference I can see here?
RE: Unbraced length of a beam
That's how I see it. If the treads were in the plane of the top flange, then it would be similar to this condition(except for the relative brace that 271828 is looking into). However, treads are a lot stiffer than plate, so it might be easier to prove torsional bracing even then.
RE: Unbraced length of a beam
RE: Unbraced length of a beam
RE: Unbraced length of a beam
I believe that it is possible to get a plate between two beams to act as a continuous torsional brace per Appendix 6.3.2b, though it might not be practicle to do so.
As an Example: Two W10x12 beams, spanning 15ft, 2'-0" apart, with a factored load of 0.80klf on each of them. Design the plate to fully brace the beams.
Condition 1: Strength Design
Mr = wl^2/8 = 0.8*(15')^2/8 = 22.5k-ft = 270 k-in.
Mbr (required brace strength) = 0.024*Mr/CbLb (from Equation A-6-9 assuming L/n=1.0 per Ap6.3.2b).
Assume Cb = 1.0. Lb = Lq per Ap6.3.2b which is the maximum unbraced length of the W10x12 that can take Mr. In this case, this equals about 9.3 ft.
Mbr = 0.024*270/(1.0)(9.3x12) = 0.058 k-in / inch length required strength.
Provide Plate Strength = (0.9*Fy=36*t^/4)*1" wide, set equal to Mbr and solve for t = 0.085 in thick plate needed for strength requirements.
The connection of the plate to the beam must also transfer Mbr, so you would need to overlap the plate a few inches over the top of the beam, and weld on the top and the bottom to form a weld couple to resist this moment.
Condition 2: Stiffness Requirements
Bsec = 3.3*29000ksi*(tw=0.190")^3 / (12)(ho=9.87-0.21) = 5.66k-in/in length of beam (this is the distortional stiffness of the W10x12 web from A-6-13).
Bt = (1/phi=0.75)(2.4*Mr=270^2)/(E=29000)(Iy=2.18)(Cb=1^2)=3.69k-in/in length of beam (brace stiffness per A-6-11 with L/n=1 per Ap6.3.2b)
So Btb (required stiffness) = Bt/(1-(Bt/Bsec)) Eq. A-6-10
= (3.69/(1-(3.69/5.66)) = 10.6k-in/inch of beam length.
The provided stiffness from the plate will be the flexural stiffness of the plate in double curvature (as the beams try to rotate, they will force the plate into double curvature). This is equal to 6EI/L for the plate spanning between the beams.
Setting 6EI/L equal to Btb = 10.6 k-in/in will yield the required plate thickness for stiffness.
[6*29,000*Ipl=(1/12)(b=1in)(treq)^3 ]/ L=24" = 10.6
treq = 0.26"
So a 1/4" plate is approximately adequate to fully brace the W10x12, which will then have a PhiMn = 32.9ft-k>22.5ft-k. OK.
Would I ever do this in practice? Probably not, but it makes for interesting discussion and illustrates what I think is a reasonable method to determine if it is possible.
RE: Unbraced length of a beam
Now, I have a question about using a torsional brace in this case.
Just repeating the problem: We have two channels 24" apart. A plate sits atop them and is welded continuously, assuming only one weld per channel at the plate edges. We don't know which way the flanges point.
How does the single weld at a channel transmit moment from the plate to the channel?
The situation seems worse to me if the channels are arranged so that the flanges are pointing toward each other. In that case, the shear centers are such that the channels will try to twist and pry the weld.
I think the torsional brace idea is possible, but dubious in many cases.
RE: Unbraced length of a beam
Now, if you do have enough torsional strength and stiffness, then it is most definitely braced. Why? Because then it becomes a built-up shape that is simply a channel, toed down, bending about its weak axis. Since the global shape is now around the weak axis, it cannot fail laterally. And since you've provided adequate torsional stiffness at the plate/channel interface, the channels can't fail in torsion independently. What can happen is the built-up channel flanges (which are the channels in this case) can start to bow outwards at high load, bending the plate in a concave fashion. They will never roll in the same direction, though, as has been proffered.
If you "heard" it on the internet, it's guilty until proven innocent. - DCS
RE: Unbraced length of a beam
I just looked in Blodgett and there is no Sw value for a single line weld about its "weak" axis. Since the line is considered to have length only, all of the area is located at the axis under consideration and has no "d", therefore can have no moment resistance.
RE: Unbraced length of a beam
RE: Unbraced length of a beam
Also, you could always bolt them together along the length and be able to put solid numbers to the problem as well...
If you "heard" it on the internet, it's guilty until proven innocent. - DCS
RE: Unbraced length of a beam
We have replaced engineering judgment with more equations!
Pretty soon all we will have is equations and no judgment.
RE: Unbraced length of a beam
RE: Unbraced length of a beam
Judgment needs some help sometimes. The OP, for instance, could place a 12GA (0.1017-inch design thickness) at the top to create a brace. Would it work? Why not, you still have a 2-foot wide inverted "U"? Do we go then immediately to a plate 1/2-inch thick from judgment when a couple of minutes show that 1/4-inch or 5/16-inch is good?
Judgment helps us identify ways to solve problems and is developed as we apply numbers and principles. If we can use numbers to solve problems, we should. Now that the solution has been verified, several times later down the road you develop judgment based on experience and could look at it and decide a 1/4-inch plate will do the job, or what-have-you.
RE: Unbraced length of a beam
Most of the time a solid engineering judgement is so obvious that makes the check with the equations just a formality.
RE: Unbraced length of a beam
"Sheesh. Some days it just doesn't pay to get out of bed."
RE: Unbraced length of a beam
Now add this to the design bending stress of the beam/plate acting compositely to get a design normal stress for the plate to determine max spacing of welds based on buckling. Use VQ/I to determine max spacing of welds based on shear flow. Since the section will be a composite section, determining the bending allowable for the full section is appropriate. The channel is continuously braced with full bending allowable stress. The b/t ratio of the plate may reduce this value.
"Sheesh. Some days it just doesn't pay to get out of bed."