Unbraced length of a beam 7

[shaneelliss]

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?

 

[miecz]

StructuralEIT

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.

 

[Lion06]

Do the treads provide torsional resraint simply because they are at an angle to the stringer and not completely parallel?
That is the only real difference I can see here?

 

[miecz]

Structural EIT-

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.

 

[Lion06]

fair enough, thanks!

 

[hokie66]

Seems to me some of you guys are making too much of this. What Shane has is an inverted U-shaped beam which is 2 ft wide. Spanning 20 ft, I don't see bracing as an issue.

 

[WillisV]

The way I see it, you cannot just look at the problem and say it is or is not braced depending on the configuration of the cross frames (or stair treads, or plate, or whatever you are assuming is acting as a brace) - you just have to run the numbers in Appendix 6.

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.

 

[271828]

Very nice Willis! A star from me.

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.

 

[swearingen]

For the weld to be an adequate torsional brace between the channels and the plate, it must provide the moment strength AND stiffness required per Appendix 6.3.2b. That's all. A single weld may indeed be good enough (although it would be a bad fatigue detail - you'd want to weld the other side too).

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

 

[Lion06]

I don't believe a single weld of the configuration under consideration will transfer any moment. It may transfer some by tension (shear) on the weld and bearing the plate on the top flange of the channel if applied in the right direction, but that is not something that can be counted on.
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.

 

[miecz]

Nicely done WillisV. I am surprised that the web flexibility doesn't dominate the system. Great observation that the double curvature of the plate would increase its stiffness. I see that the thickness required for strength in your example is only .085". It would seem that a single weld with a throat greater than the strength thickness requirement would suffice. No?

 

[swearingen]

From a direct structural load standpoint, I would agree with you, StructuralEIT. However, a weld most certainly has a moment strength along its length as jmiec implies. If its moment strength is greater than that required, you're all set. Of course, the sure fire way to take care of that is to stitch weld the plate to the channels on the top of the channels and underneath the plate.

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

 

[jike]

Maybe it is just my age, but I am in total agreement with hokie66.

We have replaced engineering judgment with more equations!
Pretty soon all we will have is equations and no judgment.

 

[Lion06]

I tend to agree with hokie66 as well, I would still check the top plate for local buckling.

 

[UcfSE]

Some confuse judgment with voodoo engineering, and the arm-waiving bracing rain-dance and so on. There is a difference. The equations help to weed-out the voodoo.

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.

 

[shin25]

Engineering judgement comes from knowledge and practicing over the years on similar kind of problems.

Most of the time a solid engineering judgement is so obvious that makes the check with the equations just a formality.

 

[darkwing888]

The plate can be considered a lateral brace (not torsional) because it will be acting as a diaphragm spanning 20' with a span/depth ratio of 10. Similar to DaveAtkins' recommendation, I recommend using 2% of the maximum vertical load as a lateral load on the plate. Evaluate the plate stress and stiffness term from this in-plane lateral load. For example with a 10 kip concentrated vertical load at midspan, the lateral load on the plate is (0.02)(10000)= 200 lbs at midspan. Calculate plate stress. For a 1/4" plate, your looking a stress of .02 ksi. Provide stitch weld spacing adequate to resist a whopping .02 ksi stress and not buckle between welds. Calculate lateral deflection to derive stiffness term #/in to compare to required. If these check out, use fully braced allowable bending stress.

"Sheesh. Some days it just doesn't pay to get out of bed."

 

[darkwing888]

Obviously, I calculated an incorrect lateral load to apply. The load should be design load of the compressive flange. Using the same example but changing the concentrated midspan vertical design load to 7 kip; 7 kip x 20'/4/0.95'moment arm = 36.8 kips. Then 2% = .736 kip lateral midspan load on plate. For a 1/4" plate the stress is .736x20'x12/4/24in3 = 1.84 ksi.
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."