×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Unbraced length of a beam
7

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

Normally, the bracing comes from either a wood or concrete diaphragm, or wood or steel purlins framed 90 degrees to the beam you are bracing.  

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

or do a custom section and call it not braced.

RE: Unbraced length of a beam

Your pair of channels with a plate welded continuously to both is not braced, but the assembly forms a beam with a top (compression) flange which is 2 feet wide, so is much stronger than just the two channels.   

RE: Unbraced length of a beam

You could check this as a beam lying on its side, 2 feet deep.  The only load the beam lying on its side needs to resist is the buckling force in the two channels (say, the summation of 2% of the force in each channel's top flange).  The stress induced by the buckling force would be added to the stress induced by the direct vertical loads.

DaveAtkins

RE: Unbraced length of a beam

I would say-tie the bottom flanges of the channels every so often and call it a box shape and forget about the lateral torsional buckling. Given, that the plate that you have welded on the top is sufficiently rigid.

RE: Unbraced length of a beam

Dave has a good answer - if in doubt, do the numbers. There are explicit rules in the code regarding what loads the braces need to take.

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

I'd say quit trying to home-brew the definition (just ribbing, guys! ;) ) of braced and read AISC 13th Ed. App. 6 and the Yura/Helwig bracing seminar notes.  The notes, especially, talk about stuff like the proposed situation.

RE: Unbraced length of a beam

It seems most people are saying this situation is unbraced.  Please see the following thread,
javascript:openindex(450,450,'http://www.tipmaster.com/includes/refinfo.cfm?w=450&h=450')
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

3
"Bracing 2 unbraced members together does not help as they could then both buckle in the same direction."

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

I agree with willisV.

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

WillisV/Shin25,

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

(OP)
Thanks for the guidance.  I guess I have not looked through the new AISC manual enough.  I didn't realize there was such a layed out explanantion of bracing in the commentary.  Has that always been there or is it new in the 13th edition?

RE: Unbraced length of a beam

Shane, it's a fairly new inclusion.  The first time it showed up was in the 3rd Ed. LRFD Manual.  The 13th Ed. Manual includes the info in Appendix 6 and (especially useful) the Commentary to Appendix 6.

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

Beware trying to use the plate as a torsional brace.  The web distortional stiffness will likely eat its lunch and make it not work.

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

I've faced this situation myself.  I don't think the plate qualifies as bracing according to Appendix 6.  It can't qualify as a lateral brace, and I don't think it can qualify as a torsional brace, for the reasons stated by 271828.  At the same time, my gut feeling is that the plate really braces the channels.  It's too bad no one has ever tested this configuration.

RE: Unbraced length of a beam

"At the same time, my gut feeling is that the plate really braces the channels.  It's too bad no one has ever tested this configuration."

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

271828-

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

Can someone please explain the difference to me between this situation and the one referenced in the thread I posted above?  I thought from the time of the first thread above that this would be an unbraced condition, but it seems the views are mixed with a consensus being reached on each thread.  People are saying this one is unbraced, but the stair stringer in the last one is not braced.  I am not following the reasoning.

RE: Unbraced length of a beam

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.

RE: Unbraced length of a beam

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?

RE: Unbraced length of a beam

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.

RE: Unbraced length of a beam

fair enough, thanks!

RE: Unbraced length of a beam

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.     

RE: Unbraced length of a beam

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.  



RE: Unbraced length of a beam

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.

RE: Unbraced length of a beam

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

RE: Unbraced length of a beam

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.

RE: Unbraced length of a beam

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?

RE: Unbraced length of a beam

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

RE: Unbraced length of a beam

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.

RE: Unbraced length of a beam

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

RE: Unbraced length of a beam

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.

RE: Unbraced length of a beam

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.

RE: Unbraced length of a beam

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."

RE: Unbraced length of a beam

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."

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources