Cb Factor in 13th Edition 1

[JAE]

We just came across a unique situation - have asked AISC for a response but thought I'd post it here to see if any of you have any views on it.

Cb in AISC beam design has been calculated using Mmax, and three 1/4 point moments along the unbraced length (M_A, M_B, and M_C). This equation:


Cb = 12.5M_max
----------------------------------------
2.5M_max + 3M_A + 4M_B + 3M_C

is given and then AISC states, "Cb is permitted to be conservatively taken as 1.0 for all cases."

This is fine - no problems here.

But now in the 13th Edition, they have introduced a "cross-section monosymmetry parameter, R_m which is based on Iyc and I.

When we use a wide flange with a cap channel for a crane beam we get a value of Rm of about .547.

Using the Rm in the Cb formula we get a Cb value less than 1.0.

So the question is - do we use 1.0 for Cb "conservatively" or should we use the lower value of Cb? Using 1.0 when we calculate Cb = .94 isn't conservative.

But AISC still has the same sentence stating that Cb can be conservatively taken as 1.0 for all cases.

 

[MarkAJohn]

Rm is 1.0 except for reverse curvature. Do you have that in a crane beam? I realize this doesn't answer the question. If you get an answer from AISC, please post it here.

MJ

 

[JAE]

We have an older crane beam that is continuous over a single support (i.e. two span condition) so the beam has reverse curvature and has a "small" bottom flange in compression with a small Iyc that causes Rm to get low and reduce the Cb to below 1.0.

 

[271828]

The question is which flange is the compression flange? Obviously, both are in compression over some parts of the span. This is a great question for AISC because it's definitely not clear from reading the Spec.

The Section F4 LTB Fcr is for the entire compression flange under uniform compression. It can't get any worse than that, so I'd use Cb=1.0 if Cb calculated to be < 1.0.

 

[JAE]

[red]"I'd use Cb=1.0 if Cb calculated to be < 1.0"[/red]

That is what I'm not sure about.

A lower-than-1 Cb is more conservative than Cb=1 obviously.

I did email AISC and got the response that Cb can never calculate BELOW 1.0 by the formula (without Rm). This may be true, but when adding on the Rm factor the Cb can be calculated below 1.0 and the code doesn't really address this...neither does the commentary.


I've re-asked them the question considering the Rm factor along with some data - we'll see what they say.

It is not too often that you get a singly symmetrical beam over two supports -but it does happen -suppose you have a two span continuous beam with an added cover plate?

 

[nutte]

This is an interesting problem. My guess is the Rm factor approximates the effect of a singly symmetric member under double curvature, and is probably conservative for most cases, which might explain why for your particular case it causes Cb to be less than 1.

I agree with 271828 that Cb shouldn't ever be less than 1, as that is the constant moment case. I would use Cb=1.

Keep us posted on AISC's response. I have found that anytime you have a question like this, where the code equations aren't clear or give you peculiar results, the guys at the Solutions Center are completely worthless. If it's not in the web site's FAQs, they don't know. I've come across a few questions that had the gurus in the office stumped, and AISC was absolutely no help.

 

[haynewp]

I was trying to figure out a way to prove that using Cb=1 is always conservative by comparing a WT with the stem in compression against the singly symmetric equations that include Cb and this Rm factor. I at first thought the Rm reduction was accounting for the shear center not aligning with the centroid of the cross section in the horizontal plane, so there is additional twisting upon lateral buckling for the singly symmetric I shape. But it says to use Rm=1.0 in the case of a doubly symmetric member so it can't be for that reason. Cb doesn't show up in the WT equations, so what do you do for a WT that is in reverse curvature? Aren't the equations still applicable? I bet they are derived using Cb=1.0.

I would like to see a graph showing a singly symmetric I shape in double curvature with the bottom flange incrementally reducing to the same width as the web, the instant the bottom flange becomes the same width as the stem, the WT equations are used instead of the Cb with Rm equation of the singly symmetric I shape. I might be able to tell if Cb=1.0 should be the lowest limit. Or better yet, just compare the WT equations against the singly symmetric I shape equations accounting for double curvature with one of the flanges barely larger than the stem. I would think the equations should give closely the same results.

 

[271828]

I think I have something helpful. The 5th Ed. SSRC GUide Page 205 shows Cb with the R factor in there and gives some background.

In short, there is an applicability limit of

0.1 <= Iyc/Iy <= 0.9

that shouldn't be violated if this equation is to be used. Your combo section might violate this and screw up the equation. The next issue is bigger, however.

They also literally call it Iy-top, not Iyc and show a mono-symm beam on the adjacent page with the bigger flange on top. I'd say this equation probably doesn't work if the smaller flange is in compression, which (I think) is the only way you can get such a small Cb outta this equation.

They also talk about tees a little on that page.

There is actually a much bigger issue with the Cb calc than R -- the load height. This is also included in the SSRC version of the equation. This could drive Cb down, which seems perfectly rational to me.

I think AISC simply accidentally left out some key info:

1. That Rm only applies if Iyc is for the bigger flange.

2. There's an applicability limit.

3. An allowance to calculate Cb using rational analysis. There are lots of Cb equations out there. One of them might fit this problem better than the AISC equation. Folks need to request this so the Spec. Committee will allow it next time.

The bottom line, regardless of what TSC says, is that the base Fcr equation is for a fictitious beam with the bottom flange with the same stress along its entire length. This is the absolute worst case moment diagram. If some of that flange has lesser compression or tension, then Fcr is really bigger than what the equation gives, hence the use of Cb>1.0. Load height is a separate issue and should definitely be considered IMO.

 

[UcfSE]

The commentary for section F4 does refer the reader to chapter 5 of the Guide when loads are not applied at the centroid or when a "more accurate solution is desired".

Unforunately, there does not appear to be much about R_m and it's source in Comm F-1. The Guide provides the equation for R as e noted above, and states it's "for beams under reverse curvature between brace points". That seems to imply you could add bracing in the right location and have single curvature between brace points and perhaps get out of the reduction, but I don't know if that's stretching too much.

 

[WillisV]

I believe 271828 has it correct in that the Rm equation is not applicable when the compression flange is smaller than the tension flange.

 

[JAE]

WillisV - how do you justify that?

When you see the equation for Rm, it includes Iyc. Iyc is further defined below that as [red]"moment of inertia about y-axis referred to the compression flange, or if reverse curvature bending, referred to the smaller flange, in⁴"[/red]

So for a span from A to B, with reverse curvature, both flanges experience compression at different points along the unbraced length. Rm is then calculated using the Iyc of the "smaller flange".

I don't see where you can simply neglect Rm (per the direct language of the AISC spec) if you have a small compression flange.

 

[WillisV]

JAE, I think 271828's reasoning is sound and I look at it as an error in the Specification - it would be difficult to justify until they (hopefully) fix it into 2010 though unless the Solution Center provides you with a straight answer. I personally would have no problem using Cb as 1.0 especially considering that the Rm factor was not even in place in any of the previous specs.

 

[JAE]

Ok - sounds reasonable. I still have the question in at AISC and we'll see how they respond.

 

[271828]

Thanks WillisV.

JAE, I hope you'll report back what AISC says.

 

[JAE]

Just got a message from AISC - here it is:

The Specification Task Committee is considering either taking the Rm factor out of Equation (F1-1) and moving it to the Commentary to discuss mono symmetry separately, or to indicate that the Cb = 1.0 does not apply to the case of reverse curvature. The issue is not resolved as to how they are going to revise it.


It would appear that however they decide to handle it, that you could have a Cb < 1.0 with the Rm factor.

So for the time being, we're using a Cb less than 1.0 for those cases...just to be conservative.

 

[MarkAJohn]

JAE,

Who was the response from in the Solutions Center? Just curious.

MJ

 

[JAE]

I'd rather not use a name on a public forum.

 

[WillisV]

Hmmm - I'll be interested in seeing how that is resolved in the final spec. Thanks for the throught-provoking post JAE.

 

[271828]

An expected response. TSC does not generally offer anything but an interpretation.