Column effective length factors for FMC frames

[Lion06]

I have a question regarding FMC frames and how you calc the effective length factors, K.
When doing a sway frame building using FMC for the lateral system, you are almost required to use the nomograph to calc the K factors. The reason being that you can't simply use any of the idealized cases shown in the AISC Manual for several reasons. First, because of the FMC you will never have an idealized fixed connection at the top of the column at the first floor or at either end of columns above that. For the lowest column level the only other option at the top is to have it free (to use the recommended K values). The top of the column cannot be assumed free, because the base is not fixed. While it may be fixed for wind moment only, I don't believe that the footing/base plate/anchor bolts designed based on the wind moment only will provide a rigid enough base to assume this is fixed.
Now you have a Gtop moving up due to the decreasing stiffness of the connecting beams (in addition to only being able to include (1) of the girders in Gtop instead of both), and a Gbottom moving up due to the decreased stiffness of the footing/BP/AB assembly.
These things add up and can significantly increase the K you would get if you assumed fixed connnections everywhere.
I have read a few papers on this and most reference using the moment-rotation curve for the connections to get the stiffness and use that in the G calcs. That is all well and good, but part of the beauty of the FMC is its economy. The economy goes away if you have to test the connections - especially if you have 10 different connection types on a building.
One paper, by Geschwindner, talks about the moment-rotation of the connection to lay the groundwork for his method, but I don't completely agree with it. He only uses (1) of the connecting girders (which I agree with), he also doubles the length of that (1) girder (to account for the far end of that girder being pinned (since the connection has reached its plastic moment capacity and is being loaded, but cannot resist any more moment), but makes no allowance for the connection being less stiff than the beam (this is the part I disagree with).
Does anyone have a method for calculating a K factor under these conditions or know of a good paper to read?
I am really trying to understand what is going on so I can come up with a method to accurately model this in RAM.
RAM has a procedure for this type of building, but I don't completely agree with it. They say to design gravity first (fine), the fix the beams and run lateral only and make sure the beams are ok (fine). Then run lateral and gravity, only sending 10% DL moment to columns with lateral loads - the problem with this step is that there are only two options for K - either RAM calcs it using the nomograph or you input it. You obviously can't input a single K for every column unless you are being VERY conservative, which again takes away from the economy of the system, but if you use the nomograph it calcs K as if the connections are fixed, which is UNconservative.

Does anyone have any suggestions?

 

[haynewp]

If you look in "The Handbook of Structural Engineering" by Chen there is a procedure and example for getting K for PR frames.

 

[Lion06]

FMC is a Flexible Moment Connection. It differs slightly from a PR (Partially Restrained). There are several papers to read - Geschwindner, and Driscoll to name two, that will explain some of the difference.

 

[swearingen]

Try breaking up your members and columns into many discrete pieces in your model. If you have your connection stiffness modeled correctly, divvying up your members will take care of any second order effects that are accounted for by k factors.




If you "heard" it on the internet, it's guilty until proven innocent. - DCS

 

[Lion06]

swearingeng-
RAM uses a P-delta analysis - that's not the issue
The problem I am having is determining an effective connection stiffnes. If I can determine an effective, reliable method for calculating an effective connection stiffness, then K can be easily calculated using the nomograph.

 

[haynewp]

but makes no allowance for the connection being less stiff than the beam (this is the part I disagree with).

Isn't that part of the simplifying assumptions of using FMC instead of PR?

Modeling PR connections and using Direct Analysis has got to be more accurate than using FMC assumptions with effective lengths. If I were you, I would assume FR connections in RAM System and find another program to do more involved analysis.

 

[Lion06]

haynewp-
I agree with you, but in an effort to understand this better I am trying to find out if anyone has a resource for finding an effective length factor in this circumstance.
Part of the economy of the FMC is supposed to be ease of construction and design. I don't think it will be economical to spend 30% more time refining the column design procedure.

 

[haynewp]

I think the resource is the Geschwindner paper. It is from 2005, so it is not old news. I am not sure I see where the paper is flawed. How do you plan on magnifying the P-delta in RAM for the FMC connections?

 

[Lion06]

I am not saying the paper is flawed. I have questions about it and would like some input. As I mentioned in the OP - He does take account for the FMC at the far end of the girder being pinned (since it will be trying to load further but has already reached its plastic moment capacity) and only accounts for one girder framing into a particular column (since one will be a "far end" connection from a girder which has already reached its plastic moment capacity and is pinned).
The problem I am having with it is that for the one girder he does consider, he is using the I/(2L) of the girder for the G calc. You would use that same value for a fully rigid connection between the two members (note that the factor of 2 in the bottom represents the fact that the far end girder connection is pinned), surely you would use a lower value if the connection is not fully rigid. G varies from 1 for a "fixed" connection to 10 for a "pinned" connection. Everything in between accounts for the relative stiffnesses of the attaching members.
I am looking at this and thinking that a W16x40 rigidly attached is going to restrain the column a lot more than a W16x40 that has a FMC. Maybe it is more along the lines of a W12x19 that is rigidly attached - either way I feel like the I/(2L) needs to be reduced to account for the connection being less stiff than the attaching beam.

 

[haynewp]

Again, isn't that one of the main assumptions of the FMC method that gets you away from PR analysis?

"Since, in the historical approach and the FMC approach,
when considering column stability, the windward connection
is assumed to be rigid"

 

[Lion06]

I agree, but it is only considered rigid for lateral loads, not gravity loads. Certainly you must consider buckling of the column under gravity loads only or as explicitly demonstrated in the paper, under combined gravity and lateral.

 

[haynewp]

That is the other assumption right? The beam ends are pinned under gravity loads so the columns must be designed with pinned beam ends under gravity combinations.

 

[Lion06]

ok, so would you use K=2.0 and design the footing as fully fixed? If the footing is not fully fixed and you assume the top is pinned (actually free because it is a sway frame, you can only assume pinned with a non-sway frame), otherwise you have a pinned base and a free top for a K=infinity.
Again, when you go back to lateral loads, you will not apply the lateral loads in a vacuum - they will be applied in combination with the gravity loads. If you assume fixed connections for lateral and pinned for gravity, then I am back to the same question of how do you marry the two and come up with an effective connection stiffness - since it is clearly less than that of a rigidly attached beam.

BTW haynewp, I appreciate the discussion.

 

[haynewp]

StructuralEIT. I need to look into all this further when I get a chance, coming up with all these K values and making sure I am still keeping with the assumptions. I don't want to mislead you.

 

[haynewp]

I looked at this paper again and it is not nearly as easy to follow as I first thought.

I don't think the assumption is that ALL beam ends are pinned under gravity, only one end of them. Or else it is not stable without fixed footings. So I should have specifically said previously that one end of the beams is assumed pinned under gravity. The other end is assumed rigid:

"Since, in the historical approach and the FMC approach,
when considering column stability, the windward connection
is assumed to be rigid"

"With the design assumption that all windward connections
behave linearly after shakedown and all leeward connections
act as plastic hinges,"


The following is what I {think} is the procedure but I may be wrong.

1. Beams are designed as pinned at each end for gravity

"When gravity load is reapplied, these positive end moments will offset the negative gravity moments in the beam so that the beam response approaches that of a simple beam. Thus, it will be appropriate to design the beam as a simple beam for gravity loads."

2. The connections are designed for lateral moments only.

3. One end of the beams are assumed as pinned when checking either gravity only or gravity plus lateral and one is assumed rigid.


So you get your K value based on this assumption and you design the column for all combinations with the same K.

If you look at the example in the paper, the whole first part looks like a gravity only combination check, but there is a lateral moment that is being brought into the column due to what I think is a locked in moment from the windward connection. So even though it is a gravity only check, there is a wind moment that is still locked into the joint from the residual deformation and kx is found to be equal to 2 with the far pinned end.

The next part of the example checks the gravity plus lateral and I think the same kx=2 shows up in the denominator of the Pe equation. (He did not calculate a new kx for the combined gravity plus lateral case). Also, I think he should have noted that the windward and leeward connections are reversible and you have to develop a kx based on the worst case.

But I am not real sure about all this.

 

[Lion06]

you are right. The paper is rather involved and not the easiest to follow.
I believe that the beams are considered pinned at both ends for gravity (at least while under lateral loading), this may not be completely true under gravity only loads since the connection can take that moment.
I don't believe that the near connection is fixed for gravity and lateral or else the connection would have to be designed for this moment, not just the lateral moment. You are right in stating that the windward connection behaves linearly after shakedown, but it....... wait, I guess that is a reasonable assumption provided that the gravity moment is less than twice the wind moment.

Let me think about this a little more!

 

[haynewp]

Look at this part, he designing the column for the lateral moment for the "gravity only case"..

"Step 3: Column design will assume that the column is
a two-story column. Selecting the lower level column for
strength under gravity load
Pu = [1.2(1.875 + 0.625) + 1.6(1.25) + (0.5(1.125)](25)
= 139 kips
The columns must also resist a moment equivalent to the
connection design moment, which is the moment that was
determined from the lateral analysis. Although the two columns
at the joint should share this moment, it will conservatively be applied fully to the column being designed. Since this is a gravity only load case it is acceptable to assume that there is no lateral translation moment thus,

Mnt = 48.0 kip-ft
Mlt = 0 kip-ft

Where Mnt is the conection moment from the lateral distribution. He then finds k to design this column for the same "gravity only case" and checks the interaction equation:

For the pinned base, the recommended value of GB =10, will be used. For the upper end of the column, account must be taken for the beam with a pinned end; thus, its length will be doubled and only one beam can be used to restrain the column.

The next load case to be checked is the lateral load case where he is using the same kx=2 (see the "2" in the "Pe2" equation):

To check strength for the lateral load case, the gravity load is taken as Pu = [1.2(1.875 + 0.625) + 0.5(1.25 + 1.125)](25) = 105 kips

For this case, the moment in the connection is due to the
lateral load, assuming that there is no moment due to the gravity load, thus,
Mnt = 0 kip-ft
Mlt = 48.0 kip-ft

....Pe2=pi*29000*209/(2.0*15*12)^2

So he is using beams that are rigid at one end and pinned at the other for both the gravity only and gravity plus lateral case, and the same moment=48.0 kip*ft is used for BOTH case checks.