Effective Length

[Calif]

I am working on a 3 bay two story steel frame for a section of a building using ETABS 9. For some of the columns, I am getting outrages columns sizes (W14x257). When I look at the print out, I see that my effective length factor is sometimes 50 and sometimes 70 for some of them. I know this is wrong because looking at the Manual of Steel Construction, effective length cant go higher than 20. Can someone who uses ETabs, explain why this happens? I have made all my end connections hinged at the based and fixed at the beam supports. The columns are only 12 feet and what is control is the compression strength of the column which ETabs says it 2 to 6 kips(this is low for a column since the manual says 2940 kips for a KL with respect to ry). Obviously, the K factor is controlling this but I do not know why the program calculates a K so high.

The resisant virtues of the structure that we seek depend on their form; it is through their form that they are stable, not because of an awkward accumulation of material. There is nothing more noble and elegant from an intellectual viewpoint than this: to resist through form. Eladio Dieste

 

[Lion06]

What kind of lateral system are you using? Are these columns part of the lateral system or are they gravity only columns?

 

[MarkAJohn]

Take this to ETABS tech support.

HTH
MJ

 

[Calif]

This is part of the lateral system for the building designed as a ordinary moment frame. I am down the road for tech support but I have been waiting for awhile for them to get back to me (almost two days).

The resisant virtues of the structure that we seek depend on their form; it is through their form that they are stable, not because of an awkward accumulation of material. There is nothing more noble and elegant from an intellectual viewpoint than this: to resist through form. Eladio Dieste

 

[271828]

How does ETABS calculate the effective length? Does it do the AISC alignment chart calcs or some kind of matrix eigenvalue calc? You really need to dig in and see what it's doing. The ETABS manuals should have info on this, probably in teh steel design chapter.

K is UNLIMITED for unbraced frames. There's no limit of 20, perhaps a *practical* limit exists, but no real limit. K = very large or infinity means that the column is unstable, like a flagpole pinned at the base with no moment restraint at the top.

I'm pretty sure ETABS will let you override Kx. If you can't figure out what it's doing then calc your own Kx and over-ride it.

You could also use the Direct Analysis Method which allows Kx=1.0 for all columns. See the 13th Ed. Spec. App. 7.

Really, the bottom line is that you MUST know how it's calculating everything.

 

[271828]

Hey, I'm going to hazard a wild guess at what's causing the problem. I betcha that you have the column base pinned and the girder Ix somehow very small, either with a property modifier or by accidentally turning the beam on its side. Other option is that the girder has the end pinned accidentally.

 

[csd72]

is the top of the column restrained against out of plane movement?

 

[alord]

Try as a test making one column pinned at the base and at the top, without sway and see if it gives you a K = 1.

 

[JLNJ]

Look at the sway and see if it's in the ballpark. Hand check the reactions and moment diagrams for busts.

Maybe you can try it with a different analysis procedure.

 

[slickdeals]

ETABS is notorious when it comes to calculating K factors. Have you meshed the column element into finite pieces explicitly? I think ETABS follows AISC alignment charts and uses the stiffness properties of the joint. I think you have a braced system because you are saying that your column bases are pinned. If that is the case, I would suggest use a K=1.0 conservatively and not spin your wheels too much unless you really have the time to.
What kind of deflections are you getting at the top. Is it reasonable? If your base is pinned and every connection is a simple connection, maybe you have an instability.

 

[Lion06]

a moment frame is a sway frame and has a minimum k of 1, it is likely higher than one. I agree with 271828 that it probably has something to do with the base conditions of the columns (being pinned) and the beams framing into them having a small Ix (relatively speaking).

 

[271828]

StrlEIT is right. If it's a frame that can have a sidesway buckling mode (Fig. C-C2.4 in the 13th Ed. Manual), the LOWER limit on Kx is 1.0. The upper limit is infinity which indicates that it's unstable.

If ETABS is using the alignment charts, then it should be checked thoroughly. It is a cakewalk to use those charts *wrong*, but it takes a little judgment (ETABS have this?!) to use them right. For example, look at the adjustments that are listed in the text on Page 16.1-242 and 243 of the 13th Ed. Manual. Some of those make a pretty big difference. What is ETABS doing with them?! There's also the tau business, but it's conservative to ignore that.

I look forward to the inevitable death of the Effective Length Method, LOL. Direct Analysis Method!!

 

[Calif]

The top and bottom is restrained in one direction in the weak axis plane and in two direction in the strong axis plane. I got a response from Etabs and they said that sometimes the program makes mistakes like this and that I can overwrite the calculate K and use the correct K factor using the Steel manual. The columns are pinned at the bottom and fixed at the joints. It could be that the beam is too small at the top since I am using a W10x17 in the weak plane and W12x14 in the strong plane. This is my first time using Etabs and I am starting to get use to all the commands but it looks like I have to dig deeper. Where can I dig up information about Direct Analysis Method?

The resisant virtues of the structure that we seek depend on their form; it is through their form that they are stable, not because of an awkward accumulation of material. There is nothing more noble and elegant from an intellectual viewpoint than this: to resist through form. Eladio Dieste

 

[271828]

So ETABS is using the alignment charts.

Which K is coming out very high, Kx or Ky? If these frames only provide lateral resistance in the strong direction, then these columns are leaner columns in the other direction and Ky=1.0 and Kx>1.0. For buckling in the weak axis, the mode is like a gravity column. Sway buckling is the behavior in the strong direction, hence Kx>1.0.

I'm not sure I'd just calc Kx and over-ride in this case because honestly, it doesn't seem like there's an understanding of what the program is doing. There might be other issues also that will go undiscovered if you over-ride.

You're right: dig deeper into what it's doing. Can't type this enough times: One MUST be able to nearly exactly duplicate manually what the program is doing (at least on a simpler version of the problem) or don't use it.

The DAM is in the 13th Ed. Manual, Appendix 7. There will be a design guide out there soon, but it's not available yet. The AISC Manual comes with a CD with examples. There might be something in there on the DAM. Other than that, there might be some papers out there or a newly updated textbook.

 

[Lion06]

271828-
I know you've mentioned before that there should be a design guide for stability out shortly. Do you have any idea when that might be? I am anxious to get my hands on a copy.
Hopefully it will be available for free download if you're a member of AISC.

 

[Calif]

It was buckling according to the weak plane of bending Ky=50 to 70. K was controlling the weak axis which is why I got W14x257.

The resisant virtues of the structure that we seek depend on their form; it is through their form that they are stable, not because of an awkward accumulation of material. There is nothing more noble and elegant from an intellectual viewpoint than this: to resist through form. Eladio Dieste

 

[slickdeals]

There is a boxed-lunch presentation on AISC's website by Shankar Nair

http://www.aisc.org/Content/Navigat...ls/A_New_Approach_to_Design_got_Stability.htm

I found that quite useful.

 

[rmurph17]

Ahh this is where you should really do things by British Standard 5950:2000, I don't know how your doing it but Le (effective length) when your considering lateral torsional buckling is dependent on if the beam or column is designed as pin ended or fixed ended etc.

So if you are looking for a factor to mulitply L with then take

Free to rotate on plan = 1
Lateral restraint only = 0.85
Lateral and torsional restraint = 0.7

I hope this helps.

Ryan