Column K Factor - Different per floor, self-check 3

[AggieYank]

Column K factors:

You have a column continuous from the foundation to the 3rd floor. Each floor is a diaphragm.

Column is fixed at the foundation.
Column is tied to 1st floor girders with pinned connections.
Column is tied to 2nd floor girders with pinned connections.
Column is tied to 3rd floor girders with pinned connections.

There is x bracing from the foundation to the first floor.
There is no bracing from the 1st to the 2nd floor, the idea is that the columns will resist lateral movement through their own bending.
There is a moment frame from the 2nd floor to the 3rd floor.


My answers. K factors are according to the latest AISC manual.
Foundation to 1st floor - K=0.8 (Fixed - pinned).
1st floor to 2nd floor - K=2.1 (Continuous up = fixed - free at top)
2nd floor to 3rd floor - K = 1.0 (pinned - pinned).

Is everyone in agreement with me, or do I have something wrong. Thanks. I was recently questioned on this, and wanted to be sure.

 

[SlideRuleEra]

AggieYank - Since assigning K values is usually a judgment call, I'll just offer this as an opinion:

With the column being continuous, restraint at each floor (both just above and just below the floor) would be the same. Taking that the foundation connection is fixed, which is very hard to guarantee, I offer the following assumptions:

Foundation to 1st floor: K=0.8 (fixed - pinned)

1st floor to 2nd floor: K=1.0 (pinned - pinned)

2nd floor to 3rd floor: K=1.2 (pinned - rotation fixed & translation free)

http://www.SlideRuleEra.net

 

[Galambos]

sliderule,

respectfully, if the second floor is pinned-pinned and not relying on stability from another frame, then this level is unstable. i would suggest that it does offer lateral resistance, and therefore the k factor would be greater than 1.0. perhaps, the unbraced length should be taken from the top level down to the top of the first level and use the nomographs for the k factor. the beams at the 2nd level offer little (theoretically zero) restraint.

ted

 

[JAE]

I agree with Ted. If you use the approximate column condition diagrams:

[blue]Foundation to 1st[/blue]
Base: rotationally and translationally fixed
Top: no sway, column top can rotate,
k = 0.8
L = distance from base to 1st

Since the second floor beams are pinned to the column and are otherwise not laterally braced then the whole second floor essentially serves no purpose, restricts no displacement in translation or rotation, so therefore should be totally ignored as it has no effect on the column other than adding vertical load.

So: [blue]1st to 3rd[/blue]
Bottom: free to rotate, no translation
Top: restricted by the beams rotationally, sway
possible
k = 2.1
L = distance from 1st to 3rd

With the added point load at the second floor, there is also a concern over P[Δ] effects at that point. With the sway condition above (at the 3rd level) the P[Δ] effects should be carefully studied as the whole arrangement seems a bit precarious.

Also, instead of using 2.1 at the upper section, you might do what Ted suggests and use the nomographs to more accurately determine k based on the beam you use.

Finally, AggieYank - you have me a bit confused...you say that the beams at the 3rd level are pinned-pinned but then you say that they are part of a moment frame. Does this mean that the moment frame is elsewhere in the story and this particular column is not part of the moment frame?

If that's the case, then one would assume (for the column diagrams) that the top level (3) is laterally restrained against translation (i.e. a leaning column dependent upon other frames to laterally support it) and we'd have:

[blue]1st to 3rd[/blue]
Bottom: free to rotate, no translation
Top: restricted by the beams rotationally, no sway
k = 1.0
L = distance from 1st to 3rd

BUT....assuming that the top level is resrained against translation isn't really all that accurate as normally you assume that moment frames are NOT restrained against sidesway...here again I see P[Δ] issues looming large!

 

[AggieYank]

Thanks for the responses. The X bracing and the moment frame are in different bays than the column we are talking about, I should have cleared that up.

SlideRuleEra:
Found - 1st: agree
1st-2nd floor: The only thing stopping the second floor from laterally translating is the column(s) which are continuous up from the first floor. The columns are laterally restrained at the first floor, and IMHO, fixed at the 1st floor because they are rotationally fixed to the column below (continous).
2nd-3rd floor: We're on different pages because I believe you thought the column was part of the moment frame. If you treat the column as part of a moment frame though, I agree.

Galambos: Even though the second floor doesn't restrain the columns from moving laterally, doesn't the second floor brace the columns from buckling in each direction, meaning the unbraced length would be floor to floor?

JAE:
Found - 1st: agree.
1st-3rd: An interesting way to look at it, similar to Galambos's. I agree the second floor doesn't laterally restrain the column.
Bottom: Is NOT free to rotate as it is continuous up from the first floor, aand is laterally restrained. It may not be perfectly fixed, as it will rotate a small degree, but it is much closer to fixed than pinned.
Top: Braced in plane by the moment frame in another bay, rotation free, so it is pinned.
k=0.8
L=1st to 3rd. Would the 2nd floor stop the column from buckling?

Also, column design is all about KL. I think both my initial factors and the factos laid about by JAE work. Using a factor of 2.1 on the column from 1st to 2nd floor gives a KL of (using 10' floor to floor heights) 21'. Using a factor of 0.8 on the column from 1st to 3rd floor gives a KL of 16. Even if you use 1.0, the KL is 20'. If you used different floor heights, this still works.

I suppose this is similar to picking any height in the column and analyzing it. Take a 50 foot column, pinned top and bottom. Put a point load at 50', what is the KL? 1.0 * 50'. Put a point load at 25', what is the KL? The column isn't ultra stiff, so at 25' it isn't exactly restrained from moving laterally. I'd use a KL of 1.8 (eng. judgement) * 25' = 45'. Put a point load at 40', what is the KL? This is fairly close to the end of the column, meaning it will be at least partially laterally restrained, depending on stiffness. I'd use a KL of 1.25 * 40'=50'.

 

[AggieYank]

Galambos, looking again at your post, the second floor girders are pinned connections, meaning the column isn't part of a braced frame. If you were to calculate K, it would be 0.8, assuming fixed at the 1st floor and pinned at the top.

If you were to pick a point in the column to analyze, you couldn't use the nomographs, and would have to use engineering judgement. The nomographs are based on fixed frame connections, even if they are very flexible fixed connections. You can figure out the column stiffness EI/L above the point you pick, but there is no fixed girder connection to cause double curvature, and you can't divide by zero in your G calculation.

If someone is skimming this thread, just read this. RISA 3D doesn't calculate K factors for any non-typical column. So for a moment frame, I believe you'll have to do the math yourself. It is probably worth checking if STAAD, RAM, or whatever you use does it as well.

 

[SlideRuleEra]

Thanks for the education on this, you have convinced me that those pins I assumed at the first and second floors would make a real unstable frame - kind of a "house of cards".

http://www.SlideRuleEra.net

 

[JAE]

AggieYank,

You said:

Bottom: Is NOT free to rotate as it is continuous up from the first floor, aand is laterally restrained. It may not be perfectly fixed, as it will rotate a small degree, but it is much closer to fixed than pinned.

. I'm not sure I agree with you. The continuation of the column below does not restrict rotation. The bottom (Fnd to 1st) WILL rotate under load and you cannot count on the column resisting itself to consider the base of 1st-3rd as fixed rotationally...think of an "S" shape from foundation to 3rd with the inflection point at the 1st level.

 

[AggieYank]

JAE, exactly. The column will be in an S shape from Found to 3rd.

The fixity of the column coming up from the first floor will depend on the stiffness of the column. Forget about the floors above for a second. Then we have a typical case of a cantilever sticking out past the support (the two supports are the found and the first floor, with the column sticking up past that).

Imagine if the column was a HSS 16x16x5/8, and the floor to floor height is 10 feet. There will be very little rotation of the column at the first floor becase the HSS 16x16 is very stiff. Now imagine if there is a HSS3x3x3/16 with 20' floor to floor heights. There will be a large rotation at the first floor, as the column is relatively flexible. Our real case is somewhere between, but I'd imagine it is more towards fixed than pinned. Doesn't using fixed vs. pinned for the K factor depend on curvature of the column? In the case of a fixed column, you have double curvature, which is what our example column has. In the case of a pinned column, there is no double curvature.

 

[JAE]

I see the point about different stiffnesses of the column (I love using the technique of taking conditions to ridiculous extremes to better visualize the situation by the way!). But when you say that the condition is "somewhere in between" the fixed and pinned condition makes me always steer towards the more conservative, especially when using the approximate AISC column curvature diagrams.

Using the nomographs - the base of the column (we're talking about the column from 1st to 3rd) would have an EI/L value for the column below and for the column from 1st to 3rd, but NO EI/L value for the girder stiffnesses - thus G for the beam is infinity at the base.

G for the top (at the 3rd level) would also have NO EI/L for the girders either side but would have an EI/L for the column from 1st to 3rd...thus G at the top = infinity.

By observation of the nomograph for a sway condition, the k would be infinity...not a good thing. Be careful here.
Again, I think there's a lot of second order effects that ought to be studied.

 

[haynewp]

"Foundation to 1st
Base: rotationally and translationally fixed
Top: no sway, column top can rotate,
k = 0.8
L = distance from base to 1st"



Why is the column fixed at the foundation if you have X-bracing from foundation to first floor? If it is part of the LFRS at the foundation to first acting with the X-bracing, shouldn't you take into account sway at the first floor level when designing the column from the foundation to the first floor?

 

[AggieYank]

haynewp: There are moment frames with fixed bases in the other direction (x-x). So the reality is that the columns are fixed at the base in both directions, but the columns aren't stiff enough to act as moment frames in the y-y direction and so x-bracing was added. The x-bracing is much much stiffer than the fixed base columns and so will take almost all of the load. For this reason, I assumed the columns were braced at the first floor.

JAE, I don't believe using the nomographs from the 1st to the 3rd floor applies, as the column from 2nd to 3rd floor is braced in plane by a portal frame in another bay. From 1st to 2nd floor, the column is part of an uninhibited sway "frame", really just cantilevering up.

If you treat the columns from 1st to 2nd as moment frames with pinned girders, you'll get a K of 20 no matter what the column sizes, as the G for each end of the column is infinity. A K of 20 seems very high. Is there some other way to do this? My gut feeling is that the K chart doesn't apply when it isn't a true moment frame, but how else would you calculate the K factor?

It's like the more I learn the less I know.

 

[UcfSE]

You mentioned RISA 3D in an above post. Are you modeling the entire frame at once or just one bay? Try taking a look at the entire frame as a whole. Even if your braced bays don't line-up from top to bottom, having a braced bay will brace the entire line of beams and columns along that floor.

Obviously, having moment frames with pinned girders doesn't make any sense. If you have pinned girders, then there's no point in using the graphs, as you can see with the outrageous results. The graphs are to help determine the effect of an elastic restraint on the buckling length of the column. If your girders are pinned, then you just have a cantilever, or a flag pole, no elastic restraint. If you don't have some LFRS in place in a different bay at this level then you have a very flexible system, probably not what you want. With a braced frame or a MRF in another bay, your columns then are restrained from sidesway and you have K=1 at most.

This can get a little more complex having addiontional framing above but you can work it out by stopping and thinking about what's going on. I would also try modeling your frame and giving it a small lateral load and checking out the plotted deflection. That may help clue you in as to how the frame is trying to behave. If all else fails, you can derive your own K by going back to mechanics of materials and stability of columns. That's the loooong way though .

 

[AggieYank]

UcfSE. It isn't that simple. It isn't just a flagpole cantilevering up from a fixed base. There is a rotation at the base of the "flagpole" which depends on the stiffness of the column. Or is it that simple? Do you usually assume a condition like that is purely fixed? I'd love to be pointed in the right direction as far as a text or reference on calculating the correct K.

You mention this is a flexible system and to check deflections. I checked deflections, and frame action is quite simple (even if unusual), and can be easily verified in RISA.

This is a retrofit type project, so your suggestions about "stopping and thinking about what is going on" don't really apply.

 

[SlideRuleEra]

Perhaps this AISC paper on "Critical Buckling Loads of Semi-Rigid Steel Frames" will be of value

http://www.aisc.org/Content/ContentGroups/Documents/Connections_IV_Proceedings/117.pdf

http://www.SlideRuleEra.net

 

[UcfSE]

Your rotation at the base will also depend on the rotation of the footing itself and the flexibility inherent in the connection, but you have to draw the line somewhere unless you want to spend weeks figuring this problem.

The text by Salmon and Johnson has some good information and derives equations used for the K charts in the AISC manual. You can take a look at these and see if you meet the assumptions to determine if you even want to use the charts. The authors also have a slope-deflection method discussed in chapter 14. If you need more information, check the references at the end of the chapter for pertinent research and information.

You may also try directly calculating the buckling load using mechanics of materials. Boresi and Schmidt have a good text on this subject. You will have to modify the differential equations most likely to account for your specific set of boundary conditions.

I don't understand how "stopping and thinking" about what's going on does not apply to a retrofit. I did not mean anything derogatory by it.

With the deflection plot, I meant to take a look at the deflected frame to give you an idea of its behvaior and perhaps give you a clue as to how the columns will try to buckle. That might help you decide which way to go when finding your K.

 

[AggieYank]

UcfSE, it's true the foundation will also rotate to a small degree. However, imo, the degree the column rotates will be a fair amount more than the foundation rotation.

Getting the K from the deflection plot is a good idea, but I'd have no confidence in it. I could tell you the column is in double curvature, that the lateral deflection at the top is fairly limited, but what would that mean. K=1.5? K=3? I don't feel comfortable doing that to be honest.

The "perfect" solution is to break out the mechanics of materials book, but as you mentioned, something like that will take "weeks". I'm still looking for the "acceptable / good" solution. I'm leaning towards treating the column as a flagpole condition from 1st to 2nd floor (k=2.1), check the additional deflection caused by rotation at the first floor, and be careful to check the P-Delta analysis as JAE mentioned. I emailed AISC and a professor who has written several papers on similar subjects. If they give a response, I'll post it on here.

 

[UcfSE]

That's about all you can really expect from looking at a deflection plot, but it helps a little to know if you're getting double or single curvature. It seems that if you do have double curvature, you might at least be able to put an upper bound on K of 2 or 2.1. Knowing that would give you a start, and if everything works with K=2 or 2.1 then you don't need to go further.

So you're columns from 1st to 2nd are cantilevering, as you say, to help with the stability of the frame as a whole?

This sounds like an interesting problem. I'm sorry I can't be of more help.

 

[miecz]

I would treat this as a two level column, braced at the first level, braced against translation at the second level. As such;

From the foundation to the first floor: K=0.8 (f-p)
From the first to the third floor: K=1.0 (p-p)

I believe JAE has stated it this way in an earlier post.

The moment frame should also be analyzed for the double height, i.e. pinned at the first floor level and extending to the top of the building.

 

[miecz]

More...

The moment frame column, I believe, would have a K of 2.1 (pinned-sway) and a length from the first floor to the top of the building.