Fixed base for a column? 1

[Lion06]

In digging into the nomograph for a previous thread I posted about FMC frames I happened upon another question that I am hoping I can get some opinions and insight on.
When using the nomograph to calc k (for either sway or non-sway frames), you can assume a G of 1 for a column rigidly attached to a properly design footing. My question is what exactly is a properly designed footing?
Initially, I thought it would be any footing designed to take the greatest load effects from the combinations assuming a fixed base. When I started thinking about this more, I don't necessarily agree with this anymore.
Now I think that the footing needs to be designed to take the full moment capacity of the column. My reasoning is this: as this column is loaded and reaches the point of buckling the COLUMN needs to be fixed at the base, not just be able to transfer some value of moment. It is likely that the moment it will see when buckling could exceed the moment it sees under combined gravity and lateral loads. I guess you could say that it won't see anything higher than gravity and lateral loads, but.. that is for strength, not stability. The lower moment resistance will likely allow more rotation, true?

 

[Lion06]

I don't believe you can use a k of 1 for pinned pinned in a sway frame. That is my point. In a sway frame, the top of the column is free to move. Therefore you either need to have a fixed base or a a fixed beam connection.
It may be leaning on the lateral system for stability under gravity loads, but when the lateral load is applied, it goes along for the ride and now you have a pinned base with a free top (k=infinity).

 

[271828]

DaveAtkins is right. It's a leaner column and K<=1.0.

The key to understanding problems like this is to visualize the potential buckled shape of the column. If the axial load is large enough to cause flexural buckling, then the buckled shape will still look exactly like it would if the top didn't move. As far as the column's concerned, it's no different. From a global viewpoint, there's a tiny rigid body rotation. The only thing that's different is the axial load which will be just a tiny bit different because now you have a tiny component of horizontal reaction at the bottom.

 

[Lion06]

271828 and dave-
Please explain this one to me. If you have a sway frame, the top can not be pinned, correct? I thought for a sway frame the top is either free to translate, but fixed for roation or it is free to translate and free to rotate (totally free condition).
I was taught in school that any column in a sway frame has a k>1.0. Is this only for the frame columns and not the gravity columns? If so, why is that? If the top is free to translate (sway frame), how do you say that the gravity columns are not subject to the same translation as the frame?

 

[271828]

This goes back to visualizing the buckled shape of the column.

A pinned-pinned column buckled shape is a half sine wave. If a column has no rotational restraint at either end, but the top can translate, the buckled shape is still a half sine wave, but with a rigid body movement also. For small translations, there's no real difference between the two cases. The only difference will be that the second column has a very slightly larger applied axial load because the reaction has a small horizontal component also.

A moment frame column has end restraint, so most likely won't buckle into a half sine wave. It'll be some combination of sine, cos, cosh, and sinh. Plotted, it'll look like one would expect from a moment frame column. This is why K>1.0 for these.

 

[Lion06]

I did just read in the commentary to use k=1 for leaners, even in sway frames.

 

[DaveAtkins]

StructuralEIT,
I agree with everything 271828 has said. I think you are wondering about something that never happens. If the column in question is PART of the rigid frame, it WILL have moment under lateral loads, and it will not be just a gravity column, and K will be greater than 1.0. But if the column in question is a leaner column, as described by 271828 (the top and bottom are pinned), then it just leans a little bit with the rigid frame, and K = 1.0 because it will buckle over its length, but its top is not free.

DaveAtkins

 

[shin25]

haynewp, correct me if I am wrong-

The magnified moment that a properly designed fixed base is going to see in a sway frame is from the p-delta effect only. When a column really buckles in a non sway frame (if the footing is truely fixed), it buckles away from the fixed connection. This means that the magnified moment due to buckling occurs in the column away from the fixed footing. This is why the code does not require the flexural members to be designed for magnified moment in a non sway frame.

 

[haynewp]

I'm confused.

 

[shin25]

haynewp,

Look at the diagrams of the buckled shapes for columns fixed at the base. At the fixed connection, the column is rotation fixed during buckling event. But, the column itself becomes increasingly more eccentric at about mid height. This means that the fixed base will not be experiencing additional moment during buckling.

 

[Lion06]

it might not be experiencing additional P-delta moment, but the moment due to the effects of buckling would be higher, no?

 

[Lion06]

just thinking out loud here, but if you have a fixed end (base) condition, how can you possibly increase teh midspan moment without increasing the end (base) moment ? The only way I can think of is if a plastic hinge has already formed at the base.

 

[miecz]

StructuralEIT

If the column buckles, then, yes, you'll get an increased moment at the bottom of a fixed end column, and it may increase up the the strength of the column. But it doesn't matter if your footing can resist that moment, because your building will collapse anyway. So, you don't design for that condition, you prevent it, by designing for the applied forces. As long as you prevent the column from buckling, the footing will never see any moment greater than that resulting from the applied forces.

 

[haynewp]

jmiec

I agree it is impractical (and is likely not even that much after all this long discussion) to try and include buckling restraint moments in the column end connections (eg footings). But there is also the restraint moment that keeps the column from buckling sooner as a pin-ended member rather than waiting to buckle as a fixed ended member.

If the column load is greater than what would cause buckling in a pin-ended column but less than what would cause buckling in a fixed ended column, then the end connections would have a restraining action and I guess technically, while restraining your hand from hitting yourself in the head with a hammer, it could be counted for in the footing design.

 

[miecz]

haynewp-

Yes, I see your point. What I don't see, is the solution.

 

[271828]

Been out of town for a couple of days, so didn't get a chance to participate in the latest volleys.

I think we're missing the boat trying to figure out what column restraint moments exist at the footing.

At the onset of buckling, regardless of the buckled shape, the amplitude of the shape will be VERY tiny, approximating the out-of-straightness tolerance from AISC.

After it buckles, the buckled shape amplitude increases without bound, so of course the base moment will grow huge.

This is really a stability bracing problem, so should be approached that way to determine the required base moment and stiffness. For example, one could conduct an analytical study. Manually calculate the buckling load one would expect using G=1.0 at the bottom, pinned at the top. Create a model of a column with curvature corresponding to AISC's sweep tolerance. Then use the model to calculate the spring one needs to restrain the column enough to get a buckling load equal to the one computed manually using G=1.0. The guys who came up with G=1.0 probably did something like this.