k value for corner column 1

[smvk3]

I have a two story wf column in a corner of a building. In the strong axis direction, the column is pinned at the base and has moment connections going into it at both the upper floor and roof levels (it is part of a two story moment frame in that direction). therefore, my k value is 2 with an unbraced length of the story to story height.

In the weak axis direction, the column is also pinned at the base, but only has simple shear connections going into it at both the upper and roof levels. However, the next column down in the weak axis direction is a moment frame.

My question is, would the k value in the weak axis direction of the column be 1 with an unbraced length of the story to story height? Or would it be 2 since the joints translate In that direction?

 

[BAretired]

It would be 1 as the corner column in the weak direction is relying on an adjacent moment frame to resist lateral forces and laterally brace the corner column at second floor and roof level. Without such frame, shear wall or similar stiffening feature, k for the corner column would be infinite and the structure would be unstable.

Horizontal reactions will result at first floor, second floor and roof which must be resisted by framing members or foundation. The magnitude depends on the vertical load and the relative horizontal displacement at first floor, second floor and roof due to frame deformation.

BA

 

[TehMightyEngineer]

Neither. You have to use the alignment charts. If you're using AISC then see AISC 360 commentary section C2 of the 2005 specification (or use the direct analysis method). The 2010 AISC specification switched chapter C with the direct analysis method in the appendix.

Two charts are presented, one for sway frames and one for braced frames.

Calculate G for the top and bottom of each of the two stories of column in the weak and strong axis. You can then use the alignment charts to calculate the K value for each direction and story.

Or you can set K = 1 and use direct analysis, assuming you're modeling this in a computer program capable of second order effects.



Maine EIT, Civil/Structural.

 

[smvk3]

BA, in the second part of your post are you talking about the lateral forces at each story level from wind and seismic and that these result in horizontal reactions that must be resisted by the foundation?

Teh, isn't conservative to just use k=2 in the strong axis and k=1 in the weak axis instead of using the alignment charts? Also, this is only a two-story frame so second order effects would be very small (although I have resisted the 10-year wind deformation to h/400 and the seismic deformations per code).

 

[TehMightyEngineer]

As far as I know K can't be any higher than 1.0 in a braced frame so, yes, you can probably use K = 1.0 for the weak axis. I'm not quite sure about the strong axis but if that too can be idealized as a braced frame then K maxes out at 1.0 as well.

Maine EIT, Civil/Structural.

 

[BAretired]

smvk3,

The corner column takes no applied lateral force from wind or seismic in the weak direction but if the adjacent moment frame moves horizontally a distance Δ under the applied load, a column hinged top and bottom carrying axial load P would produce equal and opposite horizontal hinge forces of PΔ/h where h is the story height.

The column is not truly hinged at the second floor if it is continuous from first floor to roof, but I would neglect continuity for the present purpose and treat it conservatively as hinged at each floor with a k of 1 in each story.

BA

 

[WillisV]

The direst analysis method removes the need for these kind of discussions.

 

[SAIL3]

I would tend to use a k value of 2 or close to it in the weak dir because of all of the following:
a. It is a continous col @ 1st floor level.
b. Being braced by a moment frame one can experience lateral trans @ both floor levels which may
not be equal or could be in reverse.
c. Being braced by the moment frame is not the same as being braced by a braced frame.

 

[KootK]

For the strong axis, 2.0 is the minimum value for K in the situation that you've described. K=2.0 represents pinned on one end and fixed on the other. At the second floor, you will be something less than fixed. Your K should be > 2.0.

I'd be perfectly comfortable with K=1.0 for the weak axis. Since you're modelling the column as pinned at all levels in the weak axis, it's quite impossible for you to have any modelled column moments unless there are loads applied between levels. P-delta effects will be dealt with at your moment frames and those frames should be stiff enough as to prevent the development of any serious moments in your lean-on columns.

There will of course be inherent eccentricities of load and moments due to accidental restraint in the real world. However, those are secondary effects and are likely more than offset by the fact that your columns will have some fixity at the ends and that restraint will tend to reduce K, most likeley below 1.0. If you want to be super safe, you could design your columns for K=1.0 and moments consistent with the drift that you expect in the moment frames. This should still be much less punitive than designing for K=2 (75%-ish less axial capacity).

It's important to remember that K reflects the mode of buckling, not whether or not there are column moments present (although those two things are often related). For the weak axis, lean-on columns braced by the moment frames, K=2.0 represents unbounded sway buckling up the column. And that mode of buckling can't occur because -- hopefully -- you've got a moment frame designed to stabilize all of the floor framing tributary to it.



The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[smvk3]

KootK, at the second floor (between the roof level and upper level), wouldn't 1 < K <= 2 since at both ends the rotation is restrained by the moment connections coming into at from both beams? Wouldn't that represent case (c) in table C-C2.2 (p. 16.1-240) in the AISC manual?

 

[KootK]

I assumed that we were talking about the lower floor columns as those would likely govern your design if, as is usually the case, your second floor columns are just extensions of your first floor columns.

Switching to the upper floor columns and speaking theoretically:

- Weak axis: K=1.0; same arguments as above.
- Strong axis: 1.0 (super stiff beams) < K < Infinity (super flexible beams).

I suspect that, for the upper floor columns in the strong axis direction, K probably is less than two. You can't say that for sure without looking into it however. The AISC alignment chart is probably the best way to do that.



The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[structSU10]

The nomographs are not a great way to get a K factor in my opinion. Attached is a paper that shows examples and a more refined approach to determining K, which aligns much better with the theoretical, rigorous calculations. I tend to use the direct analysis method, as WillisV said, to avoid having to worry about applying the correct K to every column - I can just set K=1 and not fiddle with K if I need to change member sizes and such.