Column Unbraced Length Question

[gregeckel]

I have a quick theoretical question.

Let’s assume we have an unbraced column 100 feet long. The top of the column is a roller (translation fixed, vertical free) and bottom of the column is a pin. There is a large point load concentric on the column close to the base (let’s assume about 15’ up).

How do I determine my K and L.

I don’t need an actual solution to the problem just a code reference or the theory is fine. The numbers I gave are just to scale up the magnitude of the idea I’m getting at.

Thanks,
Greg

 

[DaveAtkins]

Well, the simple answer is K = 1 and L = 100'. And that is what I use for design for this situation. Yes, I know that only the lower 15' of the column has compression, but the entire 100' length of column can buckle.

DaveAtkins

 

[gregeckel]

It's my understanding that the buckling shape would be different with the load at 15' than with the load at 100'. I guess what I'm really wondering is what is the K value to use if you assume L=15'.

In design this situation would likely never be this extreme and I would simply use K=1 and L=(Total Length).

-Greg

 

[Lion06]

I agre that there should be a reduction in buckling length (or increase in capacity) for this case. This is analogous to the Cb factor for bending. It is recognized that if the section is not subject to uniform compression along its length it will have greater capacity, which is why Cb>1 in almost any case. Unfortunately, I don't know of any code provision that allows you to take that into account for axial members.

 

[Lion06]

I take that back. It does say in the spec that you can use k=?? unless a smaller value can be justified by analysis.

 

[ishvaaag]

Your case is in table 3.2.4.6 of former Spain's code NBE EA-95

For 85% free of load and 15% loaded, two-hinged case, a linearly interpolated value gives K=.516. A more proper interpolation than linear by a chart would give an even lower value for K on the whole total length.

Download NBE-EA-95 from

http://tocasa.iespana.es/descargas.htm

 

[miecz]

Look for an analysis of a stepped column for an industrial building. The one I have is titled "Calculation of Effective Lengths and Effective Slenderness Ratios of Stepped Columns", by Anderson and Woodward. It was published in the AISC Journal in October of 1972. I got it from the UMI Clearinghouse.

 

[BENDOG]

See AISC Design Guide 7 Industrial Buildings, Appendix B titled Calculation of Effective Lengths of Stepped Columns. It is based on the same paper by Anderson and Woodward>

 

[paddingtongreen]

This is a very unusual stepped column, considering the end conditions.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[BAretired]

It is not a stepped column. As I understand the problem, the column is continuous from end to end and does not change cross section. The buckling load may be easily determined by hand calculations using numerical methods as proposed by Nathan Newmark and illustrated in "Theory of Elastic Stability" by Timoshenko and Gere.

You first assume a deflected shape, then proceed to determine the bending moments at intervals along the column as a result of the load and the eccentricity according to your first assumption. You then modify the contour of the column and continue iterating until you have found a deflected shape which you deem "close enough" for all practical purposes.

Finally, you determine a 'k" value based on the buckling load which you have determined by the above procedures.

BA

 

[271828]

I say put the column in SAP2000 or some other program and do an eigenvalue buckling analysis to get the elastic buckling load. Should take <10 min., maybe <5 min.

 

[paddingtongreen]

BA, I used that a number of times on damaged struts.

To clarify, divide the length into six convenient lengths, assume a deflected shape, use these deflections to calculate the moments at the nodes, use these moments to calculate new deflections, repeat until the difference is negligible. Proportion the unit load upward by the yield moment over the moment due to the unit load. Then apply a safety factor, AISC used to use 12/23.
Done on a spreadsheet it is a piece of cake.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[BAretired]

paddingtongreen,

I agree that it is a very simple and useful procedure. Any bar buckling problem can be solved using that technique, even when the EI is changing along the length of column. I am not too conversant with spreadsheet manipulations, so I have always done it by hand, but I agree that a spreadsheet would likely be a piece of cake. Also, as you iterate, you get upper and lower bounds on the value of P(critical).

With the load applied 15' from the base of a 100' column, the bending moment in the upper 85' is going to be a straight line, so the problem could easily be solved using area moment principals or conjugate beam.


BA

 

[miecz]

BAretired

Why can't this be analyzed as a stepped column, with a tiny step?

paddingtongreen-

Why is this a very unusual stepped column, given the end conditions?

 

[BAretired]

miecz,

It could be designed as a stepped column with no step and EI equal above and below the point of load application. I'm not sure I see the advantage in doing that unless you have tables for stepped columns with those features.

BA

 

[miecz]

No tables, but a step by step set of formulas in AISC DG7 Appendix B. I haven't actually cranked one, but it looks pretty straightforward. I thought you were saying it couldn't be analyzed as a stepped column.

 

[iv63]

AISC Design Guide 7 Industrial Buildings, Appendix B titled Calculation of Effective Lengths of Stepped Columns has a table for slenderness ratios for seven different end conditions.
For L1=85ft, P1=0, L2=15ft, P2=P => I1/I2=1.0, L2/LT=15/(15+85)=0.15 use 0.10, P2/PT=P2/(P1+P2)=1.0 => from Table for end-fixity type pin-pin => K1=0.0 and K2=0.517
Hope this helps,
IV

 

[BAretired]

I just did a quick approximation of the load P applied 15' from the base of a 100' long column. I came up with P(critical) = 24.5EI/L². This would be equivalent to a 40' column loaded at each end.

Can anyone confirm this?


BA

 

[BAretired]

Just found a mistake in my earlier calc. My revised buckling load is P(cr) = 28.8EI/L². This is equivalent to a 34' column with loads applied at each end.

BA

 

[271828]

LOL, you guys are nuts for wanting to approach the problem that way! Am I the only one who has a computer program that does eigenvalue buckling analysis?

Really, this problem is trivial using such a procedure.