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Column Unbraced Length Question

Column Unbraced Length Question

Column Unbraced Length Question

(OP)
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

RE: Column Unbraced Length Question

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

RE: Column Unbraced Length Question

(OP)
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

RE: Column Unbraced Length Question

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.

RE: Column Unbraced Length Question

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

RE: Column Unbraced Length Question

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

RE: Column Unbraced Length Question

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.

RE: Column Unbraced Length Question

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>

RE: Column Unbraced Length Question

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.

RE: Column Unbraced Length Question

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

RE: Column Unbraced Length Question

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.

RE: Column Unbraced Length Question

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.

RE: Column Unbraced Length Question

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

RE: Column Unbraced Length Question

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?

RE: Column Unbraced Length Question

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

RE: Column Unbraced Length Question

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.   

RE: Column Unbraced Length Question

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
 

RE: Column Unbraced Length Question

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

RE: Column Unbraced Length Question

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

RE: Column Unbraced Length Question

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.

RE: Column Unbraced Length Question

271828,

I don't have such a program.  So do it and let us know what you get for an answer.

BA

RE: Column Unbraced Length Question

CRAP!  I went to SAP2000 and tried to do this, but it kicked me out and said that I need to buy the Advanced version.  I'd forgotten that I used to have the Advanced version, but now only have the Plus version which doesn't do eigenvalue buckling.  So...I don't have such a program either at this point!  I have Ansys also, but don't know how to use it yet.

Sorry.  Now I feel stupid, LOL.  I have this stuff programmed in Mathcad, so if I get a chance, I might try that out.

Another approach is to put the column into a program that has small-displacement second-order capabilities.  Displace the column a little (like 0.1", not 0.001" and probably not 1") in the direction that it should buckle.  Run the analysis and write down the lateral displacemet near the middle.  Increase the load and repeat.  When the deflection goes the other way, that means that the geometric stiffness is larger (and negative) than the elastic stiffness and the buckling load has been exceeded.  It goes something like:

P=99kip, Delta=0.1"
P=100 kip, Delta=0.1"
P=101 kip, Delta=0.101"
P=102 kip, Delta=2"
P=103 kip, Delta=20"
p=104 kip, Delta=-6"

I know this will work, but it takes about a half hour to iterate.  If I get a chance, I'll take a shot at it.

RE: Column Unbraced Length Question

271828-

I don't have a canned program that does eigenvalue buckling analysis.  Even if I did, I would only use it to check my work.   

RE: Column Unbraced Length Question

Why not?  Do you use a program that does small-displacement 2nd-order analysis for other than checking your work?

RE: Column Unbraced Length Question

No.  I've pretty much migrated to using Mathcad for all my work, and then use some software to check the results.  So I write my own routines.  I guess that's a result of being burned by too many canned packages. That limits the complexity of the work that I do, but that's OK.

RE: Column Unbraced Length Question

I plot by hand the given values in the EA-95 code and gives K=.54 approximately for 15% of the length loaded (quite small scale so better not venture more). I was wrong in my previous post indicating that the proper interpolation would place the value below that linearly interpolated, by whatever the reason imagined in wrong way the shape of the fitting curve.

RE: Column Unbraced Length Question

Done a quick eigenvalue buckling analysis and i get that k=0.6 which is close to ishvaaags value of 0.54.

RE: Column Unbraced Length Question

If you are saying that a 100' long column, hinged top and bottom, loaded axially 15' above the lower hinge has the same buckling load as a 60' long column loaded at each end, I believe your solution is extremely conservative.  If your k value means something else, please clarify.

BA

RE: Column Unbraced Length Question

Well, I at first sight think is also a conservative value ... thinking as a child loading a stick, I would think a value closer to yours, BAretired. But that is what the code seems to have elected to set for the value. It may also correspond to exact analyses, not having done can't confirm. But no, the values mean a K factor to apply to the whole length as pinned-pinned. I have also a book with buckling factors I think taken of Pflüger but seemed to me no case applied to this, will look again.

RE: Column Unbraced Length Question

Looking at
Pandeo de Estructuras
Félix Escrig
Publicaciones de la Universidad de Sevilla
1985

case II-IIa seems to be the value given in code MV-103 (vigency prior to EA-95 till now by me quoted)
I missed it the previous time because the sliding support was placed contrarily at what here we were thinking; in the case the compressed part is the closer to the sliding end ... yet in any case laterally indesplaceable member. Making the story short (K·L)^2 for Pcr there is about 0.31·L^2 (read from middle size chart yet bigger than my hand drawn fitting curve of above) for 15% of the length in compression. That makes K=sqrt(0.31)=0.556
 

RE: Column Unbraced Length Question

I can believe K=0.55-0.6.  The bottom 15 ft is similar to a 15 ft tall column with a pin at the bottom, but with horizontal displacement allowed at the top.  Also at the top is a rotational spring that represents the upper 85 ft of the column--it has zero load, so is a restraining member in this case, like the "girder" used in the alignment charts.  If the sway uninhibited alignment chart is used (the "column"=bottom 15' and the "girder"=top 85') with Gb=infinity and Gt=(I/15) / (I/85)=5.7, then K=3.5.  KL=3.5*15=52.5 ft.  That's approximate, of course, because the chart was created for the other end of the girder to be continuous, not pinned as it is in this case.  If the chart's K is adjusted for this, then I get KL=68 ft.  From those approximate analyses, I can buy KL=60 ft.

RE: Column Unbraced Length Question

Sorry, gents.  I was having a senior moment before.  If P(crit) = 28.8 EI/L² then the corresponding length is approximately 0.59L, or in this case approximately 60'.  

For some reason, it still seems more than I would have expected.   

BA

RE: Column Unbraced Length Question

Seems to me that the top of the 15 foot section should be modeled with vertical translation, but fixed for rotation due to the continuity of the member above the 15 foot height.  This will give a higher capacity.

Mike McCann
MMC Engineering
Motto:  KISS
Motivation:  Don't ask

RE: Column Unbraced Length Question

I'm late, I know, but this case is tackled in Theory of Elastic Stability on page 98. The differential equations for the sections above and below the load are formulated and solved for a column with load P1 applied at the top and P2 applied at an intermediate point. Taking a very small load for P1 and running through with some representative numbers I also get k approximately equal to 0.6, so an effective length of 60'.

RE: Column Unbraced Length Question

Personally I would just make sure that the base plate and bolts are sufficient for the short cantilvered column restraint forces and then treat it as an effective length of 2.2 times 15 feet.

 

RE: Column Unbraced Length Question

cds72,

That would be conservative, but it does not answer the original question with hinge at the bottom and vertical roller at the top.

BA

RE: Column Unbraced Length Question

Yes but what columns actually do have a hinge top and bottom?

You normally need at least 4 bolts for health and safety anyway.

RE: Column Unbraced Length Question

The question was a theoretical one.

BA

RE: Column Unbraced Length Question

BAretired,

Scientists have abstract theories.

But as engineers these theories are meaningless until they are applied to the real world.

RE: Column Unbraced Length Question

csd72,

I don't know why we are belaboring the point.  There is nothing abstract about a hinged support and it can be achieved (almost) in the real world.  It is not meaningless.

gregeckel started with:

Quote:

I have a quick theoretical question.

Your response was conservative, but did not address the theoretical question posed.

BA

RE: Column Unbraced Length Question

I did a quick test using RISA-3D.  Assuming that the P-Delta analysis can adequately predict the elastic buckling behavior of the system.  

I used a HSS10x10x1/4, 100ft long, simply supported at both ends and divided into 40 smaller pieces (so that I can capture the P-little delta effect).  

Then I added in a load of 1/500 of the vertical load at the joint with the applied vertical load. Similar to the notional loads of AISC's direct analysis method which is used to approximate elastic buckling directly in an analysis.  

The assumption is that when the P-Delta analysis starts to diverge then the column has buckled.  This should be an easy test to perform for anyone with a program that does a good P-Delta analysis.  

I found that the elastic buckling of the column occured somewhere between 115.3 kips and 115.2 kips. That would correspond to a KL value of 591 inches or 49 feet.

If I move the notional load up to the mid-point of the column (rather than apply it at the 15ft mark) then the buckling still occurs at the same load level.  

 

RE: Column Unbraced Length Question

JoshPlum

I must be missing something.  My model diverges at 79 kips, and from that I get a KL of 43 ft.  Could you post your .r3d file?

RE: Column Unbraced Length Question

Here is the RISA-3D file.  Note a couple of things:
1) Shear deformation is turned off.

2) I'm not using the stiffness reduction associated with the Direct Analysis Method (that is done to help an elastice analysis program mimic the inelastic buckling effects).

3) I'm not using the 13th edition ASD code (which requires that loads be multiplied by 1.6 before accounting for the P-Delta effects).  

Josh  

RE: Column Unbraced Length Question

Josh has a pin at the top instead of a vertical roller.

I get 78.5 kips as well.
 

RE: Column Unbraced Length Question

That makes sense... It should be a roller rather than a full pin.  If I had reviewed the axial force in the column I would have known to change it.  Therefore, it appears that the KL value from the RISA analysis really ends up being closer to 59 feet. That's almost exactly what the theoretically derived values were from the other folks.

Note that the formula for euler buckling (without safety factors) as I understand it is Pcrit = pi^2*E*I/(KL)^2.  That's the formula I used to convert Pcrit = 79 kips to an equivalent KL of 59 feet.

Josh  

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