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

 

[BAretired]

271828,

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

BA

 

[271828]

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.

 

[miecz]

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.

 

[271828]

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

 

[miecz]

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.

 

[ishvaaag]

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.

 

[patswfc]

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

 

[BAretired]

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

 

[ishvaaag]

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.

 

[ishvaaag]

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

 

[271828]

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.

 

[BAretired]

Sorry, gents. I was having a senior moment before. If P(crit) = 28.8 EI/L[²] 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

 

[msquared48]

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

 

[ChadV]

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'.

 

[csd72]

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.

 

[BAretired]

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

 

[csd72]

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

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

 

[BAretired]

The question was a theoretical one.

BA

 

[csd72]

BAretired,

Scientists have abstract theories.

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

 

[BAretired]

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:

I have a quick theoretical question.

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

BA