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
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
DaveAtkins
RE: Column Unbraced Length Question
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
RE: Column Unbraced Length Question
RE: Column Unbraced Length Question
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
RE: Column Unbraced Length Question
RE: Column Unbraced Length Question
Michael.
Timing has a lot to do with the outcome of a rain dance.
RE: Column Unbraced Length Question
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
RE: Column Unbraced Length Question
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
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
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
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
RE: Column Unbraced Length Question
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
Can anyone confirm this?
BA
RE: Column Unbraced Length Question
BA
RE: Column Unbraced Length Question
Really, this problem is trivial using such a procedure.
RE: Column Unbraced Length Question
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
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
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
RE: Column Unbraced Length Question
RE: Column Unbraced Length Question
RE: Column Unbraced Length Question
RE: Column Unbraced Length Question
BA
RE: Column Unbraced Length Question
RE: Column Unbraced Length Question
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
RE: Column Unbraced Length Question
For some reason, it still seems more than I would have expected.
BA
RE: Column Unbraced Length Question
Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask
RE: Column Unbraced Length Question
RE: Column Unbraced Length Question
RE: Column Unbraced Length Question
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
You normally need at least 4 bolts for health and safety anyway.
RE: Column Unbraced Length Question
BA
RE: Column Unbraced Length Question
Scientists have abstract theories.
But as engineers these theories are meaningless until they are applied to the real world.
RE: Column Unbraced Length Question
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:
Your response was conservative, but did not address the theoretical question posed.
BA
RE: Column Unbraced Length Question
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
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
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
I get 78.5 kips as well.
RE: Column Unbraced Length Question
Thanks for posting the file. I can't see much difference with my model (attached), except that I had the boundary condition at the top free in the y direction, and I split the column into 40 members. I'm getting divergence at 79 kips.
RE: Column Unbraced Length Question
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