column buckling check 2

[Lion06]

I tried following the procedure in this thread (suggested by 271828) to get a column buckling load out of a program, but it just isn't working. It keeps upping the lateral displacement, but it never becomes unstable near (or above) the elastic critical buckling load. I could double the euller buckling load, but the lateral displacement is always appropriately proportional.


thread507-211026

 

[271828]

What program are you using?

What type of analysis? Small-displacement P-Delta?

 

[Lion06]

RAM Advanse. I broke a 20' column into 2' segments and modeled a half sine wave.

 

[JAE]

Does RAM Advance do a P[Δ] analysis?

 

[JAE]

I mean technically...a second order analysis since you have intermittent joints along the length.

 

[Lion06]

Advanse does do a P-big delta analysis. It doesn't do the P-little delta analysis (hence the broken member).

 

[abusementpark]

Are you modeling it as a frame element broken into 2' segements??

 

[271828]

"I could double the euller buckling load, but the lateral displacement is always appropriately proportional."

THis makes me think you are running a first-order analysis. Are you sure the second-order feature is turned on?

Also, you might try the AISC App. 7 Commentary benchmark problems to make sure RAM's second-order analysis actually works.

 

[slickdeals]

For all those who have already tried to compute p-[δ] moments using computer programs, here is a question:

Do you only have to divide a frame element into finite pieces to capture p-[δ] or should you break them into finite pieces and also introduce a "kink" in that node?

 

[271828]

It depends on exactly what you want to do. Usually, you only have to break up the member. After some load is applied, the nodes are displaced transversely, so the second order moments can be generated. This would also work for eigenvalue buckling analysis. Actually, I don't think eigenvalue buckling analysis would work correctly in most cases if the nodes were displaced initially--there would be no trivial solution, hence no eigenvalue problem. Not sure that this is always the case, though.

In StrlEIT's analysis, he is trying to compute the buckling load for a specific mode using a very specific kind of analysis. He needs to displace the modes into the shape of the mode that he's looking for. If it's simply supported on both ends, then this initial shape is something like a half-sine wave.

 

[abusementpark]

Don't you have to use brick elements to perform an eigenvalue buckling analysis??

 

[271828]

No. All you're doing is solving the following equation:

([Ke] + Lambda*[Kg]){Delta}={0} where

[Ke] = elastic stiffness matrix (the one everyone thinks of as the "stiffness matrix"

[Kg] = geometric stiffness matrix

Lambda are constants.

Of course, {Delta}={0} is a solution, but that's worthless to know.

If you set the determinant of [Ke]+Lambda*[Kg]=0, you get values of Lambda that correspond to non-trivial solution--hence the term "eigenvalue analysis." These relate to the buckling loads. Corresponding {Delta} are the modes.

[Ke] and [Kg] can be for any kind of element. [Kg] has P's in it if you have frame elements, membrane stresses if you have shells, etc.

 

[Lion06]

I modeled a 20' column (broken into 2' segments) with the half-sine wave, deformed shape of 0.01sin(pix/L). This gives an initial displacement of 0.01" at mid-height of the column.

I'm working with W8x31 (arbitrarily decided on by me). I have the buckled shape modeled for weak axis buckling. I'm coming up with a critical elasic buckling load of right around 185K (on paper). The problem is that the program is giving me results that aren't making sense to me. The lateral displacement that increases upon initial loading (as expected). What isn't expected is this - The lateral displacement stays small up until around 100k (it's around 0.17"), from there it starts jumping up faster. At 135k, the lateral displacement is 0.28", at 170k it's 0.82", at 180k (just below critical buckling) it's 1.4" (like it's already buckled, but I would have expected much higher displacements for a buckled shape), at 190k (just above critical buckling) it's 3.8" (again, possibly buckled, but I would have expected much higher displacements for the buckled shape).

Here's the real kicker. At 200k, it's buckling mode changes. It jumps from the buckled mode that I assumed to a half-sine wave in the opposite direction (i.e. my assigned half-sine wave had positive 'displacements', but the buckled shape half-sine wave has negative displacements) and of much larger magnitudes (6.9" at mid-height), more along the lines of what I would expect from a critical buckling load.

Any ideas?

 

[haynewp]

Post the file.

 

[Lion06]

Here.

 

[WillisV]

I did the same thing in SAP2000 except with a 1" initial center displacement - buckles around 185ish.

Loads and center displacements as follows:
160k -6.6"
170k -12.19"
175k -19.64"
180k -46.58"
185k -156"

 

[Lion06]

was there a reason you chose 1"? I tried to keep the initial displacements very small (thinking that the buckling behavior would be more pronounced).

 

[271828]

Check the basics of your model. I just repeated your problem and my first one screwed up because of a basic boundary condition problem.

Here's what I did. Maybe there's something in there that'll help:

Create 20' long beam of two frame elements with a node in the middle. End nodes are set up to create a pinned-pinned condition. Move the middle node 0.01" and auto submesh the members to get nodes at 1' apart. Note that the initial shape is not exactly a sine wave. This won't matter in the end. It's close enough to the mode shape we're after.

Give the W8x31 zero shear areas to ignore shear deformations (your issue might be this or something similar). Give it 1000x bigger area so axial deformations won't make the plot hard to read in its deformed shape.

Use P-Delta, which solves [Ke+Kg]{Delta}={F}, so no iteration is required. No need to use P-Delta plus large displacements option that would move nodes and iterate.

Pe = 184.352 kip

Results:

P (kip) Delta (in.)
170 0.0961
182 0.6272
184 4.229
184.32 45.68
184.349 403.0
185 -2.32

The shapes look like half sine waves, but I didn't pull numbers out and verify mathematically.

Try to see what happens between 190 kips and 200 kips. My guess is that your model has something in it that causes a little more stiffness than that used to compute Pe, and you'll see the deflection take off somewhere between 190 and 200 kips.

 

[271828]

"was there a reason you chose 1"?"

Either way should work. Like you typed, the displacement will more suddenly increase with smaller initial displacement.

 

[271828]

Oh yeah, I used SAP2000 also. I have no way to read a RAM Advanse file.