Buckling Test - Request for Help

LHartleyLHartley
I am a hobbyist who finds the interface of Mecway very user friendly.

Attached is an attempt I have made at analysing a truss member.
Geometry is a square hollow section (pipe) 65mm by 65mm by 4mm wall thickness.
Steel is 450 mpa
The member is connected to the panel point by an 18mm clevis pin at each end.
The distance from centre of pin to centre of pin is 1360mm
The member will be subject to a maximum load in compression of 32Kn.

I have four questions:
1. Have I set my model up correctly?
2. If so is the displacement of 0.233mm in Y correct?
3. If so what if any is the factor of safety?
4. If the answer to question 1 is "no" can one of you guys show me how to go about things please?

Thanks in advance

Comments

  • cwharpecwharpe
    Ok, I'll give it a try...

    1. Model Setup:
    • Constraints are correct. Since these are pinned supports, I don't think truss option is necessary.
    • There are nodes floating not attached to elements. "Delete Unused Nodes". Try the Refine X2 a few times to subdivide your column.
    2. Analysis related:
    • The displacement of 0.233mm is correct for simple axial compression. You want to run Buckling Analysis if it's Euler Critical Load you are after (more on that in a moment).
    • The software will often red-flag about not accepting a particular cross-section. Some of the members substitute CCX material cards for this situation. I find I can calculate equivalent stiffness (Ix,Iy) and area for a rectangular cross-section and get by.
    3. Safety Factor:
    • First get your Critical Load in the ballpark. Calculate the Slenderness Ratio.
    • I've attached a couple classic reference charts of column failure tests and encourage you to explore further. They illustrate transition points in curve-fitted data that mapped out critical load vs slenderness ratio (SR). Most notably, you can grossly overstate your buckling strength if you've been calculating with the Euler curve (long, slender columns) but your SR lands you in the short/intermediate column zone.
    • Most Codes have a statistically suitable safety factor built in to their (flexural buckling) equations for allowable column loads.
    • Other forms of buckling are always a possibility (Local, Lateral-Torsional) and may need investigation.
    4. Examples of 2D & Non-Linear Euler buckling attached. You'll also find some clever examples posted in Forum and YouTube by members.


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  • LHartleyLHartley
    @cwharpe - thank you very very much for taking the time to provide me with such detailed answers to my questions. I have just read through them a couple of Tim’s over my morning coffee.

    I have already learnt quite a lot from them, and I will examine them and the examples you have kindly provided, in more detail over the weekend.

    Thanks again!
  • disladisla
    edited December 2022
    I have been discussing about this recently.
    According to my new spreadsheet could be even less.
    SHS 65x4 S450.

    Pcr (Euler)=635KN
    Ncr ( EC3 )=635 KN

    Nb,Rd=261 KN (Cold Formed- Curve c )
    Nb,Rd=347 KN (Hot Formed- Curve a0)

    In any case Nc,Ed=32KN it is very far from buckling collapse.


  • LHartleyLHartley
    @disla - thank you very much also for your response. The sections I have in mind are cold formed so therefore is the factor of safety 261/32 = >8. If so I suppose this appears very conservative but in the industry I used to work in, building and construction, if there was a way to damage a system there were always certain individuals who would find a way to compromise it. I am therefore very happy with that result.

    @cwharpe - I had to have a look at your files - I have wandered into Alladin's cave! No doubt I will have more questions but the first two that come to mind are:-

    1. Why have you applied a force of 10kN to the midpoint of the member?
    2. Also a question for both you and @disla - can you give me in laymens terms an explanation of the time lapse feature. I think I understand that things won't always buckle or crush immediately ( I have seen real life examples of this) but how is this applied/set up in the software?

  • cwharpecwharpe
    @LHartley,

    You might find the following examples helpful: Reference Mecway Manual ex. 11.7,11.8,11.9. Mecway Tutorial ex. 2.11.

    1. The (10kN) lateral force is a constant, small perturbance to get the buckling mode shape started in the otherwise perfect Non-Linear model. Actually, I had mistyped. They usually want that maybe .001 to .0001 (or less) times the maximum load. Note if you suppress my perturb load and run the N-L model again, you will see no buckling at all.
    2. I only speak laymen. The N-L analysis ramps through the loading by discrete divisions, peforming a static analysis at each step to the previous step's deformed position until reaching the full load value. Proceeding this way reveals subtle differences between the full-load-slam of Static 3D vs. incremental N-L analysis. (You might test this with a simple fixed cantilever.)
    • The Buckling set-up for N-L has to do with the load. Hand-calc Euler critical load (or Buckle 2D) informs where the first mode failure load will occur. To capture this event, set the max load somewhat higher. So, for 648kN Pcrit., I used Pmax=1000 kN. You can enter the load ramping schedule here by Table, but for this case I used the simple Formula of 1000*t.
    • In Analysis select Nonlinear Static 3D/Settings, check /CCX/Quasi-Static, then enter the Time Period (usually 1.0), which is time @ completion, then Time Step, which determines the resolution (and solve-time) for the output. At any given time, the Load applied at that step is Pmax * elapsed time.
    • Occasionally, the model won't solve. As a layman, I can't speak definitively but it either doesn't like the model or it chokes while trying to converge the step Residual Force and Displacement Corrections. This is where playing with the time step, time period, etc. comes into play. Notice I set the period to .7 (still higher than .648 -- understand?) because I was crashing at time=.8 ish. Checking the Automatic time stepping box let's the solver dance around however it wants, though you can also overide the min & max step increments. So, by end of my analysis, the total Load on the column was 1000*0.7=700kN -- enough to capture the expected buckling.
    • Once your analysis solves, select your output vector from the Solution tree (say, displ-x), choose a node on the model and read the values by hovering on the chart curve, or slide-bar to a new position and refresh readout by re-selecting the node.

      ~Good Luck.
  • disladisla
    @LHartley ,

    You are absolutely right. It makes no sense to skimp one or two millimeters of thickness on such a small part. You never know where an unforeseen load may come from.
    I've seen workers on site pulling with a tractel tied to the column to position the equipment into place. :o


    It depends on the design code but usually the safety margin is normalized. That translates into the following formula.



  • LHartleyLHartley
    @cwharpe and @disla - thank you both very much. Suffice it to say it may take me some time to take all of this in. No doubt I will be back.
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