Fixed-pin connection in buckling

I am doing a buckling simulation and I can't seem to do the pin connection correctly. I have a cylindrical model under compression load, I've tried putting "displacement" support but I don't know which component should be zero. Please help. 

Comments

  • peteroznewmanpeteroznewman Member
    edited January 17
    Delete the displacement and use a Remote Displacement. Now promote to a Remote Point. On the Remote Displacement, fix X and Z to 0 leaving Y free and the rotations free. That is a pinned end. You could set Rot-Y and Rot-Z to 0 and then the pin axis is the X-axis. Apply the buckling Y component force to the Remote Point.
  • erika15erika15 Member
    edited January 18

    Thank you very much for the help! I'm not sure if I.'m doing it correctly though. I used remote displacement and promote it to remote point. On the remote point, I selected the top face of the model. 

    I then applied the force onto the remote point (which I made into a named selection), and got this result. I'm not sure if this is what it should look like. 

  • peteroznewmanpeteroznewman Member
    edited January 18
    That is normal. There is a Result scale factor in the ribbon that you can change to reduce the exaggerated deformation. The important part of the result is the Load Multiplier. That is what you multiply your applied load by to get a linear estimate of the buckling load.
  • erika15erika15 Member
    edited January 20

    Thank you very much! You are a big help! 

  • erika15erika15 Member
    edited January 25

    Hi, I'm still having problems regarding the result of the simulation. Comparing the buckling critical load I got from the simulation and the one I calculated, there is a 20% error in their values. I'm not sure if I have done something wrong in doing the remote displacement support or the force application. 

  • peteroznewmanpeteroznewman Member
    edited January 25

    The equation you used for the hand calculation might have simplifying assumptions built-in, such as Euler beam theory, while the ANSYS model is not using the same simplifying assumptions.  The agreement of 20% is adequate to show that you have not made any significant mistake such as using the wrong units in a material property, for example, which could result in an order of magnitude error.

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