Non-linear truss analysis

cingolocingolo
Good afternoon
I'm try to create a non linear simulation on truss structure for plastic deformation but i have this error:

*ERROR in cascade: zero coefficient on the dependent side of an equation dependent node: 14 direction: 1

I attach the file

best regards

Cingolo

Comments

  • VictorVictor
    Two things:

    Z Displacement constraint on the common node to constraint rigid body rotation.

    Either turn off quasi-static or make the force time-dependent (250*t). Quasi-static with no time dependence often fails.
  • cingolocingolo
    edited October 10
    It's work! Thank you very much!
    now i have try to apply isotropic hardening curve with this data:
    0, 0
    0.00125, 250
    0.025, 350
    0.15, 470

    but i display this error:

    *ERROR: increment size smaller than minimum best solution and residuals are in the frd file



    Best regads
  • cingolocingolo
    now is work! i have use quasi static optin time period 1s
    time step 0.001s
  • VictorVictor
    That's good. I still have trouble with those settings so if you do too, other things that help are:

    - Smooth the corners of the curve by adding more points near whatever stress/strain it stops at.
    - If the solution stops near final point (0.15, 470), add another point far above that, otherwise it uses constant extrapolation so the curve becomes horizontal which will likely fail.
  • cingolocingolo
    Is possible to calculate axial force in non linear problem? Only value is on node.


  • VictorVictor
    Not really. Here are some possibilities:

    Use line3 elements and take the force at the midside node. It's still not listed under Element values but isn't polluted by values from other elements sharing the node. Beware that line3 truss elements should be slender because their stiffness has an error which increases with the ratio of width to length.

    Use *SECTION PRINT to output it to a separate .dat file. Example attached. The output goes in truss_axial_force.dat in the working directory.


  • cingolocingolo
    That's works! thank you Victor!
  • JohnMJohnM


    I found it interesting to replace the force with the equivalent Y-displacement (I used 0.157399*t based on your result). It is an interesting problem - the knee in the load-deflection curve does not appear until all three spars have yielded :)
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