Steel as an Elastic/Perfectly Plastic Material

Can anyone tell me the best way to model steel so that it behaves elastically to its yield point, then plastically beyond? I think it's possible using the Calculix solver and one of the nonlinear material types, but I can't figure out how they work exactly.

Comments

  • You can use these settings:

    Analysis type: Nonlinear Static 3D
    Solver: CCX
    Material: Mechanical type = Isotropic, plastic type = bilinear

    Here's an example with a force that increases with time that causes elastic deformation initially, then plastic deformation towards the end.


  • Thanks for the help so quickly, Victor! I'll give it a try when I get back to the office tomorrow.
  • Got it. Works great, Victor!

    Does anybody know where to find material info online to get the properties necessary for inputs, like tangent modulus?
  • Hi!
    Make attention because the first point under *PLASTIC card is the yield point and so the strain must be the true plastic strain.
    About material info on plastic field you can find several models. I use the models on attached pictures for European structural steels on CCX.
    For the Standards that I know (European Standards) plastic analysis refers to elastic-perfect pastic material model or bilinear model with strain hardening. For example EN 13445 (pressure vessels) admits plastic analysis for pressure components using bilinear material models (EN13445 Part 3 - Annex B).
    Make attention to yielding criterion and to flow rules on plastic analysis


  • Thanks for the tip Andrea!
  • edited November 2017
    IN718 bolt bending in the thread. Elastic case and plastic case, 75F. True stress strain curve.

    Bending bolts is not good.
  • Hello nwedding,

    In addition, an optional true stress – true strain curve for use in elastic-plastic finite element analysis can be wholly derived according to ASME_VIII_DIV_2 ANNEX_3D.3 Stress-Strain. There is an Example in the PTB-5-2013 4.6.

    Inputs

    • Engineering Yield Strength (0.2% Offset)(Sy) per Section II-D Table Y-1
    • Engineering Tensile Strength (Su) per Section II-D Table U
    • Modulus of Elasticity (Ey) per Section II Part D
    • Material Parameter (εp) per Table 3-D.1.
  • Hello,

    I have done a comparison between the different Plastic models available in MECWAY/ccx before going to more complicated geometrical models. Isotropic hardening & Kinematic Hardening introducing the exact same values for the True Stress-True Strain curves.
    Ccx doesn’t support Elastic-Perfectly Plastic material so I have give some small Tangent Modulus at the end of the curve.
    As far as I have understood, on a unique increasing load with final release , both models should behave exactly the same ending both with the same Equivalent Plastic Strain.
    The behavior should be the same for tension than for compression.
    I have found they all 4 cases behave identical up to the point where the Plastic Strain starts to develop. They start at the expected point according to the curve but final values disagree.
    ¿ I’m I doing something wrong.?
    I would appreciate if someone could give some advice.

    Thanks




  • edited November 2021
    Thanks, John, for your fast response and link.
    I think that isotropic and kinematic show their nature and differences when we apply cyclic loading or at least tension followed by compression or vice versa.
    In my case I'm just applying a load and releasing. I have plot, as I have been suggested in the CalculiX forum, the Stress Strain curves and definitively there is some issue with my model as the hardening curves under compression are not right for any of the constitutive models. I have to check if it happens the same with bilinear hardening.
  • @disla, can you be more specific about what's not right? I'm not sure what the different cases in the screenshots are showing or what's wrong with those graphs.
  • Hi Victor ,

    I have just found my source of error. I have been comparing Principal Strain 1 against Principal Stress 1 for the whole set of models. For some reason, under compression , the principal Stress 1 is not representing the same direction as when the system is under tension. I have checked the element orientation and they are all the same but ....
    Shape is right now but I have to review the values to compare if the behavior is symmetric or not. I will redo my graphs tomorrow. Too many numbers .

    Thanks
  • Hi,
    After some sleep and coffee here are my results. I hope I can express myself better.
    I think that Plastic Strain at the end of the Load/Release process should be the same and also Compression / Tension should behave equally, and they do not.
    I have slightly reduced the model to be computed faster (00:02:37) and with smaller timestep.

    Am I assuming or doing something wrong?
    Thanks
  • I keep doing test to see what could be wrong and I have found that the solver seems is not taking Poisson's Ratio under consideration.
    I have redone the calculation again with Poisson's Ratio=0 and display show the solid is shrinking under tension and expanding under compression.

  • Oh, I understand now. It's only undergoing tension (or compression) with no load reversal so the two hardening models should be equivalent in that region.

    I don't understand the theory here so I'm not sure if Poisson's ratio is supposed to remain constant or behave the same with the two hardening models but that sounds like it might be the source of the discrepancy. The CCX manual does say "Furthermore, the plastic flow is isochoric (the volume is conserved)" in the incremental plasticity section so perhaps Poisson's ratio always becomes 0.5 for plastic deformation?

    It's normal for principal stress 1 to change direction on compression since it's the maximum (i.e. most tensile) stress.
  • Thanks Victor,

    “Without load reversal”. That's the point. I'm doing some reads to see why is this happening.

    What it is also strange is why the compressive load&release end up on a different value of plastic strain than the one under tensile/release.
    Seems physically the same problem to me. The curve is only defined in the positive region, so I do not really know what is going on at the other side but should be antisymmetric (?¿?) ¿ isn’t it?..
    Actually, seems that the areas under compression yield less than the ones under tensile.

    Thanks again for your response and patience.
  • Since Poisson effect is happening, I'd expect greater stress and plastic strain in tension due to the reduced cross-section area.

    The curve describes both tension and compression symmetrically since it uses von Mises equivalent stress and a similar kind of equivalent strain.
  • Thanks Victor,

    I have an smile in my face right now. :)
    That seems a very plausible explanation. I will change force by pressure to see what happens. It would be paradoxical that I come from reviewing the concept of True/Engineering Stress some lines above and now it kick me in the face.

  • Hi.

    These are my results after changing the force by pressure.
    I have also change in both the Poisson’s Ratio to 0.4999 just in case and to remove any source of discontinuity between elastic and plastic behavior.
    When computing with the Bilinear hardening the results agree almost perfectly at the final free stress state.
    For the calculation made when the user provides the curve the agreement it is not so good but pretty close now.
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