General Mechanical

General Mechanical

Static structural—Snap deflection analysis: Strain doesn’t correspond to stress?

    • rgarcia60
      Subscriber

      Hello,


      I ran a static structural analysis of a snap with multilinear kinematic hardening plasticity. The material at hand is a plastic resin with 30% glass fill—data sheet shows a break stress of 130MPa and a break strain of 2.5% (see engineering stress-strain curve below).



      So, in my engineering data I have Isotropic Elasticity (E=9500MPa, v=0.35) plus Multilinear Kinematic Hardening (see graph below).



      For the multilinear inputs, I made sure to transform the engineering stress-strain data given by supplier to true stress-strain data, followed by the plastic strain calculation. The model converged quickly and behaved as expected, but I got confused when checking the strain and stress results. When looking at the Max. Principal Strain, results show a max of ~1.4% of strain which is considerably below the 2.5% break strain of the material (according to engineering stress-strain curve). But when I look at the Max. Principal Stress, results show a max of ~133.3MPa which is a bit above the 130MPa break stress of the material (according to engineering stress-strain curve). 



      Why is there this discrepancy? I would think beforehand that the stress and strain results would correspond to the true stress-strain curve input (true s-s curve is very similar to engineering s-s curve anyhow). But it seems if I read strain results, I would approve this design since results are considerably below the break strain limit, whereas if I read stress results I wouldn't approve this design as it's marginal. If I read ~1.4% of strain I would expect to read ~110MPa in stress. What am I missing? I also checked Von-Mises and it's the same story, although I think Max. Principal Stress failure theory applies more due to the high percentage of glass fibre in the resin.


      Warm regards,


      Richard


       

    • kaitova
      Subscriber

      Hello rgarcia60 ,


      I have answer for you, the most probably you have correct results of your simulation. but you have settings in the analysis results representation, that are averaged in s special way, thus, they give results more than they should actually. I had this issue with Abaqus long time ago, change of setting helped. You can try to find out yourself which setting can be adapted. I will look as well and send a link if I find.


      Regards, Kamilla

    • peteroznewman
      Subscriber

      Hello Richard,


      I find it helps to make a one element model to understand what material models are doing. I made a 10 mm cube and meshed it with one linear element so there are only 8 nodes. I used symmetry on each of the three adjacent faces to hold the cube, then I applied a 0.02 mm displacement on the face normal to the X axis.


      I created a simple material with E = 100 GPa and used Bilinear Isotropic Hardening with a Yield Strength of 100 MPa and a Tangent Modulus of 0.


      All these plots are vs. Time, so at 1 the displacement is 0.02 mm on a 10 mm cube or a total strain of 2.e-3 as shown in the plot below.


       


      You know that Total Strain = Elastic Strain + Plastic Strain. In the plot of elastic strain, you can see that at a strain level of 1e-3 (time = 0.5), the elastic strain reaches the point when plastic strain begin. This is expected because Yield Strength/Modulus = 100 MPa / 100 GPa = 1e-3.



      Plastic strain is zero below the yield stress and then it increases.



      The stress increases until it reaches the yield stress.



      If I plot Normal Stress (Max) vs. Equivalent Total Strain (Max) I get the material Stress-Strain curve.



      Hope this helps.  ANSYS 2019 R3 archive is attached.

    • AkashVyas
      Subscriber

      Hello, peteroznewman I also tried this single element but with Multilinear Kinematic hardening and I have a small doubt, my results shows max Von Misses stress is 190.53Mpa which is less then yield strength(250Mpa) but this has a plastic strain, but if the max stress is less then yield strength then why I have a plastic strain.


      Stress


      plastic strain


      material


       

    • peteroznewman
      Subscriber

      The last graph shows the yield stress is 100 MPa.  That is the first point on the curve.


      If you want the plastic strain to start at 250 MPa, delete the first six lines in the multilinear table so the first number is 250.

    • AkashVyas
      Subscriber

      Okay I will try

    • AkashVyas
      Subscriber

      sir, I removed the points but my normal stress Vs total strain graph has a little tweak, can you tell me why is that


    • peteroznewman
      Subscriber

      Under Analysis Settings, with Auto Time Stepping On, set the Initial, Minimum and Maximum substeps to 200 and solve. The little tweak become much smaller.

    • AkashVyas
      Subscriber

      Sir, I have done the changes but the result are almost same 

    • AkashVyas
      Subscriber

      Sir any updates on this

    • peteroznewman
      Subscriber

      I don't know. Is it a Linear element? Please attach your project archive and say what version of ANSYS you are using.

    • AkashVyas
      Subscriber

      actually it's program-controlled, I'm using 18.2


       

    • peteroznewman
      Subscriber

      Don't leave it Program Controlled!  Use the pull down and set it to Linear.


      Here are the corrections to your one-element model.


      1) You need Frictionless support on three orthogonal faces.



      2) You must leave the two other axes Free when you apply a Displacement.



      Once you do that, you get rid of the glitch in the Stress-Strain chart.


    • AkashVyas
      Subscriber

      got it, sir, thank you

    • rgarcia60
      Subscriber

      Hi Peter,


      Apologies for my late response to this thread. I know now after reading your response that I was only looking at the max. principal elastic strain instead of looking at the max. principal total strain. I find it odd that I have to look for a user defined command, specifically EPTT1, in order to post-process max. principal total strain. Anyways, I do have a follow-up question: will EPTT1 always be the maximum strain compared to EPTT2 and EPTT3? Or is it a best practice to always look at the three EPTTs no matter what?   


      Also, I would like your opinion on using strain as failure criteria for the glass-filled polymer I'm working with. The resin supplier tells me it's best to use the break strain (2.5%) shown in the material data sheet provided by them. This is what I'm doing now by comparing the max. principal total strain (EPTT1) shown in Ansys to the 2.5% break strain of the material. The thing is that I've been refreshing my memory with failure theory lectures and readings and it's generally suggested that maximum principal strain theory is inaccurate and should be avoided if possible, so in essence I shouldn't be using EPTT1 to assess failure in my component. Perhaps the resin supplier is suggesting to use Von-Mises strain against their 2.5% break strain value, but I don't agree at all that the resin at hand is ductile! It's surely a mind-bender for me... hopefully you can shed some light over this.  


      Warm regards,


      Richard

    • peteroznewman
      Subscriber

      Hi Richard,


      I often ask someone who hijacks another member's discussion to open a New Discussion so they own the thread so the original poster is not bothered with all the new posts, but in this case, you seem to have benefited from the hijack.


      To answer your first question, EPTT1 will always be the most positive strain, EPTT3 will always be the most negative strain and EPTT2 will be between them.


      To answer your second question, look at the stress-strain curve for the glass-filled polymer. If it is basically a straight line up to fracture, that is how a brittle material behaves. In that case, the failure theory to use is Max Principal Stress is compared with Ultimate Tensile Strength. When the stress-strain curve has a different slope after yield, that is a ductile material and the failure theory to use is Total Plastic Strain (von-Mises) is compared with Elongation at break.


      The stress-strain curve you show above for temperatures at 23 C and above have a definite change in slope, so I would apply the ductile failure theory. However, I have been working with metals for many years and have not read any of the literature that might suggest a better approach.


      Warm regards,


      Peter


       

    • rgarcia60
      Subscriber

      Thank you Peter. 

Viewing 16 reply threads
  • You must be logged in to reply to this topic.