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LS Dyna

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LS-Dyna Tube Crash Simulation – message file

    • Noah


      I have a question/problem with a crash simulation of a simple pipe.

      After a certain moment of deformation the following message appears:

      Element deleted due to maximum shear  6.40633E+02 >  tauc= 6.37500E+02

      However, the program runs through without errors until the end.

      When I then look at the result in the PrePost, however, no deleted elements are visible.

      This only occurs with finer meshing; this error does not occur with coarser meshing.

      What could be the reason for this?

      Thanks a lot!

    • Jim Day
      Ansys Employee
      Generally, the element ID would be given in the failure message. View that element by itself when postprocessing d3plot to confirm whether or not that element is actually removed from the display the the reported time of failure. I would also plot a time history of stress for the element to see what happens to stress at the time of reported failure. If you have any follow-up to this reply, please state what element formulation and what material model are used for the failed element(s).
    • Noah

      Thank you very much for your answer.

      I use shell elements with the formulation ELFORM 16.
      Thereby I use a SHELL NIP of 3.
      As material model I use MAT_107 with the appropriate error parameters of Cockcroft-Latham.
      Can it be that only the outer elements that are created based on the NIP command are deleted and not the nodes?

    • Jim Day
      Ansys Employee
      To what are you referring when you say "outer elements that are created based on the NIP command"? NIP is the number of through-thickness integration points for each shell element. When an element is deleted, any nodes that are freed up will generally remain in the simulation as unattached nodes. Those unattached, free nodes will no longer participate in the contact unless the variable ENMASS in *CONTROL_CONTACT is invoked. See description of ENMASS in the User's Manual.
    • Noah

      That was my wrong interpretation.
      Attached is a picture of an element that exceeds the shear stress but is not deleted.
      Also note the diagram where the element exceeds the maximum shear stress of 637.5 MPa.
      What could be the reason for this?

      LS-DYNA keyword deck by LS-PrePost
      $#      q1        q2      type     btype    tstype      
             1.5      0.06        -1         0         0
      $#     ihq        qh  
               8      0.03
      $Shell Theory zurückgesetzt auf 02
      $#  wrpang     esort     irnxx    istupd    theory       bwc     miter      proj
            20.0         0        -1         0         2         2         1         0
      $#  endtim    endcyc     dtmin    endeng    endmas     nosol     
           0.015         0       0.0       0.01.000000E8         0
      $#  dtinit    tssfac      isdo    tslimt     dt2ms      lctm     erode     ms1st
           0.005       0.0         0       0.0       0.0         0         0         0
      $#      dt    binary      lcur     ioopt   option1   option2       
           0.001         0         0         1       0.0         0
      $#      dt    binary      lcur     ioopt     
           0.001         0         0         1
      $#      dt      lcdt      beam     npltc    psetid      
      5.00000E-4         0         0         0         0
      $#   ioopt      rate    cutoff    window      type      pset    
               0                                       0         0
      $#     cid                                                                 title
      $#   surfa     surfb  surfatyp  surfbtyp   saboxid   sbboxid      sapr      sbpr
               3                   3         0                             0         0
      $#      fs        fd        dc        vc       vdc    penchk        bt        dt
             0.2      0.18       0.0       0.0       0.0                 0.01.00000E20
      $#    sfsa      sfsb      sast      sbst     sfsat     sfsbt       fsf       vsf
             1.5       1.0                           1.0       1.0       1.0       1.0
      $#    soft    sofscl    lcidab    maxpar     sbopt     depth     bsort    frcfrq
               1       0.1         0     1.025       2.0         2         0         1
      $#  penmax    thkopt    shlthk     snlog      isym     i2d3d    sldthk    sldstf
             0.0         1         2         0         0         0       0.0       0.0
      $#    igap    ignore    dprfac    dtstif    unused    unused    flangl   cid_rcf
               1         1       0.0       0.0                           0.0         0
      $#     mid        ro         e        pr      beta       xsi        cp     alpha
               17.90000E-9  200000.0       0.3       1.0       0.95.000000E81.60000E-5
      $#   e0dot        tr        tm        t0     flag1     flag2     
             1.0     293.0    1673.0     293.0       0.0       1.0
      $#  a/siga       b/b   n/beta0   c/beta1      m/na    
           310.0    1000.0      0.65     0.007       1.0
      $#    q1/a      c1/n q2/alpha0 c2/alpha1         
      $#   dc/dc     pd/wc     d1/na     d2/na     d3/na     d4/na     d5/na     
             1.0     150.0                                                  
      $#      tc      tauc    
      $#                                                                         title
      $#     pid     secid       mid     eosid      hgid      grav    adpopt      tmid
               3         1         1         0         0         0         0         0
      $#   secid    elform      shrf       nip     propt   qr/irid     icomp     setyp
               1        16 0.8333333         3       1.0         0         0         1
      $#      t1        t2        t3        t4      nloc     marea      idof    edgset
             1.5       1.5       1.5       1.5       0.0       0.0       0.0         0


    • Jim Day
      Ansys Employee
      Set NEIPS to 9 and MAXINT=-3 in *DATABASE_EXTENT_BINARY. You should then be able to plot D, T, and taumax for all integrations points via history variables 1, 4, and 9, respectively. Those are the values to monitor to see if erosion criteria are met. Offhand, I'm not seeing in the User's Manual where it states the number of integration points that must reach the failure condition to trigger element erosion. Since there doesn't seem to be a variable that allows you to specify that number, the assumption would be either 1 integration point or all integration points. I'm not certain which it is.
    • Noah

      Thank you, this has helped me a lot.

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