General Mechanical

General Mechanical

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Spiking results

    • htozam
      Subscriber

      Respected members, 


      In applying displacement on a reinforced concrete model I am recording the correspoinding resistance force amplitude for each of the displacement (0 - 40 mm amplitude) applied, the force - displacement result graph of the analysis is attached.


      The question is how can the resistance force curve spike more than once? 


      I mean in real-life situation once a resistance force is dropped (say if a sample is cracked or damaged) it will not increase again specially if we continue to apply higher displacement amplitudes on the model. 


      Here it looks like the model is regaining strength at 4.e-3 mm (after the force drops from 37500 to approx. 6000 N) and again in 9 mm displacement amplitudes.  


      Additionlly, if the number of Sub-steps are increased the spikings increase and the whole results graph changes. For examples attached is the force - displacement curve for 1 Sub - step and 5 Sub-steps manually selected. The number of displacement steps are 14. 


      1 sub step


       


      5 sub - steps

    • Wenlong
      Ansys Employee

      Hi, 


      Before making any comments, it would be nice to how do you model your reinforced concrete? Is the reinforcement defined explicitly? If yes, What kind of contact did you specify between the reinforcement and concrete? What types of failure criteria do you use on the concrete? 


      Bests,


      Wenlong


       


       


       


       

    • htozam
      Subscriber

      Hi Wenlong, 


      Thank you for reply. 


      Yes, the reinforcement is defined explicitly. I used node merging to connect the reinforcement with concrete. 


      The concrete properties is defined as following:


       


      /PREP7 !N AND MM


       


      * GET,NMAT0,MAT,0, NUM, MAX, , 


      * GET,NELEM0,ETYPE,0, NUM, MAX,


      * GET, Nreal0, RCON, 0, NUM, MAX,


       


       


      NELEM1=NELEM0+5


      NMAT1=NMAT0+1


      Nreal1=Nreal0+5


       


      ! Define concxrete Element


      ET,NELEM1,SOLID65


      !*


      KEYOPT,NELEM1,1,0


      KEYOPT,NELEM1,3,0


      KEYOPT,NELEM1,5,0


      KEYOPT,NELEM1,6,0


      KEYOPT,NELEM1,7,0


      KEYOPT,NELEM1,8,0


       


      fc=25


      ft=fc/10


      nu=0.25


      E=0.468253*fc/0.000778


      !Define concrete Material 


      MPTEMP,,,,,,,,


      MPTEMP,1,0


      MPDATA,EX,NMAT1,,E


      MPDATA,PRXY,NMAT1,,nu


       


       


      TBDE,CONC,NMAT1,,,


      TB,CONC,NMAT1,1,9


      TBTEMP,0


      TBDATA,,0.2,1,ft,fc,,


       


      !Define concrete eral


      R,Nreal1, , , , , , , 


      RMORE, , , , , , , 


      RMORE,0.01, 


       


      ALLSEL,ALL


      CMSEL,S,Concrete,ELEM


      ALLSEL,BELOW,ELEM


      EMODIF,ALL,TYPE,NELEM1


      EMODIF,ALL,MAT,NMAT1


      EMODIF,ALL,REAL,Nreal1,


      ALLSEL,ALL


      /solu


      OUTRES,ALL,ALL


      Best regards, 


      Roman


       

    • htozam
      Subscriber

      Hi Wenlong, any comment on this regards will be very highly appreciated. Thank you for your considerations. 

    • Wenlong
      Ansys Employee

      Hi Roman,


      I would recommend you take very small sub-steps, and use only 1 or 2 iterations per sub-step, because your problem is path-dependent and bifurcation won't help you go back to the previous state. 


      You may try the following commands:


      NCNV, 0     ! simulation will keep going even without convergence


      NEQIT, 2      ! only 2 iteration per sub-step


      AUTOS, OFF   ! turn off auto time stepping


      NSUB, 1000   ! Define 1000 substeps


       


      Best Regards,


      Wenlong

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