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Re: [Question #690521]: Exponential softening shear force after peak- model implementation

 

Question #690521 on Yade changed:
https://answers.launchpad.net/yade/+question/690521

chanaka Udaya posted a new comment:
Dear Dr. Jan Stransky,

I have followed the following steps to obtain scalar shear stress
value(|sigma_S|) which is same as your sActual value magnitude


delta_us= Us.norm()- Us_old.norm() ; where, Us_old is shear displacement at previous step and delta_Us: scalar shear displacement increment

|sigma_S_trial|=Sigma_S_old+ k_s*(1-D_old)*delta_us         trial stress
calculation

F_trial=|sigma_S_trial|- Cohesion*(1-D_old)
evaluate yield function

If F_trial>0
      (dF/dsigmaS)=sign(sigmaS)=+1                                     partial derivatives to calculate plastic multiplier
      (dsigma_S/dUs_Pl)=k_s*(1-D_old)
      (dsigma_S/dD)= k_s*(Us.norm()-|Us_Pl_old|)
      (dD/dUs_pl)= Exp(-|UsPl_old|/constant)/constant

assuming associate flow rule to calculate plastic multiplier (d_lambda)
that means, (dg/dsigmaS)= (dF/dsigmaS) where g-plastic potential

      d_lambda= -F_trial/((dF/dsigmaS)*(dsigmaS/dUs_Pl)*(dg/dsigmaS)+(dF/dsigmaS)* (dsigma_S/dD)* (dD/dUs_pl)*(dg/dsigmaS))
      delta_Us_Pl=d_lambda*(dg/dsigmaS)
      |Us_pl|=|Us_Pl_old|+ delta_Us_Pl
      D=1-exp(-|Us_Pl|/constant)
      |sigma_S|=|sigma_S_trial|*(1-D)/(1-D_old)-k_s*(1-D)*delta_Us_Pl

Based on the above calculation I only know the scalar actual shear
stress,(|sigmaS|) and scalar shear plastic displacement, (|Us_Pl), at
the current step

But I need the vector value of shear plastic displacement to apply the
shear force using the below equation at the contact point

Fs=k_s*(1-D)*(Us-UsPl)*|crossSection_area| where Fs is shear force
vector

I still couldn't understand how to get that Us_Pl vector

Thanks

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