1. Use plain strain instead of plain stress;

2. If the material model has hardening,  then the problem can be with associated or non-associated plasticity (check friction and dilatency angles);

3. If the material has ideal plasticity (without hardening) and loading is kinematic, then you should get the desired result.

 Sir,

Thanks for the reply.

1. Yes sir, plane strain can be implemented easily without any such issue but for some problem cases it is mandatory to go for plane stress implementation.  

2. For Drucker-Prager Plasticity model : As far as hardening is concerned it is only the yield envelop which is either increasing (Isotropic hardening) or same (Kinematic hardening). Hardening has no role in plasticity model for being associative or non-associated. The plastic potential expression in case of non-associative model has no hardening term.    

3. The last point seems to be totally confusing, no offence  

In case if wrong, Please correct me sir

Best regards, 

Thanks

Dear Manik,

In plane stress you stress is 2D but your strain has an out of plane component which must be determined separately from the constitutive stress strain equations. For pressure independent plasticity the volume conservation condition is used for that. However when you have pressure dependent yield then the volume in plastic deformation is (mostly) not conserved. In such cases, it will not be valid to use the volume conservation condition to determine the out of plane strain in  plane stress implementation.

For plane stress situation, you still need to define a relation ep_zz = f(ep_xx,ep_yy,ep_xy). It can be any assumption

1) assume the volume to be conserved as J2 plasticity (not a good idea)

2) assume plane strain condition (ep_zz = 0). Can be used if the out of plane strain is very small compared to in-plane strains. In this case applying tension will give volume expansion and compression will result in volume dilation.

3) Use any other suitable model for ep_zz = f(ep_xx,ep_yy,ep_xy) which applies good in your specific case.

There is also another alternative to implement plane stress condition for such situations. Formulate the constitutive equations for 3D (excluding out of plane shear terms) then solve the system of equations to enforce sig_z = 0. This will give you a plane stress condition without any assumption on ep_zz = f(ep_xx,ep_yy,ep_xy). However it may be computationally expensive than a normal plane stress implementation. See the attachment.

Just a remark (may be you already know it). Generally we use nonassociative plasticity when we have pressure dependent yield. Because the plastic deformation still depends only on the deviatoric part. The plastic potential is therefore different than the yield and still depends only upon J2. When we use nonassociative plasticity we get asymmetric stiffness. So you have to keep that in  mind during the formulation/implementation.

kind regards

Muhammad

Sir,

Thanks for the reply. All the comments are really helpful and informative. With the help for the information provided by you the problem is solved finally.

I formulated the system of equations such that plane stress condition is enforced (i.e. sig_z=0). From the total strain, the total ep_zz has been evaluated by implementing Newton Raphson algorithm such that predicted value of stress sig_z=0 as mentioned in Neto et al., (2008). The elastic ep_zz has been evaluated after implemented radial return. The plastic ep_zz will be 

plastic ep_zz = (total ep_zz) - (elastic ep_zz)

I went through the attached document too. It is also very helpful. 

Best Regards

Manik

http://as.wiley.com/WileyCDA/WileyTitle/productCd-0470694521.html

I am glad your problem is solved.

Cheers!