Plasticity is volume conserving, so the sum of all plastic strains must be zero.

Dear Imanol,

When you prescribe plane strain condition then the problem is 2D. Nodal displacement is zero in direction 3 which means elastic strain 3 = 0 AND plastic strain 3 = 0. Obviously when you add these you also get total strain 3 = 0.

If the (plastic) volume is conserved then you get plastic strain 1 = -plastic strain 2. The elastic strain is (generally) not volume conserved and depends upon elastic properties of the material.

Dear Muhammad,

Thank you for your answer, but the problem is that I was considering elastic strain 3 = 0 and plastic strain 3 = 0 in an analyticial model that I was developing and the results were not the expected ones in comparison to the numerical results in Abaqus.

I defined my numerical model by a Plane Strain condition and later I realized that Abaqus is considering elastic strain 3 = - plastic strain 3 and not necesarrily equal to 0. For example, at a certain moment, I can find that plastic strain 3 = 0.00158 --> elastic strain 3 = - 0.00158

How are you defining plane strain condition? If you use a plane strain element then it only has 2 D.O.F i.e. x and y. The results shouldnt even show you strain in third direction when you use plane strain elements.

When I look to the results, I can ask for the Plastic Strain 3, Elastic Strain 3 and Total Strain 3. Even if I just define a 2D geometry, I can see the stress and strain results in the third direction and I can observe what I was explaining before.

What about the stress? Can you also get the stress in direction 3? May be you are using plane stress instead of plane strain.

Also check if the results are plotted in element (local) coordinate system.

The stress in direction 3 is not 0. I am pretty sure that I defined correctly Plane Strain