Let the yield surface be represented by
To fi nd the shear or deviatoric strains, we will assume that the resultant plastic strain increment, Dεp, for an increment of stress is normal to the yield surface ). Normally, the plastic strain increment should be normal to a plastic potential function, but we are assuming here that the plastic potential function, and the yield surface (yield function, F ) are the same. A plastic potential function is a scalar quantity that defi nes a vector in terms of its location in space. Classical plasticity demands that the surfaces defi ned by the yield and plastic potential coincide.
If they do not, then basic work restrictions are violated. However, advanced soil mechanics theories often use different surfaces for yield and potential functions to obtain more realistic stress–strain relationships. The resultant plastic strain increment has two components—a deviatoric or shear component, Dεpq , and a volumetric component, Dεpp , as shown in Figure We already found Dεpp in the previous section. Since we know the equation for the yield surface [Equation], we can fi nd the normal to it by differentiation of the yield function with respect to p9 and q. The tangent or slope of the yield surface is
Rearranging Equation , we obtain the slope as
The normal to the yield surface is
From Figure the normal, in terms of plastic strains, is dεpq /dεpp . Therefore
which leads to
The elastic shear strains can be obtained from Equation as
These equations for strain increments are valid only for small changes in stress. For example, you cannot use these equations to calculate the failure strains by simply substituting the failure stresses for p' and q. You have to calculate the strains for small increments of stresses up to failure and then sum each component of strain separately. We need to do this because the critical state model considers soils as elastoplastic materials and not linearly elastic materials.