Isotropic elasto-plasticity

The strain rate decomposition is

Using the standard definition of corotational measures, this can be written in integrated form as

The elasticity is linear and isotropic and, therefore, can be written in terms of two temperature-dependent material parameters. For the purpose of this development it is most appropriate to choose these parameters as the bulk modulus, K, and the shear modulus, G. These are computed readily from the user's input of Young's modulus, E, and Poisson's ratio, $\nu$, as

and

The elasticity can be written in volumetric and deviatoric components as follows.

Volumetric:

where

is the equivalent pressure stress.

Deviatoric:

where $\mathbf{S}$ is the deviatoric stress,

The flow rule is

where

and $d{\overline{e}}^{pl}$ is the (scalar) equivalent plastic strain rate.

The plasticity requires that the material satisfy a uniaxial-stress plastic-strain strain-rate relationship. If the material is rate independent, this is the yield condition:

where ${\sigma }^0\left({\overline{e}}^{pl},\theta \right)$ is the yield stress and is defined by the user as a function of equivalent plastic strain (${\overline{e}}^{pl}$) and temperature ($\theta$).

If the material is rate dependent, the relationship is the uniaxial flow rate definition:

where h is a known function. For example, the rate-dependent material model offers an overstress power law model of the form

where $D\left(\theta \right)$ and $n\left(\theta \right)$ are user-defined temperature-dependent material parameters and ${\sigma }^0\left({\overline{e}}^{pl},\theta \right)$ is the static yield stress.

Integrating this relationship by the backward Euler method gives

This equation can be inverted (numerically, if necessary) to give q as a function of ${\overline{e}}^{pl}$ at the end of the increment.

Thus, both the rate-independent model and the integrated rate-dependent model give the general uniaxial form

where $\overline{\sigma }={\sigma }^0$ for the rate-independent model and $\overline{\sigma }$ is obtained by inversion of Equation 6 for the rate-dependent model.

Equation 1 to Equation 7 define the material behavior. In any increment when plastic flow is occurring (which is determined by evaluating q based on purely elastic response and finding that its value exceeds ${\sigma }^0$), these equations must be integrated and solved for the state at the end of the increment. As in the general discussion in About metal plasticity models, the integration is done by applying the backward Euler method to the flow rule (Equation 4), giving

Combining this with the deviatoric elasticity (Equation 3) and the integrated strain rate decomposition (Equation 1) gives

Using the integrated flow rule (Equation 8), together with the Mises definition of the flow direction, $\mathbf{n}$ (in Equation 4), this becomes

For simplicity of notation we write

so that this equation is

Taking the inner product of this equation with itself gives

where

The Mises equivalent stress, q, must satisfy the uniaxial form defined in Equation 7, so that from Equation 11,

This is a nonlinear equation for $\mathrm{\Delta }{\overline{e}}^{pl}$ in the general case when $\overline{\sigma }$ depends on the equivalent plastic strain (that is, when the material is rate-dependent, or when there is nonzero work hardening). (It is linear in $\mathrm{\Delta }{e}^{pl}$ for rate-independent perfect plasticity.) We solve it by Newton's method:

and we iterate until convergence is achieved.

Once $\mathrm{\Delta }{\overline{e}}^{pl}$ is known, the solution is fully defined: using Equation 5,

and so, from Equation 10,

From Equation 4,

and thus, from Equation 6,

For cases where three direct strain components are provided by the kinematic solution (that is, all but plane stress and uniaxial stress cases), Equation 2 defines

so that the solution is then fully defined.

For this simple plasticity model the material stiffness matrix can be derived without the need for matrix inversion (as was needed in the general case described in Integration of plasticity models), as follows.

Taking the variation of Equation 10 with respect to all quantities at the end of the increment gives

Now, from Equation 5,

and, from Equation 11,

Combining these last two results,

where

From the definition of $\tilde{e}$ (see Equation 11),

and, hence,

Combining these results with Equation 14 gives

where $Q=\frac{2}{3}q/\tilde{e}$, $\mathrm{\Im }$ is the fourth-order unit tensor, and

For all cases where three direct strains are defined by the kinematic solution, the material stiffness is completed by

so since

and

we have