17.1.9 Fraction Model

DIANA offers the fraction model, or sublayer model, of Besseling [8] to model kinematic hardening behavior in plasticity. Basic assumption in the fraction model is that the total stress $\boldsymbol\sigma$ in a material point is the sum of a number of, say n , fraction stresses $\boldsymbol\sigma$_i

$$\boldsymbol\sigma \sum_{{i=1}}^{{n}} \phi_{{i}}^{} \boldsymbol\sigma$$ (17.163)

with $\phi_{{i}}^{}$ the weight of fraction i where the sum of the weights $\phi_{{i}}^{}$ are equal to one. The fraction model can be conceived as a parallel chain of nonlinear springs [Fig.17.16],

Figure 17.16: Concept of the fraction model

where the total strain is equal for all fractions. Each fraction has its own nonlinear material parameters, say a yield stress $\bar{{\sigma }}_{{i}}^{}$ , and by properly selecting the weights and material parameters of the fractions, the constitutive behavior can be approximated in a multilinear sense. The elastic material properties, Young's modulus E and Poisson's ratio $\nu$ , are equal for all fractions. Evidently, including more fractions results in a higher accuracy at the cost of more computing time.

The fraction model is basically a three-dimensional approach. This has some special consequences for constrained stress situations --such as occur in trusses, beams, shells and in membranes-- since the stress in the constrained directions is taken into account at fraction level. Although the sum of the fraction stresses in the constrained direction must vanish, the individual contributions of the fractions need not be zero. The stress in the constrained direction has to be considered in the calculation of the plasticity parameters of the specific plasticity model.

Example.

We will now illustrate the derivation of the fraction weights if a specified hardening behavior has to be modeled. Consider the truss element depicted in Figure 17.17.

Figure 17.17: Truss model

Suppose we have two fractions with fraction weight $\phi_{{1}}^{}$ and $\phi_{{2}}^{}$ respectively and linear elastic behavior in the first fraction and elastoplasticity governed by the Von Mises criterion in the second fraction. The stresses in the second fraction should comply with the yield condition

$$\sigma_{{xx.2}}^{} \sigma_{{yy.2}}^{} \sigma_{{yy.2}}^{} \sigma_{{zz.2}}^{} \sigma_{{zz.2}}^{} \sigma_{{xx.2}}^{} \bar{{\sigma }}^{{2}}_{}$$ (17.164)

If plastic flow occurs in fraction two, i.e., the yield condition is violated, the plastic volume change is equal to zero, or

$$\varepsilon_{{xx.2}}^{{\mathrm{p}}} \varepsilon_{{yy.2}}^{{\mathrm{p}}} \varepsilon_{{zz.2}}^{{\mathrm{p}}}$$ (17.165)

Assuming isotropic elasticity and because of the symmetry conditions we can reduce the number of unknowns as follows

$$\begin{split}\varepsilon _{xx} &= \varepsilon \\ [1ex] \varep... ...{yy.2} &= \sigma _{zz.2} = \sigma _{\mathrm{lat}.2} \end{split}$$ (17.166)

Substitution of the formulation for $\sigma_{{yy.2}}^{}$ and $\sigma_{{zz.2}}^{}$ into the yield condition results in the condition for the stresses in the second fraction

$$\sigma_{{xx.2}}^{} \bar{{\sigma }} \sigma_{{\mathrm{lat}.2}}^{}$$ (17.167)

where it is tacitly assumed that we consider tensile stresses. Applying the symmetry conditions, the stresses in the first fraction are given by

$$\begin{split}\sigma _{xx.1} &= \dfrac{(1-\nu)E}{(1+\nu)(1-2\n... ...dfrac{1}{1-\nu} \varepsilon _{\mathrm{lat}} \right) \end{split}$$ (17.168)

and in the second fraction by

$$\begin{split}\sigma _{xx.2} &= \dfrac{(1-\nu)E}{(1+\nu)(1-2\n... ...nu}{1-\nu} \varepsilon _{xx.2}^{\mathrm{p}} \right) \end{split}$$ (17.169)

The basic assumption (17.163) leads to the following equilibrium conditions

$$\begin{split}\sigma _{xx} = \phi_{1} \sigma _{xx.1} + \phi_{2... ...hrm{lat}.1} + \phi_{2} \sigma _{\mathrm{lat}.2} = 0 \end{split}$$ (17.170)

The second equilibrium condition of (17.170) results in an expression for the lateral strain $\varepsilon_{{\mathrm{lat}}}^{}$

$$\varepsilon_{{\mathrm{lat}}}^{} \nu \varepsilon {\tfrac{{1}}{{2}}} \phi_{{2}}^{} \nu \varepsilon_{{xx.2}}^{{\mathrm{p}}}$$ (17.171)

Substituting this equation in (17.168) results in

$$\begin{split}\sigma _{xx.1} &= E \left( \varepsilon - \phi_{2... ...} \dfrac{E}{1+\nu} \varepsilon _{xx.2}^{\mathrm{p}} \end{split}$$ (17.172)

The stresses in the second fraction follow from the second equilibrium condition (17.170) and the yield condition (17.165) respectively, which result in

$$\begin{split}\sigma _{\mathrm{lat}.2} &= \tfrac{1}{2}\phi_{1}... ...} \dfrac{E}{1+\nu} \varepsilon _{xx.2}^{\mathrm{p}} \end{split}$$ (17.173)

It is clear from these results that in case of a tensile loading ( $\varepsilon_{{xx.2}}^{{\mathrm{p}}}$ > 0 ) the lateral stress in the first (elastic) fraction is compressive ( $\sigma_{{\mathrm{lat}.1}}^{}$ < 0 ) and the lateral stress in the second (elastoplastic) fraction is tensile ( $\sigma_{{\mathrm{lat}.2}}^{}$ > 0 ). This results in an artificial hardening behavior which is solely determined by the three-dimensional approach of the fraction model and the different contractive behavior of the elastic and the elastoplastic fractions.

The plastic strain in the second fraction $\varepsilon_{{xx.2}}^{{\mathrm{p}}}$ can be derived from the strain decomposition assumption

$$\varepsilon_{{xx.2}}^{{\mathrm{p}}} \varepsilon \varepsilon_{{xx.2}}^{{\mathrm{e}}}$$ (17.174)

with the elastic strain $\varepsilon_{{xx.2}}^{{\mathrm{e}}}$ determined by

$$\begin{split}\varepsilon _{xx.2}^{\mathrm{e}} &= \dfrac{1}{E}... ... E}{1+\nu} \varepsilon _{xx.2}^{\mathrm{p}} \right) \end{split}$$ (17.175)

which can be elaborated to

$$\begin{split}\varepsilon _{xx.2}^{\mathrm{e}} &= \dfrac{1}{E}... ...2}\phi_{1} \dfrac{1-2\nu}{1+\nu}\varepsilon \right) \end{split}$$ (17.176)

The plastic strain $\varepsilon_{{xx.2}}^{{\mathrm{p}}}$ is finally given by

$$\varepsilon_{{xx.2}}^{{\mathrm{p}}} {\dfrac{{\varepsilon - \dfrac{\bar{\sigma}}{E}}}{{1+\tfrac{1}{2}\phi_{1} \dfrac{1-2\nu}{1+\nu}}}}$$ (17.177)

The total stress $\sigma$ finally follows from the equilibrium in x  direction

$$\sigma \phi_{{2}}^{} \bar{{\sigma}} \phi_{{1}}^{} \varepsilon {\tfrac{{1}}{{2}}} \phi_{{1}}^{} \phi_{{2}}^{} {\dfrac{{1-2\nu}}{{1+\nu}}} {\dfrac{{\varepsilon - \dfrac{\bar{\sigma}}{E}}}{{1+\tfrac{1}{2}\phi_{1} \dfrac{1-2\nu}{1+\nu}}}}$$ (17.178)

The last term of this equation is the additional hardening due to the Poisson effect of the elastic fraction. If the Poisson's ratio $\nu$ is equal to the limiting case of 0.5 , this term is equal to zero because the elastic and the plastic fraction have the same contractive behavior. In general, the effect of the different contractive behavior is not large and the fraction weights can be determined with the simplified formulation

$$\approx \phi_{{1}}^{}$$ (17.179)

Figure 17.18a shows the results for a material with two fractions with $\phi_{{1}}^{}$ = 0.1 and $\phi_{{2}}^{}$ = 0.9 .

Figure 17.18: Hardening behavior

The Young's modulus of the material E = 200000 N/mm² . A Von Mises yield condition is applied for the second fraction with a yield stress $\bar{{\sigma }}$ = 200 N/mm² . The Poisson's ratio $\nu$ is respectively set equal to 0 , 0.3 and 0.499 to show the effect of the additional hardening due to the fraction model. The kinematic hardening behavior is depicted in Figure 17.18b for a Poisson's ratio $\nu$ = 0.3 .

When we know the hardening modulus E_har , we can calculate the fraction weights for this two-fraction model. Starting from (17.178), the hardening modulus is given by the derivative

$${\frac{{\partial \sigma }}{{\partial \varepsilon }}} \phi_{{1}}^{} {\frac{{ \tfrac{1}{2}\phi_{1} \phi_{2} \dfrac{ 1 - 2 \nu }{ 1 + \nu } E }}{{ 1 + \tfrac{1}{2}\phi_{1} \dfrac{ 1 - 2 \nu }{ 1 + \nu } }}}$$ (17.180)

Substituting $\phi_{{2}}^{}$ = 1 - $\phi_{{1}}^{}$ and introducing the hardening ratio S , we arrive at

$${\frac{{ E_{\mathrm{har}} }}{{ E }}} \phi_{{1}}^{} {\frac{{ \tfrac{1}{2}\phi_{1} ( 1 - \phi_{1} ) \dfrac{ 1 - 2 \nu }{ 1 + \nu } }}{{ 1 + \tfrac{1}{2}\phi_{1} \dfrac{ 1 - 2 \nu }{ 1 + \nu } }}} {\frac{{ \phi_{1} + \tfrac{1}{2}\phi_{1} \dfrac{ 1 - 2 \nu }{ 1 + \nu } }}{{ 1 + \tfrac{1}{2}\phi_{1} \dfrac{ 1 - 2 \nu }{ 1 + \nu } }}}$$ (17.181)

Solving this equation for $\phi_{{1}}^{}$ we end at

$$\phi_{{1}}^{} {\frac{{ 2 S }}{{ 2 + \dfrac{ 1 - 2 \nu }{ 1 + \nu } ( 1 - S ) }}}$$ (17.182)

For $\nu$ = 0.5 we find again $\phi_{{1}}^{}$ = S .