Homogeneous stress–strain states computed by 3D-stress algorithms of FE-codes: application to material parameter identification

Abstract

In view of code verification of finite element implementations and for material parameter identification purposes it is of interest to make use of stress algorithms developed for three-dimensional finite element computations. In the case of homogeneous deformations various boundary-conditions for given displacements or stresses are possible and define a sub-problem of three-dimensional stress–strain states, which are either one-, two- or three-dimensional. Examples are uniaxial tension/compression, plane stress conditions or biaxial tensile problems. Caused by the fact that the stress algorithms are strain-driven, the constraints of zero stresses in a specific direction lead for elastic and inelastic constitutive models to a particular system of differential-algebraic equations. How to treat such stress algorithms and how to solve the resulting system of differential-algebraic equations, which are developed for finite element programs, for specific stress and displacement boundary conditions is discussed in this article. Additionally, it is worked out that the consistent tangent operator is required in the same manner as in 3D-FE computations. The second topic treats the extension of the entire procedure for material parameter identification procedure applied to test data for different materials such as steel, rubber material and powder. In this respect, uniaxial tensile, biaxial tensile tests, and laterally constrained loading paths are exemplarily investigated. These investigations and the proposed procedure are applied for small and finite strain problems. In this investigation measure of the quality of identification is discussed as well.

Appendix

1.1 Appendix 1: Filter matrices

In this section the filter matrices ${{\bf{\mathsf{Z}}}_1}$ and ${{\bf{\mathsf{Z}}}_2}$, see, for example, in Eq. (22), for several loading cases are assembled.

1D-tension/compression—strain control

$$ {\bf{\mathsf{Z}}}_1 = \left[\begin{array}{c} 1\\ 0\\ 0\\ 0\\ 0\\ 0 \end{array}\right] ,\quad\quad\quad {\bf{\mathsf{Z}}}_2 =\left[\begin{array}{ccccc} 0&0&0&0&0\\ 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ \end{array}\right] $$

(97)

1D-tension/compression—stress control

$$ {\bf{\mathsf{Z}}}_1 = 0,\quad\quad\quad {\bf{\mathsf{Z}}}_2 =\left[\begin{array}{cccccc} 1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1\\ \end{array}\right] $$

(98)

Torsion of thin-walled tube—small strain case

$$ {\bf{\mathsf{Z}}}_1 = \left[\begin{array}{c} 0\\ 0\\ 0\\ 1\\ 0\\ 0\\ \end{array}\right] ,\quad\quad\quad {\bf{\mathsf{Z}}}_2 =\left[\begin{array}{ccccc} 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ \end{array}\right] $$

(99)

Plane strain case

$$ {\bf{\mathsf{Z}}}_1 =\left[\begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&0\\ 0&0&1\\ 0&0&0\\ 0&0&0\\ \end{array}\right] , \quad\quad\quad{\bf{\mathsf{Z}}}_2 =\left[\begin{array}{ccc} 0&0&0\\ 0&0&0\\ 1&0&0\\ 0&0&0\\ 0&1&0\\ 0&0&1\\ \end{array}\right] $$

(100)

Plane stress case

$$ {\bf{\mathsf{Z}}}_1 =\left[\begin{array}{ccc} 0&0&0\\ 0&0&0\\ 1&0&0\\ 0&0&0\\ 0&1&0\\ 0&0&1 \end{array}\right], {\bf{\mathsf{Z}}}_2 =\left[\begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&0\\ 0&0&1\\ 0&0&0\\ 0&0&0\\ \end{array}\right] $$

(101)

Compression laterally constrained—stretch driven

$$ {\bf{\mathsf{Z}}}_1 =\left[\begin{array}{ccc} 1&0&0\\ 0&1&0\\ 0&0&1\\ 0&0&0\\ 0&0&0\\ 0&0&0\\ \end{array}\right] ,{\bf{\mathsf{Z}}}_2 =\left[\begin{array}{ccc} 0&0&0\\ 0&0&0\\ 0&0&0\\ 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{array}\right] $$

(102)

Compression laterally constrained—stress control

$$ {\bf{\mathsf{Z}}}_1 =\left[\begin{array}{ccc} 0&0&0\\ 0&1&0\\ 0&0&1\\ 1&0&0\\ 0&0&0\\ 0&0&0\\ \end{array}\right] ,{\bf{\mathsf{Z}}}_2 =\left[\begin{array}{ccc} 1&0&0\\ 0&0&0\\ 0&0&0\\ 0&0&0\\ 0&1&0\\ 0&0&1\\ \end{array}\right] $$

(103)

Equibiaxial—stress control

$$ {\bf{\mathsf{Z}}}_1 = \left[\begin{array}{cc} 1&0\\ 0&1\\ 0&0\\ 0&0\\ 0&0\\ 0&0\\ \end{array}\right] , {\bf{\mathsf{Z}}}_2 =\left[\begin{array}{cccc} 0&0&0&0\\ 0&0&0&0\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{array}\right] $$

(104)

1.2 Appendix 2: Analytical solution of parameter identification in J ₂-plasticity

To determine a possibility of material parameter identification for the J ₂-visco-plasticity model (78)–(82) in the rate-independent limit we have the three-dimensional model

$$ {\bf T} = {\bf h} ({\bf E} ,{\bf E}_{\rm p}) = K({\rm tr}\, {\bf E} ){{\rm {\bf I}}} +2G({\bf E} -{\bf E}_{\rm p})^{{\rm D}} $$

(105)

$$ {\dot{{\bf E}}}_{\rm p} = \lambda {{\bf N}} = \lambda\frac{({\bf T} - {\bf X})^{{\rm D}}} {\|({\bf T}- {{\bf X}})^{{\rm D}}\|} $$

(106)

$$ {\dot{{\bf X}} = c{\dot{{\bf E}}}_{\rm p} - b{\dot{s}}{\bf X}} = \lambda \left(c{{\bf N}} -b\sqrt{\frac{2}{3}}{{\bf X}}\right) $$

(107)

with the yield condition

$$ f({\bf T , X} ,k) = \frac{1}{2}({{\bf {T}}} -{{\bf X}})^{{\rm D}}\cdot({{{\bf T}} -{ {\bf X}}})^{{\rm D}}-\frac{1}{3}k^2 = 0. $$

(108)

For the uniaxial tensile test one has the strain

$$ {\bf E} = \varepsilon ({\bf e}_1 \otimes {\bf e}_1) + \varepsilon_Q ({\bf e}_2 \otimes {\bf e}_2 + {\bf e}_3 \otimes {\bf e}_3) $$

(109)

$$ {\bf E}_{\rm e} = \varepsilon_{\rm e} ({\bf e}_1 \otimes {\bf e}_1) - \nu \varepsilon_{\rm e} ({\bf e}_2 \otimes {\bf e}_2 + {\bf e}_3 \otimes {\bf e}_3) $$

(110)

$$ {\bf E}_{\rm p} = \varepsilon_{\rm p} ({\bf e}_1 \otimes {\bf e}_1) - {\frac{\varepsilon_{\rm p}}{2}} ({\bf e}_2 \otimes {\bf e}_2 + {\bf e}_3 \otimes {\bf e}_3) $$

(111)

and stress state T = σ (e ₁ ⊗ e ₁) and X = ξ (e ₁ ⊗ e ₁) − (ξ/2) (e ₂ ⊗ e ₂ + e ₃ ⊗ e ₃).

Under uniaxial uniaxial tension for $\sigma(t) = \dot{\sigma} t + k$, where k is the yield stress $(\dot{\sigma} =\hbox{const.} > 0)$, the following scalar equations result from Eqs. (105)–(108):

$$ \hbox{decomposition of}\,\varepsilon \quad\quad\quad \varepsilon = \varepsilon_{\rm e}+\varepsilon_{\rm p} $$

(112)

$$ \hbox{elasticity relation}\quad\quad\quad\sigma = E(\varepsilon-\varepsilon_{\rm e}) $$

(113)

$$ \hbox{yield function}\quad\quad\quad f(\sigma,\xi) = \frac{2}{3}\sigma-\xi-\frac{2}{3}k = 0 $$

(114)

$$ \hbox{flow-rule}\quad\quad\quad {\dot{\varepsilon}}_{\rm p} = \sqrt{\frac{2}{3}}\lambda,\quad\hbox{with}\quad\lambda = \sqrt{\frac{2}{3}}{\frac{\dot{\sigma}}{c-b\xi}} $$

(115)

$$ \hbox{kinematic hardening}\quad\quad\quad \dot{\xi} = \frac{2}{3}\dot{\sigma} $$

(116)

λ is obtained by the consistence condition $\dot{f} = 0$. In this stress-controlled process with the initial conditions $\varepsilon_{\rm p}(0) = 0$ and ξ(0) = 0 one calculates by Eq. (107)

$$ \xi(t) = \frac{2}{3}\dot{\sigma} t = \frac{2}{3}(\sigma(t)-k). $$

(117)

Now, the total strains can be written as

$$ \varepsilon(t) = \varepsilon_0 +\int\limits_0^t\dot{\varepsilon}_{\rm e}\,{\rm d} t+\int\limits_0^t\dot{\varepsilon}_{\rm p}\,{\rm d} t = \frac{k}{E}+\int\limits_0^t\frac{\dot{\sigma}}{E}\,{\rm d} t +\frac{2}{3}\int\limits_0^t{\frac{\dot{\sigma}}{c-b\xi(t)}}\,{\rm d} t $$

(118)

$$ ={\frac{k+\dot{\sigma} t}{E}} +\frac{1}{b}\ln\left(1-\frac{2}{3}\frac{b}{c}(\sigma(t)-k)\right) $$

(119)

for $\varepsilon=k/E$. Inversion of Eq. (119) and application to the strain-controlled case yields

$$ \sigma(\varepsilon) - \frac{E}{b}\ln\left(1-\frac{2}{3}\frac{b}{c}(\sigma(\varepsilon)-k)\right) = E\varepsilon. $$

(120)

This implicit equation can be solved for every given strain $\varepsilon$. Furthermore, it can be differentiated in respect to $\varepsilon$ to get a possibility for computing the material parameters b and c. The slopes at the yield point and at the final state σ ₀′ and σ ₁′ can be derived from the measured stress–strain relation

$$ \sigma_0' := \left.{\frac{{\rm d}\sigma}{{\rm d}\varepsilon}}\right|_{\sigma = k} = {\frac{3Ec}{3c+2E}}, \quad\quad \sigma_1' := \left.{\frac{{\rm d}\sigma}{{\rm d}\varepsilon}}\right|_{\sigma = \sigma_1} = {\frac{3c}{2(\sigma_1-k)}}-{\frac{E\sigma_1'}{(E-\sigma_1')(\sigma_1-k)}} $$

(121)

i.e. the material parameters b and c are quantifiable, see Eq. (86). Figure 7 shows the slopes and the values of a typical diagram. Obviously, b and c are not independent on each other, see Eq. (86)₂.

Fig. 7

Uniaxial stress–strain relation