Anisotropic elasto-plastic law based on HILL48 yield locus for solid elements at constant temperature.

Mechanical analysis of elasto-plastic anisotropic solids undergoing large strains. Isotropic hardening is assumed.

Prepro: LHIL3D.F
Lagamine: HILL3D.F

The inverse of the orthotropic elastic matrix is defined: $$ \begin{pmatrix} \varepsilon_{11} \\ \varepsilon_{22}\\ \varepsilon_{33}\\ \varepsilon_{12} \\ \varepsilon_{13} \\ \varepsilon_{23} \end{pmatrix} = \begin{pmatrix} \frac{1}{E_{1}} & \frac{-\nu_{12}}{E_{1}} & \frac{-\nu_{13}}{E_{1}} & 0 & 0 & 0\\ \frac{-\nu_{12}}{E_{1}} & \frac{-\nu_{12}}{E_{2}} & \frac{1}{E_{2}} & 0 & 0 & 0\\ \frac{-\nu_{13}}{E_{1}} & \frac{-\nu_{23}}{E_{2}} & \frac{1}{E_{3}} & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{2G_{12}} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2G_{13}} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2G_{23}} \end{pmatrix} \begin{pmatrix} \sigma_{11}\\ \sigma_{22}\\ \sigma_{33}\\ \sigma_{12}\\ \sigma_{13}\\ \sigma_{23}\\ \end{pmatrix}$$

The yield locus is defined:
$f = \frac{1}{2} \underline{\sigma}^{T} \underline{\underline{H}} \ \underline{\sigma} - \sigma_{F}^{2} = 0 $
with

$\underline{\underline{H}} = \begin{pmatrix} G+H &-H&-G&0&0&0\\-H&H+F&-F&0&0&0\\-G&-F&F+G&0&0&0\\0&0&0&2N&0&0\\0&0&0&0&2L&0\\0&0&0&0&0&2M\\ \end{pmatrix}$
The 6 parameters of this matrix can be computed:
$\rightarrow$ from the yield stress limits :

$\sigma_{XX}^{yield} = \sigma_{F}^{initial} =$ initial value of $\sigma_{F}$ (see Hardening form)
$\sigma_{YY}^{yield} = \sqrt{\frac{2}{H+F}}\sigma_{F}^{initial}$
$\sigma_{ZZ}^{yield} = \sqrt{\frac{2}{G+F}}\sigma_{F}^{initial}$
$\sigma_{XY}^{yield} = \sqrt{\frac{1}{N}}\sigma_{F}^{initial}$
$\sigma_{XZ}^{yield} = \sqrt{\frac{1}{L}}\sigma_{F}^{initial}$
$\sigma_{YZ}^{yield} = \sqrt{\frac{1}{M}}\sigma_{F}^{initial}$
with the additional condition:
$H + G = 2$
$\rightarrow$ for sheet metal fitting, it is more convenient to perform 3 tensile tests and use the following fitting through the Lankford coefficients:
$r_{0} = \frac{H}{G}$; $r_{90} = \frac{H}{F}$; $r_{45} = \frac{2N-F-G}{2(F+G)}$
with these additional conditions :
$H + G = 2$; $N=L=M$
And $\sigma_{F}$ is computed as above
$\rightarrow$ The 3 conditions to compute $N$, $L$ and $M$ with the yield stress limits can be replaced by the following using the tensile test along the $45^{\circ}$ direction:
$\sigma_{45^{\circ}}^{yield} = \sqrt{\frac{8}{G+F+2N}} \sigma_{F}^{initial}$ with the additional conditions $N=L=M$
It should be noticed that these 3 fittings cannot generally be fulfilled simultaneously. Indeed, each one gives enough equations to compute all the Hill coefficients.
However, for an isotropic material, the Von Mises criterion is obtained. The condition on the Lankford coefficients gives:
\[\left. \begin{array}{lll} r_{0} = 1\\ r_{90} = 1\\ r_{45} = 1 \end{array} \right\} H=G=F=1; N=L=M=3\] And the condition on the yield stresses gives :

\[\left. \begin{array}{ll} \sigma_{YY}^{yield} = \sigma_{ZZ}^{yield} = \sigma_{XX}^{yield} \\ \sigma_{45^{\circ} }^{yield} = \sigma_{XX}^{yield}\\ \end{array} \right\} H=G=F=1; N=L=M=3\]

6

The stresses are the components of CAUCHY stress tensor in global (X,Y,Z) coordinates.

8+…

• if (NTYPHP = 0), The user’s parameters CK, CW and CN are respectively $K^{W}$ , $W_{0}$ and $n^{W}$ of the hardening equation:
  $\sigma_{F} = K^{W} (W_{0} + W^{pl})^{n^{W}}$
  
• if (NTYPHP = 1), The user’s parameters CK, CW and CN are respectively K, $\varepsilon_{0}$ and n of the hardening equation:
  $\sigma_{F} = K(\varepsilon_{0} + \varepsilon^{pl})^{n}$
  

Note that in the code, only the first formulation is used and a conversion is managed if NTYPHP = 1.

• if (NTYPHP = 2), The user’s parameters CK, CW and CN are respectively K, $\sigma_{0}$ and n of the hardening equation:
  $\sigma_{F} = \sigma_{0} + K(1-exp(-n.\varepsilon^{pl}))$