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
| Plane stress state | NO |
| Plane strain state | NO |
| Axisymmetric state | NO |
| 3D state | YES |
| Generalized plane state | NO |
| Line 1 (2I5, 60A1) | |
|---|---|
| IL | Law number |
| ITYPE | 65 |
| COMMENT | Any comment (up to 60 characters) that will be reproduced on the output listing |
| Line 1 (5I5) | |
|---|---|
| NINTV | number of sub-steps used to integrate numerically the constitutive equation in a time step |
| NINTEPS | Number of sub-intervals per unit of delta epsilon |
| $\Rightarrow$ number of sub-steps = MAX(NINTV; NINTEPS*DELTA EPSILON) | |
| IKAP | 0 = Analytical compliance matrix 1 = Perturbation compliance matrix |
| MAXIT | Maximum number of iterations during stress integration |
| NTYPHP | Type of hardening law (see Hardening form) |
| Line 1 (6G10.0) | |
|---|---|
| $E_{1}$ | YOUNG's orthotropic elastic moduli |
| $E_{2}$ | YOUNG's orthotropic elastic moduli |
| $E_{3}$ | YOUNG's orthotropic elastic moduli |
| $\mbox{ANU}_{12}$ | Orthotropic POISSON's ratios |
| $\mbox{ANU}_{13}$ | Orthotropic POISSON's ratios |
| $\mbox{ANU}_{23}$ | Orthotropic POISSON's ratios |
| Line 2 (3G10.0) | |
|---|---|
| $G_{12} | COULOMB's orthotropic elastic moduli |
| $G_{13} | COULOMB's orthotropic elastic moduli |
| $G_{23} | COULOMB's orthotropic elastic moduli |
The inverse of the orthotropic elastic matrix is defined: $$ \begin{pmatrix} \varepsilon_{11} \\ \varepsilon_{22}\\ \varepsilon_{33}\\ \varepsilon_{12} \\ \varepsilon_{13} \\ \varepsilon_{23} \end{pmatrix} = \begin{pmatrix} \frac{1}{E_{1}} & \frac{-\nu_{12}}{E_{1}} & \frac{-\nu_{13}}{E_{1}} & 0 & 0 & 0\\ \frac{-\nu_{12}}{E_{1}} & \frac{-\nu_{12}}{E_{2}} & \frac{1}{E_{2}} & 0 & 0 & 0\\ \frac{-\nu_{13}}{E_{1}} & \frac{-\nu_{23}}{E_{2}} & \frac{1}{E_{3}} & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{2G_{12}} & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2G_{13}} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2G_{23}} \end{pmatrix} \begin{pmatrix} \sigma_{11}\\ \sigma_{22}\\ \sigma_{33}\\ \sigma_{12}\\ \sigma_{13}\\ \sigma_{23}\\ \end{pmatrix}$$
| Line 3 (6G10.0) | |
|---|---|
| F | Hill’s coefficients defining the yield locus |
| G | Hill’s coefficients defining the yield locus |
| H | Hill’s coefficients defining the yield locus |
| N | Hill’s coefficients defining the yield locus |
| L | Hill’s coefficients defining the yield locus |
| M | Hill’s coefficients defining the yield locus |
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 :
\[\left. \begin{array}{ll} \sigma_{YY}^{yield} = \sigma_{ZZ}^{yield} = \sigma_{XX}^{yield} \\ \sigma_{45^{\circ} }^{yield} = \sigma_{XX}^{yield}\\ \end{array} \right\} H=G=F=1; N=L=M=3\]
| Line 1 (3G10.0) | |
|---|---|
| CK | hardening factor K (see Hardening form) |
| CW | hardening coefficient W0 (see Hardening form) |
| CN | hardening exponent n (see Hardening form) |
6
The stresses are the components of CAUCHY stress tensor in global (X,Y,Z) coordinates.
| SIG(1) | $\sigma_{XX}$ |
| SIG(2) | $\sigma_{YY}$ |
| SIG(3) | $\sigma_{ZZ}$ |
| SIG(4) | $\sigma_{XY}$ |
| SIG(5) | $\sigma_{XZ}$ |
| SIG(6) | $\sigma_{YZ}$ |
8+…
| Q(1) | Yield criterion = 0 : the previous step was elastic = 1: the previous step was elasto-plastic |
| Q(2) | Accumulated plastic work ($W^{pl}$) |
| Q(3)$\rightarrow$ Q(5) | Coordinates (x,y,z) of the first local axe |
| Q(6) $ \rightarrow$ Q(8) | Coordinates (x,y,z) of the second local axe |
Note that in the code, only the first formulation is used and a conversion is managed if NTYPHP = 1.