Table of Contents

HILL3D_KI

Description

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

The model

Mechanical analysis of elasto-plastic anisotropic solids undergoing large strains.
Classic isotropic and kinematic hardening is available.
Teodosiu isotropic and kinematic hardening is available.

Files

Prepro: LHIL3D_KI.F
Lagamine: HILL3D_KI.F

Availability

Plane stress state NO
Plane strain state NO
Axisymmetric state NO
3D state YES
Generalized plane state NO

Input file

Parameters defining the type of constitutive law

Line 1 (2I5, 60A1)
ILLaw number
ITYPE 66
COMMENT Any comment (up to 60 characters) that will be reproduced on the output listing

Integer parameters

Line 1 (6I5)
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 not working
1 = Perturbation compliance matrix
MAXITMaximum number of iterations during stress integration
NTYPHP Type of hardening law (see Hardening form), no action if NTEO = 1
= 1 (Swift hardening)
= 2 (Voce hardening)
= 3 (Ludwick hardening)
INDAM 0 (no fatigue damage computation)
> 1 (fatigue damage computation)
Line 2 (4I5)
NTE00 Classic hardening
1 Teodosiu hardening
2 Teodosiu + Jauman hardening
3 Ziegler hardening
4 Armstrong-Frederick hardening with 2 terms
if nteo = 1 or nteo= 2, then
NREAD = 1 Read the 58 state variables in .f72 (in column)
KREAD= 1 Changes hardening parameters if fitted with a Von Mises yield locus and shear tests
IOPTEO= 10 Hofferlin type. SL only activated if a path change with an elastic transition occurs.
= 20 Alves type. SL activated for any path change (continuous path change or even local path change, for example simple shear where material frame rotates)
= 21 Alves type but SL activated only if a sufficiently strong path change occurs $|\Delta\mathbf{\hat{\varepsilon}}_{n+1}^{p} - \Delta\mathbf{\hat{\varepsilon}}_{n}^{p}| \geq$ prec, where n and n+1 are two successive steps and $\Delta\mathbf{\hat{\varepsilon}}^{p}$ is the increment of the local plastic strain.

Real parameters

Line 1 (6G10.0)
$E_{1}$YOUNG's orthotropic elastic moduli
$E_{2}$
$E_{3}$
$\mbox{ANU}_{12}$Orthotropic POISSON's ratios
$\mbox{ANU}_{13}$
$\mbox{ANU}_{23}$
Line 2 (3G10.0)
$G_{12} COULOMB's orthotropic elastic moduli
$G_{13}
$G_{23}

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)
FHill’s coefficients defining the yield locus, param(16, ilaw)
G Hill’s coefficients defining the yield locus, param(17, ilaw)
H Hill’s coefficients defining the yield locus, param(18, ilaw)
N Hill’s coefficients defining the yield locus, param(19, ilaw)
L Hill’s coefficients defining the yield locus, param(20, ilaw)
M Hill’s coefficients defining the yield locus, param(21, ilaw)

The yield locus is defined:

  1. Classic: $f = \frac{1}{2} \underline{\sigma}^{T} \underline{\underline{H}} \ \underline{\sigma} - \sigma_{F}^{2} = 0 $
  2. Teodosiu: $f = \sqrt{\underline{\sigma}^{T} \underline{\underline{H}} \ \underline{\sigma}^{T}} - \sigma_{F}$

Be careful! $\underline{\underline{H}}$ is divided by two for Teodosiu model

\[\underline{\underline{H}} = \begin{pmatrix} G+H &-H&-G&0&0&0\\-H&H+F&-F&0&0&0\\-G&-F&F+G&0&0&0\\0&0&0&2N&0&0\\0&0&0&0&2L&0\\0&0&0&0&0&2M\\ \end{pmatrix}\]

The 6 parameters of this matrix can be computed:
$\mathbf{\rightarrow}$ 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:

  1. Classic : H + G = 2
  2. Teodosiu: H + G = 1

$\mathbf{\rightarrow}$ 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:

  1. $H + G = 2$; $N=L=M$
  2. $H + G =1$; $N=L=M$

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:
\[\left. \begin{array}{lll} r_{0} = 1\\ r_{90} = 1\\ r_{45} = 1 \end{array} \right\} \begin{array} 11. & NTEO=0, NTEO=3 & \rightarrow & H=G=F=1 & N=L=M=3 \\ 2. & NTEO=1 & \rightarrow & H=G=F=0.5 & N=L=M=1.5 \end{array}\] And the condition on the yield stresses gives :

\[\left. \begin{array}{ll} \sigma_{YY}^{yield} = \sigma_{ZZ}^{yield} = \sigma_{XX}^{yield} \\ \sigma_{45^{\circ} }^{yield} = \sigma_{XX}^{yield}\\ \end{array} \right\} \begin{array} 11. & NTEO=0, NTEO=3 & \rightarrow & H=G=F=1 & N=L=M=3 \\ 2. & NTEO=1 & \rightarrow & H=G=F=0.5 & N=L=M=1.5 \end{array}\] 1. Classic or Ziegler hardening parameters (NTEO=0, 3 or 4)

Line 1 (3G10.0)
CK hardening factor K (see 6.3), param(28,ilaw)
CW hardening coefficient W0 or $\varepsilon_{0}$ or $\sigma_{0}$(see section hardening form), param(29,ilaw)
CN hardening exponent n (see 6.3), param(30,ilaw)

NTEO = 0

Line 1 (2G10.0)
CXkinematic hardening saturation rate, param(25,ilaw)
XSATkinematic hardening saturation value, param(26,ilaw)

NTEO = 4

Line 1 (2G10.0)
CX1kinematic hardening saturation rate 1, param(31,ilaw)
XSAT1kinematic hardening saturation value 1, param(32,ilaw)
CX2kinematic hardening saturation rate 2, param(33,ilaw)
XSAT2kinematic hardening saturation value 2, param(34,ilaw)

NTEO = 3

Line 1 (2G10.0)
$C_{A}$ initial kinematic hardening modulus, param(25,ilaw)
$G_{A}$ rate at which the kinematic hardening modulus decrease with increasing plastic deformation, param(26,ilaw)
Line 2 (1G10.0)
mkinematic-isotropic hardening balance : 0 or 1 \\(m = 1 : isotropic/mixed, m = 0 : full kinematic), param(27, ilaw)
Always use m = 1 when NTEO=0 or NTEO=3.

Amstrong-Frederick (classic) : $\underline{\dot{X}} = C_{X}(X_{sat} \underline{\dot{\varepsilon}}^{plastic} \ - \overline{\dot{\varepsilon}}^{plastic}. \ \underline{X})$
or
Ziegler kinematic hardening : $\underline{\dot{X}} = C_{A}\frac{1}{\sigma_{0}} (\underline{\sigma} - \underline{X}).\dot{\overline{\varepsilon}}^{plastic} – G_{A}. \underline{X}. \dot{\overline{\varepsilon}}^{plastic}$

(If NTEO = 4): X = X1 + X2

Parameters of fatigue law ONLY IF INDAM > 1

Line 1 (10G10.0)
SIGL fatigue limit
SIGUstress fracture stress
BETAfatigue damage parameter law
ADAMfatigue damage parameter law
bdamfatigue damage parameter law
cdamfatigue damage parameter law
dlimdamage critic value
am0fatigue damage parameter law
aprimfatigue damage parameter law
afatigue damage parameter law

2. Teodosiu hardening parameters (NTEO=1 or 2)

Line 1(2G10.0)
CPpolarity saturation rate, param (22, ilaw)
NPpolarity exponent, param(23,ilaw)
Line 2 (5G10.0)
CSD orientation saturation rate for $S_{D}$, param(24,ilaw)
CSL orientation saturation rate for$ \underline{\underline{S_{L}}}$, param(34, ilaw)
SSAT0 initial orientation saturation value, param(28, ilaw)
NLorientation exponent, param(29,ilaw)
R0 initial yield limit, param(31,ilaw)
Line 3 (2G10.0)
CXback stress saturation rate, param(25, ilaw)
XSAT0 initial back stress saturation value, param(26, ilaw)
Line 4 (2G10.0)
Minfluence of S on kinematic(m = 0)-isotropic(m = 1) hardening, param(27, ilaw)
Q $S_{D}$ - SL balance on Xsat, param(30, ilaw)
Line 5 (2G10.0)
CRisotropic hardening : saturation rate, param(32, ilaw)
RSAT isotropic hardening : saturation value, param(33, ilaw)

Stresses

Number of stresses

6

Meaning

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)$\sigma_{YZ}$

State variables

1. Classic hardening parameters (NTEO = 0, 3 or 4)

Number of state variables

9 if NTEO = 0 or 3
21 if NTEO = 4

List of state variables (if NTEO = 0 or 3)

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):Q(8)Back stress (X)
Q(9) Accumulated plastic equivalent strain divided by $\sqrt{2}$ ($\frac{\underline{\varepsilon^{pl}}}{\sqrt{2}}$ )
Q(10):Q(15)Plastic strain tensor $\underline{\varepsilon}^p$ (6 components)
Q(16):Q(18)Principal strains $\varepsilon_{I}$, $\varepsilon_{II}$, $\varepsilon_{III}$

List of state variables (if NTEO = 4)

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):Q(8)Total back stress (X)
Q(9):Q(14)First back stress (X1)
Q(15):Q(20)Second back stress (X2)
Q(21) Accumulated plastic equivalent strain divided by $\sqrt{2}$ ($\frac{\underline{\varepsilon^{pl}}}{\sqrt{2}}$ )

State variables of fatigue law (used only if nteo = 0, 3 or 4)
IF INDAM.GT.1 ONLY
Number of state variables = 32

QBCRO(1)maximum x-component of deviatoric stress over a fatigue cycle
QBCRO(2) maximum y-component of deviatoric stress over a fatigue cycle
QBCRO(3) maximum z-component of deviatoric stress over a fatigue cycle
QBCRO(4)maximum xy-component of deviatoric stress over a fatigue cycle
QBCRO(5)maximum xz-component of deviatoric stress over a fatigue cycle
QBCRO(6) maximum yz-component of deviatoric stress over a fatigue cycle
QBCRO(7) minimum x-component of deviatoric stress over a fatigue cycle
QBCRO(8) minimum y-component of deviatoric stress over a fatigue cycle
QBCRO(9) minimum z-component of deviatoric stress over a fatigue cycle
QBCRO(10)minimum xy-component of deviatoric stress over a fatigue cycle
QBCRO(11)minimum xz-component of deviatoric stress over a fatigue cycle
QBCRO(12)minimum yz-component of deviatoric stress over a fatigue cycle
QBCRO(13) maximum hydrostatic stress over a fatigue cycle
QBCRO(14) maximum of V.M stress over a fatigue cycle
QBCRO(15) number of fatigue cycles
QBCRO(16) cumul of time increment
QBCRO(17) mean stress (3D case) over a fatigue cycle
QBCRO(18) damage variable
QBCRO(19)number of cycles
QBCRO(20)minimum hydrostatic stress over a fatigue cycle
QBCRO(21) sines criterion
QBCRO(22)variable used for the damage computation (multi-blocks case)
QBCRO(23) normalised damage variable
QBCRO(24) minimum triaxiality factor over a fatigue cycle
QBCRO(25) maximum triaxiality factor over a fatigue cycle
QBCRO(26)equivalent stress amplitude (AIIa) over a fatigue cycle
QBCRO(27)maximum triaxiality function over a fatigue cycle
QBCRO(28)maximum damage equivalent stress over a fatigue cycle
QBCRO(29) number of cycle to failure for the block loading
QBCRO(30) x-component of the gradient of equivalent stress amplitude (AIIa)
QBCRO(31) y-component of the gradient of equivalent stress amplitude (AIIa)
QBCRO(32) z-component of the gradient of equivalent stress amplitude (AIIa)

Hardening form

if (NTYPHP = 0), The user’s parameters CK, CW and CN are respectively $K^{W}$ , $W_{0}$ and $n^{W}$ of the hardening equation:
$\sigma_{F} = K^{W} (W_{0} + W^{pl})^{n^{W}}$
$ K^{W}$ = param(22,ilaw)
$ W_{0}$ = param(23,ilaw)
$ n^{W}$ = param(24,ilaw)

if (NTYPHP = 1), The user’s parameters CK, CW and CN are respectively K, $\varepsilon_{0}$ and n of the hardening equation:
$\sigma_{F} = K(\varepsilon_{0} +m. \varepsilon^{pl})^{n}$
K = param(24,ilaw)
$\varepsilon_{0}$ = param(29,ilaw)
n = param(30,ilaw)
m = param(27,ilaw)

if (NTYPHP = 2), The user’s parameters CK, CW and CN are respectively K, $\sigma_{0}$ and n of the hardening equation:
$\sigma_{F} = \sigma_{0} + K(1-exp(-n.\varepsilon^{pl}.m))$
K = param(24,ilaw)
$\sigma_{0}$ = param(29,ilaw)
n = param(30,ilaw)

if (NTYPHP = 3), The user’s parameters CK, CW and CN are respectively K, $\sigma_{0}$ and n of the hardening equation:
$\sigma_{F} = \sigma_{0} + K.(\varepsilon^{pl})^{n}$
K = param(24,ilaw)
$\sigma_{0}$ = param(29,ilaw)
n = param(30,ilaw)

2. Teodosiu hardening state variables (NTEO = 1)
For details see internal report

Number of state variables

54

List of state variables

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):Q(8)Back stress (X)
Q(9) Accumulated plastic equivalent strain ($\underline{\varepsilon}^{pl}$)
Q(10:15) Polarity vector (P)
Q(16) Orientation (SD)
Q(17:52)Orientation matrix ($\underline{\underline{S}}_{D}$)
Q(53)Current elastic limit ($\sigma_{F}$)
Q(54)Isotropic hardening (R )