Issue 68
M. Sarparast et alii, Frattura ed Integrità Strutturale, 68 (2024) 340-356; DOI: 10.3221/IGF-ESIS.68.23
2
p m N N s
f
1 2
N
P
exp
0
s
2
N
A
(7)
P
0
0
f f f f
(8)
g
n
s
where k is introduced as a new material parameter for the void nucleation rate of damage in pure shear conditions, σ is the stress- function used by Nahshon and Hutchinson, S is the deviatoric stress tensor and p plastic strain rate tensor, c f is the critical volume fraction of voids and is the volume fraction of the void at the fractured moment This function is between the zero and one value, which for the axial stress state is zero and for the shear stress condition is one. Also, the ˙ f is rate of change in the voids volume fraction, g f is the void growth, n f is the nucleation rate, and s f is the shear rate of the voids. The VUMAT subroutine is implemented to utilize the modified GTN damage behavior of the material in Abaqus software. In this process stress tensor and internal state variables must be updated in the end of each increment If plastic state occurs the plastic corrections must be calculated. The corrections for ∆ε p (volumetric plastic strain increment) and ∆ε q (deviatoric plastic strain increment) are calculated using the Newton-Raphson iteration method. This iterative process helps refine the values of cp and cq, which are the corrections required for updating the internal variables. By iteratively adjusting these corrections, the algorithm achieves an accurate and reliable update of the stress tensor and internal state variables. The pseudo-code algorithm for developing the user subroutine is provided in Tab. 4 [38]. Initiate the stress tensor and internal variables at t 0 time . Calculate elastic predictor t t t t t t , , T T T p q :
1 3
3 2
: T p σ I ; T
: T e σ σ C ε ;
T q
T T
: s s
(9-11)
t
Calculate the value of the yield function:
s
T T p m
F p q
f D
Φ ( , ,
, ,
)
(12)
2
t
t
t
IF 2 0 F then k=0 While |F 1 |> tolerance or |F 2 |> tolerance do
p and
q values:
Calculate the corrections p c and q c to update the
k
k
F
F
1 F
k
1
1
c
c
p t q p q 1 1 k k t t t k k t t t t
p t q t k
1 t t k 1 t t k 3 q G T p K T p q
1
p
q
c
t
t
q
p
p
1 t
t
t
t
t
;
;
(13-15)
k
k
k
c
F
F
2 F
k
q
2
2
c
c
t
p
q
t
t
q
p
t
t
t
t
Check the yield condition:
k
k
1
1
Φ q
Φ p
p m
k
1
D
; 2 F
k
k
1
1
1 F
k
1
1 p q , k
k
k
1
1
k
k
1
1
f
Φ
,
,
,
(
(
q
)
)
(16,17)
t t
t
t
t
t
s
t t
t t
p
t t
t
t
t
t
t t
t t
t t
346
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