Computational Methods for Debonding in Composites

208 Ernest T.Y. Ng and A. Suleman
∆λ has to be positive for plastic loading, therefore the only choice is ∆e P m = 0,
which implies σm = 0. In this case, we have ∆ f = A∆ē P . But then we have arrived
an undetermined form for ∆λ which is ∆λ → 0 0 . Thus, the formula ∆λ = 3∆eP m
F ′ is
no longer valid in this case. Therefore, we expect ∆λ is no longer a function of ∆e P m
and hence, ∆λ = ∆λ(∆ē P ). Thus, we have the following theorem:
Theorem 10.2. Suppose that σ E m = 0, then we have the following form for P and
F :
t+∆t t+∆t
P = ∆λ S : t+∆t S − (1 − t+∆t f ) t+∆t ˆσy∆ē P
(10.57)
t+∆t 1t+∆t
F = S :
2
t+∆t t+∆t ˆσ 2
y
S + [2q1
3
t+∆t f ∗ − 1 − (q3 t+∆t f ∗ ) 2 ] (10.58)
Moreover, we have
∆λ = R ∆ēP
(10.59)
t+∆t ˆσy
where
3(1 −
R =
t+∆t f )
2[1 +(q3 t+∆t f ∗ ) 2 − 2q1 t+∆t f ∗ (10.60)
]
�
3q2 Proof. Sinceσm = 0, therefore cosh
t+∆t σm
2t+∆t �
= 1. Substitute these two relations
ˆσy
into the expressions of P and F of Gurson-Tvergaard model to obtain the results.
To show that ∆λ = R ∆ēP
t+∆t , simply set the expressions for P and F equal to zero
ˆσy
and combine the two expressions. That is, we have
t+∆t S : t+∆t S = 2t+∆t ˆσ 2 y
3
[1 +(q3 t+∆t f ∗ ) 2 − 2q1 t+∆t f ∗ ]
follow from the expression F = 0. Then substitute t+∆t S : t+∆t S into the expression
P = 0tosolvefor∆λ. �
However, if t f = 0andA = 0, then ∆ f = 3∆ePm 1+3∆eP and we have back to von-Mises
m
case and this leads to the following corollary of the above theorem:
Corollary 10.1. Suppose that t f = 0,A= 0 and σ E m = 0, then we have the following
form for P and F :
Moreover, we have
t+∆t P = ∆λ t+∆t S : t+∆t S − t+∆t ˆσy∆ē P
t+∆t 1t+∆t
F = S :
2
t+∆t S + 1 t+∆t ˆσ 2
y
3
∆λ = 3∆ēP
2 t+∆t ˆσy
(10.61)
(10.62)
(10.63)
Proof. The proof is by setting t+∆t f and t+∆t f ∗ equal to zero in the expression
of R. �