PSI - Issue 2_A
M. Nourazar et al. / Procedia Structural Integrity 2 (2016) 3423–3431 Author name / Structural Integrity Procedia 00 (2016) 000–000
3427
5
( j i sg y y −
)
( g y y −
)
G
e
j
i
x
sg y h
)) s x x ds −
y y
sinh( (
)) + +
}cos( (
]sin( ) θ
0
×
−
< <
j
i
j
j
i
2
2 2 ( π
2 ) ( − + −
2
2
g y y
x x
)
i
j
i
j
0 ∫ ∞
( j i sg y y −
)
−
gG sgy
sgy sgh
cosh( cosh(
) )
sinh( sinh(
)
χ χ
+ −
e
y
j
j
K
sg y h
)) s x x ds −
[
{
sinh( (
)) + −
}sin( (
=
ij
i
i
j
2
sgh
)
2
π
2
2
gG
x x −
(
)
y
i
j
]cos( ) θ
+
2
2 ) ( − + −
2
2
π
( g y y
x x
)
i
j
i
j
0 ∫ ∞
( j i sg y y −
)
−
sgy
sgy sgh
cosh( cosh(
) )
sinh( sinh(
)
+ −
χ χ
G
e
j
j
x
sg y h
)) s x x ds −
[
{
cosh( (
)) + −
}cos( (
−
i
i
j
2
sgh
)
2
π
2
2
( g y y −
)
G
j
i
x
h y y
]sin( ) θ
−
− < <
(15)
i
j
2
2 2 ( π
2 ) ( − + −
2
g y y
x x
)
i
j
i
j
The left hand side of Eq. (14) is stress components at the presumed location of the cracks with negative sign. Since stress component (11) are Cauchy singular at the dislocation location, the system of integral equations (14) for the density functions are Cauchy singular for i j = as t s → . Employing the definition of dislocation density function, the cracks opening displacement for embedded crack becomes:
s
∫
2
2
−
+
j ′
j ′
(16)
( ) w s w s − ( )
wzj b t
( ) [ ( )] [ ( )] x t +
y t dt
j
1, 2,..., . N
=
=
j
j
1
−
The integral equations must be solved under the following single-valuedness conditions:
1 ∫
2
2
j ′
(17)
wzj b t
( ) [ ( )] [ ( )] x t + ′
y t dt
j
1, 2, ..., . N
0,
=
=
j
1
−
To evaluate the dislocation density, the Cauchy singular integral equations (14) and (17) ought to be solved simultaneously. The stress fields in neighborhood of crack tips behave like 1 r where is the distance from the crack tips. Therefore, the dislocation densities are taken as
zj g t
2 ( )
(18)
wzj b t
1 1, t − ≤ ≤ = j
1, 2, ..., . N
( )
,
=
t
1
−
( ) g t zj are obtained via solution of the system of equations. The stress intensity factors may be written in terms
The parameters
of crack opening displacement takes the form
−
+
−
+
( ) w s w s − ( )
( ) w s w s − ( )
2
2
i
i
i
i
,
(19)
K
G G
K
G G
lim
,
lim
=
=
Li
y x
Ri
y x
4
4
r
r
r
r
0
0
→
→
L
Ri
L
Ri
i
i
where L and R designate, the left and right tips of a crack, respectively, and
1 2
1 2
2
2 ( 1)) ,
2
2 (1)) .
r
( ( ) x s x
y s y
r
( ( ) x s x −
y s y −
( 1)) ( ( )
(1)) ( ( ) +
=
− − +
− −
=
(20)
Li
i
i
i
i
Ri
i
i
i
i
from (20) it may easily be shown that
1 4
1 4
G G
G G
y x
y x
M
M Ri
2
2
2
2
[ ( 1)] [ ( 1)] x y ′ ′ − + −
[ (1)] [ (1)] x y ′ ′ +
K
g
K
g
( 1), −
(1).
=
= −
(21)
Li
i
i
zi
i
i
zi
2
2
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