PSI - Issue 52
J.C. Wen et al. / Procedia Structural Integrity 52 (2024) 625–646 Author name / Structural Integrity Procedia 00 (2019) 000 – 000
639 15
and
( )
( ) ,
=
(63)
II K s
K s
I
in which n u are normal discontinuous displacements of static and dynamic tangential near the crack tip, respectively. The static intensity factors can be directly determined by / u u , u , , / u u n n n u , u and
1/2
E
(
)
(
)
0
0 u t d y y
0 t u t u d + 0 x x y y
(64)
.
u t
K
0
=
+
−
I
x x
2 2(1 ) (1 ) a − + 2
t
u
4.1. Independent contour integral algorithm To avoid complicated computational processes, we divide the computation process into two steps. The first step is to obtain the coefficient solutions calculated without crack increment.
1
−
K
α
, Q α
( ) , a K Q
=
( ) a
(65)
=
1
1
Next, considering an increment of crack length a gives
1
−
K
α
, Q α
K
) , Q
(
)
(
2
a a +
a a +
(66)
=
=
2
Therefore, the variation of displacement can be obtained as
0 0 M N m n = =
0 0 M N m n = =
2 1 mn m n T T − ( ) ( ), ( , ) u u mn y u
2 1 mn m n T T − ( ) ( ). v v mn
(67)
( , ) u
=
=
x
Contour integration can be achieved by selecting the boundary of the computational domain as the contour, or by setting a contour within the region as the integration boundary. Here, we introduce the method of contour integration set within the region, which is simple and can reduce computational workload. By using reciprocal theory, it is clear that the contour path integral is independent around the crack tip for functionally graded materials. As the stress intensity factor is independent of integral contour, any integral contour ' with an enclosed domain ' can be selected to determine either static or dynamic stress intensity factor. As shown in Fig. 3a, when considering crack increment, the ( , ) P point field in the mapping domain moves to the '( , ) P + + point field, but the corresponding point ( , ) P x y in the physical domain does not change in position as the crack tip increment occurs (see Fig. 3b), which means x x x = + and y y y = + . Then
N N
8
8
8
( , ) N x i i =
( , ) N x
0,
x =
x
(68)
+
+
=
i
i
i
i
i
1
1
1
i
i
i
=
=
=
N N
8
8
8
( , ) N y i i =
( , ) N y
0.
y =
y
(69)
+
+
=
i
i
i
i
i
1
1
1
i
i
i
=
=
=
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