Crack Paths 2012

z2

A1

A2

1'1

1—»

H l

|> 2

T h einterface I is located on the sag-axis, while the crack E _ A ais located on the 0:1

axis:

E_Aa :{(ac1,0) G 9,961 g —Aa},

in particular E0 correspondsto the crack with tip at the interface I. Let ifAabe

the displacementfield solving the elasticity problem,

1 I 1 Q _ A a : Q \ E — A a 7 _ _ _ _ <1)

U l g i O ' g g i o o n\:'—Aa,:l:7

O nP ’

wheren denotes the external unit normalvector. W ehave Hooke’slaw

A 1 , 131 < 0

(2)

a : A x - e,

A at I

()

{A2, I 1 > 0 .

Furthermore, we have to assume that the external loading p is self balanced, that

is IT p,- : 0, i I 1, 2, f1. P1082 — pgxlds I O. In addition w e require the continuity of

the displacementfields and normalstresses at the interface I: on the interface I:

U ( 0 +$,2 )I ’LL(0_, IE2), U i 1 ( 0 +$,2 )I U i 1 ( 0 _$,2 ) .

The associated potential energy U can be considered as a function of the distance

Act (we use the sum convention):

1

U ( — AI a— ) Uijé‘ij — p j u J - d s .

2 92A,,

r

In order to decide whetherthe crack can reach andeventually penetratethe interface,

the following condition must be fulfilled for all (small) values of Au:

(4)

— A U: U(—Aa)— U(0) > 2'y0Aa,

where 2'y1Aa represents the energy to produce the new surface, that is 2V1 is the

critical energy release rate in the first material. Problems of this type were studied

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