13th International Conference on Fracture June 16–21, 2013, Beijing, China -5- ∫− − = + c c ist dt e p t f is p s A p s A ( , ) 1 ( , ) ( , ) * 2 1 (23) ∫ ∫ − − − + − ⋅ = − + c c ist c c ist dt e p t g is dt e p t f is e p s A p s A ( , ) 1 ) ( , ) 1 ) ( ( , ) ( ( , ) * * 0 0 4 3 κ (24) Substituting Eqs. (2) (17-19) into Eq. (10), it yields ( , ) ] 0 ( , ) [ ( , ) ] ) [ ( , ) ( 4 3 2 1 4 4 3 3 0 2 2 1 1 0 2 0 0 = + + + + h t h t h h t h t h e p s A t eps At ee eps At e p s A t e e c β β κ (25) ( , ) ] 0 [ ( , ) ( , ) ] [ ( , ) 2 4 3 2 1 4 4 3 3 0 2 2 1 1 0 = + + + h t h t h h t h t h e p s A t e p s A t e e p s A t e p s A t e e β β κ (26) One can finally obtain the following relations form Eqs. (23-26) f e t e t e t A h t h t h t 2 1 2 2 1 2 1 − − = ， f e t e t e t A h t h t h t 2 1 1 2 1 1 2 − = (27) ) ( 0 0 4 3 4 3 4 3 4 g f e e t e t e t A h t h t h t − + − − = κ ， ) ( 0 0 4 3 3 4 4 3 3 g f e e t e t e t A h t h t h t − + − = κ (28) where ∫− − = c c ist dt e p t f is f ( , ) 1 * ， ∫ − − = c c ist dt e p t g is g ( , ) 1 * (29) Then, substituting Eqs. (2) (27-29) into Eqs. (2), the basic solution expression of the electric displacement and anti-plane stresses is obtained. And then, substituting the electric displacement and stress component into Eq. (8), one arrives at an integral equation in the form [6] dt pt fptx K dt x t p t g e dt x t p t f c c c c c c ∫ ∫ ∫ − − − + − + − ( , , ) ( , ) ( , ) 1 ( , ) 1 * 11 * 0 * π π π p dt p t g p t x K e c c 0 * 12 0 ( , , ) ( , ) τ π =− − ∫ − (30) dt p t f p t x K e dt x t p t g dt x t p t f e c c c c c c ∫ ∫ ∫ − − − − − + − − ( , , ) ( , ) ( , ) ( , ) * 21 0 * 0 * 0 π π κ π p D dt p t g p t x K c c 0 * 22 0 ( , , ) ( , ) =− + ∫ − π κ (31) where ∫+∞ − − = 11 0 1 [ (, ) 1]sin( )ds x t s p s U K ， ∫ +∞ − + = 12 0 2 [ (, ) 1]sin( )ds x t s p s U K ， ∫+∞ − − = = 0 2 22 21 [ (, ) 1]sin( )ds x t s p s U K K (32) ] ) ( [ 1 2 1 2 1 4 3 4 3 2 1 1 2 1 2 0 2 0 0 4 3 3 4 3 4 0 2 0 1 h t h t h t h t h t h t h t h t e t e t e t t e t t e c e t e t e t t e t t e s U − − ⋅ − + − − ⋅ = κ κ , h t h t h t h t e t e t e t t e t t s U 4 3 4 3 4 3 3 4 3 4 2 1 − − = ⋅ (33) Because of the symmetry of fracture analysis, the dislocation density function ( , ), ( , )pupuF must be an odd function of u. Then, it is automatically satisfied (0,) 0, (0,) 0 * * = = p g p f (34) 3.3 Numerical solution Introducing rc x uc t = = , , ( , ) ( , ), ( , ) ( , ) * * puG ptg puF ptf = = , ( , , )( , 1,2) ( , , ) * = = jipru K ptx K ij ij , Eqs. (30-31) and (34) can be transformed into the standard form of the first kind Cauchy singular integral equations as
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