ICF13B

13th International Conference on Fracture June 16–21, 2013, Beijing, China 2 2. Formulation of the problem As shown in Fig.1, consider the plane problem of an arbitrarily oriented crack located in a functionally graded material. Two rectangular Cartesian coordinate systems, ( , , ) x y z and 1 1 1 ( , , ) x y z , are defined to describe the direction of the material gradient and crack orientation, respectively. The system 1 1 1 ( , , ) x y z is obtained by rotating counterclockwise to the system ( , , ) x y z with angle  .The crack is assumed to occupy the region 1 a x b   , 1 0 y  , 1z  ∣∣ . The shear modulus  and mass density  in 1 1 1 ( , , ) x y z and (, , ) x y z coordinate can be written, respectively, as follows [12] 1 1 0 1 1 2 1 1 1 0 1 1 2 1 ( , ) exp( ), ( , ) exp( ), x y x y x y x y             (1) 0 0 ( ) exp( ), ( ) exp( ), y y y y         (2) where , 0 and 0 are material constants, and 1 2 sin( ), cos( )         . The mixed boundary value problem shown in Fig.1 will be solved under the following conditions 1 1 1 1 1 1 1 1 1 1 1 1 1 ( ,0 , ) ( ,0 , ), ( ,0 , ) ( ,0 , ), , y y y y x y x y x t x t x t x t x             (3) 1 1 1 1 1 1 ( ,0,) ( ,0,), ( ,0,) ( ,0,), , or , u x t u x t v x t v x t x a x b         (4) 1 1 1 1 1 ( ,0 , ) ( ), , y y x t H t a x b       (5) 1 1 1 2 1 ( ,0 , ) ( ), , x y x t H t a x b       (6) where 1 ( ) H t  and 2 ( ) H t  are the negative of dynamic normal stress and shear stress at the crack plane under external loading in an uncracked specimen. ( ) H t is the Heaviside function of time t . When 0 t  , ( ) 0 H t  , and when 0 t  , ( ) 1 H t  . Figure 1. Geometry of the functionally graded material with an arbitrarily oriented crack 3. Method of solutions The general expressions for displacements components can be written as

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