13th International Conference on Fracture June 16–21, 2013, Beijing, China The disk is assumed to be made of isotropic elastic material and obey the small deformation hypothesis. The stress distributions in the disk relative to a polar coordinate system can be obtained based on the elasticity theory [10] as c r σ σ =− ， c σ σ θ =− ， 0=θ τ r （1） 2.2 Stress intensity factors for cracked disk under confining pressure Fig.2 Cracked Brazilian Disk under confining pressure loading The weight function method is used to calculate the stress intensity factors for the cracked disk. The cracked Brazilian disk subjected to confining pressure cσis shown in Fig.2. The x-axis lies in the crack plane. The coordinate origin is located at the center of the disk. ais the half length of the crack, Ris the disk radius and Bis the thickness. Then the ' IK and ' II K can be given as K h x a x dx a ( , ) ( ) 0 1 ' θσ ∫ = Ι （2） h x a x dx K r a ( , ) ( ) 0 2 ' θ τ ∫ = ΙΙ （3） where 1h and 2h are weight functions, θ θ τ σ r , are the stress components of the uncracked Brazilian disk under confining pressure. Specifically, 1h and 2h can be expressed as [11]: ) ] (1 1 1 1 [ 2 ( , ) 2 3 1 12 2 1 11 2 1 1 ρ ρ ρ π + − + − − = c c a h x a （4） 2 2 3 2 21 1 22 1 2 1 2 1 ( , ) [ 1 ( 1 ) ] 1 h x a c c a ρ ρ π ρ = + − + − − （5） where 8 1 13.234 24.6076 15.9344 8 4 3.8612 5 4 3 2 11 − − − + − − + = α α α α α α c （6） 8 1 12.596 24.2696 14.1232 8 4 0.6488 5 4 3 2 12 + − + − + − + − = α α α α α α c （7） 5 1 1.1028 2.376 5 2.5 1.4882 4 3 2 21 − − + − + − = α α α α α c （8） 4 1 0.7177 0.81112 4 2 0.4888 4 3 2 22 + − − + − + + = α α α α α c （9） （ a R x a/ / , 1 = = ρ α ） By substituting Eqs.(1) and (4) into Eq.(2), respectively, the following explicit formula of the stress intensity factors can be yielded as
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