13th International Conference on Fracture June 16–21, 2013, Beijing, China -3- 1 2 ,1 ,1 1 , ,1 ,1 1 , 1 (( )( ) ( )( ) ) 2 1 (( )( ) ( )( ) ) 2 aux aux aux m aux ij ij i i ik ik ik ik j j A aux aux aux m aux ij ij i i ik ik ik ik j j A J u u qdA u u q dA σ σ σ σ ε ε δ σ σ σ σ ε ε δ + = + + − + + + + + − + + ∫ ∫ (7) The interactional part is called interaction energy integral [14]. Here, it can be derived as ,1 ,1 1 , , 1 , 1 ,1 ,1 ( ) ( ) aux aux aux m ij i ij i ik ik j j A aux aux aux m aux m ij i j ij i j ij ij ij ij A I u u q dA u u qdA σ σ σ ε δ σ σ σ ε σ ε = + − + + − − ∫ ∫ (8) According to the relationship of the displacement and strain in elastic mechanics, one may write , 1 , 1 , 1 ,1 ,1 ,1 1 ( ) ( ) 2 aux aux aux t aux m th ij i j ij i j j i ij ij ij ij ij u u u σ σ σ ε σ ε ε = + = = + (9) The second integral can be obtained as , 1 , 1 ,1 ,1 , 1 ,1 ,1 ( ) ( ) aux aux aux m aux m aux aux m aux th ij i j ij i j ij ij ij ij ij i j ij ij ij ij A A u u qdA u qdA σ σ σ ε σ ε σ σ ε σ ε + − − = − + ∫ ∫ (10) If the extra strain field 0 aux tip aux ij ijkl kl S ε σ = is introduced, we can get the formulation 0 , , 1 ( ) 2 aux aux aux ij i j j i u u ε = + , Here tip ijkl S is a compliance tensor at the crack tip [15]. Thus, the interaction energy integral can be written as ,1 ,1 1 , ,1 ( ) ( ( ( )) ) aux aux aux m ij i ij i ik ik j j A tip aux th ij ijkl ijkl kl A I u u q dA S S x qdA I σ σ σ ε δ σ σ = + − + − + ∫ ∫ (11) where ,1 ,1 0 ,1 ( ) [ ( ) ] th aux th aux ij ij ii A A I qdA qdA σ ε σ α θ θ αθ = = − + ∫ ∫ (12) In the above derivation process, the domain of the integral is chosen arbitrarily around the crack-tip, so the interaction energy integral for thermal fracture problems is domain-independent. 3. Evaluation of the T-stress from the interaction energy integral The contour integral around the crack tip can be written as 0 0 1 ,1 ,1 0 lim ( ) aux aux aux ik ik j ij i ij i j I u u n d σ ε δ σ σ Γ Γ → = − − Γ ∫Ñ (13) Here we can consider only the stress parallel to the crack direction, i.e. 1 1 ij i j T σ δ δ = (14) Where T denotes the T-stress. The force f is in equilibrium can be expressed as 0 0 0 lim aux ij j f n d σ Γ Γ → =− Γ ∫Ñ (15) We can also obtain the following relationship from the kinematic equations ,1 11 1 11 11 1 ( ) t m th i i i u ε δ ε ε δ = = + (16) It can be rewritten as 11 ,1 1 ( + ) i i u C E σ α θ δ ∗ = Δ (17) where * 1 C = for generalized plane stress and * 1 ( ) C x ν = + for plane strain. Substituting Eq. (14) and Eq.(17) into the contour integral, one obtains 0 0 * * 0 lim ( + ) ( + ) aux ij j tip tip tip tip T T I C n d C f E E α θ σ α θ ∗ Γ Γ → =− Δ Γ= Δ ⋅ ∫Ñ (18)
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