13th International Conference on Fracture June 16–21, 2013, Beijing, China -4- ( ) ( ) ( ) ( ) { } ( )( )( ) 2 2 2 2 2 2 2 2 2 2 2 2 , , , d ; ' x b r x x c a E k K k a u a b c x x b x c E x a x b c x σ ⎡ ⎤ − − + ⎣ ⎦ = ⋅ ≤ ≤ − − − ∫ (6b) Having obtained the weight functions for this crack configuration, the stress intensity factors, crack opening displacements and plastic zone size of its strip yield model can be determined [8]. 2.4. A unified method for strip yield collinear cracks It is observed that the weight function method is accurate and efficient to obtain the stress intensity factor and crack opening displacement. However it is hard to obtain the weight functions for cracks in finite sheet. In this section, a simple and efficient unified method [9] is proposed to solve the strip yield model for collinear cracks in infinite and finite sheets. The key idea of this method is to treat all the cracks, the plastic zones and the remaining elastic ligament between the cracks as a single crack, which is solved using the weight function method. For example, the strip yield model for two collinear cracks in an infinite sheet shown in Fig.3a is modeled by an equivalent single center crack shown in Fig.3b. The center crack of length 2l=(4a+2b+2rB) is subjected to (i) remote uniform stress σ; (ii) segments uniform compression yield stress -σs over the plastic zones rA and rB; and (iii) continuous compression stress -σ(x) distributed along the remaining elastic ligament x [∈-(b-rA), (b-rA)], respectively. The σ(x) is further discretized by a set of uniform segment stress, as shown in Fig.3c. The requirements for the equivalent single crack are: 1) No stress singularity at the fictitious crack tip, 2) Zero crack opening displacements along the elastic ligament between the fictitious crack tips, x∈[-(b-rA), (b-rA)]. σ σ x y rA b 2a δΑ δΒ w=1 o A rB B X1 X2=b+2a l=b+2a+rB (a) σ σ x y rA b 2a w=1 rB X1 X2=b+2a l=b+2a+rB σs σs σs σs σ(x) (b) Fig.3. The concept of the unified method: (a) a strip yield model for two collinear cracks in an infinite sheet, (b) an equivalent single crack for modeling problem (a), (c) discretized stress distribution. The stress intensity factor F and crack opening displacement U(l, x) for the equivalent crack subjected to such complex loading are determined by the superposition of three component elastic solutions. Closed form solutions for the first two load cases are available in Ref.[12]. For the equivalent crack, the following equations are established to obtain the unknown variables. ( ) ( ) ( ) 1 0; , 0; 2, , ; 1,2,3... i i A A F U l x x x x b r b r i N + ⎧⎪ = ⎨ = = + − − − = ⎪⎩ (10)
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