13th International Conference on Fracture June 16–21, 2013, Beijing, China -5- 3 2 1 2, , 120 3 2 1 2, 60 , 0 1, 0 , (3) 2 (3) 1 (3) (2) 2 (2) 1 (2) (1) 2 (1) 1 (1) = =− = = = = = = = n n n n n n ο ο ο θ θ θ (7) Figure 4. Unit cell of the lattice and coordinate system Due to the requirement of symmetry with respect to the center of the unit cell, the other three springs ( 4,5,6 =b ) have the same properties as 1,2,3 =b , respectively. Every node has two degrees of freedom, and it follows that the strain energy of a unit hexagonal cell of such a lattice, under conditions of uniform strain , ) ( , 11 22 12 ε ε ε =ε , is ∑ ∑ = = ⋅ = = 6 1 () () () () () 2 6 1 ( ) ( ) 8 F u 2 1 b ij km b m b k b j b i b b b b cell n n n n l E ε ε α (8) From Eqs. (3), (5), and (8), the stiffness tensor can be obtained as ∑ = = 6 1 () () () () () 2 4 b b m b k b j b i b ijkm n n n n V l C α (9) In particular, taking all ( )bα as the same α α = ( )b , and substituting the value 3 2 2 l t V = and Eq. (7) into (9), we can get t C C C t C C 8 3 3 , 8 3 9 1212 2211 1122 2222 1111 α α = = = = = (10) It can be observed that the condition ) 2 ( 1122 1111 1212 C C C = − (11) is satisfied and there are only two independent elastic moduli, which means the modeled continuum 1 2 3 4 5 6 7 8 9 x y θ
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