13th International Conference on Fracture June 16–21, 2013, Beijing, China -7- Figure 1. Comparison between the results predicted by limit analysis and the variational approach for different volume fractions of the rigid inclusion -’line’: limit analysis with ρ=0.1, -’cross’: variational approach with ρ=0.1, -’point’: limit analysis with ρ=0.3, -’box’: variational approach with ρ=0.3, -’diamond’: limit analysis with ρ=0.6, -’circle’: variational approach with ρ=0.6 6. Strength under isotropic loading 6.1. Static approach of the limit analysis problem The theory of limit analysis teaches that a cinematic approach like the Gurson one provides an upper bound of the true strength. In order to check the accuracy of a cinematic estimate, it is therefore highly desirable to derive a static approach which in turn will deliver a lower bound of the true strength. We therefore seek the stress field solution to an isotropic loading (traction or compression) in the framework of the ’rigid core sphere model’. The macroscopic stress state is of the form 1mΣ . Accordingly the external boundary ( e r r = ) is subjected to a radial surface force m n n σ⋅ =Σ . We implement the so-called static approach of limit analysis. It consists in deriving a mesoscopic stress field σ which must be statically admissible with these boundary conditions and meet the criterion ( ) 0 meso F σ = (see (2)). Owing to the spherical symmetry, this statically admissible stress field can be sought in the form (spherical coordinates): ( ) ( ) ( ) rr r r r r e e e e e e θθ θ θ ϕ ϕ σ σ σ = ⊗ + ⊗ + ⊗ . (33) The boundary condition on the surface e r r = reads ( ) rr e m r σ =Σ . (34) The momentum balance equation 0 div σ= reduces to 2 rr rr d dr r θθ σ σ σ − = . (35) With the notation rr X θθ σ σ = − (note that 2 3 2 2 d X σ = ), it is found that 2 3 m rr X σ σ = − , so that the criterion (2) yields ( ) 2 2 2 2 3 2 2 2 2 2 1 3 2 2 2 2(1 ) (1 ) 0 3 2 3 3 rr rr f X f T X f h X f h T T σ σ + − ⎛ ⎞ ⎛ ⎞ + + + − + − − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ . (36) The values of X solutions of (36) read: 2 2 2 (3 2 ) 2 ( 1) 3 with 2 2 5 3 rr f T hT f X T f σ − − − ± Δ = , − − . (37) 2 2 2 2 2 2 (2 3)(2 3 ) 4 (2 3)( 1) 2 (5 3)( 1) rr rr f T f hT f f h T f f σ σ Δ= + − + + − + + − . (38) Recalling (35), an ordinary differential equation with respect to rrσ is obtained in the form: ( ) 2 rr rr r d X dr σ σ = . (39) Introducing (37) into (39) and integrating over the interval i e r r ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ , , one obtains
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