014 3 3 XFEM 1116 6 J kε K I K II J kε θ p J kε XFEM θ p θ p 4 θ p v 1/v lg E 1/E J O30 A doi 10.3981/j.issn.1000-7857.014.3.001 Crack Propagation Terminating at a Bimaterial Interface Studied Using Extended Finite Element Method SHI Fang, GAO Feng, GAO Yanan State Key Laboratory for Geomechanics & Deep Underground Engineering, China University of Mining & Technology, Xuzhou 1116, China Abstract A numerical method is presented for obtaining the stress intensity factors of cracks terminating at a bimaterial interface based on extended finite element method (XFEM). A new 6-term crack tip displacement enrichment function is derived. Based on the analytical solution of the stress and displacement fields around the crack tip, the expression of the path independent integral J kε and the stress intensity factors K I and K II, is established. The XFEM numerical solution is used to calculate the integral J kε, and the stress intensity factors are obtained by using the above expression. Finally, the maximum circumferential stress criterion is used to request the crack propagation angle θ p. Results of the numerical simulations show that propagation problems of vertical crack at the bimaterial interface can be solved efficiently by the combination of the integral J kε method and the XFEM. The crack propagation angle θ p is smaller when the crack propagates from a softer material into a harder material, but θ p is larger when the crack propagates from a harder material into a softer material. In the case of the four-point bending test, the crack propagation angle θ p is independent of the ratio of the Poisson's ratios (v 1/v ) of materials on both sides of the interface, but θ p and the logarithm of the ratio of elasticity modulus lg(e 1/E ) meet an exponential relation. Keywords XFEM; fracture mechanics; J integral; bimaterial interface; crack propagation - [1,] 014-04-7 014-07-11 973 011CB0105 CXZZ13_09 fshi@cumt.edu.cn fgao@cumt.edu.cn,,. XFEM [J]., 014, 3(3): 15-1. 15
014 3 3 [3] ìσ 1 θθ(r, -π) = 0 σ 1 rθ(r, -π) = 0 σ 3 θθ(r,π) = 0 îσ 3 rθ(r,π) = 0 σ k ij k extended finite element method XFEM ìσ θθ( r,π/ ) = σ 3 θθ( r,π/) [4] σ rθ( r,π/ ) = σ 3 rθ( r,π/) σ finite element method FEM 1 θθ( r, -π/) = σ θθ( r, -π/) σ 1 rθ( r, -π/) = σ rθ( r, -π/) 3 u θθ( r,π/ ) = u 3 θθ( r,π/) 1999 Belytschko [5~8] u rr( r,π/ ) = u 3 rr( r,π/) u 1 θθ( r, -π/) = u θθ( r, -π/) îu 1 rr( r, -π/) = u rr( r, -π/) XFEM Sukumar [9] Huang [10] 4 Sukumar [11] 1 Bouhala [1] Airy [13,14] [15,16] isoparametric finite elements [17] Chang [18] J kε J k J kε XFEM J kε K I K II XFEM J kε Betaxfem D 4 3 ϕ k (z) ψ k (z) k=1,, 3 ìϕ k (z) = a 1k z λ + a k z λˉ (k = 1,,3) 4 îψ k (z) = b 1k z λ + b k z λˉ λ 1 4 1 3 a 1k a 1 k b 1k b k k=1,, 3 1 λ 0.5 O Ra=0 a =[a 11,a 1,b 11,b 1,a 1,a,b 1,b,a 13,a 3,b 13,b 3 ] T (x, y) (r, θ) R 3 Ω 1 (-π θ -π/) Ω (-π/ θ π/) R = 0 Ω 3 ( π/ θ π) [ λ (α - β)( β + 1) - α + β +(1 - β )cos(λπ)] = 0 5 ϕ(z) ψ(z) 5 λ5 ìσ rr + σ θθ = [ ϕ ] (z) + -- - ϕ - (z) α β Dundurs σ θθ - σ rr + iσ rθ = e iθ [zˉϕ (z) + ψ (z)] ì G(u î rr + iu θθ ) = e [ ] -iθ κϕ(z) - z -- - ϕ - (z) - - ψ(z) --- 1 α = G 1(κ + 1) - G (κ 1 + 1) G 1 (κ + 1) + G (κ 1 + 1) β = G 1(κ - 1) - G (κ 1-1) 6 σ rr σ θθ σ rθ u rr u θθ î G 1 (κ + 1) + G (κ 1 + 1) G G κ = 3-4v κ =(3 - v)/(1 + v) α<0 Fig. 1 1 A crack terminates normally at the bimaterial interface 16
014 3 3 λ>0.5 α>0 λ<0.5 Chen [13] 7 u rr u θθ ìu k rr = r λ [(A 1 ) k rr cos(λ + 1)θ +(A ) k rr sin(λ + 1)θ + u ij = K I r λ g I ij( θ ) + K II r λ g II ij ( θ) 7 (A 3 ) k rr cos(λ - 1)θ +(A 4 ) k rr sin(λ - 1)θ] σ ij = K I r f I 1 - λ ij ( θ ) + K 14 II r f u k II θθ = r λ [(A 1 ) k θθ cos(λ + 1)θ +(A ) k θθ sin(λ + 1)θ + 1 - λ ij ( θ) 8 î (A 3 ) k θθ cos(λ - 1)θ +(A 4 ) k θθ sin(λ - 1)θ] (A i ) k rr (A i ) θθ k i=1,, 3, 4 9 1999 Belytschko [5] { u x = u rr cos θ - u θθ sin θ u y = u rr sin θ + u θθ cos θ 15 Moës [7] ìu k x = r λ [(A 1 ) k rr cos(λ + 1)θ cos θ +(A ) k rr sin(λ + 1)θ cos θ +.1 (A 3 ) k rr cos(λ - 1)θ cos θ +(A 4 ) k rr sin(λ - 1)θ cos θ - n x (A 1 ) k θθ cos(λ + 1)θ sin θ -(A ) k θθ sin(λ + 1)θ sin θ - x XFEM (A 3 ) k θθ cos(λ - 1)θ sin θ -(A 4 ) k θθ sin(λ - 1)θ sin θ] 16 u FE u enr u k y = r λ [(A 1 ) k rr cos(λ + 1)θ sin θ +(A ) k rr sin(λ + 1)θ sin θ + u h (x) = u FE + u enr (A 3 ) k rr cos(λ - 1)θ sin θ +(A 4 ) k rr sin(λ - 1)θ sin θ + 9 (A 1 ) k θθ cos(λ + 1)θ cos θ +(A ) k θθ sin(λ + 1)θ cos θ + N k ( x) ψ( x) a k î (A 3 ) k θθ cos(λ - 1)θ cos θ +(A 4 ) k θθ sin(λ - 1)θ cos θ] n = N j ( ) j = 1 m x u j + k = 1 u h (x) = é n ë ê ù N j (x)u j + é m h j = 1 ë ê ù N h (x)h(x)a h + h = 1 { F l ( r,θ) } 4 ={ r 1 - λ sin λθ,r 1 - λ cos λθ,r 1 - λ sin(λ - )θ,r 1 - λ cos(λ - )θ} l = 1 m é t ù ê N k (x) æ 11 n l ö 18 ç F l (x)b l k k = 1 ë è l = 1 ø λ=0.5 1 m h H( x) Heaviside a h m t F l ( x) n l b l k 3 J kε F l (x) 4 { F l ( )} r,θ 4 l = 1 = { } r sin θ, r cos θ, r sin θ sin θ, r sin θ cos θ H(ξ) ={ 1 ξ > 0-1 ξ 0 13. u j N j j 16 a k m { F l (r,θ),l = 1,,,6 } = ψ(x) k N k {r λ cos(λ + 1)θ cos θ,r λ sin(λ + 1)θ cos θ, N k k r λ cos(λ - 1)θ cos θ,r λ sin(λ - 1)θ cos θ, 17 9 u h ( x ) = u FE ( x ) + u H ( x ) + u tip ( x) 10 r λ cos(λ + 1)θ cos θ,r λ sin(λ + 1)θ sin θ} λ 5 u ( x) λ=0.5λ 17 u tip ( x) 17 1Sukumar [9] Huang [10] 4 Chang [18] J kε J k 1 é æ u ù J r = ( x - x tip ) + ( y - y tip ) θ = arctan[ (y - y tip )/(x - x tip )] - γ k = Γ lim 0 Γ êwn k - σ ij n i ö j ç ds k = 1, 19 ë è k ø (x, y) x (x tip, y tip ) γ W n i Γ x Heaviside H i Γ O r [18] 19 J 1 = lim 1 r 0 r λ - 1 E ( pk + qk ) I II 1 0 17
014 3 3 J = -lim r 0 r λ - 1 wk E I K II 1 1 0 1 p q w r 0 0 < λ < 0.5 J k 0.5 < λ < 1 J k 0 J ε r J kε [18] J 1ε = 1 ε λ - 1 ( pk I + qk II) E 1 ì u x = u x ε cos θ - u x sin θ u x y = u x ε sin θ + u x cos θ u y = u y ε cos θ - u y sin θ u y î y = u y ε sin θ + u y cos θ 7 J ε = - E 1 ε λ - 1 wk I K II 3 Chang [18] J kε é æ J 1ε = Γo êwn 1 - σ ij n j ç ë è u i öù ds + l1 ø + l + l 3 + l 4 é æ êwn 1 - σ ij n j ç ë è é æ u ù J ε = Γo êwn - σ ij n i ö j ç ds + ë è y s1 + s ø Wn ds u i öù ds ø 4 5 Fig. 3 3 p q w Calculation of p, q and w 4 4.1 p q w Chang [18] p q w p q w 15 u rr u θθ u x u y 6 Fig. J kε Integral paths of J kε ìσ xx = σ rr cos θ + σ θθ sin θ - σ rθ sin θ cos θ σ yy = σ rr sin θ + σ θθ cos θ + σ rθ sin θ cos θ îσ xy =(σ rr - σ θθ )sin θ cos θ + σ rθ (cos θ - sin θ) 6 3 ε N π/n ε = x + y θ = arctan( y x) ε/ = cos θ ε/ y = sin θ θ/ = -sin θ/ε θ/ y = cos θ/ε W W = 1 é u ù êσ x xx ë + σ æ u x ö xyç è y + u y u y + σ yy ø y u x J 1ε = Γ Wdy - Γ ε é ë ê ( σ xx n 1 + σ xy n ) + ( σ xyn ) 1 + σ yy n u x J ε = Γ Wdx - Γ ε é ë ê ( σ xx n 1 + σ xy n ) y + ( σ xyn ) 1 + σ yy n 8 u y ù dθ 9 u y ù y dθ 30 K I = 1 K II = 0 K I = 0 K II = 1 9 p qk I = 1 K II = 1 30 3 w 4. J kε J kε J 1ε 4 5 J kε Γ 4 4 J 1ε Fig. 4 Rectangular J 1ε integral path passes through gauss points 18
014 3 3 XFEM J 1ε n g J 1ε = δ( ξ g,η g ) I( ξ g,η g ) g = 1 ( ) 31 n g Γ ξ g,η g δ( ξ g,η g ) I( ξ g,η g ) I I(ξ g,η g ) = 1[ ] y σ xx ε xx + σ xy ε xy + σ yy ε yy η - é ù ê(σ xx n 1 + σ xy n ) u x ë +(σ xyn 1 + σ yy n ) u y æ ö ç + æ y ö è η ø è ç η ø J ε 3 4.3 p q w J 1ε J ε 3 K I K II [19] KˉI KˉI KˉI G G 1 v 1 v λ p σ θθ σ θθ Chen [13] Lin [1] θ p 0.007 0.30 0.35 0.0749 138.93 0.0195 0.019 0.0196 σ θθ θ = 0, σ θθ < 0 33 θ σ θθ 8 5 5.1 Matlab Beatxfem cm a =3 cm b=0.5 cm F=5 kn θ p D www.betaxfem.com 5 a =0.5 m b =1 m 800 6 I K II 0 p XFEM J 1ε K I Fig. 6 3.08 KˉI = K I σa 1 - λ 1 Chen [13] Lin [1] 5. 4 7 4 l=15 cm a 1= 7 4 33 θ =0 σ θθ θ p =0 Fig. 7 Geometry and loads of a 4-point bending bimaterial beam 6 Extended finite element mesh of the bimaterial plate Table 1 1 0.0433 0.30 0.35 0.1758 1.36 Normalized stress intensity factors 0.35 0.30 0.7110 0.7415 138.46 0.35 0.30 0.7335 0.6078 0.0946 4.360 4.980 0.0950 4.310 5.0010 0.0950 4.410 4.9780 Fig. 5 E =00 GPa v 1=v =0.3 E 1 10 Δa = 3 A elem A elem [5~7] E 1=E 33 θ p = arctan æ 5 ç Geometry and loads of a bimaterial plate under tension è K I - K I + 8K II 4K II ö ø 34 19
014 3 3 E /E 1=0.1 8 θ p=8 XFEM E /E 1 9 E /E 1 0.01 100 θ p 76 lg E /E 1 θ p = 10.45e -x x=lg E /E 1 9 a E/E1=0.1 b E/E1=10 10 Fig. 10 σ xx σ xx stress contours 8 Fig. 8 Extended finite element mesh and crack propagation path a E/E1=0.1 b E/E1=10 11 σ yy Fig. 11 σ yy stress contours a E/E1=0.1 9 Fig. 9 E /E 1=0.1 E /E 1=10 σ xx σ yy x 10~ 1 Fig. 1 11 11 F 1 Table Influence of Poisson's ratio on the crack E 1=E =00 GPa v =0.3 v 1 v 1 0. 0.5 0.3 lg E /E 1 Variations of crack propagation angle versus lg E /E 1 θ p/ 10.0 1 b E/E1=10 x Displacement contours in x direction propagation angle 10.04 10.0 0.35 10.09 0.4 10.05 0
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