Vol. 8 (2005 8 ) Finite Element of Composite aterials with Partial Debonding along Boundary Kohei HIGUCHI, Tetsuo IWAKUA JFE PhD 980-8579 6-6-06 Damages in composite materials are mostly governed by interfacial debonding. In order to take this into account, we employ an analytical model considering partial debonding along interface, and formulate a finite element. The model utilizes a concept where a moduli of an inclusion is replaced by an equivalent one of debonded inclusions. By several examples, eligibility of the model is shown. As an application, excavation analysis of a tunnel in jointed rock mass is carried out, and the results are compared with other observations. Key Words : composites, partial debonding, ori-tanaka theory, finite element 1. FRP, FRC 1) 2) 3 3),4) Zhao and Weng 2 Voigt Reuss 5) Eshelby 6) 7) 10),11) Eshelby 12) Hashin- Shtrikman 8) 9) Zhao and Weng
J 2 Eshelby 2. 2.1 2.2 Zhao and Weng 11),12) Hill short-hand format 13) (x 1, x 2, x 3 ) x 1 C I C i jkl σ, ε I C i jkl Voigt σ 11 = C 11 ε 11 + C 12 ε 22 + C 13 ε 33 σ 22 = C 21 ε 11 + C 22 ε 22 + C 23 ε 33 (1) σ 33 = C 31 ε 11 + C 32 ε 22 + C 33 ε 33 σ 23 = C 44 ε 23, σ 31 = C 55 ε 31, σ 12 = C 66 ε 12 x 1 σ 11 = σ 31 = σ 12 = 0 (2) (1) (2) Eshelby σ 11 = 0 ( σ 22 = C 22 C ) ( 21C 12 ε 22 + C 23 C 21C 13 σ 33 = C 11 ( C 32 C 31C 12 C 11 C 11 ) ( ε 22 + C 33 C 31C 13 C 11 ) ε 33 ) ε 33 σ 23 = C 44 ε 23, σ 31 = 0, σ 12 = 0 (3) σ = C d I ε (4) C d I C d I 10),11) 14)
(C C I ) C 1 [ I + f C (S I) {C (C C I ) S} 1 Zhao and Weng D [ I + f C (S I) {C (C C I ) S} 1 (C C I ) C 1 1 (7c) [ f C (S I) {C (C C I ) S} 1 C I E [ C (S I) {C (C C I ) S} 1 C I [ I + C (S I) {C (C C I ) S} 1 (C C I ) C 1 [ 3. I + f C (S I) {C (C C I ) S} 1 (C C I ) C 1 1 (7d) [ f C (S I) {C (C C I ) S} 1 C I 3.1 J 2 15) 15) C I (4) C d I (7) σ = A σ + D ε p (5a) σ I = B σ E ε p (5b) 16) ε = C 1 σ + F ε p (5c) 3.2 σ σ I von ises σ, ε ε p f = (J 2 ) F (ε) p (8) C C 1 = [ I + f S {C (C C I ) S} 1 (C C I ) [ C + f C (S I) {C (C C I ) S} 1 (C C I ) 1 (6) ε p C C I S Eshelby I f 4 A, B, D, E, F A [ I + f C (S I) {C (C C I ) S} 1 (C C I ) C 1 1 (7a) B [ I + C (S I) {C (C C I ) S} 1 (C C I ) C 1 1 (7b) F I [ f S {C (C C I ) S} 1 C I + [ I + f S {C (C C I ) S} 1 (C C I ) [ C + f C (S I) {C (C C I ) S} 1 (C C I ) 1 [ f C (S I) {C (C C I ) S} 1 C I (7e) (J2 ) (J2 ) 1 2 σ : σ (9) F = 1 3 σ Y + h (ε eq ) n (10) σ Y h n ε eq 2ε p : ε p (11)
1 m x 2 x 1 P ε p = 1 H σ σ 4(J 2 ) σ (12) 5 m H 1 H F σ F ε p (13) (10) (10) (13) H = 2(n/2) 1 n h ( ε p : ε p (n /2) 1 ( ) ) σ : ε p 3(J2 ) (14) 4 3.3 e (12) (5a) K e ḋ e = Ḟ e (20) ε p = 1 H σ σ 4 (J 2 ) A σ + 1 H σ σ 4 (J 2 ) D ε p (15) ε p = { I 1 σ σ } 1 1 σ σ D A H 4 (J 2 ) H 4 (J 2 ) σ (16) (5c) ε = C 1 σ + F {I 1 σ σ } 1 1 σ σ D A H 4 (J 2 ) H 4 (J 2 ) σ (17) { σ = C 1 + F I 1 σ σ } 1 1 σ σ 1 D A ε H 4 (J 2 ) H 4 (J 2 ) (18) σ = X ε (19) X 1 1 m 5 m 100 3.4 20 2000 (19) 2) 10) ḋ e Ḟ e K e K e = ( N e ) T X( N e )da (21) Ω e N e X (19) X 4. 4.1 (1) 3.0 10 3 N/ 17) 2124Al SiC
P (N) 30 (a) σ d = 100Pa 20 (b) σ d = 200Pa 10 σ d = 100 Pa σ d = 200 Pa σ d = 300 Pa (c) σ d = 300Pa 0 0 0.2 0.4 v (m) 3 2 (P = 24 N) 20% 2124Al E = 60 GPa ν = 0.3 290 Pa σ Y = 290 Pa h = 700 Pa n = 0.55 SiC E I = 450 GPa ν I = 0.2 14) σ d = 100, 200, 300 Pa 3 3 P < 5 N 1:5 (2) σ d = 100, 200, 300 Pa 3 P = 24.0 N 2 3 σ d = 100, 200, 300 Pa σ d = 100 Pa σ Y = σ d = 300 Pa P > 20 N (3) SiC 17) 1:1:5 1:5 σ d = 200 Pa
P (N) 30 (a) 20 (b) 10 (c) 4 (P = 24 N) 4 (a) P = 24 N 2.2 5 2 (b) 4.2 1 4 (b) 4 (a) 2000 20 2 4 4 (c) 2 2 1 0 0 0.2 0.4 v (m) 5 5 (1) 18) 2024Al 34% 2024Al E = 55 GPa, ν = 0.32 σ Y = 80 Pah = 827 Pa, n = 0.60 E I = 380 GPa ν I = 0.2 σ d = 50, 100, 150 Pa 3 = 150 Pa 1:5 σ d
q 10 cm 8-a q = 48 N/m 8-b q = 64 N/m r r = 1 cm 10 cm 8-c q = 80 N/m 8-d q = 104 N/m q 6 8 9-a q = 48 N/m 9-b q = 64 N/m 7 1024 9-c q = 80 N/m 9-d q = 96 N/m 6 20 cm 20 cm 1 cm q 1024 7 q = 0.016 N/m/ σ d = 150 Pa 8 q = 48, 64, 80, 104 N/m 9 q = 48 N/m q = 64 N/m q = 80 N/m q = 104 N/m
(2) 1:5 9 q = 48, 64 N/m 4 m 2 m 2 m 4 m σ v σ h σ v 10 σ h 5. 19) 5.2 BC 20) 21) 22) 5.1 4 m4 m 2 m200 m 40 m 40 m 10 11 1680 11 10 E = 20 GPa ν = 0.20 1:10 20% 2 Pa 2500 kg/m 3 σ v = 5 Pa 0.8 σ h = 4 Pa 60 30 12
0.3mm 0.2mm 0.1mm 0.3mm 0.2mm 0.1mm 60 60 60 13 (2005/02/09) 30 12 5.3 22) BC BC BC E = 20 GPa ν = 0.25 σ d = 15 Pa 60 60 60 60 13, 14 60 60 15, 16 60 60 60 14 60 6. 30 Pa (C) 15560391
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