Basic concepts and models in continuum damage mechanics

Transcription

1 Basic concepts and models in continuum damage mechanics Djimedo Kondo, Hélène Welemane, Fabrice Cormery To cite this version: Djimedo Kondo, Hélène Welemane, Fabrice Cormery. Basic concepts and models in continuum damage mechanics. Revue europã enne de gã nie civil, 2007, vol. 11, pp < /regc >. <hal > HAL Id: hal Submitted on 16 Jan 2013 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

2 Open Archive Toulouse Archive Ouverte (OATAO) OATAO is an open access repository that collects the work of Toulouse researchers and makes it freely available over the web where possible. This is an author-deposited version published in: Eprints ID: 6634 To link to this article: DOI: /regc To cite this version: Kondo, Djimédo and Welemane, Hélène and Cormery, Fabrice Basic concepts and models in continuum damage mechanics. (2007) Revue européenne de génie civil, vol. 11 (n 7-8). pp ISSN Any correspondence concerning this service should be sent to the repository administrator:

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4 S S D S D ω ω = S D S 1

5 S D = 0 ω = 0 ω S S D σ = F S σ = F F = S S D S(1 S D /S) = σ 1 ω σ σ = Eε E σ σ = (1 ω)eε ω ω = g(ε) g ω ε κ t κ(t) = max ε(τ) τ t ω = g(κ) g f(ε,κ) = ε κ

6 f 0; κ 0; κf = 0 f 0 ε κ κ κ κ σ = E s ε = (1 ω)eε E s = (1 ω)e D = 1 Es E ω = g(κ) f f(ε,κ) = ε κ D D = (1 ω)d 0 D 0 ω

7 σ = (1 ω)d 0 : ε σ σ = (1 ω) σ f f f(ε,κ) = ε eq (ε) κ ε eq ε ε eq = ε : ε ε eq = ε : D 0 : ε ε ε eq ω t ω c ω t ω c g t g c ε eq = 3 < ε I > 2 I=1

8 <. > ε I < ε I > ε I ω = α t ω t + α c ω c α t α c α t = 3 [ ] < ε t β I >< ε I > ; α c = I=1 ε 2 eq 3 [ ] < ε c β I >< ε I > α t = 1; α c = 0 ω = ω t α t = 0; α c = 1 ω = ω c I=1 ε 2 eq { 0 κ κ0 ω t = g t (κ) = 1 (1 At)κ0 κ A t exp[ B t (κ κ 0 )] κ κ 0 { 0 κ κ0 ω c = g c (κ) = 1 (1 Ac)κ0 κ A c exp[ B c (κ κ 0 )] κ κ 0 κ 0 = ft E f t E κ 0 A t B t A c B c β β

9 ω ω ω = Na 3 N a

10 ω = N i ηs 3/2 i S i η n Ò a 2c ǫ n D n n D = i ω i n i n i ω i n i i ε ω W(ε,ω) = 1 2 ε : D(ω) : ε

11 D(ω) ω = 0 D(ω) σ(ε,ω) σ(ε,ω) = W ε (ε,ω), F ω (ε,ω) F ω (ε,ω) = W ω (ε,ω) = 1 2 ε : D (ω) : ε D = D ω F ω D D = σ : ǫ Ẇ = W ω ω F ω = W ω f f(f ω,ω) = F ω κ(ω) κ ω κ ω κ(ω) = κ 0 (1 + ηω) κ 0 > 0 η > 0.

12 f ω = Λ F ω (F ω,ω) = Λ Λ = 0 f(f ω,ω) 0 f(f ω,ω) < 0 Λ > 0 f(f ω,ω) = 0 f(f ω,ω) = 0 Λ σ = L : ε L D(ω) L = D(ω) F ω ε F ω ε h f(f ω,ω) 0 f(f ω,ω) < 0 f(f ω,ω) = 0 f(f ω,ω) = 0 D(ω) h = κ 0 η F ω d D(ω) D(ω) D(ω) = 3K(ω)J + 2µ(ω)K. (K,µ) J K J = K = I J 1 I D(ω) D(ω)

13 ) 48(1 ν K(ω) = K 0 (1 2 0)ω 27(1 2ν 0 ) + 16ω(1 + ν 0 ) 2 ) 480(1 ν 0 )(5 ν 0 )ω µ(ω) = µ 0 (1 675(2 ν 0 ) + 64ω(4 5ν 0 )(5 ν 0 ) ν 0 D W ε D (ε,d) tr(ε),tr(ε 2 ),tr(ε 3 ),tr(d),tr(d 2 ),tr(d 3 ) tr(ε.d),tr(ε 2.D),tr(ε.D 2 ),tr(ε 2.D 2 ) D = 0 D W ε 2

14 W D W W = 1 2 [λ(tr(ε))2 + 2µtr(ε 2 )]+ 1 2 [ηtr(d)(tr(ε))2 + γtr(d)tr(ε 2 )] + αtr(ε)tr(ε.d) + χtr(ε 2.D) α β η γ tr(d) σ W σ = W ε = D(D) : ε D(D) = D 0 + α(1 D + D 1) + χ(d D) + η tr(d)(1 1) + γ tr(d)(1 1) D 0 = 3k 0 J + 2µ 0 K D F D = W D D D f(f D,D) = F D (a 0 + a 1 tr(d)). a 0 a 1 a 0 a 1 D Ḋ = { 0, f < 0 f = 0 f < 0 Λ f F D, f = 0 f = 0

15 Λ Λ = tr(f D Ḟ D ) a 1 tr(f D ) ω D o (ω) D cl (ω) Γ ε Γ(ε) > 0 Γ(ε) 0 D(ω) = { D o (ω) = 3K o (ω)j + 2µ o (ω)k Γ(ε) > 0 D cl (ω) = 3K cl (ω)j + 2µ cl (ω)k Γ(ε) 0 (K o,µ o ) (K cl,µ cl ) 3

16 (K o,µ o ) (K cl,µ cl ) W C 1 [D(ω)] =D o (ω) D cl (ω) = s(ω) Γ Γ (ε) (ε) ε Γ(ε) = 0 ε ε s ω D o D cl [D(ω)] = 3[K o (ω) K cl (ω)]j + 2[µ o (ω) µ cl (ω)]k [D(ω)] ω,µ o (ω) = µ cl (ω) W µ(ω) = µ o (ω) = µ cl (ω) K 0 K cl (ω) = K 0 K o (ω) = K(ω) Γ [D(ω)] [D(ω)] = 3[K o (ω) K cl (ω)]j = [K o (ω) K cl (ω)]1 1

17 Γ(ε) = tr(ε) = 0 W(ε,ω) = 1 { 3K(ω)J + 2µ(ω)K tr(ε) > 0 2 ε : D(ω) : ε D(ω) = 3K 0 J + 2µ(ω)K tr(ε) 0 K(ω) µ(ω) σ = L : ε e ε L L = 3k 1 J + 2k 2 K 2k 3 (1 e + e 1) 2k 4 e e k 2 = µ(ω) k 4 = 2(µ (ω)) 2 h { k1 = K(ω) H 1 k 3 = K(ω) H 3 tr(ε) > 0 k 1 = K 0 k 3 = 0 tr(ε) 0 H 1 = (K (ω)trε) 2 H 3 = µ (ω)k (ω) h h h h = H 0 η+k (ω)tr(ε) 1+µ (ω)e K (ω) µ (ω) K(ω) µ(ω) K (ω) µ (ω) K(ω) µ(ω)

18 K(ω) µ(ω)

19