Jacques Besson Centre des Matériaux UMR 7633 Mines ParisTech PSL Research University Institut Mines Télécom Aγνωστ oς Θεoς
Outline 1 Some definitions 2 in a linear elastic material 3 in a plastic material 4 Fracture criteria 5 Evaluation of fracture parameters: K, T, J, Q 2/60
Part I Some definitions 3/60
Small scale yielding Large scale yielding F large scale yielding small scale yielding u 4/60
Some specimen types CT / Compact tension 5/60
SENB / Single Edge Nocthed Bending 6/60
SENT / Single Edge Notched Tensile 7/60
CCP / Center Crack Panel 8/60
Part II in a linear elastic material 9/60
in a linear elastic isotropic material Geometry y M n z r θ crack T = σ. n = 0 x Material properties E ν 10/60
Crack opening modes mode I mode II mode III opening in-plane shear out-of-plane shear 11/60
Stresses Solution obtained using complex Airy s function Stresses vary asi (first term of a Taylor expansion): σ ij = K X 2πr σ X ij (θ), X = I, II, III K X : stress intensity factor (MPa m) lim r 0 σ ij = Energy is bounded e = σ ij εij = ẽ(θ)/r E = L erdrdθ = L 0 0 ẽ(θ)drdθ < 12/60
Stress field σ xx = σ yy = σ xy = K I cos θ ( 1 sin θ 3θ sin 2πr 2 2 2 K I cos θ ( 1 + sin θ 3θ sin 2πr 2 2 2 ) K I cos θ 2πr 2 sin θ 3θ cos 2 2 + K II cos θ 2πr 2 σ xz = K III 2πr sin θ 2 K II sin θ ( 2 + cos θ ) 3θ cos 2πr 2 2 2 ) + K II sin θ 2πr 2 cos θ 3θ cos 2 2 ) ( 1 sin θ 3θ sin 2 2 σ yz = σ zz = K III cos θ 2πr 2 { ν(σ xx + σ yy) plane strain 0 plane stress 13/60
Displacements Displacemens vary as: u i = K X r 2π ũx (θ), X = I, II, III u x = K I r 2µ 2π cos θ 2 u y = K I r 2µ 2π sin θ 2 u z = 2 K III r µ 2π sin θ 2 κ = ( κ 1 + 2 sin 2 θ 2 ( κ + 1 2 cos 2 θ 2 { 3 4ν plane strain (3 ν)/(1 + ν) plane stress ) + K II 2µ ) K II 2µ r 2π sin θ 2 r 2π cos θ 2 ( κ + 1 + 2 cos 2 θ 2 ( κ 1 2 sin 2 θ 2 ) ) 14/60
Energy release rate G F 1/S(a) GBda 1/S(a + da) u Determination of the the energy release rate (A = Ba) G = W A = 1 2 F 2 S A 15/60
Energy release rate and stress intensity factor (mode I) Small crack advance a σ yy y w = 1 2 σ yyu y σ θ=0 yy 1/ x u y x a u θ=π y a x x x = 0... a 16/60
Work of separation dw : Released energy with and so that Mixed mode case dw = wdx = 1 2 σyyuydx = K I 2 2µ a x I = x G a = a 0 κ + 1 a x dx 2π x wdx dx = x( a x) + a 2 arcsin I(x = a) I(x = 0) = π 2 a G = K I 2 (κ + 1) = 8µ KI 2 (1 ν 2 ) E K 2 I E plane strain plane stress G = (K I 2 + KII 2 )(1 ν 2 ) + 1 + ν E E K III 2 ( ) 2x a a 17/60
Plastic zone size Small scale yielding Calculation of the von Mises stress σ eq = K cos 2 (θ/2) (4 4ν + 4ν 2 3 cos 2 (θ/2)) = K σ eq(θ, ν) 2πr 2πr for ν = 0 or plane stress for ν = 1 2 R p(θ = 0) = 1 ( ) 2 K 2π σ 0 R p(θ = 0) = 0 18/60
Iso-values: σ eq = σ 0 for ν = 0., 0.1, 0.2, 0.3, 0.4, 0.5 0.6 0.4 0.2 0.5 0.2 0.1 0.0 πyσ 2 0/K 2 I 0 0.2 crack 0.3 0.4 0.4 0.6 0.6 0.4 0.2 0 0.2 0.4 0.6 πxσ 2 0/K 2 I 19/60
Irwin s correction Restore equilibrium when plasticity occurs σ yy σ 0 X x 20/60
Equilibrium With the condition after some maths 0 K 2πx dx = R 0 K σ 0 dx + dx R 2π(x + X) K 2π(R + X) = σ 0 X = 1 ( ) 2 K R = 1 ( ) 2 K = 2R p 2π σ 0 π σ 0 Plane strain case R 1 ( ) 2 K 3π σ 0 21/60
Validity of the toughness measurement tests according to ASTM E399 from the E399 ASTM standard so that (i.e. small scale yielding) ( ) 2 K 2.5 25R PE σ 0 R W a 22/60
FE evaluation of the plastic zone size Specific mesh ; loading using the K field solution ux = K r 2µ 2π cos θ 2 uy = K r 2µ 2π sin θ 2 ( k 1 + 2 sin 2 θ ) 2 ( k + 1 2 cos 2 θ ) 2 3 4ν plane strain k = 3 ν 1 + ν plane stress ρ J ρoy (x, y) θ Ox crack tip region L R p 23/60
Calculation for a hardening exponent equal to N = 10 ε = σ 0 E ( σ σ 0 ) N ε p < 0.2% Plane strain Plane stress 24/60
T stress Taylor expansion σ ij = K X 2πr σ ij (θ) + T δ i1 δ 1j + O(r 1/2 ) The T stress depends on the geometry of the specimens Non-dimensionnal parameter πat β = K I 25/60
Effect of the T stress of the plastic zone plastic zone for a fixed K I and various T /σ 0 values T/σ0 = +0.3 T/σ0 = +0.6 T/σ0 = 0 T/σ0 = 0.3 T/σ0 = 0.6 26/60
Part III in a plastic material 27/60
J integral Contour integral which does not depend on the contour Γ ( J = wdx 2 T. u ) x ds where Γ w = σ ij dε ij ds T = σ. n n x 2 crack Γ x 1 28/60
Proof that the J-integral is path independent (1/2) First prove that J = 0 over a closed path (Γ) ( J = wdx 2 T. u ) ( ) ( ) u k u k ds = wn 1 n j σ jk ds = wδ 1j σ jk n j ds Γ x 1 Γ x 1 Γ x 1 Using the divergence theorem: f j n j ds = J = A ( wδ 1j σ jk x j u k x Γ ) da = A A f j x j da ( w σ kj u k σ kj x 1 x j x 1 2 u k x 1 x j ) da since σ kj / x j = 0 (equilibrium σ = 0) and 2 u k / x 1 x j = ε kj / x 1 ( ) w ε kj J = σ kj da x 1 x 1 however σ kj = w/ ε kj so that σ kj ε kj x 1 A = w ε kj ε kj x 1 = w x 1 J = 0 29/60
Proof that the J-integral is path independent (2/2) Consider the following closed path Γ = Γ 1 Γ + Γ 2 Γ Γ + Γ Γ 2 Γ 1 x 2 x 1 One has (see above) 0 = J 1 + J + J 2 J so that J 1 = J 2 for any Γ 1 and Γ 2 which proves the path-independence. 30/60
Energy release rate Nonlinear material F a a + da JBda J = 1 B du da U = Pdu u 31/60
The HRR field HRR = Hutchinson+Rice+Rosengren (1968) Non linear material Small strain analysis Mode I crack ( ) N ε σ = ε 0 σ 0 32/60
Form of the HRR field for stresses ( J σ ij = σ 0 σ 0 ε 0 I N r Form of the HRR field for deformations Displacements N = 1 linear elastic case ( J ε ij = ε 0 σ 0 ε 0 I N r ( J u i = ε 0 σ 0 ε 0 I N r ) 1 N + 1 σij (θ, N) ) N N + 1 εij (θ, N) ) N N + 1 r 1 N + 1 ũ i (θ, N) 33/60
No analytical forms for σ ij (θ, N), ε ij (θ, N) and ũ i (θ, N) σ ij, ε ij and ũ i (θ, N) are obtained numerically Tables # n=10 In=4.54041 # theta Seq Srr Stt Srt Err Ett Ert Ur Ut 0. 0.66908 1.72433 2.49692-0.00000-0.01557 0.01557-0.00000-0.17129 0.00000 1. 0.66913 1.72469 2.49665 0.01622-0.01557 0.01557 0.00065-0.17125 0.00326 2. 0.66926 1.72576 2.49584 0.03241-0.01556 0.01556 0.00131-0.17114 0.00652 3. 0.66949 1.72756 2.49449 0.04857-0.01554 0.01554 0.00197-0.17096 0.00978 4. 0.66980 1.73007 2.49260 0.06466-0.01552 0.01552 0.00263-0.17069 0.01303 5. 0.67021 1.73330 2.49018 0.08067-0.01549 0.01549 0.00330-0.17036 0.01628 6. 0.67071 1.73723 2.48723 0.09657-0.01545 0.01545 0.00398-0.16995 0.01952 7. 0.67131 1.74187 2.48375 0.11236-0.01541 0.01541 0.00467-0.16946 0.02275 8. 0.67201 1.74722 2.47974 0.12799-0.01535 0.01535 0.00537-0.16890 0.02597 9. 0.67281 1.75325 2.47522 0.14346-0.01530 0.01530 0.00608-0.16826 0.02918 10. 0.67372 1.75998 2.47018 0.15875-0.01523 0.01523 0.00681-0.16755 0.03238 11. 0.67474 1.76739 2.46464 0.17383-0.01516 0.01516 0.00756-0.16676 0.03556 12. 0.67589 1.77546 2.45860 0.18869-0.01508 0.01508 0.00833-0.16590 0.03873 13. 0.67716 1.78420 2.45207 0.20331-0.01500 0.01500 0.00913-0.16495 0.04187 14. 0.67856 1.79358 2.44505 0.21766-0.01490 0.01490 0.00996-0.16393 0.04501 15. 0.68011 1.80359 2.43757 0.23174-0.01480 0.01480 0.01082-0.16283 0.04812 16. 0.68181 1.81422 2.42961 0.24551-0.01470 0.01470 0.01173-0.16165 0.05121 17. 0.68367 1.82543 2.42121 0.25897-0.01458 0.01458 0.01268-0.16038 0.05427 18. 0.68571 1.83722 2.41236 0.27210-0.01446 0.01446 0.01368-0.15904 0.05731 19. 0.68794 1.84953 2.40308 0.28487-0.01433 0.01433 0.01475-0.15761 0.06033 20. 0.69038 1.86235 2.39338 0.29728-0.01419 0.01419 0.01589-0.15609 0.06331 21. 0.69305 1.87561 2.38327 0.30931-0.01404 0.01404 0.01711-0.15449 0.06627 22. 0.69596 1.88927 2.37277 0.32095-0.01389 0.01389 0.01844-0.15280 0.06920 23. 0.69914 1.90325 2.36189 0.33218-0.01373 0.01373 0.01989-0.15101 0.07209... 34/60
Crack tip opening displacement (CTOD), crack blunting blunt while loaded CTOD Evaluation of the CTOD (α close to 1) CTOD = α J = α K I 2 σ 0 Eσ 0 Some expressions for α (McMeeking, 1977) α = 0.55 α = 0.54(1 + 1/N) α = 0.55 ( 2 3 (1 + ν)(1 + N)σ 0/E ) 1/N 35/60
T stress, Q factor and stress triaxiality T stress : +T δ i1 δ 1j Q factor : +Qσ 0 δ ij Stress triaxiality τ = 1 σ kk 3 σ eq Faster damage if T > 0, Q > 0 and high τ 36/60
Q factor Modification of the HRR field (O Dowd & Shih, 1991, 1992) ( J σ ij = σ 0 σ 0 ε 0 I N r ) 1 N + 1 σij (θ, N) + Qσ 0 δ ij Q (a non-dimensionnal factor) corresponds to an hydrostatic pressure 37/60
Part IV Fracture criteria 38/60
Fracture criterion : LEFM The released energy is used to create the crack (Griffith, 1920) G = G c = 2γ s so that the critical stress intensity factor is given by: K Ic = E plane stress 2γ se with E = E plane strain 1 ν 2 so that the failure criterion is given by K I = K Ic 39/60
Which value for G c? Surface energy: γ s 1 J/m 2 For steel (E = 200 GPa, ν = 0.3) K Ic = 2γ se = 0.7 MPa m Plasticity at the crack tip increases the apparent surface energy γ s γ s + γ p For K Ic = 50MPa m, γ p = 6000 J/m 2 γ s 40/60
Fracture criterion : NLFM J as a fracture parameter Does it work? J = J c K J = JE 41/60
Test on SENB specimens with various a/w ratios 300 250 data fit JIc (kj/m 2 ) 200 150 100 W a 50 0 0.0 0.1 0.2 0.3 0.4 a/w 0.5 0.6 0.7 0.8 SENB = Single Edge Notched Bending 42/60
J Q approach Different specimens and a/w ratios 250 200 CCP a/w = 0.63 CCP a/w = 0.77 SENB 0.05 < a/w < 0.78 fit JIc (kj/m 2 ) 150 100 50 0-1.5-1.0 Q -0.5 0.0 43/60
Effect of specimen geometry on J a curves 2000 1500 J (kj/m 2 ) 1000 SENT CT 500 X100 steel 0 0.0 1.0 2.0 3.0 a (mm) 4.0 5.0 44/60
Part V Evaluation of fracture parameters: K, T, J, Q 45/60
Calculation of K Smoothing of stresses σ yy(θ = 0) = K ( ) I KI log(σ yy) = log 1 2πr 2π 2 log(r) 6 5 log(k I / (2π)) 4 3 log(σyy) 2 1 0-1 -2-3 -2 0 2 4 6 8 10 12 14 log(x) 46/60
Smoothing of displacements u y(θ = π) = 4K I(1 ν 2 ) E 2π 1 0-1 -2 r log(uy) = log ( 4KI (1 ν 2 ) E 2π ) + 1 2 log(r) log(uy) -3-4 -5-6 ( log 4K I (1 ν 2 )/E ) (2π) -7-2 0 2 4 6 8 10 12 14 log(x) 47/60
Calculation of J and K Contour integral J = Γ ( wdx 2 T. u ) ds x 1 T = σ. n n ds x2 crack Γ x1 In fact, it is difficult to evaluate the contour integral based on FE results 48/60
Virtual crack extension method a: virtual crack extension Γ e Γ i y crack a x Γ i is translated by a Γ e is fixed Interpolation for nodes between Γ i and Γ e : x Replace the contour integral by a volume integral (Delorenzi (1985)) G = J = 1 ( σ. u ) a Ω x w1. x x dω = K 2 E 49/60
Is J path-independent? (1/2) Crack tip mesh (SENT specimen) Use of von Mises plasicity σ eq R(p) = 0 ε p = 3 2 ṗ s σ eq ε = ε e + ε p σ = E : ε e or use of non-linear elasticity σ = E : ( ε 3 2 R 1 (σ s ) eq) σ eq 50/60
Is J path-independent? (2/2) Results von Mises Non-linear elasticity 51/60
Calculation of T Simple method (σ yy σ xx) T as x 0 Contour integral T = E ( u.σ u.σ). n dc where σ and u are solutions ; σ and u are given by C σ xx = σ yy = σ xy = cos 2θ + cos 4θ 2πr 2 cos 2θ cos 4θ 2πr 2 sin 4θ 2πr 2 ux = 1 κ cos θ + cos 3θ 4πr 2G u y = 1 4πr κ sin θ + sin θ 2G Based on FE results T = E C ( u FE.σ u.σ FE ). n dc 52/60
To avoid contour integrals and use volume integral T = E (( u FE u tip FE ).σ u.σ FE ). gradq dω Ω 1.0 0.5 0.0 1.0 0.0 1.0 0.5 0.0 53/60
Example 54/60
Calculation of Q HRR r c J/σ 0 ρ 0 ρ ρ/ρ 0 > 5 finite strain zone 2r c Qσ 0 Q = σfe yy σ HRR yy σ 0 at θ = 0 and x = 2r c (O Dowd & Shih, 1991,1992) 55/60
Meshes 56/60
local fields (ρ 0 = 2µm ρ = 50µm) 1700 1 1600 Stress (MPa) 1500 1400 1300 1200 Cumulated plastic strain 0.1 0.01 1100 1000 0 0.2 0.4 0.6 0.8 1 0.001 0 0.2 0.4 0.6 0.8 1 Distance (mm) Distance (mm) 57/60
Element formulation (large plastic strains, plastic incompressibility) 58/60
Application ux = K r 2µ 2π cos θ 2 uy = K r 2µ 2π sin θ 2 ( k 1 + 2 sin 2 θ ) 2 ( k + 1 2 cos 2 θ ) 2 3 4ν plane strain k = 3 ν 1 + ν plane stress J 0 1 2 J t ρ L Interface ρoy (x, y) θ Ox crack tip region R stress triaxiality plastic strain 59/60
Characteristic lengths Plastic zone size Process zone size Relative importance R p = α 1 ( K σ 0 ) 2 α 1 = 1 3π... 1 π R PZ = α 2 K 2 Eσ 0 α 2 1 2 R PZ R p σ 0 E 1 R p R PZ crack 60/60
J. R. Rice. A path independent integral and the approximate analysis of strain concentration by notched and cracks. J. Applied Mech., 35:379, 1968. J. R. Rice and G. F. Rosengren. Plane strain deformation near a crack tip in a power-law hardening material. J. Mech. Phys. Solids, 16:1 12, 1968. J. W. Hutchinson. Plastic stress and strain fields at a crack tip. J. Mech. Phys. Solids, 16:337 342, 1968. R.M. McMeeking. Finite deformation analysis of crack-tip opening in elastic-plastic materials and implications for fracture. J. Mech. Phys. Solids, 25:357 381, 1977. C.F. Shih. Tables of Hutchinson Rice Rosengren singular field quantities. Technical report, MRL E-147, Brown University, 1983. N.P. O Dowd and C.F. Shih. Family of crack-tip fields characterized by a triaxiality parameter I. Structure of Fields. J. Mech. Phys. Solids, 39(8):989 1015, 1991. 60/60
N.P. O Dowd and C.F. Shih. Family of crack-tip fields characterized by a triaxiality parameter II. Fracture applications. J. Mech. Phys. Solids, 40(8):939 963, 1992. H.G. Delorenzi. Energy release rate calculations by the finite element method. Eng. Fract. Mech., 21(1):129 143, 1985. C.S. Chen, R. Krause, R.G. Pettit, L. Banks-Sills, and A.R. Ingraffea. Numerical assessment of T stress computation using a p version finite element method. Int. J. Frac., 107:177 199, 2001.