FRACTURE MECHANICS. Piet Schreurs. Eindhoven University of Technology Department of Mechanical Engineering Materials Technology September 5, 2013

Transcript

1 FRACTURE MECHANICS Piet Schreurs Eindhoven University of Technology Department of Mechanical Engineering Materials Technology September 5, 2013

2 INDEX back to index () 1 / 303

3 Introduction Fracture mechanisms Ductile/brittle Theoretical strength Experimental techniques Energy balance Linear elastic stress analysis Crack tip stresses Multi-mode loading Crack growth direction Crack growth rate Plastic crack tip zone Nonlinear fracture mechanics Numerical fracture mechanics Fatigue Engineering plastics () 2 / 303

4 () 3 / 303

5 back to index INTRODUCTION

6 Introduction () 5 / 303

7 Continuum mechanics V 0 A 0 u V A x 0 e 3 x e 1 O e 2 () 6 / 303

8 Continuum mechanics - volume / area V 0, V / A 0, A - base vectors { e 1, e 2, e 3 } - position vector x 0, x - displacement vector u - strains ε kl = 1 2 (u k,l + u l,k ) - compatibility relations - equilibrium equations σ ij,j + ρq i = 0 ; σ ij = σ ji - density ρ - load/mass q i - boundary conditions p i = σ ij n j - material model σ ij = N ij (ε kl ) () 7 / 303

9 Material behavior σ ε σ ε t 1 t 2 t t 1 t 2 t t 1 t 2 t t 1 t 2 t σ ε ε e ε p σ ε t 1 t 2 t t 1 t 2 t t 1 t 2 t t 1 t 2 t () 8 / 303

10 Stress-strain curves σ σ ε ε σ σ ε ε () 9 / 303

11 Fracture () 10 / 303

12 Fracture mechanics questions : when crack growth? ( crack growth criteria) crack growth rate? residual strength? life time? inspection frequency? repair required? fields of science : material science and chemistry theoretical and numerical mathematics experimental and theoretical mechanics () 11 / 303

13 Overview of fracture mechanics LEFM (Linear Elastic Fracture Mechanics) energy balance crack tip stresses SSY (Small Scale Yielding) DFM (Dynamic Fracture Mechanics) NLFM (Non-Linear Fracture Mechanics) EPFM (Elasto-Plastic Fracture Mechanics) Numerical methods : EEM / BEM Fatigue (HCF / LCF) CDM (Continuum Damage Mechanics) Micro mechanics micro-cracks (intra grain) voids (intra grain) cavities at grain boundaries rupture & disentangling of molecules rupture of atomic bonds dislocation slip () 12 / 303

14 Experimental fracture mechanics () 13 / 303

15 Linear elastic fracture mechanics () 14 / 303

16 Dynamic fracture mechanics () 15 / 303

17 Nonlinear fracture mechanics CTOD J-integral () 16 / 303

18 Numerical techniques () 17 / 303

19 Fatigue () 18 / 303

20 Objectives Insight in : crack growth mechanisms brittle / ductile energy balance crack tip stresses crack growth direction plastic crack tip zone crack growth speed nonlinear fracture mechanics numerical methods fatigue () 19 / 303

21 Outline () 20 / 303

22 back to index FRACTURE MECHANISMS

23 Fracture mechanisms shear fracture cleavage fracture fatigue fracture crazing de-adhesion () 22 / 303

24 Shearing dislocations voids crack dimples load direction () 23 / 303

25 Cleavage intra-granulair inter-granulair intra-granular HCP-, BCC-crystal T low ε high 3D-stress state inter-granular weak grain boundary environment (H2) T high () 24 / 303

26 Fatigue clam shell pattern striations () 25 / 303

27 Crazing stress whitening crazing materials : PS, PMMA () 26 / 303

28 () 27 / 303

29 back to index DUCTILE/BRITTLE

30 Ductile - brittle behavior σ ABS, nylon, PC PE, PTFE ε (%) surface energy : γ [Jm 2 ] solids : γ 1 [Jm 2 ] independent from cleavage/shearing ex.: alloyed steels; rubber () 29 / 303

31 Charpy v-notch test () 30 / 303

32 Charpy Cv-value C v fcc (hcp) metals low strength bcc metals Be, Zn, ceramics C v high strength metals Al, Ti alloys T NDT FATT FTP T t T - Impact Toughness C v - Nil Ductility Temperature NDT - Nil Fracture Appearance Transition Temperature FATT(T t ) - Nil Fracture Transition Plastic FTP () 31 / 303

33 back to index THEORETICAL STRENGTH

34 Theoretical strength f f r f x σ a λ x r S ( ) 2πx f (x) = f max sin ; x = r a 0 λ σ(x) = 1 ( ) 2πx f (x) = σmax sin S λ () 33 / 303

35 Energy balance available elastic energy per surface-unity [N m 1 ] required surface energy U i = 1 S = x=λ/2 x=0 x=λ/2 x=0 = σ max λ π f (x) dx σ max sin ( ) 2πx dx λ [Nm 1 ] energy balance at fracture U a = 2γ [Nm 1 ] U i = U a λ = 2πγ σ = σ max sin ( ) x γ σ max σ max () 34 / 303

36 Approximations linearization ( ) x σ = σ max sin γ σ max x γ σ2 max linear strain of atomic bond ε = x a 0 x = εa 0 σ = εa 0 γ σ2 max elastic modulus ( ) ( dσ dσ E = = 0) dε x=0 dx x=0 a = σ 2 max Eγ σ max = theoretical strength a 0 a 0 γ σ th = Eγ a 0 () 35 / 303

37 Discrepancy with experimental observations a 0 [m] E [GPa] σ th [GPa] σ b [MPa] σ th /σ b glass steel silica fibers iron whiskers silicon whiskers alumina whiskers ausformed steel piano wire discrepancy with experiments σ th σ b () 36 / 303

38 Griffith s experiments σ b [MPa] d [µ] DEFECTS FRACTURE MECHANICS () 37 / 303

39 Crack loading modes Mode I Mode II Mode III Mode I = opening mode Mode II = sliding mode Mode III = tearing mode () 38 / 303

40 back to index EXPERIMENTAL TECHNIQUES

41 Surface cracks dye penetration small surface cracks fast and cheap on-site magnetic particles cracks disturbance of magnetic field surface cracks for magnetic materials only eddy currents impedance change of a coil penetration depth : a few mm s difficult interpretation () 40 / 303

42 Electrical resistance () 41 / 303

43 X-ray orientation dependency () 42 / 303

44 Ultrasound piëzo-el. crystal wave sensor S in out t t () 43 / 303

45 Acoustic emission registration intern sounds (hits) () 44 / 303

46 Adhesion tests blade wedge test peel test (0 o and 90 o ) bending test scratch test indentation test laser blister test pressure blister test fatigue friction test () 45 / 303

47 back to index ENERGY BALANCE

48 () total energy balance 47 / 303 Energy balance a B = thickness A = Ba Ḃ = 0 U e = U i + U a + U d + U k [Js 1 ] d dt ( ) = da d dt da ( ) = Ȧ d da ( ) = ȧ d da ( ) du e da = du i da + du a da + du d da + du k da [Jm 1 ]

49 Griffith s energy balance no dissipation no kinetic energy energy balance energy release rate crack resistance force du e da du i da = du a da G = 1 ( due B da du ) i [Jm 2 ] da R = 1 ( ) dua = 2γ [Jm 2 ] B da Griffith s crack criterion G = R = 2γ [Jm 2 ] () 48 / 303

50 Griffith s energy balance du e = 0 a U a da needed G, R 2γ a c available a c U i () 49 / 303

51 Griffith stress σ y 2a a thickness B x σ U i = 2πa 2 B 1 σ 2 2 E G = 1 ( ) dui B da = 1 B ; U a = 4aB γ [Nm = J] ( ) dua = R 2πa σ2 da E = 4γ [Jm 2 ] Griffith stress σ gr = 2γE πa ; critical crack length a c = 2γE πσ 2 () 50 / 303

52 Griffith stress: plane stress 2γE σ gr = (1 ν 2 )πa () 51 / 303

53 Discrepancy with experimental observations σ gr σ c reason remedy neglection of dissipation measure critical energy release rate G c glass G c = 6 [Jm 2 ] wood G c = 10 4 [Jm 2 ] steel G c = 10 5 [Jm 2 ] composite design problem / high alloyed steel / bone (elephant and mouse) energy balance G = 1 ( due B da du ) i = R = G c da critical crack length a c = G ce 2πσ 2 ; Griffith s crack G = G c criterion () 52 / 303

54 Compliance change compliance : C = u/f F u a P F u P a + da F F a a + da a a + da du i du i du e fixed grips u constant load u () 53 / 303

55 Compliance change : Fixed grips fixed grips : du e = 0 Griffith s energy balance du i = U i (a + da) U i (a) (< 0) = 1 2 (F + df)u 1 2 Fu = 1 2 udf G = 1 2B udf da = 1 2B = 1 2B F 2 dc da u 2 dc C 2 da () 54 / 303

56 Compliance change : Constant load constant load du e = U e (a + da) U e (a) = Fdu Griffith s energy balance du i = U i (a + da) U i (a) (> 0) = 1 2 F(u + du) 1 2 Fu = 1 2 Fdu G = 1 2B F du da = 1 2B F 2 dc da () 55 / 303

57 Compliance change : Experiment F a 1 a 2 ap a 3 a 4 F u u G = shaded area a 4 a 3 1 B () 56 / 303

58 Example B 2h F u a u F u = Fa3 3EI = 4Fa3 EBh 3 C = u F = 2u F = 8a3 EBh 3 G = 1 [ 1 B 2 F 2 dc ] = 12F 2 a 2 da EB 2 h 3 [J m 2 ] G c = 2γ F c = B 1 a 6 γeh3 dc da = 24a2 EBh 3 () 57 / 303

59 Example h a question : which h(a) makes dc da C = u F = 2u F = 8a3 EBh 3 independent from a? dc da = 24a2 EBh 3 choice : h = h 0 a n Fa 3 u = 3(1 n)ei = 4Fa 3 (1 n)ebh 3 = 4Fa3(1 n) (1 n)ebh0 3 C = 2u F = dc da 8a3(1 n) (1 n)ebh 3 0 dc da = 24a(2 3n) EBh 3 0 constant for n = 2 3 h = h 0 a 2 3 () 58 / 303

60 back to index LINEAR ELASTIC STRESS ANALYSIS

61 Deformation P Q X + d X u x + d x Q P X e 3 e 2 x e 1 x i = X i + u i (X i ) x i + dx i = X i + dx i + u i (X i + dx i ) = X i + dx i + u i (X i ) + u i,j dx j dx i = dx i + u i,j dx j = (δ ij + u i,j )dx j ds = d x = dx i dx i ; ds = dx = dx i dx i () 60 / 303

62 Strains ds 2 = dx i dx i = [(δ ij + u i,j )dx j ][(δ ik + u i,k )dx k ] = (δ ij δ ik + δ ij u i,k + u i,j δ ik + u i,j u i,k )dx j dx k = (δ jk + u j,k + u k,j + u i,j u i,k )dx j dx k = (δ ij + u i,j + u j,i + u k,i u k,j )dx i dx j = dx i dx i + (u i,j + u j,i + u k,i u k,j )dx i dx j = ds 2 + (u i,j + u j,i + u k,i u k,j )dx i dx j ds 2 ds 2 = (u i,j + u j,i + u k,i u k,j )dx i dx j = 2γ ij dx i dx j Green-Lagrange strains γ ij = 1 2 (u i,j + u j,i + u k,i u k,j ) linear strains ε ij = 1 2 (u i,j + u j,i ) () 61 / 303

63 Compatibility 3 displacement components 9 strain components 6 dependencies 6 compatibility equations 2ε 12,12 ε 11,22 ε 22,11 = 0 2ε 23,23 ε 22,33 ε 33,22 = 0 2ε 31,31 ε 33,11 ε 11,33 = 0 ε 11,23 + ε 23,11 ε 31,12 ε 12,13 = 0 ε 22,31 + ε 31,22 ε 12,23 ε 23,21 = 0 ε 33,12 + ε 12,33 ε 23,31 ε 31,32 = 0 () 62 / 303

64 Stress unity normal vector stress vector Cauchy stress components stress cube n = n i e i p = p i e i p i = σ ij n j σ 33 σ σ 21 σ 31 σ 13 σ 32 σ 12 σ 22 1 σ 11 () 63 / 303

65 Linear elastic material behavior σ ij = C ijkl ε lk material symmetry isotropic material 2 mat.pars () 64 / 303

66 Hooke s law for isotropic materials σ 11 σ 22 σ 33 σ 12 σ 23 σ 31 σ ij = E ( ε ij + ν ) 1 + ν 1 2ν δ ijε kk i = 1, 2, 3 ε ij = 1 + ν ( σ ij ν ) E 1 + ν δ ijσ kk i = 1, 2, 3 1 ν ν ν ν 1 ν ν = α ν ν 1 ν ν ν ν α = E/[(1 + ν)(1 2ν)] ε 11 ε 22 ε 33 ε 12 ε 23 ε 31 ε 11 ε 22 ε 33 ε 12 ε 23 ε 31 = 1 E 1 ν ν ν 1 ν ν ν ν ν ν σ 11 σ 22 σ 33 σ 12 σ 23 σ 31 () 65 / 303

67 Equilibrium equations σ 12 σ 13 + σ 13,3 dx 3 σ 23 + σ 23,3 dx 3 σ 33 + σ 33,3 dx 3 σ 22 σ 32 σ 11 σ 21 1 σ 11 + σ 11,1 dx 1 σ 21 + σ 21,1 dx 1 σ 31 + σ 31,1 dx σ 13 σ 23 σ 33 σ 31 σ 12 + σ 12,2 dx 2 σ 22 + σ 22,2 dx 2 σ 32 + σ 32,2 dx 2 volume load ρq i force equilibrium σ ij,j + ρq i = 0 i = 1, 2, 3 moment equilibrium σ ij = σ ji () 66 / 303

68 Plane stress σ 33 = σ 13 = σ 23 = 0 equilibrium (q i = 0) σ 11,1 + σ 12,2 = 0 ; σ 21,1 + σ 22,2 = 0 compatibility 2ε 12,12 ε 11,22 ε 22,11 = 0 Hooke s law σ ij = E 1 + ν ( ε ij + ν ) 1 ν δ ijε kk ; ε ij = 1 + ν E Hooke s law in matrix notation ε 11 ε 22 = 1 1 ν 0 ν 1 0 E ε ν σ 11 σ 22 = E 1 ν 0 ν ν σ ν ( σ ij ν ) 1 + ν δ ijσ kk σ 11 σ 22 σ 12 ε 11 ε 22 ε 12 ε 33 = ν E (σ 11 + σ 22 ) = ν 1 ν (ε 11 + ε 22 ) ε 13 = ε 23 = 0 i = 1, 2 () 67 / 303

69 Plane strain ε 33 = ε 13 = ε 23 = 0 equilibrium (q i = 0) σ 11,1 + σ 12,2 = 0 ; σ 21,1 + σ 22,2 = 0 compatibility 2ε 12,12 ε 11,22 ε 22,11 = 0 Hooke s law ε ij = 1 + ν E (σ ij νδ ij σ kk ) ; σ ij = E ( ε ij + ν ) 1 + ν 1 2ν δ ijε kk i = 1, 2 Hooke s law in matrix notation σ 11 σ 22 E = 1 ν ν 0 ν 1 ν 0 (1 + ν)(1 2ν) σ ν ε 11 ε 22 = 1 + ν 1 ν ν 0 ν 1 ν 0 σ 11 σ 22 E ε σ 12 Eν σ 33 = (1 + ν)(1 2ν) (ε 11 + ε 22 ) = ν (σ 11 + σ 22 ) σ 13 = σ 23 = 0 ε 11 ε 22 ε 12 () 68 / 303

70 Displacement method σ ij,j = 0 σ ij = E 1 + ν E 1 + ν ( ε ij,j + ( ε ij + ν ) 1 2ν δ ijε kk,j ν 1 2ν δ ijε kk,j ) = 0 ε ij = 1 2 (u i,j + u j,i ) E ν 2 (u Eν i,jj + u j,ij ) + (1 + ν)(1 2ν) δ iju k,kj = 0 BC s u i ε ij σ ij () 69 / 303

71 Stress function method ψ(x 1, x 2 ) σ ij = ψ,ij + δ ij ψ,kk σ ij,j = 0 ε ij = 1 + ν (σ ij νδ ij σ kk ) E ε ij = 1 + ν } { ψ,ij + (1 ν)δ ij ψ,kk } E 2ε 12,12 ε 11,22 ε 22,11 = 0 2ψ, ψ, ψ,1111 = 0 (ψ,11 + ψ,22 ),11 + (ψ,11 + ψ,22 ),22 = 0 Laplace operator : 2 = 2 x x 2 2 = ( ) 11 + ( ) 22 } bi-harmonic equation 2 ( 2 ψ) = 4 ψ = 0 BC s } ψ σ ij ε ij u i () 70 / 303

72 Cylindrical coordinates z e z e t e r x e 3 e 2 e 1 θ r y vector bases { e 1, e 2, e 3 } { e r, e t, e z } e r = e r (θ) = e 1 cosθ + e 2 sinθ e t = e t (θ) = e 1 sinθ + e 2 cosθ () 71 / 303

73 Laplace operator z e z e t e r x e 3 e 2 e 1 θ r y gradient operator Laplace operator two-dimensional = er r + e 1 t r 2 = = 2 2 = 2 r r θ + e z z r r r r 2 θ 2 r + 1 r 2 2 θ z 2 () 72 / 303

74 Bi-harmonic equation bi-harmonic equation ( 2 r r r ) ( 2 ψ r 2 θ 2 r ψ r r ) ψ r 2 θ 2 = 0 stress components σ rr = 1 ψ r r ψ r 2 θ 2 σ tt = 2 ψ r 2 σ rt = 1 ψ r 2 θ 1 r ψ r θ = r ( 1 r ) ψ θ () 73 / 303

75 Circular hole in infinite plate σ y σ θ r x 2a σ rr = 1 r ( 2 r r ψ r + 1 r 2 2 ψ θ 2 ; r ) ( 2 ψ r 2 θ 2 r ψ r r ) ψ r 2 θ 2 = 0 σ tt = 2 ψ r 2 ; σ rt = 1 r 2 ψ θ 1 r ψ r θ = r ( 1 r ) ψ θ () 74 / 303

76 Load transformation σ σ θ σ rr σ rt σ rr σ rt 2a 2b equilibrium two load cases σ rr (r = b, θ) = 1 2 σ σcos(2θ) σ rt (r = b, θ) = 1 2 σsin(2θ) I. σ rr (r = a) = σ rt (r = a) = 0 σ rr (r = b) = 1 2 σ ; σ rt(r = b) = 0 II. σ rr (r = a) = σ rt (r = a) = 0 σ rr (r = b) = 1 2 σcos(2θ) ; σ rt(r = b) = 1 2 σsin(2θ) () 75 / 303

77 Load case I σ rr (r = a) = σ rt (r = a) = 0 σ rr (r = b) = 1 2 σ ; σ rt(r = b) = 0 Airy function ψ = f (r) stress components σ rr = 1 r ψ r ψ r 2 θ 2 = 1 df r dr ; σ tt = 2 ψ r 2 = d2 f dr 2 ; σ rt = r ( 1 r ) ψ = 0 θ bi-harmonic equation ( d 2 dr r ) ( d d 2 f dr dr ) df = 0 r dr () 76 / 303

78 Solution general solution stresses ψ(r) = Aln r + Br 2 ln r + Cr 2 + D σ rr = A + B(1 + 2 lnr) + 2C r2 σ tt = A + B(3 + 2 lnr) + 2C r2 strains (from Hooke s law for plane stress) ε rr = 1 [ ] A (1 + ν) + B{(1 3ν) + 2(1 ν)ln r} + 2C(1 ν) E r2 ε tt = 1 1 [ Ar ] E r (1 + ν) + B{(3 ν)r + 2(1 ν)r lnr} + 2C(1 ν)r compatibility ε rr = du dr = d(r ε tt) dr 2 BC s and b a A and C B = 0 σ rr = 1 a2 2σ(1 r 2 ) ; σ tt = 1 a2 2σ(1 + r 2 ) ; σ rt = 0 () 77 / 303

79 Load case II σ rr (r = a) = σ rt (r = a) = 0 σ rr (r = b) = 1 2 σcos(2θ) ; σ rt(r = b) = 1 2 σsin(2θ) Airy function ψ(r, θ) = g(r)cos(2θ) stress components σ rr = 1 r ψ r ψ r 2 θ 2 ; σ tt = 2 ψ r 2 σ rt = 1 r 2 ψ θ 1 r ψ r θ = r bi-harmonic equation ( 2 r r r ) {( d 2 g r 2 θ 2 dr dg r dr 4 ( d 2 dr d r dr 4 ) ( d 2 g r 2 dr dg r dr 4 r 2 g ( 1 r ψ θ ) r 2 g ) ) cos(2θ) = 0 } cos(2θ) = 0 () 78 / 303

80 Solution general solution g = Ar 2 + Br 4 + C 1 r 2 + D ( ψ = Ar 2 + Br 4 + C 1 ) r 2 + D cos(2θ) ( stresses σ rr = 2A + 6C r 4 + 4D ) r 2 cos(2θ) ( σ tt = 2A + 12Br 2 + 6C ) r 4 cos(2θ) ( σ rt = 2A + 6Br 2 6C r 4 2D ) r 2 sin(2θ) 4 BC s and b a A,B,C and D ( σ rr = 1 2 σ 1 + 3a4 r 4 4a2 r 2 σ tt = 1 2 σ ( 1 + 3a4 r 4 ) cos(2θ) ) cos(2θ) ( σ rt = 1 2 σ 1 3a4 r 4 + 2a2 r 2 ) sin(2θ) () 79 / 303

81 Stresses for total load σ rr = σ [(1 a2 2 r 2 σ tt = σ [(1 + a2 2 r 2 σ rt = σ 2 ) + (1 + 3a4 ) [1 3a4 r 4 + 2a2 r 2 r 4 4a2 r 2 (1 + 3a4 r 4 ] sin(2θ) ) cos(2θ) ) ] cos(2θ) ] () 80 / 303

82 Special points stress concentration factor σ rr (r = a, θ) = σ rt (r = a, θ) = σ rt (r, θ = 0) = 0 σ tt (r = a, θ = π 2 ) = 3σ σ tt (r = a, θ = 0) = σ K t is independent of hole diameter! K t = σ max σ = 3 [-] () 81 / 303

83 Stress gradients large hole : smaller stress gradient larger area with higher stress higher chance for critical defect in high stress area () 82 / 303

84 Elliptical hole y σ σ yy σ a b radius ρ x ( σ yy (x = a, y = 0) = σ a ) ( = σ ) a/ρ b 2σ a/ρ stress concentration factor K t = 2 a/ρ [-] () 83 / 303

85 back to index CRACK TIP STRESS

86 () 85 / 303

87 Complex plane x 2 r θ x 1 crack tip = singular point complex function theory complex Airy function (Westergaard, 1939) () 86 / 303

88 Complex variables x 2 e i r θ z e r x 1 z z = x 1 + ix 2 = re iθ ; z = x 1 ix 2 = re iθ x 1 = 1 2 (z + z) ; x 2 = 1 2i (z z) = 1 2i(z z) z = x 1 e r + x 2 e i = x 1 e r + x 2 i e r = (x 1 + ix 2 ) e r () 87 / 303

89 Complex functions complex function f (z) = φ + iζ = φ(x 1, x 2 ) + iζ(x 1, x 2 ) = f f ( z) = φ(x 1, x 2 ) iζ(x 1, x 2 ) = f φ = 1 2 {f + f } ; ζ = 1 2 i{f f } 2 φ = 2 ζ = 0 appendix!! () 88 / 303

90 Laplace operator complex function Laplacian derivatives (see App. A) g(x 1, x 2 ) = g(z, z) 2 g = 2 g x g x2 2 g = g x 1 z z + g x 1 z z = g x 1 z + g z ; 2 g x 2 1 = 2 g z g z z + 2 g z 2 g = g x 2 z z + g x 2 z z = i g x 2 z i g z ; 2 g x 2 2 = 2 g z g z z 2 g z 2 Laplacian 2 g = 2 g x = 4 z z + 2 g x 2 2 = 4 g z z () 89 / 303

91 Bi-harmonic equation Airy function ψ(z, z) bi-harmonic equation 2 ( 2 ψ(z, z) ) = 0 () 90 / 303

92 Solution of bi-harmonic equation real part of complex function f satisfies Laplace eqn. 2 ( 2 ψ(z, z) ) = 2 (φ(z, z)) = 0 φ = f + f choice Airy function 2 ψ = 4 ψ z z = φ = f + f integration ψ = 1 2 [ zω + z Ω + ω + ω ] unknown functions : Ω ; Ω ; ω ; ω () 91 / 303

93 Stresses Airy function ψ = 1 2 [ zω + z Ω + ω + ω ] stress components σ ij = σ ij (z, z) = ψ,ij + δ ij ψ,kk σ 11 = ψ,11 + ψ,γγ = ψ,22 = Ω + Ω 1 2 { zω + ω + z Ω + ω } σ 22 = ψ,22 + ψ,γγ = ψ,11 = Ω + Ω { zω + ω + z Ω + ω } σ 12 = ψ,12 = 1 2 i { zω + ω z Ω ω } () 92 / 303

94 Displacement u 2 u x 2 e i e 2 θ r u 1 e r e 1 x 1 definition of complex displacement u = u 1 e 1 + u 2 e 2 = u 1 e r + u 2 e i = u 1 e r + u 2 i e r = (u 1 + iu 2 ) e r = u e r u = u 1 + iu 2 = u 1 (x 1, x 2 ) + iu 2 (x 1, x 2 ) = u(z, z) ū = u 1 iu 2 = ū(z, z) () 93 / 303

95 Schematic u u(z, z) with int. const. M(z) z u z u ū z + ū z M(z), M( z) z u z + ū z = ε 11 + ε 22 σ 11, σ 22 M(z) u(z, z) () 94 / 303

96 Displacement derivatives u z = u x 1 x 1 z + u { x 2 u x 2 z = i u } x 1 x 2 { = 1 u1 2 + i u 2 + i u 1 u } 2 = 1 x 1 x 1 x 2 x 2 (ε 11 ε iε 12 ) 2 u z = u x 1 x 1 z + u { x 2 u x 2 z = 1 2 i u } x 1 x 2 { = 1 u1 2 + i u 2 i u 1 + u } { 2 = 1 x 1 x 1 x 2 x 2 ε 11 + ε 22 + i 2 ū z = ū x 1 x 1 z + ū { x 2 ū x 2 z = 1 2 i ū } x 1 x 2 { = 1 u1 2 i u 2 i u 1 u } 2 = 1 x 1 x 1 x 2 x 2 (ε 11 ε 22 2iε 12 ) 2 ū z = ū x 1 x 1 z + ū { x 2 ū x 2 z = i ū } x 1 x 2 { = 1 u1 2 i u 2 + i u 1 + u } 2 = 1 x 1 x 1 x 2 x 2 2 ( u2 u )} 1 x 1 x 2 { ( u2 ε 11 + ε 22 i u )} 1 x 1 x 2 () 95 / 303

97 General solution u z = 1 2 (ε 11 ε iε 12 ) Hooke s law (pl.strain) ] [σ 11 σ iσ 12 u z = ν E = 1 + ν E [z Ω + ω ] Integration u = 1 + ν E [ ] z Ω + ω + M () 96 / 303

98 Integration function u = 1 + ν E ū = 1 + ν E [ ] z Ω + ω + M u z = 1 + ν [ Ω + M ] E [ Ω + M ] [ zω + ω + M ] ū z = 1 + ν E u z + ū z = 1 + ν [ Ω + Ω + M + M ] E u z + ū z = ε 11 + ε 22 = 1 + ν [(1 2ν)(σ 11 + σ 22 )] E (1 + ν)(1 2ν) = 2 [ Ω + Ω ] E M + M = (3 4ν) [ Ω + Ω ] M = (3 4ν)Ω = κω u = 1 + ν E [ ] z Ω + ω κω () 97 / 303

99 Choice of complex functions Ω = (α + iβ)z λ+1 = (α + iβ)r λ+1 e iθ(λ+1) ω = (γ + iδ)z λ+1 = (γ + iδ)r λ+1 e iθ(λ+1) Ω = (α iβ) z λ+1 = (α iβ)r λ+1 e iθ(λ+1) Ω = (α iβ)(λ + 1) z λ = (α iβ)(λ + 1)r λ e iθλ ω = (γ iδ) z λ+1 = (γ iδ)r λ+1 e iθ(λ+1) u = 1 2µ rλ+1 [ κ(α + iβ)e iθ(λ+1) (α iβ)(λ + 1)e iθ(1 λ) (γ iδ)e iθ(λ+1)] with µ = E 2(1 + ν) displacement finite λ > 1 () 98 / 303

100 Displacement components u = 1 2µ rλ+1 [ κ(α + iβ)e iθ(λ+1) (α iβ)(λ + 1)e iθ(1 λ) (γ iδ)e iθ(λ+1)] e iθ = cos(θ) + i sin(θ) u = 1 2µ rλ+1 [ { i = u 1 + iu 2 κα cos(θ(λ + 1)) κβ sin(θ(λ + 1)) α(λ + 1)cos(θ(1 λ)) β(λ + 1)sin(θ(1 λ)) } γ cos(θ(λ + 1)) + δ sin(θ(λ + 1)) + { κα sin(θ(λ + 1)) + κβ cos(θ(λ + 1)) α(λ + 1)sin(θ(1 λ)) + β(λ + 1)cos(θ(1 λ)) + } ] γ sin(θ(λ + 1)) + δ cos(θ(λ + 1)) () 99 / 303

101 Mode I () 100 / 303

102 Mode I : displacement x 2 r θ x 1 displacement for Mode I u 1 (θ > 0) = u 1 (θ < 0) u 2 (θ > 0) = u 2 (θ < 0) } β = δ = 0 Ω = αz λ+1 = αr λ+1 e i(λ+1)θ ω = γz λ+1 = γr λ+1 e i(λ+1)θ () 101 / 303

103 Mode I : stress components σ 11 = (λ + 1) [ αz λ + α z λ { 1 2 αλ zz λ 1 + γz λ + αλz z λ 1 + γ z λ}] σ 22 = (λ + 1) [ αz λ + α z λ { αλ zz λ 1 + γz λ + αλz z λ 1 + γ z λ}] σ 12 = 1 2 i(λ + 1)[ αλ zz λ 1 + γz λ αλz z λ 1 γ z λ] with z = re iθ ; z = re iθ σ 11 = (λ + 1)r λ [ αe iλθ + αe iλθ {αλe i(λ 2)θ + γe iλθ + αλe i(λ 2)θ + γe iλθ}] 1 2 σ 22 = (λ + 1)r λ [ αe iλθ + αe iλθ + {αλe i(λ 2)θ + γe iλθ + αλe i(λ 2)θ + γe iλθ}] 1 2 σ 12 = 1 2 i(λ + 1)rλ [ αλe i(λ 2)θ + γe iλθ αλe i(λ 2)θ γe iλθ] () 102 / 303

104 with e iθ + e iθ = 2 cos(θ) ; e iθ e iθ = 2i sin(θ) σ 11 = 2(λ + 1)r λ [ α cos(λθ) {αλ cos((λ 2)θ) + γ cos(λθ)}] σ 22 = 2(λ + 1)r λ [ α cos(λθ) 1 2 {αλ cos((λ 2)θ) + γ cos(λθ)}] σ 12 = (λ + 1)r λ [αλ sin((λ 2)θ) + γ sin(λθ)] () 103 / 303

105 Stress boundary conditions σ 11 = 2(λ + 1)r λ [ α cos(λθ) {αλ cos((λ 2)θ) + γ cos(λθ)}] σ 22 = 2(λ + 1)r λ [ α cos(λθ) 1 2 {αλ cos((λ 2)θ) + γ cos(λθ)}] σ 12 = (λ + 1)r λ [αλ sin((λ 2)θ) + γ sin(λθ)] crack surfaces are stress free σ 22 (θ = ±π) = σ 12 (θ = ±π) = 0 [ ][ ] [ ] (λ 2)cos(λπ) cos(λπ) α 0 = λ sin(λπ) sin(λπ) γ 0 [ ] (λ 2)cos(λπ) cos(λπ) det = sin(2λπ) = 0 2πλ = nπ λ sin(λπ) sin(λπ) λ = 1 2, n, with n = 0, 1, 2,.. 2 () 104 / 303

106 Stress field λ = 1 2 α = 2γ ; λ = 0 α = 1 2 γ λ = 1 2 α = 2γ ; λ = 1 α = γ σ 11 = 2γr 1 2 cos( 1 2 θ)[ 1 sin( 3 2 θ)sin(1 2 θ)] + σ 22 = 2γr 1 2 cos( 1 2 θ)[ 1 + sin( 3 2 θ)sin(1 2 θ)] + σ 12 = 2γr 1 2 [ cos( 1 2 θ)cos(3 2 θ)sin(1 2 θ)] + () 105 / 303

107 Mode I : stress intensity factor definition stress intensity factor K ( Kies ) ( ) K I = lim 2πr σ22 θ=0 = 2γ 2π [ m 1 2 N m 2 ] r 0 () 106 / 303

108 Mode I : crack tip solution σ 11 = σ 22 = σ 12 = K I 2πr [ cos( 1 2 θ){ 1 sin( 1 2 θ)sin(3 2 θ)}] K I 2πr [ cos( 1 2 θ){ 1 + sin( 1 2 θ)sin(3 2 θ)}] K I 2πr [ cos( 1 2 θ)sin(1 2 θ)cos(3 2 θ)] u 1 = K I r 2µ 2π u 2 = K I r 2µ 2π [ cos( 1 2 θ){ κ sin 2 ( 1 2 θ)}] [ sin( 1 2 θ){ κ cos 2 ( 1 2 θ)}] plane stress plane strain κ = 3 ν 1 + ν κ = 3 4ν () 107 / 303

109 Mode II () 108 / 303

110 Mode II : displacement x 2 r θ x 1 displacements for Mode II u 1 (θ > 0) = u 1 (θ < 0) u 2 (θ > 0) = u 2 (θ < 0) } α = γ = 0 Ω = iβz λ+1 = iβr λ+1 e i(λ+1)θ ω = iδz λ+1 = iδr λ+1 e i(λ+1)θ () 109 / 303

111 Mode II : stress intensity factor definition stress intensity factor K ( Kies ) ( ) K II = lim 2πr σ12 θ=0 r 0 [ m 1 2 N m 2 ] () 110 / 303

112 Mode II : crack tip solution σ 11 = K II 2πr [ sin( 1 2 θ){ 2 + cos( 1 2 θ)cos(3 2 θ)}] σ 22 = K II 2πr [ sin( 1 2 θ)cos(1 2 θ)cos(3 2 θ)] σ 12 = K II 2πr [ cos( 1 2 θ){ 1 sin( 1 2 θ)sin(3 2 θ)}] u 1 = K II r 2µ 2π u 2 = K II r 2µ 2π [ sin( 1 2 θ){ κ cos 2 ( 1 2 θ)}] [ cos( 1 2 θ){ κ 1 2 sin 2 ( 1 2 θ)}] plane stress plane strain κ = 3 ν 1 + ν κ = 3 4ν () 111 / 303

113 Mode III () 112 / 303

114 Laplace equation ε 31 = 1 2 u 3,1 ; ε 32 = 1 2 u 3,2 Hooke s law σ 31 = 2µε 31 = µu 3,1 ; σ 32 = 2µε 32 = µu 3,2 equilibrium σ 31,1 + σ 32,2 = µu 3,11 + µu 3,22 = 0 2 u 3 = 0 () 113 / 303

115 Mode III : displacement general solution u 3 = f + f specific choice f = (A + ib)z λ+1 f = (A ib) z λ+1 () 114 / 303

116 Mode III : stress components σ 31 = 2(λ + 1)r λ {Acos(λθ) B sin(λθ)} σ 32 = 2(λ + 1)r λ {Asin(λθ) + B cos(λθ)} σ 32 (θ = ±π) = 0 [ ] [ ] [ ] sin(λπ) cos(λπ) A 0 = sin(λπ) cos(λπ) B 0 [ ] sin(λπ) cos(λπ) det = sin(2πλ) = 0 2πλ = nπ sin(λπ) cos(λπ) λ = 1 2, n,.. with n = 0, 1, 2,.. 2 crack tip solution λ = 1 2 A = 0 σ 31 = Br 1 2 {sin( 1 2 θ)} ; σ 32 = Br 1 2 {cos( 1 2 θ)} () 115 / 303

117 Mode III : Stress intensity factor definition stress intensity factor ( ) K III = lim 2πr σ32 θ=0 r 0 () 116 / 303

118 Mode III : crack tip solution stress components σ 31 = K III 2πr [ sin( 1 2 θ)] σ 32 = K III 2πr [ cos( 1 2 θ)] displacement u 3 = 2K III µ r [ sin( 1 2π 2 θ)] () 117 / 303

119 Crack tip stress (mode I, II, III) σ τ τ σ τ τ Mode I Mode II Mode III σ ij = K I 2πr f Iij (θ) ; σ ij = K II 2πr f IIij (θ) ; σ ij = K III 2πr f IIIij (θ) crack intensity factors (SIF) K I = β I σ πa ; K II = β II τ πa ; K III = β III τ πa () 118 / 303

120 K-zone D K-zone : D I II D II D I () 119 / 303

121 SIF for specified cases 2a σ τ 2a K I = σ ( πa sec πa W K II = τ πa ) 1/2 small a W W W () 120 / 303

122 SIF for specified cases σ K I = σ [ a 1.12 π 0.41 a W + a W ( a ) 2 ( a ) W W ( a ) ] W 1.12σ πa small a W () 121 / 303

123 SIF for specified cases a W a σ K I = σ [ a 1.12 π a W ( a ) 2 ( a ) ] W W 1.12σ πa () 122 / 303

124 SIF for specified cases W 2a σ K I /σ full 1st term a/w a W σ K I /σ full 1st term a/w plots are made with Kfac.m. () 123 / 303

125 SIF for specified cases P K I = [ PS ( a ) BW 3/2 W W a P/2 S P/2 ( a ) 3 ( 2 a ) W W ] ( a ) 7 ( 2 a ) W W () 124 / 303

126 SIF for specified cases a P W K I = [ P ( a ) BW 1/2 W ( a ) 3 ( 2 a ) W W ] ( a ) 7 ( 2 a ) W W P () 125 / 303

127 SIF for specified cases 2a p K I = p πa p per unit thickness W () 126 / 303

128 SIF for specified cases P W a P/2 S P/2 K I /P full 1st term P a W P K I /P a/w full 1st term a/w plots are made with Kfac.m. () 127 / 303

129 K-based crack growth criteria K I = K Ic ; K II = K IIc ; K III = K IIIc K Ic = Fracture Toughness calculate K I, K II, K III - analytically - literature - relation K G - numerically (EEM, BEM) experimental determination of K Ic, K IIc, K IIIc - normalized experiments (exmpl. ASTM E399) KIc 2 - correlation with C v ( KAN p. 18 : E = mcn v ) () 128 / 303

130 Relation G K I y σ yy x a a crack length a σ yy (θ = 0, r = x a) = σ a 2(x a) ; u y = 0 crack length a + a σ yy (θ = π, r = a + a x) = 0 (1 + ν)(κ + 1) σ a + a u y = E 2 a + a x plane stress : κ = 3 ν 1 + ν ; plane strain : κ = 3 4ν () 129 / 303

131 Relation G K I (continued) accumulation of elastic energy U = 2B energy release rate G = 1 B lim a 0 a+ a a ( U a ) 1 2 σ yy dx u y = B a+ a a σ yy u y dx = B f ( a) a (1 + ν)(κ + 1) = lim f ( a) = σ 2 (1 + ν)(κ + 1) aπ = KI 2 a 0 4E 4E plane stress G = K I 2 E plane strain G = (1 ν 2 ) K I 2 E () 130 / 303

132 Multi mode load G = 1 E ( c1 KI 2 + c 2 KII 2 + c 3 KIII 2 ) plane stress G = 1 E (K 2 I + K 2 II) plane strain G = (1 ν2 ) (KI 2 + K 2 E II) + (1 + ν) KIII 2 E () 131 / 303

133 The critical SIF value σ K I 2a B K Ic σ B c B K Ic = σ c πa B c = 2.5 ( ) 2 KIc σ y () 132 / 303

134 K Ic values Material σ v [MPa] K Ic [MPa m ] steel, 300 maraging steel, 350 maraging steel, D6AC steel, AISI steel, A533B reactor steel, carbon Al 2014-T Al 2024-T Al 7075-T Al 7079-T Ti 6Al-4V Ti 6Al-6V-2Sn Ti 4Al-4Mo-2Sn-0.5Si () 133 / 303

135 back to index MULTI-MODE LOADING

136 Multi-mode crack loading Mode I Mode II Mode I + II Mode I + II () 135 / 303

137 Multi-mode crack loading crack tip stresses s ij Mode I s ij = K I 2πr f Iij (θ) Mode II s ij = K II 2πr f IIij (θ) Mode I + II s ij = K I 2πr f Iij (θ) + K II 2πr f IIij (θ) () 136 / 303

138 Stress component transformation (b) p e 2 e 2 θ n e 1 e 1 = cos(θ) e 1 + sin(θ) e 2 = c e 1 + s e 2 e 2 = sin(θ) e 1 + cos(θ) e 2 = s e 1 + c e 2 e 1 stress vector and normal unity vector p = p 1 e 1 + p 2 e 2 = p1 e 1 + p 2 e [ ] [ ] [ ] p1 c s p = 1 p 2 s c p2 2 [ p 1 p 2 ] [ c s = s c ] [ p1 p 2 ] = T p p p = T T p idem : ñ = T T ñ () 137 / 303

139 Transformation stress matrix p = σñ T p = σtñ p = T T σt ñ = σ ñ σ = T T σt σ = T σ T T [ σ 11 σ 12 σ 21 σ 22 ] [ ][ ] [ ] c s σ11 σ = 12 c s s c σ 21 σ 22 s c [ ][ ] c s cσ11 + sσ = 12 sσ 11 + cσ 12 s c cσ 21 + sσ 22 sσ 21 + cσ 22 c 2 σ csσ 12 + s 2 σ 22 = csσ 11 + (c 2 s 2 )σ 12 + csσ 22 csσ 11 + (c 2 s 2 )σ 12 + csσ 22 s 2 σ 11 2csσ 12 + c 2 σ 22 () 138 / 303

140 Cartesian to cylindrical transformation σ yy σ xy e t e r σ xx e 2 r θ e r = c e 1 + s e 2 e t = s e 1 + c e 2 e 1 σ rt σ rr σ tt [ ] σrr σ rt = σ tr σ tt [ c ][ s σxx σ xy s c = ] [ c s σ xy σ yy s c c 2 σ xx + 2csσ xy + s 2 σ yy csσ xx + (c 2 s 2 )σ xy + csσ yy csσ xx + (c 2 s 2 )σ xy + csσ yy s 2 σ xx 2csσ xy + c 2 σ yy ] () 139 / 303

141 Crack tip stresses : Cartesian σ yyσxy σ xx = σ xx K I 2πr f Ixx (θ) + K II 2πr f IIxx (θ) f Ixx (θ) = cos( θ 2 )[ 1 sin( θ 2 )sin(3θ 2 )] f IIxx (θ) = sin( θ 2 )[ 2 + cos( θ 2 )cos(3θ 2 )] f Iyy (θ) = cos( θ 2 )[ 1 + sin( θ 2 )sin(3θ 2 )] f IIyy (θ) = sin( θ 2 )cos( θ 2 )cos(3θ 2 ) f Ixy (θ) = sin( θ 2 )cos( θ 2 )cos(3θ 2 ) f IIxy (θ) = cos( θ 2 )[ 1 sin( θ 2 )sin(3θ 2 )] σ yy = σ xy = K I 2πr f Iyy (θ) + K II 2πr f IIyy (θ) K I 2πr f Ixy (θ) + K II 2πr f IIxy (θ) () 140 / 303

142 Crack tip stresses : cylindrical σ rr σ rt σ rr = θ σ tt K I 2πr f Irr (θ) + K II 2πr f IIrr (θ) f Irr (θ) = [ 5 4 cos( θ 2 ) 1 4 cos(3θ 2 )] f IIrr (θ) = [ 5 4 sin( θ 2 ) sin(3θ 2 )] f Itt (θ) = [ 3 4 cos( θ 2 ) cos(3θ 2 )] f IItt (θ) = [ 3 4 sin( θ 2 ) 3 4 sin(3θ 2 )] f Irt (θ) = [ 1 4 sin( θ 2 ) sin(3θ 2 )] f IIrt (θ) = [ 1 4 cos( θ 2 ) cos(3θ 2 )] σ tt = σ rt = K I 2πr f Itt (θ) + K II 2πr f IItt (θ) K I 2πr f Irt (θ) + K II 2πr f IIrt (θ) () 141 / 303

143 Multi-mode load 2a σ 22σ12 σ 11 σ 12 σ 22 2a σ 12 σ 11 [ σ 11 σ 12 σ 21 σ 22 ] = e 2 e 2 e 1 θ e 1 σ 12 σ 22 c 2 σ csσ 12 + s 2 σ 22 csσ 11 + (c 2 s 2 )σ 12 + csσ 22 csσ 11 + (c 2 s 2 )σ 12 + csσ 22 s 2 σ 11 2csσ 12 + c 2 σ 22 crack tip stresses s ij = K I 2πr f Iij (θ) + K II 2πr f IIij (θ) with K I = β σ 22 πa ; KII = γ σ 12 πa σ 11 does not do anything () 142 / 303

144 Example multi-mode load σ σ 12 σ 22 σ 11 2a kσ 2a σ 12 θ σ 11 = c2 σ csσ 12 + s 2 σ 22 = c 2 kσ + s 2 σ σ 22 = s2 σ 11 2csσ 12 + c 2 σ 22 = s 2 kσ + c 2 σ σ 12 = csσ 11 + (c 2 s 2 )σ 12 + csσ 22 = cs(1 k)σ crack tip stresses s ij = K I 2πr f Iij (θ) + K II 2πr f IIij (θ) stress intensity factors K I = β I σ 22 πa = βi (s 2 k + c 2 )σ πa K II = β II σ 12 πa = βii cs(1 k)σ πa () 143 / 303

145 Example multi-mode load σ t 2a σ a p R t θ σ 22 σ 12 σ 11 σ t = pr = σ ; σ a = pr t 2t = 1 2 σ k = 1 2 σ 22 = s σ + c2 σ ; σ 12 = cs(1 1 2 )σ = 1 2 cs σ K I = σ 22 πa = ( 1 2 s2 + c 2 )σ πa = ( 1 2 s2 + c 2 ) pr t πa K II = σ 12 πa = 1 2 cs σ = 1 2 cs pr t πa () 144 / 303

146 back to index CRACK GROWTH DIRECTION

147 Crack growth direction criteria for crack growth direction : maximum tangential stress (MTS) criterion strain energy density (SED) criterion requirement : crack tip stresses in cylindrical coordinates () 146 / 303

148 Maximum tangential stress criterion Erdogan & Sih (1963) σ rr σ rt θ σ tt Hypothesis : crack growth towards local maximum of σ tt σ tt θ = 0 and 2 σ tt θ 2 < 0 θ c σ tt (θ = θ c ) = σ tt (θ = 0) = K Ic 2πr crack growth () 147 / 303

149 Maximum tangential stress criterion 3 2 σ tt θ = 0 K I K II [ 1 2πr 4 sin( θ 2 ) 1 4 sin(3θ 2 )] [ πr 4 cos( θ 2 ) 3 4 cos(3θ 2 )] = 0 K I sin(θ) + K II {3 cos(θ) 1} = 0 2 σ tt θ 2 < K I [ 1 2πr 4 cos( θ 2 ) 3 4 cos(3θ 2 )] [ πr 4 sin( θ 2 ) sin(3θ 2 )] < 0 K II σ tt (θ = θ c ) = K Ic 2πr 1 K I 4 K II [ 3 cos( θ c K 2 ) + cos(3θc 2 )] [ sin( θ c Ic K 2 ) 3 sin(3θc 2 )] = 1 Ic () 148 / 303

150 Mode I load K II = 0 σ tt θ = K I sin(θ) = 0 θ c = 0 2 σ tt θ 2 < 0 θc σ tt (θ c ) = K Ic 2πr K I = K Ic () 149 / 303

151 Mode II load K I = 0 σ tt θ = K II(3 cos(θ c ) 1) = 0 θ c = ± arccos( 1 3 ) = ±70.6o 2 σ tt θ 2 < 0 θ c = 70.6 o θc σ tt (θ c ) = K Ic 2πr K IIc = 3 4 K Ic τ θ c τ () 150 / 303

152 Multi-mode load K I [ sin( θ 2 ) sin(3θ 2 )] + K II[ cos( θ 2 ) 3 cos(3θ 2 )] = 0 K I [ cos( θ 2 ) 3 cos(3θ 2 )] + K II[sin( θ 2 ) + 9 sin(3θ 2 )] < 0 K I [3 cos( θ 2 ) + cos(3θ 2 )] + K II[ 3 sin( θ 2 ) 3 sin(3θ 2 )] = 4K Ic K I f 1 K II f 2 = 0 K I f 2 + K II f 3 < 0 K I f 4 3K II f 1 = 4K Ic ( KI K ( Ic KI K ( Ic KI K Ic ) ( KII f 1 ) f 2 + ) f 4 3 K ( Ic KII K ( Ic KII K Ic ) f 2 = 0 ) f 3 < 0 ) f 1 = 4 () 151 / 303

153 Multi-mode load θ c K I /K Ic K II /K Ic K /K I Ic () 152 / 303

154 Strain energy density (SED) criterion Sih (1973) σ rr σ rt θ σ tt Hypothesis : U i = Strain Energy Density (Function) = εij 0 σ ij dε ij S = Strain Energy Density Factor = ru i = S(K I, K II, θ) crack growth towards local minimum of SED S θ = 0 and 2 S θ 2 > 0 θ c S(θ = θ c ) = S(θ = 0, pl.strain) = S c crack growth () 153 / 303

155 SED U i = 1 2E (σ2 xx + σ2 yy + σ2 zz ) ν E (σ xxσ yy + σ yy σ zz + σ zz σ xx ) + 1 2G (σ2 xy + σ2 yz + σ2 zx ) σ xx = σ yy = σ xy = K I 2πr cos( θ 2 )[ 1 sin( θ 2 )sin(3θ 2 )] K II 2πr sin( θ 2 )[ 2 + cos( θ 2 )cos(3θ 2 )] K I 2πr cos( θ 2 )[ 1 + sin( θ 2 )sin(3θ 2 )] + K II 2πr sin( θ 2 )cos( θ 2 )cos(3θ 2 ) K I 2πr sin( θ 2 )cos( θ 2 )cos(3θ 2 ) + K II 2πr cos( θ 2 )[ 1 sin( θ 2 )sin(3θ 2 )] () 154 / 303

156 SED factor S = ru i = S(K I, K II, θ) = a 11 k 2 I + 2a 12 k I k II + a 22 k 2 II with a 11 = 1 16G (1 + cos(θ))(κ cos(θ)) a 12 = 1 16G sin(θ){2 cos(θ) (κ 1)} a 22 = 1 16G {(κ + 1)(1 cos(θ)) + (1 + cos(θ))(3 cos(θ) 1)} k i = K i / π S θ = 0 ki 2 16G {2 sin(θ)cos(θ) (κ 1)sin(θ)} + k Ik II 16G {2 4 sin2 (θ) (κ 1)cos(θ)} + kii 2 { 6 sin(θ)cos(θ) + (κ 1)sin(θ)} = 0 16G 2 S θ 2 > 0 k 2 I 16G {2 4 sin2 (θ) (κ 1)cos(θ)} + k Ik II { 8 sin(θ)cos(θ) + (κ 1)sin(θ)} + 16G kii 2 16G { sin2 (θ) + (κ 1)cos(θ)} > 0 () 155 / 303

157 Mode I load S = a 11 ki 2 = σ2 a {1 + cos(θ)}{κ cos(θ)} 16G S = sin(θ){2 cos(θ) (κ 1)} = 0 θ θ c = 0 or arccos ( 1 2 (κ 1)) 2 S θ 2 = 2 cos(2θ) (κ 1)cos(θ) > 0 θ c = 0 S(θ c ) = σ2 a 16G {2}{κ 1} = σ2 a (κ 1) 8G S c = S(θ c, pl.strain) = (1 + ν)(1 2ν) 2πE K 2 Ic () 156 / 303

158 Mode II load S = a 22 kii 2 = τ2 a [(κ + 1){1 cos(θ)} + {1 + cos(θ)}{3 cos(θ) 1}] 16G S = sin(θ)[ 6 cos(θ) + (κ 1)] = 0 θ 2 S θ 2 = 6 cos2 (θ) + (κ 1)cos(θ) > 0 S(θ c ) = τ2 a 16G { 1 12 ( κ2 + 14κ 1)} θ c = ± arccos ( 1 6 (κ 1)) S(θ c ) = S c τ c = 1 a 192GSc κ κ 1 () 157 / 303

159 Multi-mode load; plane strain ν= ν=0.1 ν=0.2 ν=0.3 ν=0.4 ν= θ c K /K I Ic K II /K Ic ν= ν=0.1 ν=0.2 ν=0.3 ν=0.4 ν= K /K I Ic () 158 / 303

160 Multi-mode load; plane stress ν= ν=0.1 ν=0.2 ν=0.3 ν=0.4 ν= θ c K /K I Ic K II /K Ic ν= ν=0.1 ν=0.2 ν=0.3 ν=0.4 ν= K /K I Ic () 159 / 303

161 Multi-mode load; plane strain k I = σ a sin 2 (β) ; k II = σ a sin(β)cos(β) S = σ 2 a sin 2 (β) { a 11 sin 2 (β) + 2a 12 sin(β)cos(β) + a 22 cos 2 (β) } S θ = (κ 1)sin(θ c 2β) 2 sin{2(θ c β)} sin(2θ c ) = 0 2 S θ 2 = (κ 1)cos(θ c 2β) 4 cos{2(θ c β)} 2 cos(2θ c ) > 0 σ β 2a σ θc 90 θ c ν = ν = ν = β From Gdoutos () 160 / 303

162 back to index DYNAMIC FRACTURE MECHANICS

163 Dynamic fracture mechanics impact load (quasi)static load fast fracture - kinetic approach - static approach () 162 / 303

164 Crack growth rate Mott (1948) du e da du i da = du a da + du d da + du k da σ y 2a a thickness B x du e da = 0 ; du d da = 0 U a = 4aBγ du a da = 4γB σ U i = 2πa 2 B 1 σ 2 2 E du i da = 2πaBσ2 E () 163 / 303

165 Kinetic energy U k = 1 2 ρb Ω ( u 2 x + u 2 y)dxdy material velocity u x u y = du y dt ( ) 2 U k = 1 duy 2 ρs2 B dxdy Ω da ds assumption da = 0 ( ) 2 du k da = 1 d duy 2 ρs2 B dxdy Ω da da u y = 2 2 σ E a2 ax du ( k σ ) 2 da = ρs2 B a E Ω = du y da du y da = 2 σ E da dt = du y da s 2a x a2 ax 1 x 2 (x 2a) ( σ ) 2 a 3 (a x) 2 dxdy = ρs 2 B a k(a) E () 164 / 303

166 Energy balance 2πaσ 2 E s = ( E ρ ( = 4γ + ρs 2 σ ) 2 ak E ) 1 ( ) 2 2π 1 ( 2 1 2γE ) 1 2 k πaσ 2 ( ds ) da 0!! 2π k 0.38 ; a c = 2γE E πσ 2 ; c = ρ ( s = 0.38 c 1 a c a a a c ) 1 2 s 0.38 c () 165 / 303

167 Experimental crack growth rates steel copper aluminum glass rubber E [GPa] ρ [kg/m 2 ] ν c [m/sec] s [m/sec] s/c < s c < 0.4 () 166 / 303

168 Elastic wave speeds C 0 = elongational wave speed = C 1 = dilatational wave speed = C 2 = shear wave speed = E ρ κ + 1 κ 1 µ ρ µ ρ C R = Rayleigh velocity = 0.54 C 0 á 0.62 C 0 () 167 / 303

169 Corrections ( Dulancy & Brace (1960) s = 0.38 C 0 1 a ) c a ( Freund (1972) s = C R 1 a ) c a () 168 / 303

170 Crack tip stress Yoffe (1951) : σ Dij = K D 2πr f ij (θ, r, s, E, ν) () 169 / 303

171 Crack branching Yoffe (1951) σ Dij = K ID 2πr f ij (θ, r, s, E, ν) 0.9 σ Dtt (θ) max σ Dtt (θ = 0) s c R 0.87 σ tt π π 2 θ crack branching volgens MTS Source: Gdoutos (1993) p.245 () 170 / 303

172 Fast fracture and crack arrest K D K Dc (s, T) crack growth K D < min 0<s<C R K Dc (s, T) = K A crack arrest () 171 / 303

173 Experiments Source: KAN1985 p.210 High Speed Photography : 10 6 frames/sec Robertson : CA Temperature (CAT) test (KAN1985 p.258) () 172 / 303

174 back to index PLASTIC CRACK TIP ZONE

175 () 174 / 303

176 Von Mises and Tresca yield criteria Von Mises Tresca W d = W d c (σ 1 σ 2 ) 2 + (σ 2 σ 3 ) 2 + (σ 3 σ 1 ) 2 = 2σ 2 y τ max = τ maxc σ max σ min = σ y yield surface in principal stress space () 175 / 303

177 Principal stresses at the crack tip plane stress state σ zz = σ zx = σ zy = 0 σ = σ xx σ xy 0 σ xy σ yy characteristic equation det(σ σi) = 0 σ [ σ 2 σ(σ xx + σ yy ) + (σ xx σ yy σ 2 xy )] = 0 σ 1 = 1 2 (σ xx + σ yy ) + { 1 4 (σ xx σ yy ) 2 + σ 2 xy σ 2 = 1 2 (σ xx + σ yy ) { 1 4 (σ xx σ yy ) 2 + σ 2 xy σ 3 = 0 } 1/2 } 1/2 plane strain state σ 3 = ν(σ 1 + σ 2 ) () 176 / 303

178 Principal stresses at crack tip crack tip stresses σ ij = K I 2πr f Iij (θ) σ 1(+),2( ) = K I [ cos( θ 2πr 2 )± ] { cos( θ 2 )sin( θ 2 )sin(3θ 2 )} 2 { + sin( θ 2 )cos( θ 2 )cos(3θ 2 )} 2 σ 1 = σ 2 = K I 2πr cos( θ 2 ){1 + sin( θ 2 )} K I 2πr cos( θ 2 ){1 sin( θ 2 )} σ 3 = 0 or σ 3 = 2νK I 2πr cos( θ 2 ) () 177 / 303

179 Principal stresses at crack tip plane stress σ 1 > σ 2 > σ 3 plane strain σ 1 > σ 2 > σ 3 or σ 1 > σ 3 > σ ν = ν = σ σ θ θ 1000 ν = ν = σ σ θ θ () 178 / 303

180 Von Mises plastic zone (σ 1 σ 2 ) 2 + (σ 2 σ 3 ) 2 + (σ 3 σ 1 ) 2 = 2σ 2 y plane stress σ 3 = 0 (σ 1 σ 2 ) 2 + σ σ2 1 = 2σ2 y KI 2 cos 2 ( θ 2πr 2 )[ 6 sin 2 ( θ 2 ) + 2] = 2σ 2 y y r y = K I 2 2πσ 2 y cos 2 ( θ 2 )[ sin 2 ( θ 2 )] = K I 2 [ 1 + cos(θ) + 3 4πσ 2 2 sin2 (θ) ] y plane strain σ 3 = ν(σ 1 + σ 2 ) (ν 2 ν + 1)(σ σ 2 2) + (2ν 2 2ν 1)σ 1 σ 2 = σ 2 y KI 2 cos 2 ( θ 2πr 2 )[ 6 sin 2 ( θ 2 ) + 2(1 2ν)2] = 2σ 2 y y r y = K I 2 4πσ 2 y [ (1 2ν) 2 {1 + cos(θ)} sin2 (θ) ] () 179 / 303

181 Von Mises plastic zone 1 Von Mises plastic zones pl.stress pl.strain Plot made with plazone.m. () 180 / 303

182 Tresca plastic zone σ max σ min = σ y plane stress {σ max, σ min } = {σ 1, σ 3 } K I 2πry [ cos( θ 2 ) + cos( θ 2 )sin( θ 2 ) ] = σ y r y = K I 2 [ cos( θ 2πσ 2 2 ) + cos( θ 2 )sin( θ 2 ) ] 2 y plane strain I σ 1 > σ 2 > σ 3 {σ max, σ min } = {σ 1, σ 3 } r y = K I 2 [ (1 2ν)cos( θ 2πσ 2 2 ) + cos( θ 2 )sin( θ 2 ) ] 2 y plane strain II σ 1 > σ 3 > σ 2 {σ max, σ min } = {σ 1, σ 2 } r y = K I 2 2πσ 2 sin 2 (θ) y () 181 / 303

183 Tresca plastic zone Tresca plastic zones pl.stress pl.strain sig3 = min pl.strain sig2 = min Plot made with plazone.m. () 182 / 303

184 Influence of the plate thickness B c > 25 3π ( KIc σ y ) 2 > 2.5 ( ) 2 KIc σ y () 183 / 303

185 Shear planes Source: Gdoutos p.60/61/62; Kanninen p.176 () 184 / 303

186 Plastic zone in the crack plane () 185 / 303

187 Irwin plastic zone correction σ xx σ xx σ yy σ yy σ y σ y a r y r a r y r p r θ = 0 σ xx = σ yy = K I 2πr yield σ xx = σ yy = σ y r y = 1 ( ) 2 KI 2π σ y equilibrium not satisfied correction required shaded area equal () 186 / 303

188 Irwin plastic zone correction σ xx σ xx σ yy σ yy σ y σ y a r y r a r y r p r σ y r p = ry 0 σ yy (r)dr = K I 2π ry 0 r 1 2 dr = 2K I 2π ry r p = 2K I ry r p = 1 ( KI 2π σ y π σ y ) 2 = 2 r y () 187 / 303

189 Dugdale-Barenblatt plastic zone correction y a σ y r p σ x σ load σ load σ y K I (σ) = σ π(a + r p ) K I (σ y ) = 2σ y a + rp π ( ) a arccos a + r p singular term = 0 K I (σ) = K I (σ y ) ( ) a πσ = cos r p = πk I 2 a + r p 2σ y 8σ 2 y () 188 / 303

190 Plastic constraint factor 1 2 {(σ 1 σ 2 ) 2 + (σ 2 σ 3 ) 2 + (σ 3 σ 1 ) 2 } = [ ] 1 n m + n2 + m 2 mn σ max = σ y PCF = σ max 1 = σ y 1 n m + n2 + m 2 mn () 189 / 303

191 PCF at the crack tip pl.sts n = [ 1 sin( θ 2 )] / [ 1 + sin( θ 2 )] ; m = 0 pl.stn n = [ 1 sin( θ 2 )] / [ 1 + sin( θ 2 )] ; m = 2ν/ [ 1 + sin( θ 2 )] () 190 / 303

192 PCF at the crack tip in the crack plane pl.sts n = 1 ; m = 0 PCF = 1 pl.stn n = 1 ; m = 2ν PCF = 1 1 4ν + 4ν 2 () 191 / 303

193 Plastic zones in the crack plane r y r p criterion state r y or r p (K I /σ y ) 2 ( ) 2 1 KI Von Mises plane stress π σ y ( ) 2 1 KI Von Mises plane strain π σ y ( ) 2 1 KI Tresca plane stress π σ y ( ) 2 1 KI Tresca plane strain σ 1 > σ 2 > σ π Tresca plane strain σ 1 > σ 3 > σ ( ) 2 1 KI Irwin plane stress π σ y ( ) 2 1 KI Irwin plane strain (pcf = 3) π 3σ y ( ) 2 π KI Dugdale plane stress σ y ( ) 2 π KI Dugdale plane strain (pcf = 3) σ y σ y () 192 / 303

194 Small Scale Yielding LEFM & SSY correction effective crack length a eff Irwin / Dugdale-Barenblatt correction SSY : outside plastic zone : K I (a eff )-stress a eff = a + (r y r p ) K I = β I (a eff )σ πa eff () 193 / 303

195 back to index NONLINEAR FRACTURE MECHANICS

196 Nonlinear Fracture Mechanics () 195 / 303

197 Crack-tip opening displacement crack tip displacement u y = σ πa r [ sin( 1 2µ 2π 2 θ){ κ cos 2 ( 1 2 θ)}] displacement in crack plane θ = π; r = a x u y = (1 + ν)(κ + 1) E Crack Opening Displacement (COD) δ(x) = 2u y (x) = Crack Tip Opening Displacement (CTOD) (1 + ν)(κ + 1) E δ t = δ(x = a) = 0 σ 2a(a x) 2 σ 2a(a x) () 196 / 303

198 CTOD () 197 / 303

199 CTOD by Irwin σ xx σ xx σ yy σ yy σ y σ y a r y r a r y r p r effective crack length a eff = a + r y = a + 1 2π ( ) 2 KI σ y () 198 / 303

200 CTOD by Irwin (1 + ν)(κ + 1) δ(x) = σ 2a eff (a eff x) E (1 + ν)(κ + 1) = σ 2(a + r y )(a + r y x) E (1 + ν)(κ + 1) δ t = δ(x = a) = σ 2(a + r y )r y E (1 + ν)(κ + 1) = σ 2ar y + 2ry E 2 (1 + ν)(κ + 1) σ 2ar y E plane stress : δ t = 4 = 4 G π Eσ y π σ y [ ] 1 4(1 ν 2 ) plane strain : δ t = 3 π K 2 I K 2 I Eσ y () 199 / 303

201 CTOD by Dugdale y σ y σ x a r p σ effective crack length ( ) 2 KI a eff = a + r p = a + π 8 σ y () 200 / 303

202 CTOD by Dugdale displacement from requirement singular term = 0 : ū y (x) [ ū y (x) = (a + r { p)σ y x sin 2 } { } ] 2 (^γ γ) sin(^γ) + sin(γ) ln πe a + r p sin 2 + cos(^γ) ln (^γ + γ) sin(^γ) sin(γ) ( ) x γ = arccos ; ^γ = π σ a + r p 2 σ y Crack Tip Opening Displacement δ t = lim 2ū x a y(x) = 8σ { ( )} va π πe ln σ sec 2 σ y series expansion & σ σ y plane stress : δ t = K I 2 = G Eσ y σ y [ ] 1 plane strain : δ t = 2 (1 ν 2 ) K I 2 Eσ y () 201 / 303

203 CTOD crack growth criterion δ t (G, K I ) at LEFM δ t = measure for deformation at crack tip (LEFM) δ t = measure for (large) plastic deformation at crack tip (NLFM) criterion δ t = δ tc ( ε, T) δ t calculate or measure δ tc experimental determination (ex. BS 5762) () 202 / 303

204 J-integral n t x 2 e 2 Γ Ω V S positive e 1 x 1 J k = Γ J = J 1 = ( u i Wn k t i x k Γ ( Wn 1 t i u i x 1 ) dγ ; W = specific energy = ) dγ Epq 0 σ ij dε ij [ ] N m () 203 / 303

205 Integral along closed curve J k = Γ (Wδ jk σ ij u i,k ) n j dγ inside Γ no singularities Stokes (Gauss in 3D) ( ) dw ε mn δ jk σ ij,j u i,k σ ij u i,kj dω dε mn x j Ω homogeneous hyper-elastic σ mn = W ε mn linear strain ε mn = 1 2 (u m,n + u n,m ) equilibrium equations σ ij,j = 0 Ω { 1 2 σ mn(u m,nk + u n,mk ) σ ij u i,kj } dω = ) (σ mn u m,nk σ ij u i,kj dω = 0 Ω () 204 / 303

206 Path independence x 2 Ω n e 2 e 1 Γ+ Γ n Γ B Γ A x 1 f 1 dγ + f 1 dγ + f 1 dγ + f 1 dγ = 0 Γ A Γ B Γ Γ + no loading of crack faces : n 1 = 0 ; t i = 0 on Γ + and Γ f 1 dγ + f 1 dγ = 0 Γ A Γ B J 1A + J 1B = 0 J 1A = J 1B f 1 dγ = J 1A ; f 1 dγ = J 1B Γ A Γ B () 205 / 303

207 Relation J K lin. elast. material : W = 1 2 σ mnε mn = 1 4 σ mn(u m,n + u n,m ) ( ) 1 J k = 4 σ mn(u m.n + u n,m )δ jk σ ij u i,k n j dγ Mode I + II + III Γ = Γ ( 1 2 σ mnu m,n δ jk σ ij u i,k ) n j dγ σ ij = 1 2πr [K I f Iij + K II f IIij + K III f IIIij ] u i = u Ii + u IIi + u IIIi substitution and integration over Γ = circle (κ + 1)(1 + ν) J 1 = 4E J 2 = ( K 2 I + KII 2 (κ + 1)(1 + ν) K I K II 2E ) (1 + ν) + KIII 2 E () 206 / 303

208 Relation J G Mode I J 1 = J = (κ + 1)(1 + ν) 4E K 2 I = G plane stress κ + 1 = 3 ν 1 + ν ν 1 + ν = ν J = 1 E K 2 I plane strain κ + 1 = 4 4ν J = (1 ν2 ) E K 2 I () 207 / 303

209 Relation J δ t plane stress Irwin J = π 4 σ yδ t Dugbale J = σ y δ t plane strain Irwin J = π 4 3σy δ t Dugbale J = 2σ y δ t () 208 / 303

210 Plastic constraint factor J = m σ y δ t m = a W σ u σ y () 209 / 303

211 HRR crack tip stresses and strains () 210 / 303

212 Ramberg-Osgood material law ε = σ ( ) n σ + α ε y0 σ y0 σ y0 n strain hardening parameter (n 1) n = 1 linear elastic n ideal plastic 5 Ramberg Osgood for α = 0.01 n = 1 4 n = 3 3 σ/σ y0 2 n = 5 n = 7 1 n = ε/ε y () 211 / 303

213 HRR-solution σ ij = σ y0 β r 1 n+1 σ ij (θ) ; u i = αε y0 β n 1 r n+1 ũ i (θ) [ ] J 1 n+1 with : β = (I n from num. anal.) ασ y0 ε y0 I n I n plane strain plane stress n () 212 / 303

214 J-integral crack growth criterion LEFM : J k G (K I, K II, K III ) NLFM : Ramberg-Osgood : J determines crack tip stress criterion J = J c calculate J J Ic from experiments e.g. ASTM E813 () 213 / 303

215 back to index NUMERICAL FRACTURE MECHANICS

216 Numerical fracture mechanics Methods EEM ; BEM Calculations G K δt J Simulation crack growth () 215 / 303

217 Quadratic elements ξ 2 ξ ξ 2 3 ξ isoparametric coordinates : 1 ξ i 1 shape functions for each node n ψ n (ξ 1, ξ 2 ) = quadratic in ξ 1 and ξ 2 () 216 / 303

218 Crack tip mesh bad approximation stress field results are mesh-dependent 1/ r () 217 / 303

219 Special elements enriched elements crack tip field added to element displacement field structure K and f changes transition elements for compatibility hybrid elements modified variational principle () 218 / 303

220 Quarter point elements p 3p p p Distorted Quadratic Quadrilateral (1/ r) Distorted Quadratic Triangle (1/ r) Collapsed Quadratic Quadrilateral (1/ r) Collapsed Distorted Linear Quadrilateral (1/r) good approximation stress field (1/ r or 1/r) bad approximation non-singular stress field standard FEM-programs can be used () 219 / 303

221 Crack tip rozet Quarter Point Elements : 8x Transition Elements : number is problem dependent Buffer Elements () 220 / 303

222 One-dimensional case ξ = 1 ξ = 0 ξ = 1 1 x 3 2 position displacement and strain x = 1 2 ξ(ξ 1)x ξ(ξ + 1)x 2 (ξ 2 1)x 3 = 1 2 ξ(ξ + 1)L (ξ2 1)x 3 u = 1 2 ξ(ξ 1)u ξ(ξ + 1)u 2 (ξ 2 1)u 3 du dξ = (ξ 1 2 )u 1 + (ξ )u 2 2ξu 3 du dx = du dξ dξ dx = du dξ /dx dξ () 221 / 303

223 Mid point element mid-point element : x 3 = 1 2 L ξ = 1 ξ = 0 ξ = 1 1 x 3 2 x = 1 2 ξ(ξ + 1)L (ξ2 1) 1 2 L = 1 2 (ξ + 1)L dx dξ = 1 2 L du du dx = dξ 1 2 L du dx x=0 ξ= 1 = ( 2 L ) {( 3 2 ) u1 + ( ) } 1 2 u2 + 2u 3 () 222 / 303

224 Quarter point element quarter-point element : x 3 = 1 4 L ξ = 1 ξ = 0 ξ = 1 1 x 3 2 x = 1 2 ξ(ξ + 1)L (ξ2 1) 1 4 L = 1 4 (ξ + 1)2 L ξ + 1 = dx dξ = 1 2 (ξ + 1)L = xl du du dx = dξ du xl dx singularity 1 x x=0 ξ= 1 = 4x L () 223 / 303

225 Virtual crack extension method (VCEM) u u a a + a I II fixed grips du e da = 0 G = 1 du i B da 1 U i (a + a) U i (a) B a () 224 / 303

226 VCEM : stiffness matrix variation B G = du i da = C 1 2ũT a ũ with C = C(a + a) C(a) G from analysis crack tip mesh only nodal point displacement : ± element size not possible with crack tip in interface unloaded crack plane no thermal stresses () 225 / 303

227 Stress intensity factor calculate G I and G II with VCEM calculate K I and K II from KI 2 = E G I ; KII 2 = E G II plane stress E = E plane strain E = E/(1 ν 2 ) difficult for crack propagation study () 226 / 303

228 SIF : stress field ( ) K I = lim 2πr σ22 θ=0 r 0 extrapolation to crack tip ( ) ; K II = lim 2πr σ12 θ=0 r 0 p p 4 3 p p 2 1 θ r K K p1 Kp2 K p3 K p4 r 1 r 2 r 3 r 4 r questions : which elements? how much elements? which integration points? () 227 / 303

229 SIF : displacement field crack tip displacement y-component u y = 4(1 ν2 ) r E 2π K I g ij (θ) [ ] E 2π K I = lim r 0 4(1 ν 2 u y (θ = 0) ) r more accurate than SIF from stress field () 228 / 303

230 J-integral J = Γ ( ) ε u i Wn 1 t i dγ with W = σ ij dε ij x 1 0 () 229 / 303

231 J-integral : Direct calculation J = 2 W = y [ ( u x W σ xx x + σ yx )] u y x dy 2 E 2(1 ν 2 ) (ε2 xx + 4νε xxε yy + 2(1 ν)ε 2 xy + ε2 yy ) x [( σ xy u x x + σ yy )] u y dx x path through integration points no need for quarter point elements () 230 / 303

232 J-integral : Domain integration x 2 Ω e 2 Γ + n e 1 Γ Γ B n Γ A x 1 Ω q = 0 q = 1 q J = x j Ω ( ) u i σ ij Wδ 1j dω x 1 interpolation q e = Ñ T e (ξ )q () 231 / 303

233 De Lorenzi J-integral : VCE technique ( ) u i σ i1 Wδ 1j dω x 1 q J = x j Ω u i qp i dγ x 1 Γ s rigid region Ω elongation a of crack translation δx1 of internal nodes fixed position of boundary q = δx 1 = shift function (0 < q < 1) a q(ρq i ρü i ) u i dω + x 1 Ω qσ ij ε o ij x 1 dω () 232 / 303

234 Crack growth simulation Node release Moving Crack Tip Mesh Element splitting Smeared crack approach () 233 / 303

235 Node release node collocation technique () 234 / 303

236 Moving Crack Tip Mesh () 235 / 303

237 Element splitting () 236 / 303

238 Smeared crack approach e 2 e 1 e 2 e 1 n 2 σ 2 σ 1 n1 n 2 n 1 () 237 / 303

239 back to index FATIGUE

240 Teletekst Wo 3 oktober 2007 Van de 274 stalen bruggen in ons land kampen er 25 met metaalmoeheid. Dat is de uitkomst van een groot onderzoek van het ministerie van Verkeer. Bij twaalf bruggen zijn de problemen zo groot dat noodmaatregelen nodig zijn. Ook de meer dan 2000 betonnen bruggen en viaducten zijn onderzocht. De helft daarvan moet nog nader worden bekeken. Ze gaan mogelijk minder lang mee dan was berekend, maar de veiligheid komt volgens het ministerie niet in gevaar. Verkeersbeperkende maatregelen zijn dan ook niet nodig. Die werden in april wel getroffen voor het vrachtverkeer over de Hollandse Brug bij Almere. () 239 / 303

241 Fatigue ± 1850 (before Griffith!) : cracks at diameter-jumps in axles carriages / trains failure due to cyclic loading with small amplitude Wöhler : systematic experimental examination cyclic loading : variable mechanical loads vibrations pressurization / depressurization thermal loads (heating / cooling) random external loads () 240 / 303

242 Crack surface clam shell markings (beach marks) - irregular crack growth - crack growth under changing conditions striations - sliding of slip planes - plastic blunting / sharpening of crack tip - regular crack growth () 241 / 303

243 Experiments full-scale testing a.o. train axles airplanes laboratory testing harmonic loading constant force/moment strain/deflection SIF () 242 / 303

244 Train axle D = 0.75 [m] 1 rev = πd = π [m] 1 km = 1000 m = = [c(ycles)] 1 day Maastricht - Groningen = [km] = 1000 [km] 1 day Maastricht - Groningen = [c] 1 year = [c] = [c] [c] frequency : 100 [km/h] = [c/h] = = 12.5 [c/sec] = 12.5 [Hz] () 243 / 303

245 Fatigue load (stress controlled) σ σ max σ m σ min 0 0 i i + 1 t N σ = σ max σ min ; σ a = 1 2 σ σ m = 1 2 (σ max + σ min ) ; R σ = σ min /σ max ; - frequency bending Hz tensile electric Hz mechanic < 50 Hz hydraulic 1-50 Hz - no influence frequency for ± 5000 [c/min] (metals) σ a = 1 R σ m 1 + R () 244 / 303

246 Fatigue limit (σ th ) σ σ th N σ < σ th : no increase of damage materials with fatigue limit mild steel low strength steels Ti / Al / Mg -alloys materials without fatigue limit some austenitic steels high strength steels most non-ferro alloys Al / Mg-alloys () 245 / 303

247 (S-N)-curve B.S part I 1984 : S = σ max S σ th 0 0 log(n f ) reference : R = 1 and σ m = 0 σ max = 1 2 σ fatigue life : N f at σ max (= S) fatigue limit : σ th (= σ fat ) N f = (±10 9 ) fatigue strength : σ e = σ max when N f steels : σ th 1 2 σ b () 246 / 303

248 (S a -N)-curve B.S part I 1984 : S a = 1 2 σ = σ a S a σ th 0 0 log(n f ) reference : R = 1 and σ m = 0 σ a = σ max (S a N) curve = (S N) curve () 247 / 303

249 Examples steelt1 400 σ max [MPa] steel Mgalloy Al2024T N f () 248 / 303

250 Influence of average stress σ a σ m σ th 0 0 log(n f ) () 249 / 303

251 Correction for average stress Gerber (1874) Goodman (1899) Soderberg (1939) σ a σ a = 1 σ a ( ) 2 σm σ u = 1 σ m σ a σ u σ a = 1 σ m σ a σ y0 σ u : tensile strength σ y0 : initial yield stress () 250 / 303

252 (P-S-N)-curve σ max [MPa] % prob.failure % prob.failure % prob.failure N f () 251 / 303

253 High/low cycle fatigue S a σ m = LCF 4 5 HCF log(n f ) high cycle fatigue N f > ±50000 low stresses LEFM + SSY stress-life curve Basquin relation K max = βσ max πa ; Kmin = βσ min πa ; K = β σ πa () 252 / 303

254 High/low cycle fatigue S a σ m = LCF 4 5 HCF log(n f ) low cycle fatigue N f < ±50000 high stresses EPFM strain-life curve Manson-Coffin relation () 253 / 303

255 Basquin relation 1 2 σ = σ a = σ f (2N f ) b σn b f = constant σ f = fatigue strength coefficient σ b (monotonic tension) b = fatigue strength exponent (Basquin exponent) log ( ) σ 2 log(2n f ) () 254 / 303

256 Manson-Coffin relation 1 2 εp = ε f (2N f ) c ε p N c f = constant ε f = fatigue ductility coefficient ε b (monotonic tension) c = fatigue ductility exponent ( log ) ε p 2 ( 0.5 < c < 0.7) log(2n f ) () 255 / 303

257 Total strain-life curve log( ε 2 ) log(n f ) ε 2 = εe 2 + εp 2 = 1 E σ f (2N f ) b + ε f (2N f ) c () 256 / 303

258 Influence factors load spectrum stress concentrations stress gradients material properties surface quality environment () 257 / 303

259 Load spectrum sign / magnitude / rate / history multi-axial lower f.limit than uni-axial () 258 / 303

260 Stress concentrations ρ σ th (notched) = 1 K f σ th (unnotched) ; 1 < K f < K t K f : fatigue strength reduction factor (effective stress concentration factor) K f = 1 + q(ρ)(k t 1) Peterson : q = a ρ 1 Neuber : q = 1 + b ρ with with q(ρ) = notch sensitivity factor a = material parameter b = grain size parameter () 259 / 303

261 Stress gradients full-scale experiments necessary () 260 / 303

262 Material properties grain size/structure : small grains higher f.limit at low temp. large grains higher f.limit at high temp. (less grain boundaries less creep) texture inhomogeneities and flaws residual stresses fibers and particles () 261 / 303

263 Surface quality 10µm surface extrusions & intrusions notch + inclusion of O 2 etc. bulk defect internal surfaces internal grain boundaries / triple points (high T) voids manufacturing minimize residual tensile stresses surface finish minimize defects (roughness) surface treatment (mech/temp) residual pressure stresses coating environmental protection high σ y0 more resistance to slip band formation () 262 / 303

264 Environment temperature creep - fatigue low temperature : ships / liquefied gas storage elevated temperature (T > 0.5T m ) : turbine blades creep mechanism : diffusion / dislocation movement / migration of vacancies / grain boundary sliding grain boundary voids / wedge cracks chemical influence corrosion-fatigue () 263 / 303

265 Crack growth a I II III a c a c σ a f a 1 a i N i N f N I : N < N i - N i = fatigue crack initiation life - a i = initial fatigue crack II : N i < N < N f - slow stable crack propagation - a 1 = non-destr. inspection detection limit III : N f < N - global instability - towards catastrophic failure - a = a c : failure N r N f = 1 N N f N r = rest life () 264 / 303

266 Crack growth models ( ) 2 da K striation spacing 6 dn E (Bates, Clark (1969)) da dn f (σ, a) σm a n ; m 2 7 ; n 1 2 da dn δ t ( K)2 Eσ y (BRO263) da da K dn dn K E Source: HER1976a p515 da Paris law : dn = C( K)m () 265 / 303