FRACTURE MECHANICS Piet Schreurs Eindhoven University of Technology Department of Mechanical Engineering Materials Technology September 5, 2013
INDEX back to index () 1 / 303
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
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back to index INTRODUCTION
Introduction () 5 / 303
Continuum mechanics V 0 A 0 u V A x 0 e 3 x e 1 O e 2 () 6 / 303
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
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
Stress-strain curves σ σ ε ε σ σ ε ε () 9 / 303
Fracture () 10 / 303
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
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
Experimental fracture mechanics () 13 / 303
Linear elastic fracture mechanics () 14 / 303
Dynamic fracture mechanics () 15 / 303
Nonlinear fracture mechanics CTOD J-integral () 16 / 303
Numerical techniques () 17 / 303
Fatigue () 18 / 303
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
Outline () 20 / 303
back to index FRACTURE MECHANISMS
Fracture mechanisms shear fracture cleavage fracture fatigue fracture crazing de-adhesion () 22 / 303
Shearing dislocations voids crack dimples load direction () 23 / 303
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
Fatigue clam shell pattern striations () 25 / 303
Crazing stress whitening crazing materials : PS, PMMA () 26 / 303
() 27 / 303
back to index DUCTILE/BRITTLE
Ductile - brittle behavior σ ABS, nylon, PC PE, PTFE 0 10 100 ε (%) surface energy : γ [Jm 2 ] solids : γ 1 [Jm 2 ] independent from cleavage/shearing ex.: alloyed steels; rubber () 29 / 303
Charpy v-notch test () 30 / 303
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
back to index THEORETICAL STRENGTH
Theoretical strength f f r f x σ a 0 1 2 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
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
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
Discrepancy with experimental observations a 0 [m] E [GPa] σ th [GPa] σ b [MPa] σ th /σ b glass 3 10 10 60 14 170 82 steel 10 10 210 45 250 180 silica fibers 10 10 100 31 25000 1.3 iron whiskers 10 10 295 54 13000 4.2 silicon whiskers 10 10 165 41 6500 6.3 alumina whiskers 10 10 495 70 15000 4.7 ausformed steel 10 10 200 45 3000 15 piano wire 10 10 200 45 2750 16.4 discrepancy with experiments σ th σ b () 36 / 303
Griffith s experiments 11000 σ b [MPa] 170 10 20 d [µ] DEFECTS FRACTURE MECHANICS () 37 / 303
Crack loading modes Mode I Mode II Mode III Mode I = opening mode Mode II = sliding mode Mode III = tearing mode () 38 / 303
back to index EXPERIMENTAL TECHNIQUES
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
Electrical resistance () 41 / 303
X-ray orientation dependency () 42 / 303
Ultrasound piëzo-el. crystal wave sensor S in out t t () 43 / 303
Acoustic emission registration intern sounds (hits) () 44 / 303
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
back to index ENERGY BALANCE
() 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 ]
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
Griffith s energy balance du e = 0 a U a da needed G, R 2γ a c available a c U i () 49 / 303
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
Griffith stress: plane stress 2γE σ gr = (1 ν 2 )πa () 51 / 303
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
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
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
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
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
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
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
back to index LINEAR ELASTIC STRESS ANALYSIS
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
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
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
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 23 3 2 σ 21 σ 31 σ 13 σ 32 σ 12 σ 22 1 σ 11 () 63 / 303
Linear elastic material behavior σ ij = C ijkl ε lk material symmetry isotropic material 2 mat.pars () 64 / 303
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 0 0 0 ν 1 ν ν 0 0 0 = α ν ν 1 ν 0 0 0 0 0 0 1 2ν 0 0 0 0 0 0 1 2ν 0 0 0 0 0 0 1 2ν 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 0 0 0 ν 1 ν 0 0 0 ν ν 1 0 0 0 0 0 0 1 + ν 0 0 0 0 0 0 1 + ν 0 0 0 0 0 0 1 + ν σ 11 σ 22 σ 33 σ 12 σ 23 σ 31 () 65 / 303
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 1 3 2 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
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 12 0 0 1 + ν σ 11 σ 22 = E 1 ν 0 ν 1 0 1 ν σ 2 12 0 0 1 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
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 12 0 0 1 2ν ε 11 ε 22 = 1 + ν 1 ν ν 0 ν 1 ν 0 σ 11 σ 22 E ε 12 0 0 1 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
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 1 1 + 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
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,1122 + ψ,2222 + ψ,1111 = 0 (ψ,11 + ψ,22 ),11 + (ψ,11 + ψ,22 ),22 = 0 Laplace operator : 2 = 2 x 2 1 + 2 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
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
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 2 + 1 r θ + e z z r 2 + 1 r r + 1 2 r 2 θ 2 r + 1 r 2 2 θ 2 + 2 z 2 () 72 / 303
Bi-harmonic equation bi-harmonic equation ( 2 r 2 + 1 r r + 1 2 ) ( 2 ψ r 2 θ 2 r 2 + 1 ψ r r + 1 2 ) ψ r 2 θ 2 = 0 stress components σ rr = 1 ψ r r + 1 2 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
Circular hole in infinite plate σ y σ θ r x 2a σ rr = 1 r ( 2 r 2 + 1 r ψ r + 1 r 2 2 ψ θ 2 ; r + 1 2 ) ( 2 ψ r 2 θ 2 r 2 + 1 ψ r r + 1 2 ) 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
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 + 1 2 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
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 + 1 2 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 2 + 1 r ) ( d d 2 f dr dr 2 + 1 ) df = 0 r dr () 76 / 303
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
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 + 1 2 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 2 + 1 r r + 1 2 ) {( d 2 g r 2 θ 2 dr 2 + 1 dg r dr 4 ( d 2 dr 2 + 1 d r dr 4 ) ( d 2 g r 2 dr 2 + 1 dg r dr 4 r 2 g ( 1 r ψ θ ) r 2 g ) ) cos(2θ) = 0 } cos(2θ) = 0 () 78 / 303
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
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
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
Stress gradients large hole : smaller stress gradient larger area with higher stress higher chance for critical defect in high stress area () 82 / 303
Elliptical hole y σ σ yy σ a b radius ρ x ( σ yy (x = a, y = 0) = σ 1 + 2 a ) ( = σ 1 + 2 ) a/ρ b 2σ a/ρ stress concentration factor K t = 2 a/ρ [-] () 83 / 303
back to index CRACK TIP STRESS
() 85 / 303
Complex plane x 2 r θ x 1 crack tip = singular point complex function theory complex Airy function (Westergaard, 1939) () 86 / 303
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
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
Laplace operator complex function Laplacian derivatives (see App. A) g(x 1, x 2 ) = g(z, z) 2 g = 2 g x1 2 + 2 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 2 + 2 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 2 + 2 g z z 2 g z 2 Laplacian 2 g = 2 g x 2 1 2 = 4 z z + 2 g x 2 2 = 4 g z z () 89 / 303
Bi-harmonic equation Airy function ψ(z, z) bi-harmonic equation 2 ( 2 ψ(z, z) ) = 0 () 90 / 303
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
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 + 1 2 { zω + ω + z Ω + ω } σ 12 = ψ,12 = 1 2 i { zω + ω z Ω ω } () 92 / 303
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
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
Displacement derivatives 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 + 2iε 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 = 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 2 ( u2 u )} 1 x 1 x 2 { ( u2 ε 11 + ε 22 i u )} 1 x 1 x 2 () 95 / 303
General solution u z = 1 2 (ε 11 ε 22 + 2iε 12 ) Hooke s law (pl.strain) ] [σ 11 σ 22 + 2iσ 12 u z = 1 2 1 + ν E = 1 + ν E [z Ω + ω ] Integration u = 1 + ν E [ ] z Ω + ω + M () 96 / 303
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
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
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
Mode I () 100 / 303
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
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 { + 1 2 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
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) + 1 2 {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
Stress boundary conditions σ 11 = 2(λ + 1)r λ [ α cos(λθ) + 1 2 {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
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
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
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 1 + 2 sin 2 ( 1 2 θ)}] [ sin( 1 2 θ){ κ + 1 2 cos 2 ( 1 2 θ)}] plane stress plane strain κ = 3 ν 1 + ν κ = 3 4ν () 107 / 303
Mode II () 108 / 303
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
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
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 + 1 + 2 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
Mode III () 112 / 303
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
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
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
Mode III : Stress intensity factor definition stress intensity factor ( ) K III = lim 2πr σ32 θ=0 r 0 () 116 / 303
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
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
K-zone D K-zone : D I II D II D I () 119 / 303
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
SIF for specified cases σ K I = σ [ a 1.12 π 0.41 a W + a W ( a ) 2 ( a ) 3 18.7 38.48 + W W ( a ) ] 4 53.85 W 1.12σ πa small a W () 121 / 303
SIF for specified cases a W a σ K I = σ [ a 1.12 π + 0.76 a W ( a ) 2 ( a ) ] 3 8.48 + 27.36 W W 1.12σ πa () 122 / 303
SIF for specified cases W 2a σ K I /σ 2.5 2 1.5 1 full 1st term 0.5 0 0 0.1 0.2 0.3 0.4 a/w a W σ K I /σ 2.5 2 1.5 1 full 1st term 0.5 0 0 0.1 0.2 0.3 0.4 a/w plots are made with Kfac.m. () 123 / 303
SIF for specified cases P K I = [ PS ( a ) 1 2 2.9 BW 3/2 W W a P/2 S P/2 ( a ) 3 ( 2 a ) 5 2 4.6 + 21.8 W W ] ( a ) 7 ( 2 a ) 9 2 37.6 + 37.7 W W () 124 / 303
SIF for specified cases a P W K I = [ P ( a ) 1 2 29.6 BW 1/2 W ( a ) 3 ( 2 a ) 5 2 185.5 + 655.7 W W ] ( a ) 7 ( 2 a ) 9 2 1017 + 638.9 W W P () 125 / 303
SIF for specified cases 2a p K I = p πa p per unit thickness W () 126 / 303
SIF for specified cases P W a P/2 S P/2 K I /P 100 80 60 40 20 full 1st term P a W P K I /P 0 0 0.1 0.2 0.3 0.4 a/w 200 150 100 50 full 1st term 0 0 0.1 0.2 0.3 0.4 a/w plots are made with Kfac.m. () 127 / 303
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
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
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
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
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
K Ic values Material σ v [MPa] K Ic [MPa m ] steel, 300 maraging 1669 93.4 steel, 350 maraging 2241 38.5 steel, D6AC 1496 66.0 steel, AISI 4340 1827 47.3 steel, A533B reactor 345 197.8 steel, carbon 241 219.8 Al 2014-T4 448 28.6 Al 2024-T3 393 34.1 Al 7075-T651 545 29.7 Al 7079-T651 469 33.0 Ti 6Al-4V 1103 38.5 Ti 6Al-6V-2Sn 1083 37.4 Ti 4Al-4Mo-2Sn-0.5Si 945 70.3 () 133 / 303
back to index MULTI-MODE LOADING
Multi-mode crack loading Mode I Mode II Mode I + II Mode I + II () 135 / 303
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
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
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 11 + 2csσ 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
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
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
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 ) + 3 4 sin(3θ 2 )] f Itt (θ) = [ 3 4 cos( θ 2 ) + 1 4 cos(3θ 2 )] f IItt (θ) = [ 3 4 sin( θ 2 ) 3 4 sin(3θ 2 )] f Irt (θ) = [ 1 4 sin( θ 2 ) + 1 4 sin(3θ 2 )] f IIrt (θ) = [ 1 4 cos( θ 2 ) + 3 4 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
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 11 + 2csσ 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
Example multi-mode load σ σ 12 σ 22 σ 11 2a kσ 2a σ 12 θ σ 11 = c2 σ 11 + 2csσ 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
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 2 1 2 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
back to index CRACK GROWTH DIRECTION
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
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
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 )] [ + 3 2 1 2πr 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 < 0 3 4 K I [ 1 2πr 4 cos( θ 2 ) 3 4 cos(3θ 2 )] [ + 3 1 4 2πr 4 sin( θ 2 ) + 9 4 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 )] [ + 1 4 3 sin( θ c Ic K 2 ) 3 sin(3θc 2 )] = 1 Ic () 148 / 303
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
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
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
Multi-mode load 0 10 20 30 θ c 40 50 60 70 0 0.2 0.4 0.6 0.8 1 K I /K Ic 0.9 0.8 0.7 0.6 K II /K Ic 0.5 0.4 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 K /K I Ic () 152 / 303
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
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
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 { 6 + 12 sin2 (θ) + (κ 1)cos(θ)} > 0 () 155 / 303
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
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 2 + 14κ 1 () 157 / 303
Multi-mode load; plane strain 10 0 10 ν=0.0.75 ν=0.1 ν=0.2 ν=0.3 ν=0.4 ν=0.495 20 30 θ c 40 50 60 70 80 90 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 K /K I Ic 1 0.9 0.8 0.7 K II /K Ic 0.6 0.5 0.4 ν=0.0.75 ν=0.1 ν=0.2 ν=0.3 ν=0.4 ν=0.495 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 K /K I Ic () 158 / 303
Multi-mode load; plane stress 10 0 10 ν=0.0.75 ν=0.1 ν=0.2 ν=0.3 ν=0.4 ν=0.495 20 30 θ c 40 50 60 70 80 90 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 K /K I Ic 1 0.9 0.8 0.7 0.6 K II /K Ic 0.5 0.4 0.3 0.2 ν=0.0.75 ν=0.1 ν=0.2 ν=0.3 ν=0.4 ν=0.495 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 K /K I Ic () 159 / 303
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 80 70 ν = 0.5 60 50 40 ν = 0 30 20 ν = 0.1 10 0 0 10 20 30 40 50 60 70 80 90 β From Gdoutos () 160 / 303
back to index DYNAMIC FRACTURE MECHANICS
Dynamic fracture mechanics impact load (quasi)static load fast fracture - kinetic approach - static approach () 162 / 303
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
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
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
Experimental crack growth rates steel copper aluminum glass rubber E [GPa] 210 120 70 70 20 ρ [kg/m 2 ] 7800 8900 2700 2500 900 ν 0.29 0.34 0.34 0.25 0.5 c [m/sec] 5190 3670 5090 5300 46 s [m/sec] 1500 2000 s/c 0.29 0.38 0.2 < s c < 0.4 () 166 / 303
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
Corrections ( Dulancy & Brace (1960) s = 0.38 C 0 1 a ) c a ( Freund (1972) s = C R 1 a ) c a () 168 / 303
Crack tip stress Yoffe (1951) : σ Dij = K D 2πr f ij (θ, r, s, E, ν) () 169 / 303
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) 1 0.6 0 s c R 0.87 σ tt π π 2 θ crack branching volgens MTS Source: Gdoutos (1993) p.245 () 170 / 303
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
Experiments Source: KAN1985 p.210 High Speed Photography : 10 6 frames/sec Robertson : CA Temperature (CAT) test (KAN1985 p.258) () 172 / 303
back to index PLASTIC CRACK TIP ZONE
() 174 / 303
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
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 0 0 0 0 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
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 )± ] { 1 4 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
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 2 1000 ν = 0.25 1000 ν = 0.35 800 800 600 600 σ σ 400 200 0 0 20 40 60 80 100 θ 400 200 0 0 20 40 60 80 100 θ 1000 ν = 0.45 1000 ν = 0.5 800 800 600 600 σ σ 400 200 0 0 20 40 60 80 100 θ 400 200 0 0 20 40 60 80 100 θ () 178 / 303
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 2 2 + 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 )[ 1 + 3 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 2 1 + 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)} + 3 2 sin2 (θ) ] () 179 / 303
Von Mises plastic zone 1 Von Mises plastic zones pl.stress pl.strain 0.5 0 0.5 1 0.5 0 0.5 1 1.5 Plot made with plazone.m. () 180 / 303
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
Tresca plastic zone 1 0.5 Tresca plastic zones pl.stress pl.strain sig3 = min pl.strain sig2 = min 0 0.5 1 0.5 0 0.5 1 1.5 Plot made with plazone.m. () 182 / 303
Influence of the plate thickness B c > 25 3π ( KIc σ y ) 2 > 2.5 ( ) 2 KIc σ y () 183 / 303
Shear planes Source: Gdoutos p.60/61/62; Kanninen p.176 () 184 / 303
Plastic zone in the crack plane () 185 / 303
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
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
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
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
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
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
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 0.1592 2π σ y ( ) 2 1 KI Von Mises plane strain 0.0177 18π σ y ( ) 2 1 KI Tresca plane stress 0.1592 2π σ y ( ) 2 1 KI Tresca plane strain σ 1 > σ 2 > σ 3 0.0177 18π Tresca plane strain σ 1 > σ 3 > σ 2 0 0 ( ) 2 1 KI Irwin plane stress 0.3183 π σ y ( ) 2 1 KI Irwin plane strain (pcf = 3) 0.0354 π 3σ y ( ) 2 π KI Dugdale plane stress 0.3927 8 σ y ( ) 2 π KI Dugdale plane strain (pcf = 3) 0.0436 8 3σ y σ y () 192 / 303
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
back to index NONLINEAR FRACTURE MECHANICS
Nonlinear Fracture Mechanics () 195 / 303
Crack-tip opening displacement crack tip displacement u y = σ πa r [ sin( 1 2µ 2π 2 θ){ κ + 1 2 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
CTOD () 197 / 303
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
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
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
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
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
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
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
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
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
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 + 1 + ν 1 + ν = 4 1 + ν J = 1 E K 2 I plane strain κ + 1 = 4 4ν J = (1 ν2 ) E K 2 I () 207 / 303
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
Plastic constraint factor J = m σ y δ t m = 0.111 + 0.817 a W + 1.36 σ u σ y () 209 / 303
HRR crack tip stresses and strains () 210 / 303
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 = 13 0 0 1 2 3 ε/ε y0 4 5 6 () 211 / 303
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 5 2.5 0 5 10 15 plane strain plane stress n () 212 / 303
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
back to index NUMERICAL FRACTURE MECHANICS
Numerical fracture mechanics Methods EEM ; BEM Calculations G K δt J Simulation crack growth () 215 / 303
Quadratic elements ξ 2 ξ 1 4 8 7 ξ 2 3 ξ 1 6 4 7 3 1 4 7 5 3 6 2 8 1 5 6 2 1 8 5 2 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
Crack tip mesh bad approximation stress field results are mesh-dependent 1/ r () 217 / 303
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
Quarter point elements 4 7 3 6 8 1 5 2 p 3p 3 5 6 1 4 2 4 8 1 7 p 3 6 5 2 3p 4 1 3 2 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
Crack tip rozet Quarter Point Elements : 8x Transition Elements : number is problem dependent Buffer Elements () 220 / 303
One-dimensional case ξ = 1 ξ = 0 ξ = 1 1 x 3 2 position displacement and strain x = 1 2 ξ(ξ 1)x 1 + 1 2 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 1 + 1 2 ξ(ξ + 1)u 2 (ξ 2 1)u 3 du dξ = (ξ 1 2 )u 1 + (ξ + 1 2 )u 2 2ξu 3 du dx = du dξ dξ dx = du dξ /dx dξ () 221 / 303
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
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
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
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 : ± 0.001 element size not possible with crack tip in interface unloaded crack plane no thermal stresses () 225 / 303
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
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
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
J-integral J = Γ ( ) ε u i Wn 1 t i dγ with W = σ ij dε ij x 1 0 () 229 / 303
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
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
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
Crack growth simulation Node release Moving Crack Tip Mesh Element splitting Smeared crack approach () 233 / 303
Node release node collocation technique () 234 / 303
Moving Crack Tip Mesh () 235 / 303
Element splitting () 236 / 303
Smeared crack approach e 2 e 1 e 2 e 1 n 2 σ 2 σ 1 n1 n 2 n 1 () 237 / 303
back to index FATIGUE
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
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
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
Experiments full-scale testing a.o. train axles airplanes laboratory testing harmonic loading constant force/moment strain/deflection SIF () 242 / 303
Train axle D = 0.75 [m] 1 rev = πd = π 0.75 2.25 [m] 1 km = 1000 m = 1000 2.25 = 4000 9 445 [c(ycles)] 1 day Maastricht - Groningen = 2.5 333 [km] = 1000 [km] 1 day Maastricht - Groningen = 445 10 3 [c] 1 year = 300 445 10 3 [c] = 1335 10 5 [c] 1.5 10 8 [c] frequency : 100 [km/h] = 445 10 2 [c/h] = 44500 3600 = 12.5 [c/sec] = 12.5 [Hz] () 243 / 303
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 30-80 Hz tensile electric 50-300 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
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
(S-N)-curve B.S. 3518 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 50 10 6 steels : σ th 1 2 σ b () 246 / 303
(S a -N)-curve B.S. 3518 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
Examples 600 550 500 450 steelt1 400 σ max [MPa] 350 300 250 steel1020 200 150 100 Mgalloy Al2024T4 10 4 10 5 10 6 10 7 10 8 10 9 N f () 248 / 303
Influence of average stress σ a σ m σ th 0 0 log(n f ) () 249 / 303
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
(P-S-N)-curve 550 500 450 400 σ max [MPa] 350 300 90% prob.failure 250 50% prob.failure 200 150 10% prob.failure 100 10 4 10 5 10 6 10 7 10 8 N f () 251 / 303
High/low cycle fatigue S a σ m = 0 0 0 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
High/low cycle fatigue S a σ m = 0 0 0 LCF 4 5 HCF log(n f ) low cycle fatigue N f < ±50000 high stresses EPFM strain-life curve Manson-Coffin relation () 253 / 303
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
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
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
Influence factors load spectrum stress concentrations stress gradients material properties surface quality environment () 257 / 303
Load spectrum sign / magnitude / rate / history multi-axial lower f.limit than uni-axial () 258 / 303
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 = 1 1 + a ρ 1 Neuber : q = 1 + b ρ with with q(ρ) = notch sensitivity factor a = material parameter b = grain size parameter () 259 / 303
Stress gradients full-scale experiments necessary () 260 / 303
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
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
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
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
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