INDEX Introduction (ch 1) Theoretical strength (ch 2) Ductile/brittle (ch 2) Energy balance (ch 4) Stress concentrations (ch 6) () April 30, 2018 1 / 52
back to index INTRODUCTION
Introduction () 3 / 52
Continuum mechanics V 0 A 0 u V A x 0 e 3 x e 1 O e 2 () 4 / 52
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 2ε 12,12 ε 11,22 ε 22,11 = 0 ; - 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 ) () 5 / 52
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 () 6 / 52
Stress-strain curves σ σ ε ε σ σ ε ε () 7 / 52
Fracture () 8 / 52
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 () 10 / 52
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 () 11 / 52
Approximations linearization linear strain of atomic bond elastic modulus ( ) x σ = σ max sin γ σ max x γ σ2 max ε = x a 0 x = εa 0 σ = εa 0 γ σ2 max ( ) ( dσ dσ E = = 0) dε x=0 dx x=0 a = σ 2 max Eγ σ max = a 0 a 0 γ theoretical strength σ th = Eγ a 0 () 12 / 52
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 σ th σ b () 13 / 52
Griffith s experiments 11000 σ b [MPa] 170 10 20 d [µ] DEFECTS FRACTURE MECHANICS () 14 / 52
Crack loading modes Mode I Mode II Mode III Mode I = opening mode Mode II = sliding mode Mode III = tearing mode () 15 / 52
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 () 16 / 52
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 () 17 / 52
ENERGY!! DUCTILE/BRITTLE FRACTURE back to index
Ductile - brittle behavior σ ABS, nylon, PC PE, PTFE 0 10 100 ε (%) surface energy : γ [Jm 2 ] solids : γ 1 [Jm 2 ] ex.: alloyed steels; rubber () 19 / 52
Charpy v-notch test () 20 / 52
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 () 21 / 52
Fracture mechanisms shear fracture cleavage fracture () 22 / 52
Shearing dislocations voids crack dimples load direction () 23 / 52
Dimples () 24 / 52
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 () 25 / 52
back to index ENERGY BALANCE
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 ] du e da du i da = du a da + du d da + du k da [Jm 1 ] () 27 / 52
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 ] () 28 / 52
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 () 29 / 52
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 energy balance G = 1 ( due B da du ) i = R = G c da critical crack length a c = G ce πσ 2 ; Griffith s crack criterion G = G c () 30 / 52
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 () 31 / 52
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 () 32 / 52
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 () 33 / 52
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 no fixed grips AND no constant load BUT small deviation!! () 34 / 52
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 () 35 / 52
back to index STRESS CONCENTRATIONS
Governing equations - displacements u i - strains ε kl = 1 2 (u k,l + u l,k ) - compatibility relations 2ε 12,12 ε 11,22 ε 22,11 = 0 - equilibrium equations σ ij,j + ρq i = 0 ; σ ij = σ ji - material model σ ij = C ijkl ε lk () 37 / 52
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 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 () 38 / 52
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 () 39 / 52
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 θ { e r(θ)} = e t (θ) ; θ { e t(θ)} = e r (θ) () 40 / 52
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 () 41 / 52
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 () 42 / 52
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 () 43 / 52
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) () 44 / 52
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 () 45 / 52
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 () 46 / 52
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 () 47 / 52
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) () 48 / 52
Stresses for total load σ rr = σ 2 [(1 a2 r 2 ) ) ] + (1 + 3a4 r 4 4a2 r 2 cos(2θ) σ tt = σ 2 [(1 + a2 r 2 ) ) ] (1 + 3a4 r 4 cos(2θ) σ rt = σ 2 ] [1 3a4 r 4 + 2a2 r 2 sin(2θ) () 49 / 52
Special points 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 stress concentration factor K t = σ max σ = 3 [-] K t is independent of hole diameter! () 50 / 52
Stress gradients large hole : smaller stress gradient larger area with higher stress higher chance for critical defect in high stress area () 51 / 52
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 [-] () 52 / 52