INDEX. Introduction (ch 1) Theoretical strength (ch 2) Ductile/brittle (ch 2) Energy balance (ch 4) Stress concentrations (ch 6)

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Ἐλιούδ Μπουκουβαλαίοι
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1 INDEX Introduction (ch 1) Theoretical strength (ch 2) Ductile/brittle (ch 2) Energy balance (ch 4) Stress concentrations (ch 6) () April 30, / 52

2 back to index INTRODUCTION

3 Introduction () 3 / 52

4 Continuum mechanics V 0 A 0 u V A x 0 e 3 x e 1 O e 2 () 4 / 52

5 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

6 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

7 Stress-strain curves σ σ ε ε σ σ ε ε () 7 / 52

8 Fracture () 8 / 52

9 back to index THEORETICAL STRENGTH

10 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 () 10 / 52

11 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

12 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

13 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 σ th σ b () 13 / 52

14 Griffith s experiments σ b [MPa] d [µ] DEFECTS FRACTURE MECHANICS () 14 / 52

15 Crack loading modes Mode I Mode II Mode III Mode I = opening mode Mode II = sliding mode Mode III = tearing mode () 15 / 52

16 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

17 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

18 ENERGY!! DUCTILE/BRITTLE FRACTURE back to index

19 Ductile - brittle behavior σ ABS, nylon, PC PE, PTFE ε (%) surface energy : γ [Jm 2 ] solids : γ 1 [Jm 2 ] ex.: alloyed steels; rubber () 19 / 52

20 Charpy v-notch test () 20 / 52

21 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

22 Fracture mechanisms shear fracture cleavage fracture () 22 / 52

23 Shearing dislocations voids crack dimples load direction () 23 / 52

24 Dimples () 24 / 52

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 () 25 / 52

26 back to index ENERGY BALANCE

27 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

28 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

29 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

30 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

31 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

32 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

33 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

34 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

35 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

36 back to index STRESS CONCENTRATIONS

37 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

38 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 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

39 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 () 39 / 52

40 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

41 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 () 41 / 52

42 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 () 42 / 52

43 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 () 43 / 52

44 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) () 44 / 52

45 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 () 45 / 52

46 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

47 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 () 47 / 52

48 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

49 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

50 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

51 Stress gradients large hole : smaller stress gradient larger area with higher stress higher chance for critical defect in high stress area () 51 / 52

52 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 [-] () 52 / 52

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