Fracture Mechanics, Damage and Fatigue Linear Elastic Fracture Mechanics Stress Intensity Factor (SIF)

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1 Universit of Liège Aerospace & Mechanical Engineering Fracture Mechanics, Damage and Fatigue Linear Elastic Fracture Mechanics Stress Intensit Factor (SIF) Ludovic Noels Computational & Multiscale Mechanics of Materials CM3 Chemin des Chevreuils 1, B4000 Liège Fracture Mechanics LEFM SIF

2 Asmptotic solution of LEFM Summar Mode I Mode II Mode III (opening) (sliding) (shearing) r q z Detailed epressions for the 3 modes Fracture Mechanics LEFM SIF 2

3 LEFM: Computation of SIF Computation of the SIFs Analtical methods (for LEFM) Full field solution (see net slides) Limited to infinite planes SIFs obtained from asmptotic limit Superposition (see net slides) Of eisting solutions Energetic approach (see previous lecture) Related to Griffith s work Numerical Collocation method (see Anne 2) FEM Capture asmptotic solution» Use singular (Barsoum, 1974) quadratic elements (central nodes moved to quarter edge) Energetic approach J integral Eperimental Normalized eperiments Strain Gauge Method Use of SIF handbook Fracture Mechanics LEFM SIF 3

4 SIF: Analtical methods Reminder: method in linear elasticit Problem is governed b the bi-harmonic equation One solution of this equation has the form where w(z) and W(z) are functions to be defined so The stress field The displacement field Plane s satisf the BCs Plane e Fracture Mechanics LEFM SIF 4

5 Computation of SIF b analtical method (LEFM) Full field solution of a crack submitted to traction Infinite plane Mode I:, and t (-) = t (), t (-) = -t () Westergaard approach for mode I As and as for mode I, one should have s = 0 for = 0, SIF: Analtical methods t 2a t the solution is Indeed with this choice t 2a Fracture Mechanics LEFM SIF 5

6 SIF: Analtical methods Computation of SIF b analtical method (LEFM) (2) Westergaard solution for mode I (2) The general stress/displacement fields read t 2a Using the become Onl W to be defined Fracture Mechanics LEFM SIF 6

7 SIF: Analtical methods Computation of SIF b analtical method (LEFM) (3) Full field solution of a crack submitted to traction Westergaard solution for mode I t 2a Solution W( 0) satisfing the BCs for a uniform traction (see net slides) NB: General solution (Sedov, 1972): Fracture Mechanics LEFM SIF 7

8 SIF: Analtical methods Computation of SIF b analtical method (LEFM) (4) Check solution with Smmetr with respect to O and O R (C) = R (C 1 ) = I (C) = I (C 1 ) = 0 n - n + z r a q Fracture Mechanics LEFM SIF 8

9 SIF: Analtical methods Computation of SIF b analtical method (LEFM) (5) Check solution (2) t with 2a On crack lips z = ± i e, e 0, < a n - n + z r a q Boundar condition on crack lips: t = n. s = 2 R (W ) = ± t satisfied ± Fracture Mechanics LEFM SIF 9

10 SIF: Analtical methods Computation of SIF b analtical method (LEFM) (6) Asmptotic field on crack lips n - z r q n + a For z = a-r ± i e, e 0 Crack Opening Displacement (COD) Fracture Mechanics LEFM SIF 10

11 SIF: Analtical methods Computation of SIF b analtical method (LEFM) (7) Asmptotic field ahead of crack n - z r q n + a For z = a+r ± i e, e 0, r 0 +, Fracture Mechanics LEFM SIF 11

12 SIF: Analtical methods Computation of SIF b analtical method (LEFM) (8) Asmptotic stress field ahead of crack n - z r q n + a For z = a+r ± i e, e 0, r 0 +, Fracture Mechanics LEFM SIF 12

13 SIF: Analtical methods Computation of SIF b analtical method (LEFM) (9) Crack in an infinite plate under perpendicular loading Superposition s s s 2a 2a Case 1 Case 2 s s s Case 2: solved for t = s Case 1: & & Fracture Mechanics LEFM SIF 13

14 s /s SIF: Analtical methods Computation of SIF b analtical method (LEFM) (10) Crack in an infinite plate under perpendicular loading Full field solution from W and case 1 Asmptotic solution along = 0, for > a SIF: s 100 Zone of asmptotic solution (in terms of 1/r 1/2 ) dominance s Asmptotic s True s 10 2a s Plasticit Structural response 1 asmptotic full field r/a Fracture Mechanics LEFM SIF 14

15 Computation of SIF b analtical method (LEFM) (11) Crack in an infinite plate under sliding Westergaard approach for mode II For mode II, s = 0 for = 0 SIF: Analtical methods 2a t Crack subjected to a shearing t t Appling superposition principle and t =t Fracture Mechanics LEFM SIF 15

16 SIF: Analtical methods Computation of SIF b analtical method (LEFM) (12) Crack in an infinite plate under shearing Stress field (see Anne 1) t 2a Ahead of crack tip ( = 0 and = a + r) t q 2 r 2 r 1 q 1 r q Fracture Mechanics LEFM SIF 16

17 SIF: Analtical methods Computation of SIF b analtical method (LEFM) (13) Here we have obtained the SIF using the full field solution so Wh did we develop the asmptotic solution last time instead of using this full field solution directl? Fracture Mechanics LEFM SIF 17

18 SIF: Handbooks Solutions from SIF handbooks (see references) Obtained b using a variet of methods (analtical, numerical, ) Eample: mode I for a crack in a finite plate (h/w > 3) s General formula Numerical results based on Laurent epansion» Isida, (1-a/W)^1/2 f(a/w) W 2a h Periodic crack approimation a/w» <5% error for a/w < 0.5» Irwin, 1957 Fit of Isida s values» <0.1% error» Tada, 1973 s Fracture Mechanics LEFM SIF 18

19 SIF: Numerical approaches Finite element model: etraction from stress and displacement fields Asmptotic crack tip stress and displacement fields determined b the SIF & Stress correlation FEM computation Etract stress field s(r,q) for q=0 (or an other q) Etrapolate to get the SIFs Displacement correlation Idem but with Fracture Mechanics LEFM SIF 19

20 Finite element model: etraction from stress and displacement fields (2) Advantages of the method Simple Can be used with an FE software Onl one post-processing to determine the 3 SIFs (if suitable loading) Can be used along a crack front in 3D When using displacement correlation, the field is primar solution Drawbacks SIF: Numerical approaches The accurac is strongl dependant on the mesh refinement The mesh refinement required depends on the element abilit to capture the singularit at crack tip (see net slide) Fracture Mechanics LEFM SIF 20

21 SIF: Numerical approaches Finite element model: Barsoum elements Previous method requires a fine mesh since a singularit in r ½ is not naturall captured b usual FE This singularit can be captured B enriching elements (as in XFEM for LEFM, see later) B using Barsoum (quarter-point) elements Quarter-point elements h 1 Quadratic elements with some mid-nodes located at a quarter of the edge h Mapping L/4 3L/4 Mapping L/4 3L/ Fracture Mechanics LEFM SIF 21

22 SIF: Numerical approaches Finite element model: Barsoum elements (2) Quarter-point element: 1D eample Shape functions N N N Mapping 0 L/4 L Strain field So a strain field (and so for stress) in 1/r ½ can be captured Fracture Mechanics LEFM SIF 22

23 SIF: Numerical approaches Finite element model: Barsoum elements (3) Quarter-point element: 2D eample Fine mesh at crack tip: mid-nodes of first ring are moved to quarter points Fracture Mechanics LEFM SIF 23

24 SIF: Numerical approaches Finite element model: global energ & compliance For cracks growing straight ahead (if onl one mode is involved) G=J, and in linear elasticit Prescribed loading: (in linear elasticit) Prescribed displacement: Perform 2 computations With the same loading (displacement) but with two different crack lengths So, per unit thickness: Could be rewritten using compliance C in terms of the generalized loading Q (linear elasticit) Advantages Simple Can be used with an FE software and requires onl post-processing Less mesh sensitive than correlation methods as a global variable is used Drawbacks 2 computations needed Onl one mode (so one SIF) can be considered at a time In 3D, SIF variation along crack front cannot be determined Fracture Mechanics LEFM SIF 24

25 SIF: Numerical approaches Finite element model: crack closure integral Cracks growing straight ahead r q r q a s = 0, u 0 Da s 0, u = 0 For such cracks, in linear elasticit a s = 0, u 0 Da s = 0, u Fracture Mechanics LEFM SIF 25

26 SIF: Numerical approaches Finite element model: crack closure integral (2) This integral can be computed using FE software Nodal release First computation: nodes j on both crack lips are constrained together F j-2 Second computation: constrain on node j is released j-2 SIFs can be computed from a a j-1 -F j j+1 Da Du j-1 j j+1 Da Reaction forces of first computation & Displacement jumps of second computation, & Fracture Mechanics LEFM SIF 26

27 Finite element model: crack closure integral (3) Advantage Requires onl post processing Drawbacks SIF: Numerical approaches If the crack is not a line of smmetr, the node displacements have to be constrained (using Lagrange multipliers e.g.) Two computations are needed (can be improved using modified nodal release) Fracture Mechanics LEFM SIF 27

28 SIF: Numerical approaches Finite element model: direct computation of J-integral For linear elasticit and for an contour G embedding a straight crack & Assuming smmetr & a structured meshed G 2 G 3 G 1 dl = d on G 1, dl = -d on G 1,dl = -d on G 3 As values are computed on Gauss points, ideall, the integrals will add Gauss points contributions Fracture Mechanics LEFM SIF 28

29 SIF: Numerical approaches Finite element model: J-integral b domain integration For linear elasticit and for an contour G embedding a straight crack & Let us define a contour C=G 1 +G - +G + -G G 1 Interior of this contour is region D Define q(, ) such that q = 1 on G G + q = 0 on G 1 - G - G As crack is stress free Fracture Mechanics LEFM SIF 29

30 Finite element model: J-integral b domain integration (2) Computation of SIF: Numerical approaches C is closed divergence theorem C q=0 But & =0 q=1 q is discretized using the same shape functions than the elements This integral is valid for an annular region around the crack tip As long as the crack lips are straight D Fracture Mechanics LEFM SIF 30

31 SIF: Numerical approaches Finite element model: J-integral Advantages Accurate (especiall using domain integration) Does not require Barsoum elements Drawbacks Onl one mode at the time Either modifing the FE code in order to have eas post processing or Post processing difficult when using a standard FE software As this method is reall accurate and computationall efficient, can it be modified in order to etract the SIF of each mode? Fracture Mechanics LEFM SIF 31

32 SIF: Numerical approaches Finite element model: J-integral (2) How to etract the different SIFs? Compute J for the solution u of the considered problem J Compute J for another field u au to be specialized J au Compute J for the sum of u & u au to be specialized J s Relation between the different Js? Fracture Mechanics LEFM SIF 32

33 SIF: Numerical approaches Finite element model: J-integral (3) How to etract the different SIFs (2)? Relation between the different Js can be deduced from So is rewritten The right term is called the interaction integral» What is its epression in terms of the SIFs? Fracture Mechanics LEFM SIF 33

34 SIF: Numerical approaches Finite element model: J-integral (4) How to etract the different SIFs (3)? It has been found that With & Direct substitution leads to Fracture Mechanics LEFM SIF 34

35 SIF: Numerical approaches Finite element model: J-integral (5) How to etract the different SIFs (4)? As With And as these two last relations are smmetric in K and K au B analog with One can feel that This is obtained eplicitl after substitution of f and g b their closed forms Fracture Mechanics LEFM SIF 35

36 SIF: Numerical approaches Finite element model: J-integral (6) The SIFs are deduced from the so-called interaction integral Indeed, if u au is chosen such that onl K i au 0, K i is obtained directl This method Is ver accurate compared to correlation methods But it requires More computations Etensive modifications of the FE code Fracture Mechanics LEFM SIF 36

37 Strain gauge method Asmptotic solution for mode I SIF: eperimental methods with l = -½, 0, ½, The asmptotic solution in r -½ cannot be matched accuratel using a gauge So the solution considered is limited to the 3 rd order Fracture Mechanics LEFM SIF 37

38 SIF: eperimental methods Strain gauge method (2) Position of the strain gauge Located at (r, q ) of the crack tip With an orientation a One** can show that in the referential O r q a with The gauge is located at (r, q *, a * ) such that & Mounting of the gauge is critical **Dall JW and Sanford RJ (1988), Strain gage methods for measuring the opening mode stress intensit factor KI, Eperimental Mechanics 27, Fracture Mechanics LEFM SIF 38

39 SIF: eperimental methods Strain gauge method (3) Strain gauges are now replaced b Digital Image Correlation (DIC) Markers on the surface are tracked opticall Displacements and strains can be computed These strains can be used to deduce the SIFs Fracture Mechanics LEFM SIF 39

40 Eercise 1 Flawed clinder A piston is used to increase inner pressure From 0 to 55 MPa Clinder made of Peaked-aged aluminum allo 7075-T651 Yield s p0 = 550 Mpa Toughness K IC = 30 MPa m 1/2 Malfunction Clinder burst Post failure analses Initial elliptical flaw at inner wall» 4.5 mm long Origin of burst?» 1.45 mm deep» Normal to hoop stress D in = 9 cm t = 1 cm s qq L= 20 cm a = 1.45 mm 2c = 4.5 mm s qq Fracture Mechanics LEFM SIF 40

41 Lecture notes References Lecture Notes on Fracture Mechanics, Alan T. Zehnder, Cornell Universit, Ithaca, Other references «on-line» Book Fracture Mechanics, Piet Schreurs, TUe, Fracture Mechanics: Fundamentals and applications, D. T. Anderson. CRC press, Fatigue of Materials, S. Suresh, Cambridge press, Fracture Mechanics LEFM SIF 41

42 Stress field Consider thick clinder with r in = m & r out = m Inner pressure p Eercise 1: Solution D in = 9 cm t = 1 cm L= 20 cm s qq a = 1.45 mm 2c = 4.5 mm s qq Fracture Mechanics LEFM SIF 42

43 Eercise 1: Solution SIF The wall is not perforated Use SIF for semi-elliptical crack in large plate See SIF handbook Geometrical effect Plasticit correction SSY criterion: Not full satisfied plasticit correction s qq 2c = 4.5 mm s qq a = 1.45 mm SIF with Rupture: For r = r in, p would be 100 MPa, this is the critical value Fracture Mechanics LEFM SIF 43

44 Eercise 1: Solution Rupture mode For r = r in, p would be 100 Mpa, this is the critical value This is out of the range of the piston activit So rupture should come from fatigue Cclic loading p from 0 to 55 Mpa Hoop stress from 0 to SIF Assuming a/c remains constant during crack propagation Fracture Mechanics LEFM SIF 44

45 Eercise 1: Solution Cclic loading p from 0 to 55 Mpa & Due to initial flaw Assuming curves are valid for R=0 We are in Paris regime crack propagation Rupture will happen for Number of ccles? Fracture Mechanics LEFM SIF 45

46 Eercise 1: Solution Cclic loading (2) As Assuming curves are valid for R=0 We are in Paris regime!!!life strongl depends on the maimum pressure reached during accidents Fracture Mechanics LEFM SIF 46

47 Anne 1: Analtical methods Computation of SIF b analtical method (LEFM) Crack in an infinite plate under shearing Mode III: t 2a u z is the imaginar part of a function z(z) t This function has to be found to satisf the BCs Stress field As One has And Fracture Mechanics LEFM SIF 47

48 Anne 1: Analtical methods Computation of SIF b analtical method (LEFM) (2) Crack in an infinite plate under shearing (2) Solution of the problem? t with t 2a u z = I(z) Smmetr? u z (-) = -u z () satisfied Far awa field? For ± : s z ±t satisfied Fracture Mechanics LEFM SIF 48

49 Anne 1: Analtical methods Computation of SIF b analtical method (LEFM) (3) Crack in an infinite plate under shearing (3) Solution of the problem? t with t 2a Crack lips stress free? q 2 r 2 r 1 q 1 r q satisfied Fracture Mechanics LEFM SIF 49

50 Anne 2: Semi analtical methods Computation of SIF b Boundar Collocation Method (LEFM) Eample: crack in a finite circular plate Considering mode III (can be done for I and II) Laurent series a=r R q with T z = f(q) But for the crack lies in q=p, so we use is satisfied Finite displacement n = 0, 1, Fracture Mechanics LEFM SIF 50

51 Anne 2: Semi analtical methods Computation of SIF b Boundar Collocation Method (LEFM) (2) Eample: crack in a finite circular plate (2) Unknowns a n are obtained b defining m collocation points on the boundar At these points for k=1,.., m, & assuming an epansion up to order p, p+1 m T z = f(q k ) Eample: R=1, & loading R a=r q k K III K III m=2(p+1) For accurac m = 2p+2 Collocation points onl on upper side (avoid trivial solution) Least squares resolution p Fracture Mechanics LEFM SIF 51