Introduction to fracture mechanics

Transcription

1 Introduction to fracture mechanics Prof. Dr. Eleni Chatzi Dr. Giuseppe Abbiati, Dr. Konstantinos Agathos Lecture 6-9 November, 2017 Institute of Structural Engineering, ETH Zu rich November 9, 2017 Institute of Structural Engineering Method of Finite Elements II 1

2 Outline 1 Introduction 2 Linear elastic fracture mechanics Historical background General elasticity problem Airy stress function Some relevant solutions Westergaard solution Stress Intensity Factors Energy release rate Mixed mode fracture J and interaction integrals 3 Cohesive zone models Institute of Structural Engineering Method of Finite Elements II 2

3 Learning goals Reviewing basic elasticity equations and solution methods Understanding some basic concepts of Linear Elastic Fracture Mechanics Introducing brittle and cohesive fracture Reviewing some basic methods for fracture mechanics Institute of Structural Engineering Method of Finite Elements II 3

4 Applications Fracture in brittle materials Fatigue fracture Fracture in concrete Development of numerical methods for fracture Institute of Structural Engineering Method of Finite Elements II 4

5 Resources Some useful resources: fracturemechanics.org Extended Finite Element Method for Fracture Analysis of Structures by Soheil Mohammadi, Blackwell Publishing, 2008 Institute of Structural Engineering Method of Finite Elements II 5

6 History Some research relevant to the field was conducted starting from the end of the 19th century Motivation was provided from experiments in brittle materials were the measured strength was much lower than the theoretical prediction In general very few developments were made in the field until World War II Institute of Structural Engineering Method of Finite Elements II 6

7 Motivation Due to their local nature cracks were not considered a threat for large structures During the war several failures in ships and aircraft occurred as a result of crack propagation Source:Wikipedia Research in the field was triggered as a result Institute of Structural Engineering Method of Finite Elements II 7

8 Motivation Due to their local nature cracks were not considered a threat for large structures During the war several failures in ships and aircraft occurred as a result of crack propagation Source:Wikipedia Research in the field was triggered as a result Institute of Structural Engineering Method of Finite Elements II 7

9 Motivation Due to their local nature cracks were not considered a threat for large structures During the war several failures in ships and aircraft occurred as a result of crack propagation Source:Wikipedia Research in the field was triggered as a result Institute of Structural Engineering Method of Finite Elements II 7

10 General elasticity problem General problem (in the presence of a crack): Γ 0 Ω t y t c Γ t Γc t Γc 0 Γ u x z Institute of Structural Engineering Method of Finite Elements II 8

11 Fracture mechanics Governing equations: σ + f = 0 in Ω u = ū on Γ u σ n = t on Γ t σ n = 0 on Γ 0 c σ n = t c on Γ t c where σ is the Cauchy stress tensor n is the unit outward normal f is the body force per unit volume u is the displacement field Institute of Structural Engineering Method of Finite Elements II 9

12 General elasticity problem Kinematic equations (small deformations): ɛ = s u The constitutive equations are given by Hooke s law: where D is the elasticity tensor. σ = D : ɛ Institute of Structural Engineering Method of Finite Elements II 10

13 General elasticity problem Using index notation the above can be written as: σ ji,j + f i = 0 in Ω u i = ū i on Γ u n i σ ij = t j on Γ t n i σ ij = 0 on Γ 0 c n i σ ij = t j c on Γ t c ɛ ij = 1 2 ( ui x j + u ) j x i σ ij = D ijkl ɛ kl, ɛ ij = S ijkl σ kl Institute of Structural Engineering Method of Finite Elements II 11

14 Displacement formulation Displacement formulation : Kinematic equations are substituted into the constitutive equations Both of the above are substituted into the equilibrium equations The resulting system consists of three equations in three unknowns The problem is solved in terms of the displacements Institute of Structural Engineering Method of Finite Elements II 12

15 Stress formulation Stress formulation : The equilibrium equations are directly solved in terms of stresses Equilibrium provides three equations The (symmetric) stress tensor consists of six independent components Three more equations are needed! Institute of Structural Engineering Method of Finite Elements II 13

16 Stress formulation The Saint Venant compatibility equations can be employed as additional equations: ɛ ij,kl + ɛ kl,ij = ɛ ik,jl + ɛ jl,ik In terms of stresses: (S ijmn σ mn ),kl + (S klmn σ mn ),ij = (S klmn σ mn ),jl + (S jlmn σ mn ),ik of the above only three equations are independent Institute of Structural Engineering Method of Finite Elements II 14

17 Stress formulation The compatibility equations: Guarantee that the computed strains/stresses are the symmetric derivative of a vector field Guarantee that the resulting displacement field will not exhibit any gaps or overlaps In conjunction to the equilibrium equations form a system that can be solved to provide the stress field Institute of Structural Engineering Method of Finite Elements II 15

18 Airy stress function A function Φ is introduced for the 2D case such that: σ 11 = σ xx = 2 Φ x 2 2 σ 22 = σ yy = 2 Φ x 2 1 = 2 Φ y 2 = 2 Φ x 2 σ 12 = σ xy = 2 Φ x 1 x 2 = 2 Φ x y Φ is called a stress function. Institute of Structural Engineering Method of Finite Elements II 16

19 Airy stress function In the absence of body forces f i = 0, the equilibrium equations become: σ xx,x + σ xy,y = Φ,yyx Φ,xyy = 0 σ yx,x + σ yy,y = Φ,xyx + Φ,xxy = 0 They are satisfied by default! Institute of Structural Engineering Method of Finite Elements II 17

20 Airy stress function Substituting the stress function in the compatibility equations the following can be obtained: Φ,xxxx + 2Φ,xyxy + Φ,yyyy = 0 The above is called biharmonic equation and can also be written as: 4 Φ = 0 Institute of Structural Engineering Method of Finite Elements II 18

21 Airy stress function Using stress functions: In the absence of body forces allows the reduction of the number of equations to one The problem reduces to the problem of determining an appropriate stress function for the problem at hand Solutions to several classical elasticity problems have been obtained Institute of Structural Engineering Method of Finite Elements II 19

22 Circular hole in an infinite plate We consider the problem circular in an infinite domain subjected to a far field stress: Institute of Structural Engineering Method of Finite Elements II 20

23 Circular hole in an infinite plate Kirsch solved the problem in 1898 by using the stress function: Φ = σ o r 2 cos 2 θ which results in the stress field: σ rr = σ θθ = σ rθ = σ o 2 σ o 2 σ o 2 ( ( ) ) a σ ( ( o a 1 4 r 2 r ( ( ) ) a σ ( ( ) ) o a cos 2θ r 2 r ( ( ) a 2 ( ) ) a sin 2θ r r ) 2 ( ) ) a 4 3 cos 2θ r Institute of Structural Engineering Method of Finite Elements II 21

24 Circular hole in an infinite plate At the hole (r = a): σ rr = 0 σ θθ = σ 0 (1 2 cos 2θ) σ rθ = 0 For θ = ± π 2 : σ θθ = 3σ 0 A stress concentration factor of 3 occurs! Institute of Structural Engineering Method of Finite Elements II 22

25 Circular hole in an infinite plate 3D and contour plot of the σ θθ stress component: Institute of Structural Engineering Method of Finite Elements II 23

26 Elliptical hole in an infinite plate Inglis (1913), obtained the solution for an elliptical hole in an infinite domain subjected to a far field stress: Institute of Structural Engineering Method of Finite Elements II 24

27 Elliptical hole in an infinite plate Maximum stress predicted by the solution, at [a, 0], is: ( σ max = σ o a ) b We observe that for b = a (circular hole): σ max = 3σ o For b = 0 the hole turns into a sharp crack: A singularity occurs! σ max = Institute of Structural Engineering Method of Finite Elements II 25

28 Westergaard solution We consider the problem of a sharp crack in an infinite domain subjected to a far field stress: Institute of Structural Engineering Method of Finite Elements II 26

29 Westergaard solution Westergaard solved the problem by considering the complex stress function: Φ = Re Z + y Im Z where: Z is a complex function defined as: Z = z is a complex number: z = x + iy Z denotes the integral of Z σ o ( ) a 2 1 z Institute of Structural Engineering Method of Finite Elements II 27

30 Westergaard solution For r a the stress expression of the solution is: a σ xx = σ o 2r cos θ ( 1 sin θ 2 2 a σ yy = σ o 2r cos θ ( 1 + sin θ 2 2 σ xy = σ o a 2r sin θ 2 cos θ 2 ) 3θ sin 2 sin 3θ 2 cos 3θ ) +... Institute of Structural Engineering Method of Finite Elements II 28

31 Westergaard solution 3D and contour plot of the σ yy stress component: Institute of Structural Engineering Method of Finite Elements II 29

32 Stress Intensity Factors We define the stress intensity factors (SIFs) corresponding to the three modes of fracture as: Mode I Mode II Mode III K I = lim σ yy 2πr r 0 θ=0 K II = lim σ yy 2πr r 0 θ=0 K III = lim σ yy 2πr r 0 θ=0 Institute of Structural Engineering Method of Finite Elements II 30

33 Stress Intensity Factors The mode I SIF for the Westergaard solution is: K I = σ o πa Rewritting stresses w.r.t the SIFs gives: σ xx = σ yy = σ xy = K I cos θ ( 1 sin θ 2πr 2 2 K I cos θ ( 1 + sin θ 2πr 2 2 ) 3θ sin 2 sin 3θ 2 K I sin θ 2πr 2 cos θ 3θ cos ) +... Institute of Structural Engineering Method of Finite Elements II 31

34 Energy release rate It can be observed in the solutions above that: Stresses at the tip of a crack are infinite independently of the load applied Since no material can withstand infinite stress, structures should fail even in the presence of the smallest crack Institute of Structural Engineering Method of Finite Elements II 32

35 Energy release rate To circumvent the above situation, Griffith (1920) proposed an energy based failure criterion. First the energy required to propagate the crack by a length a (in 2D) is considered: where: U Γ = 2γ s a γ s is the required energy per unit length The factor 2 is used to account for both crack surfaces Institute of Structural Engineering Method of Finite Elements II 33

36 Energy release rate Next the total change of potential energy in the system is considered: where: dπ dt = d (W U s) dt U s is the strain energy: U s = 1 2 σ : ɛ = 1 2 σ ijɛ ij W is the work produced by external forces With respect to the crack length: dπ da = d (W U s) da Institute of Structural Engineering Method of Finite Elements II 34

37 Energy release rate The total change in potential energy should be equal to the energy dissipated to propagate the crack: dπ da = d (W U s) = d (U γ) = 2γ s da da We define the energy release rate as: and critical release rate as: G = dπ da G c = 2γ s Institute of Structural Engineering Method of Finite Elements II 35

38 Energy release rate Griffith s criterion for crack propagation is written as: where: G G c The critical energy release rate (G c ) is a material property The crack will propagate if the energy release rate (change in potential energy) is greater than a critical value Institute of Structural Engineering Method of Finite Elements II 36

39 Energy release rate Computing the total potential energy for the Westergaard solution yields: Π = πσ2 oa 2 2E, E = Differentiating with respect to a: E plane stress E 1 ν 2 plane strain G = dπ da = πσ2 oa E Institute of Structural Engineering Method of Finite Elements II 37

40 Energy release rate Setting G = G c : πσ 2 ca E = G c where σ c is the critical/failure stress: Also: σ c = E Gc πa K Ic = σ c πa is defined as the critical SIF. Institute of Structural Engineering Method of Finite Elements II 38

41 Energy release rate Griffith s criterion can be written in terms of the SIFs: K I K Ic Also the mode I SIF can be related to the energy release rate by: G = K I 2 E, G c = K Ic 2 E Institute of Structural Engineering Method of Finite Elements II 39

42 Mixed mode fracture The above results are derived for pure mode I, for mixed mode fracture the following relation holds: G = K I 2 + KII 2 E ν E K III 2 Also appropriate criteria should be used which also yield the direction of propagation. Those assume the form: f (K I, K II, K Ic ) = 0 Typically mode III is not taken into account and only K Ic is experimentally measured. Institute of Structural Engineering Method of Finite Elements II 40

43 Maximum circumferential stress criterion Erdogan and Sih (1963) developed a criterion for mixed mode fracture according to which: The crack will propagate in a direction normal to the direction of the maximum circumferential stress The crack will propagate when the maximum circumferential stress reaches a critical value Institute of Structural Engineering Method of Finite Elements II 41

44 Maximum circumferential stress criterion By transforming the stresses from the Westergaard solution for modes I and II to a polar system of coordinates we obtain: σ rr = σ θθ = σ rθ = K I cos θ ( 1 + sin 2 θ ) + K ( II 2πr 2 2 2πr K I cos θ ( 1 sin 2 θ ) + K ( II 2πr 2 2 2πr K I 2πr sin θ 2 cos2 θ 2 + K II 2πr ( 1 4 sin θ sin θ sin θ ) 3θ sin 2 ) 3θ sin 2 ) 3θ sin 2 Institute of Structural Engineering Method of Finite Elements II 42

45 Maximum circumferential stress criterion The maximum circumferential stress is a principal stress, therefore the angle of crack propagation θ c is obtained by setting σ rθ = 0 in the above equation which, after some manipulations, results in: K I sin θ c + K II (3 cos θ c 1) = 0 Solving for θ c gives: θ c = 2 arctan 1 4 K I K II ± ( ) 2 KI + 8 K II Institute of Structural Engineering Method of Finite Elements II 43

46 Maximum circumferential stress criterion Setting σ θθ 2πr = KIc gives the failure criterion: K I cos 3 θ K II cos θ 2 sin θ = K Ic Institute of Structural Engineering Method of Finite Elements II 44

47 Other criteria Other criteria for mixed mode fracture have been proposed such as: The maximum energy release rate criterion The minimum strain energy criterion Institute of Structural Engineering Method of Finite Elements II 45

48 SIF evaluation With the available criteria the direction of crack propagation can be determined provided the SIFs are known Several methods are available for computing the SIFs if the displacement, stress and strain fields are available The interaction integral method is one of the most widely used and will be presented in the following Institute of Structural Engineering Method of Finite Elements II 46

49 J integral For the case of linear elasticity in the absence of body forces and crack tractions the J integral assumes the form: J = Γ ( ) 1 u σ : ɛdy t 2 x dγ where t = σ n n is the normal vector to the contour Institute of Structural Engineering Method of Finite Elements II 47

50 J integral It can be shown that the J integral is: Path independent Zero for a closed path not containing any singularities Equal to the energy release rate: J = G = K I 2 + KII 2 E Institute of Structural Engineering Method of Finite Elements II 48

51 Interaction integral We consider two states of the cracked body: The actual stress state of the body due to the applied loads with the SIFs, displacement, stress and strain fields denoted as: K I, K II, u, σ, ɛ respectively An auxiliary state where the displacement, stress and strain fields are the asymptotic fields of the Westergaard solution denoted u aux, σ aux, ɛ aux respectively, SIFs for that state are denoted as: KI aux, KII aux Institute of Structural Engineering Method of Finite Elements II 49

52 Interaction integral The J integral evaluated for the sum of the two states (denoted as J s ) is: J s = Γ [ 1 2 (σ + σaux ) : (ɛ + ɛ aux ) dy (t + t aux ) (u + ] uaux ) dγ x =J + J aux + I Where I is the interaction integral: [ I = σ aux : ɛdy (t uaux x Γ + t aux u ) ] dγ x Institute of Structural Engineering Method of Finite Elements II 50

53 Interaction integral Considering that the J integral is equal to the energy release rate, we obtain for the sum of the two states: J s = (K I + KI aux = K I 2 + 2K I K aux I = K I 2 + KII 2 E + ) 2 + (K II + K aux E II ) 2 + (KI aux ) 2 + KII 2 + 2K IIKII aux (K aux I =J + J aux + 2 K IKI aux E ) 2 + (K aux E II ) 2 + K II K aux II E + 2 K IKI aux + (K aux II ) 2 + K II K aux II E Institute of Structural Engineering Method of Finite Elements II 51

54 Interaction integral Combining the two definitions of J s we obtain: I = 2 K IKI aux + K II K aux II E Appropriate choice of the values of KI aux values for the actual state. and K aux II yields the SIF For instance K aux I = 1 and K aux II = 0 yields: K I = E I 2 Institute of Structural Engineering Method of Finite Elements II 52

55 Interaction integral - Domain form The current form of the interaction integral is a contour integral which is not very well suited for FE calculations. In order to obtain a domain integral form, we introduce a function q such that: q = 1 in Γ q = 0 in Γ 0 Institute of Structural Engineering Method of Finite Elements II 53

56 Interaction integral - Domain form Definition of the contour Γ 0 : we define C = Γ + Γ 0 + Γ + + Γ Institute of Structural Engineering Method of Finite Elements II 54

57 Interaction integral - Domain form Then the interaction integral can be substituted with: [ I = σ aux : ɛdy (t uaux + t aux u ) ] dγ q x x C By applying the divergence theorem, the domain form of the interaction integral is obtained: [ I = (σ aux : ɛ) e 1 (σ uaux + σ aux u )] qda x x A where e 1 is the unit vector along the x direction. Institute of Structural Engineering Method of Finite Elements II 55

58 Cohesive zone models Methods mentioned so far assume linear elastic material behavior and apply mostly to brittle materials When nonlinear effects are present in an area of considerable size compared to the length scale of the problem those methods will fail Next a simple method will be briefly presented which: Successfully predicts the response of structures subjected to size effects Is easy to incorporate into numerical models Institute of Structural Engineering Method of Finite Elements II 56

59 Cohesive zone models In those models some forces are introduced which resist separation of the surfaces: Those forces gradually reduce to a value of zero which corresponds to full separation Institute of Structural Engineering Method of Finite Elements II 57

60 Cohesive zone models Cohesive forces are related to the crack opening displacements through a traction-separation law: Normal traction Normal traction Normal displacement jump Linear Law Normal displacement jump Nonlinear Law Institute of Structural Engineering Method of Finite Elements II 58

61 Cohesive zone models The area under the curve is equal to the critical energy release rate G c : Normal traction Normal traction Normal displacement jump Linear Law Normal displacement jump Nonlinear Law Institute of Structural Engineering Method of Finite Elements II 58