Fracture mechanics

Fracture mechanics Nguyen Vinh Phu, PhD [email protected] Researcher at Division of Computational Mechanics Ton Duc Thang University

Nguyen Vinh Phu

Program “Master of Science, Civil Engineering, 2012, Ton Duc Thang University” 1

Sunday, September 30,

1

Textbooks • Anderson, T.L. (1995) Fracture Mechanics: Fundamentals and Applications, 2nd Edition, CRC Press, USA.

• Gdoutos E.E (2005) Fracture Mechanics: an introduction, 2nd Edition, Springer.

• Zehnder, T.A. (2007) Lecture Notes on Fracture Mechanics, Cornel University, Ithaca, New York

• imechanica.org • wikipedia 2 Sunday, September 30,

2

Outline •

Brief recall on mechanics of materials - stress/strain curves of metals, concrete

• •

Introduction Linear Elastic Fracture Mechanics (LEFM) - Energy approach (Griffith, 1921, Orowan and Irwin 1948) - Stress intensity factors (Irwin, 1960s)

•

LEFM with small crack tip plasticity - Irwin’s model (1960) and strip yield model - Plastic zone size and shape

•

Elastic-Plastic Fracture Mechanics - Crack tip opening displacement (CTOD), Wells 1963 - J-integral (Rice, 1958)

•

Mixed-mode crack propagation 3


3

Outline (cont.) Fatigue - Paris law - Overload and crack retardation - cohesive crack model (Hillerborg, 1976) - Continuum Damage Mechanics - size effect (Bazant)

Fracture of concrete Computational fracture mechanics - FEM, BEM, MMs - XFEM

4 Sunday, September 30,

4

Stress-strain curves Engineering stress and strain

δ P σe = , ✏e = A0 L0

0 0

ductile metals

Tension test E

Young’s modulus

σe = E✏e 5


elastic unloading strain hardening (tai ben) 5

Stress/strain curve fracture

Wikipedia

necking=decrease of cross-sectional area due to plastic deformation

1: ultimate tensile strength 6 Sunday, September 30,

6

Stress-strain curves True stress and true (logarithmic) strain: dL P σt = , d✏t = A L → ✏t =

Z

L L0

L (extension ratio) λ≡ L0

1 L dL = ln L L0

Plastic deformation:volume does not change L A dV = 0 → AL = A0 L0 → = L0 A0 Relationship between engineering and true stress/strain σt = σe (1 + ✏e ) = σe λ ✏t = ln(1 + ✏e ) = ln λ Sunday, September 30,

7 7

Stress-strain curve concrete pre-peak post-peak(strain softening)

strain softening=increasing strain while stress decrease 8 Sunday, September 30,

8

Concrete

fracture in concrete

ITZ

aggregates cement paste 9 Sunday, September 30,

9

Some common material models σ

σ

no hardening

σys

✏ ✏

Linear elastic

Elastic perfectly plastic

Will be used extensively in Fracture Mechanics 10 Sunday, September 30,

10

Strain energy density Consider a linear elastic bar of stiffness k, length L, area A, subjected to a force F, the work is

W =

Z

u

F du = 0

Z

u 0

1 2 1 F kudu = ku = F u 2 2

W

u This work will be completely stored in the structure in the form of strain energy. Therefore, the external work and strain energy are equal to one another σ In terms of stress/strain

1 U = W = Fu 2

σxx

1 1F u U = Fu = AL 2 2AL

Strain energy density 3

[J/m ] Sunday, September 30,

1 u = σ x ✏x 2 11

u=

Z

✏x ✏x

σx d✏x 11

Strain energy density Poisson’s ratio

1 ν 2 2 2 u= (σx + σy + σz ) − (σx σy + σy σz 2E E 1 2 2 2 +σz σx ) + (τxy + τyz + τzx ) 2µ

E µ= shear modulus 2(1 + ν)

Plane problems 

"

1 κ+1 2 2 2 u= (σx + σy ) − 2(σx σy − τxy ) 4µ 4  3 − 4ν plane strain Kolosov coefficient κ= 3−ν  plane stress 1+ν 12 Sunday, September 30,

12

Indicial notation a 3D vector

i = 1, 2, 3 √ ||x|| = xi xi

x = {x1 , x2 , x3 } q ||x|| = x21 + x22 + x23

two times repeated index=sum, summation/dummy index

||x|| =

√

xk xk

σxx ✏xx + σxy ✏xy + σyx ✏yx + σyy ✏yy

σij ✏ij

∂τxy ∂σx + = 0, ∂x ∂y

σij,j = 0

∂σy ∂τxy + =0 ∂y ∂x

σxx nx + σxy ny = tx

i: free index (appears precisely once in each side of an equation)

σyx nx + σyy ny = ty

σij nj = ti σ:✏

tensor notation


13

Engineering/matrix notation 



x1 x = x2  x3

T

||x|| =

||x|| = x x

Voigt notation 



σxx σ = σyy  σxy





✏xx ✏ =  ✏yy  2✏xy

√

xi xi

σij ✏ij

T

σ ✏ = σij ✏ij 

σxx σ= σxy

σxy σyy

"



✏xx ✏= ✏xy

✏xy ✏yy

"


14

Principal stresses

Principal direction 2τxy tan 2θp = σxx − σyy Principal stresses are those stresses that act on principal surface. Principal surface here means the surface where components of shear-stress is zero.

σxx + σyy σ1 , σ2 = ± 2 Sunday, September 30,

s✓

σxx − σyy 2 15

◆2

2 + τxy 15

Residual stresses Residual stresses are stresses that remain after the original cause of the stresses Wikipedia (external forces, heat gradient) has been removed. Residual stresses always appear to some extent during fabrication operations such as casting, rolling or welding. Causes

treatment: welding, casting processes, cooling, • Heat some parts contract more than others -> residual stresses


16

Residual stresses TOTAL STRESS = APPLIED STRESS + RESIDUAL STRESS

Welding: produces tensile residual stresses -> potential sites for cracks.

Knowledge of residual stresses is indispensable. Measurement of residual stresses: FEM packages 17 Sunday, September 30,

17

Introduction

Cracks: ubiquitous !!! 18 Sunday, September 30,

18

Conventional failure analysis before 1960s

Stresses

Failure criterion σ

f (σ, σc ) = 0

Solid mechanics, numerical methods (FEM,BEM)

Tresca, Mohr-Coulomb… critical stress: σc experimentally determined

• σσ : depends on the testing samples !!! • • Many catastrophic failures occurred during Structures: no flaws!!!

Liberty ship

cc

WWII: f (σ, σc ) = 0


19 19

New Failure analysis Stresses

1970s

f (σ, a, Kc ) = 0

σ

Flaw size a

Fracture toughness

Fracture Mechanics (FM) - FM plays a vital role in the design of every critical structural or machine component in which durability and reliability are important issues (aircraft components, nuclear pressure vessels, microelectronic devices). - has also become a valuable tool for material scientists and engineers to guide their efforts in developing materials with20improved mechanical properties. Sunday, September 30,

20

Design philosophies •

Safe life The component is considered to be free of defects after fabrication and is designed to remain defect-free during service and withstand the maximum static or dynamic working stresses for a certain period of time. If flaws, cracks, or similar damages are visited during service, the component should be discarded immediately.

•

Damage tolerance The component is designed to withstand the maximum static or dynamic working stresses for a certain period of time even in presence of flaws, cracks, or similar damages of certain geometry and size. 21


21

Definitions • Crack, Crack growth/propagation • A fracture is the (local) separation

of an object or material into two, or more, pieces under the action of stress.

• Fracture

mechanics is the field of mechanics concerned with the study of the propagation of cracks in materials. It uses methods of analytical solid mechanics to calculate the driving force on a crack and those of experimental solid mechanics to characterize the material's resistance to fracture (Wiki). 22


22

Objectives of FM

• What is the residual strength as a function of crack size? • What is the critical crack size? • How long does it take for a crack to grow from a certain initial size to the critical size? 23 Sunday, September 30,

23

Brittle vs Ductile fracture •

In brittle fracture, no apparent plastic deformation takes place before fracture, crack grows very fast!!!, usually strain is smaller than 5%.

•

In ductile fracture, extensive plastic deformation takes place before fracture, crack propagates slowly (stable crack growth).

rough surfaces


Ductile fracture is preferable than brittle failure!!! 24

Classification Fracture mechanics:

• Linear Elastic Fracture Mechanics (LEFM) - brittle-elastic materials: glass, concrete, ice, ceramic etc.

• Elasto-Plastic Fracture Mechanics (EPFM) - ductile materials: metals, polymer etc.

• Nonlinear Fracture Mechanics (NLFM) 25 Sunday, September 30,

25

Approaches to fracture • Stress analysis • Energy methods • Computational fracture mechanics • Micromechanisms of fracture (eg. atomic level) • Experiments • Applications of Fracture Mechanics

covered in the course


26

Stress concentration

load lines Geometry discontinuities: holes, corners, notches, cracks etc: stress concentrators/risers


27

Stress concentration (cont.) uniaxial

biaxial


28

Elliptic hole Inglis, 1913, theory of elasticity

ρ

✓

◆

2b σ3 = 1 + σ1 a b2 ρ= radius of curvature s !a b σ3 = 1 + 2 σ1 ρ s b !!! ∞ σ3 = 2 σ1 ρ

0 crack stress concentration factor [-]


2b σ3 =1+ KT ≡ σ1 a 29

Griffith’s work (brittle materials) FM was developed during WWI by English aeronautical engineer A. A. Griffith to explain the following observations:

• •

The stress needed to fracture bulk glass is around 100 MPa

•

experiments on glass fibers that Griffith himself conducted: the fracture stress increases as the fiber diameter decreases => Hence the uniaxial tensile strength, which had been used extensively to predict material failure before Griffith, could not be a specimen-independent material property.

The theoretical stress needed for breaking atomic bonds is approximately 10,000 MPa

Griffith suggested that the low fracture strength observed in experiments, as well as the size-dependence of strength, was due to the presence of microscopic flaws in the bulk material. 30 Sunday, September 30,

30

Griffith’s size effect experiment Size effect: ảnh hưởng kích thước

“the weakness of isotropic solids... is due to the presence of discontinuities or flaws... The effective strength of technical materials could be increased 10 or 20 times at least if these flaws could be eliminated.'' 31 Sunday, September 30,

31

Griffith’s experiment • Glass fibers with artificial cracks (much larger than natural crack-like flaws), tension tests

s

σ3 = 2


b σ1 ρ

const σc = √ a

√

σc a = const

Energy approach 32 32

Energy balance during crack growth external work

kinetic energy ˙ = U˙ e + U˙ p + U˙ k + U˙ Γ W

surface energy

internal strain energy

All changes with respect to time are caused by changes in crack size: ∂(·) ∂(·) ∂a = ∂t ∂a ∂t Energy equation is rewritten: ∂Ue ∂Up ∂UΓ ∂W = + + slow process ∂a ∂a ∂a ∂a It indicates that the work rate supplied to the continuum by the applied loads is equal to the rate of the elastic strain energy and plastic strain work plus the energy dissipated in crack propagation 33 Sunday, September 30,

33

Potential energy

Π = Ue − W ∂Up ∂UΓ ∂Π = + − ∂a ∂a ∂a

Brittle materials: no plastic deformation ∂UΓ ∂Π = − ∂a ∂a

Griffith’s through-thickness crack

γs is energy required to form a unit of new surface ∂Π = 2γs − ∂a

[J/m2=N/m]

√

Inglis’ solution 2

∂Π πσ a − = ∂a E

σc a =

πσ 2 a = 2γs → σc = E

(plane stress, constant load) Sunday, September 30,

(two new material surfaces)

r

✓

2Eγs π

◆1/2

2Eγs πa

34 34

2

[N/m2]

πσ a = 2γs → σc = E

r

2Eγs πa [N/m2]

E : MPa=N/m2

check dimension

γs : N/m a: m

App. of dimensional analysis 1 1 σ2 u = σ✏ = 2 2E Sunday, September 30,

r Dimensional Analysis 2 πσ a 2Eγs = 2γs → σc = E πa B=1 35

σ2 2 U =β a E 35

2

[N/m2]

πσ a = 2γs → σc = E

r

2Eγs πa [N/m2]

E : MPa=N/m2

check dimension

γs : N/m a: m

r Dimensional Analysis 2 πσ a 2Eγs = 2γs → σc = E πa

App. of dimensional analysis

β=π

1 1 σ2 u = σ✏ = 2 2E

σ2 2 U =β a E


B=1 35

35

Energy equation for ductile materials Plane stress r r 2Eγ ss 2Eγ σσcc = = πa πa

σc = γp

r

2E(γs + γp ) πa

Griffith (1921), ideally brittle solids

Irwin, Orowan (1948), metals

plastic work per unit area of surface created

γp ! γs γp ≈ 103 γs (metals) Sunday, September 30,

Griffith’s work was ignored for almost 20 years

36 36

Energy release rate Irwin 1956

dΠ G≡− dA

G: energy released during fracture per unit of newly created fracture surface area G = 2γ + γp | {z }

the resistance of the material that must be overcome for crack growth

energy available for crack growth (crack driving force) Energy release rate failure criterion Gc

G ≥ Gc

fracture energy, considered to be a material property (independent of the applied loads and the geometry of the body). 37


37

Energy release rate Irwin 1956

dΠ G≡− dA

Griffith

G: energy released during fracture per unit of newly created fracture surface area G = 2γ + γp | {z }

the resistance of the material that must be overcome for crack growth

energy available for crack growth (crack driving force) Energy release rate failure criterion Gc

G ≥ Gc

fracture energy, considered to be a material property (independent of the applied loads and the geometry of the body). 37


37

G from experiment a1: OA, triangle OAC=U a2: OB, triangle OBC=U

Fixed grips

dΠ G≡− dA

Dead loads 1 (OAB) G= B ∆a

B: thickness Π = Ue − W

OAB=ABCD-(OBD-OAC)

1 (OAB) G= B ∆a W =0 Sunday, September 30,

→ Ue < 0

elastic strain energy stored in the body is decreasing—is 38 being released 38

G from experiments

1 shaded area G= B a4 − a3 39 Sunday, September 30,

39

Crack extension resistance Irwin crack driving curve (R-curve) force curve dU dU dΠ

G=−

=

Γ

dA dA dUp dUΓ + R≡ dA dA

+

p

dA G=R

R-curve

Resistance to fracture increases with growing crack size in elastic-plastic materials.

R = R(a)

Irwin

Stable crack growth: fracture resistance of thin specimens is represented by a curve not a single parameter.

SLOW 40


40

R-curve shapes 2 πσ a flat R-curve G= E (ideally brittle materials)

rising R-curve (ductile metals)

slope dR dG ≤ G = R, da da Sunday, September 30,

stable crack growth 41

crack grows then stops, only grows further if there is an increase of applied load 41

G in terms of compliance P

u C= P

C

inverse of stiffness K

Fixed grips P dP

dUe = Ue (a + da) − Ue (a) 1 1 = (P + dP )u − P u 2 2 1 = dP u 2 1 dP G=− u 2B da 2 1 u dC 1 2 dC G= = P 2 2B C da 2B da

a a + da

u dA = Bda


u

dW − dUe G= dA 42

G in terms of compliance P

u C= P

C

inverse of stiffness Fixed load

1 1 dUe = P (u + du) − P u 2 2 1 = P du 2

K

u

a + da

P a

dW = P du du

1 du G= P 2B da

u

1 2 dC P G= 2B da 43 Sunday, September 30,

dW − dUe G= dA 43

G in terms of compliance Fixed grips

Fixed loads

1 u2 dC 1 2 dC G= = P 2B C 2 da 2B da

1 2 dC G= P 2B da

Strain energy release rate is identical for fixed grips and fixed loads. Strain energy release rate is proportional to the differentiation of the compliance with respect to the crack length.


44

Stress analysis of isotropic linear elastic cracked solids


45

Airy stress function for solving 2D linear elasticity problems ∂σx ∂τxy + = 0, ∂x ∂y

Equilibrium: Airy stress function Φ: Compatibility condition:

2

∂ Φ σx = , 2 ∂y

∂σy ∂τxy + =0 ∂y ∂x 2

∂ Φ σy = , 2 ∂x

4

4

2

τxy

∂ Φ =− ∂x∂y

4

∂ Φ ∂ Φ ∂ Φ r Φ= +2 2 2 + =0 4 4 ∂x ∂x ∂y ∂y 4

(*)

Bi-harmonic equation For a given problem, choose an appropriate Φ that satisfies (*) and the boundary conditions.

Φ → σij → ✏ij → ui 46 Sunday, September 30,

46

Crack modes

ar


47

Crack modes


48

Westergaard’s complex 1937 stress function for mode I 2

Z(z), z = x + iy, i = −1 Φ = ReZ¯ + yImZ¯ Z Z ¯ Z¯ = Z(z)dz, Z¯ = Z(z)dz

Kolosov  coef. κ 3 − 4ν κ= 3−ν  1+ν

plane strain plane stress

E µ= shear modulus 2(1 + ν) 49


σxx = ReZ − yImZ 0 σyy = ReZ + yImZ

0

τxy = −yReZ 0 ✏ij → ui κ−1 2µu = ReZ¯ − yImZ 2 κ+1 2µv = ImZ¯ − yReZ 2 49

Griffith’s crack (mode I) (x, y) → ∞ : σxx = σyy = σ, τxy = 0 |x| < a, y = 0 : σyy = τxy = 0

boundary conditions

σz Z(z) = √ z 2 − a2 σxx = ReZ − yImZ 0

1I

σyy = ReZ + yImZ 0 τxy = −yReZ 0

y = 0, |x| < a σx Z(z) = √ iθ is imaginary ξ= x2 z−−a2a, ξ = re √ σ(ξ + a) I σ πa Z(z) = p 2 σ Z= σa √ 0 Z(z) = p ξ(ξ + 2a) Z (z) = − → 0 2πξ 2 1 − (a/z) (z 2 − a2 )3/2

infinite plate

(x, y) → ∞ : z → ∞ Z → σ Sunday, September 30,

50 50

Griffith’s crack (mode I) (x, y) → ∞ : σxx = σyy = σ, τxy = 0 |x| < a, y = 0 : σyy = τxy = 0

boundary conditions

σz Z(z) = √ z 2 − a2 y

x

ξ = z − a, ξ = re σ(ξ + a) Z(z) = p ξ(ξ + 2a)

√ σ πa Z= √ 2πξ

infinite plate Sunday, September 30,

iθ

51 51

Griffith’s crack (mode I) (x, y) → ∞ : σxx = σyy = σ, τxy = 0 σ(ξ + a) σ(ξ + a) p p Z(z) = = |x| < a, y ξ(ξ = 0+: 2a) σyy = τxy2aξ(1 = 0 + ξ/(2a)))

p

1 + ξ/(2a)) = (1 + ξ/(2a))

−1/2

1 ξ =1− + H.O.T 2 2a =1 ξ

small

ξ

y

x

small

ξ+a=a √ √ σ√ πa σZ = πa 2πξ Z= √ 2πξ

ξ small 52


52

√ KI Z(z) = √ , KI = σ πa 2πξ KI −iθ/2 iθ Z(z) = √ e ξ = re 2πr

Recall σxx = ReZ − yImZ

0

σyy = ReZ + yImZ 0 τxy = −yReZ

0

1 KI −3/2 KI −i3θ/2 −ix Z (z) = − √ ξ =− √ e e = cos x − i sin x 2 2π 2r 2πr y = r sin θ Crack tip stress field θ θ ✓ ◆ sin θ = 2 sin cos 2 2 KI θ θ 3θ σxx = √ cos 1 − sin cos 2 2 2 2πr ✓ ◆ inverse square root KI θ θ 3θ σyy = √ cos 1 + sin cos 2 2 2 2πr θ θ 3θ KI τxy = √ sin cos sin 2 2 2 2πr r → 0 : σij → ∞ 53 0


53

√ KI Z(z) = √ , KI = σ πa 2πξ KI −iθ/2 iθ Z(z) = √ e ξ = re 2πr

Recall σxx = ReZ − yImZ

0

σyy = ReZ + yImZ 0 τxy = −yReZ

0

1 KI −3/2 KI −i3θ/2 −ix Z (z) = − √ ξ =− √ e e = cos x − i sin x 2 2π 2r 2πr y = r sin θ Crack tip stress field θ θ ✓ ◆ sin θ = 2 sin cos 2 2 KI θ θ 3θ σxx = √ cos 1 − sin cos 2 2 2 2πr ✓ ◆ inverse square root KI θ θ 3θ σyy = √ cos 1 + sin cos 2 2 2 2πr singularity θ θ 3θ KI τxy = √ sin cos sin 2 2 2 2πr r → 0 : σij → ∞ 53 0

1 √ r


53

Plane strain problems Hooke’s law ✏zz

1 = (−⌫σxx − ⌫σyy + σzz ) E

Plane strain ✏zz = 0

✓ ◆ KI θ θ 3θ σxx = √ cos 1 − sin cos 2 2 2 2πr ✓ ◆ KI θ θ 3θ σyy = √ cos 1 + sin cos 2 2 2 2πr θ θ 3θ KI sin cos sin τxy = √ 2 2 2 2πr

σz = ν(σx + σy ) θ KI σz = 2ν cos 2πr 2


54

Stresses on the crack plane ✓ ◆ KI θ θ 3θ σxx = √ 1 − sin cos cos 2 2 2 2πr ✓ ◆ KI θ θ 3θ σyy = √ 1 + sin cos cos 2 2 2 2πr θ θ 3θ KI sin cos sin τxy = √ 2 2 2 2πr

On the crack plane θ = 0, r = x σxx = σyy

KI =√ 2πx

crack plane is a principal plane with the following principal stresses

τxy = 0 55 Sunday, September 30,

σ1 = σ2 = σxx = σyy 55

Stress Intensity Factor (SIF) σxx σyy τxy

✓ ◆ √ KI θ θ 3θ K = σ πa I =√ cos 1 − sin cos 2 2 2 2πr ✓ ◆ KI θ θ 3θ =√ cos 1 + sin cos 2 2 2 2πr √ θ θ 3θ KI [MPa m] =√ sin cos sin 2 2 2 2πr

KI

SIMILITUDE

• Stresses-K: linearly proportional • K uniquely defines the crack tip stress field • modes I, II and III: K , K , K • LEFM: single-parameter I

II

III

56


56

Singular dominated zone σyy

KI θ =√ cos 2 2πr

✓

θ 3θ 1 + sin cos 2 2

(crack plane)

◆

σ∞

crack tip

σ∞ ✓ ◆ KI θ θ 3θ σxx = √ cos 1 − sin cos 2 2 2 2πr ✓ ◆ KI θ θ 3θ σyy = √ cos 1 + sin cos 2 2 2 2πr θ θ 3θ KI sin cos sin τxy = √ 2 2 2 2πr Sunday, September 30,

K-dominated zone

57 57

Mode I: displacement field KI Z(z) = √ 2πr

✓

◆

θ θ cos − i sin 2 2 Z Z¯ = Z(z)dz

Recall

κ−1 2µu = ReZ¯ − yImZ 2 KI κ+1 Z(z) = √ ¯ − yReZ 2µv = Im Z 2πξ 2 r ✓ ◆ K r θ θ I 1/2 ¯ Z(z) = 2√ ξ = 2KI cos + i sin z =ξ+a 2π 2 2 2π iθ ξ = re Displacement field −ix e = cos x − i sin x r ✓ ◆ KI r θ 2 θ u= cos κ − 1 + 2 sin 2µ 2π 2 2 r ✓ ◆ KI r θ 2 θ v= sin κ + 1 − 2 cos 2µ 2π 2 2 58 Sunday, September 30,

58

Crack face displacement y = 0, −a ≤ x ≤ a κ+1 2µv = ImZ¯ − yReZ 2 σx Z(z) = √ x 2 − a2 −a ≤ x ≤ a

√ i = −1

κ+1 p 2 v= σ a − x2 4µ

κ+1 v= ImZ¯ 4µ p ¯ Z(z) = σ x 2 − a2 ¯ Z(z) = i(σ

p

a2 − x 2 )

ellipse

Crack Opening Displacement κ+1 p 2 COD = 2v = σ a − x2 59 2µ Sunday, September 30,

59

Crack tip stress field in polar coordinates-mode I KI σij = √ fij (θ) πa

σrr

KI =√ 2πr

✓

5 θ 1 3θ cos − cos 4 2 4 2

✓

◆ ◆

KI 3 θ 1 3θ σθθ = √ cos + cos 2 4 2 2πr 4 ✓ ◆ KI 1 θ 1 3θ τrθ = √ sin + sin 2 4 2 2πr 4

stress transformation


60

Principal crack tip stresses s σxx + σyy σ1 , σ2 = ± 2

✓

◆

✓

KI θ θ σ1 = √ cos 1 + sin 2✓ 2◆ 2πr KI θ θ σ2 = √ cos 1 − sin 2 2 2πr 0 plane stress σ3 = 2νKI θ √ cos plane strain 2 2πr 61


σxx − σyy 2

◆2

2 + τxy

✓ ◆ KI θ θ 3θ σxx = √ cos 1 − sin cos 2 2 2 2πr ✓ ◆ KI θ θ 3θ σyy = √ cos 1 + sin cos 2 2 2 2πr θ θ 3θ KI sin cos sin τxy = √ 2 2 2 2πr

σ3 = ν(σ1 + σ2 ) 61

Mode II problem Boundary conditions

(x, y) → ∞ : σxx = σyy = 0, τxy = τ |x| < a, y = 0 : σyy = τxy = 0

Stress function Check BCs

iτ z Z = −√ z 2 − a2

σxx = ReZ − yImZ 0 σyy = ReZ + yImZ 0 τxy = −yReZ 0 Sunday, September 30,

62 62

Mode II problem Boundary conditions

(x, y) → ∞ : σxx = σyy = 0, τxy = τ |x| < a, y = 0 : σyy = τxy = 0

Stress function σxx σyy τxy

mode II SIF Sunday, September 30,

iτ z Z = −√ z 2 − a2 ✓ ◆ KII θ θ 3θ = −√ sin 2 + cos cos 2 2 2 2πr θ θ 3θ KII =√ sin cos cos 2 2 2 2πr ✓ ◆ KII θ θ 3θ =√ cos 1 − sin sin 2 2 2 2πr K 63 II

√ = τ πa 63

Mode II problem (cont.) iτσzxx = σyy = 0, τxy = τ (x, y) → ∞ : Z = −√ Stress function 2 − a2 z |x| < a, y = 0 : σyy = τxy = 0 (x, y) → ∞ : σxx = σyy = 0, τxy = τ iτ z Z = −√ |x| < a, y = 0 : σyy = τxy = 0 z 2 − a2

KII u= 2µ

r

◆

θ κ + 1 + 2 cos 2 r ✓ ◆ KII r θ 2 θ v= cos κ − 1 − 2 sin 2µ 2π 2 2

mode II SIF

KII

r θ sin 2π 2

✓

2

√ = τ πa


64

Mode III problem


65

Universal nature of the asymptotic stress field Westergaards, Sneddon etc. ✓ ◆ ✓ ◆ KII θ θ 3θ KI θ θ 3θ σxx = − √ sin 2 + cos cos σxx = √ 1 − sin cos cos 2 2 2 2πr 2 2 2 2πr ✓ ◆ θ θ 3θ KII KI θ θ 3θ σyy = √ sin cos cos σyy = √ 1 + sin cos cos 2 2 2 2πr 2 2 2 2πr ✓ ◆ KII θ θ 3θ θ θ 3θ KI τxy = √ cos 1 − sin sin sin cos sin τxy = √ 2 2 2 2πr 2 2 2 2πr

(mode I) Irwin

(mode II) K σij = √ fij (θ) + H.O.T 2πr 66


66

Inclined crack in tension

2

2

σ1 = σx cos θ + 2 sin θ cos θτxy + sin θσy σ2 = σy cos2 θ − 2 sin θ cos θτxy + sin2 θσx τ12 = −σx cos θ sin θ + cos 2θτxy + 0.5 sin 2θσy

Final result σ1 = (sin2 β)σ 2

σ2 = (cos β)σ

τ12 = (sin β cos β)σ Sunday, September 30,

Recall +

√ KI = σy πa √ KII = τxy πa

√ KI = σ πa cos2 β √ KII = σ πa sin β cos β

67 67

Inclined crack in tension 2

1

2

2

σ1 = σx cos θ + 2 sin θ cos θτxy + sin θσy σ2 = σy cos2 θ − 2 sin θ cos θτxy + sin2 θσx τ12 = −σx cos θ sin θ + cos 2θτxy + 0.5 sin 2θσy

Final result σ1 = (sin2 β)σ 2

σ2 = (cos β)σ

τ12 = (sin β cos β)σ Sunday, September 30,

Recall +

√ KI = σy πa √ KII = τxy πa

√ KI = σ πa cos2 β √ KII = σ πa sin β cos β

67 67

Cylindrical pressure vessel with an inclined through-thickness crack closed-ends

pR σz = 2t pR σθ = t Sunday, September 30,

R ≥ 10 thin-walled pressure t 2 (πR )p = (2πRt)σz

(∆l2R)p = (2∆lt)σθ pR √ 2 KI = πa(1 + sin β) 2t pR √ KII = πa sin β cos β 68 2t 68

Cylindrical pressure vessel with an inclined through-thickness crack σθ = 2σz This is why an overcooked hotdog usually cracks along the longitudinal direction first (i.e. its skin fails from hoop stress, generated by internal steam pressure).

Equilibrium pR σz = 2t pR σθ = t Sunday, September 30,

pR √ 2 KI = πa(1 + sin β) 2t pR √ KII = πa sin β cos β 69 2t

? 69

Computation of SIFs • Analytical methods (limitation: simple geometry) - superposition methods - weight/Green functions

• Numerical methods (FEM, BEM, XFEM) numerical solutions -> data fit -> SIF handbooks

• Experimental methods - photoelasticity 70 Sunday, September 30,

70

SIF for finite size samples Exact (closed-form) solution for SIFs: simple crack geometries in an infinite plate.

Cracks in finite plate: influence of external boundaries cannot be neglected -> generally, no exact solution


71

SIF for finite size samples KI

<

KI

dimensional analysis

√ geometry/correction K = f (a/W )σ πa I factor [-]

force lines are compressed->> higher stress concentration

a ⌧ W : f (a/W ) ⇡ 1

72


72

SIFs handbook


73

SIFs handbook


74

SIFs handbook


75

Reference stress√

KI = βσ πa

√ KI = βmax σmax πa √

KI = βxa σxa πa

βxa

βmax σmax = βmax = σxa 1 − 2a/W

Non-uniform stress distribution β for which reference stress!!! 76 Sunday, September 30,

76

Reference stress√

KI = βσ πa

√ KI = βmax σmax πa √

KI = βxa σxa πa

chosen

βxa

βmax σmax = βmax = σxa 1 − 2a/W

Non-uniform stress distribution β for which reference stress!!! 76 Sunday, September 30,

76

Superposition method A sample in mode I subjected to tension and bending: KItension

σij = √ σij =

2πr

KItension

KI =

fij (θ) +

bending KI

√ fij (θ) 2πr

bending KI

+ √ 2πr

tension KI

+

fij (θ)

bending KI

Is superposition of SIFs of different crack modes possible? 77 Sunday, September 30,

77

Determine the stress intensity factor for an edge cracked plate subjected to a combined tension and bending. a/W = 0.2 B

thickness

Solution bend KI

6M √ = fM (a/W ) πa 2 BW

ten KI

P √ = fP (a/W ) πa BW

1.12 ✓ ◆ √ 6M P KI = 1.055 + 1.12 πa BW 2 BW

1.055


78

Superposition method Centered crack under internal pressure

KId + KIe = KIb = 0 → KIe = −KId

√ = −σ πa

This result is useful for surface flaws along the internal wall of pressure vessels. 79 Sunday, September 30,

79

√ KI = σ πa


80

SIFs: asymmetric loadings Procedure: build up the case from symmetric cases and then to subtract the superfluous loadings. K A = KB + K C − K D KA = (KB + KC )/2


81

Two small cracks at a hole

a

3σ

edge crack

hole as a part of the crack


82

Photoelasticity Wikipedia Photoelasticity is an experimental method to determine the stress distribution in a material. The method is mostly used in cases where mathematical methods become quite cumbersome. Unlike the analytical methods of stress determination, photoelasticity gives a fairly accurate picture of stress distribution, even around abrupt discontinuities in a material. The method is an important tool for determining critical stress points in a material, and is used for determining stress concentration in irregular geometries.


83

K-G relationship So far, two parameters that describe the behavior of cracks: K and G. K: local behavior (tip stresses) G: global behavior (energy) Irwin: for linear elastic materials, these two params are uniquely related Crack closure analysis: work to open the crack = work to close the crack 84 Sunday, September 30,

84

Irwin

K-G relationship

B=1 (unit thickness)

G = lim

∆a→0

✓

∆U ∆a

◆

fixed load Z ∆a

work of crack closure ∆U =

dU (x)

0

1 dU (x) = 2 σyy (x)uy (x)dx 2 r (κ + 1)KI (a + ∆a) ∆aθ−=xπ uy = 2µ 2π KI (a) σyy = √ θ=0 2πx KI (a) r Z ∆a 2 (κ + 1)KI2 (κ + 1)KI ∆a − x G= G = lim dx ∆a→0 4πµ∆a 8µ 85x 0 Sunday, September 30,

85

K-G relationship (cont.) Mode I

 2 K   I E GI = 2  (1 − v 2 ) KI E

plane stress plane strain

Mixed mode 2 2 KI2 KII KIII G= 0 + 0 + E E 2µ

• • •

Equivalence of the strain energy release rate and SIF approach Mixed mode: G is scalar => mode contributions are additive Assumption: self-similar crack growth!!!

Self-similar crack growth: planar crack remains planar ( da same 86 direction as a ) Sunday, September 30,

86

SIF in terms of compliance 1 2 dC G= P 2B da KI2 GI = 0 E

B: thickness 0 2 E P dC 2 KI = 2B da

A series of specimens with different crack lengths: measure the compliance C for each specimen -> dC/da -> K and G


87

SIF in terms of compliance 1 2 dC G= P 2B da KI2 GI = 0 E

B: thickness 0 2 E P dC 2 KI = 2B da

A series of specimens with different crack lengths: measure the compliance C for each specimen -> dC/da -> K and G


87

Units


88

Example


89

1 OAi Aj G= 2B aj − ai

Gc for different crack lengths are almost the same: flat Rcurve. 90 Sunday, September 30,

90


91


92

Examples a increases -> G increase 1 2 dC load control P G= 2B da 1 u2 dC disp. control G= 2B C 2 da

load control

2

3

3u Eh G= 16a4 a increases -> G decreases!!!

Double cantilever beam (DCB) 93 Sunday, September 30,

93

Compliance-SIF r

πa √ K = sec σ πa W 2 2 P dC P dC G= = 2 dA 4B da K2 G= E 2

πa 2 W σ πa

sec P dC = 4B da E sec dC = da

C=

Z

a 0

πa 4 πa sec da + C 0 2 EBW W H δ C0 = = P EBW

πa 2 W σ πa4B P 2E

4 πa dC = πa sec 2 da EBW W Sunday, September 30,

H

⌘ πa

4 C=− ln cos EBπ W 94 ⇣

H + EBW 94

4W C/C0 = − ln cos πH W ⇣

⌘ πa

+1

compliance rapidly increases


95

K as a failure criterion √

Failure criterion

•

K = Kc

f (a/W )σ πa = Kc

fracture toughness Problem 1: given crack length a, compute the maximum allowable applied stress W, Kc

σmax

Kc √ = f (a/W ) πa

the • Problem 2: for a specific applied stress, compute σ maximum permissible crack length (critical crack √ length) ac f (ac /W )σ πac = Kc → ac

• Problem 3: compute K

c

provided crack length and

stress at fracture Kc = f (ac /W )σ πac √


96

Example


97

Example solution √ KI = 1.12σθ πa a

KI = KIc /S

pR σz = 2t pR σθ = t

problem 1

problem 2

p = 12MPa

a = 1mm


98

Example

Griffith σc =

r

Irwin 2Eγs πa

KIc σc = √ =479 Mpa πa

=5.8 Mpa


99

Mixed-mode fracture KI = KIc KIc < KIIc , KIIIc

KII = KIIc KIII = KIIIc

lowest Kc: safe

Superposition cannot be applied to SIF. However, energy can. 2 KI2 KII G= 0 + 0 E E

2 KIc Gc = 0 E

Fracture occurs when G = Gc

2 KI

+

2 KII

=

2 KIc


100

Experiment verification of the mixed-mode failure criterion 2 KI

+

2 KII

=

2 KIc

a circle in KI, KII plane

Data points do not fall exactly on the circle. ◆2 ✓ ◆2 ✓ KI KII (κ + 1)KI2 + =1 G= KIc KIIc self-similar growth 101 8µ Sunday, September 30,

101

G: crack driving force -> crack will grow in the direction that G is maximum


102

Crack tip plasticity • Irwin’s model • Strip Yield model • Plane stress vs plane strain • Plastic zone shape 103 Sunday, September 30,

103

Introduction •

Griffith's theory provides excellent agreement with experimental data for brittle materials such as glass. For ductile materials such as steel, the surface energy (γ) predicted by Griffith's theory is usually unrealistically high. A group working under G. R. Irwin at the U.S. Naval Research Laboratory (NRL) during World War II realized that plasticity must play a significant role in the fracture of ductile materials.

(SSY) Small-scale yielding: LEFM still applies with minor modifications done by G. R. Irwin

crack tip

R⌧D 104 Sunday, September 30,

104

Validity of K in presence of a plastic zone crack tip

Fracture process usually occurs in the inelastic region not the Kdominant zone.

is SIF a valid failure criterion for materials that exhibit inelastic deformation at the tip

?


105

Validity of K in presence of a plastic zone [Anderson] same K->same stresses applied on the disk stress fields in the plastic zone: the same K still uniquely characterizes the crack tip conditions in the presence of a small plastic zone. LEFM solution


106

Paradox of a sharp crack At crack tip: r = 0 → σij = ∞

An infinitely sharp crack is merely a mathematical abstraction. Crack tip stresses are finite because (i) crack tip radius is finite (materials are made of atoms) and (ii) plastic deformation makes the crack blunt. 107 Sunday, September 30,

107

Plastic correction • A cracked body in a plane stress condition σ • Material: elastic perfectly plastic with yield stress ys

On the crack plane

θ=0

stress singularity is truncated by yielding at crack tip

KI σyy = √ 2πr σyy = σys (yield occurs) σys

KI2 r1 = 2 2πσys

first order approximation of plastic zone size: equilibrium is not satisfied 108 Sunday, September 30,

108

Irwin’s plastic correction


109

Irwin’s plastic correction

plate behaves as with a longer crack stress redistribution: yellow area=hatched area σys r1 =

Z

r1

σyy dr 0

KI2 rp = 2r1 = 2 πσys 1 rp = 3π Sunday, September 30,

KI2 2 σys

plastic zone: a CIRCLE !!! Plane strain 110 110

Plane stress

Plane strain

σ1 = σ2 = σyy , σ3 = 0

σ1 = σ2 = σyy σ3 = ν(σxx + σyy ) = 2νσyy σ3 = 0.66σyy

σ1 = σys σy = σys

Tresca’s criterion

ν = 0.33

σ1 − σ3 = σys σy = 3σys


111

Irwin’s plastic correction crack tip

LEFM:

R⌧D σys r1 =

Z

rp is small

r1

σyy dr 0

1 rp = 3π

2 KI 2 σys

σys is big and KIc is small

LEFM is better applicable to materials of high yield strength and low fracture toughness 112 Sunday, September 30,

112

Plastic zone shape von-Mises criterion

σe = σys

" 1 ! 2 2 2 1/2 σe = √ (σ1 − σ2 ) + (σ1 − σ3 ) + (σ2 − σ3 ) 2

Mode I, principal stresses ✓

◆

KI θ θ σ1 = √ cos 1 + sin 2✓ 2◆ 2πr KI θ θ σ2 = √ cos 1 − sin 2 2 2πr 0 plane stress σ3 = 2νKI θ √ cos plane strain 2 2πr

KI σys

◆2 

KI σys

◆2 

1 ry (θ) = 4π

✓

1 ry (θ) = 4π

✓


3 1 + cos θ + sin2 θ 2

$

plane stress

3 2 (1 − 2µ) (1 + cos θ) + sin θ 113 2 2

$ 113

Plastic zone shape plastic zone shape (mode I, von-Mises criterion) 1 ry (θ) = 4π 0.6

rp/(KI/(N «y))2

0.4

✓

KI σys

◆2 

3 2 1 + cos θ + sin θ 2

$

plane stress plane strain

0.2 0 −0.2 −0.4 −0.6 −0.6−0.4−0.2 0 0.2 0.4 2 0.6 0.8 rp/(KI/(N «y))


114

114

Plane stress/plane strain constrained by the surrounding material

dog-bone shape

• Plane stress failure: in general, ductile • Plane strain failure: in general, brittle 115


115

Plane stress/plane strain toughness depends on thickness

Plane strain fracture toughnessKIc lowest K (safe)  # !1/2 Kc = KIc

1.4 KIc 1+ 2 B σys

4

(Irwin)


116

Fracture toughness tests • Prediction of failure in real-world applications: need the value of fracture toughness

Tests on cracked samples: PLANE STRAIN condition!!! •Compact Tension Test  $ ⌘ ⇣ ⌘ ⇣ ⌘ ⇣ ⌘ 2 3 4 a a a a a 2+ 0.886 + 4.64 − 13.32 + 14.72( − 5.6 P W W W W W KI = √ ⇣ ⌘3/2 a B W 1− W ⇣

ASTM (based on Irwin’s model) Constraint conditions 117 Sunday, September 30,

117

Compact tension test Cyclic loading: introduce a crack ahead of the notch Stop cyclic load, apply forces P Monitor maximum load and CMOD until failure (can sustain no further increase of load) P Q → KQ

check constraint conditions KIc = KQ


118

Fracture toughness test ASTM E399 plane strain

25 B > 25rp = 3π

✓

KI σys

◆2

a > 25rp

Linear fracture mechanics is only useful when the plastic zone size is much smaller than the crack size

Text Sunday, September 30,

119 119

Strip Yield Model proposed by Dugdale and Barrenblatt Infinite plate with though thickness crack 2a

• • •

Plane stress condition Elastic perfectly plastic material Hypotheses:

•

All plastic deformation concentrates in a line in front of the crack.

•

The crack has an effective length which exceeds that of the physical crack by the length of the plastic zone.

• ρ : chosen such that stress singularity at the tip disappears.


120

120

Strip Yield Model (cont.) Superposition principle KI = σys

KIσ

+

σys KI

σys

p σ KI = σ π(a + c) r ◆ ✓ a+c a σys −1 KI = −2σys cos π a+c σ ⌧ σys (derivation follows) K 1 2 σij = √ fij (θ) + H.O.T cos x = 1 − x + · · · ✓ ◆ 2πr 2! a πσ = cos KI = 0 a+c 2σys Irwin’s result 0.318 ✓ ◆ 2 ✓ ◆ 2 2 2 π σ a π KI close to 1 K I c= = r = 2 p 8σys 8 σys π σys 0.392 121 Sunday, September 30,

121

SIF for plate with normal force at crack P = −σys dx

P KA = √ πa

r

r ◆ Z c ✓r a+x σys c−x c+x σys dx + a − x KI = − √ c + x c − x πc a r P a−x KB = √ r ◆ ✓ πa a + x a+c a σys −1 cos KI = −2σys π a+c

Gdoutos, chapter 2, p40


122 122

Effective crack length

1 r1 = 2π

✓

KI σys

◆2

π rp = 8

✓

KI σys

◆2

1


123

Fracture vs. Plastic collapse σnet

P

P W = =σ W −a W −a

P σ= W

(cracked section) W = σys Yield: σ W −a

σ = σys

⇣

unit thickness

⌘ a 1− W

short crack: fracture by plastic collapse!!! high toughness materials:yielding before fracture

a

W P


124

Fracture vs. Plastic collapse σnet

P

P W = =σ W −a W −a

P σ= W

(cracked section) W = σys Yield: σ W −a

σ = σys

⇣

unit thickness

⌘ a 1− W

short crack: fracture by plastic collapse!!! high toughness materials:yielding before fracture

a

W P

LEFM applies when σc ≤ 0.66σys 124 Sunday, September 30,

124

Example Consider an infinite plate with a central crack of length 2a subjected to a uniaxial stress perpendicular to the crack plane. Using the Irwin’s model for a plane stress case, show that the effective SIF is given as follows Keff

√ σ πa = ⇣ ⌘2 $1/2 1 − 0.5 σσys

Solution: The effective crack length is a + r1 p Keff = σ π (a + r1 ) The effective SIF is thus with 125 Sunday, September 30,

2 Keff r1 = 2 2πσys 125

Example 1. Calculate the fracture toughness of a material for which a plate test with a central crack gives the following information: W=20in, B=0.75in, 2a=2in, failure load P=300kip. The yield strength is 70ksi. Is this plane strain? Check for collapse. How large is the plastic zone at the time of fracture? 2. Using the result of problem 1, calculate the residual strength of a plate with an edge crack W=5 inch, a=2inch. 3. In a toughness test on a center cracked plate one obtains the following result: W=6in, B=0.2in, 2a=2in, Pmax=41kips, σys = 50 ksi. Calculate the toughness. How large is the plastic zone at fracture? Is the calculated toughness indeed the true toughness? 126 Sunday, September 30,

126

Solution to problem 1 Stress at failure σf = 300/(20 × 0.75) = 20 ksi √ √ Toughness Kc = 1 × 20 × π × 1 = 35.4 ksi in ⇣ ⌘ a Nominal stress at collapse σ = σys 1 − W 20 − 2 σcol = 70 × = 63 ksi 20 σf < σcol

Fracture occurs before collapse. B ≥ 2.5


✓

KIc σY

◆2

= 0.64in plane strain by ASTM

127

Solution to problem 2 σf r

Kc = √ β πa

σf r

35.4 √ = = 6.73 ksi 2.1 × π × 2

σcol

5−2 = 70 × = 42 ksi 5

Residual strength 6.73 ksi


128

Solution to problem 3 Stress at failure σf = 41/(6 × 0.2) = 34.2 ksi Toughness

Kc = 1.07 × 34.2 ×

Nominal stress at collapse σcol

6−2 = 50 × = 33.3 ksi 6

σf > σcol

√

√ π × 1 = 64.9 ksi in ⇣ ⌘ a σ = σys 1 − W

β = 1.067 = 1.07

Collapse occurs before fracture

The above Kc is not the toughness!!! Sunday, September 30,

129

129

Solution to problem 3 Stress at failure σf = 41/(6 × 0.2) = 34.2 ksi Toughness

Kc = 1.07 × 34.2 ×

Nominal stress at collapse σcol

6−2 = 50 × = 33.3 ksi 6

σf > σcol

√

√ π × 1 = 64.9 ksi in ⇣ ⌘ a σ = σys 1 − W

β = 1.067 = 1.07

Collapse occurs before fracture

The above Kc is not the toughness!!! Sunday, September 30,

whole section is yielding 129

129

Elastic-Plastic Fracture Mechanics • J-integral (Rice,1958) • Crack Tip Opening Displacement (CTOD), (Wells, 1963)


130

Introduction No unloading

Monotonic loading: an elastic-plastic mater is equivalent to a nonlinear elastic material

deformation theory of plasticity can be utilized

- deformation theory plasticity models: 131 Sunday, September 30,

- flow theory 131

J-integral

Wikipedia

Eshelby, Cherepanov, 1967, Rice, 1968

Z ✓

◆

∂ui J= W n1 − ti dΓ ∂x1 Γ ◆ ◆  " Z ✓Z ✓ N ∂ui ∂ui J =J = W dxW − t2i − ti ds ds 2 dx m ∂x1 ∂x1 Γ Γ Z ✏ W = σij d✏ij strain energy density 0

ti = σij nj

surface traction

J − integral

(1) J=0 for a closed path (2) is path-independent Sunday, September 30,

notch:traction-free 132 132

Path independence of J-integral J is zero over a closed path 0 = JABCDA = JAB + JBC + JCD + JDA ◆ Z ✓ ∂ui J= W dx2 − ti ds ∂x1 Γ

AB, CD: traction-free crack faces ti = 0, dx2 = 0 (crack faces: parallel to x-axis) JAB = JCD = 0 JBC + JDA = 0

which path BC or AD should be used to compute J? JBC = JAD 133


133

Z

Π= dΠ = da

J-integral

A0

Z

A0

dW dA − da

W dA −

Z

crack grows, coord. axis move

∂ ∂ d = − da ∂a ∂x

ti ui ds Γ

Z

dui ti ds da Γ

Self-similar crack growth dΠ = da

Z

A0

✓

∂W ∂W − ∂a ∂x

@W @✏ij @W = @a @✏ij @a @✏ij 1 @ = @a 2 @a Sunday, September 30,

✓

◆

dA −

Z

ti Γ

✓

∂ui ∂ui − ∂a ∂x

◆

ds

@W = σij @✏ij ◆

@ui @uj + @xj @xi nonlinear elastic 134

∂ ∂ ∂x d = + da ∂a ∂x ∂a ∂x = −1 ∂a 134

∂W 1 ∂ = σij ∂a 2 ∂a

J-integral ✓ ◆

A:B=0 symmetric

∂ui ∂uj + ∂xj ∂xi

∂ ∂ui ∂ ∂ui ∂W = σij = σij ∂a ∂a ∂xj ∂xj ∂a Z Z ∂ ∂ui ∂ui σij dA = σij nj ds ∂xj ∂a ∂a Γ A0

Z

A0

∂W dA = ∂a

skew-symmetric

Z

∂ui ti ds ∂a Γ

Gauss theorem dΠ =− da

Z

Z

∂W ∂ui dA + ti ds ∂x A0 ∂x Γ J ◆ Z ✓ dΠ ∂ui − W dy − ti = ds da ∂x Γ

nx ds = dy


135

∂W 1 ∂ = σij ∂a 2 ∂a

J-integral ✓ ◆

A:B=0 symmetric



Z

A0

∂W dA = ∂a

skew-symmetric

Z

∂ui ti ds ∂a Γ


Z

Z

∂W ∂ui Gauss theorem, nnxxds = dy ds = dy dA + ti ds ∂x A0 ∂x Γ J ◆ Z ✓ dΠ ∂ui − W dy − ti = ds da ∂x Γ 135 Sunday, September 30,

135

∂W 1 ∂ = σij ∂a 2 ∂a

J-integral ✓ ◆

A:B=0 symmetric



Z

A0

∂W dA = ∂a

skew-symmetric

Z

∂ui ti ds ∂a Γ


Z

Z

∂W ∂ui Gauss theorem, nnxxds = dy ds = dy dA + ti ds ∂x A0 ∂x Γ J-integral is equivalent to the J ◆ Z ✓ energy release rate for a dΠ ∂ui nonlinear elastic material under − W dy − ti = ds da ∂x Γ quasi-static condition. 135


135

J-K relationship 2 2 KI2 KII KIII G= 0 + 0 + E E 2µ

dΠ − = da

Z ✓ Γ

∂ui W dy − ti ds ∂x

◆

(previous slide)

2 2 KI2 KII KIII J= 0 + 0 + E E 2µ

J-integral: very useful in numerical computation of SIFs 136 Sunday, September 30,

136

Crack Tip Opening Displacement

κ+1 p 2 v= σ a − x2 4µ

see slide 59

COD is zero at the crack tips. Sunday, September 30,

137

Crack Tip Opening Displacement Wells 1961

see slide 43 COD is taken as the separation of the faces of the effective crack at the tip of the physical crack

r

κ+1 ry uy = KI 2µ 2π ✓ ◆2 1 KI ry = 2π σys

3−ν κ= 1+ν

E 2µ = 1+ν

(Irwin’s plastic correction, plane stress) Sunday, September 30,

CTOD 4 KI2 δ = 2uy = π σys E

138 138

CTOD-G-K relation

Wells observed:

The degree of crack blunting increases in proportion to the toughness of the material

4 KI2 δ= π σys E KI2 GI = E

Fracture occurs

π GI = σys δ 4

δ = δc

material property independent of specimen and crack length (confirmed by experiments)

Under conditions of SSY, the fracture criteria based on the stress intensity factor, the strain energy release rate and the crack tip opening displacement are equivalent. 139 Sunday, September 30,

139

CTOD in design 4 KI2 δ = 2uy = π σys E

has no practical application


140

CTOD experimental determination Plastic hinge rigid

similarity of triangles

r

rotational factor [-], between 0 and 1 141


141

Governing fracture mechanism and fracture toughness


142

Example


143

Example

2a = 25.2cm


144

Fatigue crack growth • S-N curve • Constant amplitude cyclic load - Paris’ law

• Variable amplitude cyclic load - Crack retardation due to overload


145

Fatigue • •

Fatigue occurs when a material is subjected to repeated loading and unloading (cyclic loading).

•

ASTM defines fatigue life, Nf, as the number of stress cycles of a specified character that a specimen sustains before failure of a specified nature occurs.

Under cyclic loadings, materials can fail (due to fatigue) at stress levels well below their strength -> fatigue failure.

blunting

resharpening


146

Cyclic loadings σmax = −σmin

∆σ = σmax − σmin σa = 0.5(σmax − σmin ) σm = 0.5(σmax + σmin ) σmin R= σmax Sunday, September 30,

load ratio

147

147

Cyclic vs. static loadings • Static: Until K reaches Kc, crack will not grow • Cyclic: K applied can be well below Kc, crack grows still!!!

1961, Paris et al used the theory of LEFM to explain fatigue cracking successfully.

Methodology: experiments first, then empirical equations are proposed. 148 Sunday, September 30,

148

1. Initially, crack growth rate is small 2. Crack growth rate increases rapidly when a is large 3. Crack growth rate increases as the applied stress increases 149 Sunday, September 30,

149

Fatigue • •

Fatigue occurs when a material is subjected to repeated loading and unloading (cyclic loading).

•

ASTM defines fatigue life, Nf, as the number of stress cycles of a specified character that a specimen sustains before failure of a specified nature occurs.

Under cyclic loadings, materials can fail (due to fatigue) at stress levels well below their strength.

✴Stress->Nf ✴Nf->allowable S

S-N curve

scatter!!!

endurance limit (g.han keo dai)


150

Constant variable cyclic load ∆K = Kmax − Kmin

R = Kmin /Kmax ∆K = Kmax − Kmin

crack grow per cycle

da = f1 (∆K, R) dN

∆K = Kmax − Kmin = Kmax (1 − R) 151


151

Paris’ law (fatigue) Paris’ law

2≤m≤7

da da m C(∆K)m, ∆K = ==C(∆K) =K Kmax −KKmin max− min dN dN

II

(Power law relationship for fatigue crack growth in region II)

Fatigue crack growth behavior in metals

N: number of load cycles

Paris’ law is the most popular fatigue crack growth model

I

Paris' law can be used to quantify the residual life (in terms of load cycles) of a specimen given a particular crack size.

∆K ≤ ∆Kth : no crack growth

(dormant period) Sunday, September 30,

10

−8

mm/cycle

152 152

Paris’ law not depends on load ratio R da m = C(∆K) , ∆K = Kmax − Kmin dN

C

are material properties that must be C, m determined experimentally. 153 Sunday, September 30,

153

Other fatigue models Forman’s model (stage III)

Paris’ model da m = C(∆K) dN

R = Kmin /Kmax Kmax − Kmin Kc − (Kmax − Kmin ) Kmax

Kmax

da =∞ = Kc : dN

As R increases, the crack growth rate increases. 154 Sunday, September 30,

154

Fatigue life calculation • Given: Griffith crack, 2a , ∆σ, C, m, K , N √ K = σ πa • Question: compute N 0

Ic

0

f

m=4

da da √ = dN = m C(∆K) C(∆σ πa)m Z af da √ N = N0 + m C(∆σ πa) a0

1 N = N0 + C(∆σ)4 π 2

Z

af a0

da 1 = N + 0 a2 C(∆σ)4 π 2

✓

1 1 − a0 af

◆


155

Fatigue life calculation • Given: Griffith crack, 2a , ∆σ, C, m, K , N √ K = σ πa • Question: compute N 0

Ic

0

f

m=4

da da √ = dN = m C(∆K) C(∆σ πa)m Z af da √ N = N0 + m C(∆σ πa) a0

1 N = N0 + C(∆σ)4 π 2

Z

Kmax

af a0

da 1 = N + 0 a2 C(∆σ)4 π 2

✓

1 1 − a0 af

◆

√ = σmax πaf = KIc 155


155

Numerical integration of fatigue law Z af

N = N0 +

a0

da √ C(∆σf (a/W ) πa)m

tedious to compute


156

Importance of initial crack length


157

Miner’s rule for variable 1945 load amplitudes Shortcomings: 1. sequence effect not considered 2. damage accumulation is independent of stress level

∆σ1 ∆σ2 N1 a 1 N1f n X Ni =1 N if i=1

Nᵢ/Nif : damage

∆σi

Ni

number of cycles a0 to ai

Nif

number of cycles a0 to ac


158

Variable amplitude cyclic loadings

da = f2 (∆K, R, H) dN

history variables

✏

∗

three stress values

plasticity: history dependent plastic wake 159


159

Overload and crack retardation It was recognized empirically that the application of a tensile overload in a constant amplitude cyclic load leads to crack retardation following the overload; that is, the crack growth rate is smaller than it would have been under constant amplitude loading.


160

Crack retardation Point A: plastic point B: elastic

After unloading: point A and B has more or less the same strain -> point A : compressive stress. 161 Sunday, September 30,

161

Crack retardation

a large plastic zone at overload has left behind

residual compressive plastic zone close the crack->crack retards 162 Sunday, September 30,

162

Nondestructive testing Nondestructive Evaluation (NDE), nondestructive Inspection (NDI) NDT is a wide group of analysis techniques used in science and industry to evaluate the properties of a material, component or system without causing damage NDT: provides input (e.g. crack size) to fracture analysis safety factor s

K(a, σ) = Kc → ac − > at ∗ s

NDT → ao t : ao → at

(Paris)

inspection time Sunday, September 30,

163 163

Damage tolerance design (stress concentration: possible crack sites) 1. Determine the size of initial defects a0 , NDI 2. Calculate the critical crack size ac at which failure √ would occur σ πac = KIc 3. Integrate the fatigue crack growth equations to compute the number of load cycles for the crack to grow from initial size to the critical size N = N0 +

Z

ac a0

da √ C(∆σ πa)m

4. Set inspection intervals Sunday, September 30,

164

164

Examples for Fatigue

da = log C + m log ∆K log dN √ 5.6 MPa m √ ∆K = ∆σ πa √ 17.72 MPa m

log(xy) = log(x) + log(y) log(x ) = p log(x) p

[Gdoutos]

af − a0 da = dN N 165 Sunday, September 30,

165

Example


166

Example (Gdoutos p.287)

A large thick plate contains a crack of length 2a₀=10 mm and is subjected to a constant-amplitude tensile cyclic stress normal to the crack of which σmin = 100 MPa and σmax= 200 MPa. The critical SIF is KIc = 60 MPa√m. Fatigue is governed by the following equation da −11 3 = 0.42 × 10 (∆K) dN

(m/cycle)

Plot the crack growth curve--a versus N up to the point of fracture. If a lifetime of 106 cycles is required, discuss the option that the designer has for an improved lifetime. 167 Sunday, September 30,

167

√ 200 πaf = 60

af = 28.65 mm

√ da −11 3 = 0.42 × 10 (∆σ πa) dN


168

Example (Matlab) A plate of width W=6 in contains a crack of length 2a₀=0.2 in and is subjected to a constant-amplitude tensile cyclic stress normal to the crack with Δσ=12 ksi. Fatigue is governed by the following equation da −9 3.5 = 4 × 10 (∆K) dN

Given Plot the crack growth curve--a versus N up to the point of fracture at which the critical crack length 2ac = 5.6 in. For 2a₀=1 in, do the same and plot the two curves on the same figure to see the influence of a₀. 169 Sunday, September 30,

169

Summary

σres

Kc √ = f (a/W ) πa

• What is the residual strength as a function of √crack size? f (a /W )σ πa = K What is the critical crack size? • from a certain • How long does it take for a crack to grow Z da c

initial size to the critical size? N = N + 0 170


c

c

ac

a0

√

C(∆σ πa)m

170

Mixed-mode crack growth Combination of mode-I, mode-II and mode-III loadings: mixed-mode loading.

Cracks will generally propagate along a curved surface as the crack seeks out its path of least resistance. Only a 2D mixed-mode loading (mode-I and mode-II) is discussed. 171 Sunday, September 30,

171

Maximum circumferential stress criterion Erdogan and Sih

(from M. Jirasek) Sunday, September 30,

principal stress 172

τrθ = 0 172

Maximum circumferential stress criterion

τrθ = 0

KI

✓

θ 3θ sin + sin 2 2

θc = 2 arctan Sunday, September 30,

⇣ 1 4

◆

+ KII

✓

KI /KII ± 173

θ 3θ cos + 3 cos 2 2

p

◆

=0

⌘ (KI /KI I)2 + 8 173

Maximum circumferential stress criterion

Fracture criterion


Keq ≥ KIc

174

Experiment

XFEM

⌘ p 1⇣ θc = 2 arctan KI /KII ± (KI /KI I)2 + 8 4 175


175

Ductile to Brittle transition Fractures occurred in “welldesigned” steel structures in severe weather.

Titanic in the icy water of Atlantic

At low temperatures some metals that would be ductile at room temperature become brittle. This is known as a ductile to brittle transition. As a result, some steel structures are every likely to fail in winter.


176

Stress corrosion cracking corrosive environments

• •

Metals are subject to corrosion

• •

Fracture type: brittle!!!

Stress corrosion cracking (SCC): interaction of corrosion and mechanical loadings to produce a cracking failure

Stress corrosion cracking is generally considered to be the most complex of all corrosion type


177

Alternatives to LEFM • Bodies with at least one existing crack • Nonlinear zone ahead of the crack tip is negligible

crack growth

Alternatives:

discussed

• Continuum damage mechanics • Cohesive zone models • Peridynamics • Lattice models

crack initiation/ formation


178

Fracture mechanics for concrete


179

Introduction •

LEFM theory was developed in 1920, but not until 1961 was the first experimental research in concrete performed.

•

Fracture mechanics was used successfully in design for metallic and brittle materials early on; however comparatively few applications were found for concrete.

•

This trend continued up until the middle ‘70s when finally major advances were made.

•

Experimentally observed size-effect can only be explained using fracture mechanics 180


180

Tensile response of concrete • • •

Tensile behavior of concrete is usually ignored: tensile strength is small

σ − ∆L

This prevented the efficient use of concrete Tensile behavior plays a key role in understanding fracture of concrete

✏m

∆L w = = ✏0 + L L σ − ✏m

quasi-brittle


∆L 181

Fictitious crack model

Fracture Process Zone (FPZ) concrete=quasi-brittle material 182 Sunday, September 30,

182

1976

Hilleborg’s fictitious crack model

Cohesive crack/zone model

Similar to the strip yield model of Dugdale-Barenblatt 183 Sunday, September 30,

183

Cohesive crack model Fracture criterion 1 σmax ≥ ft

when

2 where (direction) G=

Z

σ([[u]])d[[u]]

Rankine criterion Sunday, September 30,

184

Cohesive crack model Governing equations (strong form)

Constitutive equations deformation separation 185 Sunday, September 30,

185

Cohesive crack model Weak form

where


186

Cohesive crack model Weak form new term where


186

Cohesive crack model Weak form new term where (1) XFEM Implementation: (2) Interface elements (to be discussed later) 186 Sunday, September 30,

186

Size effect • Experiment tests: scaled versions of real structures • The result, however, depends on the size of the specimen that was tested

• From experiment result to engineering design: knowledge of size effect required

the nominal • The size effect is defined by comparing σ

strength (nominal stress at failure) N of geometrically similar structures of different sizes.

• Classical theories (elastic analysis with allowable stress): cannot take size effect into account √ a 187 Sunday, September 30,

187

• Size effect is crucial in concrete structures (dam,

bridges), geomechanics (tunnels): laboratory tests are small

• Size effect is less pronounced in mechanical and

aerospace engineering the structures or structural components can usually be tested at full size.

σN

cN P = bD

geometrically similar structures of different sizes

b is thickness


188

Structures and tests [Dufour]


189

Size effect (cont.)

1. Large structures are softer than small structures. 2. A large structure is more brittle and has a lower strength than a small structure. 190 Sunday, September 30,

190

Bazant’s size effect law σN = ft = ft (D)

0

σN

KIc KIc =√ =√ πa πcN D

For very small structures the curve approaches the horizontal line and, therefore, the failure of these structures can be predicted by a strength theory. On the other hand, for large structures the curve approaches the inclined line and, therefore, the failure of these structures can be predicted by LEFM. 191 Sunday, September 30,

191

Bazant’s size effect law


192

Continuum damage mechanics Milan Jirasek σ nominal stress

A

σ ¯ effective stress

A¯

Equilibrium: σA = σ¯ A¯ A¯ σ= σ ¯ = (1 − ω)¯ σ, A

A¯ ω =1− A

damage variable Hook’s law: σ¯ = Eε σ = (1 − ω)Eε 193 Sunday, September 30,

193

Four point bending test


194

Single Edged Notch Beam (SEN beam)

Numerical solution with CDM

Experiment 195 Sunday, September 30,

195

Computational fracture mechanics


196

Numerical methods to solve PDEs • • BEM (Boundary Element Method) • MMs (Meshless/Meshfree methods) FEM (Finite Element Method)

MMs

FEM

BEM 197 Sunday, September 30,

197

Fracture models • Discrete crack models (discontinuous

models) - LEFM (FEM,BEM,MMs) - EPFM (FEM,MMs) - Cohesive zone models (FEM,XFEM,MMs)

• Continuous models

- Continuum damage models (FEM,XFEM) - Phase field models (FEM)

• Lattice models (FEM) • Peridynamic models (FEM,MMs) 198 Sunday, September 30,

198

FEM for elastic cracks (1) double nodes

• Developed in 1976 (Barsoum) • double nodes: crack edge • singular elements: crack tip • remeshing as crack grows (3) remeshing

(2) singular elements 1 √ behavior r


199

What’s wrong with FEM for crack problems • Element edges must conform to the crack geometry:

make such a mesh is time-consuming, especially for 3D problems.

• Remeshing as crack advances: difficult


200

However ...

Bouchard et al. CMAME 2003 201 Sunday, September 30,

Show crack growth movies 201


202

Extended Finite Element Method (XFEM)

set of enriched nodes S X X h u (x) = NI (x)uI + NJ (x)Φ(x)aJ

Belytschko et al 1999

c

I∈S

J∈S c

standard part enrichment part Partition of Unity (PUM) X

NJ (x) = 1

J

enrichment function X

NJ (x)Φ(x) = Φ(x)

J

Φ(x) known characteristics of the problem (crack tip singularity, displacement jump etc.) into the203approximate space. Sunday, September 30,

203

XFEM: enriched nodes nodal support NI (x) 6= 0 I

X

NJ (x)Φ(x) = Φ(x)

J

enriched nodes = nodes whose support is cut by the item to be enriched enriched node I: standard degrees ofuIfreedoms aI dofs 204 (dofs) and additional Sunday, September 30,

204

XFEM for LEFM crack tip with known displacement r

✓

KI r θ u= cos κ − 1 + 2 sin2 2µ 2π 2 r ✓ KI r θ v= sin κ + 1 − 2 cos2 2µ 2π 2

◆

θ 2 ◆ θ 2

√

Φ1 = f ( r, θ)

crack edge

x+

x

−

displacement: discontinuous across crack edge +

Φ2 : Φ2 (x ) 6= Φ2 (x ) −


205

XFEM for LEFM (cont.) Crack tip enrichment functions: [Bα ] =

√

✓

KI r θ cos κ − 1 + 2 sin2 2µ 2π 2 r ✓ KI r θ v= sin κ + 1 − 2 cos2 2µ 2π 2

u=



r

√ θ √ θ √ θ θ r sin , r cos , r sin sin θ, r cos sin θ 2 2 2 2

◆

θ 2 ◆ θ 2

"

Crack edge enrichment functions: H(x) =

⇢

+1 −1

S

if (x − x ) · n ≥ 0 otherwise ∗

uh (x) =

c

blue nodes

t

red nodes

S X

NI (x)uI

I∈S

+

X

NJ (x)H(x)aJ

J∈S c

+ 206 Sunday, September 30,

X

K∈S t

NK (x)

4 X

α=1

Bα bα K

! 206

Domain form of J-integral

FE mesh J-integral is a contour integral that is not well suitable to FE computations. J= Sunday, September 30,

Z

A∗



#

∂uj ∂q σij − W δ1i dA ∂x1 ∂xi 207 207

XFEM for cohesive cracks Wells, Sluys, 2001

h

u (x) =

X I∈S

NI (x)uI +

X

NJ (x)H(x)aJ

J∈S c

No crack tip solution is known, no tip enrichment!!!

S

c

not enriched to ensure zero crack tip opening!!! H(x) =

⇢

+1 −1

if (x − x∗ ) · n ≥ 0 otherwise


208

XFEM: SIFs computation Mesh

One single mesh for all angles!!! [VP Nguyen Msc. thesis]

Results

Matlab code: free Sunday, September 30,

209 209

XFEM: examples

CENAERO, M. Duflot Northwestern Univ. 210 Sunday, September 30,

210

XFEM-Crack propagation Samtech, Belgium


fracture of underwater gas-filled pipeline 211

Meshfree methods Bordas et al.

Elastic-plastic fracture Sunday, September 30,

212

Shaofan Li 2012 212

Interface elements and cohesive crack model fracture of polycrystalline material

delamination of composites 213 Sunday, September 30,

213

Dynamic fracture • Dynamics is much more difficult than static • Dynamic fracture mechanics - inertia forces (kinetic energy) - rate-dependent material behavior - reflected stress waves

• Classification - LEFM -> Elastodynamic fracture mechanics 214 Sunday, September 30,

214

Crack speed of


215

Interfacial fracture mechanics

• Thin films •


216

Fracture of composite materials


217

Fiber reinforced composites


218

Fracture mechanics

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