Fracture mechanics

Fracture mechanics

Nguyen Vinh Phu, [email protected]

Researcher at Division of Computational MechanicsTon Duc Thang University

Program “Master of Science, Civil Engineering, 2012, Ton Duc Thang University”Nguyen Vinh Phu

1

1Sunday, September 30,

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

• wikipedia2


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


Outline (cont.)

Fatigue

- Paris law- Overload and crack retardation

Fracture of concrete

Computational fracture mechanics- FEM, BEM, MMs- XFEM

- cohesive crack model (Hillerborg, 1976)- Continuum Damage Mechanics- size effect (Bazant)

4


Stress-strain curves

0

0

Engineering stress and strain

Young’s modulusstrain hardening (tai ben)

Tension test

elastic unloading

ductile metals

�e =P

A0, ✏e =

�

L0

�e = E✏eE5


Stress/strain curve

Wikipedianecking=decrease of cross-sectional

area due to plastic deformation

fracture

1: ultimate tensile strength66Sunday, September 30,

Stress-strain curvesTrue stress and true (logarithmic) strain:

Plastic deformation:volume does not change

(extension ratio)

Relationship between engineering and true stress/strain

�t =P

A, d✏t =

dL

L

! ✏t =

Z L

L0

1

LdL = ln

L

L0

� ⌘ L

L0

dV = 0 ! AL = A0L0 ! L

L0=

A

A0

�t = �e(1 + ✏e) = �e�

✏t = ln(1 + ✏e) = ln�7


Stress-strain curveconcrete

post-peak(strain softening)pre-peak

strain softening=increasing strain while stress decrease8


Concrete

aggregates

cement paste

ITZ

fracture in concrete

9


Some common material models

✏

�

�ys

Linear elastic Elastic perfectly plastic✏

� no hardening

Will be used extensively in Fracture Mechanics10


Strain energy density

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


In terms of stress/strain

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

F

u

W

�x

✏x

u =1

2�x

✏x

[J/m3]

u =

Z�x

d✏x

W =

Z u

0Fdu =

Z u

0kudu =

1

2ku2 =

1

2Fu

U = W =1

2Fu

U =1

2Fu =

1

2

F

A

u

LAL

�x

✏x

11



Plane problems

Kolosov coefficient

Poisson’s ratio

shear modulus

u =1

2E(�2

x

+ �2y

+ �2z

)� ⌫

E(�

x

�y

+ �y

�z

+�z

�x

) +1

2µ(⌧2

xy

+ ⌧2yz

+ ⌧2zx

)

=

8<

:3� 4⌫ plane strain3� ⌫

1 + ⌫plane stress

u =1

4µ

+ 1

4(�2

x

+ �2y

)� 2(�x

�y

� ⌧2xy

)

�

µ =E

2(1 + ⌫)

12


Indicial notationa 3D vector

two times repeated index=sum, summation/dummy index

tensor notation

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

x = {x1, x2, x3}

||x|| =q

x

21 + x

22 + x

23

�xx

✏xx

+ �xy

✏xy

+ �yx

✏yx

+ �yy

✏yy

@�

x

@x

+@⌧

xy

@y

= 0,@�

y

@y

+@⌧

xy

@x

= 0

�ijnj = ti

�ij,j = 0

�ij✏ij

||x|| =pxixi

||x|| =pxkxk

i = 1, 2, 3

�yx

nx

+ �yy

ny

= ty

�xx

nx

+ �xy

ny

= tx

� : ✏

13


Engineering/matrix notation

�T✏ = �ij✏ij

Voigt notation

x =

2

4x1

x2

x3

3

5

||x|| = x

Tx

� =

2

4�xx

�yy

�xy

3

5 ✏ =

2

4✏xx

✏yy

2✏xy

3

5

� =

�xx

�xy

�xy

�yy

�✏ =

✏xx

✏xy

✏xy

✏yy

�

||x|| =pxixi

�ij✏ij

14


Principal stresses

�1,�2 =�xx

+ �yy

2±

s✓�xx

� �yy

2

◆2

+ ⌧2xy

Principal stresses are those stresses that act on principal surface. Principal surface here means the surface where components of shear-stress is zero.

tan 2✓p

=2⌧

xy

�xx

� �yy

Principal direction

15


Residual stressesResidual stresses are stresses that remain after the original cause of the stresses (external forces, heat gradient) has been removed.

Wikipedia

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

Causes

Residual stresses always appear to some extent during fabrication operations such as casting, rolling or welding.

16


http://en.wikipedia.org/wiki/Stress_(physics)

http://en.wikipedia.org/wiki/Stress_(physics)

Residual stressesTOTAL 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 packages17


Introduction

Cracks: ubiquitous !!!

18


Conventional failure analysisStresses

Failure criterion

Solid mechanics,numerical methods (FEM,BEM)

• Structures: no flaws!!!

• : depends on the testing samples !!!

• Many catastrophic failures occurred during WWII:

�c

Liberty ship

before 1960s

f(�,�c) = 0

f(�,�c) = 0

Tresca, Mohr-Coulomb…critical stress: experimentally determined

�c

�

�c

19


New Failure analysis Stresses

Flaw size aFracture

toughness

Fracture Mechanics (FM)

1970s

- 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 with improved mechanical properties.

� f(�, a,Kc) = 0

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


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


http://en.wikipedia.org/wiki/Mechanics

http://en.wikipedia.org/wiki/Mechanics

http://en.wikipedia.org/wiki/Solid_mechanics

http://en.wikipedia.org/wiki/Solid_mechanics

http://en.wikipedia.org/wiki/Fracture


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


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


http://en.wikipedia.org/wiki/Brittle


http://en.wikipedia.org/wiki/Plasticity_(physics)

http://en.wikipedia.org/wiki/Plasticity_(physics)

http://en.wikipedia.org/wiki/Ductile


Classification

• 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)

Fracture mechanics:

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

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

load lines

27


Stress concentration (cont.)uniaxial

biaxial

28


Elliptic holeInglis, 1913, theory of elasticity

radius of curvature

⇢stress concentration factor [-]

0 crack

⇢ =b2

a

!!!

�3 =

✓1 +

2b

a

◆�1

�3 =

1 + 2

sb

⇢

!�1

�3 = 2

sb

⇢�1

1

KT ⌘ �3

�1= 1 +

2b

a29


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

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

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


http://en.wikipedia.org/wiki/Glass

http://en.wikipedia.org/wiki/Glass

Griffith’s size effect experimentSize 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


Griffith’s experiment• Glass fibers with artificial cracks (much larger

than natural crack-like flaws), tension tests

Energy approach

�c =constp

a�3 = 2

sb

⇢�1

�cpa = const

32


Energy balance during crack growth

external worksurface energy

kinetic energy

internal strain energyAll changes with respect to time are caused by changes in crack size:

Energy equation is rewritten:

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

slow process@W

@a=

@Ue

@a+

@Up

@a+

@U�

@a

@(·)@t

=@(·)@a

@a

@t

W = Ue + Up + Uk + U�

33


Brittle materials: no plastic deformation

Potential energy

γs is energy required to form a unit of new surface

(two new material surfaces)

(plane stress, constant load)

Inglis’ solution

Griffith’s through-thickness crack

[J/m2=N/m]

⇧ = Ue �W

�@⇧

@a=

@Up

@a+

@U�

@a

�@⇧

@a=

@U�

@a

�@⇧

@a= 2�s

�@⇧

@a=

⇡�2a

E

⇡�2a

E= 2�s ! �c =

r2E�s⇡a

�cpa =

✓2E�s⇡

◆1/2

34


⇡�2a

E= 2�s ! �c =

r2E�s⇡a

⇡�2a

E= 2�s ! �c =

r2E�s⇡a

E : MPa=N/m2

�s : N/m

a : m

check dimension

Dimensional Analysis

u =1

2�✏ =

1

2

�2

E U = ��2

Ea2

B = 1App. of dimensional analysis

[N/m2]

[N/m2]

35


⇡�2a

E= 2�s ! �c =

r2E�s⇡a

⇡�2a

E= 2�s ! �c =

r2E�s⇡a

E : MPa=N/m2

�s : N/m

a : m

check dimension

Dimensional Analysis

u =1

2�✏ =

1

2

�2

E U = ��2

Ea2

B = 1� = ⇡App. of dimensional analysis

[N/m2]

[N/m2]

35


�c =

r2E�s⇡a

Griffith (1921), ideally brittle solids

Irwin, Orowan (1948), metals

plastic work per unit area of surface created

Energy equation for ductile materials

Plane stress

(metals)

Griffith’s work was ignored for almost 20 years

�c =

r2E�s⇡a

�c =

r2E(�s + �p)

⇡a

�p ⇡ 103�s

�p � �s

�p

36


Energy release rateIrwin 1956

G: energy released during fracture per unit of newly created fracture surface area

energy available for crack growth (crack driving force)

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

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

Energy release rate failure criterionGc

G = 2� + �p| {z }

G � Gc

G ⌘ �d⇧

dA

37


http://en.wikipedia.org/wiki/Energy


http://en.wikipedia.org/wiki/Dissipation




http://en.wikipedia.org/wiki/Failure_theory_(material)


Energy release rateIrwin 1956

G: energy released during fracture per unit of newly created fracture surface area

energy available for crack growth (crack driving force)

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

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

Energy release rate failure criterionGc

Griffith

G = 2� + �p| {z }

G � Gc

G ⌘ �d⇧

dA

37










G from experimentG ⌘ �d⇧

dA

G =1

B

(OAB)

�a

G =1

B

(OAB)

�a

⇧ = Ue �W

W = 0 ! Ue < 0

Fixed grips Dead loads

B: thickness

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

a2: OB, triangle OBC=U

a1: OA, triangle OAC=U

OAB=ABCD-(OBD-OAC)

38


http://en.wikipedia.org/wiki/Strain_energy



G from experiments

G =1

B

shaded area

a4 � a3 39


Crack extension resistance curve (R-curve)

G = �d⇧

dA=

dU�

dA+

dUp

dAG = R

R ⌘ dU�

dA+

dUp

dA

R-curve

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

R = R(a) Irwin

crack driving force curve

SLOW

Irwin

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

40


R-curve shapesflat R-curve

(ideally brittle materials)rising R-curve

(ductile metals)

G = R,dG

da dR

dastable crack growth

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

G =⇡�2a

E

slope

41


G in terms of compliance

Fixed grips

inverse of stiffnessP

uK

CC =

u

P

dUe = Ue(a+ da)� Ue(a)

=1

2(P + dP )u� 1

2Pu

=1

2dPu

G = � 1

2BudP

da

G =1

2B

u2

C2

dC

da=

1

2BP 2 dC

da

a

a+ da

u

PdP

dA = Bda

G =dW � dUe

dA42



Fixed load

G =1

2BP 2 dC

da

inverse of stiffnessP

uK

CC =

u

P

dUe =1

2P (u+ du)� 1

2Pu

=1

2Pdu

dW = Pdu

G =1

2BPdu

da

P a+ da

a

u

du

G =dW � dUe

dA43



G =1

2B

u2

C2

dC

da=

1

2BP 2 dC

daG =

1

2BP 2 dC

da

Fixed grips Fixed loads

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

@�

x

@x

+@⌧

xy

@y

= 0,@�

y

@y

+@⌧

xy

@x

= 0Equilibrium:

Airy stress function :

Compatibility condition:

Bi-harmonic equation

�

x

=@

2�

@y

2, �

y

=@

2�

@x

2, ⌧

xy

= � @

2�

@x@y

r4� =@

4�

@x

4+ 2

@

4�

@x

2@y

2+

@

4�

@y

4= 0

For a given problem, choose an appropriate that satisfies (*) and the boundary conditions.

�

� ! �ij ! ✏ij ! ui

�

(*)

46


Crack modes

ar

47


Crack modes

48


Westergaard’s complex stress function for mode I

1937

shear modulus

Kolosov coef.

µ =E

2(1 + ⌫)

�xx

= ReZ � yImZ 0

�yy = ReZ + yImZ 0

⌧xy

= �yReZ 0

✏ij ! ui

=

8<

:3� 4⌫ plane strain3� ⌫

1 + ⌫plane stress 2µu =

� 1

2ReZ � yImZ

2µv =+ 1

2ImZ � yReZ

Z(z), z = x+ iy, i

2 = �1

� = Re ¯Z + yImZ

Z =

ZZ(z)dz, ¯Z =

ZZ(z)dz

49


Griffith’s crack (mode I)

Z(z) =�(⇠ + a)p⇠(⇠ + 2a)

boundary conditions

⇠ = z � a, ⇠ = rei✓

Z =�p⇡ap

2⇡⇠

infinite plate

Z(z) =�zp

z2 � a2

(x, y) ! 1 : �xx

= �

yy

= �, ⌧

xy

= 0

|x| < a, y = 0 : �yy

= ⌧

xy

= 0

Z(z) =�xp

x

2 � a

2

y = 0, |x| < a

is imaginary

Z(z) =�p

1� (a/z)2

(x, y) ! 1 : z ! 1 Z ! �

1I

IZ 0(z) = � �a2

(z2 � a2)3/2! 0

50

�xx

= ReZ � yImZ 0


⌧xy

= �yReZ 0


Griffith’s crack (mode I)

Z(z) =�(⇠ + a)p⇠(⇠ + 2a)

Z(z) =�zp

z2 � a2

(x, y) ! 1 : �xx

= �

yy

= �, ⌧

xy

= 0

|x| < a, y = 0 : �yy

= ⌧

xy

= 0

boundary conditions

x

y

⇠ = z � a, ⇠ = rei✓

Z =�p⇡ap

2⇡⇠

infinite plate 51


Griffith’s crack (mode I)(x, y) ! 1 : �

xx

= �

yy

= �, ⌧

xy

= 0

|x| < a, y = 0 : �yy

= ⌧

xy

= 0

x

y

Z =�p⇡ap

2⇡⇠⇠ small

Z =�p⇡ap

2⇡⇠

Z(z) =�(⇠ + a)p⇠(⇠ + 2a)

=�(⇠ + a)p

2a⇠(1 + ⇠/(2a)))

p1 + ⇠/(2a)) = (1 + ⇠/(2a))�1/2

= 1� 1

2

⇠

2a+H.O.T

= 1

⇠ small ⇠ + a = a

⇠ small

52


Recall

e

�ix

= cosx� i sinx

y = r sin ✓

sin ✓ = 2 sin

✓

2

cos

✓

2

Crack tip stress field

inverse square root

�xx

= ReZ � yImZ 0


⌧xy

= �yReZ 0

Z(z) =KIp2⇡⇠

,KI = �p⇡a

Z(z) =KIp2⇡r

e�i✓/2

Z 0(z) = �1

2

KIp2⇡

⇠�3/2 = � KI

2rp2⇡r

e�i3✓/2

⇠ = rei✓

�xx

=

KIp

2⇡rcos

✓

2

✓1� sin

✓

2

cos

3✓

2

◆

�yy =

KIp2⇡r

cos

✓

2

✓1 + sin

✓

2

cos

3✓

2

◆

⌧xy

=

KIp

2⇡rsin

✓

2

cos

✓

2

sin

3✓

2

r ! 0 : �ij ! 153


Recall

1pr

singularity

e

�ix

= cosx� i sinx

y = r sin ✓

sin ✓ = 2 sin

✓

2

cos

✓

2

Crack tip stress field

inverse square root

�xx

= ReZ � yImZ 0


⌧xy

= �yReZ 0

Z(z) =KIp2⇡⇠

,KI = �p⇡a

Z(z) =KIp2⇡r

e�i✓/2

Z 0(z) = �1

2

KIp2⇡

⇠�3/2 = � KI

2rp2⇡r

e�i3✓/2

⇠ = rei✓

�xx

=

KIp

2⇡rcos

✓

2

✓1� sin

✓

2

cos

3✓

2

◆

�yy =

KIp2⇡r

cos

✓

2

✓1 + sin

✓

2

cos

3✓

2

◆

⌧xy

=

KIp

2⇡rsin

✓

2

cos

✓

2

sin

3✓

2

r ! 0 : �ij ! 153


Plane strain

�z

= ⌫(�x

+ �y

)

�z = 2⌫KI

2⇡rcos

✓

2

✏zz

=1

E(�⌫�

xx

� ⌫�yy

+ �zz

)

Hooke’s law

✏zz = 0

Plane strain problems

�xx

=

KIp

2⇡rcos

✓

2

✓1� sin

✓

2

cos

3✓

2

◆

�yy =

KIp2⇡r

cos

✓

2

✓1 + sin

✓

2

cos

3✓

2

◆

⌧xy

=

KIp

2⇡rsin

✓

2

cos

✓

2

sin

3✓

2

54


Stresses on the crack plane

✓ = 0, r = x

On the crack plane

�

xx

= �

yy

=K

Ip2⇡x

⌧xy

= 0

crack plane is a principal plane with the following principal stresses

�1 = �2 = �xx

= �yy

�xx

=

KIp

2⇡rcos

✓

2

✓1� sin

✓

2

cos

3✓

2

◆

�yy =

KIp2⇡r

cos

✓

2

✓1 + sin

✓

2

cos

3✓

2

◆

⌧xy

=

KIp

2⇡rsin

✓

2

cos

✓

2

sin

3✓

2

55


Stress Intensity Factor (SIF)

KI

• Stresses-K: linearly proportional

• K uniquely defines the crack tip stress field

• modes I, II and III:

• LEFM: single-parameter

KI ,KII ,KIII

SIMILITUDE

�xx

=

KIp

2⇡rcos

✓

2

✓1� sin

✓

2

cos

3✓

2

◆

�yy =

KIp2⇡r

cos

✓

2

✓1 + sin

✓

2

cos

3✓

2

◆

⌧xy

=

KIp

2⇡rsin

✓

2

cos

✓

2

sin

3✓

2

KI = �p⇡a

[MPapm]

56


Singular dominated zone

�1

�1

crack tip

K-dominated zone

�yy =

KIp2⇡r

cos

✓

2

✓1 + sin

✓

2

cos

3✓

2

◆

(crack plane)

�xx

=

KIp

2⇡rcos

✓

2

✓1� sin

✓

2

cos

3✓

2

◆

�yy =

KIp2⇡r

cos

✓

2

✓1 + sin

✓

2

cos

3✓

2

◆

⌧xy

=

KIp

2⇡rsin

✓

2

cos

✓

2

sin

3✓

2

57


Displacement field

Mode I: displacement fieldRecall

2µu =� 1

2ReZ � yImZ

2µv =+ 1

2ImZ � yReZ

Z(z) =KIp2⇡r

✓cos

✓

2

� i sin✓

2

◆

Z(z) =KIp2⇡⇠

Z =

ZZ(z)dz

¯Z(z) = 2

KIp2⇡

⇠1/2 = 2KI

rr

2⇡

✓cos

✓

2

+ i sin✓

2

◆

u =

KI

2µ

rr

2⇡cos

✓

2

✓� 1 + 2 sin

2 ✓

2

◆

v =

KI

2µ

rr

2⇡sin

✓

2

✓+ 1� 2 cos

2 ✓

2

◆

z = ⇠ + a

⇠ = rei✓

e

�ix

= cosx� i sinx

58


Crack face displacement

2µv =+ 1

2ImZ � yReZ

Z(z) =�xp

x

2 � a

2 Z(z) = �

px

2 � a

2

v =+ 1

4µImZ

Z(z) = i(�pa

2 � x

2)

v =+ 1

4µ�

pa

2 � x

2

y = 0,�a x a

i =p�1�a x a

COD = 2v =+ 1

2µ�

pa

2 � x

2

Crack Opening Displacement

ellipse

59


Crack tip stress field in polar coordinates-mode I

stress transformation

�ij =KIp⇡a

fij(✓)

�rr =

KIp2⇡r

✓5

4

cos

✓

2

� 1

4

cos

3✓

2

◆

�✓✓ =

KIp2⇡r

✓3

4

cos

✓

2

+

1

4

cos

3✓

2

◆

⌧r✓ =KIp2⇡r

✓1

4sin

✓

2+

1

4sin

3✓

2

◆

60


Principal crack tip stresses�1,�2 =

�xx

+ �yy

2±

s✓�xx

� �yy

2

◆2

+ ⌧2xy

�1 =

KIp2⇡r

cos

✓

2

✓1 + sin

✓

2

◆

�2 =

KIp2⇡r

cos

✓

2

✓1� sin

✓

2

◆

�3 =

8<

:

0 plane stress

2⌫KIp2⇡r

cos

✓

2

plane strain

�3 = ⌫(�1 + �2)

�xx

=

KIp

2⇡rcos

✓

2

✓1� sin

✓

2

cos

3✓

2

◆

�yy =

KIp2⇡r

cos

✓

2

✓1 + sin

✓

2

cos

3✓

2

◆

⌧xy

=

KIp

2⇡rsin

✓

2

cos

✓

2

sin

3✓

2

61


Stress function

Mode II problemBoundary conditions

Z = � i⌧zpz2 � a2

(x, y) ! 1 : �xx

= �

yy

= 0, ⌧xy

= ⌧

|x| < a, y = 0 : �yy

= ⌧

xy

= 0

�xx

= ReZ � yImZ 0


⌧xy

= �yReZ 0

Check BCs

62


Stress function

Mode II problem

mode II SIF

Boundary conditions

KII = ⌧p⇡a

Z = � i⌧zpz2 � a2

(x, y) ! 1 : �xx

= �

yy

= 0, ⌧xy

= ⌧

|x| < a, y = 0 : �yy

= ⌧

xy

= 0

�xx

= � KIIp2⇡r

sin

✓

2

✓2 + cos

✓

2

cos

3✓

2

◆

�yy =

KIIp2⇡r

sin

✓

2

cos

✓

2

cos

3✓

2

⌧xy

=

KIIp2⇡r

cos

✓

2

✓1� sin

✓

2

sin

3✓

2

◆

63


Z = � i⌧zpz2 � a2

Stress function

Mode II problem (cont.)

|x| < a, y = 0 : �yy

= ⌧

xy

= 0

(x, y) ! 1 : �xx

= �

yy

= 0, ⌧xy

= ⌧

mode II SIF

Z = � i⌧zpz2 � a2

(x, y) ! 1 : �xx

= �

yy

= 0, ⌧xy

= ⌧

|x| < a, y = 0 : �yy

= ⌧

xy

= 0

KII = ⌧p⇡a

u =

KII

2µ

rr

2⇡sin

✓

2

✓+ 1 + 2 cos

2 ✓

2

◆

v =

KII

2µ

rr

2⇡cos

✓

2

✓� 1� 2 sin

2 ✓

2

◆

64


Mode III problem

65


�ij =Kp2⇡r

fij(✓) + H.O.T

Universal nature of the asymptotic stress field

Irwin

Westergaards, Sneddon etc.

(mode I) (mode II)

�xx

=

KIp

2⇡rcos

✓

2

✓1� sin

✓

2

cos

3✓

2

◆

�yy =

KIp2⇡r

cos

✓

2

✓1 + sin

✓

2

cos

3✓

2

◆

⌧xy

=

KIp

2⇡rsin

✓

2

cos

✓

2

sin

3✓

2

�xx

= � KIIp2⇡r

sin

✓

2

✓2 + cos

✓

2

cos

3✓

2

◆

�yy =

KIIp2⇡r

sin

✓

2

cos

✓

2

cos

3✓

2

⌧xy

=

KIIp2⇡r

cos

✓

2

✓1� sin

✓

2

sin

3✓

2

◆

66


Inclined crack in tension

�2 = �y

cos

2 ✓ � 2 sin ✓ cos ✓⌧xy

+ sin

2 ✓�x

�1 = �x

cos

2 ✓ + 2 sin ✓ cos ✓⌧xy

+ sin

2 ✓�y

⌧12 = ��x

cos ✓ sin ✓ + cos 2✓⌧xy

+ 0.5 sin 2✓�y

�1 = (sin2 �)�

�2 = (cos

2 �)�

⌧12 = (sin� cos�)�

KI = �yp⇡a

KII

= ⌧xy

p⇡a

Recall

KII = �p⇡a sin� cos�

KI = �p⇡a cos2 �

+

Final result

67


Inclined crack in tension

�2 = �y

cos

2 ✓ � 2 sin ✓ cos ✓⌧xy

+ sin

2 ✓�x

�1 = �x

cos

2 ✓ + 2 sin ✓ cos ✓⌧xy

+ sin

2 ✓�y

⌧12 = ��x

cos ✓ sin ✓ + cos 2✓⌧xy

+ 0.5 sin 2✓�y

�1 = (sin2 �)�

�2 = (cos

2 �)�

⌧12 = (sin� cos�)�

KI = �yp⇡a

KII

= ⌧xy

p⇡a

Recall

KII = �p⇡a sin� cos�

KI = �p⇡a cos2 �

+

Final result

12

67


Cylindrical pressure vessel with an inclined through-thickness crack

�z =pR

2t

�✓ =pR

t

KI =pR

2t

p⇡a(1 + sin2 �)

KII =

pR

2t

p⇡a sin� cos�

R

t� 10 thin-walled pressure

(�l2R)p = (2�lt)�✓

(⇡R2)p = (2⇡Rt)�z

closed-ends

68


Cylindrical pressure vessel with an inclined through-thickness crack

�z =pR

2t

�✓ =pR

t

KI =pR

2t

p⇡a(1 + sin2 �)

KII =

pR

2t

p⇡a sin� cos�

?

�✓ = 2�z

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

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


SIF for finite size samplesExact (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

force lines are compressed->> higher stress concentration

geometry/correction factor [-]

dimensional analysis

KI = f(a/W )�p⇡a a ⌧ W : f(a/W ) ⇡ 1

KI KI<

72


SIFs handbook

73


SIFs handbook

74


SIFs handbook

75


Reference stressKI = ��

p⇡a

KI = �max

�max

p⇡a

KI

= �xa

�xa

p⇡a

�xa

= �max

�max

�xa

=�max

1� 2a/W

�

Non-uniform stress distribution

for which reference stress!!!

76


Reference stressKI = ��

p⇡a

KI = �max

�max

p⇡a

KI

= �xa

�xa

p⇡a

�xa

= �max

�max

�xa

=�max

1� 2a/W

chosen

�

Non-uniform stress distribution

for which reference stress!!!

76


Superposition methodA sample in mode I subjected to tension and bending:

Is superposition of SIFs of different crack modes possible?

�ij =Ktension

Ip2⇡r

fij(✓) +Kbending

Ip2⇡r

fij(✓)

�ij =Ktension

I +Kbending

Ip2⇡r

fij(✓)

KI = Ktension

I +Kbending

I

77


KbendI = fM (a/W )

6M

BW 2

p⇡a Kten

I = fP (a/W )P

BW

p⇡a

Determine the stress intensity factor for an edge cracked plate subjected to a combined tension and bending.

a/W = 0.2

1.055 1.12

KI =

✓1.055

6M

BW 2+ 1.12

P

BW

◆p⇡a

B thickness

Solution

78


KId +KIe = KIb = 0 ! KIe = �KId = ��p⇡a

Superposition method Centered crack under internal pressure

This result is useful for surface flaws along the internal wall of pressure vessels.

79


KI = �p⇡a

80


SIFs: asymmetric loadingsProcedure: build up the case from symmetric cases and then to subtract the superfluous loadings.

KA = KB +KC �KD

KA = (KB +KC)/2

81


Two small cracks at a hole

a

3� edge crack hole as a part of the crack

82


PhotoelasticityPhotoelasticity 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.

Wikipedia

83


http://en.wikipedia.org/wiki/Stress_analysis

http://en.wikipedia.org/wiki/Stress_analysis

K-G relationshipSo 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 closethe crack

84


K-G relationshipIrwin

✓ = ⇡

✓ = 0

work of crack closure

B=1 (unit thickness)

KI(a)

G = lim�a!0

(+ 1)K2I

4⇡µ�a

Z �a

0

r�a� x

x

dx

G =(+ 1)K2

I

8µ

uy =(+ 1)KI(a+�a)

2µ

r�a� x

2⇡

�yy =KI(a)p2⇡x

dU(x) = 21

2�yy(x)uy(x)dx

�U =

Z �a

0dU(x)

G = lim�a!0

✓�U

�a

◆

fixed load

85


K-G relationship (cont.)Mode I

Mixed mode

• 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 ( same direction as )

GI =

8><

>:

K2I

Eplane stress

(1� v2)K2

I

Eplane strain

G =K2

I

E0 +K2

II

E0 +K2

III

2µ

daa 86


SIF in terms of compliance

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

B: thicknessG =1

2BP 2 dC

da

GI =K2

I

E0K2

I =E0P 2

2B

dC

da

87


SIF in terms of compliance

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

B: thicknessG =1

2BP 2 dC

da

GI =K2

I

E0K2

I =E0P 2

2B

dC

da

87


Units

88


Example

89


Gc for different crack lengths are almost the same: flat R-curve.

G =1

2B

OAiAj

aj � ai

90


91


92


Examples

Double cantilever beam (DCB)

G =1

2BP 2 dC

daload control

G =1

2B

u2

C2

dC

dadisp. control

load control

G =3u2Eh3

16a4

a increases -> G decreases!!!

a increases -> G increase

93


Compliance-SIFK =

rsec

⇡a

W�p⇡a

G =P 2

2

dC

dA=

P 2

4B

dC

da

G =K2

E

P 2

4B

dC

da=

sec ⇡aW �2⇡a

E

dC

da=

sec ⇡aW �2⇡a4B

P 2E

dC

da=

4

EBW 2⇡a sec

⇡a

W

C =

Z a

0

4

EBW 2⇡a sec

⇡a

Wda+ C0

C = � 4

EB⇡ln

⇣cos

⇡a

W

⌘+

H

EBW

C0 =�

P=

H

EBW

H

94


C/C0 = � 4

⇡

W

Hln

⇣cos

⇡a

W

⌘+ 1

compliance rapidly increases

95


K as a failure criterionFailure criterion

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

• Problem 2: for a specific applied stress, compute the maximum permissible crack length (critical crack length)

• Problem 3: compute provided crack length and stress at fracture

�

fracture toughness

ac

K = Kc f(a/W )�p⇡a = Kc

�max

=Kc

f(a/W )p⇡a

f(ac/W )�p⇡ac = Kc

Kc

Kc = f(ac/W )�p⇡ac

! ac

96


Example

97


Example solution�z =

pR

2t

�✓ =pR

t

a

KI = 1.12�✓p⇡a

KI = KIc/S

p = 12MPa

problem 1 problem 2

a = 1mm

98


Example

�c =

r2E�s⇡a

=5.8 Mpa �c =KIcp⇡a =479 Mpa

Griffith Irwin

99


Mixed-mode fractureKI = KIc

KII = KIIc

KIII = KIIIc

KIc < KIIc,KIIIc

G =K2

I

E0 +K2

II

E0

Superposition cannot be applied to SIF. However, energy can.

Gc =K2

Ic

E0

lowest Kc: safe

G = Gc

Fracture occurs when

K2I +K2

II = K2Ic

100


Experiment verification of the mixed-mode failure criterion

K2I +K2

II = K2Ic

Data points do not fall exactly on the circle.

a circle in KI, KII plane

G =(+ 1)K2

I

8µself-similar growth

✓KI

KIc

◆2

+

✓KII

KIIc

◆2

= 1101


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


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.

crack tipSmall-scale yielding:

LEFM still applies with minor modifications done

by G. R. Irwin

(SSY)

R ⌧ D

104






http://en.wikipedia.org/wiki/Steel

http://en.wikipedia.org/wiki/Steel

http://en.wikipedia.org/wiki/G._R._Irwin

http://en.wikipedia.org/wiki/G._R._Irwin

http://en.wikipedia.org/wiki/U.S._Naval_Research_Laboratory

http://en.wikipedia.org/wiki/U.S._Naval_Research_Laboratory

Validity of K in presence of a plastic zone

crack tip Fracture process usually occurs in the inelastic region not the K-dominant 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

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.

[Anderson]

LEFM solution

106


Paradox of a sharp crack

r = 0 ! �ij = 1At crack tip:

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


Plastic correction• A cracked body in a plane stress condition

• Material: elastic perfectly plastic with yield stress

On the crack plane

�ys

(yield occurs)

first order approximation of plastic zone size: equilibrium is not satisfied

stress singularity is truncated by yielding at crack tip

r1 =K2

I

2⇡�2ys

�yy =KIp2⇡r

�yy = �ys

✓ = 0

�ys

108


Irwin’s plastic correction

109


Irwin’s plastic correctionplate behaves as with a longer crack

stress redistribution: yellow area=hatched area

Plane strainplastic zone: a CIRCLE !!!

rp = 2r1 =K2

I

⇡�2ys

rp =1

3⇡

K2I

�2ys 110

�ysr1 =

Z r1

0�yydr


Plane stress Plane strain

�1 = �2 = �yy, �3 = 0 �1 = �2 = �yy

�3 = ⌫(�xx

+ �yy

) = 2⌫�yy

�3 = 0.66�yy ⌫ = 0.33

�1 � �3 = �ysTresca’s criterion�1 = �ys

�y = �ys �y = 3�ys

111


Irwin’s plastic correction

�ysr1 =

Z r1

0�yydr

rp =1

3⇡

K2I

�2ys

crack tip

R ⌧ D

LEFM:

rp is small

�ys is big and KIc is small

LEFM is better applicable to materials of high yield strength and low fracture toughness

112


Plastic zone shape

Mode I, principal stresses

von-Mises criterion

�e =1p2

�(�1 � �2)

2 + (�1 � �3)2 + (�2 � �3)

2�1/2

�e = �ys

�1 =

KIp2⇡r

cos

✓

2

✓1 + sin

✓

2

◆

�2 =

KIp2⇡r

cos

✓

2

✓1� sin

✓

2

◆

�3 =

8<

:

0 plane stress

2⌫KIp2⇡r

cos

✓

2

plane strain

ry(✓) =1

4⇡

✓KI

�ys

◆2 1 + cos ✓ +

3

2

sin

2 ✓

�

ry(✓) =1

4⇡

✓KI

�ys

◆2 (1� 2µ)2(1 + cos ✓) +

3

2

sin

2 ✓

�

plane stress

113


Plastic zone shapeplastic zone shape (mode I, von-Mises criterion)

ry(✓) =1

4⇡

✓KI

�ys

◆2 1 + cos ✓ +

3

2

sin

2 ✓

�

114

−0.6−0.4−0.2 0 0.2 0.4 0.6 0.8−0.6

−0.4

−0.2

0

0.2

0.4

0.6

rp/(KI/(/ my))2

r p/(KI/(/

my))2

plane stressplane strain


Plane stress/plane strain

dog-bone shape

• Plane stress failure: in general, ductile

• Plane strain failure: in general, brittle

constrained by the surrounding material

115


Plane stress/plane strain

Plane strain fracture toughness lowest K (safe)

KIc

toughness depends on thickness

(Irwin)

116

Kc = KIc

1 +

1.4

B2

KIc

�ys

�4!1/2


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

ASTM (based on Irwin’s model)

Constraint conditions

KI =P

BpW

⇣2 +

a

W

⌘0.886 + 4.64

a

W� 13.32

⇣ a

W

⌘2+ 14.72(

⇣ a

W

⌘3� 5.6

⇣ a

W

⌘4�

⇣1� a

W

⌘3/2

117


Compact tension testCyclic 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)

check constraint conditions

PQ ! KQ

KIc = KQ

118


Fracture toughness test

B > 25rp =25

3⇡

✓KI

�ys

◆2plane strainASTM E399

a > 25rp

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

Text119


proposed by Dugdale and Barrenblatt

Strip Yield Model• 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.

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

120


Strip Yield Model (cont.)

�ys�ys

Superposition principle

Irwin’s result

(derivation follows)

close to

KI = K�I +K

�ys

I

K�I = �

p⇡(a+ c)

K�ys

I = �2�ys

ra+ c

⇡cos

�1

✓a

a+ c

◆

c =⇡2�2a

8�2ys

=⇡

8

✓KI

�ys

◆2

a

a+ c= cos

✓⇡�

2�ys

◆

rp =1

⇡

✓KI

�ys

◆2

� ⌧ �ys

cosx = 1� 1

2!

x

2+ · · ·

KI = 00.318

0.392

�ij =Kp2⇡r

fij(✓) + H.O.T

121


SIF for plate with normal force at crack

KA =Pp⇡a

ra+ x

a� x

KB =Pp⇡a

ra� x

a+ x

Gdoutos, chapter 2, p40

K�ys

I = �2�ys

ra+ c

⇡cos

�1

✓a

a+ c

◆

K

�ys

I = � �ysp⇡c

Z c

a

✓rc� x

c+ x

+

rc+ x

c� x

◆dx

P = ��ysdx

122


Effective crack length

1

rp =⇡

8

✓KI

�ys

◆2

r1 =1

2⇡

✓KI

�ys

◆2

123


�net =P

W � a= �

W

W � a

a

W

P

� =P

W(cracked section)

�W

W � a= �ys � = �ys

⇣1� a

W

⌘Yield:

short crack: fracture by plastic collapse!!!

high toughness materials:yielding before fracture

Fracture vs. Plastic collapseP

unit thickness

124


�net =P

W � a= �

W

W � a

a

W

P

� =P

W(cracked section)

�W

W � a= �ys � = �ys

⇣1� a

W

⌘Yield:

short crack: fracture by plastic collapse!!!

high toughness materials:yielding before fracture

�c 0.66�ysLEFM applies when

Fracture vs. Plastic collapseP

unit thickness

124


ExampleConsider 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

Ke↵ =�p⇡a

1� 0.5

⇣�

�ys

⌘2�1/2

Ke↵ = �p⇡ (a+ r1)

r1 =K2

e↵

2⇡�2ys

Solution:a+ r1The effective crack length is

The effective SIF is thus

with125


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?

Example

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?


Stress at failure �f = 300/(20⇥ 0.75) = 20 ksi

Kc = 1⇥ 20⇥p⇡ ⇥ 1 = 35.4 ksi

pinToughness

Nominal stress at collapse � = �ys

⇣1� a

W

⌘

�col

= 70⇥ 20� 2

20= 63 ksi

Fracture occurs before collapse.

Solution to problem 1

�f

< �col

B � 2.5

✓KIc

�Y

◆2

= 0.64in plane strain by ASTM


�fr =Kc

�p⇡a

�fr =35.4

2.1⇥p⇡ ⇥ 2

= 6.73 ksi

�col

= 70⇥ 5� 2

5= 42 ksi

Residual strength 6.73 ksi



Stress at failure

Toughness


⇣1� a

W

⌘

Collapse occurs before fracture


� = 1.067 = 1.07

�f = 41/(6⇥ 0.2) = 34.2 ksi

Kc = 1.07⇥ 34.2⇥p⇡ ⇥ 1 = 64.9 ksi

pin

�col

= 50⇥ 6� 2

6= 33.3 ksi

�f

> �col

The above Kc is not the toughness!!!

129 129Sunday, September 30,

Stress at failure

Toughness


⇣1� a

W

⌘

Collapse occurs before fracture


� = 1.067 = 1.07

�f = 41/(6⇥ 0.2) = 34.2 ksi

Kc = 1.07⇥ 34.2⇥p⇡ ⇥ 1 = 64.9 ksi

pin

�col

= 50⇥ 6� 2

6= 33.3 ksi

�f

> �col

The above Kc is not the toughness!!! whole section is yielding


Elastic-Plastic Fracture Mechanics

• J-integral (Rice,1958)

• Crack Tip Opening Displacement (CTOD), (Wells, 1963)

130


Introduction

deformation theory of plasticity can be utilized

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

No unloading

plasticity models:

- deformation theory

- flow theory131


J-integralEshelby, Cherepanov, 1967, Rice, 1968

Wikipedia

J =

Z

�

✓Wdx2 � ti

@ui

@x1ds

◆

strain energy density

surface traction

(1) J=0 for a closed path

(2) is path-independent notch:traction-free

J =

Z

�

✓Wn1 � ti

@ui

@x1

◆d�

W =

Z ✏

0�ijd✏ij

ti = �ijnj

J � integral

J =

Z

�

✓Wdx2 � ti

@ui

@x1ds

◆ N

m

�

132


Path independence of J-integral

0 = JABCDA = JAB + JBC + JCD + JDA

J is zero over a closed path

J =

Z

�

✓Wdx2 � ti

@ui

@x1ds

◆

AB, CD: traction-free crack faces

ti = 0, dx2 = 0 (crack faces: parallel to x-axis)

JAB = JCD = 0

JBC + JDA = 0 JBC = JAD

which path BC or AD should be used to compute J?

133


J-integral

Self-similar crack growth

crack grows, coord. axis move

d

da

=@

@a

+@

@x

@x

@a

@x

@a

= �1nonlinear elastic

⇧ =

Z

A0WdA�

Z

�tiuids

d⇧

da=

Z

A0

dW

dadA�

Z

�tidui

dads

d⇧

da

=

Z

A0

✓@W

@a

� @W

@x

◆dA�

Z

�ti

✓@ui

@a

� @ui

@x

◆ds

@W

@a=

@W

@✏ij

@✏ij@a

@✏ij

@a

=1

2

@

@a

✓@ui

@xj+

@uj

@xi

◆

d

da

=@

@a

� @

@x

@W

@✏ij= �ij

134


J-integral@W

@a

= �ij1

2

@

@a

✓@ui

@xj+

@uj

@xi

◆

@W

@a

= �ij@

@a

@ui

@xj= �ij

@

@xj

@ui

@a

Z

A0

@W

@adA =

Z

�ti@ui

@ads

Z

A0�ij

@

@xj

@ui

@a

dA =

Z

��ijnj

@ui

@a

ds

d⇧

da

= �Z

A0

@W

@x

dA+

Z

�ti@ui

@x

ds

�d⇧

da

=

Z

�

✓Wdy � ti

@ui

@x

ds

◆J

A : B = 0

symmetric skew-symmetric

nx

ds = dy

Gauss theorem

135


J-integral

nx

ds = dyGauss theorem,

@W

@a

= �ij1

2

@

@a

✓@ui

@xj+

@uj

@xi

◆

@W

@a

= �ij@

@a

@ui

@xj= �ij

@

@xj

@ui

@a

Z

A0

@W

@adA =

Z

�ti@ui

@ads

Z

A0�ij

@

@xj

@ui

@a

dA =

Z

��ijnj

@ui

@a

ds

d⇧

da

= �Z

A0

@W

@x

dA+

Z

�ti@ui

@x

ds

�d⇧

da

=

Z

�

✓Wdy � ti

@ui

@x

ds

◆J

A : B = 0


nx

ds = dy

Gauss theorem

135


J-integral

J-integral is equivalent to the energy release rate for a nonlinear elastic material under quasi-static condition.

nx

ds = dyGauss theorem,

@W

@a

= �ij1

2

@

@a

✓@ui

@xj+

@uj

@xi

◆

@W

@a

= �ij@

@a

@ui

@xj= �ij

@

@xj

@ui

@a

Z

A0

@W

@adA =

Z

�ti@ui

@ads

Z

A0�ij

@

@xj

@ui

@a

dA =

Z

��ijnj

@ui

@a

ds

d⇧

da

= �Z

A0

@W

@x

dA+

Z

�ti@ui

@x

ds

�d⇧

da

=

Z

�

✓Wdy � ti

@ui

@x

ds

◆J

A : B = 0


nx

ds = dy

Gauss theorem

135


J-K relationship

G =K2

I

E0 +K2

II

E0 +K2

III

2µ

�d⇧

da

=

Z

�

✓Wdy � ti

@ui

@x

ds

◆(previous slide)

J =K2

I

E0 +K2

II

E0 +K2

III

2µ

J-integral: very useful in numerical computation of SIFs

136


v =+ 1

4µ�

pa

2 � x

2 see slide 59

Crack Tip Opening Displacement

COD is zero at the crack tips.137Sunday, September 30,

Crack Tip Opening Displacement

(Irwin’s plastic correction, plane stress)

CTOD

Wells 1961

see slide 43

uy =+ 1

2µKI

rry2⇡

ry =1

2⇡

✓KI

�ys

◆2 � = 2uy =4

⇡

K2I

�ysE

=3� ⌫

1 + ⌫

2µ =E

1 + ⌫

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

138


CTOD-G-K relation

GI =⇡

4�ys�

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.

� = �cFracture occursThe degree of crack blunting increases in proportion to the toughness of the material

Wells observed:

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

� =4

⇡

K2I

�ysE

GI =K2

I

E

139


CTOD in design

� = 2uy =4

⇡

K2I

�ysEhas no practical application

140


CTOD experimental determination

Plastic hinge

rigid

similarity of triangles

rotational factor [-], between 0 and 1r141


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).

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

• 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.

blunting

resharpening

146


http://en.wikipedia.org/wiki/ASTM_International


http://en.wikipedia.org/wiki/Structural_failure


Cyclic loadings�max

= ��min

�� = �max

� �min

�a = 0.5(�max

� �min

)

�m = 0.5(�max

+ �min

)

R =�min

�max

load ratio 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


1. Initially, crack growth rate is small2. Crack growth rate increases rapidly when a is large3. Crack growth rate increases as the applied stress increases

149


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

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

• 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.

S-N curve✴Stress->Nf✴Nf->allowable S

scatter!!!

endurance limit (g.han keo dai)150






Constant variable cyclic load

�K = Kmax

�Kmin

crack grow per cycle

R = Kmin

/Kmax

�K = Kmax

�Kmin

da

dN= f1(�K,R)

�K = Kmax

�Kmin

= Kmax

(1�R)151


Paris’ law (fatigue)

da

dN= C(�K)m, �K = K

max

�Kmin

Paris’ law

Fatigue crack growth behavior in metals

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

Paris’ law is the most popular fatigue crack growth model

N: number of load cycles

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

I

II

�K �Kth : no crack growth

da

dN= C(�K)m, �K = K

max

�Kmin

2 m 7

10�8 mm/cycle(dormant period) 152


Paris’ law

C

da

dN= C(�K)m, �K = K

max

�Kmin

C,mare material properties that must be determined experimentally.

not depends on load ratio R

153


Other fatigue modelsForman’s model (stage III)

da

dN= C(�K)m

Paris’ model

R = Kmin

/Kmax

Kmax

�Kmin

Kmax

Kc � (Kmax

�Kmin

) Kmax

= Kc :da

dN= 1

As R increases, the crack growth rate increases.

154


Fatigue life calculation• Given: Griffith crack,

• Question: compute K = �p⇡a

m = 4

dN =da

C(�K)m=

da

C(��p⇡a)m

N = N0 +

Z af

a0

da

C(��p⇡a)m

N = N0 +1

C(��)4⇡2

Z af

a0

da

a2= N0 +

1

C(��)4⇡2

✓1

a0� 1

af

◆

2a0,��, C,m,KIc, N0

Nf

155


Fatigue life calculation• Given: Griffith crack,

• Question: compute K = �p⇡a

m = 4

dN =da

C(�K)m=

da

C(��p⇡a)m

N = N0 +

Z af

a0

da

C(��p⇡a)m

N = N0 +1

C(��)4⇡2

Z af

a0

da

a2= N0 +

1

C(��)4⇡2

✓1

a0� 1

af

◆

Kmax

= �max

p⇡af = KIc

2a0,��, C,m,KIc, N0

Nf

155


N = N0 +

Z af

a0

da

C(��f(a/W )p⇡a)m

Numerical integration of fatigue law

tedious to compute

156


Importance of initial crack length

157


Miner’s rule for variable load amplitudes

��1

��2

N1

N1f

nX

i=1

Ni

Nif= 1

a1

Ni

Nif

number of cycles a0 to ai

number of cycles a0 to ac��i

1945

Shortcomings:1. sequence effect not considered2. damage accumulation is independent of stress level

Nᵢ/Nif : damage

158


Variable amplitude cyclic loadings

history variables

✏⇤ three stress values

plasticity: history dependent

plastic wake

da

dN= f2(�K,R,H)

159


Overload and crack retardationIt 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


Crack retardation

a large plastic zone at overload has left behind

residual compressive plastic zone

close the crack->crack retards162


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

(Paris)

inspection time

K(a,�) = Kc ! ac� > at ⇤ s

NDT ! ao

t : ao

! at

163


Damage tolerance design

1. Determine the size of initial defects , NDI

2. Calculate the critical crack size at which failure would occur

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

4. Set inspection intervals

�p⇡ac = KIc

ac

a0

N = N0 +

Z ac

a0

da

C(��p⇡a)m

(stress concentration: possible crack sites)


Examples for Fatigue

log

da

dN= logC +m log�K

�K = ��p⇡a

5.6 MPapm

17.72 MPapm

da

dN=

af � a0N

[Gdoutos]

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

log(x

p) = p log(x)

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

dN= 0.42⇥ 10�11(�K)3

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

(m/cycle)


168

200p⇡af = 60 af = 28.65 mm

da

dN= 0.42⇥ 10�11(��

p⇡a)3


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

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.

da

dN= 4⇥ 10�9(�K)3.5

Given


Summary

170

• 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?

f(ac/W )�p⇡ac = Kc

�res =Kc

f(a/W )p⇡a

N = N0 +

Z ac

a0

da

C(��p⇡a)m


Mixed-mode crack growth

Cracks will generally propagate along a curved surface as the crack seeks out its path of least resistance.

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

Only a 2D mixed-mode loading (mode-I and mode-II) is discussed.

171


Maximum circumferential stress criterion

Erdogan and Sih

(from M. Jirasek) principal stress ⌧r✓ = 0172



✓c = 2arctan1

4

⇣KI/KII ±

p(KI/KII)2 + 8

⌘KI

✓sin

✓

2

+ sin

3✓

2

◆+KII

✓cos

✓

2

+ 3 cos

3✓

2

◆= 0⌧r✓ = 0

173



Fracture criterion Keq � KIc


Experiment

XFEM

✓c = 2arctan1

4

⇣KI/KII ±

p(KI/KII)2 + 8

⌘

175


Ductile to Brittle transitionFractures occurred in “well-designed” steel structures in severe weather.

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.

Titanic in the icy water of Atlantic

176


Stress corrosion cracking• Metals are subject to corrosion

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

• Fracture type: brittle!!!

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

corrosive environments

177


Alternatives to LEFM• Bodies with at least one existing crack

• Nonlinear zone ahead of the crack tip is negligible

Alternatives:

• Continuum damage mechanics

• Cohesive zone models

• Peridynamics

• Lattice models

crack initiation/formation

crack growth

discussed

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


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

• This prevented the efficient use of concrete

• Tensile behavior plays a key role in understanding fracture of concrete

�L

� � ✏m

quasi-brittle

181

✏m =�L

L= ✏0 +

w

L

� ��L


Fictitious crack model

concrete=quasi-brittle material

Fracture Process Zone (FPZ)

182


Hilleborg’s fictitious crack model1976

Cohesive crack/zone model

Similar to the strip yield model of Dugdale-Barenblatt183


Cohesive crack model

G =

Z�([[u]])d[[u]]

Fracture criterion

1

2

when

where (direction)

Rankine criterion

�max

� ft


Cohesive crack model

separation

Constitutive equations

deformation

Governing equations (strong form)

185


Cohesive crack modelWeak form

where

186



new term

where

186



new term

Implementation:(1) XFEM

(2) Interface elements

(to be discussed later)

where

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 size effect is defined by comparing the nominal strength (nominal stress at failure) of geometrically similar structures of different sizes.

• Classical theories (elastic analysis with allowable stress): cannot take size effect into account

�N

pa187


• 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.

geometrically similar structures of different sizes

�N =cNP

bD 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


Bazant’s size effect law

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.

�N = ft = ft(D)0

�N =KIcp⇡a

=KIcp⇡cND

191


Bazant’s size effect law

192


Continuum damage mechanics

nominal stress

effective stress

�A = �AEquilibrium:

damage variable

Hook’s law:

Milan Jirasek

� =A

A� = (1� !)�, ! = 1� A

A

� = E"

� = (1� !)E"

�

A

A

�

193


Four point bending test

194


Numerical solution with CDM

Experiment

Single Edged Notch Beam (SEN beam)

195


Computational fracture mechanics

196


Numerical methods to solve PDEs

• FEM (Finite Element Method)

• BEM (Boundary Element Method)

• MMs (Meshless/Meshfree methods)

FEM

BEM

MMs

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


FEM for elastic cracks(1) double nodes

(2) singularelements(3) remeshing

• Developed in 1976 (Barsoum)

• double nodes: crack edge

• singular elements: crack tip

• remeshing as crack grows

1pr behavior

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


Bouchard et al. CMAME 2003

However ...

Show crack growth movies201


202


Extended Finite Element Method (XFEM)

standard part enrichment part

Partition of Unity (PUM) enrichment function

known characteristics of the problem (crack tip singularity, displacement jump etc.) into the approximate space.

Belytschko et al 1999 Sc set of enriched nodes

X

J

NJ (x) = 1X

J

NJ (x)�(x) = �(x)

�(x)

u

h(x) =X

I2SNI(x)uI +

X

J2Sc

NJ (x)�(x)aJ

203


XFEM: enriched nodes

nodal support

I

enriched nodes = nodes whose support is cut by the item to be enrichedenriched node I: standard degrees of freedoms (dofs) and additional dofs

uI

aI

NI(x) 6= 0

X

J

NJ (x)�(x) = �(x)

204


XFEM for LEFM

u =

KI

2µ

rr

2⇡cos

✓

2

✓� 1 + 2 sin

2 ✓

2

◆

v =

KI

2µ

rr

2⇡sin

✓

2

✓+ 1� 2 cos

2 ✓

2

◆

crack tip with known displacement

crack edge displacement: discontinuous across crack edge

x

+

x

�

�1 = f(pr, ✓)

�2 : �2(x+) 6= �2(x

�)205


H(x) =

⇢+1 if (x� x

⇤) · n � 0

�1 otherwise

u =

KI

2µ

rr

2⇡cos

✓

2

✓� 1 + 2 sin

2 ✓

2

◆

v =

KI

2µ

rr

2⇡sin

✓

2

✓+ 1� 2 cos

2 ✓

2

◆

XFEM for LEFM (cont.)

Sc

St

blue nodes

red nodes

Crack tip enrichment functions:

[B↵] =

pr sin

✓

2

,pr cos

✓

2

,pr sin

✓

2

sin ✓,pr cos

✓

2

sin ✓

�

Crack edge enrichment functions:

+X

K2St

NK(x)

4X

↵=1

B↵b↵K

!

u

h(x) =X

I2SNI(x)uI

+X

J2Sc

NJ(x)H(x)aJ

206


Domain form of J-integral

J-integral is a contour integral that is not well suitable to FE computations.

FE mesh

J =

Z

A⇤

�ij

@uj

@x1�W �1i

�@q

@xidA

207


XFEM for cohesive cracks

Sc

H(x) =

⇢+1 if (x� x

⇤) · n � 0

�1 otherwise

u

h(x) =X

I2SNI(x)uI +

X

J2Sc

NJ (x)H(x)aJ

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

Wells, Sluys, 2001

not enriched to ensure zero crack tip opening!!!

208


XFEM: SIFs computationMesh

Results

One single mesh for all angles!!!

Matlab code: free

[VP Nguyen Msc. thesis]

209


XFEM: examples

CENAERO, M. Duflot

Northwestern Univ.

210


XFEM-Crack propagationSamtech, Belgium

fracture of underwater gas-filled pipeline211


Meshfree methods

Elastic-plastic fracture Shaofan Li 2012

Bordas et al.

212


Interface elements and cohesive crack model

fracture of polycrystalline material

delamination of composites213


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


Crack speed of

215


Interfacial fracture mechanics

• Thin films

•

216


Fracture of composite materials

217


Fiber reinforced composites

218


Education

Fracture mechanics