Elastic-Plastic Fracture Mechanics

Introduction

• When does one need to use LEFM and EPFM?

• What is the concept of small-scale and large-scale yielding?

Contents of this Chapter

• The basics of the two criteria used in EPFM: COD (CTOD), and J-Integral (with H-R-R)

• Concept of K- and J-dominated regions, plastic zones

• Measurement methods of COD and J-integral

• Effect of Geometry

Background Knowledge

• Theory of Plasticity (Yield criteria, Hardening rules)

• Concept of K, G and K-dominated regions

• Plastic zone size due to Irwin and Dugdal

LEFM and EPFM

LEFM

• In LEFM, the crack tip stress and displacement field can be uniquely characterized by K, the

stress intensity factor. It is neither the magnitude of stress or strain, but a unique parameter that

describes the effect of loading at the crack tip region and the resistance of the material. K filed is

valid for a small region around the crack tip. It depends on both the values of stress and crack size.

We noted that when a far field stress acts on an edge crack of width “a” then

for mode I, plane strain case

σ

σ

τπ

θ

θ θ

θ θ

θ θ

x x

y y

x y

IK

r

RS|

T|

UV|

W|=

−

+

L

N

MMMMMMM

O

Q

PPPPPPP

2 2

12

3

2

12

3

2

2

3

2

c o s

s i n ( ) s i n ( )

s i n ( ) s i n ( )

s i n ( ) s i n ( )

σ σ ν σ σzz zz xx yy= = +0 for plane stress; for plane strain( )

u

u

K rk

k

x

y

IRSTUVW

=− +

+ −

L

N

MMMM

O

Q

PPPP

2

21 2

2

21 2

2

2

2µ π

θ θ

θ θ2

cos ( sin ( ))

sin ( cos ( ))

LEFM concepts are valid if the plastic zone is much smaller than the singularity zones.

Irwin estimates

Dugdale strip yield model:

rK

p

I

ys

=1

2

2

π σ( )

rK

pI

ys

=1

8

2( )σ

ASTM: a,B, W-a 2.5 , i.e. of specimen dimension.≥ ( )KI

ysσ

2 rp≤

1

50

LEFM cont.

Singularity dominated region

σ

σ

τπ

x x

y y

x y

IK

r

RS|

T|

UV|

W|

=

L

NMMM

O

QPPP2

1

1

0

For =0 θ

For =2

all ijθθ

σ, = 0

EPFM

• In EPFM, the crack tip undergoes significant plasticity as seen in the following diagram.

sharp tip

I d e a l e la s ti c b r i t t l e b e h a v i o r

c l e a v a g e f r a c tu r e

P : A p p l ie d l o a d

P : Y i e ld l o a dy D is p la c e m e n t , u

Load

rati

o,

P/P

y

1 .0

F r ac tu re

B lu n t t i p

L im ite d p la st ic it y a t c r ac k

t ip , s t ill c l e a v ag e f ra c tu r e

D isp la cem en t, u

Load

rati

o,

P/P

y

1 .0

F r ac tu re

Blunt tip

Void formation & coalescence

failure due to fibrous tearingDisp lacement, u

Load

rati

o,

P/P

y

1 .0Fracture

l a r g e s c a l e

b l u n t i n g

L a r g e s c a l e p l a s t i c i t y

f i b r o u s r a p t u r e / d u c t i l e

f a i l u r e D is p la c e m e n t , u

Load

rati

o,P/P

y

1 .0 F rac tu re

EPFM cont.

• EPFM applies to elastoc-rate-independent materials, generally in the large-scale plastic

deformation.

• Two parameters are generally used:

(a) Crack opening displacement (COD) or crack tip opening displacement (CTOD).

(b) J-integral.

• Both these parameters give geometry independent measure of fracture toughness.

δSharp crack

Blunting crack

y

x

Γds

EPFM cont.

• Wells discovered that Kic measurements in structural steels required very large thicknesses for

LEFM condition.

--- Crack face moved away prior to fracture.

--- Plastic deformation blunted the sharp crack.

δSharp crack

Blunting crack

• Irwin showed that crack tip plasticity makes the crack behave as if it were longer, say from size a to a + rp

-----plane stress

From Table 2.2,

Set ,

rK

pI

y s

=1

2

2

π σ( )

uK r

kyI= + −

2 2 21 2

2

2

µ πθ θ

s in ( ) [ c o s ( ) ]

θ π= uk

Kr

y I

y=

+ 1

2 2µ πa ry+

θ π=

δπ σ

= =24 2

2u

K

Ey

I

y s

Note:

since

k E=−+

= +3

12 1

νν

µ ν and ( )

δπ σ

= =CTOD4 G

ys

GK

E

I=2

CTOD and strain-energy release rate

• Equation relates CTOD ( ) to G for small-scale yielding. Wells proved that

Can valid even for large scale yielding, and is later shown to be related to J.

• can also be analyzed using Dugdales strip yield model. If “ ” is the opening at the end of the strip.

δπ σ

= =CTOD4 G

ys

δ δ

δ δ

δσ

ys

Consider an infinite plate with a image crack subject to a

Expanding in an infinite series,

σ σ∞ =

δσ

ππ σ

σ= = ⋅2

8u

a

Ey

y s

y s

li n s e c (2

)

δσ

ππ σ

σπ σ

σ= ⋅ + ⋅ +

8 1

12

2 4ys

ys ys

a

E[1

2(

2(

2) ) . . . ]

If , and can be given as:

In general,

δσ

π σσ

= + ⋅K

E

I

y s y s

2

2[ 11

6(

2) ]

σσ

σ σ δσ σys

ys

ysE

G≈ << =0 ( th e n =

K I

2

y s

) , δ

δσ

=G

m ys

, m =1.0 for plane stress; m = 2.0 for plane strain

Alternative definition of CTOD

δSharp crack

Blunting crack

δBlunting crack

Displacement at the original crack tip Displacement at 900 line intersection, suggested by Rice

CTOD measurement using three-point bend specimen

W

P

a

r p(W-a)

z

Vp

δp

'

'

'δ p l

p p

p

r W a V

r W a a z=

−

− + +

( )

( )

displacement

expandingδ

Elastic-plastic analysis of three-point bend specimen

δ δ δσ

= + = +−

− + +el p lI

ys

p p

p

K

m E

r W a V

r W a a z

2 ( )

( )

Where is rotational factor, which equates 0.44 for SENT specimen.δpl

• Specified by ASTM E1290-89--- can be done by both compact tension, and SENT specimen

• Cross section can be rectangular or W=2B; square W=B

KI is given by

δν

σe lI

y s

K

E=

−2 21

2

( )

KP

B Wf

a

WI= ⋅ ( )

δ p l

p p

p

r W a V

r W a a z=

−

− + +

( )

( )

load

Mouthopening

υ p υ e

V,P

CTOD analysis using ASTM standards

Figure (a). Fracture mechanism is purely cleavage, and critical CTOD <0.2mm, stable crack growth,

(lower transition).

Figure (b). --- CTOD corresponding to initiation of stable crack growth.

--- Stable crack growth prior to fracture.(upper transition of fracture steels).

Figure (c) and then ---CTOD at the maximum load plateau (case of raising R-curve).

δ c

δ i

δ i δ m

δ u

loa

d

Mouth opening

Pc

fracture

(a) (b) (c)

PiPu Pm

f racture

Pi

More on CTOD

The derivative is based on Dugdale’s strip yield model. For

Strain hardening materials, based on HRR singular field.

By setting =0 and n the strain hardening index based on

*Definition of COD is arbitrary since

A function as the tip is approached*Based on another definition, COD is the distance between upper

and lower crack faces between two 45o lines from the tip. With this

Definition

2

or ICOD

y y

K J

Eδ

σ σ= =

( )11

1 ,

n

n

ni y i

y y n

Ju r u n

Iαε θ

ασ ε

++

=

1

3

2

n

y ije

y y y

ε σσαε σ σ

−

= ( ) ( ), 0 , 0y yu x u xδ + −= −

( )1

1nx +−

COD n

y

Jdδ

σ=

Where ranging from 0.3 to 0.8 as n is varied from

3 to 13 (Shih, 1981)

*Condition of quasi-static fracture can be stated as the

Reaches a critical value . The major advantage is that this

provides the missing length scale in relating microscopic failure

processes to macroscopic fracture toughness.

*In fatigue loading, continues to vary with load and is a

function of:

(a) Load variation

(b) Roughness of fracture surface (mechanisms related)

(c) Corrosion

(d) Failure of nearby zones altering the local stiffness response

( ), ,n n yd d nα ε=

tipδ

CODδ

2

2I

y

kδ

σ ε∆

∆ =

3.2 J-contour Integral

• By idealizing elastic-plastic deformation as non-linear elastic, Rice proposed J-integral, for regions

beyond LEFM.

• In loading path elastic-plastic can be modeled as non-linear elastic but not in unloading part.

• Also J-integral uses deformation plasticity. It states that the stress state can be determined knowing

the initial and final configuration. The plastic strain is loading-path independent. True in proportional

load, i.e.

• under the above conditions, J-integral characterizes the crack tip stress and crack tip strain and

energy release rate uniquely.

• J-integral is numerically equivalent to G for linear elastic material. It is a path-independent integral.

• When the above conditions are not satisfied, J becomes path dependent and does not relates to any

physical quantities

d d d d d dk

σσ

σσ

σσ

σσ

σσ

σσ

1

1

2

2

3

3

4

4

5

5

6

6

= = = = = =

3.2 J-contour Integral, cont.

y

x

Γds

Consider an arbitrary path ( ) around the crack tip. J-integral is defined as Γ

J wdy Tu

xds w di

i

i

ij i j

ij

= −∂

∂=zz( ), σ ε

ε

0Γ

It can be shown that J is path independent and represents energy release rate for a material where

is a monotonically increasing with

σ ij

ε ij

Proof: Consider a closed contour:

Using divergence theorem:

J w dy Tu

xdsi

i

i

* ( )*

= −∂∂z

Γ

Jw

x x

u

xd xd y

i

i ji

A

*( )

*

=∂∂

−∂

∂

∂

∂z σ

where w is strain energy density, Ti is component of traction vector normal to contour.

A*

Γ*

*

*( )i

ij

jA

uwJ dxdy

x x xσ

∂∂ ∂= −

∂ ∂ ∂ ∫

Evaluate ∂

∂=

∂

∂⋅∂

∂=

∂

∂

w

x

w

x xij

ij

i j

ij

ε

εσ

ε

Note is only valid if such a potential function exists

Again,

σεij

ij

w=

∂∂

∂∂

=∂∂

+∂∂

=∂

∂∂∂

+∂

∂

∂

∂

w

x xu

xu

x

u

x x

u

x

i j i j i j

i j

j

i

i

j

1

2

1

2

σ

σ

[ ( ) ( ) ]

[ ( ) ( ) ]

, ,

Since σ σ

σ

ij ji

i j

j

i

x

u

x

=

=∂

∂∂∂

( )

Recall ∂

∂=

∂∂

∂∂

=∂

∂∂∂

σ

σ σ

i j

j

j

i

j

i ji

x

x

u

x x

u

x

0 ( e q u i l i b r i u m ) l e a d s t o

i j ( ) ( )

Evaluation of J Integral ---1

w

(equilibrium) leads to

,j i

Hence, Thus for any closed contour J*

.= 0 J*

.= 0

Now consider

Γ1

Γ2

Γ3

Γ4

1 2 3 4

0J J J J J= + + + =

Recall J w d yw

xd si

* ( )= −∂∂z τ

Γ

On crack face, (no traction and y-displacement), thus

, leaving behind

Thus any counter-clockwise path around the crack tip will yield

J; J is path independent.

τi

dy= =0 0,

J J3 4 0= = J J1 2= −

Evaluation of J Integral ---2

1 2 3 4

t i

t i

Γ '

a

y

x

2D body bounded by Γ '

In the absence of body force, potential energy Π

ΠΓ

= −z zw d A u d s

A

i i

' ' '

τ

Suppose the crack has a vertical extension, then

d

da

dw

dadA

du

dads

A

i

iΠ

Γ

= −z z' '

τ (1)

Note the integration is now over Γ '

Evaluation of J Integral ---3

t i

t i

Noting that d

da a

x

a x a x

x

a=

∂∂

+∂∂

∂∂

=∂∂

−∂∂

∂∂

= − since 1

d

da

w

a

w

xdA

w

a

u

xds

A

iiΠ

Γ

=∂∂

−∂∂

−∂∂

−∂∂z z( ) ( )

' '

τ (2)

∂∂

=∂∂

∂

∂=

∂∂

∂

∂w

a

w

a x

u

aij

iij

j

i

εε

σ ( )

Using principle of virtual work, for equilibrium, then from

eq.(1), we have

d

da

Π= 0

σ τij

jA

ii

i

x

u

adA

u

ads

∂

∂

∂

∂=

∂

∂z z( )' 'Γ

Thus, d

da

du

dxds

dw

dxdAi

i

A

Π

Γ

= −z zτ' '

Using divergence theorem and multiplying by -1

− = − = −∂∂z zd

dawn

du

dxds wdy

w

xdsx i

ii

Π

Γ Γ

( )' '

τ τ

Evaluation of J Integral ---4

j

t it i

t i

t i

t i

⋅⋅

Therefore, J is energy release rate , for linear or non-linear

elastic material

d

da

Π

In general

Potential energy; U=strain energy stored; F=work done by

external force and A is the crack area.

Π = − = −∂Π∂

U F JA

and

Π =

a

up

Evaluation of J Integral ---5

-dP

*dU dU=− d∆

Displacement

*U P UΠ = − ∆ = = Complementary strain energy = dP∆∫0

p

Load

For Load Control

For Displacement Control

The Difference in the two cases is and hence J for

both load Displacement controls are same

*

p

dUJ

da=

dUJ

da ∆

= −

1. .

2dp d dU∆ �

0 0

0

. .

.

pD

p p

J dp dpa a

or

pJ pd d

a a

∆

∆ ∆

∂ ∂∆= ∆ =

∂ ∂

∂ ∂= − ∆ = − ∆

∂ ∂

∫ ∫

∫ ∫J=G and is more general description of energy release rate

2

'

IKJ

E=

Evaluation of J-Integral

More on J Dominance

J integral provides a unique measure of the strength of the singular

fields in nonlinear fracture. However there are a few important

Limitations, (Hutchinson, 1993)

(1) Deformation theory of plasticity should be valid with small strain

behavior with monotonic loading

(2) If finite strain effects dominate and microscopic failures occur, then

this region should be much smaller compared to J dominated region

Again based on the HRR singularity

( )

1

1

,n

Iijij y

y y n

Jn

I rσ σ σ θ

ασ σ

+ =

Based on the condition (2), we would

like to evaluate the inner radius ro of J

dominance. Let R be the radius

where the J solutions are satisfied

within 10% of complete solution.

FEM shows that R

or

3o CODr δ�

•However we need ro should be greater than the forces zone

(e.g. grain size in intergranular fracture, mean spacing of voids)

•Numerical simulations show that HRR singular solutions hold

good for about 20-25% of plastic zone in mode I under SSY

• Hence we need a large crack size (a/w >0.5) . Then finite strain

region is , minimum ligament size for valis JIC is

• For J Controlled growth elastic unloading/non proportional loading

should be well within the region of J dominance

• Note that near tip strain distribution for a growing crack has a

logarithmic singularity which is weaker then 1/r singularity for a

stationary crack

3 CODδ25 IC

y

Jb

σ=

and a RdJ J

da R≥ ∆ �

Williams solution to fracture problem

Williams in 1957 proposed Airy’s stress function

As a solution to the biharmonic equation

For the crack problem the boundary conditions are

Note will have singularity at the crack tip but is single valued

Note that both p and q satisfy Laplace equations such that

( ) ( )R rψ θ θ=

2 24 2

2 2 2

1 10 where

r r r rψ

θ∂ ∂ ∂

∇ = ∇ = + +∂ ∂ ∂

0 for rθθ θσ σ θ π= = = ±

ψ

( ) ( )2, ,r p r q rψ θ θ= +

2 20p q∇ = ∇ =

Now, for the present problem.

( ) ( )

( )[ ]( )[ ]

( )[ ]

( )( ) ( )[ ]

1

2 2

1

2

1 1

2

2 2

2

1 1

2

1 12

cos sin

cos 2 sin 2

Then

cos cos 2

sin sin 2

Consider only mode I solution with

cos cos 2

1 2 cos cos 2

z

z

r

p A r A r

q B r A r

r A B

r A B

r A B

r A Br

r

λ λ

λ λ

λ

λ

λ

λθθ

θ

λθ λθ

λ θ λ θ

ψ λθ λ θ

λθ λ θ

ψ λθ λ θ

ψσ λ λ λθ λ θ

σ

+ +

+

+

+

= +

= + + +

= + +

+ + +

= + +

∂= = + + + +

∂∂

= −∂

( ) ( ) ( )[ ]1 1

1

1 sin 2 sin 2

r

r A Bλ

ψθ

λ λ λθ λ λ θ

∂ ∂

= + + + +

Williams Singularity…3

Applying boundary conditions,

( )( )

1 1

1 1

A +B cos 0

2 sin 0A B

λπ

λ λ λπ

=

+ + =

Case (i) cos 0λπ = 2 1, Z=0,1,2...

2

Zλ

+=

1 1B2

Aλ

λ= −

+or,

sin 0λπ = Zλ =1 1B A= −

Case (ii)

Since the problem is linear, any linear combination of the above two will also be

acceptable.

Thus

Though all values are mathematically fine, from the physics point of view, since 2 with Z= ... 3, 2, 1,0,1, 2,3...Zλ = − − − −

and ij ijr rλ λσ ε∝ ∝

Williams Singularity…4

( )

0

0

212

212

0

2A 2 1

0

=

ij ij

R

ij ijr

R

r

U r

rdrd

r drd

λ

π

λ

σ ε

σ ε θ

θ+

= ∝

∫ ∫

∫ ∫�

Since U should be provided for any annular rising behavior and R ,0r

0 ˆ as 0, 1 ( 1 makes 0)ijU r λ σ< ∞ → > − =− =

1

1

31 1 Z2 2 2 2

112 3

, needs > 1. Thus

=- ,0, ,1, ,2... with = Z=-1,0,

positive number.

The most dominant singular form

=- and B

ri

A

Also u r

where

λ

λ λ

λ

+∝ −

=

( )( )

( ) ( ) ( )( ) ( )

3

2

52

1 12 2

311 2 3 2

2

01

Now cos cos

+ ...

...

and

where indicates the order of

Iij ij ij ij

r A

r

r

Ar r r

θ θ

σ σ θ−

Φ = +

Φ

+ Φ

= + Φ +Φ

Φ

%

Williams Singularity…4

( )

( )

0

1

Note the second term in is a non-singular

and non-vanishing term. However, higher order vanish as r 0

with 2

(no sum on x)2

ij ij

I

IIij ij ix jx

r

KA

KT

r

σ

π

σ σ θ δ δπ

=Φ

→

=

= +%

( )( )( ) ( ) ( )

( ) ( )

32

52

1 12 2

311 2 3 2

2

01

Now

cos cos

...

...

and

where indicates the order of

I

ij ij ij ij

r A

r

r

Ar r r

θ θφ

φ

φ

σ σ θ φ φ

φ

−

= +

+

+

= + +%

Williams Singularity…5

( )

( )

0

1

Note the second term in is a non-singular

and non-vanishing term. However, higher order vanish as r 0

with 2

(no sum on x)2

ij ij

I

IIij ij ix jx

r

KA

KT

r

σ

π

σ σ θ δ δπ

=Φ

→

=

= +%

Williams Singularity…6

( ) ( )( ) ( )

For in-plane stress components,

0

0 02

I Ixx xy xx xyI

I Iyx yy yx yy

TK

r

σ σ σ θ σ θσ σ π σ θ σ θ

= +

% %

% %

I

Second-term is generally termed as "T-stress" or

"T-tensor" with

For brittle crack of length 2a in x-z plane

with & applied

K and

xx

yy xx

yy

T

a

σ

σ σ

σ π

∞ ∞

∞

=

=

T= yy xxσ σ∞ ∞−

x

y

2a

z

HRR Singularity…1

0 0 0

Hutchinson, Rice and Rosenbren have evaluated the character of crack tip

in power-law hardening materials.

Suppose the material is represented by Ramberg-Osgood model,

ε σ σ

αε σ σ

= +

0

00

Reference value of stress=yield strength

, strain at yieldE

dimensionless constant

strain-hardening exponent

n

n

σ

σε

α

−

−

−

−

1

Note if elastic strains are negligible, then

ˆ 3 3ˆ ;

2 2

n

y y

n

ij eq ij

eq ij

y ij y

ε σα

ε σ

ε σ σασ σ

ε σ σ

−

=

= =

HRR Singularity…2

( )

( ) ( )( )

40

1 2

0

Then

, , , ,

(similar to Williams expression)

s t

s

f r n

C r r

k r

φ φ σ α

φ θ ρ θ

φ σ φ θ

∇ +

= +

= ⋅ ⋅ %

( )

( )

1

1

0 2

0

10

20

Applying the appropriate boundary conditions

,

,

Integration const

n

ij ij

n

n

n

ij ij

n

n

EJn

I r

EJn

E I r

I

σ σ σ θασ

σε ε θ

ασ

+

+

=

⋅

=

⋅ −

%

%

ant

, Dimensionless functions of n and σ ε θ−%%