1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

59
1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J

Transcript of 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

Page 1: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

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plastic zone

CH-6 Elastic-Plastic Fracture Mechanics

Laurent HUMBERT

Thursday, May 27th 2010

J

Page 2: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

Non linear elastic phenomenon

Introduction

Applies when nonlinear deformation is confined to a small region surrounding the crack tip

LEFM: Linear elastic fracture mechanics

Elastic-Plastic fracture mechanics (EPFM) :

Effects of the plastic zone negligible, asymptotic mechanical field controlled by K

Generalization to materials with a non-negligible plastic zone size: elastic-plastic materials

Full yielding Diffuse dissipation

L D a

Elastic Fracture

D

L

a

, ,L a D B

B

LEFM, KIC or GIC fracture criterion

Contained yielding

L D a

EPFM, JC fracture criterion

Catastrophic failure, large deformations

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Surface of the specimen Midsection Halfway between surface/midsection

Plastic zones (light regions) in a steel cracked plate (B 5.0 mm):

Slip bands at 45°Plane stress dominant

Net section stress 0.9 yield stress in both cases.

Plastic zones (dark regions) in a steel cracked plate (B 5.9 mm):

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6.2 CTOD as yield criterion

6.1 Models for small scale yielding : - Estimation of the plastic zone size using the von Mises yield criterion - Irwin’s approach (plastic correction) - Dugdale’s model or the strip yield model

Outline

6.5 Applications of the J-integral

6.3 Elastic –plastic fracture

6.4 Elastoplastic asymptotic field (HRR theory)

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Von Mises equation:

6.1 Models for small scale yielding :

1 22 22

1 2 1 3 2 31

2e

se is the effective stress and si (i=1,2,3) are the principal normal stresses.

Recall the mode I asymptotic stresses in Cartesian components, i.e.

3cos 1 sin sin ...

2 2 22

3cos 1 sin sin ...

2 2 22

3cos sin cos ...

2 2 22

Ixx

Iyy

Ixy

K

r

K

r

K

r

yy

xy

xx

x

y

r

LEFM analysis prediction:

KI : mode I stress intensity factor (SIF)

Estimation of the plastic zone size

Page 6: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

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One has the relationships (1) ,

1 222

1,2 2 2xx yy xx yy

xy

Thus, for the (mode I) asymptotic stress field:

1,2 cos 1 sin2 22

IK

r

rrr

x

y

r

and in their polar form:Expressions are given in (4.36).

1 222

2 2rr rr

r

31 2

0 plane stress

plane strain

and

, ,rr r

3

0 plane stress

2cos plane strain

22IK

r

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Substituting into the expression of se for plane stress

1 22 2 212cos sin cos (1 sin ) cos (1 sin )

2 2 2 2 2 22 2I

eK

r

Similarly, for plane strain conditions

1 2

2 21 31 2 1 cos sin

22 2I

eK

r

1 22 2 21

6cos sin 2cos2 2 22 2

IK

r

1 221 3

1 cos sin22 2

IK

r

1 2

2 2cos 4 1 3cos2 22

IK

r

1 22 2 2 2 21

4cos sin 2cos 2cos sin2 2 2 2 22 2

IK

r

1 22cos 4 3cos

2 22IK

r

Page 8: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

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Yielding occurs when: e Y Y is the uniaxial yield strength

Using the previous expressions (2) of se and solving for r,

22 2

22 2 2

1cos 4 3cos

2 2 2

1cos 4 1 3cos

2 2 2

I

Yp

I

Y

K

rK

plane stress

plane strain

Plot of the crack-tip plastic zone shapes (mode I):

2

1

4

p

I

Y

r

K

Plane strain

Plane stress (n= 0.0)

increasing = n 0.1, 0.2, 0.3, 0.4, 0.5

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1) 1D approximation (3) L corresponding to 0pr Remarks:

221 2

2I

Y

KL

Thus, in plane strain:

2) Significant difference in the size and shape of mode I plastic zones

For a cracked specimen with finite thickness B, effects of the boundaries:

in plane stress:2

1

2I

Y

KL

- Essentially plane strain in the in the central region

Triaxial state of stress near the crack tip:

- Pure plane stress state only at the free surface

B

Evolution of the plastic zone shape through the thickness:

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4) Solutions for rp not strictly correct, because they are based on a purely elastic:

3) Similar approach to obtain mode II and III plastic zones:

Alternatively, Irwin plasticity correction using an effective crack length …

Stress equilibrium not respected

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The Irwin approach Mode I loading of a elastic-perfectly plastic material:

02

Iyy

K

r

Plane stress assumed (1)

crackx

r1

Y

r2

(1)

(2)

(2)

To equilibrate the two stresses distributions (cross-hatched region)

r2 ?

Elastic:

Plastic correction 2,yy Y r r

r1 : Intersection between the elastic distribution and the horizontal line yy Y

12I

YK

r

1

20 2

rI

Y YK

r dxx

yys

21 2

11222

I IY

Y

K Kr r

2 1r r

2

11

2I

Y

Kr

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Redistribution of stress due to plastic deformation:

Plastic zone length (plane stress):

Irwin’s model = simplified model for the extent of the plastic zone: - Focus only on the extent of the plastic zone along the crack axis, not on its shape

2

11

2 I

Y

Kr

- Equilibrium condition along the y-axis not respected

yy

real crack x

Y

fictitious crack

Stress Intensity Factor corresponding to the effective crack of length aeff =a+r1

1 1,I effK a r K a r

2

11

23

I

Y

Kr

2eff

yyK

X

X

In plane strain, increasing of sY. : Irwin suggested in place of sY 3 Y

(effective SIF)

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Application: Through-crack in an infinite plate

Effective crack length 2 (a+ry)

eff yK a r

21

2Y

effeff

KK a

Solving, closed-form solution:2

11

2Y

effa

K

(Irwin, plane stress)2

1

2I

yY

Kr

with

2a

ry ry

aeff

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More generally, an iterative process is used to obtain the effective SIF:

eff eff effK Y a a

Convergence after a few iterations…

Initial:

IK Y a a 2

01

2I

Y

Ka a

m

1 plane stress

3 plane strainm

Yes

No

1i i

Algorithm:

( )I

ieff i iK K Y a a

Application:

Through-crack in an infinite plate (plane stress):

s∞ = 2 MPa, sY = 50 MPa, a = 0.1 m KI = 1.1209982Keff= 1.1214469 4 iterations

Y: dimensionless function depending on the geometry.

Do i = 1, imax :

( )1 1I

ii iK Y a a

2( )1

2I

i

iY

Ka a

m

( ) ( 1)I I

i iK K ?

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Dugdale / Barenblatt yield strip model

2a cc

Long, slender plastic zone from both crack tips

Perfect plasticity (non-hardening material), plane stress

Assumptions:

Plastic zone extent

Elastic-plastic behavior modeled by superimposing two elastic solutions:

application of the principle of superposition (see chap 4)

Crack length: 2(a+c)

Very thin plates, with elastic- perfect plastic behavior

Remote tension + closure stresses at the crack

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2ac c

Principle:

Stresses should be finite in the yield zone:

• No stress singularity (i.e. terms in ) at the crack tip1 r

• Length c such that the SIF from the remote tension and closure stress cancel one another

SIF from the remote tension:

,IK a c a c

YY

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SIF from the closure stress Y

Closure force at a point x in strip-yield zone:

YQ dx a x a c

Total SIF at A:

IB YQ ( a c ) x

K ,a c( a c ) x( a c )

YY

a c

xAB

I A YQ ( a c ) x

K ,a c( a c ) x( a c )

Recall first,

crack tip A

crack tip B

a a c

Y YIA Y

( a c ) a

( a c ) x ( a c ) xK ,a c dx dx

( a c ) x ( a c ) x( a c ) ( a c )

By changing the variable x = -u, the first integral becomes,

1a

Y

( a c )

( a c ) u( ) du

( a c ) u( a c )

a cY

a

( a c ) udu

( a c ) u( a c )

Page 18: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

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a c

Y YIA Y

a

( a c ) x ( a c ) xK ,a c dx

( a c ) x ( a c ) x( a c ) ( a c )

a cY

a

( a c ) x ( a c ) xdx

( a c ) x ( a c ) x( a c )

2 22

a c

Ya

a c dx

( a c ) x

Thus, KIA can be expressed by

Same expression for KIB (at point B):

One denotes KI for KIA or KIB thereafter

(see eq. 6.15)

a a c

Y YIB Y

( a c ) a

( a c ) x ( a c ) xK ,a c dx dx

( a c ) x ( a c ) x( a c ) ( a c )

1a

Y

( a c )

( a c ) u( ) du

( a c ) u( a c )

a cY

a

( a c ) udu

( a c ) u( a c )

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Recall that,1

2 2

a c

a

dx acos

a c( a c ) x

12I Y Ya c a

K ,a c cosa c

The SIF of the remote stress must balance with the one due to the closure stress, i.e.

0I I YK ,a c K ,a c

12 cosYa c a

a ca c

Thus, cos

2 Y

a

a c

By Taylor series expansion of cosines,

2 4 61 1 1

1 ...2! 2 4! 2 6! 2Y Y Y

a

a c

Keeping only the first two terms, solving for c22 2

2 88I

YY

a Kc

1/ 0.318 and /8 = 0.392

Irwin and Dugdale approaches predict similar plastic zone sizes

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eff effK a

Estimation of the effective stress intensity factor with the strip yield model:

cos2eff Y

a

a

-By setting effa a c

andcos

2

eff

Y

aK

tends to overestimate Keff because the actual aeff less than a+c

- Burdekin and Stone derived a more realistic estimation:

Thus,

1/ 2

2

8lnsec

2eff YY

K a

sec2 Y

a

(see Anderson, third ed., p65)

Page 21: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

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6.2 CTOD as yield criterion

CTOD : crack tip opening displacement

Irwin plastic zone

Initial sharp crack has blunted prior to fracture

Non-negligible plastic deformation

blunted crack

LEFM inaccurate : material too tough !!!

sharp crack

Page 22: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

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Wells proposed d t (CTOD ) as a measure of fracture toughness

Estimation of d t using Irwin model :

crack length: a + ry

for plane stress conditions,

2 att y yu r rd where uy is the crack opening

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21 22 2 2 2

Iy

K ru sin - cos

Crack opening uy

12 2

IK r

uy

(see eq. 6.20)

and for plane stress, 3

1

2 1

E

One has

8

2yI

trK

E d

4

2I

yK r

uE

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From Irwin model, radius of plastic zone2

1

2I

yY

Kr

24 I

tY

K

Ed

CTOD related uniquely to KI and G.4

tY

Gd

and also,

CTOD appropriate crack-tip parameter when LEFM no longer valid

→ reflects the amount of plastic deformation

Replacing,

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J contour integral = more general criterion than K (valid for LEFM)

6.3 Elastic-plastic fracture

Derive a criterion for elastic-plastic materials, with typical stress-strain behavior:

A→B : linear

A

BC

B→C : non-linear curve

C→D : non-linear, same slope as A-B

D

non-reversibility: A-B-C ≠ C-D-A

Simplification: non-linear elastic behavior

reversibility: A-B-C = C-D-A

Material behavior is strain history dependent !Non unique solutions for stresses

elastic-plastic law replaced by the non-linear elastic law

A/D

BC

Consider only monotonic (non cyclic) loading

Page 26: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

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Definition of the J-integral

Rice defined a path-independent contour integral J for the analysis of cracked bodies

showed that its value = energy release rate in a nonlinear elastic material

J generalizes G to nonlinear materials :

→ nonlinear elastic energy release rate

As G can be used as a fracture criterion Jc

reduces to Gc in the case of linear fracture

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Fixed-grips conditions: Dead-load conditions

0

0U P( )d

Recall : Load-displacement diagram and potential energy

P 0

0

U

U*

U* : complementary energy

U : elastic strain energy

U ≠ U* (for nonlinear materials)

0

0

P*U ( P )dP

Page 28: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

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• Geometrical interpretation:

Definition of J (valid for nonlinear elastic materials):

dJ

dA

A = B a : for a cracked plate of thickness B

loading paths with crack lengths a and a+da

OA and OB :

, :P a

possible relationship between the load P and the displacement for a moving crack

Thus,

One has J dA dU dF

dU OBC OAD OAE EBCD

dF ABCD

J da

,P a

a daa

d

P

B

C

A

D

O

E

J dA OAE EAB OAB J obtained incrementally

Page 29: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

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• In particular ,

At constant force (dual form):

1 *

P

UJ

B a

At constant displacement:

1 UJ

B a

0

0

1 Pd

B a

0

0

1 P

PdP

B a

Generalization of eqs 3.38a and 3.43a to nonlinear elastic materials

Useful expressions for the experimental determination of J

dD

D D0D

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• Experimental determination of J:

Ex : multiple-specimen method - Begley and Landes (1972) :

(1) Consider identical specimens with different crack lengths ai

Procedure

(2) For each specimen, record the load-displacement P-u curve at fixed-grips conditions

(1) (2)

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(3) Calculate the potential energy for given values of displacement u

(4) Negative slopes of the – a curves :

→ critical value JIc of J at the onset of crack extension (material constant)

= area under the load-displacement curve here

Plots of J versus u for different crack lengths ai

(3)(4)

→ plots of – a curves

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J as a path-independent line integral :

ii

uJ w dy T ds

x

T : traction vector at a point M on the bounding surface G , i.e. i ij jT n

The contour G is followed in the counter-clockwise direction

n : unit outward normal

0

mn

mn ij ijw d

with strain energy density

w dy dsx

u

T

u : displacement vector at the same point MT

n

crack

M

x

y

Page 33: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

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→ Equivalence of the two definitions

• 2D solid of unit thickness and of area S

with a linear crack of length a along OX (fixed)

• Crack faces are traction-free

Imposed tractions on the part of the contour G t

Applied displacements on G u

• Total contour of the solid G 0 including the crack tip:

X x

Y

y

O

Sa

G u

u

G t

T

0 SG

Proof :

dJ

da

ii

uJ w dy T ds

x

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Recall for the potential energy (per unit thickness),

t

i iS

a wdS T u ds

Tractions and displacements imposed on G t and G u are (assumed to be) independent of a

0

ii

S

d ud d wdS T ds

d a d a d a

0

0

it

iu

dT, on

dadu

onda

G

G

ijij

w

i ij jT n

Strain energy work of the surface tractions

0 u t

u t

,G G GG G

path of integration extended to 0G

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Consider now the moving coordinate system x , y (attached to the crack tip),

x X a

d x

da a a x

a x

x X a

Thus,

0

i ii

S

u ud w wdS T ds

da a x a x

However,

ij

ij

w w

a a

ijij a

1

2ji

ijj i

uu

a x x

iij

j

u

a x

since ij ji

iij

j

u

x a

d

da: total derivative/crack length

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Using the divergence integral theorem,

iij

S S j

uwdS dS

a x a

Thus,

0

i iij ij j

S j

u udS n ds

x a a

0

ii

uT ds

a

The derivative of J reduces to,

0

i ii

S

u ud w wdS T ds

da a x a x

0 0

i i ii i

S

u u uwdS T ds T ds

x a a x

0

ii

S

uwdS T ds

x x

iij

S j

udS

x a

from equilibrium equation

Page 37: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

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Using Green Theorem, i.e. A

Q PP x,y dx Q x,y dy dx dy

x y

0

ii

S

ud wdS T ds

da x x

0

ii

uJ wdy T ds

x

J derives from a potential

End of the proof

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Properties of the J-integral

1) J is zero for any closed contour containing no crack tip.

x

y A

G

ii

uJ wdy T ds

xGG

ii

A

uwJ dxd y T ds

x x

Using the Green Theorem, i.e.

Closed contour around A

A

Q PP x,y dx Q x,y dy dx dy

x y

Consider

thus,

Again from the divergence theorem,

iij j

A

uwdxd y n ds

x x

iij j

un ds

x

i

ijA j

udxd y

x x

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However,

ij

ij

w w

x x

ijij x

1

2ji

ijj i

uu

x x x

iij

j

u

x x

since

iij

j

u

x x

ij ji

The integral becomes,

iij

A j

uwJ dxdy

x x xG

Replacing in the integral,

Invoking the equilibrium equation, 0ij

jx

0J

iji i iij ij

j j j

u u u

x x x x x x

i

ijj

u

x x

0

Measure of the energy dissipation due to the crack

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2) J is path-independent

x

y

G 3

Consider the closed contour21 3 4*GG G G G

2Γ*

G 4

and 0J 1 2 3 4

*J J J J JG G G G G One has

3G

The crack faces are traction free :

0i ij jT n on 4Gand

dy = 0 along these contours

3 40J JG G

Page 41: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

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x

y

G 3

1 3 2 4GG G G G2

Γ*

G 4

and

2 2*J JG G

Note that,

1 20J J JG G G

1 2J JG G

Any arbitrary (counterclockwise) path around a crack gives the same value of J

J is path-independent

G 2 followed in the counter-clockwise direction

Choose a convenient contour for calculation

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Evaluation of J for a circular path of radius r around the crack tip

crack

Gr

G is followed from = -q p to q = p

One has, ds rd dy r cos d

J integral becomes,

,, cos , i

iu r

J w r T r r dx

when r → 0 only the singular term remain and the integrand can be calculated

For LEFM , one obtains :2I'

KJ G

E (if mode I loading)

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6.4 HRR theory

0 0 0

n

Ramberg-Osgood equation

Power law relationship between plastic strain and stress.

derived first by Hutchinson, Rice and Rosengren assuming the constitutive law :

→ Near crack tip stress field for elastic-plastic fracture

0 0 E 0

= yield strength

For a linear elastic material n = 1.

: dimensionless constant

n : strain-hardening exponent

material properties

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1 1

1

n

ijJ

Ar

1

2

n n

ijJ

Ar

Asymptotic field given by Hutchinson Rice and Rosengren:

Ai are regular functions that depend on q and the power law parameters

J defines the amplitude of the HRR field as K does in the linear case.

1 1 13

n n niu A J r

singularity recovered for stress and strain in the linear case

1

2r

HRR stress singularity and strain singularity1

1nr

1

n

nr

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where HRR based upon small displacements is not applicable

Large strain region where crack blunting occurs :

JK

J and K dominated regions near crack tip :

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6.5 Applications the J-integral

Tyn

xn

A

B

D

C

ds

x

y

O

x

y n

ij ij ij ij xx xx yy yy xy xy

1 1= σ dε = σ ε = σ ε +σ ε +2σ ε

2 2w

Loads and geometry symmetric / Ox

for a plane stress, linear elastic problem

J-integral evaluated explicitly along specific contours

iij j

uJ w dy n ds

x

?

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2 2 2xx yy xx yy xy

1 1+ν= σ +σ -2νσ σ + σ

2E Ew

From stress-strain relation,

y yx xxx x xy y yx x yy y

u uu u=σ n ds+σ n ds+σ n ds+σ n ds

x x x x

iij j

uσ n ds

x

Expanded form for

Along AB or B´ A´ nx = -1, ny = 0 and ds=-dy ≠ 0

yxxx yx x

uu=σ dy + σ n dy

x x

Along CD or DC´

yxxx yx

uu=σ dy+σ dx

x x

nx = 1, ny = 0 and ds=dy ≠ 0

Simplification :

(2D problem)

Page 48: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

48

Along OA and A´O J is zero since dy = 0 and Ti = 0

B C Dy y yx x x

xx xy xy yy xx xy

A B C

u u uu u uJ=2 w-σ -σ dy+2 σ +σ dx+2 w-σ -σ dy

x x x x x x

Along BC or C´B´

yxxy yy

uu=-σ dx-σ dx

x x

BC : nx = 0, ny = -1 and ds=dx ≠ 0

C’B’ : nx = 0, ny = 1 and ds=-dx ≠ 0

Finally,

Page 49: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

49

Example 6.5.3

Example 6.5.2

uy

uy

2y(1-ν)Eu

J=2hw=(1+ν)(1-2ν)h

2

2 3

12P aJ

EB h

2

Page 50: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

50

Relationship between J and CTOD

ca

y

iij j

uJ n ds

xG

x

d t

Consider again the strip-yield problem,

G

the first term in the J integral vanishes because dy= 0 (slender zone)

Y

Y

iij j

un ds

x

but

yyy y

un ds

x

y

Yu

dxx

yY

uJ dx

x

t

t

Y ydu

Y t d

Page 51: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

51

Unique relationship between J and CTOD:

Y tJ m d

m : dimensionless parameter depending on the stress state and materials properties

• The strip-yield model predicts that m=1 (non-hardening material, plane stress condition)

• This relation is more generally derived for hardening materials (n >1) using the HRR displacements near the crack tip, i.e.

1 1 13

n n niu A J r

m becomes a (complicated) function of n

blunted crack 90° d tShih proposed this definition for d t :

4 Y tG

d

The proposed definition of d t agrees with the one of the Irwin model

Moreover,4

m

in this case

Page 52: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

J integral along a rectilinear bi-material interface:

x

y

Material 1

Material 2

Wa2h

and h/W = 1

a = 40 mm

W = 100 mm

h = 100 mm

• SEN specimen geometry (see Appendix III.1):

Material 1:

a/W = 0.4

1 MPa.

Remote loading:

Materials properties (Young’s modulus, Poisson’s ratio):

Plane strain conditions.

E1 = 3 GPa

n1 = 0.35

Material 2: E2 = 70 GPa

n2 = 0.2

Page 53: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

• Typical mesh using Abaqus :

Material 1= Material 2

Refined mesh around the crack tip

Number of elements used: 1376Type: CPE8 (plane strain quadratic elements)

Material 1

Material 2

Page 54: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

Isotropic Bi-material

KI KII KI KII

Annex III 0.746 0. / /

Abaqus 0.748 0. 0.747 0.183

Results:Material 1 Material 2 Bi-material

N/mm N/mm N/mm

Abaqus 0.1641 0.0077 0.0837

MPa mSIF given in

(*) same values on the contours 1-8

for the isotropic case (i =1,2).2

21I

i i

KJ

E ( )

• One checks that:

(*)

Remarks:

N/mm=J/mm2

Page 55: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

and 3 4i i 2 1

ii

i

EG

21i

ii

EE

• Relationship between J and the SIF’s for the bi-material configuration:

plane strain, i = 1,2

KI and KII are now defined as the real and imaginary parts of a complex intensity factor, such that

with

- For an interfacial crack between two dissimilar isotropic materials (plane strain),

where

Page 56: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

56

Notes for CH6

Page 57: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

57 2 2

1,21

42

a b a b c

3 3 3σe e

The 3-axis (z-axis) is considered as a principal axis:

(1) Principal stresses and invariants in plane state problems

The matrix of s in any orthogonal basis (n, t, e3) is

s : stress tensor s3 : associated principal stress

3, ,

3

0

0

0 0

a c

c b

n t eσ with the invariants,

1 3

2 22 3

23 3

1

2

det

I tr a b

I tr tr a b ab c

I ab c

2

σ

σ σ

σ

In principal coordinate system, the stress tensor is diagonal,

1 2 3

1

2, ,

3

0 0

0 0

0 0

e e eσ

and the two principal stresses in the plane 1-2 are given by,

2 21 3 1 2 3 1 3

14 (2 3 )

2I I I I

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58

22

22 2

1 31 cos sin

4 2

1 31 2 1 cos sin

4 2

I

Y

I

Y

K

K

22 2

22 2 2

1cos 4 3cos

2 2 2

1cos 4 1 3cos

2 2 2

I

Yp

I

Y

K

rK

(2) Other expression for rp :

2

4 1max( )

3 2I

pY

Kr

plane stress

plane strain

arctan (2 2)

plane stress

Page 59: 1 plastic zone CH-6 Elastic-Plastic Fracture Mechanics Laurent HUMBERT Thursday, May 27th 2010 J.

59

Along the x- axis (mode I),

1 22 22

1 2 1 3 2 31

2e

1 2 02

Ixx yy

K

x

Plane stress 3 0

(3) 1D approximation of plastic zone length

12I

e YK

r

Plane strain

3 1 2

1 22 2

1 11

2 1 2 1 22

e

1

1 22

Ie Y

K

r

1 221 1

12

2

2

11

2I

Y

Kr

22

11 2

2I

Y

Kr