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6. Elastic-Plastic Fracture Mechanics

Introduction

Applies when non-linear 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, linear asymptotic mechanical field (see eqs 4.36, 4.40).

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

crack

Plastic zone

Elastic Fracture Contained yielding Full yielding Diffuse dissipation

D

L

a

, ,L a D B

B

L D a L D a

LEFM, KIC or GIC fracture criterion

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 for some geometries (mode I loading).

6.3 The J contour integral as yield criterion.

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 312e

e is the effective stress and i (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 2 223

cos 1 sin sin ...2 2 22

3cos sin cos ...

2 2 22

Ixx

Iyy

Ixy

Kr

Kr

Kr

yy

xy

xx

x

y

Oθ

r

LEFM analysis prediction:

KI : mode I stress intensity factor (SIF)

Estimation of the plastic zone size

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We have 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

IKr

rrr

x

y

Oθ

r

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

1 222

2 2rr rr

r

31 2

0 plane stressplane strain

and

, ,rr r

3

0 plane stress2 cos plane strain

22IKr

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Substituting into the expression of e 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

1 2

2 21 31 2 1 cos sin

22 2I

eK

r

(see expression of Y2 p 6.7 )

1 22 2 21

6cos sin 2cos2 2 22 2

IKr

1 221 3

1 cos sin22 2

IKr

1 2

2 2cos 4 1 3cos2 22

IKr

1 22 2 2 2 21

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

IKr

1 22cos 4 3cos

2 22IKr

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

Using the previous expressions (2) of e 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

14

p

I

Y

r

K

Plane strain

Plane stress (= 0.0)

increasing 0.1, 0.2, 0.3, 0.4, 0.5

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

221 22

I

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

12

I

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

Kr

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

yy

21 2

11222

I IY

Y

K Kr r

2 1r r

2

11

2I

Y

Kr

11

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

112 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

112

3I

Y

Kr

2eff

yyK

X

X

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

112

Y

effaK

(Irwin, plane stress)2

12

Iy

Y

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 am

1 plane stress3 plane strain

m

YesNo

1i i

Algorithm:

( )Ii

eff i iK K Y a a

Application:

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

∞= 2 MPa, Y = 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

2Ii

iY

Ka a

m

( ) ( 1)I Ii 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 5)

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 ) xK ,a c

( a c ) x( a c )

YY

a cx

AB

I A YQ ( a c ) xK ,a c

( a c ) x( a c )

Recall first,

crack tip A

crack tip B

(see eqs 5.3)

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 )

(see equation 5.5)

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 ) u du( a c ) u( a c )

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

2 22

a cY

a

a c dx

( a c ) x

Thus, KIA can written

The same expression is obtained for KIB (at point B):

We denote 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 ) u du( a c ) u( a c )

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Recall that,1

2 2

a c

a

dx acosa c( a c ) x

12I Y Ya c aK ,a c cos

a 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 aa c

a c

Thus, cos2 Y

aa c

By Taylor series expansion of cosines,

2 4 61 1 11 ...2! 2 4! 2 6! 2Y Y Y

aa c

Keeping only the first two terms, solving for c22 2

2 88I

YY

a Kc

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

aa

-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

28 lnsec

2eff YY

K a

sec2 Y

a

(see Anderson, third ed., p65)