05_EPFM

ELASTIC-PLASTIC FRACTURE MECHANICS

Suggested readings:

Chapter 3 T.L. Anderson

1Qian X/CE5887

Concept of Small-Scale Yielding (1)

I II III IV

I: plastic zoneII: elastic region, elastic solution is influenced by the nerighbouring plastic zoneIII: K dominance zone

IV: non-singular terms become important

2

Kf

r

H.O.T2

Kf

r

SSY holds when the plastic zone of the crack tip is sufficiently small compared to the crack length and other relevant geometric length parameters, such that the elastic singular fields still give good approximation of the actual fields in an annular region surrounding the crack tip!

2Qian X/CE5887

Concept of Small-Scale Yielding (2)

ry

a W

,yr a W Under SSY conditions, the load-displacement response is still linear

pP

P

i

ii

iii

i ii iii

3Qian X/CE5887

Size of the Plastic Zone: Irwin’s Model (1)

ys

x

y

rp Plastic zone

Crack tip

i) Irwin’s modelFor plane-stress conditions, the size of the plastic zone ry, can be estimated from the stress-field equation by setting the linear-elastic stress equal to the yield strength, ys.

ry

Linear-elasticElastic-plastic

0.52 22

1 2 1 3 3 22

2ys

Based on Mises yield criterion,

31 sin sin

2 21 3

cos 1 sin sin2 2 22

3sin cos

2 2

xx

yy I

xy

Kr

4Qian X/CE5887

ry: first-order plastic zonerp: second-order plastic zone


ys

x

y

rp Plastic zone

Crack tip

ry


At = 0, f = 1 and xy = 0, therefore, 1 2yy xx

31 sin sin

2 21 3

cos 1 sin sin2 2 22

3sin cos

2 2

xx

yy I

xy

Kr

0.52 22

2ys yy yy yy

3 0zz

Plane-stress

5Qian X/CE5887

0.52 22

1 2 1 3 3 22

2ys


ys

x

y

rp Plastic zone

Crack tip

02 2

I Iyy ys

y y

K Kf

r r

21

2I

yys

Kr

ry


31 sin sin

2 21 3

cos 1 sin sin2 2 22

3sin cos

2 2

xx

yy I

xy

Kr

0.52 22

2ys yy yy yy

Therefore,Plane-stress

6Qian X/CE5887


x

y

rp

Plastic zone

Crack tip

ry

Linear-elastic

Elastic-plastic

Assumption: the effect of the plastic zone is to translate the stress distribution towards the right by the amount of distance necessary to restore equilibrium.

Equilibrium requires that Area A equals Area B

0

yr

yy ys ys p ydx r r

A

B

ys

0 2

yrI

ys ys p yK

dx r rr

0

22

yrI

ys y ys p yK

r r r r

21

2I

yys

Kr

21

2Ip y

ys

Kr r

Plane-stress

7Qian X/CE5887


For plane strain conditions,

ys

x

y

rp Plastic zone

Crack tip

ry


31 sin sin

2 21 3

cos 1 sin sin2 2 22

3sin cos

2 2

xx

yy I

xy

Kr

at = 0, f = 1 and xy = 0, therefore,

1 2yy xx 3 1 2zz v

0.52 221 2 1 3 3 2

0.52

2

2

22 1 2

2

1 2

ys

yy

yy

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Plane-strain


Assume υ= 1/3 for metals,

3yy ys

Plane-strain

ys

x

y

rp Plastic zone

Crack tip

ry


3 02 2

I Iyy ys

y y

K Kf

r r

21

6I

yys

Kr

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Corrected Crack Length (1)

To account for the crack-tip plasticity, Irwin suggested that the stress distribution at a moderately large distance from the crack tip be found by considering a “corrected crack length”,

eff ya a r

2

effI

yy yK

r rr

a ryry effI effK a

21

'2

effeff II

ys

KK a

This is an iterative process!

Plane stress

10Qian X/CE5887

Corrected Crack Length (2)2

1'

2

effeff II

ys

KK a

The difference between KIeff’ and KI

eff should be negligibly small

'eff effI IK K

221

2

eff effI I

ys

K Ka

21

12

effI

ys

aK

Plane stress

Note: KIeff > KI, the difference

becomes dramatic as ∞ → ys, whence the entire procedure of crack-tip zone correction becomes susceptible!

11Qian X/CE5887

Corrected Crack Length (3)

21

12

effI

ys

aK

For a reasonable upper limit to the applied nominal stress in a structure, say,

3

4 ys

21.18

0.8511 0.75

2

effI

a aK a

22

0.75 101 12.8 mm

2 2ysI

yys ys

Kr

For a crack with a = 10 mm in an infinite plate,

/effI IK K

/ ys

1.0

1.0

1.4

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Dugdale’s Model (aka: Strip-Yield Model)

• Irwin’s correction is limited to small-scale yielding (SSY) conditions, Dugdale’s model is not limited to SSY• Consider a crack in an infinite thin sheet of materials with very low hardening• The crack length is increased from 2a to 2(a+R), where R is the length of the plastic zone, with a closure pressure equal to ys applied at each crack tip

2a RR

R

Plastic zone

ys

p app RI I IK K K

appIK a R

?RIK

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Dugdale’s Model(1)

Dugdale argues that,

0p app RI I IK K K

i.e., when the closure pressure just closes the crack completely over the length R, the crack singularity at a + R should vanish since there is no crack there!

This relationship becomes the basis to find the distance R (plastic zone size) with respect to the ys, 2a and KI

app

Dugdale, D. S. (1960). Yielding of steel sheets containing slits. Journal of the Mechanics and Physics of Solids. Volume 8, Issue 2 , Pages 100-104

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Using the weight function technique, we can find the SIF values. Recall that KI for a crack under a pair of point forces equals,

p

x 2 / 2

/ 2Il x

K pl l x

y

x

l

2( )l a R

2 / 2 / 2

/ 2 / 2

a R aRI ys

a a R

l x l xK dx dx

l l x l x

over , and ,ysp R a a a R a

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RR2a


2 / 2 / 2

/ 2 / 2

a R aRI ys

a a R

l x l xK dx dx

l l x l x

For 2nd integral, assume x = -x’

/ 2 '2 / 2

'/ 2 / 2 '

2 / 2 / 2 ''

/ 2 / 2 '

a R aRI ys

a a R

a R a R

ysa a

l xl xK dx d x

l l x l x

l x l xdx dx

l l x l x

16Qian X/CE5887

RR2a

ys

2( )l a R


2 / 2 / 2 ''

/ 2 / 2 '

a R a RRI ys

a a

l x l xK dx dx

l l x l x

Grouping both integrals,

2

1

2

1

1

2 / 2 / 2

/ 2 / 2

2

/ 2

2 sin2 / 2

2 si

co

n2

2 s

a RRI ys

a

a R

ysa

a R

s

s

ya

y

y

s

l x l xK dx

l l x l x

l dx

l x

l x

l

a R

a R a

a

a

R

R

a

p

xy

x

l

2( )l a R

ysp

17Qian X/CE5887


12 cosRI ys

a R aK

a R

cos2 ys

a

a R

0app RI IK K Since,

12 cosysa R a

a Ra R

R

a R1 cos

2 ys

Dugdale’s paper was not accepted despite the good agreement with experiments, probably because he did not provide clear evidence that yielding was occurring in a narrow strip!Hahn and Rosenfield showed that in plane-stress conditions, the plastic zones do conform to Dugdale’s model. (Acta Metal, Vol 13, pp 296-306, 1965)

Comparison with experiments18Qian X/CE5887


Based on Taylor series,2 4

1 5sec 1 ...

2 2! 2 4! 2ys ys ys

For ∞<< ys, i,e, SSY conditions,

sec2 ys

a R

a

2 222

20.393

8 8I I

ys ysys

a K KR

Irwin’s results,2 2

10.318I I

ys ys

K KR

Close agreement

19Qian X/CE5887


The effective SIF for Dugdale’s model can be estimated using, aeff = a + R, or

seceffI

ys

K a

However, the above equation tends to over-estimate the effective KI. A more realistic estimate (See Appendix 3.1 of Anderson 2005 for the derivation) is,

2

8lnsec

2effI ys

ys

K a

0 0.2 0.4 0.6 0.8 1.00

0.5

1.0

1.5

2.0/eff

I ysK a

/ ys

Linear-elastic

Irwin’s correction

Dugdale’s correction

2a

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Dugdale’s Model: Example

A steel plate with a central through-thickness flaw of length 16 mm is subjected to a stress of 350 MPa normal to the crack plane. If the yield strength of the material is 1400 MPa, what is the plastic-zone size and the effective stress-intensity level at the crack tip?

350 8 1755 MPa mmIK a

Assume the plate is infinitely large, 2a = 16 mm

2

0.62 mm8

Ip

ys

Kr R

2

8 3501400 8 lnsec 1778 MPa mm 56.2 MPa m

2 1400effIK

8 350cos

8 2 1400R

R = 0.66 mm

Alternatively,

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Application of Dugdale’s Model

Dugdale models are often used to describe quantitatively the role of toughening mechanism in otherwise brittle materials.

The background brittle material is toughened by the ductile incldusions. These can be rubber, metal fibers, etc.

The SIF at the macroscopic crack tip thus equals,

Qian X/CE5887 22

tip app tougheningK K K

Toughening mechanism

Ktoughening

(MPa•m0.5)

Firber (glass, carbon, silica-carbon,etc.)

~ 20

Ductile network (Al2O3, AL, etc.)

12 ~ 20

For comparison, structural steels at very cold temperatures have toughness 20 – 40 MPa•m0.5

Shape of the Plastic Zone (1)

2cos 1 sin2 22

Irr

K

r

3cos22

IK

r

2sin cos2 22

Ir

K

r

• So far, we have considered the size of the plastic zone along the x-axis!• The size of plastic zone varies with • We will utilize the Mises yield criterion to investigate the shape of the plastic zone

Mises yield criterion,

0.52 22

1 2 1 3 3 22

2ys

23Qian X/CE5887

How to express the principal stresses in terms of σrr σθθ and σrθ?


2cos 1 sin2 22

Irr

K

r

3cos22

IK

r

2sin cos2 22

Ir

K

r

C

R

Based on Mohr’s circle,

1

2

rr

r

1cos

2 22I

rrK

Cr

22 1cos sin

4 2 22I

r rrK

Rr

1 C R 2 C R

24Qian X/CE5887


1 cos 1 sin2 22

IK

r

C

R

Based on Mohr’s circle,

1

2

rr

r

2 cos 1 sin2 22

IK

r

3

0 for plane stress

2 cos for plane strain22

IK

r

3 1 2

At = 0, the principal stresses 1 and 2 are equal and act in x and ydirections xx and yy are the principal stresses.At other values, the boundary of the plastic zone is obtained by setting mises = ys

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For plane stress,

0.52 22

1 2 1 3 3 2

0.52 22 2 22 2 2 2

0.52 2 2

0.52

2

2

42cos sin cos 1 sin cos 1 sin

2 2 2 2 2 2 2 2 2 2

22cos 6sin cos

2 2 2 22

2 31 sin cos

2 22

ys

I I I

I

I

K K K

r r r

K

r

K

r

2

21 31 sin cos

4 2I

yys

Kr

26Qian X/CE5887


For plane strain,

0.52 221 2 1 3 3 2

0.52 22 2 2 2

0.522

2

2

24cos sin cos 1 sin 2 cos 1 sin 2

2 2 2 2 2 2 22

2 3sin 1 cos 1 2

2 22

ys

I

I

K

r

K

r

2

221 3sin 1 cos 1 2

4 2I

yys

Kr

27Qian X/CE5887


0 0.4 0.8-0.8 -0.4-0.8

-0.4

0.4

0.8

0

21 I

ys

x

K

21 I

ys

y

K

Plane strain

Plane stress

28Qian X/CE5887

Effect of Thickness (1)

zzxx

B

crackr

yy

• Materials far away from the crack front experience a plane-stress condition• The large normal stress yy near the crack tip will try to contract in the x and zdirection, but is prevented from doing so by the surrounding materials• The crack front near the free surface of the plate experiences less constraint by the surrounding material

29Qian X/CE5887


B

z

x

y

z B/20

zz

High triaxiality

(plane strain)

Low triaxiality

plane stress

Mid-thickness

Free surface

Increasing thickness of the specimen30Qian X/CE5887


0

7075-T6 Aluminum

10

20

30

40

50

60

70

0 0.2 0.4 0.6 0.8 1.0 1.2

ksi incriticalK

Specimen thickness (in)

t

High triaxiality zone

Low triaxiality zoneB

Test specimen Surface defect in structures

31Qian X/CE5887

1 ksi in 1.1 MPa m

Crack-Tip Opening Displacement (CTOD) Developed initially in the U.K. by Wells and colleagues at The

Welding Institute

Very physically appealing approach based on critical elastic-plastic stretching at the crack front (no abstract energy concepts or advanced math!)

Plastic Deformation

X

yy

a X

yy

a CTOD is an indirect measure of severe stretching at crack tip

CTOD ( )

32Qian X/CE5887

CTOD

2

21

8 1 22 ( )

2

I

ysy I

K

v r r KE

2 24 1I

ys

K

E

CTOD increases as K2

(this is a nonlinear effect!)

( )v r

yr

plastic zone

24 1( )

2Ir

v r KE

r

( )v r

yr

Suggestion: re-derive for plane-strain conditions, verify physical units are correct

33Qian X/CE5887

Plane-stress

CTOD: Definition

A

A’

45o

45o

v

CTOD

u

Tracy, D. M. (1976). Finite element solutions for crack-tip behavior in small-scale yielding conditions. Journal of Engineering Materials and Technology. 98, 146-151.

CTOD is defined at the intercept of the two 45o lines originating from the deformed crack tip and intersecting the crack profile.

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CTOD(2)

The strip-yield model provides another means to determine CTOD.

y

x

R

A

• A: fictitious tip• R << a• No singularity at A 0app R

I IK K

02I

c

p xK dx

x

p(x)

x

y

c

020app

IR

p xK dx

x

B

At A:

35Qian X/CE5887

CTOD(3)y

x

R

A

020app

IR

pK dx

x

B

At A:

The crack-tip opening displacement,

01

2 ln2 2

ysy app R

r x xx u K dx

G x x

2

8app

ys

KR

ysp x

1ln

4

l

yc

x xu p x dx

G x x

Linear elastic fracture mechanics From weight function

1

2 2y I

ru K

G

36Qian X/CE5887

CTOD(4)

for –R < x < 0

111 ln

21

xx x x Rg

xR R R

R

where

0ln 2

R

xx xdx Rg

Rx x

2

2 2

1 1

2 2 8

1 1 1 1 1

2 8 4 8

appys ysapp app I

I Iys

app appI Iys

appys I ys

x x xKrx K Rg K g

G R G R

K Kx x xrg g

G K R G R R

2

8app

ys

KR

2

8app

ys

KR

8ys

appK R

37Qian X/CE5887

CTOD(5)

2

2

2 2

1

1 1 1

4 8

11 1 1 11 ln

4 8 21

11 1 11 ln

8 2 81

appI

ys

appI

ys

app appI I

ys ys

K x xx g

G R R

xK x x x R

xG R R R

R

xK Kx x xR g

xG R R G R

R

111 ln

21

xx x x Rg

xR R R

R

At x = -R,

2

1

8

appI

ys

KR

G

38Qian X/CE5887

CTOD(6)

21 1

8 IKB a G

G

2

1

8

appI

ys

KR

G

But we have derived:

ys

x

G

The strip-yield model assumes a non-hardening material. The actual relationship between CTOD and G depends on the strain hardening and is given by a more general form,

ys

xm

G

1.0 ≤ m ≤2.0

Try to plot (x) v.s. x to observe how the crack tip deforms!

39Qian X/CE5887

CTOD (7) Use nonlinear finite element analysis to compute CTOD as function

on applied loading

Thermal loading, residual stresses, etc. introduce NO complications

Just extract CTOD from displacements of nodes behind crack tip at each applied load level

Le

More elements than this needed to resolve variation of fields

13 nodes at this front location-collapse to a point

Extract CTOD at each front location using 45o intercept method

40Qian X/CE5887

J-Integral: Energy Release Rate

A

ds

For a 2-D body of area A, with tractions Ti

applied over the bounding curve the potential energy of the body is (assume unit thickness),

i iA

U P WdA T u ds

0 0 0 0 0... ...

ij xx xy yy zz

ij ij xx xx xy xy yy yy zz zzW d d d d d

W is the strain energy density, and ui is the displacement vector

If the 2-D body encloses a crack, the change in the potential energy with respect to crack extension becomes,

ii

A

dud dWdA T ds

da da da

Note: under linear-elastic conditions,

G = -

d= J

da41Qian X/CE5887

J-Integral (2)

“Advanced Calculus for Applications” by F.B. Hilderbrand

in our case: x = a, B = x, A = 042Qian X/CE5887

J-Integral (3)

A

ds

ii

A

dud dWdA T ds

da da da

d x

da a a x a x

a

x

yx0

0x x a 1x

a

dW

da

idu

da

d

da

Assumption: a is parallel to x

43Qian X/CE5887

J-Integral (4)

ij ijij

ij

W W

a a a

Recall the principle of virtual work, 0ij ij i i

A

dA T u ds

i i

i

A

u ud W WdA T ds

da a x a x

=0

ii

A

ud WJ dA T ds

da x x

ii

udJ Wdy T ds

da x

44Qian X/CE5887

Internal energy=external energy

J Integral: Path Independence

,

,

ii

iij j

iijA

j

i iij j ijA

j

uJ Wdy T ds

x

uWdy n ds

x

uWdxdy dA

x x

u uWdA

x x x x

A

ds

0Equilibrium

iijA

j

uWJ dA

x x x

Divergence theory

45Qian X/CE5887

J Integral: Path Independence (2)

ij ijij

ij

W W

x x x

1

2ji

ijj i

uu

x x

1

2ij ji i

ij ijij j i j

uu uW W

x x x x x x x

iijA

j

uWJ dA

x x x

ij ji use

46Qian X/CE5887


0

iijA

j

i iij ijA

j j

uWJ dA

x x x

u udA

x x x x

iij

j

uW

x x x

Path independency can be restored for the various effects by subtracting integrals for those contributions

J = 0 on the closed path when,• the path encloses no cracks or singularity• there is no body force• the material is isotropic and elastic• E, υ have no gradation in x• the stresses are the simple derivative of W (no residual stresses, thermal strains, etc.)

47Qian X/CE5887


1

2 3

4

1 2 3 4 0J J J J J

For a cracked body,

Any arbitrary (counterclockwise) path around a crack will yield the same value of J, which is path independent!

48Qian X/CE5887


• The path independence of J requires a one-to-one relationship between the stress and strain values, valid for linear elastic and nonlinear elastic materials. • For elastic-plastic materials, one strain value may correspond to one or more stress values if the material is unloaded and cyclically loaded.• However, elastic-plastic material behavior may assume the same behavior as that of the nonlinear-elastic materials, provided no unloading takes place.

Nonlinear elastic

Elastic-plastic

Rice, J. R. (1968). “A path independent integral and the approximate analysis of strain concentration by notches and cracks.” J. App. Mech., 35, 397-386.

49Qian X/CE5887


Conditions for path independence in elastic-plastic materials:1. stress must non-decreasing everywhere2. stress components remain in fixed proportion

However, plasticity induced stress redistribution may violate the 2nd condition. However, J still retains an approximate path independence if its values are obtained from a region outside the plastic deformation, i.e., a far-field J value.

50Qian X/CE5887

Example (1)

A narrow strip of material with elastic modulus E and height h is rigidly attached to parallel platens, as shown in the figure below. If the upper platen is displaced by an amount , determine the geometric stress intensity factor for an edge crack midway between the platen faces.

xy

1 2

3

45

1 2 3 4 5J J J J J J

ii

uJ Wdy T ds

x

• At 2 and 4, dy = 0 and displacements are constant, i.e., ∂ui/∂x = 0, J3 = J4 = 0• At 1 and 5 near the free surface, tractions are zero, J1 = J5 = 0

h

51Qian X/CE5887

Example (2)

xy

1 2

3

45

h

At 3, sufficiently far from the crack tip, xx = 0, xy = 0, yy ≠ 0, tractions are zero. Therefore, the contribution to J only comes from the W term. Along 3

(assume plane stress condition),

22

3 2 2

0 0 0

1 1 1

2 2 21 1

h h h

Iyy yy

KE EJ Wdy dy dy

h h h E

10xx xx yyE

/yy h

xx yy 21 1

yy yy xx yyE E

2 21 1yy yy

E E

h

22 1

IE

Kh

52Qian X/CE5887

HRR Singularity (1)

Ramberg-Osgood constitutive relationship

0 0 0

n

0

0

E

0

E

0

0

n

E E

0 0

0 0

n

E E

0

0 E

0

E

is the yield offset, equal to 0.002

for steels.

For large plastic deformation, the elastic strain is negligible:

0 0

n

53Qian X/CE5887

HRR Singularity (2)

cos

ii

ii

uJ Wdy T ds

x

ur W T d

x

dsd

Evaluate J integral over a circular path with radius r,

sindy r ds rd

cos ii

uJW T d

r x

J is independent of r! In addition, ij ij ij ijW d ii ij ij

uT

x

1

rshould be cancelled out on the left hand side and on the right hand side!

1ij ij r

0 0

n

combined with

/1

1n n

C

r /

21 1n

C

r

54Qian X/CE5887

HRR Singularity (3)

/1

1n n

C

r /

21 1n

C

r

1. Linear-elastic materials, n = 12. Elastic-perfectly-plastic, n→∞3. Stronger singularity for strain, weaker singularity for stress

Hutchinson (1969), Rice and Rosengren (1969) (HRR solution) derive independently the solution for the crack-tip stress field as,

/1 1

00 0

n

ij ijn

Jf

I r

/ 1

00 0

n n

ij ijn

Jg

I r

. . . .2 36 568 0 4744 0 040 0 001262nI n n n

. . . .2 4 34 546 0 2827 0 0175 0 4516 10nI n n n

for plane strain

for plane stress

55Qian X/CE5887

HRR Singularity (4)

/

i

n

in

jj

J

I rf

1 1

00 0

0 /2 -1

-0.5

0.0

0.5

1.0

1.5

0 /2 -1

-0.5

0.0

0.5

1.0

1.5

ijf ijf

rrf

ef

rf f

rrr

n = 13n = 3

f

rrf

rf

ef

Plane stress condition

56Qian X/CE5887

HRR Singularity (5)

/n n

ijijn

J

I rg

1

00 0

0 /2 -0.1

0.1

0.3

0.5

0.7

ijg

0 /2

-0.2

0

0.2

0.4

0.6

0.8 ijg

n = 13n = 3

Plane stress condition

rg

g

rrgg

rg

rrg

57Qian X/CE5887

J and CTOD (1)

45o

45oCTOD

/ 1

00 0

n n

ij ijn

Jg

I r

Integrating the strain expression yields the displacement vector on the traction-free crack face =

v

u

/n n

n

n

hu Jr

v I h

1 111

00 0 2

r

. CTOD r u v 0 5

The point at behind the deformed crack tip is the un-deformed location of the intercept.

r

r u v

From definition of CTOD,

58Qian X/CE5887

rrr

u

r

1r uu

r r

J and CTOD (2)

/n n

n

n

Jr r h h

I

1 1

10 1 2

0 0

n

n

n

n

Jr h h

I

11

0 1 20

n

nCTOD

n

Jv h h h

I

11

0 1 2 20

2 2

CTOD n

Jd

0

dn depends on n and 0 = 0/E

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J and CTOD (3)

CTOD n

Jd

0

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1.0

/ n1

nd

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1.0

/ n1

nd

/ E 0

0.0080.0040.0020.001

/ E 0

0.0080.0040.0020.001

Plane stress Plane strain

• The plane strain condition reduces the CTOD by about 20%• Less deformation under plane-strain conditions than that in the plane stress• More constraint in plane-strain state!

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About HRR Solutions

• The HRR solution depends on the applicability of Ramberg-Osgood constitutive model to the material. Ramger-Osgood is not a good representation for some of the mild steels which exhibit distinctive upper and lower yield points.

• HRR solution ignores the linear portion of the stress-strain relationship. The strains near the crack tip are large enough that the linear portion is negligible, but meanwhile small enough that the small strain theory is still applicable.

• The HRR stress field is not unique. There are infinite number of ways to satisfy the governing condition, ijij 1/r, corresponding to an infinite number of constitutive models.

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J Dominance (1)

• J is more than an energy parameter. It also characterizes the crack-tip stress fields.• J is also a fracture parameter: crack extension should take place if J exceeds a critical value, JIc.• Since the crack tip stress field is dependent on J everything happens near the crack tip should also depend on J a similitude concept

Boundary of K dominance

Inelastic region

For linear-elastic materials, when the plastic zone remains small compared to any dimension of the structure, K governs the stress, strain, displacement around the crack tip.

Any two cracks will be in the same state if they have equal K, regardless of the geometry, type of loading, crack size A K similitude!

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J Dominance (2)Similar to K similitude, the finite-strain zone around the crack tip should be small compared to the dimension of the structure. Hutchinson showed that the valid J dominance zone starts at,

CTODR 3

K dominance zone (outside plastic zone)

Finite-strain zone

R

J dominance zone

Stresses governed by remote loading

Nonlinear elastic

Elastic-plastic

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J Dominance (2)

For nonlinear-elastic materials, where the stress-strain curve for unloading is the same as that for loading, J remains path independent.

For real materials, crack extension causes unloading near the crack tip, which violates the assumption of deformation plasticity

Material immediately ahead of the unloading zone experiences highly non-proportional loading!

J similitude only applies to stationary cracks, or practically very small crack extensions, with the Jrising sharply!

Elastic unloading

Non-proportional plastic loading

J-dominance

For ductile materials, a small finite-strain zone restricts the crack extension in the J dominance zone.

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J From Lab Test (1)

Recall from energy principle: for a fracture specimen under tension, the strain energy and complimentary strain energy follows,

, LLDP

aU

*U

P

LLD

0

, , LLDU a P a d

*

0

, ,

P

LLD LLD LLDU a P U P P a dP Or use the complimentary energy:

65Qian X/CE5887

J From Lab Test (2)

P

LLD

dP

d daa

a

a da

Fixed load P during da

0

0

00

Total potential energy for load control:

= ,

= ,

,,

P

LLD LLD LLD

P

LLD

PP

LLDLLD P

P

U P P a dP P

P a dP

P add P a dP da dP

da a

0

,1 1P

LLD

P

P adJ dP

B da B a

J is the area between the load-displacement curves for different crack lengths, divided by the thickness (and the differential crack extension).

66

Load controlled loading

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J From Lab Test (3)

67

Displacement controlled loading

0

= , LLDU P a d

P

LLD

d

PdP da

a

a

a da

Fixed during da

J is the area between the load-displacement curves for different crack lengths, divided by the thickness (and the differential crack extension)

0

,1 1LLD

P adJ d

B da B a

The potential energy of external forces becomes zero, or Ω = 0

Qian X/CE5887

J From SE(B) Test (1)

68

• Two equivalent expressions to compute J• These expressions prompted research to find approximate, analytical

expressions for J for lab test specimens• Jim Rice (at Brown), John Merkle (at ORNL) and Herb Corten (TAM

Dept. UIUC) collaborated to derive the first simple, analytical expression for the J-integral specifically for the SE(B)

Suppose the remaining ligament, b, is fully plastic

The plastic moment is then

2L

W

a

b = W - a

P, Thickness, B 0

0

0 2Pb

F B

0 2Pb

F B

2

02 4P Pb b

M F B Qian X/CE5887


69

Pl PlUnder large deformation, the beam deforms into a mechanism with two rigid bars.The plastic rotation can be expressed as a dimensionless function,

200

, ,PPl

M Ef n

b B

20

0

, ,PlPE

M Bb h n

Thinking of the reverse form …

... is a non - dimensional functionh

2

2Pl Pl

Pl L L

From rigid-body approximation,

Plastic hinge

20

0

, ,2

4Pl

PPL E

M Bb hL

n

From equilibrium,4P

PLM

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70

20

0

, ,2

4Pl

PPL E

M Bb hL

n

20

0

24, ,PlBb h E

P nL L

0

,1LLD

P aJ d

B a

00

28 , ,

2

Pl Pl

PL EBb h n

LP

L b

P

a b

P

Assuming fixed Pl.

0Pl

b

2

Pl

PP

ba

Qian X/CE5887


71

2

Pl

PP

ba

0

,1LLD

P aJ d

B a

0 0

1 2 2LLD LLD

PJ d Pd

B b Bb

P

LLD

0

LLDPd

2 measured area under load-displacement curve

area of remaining ligamentJ

Qian X/CE5887

The Approach (1)

72

A J formula similar to that for the SE(B) can be derived for the C(T) specimen

Finite element analyses show that the formulas are surprisingly accurate – provided deformations are largely plastic

More accurate expressions may be written for other types of test specimens and through-cracked structures in the form

where the non-dimensional “eta” ( ) factor varies with specimen type, a/Wratio, W/L and, in general, material strain hardening (n)

0

UJ Pd

Bb Bb

Qian X/CE5887

The Approach (2)

73

John Sumpter and Cedric Turner in the late 1970s proposed an even more accurate framework to compute J values from measured experimental quantities

They decomposed the total (measured) load-line displacement into elastic and plastic components

Likewise, the total J value is separated into elastic and plastic parts

e plJ J J

LLD e pl

2 21I pl plK U

JE Bb

P

LLD

02pl

PU Pd

Qian X/CE5887

Begley-Landes Experiments (1)

74

Begley and Landes (1972) conceived of some simple tests they could perform to “measure” applied J values directly from lab tests.

Fabricate and test 4 identical SE(B) specimens with different crack lengths

1 2 3 4a a a a a

1. Plot the measured load-load line displacement curves from the tests

2. They used a “tough” material so that no crack growth occurs during the tests

3. Measure the area under the load-displacement curves at each prescribed

P

L L D

1a

2a

3a

4a

1 2 3 4

No fracture events

Qian X/CE5887

Begley-Landes Experiments (2)

75

1 2 3 4

1 2 3 4

1

2

At we have: , , ,

At we have: , , , , .

a a a a

a a a a

U U U U

U U U U etc

Make the following plot:

J

LLD

2IK

JE

l imitJ P U

B

4

a1

…

• Early in the loading, the contribution from plastic deformation is small and J values quadratically with LLD (since LLD varies linearly with load)

• This multi-specimen technique is very time-consuming and costly but it does work !

Begley, J.D. and Landes, J.D., “The J-integral as a Fracture Criterion” ASTM STP 514, 1972, pp. 1-24

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Resistance Curve (R-Curve)

• Many materials with high toughness do not fail catastrophically at a particular value of J or CTOD. They display a rising J or CTOD with crack extension, which is called R curve.

JR

a

JIc

Crack blunting

Crack initiation

ASTM E-1290: Standard test method for crack-tip opening displacement fracture toughness measurementASTM E-1820: Standard test method for fracture toughness

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R-Curve

JIC characterizes the material fracture toughness at the onset of ductile tearing (end of crack-tip blunting). Its value, however, is a somewhat arbitrary number since the onset of ductile tearing is nearly impossible to distinguish from continued blunting.

However, JIC is a conservative design limit since most J-R curve rises sharply beyond JIC.

The full J-a curve is more descriptive of the material fracture resistance.

At a larger a, the J dominance breaks down, J no longer characterizes the crack-tip stress-strain-displacement fields.

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R Curve: Tearing Modulus

• For ductile materials, the slope of the J-a curve indicates the relative stability of the crack growth. • The slope of the J-R curve is often given as the dimensionless tearing modulus.

RR

E dJT

da 2

0

, RJ J

Crack size

P1

P2

P3

12

3

4

• Under load control, the criterion for stable crack growth (instability occurs when the driving force curve becomes tangent to the J-R curve):

RJ J

RE dJ E dJ

da da 2 2

0 0

78Qian X/CE5887

Crack-Tip Constraints (1)

1.Is JIC independent of the specimen geometry?2.Is J-R curve unique for a given material???

J-dominance argument: if the J-R curve derives from two specimens that satisfy all the requirements for J-dominance, J for the same amount of (small) crack extension should not exhibit a dependence on the geometry, i.e., a unique J-R curve for a given material!

79Qian X/CE5887


Constraint: or sometimes termed as “plasticity constraint”, defines the

capability of the crack tip in constraining the plastic deformation around the crack tip within a localized volume of materials around the crack tip!

High-constraint: small-scale yielding configuration

Low-constraint: large-scale yielding condition

80Qian X/CE5887


600

500

400

300

200

100Specimen type

C(T)

Deep SE(B)

Shallow SE(B) SE(T)

DE(T)

A533 Grade B steel

JIC (kJ/m2)

• JIC does not indicate a clear dependence on the specimen geometry!• Reason being:

• JIC represents the energy for very small crack extension and for crack blunting• The process zone for JIC remains well within the finite strain zone• For high-constraint and low-constraint, the energies for crack blunting and crack extension within the finite strain zone are approximately the same

High-constraint specimensC(T), deep SE(B)

Low-constraint specimens:shallow SE(B), SE(T), DE(T)

81Qian X/CE5887


350

300

250

200

150

100

Tearing Modulus TR at a = 1 mm

Specimen type

C(T)

Deep SE(B)

Shallow SE(B)

SE(T)DE(T)

A533 Grade B steel

RR

E dJT

da 2

0

• The J-a curve rises more sharply for low-constraint conditions!• Since the low-constraint condition re-distributes the near-tip stresses to the adjacent materials, spreading the plastic zone to a large volume of material, the near-tip stresses in the low-constraint conditions remain relatively lower than those around a high-constraint crack tip subjected to the same remote J. Consequently, the material damage (separation) takes place at a slower rate than the high-constraint condition.

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SE(B) specimens

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Two-Parameter Crack-Tip Fields

• Under large deformation, the single-parameter approach using J or K breaks down in characterizing the crack-tip stress-strain-displacement fields• Extension of the fracture mechanics theory beyond the small-scale yielding condition often utilizes the linear-elastic T-stress or the elastic-plastic Q stress

84Qian X/CE5887


The linear-elastic T-stress is the second term in the Williams’ crack-tip stress solutions in isotropic materials, and represents a uniform stress acting parallel to the crack plane.

T-Stress

Iij ij

TK

fr

T

0 0

0 0 02

0 0

• For linear-elastic materials, the first term in the above equation exhibits the singularity.• The 3rd and higher order terms vanish near the crack tip, but the T-stress remains finite.• Research evidence reveals that the T-stress imposes a profound effect on the stresses deep within the plastic zone.

r1 /

85Qian X/CE5887


R

K-T field

10 2 3 4 5 61

2

3

4

5

HRRn = 10

/yy y

/yr J

/ yT +1.0

0-0.2

-0.4-0.6-0.8-1.0

• Negative T-stresses reduce significantly the opening stress within the plastic zone, while positive T-stresses impose very little effects on yy

• Negative T-stresses decreases the stress triaxiality near the crack tip larger plastic zone size!

Modified boundary layer model for mode I loading

86Qian X/CE5887


Biaxiality ratio :I

T a

K

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

-1

-0.5

0

0.5

1.0

a/W

SE(B)

SE(T)

DE(T)

M(T)

• Shallow cracked specimens often experience negative T-stresses• For M(T) specimens, = -1.0

87Qian X/CE5887


J-Q approach (1):

0ij ij ijT diff

Based on small-strain assumption, the crack-tip stress can be expressed as the summation of the crack-tip stress at T = 0 and the deviation from T = 0

Research evidence indicates the magnitude of the stress shift for nonzero T-stresses remains approximately constant with both distance and angular position in the forward sector of the crack-tip region.

10 2 3 4 5 61

2

3

4

5

HRRn = 10

/yy y

/yr J

/ yT

-0.8

01( )yy diff

2( )yy diff

1 2( ) ( )yy diff yy diff

( ) ( ) ( )yy diff xx diff xy diff

88Qian X/CE5887


J-Q approach (2):

O’Dowd and Shih designated the amplitude of this approximate difference field by the Q stress.

0ij ij ijT diff

00ij ij ijTQ

where,

0

0

yy yy TQ

at = 0 and 0 2

r

J

Q stress is a direct measure of the relative stress triaxiality (constraint) at the crack tip.

89Qian X/CE5887


-1 -0.5 0 0.5 1.0-2.0

-1.5

-1.0

-0.5

0.0

0.5

n = 3n = 5n = 10n = 20n = ∞

0/T

Q

Modified boundary layer model

R

K-T field

For well-contained plastic zones near the crack tip, there exists a unique relationship between Q and T-stress.

90Qian X/CE5887

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