Elastic Plastic Fracture Mechanics

8/2/2019 Elastic Plastic Fracture Mechanics

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Chapter 9: Elastic Plastic Fracture Mechanics

9.1 Crack Tip Opening Displacement

9.1.1 CTOD by Elastic Approach9.1.2 CTOD by Strip Yield Model9.1.3 Alternate Definition of CTOD9.1.4 Measurement of CTOD

Hinge Model Modified Hinge Model

9.2 J-Integral (Energy release rate)9.2.1 Definition9.2.2 Computation of J

Analytical approachExperimental approach

9.3 Crack Growth in Elastic Plastic Materials9.3.1 Criteria for Crack Growth9.3.2 J-R curves9.3.3 Stability of Crack Growth and Tearing modulus

9.4 Summary

Applies to materials that exhibit time independent nonlinear behavior.Example - Elastic-Plastic deformation


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9.1 Crack Tip Opening Displacement

A. A. Wells (1961) Cranfield, U.KFracture tests on structural steels were found to be too tough to characterize by theLEFM. Wells noticed that the crack faces move apart before fracture. Based on theseobservations, he proposed Crack Tip Opening Displacement as a fracture criteria.

9.1.1 CTOD by Elastic Approach

Assuming the effective crack length is based on the Irwins plastic correction,The opening displacement, is:

Where ry = rp* =

1

2KI

2

ys2 , for P condition.

= 3 4, for P condition

= 3

1+, for P condition

=E

2(1+ ), Shear modulu

CTOD =4

GI

= 2uy = 2 +1KI ry

2

CTOD= = 2vy =4

KI

2

ysE


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9.1.2 CTOD by Strip Yield Model

= 8

ysaE

lnSeca

2ys

Replacing the logarithmic term ln by series, after simplification we get:

= KI2

ysE1 + 1

62ys

2

+ ...

CTOD= =KI2

ysE=GI

ys, for P condition

CTOD= =KI

2

mysE=

GI

mysm is a non-dimensional factor,m =1 for p-Stress

m = 2 for P-strain


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9.1.3 Alternate Definitions of CTOD

(a) Displacement @ the original crack tip(b) Displacement at the intersection of90o points from the advancing crack tip.

These methods are commonly used in finite element analyses for estimation ofCTOD. The two methods become identical if the crack blunts in a semicircle.


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9.1.4 Measurement of CTOD using Three-Point Bend Specimen

(a) Hinge Model Extend crack flanks to an intersection point. Calculate the rotation factor, r From similar triangles

r Wa( )=

V

r W a( )+ a

Or

= r W a( )Vr W a( ) + a

r(w-a)

a

(b) Modified Hinge Model

The hinge model is inaccurate when displacements are primarily elastic.

= el +p =KI2

mysE* +

rp Wa( )V

rp W a( ) +aP Condition : m =1 & E

* =E

P Condition : m=2 & E* = E

1 2


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Eshelbys Conservation of Energy Theorem: In a singly connecteddomain (no singularities), the rate of change of potential energy () is zero.

9.2 The J-Integral

9.2.1 Definition

ds

Tx2

x1Where I, j, and k= 1,2 andUis the strain energy density defined by

U( ) = ijd ij0

J=Jx1 = Un1 ijuix1

nj

ds

J=Jx1 = Un1 ij uix1nj

ds = 0

J= Udy ijuix1

njds

= Q

Consider the crack extension in x1 or x direction, then

Cherapanov (1966) and Rice (1967) applied the concept ofconservation of energy principles to crack problems andshowed that the Jx integral is independent of the contour chosen(path independent) and it measures the severity of the crack tipif the integral is taken around a crack tip.

x1

x2A

C

B

n

ds


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9.2.2 J-integral is a variation of total potential energywith respect to direction of crack growth

JXk =Xk

, where =UV

U= Strain EnergyV = Potential Energy due to applied load

nids =

x i

dAA

+ x i

dAA

Divergence Theorem: x1

x2A

C

B

n

ds n1

n2

ds dx2

dx1

n1

n2

Consider the Jxk integral, the change of potential energy for a unittranslation in xk direction of the closed region .

Jxk =

Unk ij

uixk

nj

ds

Jxk = U xkA dA ij

xjA

ui xk

dA

ij xkA

ui xk

dA

Apply divergence theorem

Interchange dummy variable (j & k), we get

Jxk = U xkA dA

xk

ij ui xj

dA

OR

Jxk =

xkUV( )dA

A

=

xk


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9.2.3 Graphical Interpretation of J-Integral

A

B

ELoad

P

Displacement,v

P

a

a+a

J= a

Fixed P

= a

Fixed v

J=va

0

P

dP Fixed P

J= Pa

0

v

dv Fixed v

For elastic materials, J = G, strain energy release rate.


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9.2.4 Path Independence of J

Consider a contour integral D-C-B-A-F-E-D, Since it is a closed pathand does not include singularities, the total integral is zero.

x1

x2A

C

B

D

FE

1

I= QDEF + Q

FA + Q

ABC + Q

CD = 0

For traction free crack problems,crack face integrals are zero.

I= QDEF + Q

ABC = 0 = Q

DEF Q

CBA

J= QDEF = Q

CBA =

x

= a

or

Note: dx = da

Therefore, J is defined as the rate of change of total potential energywith respect to the crack length.The PE includes elastic and elastic plastic energy.

Thus J-integral is path independent.


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9.2.5 Calculation of J Integral

(a) Analytical approachExamples: Elastic (Cracked strip)

Elastic-plastic (Dugdale model)

(b) Experimental approachArea between the load-deflection curvesfor crack lengths a and a+da)


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Elastic Problem: Semi-infinite crack in an infinite strip of thickness2h subjected to uniform displacement in thickness direction.

EF oA

B C

D

x1x2

vo

v0

x1 = -x1 =

2h

Let the displacement @ x2 = h be vo.

1. Select the contour path, OA, AB, BC,

CD, DE, EF, and FO.

2. Create a table of normal vectorsdisplacements, and stresses.

3. From the table it is clear that only line integralon the path CD will contribute to the J.

First term:

Jx1 = Un1 ijuix1

nj

ds

CD

U= 121111 + 22 22 +1212( ) =Evo

2

2h2

Evo

2

2h2

CD

ds =Evo

2

2h2 ds

h

h

=Evo

2

h

Second term: 11u1x1

n1 + 12u1x1

n2 +21u2x1

n1 + 22u2x1

n2 = 0 + 0+ 0 + 0

J=Evo

2

h


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Dugdale Strip Yield Example

Consider a contour G within the yield zone and on the top and bottomsurfaces of the crack faces. That is the distance traversed in x2-directionis zero.

Jx 1 = Un1 ijui

x1nj

dsCD

The contour path is along A0, OB and BA. On the path AO and OB

Stresses: 11 = 12 = 0 and 22 = ys

Normal Vectors:n1 = 0 and n2 = -1 on AOn1 = 0 and n2 = 1 on OBn1 = -1 and n2 = 0 on BA (zero distance)

Because n1 is zero on both AO and OB, the integral ofstrain energy density is zero, only the second part of the integralcontributes to the J-integral.

2a

A

B

O

d

Contour pa


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J= 22u2x1

n2ds

J= u2

+

x1u2

x1

dx10

d

J= (u2

+ u2)

x1

dx1

0

d

Because = u2+ u2

J= d(u 2+ u2

)

0

d

= + ( )d0

d

For elastic - plastic materials, ( ) = ys, then

J= yso,

Where o is the crack tip opening displacement.

2a

A

B

O

d

Contour pa


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Experimental Measurement of J

Landes & Begleys Method

J = -1

BUa( )fixed


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9.3 Crack Growth in Elastic-Plastic Materials

Stable

crackgrowth

aaf

a

JR-curve

Rf

JR

No Fracture

FractureJR CurveCrack growth criteriaInitiation of growth:

J Jo

During the stable growth: J JR

Stability of Crack Growth

Crack growth is stable if the rate of change of J W.r.t a is less thanthe rate of change of JR with respect to a, crack length.

Crack growth is stable ifdJ

da

< dJR

da

Crack growth is unstable ifdJ

da

dJR

da

Tearing Modulus:

Paris, Tada, Zahoor & Ernst defined the J-R stability equationin a non-dimensionalized form as

Crack growth is unstable ifE

ys2

dJ

da

E

ys2

dJR

da

Tearing modulus T Tmat, Tearing resistance of the material.


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Hutchinson and Rice & Rosengren independently showed that J characterizes thecrack tip condition in a nonlinear elastic material. They assumed power lawrelationship between stress and plastic strain. If the elastic strains are included, theuniaxial stress-strain equation is given by

J as a Stress Intensity Parameter

o

= o

+o

n

Ramberg-Osgood Equation

HRR showed that for J to be path independent, stress and strain must vary as 1/r.In addition, near the crack tip elastic strains are small compare to total strains andthus the stress-strain equation reduces to a simple power law.From these 2 conditions

ij = k1J

r

1

n+1ij = k1

J

r

n

n+1

k1 and k2 are constants. The above equation reduce to square root singularity for n=1.

ij = oEJ

2Inr

1

n+1 ij (n, ) ij =

oE

EJ

2Inr

n

n+1 ij (n, )

J define s the amplitude of the HRR singularity. A structure in a small scale yielding has 2singularity dominated zones. Elastic zone and plastic zone.

Power law eq.:

o

=o

n

Under J field


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Elastic Plastic Fracture Mechanics

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