Fracture Mechanics, Damage and Fatigue Non Linear Fracture ... · PDF fileUniversity of Liège Aerospace & Mechanical Engineering Fracture Mechanics, Damage and Fatigue Non Linear

University of Liège

Aerospace & Mechanical Engineering

Fracture Mechanics, Damage and Fatigue

Non Linear Fracture Mechanics – HRR Theory

Ludovic Noels

Computational & Multiscale Mechanics of Materials – CM3

http://www.ltas-cm3.ulg.ac.be/

Chemin des Chevreuils 1, B4000 Liège

[email protected]

Fracture Mechanics – NLFM – HRR Theory

http://www.ltas-cm3.ulg.ac.be/

From elastic fracture to elastic-plastic fracture

• Small Scale Yielding assumption– The process zone, which is the

region where the inelastic

deformations• Plastic flow,

• Micro-fractures,

• Void growth, …

happen, is a small region compared to

the specimen size, and is at the crack tip

• How can plasticity be accounted for?– One approach is the cohesive model

• Non-linearities are localized at crack tip

• Dugdale assumed a perfectly plastic material

• Dugdale model’s validity is limited to thin

sheets of low-C steel or glassy polymers

– The plastic zone has actually a different shape in most cases

• When plasticity is not localized, elasto-plastic solution has to be computed

syy

x

Asymptotic syy

True syy

Zone of asymptotic

solution (in terms of

1/r1/2) dominance

Structural

responsePlasticity

a rp

dt

2a rprp

Crack tipCohesive

zone tip

x

y

sy

dT

2016-2017 Fracture Mechanics – NLFM – HRR Theory 2

From elastic fracture to elastic-plastic fracture

• Dugdale’s model findings

– Thin sheets of perfectly plastic materials

– Crack in an infinite plane

• Size of plastic zone

• J is related to CTOD

– SSY approximation

• If s∞< 40% of sp0 & a > 5 rp then first order SSY holds

• If SSY, a > 5 rp corresponds to

• Use classic LEFM

– Effective crack length method

• If s∞< 60% of sp0 then a second order SSY assumption holds

with the effective length

• For a general crack configuration

a rp

dt

2a rprp

Crack tipCohesive

zone tip

x

y

sy

dT


• Material law (small deformations)

– Yield surface

f < 0: elastic region

f = 0: plasticity– Plastic flow

• Assumption: deformations can be added

• Normal plastic flow

• Von Mises surface with isotropic plastic flow (J2-plasticity)

– Deviatoric part of the stress tensor

– Yield surface

– Normality: since

– Then the plastic flow becomes

– Path dependency (incremental equations in d )

Elastoplastic behavior

True e

Tru

es

sTS

sp0

ep ee

E


• Material law: J2-plasticity with non-linear hardening

– Hardening law

• Parameters a & n

• Term sp0/E represents the elastic

deformation before yield

– Plastic strain during plastic flow

• This law can be inverted as

• As



0 2 4 6 8 100

0.5

1

1.5

2

e E/sp

0

s/ s

p0

a = 2; n = 1

a = 4; n = 1

a = 2; n = 4

a = 4; n = 4

a = 2; n = 7

a = 4; n = 7

s/

s 0

p

e E / s 0p

• Material law: J2-plasticity with non-linear hardening (2)

– Plastic strain

– Assume no unloading & plastic flow

• The d can be omitted:

• An internal potential can be defined



• Power law– This law can be rewritten in terms of the total deformations

• Yield stress is replaced by

equivalent stress

• Plastic strain is replaced by

equivalent strain

• The governing law becomes

or

• Parameter n

– n→∞: perfect plasticity

– n→1: “elasticity”

– Doing so requires 2 assumptions• There is no unloading

• As elastic strains are assimilated to plastic strains, the material is incompressible

– Which are satisfied if• We are interested only in crack initiation and not in crack propagation• The stress components remain proportional with the loading• Elastic deformations are negligible compared to plastic ones



0 2 4 6 8 100

0.5

1

1.5

2

e E/sp

0

s/ s

p0

a = 2; n = 1

a = 4; n = 1

a = 2; n = 4

a = 4; n = 4

a = 2; n = 11

a = 4; n = 11

s/

s 0

p

e E / s 0p

• Power law (2)

– Stress increment

• Using the governing law

• The strain increment is

• Or, as there is no unloading

– Internal potential

• No unloading:



• The HRR theory

– From Hutchinson, Rice and Rosengren, 1967-1968

– Assumptions

• Semi-infinite crack

• Loading increased monotonically only for crack initiation

• Power law only if plastic deformations are large compared to elastic ones

– J-integral

• Elasto-plastic model & these assumptions = non-linear elasticity

• The J-integral can be used

• It is path independent, so one may choose

HRR theory

x

y B

n

Gr

q


• Near fields

– Evaluation of J-integral

• As J is path independent, it is the same whatever the value of r

– So the integrant should be independent of r

– As the terms involve stress times strain, it yields

– Generally speaking, the stress tensor can be expanded in a power series

where depends only on q and s’

• Let s’ be the dominant exponent near the crack tip

• Therefore as &

the strain tensor becomes

– Using theses considerations the dominant exponent is:

HRR theory


HRR theory

• Near fields (2)

– Near fields are obtained using s’ = -1 / (n+1)

• Stress field

– kn is a plastic stress intensity factor which

» Allows defining independently of the loading amplitude

» Depends on n

– “Elasticity” (n→1): LEFM solution in 1/r1/2 recovered

– Perfect plasticity (n→∞): stress remains finite


• Near fields (3)

– Near fields are obtained using s’ = -1 / (n+1) (2)

• Stress field

• Strain field

– As &

– Eventually with

• Energy

HRR theory


• Solution in terms of J-integral

– Let us assume that the solution is known

&

•

• The J-integral is rewritten

where is deduced from

• Let us define

• So stress and strain fields can be formulated in terms of J

&

HRR theory


• Solution in terms of J-integral (2)

– Stress field reads

• J is governing the stress field near the crack tip

– It is the plastic analog of K in LEFM

– Assuming SSY and if the crack tip fields are governed by K

» Then a criterion on K is still valid

– If SSY does not hold, the criterion for crack initiation may be J ≥ JC

• N.B. if n→∞ only the strains will depend on J and not the stresses

– So J plays the role of an equivalent “plastic strain intensity factor”

HRR theory


• Solution in terms of J-integral (3)

– Validity of this solution

• Stationary crack (no propagation) as unloading is prohibited

• In the region of dominance of the HRR field

– Near the crack tip

– As n increases, this region decreases (FE simulations)

• Small deformations

• Incompressible materials as we are using a power law with deviatoric strain field

– Questions

• How do we compute In, , ... ?

• What is the shape of the process zone ?

HRR theory


• Resolution for a semi-infinite crack in plane e state

– Airy functions

• Linear momentum

– If b = 0, there exists an Airy function F:

– As we are not in linear elasticity we cannot say

• In polar coordinates

• As we are seeking a solution in r -1/(n+1), we choose

• Therefore, the following stress expressions are satisfying the balance equation

Semi-infinite mode I crack in plane e state


• Resolution for a semi-infinite crack in plane e state (2)

– Stream function

• As we are incompressible &

with plane strain assumption (if not: ezz≠0 )

• So displacements derive from a stream function Y with

• In polar coordinates

– As &

the displacements become

– Strains are obtained from

• As we are seeking a solution in r -n/(n+1), we choose




– Strain tensor

• Stream function

• So the strain components are

– Function f and g are dependant on each other as they have to satisfy

• The power law

• The normality equation




– Power law with normality

• These two equations lead to (see annex 1 for details)

–

– &

– Back to HRR field

• Near the crack tip, the fields were normalized by introducing an intensity factor

&

• Therefore, the functions f and g are rewritten

&



• Resolution for a semi-infinite crack in plane e state (5)– Using the new non-dimensional functions leads to

• &

•

– These equations can be reduced to a differential equation of the 4th order in

• See appendix I

• Range-Kutta resolution

• Initial boundary conditions

– Mode I symmetry

– Stress free lips

iteration on until these 2 conditions are satisfied

– It remains to choose




– Resolution of the plane e problem

• The differential equation is valid for any multiple of

– Initial condition is chosen so that

– The problem fields can now be determined

• &

• J characterizes the loading intensity on the crack

– Problem dependant

• In is a normalizing factor allowing to express the dependency with respect to J

– Still needs to be evaluated



0 50 100 1500

0.5

1

1.5

2

2.5

q [°]

s

srr

; n = 1.1

sqq

; n = 1.1

srq

; n = 1.1

se; n = 1.1

~

~

~

~

~

q [deg.]0 50 100 150

0

0.5

1

1.5

2

2.5

q [°]

s

srr

; n = 3

sqq

; n = 3

srq

; n = 3

se; n = 3

~

~

~

~

~

q [deg.]0 50 100 150

0

0.5

1

1.5

2

2.5

q [°]

s

srr

; n = 13

sqq

; n = 13

srq

; n = 13

se; n = 13

~

~

~~

~

q [deg.]

• Determination of stress and strain fields

– We sought the approximations &

• Similarly

• As J is path independent

– With obtained from

• Let us define , which

is evaluated numerically

– Eventually



0 5 10 150

1

2

3

4

5

6

7

nI n

Plane e

n


• Process zone shape

– The elastic-plastic boundary corresponds to

• As

• In terms of ,

the boundary is obtained for



• Process zone shape (2)

– The elastic-plastic boundary corresponds to

• In terms of ,

the boundary is obtained for


-10 -5 0 5

x 10-3

-2

-1

0

1

2x 10

-3

y

n = 1.1n = 3n = 13

~

~

x x 10-3

-0.2 -0.1 0 0.1 0.2

-0.2

-0.1

0

0.1

0.2

x

y

n = 1.1n = 3n = 13

r

~

~

~

q

x x 10-3

• Process zone shape (3)

– On particular interest is the size of the

process zone

• Ahead of crack tip (q=0)

• Behind crack tip (q=p)

• At its maximum value

– SSY assumption

• For linear elasticity

– n = 1 & a = 1

–

Plastic zone size

– Here we have assumed n=1/2


100

101

102

0

0.05

0.1

0.15

0.2

0.25

max

r

n

~1/(2p)


~

100

101

102

0

0.002

0.004

0.006

0.008

0.01

r

q=0

q=p

n

r

• Stress field at crack tip

– As &

the stress field at crack tip is

• The method does not remove the stress singularity (except for n→∞)

• As the abscissa is , it appears that J is a measure of the intensity

of the singular crack tip field (except for n→∞)



r0 0.02 0.04 0.06 0.08 0.1

0

2

4

6

8

10

r

syy

/ sp0

n = 1.1n = 3n = 13

~

syy

/ s

0p

r0 1 2 3 4

0

2

4

6

8

10

rs

yy/ s

p0

n = 1.1n = 3n = 13

~s

yy/

s 0

p

• Crack opening displacement (COD)

– Displacement field on lips:

• The COD is defined by

• Using

• It can be seen that

– The COD is of course >0

– Displacement along x is also >0

– So is not the opening

measured at r

– For Dugdale’s model, a CTOD has

been defined. Quid for HRR model?


x

y

uy

ux

r


-0.1 -0.08 -0.06 -0.04 -0.02 00

0.1

0.2

0.3

0.4

0.5

x

u s

p0/J

uy; n = 1.1

uy; n = 3

uy; n = 13

ux; n = 1.1

ux; n = 3

ux; n = 13

~x

us

0p

/ J

• Crack tip opening displacement (CTOD)

– The CTOD is defined as such

that a 90°-angle is intercepted

• Let r* be the value of r satisfying this criterion:

• As

• So


x

y

uy

ux

r*


• Crack tip opening displacement (2)

– The CTOD is defined as such

that a 90°-angle is intercepted

• Let

• So the CTOD becomes

– dn depends on

» n

» but also on

• As for a given material J is uniquely

related to the CTOD, this last one can

be used as a crack initiation criterion


x

y

uy

ux

r*


0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1/n

dn

asp

0/E = 0.001

asp

0/E = 0.002

asp

0/E = 0.004

asp

0/E = 0.008

0.78

1/n

• Summary

– Assumptions

• J2-plasticity with power law description

• Small deformations

• There is no unloading and loading is

proportional in all the directions (ok for crack

initiation and not for crack propagation)

• Elastic strains are assimilated to plastic strain

(material is incompressible)

• Semi-infinite crack

– HRR results for semi-infinite mode I crack

in plane e state

• Asymptotic stress, strain and displacement fields

– The J-integral plays the role of an

equivalent “plastic strain intensity factor”

– Stress field evolves in a proportional

way, so this is applicable to incremental

plasticity as long as J increases



0 5 10 150

1

2

3

4

5

6

7

n

I n

Plane e

n

0 50 100 1500

0.5

1

1.5

2

2.5

q [°]

s

srr

; n = 3

sqq

; n = 3

srq

; n = 3

se; n = 3

~

~

~

~

~

q [deg.]

• Summary (2)

– HRR results for semi-infinite mode I crack in

plane e state (2)

• Process zone

with

– If SSY

• CTOD

– New definition

–

– Also function of J

• We did not assume SSY !!!

– Questions

• What is happening for other configurations?

• What is the range of validity?


x

y

uy

ux

r*


-0.2 -0.1 0 0.1 0.2

-0.2

-0.1

0

0.1

0.2

x

y

n = 1.1n = 3n = 13

r

~

~

~

q

x

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

1/n

dn

asp

0/E = 0.001

asp

0/E = 0.002

asp

0/E = 0.004

asp

0/E = 0.008

0.78

1/n

• Mode I crack in plane s state

– Analysis is the same with other , & fields

Other HRR solutions

0 50 100 150-0.5

0

0.5

1

1.5

q [°]

s

srr

; n = 3

sqq

; n = 3

srq

; n = 3

se; n = 3

~

~

~~

~

Plane s

q [deg.]0 50 100 150

0

0.5

1

1.5

2

2.5

q [°]

s

srr

; n = 3

sqq

; n = 3

srq

; n = 3

se; n = 3

~

~

~

~

~

Plane e

q [deg.]


0 50 100 1500

0.5

1

1.5

2

2.5

q [°]

s

srr

; n = 13

sqq

; n = 13

srq

; n = 13

se; n = 13

~

~

~~

~ Plane e

q [deg.]

0 50 100 150-0.5

0

0.5

1

1.5

q [°]

s

srr

; n = 13

sqq

; n = 13

srq

; n = 13

se; n = 13

~

~

~

~

~

Plane s

q [deg.]

• Mode I crack in plane s state (2)

– Stress field is

– Asymptotic syy stress

• So for the same J, a thin specimen is less stressed at crack tip

Other HRR solutions


0 50 100 150-0.5

0

0.5

1

1.5

q [°]

s

srr

; n = 3

sqq

; n = 3

srq

; n = 3

se; n = 3

~

~

~~

~

Plane s

q [deg.]0 50 100 150

0

0.5

1

1.5

2

2.5

q [°]

s

srr

; n = 3

sqq

; n = 3

srq

; n = 3

se; n = 3

~

~

~

~

~

Plane e

q [deg.]

0 5 10 150

1

2

3

4

5

6

7

n

I n

Plane e

Plane s

n

• Mode I crack in plane s state (3)

– Crack Tip Opening Displacement

•

Other HRR solutions

x

y

uy

ux

r*


0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1

1/n

dn

asp

0/E = 0.001

asp

0/E = 0.002

asp

0/E = 0.004

asp

0/E = 0.008

Plane e

1/n

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1

1/n

dn

asp

0/E = 0.001

asp

0/E = 0.002

asp

0/E = 0.004

asp

0/E = 0.008

Plane s

1/n

• Mode I crack in plane s state (4)– Process zone shapes

• The process zones

– Have a different shape

– Are more diffuse in plane s

• Mode III

– In SSY, the process zone has a circular

shape

• Perfectly plastic

• With hardening

Other HRR solutions


-0.2 -0.1 0 0.1 0.2

-0.2

-0.1

0

0.1

0.2

x

y

n = 3; Plane e

n = 3; Plane s

~

~

x

• Mode II and mixed mode crack

– In SSY, the solution depends on the

elastic mixity parameter

– Examples

• Process zones in plane e state

• SSY

Other HRR solutions


𝑥𝜎𝑝02

𝐾𝐼2 + 𝐾𝐼𝐼

2

𝑦𝜎𝑝02


2n=1

n=3

n=13

𝑀𝑒=0.7

𝑥𝜎𝑝02


2

𝑦𝜎𝑝02


2n=1

n=3

n=13

𝑀𝑒=0

• Compressibility

– Elastic part of the deformations in not incompressible

– Considering this effect will diffuse the plastic zone

• Example: Mode I, plane e state & SSY

• The plastic zone size rp is defined as the

length of the plastic zone ahead of the crack

Other HRR solutions


rp

𝑥𝜎𝑝02

𝐾𝐼2

𝑦𝜎𝑝02

𝐾𝐼2

n =0.4

n =0.3

n =0.1

• HRR solution in SSY can explain 3D effect

– Let us assume that the solution is

• Plane-s-like near free surfaces

• Plane-e-like near mid-plane of specimen

– Even if K and s are not constant through

the thickness, it remains that

• There is a transition in the plastic zone

shape and size

• This transition is responsible for shear lips

• Shear lips form a 45-degree-angle since

szz=0 (so max shear at 45° in plane Oyz)

What can be learned from these models

Mid

surface

Free

surface

x

y

z

x

y

z


-0.2 -0.1 0 0.1 0.2

-0.2

-0.1

0

0.1

0.2

x

y

n = 3; Plane e

n = 3; Plane s

~

~

x

• HRR solution in SSY can explain 3D effect (2)

– During a toughness test, the K measured is

an average one

• It is more important for thin specimen

• The process is never really plane e

– There are actually complex 3D effects

• In SSY, even for a thin specimen, near the

mid-plane a plane-e-state is developed

• If the load increases, the plastic zone can

be plane-s-like near the mid-plane

– Effect of T-stress should also be considered

• Recall T-stress is the 0-order term obtained

with the asymptotic solution, which is dominant

at radius rc

• In general, if the test is such that T < 0, the

measured fracture K will be larger than for

T > 0, independently of the thickness *

• For ASTM toughness tests, the thickness

is large so that T > 0


2aT>0

*D.J Smith, M.R Ayatollahi and M.J Pavier,Proc. R. Soc.

A, 2006, vol. 462, pp 415-2437

t

Kc

K rupture

Plane s

Plane e

t1

t2<<t1

t3<t2

t


• Effective crack length for SSY

– If SSY assumption holds

• J can be expressed in terms of K

• Then the plastic size can be written

• However, there are dependencies on

– Parameters n & n

– Whether it is plane e or plane s

– Rice’s model for perfectly plastic material

• Intersection of the linear elastic solution

with the yield stress leads at h rp

• But there is a redistribution of the stresses

so that the traction remains the same

– First order approximation: outside

the plastic zone, the stress

distribution is shifted


x

syy

sp0

h rp

LEFM asymptotic

syy in r-1/2

Redistributed syy

rp


rp

𝑥𝜎𝑝02

𝐾𝐼2

𝑦𝜎𝑝02

𝐾𝐼2

n =0.4

n =0.3

n =0.1

• Effective crack length for SSY (2)

– Rice’s model (2)

• 2 equations with 2 unknowns

• So the plastic zone size is &

• So everything is as if the crack had an effective length a +h rp = a + rp/2

– From HRR models, numerical simulations, etc …

• Considering blunting, compressibility, hardening, …, an estimation is


x

syy

sp0

h rp

LEFM asymptotic

syy in r-1/2

Redistributed syy

rp


• Effective crack length for SSY (3)

– If s∞< 50% of sp0 then a second order SSY assumption holds

• The cohesive zone remains small compared to crack size

• The effective crack size can be stated as a +h rp = a + rp/2

with for cracks in finite plates

• So there is an iterative procedure to follow:

– a) compute K from a

– b) compute effective crack size

– c) compute new K from aeff and back to b) if needed

– This method is a correction for linear fracture mechanics, but does not allow

considering problems with large yielding



• Validity in SSY

– We have two asymptotic solutions

• HRR field is valid in the process zone

• LEFM is still valid in the elastic zone close to the crack tip

– Conditions

• This is the case if all sizes are 25 times larger than the plastic zone

Validity of HRR field

syy

x

LEFM asymptotic

syy in r-1/2

True syy

Plasticity

HRR asymptotic

syy in r-1/(n+1)

Log

syy

LEFM

asymptotic

syy in r-1/2

HRR asymptoticsyy in r-1/(n+1)

1

2

1n+1

Log r

t

a L


• Validity in SSY (2)

– Crack initiation criteria

• Criteria based on J or dt are valid: J ≥ JC or dt ≥ dC

– J & dt depend on a, the geometry, the loading, …

• But as the LEFM solution holds, we can still use K(a)≥KC

– Might be corrected by using the effective length aeff if s∞< 50% of sp0

– Sizes for K–based toughness test

• Examples:

– Titanium alloy 6%Al-4%V

» Yield: 830 MPa

» Toughness: 55 MPa · m½

» a, t, L > 1.1 cm

– Strength steel

» Yield: 350 MPa


» a, t, L > 1.27 m !!!


t

a L


• Validity in elasto-plastic conditions– Deformations are small

– We still have one asymptotic solution valid

• HRR field is valid in the process zone

– LEFM is NOT valid in the elastic zone close to the crack

– Conditions

• This is the case if all sizes are 25 times larger than CTOD


syy

x

LEFM asymptotic

syy in r-1/2

True syy

Plasticity

HRR asymptotic

syy in r-1/(n+1)

Log

syy

LEFM

asymptotic

syy in r-1/2


1

2

1n+1

Log r

t

a L


• Validity in elasto-plastic conditions (2)

– Crack initiation criteria

• Criteria based on J or dt are valid: J ≥ JC or dt ≥ dC

– J & dt depend on a, the geometry, the loading, …

• The LEFM solution DOES NOT hold, we CANNOT use K(a)≥KC

– Sizes for J–based toughness test

• If KC is computed from JC:

• Examples:

– Titanium alloy 6%Al-4%V

» Yield: 830 MPa


» Young: 110 GPa

» a, t, L > 0.75 mm

– Strength steel

» Yield: 350 MPa


» Young: 210 GPa

» a, t, L > 1.93 cm


t

a L


• Validity in large yielding

– Example: ligament size is too small

– Small deformations assumption does not hold

– Neither HRR field nor LEFM asymptotic fields are valid

– Crack initiation criterion?

• As there is no zone of J-dominance can J still be used?


syy

x

LEFM asymptotic

syy in r-1/2

True syy

Plasticity

HRR asymptotic

syy in r-1/(n+1)

Log

syy

LEFM

asymptotic

syy in r-1/2


1

2

1n+1

Log r


• Validity in large yielding (2)

– Plastic strain concentrations depend on the experiment

– Zones near free boundaries or other cracks tend to be less stressed

– Crack initiation criterion?

• Solution is no longer uniquely governed by J

• Relation between J & dt is dependant on the configuration and on the loading

• The critical JC measured for an experiment might not be valid for another one

• A 2-parameter characterization is needed


s

s

M

M

s

s


• Thick specimen with centered crack:

– Steel

• Yield: 350 MPa

• Toughness: 250 MPa · m½

• Young: 210 GPa

• Hardening exponent >>

– Loading

• P = 25, 50 & 150 kN

– Compute the stress intensity factor

in terms of crack size a

• Compare the solutions obtained by

– SSY

– Effective crack length

• Check validity

Exercise 1: Specimen with centered crack

2a2W=1.5 in.

=3.8 cm

L = 3 in.= 7.62 cm

t = 0.5 in

=1.27 cm

P P


• A533-B1 steel:

– Exhibit brittle to ductile transition fracture

– Upper-shelf: Ductile fracture

• Observation: voids separated by 50 mm at T > 0°C

– Lower-shelf:

• Assumption: slip-induced cleavage at T< -30°C

• Critical fracture at

– 1830 MPa at strain rate = 0.01/s

– 2000 MPa at strain rate = 5/s

– Un-irradiated material

• Quasi-static, room temperature tensile test

– Yield sp0 = 481 MPa

– Ultimate sU = 642 MPa at e = 0.254

– Grain-size = 25 mm

• Yield (ksi, Rankine) in terms of temperature

and strain rate

– Questions

1) KIC in terms of temperature & strain rate in lower-shelf region

2) Validity?

Exercise 2: Nuclear reactor vessel


References

• Lecture notes

– Lecture Notes on Fracture Mechanics, Alan T. Zehnder, Cornell University,

Ithaca, http://hdl.handle.net/1813/3075

• Other references

– « on-line »

• Fracture Mechanics, Piet Schreurs, TUe, http://www.mate.tue.nl/~piet/edu/frm/sht/bmsht.html

– Book

• Fracture Mechanics: Fundamentals and applications, D. T. Anderson. CRC press, 1991.

• Fatigue of Materials, S. Suresh, Cambridge press, 1998.


http://hdl.handle.net/1813/3075

http://www.mate.tue.nl/~piet/edu/frm/sht/bmsht.html

• P= 50 kN

– Far stress field

– Solution for a = 7 mm

• SIF

with

Exercise 1: Solution

2a2W=1.5 in.

=3.8 cm

L = 3 in.= 7.62 cm

t = 0.5 in

=1.27 cm

P P


• P= 50 kN (2)

– Solution for a = 7 mm (2)

• Effective crack length

– First iteration

– Second iteration

– It has converged

• The correction is of about 1%

• Validity:

– s∞< 0.5 sp0: OK

– a = 7 mm, t = 12.7 mm, & W-a = 12 mm > 25 rp = 6.2 mm: OK



• P= 50 kN (3)

– For other a

• For a = 9.2 mm

– rp = 0.383 mm 25 rp = 9.6 mm > a : solution not valid anymore


rp > W-a


0 0.005 0.01 0.015 0.020

20

40

60

a [m]

K [

MP

a . m

1/2

]K(a); P=50kNK(a

eff); P=50kN

a [m]0 0.005 0.01 0.015 0.02

0

0.01

0.02

0.03

0.04

0.05

a [m]

[m]

25 rp; P=50kN

a; P=50kN

W-a; P=50kN

a [m]

• P= 25 kN

– Solution

• For a = 13.7 mm

– rp = 0.235 mm 25 rp = 5.9 mm > W- a = 5.3 mm: solution not valid

anymore


rp > W-a


0 0.005 0.01 0.015 0.020

5

10

15

20

25

30

35

a [m]

K [

MP

a . m

1/2

]

K(a); P=25kNK(a

eff); P=25kN

a [m]0 0.005 0.01 0.015 0.02

0

0.005

0.01

0.015

0.02

0.025

a [m]

[m]

25 rp; P=25kN

a; P=25kN

W-a; P=25kN

a [m]

[m]

• P= 150 kN

– Solution

• For this loading

– 25 rp > a always

– As s∞= 0.88 sp0 SSY theory does not hold, the size of the plastic zone

cannot be approximated by


rp > W-a


0 0.005 0.01 0.015 0.020

50

100

150

a [m]

K [

MP

a . m

1/2

]K(a); P=150kNK(a

eff); P=150kN

a [m]0 0.005 0.01 0.015 0.02

0

0.01

0.02

0.03

0.04

0.05

a [m]

[m]

25 rp; P=50kN

a; P=50kN

W-a; P=50kN

a [m]

• Hardening law

– Quasi-static, room temperature tensile test

• Yield sp0 = 481 MPa

• Ultimate sU = 642 MPa at e = 0.254

• Power law fit:

– For a = 1, n = 16

e (s = 642 MPa) = 0.244

– To use HRR theory, we take n = 10


s

ep

0.002 0.254-ee

642

481

Suresh, Subra, 3.35 Fracture and Fatigue, Fall 2003. (Massachusetts Institute of Technology: MIT OpenCourseWare),

http://ocw.mit.edu (Accessed 14 Jul, 2010). License: Creative Commons BY-NC-SA


• HRR theory

– Small-scale HRR (no ductile failure)

• Stress field

• Plane strain assumption

– in small-scale

– For a = 1, n = 10: In ~ 4.5

– Mode I

~2.5



0 50 100 1500

0.5

1

1.5

2

2.5

q [°]

s

srr

; n = 13

sqq

; n = 13

srq

; n = 13

se; n = 13

~

~

~~

~

q [deg.]

• Fracture by slip-induced cleavage

– Small-scale HRR:

– We know that

• Critical fracture from experiments at

– 1830 MPa at strain rate = 0.01/s

– 2000 MPa at strain rate = 5/s

– These values correspond to syy at a characteristic micro-scale distance

r = twice the grain size = 50 mm *

– KIC (Mpa m1/2):

• Quasi-static (strain rate = 0.01/s):

• Dynamics (strain rate = 5/s):


* R.O. Ritchie, J.F. Knott, J.R. Rice, J. Mech. Phys. Solids 21, 395-410, 1973


• Fracture by slip-induced cleavage (2)

– KIC (Mpa m1/2):



• In these expressions, sp0 should be expressed in Mpa

– In English unit

– Temperature from Rankine to Celsius

– Stress



• Fracture by slip-induced cleavage (3)

– KIC (Mpa m1/2) (2):




T [°C]

KIC [MPa m1/2]

-200 -30

3.53.1

36

32

5/s

0.01/s


• Validity

– Upper shell: voids growth small-scale HRR not valid

• We need a model including damage **

– Lower shell

• To use SIF: maximal size for max (KIC/sp0)2

– QS test and T =-30°C : (KIC/sp0)2 = 0.0037 m

– a, t, L > 2.5 (KIC/sp0)2 = 9.3 mm

• To use J: maximal size for max (KIC2/Esp

0)

– QS test and T =-30°C : (KIC2/Esp

0) = 1.1 10-5 m

– a, t, L > 25 (KIC2/Esp

0) (1-n2) = 0.25 mm


** R.O. Ritchie, A.W. Thompson, Metallurgical transactions A, 16A, 233-248, 1985


• Resolution for a semi-infinite crack in plane e state (A1)

– Power law with

• Deviatoric tensor, plane e & incompressible material:

– As or

– Out of plane stress &

– Non zero components of deviatoric tensor:

• Equivalent von Mises stress

– In terms of r and f

Annex 1: Mode I crack and HRR theory



– Power law (2)

• Equivalent deformation in terms of r and g

• So the power law becomes

• We have a differential equation in terms of functions depending on q & n

• But we are still missing a relation between the functions f & g

We will now study the normality relation




– Normality

• Strain-stress relations &

imply

• In terms of f & g:

– Back to HRR field

• Near the crack tip, the fields were normalized by introducing an intensity factor

&

• Therefore, the functions f and g are rewritten

&

• Which allows writing the differential equations in terms of &



• The two differential equations are

– &

–

– Let then, these equations become

•

•



• Theses equations can be reduced to a single ODE as

–

– After differentiating and

substituting and its derivatives, it yields the fourth order ODE



• Semi-infinite mode I crack in plane e state

– Von Mises stress (Plane e)

– Perfectly plastic material

• Von Mises stress:

• The Mohr’s circle is centered on with radius

– In Region II in frame• er, eq, we have pure shearing in the directions 0 and j=p/2

• ex, ey, we have pure shearing in the directions q and j=q+p/2

Annex 2: Slip line solution for perfectly plastic material

s

srq

2j


0 50 100 1500

0.5

1

1.5

2

2.5

q [°]

s

srr

; n = 13

sqq

; n = 13

srq

; n = 13

se; n = 13

~

~

~~

~ I IIIII

q [deg.]

• Semi-infinite mode I crack in plane e state (2)

– In Region I for q → 0: in frame

• er, eq, we have pure shearing in the

directions j=p/4 and j=-p/4

• ex, ey, we have pure shearing in the


– In Region I for q → p: in frame

• er, eq, we have pure shearing in the


• ex, ey, we have pure shearing in the

directions j=-3p/4 and j=3p/4

– Slip directions are the ones of maximal shearing


s

srq

-2j

2j


0 50 100 1500

0.5

1

1.5

2

2.5

q [°]

s

srr

; n = 13

sqq

; n = 13

srq

; n = 13

se; n = 13

~

~

~~

~ I IIIII

q [deg.]

• Mode I crack in plane s state

– Slip lines (perfectly plastic material)



Documents

Fracture Mechanics, Damage and Fatigue Non Linear Fracture ... · PDF fileUniversity of Liège Aerospace & Mechanical Engineering Fracture Mechanics, Damage and Fatigue Non Linear