Fracture Mechanics, Damage and Fatigue Linear Elastic Fracture … · 2016. 1. 28. · Resolution...

University of Liège

Aerospace & Mechanical Engineering

Fracture Mechanics, Damage and Fatigue

Linear Elastic Fracture Mechanics - Asymptotic Solution

Fracture Mechanics - LEFM - Asymptotic Solution

Ludovic Noels

Computational & Multiscale Mechanics of Materials – CM3

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

Chemin des Chevreuils 1, B4000 Liège

[email protected]





LEFM assumptions

• Balance of body B – Momenta balance

• Linear

• Angular

– Boundary conditions • Neumann

• Dirichlet

• Small deformations with linear elastic, homogeneous & isotropic material

– (Small) Strain tensor , or

– Hooke’s law , or

with

– Inverse law

with

b

T

n

2m l = K - 2m/3

2013-2014 Fracture Mechanics - LEFM - Asymptotic Solution 2

Resolution method

• Examples of problems to be solved

– Static

– 2D &

• Plane s

• Or plane e

2a 2b x

y

s∞

s∞

syy syy

y

2a x

s∞

s∞

syy syy

2a x

y

s∞

s∞

syy syy


Resolution method

• Static & 2D assumptions – Plane s ( ) or plane e ( & )

– Hooke’s law

• Plane e :

&

• Plane s :

&

– Greek subscripts substitute for x or y (Roman for x, y or z)


Resolution method

• Static & 2D assumptions (2) – Plane s ( ) or plane e ( & )

• Linear momentum

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

• Compatibility

– Plane e :

• In terms of Airy function

• Compatibility

– Plane s :

• In terms of Airy function

• Compatibility

– For both plane states:


Resolution method

• Application: infinite plate with a hole

– Cylindrical coordinates

• Boundary conditions

– At r = a

– At r = b→∞: only syy= s∞ is non zero

y

2a x

s∞

s∞

syy syy

y

s∞

s∞

r=a x

r q

r =

b→∞

Load case 1

independent

from q

Load case 2

dependent on cos 2q (symmetry with

respect to x)


Resolution method

• Application: infinite plate with a hole (2)

– For both loading cases: find potential F such that

• Bi-harmonic equation is verified

• Stress field satisfies the boundary conditions

– Superposition of cases 1 and 2 (see appendix 1)

• Stresses:

• For q = 0: syy= sqq

• Stress intensity factor: Kcircle = 3 0

0.5

1

1.5

2

2.5

3

0 2 4 6

x/a

syy

/s∞


Resolution method

• Application: infinite plate with an elliptical hole

– Elliptical coordinates

with, on the boundary,

– Resolution using complex functions of z (see next slides)

with the complex variable

• 1913, Inglis

– Stress intensity factor

with r the curvature radius at the tip

– Singularity when r → 0

2a 2b x

y

s∞

s∞

syy syy

x

y

a b


Asymptotic solution

• Fracture modes

– 1957, Irwin, 3 fracture modes

• Boundary conditions

Mode I Mode II Mode III

(opening) (sliding) (shearing)

y

x

r q


or or

y x

z

y x

z

y x

z


Asymptotic solution

• Method for linear and elastic materials – Use of

• Complex variables

• Complex functions

with a & b real functions

– If z is differentiable (analytic)

• This yields the Cauchy-Riemann relations

&

• And the functions satisfy the Laplace equation

&

2a

x

y

syy syy

y

x

r

q

z

z


Asymptotic solution

• Method for linear and elastic materials (2) – What becomes the bi-harmonic equation?

• Change of variables: x, y → z, z

• The Airy function F(x, y) = F(z, z)

– Let us define Y such that:

– The bi-harmonic equation is rewritten

– Since Y satisfies the Laplace equation, it is the real part of a function z(z):

with z = z(z) a function of z only

» But

– Let 4W’(z) = z(z) and w(z) be the unknown functions

which satisfies

N.B. W = W(z) and w = w(z) are functions of z only


Asymptotic solution

• Method for linear and elastic materials (3)

– What become the stresses?

• Since , &

• The stresses are rewritten


Asymptotic solution


– Displacements can be deduced from the stresses

• Complex form:

• Using Hooke’s law (see appendix 2)

–

– Plane s :

– Plane e:


Asymptotic solution


– Choice of complex functions

• Owing to the discontinuity for q = p we choose

Indeed, for l=-1/2 the displacement is discontinuous

• The exact solution is actually a series in u = Sl ….

• u is finite l >-1

y

x

r q


Asymptotic solution

• Application to mode I (opening)

– Mode I:

– Symmetry: &

&

or y

x

r q


Asymptotic solution

• Application to mode I (opening) (2)

– Stresses

– Boundary conditions

y

x

r q


Asymptotic solution


– Solution

• System of equations

• Non trivial solution

l = n/2, with n = -1, 0, 1, …. since l >-1

• Displacements in rl+1 and stresses in rl

the dominant term near the crack tip is obtained for l=-1/2

only this term is considered in the asymptotic solution

• Determination of the constants:

&


Asymptotic solution


– Asymptotic solution (terms of higher order, l = 0, 1/2, …, neglected)

• Displacement field

• Stress field

y

x

r q


Asymptotic solution


– Asymptotic stress field

• For mode I, the stress concentration

is obtained for syy(q = 0)

• As the stress tends to infinity,

the Stress Intensity Factor (SIF) is defined as

• C1 & so KI depend on both the geometry and loading

• Unit of the SIFs in Pa m1/2

y

x

r q


Asymptotic solution


– Asymptotic displacements


-0.1

0

0.1

-0.1

0

0.1-0.5

0

0.5

xy

ux E

/(K

I(1+

))

uxE

/(K

I(1+

))

-0.1

0

0.1

-0.1

0

0.1-0.5

0

0.5

xy

uy E

/(K

I(1+

))

opening uyE

/(K

I(1+

))

Asymptotic solution



-0.1

0

0.1

-0.1

0

0.1-5

0

5

xy

sxx

/KI

-0.1

0

0.1

-0.1

0

0.1-5

0

5

xy

sxy

/KI

-0.1

0

0.1

-0.1

0

0.1-5

0

5

xy

syy

/KI


Asymptotic solution


– Validity of the asymptotic solution

• This solution is only valid in

a restricted region

syy

x

Asymptotic syy

True syy

Zone of asymptotic

solution (in terms of

1/r1/2) dominance

Structural

response Plasticity


Asymptotic solution

• Application to mode II (sliding)

– Mode II:

• Proceeding as for mode I, with new BCs (C1 = C3 = 0)

&

or

y

x

r q


Asymptotic solution

• Application to mode II (sliding) (2)

– Asymptotic displacements


-0.1

0

0.1

-0.1

0

0.1-0.5

0

0.5

xy

ux E

/(K

II(1

+

))

sliding uxE

/(K

II(1

+

))

-0.1

0

0.1

-0.1

0

0.1-0.5

0

0.5

xy

uy E

/(K

II(1

+

))u

yE/(

KII(1

+

))

Asymptotic solution

• Application to mode II (sliding) (3)


-0.1

0

0.1

-0.1

0

0.1-5

0

5

xy

sxx

/KII

-0.1

0

0.1

-0.1

0

0.1-5

0

5

xy

syy

/KII

-0.1

0

0.1

-0.1

0

0.1-5

0

5

xy

sxy

/KII


Asymptotic solution

• Application to mode III (shearing)

– Mode III:

– Equations

• Hooke’s law

• Equilibrium

– Laplace equation satisfied uz is the imaginary part of a function z(z)

• Choice of function

is anti-symmetric

• Cylindrical CV

y

x

r q


Asymptotic solution

• Application to mode III (shearing) (2)

– Strain field

– Stress field

• Hooke’s law

y

x

r q


Asymptotic solution


– Displacement field

•

• Should be finite l > 0


•

• Crack is stress free , n = 1, …

– Asymptotic solution for l = ½

&

y

x

r q


Asymptotic solution


– Asymptotic fields

-0.1

0

0.1

-0.1

0

0.1-5

0

5

xy

sxz

/KII

I

-0.1

0

0.1

-0.1

0

0.1-5

0

5

xy

syz

/KII

I


-0.1

0

0.1

-0.1

0

0.1-0.5

0

0.5

xy

uz E

/(K

III(1

+

))

shearing uzE

/(K

III(

1+

))

Asymptotic solution

• Summary in 2D

– Principle of superposition holds as linear responses have been assumed

• Ki depends on

– The geometry and

– The loading

• u, s can be added for

– ≠ modes: u = u mode I + u mode II, s = s mode I + s mode II

– ≠ loadings u = u loading 1 + u loading 1, s = s loading 1 + s loading 2

• Ki can be added

– For ≠ loadings of the same mode Ki = Kiloading 1 + Ki

loading 2

• But since f and g depend on the mode K ≠ Kmode 1 + Kmode 2


(opening) (sliding) (shearing)

y x

z


• New failure criterion

– 1957, Irwin, crack propagation

• smax → ∞ s is irrelevant

• If Ki = KiC crack growth

– Toughness (ténacité) KIc

• Steel, Al, … : see figures

• Concrete : 0.2 - 1.4 MPa m1/2

(brittle failure)

Fracture toughness

brittle

ductile

brittle

ductile


• Measuring KIc

– Done by strictly following the ASTM E399 procedure

• A possible specimen is the Single Edge

Notch Bend (SENB)

– Plane strain constraint (thick enough

specimen) conservative (see next slide)

– Specimen machined with a V-notch in

order to start a sharp crack

• Cyclic loading to initiate a fatigue crack

• Toughness test performed with

– A calibrated P, d recording equipment

– The Crack Mouth Opening Displacement

(CMOD=v) is measured with a clipped gauge

– Pc is obtained on P-v curves

» Either the 95% offset value or

» The maximal value reached before

– KIc is deduced from Pc using

» f(a/W) depends on the test (SENB, …)

Fracture toughness

P, d

v/2 v/2

L=4W

W Thickness t

CMOD,

v

v

P

tg 0.95 tg

Pc

P

tg 0.95 tg

Pc


• Only 2D solutions have been considered but a real crack is clearly 3D

– At any point of the crack line

• A local referential can be defined

• Since the asymptotic solutions hold for

r → 0, at this distance the crack line seems

straight, and the problem is locally 2D

• The crack tip field can be broken into

3 2D problems (3 2D modes)

3D problems

y

x

z


• 3D effects

– Near the border of a specimen the problem

is plane s, while it is plane e near the center

at the center there is a triaxial state

• The SIF is larger at the center as

no lateral deformations are possible

(see next lecture)

• 2 consequences

– The front will first propagate at the center

– The toughness decreases with the thickness

» Crude approximation

» Later on we will see how to evaluate this effect

– The practical toughness Kc is the plane strain one

3D problems

t K rupture

t

Kc


• Computation of SIFs (see next lecture) – Computation of the SIFs Ki

• Analytical (crack 2a in an infinite plane)

• For other geometries or loadings

• bi obtained by

– Superposition

– FEM

– Energy approach

» Related to Griffith’s work

» See next lecture

– For 2 loadings a & b: KI = KIa+ KI

b

– BUT for 2 modes K ≠ KI+KII

SIF

s∞ s∞

y

2a

x

s∞

s∞

y

2a

x

t∞

t∞

y

2a

x

t∞

t∞


• Exercise 1

– Let us consider a mode I state

around a notch of opening p-b

– What are the BCs ?

– What is the asymptotic behavior ?

– When does the SIF concept make sense ?

Exercise

x

y

b

s∞

s∞


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.


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

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

• One can use the developments of mode I


&

– Change of coordinates

– System of equations

Exercise 1: Solution

x

y

b

s∞

s∞


• Non trivial solution

– Determinant is equal to zero

– For b=p , it becomes

• Since l>-1 (for the displacements to be finite)

l = -½ , 0, ½, ..

singularity at crack tip

– There is no singularity if the first l >-1 is ≥ 0

• For b=p/2, the equation becomes

l = 0, 1 , ..

no singularity at crack tip

• For b < p/2: no singularity

• For b > p/2: there is a singularity

– Solving the equation with NR

• Considering the minimum l >-1

Exercise 1: Solution

x

y

b

s∞

s∞


60 80 100 120 140 160 180-0.5

0

0.5

b

lmin

b [deg.]

lm

in

Annex 1: Plate with a hole

• Cylindrical coordinates


• At r = a

• At r = b→∞: only syy= s∞ is non zero

y

2a x

s∞

s∞

syy syy

y

s∞

s∞

r=a x

r q

r =

b→∞

Load case 1

independent

from q

Load case 2

dependent on cos 2q (symmetry with

respect to x)


• Cylindrical coordinates (2)

– Bi-harmonic equation

with

– Stresses from Airy function

with , &

• Eventually, after substitutions

y

s∞

s∞

r=a x

r q

r =

b→∞



• Load case 1

– Independent from q F(r)

• Bi-harmonic equation

• General solution:

• Stresses:

• Boundary conditions:

&

, &

• Solution: , &



• Load case 2

– Dependent on cos 2q F(r, cos 2q) = g(r) cos 2q

• Bi-harmonic equation

• General solution:

• Stresses:



• Load case 2 (2)

– Stresses

– Boundary conditions:

, , &



• Load case 2 (2)

– Solution:



• Superposition of case 1 and 2

– Stresses:

– For q = 0: syy= sqq

– Stress intensity factor: Kcircle = 3

0

0.5

1

1.5

2

2.5

3

0 2 4 6

x/a

syy

/s∞



Annex 2: Displacement of asymptotic solution

• Displacements for linear and elastic materials

– Complex form:

– In terms of strains:

– Hooke’s law (Plane s or e):

• With m(z) such that the plane s or e condition is satisfied:

• But


• Displacements for linear and elastic materials (2)

– Out-of-plane state effect

• Plane s :

with

• Plane e:

with

– For both plane states:

Annex 2: Displacement of asymptotic solution


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Fracture Mechanics, Damage and Fatigue Linear Elastic Fracture … · 2016. 1. 28. · Resolution...

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