Fracture Mechanics, Damage and Fatigue Linear … solution directly? SIF: Analytical methods...

University of Liège

Aerospace & Mechanical Engineering

Fracture Mechanics, Damage and Fatigue

Linear Elastic Fracture Mechanics – Stress Intensity Factor (SIF)

Fracture Mechanics – LEFM – SIF

Ludovic Noels

Computational & Multiscale Mechanics of Materials – CM3

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

Chemin des Chevreuils 1, B4000 Liège

[email protected]

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

Asymptotic solution of LEFM

• Summary

– Detailed expressions for the 3 modes

y

x

rq

Mode I Mode II Mode III

(opening) (sliding) (shearing)

y x

z

2016-2017 Fracture Mechanics – LEFM – SIF 2

• Computation of the SIFs– Analytical methods (for LEFM)

• Full field solution (see next slides)

– Limited to infinite planes

– SIFs obtained from asymptotic limit

• Superposition (see next slides)

– Of existing solutions

• Energetic approach (see previous lecture)

– Related to Griffith’s work

– Numerical

• Collocation method (see Annex 2)

• FEM

– Capture asymptotic solution

» Use singular (Barsoum, 1974) quadratic elements (central nodes moved to quarter edge)

– Energetic approach

– J integral

– Experimental

• Normalized experiments

• Strain Gauge Method

– Use of SIF handbook

LEFM: Computation of SIF


• Reminder: method in linear elasticity

– Problem is governed by the bi-harmonic equation

– One solution of this equation has the form

where w(z) and W(z) are functions to be defined so

• The stress field

• The displacement field

satisfy the BCs

SIF: Analytical methods

Plane e

Plane s


• Computation of SIF by analytical method (LEFM)

– Full field solution of a crack submitted to traction

• Infinite plane

• Mode I: , and ty (-x) = ty (x), ty (-y) = -ty (y)

– Westergaard approach for mode I

• As

and as for mode I, one should have

sxy = 0 for y = 0,

the solution is

• Indeed with this choice


2a

x

y

ty

tx

2a

y

ty

x



• Computation of SIF by analytical method (LEFM) (2)

– Westergaard solution for mode I (2)

• The general stress/displacement fields read

• Using they become

2a

y

ty

x

Only W to be defined



– Full field solution of a crack submitted to traction

• Westergaard solution for mode I

• Solution W(x≥0) satisfying the BCs for a uniform traction (see next slides)

• NB: General solution (Sedov, 1972):


2a

y

ty

x



– Check solution

• with

– Symmetry with respect to Ox and Oy

R (C) = R (C1) = I (C) = I (C1) = 0


a

y

x

qr

z

n+

n-




– Check solution (2)

• with

• On crack lips z = x ± i |e|, e → 0, |x| < a

Boundary condition on crack lips: t = n . s = 2 R (W’) = ± ty satisfied

2a

y

ty

x

a

y

x

qr

z

n+

n-

±




– Asymptotic field on crack lips

• For z = a-r ± i |e|, e → 0

– Crack Opening Displacement (COD)

a

y

x

qr

z

n+

n-



– Asymptotic field ahead of crack

• For z = a+r ± i |e|, e → 0, r →0+

,


a

y

x

qr

z

n+

n-



– Asymptotic stress field ahead of crack

• For z = a+r ± i |e|, e → 0, r →0+

,


a

y

x

qr

z

n+

n-



– Crack in an infinite plate under perpendicular loading

• Superposition

• Case 2: solved for ty = s∞

• Case 1:

&

&


2ax

ys∞

s∞

s∞

s∞

x

y

2ax

y

s∞

s∞Case 1 Case 2



– Crack in an infinite plate under perpendicular loading

• Full field solution from W and case 1

• Asymptotic solution along y = 0, for x > a

• SIF:


syy

x

Asymptotic syy

True syy

Zone of asymptotic

solution (in terms of

1/r1/2) dominance

Structural

responsePlasticity

∞

2ax

ys∞

s∞

0.1

1

10

100

0.001 0.01 0.1 1 10 100

r/a

syy

/ sasymptotic

full field



– Crack in an infinite plate under sliding

• Westergaard approach for mode II

– For mode II, syy = 0 for y = 0

– Crack subjected to a shearing tx

• Applying superposition principle and tx =t∞


2ax

y t∞

t∞




– Crack in an infinite plate under shearing

• Stress field (see Annex 1)

• Ahead of crack tip (y = 0 and x = a + r)

2ax

y t∞

t∞

x

y

rq

q1r1

r2

q2



– Here we have obtained the SIF using the full field solution so

– Why did we develop the asymptotic solution last time instead of using this full field solution directly?



• Solutions from SIF handbooks (see references)

– Obtained by using a variety of methods (analytical, numerical, …)

• Example: mode I for a crack in a finite plate (h/W > 3)

– General formula

– Numerical results based on Laurent expansion

» Isida, 1973

– Periodic crack approximation

» <5% error for a/W < 0.5

» Irwin, 1957

– Fit of Isida’s values

» <0.1% error

» Tada, 1973

SIF: Handbooks

2ax

ys

s

2Wh

0.8

0.85

0.9

0.95

1

0 0.5 1

a/W

(1-a

/W)^

1/2

f(a

/W)


• Finite element model: extraction from stress and displacement fields

– Asymptotic crack tip stress and displacement fields determined by the SIF

&

– Stress correlation

• FEM computation

• Extract stress field s(r,q) for q=0 (or any other q)

• Extrapolate to get the SIFs

– Displacement correlation

• Idem but

with

SIF: Numerical approaches


• Finite element model: extraction from stress and displacement fields (2)

– Advantages of the method

• Simple

• Can be used with any FE software

• Only one post-processing to determine the 3 SIFs (if suitable loading)

• Can be used along a crack front in 3D

• When using displacement correlation, the field is primary solution

– Drawbacks

• The accuracy is strongly dependant on the mesh refinement

• The mesh refinement required depends on the element ability to capture the

singularity at crack tip (see next slide)



• Finite element model: Barsoum elements

– Previous method requires a fine mesh since a singularity in r½ is not naturally

captured by usual FE

– This singularity can be captured

• By enriching elements (as in XFEM for LEFM, see later)

• By using Barsoum (quarter-point) elements

– Quarter-point elements

• Quadratic elements with some mid-nodes located at a quarter of the edge


x

h

-1 1

1

-1x

h

0 1

1

x

y

3L/4L/4Mapping

x

y

3L/4L/4

Mapping


• Finite element model: Barsoum elements (2)

– Quarter-point element: 1D example

• Shape functions

• Mapping

• Strain field

• So a strain field (and so for stress) in 1/r½ can be captured


x

-1 10

-0.2

0

0.2

0.4

0.6

0.8

1

-1 -0.5 0 0.5 1

x

N

N

N

1

2

3

x

0 LL/4


• Finite element model: Barsoum elements (3)

– Quarter-point element: 2D example

• Fine mesh at crack tip: mid-nodes of first ring are moved to quarter points



• Finite element model: global energy & compliance

– For cracks growing straight ahead (if only one mode is involved)

• G=J, and in linear elasticity

• Prescribed loading: (in linear elasticity)

• Prescribed displacement:

– Perform 2 computations

• With the same loading (displacement) but with two different crack lengths

• So, per unit thickness:

• Could be rewritten using compliance C in terms of the generalized loading Q

(linear elasticity)

– Advantages• Simple

• Can be used with any FE software and requires only post-processing

• Less mesh sensitive than correlation methods as a global variable is used

– Drawbacks• 2 computations needed

• Only one mode (so one SIF) can be considered at a time

• In 3D, SIF variation along crack front cannot be determined



• Finite element model: crack closure integral

– Cracks growing straight ahead

– For such cracks, in linear elasticity


y

x

a Da

r

q

syy = 0,

uy ≠ 0

syy ≠ 0,

uy = 0

y

x

a Da

r q

syy = 0,

uy ≠ 0

syy = 0,

uy ≠ 0


• Finite element model: crack closure integral (2)

– This integral can be computed using FE software

• Nodal release

– First computation: nodes j on both crack lips are constrained together

– Second computation: constrain on node j is released

• SIFs can be computed from

– Reaction forces of first computation &

– Displacement jumps of second computation

– , &


j-1 j j+1j-2

a Da

Du

j-1 j j+1j-2

a Da

F

-F


• Finite element model: crack closure integral (3)

– Advantage

• Requires only post processing

– Drawbacks

• If the crack is not a line of symmetry, the node displacements have to be

constrained (using Lagrange multipliers e.g.)

• Two computations are needed (can be improved using modified nodal release)



• Finite element model: direct computation of J-integral

– For linear elasticity and for any contour G embedding a straight crack

&

– Assuming symmetry & a structured meshed

• dl = dy on G1, dl = -dx on G1,dl = -dy on G3

• As values are computed on Gauss points,

ideally, the integrals will add Gauss points’

contributions


G1

G2

G3

x

y


• Finite element model: J-integral by domain integration

– For linear elasticity and for any contour G embedding a straight crack

&

– Let us define a contour C=G1+G-+G+-G

• Interior of this contour is region D

• Define q(x, y) such that

– q = 1 on G

– q = 0 on G1

• As crack is stress free


x

y

G

G1

G+

G-

-


• Finite element model: J-integral by domain integration (2)

– Computation of

• C is closed divergence theorem

• But & =0

• q is discretized using the same

shape functions than the elements

– This integral is valid for any annular region around the crack tip

• As long as the crack lips are straight


x

y

C

D

q=1

q=0


• Finite element model: J-integral

– Advantages

• Accurate (especially using domain integration)

• Does not require Barsoum elements

– Drawbacks

• Only one mode at the time

• Either modifying the FE code in order to have easy post processing or

• Post processing difficult when using a standard FE software

– As this method is really accurate and computationally efficient, can it be

modified in order to extract the SIF of each mode?



• Finite element model: J-integral (2)

– How to extract the different SIFs?

• Compute J for the solution u of the considered problem J

• Compute J for another field uaux to be specialized Jaux

• Compute J for the sum of u & uaux to be specialized Js

• Relation between the different Js?




– How to extract the different SIFs (2)?

• Relation between the different Js can be deduced from

–

–

• So is rewritten

– The right term is called the interaction integral

» What is its expression in terms of the SIFs ?





• It has been found that

• With &

• Direct substitution leads to

–

–





• As

– With

• And as these two last relations are symmetric in K and Kaux

– By analogy with

– One can feel that

• This is obtained explicitly after substitution of f and g by their closed forms




– The SIFs are deduced from the so-called interaction integral

•

• Indeed, if uaux is chosen such that only Kiaux ≠ 0, Ki is obtained directly

– This method

• Is very accurate compared to correlation methods

• But it requires

– More computations

– Extensive modifications of the FE code



• Strain gauge method

– Asymptotic solution for mode I

• with l = -½, 0, ½, …

• The asymptotic solution in r-½ cannot be matched accurately using a gauge

• So the solution considered is limited to the 3rd order

SIF: experimental methods


• Strain gauge method (2)

– Position of the strain gauge

• Located at (r, q ) of the crack tip

• With an orientation a

– One** can show that in the referential O’x’y’

with

• The gauge is located at (r, q*, a* ) such that

&

– Mounting of the gauge is critical


y

x

rq

y’

x’

a

**Dally JW and Sanford RJ (1988), Strain gage methods for measuring the opening mode stress intensity

factor KI , Experimental Mechanics 27, 381–388.


• Strain gauge method (3)

– Strain gauges are now replaced by Digital Image Correlation (DIC)

• Markers on the surface are tracked optically

• Displacements and strains can be computed

– These strains can be used to deduce the SIFs



• Flawed cylinder

– A piston is used to increase inner pressure

• From 0 to 55 MPa

– Cylinder made of

• Peaked-aged aluminum alloy

– 7075-T651

• Yield sp0 = 550 Mpa

• Toughness KIC = 30 MPa m1/2

– Malfunction

• Cylinder burst

• Post failure analyses

– Initial elliptical flaw at inner wall

» 4.5 mm long

» 1.45 mm deep

» Normal to hoop stress

– Origin of burst?

Exercise 1

L=

20

cm

t = 1 cm

Din = 9 cm

2c = 4.5 mm

a = 1.45 mm

sqq

sqq


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

• Stress field

– Consider thick cylinder with

• rin = 0.045 m & rout = 0.055 m

• Inner pressure p

Exercise 1: Solution

L=

20

cm

t = 1 cm

Din = 9 cm

2c = 4.5 mm

a = 1.45 mm

sqq

sqq


• SIF

– The wall is not perforated

• Use SIF for semi-elliptical crack in large plate

• See SIF handbook

• Geometrical effect

• Plasticity correction

– SSY criterion:

– Not fully satisfied plasticity correction

• SIF with

– Rupture:

• For r = rin, p would be 100 MPa, this is the critical value


2c = 4.5 mm

a = 1.45 mm

sqq

sqq


• Rupture mode

– For r = rin, p would be 100 Mpa, this is the critical value

– This is out of the range of the piston activity

• So rupture should come from fatigue

• Cyclic loading

– p from 0 to 55 Mpa

• Hoop stress from 0 to

– SIF

• Assuming a/c remains constant during crack propagation



• Cyclic loading

– p from 0 to 55 Mpa

&

– Due to initial flaw

• Assuming curves are valid for R=0

• We are in Paris regime

crack propagation

– Rupture will happen for

• Number of cycles?



• Cyclic loading (2)

– As

• Assuming curves are valid for R=0

• We are in Paris regime

• !!!Life strongly depends on the maximum pressure reached during accidents



• Computation of SIF by analytical method (LEFM)

– Crack in an infinite plate under shearing

• Mode III:

uz is the imaginary part of a function z(z)

• This function has to be found to satisfy the BCs

• Stress field

– As

– One has

– And

Annex 1: Analytical methods

2ax

y t∞

t∞




– Crack in an infinite plate under shearing (2)

• Solution of the problem?

–

– with

uz = I(z)

• Symmetry?

– uz(-y) = -uz(y) satisfied

• Far away field?

– For y→±∞: syz → ±t∞ satisfied

2ax

y t∞

t∞




– Crack in an infinite plate under shearing (3)

• Solution of the problem?

–

– with

• Crack lips stress free?

satisfied

2ax

y t∞

t∞

x

y

rq

q1r1

r2

q2


• Computation of SIF by Boundary Collocation Method (LEFM)

– Example: crack in a finite circular plate

• Considering mode III (can be done for I and II)

• Laurent series

with

• But for the crack lies in q=p, so we use

is satisfied

– Finite displacement n = 0, 1, …

Annex 2: Semi analytical methods

a=R

x

y

R

q

Tz = f(q)


• Computation of SIF by Boundary Collocation Method (LEFM) (2)

– Example: crack in a finite circular plate (2)

• Unknowns an are obtained by defining m collocation points on the boundary

– At these points

for k=1, .., m, & assuming an expansion up to order p, p+1 ≤ m

Annex 2: Semi analytical methods

a=R

x

y

R

qk

Tz = f(qk)

• Example: R=1, & loading

– For accuracy m = 2p+2

– Collocation points only on upper

side (avoid trivial solution)

– Least squares resolution


100

101

102

1.5

2

2.5

3

p

KIII

m=2(p+1)

p

KII

I

Fracture Mechanics, Damage and Fatigue Linear … solution directly? SIF: Analytical methods...

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