fracture mechanics: linear elastic fracture mechanics...

Task 6 - Safety Review and LicensingOn the Job Training on Stress Analysis

Pisa (Italy)June 15 – July 14, 2015

Fracture Mechanics: Linear Elastic Fracture Mechanics 2/2

Davide Mazzini – Ciro Santus

2

Content

• Stress singularity

- Notch degenerating into a crack

- Multi-axial stress at notch root/ crack tip

- The Williams problem

• Linear Elastic Fracture Mechanics (LEFM)

- The Westergaard stress function

- Definition and calculation of the Stress Intensity Factors (SIFs)

- LEFM Validity limitations

Table of content – Class VI.b.2

Pisa, June 15 – July 14, 2015

3

In-plane stresses

From notch to crack

x/a

y/a

y

x

x/a

y/a

y

x

2a

0

0,at the notchroot freesurface

x

In mid region

x y

n

0y

x


4

Westergaard stress function

From notch to crack

r

s0

1 1/2

1/ 21

ij r rr

00

...

...

...

nonsingularOtherterms


5

Cartesian/ Cylindrical coordinates

Westergaard/ Irwin stresses

r

xx

yyxy

r

rr r

5 1 3cos cos4 2 4 223 1 3cos cos4 2 4 22

1 1 3sin sin4 2 4 22

Irr

I

Ir

Kr

Kr

Kr

I

I

I

3cos 1 sin sin2 2 22


3cos sin cos2 2 22

xx

yy

xy

Kr

Kr

Kr

Crack Crack


6

Stresses ahead of the crack tip

Westergaard stresses

r0

xx

yyxy

I

I

0 :

2

20

xx rr

yy

xy r

Kr

Kr

Crack

Notes:even at the notch root

0 symmetryof the problemxx yy

xy


7

Stresses ahead of the crack tip


r

I

2x yK

r

Crack

y

x

n

0y

x

Singularitydominated zone

Far field zone


8

Stress Intensity Factor (SIF)


rCrack

I

2yK

r

Different KI values,weaker and strongersingularities

I is the (first)Stress Intensity FactorK

I I (sameintensification meaning)

What is the reason of ?Griffith Energy releas

2

e rate"2

"

K kr r

G

Now thisequation holdswithout any 2 term


9

The Stress Intensity Factor is NOT the Stress Concentration Factor

Stress Intensity Factor

rCrack

I

2yK

r

x/a

y/a

max t nK


10

Units, two options


I

2yK

r

MPa

m

MPa m

I

2yK

r

MPa

mm

32

2

MPa mm

N mm N mmmm

Conversion:

MPa m MPa 1000 mm 1000 MPa mm 31.6 MPa mmExample:

8.0 MPa m 252.8MPa mm

x x x x


11

Logarithm scale


rCrack

I

2yK

r

(log)rCrack

(log)y

0goes tor

I

Let's log both sides...1log( 1) log( ) log(2 ) log( )2 2y K r

12

1


12


ANSYS Apdl (classic) – MATLAB:

Verify the -1/2 slope (log-log) and calculate the SIF for a plate with a lateral crack

100mmb

2 70 mmh 15mma

100MPa

Element type to be used


13


- Half model with Symmetry- Full model


14


- Half model with Symmetry- Full model

“Kidney bean” shape


15


MATLAB elaboration

0 2 4 6 8 10-1000

0

1000

2000

3000

4000

5000

r, mm

Stre

ss c

ompo

nent

s, M

Pa

x

y

xy

10-2 10-1 100 101102

103

104

r, mm

y, M

Pa

KI = 1047 MPa mm1/2

Linear scales

log-log scales

2 10

1 2

11 log (2 )2

I

MATLAB "polyfit": , 0.5(good approx.)

10P

P PP

K


16


Homework:

Verify the angle dependency of the stress distribution at the crack tip

0

Path at any differentangle

I

I

I



3cos sin cos2 2 22

xx

yy

xy

Kr

Kr

Kr


17

There are two distinct ways to apply in-plane loading

In-plane Stress Intensity Factors

II

II

I

I

I

3 3cos 1 sin sin sin 2 cos cos2 2 2 2 2 22 2

3 3cos 1 sin sin sin cos cos2 2 2 2 2 22 2

2

xx

yy

xy

K

K

K

K

K

r r

r r

II3 3cos sin cos cos 1 sin sin2 2 2 2 2 22 r

Kr

r

xx

yyxy

Crack

Mode I and Mode IIStress Intensity Factors

Symmetrical andNonsymmetrical stress components

Car

tesi

an c

oord

inat

es


18

There are two distinct ways to apply in-plane loading

In-plane Stress Intensity Factors

I

I

I

II

I I5 1 3 5 3 3cos cos sin sin4 2 4 2 4 2 4 22 23 1 3 3 3 3cos cos sin sin4 2 4 2 4 2 4 22 2

1 1 3sin sin4 2 42

rr

r

K

K

K

r r

r r

K

K

r

II 1 3 3cos cos2 4 2 4 22

Kr

rCrack

rr

rMode I and Mode IIStress Intensity Factors

Symmetrical andNonsymmetrical stress components

Cyl

indr

ical

coo

rdin

ates


19

Same solution In-plane

Plane Stress/ Strain

I II

I II

I

3 3cos 1 sin sin sin 2 cos cos2 2 2 2 2 22 2

3 3cos 1 sin sin sin cos cos2 2 2 2 2 22 2

2

xx

yy

xy

K Kr r

K Kr r

K

II3 3cos sin cos cos 1 sin sin2 2 2 2 2 22

Kr r

Similarly to the notch,any point even verycloseto thesingularity:

0 plane stress

plane strainzzxx yy


20

Plane strain, transversal stress/ strain


Plane stress, transversal stress/ strain

I II( ) cos 2 sin2 22 2

0

zz xx yy

zz

K Kr r

I II

0

( ) cos 2 sin2 22 2

zz

zz xx yyK K

E E r r


21

How plane stress can be possible at the crack tip?


a

B

0

0

Plane stress solution if a>>B

Plane strain if a<<B

t0

Any thickness, the local radius is “infinitely” smaller being just zero.Why talking about plane stressfor a crack?Plastic zone need to introduced.

:thickness:radius

t


22

The three crack loading modes and related Stress Intensity Factors

Fully three-dimensional problem

I II

In-plane:,K K III

Out-of-plane:K


23

Many figures available to show the three crack modes

Fully three-dimensional problem


24

Out of plane shear stresses

Mode III stress distribution

III

III

sin22

cos22

other stress component are zero

xz

yz

Kr

Kr

, or alternatively ,

were zero for Mode I and Mode IIeither plane Stress or plain Strain

xz yz xz yz


25

How to calculate SIFs for structures

Different approaches

Dimensional approach, with graph and tabular data

Weight function

Finite elements with different techniques, previously an example with the

stress asymptotic approach has been shown


26

Dimensional approach


I

t

and are required just for thedimensionalanalysis

is the "shape function" it isdimensionless and depends on theloading configuration and relativedimensions (similarly to )

For thisspecific problem:

K FS a

a S

F

K

F

1.0(theoretical result)

2a

S

Far boundaries

IK S a


27

Dimensional approach


I g

g

n

Usually grossstress is used

in the formula rather thanwhich is the net stress

For thisspecificcase:dependson / ratio

... tabular cases available on textbooksand atlas to find (approximated)values f

K FS a

S

S

F a b

or F

2a

gS

Finite with, far stress

2b


28

Crack on plate cases


for a single-edge-crackedplate with a small

1.12

crack

F


29

Previous example solved with FE simulation


12

I12

I

Here is ,however,approximately:

/ 0.151.28(not much larger than1.12)

880 MPa mm

(ANSYS) 1047 MPa m

no

m

t satisfied

16%( )

h b

a bF

K FS a

K

100mmb

2 70 mmh 15mma

100MPa


30

Previous example modified, larger height


12

I12

I

Now is :/ 0.15

1.28(not much larger than1.12)

880 M

sati

Pa mm

sfied

2%(ANSYS) 899 MPa mm ( )

h ba b

F

K FS a

K

100mmb

2 300 mmh 15mma

100MPa


31

Three-dimensional elliptical or circular (penny-shaped) cracks


Infinitespace

Halfspace

Ellipt. Circ.Surf. Ellipt.


32

Many cases available on books and atlas


T.L. Anderson, Fracture Mechanics: Fundamentals and Applications,third edition. CRC Press 2005.

S.A. Laham, R.A. Ainsworth, Stress Intensity Factor and Limit Load Handbook. EPD/GEN/REP/0316/98, ISSUE 2, 1998.

Y. Murakami, Stress Intensity Factors Handbook. Pergamon, 1986.

ASTM standards (to be shown next)

… and others


33Pisa, June 15 – July 14, 2015


S.A. Laham, R.A. Ainsworth, Stress Intensity Factor and Limit Load Handbook. EPD/GEN/REP/0316/98, ISSUE 2, 1998.

34

Homework:


Calculate theSIF for asurfacecrack in a half-spaceat A and B points:

2 mm, 2.5mm100MPa (uniform)

Compare the resultsAndreson vs. Laham

a ta c


35

Crack tip plastic zone

Irwin 1961

YS

Material model:elastic perfectly plastically

Y

2

IY

Y

Plane :

0

stress

12

y

y

S

KrS

2

Ip

Y

p Y

After equlibriumcorrection:

( 2 )

1 KrS

r r


36


Irwin 1961

YS

Material model:elastic perfectly plastically

Y

2

IY

Y

Plane :

0

strain

16

y

y

S

KrS

2

Ip

Y

p Y

After equlibriumcorr

13

ection:

( 2 )

KrS

r r


37

Plane Stress/ Plain Strain


Always Pl. Stress at the surface

“Kidney bean” shapePl. Strain at intermediate thickness, only for a thick plate

3 times size


38



0

Within the Linear Elastic assumption, always plain strain at the singularity, local radius = 0

pr

After considering the plastic zone, the size to be compared to the thickness is To distinguish between plane Stress/ Strain

pr


39



2

I

Y

2

I

Y

2

I

Y

Y Y

To havefully developedplanestrain:

2.5

2.5

1(2.5 6 )6

(2.5 6 ) 50

KBS

KS

KS

r r


40

Small Scale Yielding for LEFM validity


p

Y

Y

To have LEFM validity:

this way the plastic zoneis dominated by the -field.Usually it is stated that:

8 (for pl.stress)8 (for pl.stress)

a r

K

a ra r

a

2 2

I I

Y Y

for Pl. Stress: for Pl. Strain:

4 43

K Ka aS S


41

Small Scale Yielding for LEFM validity


Y p( ) has to be limited also

with respect to the geometry boundaries

r r2 2

I I

Y Y

for Pl. Stress: for Pl. Strain:

4 4, ( ), , ( ),3

K Ka b a h a b a hS S


42


ANSYS Wb – ASTM CT specimen: At surface (Pl. Stress)

Interior section (Pl. Strain)

Elastic-Plastic material


43


ANSYS Wb – ASTM CT specimen:Plane Stress (small thickness with respect to the plastic zone)

Same F/thick.


44


ANSYS Wb – ASTM CT specimen: Plane Strain, interior thick specimen

Plane Stress, small thickness specimen

Plane Stress, at thick specimen surface


fracture mechanics: linear elastic fracture mechanics...

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