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7.6

An Al2O3 specimen is being pulled in tension. The specimen contains flaws

having a size of 100 m . If the surface energy of Al2O3is 0.8 J/m2, what is the

fracture stress? Use Griffiths criterion. E = 380 GPa.

According to Griffiths criterion, the critical stress required for the crack to

propagate in the plane-stress situation.

MPac

c

mlengthcrackthehalfisa

mJenergysurfacetheiss

where

a

sE

c

2.62

61050

8.09103802

50

28.0

2

=

=

=

=

=

7.7

A thin plate is rigidly fixed at its edges (see Figure Ex. 7.7). The plate has aheight L and thickness t (normal to the plane of the figure). A crack moves from

left to right through the plate. Every time the crack moves a distance s, twothings happen:

1. Two new surfaces (with specific surface energy) are created.

2. The stress falls to zero behind the advancing crack front in acertain volume of the material.

Obtain an expression for the critical stress necessary for crack propagation in thiscase. Explain the physical significance of this expression.

Strain energy per unit volume =E2

2


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Total strain energy released when the crack moves a distance x =E2

2. L.t.x.

At the same time as the strain energy is released due to the crack propagation, two

new surfaces are created which result in an increase in surface energy equal to

x.2. .t.

If U represents the change in energy, then for crack propagation to occur, we musthave

0=

x

U

U = (L.x.t) txE

..22

2

+

Therefore,

022

2

=+=

t

ELt

x

U

Or, the critical stress for crack propagation is

L

E

L

E 2

4==

Significance: The thinner is the plate (i.e., the smaller is the L), the larger is thestress necessary for crack propagation. The situation is akin to that obtained in an

adhesive joining to two parts. The thinnest possible adhesive layer will lead to the

strongest possible bond.

7.8 A central through-the-thickness crack, 50 mm long, propagates in a thermoset

polymer in an unstable manner at an applied stress of 5 MPa. Find Kc.

For a central through-the-thickness crack, we take Y = 1. Thus

K = a


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Kc= 5 ( )05.0

= 19.8 MPa m

7.10

An AISI 4340 steel plate has a width W of 30 cm and has a central crack 2a of 3

mm. The plate is under a uniform stress, . This steel has a KIc value of 50

mMPa and a services stress of 1500 MPa. Compute the maximum crack size

that the steel may have without failure.

mmma

m

a

w

af

w

aFor

mMPaK

MPaw

a

aw

afK

c

c

Ic

cIc

7.00007.02

00035.0

1

1500

50

1,074.0

50

1500

005.0

2

==

=

=

=

=

=

=

=

7.11 A microalloyed steel, quenched and tempered at 250C, has a yieldstrength (y) of 1750 MPa and a plane-strain fracture toughness KIcof 43.50

mMPa . What is the largest disk-type inclusion, oriented most unfavorably, that

can be touched in this steel at an applied stress f 0.5 y?

mMPaK

MPa

Ic

y

5.43

1750

=

=


6/29

A disk type inclusion, oriented most unfavorably, can be likened to a penny-

shaped internal crack. In the most general case, such a crack will have an ellipticalform. The stress intensity factor for an elliptical crack is given by

( ) (

4/12222

cossin

bakEaKI +=

)

where a and b the semi-major and semi-minor axes of the ellipse, respectively,

and E(k) is the complete elliptical integral of the second kind. For the present

case, we can consider the inclusion to be a circular one of radius, a. Then, puttinga = b, we have (see p. 426 in the text)

mm

m

K

aFor

aK

y

Ic

cy

I

8.7

0078.0875

5.43

5.0

2

2

2

,

=

=

=

==

=

The diameter of the largest inclusion tolerable isd = 2ac = 15.6 mm

7.12 A 25-mm2bar of cast iron contains a crack 5 mm long and normal to one face.What is the load required to break this bar if it is subjected to three- point bending with

the crack toward the tensile side and the supports 250 mm apart?

A = 5 mm

B = W = 25 mm

L = 250 mm

For three point bending situation, we have

+

+

=

2/32/72/52/32/1

2/37.386.378.216.49.2

W

a

W

a

W

a

W

a

W

a

BW

PLK


7/29


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,69.0.,.,1sec

0138.050

69.0

50

69.01

1500

701

1sec,1

sectan

22

2/12/1

mmaandrightisaboveassumptionoureiw

a

w

a

mmW

mmK

aor

aK

W

a

W

aAssume

W

aaKor

W

aWK

c

Icc

cIc

II

==

==

=

=

=

=

=

=

=

Thus, no iteration is required.

W =5 mm

102.1sec

138.06

69.0

=

==

W

a

W

a

which is not equal 1.

In this case, we must use an iterative process to calculate the a cvalue until successive

values are close enough. Thus

(i) 1.1sec =

W

a

(ii) 126.05

630.0==

W

a

( )

mma 639.0084.1

11

1500

70

084.1126.0sec2

2 =

=

=


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(iii)

mmaTherefore

mma

w

a

W

a

c 638.0

0638.086.1

11

1500

70

086.1sec

1278.05

639.0

2

3

=

=

=

=

==

Single Edge Notch

6.00

85.5348.38

7.1841.099.1

43

2

+

+

=

=

W

afor

W

a

W

a

W

a

W

a

W

afwhere

aW

afKI

( )

( )

99.1011.0,5.0

50

55.0

00055.0

1500

70

99.1

1

99.112.1

99.1,

1

2

2

2

=

==

=

=

=

=

=

=

=

W

afand

W

ammafor

mmW

mma

mm

aThen

thatNote

W

afthensmallbeto

W

aAssume

W

af

Ka

c

c

Icc


10/29

Therefore, ac= 0.55 mm

For W = 5 mm

Let a = 0.55 m, 18.2,11.0555.0 =

==Waf

Wa

( ) mma 459.0

1500

70

18.2

12

2 =

=

Taking this value of a, we estimate the new value of

W

af until two successive

values are about the same. Thus,

a (mm)

W

a

W

af

( )mmac

0.459 0.0918 2.08 0.503

0.503 0.1006 2.10 0.494

0.494 0.0988 2.10 0.494

ac= 0.494 mm

7.14 An infinitely large plate containing a central crack of length 2a = 50/mm is

subjected to a nominal stress of 300 MPa. The material yields at 500 MPa.

Compute:

(a) The stress intensity factor at the crack.(b) The size of the plastic zone at the crack up.

Comment on the validity of Irwins correction for the size of the plastic zone in this

case.

2a = 50/mmy = 500 MPa

= 300 MPa

(a)For an infinitely large plate containing a central crack,


11/29

mMPaK

mMPa

aK

4.47

1025

3003

=

=

=

(b)Plastic zone size at crack tip,

mmr

m

Kr

y

y

y

43.1

1043.1500

4.47

2

1

2

1

3

2

2

=

=

=

=

Comment: It is valid to use Irwins correction for the plastic zone when ryis small.

Specifically, in the present case,( )

,18.0/25

43.1==

a

ryi.e., the condition

50

ary is not

satisfied. Thus, it would be valid to use Irwins correction for plastic zone.


12/29

7.15 A steel plate containing a through-the-thickness central crack of length 15 mm is

subjected to a stress of 350 MPa normal to the crack plane. The yield stress of thesteel is 1500 MPa. Compute the size of the plastic zone of the plastic zone and the

effective stress intensity factor.

( ) (

( )

)

( )

( )

aK

Kr

raa

mMPa

K

mm

Kr

mma

MPa

MPa

I

y

I

y

yef

ef

y

y

y

=

=

+=

=

+=

=

=

=

=

=

=

2

2

22

2

1

2)2(

44.54

000204.00075.0350

204.0

1500

0075.0350

2

1

2

1

152

350

1500

Taking the given value of crack length as aef, we compute Kef and ry. This gives us

a new value of aef. We recalculate Kefand ryand repeat the process until the successivevalues are close enough. Thus,

(i)

mmr

mMPa

K

y

I

204.01500

7.53

2

1

7.53

)105.7(350

2

3

=

=

=

=


13/29

(ii)

mmr

mMPa

K

y

I

209.01500

45.54

2

1

45.54

)10204.0105.7(350

2

33

=

=

=

+=

(iii)

mmr

mMPa

K

y

I

209.01500

46.54

2

1

46.54

)10209.0105.7(350

2

33

=

=

=

+=

Thus, ry= 0.209 mm and Kef = mMPa46.54


14/29

7.16 The size of the plastic zone at the crack tip in the general plane stress case is

given by

( )2222

2

2/cos342/cos2

=y

I

y

Kr

(a)Determine the radius if the plastic zone in the direction of the crack.

(b)Determine the angle at which the plastic zone is largest.

(a)Plastic zone in the crack direction

= 0 in the crack direction, therefore,

2

2

2 y

I

y

Kr

=

(b)Angle at which the plastic zone is the largest.

In order to find this angle, we differentiate the expression for ry with respect to and

equate it to zero. Let A = 22 y

IK

2

, then

0]22/cos3[2/sin

2/sin2/cos62/sin2/cos4/ddr

3

3

y

==

+=

A

AA

Now, sin= 0 gives = 0, which corresponds to a minima in ry . 3 cos2 /2 = 0 gives =

70.5o.


15/29

7.17 For the plane-strain case, the expression for the size of the plastic zone is

( )

=

2cos3114

2cos

2

22

2

2

vv

Kr ly

(a) Show that this expression reduces to the one for plane stress for = 0.

(b) Make plots of size of the plastic zone as a function of for = 0, = 1/3, and =. Comment on the size and form of the zone in the three cases.

Plane strain case

( ) strainplanevvy

Kr Iy

=

2

cos3114

2

cos

2

22

2

2

(a)For plane stress, = 0 and the above expression reduces to

=

2cos34

2cos

2

22

2

2

y

I

y

Kr

which is the expression for the plane stress case.

(b)Plots of plastic zone sizes as function of for = 0, 1/3, and are given in

the figure below.


16/29

= 0.5 X

Y

0

= 0

= 0.33

Comments

For =0, we have the case of plane stress and the size of the plastic zone is the largest.

For =1/3 and = , the constraint in the thickness direction increases and the plastic

zone reduces in size accordingly.

For =1/2, we have the extreme case of an incompressible material which has ry=0 at = 0, as there will be equal triaxial tension in this case.


17/29

7.18 A sheet of polystyrene has a thin central crack with 2a = 50 mm. The crack

propagates catastrophically at an applied stress of 10 MPa. Youngs modulus polystyreneis 3.8 GPa, and the Poissons ratio is 0.4. Find G Ic.

( )( )

( ) ( )( )

2

9

226

22

2

1736

108.3

4.01025.01010

1

1

4.0

8.3

10

025.0

25

502

=

=

=

=

=

=

=

=

=

=

mJ

E

aG

a

GE

GPaE

MPa

m

mma

mma

Ic

Ic


18/29


19/29

7.20 300-M steel, commonly used for airplane landing gears, has a G cvalue of 10

kN/m. A nondestructive examination technique capable of detecting cracks that

are 1 mm long is available. Compute the stress level that the landing gear can

support without failure.

mMPaK

GEK

mkNG

c

cc

c

8.45

102100101010210

/10

12392

=

===

=

The nondestructive examination technique can detect 1 mm long cracks, i.e., ac=1 mm. In other words, we are assuming that when the cracks become detectable,

the landing gear must be substituted. Assume that the situation in practice

corresponds to a single edge notch, i.e.,

MPa

a

K

aK

c

c

cc

6.729

10112.1

8.45

12.1

12.1

3

=

==

=


20/29

7.25 An engineering ceramic has a flexure strength that obeys Weibull statistics with m

= 10. If the flexure strength is equal to 200 MPa at 50 % survival probability, what is the

flexure strength level at which the survival probability is 90%?

( )

( )[ ] [ ]

( )[ ] [[ ] [

[

( )

( ) [ ]MPa

nnn

VP

n

nn

nnnn

nnmVPnn

VPn

MPaVP

o

o

o

o

m

o

16533.5109.0/1

?9.0

33.5

2001037.0

200105.0/1

/1

//1

2005.0

=

=

==

=

=

=

=

=

==

]]

]

lll

l

ll

llll

llll

l


21/29

7.26 What would be the flexure strength, at 40 % survival probability, if the ceramic in

the preceding problem is subjected to a hot isostatic processing (HIP) treatment that

greatly reduces the population of flaws and increases m to 60? Assume that the flexure

strength at 50 % survival probability is unchanged.

Weibull modulus, m = 10

HIP treatment increases m to 60

= 200 MPa at 50 % survival probability

Weibull statistics:

( )

( )( )

( )( )[ ]

( )( )[ ] MPaMPavPn

yprobabilitsurvivalat

MPavPn

vPn

vP

m

oo

m

o

o

m

o

o

m

o

o

8.1938.193/1

?%90

2.201/1

exp

/1

/1

===

=

==

=

=

l

l

l

The population of flaws is reduced after HIP treatment. Thus, the flexure strength isincreased by about 17% in this case.


22/29


23/29

7.29 Aluminum has a surface energy of 0.5 Jm-2

and a Youngs modulus of 70 GPa.

Compute the stress at the crack tip for two different crack lengths: 1 mm and 1 cm.

a

Ec

2= GPaE

mJ70

/5.02

==

For 2a = 1 mm

( )( )( )

MPac

c

68.6

105

5.0107024

9

=

=

For 2a = 1 cm

( )( )( )

MPac

c

11.2

105

5.0107023

9

=

=


24/29

7.30 Determine the stress for crack propagation under plane strain for a crack length

equal to 2 mm in aluminum. Take the surface energy equal to 0.018 J/m2, Poissons ratio

to be 0.345, and the modulus of E = 70.3 GPa.

( )

( )( )( )kPa

a

E

ma

GPaE

m

J

c

c

c

2.954

345.01101018.0103.702

1

2

1012

1102

345.0

3.70

018.0

23

9

2

33

2

=

=

=

==

=

=

=


25/29

7.31 Calculate the maximum load that a 2024-T851 aluminum

alloy (10 cm x 2 cm) with a central through-the-thickness crack

(length 0.1 mm) can withstand without yielding. Given: y =500

MPa and KIc=30 MPa m .

011.1

1.0

1052

1

100

1

200,13152.1256.01

3

32

=

=

==

+

+=

Y

mW

ma

W

a

W

a

W

aY

( )

( )(

kNP

MPaPA

P

MPa

MPa

aYKIc

5.473

02.01.08.236

8.236

105011.1

30

3

=

=

=

=

=

=

)


26/29

7.32 An infinitely large sheet is subjected to a far-field stress of 300 MPa. The

material has yield strength of 600 MPa, and there is a central crack

/7 cm long.

a)

Calculate the stress intensity factor at the tip of the crack.b) Estimate the size of the plastic zone size at the tip of the crack.

a) aK =

mMPaK

K

12.56

100

1

2

1710300 6

=

=

b) stressplaneforr

K

yy =

2

mxr

mr

strainplanefory

Kr

mxr

MPa

mMPar

Kr

y

y

y

y

y

y

y

4

2

2

3

2

2

2

1064.4

600

12.56

6

1

6

1

1039.1

)600(2

)12.56(

2

=

=

=

=

=

=


27/29

7.33

What is the maximum allowable crack size for a material that

has MPaandmMPaK vIc 380,155 == ? Assume a plane-strain condition

and a central crack.

We take the design stress to be half the yield stress.

MPadesign 6902

1380==

mma

ma

a

Ka

aK

Ic

Ic

42

1002.2

1

690

55

1

3

2

2

=

=

=

=

=

7.35 An Al2 O3specimen is being pulled in tension. The specimen contains flaws

having a size of 100 m.

a) If the surface energy of Al2O3is 0.8 J/m2, what is the fracture stress? Use

Griffith criterion. E = 380 GPab) Estimate the fracture stress if the fracture toughness is 4 MPa m

0.5. Assume two

positions for flaws :

1) in the center of an infinite body

2) in the edge of an infinite body.

(a) According to Griffiths criterion, the critical stress required for the crack to

propagate in the plane-stress situation.


28/29

MPac

c

mlengthcrackthehalfisa

mJenergysurfacetheiss

where

a

sE

c

2.62

61050

8.09103802

50

28.0

2

=

=

=

=

=

(b)

1) For center cracked infinite body, Y = 1

( )MPa

MPa

aYKIc

2.319

10500.1

4

6

=

=

=

2)

For single edge cracked infinite body, Y = 1.12

( )MPa

MPa

aYKIc

201

1010012..1

4

6

=

=

=


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