mekanika retak sedikit
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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
=
=
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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
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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
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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,
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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.
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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
=
=
=
=
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(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
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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.
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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.
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= 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.
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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
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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
=
==
=
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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
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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.
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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
=
=
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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
=
=
=
==
=
=
=
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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
=
=
=
=
=
=
)
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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
=
=
=
=
=
=
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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.
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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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