Chapter 1 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
CHAPTER 1
THEORY OF ELASTICITY
1.1 A thin sheet made of an aluminum alloy having E = 67 GPa, G = 256 GPa and also v = 1/3
was used for two dimensional surface strain measurements. The measurements provided
 xx  10.5 x10 5 ,  yy  20 x10 5 , and  xy  240 x10 5. Determine the corresponding stresses.
Solution:
 xx 
E  xx  v yy   2.89 MPa 

  

2 
 yy  (1  v )  yy  v xx   12.44 MPa 
 xy 
E xy
2(1  v)
 60.45 MPa
1.2 Determine (a) the principal stresses and strains and (b) the maximum shear stress for the case
described in Problem 1.1.
Solution:
(a)  1, 2 
 xx   yy
2
  xx   yy
 
2

2

   xy2  4.28 MPa  60.60 MPa

 1  56.32 MPa
 2  64.88 MPa
The principal strains are
 1, 2 
 xx   yy
2
  xx   yy
 
2

   xy 
  

  2 
2
2
 4.75 x10 5  1.20 x10 3
 1  1.15 x10 3
 2  1.25 x10 3
(b) The maximum
 max
  xx   yy
 
2

2

   xy2  60.60 MPa

1
Chapter 1 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
1.3 Calculate the diameter of a 1-m long wire that supports a weight of 200 Newton. If the wire
stretches 2 mm, calculate the strain and the stress induced by the weight. Let E = 207 GPa.
Solution:

2 mm
l
 3
 2 x10 3  0.20%
l o 10 mm
4P
 0.78 mm

  E  414 MPa
P  A  ( / 4)d 2  200 N
d
1.4 Derive an expression for the local uniform strain across the neck of a round bar being loaded
in tension. Then, determine its magnitude if the original diameter is reduced 80%.
Solution:
Volume constancy:
AL  Ao Lo
 zz 
L
dL
2
 lnL / Lo   ln d o / d 
L
Lo

 zz  2 ln d o / d 
If d  0.80d o , then
 zz  2 ln 1 / 0.80  0.4463
 zz  44.63%
1.5 The torsion of a bar containing a longitudinal sharp groove may be characterized by a
warping function of the type [after F.A McClintock, Proc. Inter. Conf. on Fracture of Metals,
r
Inst. of Mechanical Eng., London, (1956) 538] w   z    ( ydx  xdy ) . The displacements are
0
 x  0 and  y  rz , where  and r are the angle of twist per unit length and the crack tip
radius, respectively. The polar coordinates have the origin at the tip of the groove, which has a
radius (R). Determine w, the shear strains  rz and   z . In addition, predict the maximum of the
shear strain   z .
2
Chapter 1 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Solution:
 y  1   z
 
z  r  
R 2
 1
 r    R 2   r 
r
r
Let x  R cos and y  R sin  so that
 z 
dx   R sin d and dy  R cos d
 z
r

0
0
 z    ( ydx  xdy)    R 2 d
If r  0 then   z   max  
 rz  0
1.6 A cantilever beam having a cross-sectional area of 1.5 cm2 is fixed at the left-hand side and
loaded with a 100 Newton downward vertical force at the extreme end as shown in the figure
shown below. Determine the strain in the strain gage located at 8 cm from the fixed end of the
shown steel cantilever beam. The steel modulus of elasticity of is E = 207 GPa.
-
+
Strain gage
P = 100 N
h = 0.5 cm
b = 3 cm
8 cm
X = 20 cm
Solution:

M   0
M
FBD:
( Moments )
P X  X 1   M  0
M  P X  X 1 
X – X1
Thus,
6P X  X 1 
  P( X  X13)( h / 2 )   2 
nh
bh
12
Using Hooke’s law yields the elastic strain



E
6P X  X

Ebh 2
1

(6)(100 N )(20 m  8 m) x10  2
(207 x10 6 N / m)(3 x10  2 m)(0.50 x10  2 m) 2
9
3
 4.64 x10  4
  MC
I
Ch
2
3
bh
I
12
Chapter 1 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
1.7 The stress-strain behavior of an annealed low-carbon steel (σys = 200 MPa and E = 207 GPa)
obeys the Hollomon equation with k = 530 MPa and n =0.25. (a) Plot the true and engineering
stress-strain curves. Calculate (b) the tensile strength (  ts ) and (c) the strain energy density up to
the instability point.
Solution:
(a) Plots
 t  1   
 t  ln 1   
 t  k  t n
1     k ln1   n

k ln 1   
1   
n
 t  n  0.25
@ Instability point
1000
True
 t  1200 t 
0.25
Stress
750
Engineering

500
1200ln 1   
1  0.25
0.25
Yield Point
250

ys ,  ys   0.0043, 305.81 MPa 
 ts  0.25
00
00
0.05
0.05
0.10
0.1
0.15
0.15
0.20
0.2
0.25
0.25
0.30
0.3
Strain
(b) The engineering yield strain and the true strain at the instability point on the true stress-strain
curve are, respectively
 ys
900 MPa

 4.3478 × 10- 3
3
E
207 x10 MPa
 t  n  0.25 @ Instability point
 ys 
At the instability or ultimate tensile strength point (maximum),
 max  k tn  1,200 MPa 0.250.25  848.53 MPa
 t  ln 1   
  1  exp( t )  1  exp(0.25)  0.28403
4
Chapter 1 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
 max  1    ts
 ts   max / 1   
 ts  374.77 MPa  / 1  0.28403  291.87 MPa
(c) The strain energy density (W) is just the area under the stress-strain curve. In general, the true
and engineering strain energy densities are

  ts

  ys


Wt   d
   d 
0




 Elastic   ys
 Plastic
 ys
 ts

E ys2
k
 ts 1 n   ys 1 n
Wt   Ed   k d 

2
1 n
0
 ys
n
Wt
2

207,000 4.3478 × 10- 3 





1,200
0.284031 0.25  4.3478 × 10-3 1 0.25  335.50 MPa
1  0.25
2
MN m
Wt  335.50 2
 335.50 MJ /m3
m m
and

  ts

  ys


W    d
    d 
0




 Elastic   ys
 Plastic
n
E ys2
k ln 1   
k
n 1
W   E d    
d 

ln  ts  1   ys  1
1   
2
n 1
0
 ys
 ys
 ts

2

207,000 4.3478 × 10- 3  
W




1,200 
- 3 0.25 1

 ln 0.28403 - 4.3478 × 10 
 0.25  1 
2
W  0.99932  0.95718i
Therefore, the engineering strain energy density can not be determined analytically using the
modified Hollomon equation. Instead, the solution can be achieved numerically. Thus,
1200ln 1   
0 207000 d  0.0043478

1   
0.0043478
W 
0.28403
0.25
d  170.60 MPa
W  170.60 MJ /m 3
Wt  2W
5
Chapter 1 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
1.8 The figure below shows a schematic cross-sectional view of a pressure vessel (hollow
cylinder) subjected to internal and external pressures. Determine the stresses at a point P r,   in
polar coordinates when (a) Pi  0 and Pi  0, (b) Po  0, (c) Pi  0 and (d) a  0 so that the
hollow cylinder becomes a solid cylinder. (e) Plot  rr  f (r ) and  rr  f (r ) . Let a  450 mm,
b  800 mm, Pi  0 and P  40 MPa. The valid radius range must be 0.45 m  r  0.80 m . Use
the following Airy’s stress function
  c1  c2 ln r  c3 r 2
Along with the boundary conditions
 rr   Pi
 r  0 @ r  a
 rr   Po
 r  0
Solution:
(a) Derivatives of   c1  c2 ln r  c3 r 2
 c2

0

 2 c3 r
 2
0
r
r

r
2
2
c


0
  22  2c3
2
r
r
 2
From eq. (1.58),
1  1  2 c2


 2 c3
r r r 2  2 r 2
 2
c
   2   22  2c3
r
r
1  1  2
 r  2

0
r  r r
 rr 
6
@r b
Chapter 1 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Using the boundary conditions yields
c
c
 rr   Pi  22  2c3
 Pi  22  2c3
r
r
c
c
 rr   Po  22  2c3
 Po  22  2c3
r
r
Replace r for the proper radius and solve eq. (a) for c2 and c3
(a)
c2
 2 c3
r2
c
 Po  22  2c3
r
from which
c
 Pi  22  2c3
a
c
2c3   Pi  22
a
and
c
 Po  22  2c3
b
c
2c3   Po  22
b
 Pi 
Then,
c
c
Pi  22  Po  22
a
b
c2 c 2

 Po  Pi
a 2 b2
o  Pi
c2
1 a 2b2
a 2b2


Pi  Po 


P

P

P


P

o
o
i
o
b2
b2 b2  a 2
b2  a 2
 a2

a2P
a2P
a2P
& 2c3  2 i 2  2 i 2  Po  2 i 2  Po  2
 1
2
b a
b a
b a
b a

a 2b2
Po  Pi 
c2  2
b  a2
a 2b2
Po  Pi  &
c2  2
b  a2
2c3   Po 
2 c3 
c3 
a 2 Pi
b 2 Po

b2  a 2 b2  a 2
a 2 Pi  b 2 Po
2b 2  a 2 
Thus,
a 2 Pi  b 2 Po a 2 b 2 Po  Pi 
 rr 

b 2  a 2 r 2
b2  a 2
a 2 Pi  b 2 Po a 2 b 2 Po  Pi 

b 2  a 2 r 2
b2  a 2
0
  
 r
7
Chapter 1 Fracture Mechanics, 2nd ed. (2015)
(b) If Po  0, then
Solution Manual
(c) If Pi  0, then
a P
 rr  2 i 2
b a
 b2 
1  2 
r 

b 2 Po
 rr  2
b  a2
 a2

 2  1
r

a 2 Pi
b2  a 2
 b2 
1  2 
r 

  
b 2 Po
b2  a 2
 a2

 2  1
r

2
  
 r  0
 r  0
(d) If a  0, then
 rr   Po
    Po
 r  0
(e) The plot
100
75

(MPa)
 
50
25
0
 rr
-25
-50
0.4
0.5
0.6
Radius r (m)
8
0.7
0.8
Chapter 1 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
1.9 Consider an infinite plate with a central hole subjected to a remote uniform stress as shown in
Example 1.5. The boundary conditions for this loaded plate are (1)  x   x  S and σy = τxy = 0
at r   and (2)  r   r  0 at r  a. Use the following complex potentials [12]
  z 
S
4
1
2a 2
z2
and   z   S2 1 
a2
z2

3a 4
z4
to determine σr, σθ and τrθ (in polar coordinates).
Solution: The infinite plate with a central hole subjected to a remote uniform stress is
Use the following stress expressions
 r     2  z  2  z  4 Re   z
    r  i2 r  2z  z    ze i2
Recall that the Euler's formula for z and z² are
z  re i  rcos   i sin 
z  re i  rcos   i sin 
z m  r m e im  r 3 cos m  i sin m
and their real and imaginary parts are
Re z  r cos  & Re z m  r m cos m
Im z  r sin  & Im z m  r m sin m
Apply the Euler's formula to the above complex potentials so that
2
2
  z  S 1  2a2
 S 1  2a
2
i2
4
4
z
r e
2
S
2a
  z 
1  2 e i2
4
r
2
4 Re   z  S 1  2a2 cos 2
r
(a) The first stress equation:
(a)
 r     4 Re  z  S 1 

2a 2
r2
cos 2
Then,
9
Chapter 1 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
2
 z  S  a 2
4
2z
2
2
2
2
 z   2Sa3  Sa3  Sa
 Sa3 e i3
3 i3
2z
z
r e
r
2
Sa

 z  3 cos 3 (Real part)
r
Thus,
z  z  re i 
Sa 2
r 3 e i3

Sa 2
r2
e i4
(b) The second potential:
  z   S2 1 
a2
z2

3a 4
z4
  S2 1 
a2
r 2 e i2

3a 4
r 4 e i4
Hence,
    r  i2 r  2z  z    ze i2
2
2 i2
4 i2
    r  i2 r  2 Sa2 e i4 e i2  S e i2  Sa2 e i2  3Sa4 ei4
2
r
2r e
2r e
2
2
4
    r  i2 r  S 2a2 e i2  e i2  a2  3a4 e i2
r
r
r
Real parts:
2
2
4
    r  S 2a2 cos 2  cos 2  a2  3a4 cos 2
r
r
(b)
(c)
r
Add eqs. (a) and (c) and solve for σθ
2
4
   S2 1  a2  S2 1  3a4 cos 2
(d1)
Use eq. (a) to solve for σr
2
 r  S2 1  a2  S2 1 
(d2)
r
r
r
4a 2
r2

3a 4
r4
cos 2
Using the imaginary part of eq. (b) yields shear stress τrθ as
2
4
i2 r  S i 2a2 sin 2  i sin 2  i 3a4 sin 2
r
r
2
4
 r   S 1  2a2  3a4 sin 2
2
r
r
(d3)
At r = a
r  0
(f1)
   S  2S cos 2  S1  2 cos 2
(f2)
 r  0
(f3)
10
Chapter 1 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
1.10 Use the Cauchy-Riemann condition to show that (a) f(z) = 1/z is analytic and (b) its
derivative is f(z)= -1/z².
Solution:
(a) If z = x + iy, then
1
z

1
z

1 xiy
xiy xiy

1
xiy
x
x 2 y 2

xiy

x 2 y 2
iy
x 2 y 2
Let
u
v 
x
x 2 y 2
y
x 2 y 2
Then
x 2 y 2 x  x x 2 y 2
u
x

u
y

x 2 y 2

x 2 y 2 2x 2

2
2
x 2 y 2

y 2 x 2
x 2 y 2
2xy
x 2 y 2
2
and
2xy
v 
2
2
x
x  y 2 
v  x 2  x 2
2
y
x 2  y 2 
Thus,
u

y
 v
x
Therefore, f(z) = 1/z is analytic because ∂u/∂y = ∂v/∂x.
(b) The derivative of f(z) = 1/z is
f  z 
f  z 

d
dz
 1z  
y 2 x 2
x 2 y 2
2
 
z2
zz 2
2
 i v
x
2xy
i
iiyx  2
i x 2 y 2
u
x
x 2 y 2
x 2 y 2
xiy 2


2
yix  2

x 2 y 2
2
2

1
z2
11
z2
xiyxiy 2
2
Chapter 1 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
1.12 Evaluate the Cauchy integral formula given below for the complex function when zo = π.
fz o  
1
2i

cos z
z 2 1
dz
Solution:
The integral has to be modified as
1
2i
fz o   1
2i
fz o  
fz o   1
z dz
 zcos
2
1
cos z
dz
 z  1z
 1
dz
 zfz
1
dz  2ifz o 
 zfz
1
where
fz 
cos z
z1
Then,
dz  2ifz o 
 zfz
1
dz  2i
 zfz
1
cos z o
zo  1
dz  2i
 zfz
1
cos 
1
z o 
 1. 52i
12
Chapter 2 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
CHAPTER 2
INTRODUCTION TO FRACTURE
MECHANICS
2.1 Show that the applied stress   0 as the crack tip radius   0. Explain
Solution:
From eq. (2.28),
a
 max  2

 
and
1

 max
2
a
Therefore,   0 as   0. This means that the applied stress   0 since the material has
fractured or separated, at least into two pieces, and there is not a crack present so that   0. On
the other hand,   0 means that the existing crack is very sharp as in the case of a fatigue
crack.
2.2 For a Griffith crack case, crack propagation takes place if the strain energy satisfies the
inequality 
U ( a )  U ( a   a ) 2  a , where a  crack extension and  is the surface energy.
Show that the crack driving force or the strain energy release rate at instability is G 
U(a)
a
Solution:
Use the Taylor’s series to expand the left side term of the inequality. Thus,
2
3
1  U (a ) 2 1  U ( a ) 3
1 U ( a )
U (a   a )  U ( a ) 
a 
a 
 a  .......... ....
1! a
2! a 2
3! a 3
(2.1)
Using the first two terms yields
U (a   a )  U (a ) 
U ( a )
U (a )  U (a   a )  
a
U ( a )
a
(2.2)
a
a
(2.2)
1
.
Chapter 2 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Then,


U ( a )
a
U ( a )
a
 a  2 a
(2.3)
 2
(2.4)
 G
Therefore,
G
U ( a )
a
(2.5)
2.3 A 1mmx15mmx100mm steel strap has a 3-mm long central crack is loaded to failure.
Assume that the steel is brittle and has E  207,000 MPa,  ys  1500 MPa, and
K IC  70 MPa m . Determine the critical stress ( c ) and the critical strain energy release rate.
Solution:
a = 1.5 mm and a / w  0.10
The geometry correction factor:
  f (a / w) 
w
a 
tan    1.07
a  w 
The stress intensity factor
K IC    f a
c 
2
K IC

a
 1020 MPa
The strain energy release rate
2
K IC
E
GIC  23.70 kPa m
GIC 
GIC  23.70 kN / m 2
2.4 Suppose that a structure made of plates has one cracked plate. If the crack reaches a
critical size, will that plate fracture or the entire structure collapse? Explain.
Answer: If the structure does not have crack stoppers, the entire structure will collapse since the
cracked plate will be the source of structural instability.
2
Chapter 2 Fracture Mechanics, 2nd ed. (2015)
2.5
Solution Manual
What is crack instability based on according to Griffith criterion?
Answer: It is based on the energy consumption creating new surfaces or in developing crack
extension. This energy is the surface energy 2.
2.6 Can the Griffith Theory be applied for a quenched steel containing 1.2%C, if a pennyshaped crack is detected?
Answer: Despite the quenched steel is brittle, the Griffith Theory applies only if a plastic zone
does not form at the crack tip.
2.7 Will the Irwin Theory (or modified Griffith Theory) be valid for a changing plastic zone size
as the crack advances?
Answer: No. It is valid if the plastic zone remains constant as it the case in many materials
under plane strain conditions.
2.8 What are the major roles of the surface energy and the stored elastic energies in a crack
growth situation?
Answer: The surface energy acts as a retarding force for crack growth and the stored elastic
energy acts to extend the crack. Thus, concurrent with crack growth is the recovery of elastic
strain energy by the relaxation of atomic bonds above and below the fracture plane.
2.9
What does happen to the elastic strain energy when crack growth occur?
Answer: It is released as the crack driving force for crack growth.
2.10 What does U/a = 0 mean?
Answer: Crack growth occurs at    f , as shown in eq. (2.23), since the magnitude of crack
growth and the crack resistance force become equal. That is,
2s 
a 2
E
2
If 2s  a , then fracture occurs and the fracture toughness without any plastic zone
E
deformation is G  2γ at    f , as shown in eq. (2.31) with  p  0. Thus, G  2γ is related
c
c
s
to an irreversible process of fracture.
3
s
Chapter 2 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
2.11 Derive eq. (2.28) starting with eq. (2.20).
Solution:
From eq. (2.20),
Kt 
 max
a
 2


1
 max 
2
a


Multiplying this expression by
eq. (2.28)
1
 max  
2
 yields the stress intensity factor as per
a
    a

K I   a
4
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
CHAPTER 3
LINEAR ELASTIC FRACTURE
MECHANICS
3.1 A steel strap 1-mm thick and 20-mm wide with a through-the-thickness central crack 4-mm
long is loaded to failure. (a) Determine the critical load if KIC = 80 MPa m for the strap
material. (b) Use the available correction factor,   f ( a / w) , for this crack configuration. Now,
do the following three comparisons to have a better understanding: Feddersen:
 c  ( fraction)  , Irwin:  c  ( fraction)  and Koiter-Benthem:  c  ( fraction)  .
2a
P
w
P
B=1 mm
Solution:
B = 1mm, w = 20mm, 2a = 4 mm, a/w = 0.10, KIC = 80 MPa m , K I   c a f a / w ,
(a) Finite plate:
From eqs. (3.31) and (3.32) along with a / w  0.10 rad .  (180o )( 0.10) /   5.7296o
.
  seca w  1025
w
a 
Irwin [13]:
f  a w 
tan   1017
.
a  w 
1
1  a  1304
 .
a
Koiter-Benthem [14]: f  a w 
  w .  w   1021
a
1 2
f aw 
Feddersen [12]:
2
 c  984.64 MPa
 c  992.38 MPa
 c  988.49 MPa
w
(b) Infinite plate approach [a/w  0 since f(a/w)  1].
K
80MPa m
 c  IC 
 1009.25MPa
a
 2  10 3 m

Feddersen:
Irwin:
Koiter:

 c  97.56% 
 c (infinite)
 c  98.33% 
 c (infinite)
 c  97.94% 
 c (infinite)
Therefore, the calculated  c values do not differ much since the plate dimensions are large
enough and the initial crack size is relatively small.
1
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
3.2 A steel tension bar 8-mm thick and 50-mm wide with an initial single-edge crack of 10-mm
long is subjected to a uniaxial stress  = 140 MPa. (a) Determine the stress intensity factor KI. If
KIC = 60 MPa m , is the crack stable? (b) Determine the critical crack size, and (c) determine the
critical load.
a
w
P
P
B
Solution:
 = P/A= 140 MPa, KIC = 60 MPa m ,
B = 8 mm,
w = 50 mm,
a = 10 mm,
a 
K I   a f  
w 
x = a/w = 0.20,
A = Bw
From Table 3.1, K I    a and
  f  x   1.12  0.231x   10.55 x   21.71 x   30.38x 
  f  x  0.20  1.37
2
3
4
(a) The applied stress intensity factor
KI = 34 MPa m
KIC = 60 MPa m
Therefore, the crack is stable because KI < KIC.
(b) K IC   a c . f  x 
2
1  K IC 
ac  
 0.0311 m  31.10 mm
  f x  
 
(c) The critical load
P
 
A
A  Bw  8 mm 50 mm   400 mm 2  4  10 4 m 2
 MN 
P  A  4  10  4 m 2 140 2  0.056MN  56 kN (Applied)
m 

 MN


4  10 4 60 2 m 
AK IC
 m
  0.0988 MN  98.84 kN (Critical)
Pc 

3
f  x  a
1.37  10  10 m 
2
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
3.3 A very sharp penny-shaped crack with a diameter of 22-mm is completely embedded in a
highly brittle solid. Assume that catastrophic fracture occurs when a stress of 600 MPa is
applied. a) What is the fracture toughness for this solid? (Assume that this fracture toughness is
for plane strain conditions). b) If a sheet 5-mm thick of this solid is prepared for fracturetoughness testing. Would the fracture-toughness value [(calculated in part a)] be an acceptable
number according to the ASTM E399 standard? Use  ys = 1342 MPa. c) What thickness would
be required for the fracture-toughness test to be valid?
Solution:


2
2
 a  600 MPa   11  10 3 m  71 MPa m


a) K IC 
K
b) B  2.5 IC

 ys
c) B  7 mm
2
2



  2.571 MPa m   7 mm > 5 mm (Not valid)
1342 MPa 




3.4 (a) One guitar steel string has a miniature circumferential crack of 0.009 mm deep. This
implies that the radius ratio is almost unity, d / D  1 . b) Another string has a localized miniature
surface crack (single-edge crack like) of 0.009 mm deep. Assume that both
strings are identical with an outer diameter of 0.28 mm. If a load of 49 N is

applied to the string when being tuned, will it break? Given properties:
K IC  15MPa m ,  ys  795MPa.

Solution:
(a) Circumferential crack
Dd
If a 
 0.009 mm , then d  0.262 mm
2
4P
(4)(49 N )
 

 908.87 MPa
2
d
 (0.262 x10 3 m)
  f (d / D) 
1
2
2
3
D  1 D 3 d 
d 
d  



0
.
36

0
.
73








d  2 d 8 D 
D 
D  
For d / D  0.9357,
  1.14
3
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
K I    a
Solution Manual




K I  (1.14)908.87 MPa   0.009  10 3 m
K I  5.51 MPa m
(b) Surface edge crack
  1.12
K I    a
K I  (1.12)908.87 MPa   0.009  10 3 m
K I  5.41 MPa m
Therefore, the strings will not break since K I  K IC in both cases. When d / D  1 and the crack
size is very small, either approach gives similar results.
3.5 A 7075-T6 aluminum alloy is loaded in tension. Initially the 10mmx100mmx500mm plate
has a 4-mm single-edge through–the-thickness crack. (a) Is this test valid? (b) Calculate the
maximum allowable tension stress this plate can support, (c) Is it necessary to correct KI due to
crack-tip plasticity? Why? or Why not? (d) Calculate the design stress and stress intensity factor
if the safety factor is 1.5. Data:  ys = 586 MPa and KIC = 33 MPa m .
Solution:
2


(a) B  2.5K IC
  7.93 mm and a w  0.0375

ys


Plane strain condition holds because the actual thickness is greater than the required value.
(b) K I    a ; a = 4 mm and w = 100 mm
 
f a  
w
  f a
 
2
3
f 0.04  1.13
Thus,
 
K IC

4
a 
a 
a 
a 
 1.12  0.231  10.55   21.71   30.38 
w
w 
w 
w 
w 
a
 261 MPa
4
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
(c) Plastic zone correction:
2
1 K IC

r
    0.17mm
ys 
6 
K eff    a eff
with a eff  a  r = 4.17 mm
K eff  1.13261 MPa   4.17 x10 3 m 
K eff  33.76 MPa m
Therefore, there is no need for crack tip plasticity since K I ,eff  K IC .

261 MPa

 174 MPa
SF
1.5
K
33 MPa m
K Id  IC 
 22 MPa m
SF
1.5
(d)  d 
3.6 A steel ship deck (30mm thick, 12m wide, and 20m long) is stressed in the manner shown
below. It is operated below its ductile-to-brittle transition temperature (with KIc = 28.3 MPa m ).
If a 65-mm long through–the-thickness central crack is present, calculate the tensile stress for
catastrophic failure. Compare this stress with the yield strength of 240 MPa for this steel.
Solution:
2a = 65mm
w = 12 m
L = 20m
  f a / w  1 since a/w  0
P
P
2a
KI
Transition Range.
(MPa m )
or
CVN
Energy
(Joules)
T
5
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
K IC    a
 
K IC
a

Solution Manual
K IC  28.3 MPa m and  ys  240 MPa
28.3 MPa m
65

   10 3 m 
2

 88.57 MPa
Therefore,    ys and to assure structural integrity, the safety factor (SF) that can be used to
avoid fracture or crack propagation is in the order of
SF 
 ys


240 MPa
 2.71
88.57 MPa
3.7 Show that
KI
 0 for crack instability in a large plate under a remote tensile external stress.
da
Solution:
K I   a
dK I 1

 
 0 since   0 and a  0
da
2
a
3.8 The plate below has an internal crack subjected to a pressure P on the crack surface. The
stress intensity factors at points A and B are
KA 
KB 

P

P
a
a
ax
dx
ax
ax
dx
ax
2a
P
A
B
Use the principle of superposition to show that the total stress intensity factor is of the form
K I  P a
Solution:
According to the principle of superposition, the total stress intensity factor is
KI  KA  KB 
P
 ax
ax
a
dx
dx  2 P



ax

a2  x2
 ax

a 
6
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Furthermore,
ax

ax
ax

ax
ax
ax
dx
x
 ar c sin
2
2
a
a x


ax
ax

(a  x)(a  x)
( a  x )( a  x )

2a
a2  x2
and
a
x
arcsin

a
K I  2P
If x = a on the crack surface, then arcsin 1   / 2  90 o and
a
x 2P

ar sin 
a   P a

a 
2
K I  2P
3.9 A pressure vessel is to be designed using the leak-before-break criterion based on the
circumferential wall stress and plane strain fracture toughness. The design stress is restricted by
the yield strength  ys and a safety factor (SF). Derive expressions for (a) the critical crack size
and (b) the maximum allowable pressure when the crack size is equals to the vessel thickness.
Solution:
(a) The critical
crack size
K IC    ac
 ys
SF
d 
Thus,
K IC 
  ys
ac
SF
1 S  K
ac   F   IC
   
 ys
2
2




Select some alloy having known K IC and  ys values. The alloy with the highest
K IC


 ys
2

 values should be used for constructing the pressure vessel.


7
2
K IC
and
 ys
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
3.10 A stock of steel plates with GIC = 130 kJ/m², σys = 2,200 MPa, E = 207 GPa, v = 1/3 are
used to fabricate a cylindrical pressure vessel (di = 5 m and B = 25.4 mm). The vessel fractured
at a pressure of 20 MPa. Subsequent failure analysis revealed an internal semi-elliptical surface
crack of a = 2.5 mm and 2c = 10 mm. (a) Use a fracture mechanics approach to predict the
critical crack length this steel would tolerate. (b) Based on this catastrophic failure, another
vessel was constructed with do = 5.5 m and di = 5 m. Will this new vessel fracture at a pressure
of 20 MPa if there is an internal semi-elliptical surface crack having the same dimensions as in
part (a)?
Solution:
(a) The vessel is based on the thin-wall theory since the thickness is
d
5 x103 mm
B i 
 250 mm (Requirement)
20
20
B  25.4 mm (Given) is less than the required thickness. So use the thin-wall theory.
Thus, the fracture hoop stress is
h  Pi di 
2B
20 MPa
5x10 3 mm
2 25. 4 mm
 h  1, 968. 50 MPa
Using eq. (2.34) yields the critical crack length
ac 
ac 
EGIC
 1  v 2 2h
207x10 3 MPa
130x10 3 MPa. m
 1  1/3 2  1, 968. 50 MPa
2
a c  2. 49 mm  2. 5 mm
(b) For the new vessel,
5. 6 m  5 m
B  do  di 
 0. 30 m
2
2
d o  5. 6  1. 12
5
di
Also,
h 
d o /d i  2  1
d o /d i  2  1
P i  8. 86Pi
 h  8. 8620 MPa  177. 23 MPa
   h  P i  9. 86P i  197. 23 MPa
8
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Then,
a/2c  2. 5/10  0. 25
/ys  197. 23/2, 200  0. 09
From eq. (3.44),

2
Q 
2
Q  1. 63
Q
2
2
3  a 2
4
2c
3  0. 25 2
4
2
 7 ys
33
2
 7 0. 09 2
33
2
Thus, the applied stress intensity factor is
KI    a
Q
KI  197. 23 MPa
 2. 5x10 3 
1. 63
KI  13. 69 MPa m
From eq. (2.33) at fracture,
EGIC
1  v2
KIC 
207x10 3 MPa
KIC 
130x10 3 MPa. m
1  1/3 2
KIC  174 MPa m
Therefore, the new vessel will not fracture because KI < KIC.
3.11 A cylindrical pressure vessel with B = 25.4 mm and di = 800 mm is subjected to an internal
pressure Pi. The material has K IC  31 MPa m and σys = 600 MPa. (a) Use a safety factor to
determine the actual pressure Pi. (b) Assume there exists a semi-elliptical surface crack with a =
5 mm and 2c = 25 mm and that a pressure surge occurs causing fracture of the vessel. Calculate
the fracture internal pressure Pf. (c) Calculate the critical crack length.
Solution:
(a) Let's figure out which pressure vessel wall theory should be used.
B
di
20

800 mm
20
 40 mm
9
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Therefore, the thin-wall theory should be used. The actual or applied pressure is
 ys
h  P i d i 
2B
SF
2 25. 4 mm 600 MPa
2B ys
Pi 

SFdi
2. 5 800 mm
P i  15. 24 MPa
(b) Using the principle of superposition yields
  h  Pi  Pi di  Pi
2B
d
800  1 P i
i

 1 Pi 
2B
2x25. 4
  16. 75P i  16. 75 15. 24 MPa
  255. 24 MPa (Actua l)
Also,
a  5  0. 20
2c
25
  255. 24  0. 43
 ys
600
a  5  0. 20
B
25. 4
M  Mk  1 [
See eq.(3.46) & Figure 3.6]
From eq. (3.44),

2
Q 
2
Q  1. 50
Q
2
2
3  a
4
2c
3  0. 2 2
4
2
2
 7 ys
33
2
 7 0. 43 2
33
2
From eq. (3.41),
K IC  MM k 
a
Q
 16. 75P f
a
Q
Now, the surge pressure is
Pf 
K IC
16. 75
31 MPa m
Q
a 
16. 75
1. 50
 5x10 3 m
Pf  18. 09 MPa
10
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
(c) The critical crack length is
KIC    a  16. 75P i
Q
Q
ac  
KIC
16. 75Pi
2
 ac
Q
 1.5
31 MPa m
2
16. 75  15. 24 MPa
a c  7. 04 mm
3.12 This is a problem that involves strength of materials and fracture mechanics. An AISI 4340
steel is used to design a cylindrical pressure having an inside diameter and an outside diameter of
6.35 cm and 12.07 cm, respectively. The hoop stress is not to exceed 80% of the yield strength of
the material. (a) Is the structure a thin-wall vessel or a thick-wall pipe? (b) What is the internal
pressure? (c) Assume that an internal semi-elliptical surface crack exist with a = 2 mm and 2c =
6 mm. Will the vessel fail? (d) Will you recommend steel for the pressure vessel? Why? or Why
not? (e) What is the maximum crack length the AISI 4340 steel can tolerate? Explain.
Solution:
Internal Crack
d i  6.35 m
External Surface
d O  12.07 m
a
2c
a  2 mm and 2c  6 mm
From Table 3.2, σys = 1,476 MPa and KIC = 81 MPa for AISI 4340 steel.
(a) Using the diameters yields
d o  12. 07  1. 09
6. 35
di
12. 07 cm  6. 35 cm
B  do  di 
2
2
B  2. 86 cm
Therefore, the pressure vessel is based on the thick-wall theory because
d o  1. 1
di
B  d i  0. 32
20
(b) From eq. (3.43e),
h 
d o /d i  2  1
d o /d i  2  1
P i  0. 8 ys
Pi  0. 8 ys
d o /d i  2  1
d o /d i  2  1
11
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Thus, the internal pressure is
P i  0. 8 1, 476 MPa
1. 09 2  1
1. 09 2  1

P i  101. 51 MPa
(c) Fracture mechanics:
   h  P i  0. 8 ys  P i
  0. 81, 476 MPa  101. 51 MPa
  1, 282. 30 MPa
Thus, a/2c = 2/6 = 0.33, σ/σys = 0.87 and the shape factor becomes

2
Q 
2
Q  1. 66
Q
3  a 2
4
2c
3  0. 33 2
4
2
2
2
 7 ys
33
2
 7 0. 87 2
33
2
Then,
K I    a  1, 282. 30 MPa
Q
 2x10 3 m
1. 66
K I  78. 89 MPa m
Therefore, the pressure vessel will not fail because KI < KIC.
(d) Denote that KI is slightly below the KIC value and safety precautions should be taken because
a minor increase in pressure due to a pressure surge will cause fracture. The margin of safety is
very small and therefore, a new steel should be used having a higher KIC value.
(e) The critical crack length is
KIC    a c
Q
Q
ac  
KIC

2
 1.66
81 MPa m
1, 282. 30 MPa
a c  2. 11 mm
This result implies that the pressure vessel may fracture because the existing crack of 2 mm in
length is approximately 5% smaller than its critical value. Therefore, a new steel should be used
to design the pressure vessel having the same dimensions and being subjected to the design
conditions.
12
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
3.13 A simple supported beam made of soda glass (E =71 MPa and GIC = 12 J/m²) is subjected to
a uniformly distributed bending load as shown in the figure below. Assume an initial crack
length of 0.1 mm due to either stress corrosion cracking (SCC) or a mechanical defect introduced
during fabrication or handling. The design stress (working stress) and the crack velocity equation
.83
are 0.10 MPa and v  K Im  6.24 m/s K 15
, respectively. Use this information to calculate the
I
lifetime of the beam if the maximum bending stress is given by σ = (MB/2)/I and M = σdL²/8,
where the second moment of area is I = wB³/12.
P
B  15 mm
w  200 mm
L 2m
a
Solution:
GI  12 J / m 2  12 Pa.m, E  71 GPa
B  20 mm, w  200 mm, L  2 m
Given data:
The applied stress
 
3 d L2 30.10 MPa 2,000 mm 

 6.67 MPa
4 wB 2
4200 mm15 mm2
2
Using Griffith expression, eq. (2.34), gives the critical crack length
ac 



EGIC
71x103 MPa 12 x10 6 MPa.m

 6.10 mm
 2
 6.67 MPa 2
This is a large crack length for a brittle material such as soda glass. The applied and the critical
stress intensity factors, eq. (3.29), are


K I    a  1.126.67 MPa   10 4 m  0.13 MPa m


K IC    ac  1.126.67 MPa   6.10 x10 3 m  1.03 MPa m
The crack velocity equation can be rearranged using the Chain Rule. Thus,
v
da
da dK I

dt dK I dt
13
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
But,
K I    a
a
K I2
 2 2
Then,
da
2K I

dK I  2 2
and
v
da
2 K I dK I
& v  K Im  6.24 m/s K I15.83

dt  2 2 dt
K Im 
2 K I dK I
 2 2 dt
Rearranging this equation and integrating with respect to time yields
2K I
dt 
dK I
2 2
  K Im
tf
2
o dt   2 2 
K IC
where
K IC
K
1 m
I
1 m
K I dK I  
KI
dK I
KI
1
K 2m
m2
K IC
KI
 

1
2 m
K IC
 K I2 m
m2

Then,
tf  

2
1
2m
K IC
 K I2m
2
   m  2
2



 1
2

215.83
2 15.83
t f  
 0.13
1.03

2
2
  1.12  6.67 MPa  6.24 m/s  15.83  2 
t f  2.3733x108 seconds  7.52 years
Therefore, the beam will last 7.53 years prior to fracture.
14

Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
3.14 Plot the given data for a hypothetical solid SE(T) specimen and determine the plane strain
fracture toughness KIC. Here,   f ( a / w)  is the modified geometry correction factor.
a 1/2 m 1/2  6 9.8 15 18 22
 MPa
25
35
38
30 50 70 90 116 122 175 190
Solution:
The plot
200
  150
MPa 
100
Slope  5.0144 MPa m
50
0
0
10
20
a 1/ 2 m1/ 2 
Linear curve fitting gives
   5.0144a -1/2 - 0.42832
Slope 
   
 5.0144 MPa m
 a 1 / 2 
K IC    a  5.0144 MPa m
15
30
40
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
3.15 A pressure vessel (L >> B = 15 mm, di = 2 m) is to be made out of a weldable steel alloy
having σys = 1,200 MPa and KIC = 85 MPa. If an embedded elliptical crack (2a = 5 mm and 2c =
16 mm) as shown below is perpendicular to the hoop stress, due to welding defects, the given
data correspond to the operating room temperature and the operating pressure is 8 MPa, then
calculate the applied stress intensity factor. Will the pressure vessel explode?
2a
B
2c
Solution:
d o  d i  2B  2 m  2  15  10 3 m  2. 03 m
d o  2. 03  1. 02
2
di
2  10 3 mm
B  di 
 100 mm
20
20
The pressure vessel is of thin-wall type because d o /d i  1. 1 and B  d i /20. The hoop stress and
the stress intensity factor for the embedded elliptical crack are, respectively
where
Then,
The pressure vessel will not explode because KI < KIC.
16
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
3.16 A steel pressure vessel subjected to a constant internal pressure of 20 MPa contains an
internal semi-elliptical surface crack with dimensions shown below. Calculate K I when (a) Q
and σ/σ ys in eq. (3.44) and (b) Q  1  1.464a/c 
1.65
as per reference [31]. (c) Compare results
and explain. (d) If fracture does not occur, calculate the safe factors S F( a ) and S F(b ) . Explain the
results. Data: K IC  92 MPa m ,  ys  900 MPa , v  0.3 and B  30 mm.
2c = 8 mm
di = 1 m
do = 1.04 m
B
a
a = 2 mm
Solution:
For d o /d i  1.04 and B  d o  d i / 2  1.04 m  1 m/ 2  0.02 m  20 mm
Pi  20 MPa , a / 2c  2 / 8  1 / 4
Pi d i
1m 
 d  

 Pi  1  i Pi  
1

20 MPa   520 MPa  σ ys
2B
 2 B   2 x 0.02 m 
 520 / 900  0.57778
(a) σ  σ h  Pi 
 /  ys
2
2
2
2
7  
π   3 a  
Q         

2   4 2c   33 
 ys 
2
2
2
7
π   3 1  
2
Q          0.57778  1.5581
2   4 4   33
a
 2 x10 3 
K I  1.12
 1.12520 MPa 
 36.98 MPa m
Q
1.5581
Validity as per ASTM E399
2
K IC 


  2.592 MPa m   2.6123x10 2 m  26.12 mm
BASTM  2.5
 900 MPa 
σ 


 ys 
B  30 mm  BASTM OK!
2
(b) For Q  1  1.464a / c 
1.65
K I  1.12
 1  1.4642 / 4 
1.65
 1.4665 ,
a
 2 x10  3 
 1.12520 MPa 
 38.12 MPa m
Q
1.4665
17
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
(c) Comparing these results yields K I( a )  0.97 K I( b ) and the percent error is
% Error 
38.12  36.98
x100  3%
38.12
This indicates that the Q equations give dissimilar results despite the different mathematical
approaches being used. Nonetheless, both calculated K I values are below K IC and therefore,
fracture does not occur.
(d) Since K I  K IC the safety factor can be calculated. Thus,
S F( a )  K IC / K I( a )  92 / 36.98  2.49
S F( b )  K IC / K I( b )  92 / 38.12  2.41
These are reasonably high values and therefore, the pressure vessel can be kept in service, but a
nondestructive technique must be used to monitor any crack growth that may lead to a  a c and
K I  K IC .
3.17 An aluminum-alloy plate has a plane strain fracture toughness of 30 MPa m . Two
identical single-edge cracked specimens are subjected to tension loading. (a) One specimen
having a 2-mm crack fractures at a stress level of 330 MPa. Calculate the geometry correction
factor   f a / w . (b) Will the second specimen having a 1-mm crack fracture when loaded at
430 MPa?
Solution:
(a) The geometry correction factor is
K IC
30 MPa m
  f a / w  

 1.1469
 a 330 MPa   2 x10 3 m
(b) The stress intensity factor for the second specimen is
K I    a  1.1469430 MPa   1x10 3 m  27.64 MPa m
Therefore, fracture will not occur because K I  K IC .
18
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
EXTRA PROBLEM NOT INCLUDED IN THE BOOK
A thin-walled cylindrical pressure vessel contains gas at a pressure of 100 MPa. During the
initial pressurization of the vessel, the axial and tangential strains were measured on the inside
surface as xx = 500x10-6 and yy = 750x10-6, respectively. Calculate a) the axial and hoop
stresses associated with these strains. Assume that an undetected internal semi-elliptical surface
crack (a = 1 mm deep and 4 mm long) grows 0.5 mm/year and that the pressure remains
constant. b) Deduce whether or not the vessel will fracture by calculating the applied stress
intensity factor using the constant hoop stress (tangential stress) and c) if the vessel will not
fracture, then calculate the time it takes for such unpleasant occurrence.
Note: Only the shape factor should be used to
correct the stress intensity factor KI.
Given data:  ys  600 MPa
K IC  30 MPa m
E  207 GPa
v  0. 3
B  2 cm
d  0. 50 m (Diameter)
19
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Solution:
a) From eq. (1.6) along with  zz  P, the inner strains are
xx

yy
 xx  v yy  P
1
E
(1.6)
 yy  v xx  P
Solving these two equations simultaneously yields
 xx 
E
1v
yy 
 xx  yy
1v

 yy  Eyy  v xx  P
vP
1v

E
xx
1v 2
 vyy  
vP
1v
(a)
(b)
Given data:
xx  500  10 6
 ys  600 MPa
yy  750  10 6
K IC  30 MPA m
E  207, 000 MPa
Thus,
 xx 
207,000 MPa
10.3 2
500  10 6  0. 3 750  10 6

0.3
10.3
100 MPa
 xx  122. 06 MPa (Axial Stress)
 yy  207, 000 MPa 750  10 6  0. 3 122. 06 MPa  100 MPa
 yy  161. 87 MPa
(Hoop Stress or Tangential Stress)
b) ASTM E399 thickness requirement, eq. (3.30):
2
30
a, BASTM  2. 5  10 3 600
 6. 25 mm
B  BASTM , but a is not met for use LEFM.
The valid average tangential or hoop stress
regardless of the wall thickness is
 av   yy,av 
Pd
2B

1000.510 2 cm
22 cm
 1, 250 MPa
The applied stress intensity at initial pressurization hoop stress is
E  207 GPa
K IC  30 MPa m
v  0. 3
   yy  161. 87 MPa
 ys  600 MPa
/ ys  161. 87/600  0. 269 78
a  1 mm
2c  4 mm
a/2c  0. 25
Q

2
2
3
4

a
2c
2
2

7
33
K I   a/Q  161. 87 MPa

 ys
2
 1. 6134
 110 3 m
1.6134
20
 7. 14 MPa m
Chapter 3 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
 The the initial pressurization process of the vessel will not cause
fracture because K I  K IC at    yy  161. 87 MPa.
 If a  B  20 mm (leak-before-fracturecondition), then
 2010 3 m
K I   yy a/Q  161. 87 MPa
1.6134
 31. 94 MPa m
Therefore, fracture will occur because K I  31. 94 MPa m  K IC
 When  yy   av fracture will occur because
K I   av a/Q  1, 250 MPa
 110 3 m
1.6134
 55. 16 MPa m
K I  K IC  30 MPa m
c) If the tangential stress,  yy  161. 87 MPa, is assumed constant
thoughtout the test, then the critical crack length becomes
K IC   a c /Q
Q
2
a c   K IC

yy
a c  B  20 mm
1.6134

10 3
30
161.87
2
 17. 64 mm
Thus, the life of the vessel is approximately
t  a c  a/da/dt  17. 64 mm  1 mm / 0. 5 mm/year
t  33. 28 years
This is an unrealistic result because  yy   av after the initial pressurization
process is completed. Consequently, the vessel fractures immediately and its
lifetime is basically t  0.
21
Chapter 4 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
CHAPTER 4
ELASTIC STRESS FIELD EQUATIONS
4.1 (a) Calculate KI using the singularity and non-singularity stresses for a single-edge cracked
plate having a = 4 mm, x = 0.1, and L/w ≥ 1.5, and subjected to 300 MPa in tension. (b) Plot the
stress ratio Tx/ and the biaxiality ratio  as functions of x = a/w.
Solution:
2) Single-edge Crack (SET)
K I   a

L / w  1.5 and x  a/w
w
L

  1.12 - 0.23a/w   10.55 a / w  21. 71a / w
2
a
 30.38 a / w

3
4
Given data:   300 MPa, x  0.1, a  4 x10 3 m
The stress ratio
Tx
1

 0.526  0.641x  0.2049 x 2  0.755 x 3  0.7974 x 4  0.1966 x 5
2
 1  x 


Tx
 0.56688

Tx   0.56688   0.56688300 MPa 
Tx  170.06 MPa
The biaxiality ratio
1

 0.469  0.1414 x  1.433x 2  0.0777 x 3  1.6195 x 4  0.859 x 5
1/ 2
1  x 
  0.46444

1

Chapter 4 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Plots:
Parameter
Parameter


Tx


x
Stress ratio approach:
T
  170.06 MPa 
3
K I( 2)  x a  
  4 x10 m

  0.46444 


K I( 2)  41.05 MPa m
Classical approach:
The geometry correction factor
  1.12  0.23a/w  10.55a/w2  21.71a/w3  30.38a/w4
  1.12  0.230.1  10.550.1  21.710.1  30.380.1
  1.1838
2
The stress intensity factor
3

4
K I(1)   a  1.1838300 MPa   4 x10 3 m
K
( 2)
I

 39.81 MPa m
Therefore, the Tx stress approach gives slight higher values for K I .
2
Chapter 4 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
4.2 Using the information given in problem 4.1, calculate the stress intensity factor K I when the
stress ratio is Tx  0.
Solution:
2) Single-edge Crack (SET)

w

  1.12 - 0.23a/w   10.55 a / w  21. 71a / w
2
a

a  4 x10 3 m
 30.38 a / w
3
4
  300 MPa
x  a / w  0.1
  1.12  0.23x  10.55x 2  21.71x 3  30.38 x 4
  1.12  0.230.1  10.550.12  21.710.13  30.380.14
  1.1838


K I   a  1.1838300 MPa   4 x10 3 m 
K I  39.81 MPa m
4.3 Assume that a single-edge crack in a plate is loaded in tension. Derive the dominant near
crack-tip stresses in Cartesian coordinates using the Westergaard complex function
Z z  
KI
where K I   a
2z
Solution:
Polar coordinates:
x  r cos , y  r sin  , r  x 2  y 2 ,
1


sin   cos sin
2
2
2
From Euler’s formulas,
z  reiθ / 2  r cos  i sin  



z  r 1/ 2 e iθ / 2  r 1/ 2  cos  i sin 
2
2

3
3 

z  r 3 / 2 e i 3θ / 2  r 3 / 2  cos  i sin 
2
2 

3
Chapter 4 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Find the derivative of the Westergaard complex function
KI
KI
KI
KI
Z z  
& Z ' z   

z 3 / 2  
r 3 / 2 e i 3 / 2
3/ 2
2z
2 2 z
2 2
2 2
Using the Euler’s formula gives the Westergaard complex function in terms of trigonometric
functions
KI
K
K



Z z  
 I r 1 / 2 e iθ / 2  I r 1/ 2  cos  i sin 
iθ / 2
2
2
2π
2π

2πre
Z ' z   
KI
KI
3
3 

r 3 / 2 e i 3 / 2  
r 3 / 2  cos  i sin 
2
2 
2 2
2 2

The real and imaginary parts of the trigonometric functions are
K
K


Re Z  z   I r 1 / 2 cos
Im Z  z    I r 1/ 2 sin
2
2
2π
2π
K
K
3
3
Re Z '  z    I r 3 / 2 cos
Im Z '  z    I r 3 / 2 sin
2
2
2π
2π
From eq. 4.11), the stress in the x-direction becomes
K

3 
 K I 3 / 2
 x  Re Z  z   y Im Z ' z   I r 1/ 2 cos  r sin  
r
sin 
2
2 
2π
 2 2π
K
 1
3  K



3 


 x  I r 1/ 2  cos  sin  sin   I r 1 / 2  cos  sin cos sin 
2 2
2 
2
2
2
2 
2π
2π


KI


3 
x 
cos 1  sin sin 
2
2
2 
2πr
Similarly,
 y  Re Z  z   y Im Z '  z  
KI
2π
r 1/ 2 cos

3 
 K I 3 / 2
 r sin  
r
sin 
2
2 
 2 2π
K I 1/ 2 
 1
3  K



3 

r  cos  sin  sin   I r 1 / 2  cos  sin cos sin 
2 2
2 
2
2
2
2 
2π
2π


KI


3 
y 
cos 1  sin sin 
2
2
2 
2πr
y 
and
K I 3 / 2
3  K
1
3

 xy   y Re Z '  z   r sin   
r
cos   I r 1 / 2 sin  cos
2 
2
2
2π
 2 2π
KI


3
 xy 
cos sin cos
2
2
2
2πr
4
Chapter 4 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
4.4 Consider a unit circle with its center at the origin and let function F  z   Pz   iQz  be
holomorphic inside the contour C. Also let the function F   take definite boundary values
______
where   e i gives the points on C. If the boundary condition is F    F    f   , then
determine the Cauchy Integral formulae and the integral for F(z).
Solution:
Multiply the equation for the boundary condition by
1 d
2i z
and integrate around C to get the Cauchy Integral formulae as
1
2i
C
Fd
z

C
1
2i
Fd
z

1
2i
C
fd
z
The first integral is
Fz 
1
2i
C
Fd
z
The second integral becomes
F0  a o  ia 1
which are unknown so far. Then,
Fz 
1
2i
C
fd
z
 a o  ia 1
Let z = 0 so that
ao 
1
2i
C
Fd


1
2
2
 o fd
The remaining work to completely solve this problem can be found in Muskhelishvili's book,
1977 published edition, page 309-311.
4.5 Show that the fundamental combination of stresses can be defined in terms of Cauchy
Integral formulae.
hd
 1
i
  z 2
C
hd
 1
i
  z 2
hd
 1
3
i

z

hd
 12 
iz C  2
C
hd
 12
2
iz

z

x  y  1
i
 y   x  2i xy  1
i
C
C
5
C
hd
2
C
hd
  z 2
Chapter 4 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Solution:
z 
1
2i
C
h d  a z
1
z
  z
h
a
d  z  z1
z
h
a1  a1  1 
d
2i C  2
a 2  1  h3 d
2i C 
z 
C
(4.124)
The resultant expressions are
 x   y  2   z    z
(4.95a)
 1
i
C
h
d  2a 1  1
i
  z 2
 1
i
C
hd
 1
2
i

z

C
C
h
d  2a 1
  z 2
hd
 1
2
i

z

C
hd
2
 y   x  2i xy  2z  z    z
 1
i
C
hd
 1
i
  z 3
C
2  z
hd

 2a21
2
2
z
z
  z
 1
i
C
hd
 1
i
  z 3
C
hd
 12
iz
  z 2
C
hd
2a
2a
 21  21
z
z
  z 2
hd
 1
i
  z 3
hd
 12 
iz C  2
C
hd
 12
iz
  z 2
C
hd
  z 2
 1
i
C
4.6 Suppose that a plate containing a single unstressed crack (Figure 4,15) is deformed by the
unknown stresses at infinity and assume that the complex potential, f ' z     i xy , represents
the stress distributions across the crack contour C. (a) Determine the stress distribution on y = 0
outside the crack and (b) the upper and lower boundary functions for x  a.
Solution:
(a) Use eq. (4.124d) for z = x + iy = x since y = 0.
 yy  i xy  

 
y i xy 
2
1
z
z 2 a 2
In order to cover the stress distribution on the upper and lower planes the boundary function χ(z)
has to be rewritten as   z    x2  a 2 for x > a and x < a, respectively. Therefore, |x| ≥ a and
the stress distribution becomes
 yy  i xy 

 
y i xy 
2
6
|x |
x 2 a 2
1
Chapter 4 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
(b) The upper and lower boundary functions for |x| ≤ a are
  z    z  i a 2  x 2
since z – a = reiθ and z + a = reiθ so that for a unit disk with z = x + iy = x
  z 
x  ax  a 
x 2  a2  i a2  x2
  z  i a 2  x 2
and
  z  i a 2  x 2
   z  i a 2  x 2
4.7 Consider an infinite plate subjected to a tensile remote stress (S) normal to the direction of a
central crack. If the complex potentials for this type of crack are
z  S 2 z 2  a 2  z
4
  z  z  S
2
z
a2
z2  a2
then determine the crack tip stresses using the Westergaard stress function Z  z   2 '  z  when
x  a.
Solution:
  z  S
4
2z
1
z  a2
2
Zz  2  z  S
2
Z  z  Z   S
2
z
1
z  a2
2
1
z2

3
z2  a2
z 2  a 2  2
The Westergaard stress functions, eq. (3.20) or (4.11), are
 x  Re Z  y ImZ 
 y  Re Z  y ImZ 
 xy  y Re Z 
7
Chapter 4 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
At y = 0 and z = x + iy = x > a, these stress functions become
 x  Re Z  S
2
z
1
z  a2
 y  Re Z  S
2
x
1
x  a2
2
 S
2
x
1
x  a2
2
#
2
 xy  0
For large z = x + iy = x since y = 0, the elastic stresses far from the crack tip vanish when x >> a
along the x-axis. Therefore,
x  0
y  0
 xy  0
4.8 Consider the elliptical crack shown below and assume that the crack is in the z-plane where
z    a and p z   p . Derive the stress equations using the given crack geometry and the
Westergaard complex method.
p z   p  
iy
z
ζ θ
x
a
Solution:
Let us move the origin to crack tip. Then the coordinates of P can be expressed in complex
variable z    a . Use the Westergaard complex equation so that
Zz 


1  a/z 2
z
z  a2
Z 
  a

  2a
a 1 
8
2
2a 1 

a

2a
Chapter 4 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
For |ζ| << a,
a   a
2a
2
Z 
Z    
 a
2 3/2
But ζ = reiθ and
Zr,   
Zr,  
Z  r,  
Z  r,  
 a
i 3/2
2re 
 a
2r
 a

 a
2r 3/2
e i/2
cos   i sin 
2
2
i 3/2
2re 
 a
i 3/2
2re 


 a
2r 3/2
 a
2r 3/2
e i3/2
cos 3  i sin 3
2
2
Then,
 y  Re Z  y Im Z 
 a
y 
cos  1  sin  sin 3
2
2
2
2r
and
 xy
 a
cos  1  sin  sin 3
2
2
2
2r
 a

cos  sin  sin 3
2
2
2
2r
x 
4.9 Assume that an infinite plate contains a through-central crack along the x-axis. If the plate is
subjected to a remote or infinite stress loading condition,  y   y  S ,  x   xy  0, then use
the Sanford [54] modified Irwin-Sih complex potential
2  z  Zz    z
Here, γ′(z) may be defined by a Cauchy integral, Z(z) is the Westergaard complex function,
ψ∗(z) is a complex polynomial and z is the complex variable define as z = x + iy. Based on this
information, determine a function for the stress intensity factor KI and expand ψ∗(z) when n = 0
and 1, and z = a, half the crack length.
9
Chapter 4 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Solution:
The Westergaard complex function and the complex polynomial are
KI
2z
Zz 
m
  z 
 b n zn/2
n0
The given complex function yields
KI
2z
 2  z   n0 b n z n/2
m
For n = 0 and 1 and z = a,
KI 
2a 2  z  b o  b 1 z 1/2 
10
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
CHAPTER 5
CRACK TIP PLASTICITY
5.1 Use the inequality K IC  K I as a criterion for crack instability where K I is defined by
Irwin’s plastic zone corrected expression for a finite size, to determine if a steel pressure vessel
is susceptible to explode under   200 MPa hoop stress. The vessel contains an internal
circular crack perpendicular to the hoop stress. If K IC  60 MPa m ,  ys  700 MPa, and the
crack size is 20 mm, (a) determine the ASTM E399 thickness requirement and the minimum
thickness to be used to prevent explosion, (b) Will crack propagation occur at 200 MPa? (c) Plot
B (thickness) vs.  /  ys for a = 10, 20, and 30 mm, and (d) Will the pressure vessel explode
when the crack size is 30 mm? Why? or Why not? and e) When will the pressure vessel
explode?
Solution:
(a) From eq. (5.12) along with  

K I   a 1  0.5( /  ys ) 2

2
,

Let K IC  K I so that
 K IC


 ys
2


   KI



 ys
K
2.5 IC

 ys




2
2

K

 2.5 I


 ASTM
 ys
2



 Actual
Thus,
2
B ASTM
K
 2.5 I

 ys



 Actual
B ASTM
  a 1  0.5( /  ) 2
ys
 2.5
 YS



 
2



2 

2
.
5

a





 ys

1




2
 1 
1  
 2   ys





2

 


Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
If the equality is used, then B ASTM  Bmin is the minimum thickness required by the ASTM E399
standard testing method. Substituting values, such as  /  ys  0.285714, a = 20 mm, and
  2 /  in the derived thickness expression, yields B ASTM  Bmin  5.41 mm .

(b) K I   a 1  0.5( /  ys 
2
  32.56 MPa
m . Therefore, crack propagation will not
occur because K I  K IC . In this case, the crack is stable.
.
(c) Plot

 
2
B ASTM
  a 1  0.5( /  ys ) 2
 2.5

 YS

B ASTM
2
2


  1     

2  
 
1  
 2.5a
    2    

 ys  
 ys   


160
140
120
B ASTM
a  30 mm
100
20
80
60
10
40
20
0
0.2
0.4
0.6
0.8
1
 /  ys

(d) It will not explode since K I   a c 1  0.5( /  ys 

(e) Using K IC   a c 1  0.5( /  ys 2

2
  40 MPa
m and K I  K IC .
yields the critical or maximum crack size as
a c  67.91 mm . Therefore, the pressure vessel will explode when a  a c  67.91 mm.
2
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
5.2 A project was carried out to measure the elastic-strain energy release rate as a function of
normalized stress  / ys of large plates made out a hypothetical brittle solid. All specimens had
a single-edge crack of 3-mm long. Plot the given data and do regression analysis on this data set.
Determine (a) the maximum allowable  /  ys ratio for G IC  30 kPa.m and (b) K IC in
MPa m . Given data: v  0.3 ,  ys  900 MPa and E  207 GPa .
 / ys
G IC kPa.m 
0
0.10 0.20 0.30 0.40 0.50
0.60
0.70
0.80
0.90
0
0.40 1.70 1.90 7.00 12.00 19.00 26.00 36.00 48.00
Solution: The required regression equation and the plot are given below and the
2
  




  53.4440    14.9570  
G IC  0.2322  7.3168
 
 
 
 ys 
 ys 
 ys 
3
50
G IC
kPa.m 
40
30
20
10
0
0.2
0.4
0.6
0.8
1
 / ys
(a) Combining eqs. (3.5) and (5.12) yields


 2 1  v 2 a ys  /  ys 2 1  0.5 /  ys 2
E
2
2
 1.12  1  0.3 3 x10  3 m 900 MPa x 2 1  0.5 x 2 
G IC 
 42.0980 x 2  21.0940 x 4
207,000 MPa
where x   /  ys  0.75 at G IC  30 kPa.m.
G IC 
(b) The fracture stress and the plane strain fracture toughness
  0.75900 MPa   675 MPa
EG IC
207,000 MPa 0.03 MPa.m   82.61 MPa m

2
1 v
1  0.32
--------------------------------------------------------------------------------------------------------------------K IC 
3
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
5.3 Determine the critical crack length of problem 5.2.
Solution:
 / ys
G IC kPa.m 
0
0.10 0.20 0.30 0.40 0.50
0.60
0.70
0.80
0.90
0
0.40 1.70 1.90 7.00 12.00 19.00 26.00 36.00 48.00
50
G IC
kPa.m 
40
30
20
10
0
0.2
0.4
0.6
0.8
1
 / ys
Combining eqs. (3.5) and (5.12) yields


 2 1  v 2 a ys  /  ys 2 1  0.5 /  ys 2
G IC 
E
2
2
 1.12  1  0.3 3x10 3 m 900 MPa x 2 1  0.5 x 2 
G IC 
 42.098 x 2  21.094 x 4
207,000 MPa
where x   /  ys  0.75 at G IC  30 kPa.m. The fracture stress and the plane strain fracture
toughness
  0.75900 MPa   675 MPa
K IC 
EG IC

1  v2
207,000 MPa 0.03 MPa.m   82.61 MPa
1  0.32
m
Thus,
K IC
2

   
 
  a c 1  0.5
  

ys

 

1
2
2
2
   
1  K IC  
1  82.61 MPa m 
2 1


  
ac  
 3 mm
 1  0.5
 1  0.50.75

    
 ys  
  1.12 675 MPa 



---------------------------------------------------------------------------------------------------------------------

4

Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
5.4 A large brittle plate containing a central crack 4-mm long is subjected to a tensile stress of
800 MPa. The material has KIC = 80 MPa m , ys = 1200 MPa and v = 0.30. Calculate (a) the
applied KI, (b) the plastic zone size using the Von Mises yield criterion and prove that r = rmax
when    o . Consider all calculations for plane stress and plane strain conditions, and (c) draw
the entire plastic zone contour where the crack tip is the origin of the coordinates.
Solution:
(a) K I   a  (800 MPa)  2  10 3 m   63.41 MPa m with   f (a / w)  1
The crack is stable since KI < KIC.
(b) From eq. (5.52),
2
KI 3 2

r
sin   h1  cos   ;
2 
4 ys  2

For plane stress 
 1
h

2
1  2v  For plane strain 
If  = 0, then
r
hK I 2
2 ys
2
0.44 mm plane stress 
and r  

0.07 mm plane strain 
Using eq. (5.52) yields
2
r 1  K I 

sin  3 cos  h
   2 ys 
For maximum KI and critical plastic zone size r = rc, let
r
 0 so that

sin  o 3 cos o  h   0
h
3
70.53o
o  
o
86.94
cos o 
For plane stress 

For plane strain 
Thus,
2
rmax 
KI
2
4 ys
3 2
 0.59 mm


sin


h
1

cos

o
o
 2
   0.37 mm

Furthermore,
KI
r

sin  3 cos   h
 4 ys 2
5
For plane stress 

For plane strain 
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual



KI
KI
 2r

3 cos  3 sin 2   h cos 
3  h  cos  3 sin 2 
2
2
2

4 ys
4 ys
2
 2r
 2
o
2

 0.44 mm plane stress 


 0.63 mm plane strain 
Therefore, r = rmax at o since
 2r
 0 in both cases.
 2
(c) The contours in polar coordinates are as per r 
KI
2
4 ys
2
3 2

 2 sin   h1  cos  


r mm 
0.4
Crack Tip
0.2
-0.1
0
0.1
0.2
0.3
0.4
 Radians 
-0.2
-0.4
Plane Strain
Plane stress
---------------------------------------------------------------------------------------------------------------------
5.5 (a) Use the data given in Example 3.3 for a pressure vessel containing a semi-elliptical crack
(Figure 3.6) to calculate Irwin’s and Dugdale’s (a) plastic zones, (b) K I using Kabayashi’s finite
size correction factor ( ) and plasticity correction factor. (c) Compare results and determine the
percent error against each case. (d) Is it necessary to include a plastic correction factor? Explain.
Solution:
Data from Example 3.3:
a = 3 mm, 2c = 10 mm, B = 6 mm,   420 MPa,  ys  700 MPa, a/(2c) = 0.30, a/B = 0.50,
K IC  60 MPa m ,  /  ys  0.60, M = 1, Mk = 1.12, and Q = 1.70 .
The correction factor and the stress intensity factor are, respectively
  1.12 M k M / Q  0.127
K I   a  5.18 MPa m
6
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Comparison:
(a) Irwin’s Approach:
  1.12 M k M / Q  0.127
a
2
r   /  ys   0.54 mm Eq. (5.13)
2
Dugdale’s Approach:
  1.12 M k M / Q  0.127
a
2
r   / 2 ys   1.33 mm
2
(b) Using a finite correction factor
K I    (a  r )

2

2
K I   a 1  0.5 /  ys 
K I   a 1  0.5 /  ys 

Eq. (5.24)
K I    (a  r )
Eq. (5.12)

K I ( Irwin' s )  5.63 MPa m

 
K I   a 1  0.5
 2

 ys





2

 
K I   a 1  0.5
 2

 ys





2




Eq.(5.25)




K I ( Dugdale' s )  6.22 MPa m
Comparisons:
K I ( eq. 3.29)  5.18 MPa m < K I ( Irwin' s ) < K I ( Dugdale' s )  6.22 MPa m
Dugdale’s - Irwin’s Error  100%(6.22  5.63) / 5.63  10.48%
Irwin’s - eq. (3.29)
Error  100%(5.63  5.18) / 5.18  8.69%
Dugdale’s - eq. (3.29) Error  100%(6.22  5.18) / 5.18  20.08%
Observe that eq. (3.29) does not include plasticity correction and it yields a smaller value than
both Dugdales’s and Irwin’s expressions. The latter expression gives an error of 26.5%, which is
slightly large. On the other hand, 8.3% error seems to be more acceptable for comparison
purposes. This implies that a plastic zone correction may not be necessary.
(c) Using Irwin’s approach yields a plastic zone of 0.54 mm for one side of the semi-ellipse and
as a result, r << a and the plastic zone size could have been excluded. On the other hand,
Dugdale’s approach gives a plastic zone of 1.33 mm, which is large enough to be excluded;
therefore, a plasticity correction is needed to obtain more a accurate KI- result.
---------------------------------------------------------------------------------------------------------------------
7
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
5.6 A 50- mm thick pressure vessel is to support a hoop stress of 300 MPa at room temperature
under no action of corrosive agents. Assume that a semi-elliptical crack (Figure 3.6) is likely to
develop on the inner surface with the major axis 2c = 40 mm and semi-minor axis a = 10 mm. A
300-M steel, which is normally used for airplane landing gear, is to be considered. Will crack
propagation occur at 300 MPa hoop stress? Make sure you include the Irwin’s plastic zone
correction in your calculations and explain if it is necessary to do so. Use the data below and
select the suitable tempered steel.
 ys
300-M Steel
K IC
(MPa )
o
650 Temper
300o Temper
( MPa m)
1070
1740
152
65
Solution:
a = 10 mm, 2c = 40 mm, B = 50 mm, a/2c = 0.25, a/B = 0.20
Mk = 1.02 (From Figure 3.6b)
  1.12M k / Q since M < 0.5 [lower limit according to eq. (3.46)]
From eq. (3.42),
2
Q   2  0.212 /  ys 
 /2
 /2
 c2  a2  2


1

sin


d


1  0.75 sin 2   d
0

 c2 


0
  0.13169 (From a Table of Elliptic Integral of the Second Kind)

Thus,

K I   a 1  0.50 /  ys 
2
a
 /  ys 2
2
300-M Steel (650o C Temper)
 /  ys  0.28
Q  1.72
  0.87
K I  47.16 MPa m  K IC
r = 0.39 mm
r << a = 10 mm

r
300-M Steel (300o C Temper)
 /  ys  0.17
Q  1.67
  0.88
K I  47.13 MPa m  K IC
r = 0.15 mm
r << a = 10 mm
These results are similar and crack propagation will not occur since both steels have KI < KIC.
Select 300-M Steel (650o C Temper) since it has a larger K IC value. The plastic zone correction
was not necessary. Thus, K I   a  46.26 MPa m  K IC . Both steels can be used for
the sought application. In fact, the calculated K I values are similar for both steels. This is purely
accidental. Anyway, the plastically corrected KI values are approximately 2% higher.
8
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
5.7 If localized plasticity is to be considered, explain the physical meaning of the following
a 2
inequality
  t  ys .
E
Solution:
If the applied stress (  ) is unchanged, but causes crack growth, then  t  ys is constant since
 ys  constant and  t   c . In this case, the crack length is the only changing variable that
increases; therefore, the inequality holds.
--------------------------------------------------------------------------------------------------------------------5.8 Show that r   t /( ys ), where r is the plastic zone due to dislocation networks within the
plastic zone area ahead of the crack tip.
Solution:
K
 ys
K I2
If  t 
and E 
, then,  t   ys  I

 ys E
 ys
 ys
2

2 ys
 

2

 KI


 ys
2

  2 ys  r


t
.
2 ys
--------------------------------------------------------------------------------------------------------------------Thus, r 
5.9 Show that  t /  ys  K I /  ys  and give a reasonable interpretation of this equality.
2
Solution: Apparently,  t /  ys is related to the crack size [  t /  ys  a] since K I /  ys  is
2
associated with the plastic zone [r  K I /  ys  ]. Therefore, r   t /  ys . Here  means
“proportional.”
--------------------------------------------------------------------------------------------------------------------2
5.10
A large plate containing a through-the-thickness central crack of 2 a c  20 mm has
E  207 GPa,  ys  1,275 MPa, and  c  9.47 m at service temperature. Determine (a) the
plane strain fracture toughness, (b) the design stress intensity factor for a safety factor (SF) of 2,
(c) the critical fracture stress, and (d) the design service stress.
Solution:
2
K IC
(a)  c 
 ys E
K IC  50 MPa m
K
(b) K ID  IC  25 MPa m
SF
(c)  c  K IC / a c  282 MPa
(d)  d   c / SF  141 MPa
9
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
4
a2  x2
E
Solution: Glass is a brittle material and there is no need for plasticity correction; therefore,
 t  0 @ x  a.
---------------------------------------------------------------------------------------------------------------------
5.11 Predict  t for glass using  
5.12 Develop a  t  expression for a von Mises material. Compare it with  t for a Tresca
material under plane strain condition. Assume that crack growth occurs along the crack plane.
Solution:
von Mises  t  expression:
From eq. (5.53),
hK I2
K r
r
and  t  I
2
2 ys
E 2
 t von Mises  
h1 / 2 K I2
2E ys
Tresca  t  expression:
From eq. (5.59) with   0 ,
K I2
K r
r
and  t  I
2
2 ys
E 2
 t Tresca  
Combining  t Mises  and  t Tresca  yields
K I2
2E ys
 t Mises   h1 / 2 t Tresca 
 t Mises   1  2v  t Tresca 
For Poisson’s ratio v  1,  t Mises    t Tresca  because r Mises  r Tresca
--------------------------------------------------------------------------------------------------------------------5.13 A material has E  70 GPa,  ys  500 MPa and v  1 / 3. It has to be used as a plate in a
large structure. Non-destructive evaluation detects a central crack of 50 mm long. If the
displacement at fracture is 0.007 mm and the plate width is three times the thickness, calculate
(a) the crack tip opening displacement, (b) the plane strain fracture toughness, (c) the plane
strain energy release rate, (d) the plate thickness and (e) What’s the safety factor being indirectly
included in this elastic-plastic fracture mechanics approach? Assume plane strain conditions as
per eq. (5.49) and a fracture load of 200 kN.
Solution:
Given data:
a = 25 mm,  f  0.007 mm ,   3 and v  1 / 3.  ys  500 MPa, P = 200 kN and E  70 GPa .
(a)  tc  2 f  0.014 mm  1.40 x10 5 m
(b) From eq. (5.49) with   3  1.7321 , which is defined below eq. (5.5), the plane strain
fracture toughness ( K IC ), as per Irwin’s model, is
10
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
K IC 
G IC
Solution Manual
 3 1.40 x10 m70 x10 MPa 500 MPa   25.82 MPa
1  1 / 925.82 MPa m   8.46 kJ / m

1
1
 tc E ys 

2
2
5
3
2
2
(1  v 2 ) K IC

E
2
3
70 x10 MPa
(c)  c  K IC / a  92.13 MPa m and  c 
B
Pc

3 c
m
200 kN
3 92.13x10 3 kN / m 2


Pc
P
 c2
wB 3B
 26.90 mm and w  3B  80.70 mm (Width)
(d) SF   ys /  c  500 / 92.13  5.43
--------------------------------------------------------------------------------------------------------------------5.14 Repeat problem 5.13 using eq. (5.41). Compare results.
Solution:
Data: a = 25 mm,  f  0.007 mm ,   3 and v  1 / 3.
 ys  500 MPa, P = 200 kN, E  70 GPa and  c  2  f  0.014 mm  1.4 x10 5 m
  3  4v  5 / 3 For plane strain
(a) c =
2
2
(  1)(1  v) K IC
8 K IC

4E ys
9E ys
K IC 
(b) G IC
9 c E ys
8

9 1.4 x10 5 m 70 x10 3
8

2
(1  v 2 ) K IC
1  1 / 9 41.61 MPa m


E
70 x10 3 MPa
(c)  c 
K IC
a


MPa 500 MPa 

 41.61 MPa m
2
 0.022 MJ / m 2  22 kJ / m 2
41.61 MPa m   148.48 MPa
 25 x10 3 m 
Pc
P
 c2
wB 3B
Pc
200 kN
B

3 c
3 148.48 x10 3 kN / m 2
c 

(d) SF 
 ys
c
Problem 5.13
Problem 5.14

 21.19 mm and w  3B  63.57 mm (Width)
 3.37
K IC  41.61 MPa m
 c  148.48 MPa
B  21.19 mm
SF  5.43
K IC  25.82 MPa m
 c  92.13 MPa
B  26.90 mm
SF  3.37
11
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
5.15 A hypothetical large metallic plate containing a 10-mm central crack is 30-mm wide and 5mm thick and mechanically loaded in tension. This plate has E  69 GPa,  ys  500 MPa ,
v  1 / 3, and   0.3% (plane stress strain). Determine (a)  t , (b) G , and c)  as per Irwin,
Dugdale, Burdekin, Rice, and Average equations, and compare the results.
Solution:
Data: w  30 mm, B  5 mm, v  1 / 3, a  5 mm ,   0.3%,  ys  500 MPa, E  69 GPa
Calculations:
  B  5 mm 0.003  0.015 mm and K I   a
G I   ys  1.5 x10 5 m 500 MPa   0.0075 m.MPa  7.5  kJ / m 2
(a) In general,
t 
 t E ys
K I2
a c2

and  
E ys
E ys
a
Irwin–eq. (5.31):
 E
  t ys 
4a
Dugdale–eq. (5.32):
 
Burdekin–eq. (5.40):  
Rice–eq. (5.41):
 
 t E ys
2a
 t E ys
a
 t E ys
a



0.015 mm 69 x103 MPa 500 MPa   161 MPa
45 mm 
0.015 mm69 x10 3 MPa 500 MPa   128 MPa
2 5 mm
0.015 mm 69 x10 3 MPa 500 MPa   182 MPa
 5 mm 
0.015 mm 69 x10 3 MPa 500 MPa   322 MPa
5 mm 
Rice equation yields  Rice   2 Irwin  .
12
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Determine (a) the critical crack tip opening displacement (  c ), (b) the plastic zone size
5.16
(r) and (c) the fracture stress ( c ) for a large aluminum alloy plate containing a central crack of
5-mm long. Use the data given below and assume that plane strain conditions exist. Data:
K IC  25 MPa m ,  ys  500 MPa, and E  70 GPa.
Solution:
(a) From eq. (5.40),


2
K2
25 MPa m
 c  IC 
 0.018 mm
E ys 70 x103 500 MPa 
(b) For plane strain condition,
1
r
6
 K IC


 ys
2
2



  1  25 MPa m   0.13 mm


6  500 MPa 

From eq. (3.24),
c 
K IC
a
25 MPa m

 2.5 x10 3 m
 282 MPa
--------------------------------------------------------------------------------------------------------------------2
 E  2
5.17 Show that    t 1  
  t for plane stress conditions, where  is the crack
 8a 
opening displacement (COD) and  t is the crack tip opening displacement (CTOD).
Schematically, plot   f  t  for various  and fixed a values.
Solution:
From eq. (5.24),
4
t 
2ar
E
2
 E t 
2ar  

 4 
(5.30)
(a)
 
1  E t 
r


2a  4 
(b)
If a  x, then
(c)
 
r2 
1
2a 2
From eq. (5.23a),
2
4
 

a2  r  x2
E
4
E
a 2  2ar  r 2  x 2
(5.29)
(d)
2
 E t 


 4 
4
13
4
E
2ar  r 2
(e)
(f)
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Inserting eqs. (b) and (c) into (f) yields
4

E
 E  2  1  E  4

 t  
 t
2 
 4 
 2a   4 
2
4
 1  E  4

   t2  
 t
2 
 2a   4 
2
2
 E  2
  t 1  
 t
 8a 
The plot is


a  a o  fixed
Increasing
t
--------------------------------------------------------------------------------------------------------------------5.18 If the plane strain fracture toughness (KIC) and the yield strength (ys) of a 12-mm thick
C(T) steel specimen are 71 MPa.m1/2 and 1,896 MPa, respectively, determine (a) the validity of
the fracture mechanics tension test as per ASTM E399 for the plate containing a single-edge
crack of 10-mm long at fracture, (b) the fracture stress if the plate is 20-mm wide, (c) the critical
crack tip opening displacement, (d) the plastic zone size, and (e) interpret the results with regard
to plane strain condition. Use a Poisson's ratio of 1/3 and assume that the elastic modulus of the
steel 207 GPa.
Solution:
14
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
(a) Using eq. (3.5) yields the critical stress intensity factor
The minimum size requirements can be computed using eq. (3.30). Thus,
Therefore, the test is valid because a, B10, 12   amin , Bmin 3.51 mm and a/w  0.5 is within the
0.2  a/w  1 valid range. So proceed with the required calculations.
(b) The fracture stress is determined from eq. (3.29) along with α=9.6591 since a/w=0.5 (Consult
Table 3.1).
Thus,
(c) The crack-tip opening displacement is calculated using Rice’s equation, eq. (5.41), with
  3  4v  5 / 3 and   11  v   32 / 9.
15
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
(d) The plastic zone can be calculated using eq. (5.53) along with the constant h = (1 - 2v)2 = 1/9
(e) The above results suggests that the plate met the ASTM E399 size requirements because
a, B  amin , Bmin and 0.2  a/w  1 . The C(T) specimen breaks in a brittle manner because both
the plastic zone size and the crack-tip opening displacement are very small.
---------------------------------------------------------------------------------------------------------------------
5.19 Assume that isotropic solid material having a single-edge crack is subjected to a remote
tensile stress at room temperature. Let the properties of the material be  ys  500 MPa,
Poisson’s ratio v  1 / 3 and E  72 MPa. Let the applied stress intensity factor for mode I
loading be K I  20 MPa m . Excluding microstructural details and microscale defects, use the
Tresca yielding criterion to derive (a) an expression for the critical plastic zone angle (  c ) and its
magnitude when minimum principal stresses are equal (  2   3 ), (b) determine when  min   2
and  min   3 by knowing the value of  c , (c) the plastic zone size at  c and K I  20 MPa m .
The Tresca yielding criterion is based on the maximum shear stress reaching a critical or failure
level. Hence, the definition of the maximum shear stress for this criterion is
 max  0.5 max   min   0.5 ys
Here  max and  min are principal stresses and  ys is the monotonic tensile yield strength of a
solid material. Let  max   1 and  min   2 or  min   3 .
Solution:
Needed equations for solving the problem. The principal stresses are
Hence,
16
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
(a) The expression for the critical plastic zone angle (  c ). For plane strain and  c   conditions,
If v = 1/3, then
and the minimum principal stresses become
(b) The minimum stresses. Let's calculate the values of  2 and  3 at    c and    c .
For    c  0.67967 rad  38.942o , say   35o so that
For   20o ,
17
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Therefore, the calculated values of the principal stress  2 and  3 at    c and fixed
r
indicate that  min   3 .
For    c  0.67967 rad  38.942o , say   40o so that
For   60o ,
Therefore, the calculated values of the principal stress  2 and  3 at    c and fixed
indicate that  min   2 .
18
r
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
(c) The plastic zone sizes for plane stress and plane strain at  c  38.942o are
As expected, r for plane stress is greater than that for plane strain.
19
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
5.20 Consider a ductile steel plate containing a 50-mm through-thickness central crack subjected
to a remote tensile stress of 40 MPa. If the yield strength of the steel is 300 MPa, then calculate
K I as per LEFM, Irwin’s approximation and the Dugdale’s strip yield criterion. Compare
results. Repeat all calculations for 290 MPa remote stress. Explain.
Solution:
Given data:   40 MPa and  ys  300 MPa
LEFM:
Irwin’s:


 50 
K I   a  40 MPa     10 3 m  11.21 MPa m
 2
KI 
Dugdale’s: K I 
40 MPa 
 a
1  0.5 /  ys 
2

 a
8    
ln sec

 2   2 ys 
 50 
  10 3 m 
 2
1  0.540 / 300
2
 11.26 MPa m
 50 
  10 3 m 
 2
 11.25 MPa m
8   x 40 
ln sec

 2   2 x300 
40 MPa 

These results are similar. However, if   290 MPa, then K I
LEFM:
K I  81.27 MPa m
Irwin’s:
K I  115.18 MPa m
Dugdale’s: K I  130.01 MPa m
Therefore, K I values are dubious. See Figure 5.4 for comparing Irwin’s and Dugdale’s
approximation schemes.
20
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
5.21 A single-edge SE(B) specimen with B = 20 mm, w = 40 mm is used to determine the critical
CTOD  tc and J IC . See Example 5.4 for the specimen configuration and the load-displacement
plot. Assume plane strain conditions and let
Pmax  38 kN ,  ys  800 MPa, E  207 GPa, v  0.3, K IC  60 MPa m
B  20 mm; w  40 mm;  p  1.00 mm; S  150 mm; z  1.5 mm; a  12 mm
Solution:
From Example 5.4,
The geometry correction factor  :
x  a / w  12 / 40  0.3
3x1 / 2 1.99  x1  x  2.15  3.93 x  2.7 x 2

3/ 2
21  2 x 1  x 



30.3
1/ 2

1.99  0.31  0.32.15  3.930.3  2.70.3   1.5212
2
21  2 * 0.31  0.3
3/ 2
The applied stress intensity factor as per LEFM is
P S 1.521238000 N  150 x103 m
K I  max

 54.19 MPa m
3/ 2
Bw3 / 2
20 x103 m 40 x10 3 m


Then, CTOD becomes





2
K I2
54.19 MPa m
 te 

 8.866 x103 m  0.009 mm
m ys E 2800 MPa 207000 MPa 
 tp 
0.44w  a  p

0.4440 mm  12 mm1 mm
 0.47715 mm
0.4440 mm  12 mm  12 mm  1.5 mm
0.44w  a   a  z
 tc   te   tp  0.008866  0.47715  4.78 mm
Thus, the J-integral is
J IC  m ys tc  2 800 MPa  4.78 x103 m  7.65 MPa.m  7.65MN/m 2


21
Chapter 5 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
22
Chapter 6 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
CHAPTER 6
THE ENERGY PRINCIPLE
6.1 Use Dugdale’s model for a fully developed plane stress yielding confined to a narrow
plastic zone. Yielding is localized to a narrow size roughly equal to the sheet thickness (B). This
is a fully elastic case, in which the plastic strain may be defines as ε  δt /B, where t is the crack
tip opening displacement. If the J-integral is defined by dJ  d , then show that
a 2 2
t 
E ys
Solution:
If the strain energy density and the J-integral at yielding are defined by


0
0
W   d    ys d

J  BW  B   ys d   t  ys
o
Thus,

t
o
o
J  B   ys d    ys d
Solving the J-integral yields
t
J    ys d   t  ys
o
Then,
K I2
K I2
J
and
 t ys 
E
E
2
2 2
KI
a 
t 

E ys
E ys
Here,  is a correction factor.
1
Chapter 6 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
6.2 The crack tip opening displacement (t) for perfectly plastic solution to the Dugdale model
was derived by Rice in 1966 [21] as
k  1a 0    
t 
log sec

G
  2 0 
where  o   ys and   Remote tensile stress (    o ), a  Crack size, and G = Shear
modulus. Show that the path-independent J-integral is defined by
J
k  11   a 2
E
 f a,  
Solution:
   
 yields
sec
2

0



2
2
y
5 4
y
sec y   1 

y  ...  1 
2 24
2
  
 and on log
Using Taylor’s series on sec
 2 0 
y

2 0
and
Let x  sec y  so that
 x  1 1  x  1 3

log x   2
 
  ...

 x  1 3  x  1 

1 
 sec y   1
 x 1
log x   2

2

2


 sec y   1
 x 1


1 

For x  0
    2 

y2
 
 
 1
 y2 
  2 o  
2
 2
  2
2
2
y2
4

y


 1
 4     

 2  
2

 o 
   
  

 and log sec
 
If
 1, then 4  
o
2

o
 2 o 



2
Thus,  t 
k  1a 0  
2G

 2 
 o
1  

2  2 o



2
2
From eq. (6.66),
J   t 0 
k  1a o2  
2G

k  1a 2


 2 
8G
 o
2
The elastic shear modulus of elasticity is G 
J
k  11  v a 2
4E
E
. Thus, J becomes
21  v 
 f a,  
2
Chapter 6 Fracture Mechanics, 2nd Ed. (2015)
6.3
Solution Manual
A bending test specimen made out of carbon steel showed a load-displacement behavior
A = Area =10 Joules
E = 207 GPa
ys = 1,500 MPa
B = 25 mm
b = 25 mm
P
Area

If the area under the P vs.  curve is 10 joules at the onset of crack growth, determine (a) the
fracture toughness in terms of J IC as per ASTM E-813 Standard, (b) K IC and its validity as per
ASTM E399 testing method, and (c)  tc [according to eq. (5.31) with   3 ] ,  f and r .
Explain the meaning of the results.
Solution:
(a) J IC 
210 joules 
2A

 32 kJ/m 2  32 x10  3 m. MPa
Bb 25  10  3 m25  10 3 m 
207 x10 MPa32 x10 m.MPa  81.39 MPa
 2.5K /    7.36 mm  B
and B
B
(b) K IC  EJ IC 
Validity: BASTM
3
3
m
2
IC
ys
min
Actual
ASTM
Therefore, K IC  81.39 MPa m is valid.
(c) From eq. (5.31),


2
4 81.39 MPa m
4 K IC
 tc 

 0.016 mm
E ys   3 207 x103 MPa 1,500 MPa 
 
2

Using eqs. (5.44) and (5.13) yields the fracture strain and the plastic zone size, respectively
f 
 c 0.016 mm

 6.4 x10 4 or 0.064%
B
25 mm
The plastic zone size:
1
r
6
 K IC


 ys
2
2



  1  81.39 MPa m   0.16 mm  a

6  1,500 MPa 

Therefore, the material behaved in a brittle manner implying that very little plastic deformation
took place at the crack tip. These are very small values indicating that the material is brittle and
undergoes very little plastic deformation. With regard to the critical crack tip displacement of
0.016 mm, it indicates that the crack tip displacement is very small; that is,
 y   c / 2  0.008 mm .
3
Chapter 6 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
6.4 If J I   ys  t is used to determine the fracture toughness, will  t be a path-independent
entity? Explain.
Solution:
Using Dugdale’s model, Figure 5.3 and eq. (6.33) yields
y
P
Plastic Zone
2 y
x
P
2a
JI  
ar
a
r
  

ds 
Wdy  T
x 

According to Figure 5.3, dy  0 from a to a  r and T   y   ys . Thus,
J I  
ar
a
 ys

  y   y
x
 dx 

t
o
 ys d  y   y dx  t  ys
Therefore, both J I and  t are path-independent. Only the initial and final values of  t are
needed to solve the integral. In fact,  t is just an opening as well as a path-independent entity.
2
6.5 Assume that crack growth occurs when J I  J IC , where J IC  (1   2 ) K IC
/ E . If a welldeveloped plastic flow occurs, will these expressions be valid? Explain.
Solution: It is valid in the elastic regime where a  r.
4
Chapter 6 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
6.6 A double cantilever beam (DCB) is slowly loaded in tension up 10 MN as shown
schematically below. Assume that there is no rotation at the end of the beam and that the beam is
made of isotropic steel having the following properties: K IC  47 MPa m and E  207 GPa.
Will fracture occur?
AokN
 B h
PP  10
M  Pa
M
B
a = 20 mm
B = 20 mm
h = 10 mm
h
h
  2
M
a
P
Solution:
From eq. (6.42) along with E '  E / 1  v 2  and I  Bh 3 / 12,
P 2 a 2 121  v 2 P 2 a 2
121  0.32 10 x10 3 MN  20 x10 3 m 
GI 


BE ' I
EB 2 h 3
207,000 MN / m 2 20 x103 m 2 10 x103 m 3
2
2
G I  5.28 x10 3 MPa.m
EG I
207,000 MPa 5.28 x10 3 MPa.m 

1  v2
1  0.32
K I  34.66 MPa m
KI 
Therefore, fracture will not occur because K I  34.66 MPa m  K IC  47 MPa m
5
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
CHAPTER 7
PLASTIC FRACTURE MECHANICS
7.1 Determine (a) the J-integral (J) and (b) the dJ/da for a hypothetical steel plate containing a
central crack of 114 mm long loaded at 276 MPa and exposed to room temperature air. What
does TJ < TR mean? Assume plane strain conditions and that the stainless steel obeys the
Ramberg-Osgood relation with curve fitting parameters such as  ' 1.69 and n  5.42. Explain
the results based on the strain hardening effect. Data: E  206,850 MPa ,  ys  207 MPa,
  0.3 and J IC  130 kJ/m 2 . Dimensions: w  4a  228 mm, L  2w

2w

2a
2L
Solution:
(a) For a central crack, use Table 3.1 for 
x  a/w  57 / 228  0.25
  1  0.5 x 2  20.46 x 4  81.72 x 6  1  0.50.25  20.460.25  81.720.25  1.0635
2
4
6
  sec 0.25  1.1892
K I   ae
The effective crack length needed for calculating K I  K I ae  is given below.
 1
ry  
 
 n  1  K I


 n  1   o
2
  1  n  1  
   

    n  1   o
2

 ae

&  

 1  n  1  
1
ae  a  w ry  a  


2 
1  P / Po     n  1   o
With   sec 0.25  1.1892
1
2
1  P / Po 
2

 ae


 1  5.42  1   1.1892276 
1
ae  57 mm  

2  
 ae
207
1  125.86 / 81.75  6  5.42  1  
ae  57 mm  8.5597 x10 2 ae
ae  62.336 mm
2
1
 6
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
K I   ae  1.1892276 MPa   62.336 x10 3 m
K I  145.25 MPa m
The elastic J-integral values that include the effects strain hardening is
K I2 (1   2 ) K I2 (1  0.32 )145.25


 9.2815x10 2 MPa  m
E'
E
206,850
J e  92.82 kPa.m  92.82 kJ/m 2
2
Je 
Plastic Part (continued): Table 7.2 for a central crack
εo  σ o /E  0.001 and   0.001
Po  4(w  a)σ o / 3  81.75 MN/m & P  2 wσ  2228 x103 m276 MPa   125.86 MN/m
w  a  228  57  171 mm
The Plastic zone:
2
 1  5.42  1  145.25 
3
r  ry   

 10   17.855 mm
 6  5.42  1  207 
From eq. (7.25) and Table 7.1 with n = 5.42,
I n  6.2511  0.29381n  1.1724 x10 2 n 2  5.0031
 
J p   ' o o I n r  
o 
J p  87.17 kJ/m 2
n 1
 276 
 1.69 207 MPa 0.0015.003117.855 x10 m

 207 
3
Thus, eq. (7.36) gives
J  J e a e   J p a, n   92.82 kJ/m 2  87.17 kJ/m 2
J  J I  180 kJ/m2
Therefore, crack grows occurs since J  290.96 kJ/m 2  J IC  130 kJ/m 2
(b) Crack instability
K I2    
2
Je 
K I2 (1   2 ) K I2  (1   2 ) a


E'
E
E
2
5.42 1
10 
3
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
 1
r  ry  
 
 n  1  K I


 n  1   o
 
J p   ' o o I n r  
o 
n 1
Solution Manual
2
  1  n  1  
   

    n  1   o
2

 a

 1  n  1  
  '  o o I n  

   n  1   o



2
 
 
o 
n 1
 1  n  1   
 (1   2 ) a
J  Je  J p 
  ' o o I n 2  
 
E
   n  1   o 
2
 1  n  1   
dJ  (1   2 ) 

  ' o o I n 2  
 
da
E
   n  1   o 
a
n 3
a
n3
0
dJ  (1  0.32 )1.1892 x 276 MPa 

da
206,850 MPa
2
 1  5.42  1  276 
 1.69207 MPa 0.0015.00311.0635  


 6  5.42  1  207 
2
dJ
 4.05 MPa  4.05 MJ/m3
da
From eq. (6.70), the tearing modulus is
E dJ 206,850 MPa 

4.05 MPa   19.55
 o2 da
207 MPa 2
TJ  19.55
TJ 
Therefore, crack growth is stable if TJ  TR .
3
5.42  3
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
7.2 Plot the given uniaxial stress-strain data for an ASTM A533B steel at 93ºC, and perform a
regression analysis based on the Ramberg-Osgood equation. A single-edge-cracked plate made
out of this steel is loaded in tension at 93ºC. Determine the total elastic-plastic J-integral when
the applied load is   500 MPa . Will the crack grow in a stable manner? Why? or Why not?
Assume plane strain conditions and necessary assumptions. Let a  100 mm, v = 0.30, E = 207
GPa, L = 3w, w = 400 mm, JIC = 1.20 MPa.m, h1  0.523 and B = 150 mm.
Strain
(x10-3)
0
1.0
0
2.2
4
2.3
0
2.3
5
5.0
0
7.5
0
10
20
40
Stress
(MPa)
0
381
414
415
416
450
469
483
519
557
Solution:
Single-edge cracked specimen
L
w


a
Stress v s. Stra in
Stress (M Pa )
800
600
400
200
0
0
0.05
0.1
0.15
0.2
0.25
0.3
Stra in
Data for plane strain condition:
 ys   o  414 MPa (From the above Figure)
v  0.30
E  207 GPa
in  kips
(From Ref. [8])
J IC  1.20 MPa  m  6.8
in 2
  500 MPa (Applied)
4
a  100 mm
w  400 mm
B  150 mm
L  3w  1,200 mm
b  w  a  300 mm (Ligament)
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
Elastic Part:
n
 
Nonlinear regression using     0   yields  '  1.12 and n  9.71 . From Table 7.2,
 0 
'
P  wσ  400 x103 m500 MPa   200 MN/m
1/ 2
1/ 2
  a 2 
  100 2 
a
100
x  1  
 1  
100  0.82137
  
  
w  a   400  100  
400  a
  w  a  
P0  1.455x w  a  0  1.4550.82137300 x103 m 414 MPa   148.43 MN/m
1  w  a a / w   400  100100 / 400  75 mm
Also,   6 and
 n  1  K I


n

1

  0

1
w
2
1  P/P0 
 1
ry  
 
  1  n  1   ae
  



n

1

  0

 
2
2
  ae  n  1   
   
 
    n  1   
 0

2
The crack size can be calculated using
ae  a  w ry
2

 1  n  1   

 1  9.71  1  500 
1
1
ae  a  
 
  ae  a  

ae 
2 
2  
1  200 / 148.43  6  9.71  1  414 
1  P/P0     n  1   0 
ae  a  0.05814ae
ae  1.0617a
a  100 mm
ae  1.0617100 mm  106.17 mm
For a / w  100 / 400  0.25,
  1.12  0.23a / w   10.55a / w   21.71a / w   30.38a / w   1.5013
2
3
4
K I   ae  1.5013500 MPa   106.17 x10  3 m  469.10 MPa m

K 2 1  ν 2 K I2 1  0.32  469.10 MPa m
J e  I' 

E
E
207 x103 MPa

5
2
 0.97 m.MPa  0.97 MJ/m 2
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
Plastic part:
2
 a  n  1     106.17 mm  9.71  1  500 
r  ry   e 
   


  21 mm
6
 9.71  1  414 
   n  1   0  
2
From eq. (7.51) and Table 7.2,
h1  0.523 ,  '  1.12 and 1  75 mm  75 x103 m
P
J p    0 01h1  
 P0 
'
n 1
with ε0  σ 0 /E  0.002
1 9.71
 200 
J p  1.12414 MPa 0.00275 x10 m0.523

 148.43 
J p  0.89 MJ/m 2
3
Thus, the total value of the J-integral is
J  J e  J p  0.97  0.89  1.86 MJ/m2
Therefore, crack will grow because J  J IC  1.20 m. MPa. Stable? Unstable? Let’s find out how
the crack will grow. Let




K 2 1   2 K I2 ae 1   2  
K I   ae ;
J e  I' 

E
E
E
so that the total value of the J-integral can be computed by
P
a 1   2  2
J  Je  J p  e
  ' 0 01h1  
E
 P0 
2
and
P
J p    0 01h1  
 P0 
'
n 1
If ae  1.0617a and 1  w  a a / w  , then


n 1

 1   2  2 
P 
'
J  
 1.0617a     0 0 h1   w  a a / w 
E

 P0  


Thus, at constant load dJ/da is
n 1
 1   2  2   '
 P   w  a a 
dJ
 1.0617
  1  26 MPa
    0 0 h1    
da
E
w 
 P0    w

 
E dJ 207 x103 MPa 
TJ  2

26 MPa   31.40
 o da
414 MPa 2
Therefore, the crack will grow in a stable manner TJ  31.40  TR
6
n 1
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
7.3 Repeat Problem 7.2 using the Hollomon equation   K n for the plastic region. Curve
fitting should be performed using this equation for obtaining K and n. Assume plane strain
conditions and make the necessary assumptions. Compare the result from Problem 7.2. What can
you conclude from these results? Assume a contour as shown in Figure 7.6 with y23  70 mm.
Solution:
Single-edge cracked specimen
Stress v s. Stra in
Stress (M Pa )
800
600
400
200
0
0
0.05
0.1
0.15
0.2
0.25
0.3
Stra in
Data for plane strain condition:
 ys   o  414 MPa (From the above Figure)
a  100 mm
E  207 GPa
v  0.30
w  400 mm
B  150 mm
J IC  1.20 MPa  m  6.8
  500 MPa (Applied)
in  kips
(From Ref. [8])
in 2
L  3w  1,200 mm
b  w  a  300 mm (Ligament)
Elastic Part:
For a / w  100 / 400  0.25,
  1.12  0.23a / w   10.55a / w   21.71a / w   30.38a / w   1.5013
2
3
4
K I   ae  1.5013500 MPa   106.17 x10  3 m  469.10 MPa m

K 2 1  ν 2 K I2 1  0.32  469.10 MPa m
J e  I' 

E
E
207 x103 MPa

7
2
 0.97 m.MPa  0.97 MJ/m 2
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
Plastic part: Nonlinear regression using   k n yields k  776 MPa and n  0.103. The
plastic strain energy density can be defined using the Hollomon equation as
1/ n
 
  k n and    
k


0
0
Wp     d   k n d
k  776 MPa and
k n 1  k   
Wp 
 
 
n 1
 n  1  K 
n 1
n
n  0.103
10.7087
 776 MPa  500 MPa 



 1.103  776 MPa 
W p  6.35 MPa  6.35 MJ/m3
Assume a contour as shown in Figure 7.6 with y23  70 mm so that J p  2 y23W p as per eq. (7.43),
which represents a near-field condition. Thus,


J p  2 y23W p  2  70 x103 m 6.35 MPa   0.89 m. MPa
J I  J e  J p  0.97  0.89  1.86 MPa  m  1.86 MJ/m 2
Crack growth occurs since J  J IC . Note that if the value of y23 changes so will J p and J I .
Anyway, both problems 7.2 and 7.3 (P7.2 and P7.3) give the same result as per chosen contour
segment y23  70 mm; otherwise J p P7.2   J p P 7.3 or J I P7.2   J I P7.3 .
Then, K I   a and K I2  a  
2
K I2  1   2   a
Je  ' 
E
E
2
K n 1  K   
Wp 
 
 
n 1
 n  1  K 
n 1
n
K n 1
 K   
J p  2 y23W p 
  2 y23 
 
n 1
 n  1  K 
n 1
n
 1   2  2 a
 K   
JI  Je  J p 
 2 y23 
 
E
 n  1  K 
2
dJ I  1   2  2 2  1  0.32 1.5013x500 


da
E
207 x103
n 1
n
dJ I
 7.78 MPa  7.78 MJ/m3
da
E dJ 207 x103 MPa 
TJ  2

7.78 MPa   9.40
 o da
414 MPa 2
Therefore, the crack will grow in a stable manner TJ  9.40  TR .
8
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
7.4 A steel plate having a single-edge crack is loaded as shown below. Calculate (a) the loadline displacement (  ) and (b) the crack opening displacement ( ) that corresponds to a point on
the resistance curve (not included). Assume plane strain conditions and use the following data:
v  0.3 , h1  0.523, h2  1.93, h3  3.42, B  0.15 m, w  0.40 m, L  1.2 m and a  0.1 m.
J IC  1.20 kJ/m 2 ,   500 MPa and E  206,850 MPa
Solution:
L
w


a
(a) From Table 7.2,
P
J p   ' 0 0 w  a h1  
 P0 
P
 p   ' a 0 h3  
 P0 
n
n 1
1
 P   p n
so that    

 P0   ' a 0 h31 
Eliminating P/Po yields
 p 

J p   ' 0 0 w  a h1 
  ' a 0 h3 
 h w  a    
J p  0 1 1 / n  p 
 '  0   ah3 
 n  1 / n
n 1 / n
n
 J p  n 1

 p  ah3  '  0 1 /( n 1) 
  01h1 
Similarly,
n
n
P
 J p  n 1

   ' a 0 h2    ah2  '  0 1 /( n 1) 
 P0 
  01h1 
These are the equations to be used in this problem. However, we need the value of J p at
  500 MPa. Let’s get.
9
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
Data for plane strain condition:
 ys   o  414 MPa (From the above Figure)
E  207 GPa
a  100 mm
w  400 mm
v  0.30
J IC  1.20 MPa  m  6.8
  500 MPa (Applied)
in  kips
(From Ref. [8])
in 2
B  150 mm
L  3w  1,200 mm
b  w  a  300 mm (Ligament)
Elastic Part:
n
 
Nonlinear regression using     0   yields  '  1.12 and n  9.71 . From Table 7.2,
 0 
P  wσ  400 x103 m500 MPa   200 MN/m
'
1/ 2
1/ 2
  a 2 
x  1  
 
  w  a  
  100 2 
a
100

 1  
100  0.82137
  
w  a   400  100  
400  a
P0  1.455x w  a  0  1.4550.82137 300 x103 m 414 MPa   148.43 MN/m


1  w  a a / w   400  100100 / 400  75 mm
Also,   6 and
 1  n  1  K I
r  ry  


   n  1   0
1
w
2
1  P/P0 
2
  1
  
  
 n  1   ae


 n  1   0
2
  ae  n  1   
   
 
    n  1   
 0

2
The crack size can be calculated using
ae  a  w ry
2

 1  n  1   

 1  9.71  1  500 
1
1
 
ae  a  
  ae  a  

ae 
2 
2  

n

1

6
9
.
71

1
414




1

P/P
1

200
/
148
.
43












 0
0


ae  a  0.05814ae
ae  1.0617a
a  100 mm
ae  1.0617100 mm  106.17 mm
10
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
For a / w  100 / 400  0.25,
  1.12  0.23a / w   10.55a / w   21.71a / w   30.38a / w   1.5013
2
3
4
K I   ae  1.5013500 MPa   106.17 x10  3 m  469.10 MPa m

K 2 1  ν 2 K I2 1  0.32  469.10 MPa m
J e  I' 

E
E
207 x103 MPa

2
 0.97 m.MPa  0.97 MJ/m 2
J e  0.97 MJ/m2
From Problem 7.2,
2
 a  n  1     106.17 mm  9.71  1  500 
r  ry   e 
   


  21 mm
6
 9.71  1  414 
   n  1   0  
2
From eq. (7.51) and Table 7.2,
P  wσ  400 x103 m500 MPa   200 MN/m
1/ 2
  a 2 
x  1  
 
  w  a  
1/ 2
  100 2 
a

 1  
 
w  a   400  100  


100
100  0.82137
400  a

P0  1.455x w  a  0  1.4550.82137 300 x103 m 414 MPa   148.43 MN/m
1  w  a a / w   400  100100 / 400  75 mm
h1  0.523 ,  '  1.12 and 1  75 mm  75 x103 m
P
J p    0 01h1  
 P0 
'
n 1
with ε0  σ 0 /E  0.002
1 9.71
 200 
J p  1.12414 MPa 0.00275x10  3 m0.523

 148.43 
J p  0.89 MJ/m2
J  J e  J p  0.97 MJ/m2  0.89 MJ/m2  1.86 MJ/m2
11
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
Thus,
n
 J p  n1

 p  ah3  '  0 1/( n1) 
  01h1 
 p  0.1 m 3.421.120.002
0.0934


0.89 m.MPa
 414 MPa 0.3 m 0.523 


0.9066
 p  3.90 mm
(b) The crack opening displacement is
n
n
P
 J p  n 1

   ' a 0 h2    ah2  '  0 1 /( n 1) 
 P0 
  01h1 
  0.1 m1.931.12 0.002 
0.0934
  2.20 mm


0.89 m. MPa
 414 MPa 0.3 m0.523 


0.9066
Therefore,  p  3.90 mm and   2.20 mm at the onset of crack growth since J IC  1.2 MN/m2
and crack blunting required these values.
7.5 Determine the strain hardening exponent (n) for a steel with  ys = 400 MPa, E = 207 GPa.
Assume that it obeys the Hollomon equation  p  k n . Consider the maximum plastic stress in
your calculations.
Solution:
and  max  k  700 MPa
 p  max @    p  1   max
n
@ yielding,  ys  k ysn ,  max  k max
. and  ys 
 ys
k ysn

  ysn
 max k max
  @  ys
ln 
 max
n 
ln  ys 




 400 
ln

 700   0.0895
ln 1.93  10 3


12
 ys
E
 1.93  10 3
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
7.6 (a) Derive an expression for the J-integral J p /J e  f(σ(σo ) ratio as per Rice model. (b) Plot
the resultant expression for a remote stress to yield stress ratio range 0   /  o  1 . Here,
 o   ys . Explain the resultant trend.
Solution:
(a) The plastic case:
k  1a o    
t 
log  sec

G
  2 o 
J p   o t 
k  1a o2
G
   

log  sec
2

o

 
The elastic case:


k  1K I2  k  1a 2

 Je 
8G
8G




   



2 log sec

 Jp
 2 o 

 f ( /  o ) 
J 
2
  

 e




 2 o 


From the plot, J p /J e   as σ/σ o  1 since secπ/ 2    and σ  σ o .
5
They significantly differ from each
other due to a large plasticity.
At σ/σ o  0.8 , JP dominates.
4
J p /J e
3
2
1
0
0.2
0.4
0.6
0.8
1
σ/σ o
13
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
7.7 Calculate the total J-integral (J) for a 2024 Al-alloy plate containing a single-edge crack
under plane stress conditions. Plot b) the   f   and J I  f  . Use the following data to
carry out all calculations: a = 1.40 mm, w = 19 mm, B = 0.8 mm, L = 10 cm,  o  64 MPa,
E  72,300 MPa, α′ = 0.35, n = 5 and F = 1.01 kN.
Solution:
Elastic part:
a) The following calculations are self-contained. Thus,
64
o   o 
 8. 852  10 4
A  wB  15. 20x10 6 m 2
E
72, 300
1. 01x10 3 MN
  F 
 66. 45 MPa   ys
A
15. 20x10 6 m 2
For a single-edge crack,
x  a/w  1. 4/19  0. 073684
  1. 12  0. 230. 073684  10. 550. 073684 2  21. 710. 073684 3  1. 1516
  1. 12  0. 23x  10. 55x 2  21. 71x 3  1. 1516
Je 
2 2
1. 40x10 3 m1. 1516 2 66. 45 MPa
K 2I
 a  
E
E
72, 300 MPa
2
J e  3. 5623  10 4 MPa. m  356. 23 J/m 2
J e  356. 23 J/m 2
Plastic part:
From Table 7.2 and eqs. (7.60),
 1  w  aa/w  19  1. 41. 4/19  1. 2968 mm
 
a
1   wa
2 
a
wa

1
1.4
191.4
2

1.4
191.4
 0. 92361
P o  1. 072w  a o  1. 072  0. 92361  19  1. 4  10 3  64  1. 1153 MN/m
P  w  19  10 3 m66. 45 MPa  1. 2626 MN/m
h1 
h1 
3a n
4 1n  1
 o 
31.4 5
41 1n 1.2968
n1
Po
P
66.45
64
n1
51
1.1153
1.2626
51
 3. 3853
From eq. (7.54a),
J p    o o  1 h 1
P
Po
n1
 0. 35  64  8. 852  10 4  1. 2968  10 3   3. 3853
J p  183. 23 J/m 2
J I  J e  J p  356. 23 J/m 2  183. 23 J/m 2  539. 46 J/m 2
14
1.2626
1.1153
51
10 6 
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
(b) These results indicate that J p  0.34 J and J e  0.66 J . Therefore, J e  J p and contributes
66% to J.
(c) The required stress-strain curve can be determined using eq. (7.30). Thus,
   ys

 ys
1/n
 322. 08 MPa 5 
80
80
70
Stress 
(MPa)
60
60
50
40
40
Load Point
30
20
20
10
0
0
0
0.0002
0.0003
10.0001
2
3
-4
Strain

Strain  x10


0.0004
4
The J-integral plot along with the load point is based on the following equation:
J I  2. 1247  10 2  2  1. 0638  10 9  6
200
150
JI
2
(J/m )
Loading
Point
100
50
0
0
20
40
 MPa 
60
80
The trend of the J I  f   resembles the trend shown in Figure 7.5.
15
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
7.8 The plastic J-integral (Jp) for some configurations can be defined by J  ηW/(Bb), where
W  absorbed energy, Bb = cross sectional area, b = (w – a) = ligament and   constant. This
integral can then be separated into elastic and plastic components. For pure tension,
J  Je  J p 
K I2  pWp

E'
Bb
(1)
Consider a strain hardenable material and a specimen with unit thickness B. If the plastic load is
defined by P  C pn , where C is the compliance and n is the strain hardening exponent, then the
load-displacement and J-P profiles are schematically shown below.
Jlimit
J
P
W

P
Po = Plimit
Derive an expression for  p .
Solution:
The area under the curve is the strain energy density define by
Wp  
p
o
p
Pd p   C np d p 
o
n
n
C np 1 
P p since P  C pn
n 1
n 1
(2)
Since W p is a maximum energy then P reaches Po (the load limit). Thus,
Wp 
n
Po  p
n 1
(3)
n
P
From eq. (7.56),  p   ' a o h3   and
 Po 
P
n
Wp 
Po ' a o h3  
n 1
 Po 
n
(4)
Substituting this expression into the J-integral equation above yields the plastic part as
 n
P
Jp  p
Po ' a o h3  
Bb n  1
 Po 
n
(5)
16
Chapter 7 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
Equate eq. (7.54) and (5) and solve for  p
p n
P
P
Po ' a o h3     '  o o1h1  
b n 1
 Po 
 Po 
n
p 
n
n  1 b 2  o h1
n a Po h3
17
Chapter 8 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
CHAPTER 8
MIXED-MODE FRACTURE MECHANICS
8.1 A large plate (2024-0 Al-alloy) containing a central crack is subjected to a combined mode III loading. The internal stresses are  y = 138 MPa and xy = 103 MPa. Use the Maximum
Principal Stress Criterion ( -criterion) to calculate the fracture toughness (KIC). Use the
following data: crack length 2a = 76 mm,  = 1/3, and E = 72,300 MPa.
Solution:
a = 38 mm,  = 138 MPa,  = 103 MPa
K I   y a  47.68 MPa m
K II   xy a  35.60 MPa m
K I / K II  1.3393
Using eq. (8.29a) yields
1  x x  3 x2  8
0   ar cos 1 
x2  9
 3 


  48.60


From eq. (8.33),



K IC  K I cos 3 0  3K II cos 2 0 sin 0
2
2
2
K IC  72.50 MPa m
o
From eq. (8.34),
3
K IIC 
K IC  62.77 MPa m
4
1
Chapter 8 Fracture Mechanics, 2nd Ed. (2015)
8.2
Solution Manual
Repeat problem 8.1 using the Strain Energy Density Factor (S) Criterion (S-criterion).
Solution:
K I   a  47.68 MPa m
K I   xy a  35.60 MPa m
K II / K I  0.75
Use eq. (8.53) to solve for the fracture angle


sin  0 2 cos 0  k  1K I2  4 cos 2  0  k  1 cos  0  2 K I K II  sin  0 6 cos 0  k  1K II2  0
  48.10 o
Using Equation (8.54) yields
4E
2
K IC

a K 2  2a12 K I K II  a 22 K II2
k  11    11 I


K IC  59.35MPa m
From eq. (8.57) and (8.51),
K IIC  0.9045K IC  53.68 MPa m
S c  14.26 kPa m  14.26 kJ / m 2
8.3 Determine (a) the incline angle  and (b) the applied tension  in Example 8.1.
Solution:
(a) K I   a 
sin 2 
(b) K I   a sin 2 
K II   a 
sin  cos 
 
KI
47.68MPa m
 tan  
K II
35.60MPa m
  53.25º
  214.95 MPa
KI
a 
sin 
2

a = 38 mm
47.68MPa m
 38  10 3 m sin 2 53.25º 
8.4 Calculate the critical stress (fracture stress) for the problem described in Problem 8.1
according to the   - criterion. Will crack propagation take place?
Solution:
From problem 8.1 & 8.2:
From Example 8.4:
  53.25º
  215MPa (Applied)
a = 38 mm
E = 72,300 MPa
2
o = -48.60º
 = 1/3
Chapter 8 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
From eq. (8.35),
c 
K IC
sin  a
b
11

1 / 2
 2b12 cos   b22 cos 2    0  492 MPa
where
1
3
b11  1  cos 0   0.57314
8
3
2
b12  sin  0 1  cos  0   0.77635
8
9
b22  sin 2  0 1  cos 0   1.05161
8
Crack propagation will not take place since    c .
8.5 Repeat problem 8.4 using the S-criterion with   215 MPa.
Solution:
K I   a 
sin 2 
K I   2a 
sin 4 
K II   a 
sin  cos 
K II   2a 
sin 2  cos 2 
K I K II   2a 
sin 2  cos 
From eq. (8.54),
c 
K IC
sin  a

(   1 )( 1   )
a11 sin 2   2a12 cos   a 22 cos 2 
4E
where

1 / 2
  o
1
1  cos k  cos   1  v  1  cos  k  cos  3.84 x10 6
16G
8E
1
a12 
sin  
2 cos   k  1 1  v  sin 2 cos   k  1 1.15 x10 6
16G
8E

1  v
a 22 
k  11  cos    1  cos 3 cos  1 5.90 x10 6
8E
a11 
Thus,
 c  204 MPa
  215 MPa (Applied)
Therefore, crack propagation will occur because    c .
3
Chapter 8 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
8.6 Show that the stress intensity factor is K I 
Poisson’s ration is v  1 / 3 .
Solution:


From eq. (8.13) with E '  E / 1  v 2 and K II  0,
E ' (1   ) 2
2
K IC
 K I2 
K III with K I  2 K III
E
E (1   ) 2
(1   )
2
2
K IC
 K I2 
K III  K I2 
K III
2
1  v 1  v 
E 1  v 
2
K IC
 K I2 
2
K III
K I2
 K I2 

1  v 
41  v 

1  2
1  41  v  K I



 2 11 2
1
2
K IC
 1 
K I   8 K I
 
 41  1 / 3
Thus,
KI 
8
K IC
11
4
8 / 11K IC when K I  2 K III and that the
Chapter 8 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
8.7 Use the experimental mixed mode fracture data or Solid A and Solid B given below to
compare the mixed mode fracture criteria discussed in this chapter. Determine which criterion is
the most suited to predict the mixed mode fracture behavior of these solids.
Solid A
K II
Solid B
KI
K II
KI
(MPa m  (MPa m  (MPa m  (MPa m 
0.49
0
0.42
0
0.43
0.52
0.41
0.10
0.33
1.02
0.43
0.21
0.31
1.10
0.38
0.40
0.28
1.20
0.40
0.50
0.26
1.25
0.37
0.60
0.16
1.40
0.39
0.65
0
1.50
0.37
0.80
0.31
1.20
0.27
1.26
0.20
1.55
0
1.65
Solution:
Plot the given data and use all possible mixed mode criteria in order to determine the most suited
criterion.
2
KI
MPa m 
1.5
Solid A
1
Solid B
0.5
0
0
0.125
0.25

K II MPa m
5

0.375
0.5
Chapter 8 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
From the given data, Solid A:
K IC  1.50 MPa m & K IIC  0.49 MPa m
Solid B:
K IC  1.65 MPa m & K IIC  0.42 MPa m
For Solid A,
KI
2
KI
MPa m 
K I

K
 IC
1.5
2
 K II

 

 K IIC
2


  1

1
2
K I K II

K IC 
K IIC
0.5


  1

0
0
0.125
0.25
0.375

K II MPa m

0.5
KII
For Solid B,
KI
2
KI
MPa m 
1.5
2
K I

K
 IC
 K II

 

 K IIC

0.375
2


  1

1
K I K II

K IC 
K IIC
0.5
0
0
0.125
2


  1


0.25
K II MPa m
K I
Therefore, the criterion 
K
 IC
2
 K II

 

 K IIC
2
0.5
KII

K I K II

  1 fits both data sets better than K  
K
IC

 IIC
6
2


  1.

Chapter 8 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
8.8 A large and wide plate has a small through-the-thickness central crack as shown in the figure
below. Use the free-body diagram (FBD) to derive expressions for (a) the normal (σN) and shear
(τ) stresses and (b) the intensity factors KI and KII as functions of β.
FBD


N

A
2a
Ao
cos 
B

P
Ao
 



Ao  Area
Solution:
(a) Using the FBD yields the normal and shear stresses
F
N
N
y
F
  N A   Ao cos   0
x
  A   Ao cos   0
Ao
  Ao cos   0
cos 
   cos  cos    sin  . cos 
Ao
  Ao cos   0
cos 
  cos 2    sin 2 

(b) The stress intensity factor equations are
K I   N a 
K II   a
K I   a sin 2  
K II   a sin  cos 
7
Chapter 8 Fracture Mechanics, 2nd Ed. (2015)
Solution Manual
8.9 Two identical cracked plates as shown in problem 8.8 are to be tested to determine K IC
and K IIC . The observed critical tension loads and the incline angles were 120 MPa @   0 and
130 MPa @    / 4, respectively. Use the equations given below to determine the fracture
toughness for mode I and II.
K I K II

K IC 
K IIC
2


  1

K I

K
 IC
and
2
 K II

 

 K IIC
2


  1

Solution:
For case   120 MPa @   0 and    / 2  90o ,


K I   a sin 2    120 MPa   20 x10 3 m 1  30 MPa m
K II   a sin   cos   0 MPa m
0
K I K II

K IC 
K IIC
2


  1

K IC  K I  30 MPa m
For case   130 MPa @    / 4 and    / 4  45o ,

K I   a sin    130 MPa   20 x10
2
3
2
 2

m
 2   16.30 MPa m
 

 2  2 
 
K II   a sin   cos   130 MPa   20 x10 3 m 
 2  2  16.30 MPa m
  

K I K II

K IC 
K IIC

2


  1

K IC  30 MPa m
K IIC


K IIC  16.30 MPa m 1  16.30 / 30
1 / 2
2
 K II

 

 K IIC
K IIC
 K
I
 K II 1  
K
  IC
1 / 2
 KI 
 K II 
1  K 

IC 

2
K I

K
 IC



  1; K IC  30 MPa m




 
2
1 / 2

K IIC  16.30 MPa m 1  16.30 / 30 
K IIC  24.12 MPa m

2 1 / 2
K IIC  19.42 MPa m
These results indicate that the care must be taken in selecting a particular fracture criterion.
K I
Nevertheless, the fracture criterion 
K
 IC
K II
K
when compared to I  
K IC 
K IIC
2
 K II

 

 K IIC
2


  1

8
2


  1 gives conservative values for K IIC

Chapter 9 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
CHAPTER 9
FATIGUE CRACK GROWTH
 E ys 
da
  t n / 2 , where  t is the
9.1 Show that Paris equation can take the form
 A
2 
dN
1 v 
change of crack tip opening displacement, E is the modulus of elasticity, and A is a constant.
n/2
Solution:
From eq. (7.45) or (8.10) with KII = KIII = 0
J


K I2
1  v 2 K I2

E'
E
KI 
EJ
1 v2
K 
EJ
1 v2
Substituting K into Paris expression, eq. (9.6) yields
 EJ 
da
n
 AK   A
2 
dN
 1 v 
n
From eq. (6.66), J   ys t and J   ys  t . Thus,
 E ys 
da

 A
2 
dN
1

v


n/2
 x n / 2
da
A
n

and K e  K max 1  R  is
 C K e  , where C 
n( 1 )
dN
1  R 
da
da
n
n
Walker’s effective stress intensity factor range. (b) Plot
 C K  and
 AK  for a
dN
dN
4340 steel having  ys  1,254 MPa,  ts  1,296 MPa, K IC  130 MPa m ,   0.42, n  3.24 ,
9.2
(a) Show that
R  0.7 , and A  5.11x10 11 . .
Solution
(a) K e  K max (1  R )
R
K min K max  K max  K min K max  K
K


1
K max
K max
K max
K max
1
Chapter 9 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
K  K max (1  R )
K
K max 
1 R
Then,
K e 
K
and
(1  R)1
n
 K

da
A
 A( K e ) n  A

K n
1 
n( 1 )
dN
(1  R )
(1  R ) 
where
C
A
( 1  R ) n( 1 )
For R  1
Thus,
da
n
 C K 
dN
If R  0 , then C  A.
(b) If  ys  1,254 MPa,  ts  1,296 MPa, K IC  130 MPa m ,   1 / 3, n  3, R  0.7 and
A  5.11x10 11
MN n  mm
cycles
2n 2
2
, then n(1   )  31  1 / 3  2
Thus,
C
A
5.11x10 11 MN 3  mm 4 / cycles

 5.68 x1010 MN 3  mm 4 / cycles
2
n (1 )
(1  R)
1  0.7 
Comparison:
Walker:
Paris:
da 
MN 3  mm 4 
K 3
  5.68  10 10
dN 
cycles 
da 
MN 3  mm 4 
K 3
  5.11  10 11
dN 
cycles 
2
Chapter 9 Fracture Mechanics, 2nd ed. (2015)
Plot
Solution Manual
da
vs . K for 20 MPa m  K  100 MPa m
dN
2.5e-08
Walker
2e-08
1.5e-08
da
mm / cycles 
dN
1e-08
Paris
5e-09
0
20
40

60
K mpA M

80
100
9.3 Suppose that a single-edge crack in a plate grows from 2 mm to 10 mm at a constant
loading frequency of 0.02 Hz. The applied stress ratio and the maximum stress are zero and 403
MPa, respectively. The material has a plane strain of 80 MPa.m1/2 and a crack growth behavior
described by
da
4
 3.68 x10 12 K I 
dN
Here, da / dN is in m/cycle and K I is in MPa m . Determine the time it takes for rupture to
occur.
a


Solution:
  1.12
K IC  80 MPa m
a o  2 mm
a f  10 mm
R0
A  3.68 x10 12
n4
 min  0
 max  403 MPa
Using K IC   max a c yields
a c  a f  10 mm
3
f  20 Hz
  403 MPa
Chapter 9 Fracture Mechanics, 2nd ed. (2015)

da
n
 AK I   A  a
dN

af
2
 a da  A  
t
4
Nf
f

278.31 cycles
 13,916 sec
0.020 cycles / sec
Nf
t  3.87 hours
0
a a
Nf
4
  dN
ao
1
o

Solution Manual
1
f
 1.51N f
2 x10

m   10 x10 3 m 
1.51 m 1 .cycles 1
3
1
1
N f  278.31 cycles
9.4 If a large component is subjected to a cyclic loading under   300 MPa and R  0.
The material behaves according to Paris Law
da
2.45
 2 x10 8 K I 
dN
where da / dN is in m/cycles and K I is in MPa MPa m . Determine the plane strain fracture
toughness for the component to endure 37,627 cycles so that a single-edge crack grows from 2
mm to a c .
Solution:

da
2.45
 2 x10 8 K I   A 1.12 
dN
ac
Nf
ao
0

2.45
K IC  1.12 a c
a 2.45
K IC  1.12 300 MPa   7 x10 3 m 
 2.45
 a da  0.12577  dN  4 ,732.30
K IC  50 MPa m
ao1.45  a c1.45
 4,732.30
1.45
a c  a f  7 mm
9.5 Consider a part made of a polycrystalline metal that is stresses in the elastic stress range. If
the metal contains inclusions, has an imperfectly smooth exterior surface, and natural
dislocation, would the metal experience irreversible changes in a micro-scale? Explain.
Answer: Many dislocations would move and contribute to a small irreversible process since the
dislocation configurations would eventually change. This is one mechanism would lead to
fatigue cracking due to dislocations.
4
Chapter 9 Fracture Mechanics, 2nd ed. (2015)
9.6
Solution Manual
Why most service fatigue fractures are normally not clear?
Answer: Most service fatigue fractures experience varying stress amplitude, leading to
irregularly spaced striations, if any.
9.7 What is the physical meaning of the slope n of the Stage II line according to the Paris
model?
da
n
 AK I  , the higher the value of the slope n , the
dN
n
higher the crack growth rate da / dN . For instance, if K  0 and n   , then K I   
and da / dN   since A is a constant.
Answer: According to Paris law,
9.8 Suppose that d ( 2a ) / dN  0.001 mm / cycle and n  4 in the Paris equation for 7075-T6
(FCC), 2024-T3 (FCC), Mo (BCC), and steel (BCC). Determine a) the constant A and its units,
and b) which of these materials will have the higher crack growth, rate?
Solution:
From Figure 9.6,
10 3 mm / cycle
d ( 2a )
4
 AK I  and A 
dN
K I 4
Material
7075-T6 (FCC)
2024-T3 (FCC)
Mo (BCC)
Steel (BCC)
K I
10
15
20
30
MPa
m)

A
MN
4
 m 5 / cycle

1.00 x10 10
1.98 x10 11
6.25 x10 12
1.23 x10 12
Therefore, 7075-T6 (FCC) alloy has the highest fatigue crack growth rate because it has the
highest constant A.
5
Chapter 9 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
9.9 A Ti-6Al-4V large plate containing a 4-mm long central crack is subjected to a steady
cyclic loading R = 0.10. The plate plane strain and the threshold fracture toughness are 70
MPa m and 14.7 MPa m . Determine (a) the minimum stress range, (b) the maximum applied
stress range for a fatigue life of 3,000 cycles, and (c) the critical crack size for 3,000 cycles. Let
the Paris equation be applicable so that n  4 and A  10 12 MN 4  m12 / cycles .

Solution:

2a
2a o  4 mm
N  10 3 cycles
K IC  70 MPa m
a o  2 mm
R  0.40
K th  14.7 MPa m
(a) Using eq. (9.2b) yields
K th
14.7 MPa m
 th 

 185.45 MPa
 a o
 2 x10 3 m


 min   th  185.45 MPa
(b) From eq. (9.9) with N o  0 since ao exists,
C1    C2  
n
n 2
 C3  0
( n  2)
C1  A N  N o n  2K IC   0.50 m
n
C 2  ( 2 /  )( K IC /  ) 2  3.12 x10 3 MPa 2  m
C 3  ( 2a 1o n / 2 )( 1 /  )n / 2 ( K IC /  )n  2.43 x10 9 MPa 2  m
0.50 4  3.12 x10 3  2  2.43x10 9  0
Thus,
C2
1

C 22  4C1C 3   3.12 x10 4 MPa 2  3.81x10 5 MPa 2
2C1 2C1
  642 MPa   max (Positive real root)
 2  
 min
and    max   min
 max
R max   min and  min   max  
R max   max  
642.31 MPa

 max 

 713 MPa
1 R
1  0.1
Then,
2
1  K IC 
ac  
 3.1 mm and 2a c  6.2 mm
   max 
(c) R 
6
 1
Chapter 9 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
9.10 Plot da / dN vs . K for 403 S.S. using the Paris, Forman and Broek/Schijve equations. Use
the data given in Table 9.2 and a (20 mm)x(300 mm)x(900 mm) plate containing a single-edge
crack of 2-mm long. Let 20 MPa m  K  80 MPa m .
Solution:
a  2 x10 3 m
a / w  0.007 and
B  20 mm
w  300 mm
  f ( a / w )  1.12
L  900 mm
Using data for AISI 403 stainless steel yields
K IC  36 MPa m
R   min /  max
R max   min and
K th  1.37 MPa m
 ys  1172 MPa
R  0.1
and
and
   max   min
 min   max  
R max   max  
 max   / 1  R 
n  3.5
for R  1
A  5.98 x10 13
Need to calculate K max for each given K . For instance, using the upper limit in the range
20 MPa m  K  80 MPa m , the procedure is as follows:
K max   max a 

a and K   a  80 MPa m
1 R
Solve for the stress range,
  901.12 MPa (Maximum)

  max   / 1  R   1001 MPa (upper limit)

 K max   max a  89 MPa m (upper limit)

 a c 
1  K IC
   max
2

  48.30 mm


It is suggested that you use a computer program to handle all the calculations needed in order to
plot smooth curves.
Paris :
Forman :
Broek/Schijve :
da
n
 A K 
dN
n
da
AK 
=
dN 1  R K IC  K
da
2
 AK max
 K n
dN
(9.6)
Plane strain 
(9.10)
(9.11)
7
Chapter 9 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
9.11 Plot the data given below and use eq. (9.6) as the model to draw a curve fitting line on loglog scales. Determine the constants in such an equation.
No
.
1
2
3
4
5
6
7
K
MPa m 
da / dN
m / cycle
20
30
40
50
60
70
80
2.14x10-8
8.84x10-8
2.42x10-7
5.29x10-7
1.00x10-6
1.72x10-6
2.74x10-6
Solution:
Since the Stage II data is for a linear behavior in a
log-log scale, the constants in eq. (9.6) can be
estimated as follows:
 da 
n

  AK 1 
 dN 1
 da 
n


 dN 1  K 1 


 da 
 K 2 


 dN  2
8
lnda / dN 1 / da / dN 2  ln 2.14 x10 / 8.84 x10 8
n

lnK 1 / K 2 
ln20 / 30
n  3.50

da / dN 1
A
K1 n

 da 
n

  AK 2 
 dN  2
and

2.14 x10 8 m / cycles
20 MPa m 
3.50
A  5.98 x10 13 MPa 3.50  m 8 / cycles
Then, the equation for curve fitting is of the following nature
da
3.50
 5.98 x10 13 K 
dN
Perhaps, the curve fitting procedure gives slightly different results.
4e-06
3e-06
da
mm / cycles 
dN
2e-06
1e-06
0
20
40
60

K MPa m
8
80

100
Chapter 9 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
9.12 A steel plate containing a single edge crack was subjected to a uniform stress range  at a
stress ratio of zero. Fatigue fracture occurred when the total crack length was 0.03 m. Subsequent
fatigue failure analysis revealed a striation spacing per unit cycle of 7.86x10-8 m. The
hypothetical steel has a modulus of elasticity of 207 GPa. Predict (a) the maximum cyclic stress
for a crack length of 0.01 m, (b) the striation spacing per unit cycle when the crack length is 0.02
m, (c) the Paris equation constants and (d) the plane strain fracture toughness and e) the fatigue
crack growth rate assuming that the Paris equation is applicable nearly up to fracture.
Solution:
a


(a) x  7.86 x10 8 m
a
 da 
 7.86 x10 8 m / cycles

 
 dN  a N
K a  E
x
7.86 x10 8 m
 207,000 MPa 
 23.69 MPa m  K max
6
6
Thus,
 
K 23.69 MPa m

 133.66 MPa   max
a
 0.01 m 
(b) For a  0.02 m ,
K b  K max   a  133.69 MPa   0.02 m   33.50 MPa m
Then,
2
 33.50 MPa m 
 K 
  1.57 x10 7 m
x  6
  6

 E 
 207,000 MPa 
a
x
1.57 x10 7 m
 da 

 1.57 x10 7 m / cycle

  b 
dN

N

N
1
cycle

b
2
(c) The exponent in the Paris equation is
 da 
n

  AK a 
 dN  a
 da 
n

  AK b 
 dN  b
 da / dN a 
 7.86 x10 8 
ln 
ln 
da / dN b 
1.57 x10  7 


n

2
 K a 
 23.69 
ln 
ln 

 33.50 
 K b 
9
Chapter 9 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
and the constant is
A
da / dN a
K a n

7.86 x10 8 m / cycle
23.69 MPa m 
2
 1.40 x10 10 MPa  2 / cycle
da
2
 1.40 x10 10 K 
dN
(d) For ac = 0.03 m,
K IC   max a c  133.66 MPa   0.03 m 
K IC  41 MPa m
(e) Finally,
A
da / dN a
K a n

7.86 x10 8 m / cycle
23.69 MPa m 
2
 1.40 x10 10 MPa  2 / cycle
da
2
2
 1.40 x10 10 K IC   1.40 x10 10 41  2.35 x10 7 m / cycle
dN
9.13 A 2-cm thick pressure vessel made of a high strength steel welded plates burst at an
unknown pressure. Fractographic work using a scanning electron microscope (SEM) revealed a
semielliptical fatigue surface crack (a = 0.1 cm and 2c = 0.2 cm) located perpendicular to the
hoop stress and nearly in the center of one of the welded plates. The last fatigue band exhibited
three striations having an average length of 0.34 mm at 10,000 magnification. The vessel internal
diameter was 10 cm. Calculate a) the pressure that caused fracture and b) the time it took for
fracture to occur due to pressure fluctuations. Assume a pressure frequency of 0.1 cycles per
minute (cpm). Given data: ys = 600 MPa, E = 207 GPa, K IC  75 MPa m and
da/dN = 4.50x10-7(K)2.
Solution:
a 0.1

 0.5
2c 0.2
a 0.1

 0.05
B
2
da a 0.00034 m


 1.13x10  4 m / cycle
dN N
3 cycle
10
Chapter 9 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Using the given crack growth rate equation, K becomes
1/ 2
 da / dN 
K  

 A 
1/ 2
 1.13x10  4 m / cycle 

 
2
4
.
50
x
10
MPa
/
cycle


K  15.85 MPa m
From eq. (9.23),
x
0.00034 m
K  E
 207,000 MPa 
610,000
6
K  15.58 MPa m
which is similar to the previous K value. The average becomes K  15.72 MPa m . Assume
the following stress ratio  /  ys  0.64 so that   384 MPa , which is verified below. Thus, the
shape factor becomes
 
Q 
2
2
2
2
2
2
2
 3  a 2 
7        3  0.1  
7
2
   
   0.64   2.38
     

 4  2c   33   ys   2   4  0.2   33
The hoop stress range along with the average value of K is
 
K
15.72 MPa m

 386.32 MPa
 a / Q 1.12  0.001 m  / 2.38
Thus,
 386.32

 0.64
 ys
600
Therefore, the assumed stress ratio is reasonably correct and fortunately, a single iteration is
sufficient. The pressure is estimated as
P
2 B 2 2 cm 384 MPa 

 154.40 MPa
d
10 cm
(b) The critical crack length is
2
2
1 K 
1  75 MPa m 
  9.56 x10 3 m  0.956 cm
ac   IC   
      1.12 x386.32 MPa 
11
Chapter 9 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Integrating eq. (9.15) along with N o  0 yields
c
c
da
da
1
da ln a / a 
dN



N
a AK 2 a A  a 2 A 2 a a  Aco2
o
o
o
o
ln ac / ao 
ln(0.956 / 0.1)
N

 8.53 cycles
2
A   4.50 x10 7 1.12 x386.322
Then,
ac
N
a
a



t

N
8.53 cycles

 85.30 min
cpm 0.1 cycle / min
9.14 Show that ac  aK IC / K max 
2
Solution:
From,
K IC   max a c and K max   max a
 max a c
K IC


K max
 max a
 K IC

 K max
ac
a
2

a
  c
a

Then,
 K
a c  a  IC
 K max



2
12
Chapter 10 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
CHAPTER 10
FRACTURE TOUGHNESS CORRELATIONS
10.1 (a) Plot the given CVN–T data for a carbon steel. (b) Calculate KIC using the CVN values up
to zero oC and Plot KIC-cal and KIC-exp. vs. Temperature. Is there a significant difference? If so,
explain.
T (oC)
CVN (J)
KIC-exp.
(MPa.m1/2)
-125
12
-100
18
-60
20
-40
28
-12
40
0
78
40
50
80
88
50 210
10
98
25
110
40
125
Solution:
(a) Plots
140
120
100
CVN J 
80
60
40
20
0
-100
-50
0
T
1
 C
o
50
100
45
126
60
28
80
30
Chapter 10 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
K IC  205.05 + 4. 5721T + 5.0958x10 -2 T 2 + 2.0082x10 -4 T 3
240
220
200
K IC
MPa m 
180
160
140
120
100
80
60
40
20
-120
-100
-80
-60
T
-40
 C
-20
0
o
(b) Do nonlinear curve fitting on K IC and CVN data set as per eqs. (10.66) and (10.67). Then, use
eq. (10.68) for calculating K IC values. Thus, plot KIC-cal and KIC-exp. vs. Temperature. Below
is a linear curve fitting result for plane strain fracture toughness and Charpy impact energy.
250
K IC  17.963  2.603U
K IC
MPa m 
200
150
100
50
0
10
20
30
40
U (J )
2
50
60
70
80
Chapter 10 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
100
K IC
MPa m 
80
60
40
Exp.
20
0
-120
-100
-80
-60
T
-40
-20
0
 C
o
10.2 A mild steel plate has a through the thickness single-edge crack, a yield strength of 800
MPa and a static fracture strength is  f  3.2 ys . If the plate is loaded in tension and fractures at
600 MPa, calculate the plane strain fracture toughness of the steel plate and the critical crack
length.
Solution:
 f  3.2 ys

f
 ys
and
 3.2 and
 ys  800 MPa
  20 m 1/ 2
From eq. (10.12),
K IC 
  1 ys 3.2  1800 MPa 

 88 MPa

20 m 1 / 2
m
If   600 MPa, then
K IC  1.12 a c
2
2
1  K IC 
1  88 MPa m 
ac  
  
  5.46 mm
  1.12    1.12 600 MPa 
3
Chapter 10 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
10.3 A standard Charpy specimen with B = 5 cm, w = 5 cm and S/w = 4 was tested at a room
temperature. The measured impact energy was 30 joules. The tested material has a modulus of
elasticity of 70,000 MPa. Calculate the KIC for this hypothetical specimen having x = a/w = 0.2.
Solution:
Data: x = a/w = 0.2
E = 70,000 MPa
w = 5×10-2 m B = 5×10-2 m
U = 30×10-6 MJ = 30×10-6 MN.m
From eqs. (10.57) and (10.43), respectively,
2
2


 0.45473
7x 7 0.2 
K IC  K I 
UE

Bw
30 x10
MN .m70,000 MPa 
0.454735x10 2 m5x10 2 m
6
K IC  42.98 MPa m
10.4 Suppose that a design code calls for a Charpy impact energy of 22 Joules for building a
large pressure vessel containing an inert gas. If A533B and A723 (  ys  1,100 MPa ) steel plates
are available for such a purpose, then (a) select the steel that will tolerate the largest critical
crack length (depth of a surface semi-elliptical crack) when the hoop stress is 500 MPa and (b)
determine the minimum plate thickness as per ASTM E399 standard for the selected steel.
Solution:
For   500 MPa, U  22 J and  ys  1,100 MPa (A723),
 196 ) 
K IC A533B  20+139
-1
 U

- 0 .54
 65.50 MPa m
K IC A723  0.644U ys  0.006 ys2  91.24 MPa m
eq. (10.71)
eq. (10.74)
(a) The critical length
2
2
1K 
1  65.50 MPa m 
  5.46 mm
ac (A533B)   IC   
      500 MPa 
2
2
1  K IC 
1  91.24 MPa m 
  10.60 mm
ac (A723)  
  
      500 MPa 
The ASTM A723 steel allows a larger critical crack length; therefore, this material is selected for
such an application.
(b) The plate thickness for the selected A723 steel is
2
2
K 
 91.24 MPa m 
  17.20 mm  0.6772 inches
B  2.5 IC   2.5

 
1
,
100
MPa
ys




4
Chapter 10 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
10.5 Plot the fracture toughness data for a hypothetical polymer and determine the brittle-ductile
transition temperature.
T (oC)
K IC ( MPa m )
-160
4.10
-120
4.11
-90 -80 -75 -70 -60 -50 -30
4.00 4.05 4.10 4.50 5.50 6.60 11.00
Solution:
20
20
K IC
15
15
MPa m 
10
10
 80o C
55
Brittle
Ductile
00
-200
-200
-150
-150
-100
-100
T
 C
-50
-50
00
O
10.6 For linear motion during Charpy impact tests, the energy lost by the striker ( U S ) and the
kinetic energy ( mv 2 ) data for some polymers with different masses and spans to depth (S/w)
ratios [47] are given below. Let m/M  1 and a) plot U S  f(mv²) and estimate the coefficient
of restitution (e) for these data. What does 0  e  1 mean? Theoretically, determine b) the
U S  f(mv²) ratio for the first bound which is transformed into specimen strain energy and c) the
U S  f(mv²) ratio when there is no bouncing.
mv 2 mJ 0 50 100 150 200
U S mJ
0 75 150 222 297
5
Chapter 10 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
Solution:
(a) The plot and the coefficient of restitution are
y  0. 6  1. 482x or U S  0. 6  1. 482mv 2 
Polynomial fit:
350
300
y
US
250
(mJ )
200
150
100
50
0
0
50
100
150
200
250
2
mv ( mJ )
x
From eq. (10.39) along with m/M << 1,
U S  1  e mv²
Slope  1  e  15
e05
This result, e  0.5, means that a repeated impact of converging energy into strain energy occurs
until the last contact is achieved.
(b) The U S  f(mv²) ratio for the first bound which is transformed into specimen strain energy
requires that the coefficient of restitution be e = 1. Thus,
U S / mv²   1  e  2
c) The U S  f mv²  ratio when there is no bouncing means that e  0 . Thus,
U S / mv²   1  e  1
6
Chapter 10 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
10.7 A large and thick ASTM A533B-1 steel plate containing a 4-mm long through-the-thickness
central crack fractures when it is subjected to a tensile stress of 7 MPa. Plot U, KIC, and GIC at
 200o C  T  100o C . Which of the plots is more suitable for determining the transition
temperature? Explain. Data: E = 207,000 MPa and v = 1/3.
Solution:
From eq. (10.72) and (10.73),
U
196
1  exp(0.0297T )
 196 
K IC  20  139
 1
 U

in Joules
 0.54
in MPa m
200
K IC
U ( JU
)
K IC
MPa m 
U
150
100
To  65o C
50
0
-200
-150
-100
-50
T
 C
o
0
50
100
T
Denote that the transition temperature ( To ) is easily determined using the U  f (T ) plot. This is
shown with dash lines.
GIC 
1  v  K
2
2
IC

10 1  1/ 3 20  95.50 exp0.016T 
6
2
E
207,000
GIC  4.294120  95.50 exp0.016T 
7
in Joules
Chapter 10 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
4
GICG(J)
IC
3
(J )
2
T0  55o C
1
0
-200
-150
-100
-50
T C
 
0
50
100
o
T (C)
It is clearly shown that the U IC  f (T ) plot is more suitable for determining the transition
temperature T0  65o C .
10.8 A large plate made of 18Ni-8Co-3Mo Grade 200 alloy is part of structure exposed to
relatively high temperature. Charpy impact tests were carried out and the average impact energy
is 60 J. Use this information to calculate (a) the plane strain fracture toughness, (b) the minimum
thickness ASTM requirement. The plate width is at least twice the thickness. Is this thickness
practical? (c) Assume that a single-edge through the thickness crack develops. What will the
critical crack length be? Will its value be reasonable?
Solution:
(a) If  60 J and  ys  1,310 MPa (From Table 10.1), then eq. (10.76) yields
K IC
 ys
2
 4. 69 Uys
 0. 20
K IC   ys 4. 69 Uys
K IC  1, 310 MPa
 0. 20
4. 69
60
1, 310
 0. 20
K IC  159. 42 MPa m
8
Chapter 10 Fracture Mechanics, 2nd ed. (2015)
Solution Manual
(b) From eq. (3.30), the minimum plate thickness is
IC
B  2. 5 K
 ys
2
 2. 5 159. 42
1310
B  0. 30424 m  304. 24 mm  12 inches
This is a very impractical thick for plate, but it complies with the ASTM plane strain conditions.
(c) Now if a single-edge crack develops during service, the critical crack length under a stress
half the yield strength will be
 ys
a c
2
1 2K IC
ac  
 ys
K IC 
2
3
 10

2  159. 42
1310
2
a c  19 mm
This is a reasonable result for a plate so thick.
9