2 Fracture Mechanics: Fundamentals and Applications
Solutions Manual for Fracture Mechanics Fundamentals and
Applications, 4th Edition by Anderson| Latest 2026/2027
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CHAPTER 1
1.2 A flat plate with a through-thickness crack (Fig. 1.8) is subject to a 100 MPa (14.5 ksi)
tensile stress and has a fracture toughness (KIc) of 50.0 MPa m (45. ksi in ). Determine
the critical crack length for this plate, assuming the material is linear elastic.
Ans:
At fracture, KIc = KI = . Therefore,
50 MPa = 100 MPa
ac = 0.0796 m = 79.6 mm
Total crack length = 2ac = 159 mm
1.3 Compute the critical energy release rate (Gc) of the material in the previous problem for E =
207,000 MPa (30,000 ksi)..
Ans:
(50 MPa m )
2
KIc
Gc = = = 0.0121 MPa mm = 12.1 kPa m
E 207,000 MPa
= 12.1 kJ/m2
Note that energy release rate has units of energy/area.
1.4 Suppose that you plan to drop a bomb out of an airplane and that you are interested in the
time of flight before it hits the ground, but you cannot remember the appropriate equation
from your undergraduate physics course. You decide to infer a relationship for time of flight
of a falling object by experimentation. You reason that the time of flight, t, must depend on
the height above the ground, h, and the weight of the object, mg, where m is the mass and g
is the gravitational acceleration. Therefore, neglecting aerodynamic drag, the time of flight
is given by the following function:
t = f (h, m, g)
Apply dimensional analysis to this equation and determine how many experiments would
be required to determine the function f to a reasonable approximation, assuming you know
the numerical value of g. Does the time of flight depend on the mass of the object?
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,4 Fracture Mechanics: Fundamentals and Applications
Ans:
Since h has units of length and g has units of (length)(time)-2, let us divide both
sides of the above equation by :
t f (h, m, g )
=
h g h g
The left side of this equation is now dimensionless. Therefore, the right side must
also be dimensionless, which implies that the time of flight cannot depend on the
mass of the object. Thus dimensional analysis implies the following functional
relationship:
h
t=
g
where is a dimensionless constant. Only one experiment would be required to
estimate , but several trials at various heights might be advisable to obtain a
reliable estimate of this constant. Note that = according to Newton's laws of
motion.
CHAPTER 2
2.1 According to Eq. (2.25), the energy required to increase the crack area a unit amount is equal
to twice the fracture work per unit surface area, wf. Why is the factor of 2 in this equation
necessary?
Ans:
The factor of 2 stems from the difference between crack area and surface area.
The former is defined as the projected area of the crack. The surface area is twice
the crack area because the formation of a crack results in the creation of two
surfaces. Consequently, the material resistance to crack extension = 2 wf.
2.2 Derive Eq. (2.30) for both load control and displacement control by substituting Eq. (2.29)
into Eqs. (2.27) and (2.28), respectively.
Ans:
(a) Load control.
P d CP
G = 2B da = 2B da = 2B
P d P dC
da
P P
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(b) Displacement control.
dP
G =−
2B da
dP ( )
d 1C dC
= =−
da da C 2 da
G = ( C ) dC = P
2
2
dC
2B da 2B da
2.3 Figure 2.10 illustrates that the driving force is linear for a through-thickness crack in an
infinite plate when the stress is fixed. Suppose that a remote displacement (rather than load)
were fixed in this configuration. Would the driving force curves be altered? Explain. (Hint:
see Section 2.5.3).
Ans:
In a cracked plate where 2a << the plate width, crack extension at a fixed remote
displacement would not effect the load, since the crack comprises a negligible
portion of the cross section. Thus a fixed remote displacement implies a fixed load,
and load control and displacement control are equivalent in this case. The driving
force curves would not be altered if remote displacement, rather than stress, were
specified.
Consider the spring in series analog in Fig. 2.12. The load and remote
displacement are related as follows:
T = (C + Cm) P T = (C + Cm ) P
where C is the “local” compliance and Cm is the system compliance. For the present
problem, assume that Cm represents the compliance of the uncracked plate and C is
the additional compliance that results from the presence of the crack. When the
crack is small compared to the plate dimensions, Cm >> C. If the crack were to
grow at a fixed T, only C would change; thus load would also remain fixed.
2.4 A plate 2W wide contains a centrally located crack 2a long and is subject to a tensile load,
P. Beginning with Eq. (2.24), derive an expression for the elastic compliance, C (= /P) in
terms of the plate dimensions and elastic modulus, E. The stress in Eq. (2.24) is the nominal
value; i.e., = P/2BW in this problem. (Note: Eq. (2.24) only applies when a << W; the
expression you derive is only approximate for a finite width plate.)
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Solutions Manual for Fracture Mechanics Fundamentals and
Applications, 4th Edition by Anderson| Latest 2026/2027
Update
@
@SSeeisismmicicisisoolalatitoionn
, Solutions Manual 3
CHAPTER 1
1.2 A flat plate with a through-thickness crack (Fig. 1.8) is subject to a 100 MPa (14.5 ksi)
tensile stress and has a fracture toughness (KIc) of 50.0 MPa m (45. ksi in ). Determine
the critical crack length for this plate, assuming the material is linear elastic.
Ans:
At fracture, KIc = KI = . Therefore,
50 MPa = 100 MPa
ac = 0.0796 m = 79.6 mm
Total crack length = 2ac = 159 mm
1.3 Compute the critical energy release rate (Gc) of the material in the previous problem for E =
207,000 MPa (30,000 ksi)..
Ans:
(50 MPa m )
2
KIc
Gc = = = 0.0121 MPa mm = 12.1 kPa m
E 207,000 MPa
= 12.1 kJ/m2
Note that energy release rate has units of energy/area.
1.4 Suppose that you plan to drop a bomb out of an airplane and that you are interested in the
time of flight before it hits the ground, but you cannot remember the appropriate equation
from your undergraduate physics course. You decide to infer a relationship for time of flight
of a falling object by experimentation. You reason that the time of flight, t, must depend on
the height above the ground, h, and the weight of the object, mg, where m is the mass and g
is the gravitational acceleration. Therefore, neglecting aerodynamic drag, the time of flight
is given by the following function:
t = f (h, m, g)
Apply dimensional analysis to this equation and determine how many experiments would
be required to determine the function f to a reasonable approximation, assuming you know
the numerical value of g. Does the time of flight depend on the mass of the object?
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@SSeeisismmicicisisoolalatitoionn
,4 Fracture Mechanics: Fundamentals and Applications
Ans:
Since h has units of length and g has units of (length)(time)-2, let us divide both
sides of the above equation by :
t f (h, m, g )
=
h g h g
The left side of this equation is now dimensionless. Therefore, the right side must
also be dimensionless, which implies that the time of flight cannot depend on the
mass of the object. Thus dimensional analysis implies the following functional
relationship:
h
t=
g
where is a dimensionless constant. Only one experiment would be required to
estimate , but several trials at various heights might be advisable to obtain a
reliable estimate of this constant. Note that = according to Newton's laws of
motion.
CHAPTER 2
2.1 According to Eq. (2.25), the energy required to increase the crack area a unit amount is equal
to twice the fracture work per unit surface area, wf. Why is the factor of 2 in this equation
necessary?
Ans:
The factor of 2 stems from the difference between crack area and surface area.
The former is defined as the projected area of the crack. The surface area is twice
the crack area because the formation of a crack results in the creation of two
surfaces. Consequently, the material resistance to crack extension = 2 wf.
2.2 Derive Eq. (2.30) for both load control and displacement control by substituting Eq. (2.29)
into Eqs. (2.27) and (2.28), respectively.
Ans:
(a) Load control.
P d CP
G = 2B da = 2B da = 2B
P d P dC
da
P P
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(b) Displacement control.
dP
G =−
2B da
dP ( )
d 1C dC
= =−
da da C 2 da
G = ( C ) dC = P
2
2
dC
2B da 2B da
2.3 Figure 2.10 illustrates that the driving force is linear for a through-thickness crack in an
infinite plate when the stress is fixed. Suppose that a remote displacement (rather than load)
were fixed in this configuration. Would the driving force curves be altered? Explain. (Hint:
see Section 2.5.3).
Ans:
In a cracked plate where 2a << the plate width, crack extension at a fixed remote
displacement would not effect the load, since the crack comprises a negligible
portion of the cross section. Thus a fixed remote displacement implies a fixed load,
and load control and displacement control are equivalent in this case. The driving
force curves would not be altered if remote displacement, rather than stress, were
specified.
Consider the spring in series analog in Fig. 2.12. The load and remote
displacement are related as follows:
T = (C + Cm) P T = (C + Cm ) P
where C is the “local” compliance and Cm is the system compliance. For the present
problem, assume that Cm represents the compliance of the uncracked plate and C is
the additional compliance that results from the presence of the crack. When the
crack is small compared to the plate dimensions, Cm >> C. If the crack were to
grow at a fixed T, only C would change; thus load would also remain fixed.
2.4 A plate 2W wide contains a centrally located crack 2a long and is subject to a tensile load,
P. Beginning with Eq. (2.24), derive an expression for the elastic compliance, C (= /P) in
terms of the plate dimensions and elastic modulus, E. The stress in Eq. (2.24) is the nominal
value; i.e., = P/2BW in this problem. (Note: Eq. (2.24) only applies when a << W; the
expression you derive is only approximate for a finite width plate.)
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