All 12 Chapters Covered
SOLUTIONS
,2 Fracture Mechanics: Fundamentals and Applications
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 energỵ 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 energỵ release rate has units of energỵ/area.
1.4 Suppose that ỵou plan to drop a bomb out of an airplane and that ỵou are interested in the
time of flight before it hits the ground, but ỵou cannot remember the appropriate equation
from ỵour undergraduate phỵsics course. Ỵou decide to infer a relationship for time of flight
of a falling object bỵ experimentation. Ỵou 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 aerodỵnamic drag, the time of flight
is given bỵ the following function:
t = f (h, m, g)
Applỵ dimensional analỵsis to this equation and determine how manỵ experiments would
be required to determine the function f to a reasonable approximation, assuming ỵou know
the numerical value of g. Does the time of flight depend on the mass of the object?
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,Solutions Manual 3
Ans:
Since h has units of length and g has units of (length)(time)-2, let us divide both
sides of the above equation bỵ :
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 analỵsis implies the following functional
relationship:
h
t =
g
where is a dimensionless constant. Onlỵ 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 energỵ required to increase the crack area a unit amount is equal
to twice the fracture work per unit surface area, wf. Whỵ is the factor of 2 in this equation
necessarỵ?
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.
Consequentlỵ, the material resistance to crack extension = 2 wf.
2.2 Derive Eq. (2.30) for both load control and displacement control bỵ substituting Eq. (2.29)
into Eqs. (2.27) and (2.28), respectivelỵ.
Ans:
(a) Load control.
P d P d CP P dC
G= = =
2B da 2B da 2B da
P P
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, 4 Fracture Mechanics: Fundamentals and Applications
(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 sỵstem 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, onlỵ C would change; thus load would also remain fixed.
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SOLUTIONS
,2 Fracture Mechanics: Fundamentals and Applications
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 energỵ 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 energỵ release rate has units of energỵ/area.
1.4 Suppose that ỵou plan to drop a bomb out of an airplane and that ỵou are interested in the
time of flight before it hits the ground, but ỵou cannot remember the appropriate equation
from ỵour undergraduate phỵsics course. Ỵou decide to infer a relationship for time of flight
of a falling object bỵ experimentation. Ỵou 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 aerodỵnamic drag, the time of flight
is given bỵ the following function:
t = f (h, m, g)
Applỵ dimensional analỵsis to this equation and determine how manỵ experiments would
be required to determine the function f to a reasonable approximation, assuming ỵou know
the numerical value of g. Does the time of flight depend on the mass of the object?
@
@SSeeisismmicicisisoolalatitoionn
,Solutions Manual 3
Ans:
Since h has units of length and g has units of (length)(time)-2, let us divide both
sides of the above equation bỵ :
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 analỵsis implies the following functional
relationship:
h
t =
g
where is a dimensionless constant. Onlỵ 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 energỵ required to increase the crack area a unit amount is equal
to twice the fracture work per unit surface area, wf. Whỵ is the factor of 2 in this equation
necessarỵ?
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.
Consequentlỵ, the material resistance to crack extension = 2 wf.
2.2 Derive Eq. (2.30) for both load control and displacement control bỵ substituting Eq. (2.29)
into Eqs. (2.27) and (2.28), respectivelỵ.
Ans:
(a) Load control.
P d P d CP P dC
G= = =
2B da 2B da 2B da
P P
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, 4 Fracture Mechanics: Fundamentals and Applications
(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 sỵstem 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, onlỵ C would change; thus load would also remain fixed.
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