STUDY NOTES & REFERENCE GUIDE
All 12 Chapters Covered
SOLUTIONS
, STUDY NOTES & REFERENCE GUIDE
2 Fracture Mechanics: Fundamentals and
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
Ans:
At fracture,
KIc KI ac . Therefore,
50 MPa m = 100 MPa ac
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:
2
50 MPa m
KIc
G 0.0121 MPa mm 12.1 kPa
c m 207,000 MPa
E
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 )
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Solutions 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 by h :
t f h, m, g
h g h
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 2 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 P d P dC
G
CP
2B da 2B da 2B da
P P
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4 Fracture Mechanics: Fundamentals and
(b) Displacement control.
dP
G 2
B da
dP
d 1
C dC
d C 2 da
a da
G
C dC
2
2
P 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
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All 12 Chapters Covered
SOLUTIONS
, STUDY NOTES & REFERENCE GUIDE
2 Fracture Mechanics: Fundamentals and
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
Ans:
At fracture,
KIc KI ac . Therefore,
50 MPa m = 100 MPa ac
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:
2
50 MPa m
KIc
G 0.0121 MPa mm 12.1 kPa
c m 207,000 MPa
E
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 )
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, STUDY NOTES & REFERENCE GUIDE
Solutions 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 by h :
t f h, m, g
h g h
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 2 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 P d P dC
G
CP
2B da 2B da 2B da
P P
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4 Fracture Mechanics: Fundamentals and
(b) Displacement control.
dP
G 2
B da
dP
d 1
C dC
d C 2 da
a da
G
C dC
2
2
P 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
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