Basic Fracture Mechanics
and Its Applications
ST
1st Edition
UV
SOLUTIONS
IA
_A
MANUAL
PP
RO
Ashok Saxena
VE
────────────────────────────────────────────────────
Comprehensive Solutions Manual for Instructors and
D?
Students
© Ashok Saxena. All rights reserved. Reproduction or distribution without permission is
prohibited.
?
© DreamsHub
, Solutions Manual for Basic Fracture Mechanics and Its Applications (1st
Edition)
Ashok Saxena
ISBN: 9781032267197
ST
UNIT 1: INTRODUCTION AND EARLY FRACTURE THEORIES
1. Fracture in Structural Components
2. Early Theories of Fracture
UV
UNIT 2: LINEAR ELASTIC FRACTURE MECHANICS
3. Theoretical Basis for Linear Elastic Fracture Mechanics
4. Crack Tip Plasticity
IA
UNIT 3: FRACTURE TOUGHNESS AND FATIGUE
5. Fracture Toughness and Its Measurement
6. Fatigue Crack Growth
_A
UNIT 4: ENVIRONMENTAL AND MIXED-MODE FRACTURE
7. Environment-Assisted Cracking
8. Fracture under Mixed-Mode Loading
PP
UNIT 5: ELASTIC-PLASTIC AND CREEP FRACTURE
9. Fracture and Crack Growth under Elastic/Plastic Loading
10. Creep and Creep-Fatigue Crack Growth
RO
UNIT 6: APPLICATIONS AND CASE STUDIES
11. Case Studies in Applications of Fracture Mechanics
VE
D?
?
© DreamsHub
, Solutions Manual for Basic Fracture Mechanics and its
Applications, 1e by Ashok Saxena (All Chapters, Ch 1 Missing)
Chapter 2
1. According to the Griffith’s theory for brittle fracture, the energy required to increase
the crack area by a unit amount is equal to twice the energy required per unit area
for creating new surfaces. Why is the factor of 2 needed?
ST
Let the crack surface extend by an incremental area, ∆A = B∆a, where B = thickness
of the planar body and ∆a be the increase in the crack length. The area of new
surfaces thus created is twice the area of crack extension = 2B∆A and therefore the
surface energy required to create the new surfaces = 2𝛾𝛾∆A, where, 𝛾𝛾 = surface energy
per unit area of surface.
UV
2. What is the Griffith’s crack extension force, G?
For cracks to extend, energy is needed for forming new surfaces and for plastic
IA
deformation that accompanies crack growth. The sum of the surface energy and
the energy for plastic deformation can be expressed as energy required for unit
area of crack extension. Griffith’s crack extension force, G, is the energy available
for release that can supply the energy needed for crack growth. If ∆F = work done
_A
by the external forces when the crack extends by an area, ∆A, then energy balance
requires that:
∆𝐹𝐹 ∆Π Δ𝑊𝑊𝑠𝑠
∆𝐴𝐴
= Δ𝐴𝐴 + (1)
Δ𝐴𝐴
Where, ΔΠ = change in potential energy of the cracked body due to crack growth,
PP
and Δ𝑊𝑊𝑠𝑠 = energy dissipated in the form of surface energy and plastic deformation
due to crack growth. The change in potential energy during fracture under
isothermal conditions is equal to the change in the elastic strain energy, ∆U, of the
cracked body. Thus equation (1) can be written as:
RO
∆𝐹𝐹 ∆U Δ𝑊𝑊𝑠𝑠
∆𝐴𝐴
= Δ𝐴𝐴 + (2)
Δ𝐴𝐴
The energy available for crack extension is given by,
VE
∆𝑊𝑊𝑠𝑠 ∆𝐹𝐹 ∆U
𝐺𝐺 = lim = ∆𝐴𝐴 − Δ𝐴𝐴 (3)
Δ𝐴𝐴→0 ∆𝐴𝐴
For a planar body of constant thickness, B, equation (3) can be written as:
1 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝐺𝐺 = 𝐵𝐵 �𝑑𝑑𝑑𝑑 − 𝑑𝑑𝑑𝑑 � (4)
D?
If the crack extends under conditions of constant deflection, dF = 0
2
?
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, 1 𝑑𝑑𝑑𝑑
� �
𝐺𝐺 = −
𝐵𝐵 𝑑𝑑𝑑𝑑
and if the crack extension occurs under constant load, it can be shown that
1 𝑑𝑑𝑑𝑑
𝐺𝐺 = � �
𝐵𝐵 𝑑𝑑𝑑𝑑
Thus, in both cases, G is directly related to the change in strain energy due to crack
extension and therefore, G is also known as the strain energy release rate, where the
rate is with respect to unit area of crack extension.
ST
3. Show on your own that the Griffith’s Crack Extension Force G, for crack extension
under both displacement-control and under load-control conditions is given by
UV
P 2 dC
G=
2 B da
If the crack grows by an incremental amount, ∆a, under the conditions of constant load,
P, the load versus displacement diagram will be as shown below:
IA
∆Π = Δ𝑈𝑈 = (𝑈𝑈𝑎𝑎+Δ𝑎𝑎 − 𝑈𝑈𝑎𝑎 )
𝑃𝑃
_A
= (Δ − Δ𝑎𝑎 )
2 𝑎𝑎+Δ𝑎𝑎
Load, P
a ∆𝐹𝐹 = 𝑃𝑃(∆𝑎𝑎+∆𝑎𝑎 − ∆𝑎𝑎 )
a+∆a 1 ∆𝐹𝐹 ∆𝑈𝑈 𝑃𝑃 𝑑𝑑∆
lim 𝐺𝐺 = � − �= � �
Δ𝑎𝑎→0 𝐵𝐵 ∆𝑎𝑎 ∆𝑎𝑎 2𝐵𝐵 𝑑𝑑𝑑𝑑 𝑃𝑃
PP
𝑃𝑃 𝑑𝑑(∆/𝑃𝑃) 𝑃𝑃2 𝑑𝑑𝑑𝑑
= 𝑃𝑃 � �=
2𝐵𝐵 𝑑𝑑𝑑𝑑 2𝐵𝐵 𝑑𝑑𝑑𝑑
Displacement, ∆
RO
If the crack grows by an incremental amount, ∆a, under the conditions of constant
displacement, the load versus displacement diagram will be as shown below:
∆Π = Δ𝑈𝑈 = (𝑈𝑈𝑎𝑎+Δ𝑎𝑎 − 𝑈𝑈𝑎𝑎 )
1 𝑑𝑑𝑑𝑑
𝐺𝐺 = − , and −∆𝑈𝑈 = (𝑈𝑈𝑎𝑎 − 𝑈𝑈𝑎𝑎+∆𝑎𝑎 )
VE
𝐵𝐵 𝑑𝑑𝑑𝑑
𝜕𝜕𝜕𝜕
𝑃𝑃∆ �𝑃𝑃 + � 𝜕𝜕𝜕𝜕 �∆ ∆𝑎𝑎� ∆
Load, P
= −
a 2 2
1 𝑑𝑑𝑑𝑑 1 𝜕𝜕𝜕𝜕
a+∆a
D?
lim 𝐺𝐺 = − =− � � ∆
Δ𝑎𝑎→0 𝐵𝐵 𝑑𝑑𝑑𝑑 2𝐵𝐵 𝜕𝜕𝜕𝜕 ∆
𝑃𝑃 1
1 𝜕𝜕� � ∆2 𝜕𝜕� � 𝑃𝑃2 𝑑𝑑𝑑𝑑
= −∆2 2𝐵𝐵 � ∆
𝜕𝜕𝜕𝜕
� = − 2𝐵𝐵 � 𝐶𝐶
𝑑𝑑𝑑𝑑
� = 𝐺𝐺 = 2𝐵𝐵 𝑑𝑑𝑑𝑑
∆
Displacement, ∆
3
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and Its Applications
ST
1st Edition
UV
SOLUTIONS
IA
_A
MANUAL
PP
RO
Ashok Saxena
VE
────────────────────────────────────────────────────
Comprehensive Solutions Manual for Instructors and
D?
Students
© Ashok Saxena. All rights reserved. Reproduction or distribution without permission is
prohibited.
?
© DreamsHub
, Solutions Manual for Basic Fracture Mechanics and Its Applications (1st
Edition)
Ashok Saxena
ISBN: 9781032267197
ST
UNIT 1: INTRODUCTION AND EARLY FRACTURE THEORIES
1. Fracture in Structural Components
2. Early Theories of Fracture
UV
UNIT 2: LINEAR ELASTIC FRACTURE MECHANICS
3. Theoretical Basis for Linear Elastic Fracture Mechanics
4. Crack Tip Plasticity
IA
UNIT 3: FRACTURE TOUGHNESS AND FATIGUE
5. Fracture Toughness and Its Measurement
6. Fatigue Crack Growth
_A
UNIT 4: ENVIRONMENTAL AND MIXED-MODE FRACTURE
7. Environment-Assisted Cracking
8. Fracture under Mixed-Mode Loading
PP
UNIT 5: ELASTIC-PLASTIC AND CREEP FRACTURE
9. Fracture and Crack Growth under Elastic/Plastic Loading
10. Creep and Creep-Fatigue Crack Growth
RO
UNIT 6: APPLICATIONS AND CASE STUDIES
11. Case Studies in Applications of Fracture Mechanics
VE
D?
?
© DreamsHub
, Solutions Manual for Basic Fracture Mechanics and its
Applications, 1e by Ashok Saxena (All Chapters, Ch 1 Missing)
Chapter 2
1. According to the Griffith’s theory for brittle fracture, the energy required to increase
the crack area by a unit amount is equal to twice the energy required per unit area
for creating new surfaces. Why is the factor of 2 needed?
ST
Let the crack surface extend by an incremental area, ∆A = B∆a, where B = thickness
of the planar body and ∆a be the increase in the crack length. The area of new
surfaces thus created is twice the area of crack extension = 2B∆A and therefore the
surface energy required to create the new surfaces = 2𝛾𝛾∆A, where, 𝛾𝛾 = surface energy
per unit area of surface.
UV
2. What is the Griffith’s crack extension force, G?
For cracks to extend, energy is needed for forming new surfaces and for plastic
IA
deformation that accompanies crack growth. The sum of the surface energy and
the energy for plastic deformation can be expressed as energy required for unit
area of crack extension. Griffith’s crack extension force, G, is the energy available
for release that can supply the energy needed for crack growth. If ∆F = work done
_A
by the external forces when the crack extends by an area, ∆A, then energy balance
requires that:
∆𝐹𝐹 ∆Π Δ𝑊𝑊𝑠𝑠
∆𝐴𝐴
= Δ𝐴𝐴 + (1)
Δ𝐴𝐴
Where, ΔΠ = change in potential energy of the cracked body due to crack growth,
PP
and Δ𝑊𝑊𝑠𝑠 = energy dissipated in the form of surface energy and plastic deformation
due to crack growth. The change in potential energy during fracture under
isothermal conditions is equal to the change in the elastic strain energy, ∆U, of the
cracked body. Thus equation (1) can be written as:
RO
∆𝐹𝐹 ∆U Δ𝑊𝑊𝑠𝑠
∆𝐴𝐴
= Δ𝐴𝐴 + (2)
Δ𝐴𝐴
The energy available for crack extension is given by,
VE
∆𝑊𝑊𝑠𝑠 ∆𝐹𝐹 ∆U
𝐺𝐺 = lim = ∆𝐴𝐴 − Δ𝐴𝐴 (3)
Δ𝐴𝐴→0 ∆𝐴𝐴
For a planar body of constant thickness, B, equation (3) can be written as:
1 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝐺𝐺 = 𝐵𝐵 �𝑑𝑑𝑑𝑑 − 𝑑𝑑𝑑𝑑 � (4)
D?
If the crack extends under conditions of constant deflection, dF = 0
2
?
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, 1 𝑑𝑑𝑑𝑑
� �
𝐺𝐺 = −
𝐵𝐵 𝑑𝑑𝑑𝑑
and if the crack extension occurs under constant load, it can be shown that
1 𝑑𝑑𝑑𝑑
𝐺𝐺 = � �
𝐵𝐵 𝑑𝑑𝑑𝑑
Thus, in both cases, G is directly related to the change in strain energy due to crack
extension and therefore, G is also known as the strain energy release rate, where the
rate is with respect to unit area of crack extension.
ST
3. Show on your own that the Griffith’s Crack Extension Force G, for crack extension
under both displacement-control and under load-control conditions is given by
UV
P 2 dC
G=
2 B da
If the crack grows by an incremental amount, ∆a, under the conditions of constant load,
P, the load versus displacement diagram will be as shown below:
IA
∆Π = Δ𝑈𝑈 = (𝑈𝑈𝑎𝑎+Δ𝑎𝑎 − 𝑈𝑈𝑎𝑎 )
𝑃𝑃
_A
= (Δ − Δ𝑎𝑎 )
2 𝑎𝑎+Δ𝑎𝑎
Load, P
a ∆𝐹𝐹 = 𝑃𝑃(∆𝑎𝑎+∆𝑎𝑎 − ∆𝑎𝑎 )
a+∆a 1 ∆𝐹𝐹 ∆𝑈𝑈 𝑃𝑃 𝑑𝑑∆
lim 𝐺𝐺 = � − �= � �
Δ𝑎𝑎→0 𝐵𝐵 ∆𝑎𝑎 ∆𝑎𝑎 2𝐵𝐵 𝑑𝑑𝑑𝑑 𝑃𝑃
PP
𝑃𝑃 𝑑𝑑(∆/𝑃𝑃) 𝑃𝑃2 𝑑𝑑𝑑𝑑
= 𝑃𝑃 � �=
2𝐵𝐵 𝑑𝑑𝑑𝑑 2𝐵𝐵 𝑑𝑑𝑑𝑑
Displacement, ∆
RO
If the crack grows by an incremental amount, ∆a, under the conditions of constant
displacement, the load versus displacement diagram will be as shown below:
∆Π = Δ𝑈𝑈 = (𝑈𝑈𝑎𝑎+Δ𝑎𝑎 − 𝑈𝑈𝑎𝑎 )
1 𝑑𝑑𝑑𝑑
𝐺𝐺 = − , and −∆𝑈𝑈 = (𝑈𝑈𝑎𝑎 − 𝑈𝑈𝑎𝑎+∆𝑎𝑎 )
VE
𝐵𝐵 𝑑𝑑𝑑𝑑
𝜕𝜕𝜕𝜕
𝑃𝑃∆ �𝑃𝑃 + � 𝜕𝜕𝜕𝜕 �∆ ∆𝑎𝑎� ∆
Load, P
= −
a 2 2
1 𝑑𝑑𝑑𝑑 1 𝜕𝜕𝜕𝜕
a+∆a
D?
lim 𝐺𝐺 = − =− � � ∆
Δ𝑎𝑎→0 𝐵𝐵 𝑑𝑑𝑑𝑑 2𝐵𝐵 𝜕𝜕𝜕𝜕 ∆
𝑃𝑃 1
1 𝜕𝜕� � ∆2 𝜕𝜕� � 𝑃𝑃2 𝑑𝑑𝑑𝑑
= −∆2 2𝐵𝐵 � ∆
𝜕𝜕𝜕𝜕
� = − 2𝐵𝐵 � 𝐶𝐶
𝑑𝑑𝑑𝑑
� = 𝐺𝐺 = 2𝐵𝐵 𝑑𝑑𝑑𝑑
∆
Displacement, ∆
3
?
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