Solutions Manual for Basic Fracture Mechanics and its
Applications, 1e by Ashok Saxena (All Chapters)
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?
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.
2. What is the Griffith’s crack extension force, G?
For cracks to extend, energy is needed for forming new surfaces and for plastic
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
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,
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:
∆𝐹𝐹 ∆U Δ𝑊𝑊𝑠𝑠
∆𝐴𝐴
= Δ𝐴𝐴 + (2)
Δ𝐴𝐴
The energy available for crack extension is given by,
∆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)
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.
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
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:
∆Π = Δ𝑈𝑈 = (𝑈𝑈𝑎𝑎+Δ𝑎𝑎 − 𝑈𝑈𝑎𝑎 )
𝑃𝑃
= (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 𝑑𝑑(∆/𝑃𝑃) 𝑃𝑃2 𝑑𝑑𝑑𝑑
= 𝑃𝑃 � �=
2𝐵𝐵 𝑑𝑑𝑑𝑑 2𝐵𝐵 𝑑𝑑𝑑𝑑
Displacement, ∆
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 )
𝐵𝐵 𝑑𝑑𝑑𝑑
𝜕𝜕𝜕𝜕
𝑃𝑃∆ �𝑃𝑃 + � 𝜕𝜕𝜕𝜕 �∆ ∆𝑎𝑎� ∆
Load, P
= −
a 2 2
1 𝑑𝑑𝑑𝑑 1 𝜕𝜕𝜕𝜕
a+∆a lim 𝐺𝐺 = − =− � � ∆
Δ𝑎𝑎→0 𝐵𝐵 𝑑𝑑𝑑𝑑 2𝐵𝐵 𝜕𝜕𝜕𝜕 ∆
𝑃𝑃 1
1 𝜕𝜕� � ∆2 𝜕𝜕� � 𝑃𝑃2 𝑑𝑑𝑑𝑑
= −∆2 2𝐵𝐵 � ∆
𝜕𝜕𝜕𝜕
� = − 2𝐵𝐵 � 𝐶𝐶
𝑑𝑑𝑑𝑑
� = 𝐺𝐺 = 2𝐵𝐵 𝑑𝑑𝑑𝑑
∆
Displacement, ∆
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,Note that the compliance C is only a function of crack size, a, so the partial derivative of
C with respect to a is converted into a total derivative without loss of generality. Also note
that the expression for G is the same for crack extension under constant load and
constant displacement conditions making the result general for any loading conditions.
4. A material exhibits the following crack growth resistance behavior:
R = 50 +200(a – a0) 0.5
Where a0 = initial crack size. R has units of kJ/m2 and crack size is in mm. The
elastic modulus of the material is 207,000 MPa. Consider a very wide plate and
assume the crack size is always much smaller than the width of the plate and the
crack is located at the center of the plate; a represents half crack size.
a) If the plate fractures at 138 MPa, compute the following:
I. The half crack size at failure
II. The amount of stable crack growth (at each crack tip) that precedes
failure.
b) If this plate has an initial crack length (2a0) of 50.8 mm and the plate is loaded
to failure, compute the following:
I. The stress at failure
II. The half crack size at failure
III. The stable crack growth at each crack tip
To be consistent in units, we take stress and E in N/m2 or in Pa, R in Joules/m2 and crack
size in m, the crack growth resistance equation can be re-written as:
𝑅𝑅 = 5𝑥𝑥104 + 200(1000)0.5 (𝑎𝑎 − 𝑎𝑎0 )0.5 𝑥𝑥1000
= 5𝑥𝑥104 + 6.324𝑥𝑥106 (𝑎𝑎 − 𝑎𝑎0 )0.5
For unstable fracture to occur,
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝐺𝐺 = 𝑅𝑅 and 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑
(a) Thus, G at fracture is given by,
𝜎𝜎𝑓𝑓2 𝜋𝜋𝜋𝜋
𝐺𝐺 = 𝑅𝑅 =
𝐸𝐸
where, 𝜎𝜎𝑓𝑓 = stress at fracture. Thus,
(138𝑥𝑥106 )2 𝜋𝜋𝑎𝑎𝑓𝑓
= 5𝑥𝑥104 + 6.324𝑥𝑥106 (𝑎𝑎 − 𝑎𝑎0 )0.5
207𝑥𝑥109
5𝑥𝑥104 + 6.324𝑥𝑥106 (𝑎𝑎 − 𝑎𝑎0 )0.5
𝑎𝑎𝑓𝑓 =
(138𝑥𝑥106 )2 𝜋𝜋
207𝑥𝑥109
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, 𝑎𝑎𝑓𝑓 = 0.173 + 21.94(𝑎𝑎 − 𝑎𝑎0 )0.5
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
Also, 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑
𝜎𝜎𝑓𝑓2 𝜋𝜋 −.5
= 3.162𝑥𝑥106 �𝑎𝑎𝑓𝑓 − 𝑎𝑎0 � = 288.9𝑥𝑥103
𝐸𝐸
−.5
�𝑎𝑎𝑓𝑓 − 𝑎𝑎0 � = 9.1𝑥𝑥10−2
�𝑎𝑎𝑓𝑓 − 𝑎𝑎0 � = 120.22 𝑚𝑚
or 𝑎𝑎𝑓𝑓 = 0.173 + 21.94(120.22)0.5 = 240.73
and 𝑎𝑎0 = 120.51 𝑚𝑚
Thus, the amount of stable crack extension = 120.51 m. These numbers are very high
indicating that fracture in the component at a stress of 138 MPa is very unlikely to occur.
(b) If 2𝑎𝑎0 = 50.8 𝑚𝑚𝑚𝑚 = 0.0254𝑚𝑚
Invoking the two conditions for fracture for this situation,
2
�𝜎𝜎𝑓𝑓 � 𝜋𝜋𝑎𝑎𝑓𝑓 0.5
9
= 5𝑥𝑥104 + 6.324𝑥𝑥106 �𝑎𝑎𝑓𝑓 − 𝑎𝑎0 �
207𝑥𝑥10
𝜎𝜎𝑓𝑓2 𝜋𝜋 −.5
and 𝐸𝐸
= 3.162𝑥𝑥106 �𝑎𝑎𝑓𝑓 − 𝑎𝑎0 �
−.5 0.5
thus, 3.162𝑥𝑥106 �𝑎𝑎𝑓𝑓 − 𝑎𝑎0 � 𝑎𝑎𝑓𝑓 = 5𝑥𝑥104 + 6.324𝑥𝑥106 �𝑎𝑎𝑓𝑓 − 𝑎𝑎0 �
0.5
or 𝑎𝑎𝑓𝑓 = 0.0158�𝑎𝑎𝑓𝑓 − 𝑎𝑎0 � + 2�𝑎𝑎𝑓𝑓 − 𝑎𝑎0 �
0.5
or −𝑎𝑎𝑓𝑓 = 0.0158�𝑎𝑎𝑓𝑓 − 𝑎𝑎0 � − 2𝑎𝑎0
0.5
or 0.0158�𝑎𝑎𝑓𝑓 − 0.0254� + 𝑎𝑎𝑓𝑓 − 0.0508 = 𝐹𝐹�𝑎𝑎𝑓𝑓 � = 0
We can solve this equation iteratively as follows:
af F(af)
0.03 -0.0197
0.04 -0.0089
0.045 -0.003588
0.048 -0.00042
0.0485 0.000101
5
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Applications, 1e by Ashok Saxena (All Chapters)
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?
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.
2. What is the Griffith’s crack extension force, G?
For cracks to extend, energy is needed for forming new surfaces and for plastic
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
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,
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:
∆𝐹𝐹 ∆U Δ𝑊𝑊𝑠𝑠
∆𝐴𝐴
= Δ𝐴𝐴 + (2)
Δ𝐴𝐴
The energy available for crack extension is given by,
∆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)
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.
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
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:
∆Π = Δ𝑈𝑈 = (𝑈𝑈𝑎𝑎+Δ𝑎𝑎 − 𝑈𝑈𝑎𝑎 )
𝑃𝑃
= (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 𝑑𝑑(∆/𝑃𝑃) 𝑃𝑃2 𝑑𝑑𝑑𝑑
= 𝑃𝑃 � �=
2𝐵𝐵 𝑑𝑑𝑑𝑑 2𝐵𝐵 𝑑𝑑𝑑𝑑
Displacement, ∆
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 )
𝐵𝐵 𝑑𝑑𝑑𝑑
𝜕𝜕𝜕𝜕
𝑃𝑃∆ �𝑃𝑃 + � 𝜕𝜕𝜕𝜕 �∆ ∆𝑎𝑎� ∆
Load, P
= −
a 2 2
1 𝑑𝑑𝑑𝑑 1 𝜕𝜕𝜕𝜕
a+∆a lim 𝐺𝐺 = − =− � � ∆
Δ𝑎𝑎→0 𝐵𝐵 𝑑𝑑𝑑𝑑 2𝐵𝐵 𝜕𝜕𝜕𝜕 ∆
𝑃𝑃 1
1 𝜕𝜕� � ∆2 𝜕𝜕� � 𝑃𝑃2 𝑑𝑑𝑑𝑑
= −∆2 2𝐵𝐵 � ∆
𝜕𝜕𝜕𝜕
� = − 2𝐵𝐵 � 𝐶𝐶
𝑑𝑑𝑑𝑑
� = 𝐺𝐺 = 2𝐵𝐵 𝑑𝑑𝑑𝑑
∆
Displacement, ∆
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,Note that the compliance C is only a function of crack size, a, so the partial derivative of
C with respect to a is converted into a total derivative without loss of generality. Also note
that the expression for G is the same for crack extension under constant load and
constant displacement conditions making the result general for any loading conditions.
4. A material exhibits the following crack growth resistance behavior:
R = 50 +200(a – a0) 0.5
Where a0 = initial crack size. R has units of kJ/m2 and crack size is in mm. The
elastic modulus of the material is 207,000 MPa. Consider a very wide plate and
assume the crack size is always much smaller than the width of the plate and the
crack is located at the center of the plate; a represents half crack size.
a) If the plate fractures at 138 MPa, compute the following:
I. The half crack size at failure
II. The amount of stable crack growth (at each crack tip) that precedes
failure.
b) If this plate has an initial crack length (2a0) of 50.8 mm and the plate is loaded
to failure, compute the following:
I. The stress at failure
II. The half crack size at failure
III. The stable crack growth at each crack tip
To be consistent in units, we take stress and E in N/m2 or in Pa, R in Joules/m2 and crack
size in m, the crack growth resistance equation can be re-written as:
𝑅𝑅 = 5𝑥𝑥104 + 200(1000)0.5 (𝑎𝑎 − 𝑎𝑎0 )0.5 𝑥𝑥1000
= 5𝑥𝑥104 + 6.324𝑥𝑥106 (𝑎𝑎 − 𝑎𝑎0 )0.5
For unstable fracture to occur,
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝐺𝐺 = 𝑅𝑅 and 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑
(a) Thus, G at fracture is given by,
𝜎𝜎𝑓𝑓2 𝜋𝜋𝜋𝜋
𝐺𝐺 = 𝑅𝑅 =
𝐸𝐸
where, 𝜎𝜎𝑓𝑓 = stress at fracture. Thus,
(138𝑥𝑥106 )2 𝜋𝜋𝑎𝑎𝑓𝑓
= 5𝑥𝑥104 + 6.324𝑥𝑥106 (𝑎𝑎 − 𝑎𝑎0 )0.5
207𝑥𝑥109
5𝑥𝑥104 + 6.324𝑥𝑥106 (𝑎𝑎 − 𝑎𝑎0 )0.5
𝑎𝑎𝑓𝑓 =
(138𝑥𝑥106 )2 𝜋𝜋
207𝑥𝑥109
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, 𝑎𝑎𝑓𝑓 = 0.173 + 21.94(𝑎𝑎 − 𝑎𝑎0 )0.5
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
Also, 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑
𝜎𝜎𝑓𝑓2 𝜋𝜋 −.5
= 3.162𝑥𝑥106 �𝑎𝑎𝑓𝑓 − 𝑎𝑎0 � = 288.9𝑥𝑥103
𝐸𝐸
−.5
�𝑎𝑎𝑓𝑓 − 𝑎𝑎0 � = 9.1𝑥𝑥10−2
�𝑎𝑎𝑓𝑓 − 𝑎𝑎0 � = 120.22 𝑚𝑚
or 𝑎𝑎𝑓𝑓 = 0.173 + 21.94(120.22)0.5 = 240.73
and 𝑎𝑎0 = 120.51 𝑚𝑚
Thus, the amount of stable crack extension = 120.51 m. These numbers are very high
indicating that fracture in the component at a stress of 138 MPa is very unlikely to occur.
(b) If 2𝑎𝑎0 = 50.8 𝑚𝑚𝑚𝑚 = 0.0254𝑚𝑚
Invoking the two conditions for fracture for this situation,
2
�𝜎𝜎𝑓𝑓 � 𝜋𝜋𝑎𝑎𝑓𝑓 0.5
9
= 5𝑥𝑥104 + 6.324𝑥𝑥106 �𝑎𝑎𝑓𝑓 − 𝑎𝑎0 �
207𝑥𝑥10
𝜎𝜎𝑓𝑓2 𝜋𝜋 −.5
and 𝐸𝐸
= 3.162𝑥𝑥106 �𝑎𝑎𝑓𝑓 − 𝑎𝑎0 �
−.5 0.5
thus, 3.162𝑥𝑥106 �𝑎𝑎𝑓𝑓 − 𝑎𝑎0 � 𝑎𝑎𝑓𝑓 = 5𝑥𝑥104 + 6.324𝑥𝑥106 �𝑎𝑎𝑓𝑓 − 𝑎𝑎0 �
0.5
or 𝑎𝑎𝑓𝑓 = 0.0158�𝑎𝑎𝑓𝑓 − 𝑎𝑎0 � + 2�𝑎𝑎𝑓𝑓 − 𝑎𝑎0 �
0.5
or −𝑎𝑎𝑓𝑓 = 0.0158�𝑎𝑎𝑓𝑓 − 𝑎𝑎0 � − 2𝑎𝑎0
0.5
or 0.0158�𝑎𝑎𝑓𝑓 − 0.0254� + 𝑎𝑎𝑓𝑓 − 0.0508 = 𝐹𝐹�𝑎𝑎𝑓𝑓 � = 0
We can solve this equation iteratively as follows:
af F(af)
0.03 -0.0197
0.04 -0.0089
0.045 -0.003588
0.048 -0.00042
0.0485 0.000101
5
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