Physical Metallurgy Principles,
5th Edition by Reza Abbaschian
Complete Chapter Solutions Manual
are included (Ch 1 to 21)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
1. The Structure of Metals.
2. Characterization Techniques.
3. Crystal Binding.
4. Introduction to Dislocations.
5. Dislocations and Plastic Deformation.
6. Elements of Grain Boundaries.
7. Vacancies and Thermodynamics.
8. Annealing.
9. Solid Solutions.
10. Phases.
11. Binary Phase Diagrams.
12. Diffusion in Substitutional Solid Solutions.
13. Interstitial Diffusion.
14. Solidification of Metals.
15. Nucleation and Growth Kinetics.
16. Precipitation Hardening.
17. Deformation Twinning and Martensite Reactions.
18. The Iron-Carbon Alloy System.
19. The Hardening of Steel.
20. Selected Nonferrous Alloy Systems.
21. Failure of Metals.
,Solutions Manual organized in reverse order, with the last chapter
displayed first, to ensure that all chapters are included in this document.
(Complete Chapters included Ch21-1)
Solution and Answer Guide
ABBASCHIAN, PHYSICAL METALLURGY PRINCIPLES, 5E, 2025, 9798214001661;
CHAPTER 21: FAILURE OF METALS
PROBLEMS
21.1 (a) Make a list of the different types of fracture that occur in single crystals.
(b) Now consider a polycrystalline metal. Indicate how grain boundaries add to the
fracture possibilities.
Solution:
(a) [1] Cleavage – It is possible for some single crystals to undergo a brittle type of
fracture, in which they split apart along a crystallographic plane of low Miller
indices, such as the basal plane in zinc or the cube plane in bcc metals.
[2] Failure by easy glide – At relatively high temperatures, it has been observed
that the two parts of a crystal can be effectively sheared apart along a slip
plane. [3] Rupture by multiple glide – If multiple glide along several intersecting
slip plane occurs is concentrated in a restricted area, it is possible to obtain a
necking rupture. If only two slip planes are involved, this may produce a chisel
edge on each half of the crystal. If more than three slip planes are involved, the
two ends may pull down to a point. [4] Twinning – Twinning may also lead to
fracture of a single crystal. This is particularly true if the reoriented lattice in
the twin is more conducive to slip than the matrix in which it forms. The
resulting large plastic deformation in the twin may result in a tearing fracture
within the boundaries of the twin.
(b) [1] The grain boundaries themselves may form a path along which fracture
occurs. Such a fracture is called an intergranular fracture. Grain boundaries
attract impurity atoms and this often leads to embrittlement. In some cases,
brittle precipitates may form along the boundaries. A typical example is
bismuth, which tends to collect along the boundaries of copper, making the
copper subject to a brittle intergranular fracture. On the other hand, when FeS
precipitates along the grain boundaries of polycrystalline iron and steel
specimens, it can remain liquid well after the iron or steel has solidified. In such
a case, the metal becomes “hot short” and cannot be hot worked. [2]
Intergranular fractures are common in metal specimens, deformed at high
temperature, under creep conditions. In this temperature range, plastic
deformation tends to concentrate in the parts of the crystals adjacent to the
grain boundaries. This concentrated plastic flow can cause a large localized
increase in the vacancy concentrations, at the boundaries, so that the pores
nucleate and form on the boundaries. These pores can then grow and
interconnect to form intergranular fractures. [3] In general, the effect of a grain
boundary is to make slip more difficult. This is because the slip planes in a given
crystal are normally poorly aligned with respect to those of its neighbors. As a
result, the stress level required to deform a metal increases as its grain size
, decreases and its grain boundary area per unit volume increases. This also
increases the probability of fracture in polycrystalline metals.
21.2 It is possible for zinc crystals to undergo very large plastic strains at temperatures
below room temperature. This is not true in polycrystalline zinc, which normally
shows a very low ductility at sub-ambient temperatures. Give an explanation for this
fact based on the mechanical properties of zinc.
Solution:
Zinc is a metal that deforms very easily by basal slip at a very low level of stress,
even at low temperatures. In other words, it deforms readily by easy glide and the
work hardening rate is low. At the same time, slip on other slip planes in zinc
becomes increasingly more difficult the lower the temperature. In a single crystal of
zinc, easy glide is effectively unrestricted because the shears along the slip planes
do not end at a free surface, but rather at a grain boundary where the basal plane of
the next grain is not normally aligned so as to pass the shear into the next grain.
Compounding the problem is the fact that zinc cleaves easily along its basal at low
temperatures. Thus, the high stresses that develop at the grain boundaries, due to
the inability of the basal slip to be accommodated between grains, leads to brittle
cleavage or intergranular fracture.
21.3 S. S. Brenner, (Jour. Appl. Phys., 27 1484 [1956]), has reported that iron whisker single
crystals may have a tensile strength of 13,130 MPa. Assume that the iron whiskers
have a 〈 100〉 orientation; that is, a 〈 100〉 direction of the crystal lies along its central
axis.
(a) Compute the magnitude of the shear stress on the most highly stressed
{110} 111 slip systems when the ultimate tensile stress is attained. How
many of these slip systems are there?
(b) The shear modulus, µ , along a 110 direction of a {110} plane is 59,810 MPa.
Compute the ratio of τ max µ . .
(c) According to theoretical calculations, the maximum shear strength of a crystal
should be of the order of µ 10 to µ 30 . How does the value obtained in (b) of
this problem compare to the theoretical predictions?
(d) Now assume that one of the slip planes is ideally oriented so as to have both θ
and φ in the resolved shear stress equation equal to 45 degrees, and compute
τ max µ .
.
(e) On the basis of these calculations, what can one conclude about the occurrence
of slip in the iron whiskers? Does this signify anything about the perfection of
the crystal structure in the iron crystal whiskers?
5th Edition by Reza Abbaschian
Complete Chapter Solutions Manual
are included (Ch 1 to 21)
** Immediate Download
** Swift Response
** All Chapters included
,Table of Contents are given below
1. The Structure of Metals.
2. Characterization Techniques.
3. Crystal Binding.
4. Introduction to Dislocations.
5. Dislocations and Plastic Deformation.
6. Elements of Grain Boundaries.
7. Vacancies and Thermodynamics.
8. Annealing.
9. Solid Solutions.
10. Phases.
11. Binary Phase Diagrams.
12. Diffusion in Substitutional Solid Solutions.
13. Interstitial Diffusion.
14. Solidification of Metals.
15. Nucleation and Growth Kinetics.
16. Precipitation Hardening.
17. Deformation Twinning and Martensite Reactions.
18. The Iron-Carbon Alloy System.
19. The Hardening of Steel.
20. Selected Nonferrous Alloy Systems.
21. Failure of Metals.
,Solutions Manual organized in reverse order, with the last chapter
displayed first, to ensure that all chapters are included in this document.
(Complete Chapters included Ch21-1)
Solution and Answer Guide
ABBASCHIAN, PHYSICAL METALLURGY PRINCIPLES, 5E, 2025, 9798214001661;
CHAPTER 21: FAILURE OF METALS
PROBLEMS
21.1 (a) Make a list of the different types of fracture that occur in single crystals.
(b) Now consider a polycrystalline metal. Indicate how grain boundaries add to the
fracture possibilities.
Solution:
(a) [1] Cleavage – It is possible for some single crystals to undergo a brittle type of
fracture, in which they split apart along a crystallographic plane of low Miller
indices, such as the basal plane in zinc or the cube plane in bcc metals.
[2] Failure by easy glide – At relatively high temperatures, it has been observed
that the two parts of a crystal can be effectively sheared apart along a slip
plane. [3] Rupture by multiple glide – If multiple glide along several intersecting
slip plane occurs is concentrated in a restricted area, it is possible to obtain a
necking rupture. If only two slip planes are involved, this may produce a chisel
edge on each half of the crystal. If more than three slip planes are involved, the
two ends may pull down to a point. [4] Twinning – Twinning may also lead to
fracture of a single crystal. This is particularly true if the reoriented lattice in
the twin is more conducive to slip than the matrix in which it forms. The
resulting large plastic deformation in the twin may result in a tearing fracture
within the boundaries of the twin.
(b) [1] The grain boundaries themselves may form a path along which fracture
occurs. Such a fracture is called an intergranular fracture. Grain boundaries
attract impurity atoms and this often leads to embrittlement. In some cases,
brittle precipitates may form along the boundaries. A typical example is
bismuth, which tends to collect along the boundaries of copper, making the
copper subject to a brittle intergranular fracture. On the other hand, when FeS
precipitates along the grain boundaries of polycrystalline iron and steel
specimens, it can remain liquid well after the iron or steel has solidified. In such
a case, the metal becomes “hot short” and cannot be hot worked. [2]
Intergranular fractures are common in metal specimens, deformed at high
temperature, under creep conditions. In this temperature range, plastic
deformation tends to concentrate in the parts of the crystals adjacent to the
grain boundaries. This concentrated plastic flow can cause a large localized
increase in the vacancy concentrations, at the boundaries, so that the pores
nucleate and form on the boundaries. These pores can then grow and
interconnect to form intergranular fractures. [3] In general, the effect of a grain
boundary is to make slip more difficult. This is because the slip planes in a given
crystal are normally poorly aligned with respect to those of its neighbors. As a
result, the stress level required to deform a metal increases as its grain size
, decreases and its grain boundary area per unit volume increases. This also
increases the probability of fracture in polycrystalline metals.
21.2 It is possible for zinc crystals to undergo very large plastic strains at temperatures
below room temperature. This is not true in polycrystalline zinc, which normally
shows a very low ductility at sub-ambient temperatures. Give an explanation for this
fact based on the mechanical properties of zinc.
Solution:
Zinc is a metal that deforms very easily by basal slip at a very low level of stress,
even at low temperatures. In other words, it deforms readily by easy glide and the
work hardening rate is low. At the same time, slip on other slip planes in zinc
becomes increasingly more difficult the lower the temperature. In a single crystal of
zinc, easy glide is effectively unrestricted because the shears along the slip planes
do not end at a free surface, but rather at a grain boundary where the basal plane of
the next grain is not normally aligned so as to pass the shear into the next grain.
Compounding the problem is the fact that zinc cleaves easily along its basal at low
temperatures. Thus, the high stresses that develop at the grain boundaries, due to
the inability of the basal slip to be accommodated between grains, leads to brittle
cleavage or intergranular fracture.
21.3 S. S. Brenner, (Jour. Appl. Phys., 27 1484 [1956]), has reported that iron whisker single
crystals may have a tensile strength of 13,130 MPa. Assume that the iron whiskers
have a 〈 100〉 orientation; that is, a 〈 100〉 direction of the crystal lies along its central
axis.
(a) Compute the magnitude of the shear stress on the most highly stressed
{110} 111 slip systems when the ultimate tensile stress is attained. How
many of these slip systems are there?
(b) The shear modulus, µ , along a 110 direction of a {110} plane is 59,810 MPa.
Compute the ratio of τ max µ . .
(c) According to theoretical calculations, the maximum shear strength of a crystal
should be of the order of µ 10 to µ 30 . How does the value obtained in (b) of
this problem compare to the theoretical predictions?
(d) Now assume that one of the slip planes is ideally oriented so as to have both θ
and φ in the resolved shear stress equation equal to 45 degrees, and compute
τ max µ .
.
(e) On the basis of these calculations, what can one conclude about the occurrence
of slip in the iron whiskers? Does this signify anything about the perfection of
the crystal structure in the iron crystal whiskers?
