SOLUTION MANUAL FOR Physical Metallurgy Principles 5th Edition Abbaschian

, SOLUTION MANUAL FOR Physical Metallurgy Principles 5th Edition
Abbaschian

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, Solution and Answer Guide: Abbaschian, Physical Metallurgy Principles, 5e, 2025, 9798214001661;
Chapter 1: The Structure of Metals



Solution and Answer Guide
ABBASCHIAN, PHYSICAL METALLURGY PRINCIPLES, 5E, 2025, 9798214001661;
CHAPTER 1: THE STRUCTURE OF METALS


PROBLEMS
1.1




Determine the direction indices for (a) line om, (b) line on, and (c) line op in the
accompanying drawing of a cubic unit cell.

Solution:
(a) The direction indices for line om are [111].

1
(b) The vector components of line on, in units of, a = 1, are , 1 and 0 along the x, y
2
and z axes respectively. The corresponding direction indices are accordingly [120].

2 1 1
(c) The components of line op are , , and . Thus the indices are [436].
3 2 1




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, Solution and Answer Guide: Abbaschian, Physical Metallurgy Principles, 5e, 2025, 9798214001661;
Chapter 1: The Structure of Metals

1.2




Determine direction indices for lines (a) qr, (b) qs, and (c) qt.




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posted to a publicly accessible website, in whole or in part.

, Solution and Answer Guide: Abbaschian, Physical Metallurgy Principles, 5e, 2025, 9798214001661;
Chapter 1: The Structure of Metals

Solution:
(a) The direction indices of line qr are [001].

(b) The direction indices of line qs are [101].

1 1 1
(c) The direction indices of line qt along the x, y and z axes are , , and ,
2 4 1
yielding the indices 2 1 4  .



1.3




In this figure, plane pqr intercepts the x, y, and z axes as indicated. What are the
Miller indices of this plane?

Solution:
1 3 2
The intercepts of plane pqr with the three axes are , , and , with the
2 4 3
1 1
reciprocals , , so that the Miller indices of the plane are (12 8 9).
1/ 2




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posted to a publicly accessible website, in whole or in part.

, Solution and Answer Guide: Abbaschian, Physical Metallurgy Principles, 5e, 2025, 9798214001661;
Chapter 1: The Structure of Metals

1.4




What are the Miller indices of plane stu?

Solution:
1 1
This plane has intercepts , –1 and , so that the Miller indices are 2 12 .
2 2  



1.5




Write the Miller indices for plane vwx.

Solution:
On the assumption that parallel planes can be described by the same set of Miller
indices, this plane should have the same indices as plane (C) of Fig. 1.16, which are (111).




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, Solution and Answer Guide: Abbaschian, Physical Metallurgy Principles, 5e, 2025, 9798214001661;
Chapter 1: The Structure of Metals

1.6 Linear density in a given crystallographic direction represents the fraction of a line
length that is occupied by atoms whereas linear mass density is mass per unit
length. Similarly, planar density is the fraction of a crystallographic plane occupied
by atoms. The fraction of the volume occupied in a unit cell, on the other hand, is
called the atomic packing factor. The latter should not be confused with bulk
density, which represents weight per unit volume.

(a) Calculate the linear density in the [100], [110], and [111] directions in body-
centered cubic (BCC) and face-centered cubic (FCC) structures.

(b) Calculate planar densities in (100) and (110) planes in bcc and fcc structures.

(c) Show that atomic packing factors for BCC, FCC, and hexagonal close-packed (HCP)
structures are 0.68, 0.74, and 0.74, respectively.

Solution:
(a) Referring to the hard ball model of BCC structure shown in Fig 1.1c, it can be
seen that the atoms touch each other along the diagonal, [111] direction, of the
unit cell. Taking R as radius of the atom, its length is equal to 4R. The lattice
4R
parameter, is therefore a = . The edge of the unit cell, [100], is occupied by
3
two half atoms. The linear density in [100] direction is:
2R
LD[ 100] = = 0.866
4R 3

For the [110] direction, the length of the diagonal of the cube face is 2 a, which
is again occupied by two half atoms.

2R 2R
LD110 = = = 0.612
 
2a 2 × 4R 3

The linear density along [111] direction of BCC is obviously 1.
For FCC unit cell, the atoms touch each other along [110] direction, as can be
seen in Fig 1.2B. The lattice parameter is therefore a = 4R 2 , and the unit cell

4 3R
diagonal is 3a = . Therefore:
2

2R
LD[100]
= = 0.707
4R 2
LD[110] = 1
2R
LD[111]
= = 0.408
4 3R 2

(b) The (100) face of the BCC unit cell, with an area of a2 , is occupied by 4 quarter
circle areas. Therefore:




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, Solution and Answer Guide: Abbaschian, Physical Metallurgy Principles, 5e, 2025, 9798214001661;
Chapter 1: The Structure of Metals


1 2
4 x x πR
4 πR 2
PD[100]
= = = 0.589
a2
( )
2
4R 3


The [110] plane of BCC has an area of 2 a2 . This plane is occupied by two
circles (4 quarter circles + 1). Therefore:
2
2πR
PD[110] = = 0.833
( )
2
2 4R 3

For FCC structure, the [100] face is occupied by 2 circles (4 quarter circles + 1).
2 2
2πRπR 2
PD
= = = 0.785
( )
 100
  a2 4R 2
2




The [110] plane of FCC again has 2 circles (4 quarter circles + 2 half circles). It’s
planar density is therefore:
2 2
2πRπR 2
PD110 = = = 0.555
( )
2
 
a × 2a 2 4R 2


( )
3
3
64R
(c) The volume of a BCC unit cell
= is a3 . The unit cell contains
4R 3
=
3 3
2 atoms (1 center atom + 8 quarter corner atoms). The packing factor is therefore:
3
2 × 4 3 πR
PFBCC  = = 0.68
3
 
64R 3 3
The unit cell of FCC structure contains 4 atoms (8 quarter atoms in corners + 6
half atoms at the center of the faces).
3 3
4 ×4 3 πRπR 4 × 4 3
PFFCC  = = = 0.74
a3
( )
  3
4R3 2
Since HCP has the same packaging as FCC, its packaging factor is also 0.74. this
can also be calculated following Problem 1.7, by calculating the volume of the
unit cell and considering that each unit cell contains 6 atoms (3 central atoms +
2 half atoms at the center of each base + 12 one sixth atoms on the corners).


1.7 Show that the c/a ratio (see Fig. 1.16) in an ideal hexagonal close-packed (HCP)
structure is 1.63. (Hint: Consider an equilateral tetrahedron of four atoms which
touch each other along the edges.)

Solution:
Atom A depicts the one shown in the center of the top plane in Fig. 1.16, whereas the
other three atoms are the central ones within the unit cell.



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, Solution and Answer Guide: Abbaschian, Physical Metallurgy Principles, 5e, 2025, 9798214001661;
Chapter 1: The Structure of Metals


= sin 60
BC a= 3a a
BO 2=
= 3 BC 3a 3
AO = AB − AO =
2 2 2 a
3

= AO 2 2
C 2= a
3
c
= 2
2= 1.63
a 3



1.8 Iron has a BCC structure at room temperature. When heated, it transforms from BCC
to FCC at 1185 K. The atomic radii of iron atoms at this temperature are 0.126 and
0.129 nm for bcc and fcc, respectively. What is the percentage volume change upon
transformation from BCC to FCC?

Solution:
Each unit cell of BCC iron contains 2 atoms. The volume of the unit cell is
3 3
 4R  3 1  4R 
a3 =   . The volume occupied by each atom is therefore a 2 =   .
 3 2  3
Substituting 0.126 nm for R results in:
Volume occupied per atom in BCC = 0.01232 nm3

In FCC, the lattice parameter is equal to 4R 2 , and each unit cell contains four
atoms. Therefore:

a3
Volume occupied per atom is FCC = = 0.01214 nm3
4


% volume change
=
(0.0121 − 0.0123) 100 = −1.4%
0.0123
Iron shrinks by 1.4% as it transforms from BCC to FCC at 1185k.




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posted to a publicly accessible website, in whole or in part.

, Solution and Answer Guide: Abbaschian, Physical Metallurgy Principles, 5e, 2025, 9798214001661;
Chapter 1: The Structure of Metals


1.9




This diagram shows the Thompson Tetrahedron, which is a geometrical figure formed by
the four cubic {111} planes. It has special significance with regard to plastic deformation
in face-centered cubic metals. The corners of the tetrahedron are marked with the
letters A, B, C, and D. The four surfaces of the tetrahedron are defined by the triangles
ABC, ABD, ACD, and BCD. Assume that the cube in the above figure corresponds to a
face-centered cubic unit cell and identify, with their proper Miller indices, the four
surfaces of the tetrahedron.

Solution:
(a) The indices of plane ABD are (111).
( )
(b) The indices of plane ABC are 111 .

(c) The indices of plane ADC are ( 111 ) .

(d) The indices of plane BCD are ( 111 ) .


1.10




The figure accompanying this problem is normally used to represent the unit cell of a
close-packed hexagonal metal. Determine Miller indices for the two planes, defg and
dehj, that are outlined in this drawing.




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