ETM 307 MSU Crystalline Materials & Atomic Packing Factor Questions

Lesson Note # 4
Crystal Structure
From our previous learning, we know that chemical bonds determine to a great extent the physical, chemical and
mechanical properties of materials. Materials strength is directly related to the atomic binding force and energy. Also
from our previous learning, we know that materials ductility is a cumulative result of discrete atomic slips driven by
shear stress. For a material, how an atomic slip occurs on what plane and direction is determined by its crystal
structure. The crystal structures and their associated defects play a pivotal role at deciding the ductility and fracture
toughness of engineering materials. This lesson note assists you in gaining knowledge of crystal structures.
Crystalline and amorphous
An atom can be modeled as a rigid sphere. For the equilibrium-solidified metallic materials and most of ceramic
materials, atoms are not packed in a random, irregular manner but are arranged into a precise, periodical, threedimensional pattern over a long distance (hundreds and thousands of atomic sizes long). The long-distance regularity
in atomic arrangement is called a long-range order. The periodical pattern of atomic arrangement is called a crystal
structure. Metallic materials have very simple crystal structures. Ceramic materials are composed of different-size
atoms in either covalent bond, ionic bond or mixed covalent/ionic bond. Depending on the complexity of their
compositions, some of them have relatively simple crystal structures while others have quite complex crystal
structures. The materials with regular atomic arrangement are all called crystalline materials.
For some materials like glasses, their atoms are very difficult to get into a periodical arrangement because of
thermodynamic barriers faced in the transformation. The irregularity in atomic arrangement is called an amorphous or
non-crystalline structure. The materials with irregular atomic arrangement are called amorphous materials or noncrystalline materials. Amorphous metallic materials often have unique properties. The methods of rapid solidification
and severe mechanical disruption are used to intentionally realize amorphous metallic materials.
Polymeric materials are made of either large covalent-networking molecules or long-chain covalent molecules. They
cannot be produced in a fully-crystallized form. They are either amorphous or partially crystallized.
Unit cell and crystal constant
Figure 1a shows a two-dimensional regular atomic arrangement. Given its repeatability and symmetry, the description
of the structural features of this atomic arrangement does not necessarily include all the atoms. If we find a small
structural unit that can generate the whole atomic arrangement by simply translating the unit in space, this unit indeed
represents all the features of the arrangement, no more and no less. Although there are many choices as illustrated
in Figure 1a, there is one that is the simplest and smallest but the best to represent the geometrical symmetry. Such
a structural subdivision, like the no. 1 choice in Figure 1a, is called a unit cell. A real crystal structure is threedimensional. If the atomic arrangement in the other direction is the same, as shown in Figure 1b, we have a threedimensional unit cell, like Figure 1c, after connecting the centers of the eight atoms. The edge length of a unit cell is
called the crystal constant. A cubic unit cell has only one crystal constant.
1
2
3
6
4
(c )
5 (a )
(b )
Fig.1: Subdivision of a periodic atomic arrangement.
ETM 307 Materials Science © Q. Xu, MSU
1
Lattice structure
In Figure 1, the physical locations of individual atoms are used to develop the concepts of crystal structure and unit
cell and each corner of the unit cell contains one actual atom. Now we slightly expand the concepts. The ionic and
covalent materials and intermetallic compounds contain atoms of different species. The atoms of different species are
associated by ionic or covalent bonds to form discrete groups of atoms at a certain atomic ratio that is decided by
chemical bonds. In the crystal structures of these materials, the discrete atomic groups are arranged in a periodic
+
manner. Figure 2 shows one example in which one Cl ion and one Cs ion always appear together in the crystal
structure of the ionic CsCl compound. In this case, one discrete atomic group is viewed as one entity that is
represented by one dot. Note that the dot of this kind does not represent the size of an actual atom or atomic group
but just its location. The obtained structure is called a lattice structure; a dot in a lattice structure is called a lattice
point; and the edge length is the lattice constant. By using the concept of lattice, the atom arrangements in Figure 1b
and Figure 2a, although strikingly different, can actually be represented by the same simple cubic unit cell.
Cl
Cs
(b )
(a ) CsCl
Fig.2: Lattice structure of CsCl.
The concept of lattice is of great significance for crystallographic analysis. The atoms of different species in discrete
atomic groups react exactly in the same way to incident X-ray and electron waves. The resulted diffraction patterns of
X-ray and electron waves are determined by the periodicity of the lattice structure instead of how the atoms of
different species are packed in discrete atomic groups.
For metallic materials, one lattice point represents one actual atom and, thus, their lattice structures are equivalent to
their crystal structure. For ionic and covalent materials and intermetallic compounds, one lattice point often
represents more than one atom and their lattice structures are not equal to the actual atom-packing situations.
Seven crystal systems and fourteen Bravais lattices
In the above learning, we know that a crystal structure containing a large number of atoms can be fully represented
by a unit cell. A unit cell is a structural unit that can seamlessly re-construct the crystal structure by repeating itself.
Given such geometrical requirements, not any geometry can serve as an effective unit cell. Figure 1c and 2b
introduced a simple unit cell. In general, a parallelepiped, like Figure 3, can serves as a unit cell.
b
a
α
β
c
γ
Fig.3: A general unit cell.
ETM 307 Materials Science © Q. Xu, MSU
2
Based on the differences in the angles and edge lengths, the general unit cell in Figure 3 can be divided into seven
lattice systems without regard to actual atoms. According to the atomic arrangements of actual materials in nature,
the seven lattice systems are further divided into fourteen Bravais lattices. They are summarized as follows:
7 Systems
Cubic
Tetragonal
Length & Angle
Cell Geometry
14 Bravais Lattices
a=b=c
α = β = γ = 90 o
a=b≠c
α = β = γ = 90 o
a≠b≠c
Orthorhombic
α = β = γ = 90 o
a=b=c
Rhombohedral
α = β = γ ≠ 90 o
a=b≠c
Hexagonal
α = β = 90 o
γ = 120 o
a≠b≠c
Monoclinic
α = β = 90 o
γ ≠ 90 o
a≠b≠c
Triclinic
α ≠ β ≠ γ ≠ 90 o
ETM 307 Materials Science © Q. Xu, MSU
3
Coordination number
The coordination number of an atom in a lattice is the number of atoms that are either in touch with this atom or
nearest to this atom. Because metallic and ionic bonds have no directionality, the atoms in the crystal structures of
the metallic and ionic materials are typically characteristic of a large coordination number. Since covalent bond has
directionality, the atoms in the crystal structures of the covalent materials have a small coordination number. Let us
use the simple cubic cell to show how to decide the coordination number. When counting the coordination number,
the important thing is to keep in mind that a unit cell is a subdivision of a crystal structure and do not miss the atoms
outside the unit cell, like Figure 4. The coordination number of the atom at the left bottom corner of the unit cell is six.
Fig.4: Coordination number of the simple unit cell.
Atomic packing factor
Eight atoms are located at the eight corners of the cubic unit cell in Figure 1b. Let us look at one single complete
atom at its right top corner. By comparing Figures 4a and 4b, it is found that only one eighth of the atom’s volume is
really located inside the unit cell. Thus, there is a total atomic volume of 8 × 1/8 = 1 atom for this unit cell. In other
words, with respect to the atomic volume, a cubic unit cell contains only one atom.
(a )
(b )
(c )
Fig.4: Simple cubic unit cell.
Now it is ready to introduce a concept of atomic packing factor (APF). The APF is the ratio of the total volume of
atoms within a unit cell over the volume of the unit cell. It represents the atomic packing density of a material. An APF
is expressed as:
APF =
Volume of atoms i n unit cell
Volume of unit cell
Assuming the adjacent atoms contact each other. For the cubic unit cell, its edge length will be 2R if the radius of an
atom is R. The atomic packing factor of the cubic unit cell is calculated as:
ETM 307 Materials Science © Q. Xu, MSU
4
4 3
πR
π
APF = 3 3 = = 52.36%
(2 R ) 6
Interstitial size
An atom is viewed as a rigid sphere. Accordingly, there are open spaces located between the crystal host atoms. The
size of an interstitial is defined by the size of the biggest rigid ball that can be placed into the interstitial site often at a
form of radius ratio between the interstitial atom and the host atom. Figure 5 uses the cubic unit cell as an example to
demonstrate the process of determining the size of an interstitial site. The size of the interstitial site in the cubic unit
cell is r / R = ( 3 − 1) =0.732.
(a )
(
)
2 3 −1 R
(b )
2R
2 3R
2 2R
Fig.5: The interstitial site of cubic unit cell.
Lattice point and Miller indexes of lattice direction and plane
1. Lattice points
A common Cartesian coordinate system is used to describe a lattice structure. The positive x axis is usually coming
out of the paper. The position of a lattice point or an atom is measured by using the unit of lattice constant instead of
using the absolute dimensions along the x, y and z axes. Thus, the coordinates per unit distance is one. After the unit
decided, the coordinate of any point can be easily identified through projecting this point to the three axes. Figure 6
shows a cubic unit cell in such a coordinate system. It will be used in the following to demonstrate how to identify
lattice directions and planes.
z
0,0,1
y
0,1,0
1,0,0
x
Fig.:6: Coordination system for a lattice structure.
ETM 307 Materials Science © Q. Xu, MSU
5
2. Miller index of lattice direction
The determination of a lattice direction is straight forward:
1)
2)
3)
4)
Figure out the coordinates of two points;
Subtract the coordinates of the end point and the start point;
Transfer the fractions, if happening, into full numbers;
Enclose the numbers in square brackets [ ]. If there are negative numbers, put bars on top of the numbers.
The obtained number combination is called a Miller index of lattice direction.
z
1,1,1
1
4 C
0 ,0 ,
D
y
B
0,1,0
1,0,0
A
x
Fig.7: Miller index of lattice direction.
Figure 7 shows two lines AB and CD. For line AB, 0,1,0 − 1,0,0 = −1,1,0 . So, the Miller index of line AB is [ 1 10] . For
line CD, 1,1,1 − 0,0,
3
1
= 1,1, = 4,4,3 . Thus, the Miller index of line CD is [443] .
4
4
3. Miller index of lattice plane
Miller index of lattice plane is calculated by using the reciprocals of the intercepts of a plane with the three axes:
1)
2)
3)
4)
Identify the intercepts of the plane with the three axes;
Calculate the reciprocal of the three intercepts;
Convert the fractions, if occurring, into full integers;
Put the numbers into parentheses ( ). If there are negative numbers, put bars on the numbers. The obtained
number combination is called a Miller index of lattice plane.
Figure 8 shows a plane having three intercepts, 1, 1, and 1/2, respectively, with the x, y and z axes. Their reciprocals
are 1, 1 and 2. Thus, the Miller index of this plane is (112).
z
0, 0,
1
2
y
0,1,0
x
1,0,0
Fig.8: Miller index of lattice plane.
ETM 307 Materials Science © Q. Xu, MSU
6
The plane in Figure 9 has only one intercept with the x axis. Its intercepts with the y and z axes are considered to be
infinity. Thus, the reciprocals of the three intercepts are:
1 1 1
, , = 1,0,0 . The Miller index of this plane is (100).
1 ∞ ∞
z
y
x
1, 0 , 0
Fig.9: Miller index of lattice plane.
Figure 10 shows a lattice plane that does not intercept the y axis within the unit cell. In this case, the plane needs to
be extended to get an intercept. After the three intercepts are obtained, their reciprocals are:
1 1 1
,
, = 2,−1,1 .
1/ 2 − 1 1
So, the miller index of this plane is (2 1 1) .
z
z
0,0,1
0,0,1
y
0, −1,0
y
1
, 0, 0
2
1
, 0, 0
2
1,1,0
x
1,1,0
x
Fig.10: Miller index of lattice plane.
Figure 11 is another case in which the lattice plane has no intercepts with the three axes within the unit cell. A parallel
plane that intercepts with the three axes is drawn. Then, the Miller index can be easily decided to be (111).
z
z
0,0,1
0,0,1
y
1,0,0
x
0,1,0
y
1,0,0
0,1,0
x
Fig.11: Miller index of lattice plane.
ETM 307 Materials Science © Q. Xu, MSU
7
Inter-planar spacing
A unit cell is one representative unit of a whole crystal structure. A crystal structure may contain more than thousands
of unit cells. For one lattice direction or plane, there are thousands of such lattice directions or planes inside a crystal
structure. Therefore, a Miller index does not represent only one lattice direction or plane but thousands of lattice
directions or planes in the whole crystal structure taking the same Miller index. Figure 12 shows an arbitrary lattice
plane of (hkl) in a unit cell. Actually, there are a series of such lattices planes in the crystal structure.
(hkl )
z
(hkl )
(hkl )
(hkl )
(hkl )
y
plane #3
plane # 2
plane #1
x
plane #−1
plane #−2
Fig.12: a series of lattice planes taking the same Miller index.
For a cubic lattice system, the inter-spacing of lattice planes in Figure 12 can be calculated by:
d hkl =
a
h + k2 + l2
2
where a is the lattice constant. For the crystal structure of a material, a specific lattice plane has a unique interspacing. Once the value of inter-spacing is identified, people can immediately judge what material this crystal
structure belongs to. This is the reason why x-ray and electron diffraction techniques can be used to identify materials
composition, phase and structure.
Families of lattice directions and planes
A cubic has a high degree of symmetry. Many of its lattice directions and planes, even though pointing to and laying
along different directions, have no difference. For example, the four directions in Figure 13a and the three hatched
planes in Figure 13b are the same thing, respectively. We say that the lattice directions belong to a family of lattice
directions and the lattice planes belong to a family of lattice planes.
(100 )
[110 ]
[1 0 1 ]
[10 1 ]
( 010 )
[ 01 1 ]
( 001 )
(a )
(b )
Fig.13: Families of lattice directions and planes
ETM 307 Materials Science © Q. Xu, MSU
8
Let us look at the Miller indexes of lattice directions in Figure 13a. It is obvious that the Miller indexes have the same
three integers but in different orders and different signs (+/-). It is the same situation with the Miller indexes of lattice
planes in Figure 13b. A family of lattice directions or planes is defined as a complete group of lattice directions or
planes that have the same index numbers but either in different orders or with different sign (+/-). A family of crystal
directions is expressed by angular brackets < > and a family of crystal planes is denoted by braces { }.
The lattice directions in Figure 12a belong to the [110] family. The complete list of lattice directions in the family is:
[1 1 0], [10 1 ], [0 1 1], [ 1 1 0], [ 1 0 1 ], [0 1 1 ], [ 1 10], [ 1 01], [0 1 1], [110], [101], [011]
The lattice planes in Figure 12b belong to the {100} family. The complete list is the three planes:
(100), (010), (001)
Lattice directions or planes in one family has the same characteristics of atomic arrangement and are equivalent
crystallographically. The families of crystal directions and planes are natural results of symmetrical geometries of
crystals. An important relationship for a cubic lattice system is that the Miller index of a direction has the same Miller
index of a plane that it is perpendicular to. For example, the [100] direction is perpendicular to the (100) plane.
Crystal structures of metals
With the basic but necessary crystallographic knowledge, we are ready to learn the three common crystal structures
of actual metallic materials: body-centered cubic (BCC); face-centered cubic (FCC) and hexagonal close-packed
(HCP). When learning the three crystal structures, special attention is needed to pay to atomic packing factor,
interstitial size, close-packed crystal direction, and close-packed crystal plane. When we learn dislocations, we will
know that dislocations preferably slide along close-packed crystal directions and on close-packed crystal planes.
Close-packed crystal directions and close-packed crystal planes refer to as crystal directions and planes whose
atoms are in direct contact. They are important for understanding materials ductility.
1. Body-centered cubic (BCC)
Based on the name, a body-centered cubic crystal structure has one atom at the center of a cubic unit cell, as shown
in Figure 14. When we learned a simple cubic cell in Figure 4, we knew that it contained one atom in terms of atomic
volume. Now, the BCC crystal structure has an extra atom at its center. Immediately, we conclude that the BCC
crystal structure contains two atoms within its unit cell.
(a )
(b )
Fig.14: Body-centered cubic crystal structure: a) actual atomic arrangement; b) lattice structure.
If atoms are assumed to touch along the cubic diagonal in the BCC, the lattice constant is related to the atom radius:
ETM 307 Materials Science © Q. Xu, MSU
9
a=
4R
3
Thus, the atomic packing factor is:
4 3
πR
π 3
3
==
= 67.98%
APF =
3
8
⎛ 4R ⎞
⎜⎜
⎟⎟
⎝ 3⎠
2×
The BCC crystal structure has two types of interstitial: tetrahedral site and octahedral site. The tetrahedral interstitial
gets its name because it has four surfaces, as shown in Figure 15a. There are a number of tetrahedral interstitial
sites at equivalent positions in this unit cell. On the top crystal plane in Figure 15a, there are four tetrahedral
interstitial sites; and, in total, there are twenty-four tetrahedral interstitial sites in the unit cell in terms of number (note:
not in terms of volume). Figure 15b shows the position of the tetrahedral interstitial site in Figure 15a, relative to its
nearby crystal atoms. If the radius of a crystal atom is R and the radius of the largest interstitial atom is r, there is the
following equation:
Atom #1
1 1
, ,1
2 4
Atom #1
a
Atom # 2
Atom # 2
(a )
(b )
a
4
Fig.15: a) Tetrahedral interstitial in body-centered cubic crystal structure; b) the position of the tetrahedral interstitial site
relative to the nearby crystal atoms.
2
⎛a⎞ ⎛a⎞
(R + r) = ⎜ ⎟ + ⎜ ⎟
⎝2⎠ ⎝4⎠
2
2
The crystal constant a = 4 R / 3 . Substitute it to the above equation, the size of the interstitial site is figured out to
be r / R = ( 5 / 3 − 1) = 0.291.
The octahedral interstitial gets its name because it has eight surfaces, as shown in Figure 16a. There are a total of
fourteen octahedral interstitial sites in a BCC crystal unit cell in terms of number. Figure 16b shows the position of the
octahedral interstitial site in Figure 16a, relative to its nearby crystal atoms. This interstitial site is not symmetrical.
This means that an interstitial atom can not touch all eight atoms simultaneously. The interstitial atom will first get into
contact with Atoms #1&2. From the sketch in Figure 16b, there is the following relationship:
ETM 307 Materials Science © Q. Xu, MSU
10
Atom #1
1 1
, ,1
2 2
Atom #1
Atom # 4
a
Atom #3
Atom #3
Atom # 4
Atom # 2
Atom # 2
(b )
(a )
Fig.16: a) Octahedral interstitial in body-centered cubic crystal structure; b) the position of the Octahedral interstitial site
relative to the nearby crystal atoms.
2( R + r ) = a =
4R
3
After solving this equation, the size of the octahedral interstitial site is r / R = ( 2 3 / 3 − 1) = 0.155 . The above
results show that, in the BCC crystal structure, a tetrahedral interstitial is larger than an octahedral interstitial.
It is important to know the close packed directions and close packed planes of the BCC crystal structure. If they do
not exist, it is important to figure out the closest packed directions and closest packed planes. In Figure 17a, a crosssectional plane is made cross the diagonals of the BCC’s top and bottom planes. After this cross-sectional plane is
laid out on the paper surface with actual atoms, it looks like Figure 17b. Apparently, the family of crystal
directions are the close packed directions for the BCC crystal structure. However, on the plane of (1 1 0) , the atoms
are not closely packed. Thus, the BCC has no close packed planes. Its closest packed planes are the family of {110}.
z
[ 1 1 1]
[111]
[111]
[ 1 1 1]
(1 1 0)
y
(1 1 0)
x
[1 1 1 ]
(a )
[11 1 ]
[1 1 1]
(b )
[11 1 ]
Fig.17: Crystal directions and planes in BCC.
2. Face-centered cubic (FCC)
According to the name, the face-centered cubic (FCC) crystal structure has atoms in the centers of its six surface
planes. On the basis of the simple cubic cell in Figure 4, when six of half atoms are added on the six surface planes,
we obtain a FCC crystal structure, like in Figure 18a. As we knew, a simple cubic cell contains one atom. Now six half
atoms are incorporated. Thus, a FCC unit cell contains four atoms in terms of atomic volume. In the FCC, the atoms
are in direct touch along the diagonals of the surface planes, as shown in Figure 18a. So, the crystal constant and
atomic radius has the following relationship:
ETM 307 Materials Science © Q. Xu, MSU
11
(a)
(b )
Fig.18: Face-centered cubic crystal structure.
a = 4R / 2
Then, the atomic packing density is:
4 3
πR
π 2
3
=
= 74.05%
APF =
3
6
⎛ 4R ⎞
⎟⎟
⎜⎜
⎝ 2⎠
4×
Similar to the BCC crystal structure, the FCC also has two types of interstitial: tetrahedral site and octahedral site.
The tetrahedral interstitial of the FCC is depicted in Figure 19a. There are eight tetrahedral interstitial sites in the FCC
unit cell in terms of number (note again: not in terms of volume). Figure 19b shows the position of the tetrahedral
interstitial site in Figure 19a, relative to its nearby crystal atoms. If the radius of a crystal atom is R and the radius of
the largest interstitial atom is r, there is the following equation:
3 1 3
, ,
4 4 4
R
(a )
R
a
4
(b )
\
Fig.19: Tetrahedral interstitial in face-centered cubic crystal structure.
2
⎛a⎞
(R + r) = ⎜ ⎟ + R 2
⎝4⎠
2
Substituting a = 4R /
2 to the equation, the size of the tetrahedral interstitial in the FCC is calculated to be
r / R = ( 3 / 2 − 1) = 0.225 .
Figure 20a shows an octahedral interstitial. Unlike the octahedral interstitial in the BCC, the octahedral interstitial here
is symmetrical. That means that an interstitial atom can get into contact simultaneously with all the eight surrounding
atoms. There are nine octahedral interstitial sites in a FCC unit cell in terms of number. Figure 20b shows the relative
ETM 307 Materials Science © Q. Xu, MSU
12
position of this interstitial with the surrounding crystal atoms. The radius of the interstitial and the radius of the crystal
atoms have such a relationship:
1 1 1
, ,
2 2 2
a
(a)
(b )
Fig.20: Octahedral interstitial in face-centered cubic crystal structure.
a = 2( R + r ) = 4 R / 2
The size of the octahedral interstitial is calculated as r / R = ( 2 − 1) = 0.414 .
It is important to figure out the close packed directions and planes in the FCC crystal structure. Make a cut like Figure
21a and expand the cut on the paper plane with full-size atoms. We obtain an atomic arrangement like Figure 21b.
Obviously, the family of crystal directions are the close packed directions and the family of {111} crystal planes
are the close packed planes for the FCC crystal structure.
z
[ 1 01]
[0 1 1]
[ 1 10]
(111)
[ 1 01]
[0 1 1]
[1 1 0]
[1 1 0]
[ 1 10]
y
[01 1 ]
x
[10 1 ]
[01 1 ]
[10 1 ]
(a)
(b )
Fig.21: Crystal directions and planes in face-centered cubic crystal structure.
3. Hexagonal close-packed (HCP)
A hexagonal crystal structure is a crystal structure that has a hexagonal symmetry. Its unit cell is shown in Figure 22a.
The cell has two interior angles of 120° and is not symmetrical against the z axis, different from BCC and FCC’s cubic
geometries. The unit cell in Figure 22b is simple and contains only one atom in terms of atomic volume. However, it
cannot explicitly reveal the hexagonal symmetry. As a result, on the basis of this unit cell, people develop a large cell,
like Figure 22b. The crystal structure is a simple hexagonal structure with a low atomic packing density. Actual
materials crystallize to a more close packed crystal structure like Figure 22c, with three additional atoms inside. This
crystal structure is stable. It is the hexagonal close-packed crystal structure we are going to learn.
ETM 307 Materials Science © Q. Xu, MSU
13
120o
120o
(a )
Fig.22:
120o
(b )
(c )
a) the unit cell of a hexagonal crystal system; b) the simple hexagonal crystal structure; and c) the hexagonal closepacked crystal structure.
Figure 23a is the hexagonal close-packed crystal structure with actual atoms. Adjacent atoms are in direct contact,
indicating the nature of close atomic packing. On each corner of the hexagonal structure, one-sixth of an atomic
volume is located inside the cell. The hexagonal structure contains three layers of atoms. At the center of the
top/bottom plane, a half of an atomic volume is located inside the cell. The top layer contains one and half of atoms in
terms of atomic volume, so does the bottom layer. In order to make it easy to identify how many atoms are in the
middle layer, the layer is taken out and is flatten like Figure 23b, where the cross indicates the center line and the
hexagram reveals the border of the hexagonal cell. Part of the three atoms in the center stretches out of the cell; part
of the three nearby atoms stays inside the cell. The in and out parts are equal. After counting in and counting out
these two parts, the middle layer has three atoms inside the cell. In total, the hexagonal close-packed crystal
structure contain six atoms in terms of atomic volume.
(a )
(b )
Fig.23: Hexagonal close-packed crystal structure.
The calculation of atomic packing factor for the hexagonal close-packed crystal structure needs to know the
relationship between its height and atom radius. In order to facilitate doing it, five atoms along the HCP central line
are depicted in Figure 24a. These five atoms are in direct touch. Connecting the centers of these five atoms produces
two perfect tetragons whose bottom planes are joined together. The surfaces of the tetragons are equilateral triangles
of 2R (R is the atom radius of the HCP). The height of one tetragon is half of the HCP’s height. Figure 24b shows one
tetragon with its relevant dimensions labeled. From Figure 24b, we have the following equation:
(R 3 )
2
2
2
⎛R 3⎞
⎟ + ⎛⎜ c ⎞⎟
=⎜
⎜ 3 ⎟
⎝2⎠
⎝
⎠
Solving this equation, obtain: c = 4R 2 / 3
ETM 307 Materials Science © Q. Xu, MSU
14
2R
c
2
R 3
c
(b )
R
(a )
R 3
3
Fig.24: The relationship between the height and atomic radius of hexagonal close-packed crystal structure.
As showed in Figure 24, the top/bottom planes of the HCP structure consist of six equilateral triangles of 2R. Thus,
the area of the top or bottom plane is:
⎛1
⎞
A = 6 × ⎜ × 2 R × R 3 ⎟ = 6 3R 2
2
⎝
⎠
4 3
πR
3
=
A× c
6×
Therefore, the atomic packing factor of the HCP is: APF =
6×
4 3
πR
3
=
2
6 3R × 4 R
3
2
π
= 74.05%
3 2
As showed above, the face-centered cubic crystal structure also has an atomic packing factor of 74%. So, there
should be some similarity between them. As illustrated in Figure 21 and Figure 23, both FCC and HCP are formed by
stacking close-packed atomic planes. This is why they have the same atomic packing factor.
Then, what is the difference between them? In order to answer this question, we need to figure out the different
relative positions of close-packed atomic planes when stacking them. Figure 25a shows one close-packed atom
plane. It has two types of void between atoms: standing triangle and Inverting triangle. When the second atomic layer
is stacked on this first atom layer, it has two choices: either go to the standing triangle valleys or go to the Inverting
triangle valleys. Plus the first atom layer, this produces three atom layers in terms of their relative positions. The three
atom layers are named A-plane; B-plane and C-plane.
A-plane
C-plane
Inverting
triangle
Standing
triangle
B-plane
(a )
(b )
Fig.25: a) two types of voids; b) three atom layers
ETM 307 Materials Science © Q. Xu, MSU
15
After knowing the differences in the positions of the close-packed atom planes, it is easy to understand the difference
between FCC and HCP. As illustrated in Figure 26, the FCC crystal structure is formed by stacking close-packed
atom planes in a sequence of A, B, C, A, B, C, …; while the HCP crystal structure is generated in an atomic layer
order of A, B, A, B, ….
B-Plane
C-Plane
A-Plane
A-Plane
B-Plane
A Plane
A-Plane
(a )
(b )
Fig.26: The stacking sequence of close-packed atom planes in a) FCC; b) HCP.
The difference between the FCC and HCP crystal structures looks very minor. In reality, the difference causes
tremendous differences in plastic deformation of FCC and HCP materials. Here, give a few examples. The FCC has
many more easy slip systems than the HCP. As a result, dislocation cross-slip is a common phenomenon in FCC but
is difficult to happen in HCP. Because of the same reason, deformation twin is a common phenomenon in HCP but
seldom occurs in FCC under rapid plastic deformation. FCC materials rarely experience low-temperature
embrittlement but HCP materials are prone to low-temperature brittle fracture.
The HCP crystal structure also has two types of interstitial: tetrahedral and octahedral interstitials. Their locations are
shown in Figure 27. Note that Figure 27 just gives one example of each and there are a number of them in equivalent
position. Because the HCP has the same atomic packing density as the FCC, the sizes of their tetrahedral and
octahedral interstitials are exactly identical. The size of tetrahedral interstitial in the HCP is r / R = 0.225 and the
size of octahedral interstitial in the HCP is r / R = 0.414 . Both are the same as those in the FCC.
(a )
(b )
Fig.27: a) Tetrahedral and b) octahedral interstitials in HCP.
ETM 307 Materials Science © Q. Xu, MSU
16
In theory, the same indexing method as the ones used for the BCC and FCC can be used for the HCP. Figure 28a
shows such a method being used for indexing two crystal directions. Based on the index numbers, the two crystal
directions belong to one family of crystal directions. After we put the two directions into the actual HCP atomic
structure like Figure 28b, we immediately find these two directions have totally different crystallographic
characteristics. The three-axis coordinate system does not work for the HCP. This discrepancy is caused by the
angle of 120°, which results in the asymmetry of the HCP unit cell around the z axis.
z
[ 1 10]
[110]
y
120 o
x
(a )
(b )
Fig. 28: Indexing the HCP by a three-axis system.
In order to solve this discrepancy and to better reveal the hexagonal symmetry of the HCP, people develop a fouraxis coordinate system, as shown Figure 29. However, for a point on a plane, two axes are sufficient to locate it. Now,
there are three axes on the basal plane of the HCP. One axis should be redundant. This problem can be solved as
follows. If the index numbers, h, k, i and l is corresponding to the values on the a1, a2, a3 and c axes, we
deliberately assign: -i = h + k. The indexes obtained by using this four-axis coordinate system are called the MillerBravais indexes.
c
a3
a1
a2
Fig.29: A four-axis coordinate system for the HCP.
The crystal directions in the HCP are decided by the following two steps:
1)
Use the three axes, a1, a2 and c, to decide the three index numbers: h’, k’, and l’.
2)
Then, use the following the equation to covert the three index numbers: h’, k’, and l’ to the four index
numbers: h, k, i and l.
ETM 307 Materials Science © Q. Xu, MSU
17
1
(2h ′ − k ′)
3
1
k = (2k ′ − h ′)
3
i = −( h + k )
l = l′
h=
Let us use the above method to the two crystal directions in Figure 28a. The result is shown in Figure 30. Now the
two directions have different index numbers, reflecting their different crystallographic properties.
[h′k ′l ′]
[hkil ]
z
[ 1 100]
[ 1 10]
[11 20]
[110]
y
120 o
x
(a )
(b )
Fig.30: Crystal directions in the HCP.
Let us use the above method to decide another three directions in Figure 31a. The three directions have the same
index numbers but in different orders and sign (+/-). As shown in Figure 31b, the three directions have the same
crystallographic characteristic. So, they belong to a family of [1120] crystal directions. This example demonstrates
that the four-axis coordinate system is good at revealing the characteristics of the HCP.
c
[ 1 2 1 0]
[11 20]
[ 2 1 1 0]
a3
a2
a1
(a )
(b )
Fig.31: Crystal directions in the HCP.
The procedure of deciding crystal plane indexes in the HCP is the same as those for the BCC and FCC, except that
the HCP has one more intercept on the a3 axis. That is, figure out the intercepts and calculate their reciprocals. For
the HCP crystal plane indexes, the relationship of -i = h + k. is naturally satisfied. Figure 32 shows three crystal
planes in the HCP. The plane in Figure 32a is an example to show how to get the plane index. It has four intercepts 1,
1, -1/2 and 1. The reciprocals of the intercepts are 1, 1, -2 and 1. Thus, the plane index is (11 2 1) . The indexes of the
other two planes can be obtained similarly.
ETM 307 Materials Science © Q. Xu, MSU
18
c
c
c
(0001)
(11 2 1)
a3
a3
a2
a2
a1
(a )
a1
(10 1 0)
a3
(b )
a2
a1
(c )
Fig.31: Crystal planes in the HCP.
Allotropic and polymorphic
Materials that have more than one crystal structure are called allotropic or polymorphic. When an elemental material
changes from a crystal structure to another, it called an allotropic transformation. The representative materials that
have an allotropic transformation include iron and titanium. Iron has BCC and FCC structures and Titanium has BCC
and HCP structures at different temperatures.
When a compound changes from a crystal structure to another, it called a polymorphic transformation. Most ceramic
materials have polymorphic phenomena. For example, ZrO2 has monoclinic, tetragonal and cubic structures at
different temperatures.
As learned from the above, different crystal structures often have different atomic packing factors. The change in
crystal structure involves the change in volume. Without careful control, the volumetric change may cause the
material to fracture.
ETM 307 Materials Science © Q. Xu, MSU
19
Name: ___________________________
HW of ETM 307 Chapter 4 of ETM307: Crystal Structure
80 points in total
1.
What are crystalline materials? What are non-crystalline materials (5 points)
2.
What is allotropic? What is polymorphic? (5 points)
3.
What are unit cell, lattice constant and the number of atoms per unit cell? What is the number of atoms of BCC,
FCC and HCP? (10 points)
4.
What is the atomic packing factor? What are the atomic packing factors of BCC, FCC and HCP? Does the
volume increase or decrease when FCC iron changes to BCC iron? Why? (10 points)
5.
What are the families of crystal directions and planes? (10 points)
6.
What are the closest packed crystalline direction and plane of BCC and FCC? (10 points)
7.
What are the similarity and difference between FCC and HCP crystal structures? (10 points)
8.
What are the Miller indexes of two crystal directions and one crystal plane in the following figure? (10 points)
z
z
0,0,0.5
y
x
9.
0.5,0,0
y
x
Please identify the Miller-Bravais’ indexes of the crystal direction and plane indicated in the following hexagonal
crystal structure. (10 points)
c
c
a3
a3
a2
a1
ETM307 Materials Science, © Q. Xu, MSU
a2
a1