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

Don't use plagiarized sources. Get Your Custom Essay on

ETM 307 MSU Crystalline Materials & Atomic Packing Factor Questions

Just from $13/Page

The price is based on these factors:

Academic level

Number of pages

Urgency

Basic features

- Free title page and bibliography
- Unlimited revisions
- Plagiarism-free guarantee
- Money-back guarantee
- 24/7 support

On-demand options

- Writer’s samples
- Part-by-part delivery
- Overnight delivery
- Copies of used sources
- Expert Proofreading

Paper format

- 275 words per page
- 12 pt Arial/Times New Roman
- Double line spacing
- Any citation style (APA, MLA, Chicago/Turabian, Harvard)

Delivering a high-quality product at a reasonable price is not enough anymore.

That’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe.

You have to be 100% sure of the quality of your product to give a money-back guarantee. This describes us perfectly. Make sure that this guarantee is totally transparent.

Read moreEach paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in.

Read moreThanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result.

Read moreYour email is safe, as we store it according to international data protection rules. Your bank details are secure, as we use only reliable payment systems.

Read moreBy sending us your money, you buy the service we provide. Check out our terms and conditions if you prefer business talks to be laid out in official language.

Read more