2 Learning Objectives Crystal Structuresnhayati/chapter 2 crystal structure.pdf · 2014-09-18 ·...

1

22

Crystal StructuresCrystal Structures

22Learning Objectives

� Distinguish between crystal structure and crystalsystem

� Relationship between atomic radius (R) and latticeparameter (a)

� Calculate Atomic Packing Factor (APF) and Volumedensity (ρ)

� Determine Indices for ‘Directions’ and ‘Planes’ in acrystal structure.

2-2

Chapter Outline

� Space lattice, crystal structures and crystal systems

� Atomic Packing Factor (APC), Coordination number (CN) & Volume Density

� Crystal Planes and Directions

2-3

Crystalline & Amorphous

� Materials are either:

1. Crystalline2. Non-crystalline – Amorphous

What is crystalline material?

� Crystalline solid must have an ordered arrangementof atoms (occupy specific or predictable positions ina 3D array).

� Crystalline materials include metals, many ceramicsand some polymers


� In an amorphous solid atoms do not possess well-defined arrangement (atoms have no periodicpacking)

� Amorphous structures form in complex structures,polymers, and rapidly cooled materials.

2-5


2-6

(c) 2003 Brooks/Cole Publishing / Thomson

Learning™ Atomic arrangements in (a) amorphous silicon and (b) crystalline

silicon. Note the variation in the inter-atomic distance for amorphoussilicon.

2

Space Lattice & Unit Cell

Why do we need to know the crystal structure?

� Crystal structure of a material affects its mechanicalproperties (strength, toughness, ductility)

� Understanding crystal structures is essential inunderstanding the crystalline materials.

2-7 2-8

(c) 2003 Brooks/Cole Publishing / Thomson Learning™

Classification of materials based on the type of atomic order.


� Lattice - A collection of points that divide space intosmaller equally sized segments.

� Basis - A group of atoms associated with a lattice point.

� Unit cell - A subdivision of the lattice that still retains theoverall characteristics of the entire lattice.

� Atomic radius - The apparent radius of an atom, typicallycalculated from the dimensions of the unit cell, usingclose-packed directions (depends upon coordinationnumber).

� Packing factor - The fraction of space in a unit celloccupied by atoms.

2-9


� A crystal is described as consisting of two parts:

� The lattice: a 3-D array of points in space. Eachpoint must have identical surroundings.

� The basis (motif): the identical group of atoms whichsurround each point in the lattice (repeated bysymmetry) to construct the crystal.

2-10


2-11

Lattice point

Unit cell


2-12

3

� There are 14Bravais latticesgrouped into 7crystal systems(7 types of unitcells)

2-13


Crystal Systems and Bravais Lattices

2-14

2-15

Lattice Parameters

2-16


Principals Metallic Crystal Structures

� Most important engineering metals have one of thefollowing crystal structures:

� BCC - body centered cubic

� FCC - face centered cubic

� HCP – hexagonal close packed

2-17


2-18

BCC Crystal Structure FCC Crystal Structure

4


2-19

HCP Crystal StructureHCP Crystal Structure

2 atoms/ unit cell6 atoms/ unit cell

Number of Atoms/ Unit Cell

� Each corner atom contributes as1/8

� There are 8 corner atoms/ BCCunit cell

2-20

Number of atoms in a BCC unit cell

cellunitatomsatomsofnumber /21188

1=×+×=

• There is 1 atom in the centre(non shared)

Number of Atoms/ Unit Cell

� Each corner atom contributesas 1/8

� There are 8 corner atoms/BCC unit cell

2-21

Number of atoms in a FCC unit cell

cellunitatomsatomsofnumber /42

168

8

1=×+×=

• Each face atom contributesas ½

• There are 6 face atoms

Relationship between Atomic Radius and

Lattice Parameter

2-22

Determine the relationship between the atomic radius and the

lattice parameter in SC, BCC, and FCC structures when one atomis located at each lattice point.

3

4ra =

2

4ra =ra 2=

(c) 2003 Brooks/Cole Publishing / Thomson

Learning™

Atomic Packing Factor

2-23

The atomic packing factor (APF) is defined as the ratio of the volume of atoms in the unit cell to the volume of the unit cell

APF Depends on:

• Crystal structure.• How “close” packed the atoms are.

APF =Vol. of atoms in unit cell

Vol. of unit cell

Atomic Packing Factor (APF) for BCC

2-24

=

3

3

42 RVatoms π

33

64)

3

4(

333

_

RRaV cellunit ===

68.08

3

33

64

3

8

3

3

_

==

==π

π

R

R

V

VAPF

cellunit

atoms

a

a2

R43

4

)4()2( 222

Ra

Raa

=

=+

2 atoms/unit

cell

5

Atomic Packing Factor (APF) for FCC

2-25

2R

R22

R22

a 4R

a

=

3

3

44 RVatoms π

333

_ 216)22( RRaV cellunit ===

74.023216

3

16

3

3

_

==

==π

π

R

R

V

VAPF

cellunit

atoms

Coordination Number (CN) for SC

2-26

• CN is the number of nearest

neighbor atoms

• It indicates how efficient atoms are

packed together

• Higher CN means a more efficient

structure in forming bonds to thelargest number of neighboring atoms

For simple cubic: CN = 6

Coordination Number for BCC

2-27

Total 8 nearest neighbor atomsCoordination number = 8

4

8

7

32

6

5

1

Coordination Number for FCC

2-28

1 4

32

8

7

6

5

12 11

1092R

R22

R22

Total 12 nearest neighbor atomsCoordination number = 12

Theoretical Density, ρ

2-29

The theoretical density of a material is given by:

ρ = n A

VcNA

# atoms/unit cell Atomic weight (g/mol)

Volume/unit cell

(cm 3/unit cell)

Avogadro's number

(6.023 x 10 23 atoms/mol)

Theoretical Density, ρ

3

2324/882.7

)1002.6)(1054.23(

)847.55)(2(cmg=

××=

−ρ

2-30

Example:

Determine the density of BCC iron, which has a lattice parameter of

0.2866 nm.

Solution:

Atoms/cell = 2, a = 0.2866 nm = 2.866 ×××× 10-8 cm

Atomic mass = 55.847 g/mol

Volume of unit cell = = (2.866 ×××× 10-8 cm)3 = 23.54 ×××× 10-24 cm3/cell

Avogadro’s number NA = 6.02 ×××× 1023 atoms/mol

3a

Compare this with:

ρFe = 7.87 g/cm3

6

Atom Positions in Cubic Unit Cell

2-31

• Cartesian coordinate system is used to locate atoms.

• In a cubic unit cell

� y axis is the direction to the right.

� x axis is the direction coming out of the paper.

� z axis is the direction towards top.

� Negative directions are to the opposite of positive directions

Miller Indices of Directions

011

2-32

To determine the Miller indices of a direction (define a vector)

1. Determine coordinates of points of direction end and origin

2. Subtract coordinates of point of origin from the point of end

3. Enclose in square brackets: [u v w] where u, v, and w are integersalong unit vectors a, b, c respectively

4. Multiply or divide by a common factor to reduce the lengths to the

smallest integer values.5. Enclose in square brackets: [u v w] where u, v, and w are integers

along x, y and z.

In any of the 3 directions, there are positive and negative directions

Example: [ ]

221−−

2-33

Example:

Determine the Miller indices of directions A, B, and C

(c) 2003 Brooks/Cole Publishing /

Thomson Learning™

Direction A

1. Two points are 1, 0, 0, and 0, 0, 0

2. 1, 0, 0, -0, 0, 0 = 1, 0, 0

3. No fractions to clear or integers to reduce

4. [100]

Direction B

1. Two points are 1, 1, 1 and 0, 0, 0

2. 1, 1, 1, -0, 0, 0 = 1, 1, 1

3. No fractions to clear or integers to reduce

4. [111]

Direction C

1. Two points are 0, 0, 1 and 1/2, 1, 0

2. 0, 0, 1 -1/2, 1, 0 = -1/2, -1, 1

3. 2(-1/2, -1, 1) = -1, -2, 2 = [ ]


2-34

• A direction and its negative are not identical

• A direction and its multiple are identical

• All equivalent directions are represented by a set of Miller indices

in < >

[ 1 0 0 ]

[ 1 0 0 ]

_

[ 0 0 1 ]

[ 0 0 1 ]

_[ 0 1 0 ]_

[ 0 1 0 ]

All directions belong to

<100>


2-35

How to draw a direction?

1. Select origin

2. Determine length of vector

projection in each of the 3 axes inunits (or fractions) of a1, a2, and a3.

3. Multiply or divide by a commonfactor to reduce the lengths to the

smallest integer values.

4. Enclose in square brackets: [u v w]

where: u, v, and w are integers

along a1, a2, and a3

a2

a3

a1


2-36

Example:

• Draw the direction [110]

1. We select the origin at 0,0,0

2. The lengths of vector projection in x,

y, z axes are: 1, 1, 0

3. Since the lengths are integers, no

fraction to clear

4. The direction is [110]

0, 0, 0

1, 1, 0

a2

a3

a1

7

Miller Indices of Planes

2-37

Why do we need to draw planes?

• Metal deform along planes of atoms that are most tightly packed together

• Therefore it is important to identify these planes in a crystal

• This is done by labeling each face (plane) by its Miller Indices

To find the Miller indices of a plane (Cubic Structures)

1. Find its intercept with the a1, a2, and a3 axes2. Take the reciprocals of these intercepts

3. Reduce all fractions to the lowest denominators

4. Enclose the 3 numbers in round ( ) to represent the specific plane


2-38

Example

Indexing the (111) plane

1. Intercepts: 1, 1, 1

2. Reciprocals: 1/1, 1/1, 1/1

3. Clear fractions: 1, 1, 1

4. Miller indices: (111)

a2

a3

a1


2-39

(100) (110) (111)

All equivalent planes are represented by a set of Miller indices in { }

� For example: the faces of the cube belong to the family of {100}

planes which contains 6 planes: (100), (010), (001), (-100), (0-10), (00-1)


2-40

Zone axis and zone law

Any two non-parallel planes will intersect in a line

In a crystal, a plane (h1k1l1) will intersect the plane (h2k2l2) along a line

with direction [uvw] where:

[uvw] is known as the “zone axis” of the two planes

u = k1l2 - k2l1

v = l1h2- l2h1

w = h1k2 - h2k1


2-41

x

y

z

-y

Zone axis

Zone axis


2-42

Zone axis and zone law

To remember these relationships, the following can be used

h1 k1 l1 h1 k1 l1

h2 k2 l2 h2 k2 l2u v w

The planes of a zone axis [uvw] (if the plane lies in the zone axis) must satisfy the “Weiss Zone Law”

hu + kv + lw = 0

+

+_

+_

+_

8

Crystallographic Planes in HCP

2-43

� Directions in HCP unit cells are denoted with either the 3-axis or 4-

axis system

With the 3-axis system, the procedure is the same as for the cubicMiller indices explained above but ignoring the a3 axis

Directions in the 4-axis system

• Symmetry about the c axis in the HCP unit cell makes a3 equivalent to

a1 and a2


2-44

Directions in the 4-axis system

• To determine the Miller indices for adirection in HCP unit cell [u v i w],

these vectors are translated parallel

to the 4 axes in the requireddirection.

• We determine the “end” and “start”

coordinates.

• We must also maintain the “relation”

i = - (u + v)


2-45

Determine the Miller-Bravais indices for planes A and B and directions C and D


Thomson Learning™

Plane A

1. a1 = a2 = a3 = ∞ , c = 12. 1/a1 = 1/a2 = 1/a3 = 0, 1/c = 1

3. No fractions to clear

4. (0001)

Plane B

1. a1 = 1, a2 = 1, a3 = -1/2, c = 12. 1/a1 = 1, 1/a2 = 1, 1/a3 = -2, 1/c = 1

3. No fractions to clear

4.

∞

)1211(


]011[

2-46

Determine the Miller-Bravais indices for planes A and B and directions C and D


Thomson Learning™

Direction C

1. Two points are 0, 0, 1 and 1, 0, 0.2. 0, 0, 1, -1, 0, 0 = -1, 0, 1

3. No fractions to clear or integers to

reduce.4.

Direction D

1. Two points are 0, 1, 0 and 1, 0, 0.2. 0, 1, 0, -1, 0, 0 = -1, 1, 0

3. No fractions to clear or integers to reduce.

4. ]101[


]011[

2-47

Conversion from 3-axis to 4-axis for Directions


Thomson Learning™

Direction C

Convert into 4-axis direction

[u’v’w’] [uvtw]

( )''23

vun

u −= ( )''23

uvn

v −=

'nww =( )vut +−=

]011[

2-48

Conversion from 3-axis to 4-axis for Directions


Thomson Learning™

Direction C

Convert into 4-axis direction

If n = 3, thenu = -2, v = -1, t = 1, w = 3

( ) )2(3

0123

''23

−−

=

−∗=−=

nnvu

nu

( )

=

−∗=−=

−−

13

1023

''23

nnuv

nv

'nww =

( ) 101 =

+−=+−=

−

vut

= 1312 ]011[

__

2 Learning Objectives Crystal Structuresnhayati/chapter 2 crystal structure.pdf · 2014-09-18 ·...

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Transcript of 2 Learning Objectives Crystal Structuresnhayati/chapter 2 crystal structure.pdf · 2014-09-18 ·...