crystal structure - iowa state universitynano.engineering.iastate.edu/courses/mate271/week2.pdf ·...

Material Sciences and Engineering, MatE271 1

Material Sciences and Engineering MatE271 1

Crystal Structure

Ashraf Bastawros

www.public.iastate.edu\~bastaw\courses\Mate271.html

Week 2

Material Sciences and Engineering MatE271 Week 2 2

- Define basic terms in crystallography

- Be able to identify 7 crystal systems and 14Bravais lattices

- To compare and contrast the structures of metal, ceramics and polymer materials

- Explain the three most important structures for metals

- Calculate atomic structure densities

Goals for this unit



Crystal Structure and Periodicity

� Crystalline Materials atoms are in an ordered 3-D periodic arraySingle crystal or polycrystalline solids (metals, ceramics, semiconductors, some polymers )

� Amorphous Materialsshort range order

( glasses, many polymers)� Intermediates


Crystal Structure and PeriodicityLattice - a 3-D array of positions in space. A geometric framework on which to hang atoms

Unit Cell - the smallest repeat unit which defines the crystal structure

2-D 3-D

2-D

Note - In these cases, there is one atom centered on each lattice site



Simple Cubic (SC)

Lattice

Crystal Structure

Hang 1 atom on each lattice point

SC lattice and crystal structure

a = 2RWhere:

R = atomic radius atom a = lattice parameter


Axes Labels

Lattice constant- a, b and c are lengths of edges- α, β, γ are angles (α is across from a, etc.)

a

b

c

αβγ

By convention origin - 0,0,0 z

y

x



Crystal Systems (7-systems)

Tetragonala=b≠c

α=β=γ=90

Cubica=b=c

α=β=γ=90

Orthorhombica≠b≠c

α=β=γ=90


�Squished�tetragonal

�Pushed over�cube

Rhombohedrala=b=c

α=β=γ 90≠

Hexagonala=b≠c

α=β=90 γ=120




�Pushed over� orthorhombic(in one direction)

�Pushed over� orthorhombic(in two directions)


Orthorhombica≠b≠c

α=β=γ=90

Monoclinica≠b≠c

α=β=90, β ≠ 90

Triclinica≠b≠c

α ≠ β ≠ γ ≠ 90


Fourteen Crystal (Bravais) Lattices

- Lattices are merely geometric constructs forming a framework of points in space on (or around) which you can position atoms

- In the simplest case, you can apply one atom centered at each lattice point�Most METALLIC materials fall into this class

- Represent the entire lattice with one unit cell



Cubic Systems

- Based on a Cubic unit Cell

� Simple Cubic (SC)� One atom on each corner� Coordination number of 6

� Body-centered Cubic (BCC)� One atom on each corner and one in the center� Coordination number of 8

� Face-centered Cubic (FCC)� One atom on each corner and on each face� Coordination number of 12

a=b=c, α=β=γ=90°


Simple Cubic (SC)

Lattice

Crystal Structure

Hang 1 atom on each lattice point

- SC lattice and crystal structure

a = 2RWhere:

R = atomic radius atom a = lattice parameter



Body Centered Cubic (BCC)

- BCC lattice and crystal structure

a = 4R3

where:R = atomic radius atom a = lattice parameter

A

B

Staking order A-B-A-B


Cubic Packing - BCC

a

a

√2 a √2 a

a √3a

√3a=4Ra=4R/√3



Face Centered Cubic (FCC)

a = 2R 2

a = 4R2

where:R = atomic radius atom a = lattice parameter

A

B

Staking order A-B-A-B

Stacking Direction[100]


Cubic Packing - FCC

a

a

√2 a

4R=√2aa=2R√2



Atomic Packing Factor

- Fraction of solid sphere volume in a unit cell

APF = volume of atoms in unit celltotal cell volume

- Example for FCCHow many atoms are in the unit cell?What is the cell volume?


Atomic Packing Factor - FCC

- There are 4 spheres in the cell

- The volume of the spheres:

4 ×43

πR3 =163

πR3

4R




- What is the volume of the cube?� a3 (length of side

cubed)- What is that in terms

of R? (sphere radius)� a = 2R√2� (2R√2)3 = 16R3√2



APF = volume of atoms in unit celltotal cell volume

= Vs/Vc

163

πR 3

16R3 2 =

π3 2

= 0.74

The atomic packing factor for FCC is 0.74(74% of the space is filled)



HCP Crystal Structure

HCP crystal structure = HexagonalBravais lattice with 2 atoms per lattice site.

1st atom at 0,0,0 (i.e. lattice point) 2nd atom at 2/3, 1/3, 1/2

Note - 2nd atom environment different than the 1st atom

This is not a lattice point!

For �ideal� HCP onlyc = 1.633 a

a

c

0,0,0


HCP Crystal Structure

This sphere is not totally inside unit cell although the partial spheres all add up so that there are two spheres per unit cell!



ABA vs ABC Packing


FCC and HCP Compared



Structure of Selected Metals

Metal

Crystal Structure

Atomic Radius (nm)

Aluminum FCC 0.1431 Chromium BCC 0.1249

Cobalt HCP 0.1253 Copper FCC 0.1278

Gold FCC 0.1442 Lead FCC 0.1750


Example Problem

o If you know the crystal structure, the atomic radius and the atomic weight, you can calculate the density of a particular material.

o Copper has an atomic radius 0.128 nm an FCC crystal structure and an atomic weight of 63.5 g/mol. Calculate its density.



Crystalline & Noncrystalline Materials

o Single crystals� repeated arrangement of atoms extends

throughout the specimen� all unit cells have the same orientation� exist in nature� can also be grown (Si)

� without external constraints, will have flat, regular faces


Polycrystalline Materials

o Crystals of different� sizes� orientations� shapes

o Grain Boundaries� mismatch between

two neighboring crystals



Polycrystalline Materials

o Most crystalline materials are composed of many small crystals called grains

o Crystallographic directions of adjacent grains are usually random

o There is usually atomic mismatch where two grains meet - this is called a grain boundary

o Most powdered materials have a many randomly oriented grains


Reading Assignment

Shackelford 2001(5th Ed)

� Read Chapter 3, pp 59-64

Read ahead to page 88, 101-110

Check class web site:

www.public.iastate.edu\~bastaw\courses\Mate271.html 2

crystal structure - iowa state universitynano.engineering.iastate.edu/courses/mate271/week2.pdf ·...

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