Engineering Materials ME Chap 3 10th Ed Part (I) Aoh

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Engineering MaterialsChapter 3 (I)

The Structure of Crystalline Solids

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Prof. J.N. Aoh CCU ME 10.12.2021

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Crystalline Solids

• Periodic atom sites• Crystal directions

• Crystal planes

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SiO4 Crystal

Figun_03_p047a

Figun_03_p047b

Figun_03_p047c Figun_03_p047d

Fundamental of CrystallographyPart (I) 2021.10.12.

矽酸鹽類

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Crystal structures

• Crystal structure :the atoms are situated in a (1) repeating or periodic 3 D array over (2) large atomic distances, or “________________”.

• Crystalline amorphous• Single crystal Polycrystalline• Crystal lattice

• Unit cell

What is crystallography?

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Unit cell

The smallest portion of a crystal lattice that shows the three-dimensional pattern of the entire crystal.

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Crystallineamorphous• Amorphous: noncrystalline, “without form”• Amorphous metals“_____________”• Glass is noncrystallineamorphous.• occurs at extreme rapid cooling

Fig_03_11 SiO2

Metal Glass

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Unit cell & crystal systems

Fig_03_02

• The unit cell is the basic structural unit of the crystal structure and defines the crystal structure by its geometry and the atomic positions within.

• Lattice parameter a, b, c• Angles between axes: α, β, γ• Crystal system: axes, a, b, c and α, β, γ

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14 Bravais Crystal lattice

7 Crystal systems

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Metallic crystal structures

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Solidification and Crystallization3.14/3.15 Single vs Polycrystalline

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• Single Crystals-Properties vary with

direction: anisotropic.-Example: the modulus

of elasticity (E) in BCC iron:

• Polycrystalline-Properties may/may not

vary with direction.-If grains are randomly

oriented: isotropic.(Epoly iron = 210 GPa)

-If grains are textured,anisotropic.

200 μm

3.14/3.15 Single/Polycrystalline & AnisotropyE (diagonal) = 273 GPa

E (edge) = 125 GPa

Unit cell

Iron microstructure, magnification: x100


Elastic modulus along different orientationCrystallographic Directions

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Growth of silicon single crystal

Single crystal growthSi single crystal Silicon wafer 晶圓

Silicon chip晶元

1420°C

Homework I – Short report

Indicate crystallographic direction of the wafer

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3.4 Metallic crystal structures• Tend to be densely packed.

• Reasons for dense packing:- For element metals, all atoms have the same

radius.- Metallic bonding is not directional.- Nearest neighbor distances tend to be small in

order to lower bond energy.- Electron cloud shields cores from each other

• Have the simplest crystal structures.

We will examine three such structures...

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• Coordination number CN= 12

• Atoms touch each other along face diagonals.--Note: All atoms are identical; the face-centered atoms are shaded

differently only for ease of viewing.

3.4.1.Face Centered Cubic Structure (FCC)

ex: Al, Cu, Au, Pb, Ni, Pt, Ag

atoms per unit cell: 6 face x 1/2 + 8 corners x 1/8= _____

• Close packed directions are face diagonals.

Sphere model site model


Hard sphere model Atomic site model

•long range order”• long range repetition.

∞

∞

∞

3.4.1.Face Centered Cubic Structure (FCC)

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APF = a3

4

3π ( 2a/4)34

atoms

unit cell atomvolume

unit cell

volume

Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell

Close-packed directions: length = 4R

= 2 a

ATOMIC PACKING FACTOR (APF): FCC

a

Face diagonal 4R

• APF for a face-centered cubic structure = _______

a is lattice constant

a

APF = Volume of atoms in unit cell*

Volume of unit cell*assume hard spheres


Examples• Determine the FCC unit cell volume

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CN = 8

(Courtesy P.M. Anderson)

• Atoms touch each other along cube diagonals.--Note: All atoms are identical; the center atom is shaded

differently only for ease of viewing.

ex: Cr, W, Fe (α), Ta, Mo

atoms/unit cell: 1 center + 8 corners x 1/8 = ____coordination number CN = ______

3.4.2. Body Centered Cubic Structure (BCC)


aR

Close-packed directions: length = 4R

= 3 a

Unit cell contains: 1 + 8 x 1/8 = 2 atoms/unit cell

ATOMIC PACKING FACTOR: BCC

APF = a3

4

3π ( 3a/4)32

atoms

unit cell atomvolume

unit cell

volume

APF for a body-centered cubic structure = ______

body diagonals

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• Rare due to low packing density (only Polonium, Po 釙)• Close-packed directions are cube edges.

• Coordination number =CN= ____(No. of nearest neighbor atoms)

3.4.2. Simple Cubic Structure (SC)

Sphere model lattice model

or

Site model


• APF for a simple cubic structure = ____

APF = a3

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π (0.5a) 31atoms

unit cellatom

volume

unit cellvolume

Atomic Packing Factor for SC

APF = Volume of atoms in unit cell*

Volume of unit cell

.

close-packed directions

a

R=0.5a

contains 8 x 1/8 = 1 atom/unit cellLattice const.

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3.4.3. HEXAGONAL close-packed structure (HCP)

a and c are lattice constants in HCP

ex: Cd, Mg, Ti, Zn

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• =

HEXAGONAL CLOSE-PACKED STRUCTURE (HCP)

CN= 126 atoms/unit cell

APF= 6Vsphere/3Vparall = 0.74

Example 3.3. Determine unit cell volume HCP

36

22

33

)2

3(3A

2

2

cRV

Ra

caAcV

aa

parall

=

=

==

=

Figun_04_p080b

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c /a =1.633R

HCP c/a ratio

c

a

h=1/2 C

a


Example: Copper

ρ = n AVcNA

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

Volume/unit cell

(cm3/unit cell)Avogadro's number

(6.023 x 1023 atoms/mol)

Data from Table inside front cover of Callister:• crystal structure = FCC: 4 atoms/unit cell• atomic weight = 63.55 g/mol (1 amu = 1 g/mol) • atomic radius R = 0.128 nm (1 nm = 10 cm)-7

Vc = a3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x 10-23cm3

Compare to actual: ρCu = 8.94 g/cm3Result: theoretical ρCu = 8.89 g/cm3

3.5 THEORETICAL DENSITY, ρ

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• Ex: Cr (BCC) A = 52.00 g/molR = 0.125 nmn = 2

ρtheoretical

a = 4R/ 3 = 0.2887 nm

ρactual

aR

ρ = a3

52.002atoms

unit cell molg

unit cellvolume atoms

mol

6.023 x 1023

= 7.18 g/cm3

= 7.19 g/cm3

3.5 THEORETICAL DENSITY, ρ


3.12. Close-packed crystal structure--plane stacking sequence for HCP and FCC

A site

B site

C site

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• 3D Projection • 2D Projection

Hexagonal Close-Packed Structure (HCP)

c

a

A sites

B sites

A sitesBottom layer

Middle layer

Top layer


Stacking Sequence ABAB... observed from z-direction

Close-packed stacking sequence for HCP

A sites

B B

B

BB

B BC C

CA BB sites B B

B

BB

B BB sites

A

A A

AA

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Stacking Sequence ABCABC... Along body diagonal

• FCC Unit Cell

A sites

B B

B

BB

B B

C sites

C C

CA BB sites B B

B

BB

B BB sites

C C

CAC C

CA

AB

C

http://www.dawgsdk.org/crystal/index.en

Close-packed stacking sequence for FCC


Close-packed stacking sequence for FCC

AB

C

Body diagonal

Body diagonal

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Coordination Numbers for various radius ratiosusually refers to ceramic structures metallic ionscations, rc , nonmetallic ionsanions, rA

12.2 Ceramic crystal structure


Table 12. 2 Coordination Numbers andGeometries for Various Cation–AnionRadius Ratios (rC/rA)

(rC/rA)

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frRR

−+=

3

30cos

RrR f =+ − 30cos30cos 3

30cos30cos 3 frRR −=−30cos)30cos1( 3 frR −=−

155.030cos

30cos13 =−=−

Rr f

Ex.12.1 Coordination Numbers (CN)

r3-f R


• r4-f = 0.225R

Coordination Numbers (CN)

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12.2 Ceramic crystal structureClose-packed anion

Spheres with different radius, R and r

CN=4 CN=6

Tetrahedral and octahedral positions

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12.2 Rock salt structure

• Anion (green)• Cation (red)-octahedral positions

Fig_04_27

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• Polymorphism - metals and nonmetals having more than one crystal structure. Water, steel, etc.

• Allotropy - polymorphism in elemental solid, e.g. C, S,

3.6 Polymorphism and Allotropy

BCC

FCC

BCC

1538ºC

1394ºC

912ºC

δ-Fe

γ-Fe

α-Fe

liquidiron system


Tin and its allotropic transformation(Tin desease)

Increase in volume 27% and decrease in density from 7.3 to 5.77 g/ccm

Body Centered Tetragonal

Diamond cubic

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Exercise Crystal structure of Tin• The unit cell for tin has tetragonal symmetry, with a and b

lattice parameters of 0.583nm and 0.318nm, respectively. If its density=7.30 g/cm3, atomic weight=118.69 g/mol, and atomic radius=0.152 nm, compute the atomic packing factor.

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ρVC NA

A=n =

APF = = 0.544

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Engineering Materials ME Chap 3 10th Ed Part (I) Aoh