crystal metallic structure

7/27/2019 Crystal Metallic Structure

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Fundamental Structure

of Crystalline Solids

2

Non dense, random packing

Dense, ordered packing

Dense, ordered packed structures tend to havelower energies.

Energy and

Packing

Energy

r

typical neighborbond length

typical neighborbond energy

Energy

r

typical neighborbond length

typical neighborbond energy


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3

atoms pack in periodic, 3D arraysCrystalline materials...

metalsmany ceramicssome polymers

atoms have no periodic packing

Noncrystalline materials...

complex structuresrapid cooling

crystalline SiO2

"Amorphous " = Noncrystalline

Materials and Packing

Si Oxygen

typical of:

occurs for:

noncrystalline SiO2

Crystal Terminalogy

Crystal : Regular, periodic and repeated arrangement

Lattice: Three dimensional array of points coinciding with atom positions.

Unit cell: smallest repeatable entity that can be used to

completely represent a crystal structure. It is the building block of crystal structure.


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5

7 crystal systems

14 Bravais /space lattices

A small group of lattice points which formsa repetitive pattern is called a unit cell .

a, b, and c are the lattice constants

Classification of Crystal Systems:

The angles ( ,, ) between the lattice translation vectors and thedistance(a,b,c) between two successive points along these vectorsare used to classify the crystal structures into 7 systems


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9

Rare

due

to

low

packing

density (only

Polonium

(Po)

has

this

structure)

Coordination # = 6(# nearest neighbors)

(Courtesy P.M. Anderson)

Simple Cubic Structure (SC)

10

APF for a simple cubic structure = 0.52

APF = a 3

4

3 (0.5 a ) 31

atoms

unit cellatom

volume

unit cell

volume

Atomic Packing

Factor

(APF)

APF = Volume of atoms in unit cell*

Volume of unit cell

*assume hard spheres

a

R=0.5 a

contains 8 x 1/8 =

1 atom/unit cellEffective no of atoms / UC


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11

Coordination # = 8


Atoms touch each other along cube diagonals.Note: All atoms are identical; the center atom is shadeddifferently only for ease of viewing.

Body Centered Cubic Structure (BCC)

ex: Cr, W, Fe ( ), Tantalum, Molybdenum

2 atoms/unit cell: 1 center + 8 corners x 1/8

12

Atomic Packing

Factor:

BCC

a

APF =

4

3 ( 3 a /4 ) 32

atoms

unit cell atom

volume

a 3

unit cell

volume

length = 4R =Close packed directions:

3 a

APF for a body centered cubic structure = 0.68

aR

a2

a3


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13

Coordination # = 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.

Face Centered Cubic Structure (FCC)

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

4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8

14

APF for a face centered cubic structure = 0.74Atomic Packing

Factor:

FCC

maximum achievable APF

APF =

4

3 ( 2 a /4 ) 34

atoms

unit cell atomvolume

a 3

unit cell

volume

Close packed directions:

length = 4R = 2 a

Unit cell contains:6 x1/2 + 8 x1/8

= 4 atoms/unit cella

2 a


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Coordination # = 12

APF = 0.74

3D Projection

2D Projection

Hexagonal Close Packed Structure

(HCP)

6 atoms/unit

cell

ex: Cd, Mg, Ti, Zn

c/ a = 1.633

c

a

A sites

B sites

A sites Bottom layer

Middle layer

Top layer

Possible Bravais lattices:

Cubic : Simple, BC, FCHexagonal : SimpleTetragonal : Simple, BCOrthorhombic : Simple, Base C, Body-C, FCMonoclinic : Simple, BCTriclinic, Trigonal : Simple

Crystal System Axial relationships Interaxial Angles

Cubic a=b=c == =90 o

tetragonal a=b c == =90 o

Hexagonal a=b c ==90 =120

Rhomohedral/trygonal

a=b=c == 90o

Orthorhombic a b c == =90 o

Monoclinic a b c = =90 o

triclinic a b c 90o


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Allotropy


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Example: Copper

= n AVc NA

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

Volume/unit cell(cm 3/unit cell)

Avogadro's number(6.023 x 10 23 atoms/mol)

Data from Table inside front page 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 -7 cm)

Vc = a3

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

cm3

Compare to actual: Cu = 8.94 g/cm 3Result: theoretical Cu = 8.89 g/cm 3

THEORETICAL DENSITY,

15

Element AluminumArgonBariumBerylliumBoronBromineCadmiumCalciumCarbonCesiumChlorineChromium

Cobalt CopperFlourineGalliumGermaniumGoldHeliumHydrogen

SymbolAlArBaBeBBrCdCaCCsClCr

CoCu FGaGeAuHeH

At. Weight (amu)26.9839.95137.339.01210.8179.90112.4140.0812.011132.9135.4552.00

58.9363.55 19.0069.7272.59196.974.0031.008

Atomic radius(nm)0.143------0.2170.114------------0.1490.1970.0710.265------0.125

0.1250.128 ------0.1220.1220.144------------

Density(g/cm 3 )2.71------3.51.852.34------8.651.552.251.87------7.19

8.98.94 ------5.905.3219.32------------

CrystalStructureFCC------BCCHCPRhomb------HCPFCCHexBCC------BCC

HCPFCC ------Ortho.Dia. cubicFCC------------

Characteristics of Selected Elements at 20C


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metals ceramics polymers

16

(

g / c m

3 )

Graphite/Ceramics/Semicond

Metals/Alloys Composites/ fibersPolymers

1

2

2 0

30B ased on data in Table B1, Callister

*GFRE, CFRE, & AFRE are Glass,Carbon, & Aramid Fiber-ReinforcedEpoxy composites (values based on

60% volume fraction of aligned fibersin an epoxy matrix).10

345

0.30.40.5

Magnesium

Aluminum

Steels

Titanium

Cu,Ni

Tin, Zinc

Silver, Mo

TantalumGold, WPlatinum

G raphiteSiliconGlass - sodaConcrete

Si nitrideDiamondAl oxide

Zirconia

H DPE, PSPP, LDPEPC

PTFE

PET

PVCSilicone

Wood

AFRE *

CFRE *GFRE*Glass fibers

Carbon fibers

A ramid fibers

DENSITIES OF MATERIAL CLASSES

a) crystal System? b) Crystal Structure? c) Density ? if Aw=141 g/mol

3.2

Tetragonal

BCT


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Crystallographic Points Coordinates

Fractional multiples of unit cell, edge lengths. Less than or equal 1 With out separation with commas (,)

24

Point CoordinatesPoint coordinates for unit cell

center area /2, b /2, c/2

Point coordinates for unit cell corner are 111

z

x

y

a b

c

000

111


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Crystallographic Directions1. Vector repositioned (if necessary) to pass

through origin.2. Read off projections in terms of

unit cell dimensions a , b , and c3. Adjust to smallest integer values4. Enclose in square brackets, no commas

[uvw ]

ex: 1 0 => 2 0 1 => [ 201 ]

-1 1 1

families of directions < uvw >

z

x

Algorithm

where overbar represents a negative index[ 111 ]=>

y


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3.14

A)

B)

C)

D)

How to draw the direction of [uvw]

1.Divide with the biggest integer among them.

2. Multiply with edge lengths.

3. Take corresponding length on ref. Axis/edges.

4. Farm a line by projection.


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3.9: Draw a [211] direction in Orthorhombic unit cell?

4. Move along the edges / axis

1 2 -1 1

2 2/2 -1/2 1/2

3 a - b/2 c/2

U V W

_

3.9: Draw a [211] direction in Orthorhombic unit cell?

4. Move along the edges / axis

1 2 -1 1

2 2/2 -1/2 1/2

3 a - b/2 c/2

U V W

_

a-b/2c/2


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HCP Crystallographic Directions

32

1. Vector repositioned (if necessary) to passthrough origin.

2. Read off projections in terms of unitcell dimensions a 1, a 2, a 3, or c

3. Adjust to smallest integer values4. Enclose in square brackets, no commas

[uvtw ]

[ 1120 ]ex: , , -1, 0 =>

dashed red lines indicateprojections onto a 1 and a 2 axes a 1

a 2

a 3

-a 32

a 2

2a1

-a 3

a 1

a 2

z Algorithm


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HCP Crystallographic Directions Hexagonal Crystals

4 parameter Miller Bravais lattice coordinates are related to the direction indices (i.e., u 'v 'w ') as follows.

==

=

'wwt

v

u

)vu( +-

)'u'v2(3

1-

)'v'u2(3

1-=

]uvtw[]'w'v'u[

-a 3

a 1

a 2

z

3.17


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Crystallographic Planes

Crystallographic planes are specified byMiller indices (hkl)

Indices are smallest Reciprocal of planeIntercepts with edges (Axis) of unit cell.

Any two planes parallel to each other areequivalent and have identical indices.

Procedure:1.Determine the intercepts made by the plane withthe three axes. If the plane passes through theorigin pick a parallel plane within the unit cell.2. Calculate the reciprocals of the intercepts.3. Divide by a common factor to convert tosmallest possible integers.4. Enclose in parentheses no

commas i.e., (hkl)


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4. Miller Indices (110)

1. Intercepts 1 1

2. Reciprocals 1/1 1/1 1/ 1 1 0

3. Reduction 1 1 0

example a b c


1. Intercepts 1 1 12. Reciprocals 1/1 1/1 1/1

1 1 13. Reduction 1 1 1

Example 2 a b c

z

x

ya b

c

38

z

x

ya b

c


example

1. Intercepts 1/2 1 3/4a b c

2. Reciprocals 1/ 1/1 1/

2 1 4/3

3. Reduction 6 3 4

(001)(010),

Family of Planes { hkl}Same atomic arrangement

( 100), (010),(001),Ex: {100} = (100),


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3.22: Determine miller indices for the planes?

2. Intercepts (in a,b,c) 1 1 -1

Plane A

3. Reciprocals 1 1 -1

Miller Indices (111) _

Plane B

1. Intercepts a b -c

h k l

3.19: a) Draw (021) plane for orthogonal unit cell? _

3 Intercept par b/2 -c

1. Indices 0 2 -1

2. Reciprocals 1/2 -1

h k l


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3.19: b) (200) monoclinic

3.21: for cubic unit cell

a) (101), b) (211) c) (012) d) (313) e) (111) f) (212) g) (312) h) (301) _ _ _ _ _ _ _


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4.3) c/a = 1.633in HCP

Atom at M is midway between the top andbottom so

Atom at M is midway between the top andbottom so


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(110) For FCC

Atomic Arrangements in planes

(110) For BCC

46

ex: linear density of FCC in [110]direction

Linear Density

a

[110]vector directionof length

vector directiononcentered atomsof number DensityLinear =

# atoms

length

12/a nm

a2

2LD

==


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[111] 2/ 3 a

[100] 1/a[110] 1/ 2 a

BCC

aR

a

2 a

FCC

[111] 1/ 3 a

[100] 1/a[110] 2/ a

Planar Densities

planeof area planeaoncentered atomsof number

DensityPlanar =

# atoms

length

22 /a 2 nm

a 22

PD==

(001) plane


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(111) 1/ 3a 2

(100) 1/a 2

(110) 2 /a 2

BCC

aR

a

2 a

FCC

(111) 4/ 3 a 2

(100) 2/a 2

(110) 2/ a 2

50

Planar Density of (100) IronSolution: At T < 912 C iron has the BCC structure.

Radius of iron R = 0.1241 nm

R3

4a =

2D repeat unit

=Planar Density =a 2

1

atoms2D repeat unit

=nm 2

atoms12.1

m2atoms

= 1.21 x 10 191

2

R34area

2D repeat unit

(100) 1/a 2

aR


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Linear Density of [111] Iron

[111] 2/ 3a

aR

Radius of iron R = 0.1241 nm

R3

4a =

2= =

nmatoms4.029

matoms4.029 x 10 9

R3linear Density =

atoms2D repeat unit

length2D repeat unit 3

4

a

2 a

[110] 2/ a

Highest planar & Linear density for FCC / CCP(Cubic Closed Pack)

(111) 4/ 3 a 2

Closed packed directions

Closed packed planes


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ABCABC... Stacking Sequence 2D Projection

A sites

B sitesC sites

B B

B

BB

B BC C

CA

A

FCC Unit Cell AB

C

FCC STACKING SEQUENCE

FCC Stacking sequence

A

B

C

A

B

C


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ABAB... Stacking Sequence

3D Projection 2D Projection

A sites

B sites

A sites

Bottom layer

Middle layer

Top layer

HEXAGONAL CLOSEPACKED STRUCTURE (HCP)


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A

B

A

B

A

A

HCP Stacking sequence


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3.27: fallowing are the planes

a) What is crystal system?

b) What is crystal lattice?

Tetragonal

BCT

BCT


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4.20: fallowing are the planes

a) What is crystal system?

b) What is crystal lattice?

Orthorambic

FCO


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c) Determine atomic wt if = 18.91 ?

= n AVc NA

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

Volume/unit cell(cm 3/unit cell)

Avogadro's number(6.023 x 10 23 atoms/mol)

4.10

Titanium is HCP crystal and density 4.51 gr / cm 3

1) Vol of unit cell in m 3 ?2) If c/ a ratio is 1.58 value of a, c ?


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4.12

Niobium atomic radius 0.143nm, density 8.57 g/ccDetermine whether it is a FCC or BCC

4.15


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Polymorphism Two or more distinct crystal structures for the

same material (allotropy/polymorphism)

Titanium , Ti

carbondiamond, graphite

BCC

FCC

BCC

1538C

1394C

912C

-Fe

-Fe

-Fe

liquid

iron system

"Existence of an elemental solid into more thanone physical forms is known as ALLOTROPY "

Existence of substance into more than onecrystalline forms is known as"POLYMORPHISM".

Example:1) Mercuric iodide (HgI 2 ) forms two types of crystals.

a . Orthorhombicb . Trigonal

2) Calcium carbonate (CaCO3) exists in two types of

crystalline forms.a. Orthorhombic (Aragonite)b. Trigonal


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Allotropy

So me engineering applications require single crystals:

Crystal properties reveal featuresof atomic structure.

--Ex: Certain crystal planes in quartzfracture more easily than others.

--diamond singlecrystals for abrasives

--turbine blades

CRYSTALS AS BUILDING BLOCKS


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Most engineering materials are polycrystals.

Nb-Hf-W plate with an electron beam weld. Each "grain" is a single crystal. If grains are randomly oriented,

overall component properties are not directional. Grain sizes typ. range from 1 nm to 2 cm

(i.e., from a few to millions of atomic layers).

1 mm

Polycrystals

Isotropic

Anisotropic


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Single Crystals

-Properties vary withdirection: anisotropic .

-Example: the modulusof elasticity (E) in BCC iron:

Polycrystals

-Properties may/may notvary with direction.

-If grains are randomlyoriented: isotropic .

(E poly iron = 210 GPa)-If grains are textured ,anisotropic.

200 m

Single vs PolycrystalsE (diagonal) = 273 GPa

E (edge) = 125 GPa

22

Demonstrates " polymorphism " The same atoms canhave more than onecrystal structure.

DEMO: HEATING ANDCOOLING OF AN IRON WIRE

Temperature, C

BCC Stable

FCC Stable

914

1391

1536

shorter

longer!shorter!

longer

Tc 768 magnet falls off

BCC Stable

Liquid

heat up

cool down


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Non crystalline solid

atoms have no periodic packing

Noncrystalline materials...

complex structuresrapid cooling

"Amorphous " = Noncrystalline = super cooled liquid

occurs for:

crystal metallic structure

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