Crystal Metallic Structure
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Transcript of 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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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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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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Coordination # = 8
(Courtesy P.M. Anderson)
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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Coordination # = 12
(Courtesy P.M. Anderson)
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
4. Miller Indices (111)
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
4. Miller Indices (634)
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: