4 – Crystal Structure and Defects in Metals

Outline

• Space Lattice & Unit Cell– BCC– FCC– HCP

• Miller Indices – Directions– Planes

• Defects – 0-D: solid solutions, vacancies– 1-D: dislocations– 2-D: Grain boundaries, twin boundaries

The Space Lattice and Unit Cells

• Atoms, arranged in repetitive 3-Dimensional pattern, in long range order (LRO) give rise to crystal structure.

• Properties of solids depends upon crystal structure and bonding force.

• An imaginary network of lines, with atoms at intersection of lines, representing the arrangement of atoms is called space lattice.

Unit Cell

Space Lattice

• Unit cell is that block of

atoms which repeats itself

to form space lattice.

3-2

• Materials arranged in shortrange order are calledamorphous materials

Crystal Systems and Bravais Lattice

• Only seven different types of unit cells are necessary to create all point lattices.

• According to Bravais (1811-1863) fourteen standard unit cells can describe all possible lattice networks.

• The four basic types of unit cells are Simple Body Centered Face Centered Base Centered

3-3

Types of Unit Cells

• Cubic Unit Cell a = b = c α = β = γ = 900

• Tetragonal a =b ≠ c α = β = γ = 900

Simple Body Centered

Face centered

Simple Body Centered

3-4 After W.G. Moffatt, G.W. Pearsall, & J. Wulff, “The Structure and Properties of Materials,” vol. I: “Structure,” Wiley, 1964, p.47.)

Types of Unit Cells (Cont..)

• Orthorhombic a ≠ b ≠ c α = β = γ = 900

• Rhombohedral a =b = c α = β = γ ≠ 900

Simple Base Centered

Face CenteredBody Centered

Simple

3-5 After W.G. Moffatt, G.W. Pearsall, & J. Wulff, “The Structure and Properties of Materials,” vol. I: “Structure,” Wiley, 1964, p.47.)

Figure 3.2

Types of Unit Cells (Cont..)

• Hexagonal a ≠ b ≠ c α = β = γ = 900

• Monoclinic a ≠ b ≠ c α = β = γ = 900

• Triclinic a ≠ b ≠ c α = β = γ = 900

Simple

Simple

Simple

BaseCentered

3-6 After W.G. Moffatt, G.W. Pearsall, & J. Wulff, “The Structure and Properties of Materials,” vol. I: “Structure,” Wiley, 1964, p.47.)

Figure 3.2

Principal Metallic Crystal Structures

• 90% of the metals have either Body Centered Cubic (BCC), Face Centered Cubic (FCC) or Hexagonal Close Packed (HCP) crystal structure.

• HCP is denser version of simple hexagonal crystal structure.

BCC Structure FCC Structure HCP Structure

3-7

Body Centered Cubic (BCC) Crystal Structure

• Represented as one atom at each corner of cube and one at the center of cube.

• Each atom has 8 nearest neighbors.

• Therefore, coordination number is 8.

• Examples :- Chromium (a=0.289 nm) Iron (a=0.287 nm) Sodium (a=0.429 nm)

3-8

BCC Crystal Structure (Cont..)

• Each unit cell has eight 1/8

atom at corners and 1

full atom at the center.

• Therefore each unit cell has

• Atoms contact each

other at cube diagonal

(8x1/8 ) + 1 = 2 atoms

3

4RTherefore, lattice constant a =

3-9

Atomic Packing Factor of BCC Structure

Atomic Packing Factor = Volume of atoms in unit cell

Volume of unit cell

Vatoms = = 8.373R3

3

3

4

R= 12.32 R3

Therefore APF = 8.723 R3

12.32 R3 = 0.68

V unit cell = a3 =

3

4.2

3R

3-10

Face Centered Cubic (FCC) Crystal Structure

• FCC structure is represented as one atom each at the corner of cube and at the center of each cube face.

• Coordination number for FCC structure is 12

• Atomic Packing Factor is 0.74

• Examples :- Aluminum (a = 0.405) Gold (a = 0.408)

3-11

FCC Crystal Structure (Cont..)

• Each unit cell has eight 1/8

atom at corners and six ½

atoms at the center of six

faces.

• Therefore each unit cell has

• Atoms contact each other

across cubic face diagonal

(8 x 1/8)+ (6 x ½) = 4 atoms

2

4RTherefore, lattice constant a =

3-12

Hexagonal Close-Packed Structure

• The HCP structure is represented as an atom at each of 12 corners of a hexagonal prism, 2 atoms at top and bottom face and 3 atoms in between top and bottom face.

• Atoms attain higher APF by attaining HCP structure than simple hexagonal structure.

• The coordination number is 12, APF = 0.74.

3-13 After F.M. Miller, “Chemistry: Structure and Dynamics,” McGraw-Hill, 1984, p.296

HCP Crystal Structure (Cont..)

• Each atom has six 1/6 atoms at each of top and bottom layer, two half atoms at top and bottom layer and 3 full atoms at the middle layer.

• Therefore each HCP unit cell has

• Examples:- Zinc (a = 0.2665 nm, c/a = 1.85) Cobalt (a = 0.2507 nm, c.a = 1.62)

• Ideal c/a ratio is 1.633.

(2 x 6 x 1/6) + (2 x ½) + 3 = 6 atoms

3-14 After F.M. Miller, “Chemistry: Structure and Dynamics,” McGraw-Hill, 1984, p.296

Atom Positions in Cubic Unit Cells

• Cartesian coordinate system is use 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.

• Atom positions are located using unit distances along the axes.

3-15

Directions in Cubic Unit Cells

• In cubic crystals, Direction Indices are vector components of directions resolved along each axes, resolved to smallest integers.

• Direction indices are position coordinates of unit cell where the direction vector emerges from cell surface, converted to integers.

3-16

Procedure to Find Direction Indices

(0,0,0)

(1,1/2,1)

zProduce the direction vector till it emerges from surface of cubic cell

Determine the coordinates of pointof emergence and origin

Subtract coordinates of point of Emergence by that of origin

(1,1/2,1) - (0,0,0) = (1,1/2,1)

All areintegers?

Convert them to smallest possible

integer by multiplying by an integer.

2 x (1,1/2,1) = (2,1,2)

Are any of the directionvectors negative?

Represent the indices in a square bracket without comas with a over negative index (Eg: [121])

Represent the indices in a square bracket without comas (Eg: [212] )

The direction indices are [212]

x

y

YES

NO

YESNO

3-17

Direction Indices - Example

• Determine direction indices of the given vector. Origin coordinates are (3/4 , 0 , 1/4). Emergence coordinates are (1/4, 1/2, 1/2). Subtracting origin coordinates from emergence coordinates, (1/4, 1/2, 1/2)-(3/4 , 0 , 1/4) = (-1/2, 1/2, 1/4) Multiply by 4 to convert all fractions to integers 4 x (-1/2, 1/2, 1/4) = (-2, 2, 1) Therefore, the direction indices are [ 2 2 1 ]

3-18

Miller Indices

• Miller Indices are are used to refer to specific lattice planes of atoms.

• They are reciprocals of the fractional intercepts (with fractions cleared) that the plane makes with the crystallographic x,y and z axes of three nonparallel edges of the cubic unit cell.

z

x

y

Miller Indices =(111)

3-19

Miller Indices - Procedure

Choose a plane that does not pass through origin

Determine the x,y and z intercepts

of the plane

Find the reciprocals of the intercepts

Fractions?Clear fractions by

multiplying by an integerto determine smallest set

of whole numbers

Enclose in parenthesis (hkl)where h,k,l are miller indicesof cubic crystal plane

forx,y and z axes. Eg: (111)

Place a ‘bar’ over theNegative indices

3-20

Miller Indices - Examples

• Intercepts of the plane at x,y & z axes are 1, ∞ and ∞

• Taking reciprocals we get (1,0,0).

• Miller indices are (100).

*******************

• Intercepts are 1/3, 2/3 & 1.

• taking reciprocals we get (3, 3/2, 1).

• Multiplying by 2 to clear fractions, we get (6,3,2).

• Miller indices are (632).

xx

y

z

(100)

3-21

Miller Indices - Examples

• Plot the plane (101)

Taking reciprocals of the indices we get (1 ∞ 1).

The intercepts of the plane are x=1, y= ∞ (parallel to y) and z=1.

******************************

• Plot the plane (2 2 1)

Taking reciprocals of the indices we get (1/2 1/2 1).

The intercepts of the plane are x=1/2, y= 1/2 and z=1.

3-22

Figure EP3.7 a

Miller Indices - Example

• Plot the plane (110) The reciprocals are (1,-1, ∞) The intercepts are x=1, y=-1 and z= ∞ (parallel to z

axis)

To show this plane a single unit cell, the origin is moved along the positive direction of y axis by 1 unit. x

y

z(110)

3-23

Miller Indices – Important Relationship

• Direction indices of a direction perpendicular to a crystal plane are same as miller indices of the plane.

• Example:-

• Interplanar spacing between parallel closest planes with same miller indices is given by

[110](110)

x

y

z

lkhd

ahkl 222

3-24

Comparison of FCC and HCP crystals

• Both FCC and HCP are close packed and have APF 0.74.

• FCC crystal is close packed in (111) plane while HCP is close packed in (0001) plane.

3-28 After W.G. Moffatt, G.W. Pearsall, & J. Wulff, “The Structure and Properties of Materials,” vol. I: “Structure,” Wiley, 1964, p.51.)

Structural Difference between HCP and FCC

Consider a layerof atoms (Plane ‘A’)

Another layer (plane ‘B’)of atoms is placed in ‘a’

Void of plane ‘A’

Third layer of Atoms placed in ‘b’ Voids of plane ‘B’. (Identical to plane ‘A’.) HCP crystal.

Third layer of Atoms placed in ‘a’ voids of plane ‘B’. ResultingIn 3rd Plane C. FCC crystal.

Plane A‘a’ void‘b’ void

Plane APlane B

‘a’ void‘b’ void

Plane APlane B

Plane A

Plane APlane B

Plane C

3-29

Figure 3.20

Volume Density

• Volume density of metal =

• Example:- Copper (FCC) has atomic mass of 63.54 g/mol and atomic radius of 0.1278 nm.

v Mass/Unit cellVolume/Unit cell

=

a=2

4R=

2

1278.04 nm= 0.361 nm

Volume of unit cell = V= a3 = (0.361nm)3 = 4.7 x 10-29 m3

v

FCC unit cell has 4 atoms.

Mass of unit cell = m =

g

Mg

molatmos

molgatoms 6

23

10

/10023.6

)/54.63)(4(= 4.22 x 10-28 Mg

33329

28

98.898.8107.4

1022.4

cm

g

m

Mg

m

Mg

V

m

3-30

Metallic Solid Solutions

• Alloys are used in most engineering applications.

• An Alloy is a mixture of two or more metals and nonmetals.

• Example: Cartridge brass is binary alloy of 70% Cu and 30% Zinc. Inconel is a nickel based superalloy with about 10 elements.

• Solid solution is a simple type of alloy in which elements are dispersed in a single phase.

4-14

Substitutional Solid Solution

• Solute atoms substitute for parent solvent atom in a crystal lattice.

• The structure remains unchanged.

• Lattice might get slightly distorted due to change in diameter of the atoms.

• Solute percentage in solvent

can vary from fraction of a

percentage to 100%

Solvent atoms

Solute atoms

4-15

Figure 4.14

Substitutional Solid Solution (Cont..)

• The solubility of solids is greater if The diameter of atoms does not differ by more than 15% Crystal structures are similar. Not much difference in electronegativity (or compounds

will be formed). Have same valence.

• Examples:-

System

Atomic radius

Difference

Electro-negativity difference

Solid

Solubility

Cu-Zn 3.9% 0.1 38.3%

Cu-Pb 36.7% 0.2 0.17%

Cu-Ni 2.3% 0 100%

4-16

Interstitial Solid Solution

• Solute atoms fit in between the voids (interstices) of solvent atoms.

• Solvent atoms in this case should be much larger than solute atoms.

• Example:- between 912 and 13940C, interstitial solid solution of carbon in γ iron (FCC) is formed.

• A maximum of 2.8%

of carbon can dissolve

interstitially in iron.

Carbon atoms r=0.075nm

Iron atoms r00.129nm

4-17

Crystalline Imperfections

• No crystal is perfect.

• Imperfections affect mechanical properties, chemical properties and electrical properties.

• Imperfections can be classified as Zero dimensional point defects. One dimensional / line defects

(dislocations). Two dimensional defects. Three dimensional defects (cracks).

4-18

Point Defects – Vacancy

• Vacancy is formed due to a missing atom.

• Vacancy is formed (one in 10000 atoms) during crystallization or mobility of atoms.

• Energy of formation is 1 eV.

• Mobility of vacancy results in cluster of vacancies.

• Also caused due

to plastic defor-

-mation, rapid

cooling or particle

bombardment.

Figure: Vacancies moving to form vacancy cluster

4-19

Point Defects - Interstitially

• Atom in a crystal, sometimes, occupies interstitial site.

• This does not occur naturally.

• Can be induced by irradiation.

• This defects caused structural distortion.

4-20

Figure 4.16b

Point Defects in Ionic Crystals

• Complex as electric neutrality has to be maintained.

• If two oppositely charged particles are missing, cation-anion di-vacancy is created. This is Scohttky defect.

• Frenkel defect is created when cation moves to interstitial site.

• Impurity atoms are

also considered as

point defects.

4-21

Yield Stress in Crystalline Materials

Initial Position

Saddle Point

G

a

a

2

1

2At Saddle

Point

For Mg Single Crystal predicts a yield stress of 17.2*0.5=8.6GPa

Measured yield stress is 0.7MPa

10,000 times less than expected!

Line Defects – (Dislocations)

• Discrepancy is due to dislocations.

• Dislocations are lattice distortions centered around a line.

• Formed during Solidification Permanent Deformation Vacancy condensation

• Different types of line defects are Edge dislocation Screw dislocation Mixed dislocation

4-22

Edge Dislocation

• Created by insertion of extra half planes of atoms.

• Positive edge dislocation

• Negative edge dislocation

• Burgers vector

Shows displa-

cement of

atoms (slip).

4-23

Burgers vector

After A.G. Guy , “Essentials of Materials Science,” McGraw-Hill, 1976, p.153After M. Eisenstadt, “Introduction to Mechanical Properties of Materials,” Macmillan, 1971, p.117

Screw Dislocation

• Created due to shear stresses applied to regions of a perfect crystal separated by cutting plane.

• Distortion of lattice in form of a spiral ramp.

• Burgers vector is parallel to dislocation line.

4-24 After M. Eisenstadt, “Introduction to Mechanical Properties of Materials,” Macmillan, 1971, p.118

Mixed Dislocation

• Most crystal have components

of both edge and screw

dislocation.

• Dislocation, since have

irregular atomic arrangement

will appear as dark lines

when observed in electron

microscope.

(After John Wolff et al., “Structure and Properties of Materials,” vol 3: “Mechanical Properties,” Wiley, 1965, p.65.(After “Metals Handbook” vol. 8, 8th ed., American Society of Metals, 1973, p.164)

Dislocation structure of iron deformed14% at –1950C

Transmission Electron Microscope

• Electron produced by heated tungsten filament.

• Accelerated by high voltage (75 - 120 KV)

• Electron beam passes through very thin specimen.

• Difference in atomic arrangement change directions of electrons.

• Beam is enlarged and focused on fluorescent screen.

Collagen Fibrilsof ligament asseen in TEM

(After L.E. Murr, “ Electron and Ion Microscopy and Microanalysis, “ Marcel Decker, 1982, p.105)

TEM (..Cont)

• TEM needs complex sample preparation• Very thin specimen needed ( several hundred

nanometers)• High resolution TEM (HRTEM) allows

resolution of 0.1 nm.• 2-D projections of a crystal with accompanying

defects can be observed. Low angle boundaryAs seenIn HTREM

The Scanning Electron Microscope

• Electron source generates electrons.

• Electrons hit the surface and secondary electrons are produced.

• The secondary electrons are collected to produce the signal.

• The signal

is used to

produce

the image.

SEM of fractured metal end

Figure 4.31

After V.A. Phillips, “Modern Photographic techniques and Their Applications,” Wiley, 1971, p.425

Grain Boundaries

• Grain boundaries separate grains.

• Formed due to simultaneously growing crystals meeting each other.

• Width = 2-5 atomic diameters.

• Some atoms in grain boundaries have higher energy.

• Restrict plastic flow and prevent dislocation movement.

4-27

3D view ofgrains

Grain BoundariesIn 1018 steel

(After A.G. Guy, “ Essentials of materials Science,” McGraw-Hill, 1976.)

Planar Defects

• Grain boundaries, twins, low/high angle boundaries, twists and stacking faults

• Free surface is also a defect : Bonded to atoms on only one side and hence has higher state of energy Highly reactive

• Nanomaterials have small clusters of atoms and hence are highly reactive.

Twin Boundaries

• Twin: A region in which mirror image pf structure exists across a boundary.

• Formed during plastic deformation and recrystallization.

• Strengthens the metal.

Twin

Twin Plane

Observing Grain Boundaries - Metallography

• To observe grain boundaries, the metal sample must be first mounted for easy handling

• Then the sample should be ground and polished with different grades of abrasive

paper and abrasive solution.• The surface is then etched chemically. • Tiny groves are produced at grain boundaries. • Groves do not intensely reflect light. Hence observed by optical microscope.

4-28 After M. Eisenstadt, “Introduction to Mechanical Properties of Materials,” Macmillan, 1971, p.126

Grain Size

• Affects the mechanical properties of the material

• The smaller the grain size, more are the grain boundaries.

• More grain boundaries means higher resistance to slip (plastic deformation occurs due to slip).

• More grains means more uniform the mechanical properties are.

4-30

Measuring Grain Size

• ASTM grain size number ‘n’ is a measure of grain size.

N = 2 n-1 N = Number of grains per

square inch of a polished

and etched specimen at 100 x.

n = ASTM grain size number.

200 X 200 X

1018 cold rolled steel, n=10 1045 cold rolled steel, n=8

4-31

N < 3 – Coarse grained4 < n < 6 – Medium grained7 < n < 9 – Fine grainedN > 10 – ultrafine grained