Transcript Foundations of Materials Science and Engineering Third Edition

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.
Space Lattice
• Unit cell is that block of
atoms which repeats itself
to form space lattice.
• Materials arranged in short
range order are called
amorphous materials
Unit Cell
3-2
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
Simple
Body Centered
Face centered
• Tetragonal
 a =b ≠ c
 α = β = γ = 900
Simple
3-4
Body Centered
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
Simple
Base Centered
Body Centered
Face Centered
• Rhombohedral
Figure 3.2
 a =b = c
 α = β = γ ≠ 900
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.)
Types of Unit Cells (Cont..)
•
Hexagonal
 a≠ b≠ c
 α = β = γ = 900
Simple
• Monoclinic
 a≠ b≠ c
 α = β = γ = 900
Base
Centered
Simple
• Triclinic
Figure 3.2
 a≠ b≠ c
 α = β = γ = 900
Simple
3-6
After W.G. Moffatt, G.W. Pearsall, & J. Wulff, “The Structure and Properties of Materials,” vol. I: “Structure,” Wiley, 1964, p.47.)
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
3-7
FCC Structure
HCP Structure
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
(8x1/8 ) + 1 = 2 atoms
• Atoms contact each
other at cube diagonal
Therefore, lattice
constant a =
4R
3
3-9
Atomic Packing Factor of BCC Structure
Atomic Packing Factor =
Vatoms =
 4 R 3 

2.


 3 
V unit cell = a3 =
Therefore APF =
3-10
Volume of atoms in unit cell
Volume of unit cell
= 8.373R3
 4R 




 3
3
= 12.32 R3
8.723 R3
12.32 R3 = 0.68
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
(8 x 1/8)+ (6 x ½) = 4 atoms
• Atoms contact each other
across cubic face diagonal
Therefore, lattice 4 R
constant a
3-12
=
2
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
(2 x 6 x 1/6) + (2 x ½) + 3 = 6 atoms
• Examples: Zinc (a = 0.2665 nm, c/a = 1.85)
 Cobalt (a = 0.2507 nm, c.a = 1.62)
•
3-14
Ideal c/a ratio is 1.633.
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
Produce the direction vector till it
emerges from surface of cubic cell
z
(1,1/2,1)
Determine the coordinates of point
of emergence and origin
Subtract coordinates of point of
Emergence by that of origin
NO
All are
integers?
YES
Are any of the direction
vectors negative?
YES
Represent the indices in a square
bracket without comas with a
over negative index (Eg: [121])
3-17
(1,1/2,1) - (0,0,0)
= (1,1/2,1)
y
(0,0,0)
x
2 x (1,1/2,1)
= (2,1,2)
The direction indices are [212]
Convert them to
smallest possible
integer by multiplying
by an integer.
NO
Represent the indices in a square
bracket without comas (Eg: [212] )
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
Miller Indices =(111)
y
x
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?
Place a ‘bar’ over the
Negative indices
Clear fractions by
multiplying by an integer
to 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)
3-20
Miller Indices - Examples
•
z
•
(100)
y
x
x
•
•
•
•
•
3-21
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).
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.
(110)
y
x
3-23
z
Miller Indices – Important Relationship
•
Direction indices of a direction perpendicular to a
crystal plane are same as miller indices of the plane.
z
• Example:-
y
(110)
[110]
x
•
Interplanar spacing between parallel closest planes
with same miller indices is given by
d
3-24
hkl
a

2
h
k l
2
2
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
Plane A
‘a’ void
‘b’ void
Consider a layer
of atoms (Plane ‘A’)
Another layer (plane ‘B’)
of atoms is placed in ‘a’
Void of plane ‘A’
Plane A
Plane B
‘a’ void
‘b’ void
Third layer of Atoms placed
Third layer of Atoms placed
in ‘b’ Voids of plane ‘B’. (Identical in ‘a’ voids of plane ‘B’. Resulting
to plane ‘A’.)
HCP crystal. In 3rd Plane C.
FCC crystal.
Plane A
Plane B
Plane A
Plane B
Plane A
Plane C
Figure 3.20
3-29
Volume Density
•
v
Volume density of metal =
=
Mass/Unit cell
Volume/Unit cell
• Example:- Copper (FCC) has atomic mass of 63.54
g/mol and atomic radius of 0.1278 nm.
a=
4R
=
4  0.1278nm
2
= 0.361 nm
2
Volume of unit cell = V= a3 = (0.361nm)3 = 4.7 x 10-29 m3
FCC unit cell has 4 atoms.
(4atom s)(63.54g / m ol)  106 Mg 

 = 4.22 x 10-28 Mg
Mass of unit cell = m =
23
6.02310 atm os/ m ol  g 
v

m
V
3-30

4.22  1028 Mg
4.7  1029 m 3
 8.98
Mg
m3
 8.98
g
cm 3
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.
•
4-14
Solid solution is a simple type of alloy in which elements
are dispersed in a single phase.
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
Figure 4.14
4-15
Solute atoms
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
4-16
Atomic
ElectroSolid
radius
negativity Solubility
Difference difference
Cu-Zn
3.9%
0.1
38.3%
Cu-Pb
36.7%
0.2
0.17%
Cu-Ni
2.3%
0
100%
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.
Iron atoms r00.129nm
Carbon atoms r=0.075nm
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.
Figure 4.16b
4-20
Point Defects in Ionic Crystals
•
•
Complex as electric neutrality has to be maintained.
If two oppositely charged particles are missing, cationanion 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
t
Initial Position
t
t
Saddle Point
t
At Saddle
Point
a 1


2a 2
t
G

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
•
4-23
Burgers vector
Shows displacement of
atoms (slip).
After M. Eisenstadt, “Introduction to Mechanical Properties of Materials,” Macmillan, 1971, p.117
After A.G. Guy , “Essentials of Materials Science,” McGraw-Hill, 1976, p.153
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.
Dislocation structure of iron deformed
14% at –1950C
(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)
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 Fibrils
of ligament as
seen 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
boundary
As seen
In 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
After V.A. Phillips, “Modern Photographic techniques and Their Applications,” Wiley, 1971, p.425
Figure 4.31
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.
3D view of
grains
Grain Boundaries
In 1018 steel
4-27
(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
Plane
Twin
Observing Grain Boundaries - Metallography
•
•
•
•
•
4-28
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.
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
N < 3 – Coarse grained
4 < n < 6 – Medium grained
7 < n < 9 – Fine grained
N > 10 – ultrafine grained
200 X
1018 cold rolled steel, n=10
4-31
square inch of a polished
and etched specimen at 100 x.
n = ASTM grain size number.
200 X
1045 cold rolled steel, n=8