5. Recommended
Books
1. Callister
W.D.,
Materials
Science
and
Engineering
an
Introduc&on
2. Askeland
D.R.,
The
Science
and
Engineering
of
Materials
3. Raghavan
V., Materials
Science
and
Engineering-‐
A
first
Course,
4. Avener
S.H,
IntroducBon
to
Physical
Metallurgy,
5
6. The
Structure
of
Crystalline
Solids
CRYSTALLINE
STATE
•
Most
solids
are
crystalline
with
their
atoms
arranged
in
a
regular
manner.
•
Long-‐range
order:
the
regularity
can
extend
throughout
the
crystal.
•
Short-‐range
order:
the
regularity
does
not
persist
over
appreciable
distances.
Ex.
amorphous
materials
such
as
glass
and
wax.
•
Liquids
have
short-‐range
order,
but
lack
long-‐range
order.
•
Gases
lack
both
long-‐range
and
short-‐range
order.
•
Some
of
the
properBes
of
crystalline
solids
depend
on
the
crystal
structure
of
the
material,
the
manner
in
which
atoms,
ions,
or
molecules
are
arranged.
6
7. Lace
•
SomeBmes
the
term
lace
is
used
in
the
context
of
crystal
structures;
in
this
sense
lace
means
a
three-‐
dimensional
array
of
points
coinciding
with
atom
posiBons
(or
sphere
centers).
A
point
la*ce
7
8. Unit
Cells
• The unit cell is the basic structural unit or building block of the crystal
structure and defines the crystal structure by virtue of its geometry and
the atom positions within.
• This size and shape of the unit cell can be described in terms of their
lengths (a,b,c) and the angles between then (α,β,γ). These lengths and
angles are the lattice constants or lattice parameters of the unit cell.
A
point
la*ce
A
unit
cell
8
9. Bravais
Lace
Table 1: Crystal systems and Bravais Lattices
Crystal systems and Bravais Lattice
9
10. Types of crystals
Three relatively simple crystal structures are found for most
of the common metals; body-centered cubic, face-centered
cubic, and hexagonal close-packed.
1. Body Centered Cubic Structure (BCC)
2. Face Centered Cubic Structure (FCC)
3. Hexagonal Close Packed (HCP)
10
11. 1. Body Centered Cubic Structure (BCC)
In these structures, there are 8 atoms at the 8 corners and
one atom in the interior, i.e. in the centre of the unit cell with
no atoms on the faces.
11
12. 2. Face Centered Cubic Structure (FCC)
In these structures, there are 8 atoms at the 8 corners,
6 atoms at the centers of 6 faces and no interior atom.
12
13. 3. Hexagonal Close Packed (HCP)
In these structures, there are 12 corner atoms (6 at the bottom
face and 6 at the top face), 2 atoms at the centers of the
above two faces and 3 atoms in the interior of the unit cell.
13
14. Average Number of Atoms per Unit Cell
Since the atoms in a unit cell are shared by the neighboring
cells it is important to know the average number of atoms per
unit cell. In cubic structures, the corner atoms are shared by 8
cells (4 from below and 4 from above), face atoms are shared
by adjacent two cells and atoms in the interior are shared by
only that one cell. Therefore, general we can write:
Nav = Nc / 8 + Nf / 2 + Ni / 1
Where,
Nav = average number of atoms per unit cell.
Nc = Total number of corner atoms in an unit cell.
Nf = Total number of face atoms in an unit cell.
Ni = Centre or interior atoms.
14
15. • Simple cubic (SC) structures: In these structures there are
8 atoms corresponding to 8 corners and there are no atoms
on the faces or in the interior of the unit cell. Therefore,
Nc = 8, Nf = 0 and Ni = 0
Using above eqn. we get, Nav = 8/8 + 0/2 + 0/1 = 1
15
16. 2. Body centered cubic (BCC) structures: In these
structures, there are 8 atoms at the 8 corners and one
atom in the interior, i.e. in the centre of the unit cell with
no atoms on the faces. Therefore Nc = 8, Nf = 0 and Ni = 1
Using above eqn. we get, Nav = 8/8 + 0/2 + 1/1 = 2
16
17. 3. Face Centered Cubic Structure (FCC): In these structures,
there are 8 atoms at the 8 corners, 6 atoms at the centers
of 6 faces and no interior atom
Therefore Nc = 8, Nf = 6 and Ni = 0
Using above eqn. we get, Nav = 8/8 + 6/2 + 0/1 = 4
17
18. 4. Hexagonal Close Packed (HCP) Structures:
In these structures, there are 12 corner atoms (6 at the bottom face and 6 at
the top face), 2 atoms at the centers of the above two faces and 3 atoms in
the interior of the unit cell.
For hexagonal structures, the corner atoms are shared by 6 cells (3 from
below and 3 from above), face atoms are shared by adjacent 2 cells and
atoms in the interior are shared by only one cell. Therefore, in general the
number of atoms per unit cell will be as: Nav = Nc / 6 + Nf / 2 + Ni / 1
Here Nc = 12, Nf = 2 and Ni = 3
Hence, Nav = 12 / 6 + 2 / 2 + 3 / 1 = 6
18
19. Co-‐ordina&on
Number
Co-‐ordinaBon
number
is
the
number
of
nearest
equidistant
neighboring
atoms
surrounding
an
atom
under
consideraBon
1.
Simple
Cubic
Structure:
Simple
cubic
structure
has
a
coordinaBon
number
of
6
19
20. 2.
Body
Centered
Cubic
Structure:
Body
centered
cubic
structure
has
a
coordinaBon
number
of
8
20
21. 3.
Face
Centered
Cubic
Structure:
Face
centered
cubic
structure
has
a
coordinaBon
number
of
12
21
22. 4.
Hexagonal
Close
Packed
Structure:
Hexagonal
close
packed
structure
has
a
coordinaBon
number
of
12
22
23. Stacking
Sequence
for
SC,
BCC,
FCC
and
HCP
•
Lace
structures
are
described
by
stacking
of
idenBcal
planes
of
atoms
one
over
the
other
in
a
definite
manner
•
Different
crystal
structures
exhibit
different
stacking
sequences
1. Stacking
Sequence
of
Simple
Cubic
Structure:
Stacking
sequence
of
simple
cubic
structure
is
AAAAA…..since
the
second
as
well
as
the
other
planes
are
stacked
in
a
similar
manner
as
the
first
i.e.
all
planes
are
stacked
in
the
same
manner.
A
A
A
23
24. 2.
Stacking
Sequence
of
Body
Centered
Cubic
Structure:
•
Stacking
sequence
of
body
centered
cubic
structure
is
ABABAB….
•
The
stacking
sequence
ABABAB
indicates
that
the
second
plane
is
stacked
in
a
different
manner
to
the
first.
•
Any
one
atom
from
the
second
plane
occupies
any
one
intersBBal
site
of
the
first
atom.
•
Third
plane
is
stacked
in
a
manner
idenBcal
to
the
first
and
fourth
plane
is
stacked
in
an
idenBcal
manner
to
the
second
and
so
on.
A
This
results
in
a
bcc
structure.
B
A
B
24
25. 3.
Stacking
Sequence
of
Face
Centered
Cubic
Structure:
•
Stacking
sequence
of
face
centered
cubic
structure
is
ABCABC….
•
The
close
packed
planes
are
inclined
at
an
angle
to
the
cube
faces
and
are
known
as
octahedral
planes
•
The
stacking
sequence
ABCABC…
indicates
that
the
second
plane
is
stacked
in
a
different
manner
to
the
first
and
so
is
the
third
from
the
second
and
the
first.
The
fourth
plane
is
stacked
in
a
similar
fashion
to
the
first
25
27. 4.
Stacking
Sequence
of
Hexagonal
Close
Packed
Structure:
•
Stacking
sequence
of
HCP
structure
is
ABABAB…..
•
HCP
structure
is
produced
by
stacking
sequence
of
the
type
ABABAB…..in
which
any
one
atom
from
the
second
plane
occupies
any
one
intersBBal
site
of
the
first
plane.
•
Third
plane
is
stacked
similar
to
first
and
fourth
similar
to
second
and
so
on.
27
28. Atomic Packing Factor (APF)
Atomic packing factor is the fraction of volume or
space occupied by atoms in an unit cell. Therefore,
APF = Volume of atoms in unit cell
Volume of the unit cell
Since volume of atoms in a unit cell = Average number
of atoms/cell x Volume of an atom
APF = Average number of atoms/cell x Volume of an atom
Volume of the unit cell
28
29. 1.
Simple
Cubic
Structures:
In
simple
cubic
structures,
the
atoms
are
assumed
to
be
placed
in
such
a
way
that
any
two
adjacent
atoms
touch
each
other.
If
a
is
the
lace
parameter
of
the
simple
cubic
structure
and
r
is
the
radius
of
atoms,
it
is
clear
from
the
fig
that:
r
=
a/2
APF = Average number of atoms/cell x Volume of an atom
Volume of the unit cell
= 1 x 4/3 π r3 = 4/3 π r3 = 0.52
a3 (2r)3
APF
of
simple
cubic
structure
is
0.52
or
52%
29
30. 2.
Body
Centered
Cubic
(BCC)
Structures:
In
body
centred
cubic
structures,
the
centre
atom
touches
the
corner
atoms
as
shown
in
fig.
If
a
is
the
lace
parameter
of
BCC
structure
and
r
is
the
radius
of
atoms,
we
can
write
(DF)2
=
(DG)2
+
(GF)2
Now
(DG)2
=
(DC)2
+
(CG)2
and
DF
=
4r
Therefore,
(DF)2
=
(DC)2
+
(CG)2
+
(GF)2
30
31.
(4r)2
=
a2
+
a2
+
a2
Therefore,
r
=
a√3
/
4
APF
=
Average
number
of
atoms/cell
x
Volume
of
an
atom
Volume
of
the
unit
cell
2
x
4/3
π
(a√3
/
4)3
=
0.68
a3
APF
of
body
centered
cubic
structure
is
0.68
or
68%
31
32. 3.
Face
Centered
Cubic
(FCC)
Structures:
In
face
centred
cubic
structures,
the
atoms
at
the
centre
of
faces
touch
the
corner
atoms
as
shown
in
figure.
If
a
is
the
lace
parameter
of
FCC
structure
and
r
is
the
atomic
radius
(DB)2
=
(DC)2
+
(CB)2
i.e.
(4r)2
=
a2
+
a2
Therefore,
r
=
a
/
2√2
APF
=
Average
number
of
atoms/cell
x
Volume
of
an
atom
Volume
of
the
unit
cell
=
4
x
4
/
3
x
π
(a/2√2)3
=
0.74
a3
APF
of
face
centered
cubic
structure
is
0.74
or
74%
32
33. 4.
Hexagonal
Close
Packed
(HCP)
Structures
The
volume
of
the
unit
cell
for
HCP
can
be
found
by
finding
out
the
area
of
the
basal
plane
and
then
mulBplying
this
by
its
height
This
area
is
six
Bmes
the
area
of
equilateral
triangle
ABC
Area
of
triangle
ABC
=
½
a2
sin
60
Total
area
ABDEFG
=
6
x
½
a2
sin
60
=
3
a2
sin
60
Now
volume
of
unit
cell
=
3
a2
sin
60
x
c
For
HCP
structures,
the
corner
atoms
are
touching
the
centre
atoms,
i.e.
atoms
at
ABDEFG
are
touching
the
C
atom.
Therefore
a
=
2r
or
r
=
a
/
2
33
34. APF
=
Average
number
of
atoms/cell
x
Volume
of
an
atom
Volume
of
the
unit
cell
APF
=
6
x
4π/3
r3
3
a2
sin
60
x
c
APF
=
6
x
4π/3
(a/2)3
3
a2
sin
60
x
c
APF
=
π
a
3
c
sin
60
The
c/a
raBo
for
an
ideal
HCP
structure
consisBng
of
uniform
spheres
packed
as
Bghtly
together
as
possible
is
1.633.
Therefore,
subsBtuBng
c/a
=
1.633
and
Sin
60o
=
0.866
in
above
equaBon
we
get:
APF
=
π
/
3
x
1.633
x
0.899
=
0.74
APF
of
face
centered
cubic
structure
is
0.74
or
74%
34
35. Atomic
Packing
Factor
1. Simple
cubic
structure:
0.52
2.
Body
centered
cubic
structure:
0.68
3.
Face
centered
cubic
structure:
0.74
4.
Hexagonal
close
packed
structure:
0.74
35
36. Crystallographic
Points,
Planes
and
DirecBons
1. Point
Coordinates
When
dealing
with
crystalline
materials
it
olen
becomes
necessary
to
specify
a
parBcular
point
within
a
unit
cell.
The
posiBon
of
any
point
located
within
a
unit
cell
may
be
specified
in
terms
of
its
coordinates
as
fracBonal
mulBples
of
the
unit
cell
edge
lengths.
36
38. 2.
Plane
Coordinates
1. Find
out
the
intercepts
made
by
the
plane
at
the
three
reference
axis
e.g.
p,q
and
r.
2. Convert
these
intercepts
to
fracBonal
intercepts
by
dividing
with
their
axial
lengths.
If
the
axial
length
is
a,
b
and
c
the
fracBonal
intercepts
will
be
p/a,
q/b
and
r/c.
3.
Find
the
reciprocals
of
the
fracBonal
intercepts.
In
the
above
case
a/p,
b/q
and
c/r.
4. Convert
these
reciprocals
to
the
minimum
of
whole
numbers
by
mulBplying
with
their
LCM.
5. Enclose
these
numbers
in
brackets
(parenthesis)
as
(hkl)
Note:
If
plane
passes
through
the
selected
origin,
either
another
parallel
plane
must
be
constructed
within
the
unit
cell
by
an
appropriate
translaBon
or
a
new
origin
must
be
established
at
the
corner
of
the
unit
cell.
38
39. 1.
Intercepts:
p,q
and
r.
2. FracBonal
intercepts:
p/a,
q/b
and
r/c.
3.
Reciprocals:
a/p,
b/q
and
c/r.
4. Convert
to
whole
numbers
5. Enclose
these
numbers
in
brackets
(parenthesis)
as
(hkl)
39
40. Step
1
:
IdenBfy
the
intercepts
on
the
x-‐
,
y-‐
and
z-‐
axes.
In
this
case
the
intercept
on
the
x-‐axis
is
at
x
=
1
(
at
the
point
(1,0,0)
),
but
the
surface
is
parallel
to
the
y-‐
and
z-‐axes
so
we
consider
the
intercept
to
be
at
infinity
(
∞
)
for
the
special
case
where
the
plane
is
parallel
to
an
axis.
The
intercepts
on
the
x-‐
,
y-‐
and
z-‐axes
are
thus
Intercepts
:
1
,
∞
,
∞
Step
2
:
Specify
the
intercepts
in
fracBonal
co-‐ordinates
Co-‐ordinates
are
converted
to
fracBonal
co-‐ordinates
by
dividing
by
the
respecBve
cell-‐dimension
-‐
This
gives
FracBonal
Intercepts
:
1/1
,
∞/1,
∞/1
i.e.
1
,
∞
,
∞
Step
3
:
Take
the
reciprocals
of
the
fracBonal
intercepts
This
final
manipulaBon
generates
the
Miller
Indices
which
(by
convenBon)
should
then
be
specified
without
being
separated
by
any
commas
or
other
symbols.
The
Miller
Indices
are
also
enclosed
within
standard
brackets
(….).
The
reciprocals
of
1
and
∞
are
1
and
0
respecBvely,
thus
yielding
Miller
Indices
:
(100)
So
the
surface/plane
illustrated
is
the
(100)
plane
of
the
cubic
crystal.
40
51. Miller
Indices
of
Planes
for
Hexagonal
Crystals
•
Crystal
Plane
in
HCP
unit
cells
is
commonly
idenBfied
by
using
four
indices
instead
of
three.
• The
HCP
crystal
plane
indices
called
Miller-‐Bravis
indices
are
denoted
by
the
lepers
h,
k,
i
and
l
are
enclosed
in
parentheses
as
(hkil)
• These
four
digit
hexagonal
indices
are
based
on
a
coordinate
system
with
four
axes.
• The
three
a1,
a2
and
a3
axes
are
all
contained
within
a
single
plane
(called
the
basal
plane),
and
at
1200
angles
to
one
another.
The
z-‐axis
is
perpendicular
to
the
basal
plane.
• The
unit
of
measurement
along
the
a1,
a2
and
a3
axes
is
the
distance
between
the
atoms
along
these
axes.
• The
unit
of
measurement
along
the
z-‐
axis
is
the
height
of
the
unit
cell.
•
The
reciprocals
of
the
intercepts
that
a
crystal
plane
makes
with
the
a1,
a2
and
a3
axes
give
the
h,
k
and
I
indices
while
the
reciprocal
of
the
intercept
with
the
z-‐axis
gives
the
index
l.
51
54. Miller Indices of Directions for Cubic Crystals
• A
vector
of
convenient
length
is
posiBoned
such
that
it
passes
through
the
origin
of
the
coordinate
system.
•
The
length
of
the
vector
projecBon
on
each
of
the
three
axes
is
determined.
•
These
three
numbers
are
mulBplied
or
divided
by
a
common
factor
to
reduce
them
to
the
smallest
integer
values.
•
The
three
indices,
not
separated
by
commas,
are
enclosed
in
square
brackets
[uvw]
•
If
a
negaBve
sign
is
obtained
represent
the
–ve
sign
with
a
bar
over
the
number
54
59. Family
of
Symmetry
Related
Planes
_
(
1
1
0
)
(1
1
0)
_
(
1
0
1
)
(
1
0
1
)
_
(
0
1
1
)
(
0
1
1
)
{
1
1
0
}
{
1
1
0
}
=
Plane
(
1
1
0
)
and
all
other
planes
related
by
symmetry
to
(
1
1
0
)
59
60. Family
of
Symmetry
Related
DirecBons
[
0
0
1
]
IdenBcal
atomic
density
IdenBcal
properBes
_
[
1
0
0
]
〈
1
0
0
〉
[
0
1
0
]
_
[
1
0
0
]
[
0
1
0
]
〈1
0
0〉=
[
1
0
0
]
and
all
other
z _
[
0
0
1
]
direcBons
related
to
[
1
0
0
]
by
symmetry
y
x 60
61. SUMMARY OF MEANINGS OF PARENTHESES
q r s represents a point
(hkl) represents a plane
{hkl} represents a family of planes
[hkl] represents a direction
<hkl> represents a family of directions
61
63. Anisotropy
of
crystals
(contd.)
Different
crystallographic
planes
have
different
atomic
density
And
hence
different
properBes
Si
Wafer
for
computers
63
64. Linear
and
Planar
DensiBes
Linear
Density
•
Linear
density
(LD)
is
defined
as
the
number
of
atoms
per
unit
length
whose
centers
lie
on
the
direcBon
vector
LD
=
number
of
atoms
centered
on
direcBon
vector
length
of
direcBon
vector
The
[110]
linear
density
for
FCC
is:
LD110
=
2
atoms/4R
=
1/2R
64
65. Planar
Density
•
Planar
density
(PD)
is
defined
as
the
number
of
atoms
per
unit
area
that
are
centered
on
a
parBcular
crystallographic
plane
•
PD
=
number
of
atoms
centered
on
a
plane
area
of
plane
Planar
density
on
(110)
plane
in
a
FCC
unit
cell
•
Number
of
atoms
on
(110)
plane
is
2
•
Area
of
(110)
plane
(rectangular
secBon)
is
4R
(length)
x
2√2R
(height)
=
8R2√2
PD
=
2
atoms
/
8R2√2
=
1
/
4R2√2
65
66. Planar
density
on
(100)
plane
in
a
Simple
Cubic
Structure:
•
Number
of
atoms
on
(100)
plane
is
1
•
Area
of
(100)
plane
(square
secBon)
is
a
x
a
=
a2
PD
=
1
atom
/
a2
=
=
1
/
a2
Planar
density
on
(110)
plane
in
a
Simple
Cubic
Structure:
•
Number
of
atoms
on
(110)
plane
is
1
•
Area
of
(110)
plane
(rectangular
secBon)
is
√2a2
PD
=
1
atom
/
√2
a2
=
=
1
/
√2
a2
66
67. Planar
density
on
(111)
plane
in
a
Simple
Cubic
Structure:
•
Number
of
atoms
on
(111)
plane
is
1/6
x
3
=
0.5
•
Area
of
(111)
plane
(triangle
DEF)
is
1/2
x
(√2a)
x
(0.866
x
√2a)
=
0.866a2
PD
=
0.5
atom
/
0.866a2
=
=
0.577
/
a2
Planar
density
on
(100)
plane
in
a
Body
Centred
Cubic
Structure:
•
Number
of
atoms
on
(100)
plane
is
1
•
Area
of
(100)
plane
(square
secBon)
is
a
x
a
=
a2
PD
=
1
atom
/
a2
=
1
/
a2
67
68. Planar
density
on
(110)
plane
in
a
Body
Centered
Cubic
Structure:
•
Number
of
atoms
on
(110)
plane
is
1/4
x
4
+
1
=
2
•
Area
of
(110)
plane
(rectangle
AFGD)
is
a
x
√2a
=
√2a2
PD
=
2
atoms
/
√2a2
=
=
√2
/
a2
=
1.414
/
a2
Planar
density
on
(111)
plane
in
a
Body
Centered
Cubic
Structure:
•
Number
of
atoms
on
(111)
plane
is
1/6
x
3
+
1
=
1.5
•
Area
of
(111)
plane
(triangle
DEG)
is
½
x
√2a
√2a
sin60o
=
0.866
a2
PD
=
1.5
atoms
/
0.866a2
=
=
1.732
/
a2
68
69. Voids
in
crystalline
structures
We have already seen that as spheres cannot fill entire space → the atomic
packing fraction (APF) < 1 (for all crystals)
This implies there are voids between the atoms. Lower the PF, larger the
volume occupied by voids.
These voids have complicated shapes; but we are mostly interested in the
largest sphere which can fit into these voids
The size and distribution of voids in materials play a role in determining
aspects of material behaviour → e.g. solubility of interstitials and their
diffusivity
The position of the voids of a particular type will be consistent with the
symmetry of the crystal
In the close packed crystals (FCC, HCP) there are two types of voids →
tetrahedral and octahedral voids (identical in both the structures as the voids
are formed between two layers of atoms)
The tetrahedral void has a coordination number of 4
The octahedral void has a coordination number 6
69
73. Voids:
Tetrahedral
and
Octahedral
Sites
•
Tetrahedral
and
octahedral
sites
in
a
close
packed
structure
can
be
occupied
by
other
atoms
or
ions
in
crystal
structures
of
alloys.
•
Thus,
recognizing
their
existence
and
their
geometrical
constrains
help
in
the
study
and
interpretaBon
of
crystal
chemistry.
•
The
packing
of
spheres
and
the
formaBon
of
tetrahedral
and
octahedral
sites
or
holes
are
shown
below.
73
75. What is the radius of the largest sphere that can be placed in a tetrahedral
void without pushing the spheres apart?
To solve a problem of this type, we need to construct a model for the analysis.
Use the diagram shown here as a starting point, and construct a tetrahedral
arrangement by placing four spheres of radius R at alternate corners of a cube.
• What is the length of the face diagonal fd of this cube in terms of R?
Since the spheres are in contact at the centre of each cube face, fd = 2 R.
• What is the length of the edge for such a cube, in terms of R?
Cube edge length a = √2 R
• What is the length of the body diagonal bd of the cube in R?
bd = √6 R
• Is the center of the cube also the center of the tetrahedral hole?
Yes
• Let the radius of the tetrahedral hole be r, express bd in terms
of R and r
If you put a small ball there, it will be in contact with all four spheres.
bd = 2 (R + r). r = (2.45 R) / 2 - R
= 1.225 R - R
= 0.225 R
• What is the radius ratio of tetrahedral holes to the spheres?
r / R = 0.225 75
76. Derive the relation between the radius (r) of the octahedral void and the
radius (R) of the atom in a close packed structure
(Assume largest sphere in an octahedral void without pushing the parent atom)
A sphere into the octahedral void is shown
in the diagram. A sphere above and a
sphere below this small sphere have not
been shown in the figure. ABC is a right
angled triangle. The centre of void is A.
Applying Pythagoras theorem.
BC2 = AB2 + AC2
(2R)2 + (R + r)2 + (R + r)2 = 2(R + r)2
4R2/2 = (R + r)2
2R2
=
(R
+
r)2
√2R
=
R
+
r
r
=
√2R
–
R
=
(1.414
–1)R
r
=
0.414
R
76
77. Single
Crystal
and
Polycrystalline
Stages
of
solidificaBon
of
a
polycrystalline
material
Single
Crystal
77
81. Ceramic
Crystal
Structures
•
Ceramics
are
compounds
between
metallic
&
nonmetallic
elements
e.x.
Al2O3,
FeO,
SiC,
TiN,
NaCl
•
They
are
hard
and
briple
•
Typically
insulaBve
to
the
passage
of
electricity
&
heat
Crystal
Structures
•
Atomic
bonding
is
mostly
ionic
i.e.
the
crystal
structure
is
composed
of
electrically
charged
ions
instead
of
atoms.
•
The
metallic
ions,
or
caBons
are
posiBvely
charged
because
they
have
given
up
their
valence
electrons
to
the
nonmetallic
Ions
or
anions,
which
are
negaBvely
charged
81
83. •
In
a
ceramic
material
two
characterisBcs
of
the
component
ions
influence
the
crystal
structure:
1. Charge
neutrality
2.
The
relaBve
sizes
of
the
caBons
and
anions
83
84.
1.
Charge
neutrality:
each
crystal
should
be
electrically
neutral
e.x.
NaCl
and
CaCl2
84
85. 2.
The
relaBve
sizes
of
the
caBons
and
anions
•
Because
the
metallic
elements
give
up
electrons
when
Ionized,
caBons
are
smaller
than
anions
•
Hence
rc
/
ra
is
less
than
unity
•
Stable
ceramic
crystal
structures
form
when
those
anions
surrounding
a
caBon
are
all
in
contact
with
the
that
caBon
85
86. •
CoordinaBon
number
is
related
to
the
caBon-‐anion
raBo
•
For
a
specific
coordinaBon
number
there
is
a
criBcal
or
minimum
rc
/
ra
raBo
86
91. AX-‐TYPE
STRUCTURES
•
Equal
number
of
caBons
and
anions
referred
to
as
AX
compounds
A
denotes
the
caBon
and
X
denotes
the
anion
rNa
=
0.102
nm
rCl
=
0.181
nm
r
Na
/
rCl
=
0.564
CaBons
prefer
octahedral
sites
Rock
Salt
Structure
91
92. AX-‐TYPE
STRUCTURES
conBnued…
MgO
also
has
a
NaCl
type
structure
rO
=
0.140
nm
rMg
=
0.072
nm
rMg
/
rO
=
0.514
CaBons
prefer
octahedral
sites
92
94. AmXp-‐TYPE
STRUCTURES
•
number
of
caBons
and
anions
are
different,
referred
to
as
AmXp
compounds
Calcium
Fluorite
Structure
94
95. AmBnXp-‐TYPE
STRUCTURES
•
Ceramic
compound
with
more
than
two
types
of
caBons,
referred
to
as
AmBnXp
compounds
95
96. Crystal
defects
(ImperfecBons
in
Solids)
• Perfect order does not exist throughout a crystalline material
on an atomic scale. All crystalline materials contain large
number of various defects or imperfections.
• Defects or imperfections influence properties such as
mechanical, electrical, magnetic, etc.
• Classification of crystalline defects is generally made
according to geometry or dimensionality of the defect
i.e. zero dimensional defects, one dimensional defects and
two dimensional defects.
96
97. Crystal defects / imperfections are broadly classified
into three classes:
1. Point
defect
(zero
dimensional
defects)
Vacancy,
Impurity
atoms
(
subs&tu&onal
and
inters&&al)
Frankel
and
Scho]ky
defect
2.
Line
defect
(one
dimensional
defects)
Edge
disloca&on
Screw
disloca&on,
Mixed
disloca&on
3.
Surface
defects
or
Planer
defects
(two
dimensional
defects)
Grain
boundaries
Twin
boundary
Stacking
faults
97
99. Vacancy
• If an atom is missing from its regular site, the defect produced
is called a vacancy
• All crystalline solids contain vacancies and their number
increases with temperature
• The equilibrium concentration of vacancies Nv for a given
quantity of material depends on & increases with temperature
according to
Where:
N is the total number of atomic sites
Qv is the energy required for the formation of a vacancy
T is the absolute temperature &
k is the gas or Boltzmann s constant i.e. 1.38 x 10-23 J/atom-K or
8.62 X 10-5 eV/atom-K
99
103. • Impurity point defects are of two types
1. Substitutional
2. Interstitial
• For substitutional, solute or impurity atoms replace or
substitute for the host atoms
• For interstitial, solute or impurity atoms fill the void or
interstitial space among the host atoms
• Both the substitutional and interstitial impurity atoms
distort the crystal lattice affecting the mechanical and
electrical / electronic properties
103
104. • Impurity atoms generate stress in the lattice by distorting the
lattice
• The stress is compressive in case of smaller substitutional
atom and tensile in case of larger substitutional atom
• These stresses act as barriers to movement of dislocations and
thus improve the strength / hardness of a material
• These stresses also act as barriers to the movement of
electrons and lower the electrical conductivity (increases
resistivity) of the material
104
106. • Frenkel and Schottky defects occur in ionic solids like ceramics
• An atom may leave its regular site and may occupy nearby
interstitial site of the matrix giving rise to two defects
simultaneously i.e. one vacancy and the other self interstitial.
These two defects together is called a Frenkel defect. This can
occur only for cations because of their smaller size as
compared to the size of anions.
• When cation vacancy is associated with an anion vacancy, the
defect is called Schottky defect.
Schottky defects are more
common in ionic solids because
the lattice has to maintain
electrical neutrality
106
107. 2. Line defects
DislocaBons
•
A
missing
line
or
row
of
atoms
in
a
regular
crystal
lace
is
called
a
dislocaBon
•
DislocaBon
is
a
boundary
between
the
slipped
region
and
the
unslipped
region
and
lies
in
the
slip
plane
•
Movement
of
dislocaBon
is
necessary
for
plasBc
deformaBon
•
There
are
mainly
two
types
of
dislocaBons
(a)
Edge
dislocaBons
and
(b)
Screw
dislocaBons
107
115. DislocaBons
as
seen
under
Transmission
Electron
Microscope
(TEM)
115
116. 3. Surface defects
Grain
Boundary
• Grain boundary is a defect which separates grains of different
orientation from each other in a polycrystalline material.
• When this orientation mismatch is slight, on the order of a few
degrees (< 15 degrees) then the term small- (or low- ) angle
grain boundary is used. When the same is more than 15
degrees its is know as a high angle grain boundary.
• The total interfacial energy is lower in large or coarse-grained
materials than in fine-grained ones, since there is less total
boundary area in the former.
• Mechanical properties of materials like hardness, strength,
ductility etc are influenced by the grain size.
• Grains grow at elevated temperatures to reduce the total
boundary energy.
116
118. Coarse and fine grain structure
Grain boundaries acting as barriers
to the movement of dislocations
Deformation of grains during 118
cold
working (cold rolling in this case)
119. Twin
Boundary
Twin
boundary
Atoms
on
one
side
of
the
boundary
are
located
in
Mirror
image
posiBons
of
the
atoms
on
the
other
side
119
120. A twin boundary is a special type of grain boundary across which there is
a specific mirror lattice symmetry; that is, atoms on one side of the
boundary are located in mirror-image positions of the atoms on the other
side.
The region of material between these boundaries is appropriately termed
a twin.
Twins result from atomic displacements that are produced from applied
mechanical shear forces (mechanical twins), and also during annealing
heat treatments following deformation (annealing twins).
Twinning occurs on a definite crystallographic plane and in
a specific direction, both of which depend on the crystal structure.
Annealing twins are typically found in metals that have the FCC crystal
structure, while mechanical twins are observed in BCC and HCP metals.
Twins contribute to plastic deformation in a small way
120
121. Stacking fault
•
Occurs
when
there
is
a
flaw
in
the
stacking
sequence
•
Stacking
fault
results
from
the
stacking
of
one
atomic
plane
out
of
sequence
on
another
and
the
lace
on
either
side
of
the
fault
is
perfect
•
BCC
and
HCP
stacking
sequence:
ABABABAB……
with
stacking
fault:
ABABBABAB……or
ABABAABABAB……..
•
FCC
stacking
sequence:
ABCABCABC….
with
stacking
fault:
ABCABCABABCABC……
Stacking
fault
FCC
Stacking 121
123. Principles
of
Alloy
FormaBon
Solid
Solu&on:
•
A
homogeneous
crystalline
phase
that
contains
two
or
more
chemical
species
•
It
is
an
alloy
in
which
the
atoms
of
solute
are
distributed
in
the
solvent
and
has
the
same
structure
as
that
of
the
solvent
Types
of
Solid
Solu&ons:
1.
IntersBBal
solid
soluBon,
ex.
Fe-‐C
IntersBBal
Solid
Soln
2.
SubsBtuBonal
solid
soluBon,
ex.
Au-‐Cu
123
SubsBtuBonal
Solid
Soln
124. 1.
Inters&&al
Solid
Solu&on
Alloys
•
Parent
metal
atoms
are
bigger
than
atoms
of
alloying
metal.
•
Smaller
atoms
fit
into
spaces,
(IntersBces),
between
larger
atoms.
124
129. 1. Atomic
size
factor:
If
the
atomic
sizes
of
solute
and
solvent
differ
by
less
than
15%,
it
is
said
to
have
a
favourable
size
factor
for
solid
soluBon
formaBon.
If
the
atomic
size
difference
exceeds
15%
solid
solubility
is
limited
2.
Crystal
Structure
factor:
Metals
having
same
crystal
structure
will
have
greater
solubility.
Difference
in
crystal
structure
limits
the
solid
solubility
+
A
(fcc)
B
(fcc)
AB
solid
solu&on
(fcc)
129
130. 3.
Electronega&vity
factor:
The
solute
and
solvent
should
have
similar
electronegaBvity.
If
the
electronegaBvity
difference
is
too
great,
the
metals
will
tend
to
form
compounds
instead
of
solid
soluBons.
If
electronegaBvity
difference
is
too
great
the
highly
electroposiBve
element
will
lose
electrons,
the
highly
electronegaBve
element
will
acquire
electrons,
and
compound
formaBon
will
take
place.
4.
Rela&ve
Valency
factor:
Complete
solubility
occurs
when
the
solvent
and
solute
have
the
same
valency.
If
there
is
shortage
of
electrons
between
the
atoms,
the
binding
between
them
will
be
upset,
resulBng
in
condiBons
unfavourable
for
solid
solubility
130
132. Phase
Diagrams
Phase
diagrams:
Phase
or
equilibrium
diagrams
are
diagrams
which
indicate
the
phases
exisBng
in
the
system
at
any
temperature,
pressure
and
composiBon.
Why
study
Phase
Diagrams?
•
Used
to
find
out
the
amount
of
phases
exisBng
in
a
given
alloy
with
their
composiBon
at
any
temperature.
•
From
the
amount
of
phases
it
is
possible
to
esBmate
the
approximate
properBes
of
the
alloy.
•
Useful
in
design
and
control
of
heat
treatment
procedures
132
133. Terms:
System:
A
system
is
that
part
of
the
universe
which
is
under
consideraBon.
Phase:
A
phase
is
a
physically
separable
part
of
the
system
with
disBnct
physical
and
chemical
properBes.
(In
a
system
consis6ng
of
ice
and
water
in
a
glass
jar,
the
ice
cubes
are
one
phase,
the
water
is
a
second
phase,
and
the
humid
air
over
the
water
is
a
third
phase.
The
glass
of
the
jar
is
another
separate
phase.)
Variable:
A
parBcular
phase
exists
under
various
condiBons
of
temperature,
pressure
and
concentraBon.
These
parameters
are
called
as
the
variables
of
the
phase
Component:
The
elements
present
in
the
system
are
called
as
components.
For
ex.
Ice,
water
or
steam
all
contain
H2O
so
the
number
of
components
is
2,
i.e.
H
and
O.
133
134. Gibb s
Phase
Rule:
The
Gibb s
phase
rule
states
that
under
equilibrium
condiBons,
the
following
relaBon
must
be
saBsfied:
P
+
F
=
C
+
2
Where,
P
=
number
of
phases
exisBng
in
a
system
under
consideraBon.
F
=
degree
of
freedom
i.e.
the
number
of
variables
such
as
temperature,
pressure
or
composiBon
(concentraBon)
that
can
be
changed
independently
without
changing
the
number
of
phases
exisBng
in
the
system.
C
=
number
of
components
(i.e.
elements)
in
the
system,
and
2
=
represents
any
two
variables
out
of
the
above
three
i.e.
temperature
pressure
and
composiBon.
134
135. Most
of
the
studies
are
done
at
constant
pressure
i.e.
one
atmospheric
pressure
and
hence
pressure
is
no
more
a
variable.
For
such
cases,
Gibb s
phase
rule
becomes:
P
+
F
=
C
+
1
In
the
above
rule,
1
represents
any
one
variable
out
of
the
remaining
two
i.e.
temperature
and
concentraBon.
Hence,
Degree
of
Freedom
(F)
is
given
by
F
=
C
–
P
+
1
135
136. ApplicaBon
of
Gibbs
Phase
Rule
•
C
At
point
A
P
=
1,
C
=
2
F
=
C
–
P
+
1
F
=
2
–
1
+1
F
=
2
The
meaning
of
F
=
2
is
that
both
temperature
and
concentraBon
can
be
varied
independently
without
changing
the
liquid
phase
exisBng
in
the
system
At
point
C
At
point
B
P
=
1,
C
=
2
P
=
2,
C
=
2
F
=
C
–
P
+
1
F
=
C
–
P
+
1
F
=
2
–
1
+1
F
=
2
–
2
+1
F
=
2
F
=
1
The
meaning
of
F
=
2
is
that
both
temperature
The
meaning
of
F
=
1
is
that
any
one
variable
and
concentraBon
can
be
varied
independently
out
of
temperature
and
composiBon
can
be
without
changing
the
liquid
phase
exisBng
in
changed
independently
without
altering
the
the
system
liquid
and
solid
phases
exisBng
in
the
system
136
138. 1. Unary
Phase
diagram
(one
component)
The
simplest
phase
diagrams
are
pressure-‐temperature
diagrams
of
a
single
simple
substance,
such
as
water.
The
axes
correspond
to
the
pressure
and
temperature.
138
139. 2.
Binary
Phase
diagram
(two
components)
•
A
phase
diagram
plot
of
temperature
against
the
relaBve
concentraBons
of
two
substances
in
a
binary
mixture
called
a
binary
phase
diagram
•
Types
of
binary
phase
diagrams:
1.
Isomorphous
2.
EutecBc
3.
ParBal
EutecBc
139
140. 3.
Ternary
Phase
diagram
(three
components)
•
A
ternary
phase
diagram
has
three
components.
•
It
is
three
dimensional
put
ploped
in
two
dimensions
at
constant
temperature
•
Stainless
steel
(Fe-‐Ni-‐Cr)
is
a
perfect
example
of
a
metal
alloy
that
is
represented
by
a
ternary
phase
diagram.
140
141. Binary
phase
diagram
The
binary
phase
diagram
represents
the
concentraBon
(composiBon)
along
the
x-‐axis
and
the
temperature
along
the
y-‐axis.
These
are
ploped
at
atmospheric
pressure
hence
pressure
is
constant
i.e.
1
atm.
pressure.
These
are
the
most
widely
used
phase
diagrams.
Types
of
binary
phase
diagrams:
• Binary
isomorphous
system:
Two
metals
having
complete
solubility
in
the
liquid
as
well
as
the
solid
state.
• Binary
eutecBc
system:
Two
metals
having
complete
solubility
in
the
liquid
state
and
complete
insolubility
in
the
solid
state.
• Binary
parBal
eutecBc
system:
Two
metals
having
complete
solubility
in
the
liquid
state
and
parBal
solubility
in
the
solid
state.
• Binary
layer
type
system:
Two
metals
having
complete
insolubility
in
the
liquid
as
well
as
in
the
solid
state.
141
146. Binary
isomorphous
system:
• These
phase
diagrams
are
of
loop
type
and
are
obtained
for
two
metals
having
complete
solubility
in
the
liquid
as
well
as
solid
state.
•
Ex.:
Cu-‐Ni,
Au-‐Ag,
Au-‐Cu,
Mo-‐W,
Mo-‐Ti,
W-‐V.
146
147. Lever
rule
Finding
the
amounts
of
phases
in
a
two
phase
region
:
1.
Locate
composiBon
and
temperature
in
phase
diagram
2.
In
two
phase
region
draw
the
Be
line
or
isotherm
3.
FracBon
of
a
phase
is
determined
by
taking
the
length
of
the
Be
line
to
the
phase
boundary
for
the
other
phase,
and
dividing
by
the
total
length
of
Be
line
147
148. %
of
Solid
=
LO
/
LS
X
100=
(Wo-‐Wi)
/
(Ws-‐Wi)
X
100
%
of
Liquid
=
OS
/
LS
X
100=
(Ws-‐Wi)
/
(Ws-‐Wi)
X
100
or
simply
%
Liquid
=
100
-‐
%
of
Solid
or
vice
versa
148
150. ProperBes
of
alloys
in
Isomorphous
systems
with
variaBon
in
composiBon
(a) Phase diagram of the Cu-Ni alloy system.
Above the liquidus line only the liquid phase
exists. In the L + S region, the liquid (L) and
solid (S) phases coexist whereas below the
solidus line, only the solid phase (a solid
solution) exists.
(b) The resistivity of the Cu-Ni alloy as a
Function of Ni content (at.%) at room
temperature
150
152. Binary
EutecBc
System:
These
diagrams
are
obtained
for
two
metals
having
complete
solubility
(i.e.
miscibility)
in
the
liquid
state
and
complete
insolubility
in
the
solid
state.
Examples:
Pb-‐As,
Bi-‐Cd,
Th-‐Ti,
and
Au-‐Si.
152
153. What
is
a
EutecBc?
•
A
eutec6c
or
eutec6c
mixture
is
a
mixture
of
two
or
more
phases
at
a
composiBon
that
has
the
lowest
melBng
point
•
EutecBc
ReacBon:
Liquid
↔
Solid
A
+
Solid
B
153
156.
Binary
Par&al
Eutec&c
System
These
diagrams
are
obtained
for
two
metals
having
complete
solubility
(i.e.
miscibility)
in
the
liquid
state
and
parBal
solubility
in
the
solid
state.
Examples:
Pb-‐Sn,
Ag-‐Cu,
Sn-‐Bi,
Pb-‐Sb,
Cd-‐Zn
and
Al-‐Si.
156
160. 3.
SolidificaBon
of
composiBons
that
range
between
the
room
temperature
solubility
limit
and
the
maximum
solid
solubility
at
the
eutecBc
temperature
160
161. Uses
of
Eutec&c
/
Par&al
Eutec&c
Alloys
Alloys
of
eutecBc
composiBons
have
some
specific
properBes
which
make
them
suitable
for
certain
applicaBons:
• Since
they
fuse
at
constant
temperature,
they
are
used
for
electrical
and
thermal
fuses.
• They
are
used
as
solders
due
to
their
lower
melBng
temperature.
• Since
eutecBc
alloys
have
low
melBng
points,
some
of
them
are
used
coaBngs
by
spraying
techniques
• Since
they
melt
at
constant
temperature
they
can
be
used
for
temperature
measurement.
•
Majority
of
the
eutecBc
alloys
are
superplasBc
in
character.
SuperplasBcity
is
the
phenomenon
by
which
an
alloy
exhibits
large
extension
(ducBlity)
when
deformed
with
certain
rate
at
some
temperature.
The
alloy
behaves
like
plasBc
and
can
be
formed
into
many
shapes.
161
162. The
Iron
–
Carbon
System
Allotrophic
TransformaBons
in
Iron
162
164. Phases
in
Iron-‐Carbon
Phase
Diagram
1.
Ferrite:
Solid
soluBon
of
carbon
in
bcc
iron
2.
Austenite:
Solid
soluBon
of
carbon
in
fcc
iron
3.
δ-‐iron:
Solid
soluBon
of
carbon
in
bcc
iron
4.
Cemen&te
(Fe3C):
Intermetallic
compound
of
iron
and
carbon
with
a
fixed
carbon
content
of
6.67%
by
wt.
5.
Pearlite:
It
is
a
two
phased
lamellar
(or
layered)
structure
composed
of
alternaBng
layers
of
ferrite
and
cemenBte
164
167. The
iron-‐carbon
system
exhibits
three
important
transformaBons
/
reacBons
as
described
below:
Eutectoid
Reac&on:
Solid1
↔
Solid2
+
Solid3
Austenite
↔
Ferrite
+
CemenBte
Eutec&c
Reac&on:
Liquid
↔
Solid1
+
Solid2
Liquid
↔
Austenite
+
CemenBte
Peritec&c
Reac&on:
Solid1
+
Liquid
↔
Solid2
δ-‐iron
+
Liquid
↔
Austenite
167
168. What
is
Pearlite?
Pearlite
is
a
two
phased
lamellar
(or
layered)
structure
composed
of
alternaBng
layers
of
ferrite
and
cemenBte
that
occurs
in
some
steels
and
cast
irons
100%
pearlite
is
formed
at
0.8%C
at
727oC
by
the
eutectoid
reacBon
/
PearliBc
transfromaBon
Eutectoid
Reac&on:
Solid1
↔
Solid2
+
Solid3
Austenite
↔
Ferrite
+
CemenBte
168
172. Non-‐Equilibrium
Cooling
•
Non-‐equilibrium
cooling
leads
to
shil
in
the
transformaBon
temperatures
that
appear
on
the
phase
diagram
•
Leads
to
development
of
non-‐equilibrium
phases
that
do
not
appear
on
the
phase
diagram
172
