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MT-201B MATERIALS SCIENCE
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Why Study Materials Science?
1. Application oriented Properties
2. Cost consideration
3. Processing route
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Classification of Materials
1. Metals
2. Ceramics
3. Polymers
4. Composites
5. Semiconductors6. Biomaterials
7. Nanomaterials
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1. Introduction to Crystallography
2. Principle of Alloy Formation
3. Binary Equilibria
4. Mechanical Properties
5. Heat Treatments
6. Engineering Materials
7. Advanced Materials
Syllabus
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Recommended Books
1. Callister W.D., Materials Science andEngineering an Introduction
2. Askeland D.R., The Science andEngineering of Materials
3. Raghavan V.,Materials Science and
Engineering- A first Course,4. Avener S.H, Introduction to Physical
Metallurgy,
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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 properties of crystalline solids depend on the
crystal structure of the material, the manner in which atoms,
ions, or molecules are arranged.
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Sometimes the term lattice is used in the context of crystalstructures; in this sense lattice means a three-
dimensional array of points coinciding with atom positions
(or sphere centers).
A point lattice
Lattice
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Unit Cells
The unit cell is the basic structural unit or building block of the crystalstructure and defines the crystal structure by virtue of its geometry andthe atom positions within.
A point lattice A unit cell
This size and shape of the unit cell can be described in terms of theirlengths (a,b,c) and the angles between then (,,). These lengths and
angles are the lattice constants or lattice parameters of the unit cell.
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Table 1: Crystal systems and Bravais Lattices
Crystal systems and Bravais Lattice
Bravais Lattice
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Types of crystals
Three relatively simple crystal structures are found for mostof the common metals; body-centered cubic, face-centeredcubic, and hexagonal close-packed.
1. Body Centered Cubic Structure (BCC)
2. Face Centered Cubic Structure (FCC)
3. Hexagonal Close Packed (HCP)
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1. Body Centered Cubic Structure (BCC)
In these structures, there are 8 atoms at the 8 corners andone atom in the interior, i.e. in the centre of the unit cell withno atoms on the faces.
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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.
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3. Hexagonal Close Packed (HCP)
In these structures, there are 12 corner atoms (6 at the bottomface and 6 at the top face), 2 atoms at the centers of theabove two faces and 3 atoms in the interior of the unit cell.
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Average Number of Atoms per Unit CellSince the atoms in a unit cell are shared by the neighboring
cells it is important to know the average number of atoms perunit cell. In cubic structures, the corner atoms are shared by 8cells (4 from below and 4 from above), face atoms are sharedby 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.
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Simple cubic (SC) structures: In these structures there are8 atoms corresponding to 8 corners and there are no atomson the faces or in the interior of the unit cell. Therefore,Nc = 8, Nf = 0 and Ni = 0Using above eqn. we get, Nav = 8/8 + 0/2 + 0/1 = 1
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2. Body centered cubic (BCC) structures: In thesestructures, there are 8 atoms at the 8 corners and one
atom in the interior, i.e. in the centre of the unit cell withno atoms on the faces. Therefore Nc = 8, Nf = 0 and Ni = 1Using above eqn. we get, Nav = 8/8 + 0/2 + 1/1 = 2
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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 atomTherefore Nc = 8, Nf = 6 and Ni = 0Using above eqn. we get, Nav = 8/8 + 6/2 + 0/1 = 4
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4. Hexagonal Close Packed (HCP) Structures:In these structures, there are 12 corner atoms (6 at the bottom face and 6 atthe 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 frombelow and 3 from above), face atoms are shared by adjacent 2 cells andatoms in the interior are shared by only one cell. Therefore, in general thenumber of atoms per unit cell will be as: Nav = Nc / 6 + Nf / 2 + Ni / 1
Here Nc = 12, Nf = 2 and Ni = 3Hence, Nav = 12 / 6 + 2 / 2 + 3 / 1 = 6
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Co-ordination Number
Co-ordination number is the number of nearest equidistant
neighboring atoms surrounding an atom under consideration
1. Simple Cubic Structure:
Simple cubic structure has a coordination number of6
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2. Body Centered Cubic Structure:
Body centered cubic structure
has a coordination number of8
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3. Face Centered Cubic Structure:
Face centered cubic structure has a coordination number of12
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4. Hexagonal Close Packed Structure:
Hexagonal close packed structure has a coordination number of12
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Stacking Sequence for SC, BCC, FCC and HCP
Lattice structures are described by stacking of identical 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 manneras the first i.e. all planes are stacked in the same manner.
A
A
A
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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 interstitial
site of the first atom. Third plane is stacked in a manner identical to the first and fourth
plane is stacked in an identical
manner to the second and so on.
This results in a bcc structure.
A
B
A
B
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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 facesand 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 similarfashion to the first
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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 interstitial site of the first plane.
Third plane is stacked similar to first and fourth similar to
second and so on.
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Atomic packing factor is the fraction of volume orspace occupied by atoms in an unit cell. Therefore,
APF = Volume of atoms in unit cellVolume of the unit cell
Atomic Packing Factor (APF)
APF = Average number of atoms/cell x Volume of an atomVolume of the unit cell
Since volume of atoms in a unit cell = Average number
of atoms/cell x Volume of an atom
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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 isthe lattice parameter of the simple cubic structure and r is theradius of atoms, it is clear from the fig that: r = a/2
APF = Average number of atoms/cell x Volume of an atomVolume of the unit cell
= 1 x 4/3 r3 = 4/3 r3 = 0.52a3 (2r)3
APF of simple cubic structure is 0.52 or 52%
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In body centred cubic structures, the centre atom touches the
corner atoms as shown in fig.
If a is the lattice 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
2. Body Centered Cubic (BCC) Structures:
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APF = Average number of atoms/cell x Volume of an atom
Volume of the unit cell
2 x 4/3 (a3 / 4)3 = 0.68a3
(4r)2 = a2 + a2 + a2
Therefore, r = a3 / 4
APF of body centered cubic structure is 0.68 or 68%
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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 lattice 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 / 22APF = Average number of atoms/cell x Volume of an atom
Volume of the unit cell
= 4 x 4 / 3 x (a/22)3 = 0.74
a3
APF of face centered cubic structure is 0.74 or 74%
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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 multiplying this by its height
This area is six times the area of
equilateral triangle ABC
Area of triangle ABC = a2
sin 60Total area ABDEFG = 6 x 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
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APF = Average number of atoms/cell x Volume of an atom
Volume of the unit cell
APF = 6 x 4/3 r33 a2 sin 60 x c
APF = 6 x 4/3 (a/2)3
3 a2 sin 60 x c
APF = a3 c sin 60
The c/a ratio for an ideal HCP structure consisting of uniform spheres packed as
tightly together as possible is 1.633.
Therefore, substituting c/a = 1.633 and Sin 60o = 0.866 in above equation we get:
APF = / 3 x 1.633 x 0.899 = 0.74
APF of face centered cubic structure is 0.74 or 74%
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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
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Crystallographic Points, Planes and Directions
1. Point Coordinates
When dealing with crystalline materials it often becomes necessary to
specify a particular point within a unit cell.
The position of any point located within a unit cell may be specified in
terms of its coordinates as fractional multiples of the unit cell edge lengths.
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2. Plane Coordinates
1. Find out the intercepts made by the plane at the threereference axis e.g. p,q and r.
2. Convert these intercepts to fractional intercepts by dividingwith their axial lengths. If the axial length is a, b and c thefractional intercepts will be p/a, q/b and r/c.
3. Find the reciprocals of the fractional intercepts. In the abovecase a/p, b/q and c/r.
4. Convert these reciprocals to the minimum of whole numbersby multiplying with their LCM.
5. Enclose these numbers in brackets (parenthesis) as (hkl)Note: If plane passes through the selected origin, either anotherparallel plane must be constructed within the unit cell by anappropriate translation or a new origin must be established at thecorner of the unit cell.
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1. Intercepts: p,q and r.
2. Fractional 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 inbrackets (parenthesis) as (hkl)
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Step 1 : Identify 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 fractional co-ordinatesCo-ordinates are converted to fractional co-ordinates by dividing by the respective
cell-dimension - This gives
Fractional Intercepts : 1/1 , /1, /1 i.e. 1 , ,
Step 3 : Take the reciprocals of the fractional intercepts
This final manipulation generates the Miller Indices which (by convention) shouldthen 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 respectively, thus yielding
Miller Indices : (100) So the surface/plane illustrated is the (100) plane of the
cubic crystal.
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Intercepts : 1 , 1,
Fractional intercepts : 1 , 1 ,
Reciprocal: 1,1,0
Miller Indices : (110)
Intercepts : 1 , 1 , 1
Fractional intercepts : 1 , 1 , 1Reciprocal: 1,1,1
Miller Indices : (111)
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Intercepts : , 1,
Fractional intercepts : , 1 ,
Reciprocal: 2,1,0
Miller Indices : (210)
Intercepts : 1/3 , 2/3 , 1
Fractional intercepts : 1/3 , 2/3 , 1Reciprocal: 3, 3/2, 1
Miller Indices : (632)
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Exercise
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Exercise
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Exercise
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Exercise
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Exercise
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http://en.wikipedia.org/wiki/File:Miller_Indices_Felix_Kling.svg -
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If the plane passes through the origin, the origin
has to be shifted for indexing the plane
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Miller Indices of Planes for Hexagonal Crystals
Crystal Plane in HCP unit cells is commonly identified by using four indices
instead of three.
The HCP crystal plane indices called Miller-Bravis indices are denoted by the
letters 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.
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Miller Indices of Directions for Cubic Crystals
A vector of convenient length is positioned such that it
passes through the origin of the coordinate system.
The length of the vector projection on each of the three axes
is determined.
These three numbers are multiplied 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 negative sign is obtained representtheve sign with a
bar over the number
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For direction not originating from origin the origin has to be shifted
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Examples of directions with shift of origin
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F il f S t R l t d Pl
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Family of Symmetry Related Planes
(1 1 0)_
( 1 1 0 )
( 1 0 1 )
( 0 1 1 )
_
( 0 1 1 )
( 1 0 1 )
_
{ 1 1 0 }
{ 1 1 0 } = Plane ( 1 1 0 ) and all other planes related by
symmetry to ( 1 1 0 )
Family of Symmetry Related Directions
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Family of Symmetry Related Directions
x
y
z[ 1 0 0 ]
[ 1 0 0 ]
_
[ 0 0 1 ]
[ 0 0 1 ]_
[ 0 1 0 ]
_[ 0 1 0 ]
Identical atomic density
Identical properties
1 0 0 1 0 0= [ 1 0 0 ] and all otherdirections related to [ 1 0 0 ]
by symmetry
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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
represents a family of directions
A i t f t l
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Anisotropy of crystals
66.7 GPa
130.3 GPa
191.1 GPa
Youngs modulus
of FCC Cu
A i t f t l ( td )
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Anisotropy of crystals (contd.)
Different crystallographicplanes have different
atomic density
And hence
different
properties
Si Wafer for
computers
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Linear and Planar Densities
Linear density (LD) is defined as the number of atoms per
unit length whose centers lie on the direction vector
LD = number of atoms centered on direction vector
length of direction vector
Linear Density
The [110] linear density forFCC is:
LD110 = 2 atoms/4R = 1/2R
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Planar Density
Planar density (PD) is defined as the number of atoms per
unit area that are centered on a particular 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 section) is4R (length) x 22R (height) = 8R22
PD = 2 atoms / 8R22 =
1 / 4R22
Planar density on (110) plane in a FCC unit cell
Number of atoms on (110) plane is 2
Area of (110) plane (rectangular section) is4R (length) x 22R (height) = 8R22
PD = 2 atoms / 8R22 =
1 / 4R22
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Planar density on (100) plane in a Simple Cubic
Structure:
Number of atoms on (100) plane is 1 Area of (100) plane (square section) 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
section) is 2a2
PD = 1 atom / 2 a2 =
= 1 / 2 a2
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Planar density on (111) plane in a
Simple Cubic Structure:
Number of atoms on (111) plane is1/6 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) planeis 1
Area of (100) plane (square
section) is a x a = a2
PD = 1 atom / a2 = 1 / a2
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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 x 2a
2a sin60o = 0.866 a2
PD = 1.5 atoms / 0.866a2 =
= 1.732 / a2
Voids in crystalline structures
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Voids in crystalline structures
We have already seen that as spheres cannot fill entire space the atomicpacking 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 determiningaspects of material behaviour e.g. solubility of interstitials and theirdiffusivity
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 voidstetrahedral 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
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Interstitial sites / voids
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Tetrahedral sites in HCP
Octahedral sites in HCP
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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 interpretation of crystal chemistry.
The packing of spheres and the formation of tetrahedral and
octahedral sites or holes are shown below.
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What is the radius of the largest sphere that can be placed in a tetrahedral
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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 tetrahedralarrangement by placing four spheres of radius Rat alternate corners of a cube.
What is the length of the face diagonal fdof 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 bdof 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 bdin termsof Rand rIf 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
Derive the relation between the radius (r) of the octahedral void and the
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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
radius (R) of the atom in a close packed structure
(Assume largest sphere in an octahedral void without pushing the parent atom)
2R2 = (R + r)2
2R = R + r
r = 2R R = (1.4141)Rr = 0.414 R
Si l C l d P l lli
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Single Crystal and Polycrystalline
Stages of solidification of a polycrystalline
material
Single Crystal
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silicon single crystal
Micrograph of a polycrystallinestainless steel showing grainsand grain boundaries
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P l hi
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Polymorphism
Ceramic Crystal Structures
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Ceramic Crystal Structures
Ceramics are compounds between metallic & nonmetallicelements e.x. Al2O3, FeO, SiC, TiN, NaCl
They are hard and brittle
Typically insulative 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 cations are positively charged becausethey have given up their valence electrons to the
nonmetallic Ions or anions, which are negatively charged
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Ionic bonding
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In a ceramic material two characteristics of the
component ions influence the crystal structure:
1. Charge neutrality
2. The relative sizes of the cations and anions
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1. Charge neutrality: each crystal should be
electrically neutral e.x. NaCl and CaCl2
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2. The relative sizes of the cations and anions
Because the metallic elements give up electrons when
Ionized, cations are
smaller than anions
Hence rc / ra is less than unity
Stable ceramic crystal structures form when those
anions surrounding a cation are all in contact with thethat cation
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Coordination number is related to the cation-anion ratio
For a specific coordination number there is a critical
or minimum rc / ra ratio
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Predicting Structure of FeO
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Predicting Structure of FeO
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AX-TYPE STRUCTURES
Equal number of cations and anions referred to asAX compounds
A denotes the cation and
X denotes the anion
rNa = 0.102 nm
rCl = 0.181 nm
rNa / rCl = 0.564
Cations prefer octahedral sites
Rock Salt Structure
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rO = 0.140 nm
rMg = 0.072 nm
rMg / rO = 0.514
Cations prefer octahedral sites
MgO also has a NaCl type structure
AX-TYPE STRUCTURES continued
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AX-TYPE STRUCTURES continued
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AmXp-TYPE STRUCTURES
number of cations and anions are different,
referred to as AmXp compounds
Calcium Fluorite Structure
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AmBnXp-TYPE STRUCTURES
Ceramic compound with more than two typesof cations, referred to as AmBnXp compounds
Crystal defects (I f ti i S lid )
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Crystal defects (Imperfections in Solids)
Perfect order does not exist throughout a crystalline materialon an atomic scale. All crystalline materials contain largenumber of various defects or imperfections.
Defects or imperfections influence properties such asmechanical, electrical, magnetic, etc.
Classification of crystalline defects is generally madeaccording to geometry or dimensionality of the defecti.e. zero dimensional defects, one dimensional defects andtwo dimensional defects.
Crystal defects / imperfections are broadly classified
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1. Point defect (zero dimensional defects)Vacancy,Impurity atoms ( substitutional and interstitial)
Frankel and Schottky defect
2. Line defect (one dimensional defects)
Edge dislocation
Screw dislocation,
Mixed dislocation
3. Surface defects or Planer defects (two dimensionaldefects)
Grain boundaries
Twin boundary
Stacking faults
y p yinto three classes:
Vacancy1. Point defects
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Vacancy
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If an atom is missing from its regular site, the defect producedis called a vacancy
All crystalline solids contain vacancies and their numberincreases with temperature
The equilibrium concentration of vacancies Nv for a given
quantity of material depends on & increases with temperatureaccording to
Where:N is the total number of atomic sites
Qv is the energy required for the formation of a vacancyT is the absolute temperature &k is the gas or Boltzmanns constant i.e. 1.38 x 10-23 J/atom-K or8.62 X 10-5 eV/atom-K
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Vacancies aid in the movement (diffusion) of atoms
Impurity atoms ( substitutional and interstitial)
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I it i t d f t f t t
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Impurity point defects are of two types1. Substitutional
2. Interstitial
For substitutional, solute or impurity atoms replace orsubstitute for the host atoms
For interstitial, solute or impurity atoms fill the void orinterstitial space among the host atoms
Both the substitutional and interstitial impurity atomsdistort the crystal lattice affecting the mechanical andelectrical / electronic properties
Impurity atoms generate stress in the lattice by distorting the
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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 (increasesresistivity) of the material
Frankel and Schottky defects
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Frankel and Schottky defects
Frenkel and Schottky defects occur in ionic solids like ceramics
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An atom may leave its regular site and may occupy nearbyinterstitial 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 canoccur only for cations because of their smaller size ascompared to the size of anions.
When cation vacancy is associated with an anion vacancy, thedefect is called Schottky defect.Schottky defects are more
common in ionic solids becausethe lattice has to maintainelectrical neutrality
Dislocations2. Line defects
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A missing line or row of atoms in a regular crystal
lattice is called a dislocation Dislocation is a boundary between the slipped region
and the unslipped region and lies in the slip plane
Movement of dislocation is necessary for plastic
deformation There are mainly two types of dislocations (a) Edge
dislocations and (b) Screw dislocations
Edge Dislocation
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Dislocation line and b are perpendicular to each other
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Movement of edge dislocation
Elastic stress field responsible for electron scattering and
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increase in electrical resistivity
lattice strain around
dislocation
Screw Dislocation
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Dislocation line and b areparallel to each other
Movement of Screw Dislocation
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When Dislocations Interact
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Mixed Dislocations
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By resolving, the contribution
from both types of
dislocations can be
determined
i l i
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Dislocations
as seen under
TransmissionElectron Microscope
(TEM)
3. Surface defectsGrain Boundary
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Grain boundary is a defect which separates grains of differentorientation from each other in a polycrystalline material.
When this orientation mismatch is slight, on the order of a fewdegrees (< 15degrees) then the term small- (or low- ) anglegrain boundary is used. When the same is more than 15degrees its is know as a high angle grain boundary.
The total interfacial energy is lower in large or coarse-grainedmaterials than in fine-grained ones, since there is less totalboundary 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 totalboundary energy.
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Coarse and fine grain structure
Grain boundaries acting as barriers
to the movement of dislocations
Deformation of grains during cold
working (cold rolling in this case)
Twin Boundary
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Twin boundary
Atoms on one side of the boundary are located inMirror image positions of the atoms on the other side
A twin boundary is a special type of grain boundary across which there is
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a specificmirror lattice symmetry; that is, atoms on one side of theboundary are located in mirror-image positions of the atoms on the other
side.
The region of material between these boundaries is appropriately termeda twin.
Twins result fromatomic displacements that are produced from appliedmechanical shear forces (mechanical twins), and also during annealingheat treatments following deformation (annealing twins).
Twinning occurs on a definite crystallographic plane and ina specific direction, both of which depend on the crystal structure.
Annealing twins are typically found in metals that have the FCC crystalstructure, while mechanical twins are observed in BCC and HCP metals.
Twins contribute to plastic deformation in a small way
Stacking fault Occurs when there is a flaw in the stacking sequence
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g q
Stacking fault results from the stacking of one atomic plane out of
sequence on another and the lattice on either side of the fault is
perfect
BCC and HCP stacking sequence: ABABABAB
with stacking fault: ABABBABABor ABABAABABAB..
FCC stacking sequence: ABCABCABC.
with stacking fault: ABCABCABABCABC
Stacking fault
FCCStacking
Plastic Deformation
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Principles of Alloy Formation
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Solid Solution:
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 thesolvent
Types of Solid Solutions:
1. Interstitial solid solution, ex. Fe-C2. Substitutional solid solution, ex. Au-Cu
Interstitial Solid Soln
Substitutional Solid Soln
1. Interstitial Solid Solution Alloys
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Parent metal atoms are bigger than atoms of alloying metal.
Smaller atoms fit into spaces, (Interstices), between larger
atoms.
Interstitial sites
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2. Substitutional Solid Solution Alloys
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y Atoms of both metals are of almost similar size.
Direct substitution takes place.
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Alloy Unit Cell Structure
Copper - Nickel FCC
Copper - Gold FCCGold - Silver FCC
Nickel - Platinum FCC
Molybdenum - Tungsten BCC
Iron - Chromium BCC
Some Solid Solution Alloys
Hume-Rotherys Rules of Solid Solubility
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Hume Rothery s Rules of Solid Solubility
1. Atomic size factor
2. Crystal structure factor
3. Electronegativity factor
4. Relative valency factor
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
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y ,
factor for solid solution formation. 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 solution (fcc)
3. Electronegativity factor:
h l d l h ld h l l f
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The solute and solvent should have similar electronegativity. If
the electronegativity difference is too great, the metals will tend
to form compounds instead of solid solutions.
If electronegativity difference is too great the highly electropositive
element will lose electrons, the highly electronegative element will
acquire electrons, and compound formation will take place.
4. Relative 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, resulting in conditions unfavourable for
solid solubility
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Phase Diagrams
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Phase diagrams:Phase or equilibrium diagrams are diagrams which indicate the
phases existing in the system at any temperature, pressure and
composition.
Why study Phase Diagrams?
Used to find out the amount of phases existing in a given alloy
with their composition at any temperature.
From the amount of phases it is possible to estimate the
approximate properties of the alloy.
Useful in design and control of heat treatment procedures
Terms:
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Terms:
System: A system is that part of the universe which is under
consideration.Phase: A phase is a physically separable part of the system
with distinct physical and chemical properties. (In a system
consisting 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 particular phase exists under various conditions
of temperature, pressure and concentration. These
parameters are called as the variables of the phaseComponent: 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.
Gibbs Phase Rule:
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Gibb s Phase Rule:
The Gibbs phase rule states that under equilibrium conditions,
the following relation must be satisfied:P + F = C + 2
Where,
P = number of phases existing in a system under consideration.
F = degree of freedom i.e. the number of variables such as
temperature, pressure or composition (concentration) that can
be changed independently without changing the number of
phases existing 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 composition.
Most of the studies are done at constant pressure i.e. one
atmospheric pressure and hence pressure is no more a variable
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atmospheric pressure and hence pressure is no more a variable.
For such cases, Gibbs 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 concentration.
Hence, Degree of Freedom (F) is given by
F = C P + 1
Application of Gibbs Phase Rule
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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 concentration can be varied independently
without changing the liquid phase existing in
the system
At point B
P = 2, C = 2
F = C P + 1
F = 2 2 +1
F = 1
The meaning of F = 1 is that any one variable
out of temperature and composition can be
changed independently without altering the
liquid and solid phases existing in the system
C
At point C
P = 1, C = 2
F = C P + 1
F = 2 1 +1F = 2
The meaning of F = 2 is that both temperature
and concentration can be varied independently
without changing the liquid phase existing in
the system
Types of Phase Diagrams:
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Unary phase diagram
Binary phase diagram
Ternary phase diagram
Types of Phase Diagrams:
1. Unary Phase diagram (one component)
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The simplest phase diagrams are pressure-temperature
diagrams of a single simple substance, such as water. Theaxes correspond to the pressure and temperature.
2 Binary Phase diagram (two components)
http://en.wikipedia.org/wiki/Water_(molecule)http://en.wikipedia.org/wiki/Cartesian_coordinate_systemhttp://en.wikipedia.org/wiki/Pressurehttp://en.wikipedia.org/wiki/Temperaturehttp://en.wikipedia.org/wiki/Temperaturehttp://en.wikipedia.org/wiki/Pressurehttp://en.wikipedia.org/wiki/Cartesian_coordinate_systemhttp://en.wikipedia.org/wiki/Water_(molecule) -
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2. Binary Phase diagram (two components)
Aphase diagram plot of temperature against therelative concentrations of two substances in a binary
mixture called a binary phase diagram
Types of binary phase diagrams:
1. Isomorphous
2. Eutectic
3. Partial Eutectic
3. Ternary Phase diagram (three components)
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A ternary phase diagram has three components.
It is three dimensional put plotted in two dimensions atconstant temperature
Stainless steel (Fe-Ni-Cr) is a perfect example of a metal alloy
that is represented by a ternary phase diagram.
Binary phase diagram
The binary phase diagram represents the concentration (composition)
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The binary phase diagram represents the concentration (composition)
along the x-axis and the temperature along the y-axis. These are
plotted 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 inthe liquid as well as the solid state.
Binary eutectic system: Two metals having complete solubility in the
liquid state and complete insolubility in the solid state.
Binary partial eutectic system: Two metals having complete solubility
in the liquid state and partial solubility in the solid state.
Binary layer type system: Two metals having complete insolubility in
the liquid as well as in the solid state.
Cooling curve for Pure Metal (one component)
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Cooling curve for an alloy / solid solution
(two components)
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(two components)
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Plotting of Phase Diagrams
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These phase diagrams are of loop type and are obtained for
Binary isomorphous system:
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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.
Fi di h f h i h i
Lever rule
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Finding the amounts of phases in a two phase region :
1. Locate composition and temperature in phase diagram
2. In two phase region draw the tie line or isotherm
3. Fraction of a phase is determined by taking the length of the
tie line to the phase boundary for the other phase, and dividing
by the total length of tie line
% of Solid = LO / LS X 100= (Wo-Wi) / (Ws-Wi) X 100
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% of Liquid = OS / LS X 100= (Ws-Wi) / (Ws-Wi) X 100
or simply % Liquid = 100 - % of Solid or vice versa
Development of Microstructure during slow cooling in
isomorphous alloys
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Properties of alloys in Isomorphous systems
with variation in composition
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(a) Phase diagram of the Cu-Ni alloy system.Above the liquidus line only the liquid phase
exists. In theL + 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
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These diagrams are obtained for two metals having complete
Binary Eutectic System:
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g g p
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.
What is a Eutectic?
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A eutectic or eutectic mixture is a mixture of two or more phases
at a composition that has the lowest melting point Eutectic Reaction:
Liquid Solid A + Solid B
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Cooling Curves in Eutectic System
Plotting of Eutectic Phase Diagrams
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These diagrams are obtained for two metals having complete
Binary Partial Eutectic System
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solubility (i.e. miscibility) in the liquid state and partial solubility
in the solid state.Examples: Pb-Sn, Ag-Cu, Sn-Bi, Pb-Sb, Cd-Zn and Al-Si.
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Development of microstructure in binary partial eutectic alloys
during equilibrium cooling
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1. Solidification of the eutectic composition
2. Solidification of the off - eutectic composition
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3. Solidification of compositions that range between the room
temperature solubility limit and the maximum solid solubility at
th t ti t t
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the eutectic temperature
Uses of Eutectic / Partial Eutectic Alloys
Alloys of eutectic compositions have some specific properties
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Alloys of eutectic compositions have some specific properties
which make them suitable for certain applications:
Since they fuse at constant temperature, they are used for
electrical and thermal fuses.
They are used as solders due to their lower melting temperature.
Since eutectic alloys have low melting points, some of them are
used coatings by spraying techniquesSince they melt at constant temperature they can be used for
temperature measurement.
Majority of the eutectic alloys are superplastic in character.
Superplasticity is the phenomenon by which an alloy exhibits large
extension (ductility) when deformed with certain rate at some
temperature. The alloy behaves like plastic and can be formed into
many shapes.
The Iron Carbon System
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Allotrophic Transformations in Iron
Iron Carbon Phase Diagram
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Phases in Iron-Carbon Phase Diagram
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1. Ferrite: Solid solution of carbon in bcc iron
2. Austenite: Solid solution of carbon in fcc iron
3. -iron: Solid solution of carbon in bcc iron
4. Cementite (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 alternating layers of ferrite andcementite
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Ferrite and -iron
Austenite
Cementite
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The iron-carbon system exhibits three important
transformations / reactions as described below:
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Eutectoid Reaction:
Solid1 Solid2 + Solid3Austenite Ferrite + Cementite
Eutectic Reaction:Liquid Solid1 + Solid2Liquid Austenite + Cementite
Peritectic Reaction:Solid1+ Liquid Solid2
-iron + Liquid Austenite
What is Pearlite?
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Pearlite is a two phased lamellar (or layered) structure composed
of alternating layers of ferrite and cementite that occurs in somesteels and cast irons
100% pearlite is formed at 0.8%C at 727oC by the eutectoid reaction /
Pearlitic transfromation
Eutectoid Reaction:
Solid1 Solid2 + Solid3Austenite Ferrite + Cementite
Development of microstructures in steel during
slow cooling
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Eutectoid Steel
Hypoeutectoid Steel
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Hypereutectoid Steel
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Non-Equilibrium Cooling
Non-equilibrium cooling leads to shift in the transformation
temperatures that appear on the phase diagram
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temperatures that appear on the phase diagram
Leads to development of non-equilibrium phases that do notappear on the phase diagram
Some common binary phase diagrams and
important alloys belonging to these systems
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Copper Nickel (Cu-Ni)
Cooper Zinc (Cu-Zn)
Cooper Tin (Cu-Sn)
Aluminum Silicon (Al-Si)
Lead Tin (Pb-Sn)
Copper and copper alloys
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Properties:1. It has good ductility and malleability
2. It has high electrical and thermal conductivity
3. It is non-magnetic and has a pleasing reddish colour
4. It has fairly good corrosion resistance
5. It has good ability to get alloyed with other elements
Major copper alloys
1. Brass: Alloys of copper and zinc
2. Bronzes: Alloys of copper containing elements other than zincex. Copper-Tin alloys
Copper-Nickel alloys
Cooper Zinc (Cu-Zn)
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Cu-Zn alloys exhibit good
ductility at lower amounts
of Zn.
These alloys are mostly
cast and formed
Widely used for tubes in
heat exchangers, cartridge
cases, fixtures, springs,
utensils, pump parts,
propeller shafts, etc
Cooper Tin (Cu-Sn)
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Cu-Sn alloys exhibit good
ductility and malleability
along with good corrosion
resistance.
Widely used for pumps,
gears, marine fittings,
bearings, coins etc
Copper Nickel (Cu-Ni)
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Cu-Ni Complete solubility ineach other
Copper alloy containing about
45% Nickel has very high
electrical resistivity
Hence used for resistors and
thermocouple wires
Aluminium and aluminium alloys
Properties:
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Properties:
1. It is ductile and malleable2. It is light in weight
3. It has good thermal and electrical conductivity
4. It has excellent ability to get alloyed with other elements
like Cu, Si, Mg, etc.
5. It has excellent corrosion and oxidation resistance
6. It is non-magnetic and non-sparking
Major Aluminium Alloys:
1. Aluminium- silicon2. Aluminium copper
3. Aluminium- Magnesium
Aluminum Silicon (Al-Si)
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Al-Si alloys widely usedfor castings due to their
excellent fluidity and casting
characteristics.
Higher silicon content givesbetter mechanical properties,
better corrosion resistance,
Improved fludity
Widely used for automobilecastings like engine block etc
Lead Tin (Pb-Sn)
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Pb-Sn alloys form a eutectic
at 61.9% Sn at 183oC.
These alloys widely used
as solders because of theirlow melting point and flow
characteristics