mt-201b material science new

8/2/2019 Mt-201b Material Science New

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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


= 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


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



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Intercepts : , 1,

Fractional intercepts : , 1 ,

Reciprocal: 2,1,0


Intercepts : 1/3 , 2/3 , 1

Fractional intercepts : 1/3 , 2/3 , 1Reciprocal: 3, 3/2, 1



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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


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:


Area of (110) plane (rectangular

section) is 2a2

PD = 1 atom / 2 a2 =

= 1 / 2 a2


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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


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


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

mt-201b material science new

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