mt 201 b material science new

MT-‐201B MATERIALS SCIENCE

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Why Study Materials Science?

1. ApplicaBon oriented Proper&es 2. Cost consideraBon 3. Processing route

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ClassificaBon of Materials

1. Metals 2. Ceramics 3. Polymers 4. Composites 5. Semiconductors 6. Biomaterials 7. Nanomaterials 3

1.  IntroducBon to Crystallography 2.  Principle of Alloy FormaBon 3.  Binary Equilibria 4.  Mechanical ProperBes 5.  Heat Treatments 6.  Engineering Materials 7.  Advanced Materials

Syllabus

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Recommended Books

1.   Callister W.D., “Materials Science and Engineering an Introduc&on”

2.  Askeland D.R., “The Science and Engineering of Materials”

3.  Raghavan V.,”Materials Science and Engineering-‐ A first Course,”

4.  Avener S.H, “IntroducBon to Physical Metallurgy,”

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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 properBes of crystalline solids depend on the crystal structure of the material, the manner in which atoms, ions, or molecules are arranged. 6

•  SomeBmes the term la\ce is used in the context of crystal structures; in this sense “la\ce” means a three-‐ dimensional array of points coinciding with atom posiBons (or sphere centers).

A point la*ce

La\ce

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Unit Cells

•  The unit cell is the basic structural unit or building block of the crystal structure and defines the crystal structure by virtue of its geometry and the atom positions within.

A point la*ce A unit cell

•  This size and shape of the unit cell can be described in terms of their lengths (a,b,c) and the angles between then (α,β,γ). These lengths and angles are the lattice constants or lattice parameters of the unit cell.

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Table 1: Crystal systems and Bravais Lattices

Crystal systems and Bravais Lattice

Bravais La\ce

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Types of crystals Three relatively simple crystal structures are found for most of the common metals; body-centered cubic, face-centered cubic, and hexagonal close-packed.

1. Body Centered Cubic Structure (BCC) 2. Face Centered Cubic Structure (FCC) 3. Hexagonal Close Packed (HCP)

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1.  Body Centered Cubic Structure (BCC) In these structures, there are 8 atoms at the 8 corners and one atom in the interior, i.e. in the centre of the unit cell with no atoms on the faces.

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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 bottom face and 6 at the top face), 2 atoms at the centers of the above two faces and 3 atoms in the interior of the unit cell.

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Average Number of Atoms per Unit Cell Since the atoms in a unit cell are shared by the neighboring cells it is important to know the average number of atoms per unit cell. In cubic structures, the corner atoms are shared by 8 cells (4 from below and 4 from above), face atoms are shared by adjacent two cells and atoms in the interior are shared by only that one cell. Therefore, general we can write: Nav = Nc / 8 + Nf / 2 + Ni / 1 Where, Nav = average number of atoms per unit cell. Nc = Total number of corner atoms in an unit cell. Nf = Total number of face atoms in an unit cell. Ni = Centre or interior atoms.

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•  Simple cubic (SC) structures: In these structures there are 8 atoms corresponding to 8 corners and there are no atoms on the faces or in the interior of the unit cell. Therefore, Nc = 8, Nf = 0 and Ni = 0 Using above eqn. we get, Nav = 8/8 + 0/2 + 0/1 = 1

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2. Body centered cubic (BCC) structures: In these structures, there are 8 atoms at the 8 corners and one atom in the interior, i.e. in the centre of the unit cell with no atoms on the faces. Therefore Nc = 8, Nf = 0 and Ni = 1 Using above eqn. we get, Nav = 8/8 + 0/2 + 1/1 = 2

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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 atom Therefore Nc = 8, Nf = 6 and Ni = 0 Using 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 at the top face), 2 atoms at the centers of the above two faces and 3 atoms in the interior of the unit cell. For hexagonal structures, the corner atoms are shared by 6 cells (3 from below and 3 from above), face atoms are shared by adjacent 2 cells and atoms in the interior are shared by only one cell. Therefore, in general the number of atoms per unit cell will be as: Nav = Nc / 6 + Nf / 2 + Ni / 1 Here Nc = 12, Nf = 2 and Ni = 3 Hence, Nav = 12 / 6 + 2 / 2 + 3 / 1 = 6

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Co-‐ordina&on Number

Co-‐ordinaBon number is the number of nearest equidistant neighboring atoms surrounding an atom under consideraBon

1. Simple Cubic Structure:

Simple cubic structure has a coordinaBon number of 6 19

2. Body Centered Cubic Structure:

Body centered cubic structure has a coordinaBon number of 8

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3. Face Centered Cubic Structure:

Face centered cubic structure has a coordinaBon number of 12

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4. Hexagonal Close Packed Structure:

Hexagonal close packed structure has a coordinaBon number of 12

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Stacking Sequence for SC, BCC, FCC and HCP •  La\ce structures are described by stacking of idenBcal planes of atoms one over the other in a definite manner

•  Different crystal structures exhibit different stacking sequences 1.  Stacking Sequence of Simple Cubic Structure:

Stacking sequence of simple cubic structure is AAAAA…..since the second as well as the other planes are stacked in a similar manner as the first i.e. all planes are stacked in the same manner.

A

A

A 23

2. Stacking Sequence of Body Centered Cubic Structure: •  Stacking sequence of body centered cubic structure is ABABAB….

•  The stacking sequence ABABAB indicates that the second plane is stacked in a different manner to the first.

•  Any one atom from the second plane occupies any one intersBBal site of the first atom.

•  Third plane is stacked in a manner idenBcal to the first and fourth plane is stacked in an idenBcal manner to the second and so on. This results in a bcc structure.

A B

AB

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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 faces and are known as octahedral planes

•  The stacking sequence ABCABC… indicates that the second plane is stacked in a different manner to the first and so is the third from the second and the first. The fourth plane is stacked in a similar fashion to the first

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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 intersBBal 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 or space occupied by atoms in an unit cell. Therefore, APF = Volume of atoms in unit cell Volume of the unit cell

Atomic Packing Factor (APF)

APF = Average number of atoms/cell x Volume of an atom Volume 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” is the la\ce parameter of the simple cubic structure and “r” is the radius of atoms, it is clear from the fig that: r = a/2

APF = Average number of atoms/cell x Volume of an atom Volume of the unit cell

= 1 x 4/3 π r3 = 4/3 π r3 = 0.52 a3 (2r)3

APF of simple cubic structure is 0.52 or 52% 29

In body centred cubic structures, the centre atom touches the corner atoms as shown in fig.

If “a” is the la\ce 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 π (a√3 / 4)3 = 0.68 a3

(4r)2 = a2 + a2 + a2 Therefore, r = a√3 / 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 la\ce parameter of FCC structure and “r” is the atomic radius (DB)2 = (DC)2 + (CB)2 i.e. (4r)2 = a2 + a2 Therefore, r = a / 2√2 APF = Average number of atoms/cell x Volume of an atom Volume of the unit cell = 4 x 4 / 3 x π (a/2√2)3 = 0.74 a3

APF of face centered cubic structure is 0.74 or 74% 32

4. Hexagonal Close Packed (HCP) Structures

The volume of the unit cell for HCP can be found by finding out the area of the basal plane and then mulBplying this by its height

This area is six Bmes the area of equilateral triangle ABC Area of triangle ABC = ½ a2 sin 60 Total area ABDEFG = 6 x ½ a2 sin 60 = 3 a2 sin 60 Now volume of unit cell = 3 a2 sin 60 x c

For HCP structures, the corner atoms are touching the centre atoms, i.e. atoms at ABDEFG are touching the C atom. Therefore a = 2r or r = a / 2

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APF = Average number of atoms/cell x Volume of an atom Volume of the unit cell APF = 6 x 4π/3 r3 3 a2 sin 60 x c APF = 6 x 4π/3 (a/2)3 3 a2 sin 60 x c APF = π a 3 c sin 60

The c/a raBo for an ideal HCP structure consisBng of uniform spheres packed as Bghtly together as possible is 1.633.

Therefore, subsBtuBng c/a = 1.633 and Sin 60o = 0.866 in above equaBon we get: APF = π / 3 x 1.633 x 0.899 = 0.74

APF of face centered cubic structure is 0.74 or 74% 34

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 DirecBons

1.  Point Coordinates When dealing with crystalline materials it olen becomes necessary to specify a parBcular point within a unit cell.

The posiBon of any point located within a unit cell may be specified in terms of its coordinates as fracBonal mulBples of the unit cell edge lengths.

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2. Plane Coordinates

1.  Find out the intercepts made by the plane at the three reference axis e.g. p,q and r.

2.  Convert these intercepts to fracBonal intercepts by dividing with their axial lengths. If the axial length is a, b and c the fracBonal intercepts will be p/a, q/b and r/c.

3.  Find the reciprocals of the fracBonal intercepts. In the above case a/p, b/q and c/r.

4.  Convert these reciprocals to the minimum of whole numbers by mulBplying with their LCM.

5.  Enclose these numbers in brackets (parenthesis) as (hkl) Note: If plane passes through the selected origin, either another

parallel plane must be constructed within the unit cell by an appropriate translaBon or a new origin must be established at the corner of the unit cell.

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1.  Intercepts: p,q and r. 2.  FracBonal intercepts: p/a, q/b and r/c. 3.  Reciprocals: a/p, b/q and c/r. 4.  Convert to whole numbers 5.  Enclose these numbers in brackets (parenthesis) as (hkl)

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Step 1 : IdenBfy the intercepts on the x-‐ , y-‐ and z-‐ axes. In this case the intercept on the x-‐axis is at x = 1 ( at the point (1,0,0) ), but the surface is parallel to the y-‐ and z-‐axes so we consider the intercept to be at infinity ( ∞ ) for the special case where the plane is parallel to an axis. The intercepts on the x-‐ , y-‐ and z-‐axes are thus Intercepts : 1 , ∞ , ∞ Step 2 : Specify the intercepts in fracBonal co-‐ordinates Co-‐ordinates are converted to fracBonal co-‐ordinates by dividing by the respecBve cell-‐dimension -‐ This gives FracBonal Intercepts : 1/1 , ∞/1, ∞/1 i.e. 1 , ∞ , ∞ Step 3 : Take the reciprocals of the fracBonal intercepts This final manipulaBon generates the Miller Indices which (by convenBon) should then be specified without being separated by any commas or other symbols. The Miller Indices are also enclosed within standard brackets (….). The reciprocals of 1 and ∞ are 1 and 0 respecBvely, thus yielding Miller Indices : (100) So the surface/plane illustrated is the (100) plane of the cubic crystal.

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Intercepts : 1 , 1 , ∞ FracBonal intercepts : 1 , 1 , ∞ Reciprocal: 1,1,0 Miller Indices : (110)

Intercepts : 1 , 1 , 1 FracBonal intercepts : 1 , 1 , 1 Reciprocal: 1,1,1 Miller Indices : (111)

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Intercepts : ½ , 1 , ∞ FracBonal intercepts : ½ , 1 , ∞ Reciprocal: 2,1,0 Miller Indices : (210)

Intercepts : 1/3 , 2/3 , 1 FracBonal intercepts : 1/3 , 2/3 , 1 Reciprocal: 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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If the plane passes through the origin, the origin has to be shiled for indexing the plane

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Miller Indices of Planes for Hexagonal Crystals •  Crystal Plane in HCP unit cells is commonly idenBfied by using four indices instead of three.

• The HCP crystal plane indices called Miller-‐Bravis indices are denoted by the lepers h, k, i and l are enclosed in parentheses as (hkil)

• These four digit hexagonal indices are based on a coordinate system with four axes.

• The three a1, a2 and a3 axes are all contained within a single plane (called the basal plane), and at 1200 angles to one another. The z-‐axis is perpendicular to the basal plane.

• The unit of measurement along the a1, a2 and a3 axes is the distance between the atoms along these axes.

• The unit of measurement along the z-‐ axis is the height of the unit cell.

•  The reciprocals of the intercepts that a crystal plane makes with the a1, a2 and a3 axes give the h, k and I indices while the reciprocal of the intercept with the z-‐axis gives the index l.

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Miller Indices of Directions for Cubic Crystals •  A vector of convenient length is posiBoned such that it passes through the origin of the coordinate system.

•  The length of the vector projecBon on each of the three axes is determined.

•  These three numbers are mulBplied or divided by a common factor to reduce them to the smallest integer values.

•  The three indices, not separated by commas, are enclosed in square brackets [uvw]

•  If a negaBve sign is obtained represent the –ve sign with a bar over the number

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For direcBon not originaBng from origin the origin has to be shiled

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Examples of direcBons with shil of origin

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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 )

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Family of Symmetry Related DirecBons

x

y

z[ 1 0 0 ]

[ 1 0 0 ] _

[ 0 0 1 ]

[ 0 0 1 ] _

[ 0 1 0 ] _

[ 0 1 0 ]

IdenBcal atomic density

IdenBcal properBes

〈 1 0 0 〉

〈1 0 0〉= [ 1 0 0 ] and all other direcBons 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

<hkl> represents a family of directions

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Anisotropy of crystals

66.7 GPa

130.3 GPa

191.1 GPa

Young’s modulus of FCC Cu

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Anisotropy of crystals (contd.)

Different crystallographic planes have different atomic density

And hence different properBes

Si Wafer for computers

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Linear and Planar DensiBes

•  Linear density (LD) is defined as the number of atoms per unit length whose centers lie on the direcBon vector LD = number of atoms centered on direcBon vector length of direcBon vector

Linear Density

The [110] linear density for FCC 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 parBcular crystallographic plane

•  PD = number of atoms centered on a plane area of plane

Planar density on (110) plane in a FCC unit cell •  Number of atoms on (110) plane is 2 •  Area of (110) plane (rectangular secBon) is 4R (length) x 2√2R (height) = 8R2√2 PD = 2 atoms / 8R2√2 = 1 / 4R2√2

Planar density on (110) plane in a FCC unit cell •  Number of atoms on (110) plane is 2 •  Area of (110) plane (rectangular secBon) is 4R (length) x 2√2R (height) = 8R2√2 PD = 2 atoms / 8R2√2 = 1 / 4R2√2

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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 secBon) is a x a = a2 PD = 1 atom / a2 = = 1 / a2

Planar density on (110) plane in a Simple Cubic Structure: •  Number of atoms on (110) plane is 1 •  Area of (110) plane (rectangular secBon) is √2a2 PD = 1 atom / √2 a2 = = 1 / √2 a2

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Planar density on (111) plane in a Simple Cubic Structure: •  Number of atoms on (111) plane is 1/6 x 3 = 0.5 •  Area of (111) plane (triangle DEF) is 1/2 x (√2a) x (0.866 x √2a) = 0.866a2 PD = 0.5 atom / 0.866a2 = = 0.577 / a2

Planar density on (100) plane in a Body Centred Cubic Structure: •  Number of atoms on (100) plane is 1 •  Area of (100) plane (square secBon) is a x a = a2 PD = 1 atom / a2 = 1 / a2

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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 ½ x √2a √2a sin60o = 0.866 a2 PD = 1.5 atoms / 0.866a2 = = 1.732 / a2

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Voids in crystalline structures   We have already seen that as spheres cannot fill entire space → the atomic

packing fraction (APF) < 1 (for all crystals)   This implies there are voids between the atoms. Lower the PF, larger the

volume occupied by voids.   These voids have complicated shapes; but we are mostly interested in the

largest sphere which can fit into these voids   The size and distribution of voids in materials play a role in determining

aspects of material behaviour → e.g. solubility of interstitials and their diffusivity

  The position of the voids of a particular type will be consistent with the symmetry of the crystal

  In the close packed crystals (FCC, HCP) there are two types of voids → tetrahedral and octahedral voids (identical in both the structures as the voids are formed between two layers of atoms)

  The tetrahedral void has a coordination number of 4   The octahedral void has a coordination number 6

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Inters&&al 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 interpretaBon of crystal chemistry.

•  The packing of spheres and the formaBon 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 void without pushing the spheres apart? To solve a problem of this type, we need to construct a model for the analysis. Use the diagram shown here as a starting point, and construct a tetrahedral arrangement by placing four spheres of radius R at alternate corners of a cube.

•  What is the length of the face diagonal fd of this cube in terms of R? Since the spheres are in contact at the centre of each cube face, fd = 2 R. •  What is the length of the edge for such a cube, in terms of R? Cube edge length a = √2 R

•  What is the length of the body diagonal bd of the cube in R? bd = √6 R

•  Is the center of the cube also the center of the tetrahedral hole? Yes •  Let the radius of the tetrahedral hole be r, express bd in terms of R and r If you put a small ball there, it will be in contact with all four spheres. bd = 2 (R + r). r = (2.45 R) / 2 - R = 1.225 R - R = 0.225 R •  What is the radius ratio of tetrahedral holes to the spheres? r / R = 0.225 75

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

Derive the relation between the radius (r) of the octahedral void and the radius (R) of the atom in a close packed structure (Assume largest sphere in an octahedral void without pushing the parent atom)

2R2 = (R + r)2 √2R = R + r r = √2R – R = (1.414 –1)R r = 0.414 R 76

Single Crystal and Polycrystalline

Stages of solidificaBon of a polycrystalline material

Single Crystal

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silicon single crystal

Micrograph of a polycrystalline stainless steel showing grains and grain boundaries

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Polymorphism

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Ceramic Crystal Structures

•  Ceramics are compounds between metallic & nonmetallic elements e.x. Al2O3, FeO, SiC, TiN, NaCl •  They are hard and briple •  Typically insulaBve to the passage of electricity & heat

Crystal Structures •  Atomic bonding is mostly ionic i.e. the crystal structure is composed of electrically charged ions instead of atoms. •  The metallic ions, or caBons are posiBvely charged because they have given up their valence electrons to the nonmetallic Ions or anions, which are negaBvely charged

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Ionic bonding

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•  In a ceramic material two characterisBcs of the component ions influence the crystal structure: 1.  Charge neutrality 2. The relaBve sizes of the caBons and anions

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1. Charge neutrality: each crystal should be electrically neutral e.x. NaCl and CaCl2

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2. The relaBve sizes of the caBons and anions

•  Because the metallic elements give up electrons when Ionized, caBons are smaller than anions

•  Hence rc / ra is less than unity

•  Stable ceramic crystal structures form when those anions surrounding a caBon are all in contact with the that caBon

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•  CoordinaBon number is related to the caBon-‐anion raBo

•  For a specific coordinaBon number there is a criBcal or minimum rc / ra raBo

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PredicBng Structure of FeO

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AX-‐TYPE STRUCTURES

•  Equal number of caBons and anions referred to as AX compounds

A denotes the caBon and X denotes the anion

rNa = 0.102 nm

rCl = 0.181 nm

rNa / rCl = 0.564

CaBons prefer octahedral sites

Rock Salt Structure 91

rO = 0.140 nm

rMg = 0.072 nm

rMg / rO = 0.514

CaBons prefer octahedral sites

MgO also has a NaCl type structure

AX-‐TYPE STRUCTURES conBnued…

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AX-‐TYPE STRUCTURES conBnued…

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AmXp-‐TYPE STRUCTURES •  number of caBons 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 types of caBons, referred to as AmBnXp compounds

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Crystal defects (ImperfecBons in Solids) •  Perfect order does not exist throughout a crystalline material on an atomic scale. All crystalline materials contain large number of various defects or imperfections. •  Defects or imperfections influence properties such as mechanical, electrical, magnetic, etc. •  Classification of crystalline defects is generally made according to geometry or dimensionality of the defect i.e. zero dimensional defects, one dimensional defects and two dimensional defects.

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1. Point defect (zero dimensional defects) Vacancy, Impurity atoms ( subs&tu&onal and inters&&al) Frankel and Scho]ky defect

2. Line defect (one dimensional defects) Edge disloca&on Screw disloca&on, Mixed disloca&on

3. Surface defects or Planer defects (two dimensional defects)

Grain boundaries Twin boundary Stacking faults

Crystal defects / imperfections are broadly classified into three classes:

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Vacancy 1. Point defects

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Vacancy •  If an atom is missing from its regular site, the defect produced is called a vacancy

•  All crystalline solids contain vacancies and their number increases with temperature

•  The equilibrium concentration of vacancies Nv for a given

quantity of material depends on & increases with temperature according to Where: N is the total number of atomic sites Qv

is the energy required for the formation of a vacancy T is the absolute temperature & k is the gas or Boltzmann’s constant i.e. 1.38 x 10-23

J/atom-K or 8.62 X 10-5 eV/atom-K

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Vacancies aid in the movement (diffusion) of atoms

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Impurity atoms ( subs&tu&onal and inters&&al)

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•  Impurity point defects are of two types 1. Substitutional 2. Interstitial •  For substitutional, solute or impurity atoms replace or substitute for the host atoms •  For interstitial, solute or impurity atoms fill the void or interstitial space among the host atoms •  Both the substitutional and interstitial impurity atoms distort the crystal lattice affecting the mechanical and electrical / electronic properties

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•  Impurity atoms generate stress in the lattice by distorting the lattice •  The stress is compressive in case of smaller substitutional atom and tensile in case of larger substitutional atom •  These stresses act as barriers to movement of dislocations and thus improve the strength / hardness of a material •  These stresses also act as barriers to the movement of electrons and lower the electrical conductivity (increases resistivity) of the material

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Frankel and Scho]ky defects

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•  Frenkel and Schottky defects occur in ionic solids like ceramics •  An atom may leave its regular site and may occupy nearby interstitial site of the matrix giving rise to two defects simultaneously i.e. one vacancy and the other self interstitial. These two defects together is called a Frenkel defect. This can occur only for cations because of their smaller size as compared to the size of anions. •  When cation vacancy is associated with an anion vacancy, the defect is called Schottky defect. Schottky defects are more common in ionic solids because the lattice has to maintain electrical neutrality

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DislocaBons

•  A missing line or row of atoms in a regular crystal la\ce is called a dislocaBon •  DislocaBon is a boundary between the slipped region and the unslipped region and lies in the slip plane •  Movement of dislocaBon is necessary for plasBc deformaBon •  There are mainly two types of dislocaBons (a) Edge dislocaBons and (b) Screw dislocaBons

2. Line defects

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DislocaBon line and b are perpendicular to each other

Edge Dislocation

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Movement of edge dislocaBon

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ElasBc stress field responsible for electron scapering and increase in electrical resisBvity

la\ce strain around dislocaBon

110

DislocaBon line and b are parallel to each other

Screw Dislocation

111

Movement of Screw Dislocation

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When Dislocations Interact

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By resolving, the contribuBon from both types of dislocaBons can be determined

Mixed Dislocations

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DislocaBons as seen under Transmission Electron Microscope (TEM)

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3. Surface defects Grain Boundary

•  Grain boundary is a defect which separates grains of different orientation from each other in a polycrystalline material.

•  When this orientation mismatch is slight, on the order of a few degrees (< 15 degrees) then the term small- (or low- ) angle grain boundary is used. When the same is more than 15 degrees its is know as a high angle grain boundary.

•  The total interfacial energy is lower in large or coarse-grained materials than in fine-grained ones, since there is less total boundary area in the former.

•  Mechanical properties of materials like hardness, strength, ductility etc are influenced by the grain size.

•  Grains grow at elevated temperatures to reduce the total boundary energy.

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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)

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Twin boundary Atoms on one side of the boundary are located in Mirror image posiBons of the atoms on the other side

Twin Boundary

119

A twin boundary is a special type of grain boundary across which there is a specific mirror lattice symmetry; that is, atoms on one side of the boundary are located in mirror-image positions of the atoms on the other side. The region of material between these boundaries is appropriately termed a twin. Twins result from atomic displacements that are produced from applied mechanical shear forces (mechanical twins), and also during annealing heat treatments following deformation (annealing twins). Twinning occurs on a definite crystallographic plane and in a specific direction, both of which depend on the crystal structure. Annealing twins are typically found in metals that have the FCC crystal structure, while mechanical twins are observed in BCC and HCP metals. Twins contribute to plastic deformation in a small way 120

Stacking fault •  Occurs when there is a flaw in the stacking sequence •  Stacking fault results from the stacking of one atomic plane out of sequence on another and the la\ce on either side of the fault is perfect •  BCC and HCP stacking sequence: ABABABAB…… with stacking fault: ABABBABAB……or ABABAABABAB…….. •  FCC stacking sequence: ABCABCABC…. with stacking fault: ABCABCABABCABC……

Stacking fault

FCC Stacking 121

PlasBc DeformaBon

122

Principles of Alloy FormaBon

Solid Solu&on: •  A homogeneous crystalline phase that contains two or more chemical species •  It is an alloy in which the atoms of solute are distributed in the solvent and has the same structure as that of the solvent

Types of Solid Solu&ons: 1. IntersBBal solid soluBon, ex. Fe-‐C 2. SubsBtuBonal solid soluBon, ex. Au-‐Cu

IntersBBal Solid Soln

SubsBtuBonal Solid Soln 123

1. Inters&&al Solid Solu&on Alloys •  Parent metal atoms are bigger than atoms of alloying metal. •  Smaller atoms fit into spaces, (IntersBces), between larger atoms.

124

Inters&&al sites

125

2. Subs&tu&onal Solid Solu&on Alloys •  Atoms of both metals are of almost similar size. •  Direct subsBtuBon takes place.

126

Alloy Unit Cell Structure Copper -‐ Nickel FCC Copper -‐ Gold FCC Gold -‐ Silver FCC Nickel -‐ PlaBnum FCC Molybdenum -‐ Tungsten BCC Iron -‐ Chromium BCC

Some Solid Solution Alloys

127

Hume-‐Rothery’s Rules of Solid Solubility

1. Atomic size factor

2. Crystal structure factor

3. ElectronegaBvity factor

4. RelaBve valency factor

128

1.   Atomic size factor: If the atomic sizes of solute and solvent differ by less than 15%, it is said to have a favourable size factor for solid soluBon formaBon. If the atomic size difference exceeds 15% solid solubility is limited

2. Crystal Structure factor: Metals having same crystal structure

will have greater solubility. Difference in crystal structure limits the solid solubility

+

A (fcc) B (fcc) AB solid solu&on (fcc) 129

3. Electronega&vity factor: The solute and solvent should have similar electronegaBvity. If the electronegaBvity difference is too great, the metals will tend to form compounds instead of solid soluBons. If electronegaBvity difference is too great the highly electroposiBve element will lose electrons, the highly electronegaBve element will acquire electrons, and compound formaBon will take place. 4. Rela&ve Valency factor: Complete solubility occurs when the solvent and solute have the same valency. If there is shortage of electrons between the atoms, the binding between them will be upset, resulBng in condiBons unfavourable for solid solubility

130

Phase diagrams: Phase or equilibrium diagrams are diagrams which indicate the phases exisBng in the system at any temperature, pressure and composiBon. Why study Phase Diagrams? •  Used to find out the amount of phases exisBng in a given alloy with their composiBon at any temperature.

•  From the amount of phases it is possible to esBmate the approximate properBes of the alloy.

•  Useful in design and control of heat treatment procedures

Phase Diagrams

132

Terms: System: A system is that part of the universe which is under consideraBon. Phase: A phase is a physically separable part of the system with disBnct physical and chemical properBes. (In a system consis6ng of ice and water in a glass jar, the ice cubes are one phase, the water is a second phase, and the humid air over the water is a third phase. The glass of the jar is another separate phase.) Variable: A parBcular phase exists under various condiBons of temperature, pressure and concentraBon. These parameters are called as the variables of the phase Component: The elements present in the system are called as components. For ex. Ice, water or steam all contain H2O so the number of components is 2, i.e. H and O.

133

Gibb’s Phase Rule: The Gibb’s phase rule states that under equilibrium condiBons, the following relaBon must be saBsfied: P + F = C + 2 Where, P = number of phases exisBng in a system under consideraBon. F = degree of freedom i.e. the number of variables such as temperature, pressure or composiBon (concentraBon) that can be changed independently without changing the number of phases exisBng in the system. C = number of components (i.e. elements) in the system, and 2 = represents any two variables out of the above three i.e. temperature pressure and composiBon.

134

Most of the studies are done at constant pressure i.e. one atmospheric pressure and hence pressure is no more a variable. For such cases, Gibb’s phase rule becomes: P + F = C + 1 In the above rule, 1 represents any one variable out of the remaining two i.e. temperature and concentraBon. Hence, Degree of Freedom (F) is given by

F = C – P + 1

135

At point A P = 1, C = 2 F = C – P + 1 F = 2 – 1 +1 F = 2 The meaning of F = 2 is that both temperature and concentraBon can be varied independently without changing the liquid phase exisBng in the system At point 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 composiBon can be changed independently without altering the liquid and solid phases exisBng in the system

•  C

At point C P = 1, C = 2 F = C – P + 1 F = 2 – 1 +1 F = 2 The meaning of F = 2 is that both temperature and concentraBon can be varied independently without changing the liquid phase exisBng in the system

ApplicaBon of Gibbs Phase Rule

136

•  Unary phase diagram

•  Binary phase diagram

•  Ternary phase diagram

Types of Phase Diagrams:

137

1.   Unary Phase diagram (one component)

The simplest phase diagrams are pressure-‐temperature diagrams of a single simple substance, such as water. The axes correspond to the pressure and temperature.

138

2. Binary Phase diagram (two components)

•  A phase diagram plot of temperature against the relaBve concentraBons of two substances in a binary mixture called a binary phase diagram •  Types of binary phase diagrams: 1. Isomorphous 2. EutecBc 3. ParBal EutecBc

139

3. Ternary Phase diagram (three components)

•  A ternary phase diagram has three components. •  It is three dimensional put ploped in two dimensions at constant temperature •  Stainless steel (Fe-‐Ni-‐Cr) is a perfect example of a metal alloy that is represented by a ternary phase diagram.

140

Binary phase diagram The binary phase diagram represents the concentraBon (composiBon) along the x-‐axis and the temperature along the y-‐axis. These are ploped at atmospheric pressure hence pressure is constant i.e. 1 atm. pressure. These are the most widely used phase diagrams. Types of binary phase diagrams: • Binary isomorphous system: Two metals having complete solubility in the liquid as well as the solid state. • Binary eutecBc system: Two metals having complete solubility in the liquid state and complete insolubility in the solid state. • Binary parBal eutecBc system: Two metals having complete solubility in the liquid state and parBal solubility in the solid state. • Binary layer type system: Two metals having complete insolubility in the liquid as well as in the solid state.

141

Cooling curve for Pure Metal (one component)

142

Cooling curve for an alloy / solid soluBon (two components)

143

Plo\ng of Phase Diagrams

145

• 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.

Binary isomorphous system:

146

Finding the amounts of phases in a two phase region : 1. Locate composiBon and temperature in phase diagram 2. In two phase region draw the Be line or isotherm 3. FracBon of a phase is determined by taking the length of the Be line to the phase boundary for the other phase, and dividing by the total length of Be line

Lever rule

147

% of Solid = LO / LS X 100= (Wo-‐Wi) / (Ws-‐Wi) X 100 % of Liquid = OS / LS X 100= (Ws-‐Wi) / (Ws-‐Wi) X 100 or simply % Liquid = 100 -‐ % of Solid or vice versa

148

Development of Microstructure during slow cooling in isomorphous alloys

149

ProperBes of alloys in Isomorphous systems with variaBon in composiBon

(a)  Phase diagram of the Cu-Ni alloy system. Above the liquidus line only the liquid phase exists. In the L + S region, the liquid (L) and solid (S) phases coexist whereas below the solidus line, only the solid phase (a solid solution) exists.

(b) The resistivity of the Cu-Ni alloy as a Function of Ni content (at.%) at room temperature

150

These diagrams are obtained for two metals having complete solubility (i.e. miscibility) in the liquid state and complete insolubility in the solid state. Examples: Pb-‐As, Bi-‐Cd, Th-‐Ti, and Au-‐Si.

Binary EutecBc System:

152

What is a EutecBc? •  A eutec6c or eutec6c mixture is a mixture of two or more phases at a composiBon that has the lowest melBng point •  EutecBc ReacBon: Liquid ↔ Solid A + Solid B

153

Cooling Curves in EutecBc System

154

Plo\ng of EutecBc Phase Diagrams

155

These diagrams are obtained for two metals having complete solubility (i.e. miscibility) in the liquid state and parBal solubility in the solid state. Examples: Pb-‐Sn, Ag-‐Cu, Sn-‐Bi, Pb-‐Sb, Cd-‐Zn and Al-‐Si.

Binary Par&al Eutec&c System

156

Development of microstructure in binary par&al eutec&c alloys during equilibrium cooling

1. SolidificaBon of the eutecBc composiBon

158

2. SolidificaBon of the off -‐ eutecBc composiBon

159

3. SolidificaBon of composiBons that range between the room temperature solubility limit and the maximum solid solubility at the eutecBc temperature

160

Uses of Eutec&c / Par&al Eutec&c Alloys

Alloys of eutecBc composiBons have some specific properBes which make them suitable for certain applicaBons: • Since they fuse at constant temperature, they are used for electrical and thermal fuses. • They are used as solders due to their lower melBng temperature. • Since eutecBc alloys have low melBng points, some of them are used coaBngs by spraying techniques • Since they melt at constant temperature they can be used for temperature measurement. •  Majority of the eutecBc alloys are superplasBc in character. SuperplasBcity is the phenomenon by which an alloy exhibits large extension (ducBlity) when deformed with certain rate at some temperature. The alloy behaves like plasBc and can be formed into many shapes.

161

The Iron – Carbon System

Allotrophic TransformaBons in Iron

162

Iron – Carbon Phase Diagram

163

Phases in Iron-‐Carbon Phase Diagram

1. Ferrite: Solid soluBon of carbon in bcc iron

2. Austenite: Solid soluBon of carbon in fcc iron

3. δ-‐iron: Solid soluBon of carbon in bcc iron

4. Cemen&te (Fe3C): Intermetallic compound of iron

and carbon with a fixed carbon content of 6.67% by wt. 5. Pearlite: It is a two phased lamellar (or layered) structure composed of alternaBng layers of ferrite and cemenBte

164

Ferrite and δ-‐iron

Austenite

Cemen&te 165

The iron-‐carbon system exhibits three important transformaBons / reacBons as described below:

Eutectoid Reac&on: Solid1 ↔ Solid2 + Solid3 Austenite ↔ Ferrite + CemenBte

Eutec&c Reac&on: Liquid ↔ Solid1 + Solid2 Liquid ↔ Austenite + CemenBte

Peritec&c Reac&on: Solid1 + Liquid ↔ Solid2 δ-‐iron + Liquid ↔ Austenite 167

What is Pearlite?

Pearlite is a two phased lamellar (or layered) structure composed of alternaBng layers of ferrite and cemenBte that occurs in some steels and cast irons 100% pearlite is formed at 0.8%C at 727oC by the eutectoid reacBon / PearliBc transfromaBon

Eutectoid Reac&on: Solid1 ↔ Solid2 + Solid3 Austenite ↔ Ferrite + CemenBte

168

Development of microstructures in steel during slow cooling

Eutectoid Steel

169

Hypoeutectoid Steel

170

Hypereutectoid Steel

171

Non-‐Equilibrium Cooling •  Non-‐equilibrium cooling leads to shil in the transformaBon temperatures that appear on the phase diagram •  Leads to development of non-‐equilibrium phases that do not appear on the phase diagram

172

Some common binary phase diagrams and important alloys belonging to these systems

Copper – Nickel (Cu-‐Ni)

Cooper – Zinc (Cu-‐Zn)

Cooper – Tin (Cu-‐Sn)

Aluminum – Silicon (Al-‐Si)

Lead – Tin (Pb-‐Sn)

173

Copper and copper alloys

Proper&es: 1.  It has good ducBlity and malleability 2.  It has high electrical and thermal conducBvity 3.  It is non-‐magneBc 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 zinc ex. Copper-‐Tin alloys Copper-‐Nickel alloys

174

Cooper – Zinc (Cu-‐Zn)

•  Cu-‐Zn alloys exhibit good ducBlity 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 shals, etc

175

Cooper – Tin (Cu-‐Sn)

•  Cu-‐Sn alloys exhibit good ducBlity and malleability along with good corrosion resistance.

•  Widely used for pumps, gears, marine fi\ngs, bearings, coins etc

176

Copper – Nickel (Cu-‐Ni)

•  Cu-‐Ni Complete solubility in each other

•  Copper alloy containing about 45% Nickel has very high electrical resisBvity

•  Hence used for resistors and thermocouple wires

177

Aluminium and aluminium alloys

Proper&es: 1. It is ducBle and malleable 2. It is light in weight 3. It has good thermal and electrical conducBvity 4. It has excellent ability to get alloyed with other elements like Cu, Si, Mg, etc. 5. It has excellent corrosion and oxidaBon resistance 6. It is non-‐magneBc and non-‐sparking Major Aluminium Alloys: 1. Aluminium-‐ silicon 2. Aluminium – copper 3. Aluminium-‐ Magnesium 178

Aluminum – Silicon (Al-‐Si)

•  Al-‐Si alloys widely used for casBngs due to their excellent fluidity and casBng characterisBcs.

•  Higher silicon content gives beper mechanical properBes, beper corrosion resistance, Improved fludity

•  Widely used for automobile casBngs like engine block etc

179

Lead – Tin (Pb-‐Sn)

•  Pb-‐Sn alloys form a eutecBc at 61.9% Sn at 183oC.

•  These alloys widely used as solders because of their low melBng point and flow characterisBcs

180

mt 201 b material science new

Education

interior atoms

unit cell withno atoms

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cubic structures

total number of face

total number of corner

general thenumber of

simple cubic structure