MT-201B MATERIAL SCIENCE NEW - Copy.ppt

8/10/2019 MT-201B MATERIAL SCIENCE NEW - Copy.ppt

http://slidepdf.com/reader/full/mt-201b-material-science-new-copyppt 1/295

MT-201B MATERIALS SCIENCE



Why Study Materials Science?

1. Application oriented Properties

2. Cost consideration

3. Processing route



Classification of Materials

1. Metals

2. Ceramics

3. Polymers

4. Composites

5. Semiconductors6. Biomaterials

7. Nanomaterials



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



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,”



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.



• 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



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 and

the atom positions within.

A point lattice 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.



Table 1: Crystal systems and Bravais Lattices

Crystal systems and Bravais Lattice

Bravais Lattice



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)



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.



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.



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.



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 perunit 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.



• 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



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 withno atoms on the faces. Therefore Nc = 8, Nf = 0 and Ni = 1




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 = 0




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 = 3Hence, Nav = 12 / 6 + 2 / 2 + 3 / 1 = 6



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 of 6



2. Body Centered Cubic Structure:

Body centered cubic structure

has a coordination number of 8



3. Face Centered Cubic Structure:

Face centered cubic structure has a coordination number of 12



4. Hexagonal Close Packed Structure:

Hexagonal close packed structure has a coordination number of 12



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



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



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



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.



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 atom

Volume of the unit cell

= 1 x 4/3 π r 3 = 4/3 π r 3 = 0.52

a3 (2r)3

APF of simple cubic structure is 0.52 or 52%





2 x 4/3 π (a√3 / 4)3 = 0.68a3

(4r)2 = a2 + a2 + a2

Therefore, r = a√3 / 4

APF of body centered cubic structure is 0.68 or 68%



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 / 2√2 APF = Average number of atoms/cell x Volume of an atom


= 4 x 4 / 3 x π (a/2√2)3 = 0.74

a3

APF of face centered cubic structure is 0.74 or 74%



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 = ½ a2

sin 60Total 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





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 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 = π / 3 x 1.633 x 0.899 = 0.74

APF of face centered cubic structure is 0.74 or 74%



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



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.



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.



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)



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.



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




Intercepts : ½ , 1 , ∞

Fractional intercepts : ½ , 1 , ∞

Reciprocal: 2,1,0


Intercepts : 1/3 , 2/3 , 1

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




Exercise



http://en.wikipedia.org/wiki/File:Miller_Indices_Felix_Kling.svg



If the plane passes through the origin, the origin

has to be shifted for indexing the plane



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.



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 representthe –ve sign with a

bar over the number



For direction not originating from origin the origin has to be shifted



Examples of directions with shift of origin



F il f S t R l t d Pl



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



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



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

A i t f t l



Anisotropy of crystals

66.7 GPa

130.3 GPa

191.1 GPa

Young’s modulus

of FCC Cu

A i t f t l ( td )



Anisotropy of crystals (contd.)

Different crystallographicplanes have different

atomic density

And hence

different

properties

Si Wafer for

computers



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 2√2R (height) = 8R2√2

PD = 2 atoms / 8R2√2 =

1 / 4R2√2

Planar density on (110) plane in a FCC unit cell


• Area of (110) plane (rectangular section) is4R (length) x 2√2R (height) = 8R2√2

PD = 2 atoms / 8R2√2 =

1 / 4R2√2



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




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



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) is½ x √2a

√2a sin60o = 0.866 a2

PD = 1.5 atoms / 0.866a2 =

= 1.732 / a2

Voids in crystalline structures



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 determiningaspects 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



Interstitial sites / voids



Tetrahedral sites in HCP

Octahedral sites in HCP



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.



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

Derive the relation between the radius (r) of the octahedral void and the



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

4R 2/2 = (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 = (1.414 –1)R

r = 0.414 R

Si l C l d P l lli



Single Crystal and Polycrystalline

Stages of solidification of a polycrystalline

material

Single Crystal



silicon single crystal

Micrograph of a polycrystallinestainless steel showing grains

and grain boundaries



Ceramic Crystal Structures



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



Ionic bonding



1. Charge neutrality: each crystal should be

electrically neutral e.x. NaCl and CaCl2



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



• Coordination number is related to the cation-anion ratio

• For a specific coordination number there is a critical

or minimum rc / ra ratio



Predicting Structure of FeO



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



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…



AX-TYPE STRUCTURES continued…



AmXp-TYPE STRUCTURES

• number of cations and anions are different,

referred to as AmXp compounds

Calcium Fluorite Structure



AmBnXp-TYPE STRUCTURES

• Ceramic compound with more than two typesof cations, referred to as AmBnXp compounds

Crystal defects (I f ti i S lid )



Crystal defects (Imperfections in Solids)

• Perfect order does not exist throughout a crystalline materialon an atomic scale. All crystalline materials contain large

number of various defects or imperfections.

• Defects or imperfections influence properties such asmechanical, 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.

Crystal defects / imperfections are broadly classified



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 y

into three classes:



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 temperatureaccording 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



Vacancies aid in the movement (diffusion) of atoms

Impurity atoms ( substitutional and interstitial)



I it i t d f t f t t



• 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 atomsdistort the crystal lattice affecting the mechanical and

electrical / electronic properties

• 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 (increasesresistivity) of the material

Frankel and Schottky defects



Frankel and Schottky defects

• 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

Dislocations2. Line defects



• 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



Dislocation line and b are perpendicular to each other



Movement of edge dislocation



Screw Dislocation



Dislocation line and b are parallel to each other

Movement of Screw Dislocation



When Dislocations Interact



Mixed Dislocations



By resolving, the contribution

from both types of

dislocations can be

determined

i l i



Dislocations

as seen under

TransmissionElectron Microscope

(TEM)

3. Surface defectsGrain 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.



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)

Grains and grain boundaries as seen under a microscope



Stacking fault• Occurs when there is a flaw in the stacking sequence



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: ABABBABAB……or ABABAABABAB……..

• FCC stacking sequence: ABCABCABC….

with stacking fault: ABCABCABABCABC……

Stacking fault

FCC

Stacking

Plastic Deformation



Plastic Deformation

Mechanisms of Plastic Deformation



1. Slip

Concept of slip plane and slip direction



Dislocations aid plastic deformation



Plastic deformation

by slip

Plastic deformationby movement of dislocation

Slip System (slip planes and slip directions)



• Plastic deformation takes place by sliding (slip)of close- packed planes over one another.

• The combination of planes and directions on

which slip takes place constitutes the slipsystems of the material.

• Slip plane is generally taken as the closest

packed plane in the system.

• Slip direction is taken as the direction on the slip

plane with the highest linear density.



Slip in Single Crystals

Even if an applied stress is purely tensile there are shear



Even if an applied stress is purely tensile, there are shear

components to it in directions at all but the parallel and

perpendicular directions.

Classified as resolved shear stresses magnitude depends on applied stress, as well as its

orientation with respect to both the slip plane and slip

direction

Polycrystalline Deformation

Slip in polycrystalline systems is more complex



y y y

direction of slip will vary from one crystal to another in the system

Polycrystalline slip requires higher values of applied stresses than

single crystal systems.Because even favorably oriented grains cannot slip until the less

favorably oriented grains are capable of deformation.

Influence of grain boundary on dislocation motion

(Grain boundary strengthening)



(Grain boundary strengthening)

σys is the yield strength or hardness

σo is the yield strength or hardness of single crystal of thesame material K is a constant and

d is the mean grain size

Hall Petch Relationship



Influence of solute atoms on dislocation motion

(Solid solution strengthening)



(Solid solution strengthening)



Influence of dispersoids on dislocation motion

(Dispersion strengthening)



(Dispersion strengthening)

dispersoid

dislocation

Effect of Cold Working and Annealing



Annealing> 0.5 Tm

Cold worked structure

Equiaxed structure• Cold working is the working or deformation ofa material at room temperature or below its

recrystallization temperature

• Annealing is a heat treatment in which the material

is heated above 0.5 of its absolute temperature

Equiaxed structure Cold worked structure

Recovery, Recrystallization and Grain growth



Recovery: Recrystallization: Grain growth:



Cold Working



Hot Working

Strain Hardening



Hardening of a metal due to cold working



1. Interstitial Solid Solution Alloys • Parent metal (solvent )atoms are bigger than atoms of



• Parent metal (solvent )atoms are bigger than atoms of

alloying metal (solute).

• Smaller atoms (solute) fit into spaces, (Interstices), between

larger atoms (solvent).

Interstitial sites



2. Substitutional Solid Solution Alloys At f b th t l f l t i il i



• Atoms of both metals are of almost similar size.

• Direct substitution takes place.

Some Solid Sol tion Allo s



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

MT-201 B, MATERIAL SCIENCE B

(Syllabus for Midterm test 1)



Introduction to Crystallography:

Ionic, covalent, and metallic bonding, van der walls forces Amorphous and Crystalline materials in terms of short range and long

range order

Single crystal and polycrystalline material

Lattice, unit cell, Bravais lattice Types of crystals, SC, BCC, FCC, HCP, Avg. number of atoms per unit

cell, Co-ordination number, Stacking sequence, Atomic packing factor

Crystallographic points, planes and directions (Millers indices), concept

of family of planes

Linear and planer densities Voids in crystalline structures: tetrahedral and octahedral voids

Ceramic crystal structures: cation-anion ratio, AX, AmX p, AmBnX p type

structures

H R th ’ R l f S lid S l bilit



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

factor for solid solution formation If the atomic size difference



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:

The solute and solvent should have similar electronegativity. If



g y

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 similar / 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



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.

Gibb’s Phase Rule:



The Gibb’s 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.



Binary phase diagram

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.

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)



Cooling curve for an alloy / solid solution

(two components)



Plotting of Phase Diagrams



•These phase diagrams are of loop type and are obtained for

l h l l b l h l d ll

Binary isomorphous system:



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.

Finding the amounts of phases in a two phase region :

Lever rule



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



% 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



These diagrams are obtained for two metals having complete

l bili (i i ibili ) i h li id d l

Binary Eutectic System:



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?

• A eutectic or eutectic mixture is a mixture of two or more phases



• 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



Plotting of Eutectic Phase Diagrams



These diagrams are obtained for two metals having complete

solubility (i e miscibility) in the liquid state and partial solubility

Binary Partial Eutectic System



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.



Development of microstructure in binary partial eutectic alloys

during equilibrium cooling



1. Solidification of the eutectic composition



3. Solidification of compositions that range between the room

temperature solubility limit and the maximum solid solubility at

the eutectic temperature



p



Example:

From the data given below for Bi-Cd system, plot the equilibrium diagram to scale

and find:

(a) Amount of eutectic in 20% Cd alloy at room temperature



(b) Free Cd in 70% Cd alloy at room tempearture

Given:Melting temperature of Bi = 271oC

Melting temperature of Cd = 321oC

Eutectic temperature = 144oC

Eutectic composition = 39.7% Cd



The Iron – Carbon System



Allotrophic Transformations in Iron

Iron – Carbon Phase Diagram



Phases in Iron-Carbon Phase Diagram

i l d l f b b



1. Ferrite: Solid solution of carbon in bcc iron

2. Austenite: Solid solution of carbon in fcc iron

3. δ-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



Ferrite and δ-iron

Austenite

Cementite



What is Pearlite?

Pearlite is a two phased lamellar (or layered) structure composed



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 + Solid3

Austenite ↔ Ferrite + Cementite

Development of microstructures in steel during

slow cooling

Eutectoid Steel



Eutectoid Steel

Hypoeutectoid Steel



Hypereutectoid Steel



Mechanical Properties



1. Hardness Test

2. Tensile Test

- Ductile and brittle fracture

3. Impact Test- Ductile-to-brittle transition

4. Fatigue Test

5. Creep Test

Hardness Test

H d i f t i l i t



Hardness is a measure of a materials resistanceto localized plastic deformation.

Types of hardness tests:

1. Brinell hardness test

2. Vickers hardness test

3. Rockwell hardness test

4. Microhardness test

Hardness tests are preformed more frequently than

any other mechanical test for several reasons:



1. They are simple and inexpensive – no special

specimen required and testing apparatus is relatively

inexpensive

2. The test is non-destructive – the specimen is

neither fractured nor excessively deformed; a small

indentation is the only deformation

3. Other mechanical properties may be estimated

from hardness data, such as tensile strength



Brinell Hardness Test

Uses a hard spherical indentor of 10mm diameter



made up of hardened steel or tungsten carbide• Standard loads range between

500 to 3000 Kg and during

the test

• The load to applied for aspecified time 10 – 30s

where:

P = applied force (kgf )

D = diameter of indenter (mm)

d = diameter of indentation (mm)

http://en.wikipedia.org/wiki/Kgf




where:

P = applied force (kgf )

D = diameter of indenter (mm)

d = diameter of indentation (mm)

Advantages and Limitation of Brinell Hardness

Test





Limitations:

1. Ball is likely to deform

2. Thin materials cannot be tested

3. For some materials ridging or piling up is observed4. Test requires more time and reading are subject to

personal errors

Advantages:Useful for measuring hardness of heterogeneous

materials

Vickers Hardness Test

Uses a square based pyramid with an included angleo



of 136o

between opposite faces

• Loads range between

1 to 120 Kg

• The load to applied for a

specified time of 10s



where d is the average length ofthe diagonal left by the indenter

in mm.

F is kgf

Microhardness Test

Uses a square based pyramid indentor as in

http://en.wikipedia.org/wiki/Kilograms-force

http://en.wikipedia.org/wiki/Kilograms-force



q py

Vickers as well as a Knoop indentor

Load applied is between 1 to 1000gm



Rockwell Hardness Test

In this method the hardness is correlated with the depth



of the indentation and not with the area as in Brinell andVickers hardness tests

Indentors used:

1. Hardened steel balls of 1/16”, ⅛”, ¼”, and ½” 2. Conical diamond (Brale indentor)

Load applied:

Minor load: 10 Kg

Major load: 60, 100 and 150 Kgs.



Tensile Test



Concept of stress and strain

Tensile Compression Shear Torsion

Engineering stress = Force

Original cross sectional area



F is the instantaneous load applied perpendicular to the specimen cross

section, in units of newtons (N) or pounds force , and is the original cross sectional

area before any load is applied (m2 or in.2)

Engineering strain = Instantaneous length – original lengthoriginal length

in which lo is the original length before any load is applied, and is the instantaneouslength. Sometimes the quantity li - iois denoted as Δl and is the deformation

elongation or change in length at some instant, as referenced to the original

length. Engineering strain (subsequently called just strain) is unitless, but meters

per meter or inches per inch are often used



Standard tensile specimen

Stress strain test



Elastic deformation



Plastic deformation



3. Ultimate Tensile Stress (U.T.S): It is the highest value of

the stress that the material can bear or sustain without

fracture.



4. Fracture stress or Breaking stress: It is the stress value

at the point of fracture or failure.



Determination of Proof Stress



Ductility

It is the ability of a material to exhibit plastic deformation prior

to fracture under tensile loading conditions.



to actu e u de te s e oad g co d t o s

It is determined by two ways: (a) % elongation (b) % reduction

in cross sectional area.

Resilience

It is the total energy absorbed by the material during its

elastic defromation.



It is the area up to the elastic stress / limit

in a stress-strain diagram

Toughness

It is the total energy absorbed by the material prior to its fracture



It is the sum of elastic and plastic energy

It is the total area under the

stress-strain curve

Toughness of ductile and

brittle materials



In general ductile materials

exhibit more toughness and

less strength while brittlematerials exhibit more

strength and less toughness

True Stress – Strain Curve



True stress-strain curve

Engineering stress-strain curve

Engineering stress = Force

Original cross sectional area



F is the instantaneous load applied perpendicular to the specimens original

cross sectional area

True stress = Force

Instantaneous cross sectional area

True stress is defined as the load F divided by the instantaneous cross-

sectional area over which deformation is occurring (i.e., the neck, past

the tensile point)



Ductile and Brittle Fracture



Ductile Fracture Brittle Fracture

Substantial amount of plastic

deformation prior to fracture

Takes place over a considerable

Negligible or small amount of plastic

deformation prior to fracture

Takes place instantaneously



p

length of time

High amount of energy is

absorbed prior to fracture

Crack is stable and proceeds

relatively slowly

Fracture surface is usually cup

and cone type or with a rough

surface

p y

Low amount of energy is absorbed

prior to fracture

Crack is unstable and proceeds

rapidly

Fracture surface is usually flat

and smooth

Impact Test(Toughness test)



• Toughness is the ability of a material to withstand shock orimpact.

• Toughness is the total energy absorbed by a material prior

to fracture.

• Toughness is determined by the impact test.

• Toughness can also be determined by the total area under

the stress – strain diagram.

• Depending upon the standardized tests using two different

standard specimens, the Charpy and Izod impact tests

were developed and are in use.

Initial

position of

hammer

Final

position of

hammer



Initial energy of = Energy necessary to + Swing energy of

pendulum fracture the specimen of pendulum

mgh = toughness + mgh’



Toughness = mg (h – h’)

W (h – h’)

W = Weight of pendulum (kg)

h = Original height of pendulum (m)h’ = Swing height of pendulum (m)

h

h’

Charpy Specimen



Charpy Specimen after

Impact test

Charpy specimen



Izod specimen

Placement of Charpy and

Izod specimen with respect

to the striker / pendulum

Ductile to Brittle Transition

The transition from ductile to brittle behavior with decrease in

temperature exhibited by metals is know as the ductile to



brittle transition.

• Not all metals / alloys display a ductile to brittle transition. Those

having FCC crystal structure (like aluminium and copper) remain

ductile even at extremely low temperature. However BCC and HCP

metals experience this transition.



• In plain carbon steels the transition temperature increases with

the Increase in the carbon content

• Alloying elements like Ni and Mn effectively lower the transition

temperature while phosphorus lowers the transition temperaturein steels

• Ductile to brittle transition temperature depends on crystal

structure, grain size, alloying elements, impurities, strain hardening

rate and microstructure

• Most ceramics and polymers also experience a ductile to brittle

transition.

Effect of carbon content on the ductile-to-brittle transition

temperature / temperature range in plan carbon steels



Temperatures at which impact test was carried out in oC



Impact test fracture surface at lower temperatures (brittle fractures)

and at higher temperatures (ductile fractures) due to ductile-to-brittletransition

Design Strategy for Toughness:

Stay Above The Ductile to Brittle Transition Temperature!



239

• Pre-WWII: The Titanic • WWII: Liberty ships

• Problem: Used a type of steel with a DBTT ~ Room temp.

Fatigue Test

Fatigue is a form of failure that occurs in components / structures



Subjected to dynamic and cyclic stresses.

Under these circumstances it is possible for failure to occur at a

stress level considerably lower than the tensile or yield strength

for a static load.

The term “fatigue” is used because this type of failure normally

occurs after a lengthy period of repeated stress or strain cycling.

It comprises of approximately 90% of all metallic failures

Alternate tensile and compressive stresses developed in a

rotating shaft leading to fatigue failure



→ ←

Compressive stress

← →

Tensile stress

Variation of stress with time that accounts for fatigue failures

Reversed stress cycle, in which the

stress alternates from a maximum



tensile stress (+) to a maximumCompressive stress (-) of equal

magnitude

Repeated stress cycle, in which

Maximum and minimum stressesare asymmetric relative to the zero

stress level

Random stress cycle

Fatigue Testing Apparatus



Stress vs. Number of cycles curve

(S – N Curve)



Fatigue Fracture

Fatigue failure is brittle like in nature even in normally ductile metals

i.e. associated with very little plastic deformation



Striations or dimples

Brittle fracture

The process of fatigue failure is characterized by three distinct steps:

1. Crack initiation, wherein a small crack forms at some point of high

concentration

2. Crack propagation, during which this crack advances incrementally

with each stress cycle and

3. Final failure, which occurs very rapidly once the advancing crackhas reached a critical size.



Striations as seen

under electron microscope

Brittle fracture

1. Impose a compressive surface stress

(t f k f i )

Improving Fatigue life Small, hard

particles (shot) having diamet

at high velocities onto the surf

compressive stresses to a de

diameter. The influence of sho



247

(to suppress surface cracks from growing)

--Method 1: shot peening

putsurface

intocompression

shot --Method 2: carburizing

C-rich gas

2. Remove stress concentrators. bad

bad

better

better

Case hard

life are enh

process wh

atmospherelayer (or “c

case is nor

material. (T

in Figure 10

increased h

stresses th

Creep Test

Creep is the slow and progressive deformation of a material

ith ti d t t t t t t i t l



with time under a constant stress at temperatures approximatelyabove 0.4 Tm (where Tm is the melting point of the metal or alloy

in degrees Kelvin)

Creep strength of a material is the highest stress that a materialcan withstand for a specified length of time without exceeding

the specified deformation at a given tempearture.

Creep is a thermally activated process and hence is a function

of temperature and time.

The Creep Test



The Creep Curve



Effect of temperature on creep curves at constant stress



Creep Fracture

At low temperature grain boundaries are stronger than the grains and at

high temperatures grains are stronger than grain boundary.

The temperat re at hich the strength of the grain bo ndar is eq al to



Transgranular fracture

at room / low temperature

Intergranular fracture

at high temperature

(creep fracture)

The temperature at which the strength of the grain boundary is equal tothe strength of the grain is called the “Equicohesive temperature”

A crack always moves through weak regions and hence below the

equicohesive temperature, fracture is transgranular i.e. is moves through

the grains. Above equicohesive

temperature, fracture is

Intergranular i.e. it moves along

the grain boundaries.

Creep is a high temperature

process and hence creepfractures are always

intergranular

Intergranular Creep Fracture



Material consideration for high temperature use

(for high creep strength)



• It should have a high melting point

• It should have coarse grained structure

• It should have high oxidation resistance

• Dispersion strengthening improves creep resistance



Heat Treatments

Eutectoid Reaction:


Equilibrium and Non-Equilibrium Cooling




Austenite ↔ Ferrite + Cementite

(Perlite)

Austenite (at 0.8% C) on

equilibrium cooling (slow cooling)

transforms to Perlite (as per the

Iron-Iron carbide phase diagram)

while on non-equilibrium cooling

(rapid cooling) transforms to otherphases like Martensite and Bainite

Slow Cooling

Interrupted Cooling

Perlite



Austenite

Rapid Cooling

Interrupted CoolingBainite

Martensite

Perlite: A two phase microstructure consisting of alternating layers of ferrite

and cementiteBainite: Extremely fine mixture of ferrite and cementite

Martensite: Supersaturated solid solution of carbon in BCC iron having

BCT structure (body centered tetragonal).



Pearlite(soft phase)

Bainite

(hard phase)

Martensite(hardest phase)



Austenite

Slow Cooling

Rapid Cooling

Interrupted Cooling

Perlite

Bainite

Martensite

Time Temperature Transformation (TTT) Diagrams



TTT curves of hypoeutectoid and hypereutectoid steels



Hypoeutectoid steel Hypereutectoid steels

Continuous Cooling Transformation (TTT) Diagrams



Objectives of various heat treatments:

1. To modify / adjust its mechanical, physical or chemical

properties such as hardness strength ductility electrical



properties such as hardness, strength, ductility, electrical

& magnetic properties, or corrosion resistance

2. To increase hardness, wear and abrasion resistance

3. To resoften the steel after it has been hardened by heattreatments or cold working

4. To reduce or eliminate residual stresses

5. To increase or decrease the grain size

Common Heat Treatments in Steel



1. Annealing

2. Normalizing

3. Hardening (Quenching)

4. Tempering

Annealing

Purpose of Anneling:

1 To relieve the internal stresses induced due to cold



1. To relieve the internal stresses induced due to coldworking, welding, etc.

2. To reduce hardness (i.e. to make the steel soft) and to

increase ductility

3. To increase the uniformity of phase distribution and to

make the material isotropic in respect of mechanical

properties

4. To refine the grain size5. To make the material homogeneous in respect of

chemical composition

Process of Annealing:

• The process consists of heating the steel to above A3

temperature for hypoeutectoid steels and above A1 for

hypereutectoid steels by 30-50oC holding at this



hypereutectoid steels by 30 50 C, holding at thistemperature for a definite period and slow cooling to below

A1 or room temperature usually in a furnace.

• Due to slow cooling (furnace cooling), eutectoid reaction

occurs very nearly in accordance with conditionsrepresented in the Fe-C phase diagram i.e. phase changes

follow the Fe-C phase diagram.

The Annealing (Blue region) heating temperature ranges for

hypoeutectoid and hypereutectoid steels on the Fe-C diagram



Annealing on the basis of TTT diagram



Annealing

Effect of Annealing on Cold Working

Equiaxed structure

• Cold working is the working or deformation of

a material at room temperature or below its

recrystallization temperature



Annealing> 0.5 Tm

Cold worked structure

recrystallization temperature• Annealing leads to a equiaxed, stress free

structure which is softer and ductile

• Effect of strain hardening is lost and the material

can be further cold worked

Equiaxed structure Cold worked structure

Recovery, Recrystallization

and Grain growth occur

during Annealing



Normalising

Purpose of Annealing:

• Similar to annealing



Process of Normalizing:

• The process consists of heating the steel to above A3

temperature for hypoeutectoid steels and above Acm for

hypereutectoid steels by 30-50oC, holding at this

temperature for a definite period and slow cooling to below

A1 or room temperature usually in a furnace.

• Due to air cooling which is slightly fast as compared to

furnace cooling employed in annealing, normalised

components show slightly different structure and properties

than annealed components.

Normalising on the basis of TTT diagram



Normalising



Hardening (Quenching)Purpose of Hardening:

1. To harden the steel to the maximum level by austenite to

martensite transformation.



a e s e a s o a o

2. To increase the wear resistance and cutting ability of steel

Process of Hardening:

• The process consists of heating to above the upper critical

temperature (A3) for hypoeutectoid steels and above A1 for

hypereutectoid steels by 50oC, holding at this temperature

for a definite period for complete austenitizing for a

sufficient time and cooling with a rate just exceeding thecritical cooling rate of that steel to room temperature or

below room temperature.

Critical Cooling Rate (CCR)



Hardening on the basis of TTT diagram



Hardening

Tempering

Purpose of Tempering:

1. To relieve the internal stresses developed due to rapid

cooling during hardening process (i e austenite to



cooling during hardening process (i.e. austenite to

martensite transformation) .

2. To reduce hardness and to increase ductility and

toughness i.e. to reduce brittleness.

Process of Hardening:

• The process consists of heating the hardened components

to temperature between 100

o

C and 700

o

C (below A1)holding at the temperature for specific period (1-2 hrs) and

cooling to room temperature usually in air.



Tempering range

100oC to 700oC

Hardening and Tempering on the basis of TTT diagram



Tempering

Martempering and Austempering

Martempering

In this process the austenitized steel is cooled rapidly avoiding

the nose of the TTT curve to a temperature between the nose



the nose of the TTT curve to a temperature between the nose

and Ms, soaked at this temperature for a sufficient time for the

equalization of temperature but not for long enough to permit

the formation of bainite and then cooled to room tempearture in

air or oil.

The process has the following advantages:

1. It results in less distortion and warping, since the martensite

formation occurs at the almost the same time throughout the

cross section of the component.

2. There is less possibility of quenching cracks appearing in

the component.



Austempering

• Austempering consists of cooling the austenitised steel with a rate

exceeding the critical cooling rate and then held at some constanttemperature between the nose of TTT curve and M temperature



temperature between the nose of TTT curve and Ms temperature

i.e. in the bainitic region, holding at this temperature for a sufficient

period for the completion of bainitic transformation and cooling to

room temperature at any desired rate.

• Here austenite transforms to bainite.

• Properties of bainite are Intermediate to those of marteniste and

pearlite and are very much similar to that of tempered martensite.

• The most important advantage of austempering is that it produces

structures and properties very much similar to tempered martensitewithout involving the martensitic transformation as in the hardening

heat treatment:



Surface Hardening• In many applications such as gears, crankshafts, etc.

a hard and wear resistant surface (case) is required

with a tough centre (core) to withstand impact loads.

• Such a requirement is difficult to achieve by using a steel of uniformcomposition



composition.

• Low C steels are ductile and tough but will wear due to less

hardness and wear resistant while high C steels are hard and wear

resistant but have less toughness. Medium C steels are Intermediatein properties to those of low and high C steels but do not satisfy the

requirements to the optimum level

Case (hard and

wear resistant)

Core (ductile

and tough)

• Such a problem can be solved by:

1. Increasing the carbon in the surface of a low C steel and

subsequently heat treating the component in a specific manner to

produce hard and wear resistant surface (case) and a tough

centre (core)



2. Introducing nitrogen in the surface of a tough steel so as to produce

hard nitrided case with no subsequent heat treatment.

3. By selectively hardening the surface only by the hardening heat

treatment

• Surface hardening treatments like those mentioned below are carried

out to selectively harden the surface

1. Carburizing2. Nitriding

3. Flame hardening

4. Induction hardening

Carburizing

• Method of increasing the carbon on the surface of a steel is know

carburizing.

• It consists of heating the steel to the austenitic region in contact witha carburizing medium holding at this temperature for a sufficient



a carburizing medium, holding at this temperature for a sufficient

period and cooling to room temperature.

• The carburized steel may be directly quenched / hardened from the

carburizing temperature or may be given a quenching / hardening

heat treatment later i.e. after cooling from the carburizing

temperature.

• Depending on the medium employed

carburizing can be classified into:

1. Pack carburizing2. Liquid carburizing

3. Gas carburizing

1. Pack Carburizing

• The components to be carburized are packed with a carbonaceous

material in steel or cast iron boxes and sealed with clay.

• The usual carbonaceous material consists of charcoal and coke.



• The sealed boxes are heated to the austenitic region and kept at

the temperature until the desired degree of penetration is obtained.

2. Liquid Carburizing• In this method, carburizing is done by immersing the steel

component in a carbonaceous fused salt bath medium at a

temperature in the austenitic region.

• The bath is composed of sodium cynide, sodium carbonate andsodium chloride



sodium chloride.

3. Gas Carburizing

• Here the components are heated to the austenitic region in the

presence of carbonaceous gas such as methane, ethane or butanediluted with a carrier gas.

CH4 C + 2H2

Nitriding

• Nitriding is accomplished by heating the steel in contact with a

source of atomic nitrogen at a temperature of about 550oC.

• The atomic nitrogen diffuses into the steel and combines with iron

and certain alloying elements present in the steel and forms



y g p

respective nitrides.

• The atomic nitrogen source can be a

molten salt bath for liquid nitriding

containing NaCN similar to that used inliquid carburizing or dissociated ammonia

2NH3 2N + 3H2

Flame Hardening

• Flame hardening is a process heating the surface layer of a

hardenable steel to its austenitic range by means of a

oxyacetylene flames followed by water spray quenching totransform austenite to martensite



transform austenite to martensite.

Induction Hardening

• In induction hardening the surface layer of a material is heated by

induction heating by passing a high frequency current through

a induction coil.

• The high frequency alternating currents flowing through the inductor



generates alternating magnetic field which inturn induces eddy

currents in the surface layer which rapidly heats the surface

MT-201B MATERIAL SCIENCE NEW - Copy.ppt

Documents

Transcript of MT-201B MATERIAL SCIENCE NEW - Copy.ppt