Unit I
1.1HISTORICAL USE OF MATERIALS
Stone Age
•
~500,000 yrs ago, early man used flint, bones and stones
• ~100,000 yrs ago, Piltdown man used stone for knives, axes, and this period ended ~6000 yrs ago but
continues in some restricted places still today (S. America, Africa and Indonesia)
Metal Age
•
the real history of metals and man starts ~35,000 yrs ago when homo sapiens displaced Neanderthals
•
gold was first to be discovered and used for ornaments.
•
iron from meteorites was used for tools.
It took the longest for iron to be produced by man into useful tools. Learning how to produce iron or steel still
continues today. Today 90% of man-made material is iron-based
1.2. INTER-RELATIONSHIP BETWEEN STRUCTURE PROPERTIES & PERFORMANCE
Four basic elements :
Structure, Property, Processing, Performance
Processing: different ways for shaping materials into useful components or changing their properties.
Performance: depends upon material‟s properties.
Internal
Structure
Properties
Performance
1.3 METALLURGY:The metallurgy is a science of study of metals. The main classe of metallurgy are
as under:
A) EXCTRACTIVE
METALLURGY
Production of metals by various
chemical processes from ores in
which they are found.
- Ferrous Metallurgy
-Non-Ferrous Metallurgy
B) PHYSICAL
METALLURGY
Deals with study of metals &
alloys with aim of producing
such structures to get desired
properties through various heat
treatment processes like
annealing, normalizing ,
hardening.
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C) MECHANICAL
METALLURGY
-Deals with the properties or
behavior of metal in mechanical
working.
-Relationship between properties
or performance of materials
during manufacturing & in
service.
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1.4 CLASSIFICATION OF MATERIALS:
A) According to Nature:
1Metals& alloys: Fe, Al, Cu
2 Ceramics and Glasses: Borides, Oxides, Nitrides, Carbides, Silicates of metals & non Metals
eg. Al2O3
3. Organic polymers: Plastics
4. Composites of above three: Boride reinforced steels, metal reinforced glass, Glass fiber
reinforced plastic
B) According to major usage/application:
1 Materials used for Structures,
2 Materials used for machines,
3 Materials used for devices
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A) According to Nature:
B. According to major usage/application:
1. Materials used for structures: Non moving parts. e.g. Bridges, Dams
2. Materials used for machines: Moving parts. e.g. turbines, engine ,lathes.
3. Materials used for devices: eg. Transistors, piezoelectric pressure gauge.
A.
Metals: Ferrous & Non-Ferrous
• Metals are composed of elements which readily give up electrons to provide metallic
bonds & electrical conductivity.
• Important properties: Luster, hardness, low specific heat, deformability, good thermal &
electrical conductivity, ductility, malleability & machinability.
Composed of one or more metallic elements and nonmetallic elements
 Good conductors of electricity and heat
 Strong but deformable
 Of all the metallic alloys in use today, ferrous alloys make up the largest proportion both by
quantity and commercial value.
Ferrous alloys Nonferrous alloys
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B. Ceramic Materials: The term “ceramic” comes from the Greek word keramikos, which means
burned stuff,” indicating that desirable properties of these materials are normally achieved through a hightemperature heat treatment process called firing.
• Oxides, Nitrides, Carbides, Silicates or borides of various materials.
• Any inorganic, non metallic solids processed or used at high temperature.
• Contain compounds of metallic or non-metallic elements, such as, MgO, SiC, Glasses.
• They Contain both ionic & metallic bonds.
• Important properties: Brittleness, Hardness, Resistance to high temperature, Abrasiveness,
Corrosion resistance.
i) Traditional ceramics: clay , porcelain, bricks, tiles, glasses，high-temperature ceramics
ii)Advanced ceramics : oxides (Al2O3), nitrides (Si3N4), carbides (SiC), and many other materials including
the superconductors have a rather dramatic effect on our lives electronics, computers,
communication, aerospace
C. Organic Materials:
 Polymeric materials composed of carbon compounds.
 Substances having large molecules consisting of repeated structural units
 very large molecule structure, chain-like in nature
 low density, extremely flexible
 eg. Papers, plastics, woods, rubber, lubricants, textiles.
D. Composites
Composed of two or more individual materials (metals, ceramics, polymers)
designed to display a combination of the best characteristics of each of the component materials
eg. Carbon fiber composites consisting of strength due to carbon fibres and flexiblility due to
epoxiy resin
1.5 FACTORS FOR SELECTION OF MATERIALS:
a.Manufacturing Process: Plasticity, Malleability, Ductility, Machinability, Castability,
Weldability, Heat Treatability
b. Function Requirement:
I)Mechanical Properties:
i) Tensile Strength- Ability to withstand tensile forces. ii)Hardness- Resistance to penetration.
iii)Ductility- Ability to withstand elongation. iv)Impact Strength- Ability to absorb impact.
v)Wear/Corrosion Resistance- Ability to resist wear or corrosion. vi)Elasticity- Stress
disappears completely after removal of load. vii)Plasticity- Property that enables the formation
of permanent deformation.
b)Thermal Properties:
i)Specific Heat: Amount of heat required to change the temperature of unit mass of the body by
one degree Celsius. ii)Thermal Conductivity: Ability to conduct heat. iii)Thermal Expansion:
Tendency of matter to change in volume in response to change in temperature
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c)Electrical Properties:
i)Conductivity: Ability to conduct electric current. ii)Resistivity: Ability to oppose electric
current.
d) Magnetic Properties: This property demonstrates the response of a material to the
application of a magnetic field. Magnetic moment, magnetic permeability
E. Chemical Properties: This property relate to the chemical reactivity of materials. Atomic
weight, Atomic Number, Molecular Number, Valancy, Chemical Composition
F. Optical Properties :This property means a material‟s response to exposure to
electromagnetic radiation and, in particular, to visible light Refractive Index, Reflectivity,
Absorption Coefficient.
c.Cost Consideration: Raw Material, Processing, Storage,Special Tratment, Packaging,
Inventory, Taxes
d. Operating Parameters: Pressure, Flow, Type of material, Corrosion Requirement,
Environmental, Fire Hazard, Weathering, Biological Effect.
d) Magnetic Properties: This property demonstrates the response of a material to the
application of a magnetic field.
Magnetic moment, magnetic permeability
E. Chemical Properties: This property relate to the chemical reactivity of materials.
weight, Atomic Number, Molecular Number, Valancy, Chemical Composition
Atomic
F. Optical Properties :This property means a material‟s response to exposure to
electromagnetic radiation and, in particular, to visible light Refractive Index, Reflectivity,
Absorption Coefficient.
ILLUSTRATIVE EXAMPLE OF SELECTION OF MATERIALS FOR HEATING PAN
•
Functional Requirement: Thermal Conductivity (Steel, Al, Cu)
•
Operating Parameters: Withstand strength at high temp.
•
Manufacturing Process: Malleability, Weldability.
•
Cost: Low- Steel, Aluminum. High- Copper bottom.
Handle Requirements:
1.
Poor thermal conductivity: Wood, Bakelite.
2.
Aesthetics.
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1.6 THE STRUCTURE OF A MATERIAL: The structure of a material usually relates to the
arrangement of its internal components.
i)Subatomic: Electrons within individual atom and interaction with nuclei .
ii) Atomic: Organization of atoms or molecules related to one another
iii) Microscopic: Large groups of atoms, direct observation by microscope.
iv) Macroscopic: Structural elements may be viewed with naked eye.
i)Subatomic structure ii) Atomic structure
iii) Microscopic str.
iv) Macroscopic str.
Solids may present in the nature in the following forms:
1. Crystalline: Atoms pack in periodic 3D arrays . typical of: metals, many ceramics, few
polymers eg.
Crystalline SiO2
(Amorphous)Non Crystalline SiO2
a. Single Crystal: Anisotropic having directional properties. The modulus of elasticity (E) in
BCC iron is different in different directions as shown in the figure.e.g. Sugar, NaCl, Diamond.
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Iron Crystal BCC
Isotropic grains(Epoly iron =210 GPa)
Anisotropic grains( Textured)
b.Polycrystalline: Isotropic having same properties in all directions. e.g. Sugar, NaCl, Diamond.
Most engineering metals are polycrystals. Each "grain" is a single crystal. If crystals are
randomly oriented, overall component properties are not directional. Crystal sizes range from
1 nm to 2 cm. (i.e. From a few to millions of atomic layers).
2. Amorphous: Non-crystalline. Atoms have no periodic packing. typical for complex
structures, during rapid cooling . e.g. ordinary glass, selenium, glycerin.
3. A combination of two.
METALLIC CRYSTALS: These crystals have the simplest crystal structures. They tend to be
densely packed and have several reasons for dense packing:
i)Typically, only one element is present, so all atomic radii are the same.
ii)Metallic bonding is not directional.
iii)Nearest neighbor distances tend to be small in order to lower bond energy.
Some Definitions
Space Lattice:
A regular distribution of points in space that has adentical surrounding.
Crystal Lattice:
Whrn atoms actually occupy the positions on or in relation to these lattice points, we get a crystal lattice.
Unit Cell:
It is the smallest regular shape in the space which when repeated, generates the whole crystal lattice.
Crystal Systems
The shape & size of a unit cell can be defined by help of three vectors a, b, c: specifying the vextor
lengths & the angles Alpha(α), Beta(β) and Gamma(γ) between them.
On the basis of different vector lengths & angles seven crystal systems have been recognized. Depending
on the arrangements of atoms in these systems 14 Bravais space lattices have been defined.
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14 Bravais Lattices in 7 crystal systems
Crystal Systems
Cubic
Monoclinic
Triclinic
Tetragonal
Orthogonal
Trigonal
Hexagonal
Lattice Type
P,F,C
P,B
P
P,C
P,B,F,C
P or R
P
No. of Lattices
3
2
1
2
4
1
1
Vector lengths
a=b=c
a≠b≠c
a≠b≠c
a=b≠c
a≠b≠c
a=b=c
a=b≠c
Angles Relation
α=β=γ≠90°
α=β=90°≠γ
α≠β≠γ≠90°
α=β=γ=90°
α=β=γ=90°
α=β=γ≠90°
α=β=90°
γ=120°
ATOMIC PACKING FACTOR
• APF for a simple cubic structure = 0.52
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Body Centered Cubic Structure (BCC)
Atomic Packing Factor: BCC
• Atoms touch each other along cube diagonals.
• APF for a body-centered cubic structure = 0.68
--Note: All atoms are identical; the center atom is shaded
differently only for ease of viewing.
3a
ex: Cr, W, Fe (), Tantalum, Molybdenum
a
• Coordination # = 8
2a
R
a
Adapted from
Fig. 3.2(a), Callister 7e.
atoms
unit cell
Adapted from Fig. 3.2,
Callister 7e.
APF =
4
2
p ( 3 a/4 ) 3
volume
atom
3
a3
2 atoms/unit cell: 1 center + 8 corners x 1/8
(Courtesy P.M. Anderson)
Close-packed directions:
length = 4R = 3 a
volume
Unit cell contains:
1 + 8 x 1/8
= 2 atoms/unit cell
unit cell
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ATOMIC PACKING FACTOR: FCC
Face Centered Cubic Structure (FCC)
• APF for a body-centered cubic structure = 0.74
• Atoms touch each other along face diagonals.
--Note: All atoms are identical; the face-centered atoms are shaded
differently only for ease of viewing.
ex: Al, Cu, Au, Pb, Ni, Pt, Ag
•
Coordination # = 12
a
Unit cell contains:
6 x 1/2 + 8 x 1/8
= 4 atoms/unit cell
Adapted from
Fig. 3.1(a),
Callister 6e.
Adapted from Fig. 3.1, Callister 7e.
4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8
(Courtesy P.M. Anderson)
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MILLER INDICES
Lattice planes: The crystal lattice may be regarded as made up of an infinite set of parallel
equidistant planes passing through the lattice points . In simple terms, the planes passing
through lattice points are called „lattice planes‟. For a given lattice, the lattice planes can be
chosen in a different number of ways.
The orientation of planes or faces in a crystal can be described in terms of their intercepts on the
three axes.
Miller introduced a system to designate a plane in a crystal by a set of three numbers known as
„Miller Indices‟ of the concerned plane. Miller indices is defined as the reciprocals of the
intercepts made by the plane on the three axes.
Procedure for finding Miller Indices
Step 1: Determine the intercepts of the plane along the axes X,Y and Z in terms of the lattice
constants a, b and c.
Step 2: Determine the reciprocals of these numbers.
Step 3: Find the least common denominator (lcd) and multiply each by this lcd.
Step 4:The result is written in parenthesis. This is called the `Miller Indices‟ of the plane in the
form (h k l).
Plane ABC has intercepts of 2 units along X-axis, 3 units along Y-axis and 2 units along Z-axis.
DETERMINATION OF ‘MILLER INDICES’
Step 1:The intercepts are 2,3 and 2 on the three axes.
Step 2:The reciprocals are 1/2, 1/3 and 1/2.
Step 3:The least common denominator is „6‟. Multiplying each reciprocal by lcd, we get, 3,2
and 3.
Step 4:Hence Miller indices for the plane ABC is (3 2 3)
IMPORTANT FEATURES OF MILLER INDICES
For the cubic crystal especially, the important features of Miller indices are,
A plane which is parallel to any one of the co-ordinate axes has an intercept of infinity
(). Therefore the Miller index for that axis is zero; i.e. for an intercept at infinity, the
corresponding index is zero.
A plane passing through the origin is defined in terms of a parallel plane having non zero
intercepts.
All equally spaced parallel planes have same „Miller indices‟ i.e. The Miller indices do
not only define a particular plane but also a set of parallel planes. Thus the plane whose
intercepts are 1, 1, 1; 2,2,2; -3,-3,-3 etc., are all represented by the same set of Miller
indices.
It is only the ratio of the indices which is important in this notation. The (6 2 2) planes are
the same as (3 1 1) planes.
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-If a plane cuts an axis on the negative side of the origin, corresponding index is negative. It is
represented by a bar, like (1 0 0). i.e.Miller indices (1 0 0) indicates that the plane has an
intercept in the –ve X –axis.
Worked Example:
A certain crystal has lattice parameters of 4.24, 10 and 3.66 Å on X, Y, Z axes
respectively. Determine the Miller indices of a plane having intercepts of 2.12, 10 and 1.83
Å on the X, Y and Z axes.
Lattice parameters are = 4.24, 10 and 3.66 Å
The intercepts of the given plane = 2.12, 10 and 1.83 Å
i.e. The intercepts are, 0.5, 1 and 0.5.
Step 1:The Intercepts are 1/2, 1 and 1/2.
Step 2:The reciprocals are 2, 1 and 2.
Step 3:The least common denominator is 2.
Step 4:Multiplying the lcd by each reciprocal we get, 4, 2 &4.
Step 5:By writing them in parenthesis we get (4 2 4)
Therefore the Miller indices of the given plane is (4 2 4) or (2 1 2).
MILLER INDICES FOR DIRECTIONS
To find the Miller indices of a direction, Choose a perpendicular plane to that direction.
Find the Miller indices of that perpendicular plane.
The perpendicular plane and the direction have the same Miller indices value.
Therefore, the Miller indices of the perpendicular plane is written within a square
bracket to represent the Miller indices of the direction like [ ].
•
i)
ii)
iii)
iv)
v)
vi)
Procedure for finding out Miller indices for directions
Draw a line parallel to given direction starting from origin
For any selected point on this line draw perpendiculars on the three reference planes xy,y-z, & z-x
Find the intercepts in terms of axial units.
Convert these intercepts to smallest integers by multiplication or division throughout.
Enclose these in [h k l]
<010> means family of directions [010], [001], [001],[ 100], [1bar00], 11bar0] 001bar]
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vii)
All parallel directions have same miller indices.
SOLID SOLUTIONS
Alloys : A macroscopically homogeneous mixture of two or more elements, one of which is
essentially a metal is called alloy. The alloy exhibits metallic properties. Eg. An alloy of Fe and
C is plain carbon steel, An alloy of Cu and Zn is Brass.
Solid solutions: A microscopically homogeneous (mixture of two or more elements, one of
which is essentially a metal) alloy is called solid solution. Thus all solid solutions are alloys but
all alloys need not be homogeneous at microscopic level and hence cannot be solid solutions. In
the solid solution both solute and solvent atoms are solid.
The solid solutions may be Substitutional when both solute & solvent atoms are having almost
same size, Crystal structure, valances and electro negativity (Cu & Ni). Or Interstial when the
size of solute atom is relatively very small as compared to solvent atoms ( E.g. Fe & C). The
conditions in which the different types of solid solutions are formed is governed by the Hume
Rothery rules .
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HUME-ROTHERY RULES
The Hume-Rothery rules are a set of basic rules describing the conditions under which
an element could dissolve in a metal, forming a solid solution. There are two sets of rules,
one which refers to substitutional solid solutions, and another which refers to interstitial
solid solutions.
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INTERMETALLIC COMPOUNDS:
i)For substitutional solid solution to be formed the solute and solvent should have similar
electronegativity. If the electronegativity difference is too great, the metals will tend to form
intermetallic compounds instead of solid solutions.
ii)These phases range between the ideal solid solutions and the ideal chemical compunds .
iii)There are two types of intermetallic compounds. Viz
a) intermetallic compounds of fixed composition which obeys the valence laws of ideal chemical
compound ( like Nacl), eg Mg2 Sn which has 29.02% Mg, and a fixed M. P.
b) intermetallic compounds of variable composition which do not obey the valence laws of ideal
chemical compound
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Solidification of pure metal
The nucleus is a cluster of atoms having right crystalline arrangement (structure).
The nuclei are formed upon under cooling in many parts at a time.
The nuclei then grow into grains.
The number of nuclei depends on the degree of under cooling.
NUCLEATION & GROWTH DURING SOLIDIFICATION
1 Positive temp gradient at liquid solid interface( pure metal): During solidification for
pure metal the positive temp gradient exist between solid interface and liquid ( i. e. temp
of solid<temp of interface< temp of liquid. ) . Under such positive temp gradient
condition as the pure molten metal touch the side walls of the mould they nuclei grow
preferentially normal to mould walls. These elongated grains look like columns and
hence they are called “Columnar Grains”. During the growth of these grains the
temperature goes down below its melting point( Tm).
2 Inverse (Negative) temp gradient at liquid solid interface ( Impure metal): During
solidification for impure metal the Inverse temp gradient exist between solid interface
and liquid ( i. e. temp of solid< temp of liquid < temp of interface. ) . Under such
Negative temp gradient condition as the relatively impure molten metal touch the side
walls of the mould the nuclei grow preferentially faster in certain directions than in
other giving rise to branch like structure called dendrites. In highly pure metal however
the dendrites are not formed as the Inverse temp gradient is never present.
Fig b Inverse temp gradient in impure metal
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Fig c Dendrites in impure metal
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a) Formation of nuclei.
b) Small Dendrites begin to grow from the nuclei.
c) The Dendrites continue to grow from the nuclei along all directions and occupying
the space between them starts filling.
d) Solidification is complete and shows no existence of dendrites but only different grains
having different crystal orientation.
1 Chilled grain: When liquid is poured into ingot mould, the liquid in contact with the
bottom wall of mould is chilled. A large number of heterogeneous nuclei are formed (due
to the heat extraction by the mould walls & suspended impurity particles in the liquid)
on the mould and these nuclei grow on this bottom wall to small grains called “chilled
grains.”
2. Columnar Grains: As the molten metal starts touching the mould walls the chilled
grains are formed along it and they grow preferentially normal to mould walls. These
elongated grains look like columns and hence they are called “Columnar Grains”. During
the growth of these grains the temperature goes down below its melting point.
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QUESTION BANK
Q.1 What is metallurgy? What are its main branches? Explain.
Q.2 How are the engineering materials classified? State the main characteristics and
applications of each class.
Q.3 What is the significance of internal structure, properties and performance
of material? Explain.
Q.4 What are the factors to be considered for selecting material for a particular purpose?
With a suitable example explain the process.
Q. 5 State different structures of metal and alloys.
Q.6Define atomic packing factor. Describe the procedure of calculating the value of APF
for BCC and FCC structure.
Q.7Define co-ordination number. Explain how the co-ordination number for BCC and
FCC are calculated?
Q.8 Define solid solution and explain „Hume-Rothery‟ rules applicable to formation of
substitutional solid solution.
Q.9Define the following term
i ) Space lattice ii ) Crystal lattice iii) Unit cell iv)coordination number v)Atomic packing
factor
Q.10 What is the procedure of finding Miller Indices for crystallographic planes and
also draw (0 1 1) and (123) planes.
Q.11 What is the procedure of finding Miller Indices for crystallographic directions and
also draw [0 1 1] and [1 0 1] planes. What do <100> mean?
Q. 12 A certain crystal has lattice parameters of 4.24, 10 and 3.66 Å on X, Y, Z axes
respectively. Determine the Miller indices of a plane having intercepts of 2.12, 10 and
1.83 Å on the X, Y and Z axes.
Q.13Describe how solidification of pure metal occurs? Also explain the phenomenon of
nucleation and growth in pure metal
Q. 14 Explain in brief solidification of ingot in a mould. Also show the different types of
grains formed during solidification with neat sketch.
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