Mec281 lecture

This course covers some fundamentals of materials science, which are
necessary for the understanding of materials properties for their
appropriate applications. The major families of materials such as
metals, ceramics, polymers and composite are discussed for their
structures, properties and applications.
Course Description

• CO1 :Explain basic concepts of structure, mechanical and physical
properties of engineering materials. [PO1, LO1].
• CO2 ;Apply the basics concepts to identify the relationships
between properties and structure of materials. [PO3, LO3, SS1].
• CO3 :Choose the suitable material for appropriate engineering
applications. [PO3, LO3, SS1].
Course Outcome

• Course Work : 40%
Test 1 : 15%
Test 2 : 15%
Quiz (x4) : 10% (i-Learn)
• Final Examination : 60%
• Total : 100%
ASSESSMENT

Syllabus content
CHAPTER CONTENT/SUB-CHAPTER
1
STRUCTURE
(10 HOURS)
1. Atomic Structure.
2. Interatomic Bonding Amorphous and Crystalline Solid.
3. Crystal Structures.
4. Efficiency of Atomic Packing, Density Computation, Miller Indices.
5. Relationship between Atomic Structure, Crystal Structures and
Properties of Material.
2
METALLIC MATERIALS
(14 HOURS)
1. Solidification Of Pure Metal And Alloys
2. Phase Diagram: Microstructure Development, Microconstituent of
Phases.
3. Fe-Fe3C System: Microstructure Development, Microconstituent of
Phases.
4. Ferrous and Non-Ferrous Metals
3
THERMAL
TREATMENT OF
METALLIC MATERIALS
(8 HOURS)
1. Heat Treatment of Ferrous Metals
2. Hardenability
3. Isothermal Transformation Diagram (TTT Diagram)
4
ENGINEERING
MATERIALS
(10 HOURS)
1. Classification of Engineering Materials
2. Plastics And Elastomer: Molecular Structures, Properties and
Applications
3. Ceramic: Structure, Properties, and Applications
4. Composite Materials: Types, Properties and Applications.

REFERENCES
 William D. Callister, Jr., Materials Science and
Engineering and Engineering, An Introduction, John
Wiley & Sins, Inc., 2011.
 William F. Smith, Foundations of Materials Science and
Engineering, Second Edition. McGraw-Hill, 2000.
 William F. Smith, Structure and Properties of
Engineering Alloys, Second Edition, McGraw-Hill, 1993.
 K.R. Tretheway and J. Chamberlain, Corrosion,
Longman Scientific Technical, 1998.

Teaching Plan (June – October 2014)
Week Date Syllybus Activity(s)
1* 9/6/14 – 15/6/14
CHAPTER 12* 16/6/14 – 22/6/14 Tutorial 1
3 23/6/14 – 29/6/14 Tutorial 2
4 30/6/14 – 6/7/14
CHAPTER 2
Tutorial 3
5 7/7/14 – 13/7/14 Tutorial 4 & QUIZ 1’
6 14/7/14 – 20/7/14 Tutorial 5
7 21/7/14 – 25/7/14
26/7/14 – 3/8/14 Mid Term Break
8 4/8/14 – 10/8/14 TEST 1 (8/8/14)
9 11/8/14 – 17/8/14
CHAPTER 3
Tutorial 6 & QUIZ 2’
10 18/8/14 – 24/8/14 Tutorial 7
11 25/8/14 – 31/8/14 Tutorial 8 & QUIZ 3’
12 1/9/14 – 7/9/14
CHAPTER 4
Tutorial 9
13** 8/9/14 – 14/9/14 TEST 2 (12/9/14)
14** 15/9/14 – 21/9/14 Tutorial 10 & QUIZ 4’
2 days 22/9/14 – 23/9/14 Revision
3 weeks 24/9/14 – 17/10/14 Examination
Remarks: (online –iLearn )
*Entrance Survey
**Exit Survey
Quiz’
SuFO (18 Ogos – 3 September 2014)
Lecture Notes, Tutorial, etc. - FOLDER (June – October 2014)

Tutorial
Tutorial Topic
1 Atomic structure
2 Interatomic bonding, Crystal Structures
3 APF, Density Computation, Miller Indices.
4 Phase diagram
5 Iron-Iron Carbide Phase Diagram
6 Ferrous and Non-Ferrous Metal
7 TTT- Diagram
8 Heat Treatment of Ferrous Metal
9 Plastic and Elastomer
10 Ceramic and Composite materials

CHAPTER 1 : STRUCTURE
(10 hours)

SUBCONTENT :
1.1 ATOMIC STRUCTURE.
1.2 INTERATOMIC BONDING AMORPHOUS AND CRYSTALLINE SOLID.
1.3 CRYSTAL STRUCTURES.
1.4 EFFICIENCY OF ATOMIC PACKING, DENSITY COMPUTATION,
MILLER INDICES.
1.5 RELATIONSHIP BETWEEN ATOMIC STRUCTURE, CRYSTAL STRUCTURES
AND PROPERTIES OF MATERIAL.

You should be able:
Describe an atomic structure
Configure electron configuration
Differentiate between each atomic bonding
Briefly describe ionic, covalent, metallic, hydrogen
and Van der Waals bonds
Relate the atomic bonding with material
properties
LEARNING OBJECTIVE

4
1.1 ATOMIC STRUCTURE
All matter is made up of tiny particles called atoms.
What are ATOMS?
Since the atom is too small to be seen even with the most powerful
microscopes, scientists rely upon on models to help us to understand the
atom.
Even with the world’s best
microscopes we cannot clearly
see the structure or behavior
of the atom.

5
Even though we do not know what an
atom looks like, scientific models
must be based on evidence. Many of
the atom models that you have seen
may look like the one below which
shows the parts and structure of the
atom.
Is this really an ATOM?
This model represents the
most modern version of the
atom.
Bohr Theory
Wave Mechanical Atomic Model

6
Protons and neutrons join together
to form the nucleus – the central
part of the atom
+
+ ‐‐
Electrons
move
around the
nucleus
Neutron
Proton
Electron
Nucleon or
Nucleus
Fig. : A simplified diagram of atom
Shell @ Orbital @ Energy level
•Atoms are made of a nucleus that contains protons, neutrons and electrons that
orbit around the nucleus at different levels, known as shells.
What does an ATOM look like?

7
•These particles have the following properties:
Particle Charge Location Mass (amu) Symbol
Proton Positive (+ve) Nucleus 1.0073
Neutron Neutral Nucleus 1.0087
Electron Negative (-ve) Orbital 0.000549
‐
To describe the mass of atom, a unit of mass called the atomic mass unit (amu) is used.
•The number of protons, neutrons and electrons in an atom completely determine
its properties and identity. This is what makes one atom different from another.
+

8
Most atoms are electrically neutral, meaning that they have an equal number of
protons and electrons. The positive and negative charges cancel each other out.
Therefore, the atom is said to be electrically neutral.
Why are all ATOMS are ELECTRICALLY
NEUTRAL?
+
‐
Neutron
Proton
Electron
++
+‐
‐
+ ‐
Fig. : Beryllium atom
Proton = 4
Electron = 4
NEUTRAL
CHARGE

9
cation ‐ ion with a positive charge
‐ If a neutral atom loses one or more electrons, it becomes a cation.
anion ‐ ion with a negative charge
‐ If a neutral atom gains one or more electrons, it becomes an anion.
Na
11 protons
11 electrons Na+ 11 protons
10 electrons
Cl
17 protons
17 electrons
Cl‐
17 protons
18 electrons
Cations are smaller than their “parent atom” because
there is less e‐e repulsion
Anions are larger than their “parent atom” because there is
more e‐‐ e repulsion
If an atom gains or loses electrons, the atom is no longer neutral and
it become electrically charged . The atom is then called an ION.

10
periodic: a repeating pattern
table: an organized collection of information
Periodic Table (P.T.)
An arrangement of elements in
order of atomic number;
elements with similar
properties are in the same
group.
Basics of the PERIODIC TABLE

11
The periodic table below is a simplified representation which
usually gives the :
1) period: horizontal row on the P.T.
•Designate electron energy levels
2) group or family: vertical column on the P.T.
Two main classifications in P.T.

12
ATOMIC NUMBER and ATOMIC MASS
1) ATOMIC NUMBER 2) ATOMIC MASS
Atom can be described using :
The element helium has the atomic number 2, is represented by
the symbol He, its atomic mass is 4 and its name is helium.
ATOMIC MASS , A =
no. of protons (Z) + number of neutrons (N)
SYMBOL
ATOMIC NUMBER, Z = no. of protons

14
ATOMIC NUMBER tells how many PROTONS (Z) are in its atoms which determine the
atom’s identity.
The list of elements (ranked according to an increasing no. of protons) can be looked up
on the Periodic Table. So, if an atom has 2 protons (atomic no. = 2), it must be helium(He).
ATOMIC MASS tells the sum of the masses of  PROTONS (Z) and NEUTRONS (N) within the
nucleus  E.g :
Lithium:
Atomic number = 3
3 protons, Z
4 neutrons, N
Atomic mass, A = 3 + 4 = 7
BUT...  although each element has a defined number of protons, the number of neutrons
is not fixed   isotopes

15
•Atoms which have the same
number of protons but different
numbers of neutrons.
•Atoms which have the same
atomic number but different atomic
mass .
•Eg : Hydrogen has 3 isotopes.
Natural
Isotope
Proton Neutron Atomic
Mass
Hydrogen 1
(hydrogen)
1 0 1
Hydrogen 2
(deuterium)
1 1 2
Hydrogen 3
(tritium)
1 2 3
H1
1 H (D)2
1 H (T)3
1
Same atomic no. @ no. of protons
Different mass number
ISOTOPES

Exercise of isotopes :
Element Name
Number of
Proton
Nucleon
Number
Number of
Neutron
Hydrogen
Hydrogen
Deuterium
Tritium
Oxygen
Oxygen‐16
Oxygen‐17
Oxygen‐18
Carbon
Carbon‐12
Carbon‐13
Carbon‐14
Chlorine
Chlorine‐35
Chlorine‐37
Sodium
Sodium‐23
Sodium‐24

17
Naturally occurring carbon consists of three isotopes,
12C, 13C, and 14C.  State the number of protons,
neutrons, and electrons in each of these carbon atoms.
12C 13C 14C
6 6                        6
#p   _______         _______                _______
#n   _______         _______             _______
#e   _______         _______            _______
EXERCISE

18
The electron cloud that surrounded the nucleus is divided into 7 shells (a.k.a energy level)
 K (1st shell, closest to nucleus) followed by L, M, N, O, P, Q.
Each of the shell, hold a limited no. of electrons.
 E.g : K (2 electrons), L (8 electrons), M (18 electrons), N (32 electrons).
3rd shell
4th shell
2nd shell
1st shell
K (2 electrons)
L (8 electrons)
M (18 electrons)
N (32 electrons)
ELECTRON SHELLS

• Within each shell, the electrons occupy sub shell (energy sublevels)
– s, p, d, f, g, h, i. Each sub shell holds a different types of orbital.
• Each orbital holds a max. of 2 electrons.
• Each orbital has a characteristic energy state and characteristic shape.
• s - orbital
–Spherical shape
–Located closest to nucleus (first energy level)
–Max 2 electrons
• p - orbital
- There is 3 distinct p - orbitals (px, py, pz)
- Dumbbell shape
- Second energy level
- 6 electrons
ORBITAL

20
d- orbital
- There is 5 distinct d – orbitals
- Max 10 electrons
- Third energy level

Table : The number of available electron states in some of the electrons
shells and subshells.
The max. no. of electrons that can occupy a specific shell can be found
using the following formula:
Electron Capacity = 2n2

• The following representation is used :
• Example: it means that there are two electrons in the ‘s’ orbital of the
first energy level. The element is helium.
ELECTRON CONFIGURATIONS
Electron configuration – the ways in which electrons are arranged
around the nucleus of atoms. The following representation is used :
1s2
Energy level @
Principal
quantum no.
Orbital
No. of electrons
in the orbital

Based on the Aufbau principle, which assumes that electrons
enter orbital of lowest energy first.
The electrons in their orbital are represented as follows :
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p6
The sequence of addition of the
electrons as the atomic number
increases is as follows with the
first being the shell number the
s, p, d or f being the type of
subshell, the last number being
the number of electrons in the
subshell.

24
e-e- e-
2nd shell
(energy
level)
Lithium (3 electrons)
How to Write the Electron Configuration of the Element?
e-e- e- e-
e-
e-
e-
e-
e-
3rd shell
(energy
level)
Magnesium (12 electrons)
e-
e-
e-

25
Exercise: Electron Configurations
Atom Symbol Atomic Number Electron configuration
Hydrogen H
Helium He
Lithium Li
Beryllium Be
Chlorine Cl
Argon Ar
Potasium K
Calcium Ca

TRANSITION ELEMENT
Cr [Z = 24] 1s2 2s2 2p6 3s2 3p6 4s1 3d5 (correct) halfly filled
Mo [Z = 42] … 5s1 4d5 (correct) halfly filled
Cu [Z = 29] 1s2 2s2 2p6 3s2 3p6 4s1 3d10 (correct) completely filled
Ag [Z = 47] … 5s1 4d10 (correct) completely filled
Au [Z = 79] …6s1 5d10 (correct) completely filled

Exercise
Write the electron configuration for below element.
a) K
b) K1+
c) Fe
d) Fe3+
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2
4f14 5d10 6p6 7s2 5f14 6d10 7p6

Answer: TEST 1 [July 2011]
1a] With the aid of sketches, describe the Bohr Model of the
sodium [Na] and its ion in terms of valence electron , number of
electron and shell.
[4 marks]

1.2 INTERATOMIC BONDING AMORPHOUS
AND CRYSTALLINE SOLID
2) Secondary Atomic Bonding
Van der Waals
1) Primary Interatomic Bonding
Metallic, ionic and covalent
• The forces of attraction that hold atoms together are called chemical bonds which can
be divided into 2 categories :
• Chemical reactions between elements involve either the releasing/receiving or sharing of
electrons .

How is ionic bonding formed??
30
1) IONIC BONDING
PRIMARY INTERATOMIC BONDING
•Often found in compounds composed of electropositive
elements (metals) and electronegative elements (non metals)
•Electron are transferred to form a bond
•Large difference in electronegativity required

• Properties :
 Solid at room temperature (made of ions)
 High melting and boiling points
 Hard and brittle
 Poor conductors of electricity in solid state
 Good conductor in solution or when molten
IONIC BONDING

• Predominant bonding in Ceramics
Give up electrons Acquire electrons
He
-
Ne
-
Ar
-
Kr
-
Xe
-
Rn
-
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
Li
1.0
Na
0.9
K
0.8
Rb
0.8
Cs
0.7
Fr
0.7
H
2.1
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
Ra
0.9
Ti
1.5
Cr
1.6
Fe
1.8
Ni
1.8
Zn
1.8
As
2.0
CsCl
MgO
CaF2
NaCl
O
3.5
EXAMPLE : IONIC BONDING

Exercise : Final Exam [April 2008]
1c] With the aid of sketches, describe how Sodium and Chlorine
atoms are joined.
[3 marks]

EXERCISE
Draw the following ionic bonding?
IONIC BONDING :
Group 1 metal + Group 7 non metal, eg : NaCl
Group 2 metal + Group 7 non metal, eg : MgF₂, BeF₂, MgBr₂, CaCl₂ or CaI₂
Group 2 metal + Group 6 non metal, eg : CaO, MgO, MgS, or CaS

• Electrons are shared to form a bond.
• Most frequently occurs between atoms with similar electronegativities.
• Often found in:
2) COVALENT BONDING
How is covalent bonding formed??
• Molecules with nonmetals
• Molecules with metals and nonmetals
(Aluminum phosphide (AlP)
• Elemental solids (diamond, silicon, germanium)
• Compound solids (about column IVA)
(gallium arsenide - GaAs, indium antimonide - InSb
and silicone carbide - SiC)
• Nonmetallic elemental molecules (H₂, Cl₂, F₂, etc)

Properties
• Gases, liquids, or solids (made of molecules)
• Poor electrical conductors in all phases
• Variable ( hard , strong, melting temperature, boiling point)
2) COVALENT BONDING

• Molecules with nonmetals
• Molecules with metals and nonmetals
• Elemental solids
• Compound solids (about column IVA)
He
-
Ne
-
Ar
-
Kr
-
Xe
-
Rn
-
F
4.0
Cl
3.0
Br
2.8
I
2.5
At
2.2
Li
1.0
Na
0.9
K
0.8
Rb
0.8
Cs
0.7
Fr
0.7
H
2.1
Be
1.5
Mg
1.2
Ca
1.0
Sr
1.0
Ba
0.9
Ra
0.9
Ti
1.5
Cr
1.6
Fe
1.8
Ni
1.8
Zn
1.8
As
2.0
SiC
C(diamond)
H2O
C
2.5
H2
Cl2
F2
Si
1.8
Ga
1.6
GaAs
Ge
1.8
O
2.0
columnIVA
Sn
1.8
Pb
1.8
EXAMPLE : COVALENT BONDING

Draw the following covalent bonding?
SINGLE BOND :
Hydrogen
Fluorine
Water
DOUBLE BOND :
Oxygen
TRIPLE BOND :
Nitrogen
EXERCISE

• Occur when some electrons in the valence shell separate
from their atoms and exist in a cloud surrounding all the
positively charged atoms.
• The valence electron form a ‘sea of electron’.
• Found for group IA and IIA elements.
• Found for all elemental metals and its alloy.
3) METALLIC BONDING
How is metallic bonding formed??

Properties:
 Good electrical conductivity
 Good heat conductivity
 Ductile
 Opaque
3) METALLIC BONDING

Explain why are metals ductile and can conduct
electricity?

• Arise from atomic or molecular dipoles
VAN DER WAALS
SECONDARY INTERATOMIC BONDING

• Three bonding mechanism
– Fluctuating Induced Dipole Bonds
• Eg: Inert gases, symmetric molecules (H2, Cl2)
– Polar molecule‐Induced Dipole Bonds
• Asymmetrical molecules such as HCl, HF
– Permanent Dipole Bonds
• Hydrogen bonding
• Between molecules
• H‐F, H‐O, H‐N

• Molecule is considered the smallest particle of a pure
chemical substance that still retains its composition
and chemical properties.
• Most common molecules are bound together by
strong covalent bonds.
• E.g. : F2, O2, H2.
• The smallest molecule : Hydrogen molecule .
MOLECULE

48
Summary of BONDING
* Directional bonding – Strength of bond is not equal in all directions
* Nondirectional bonding – Strength of bond is equal in all directions
Type Bond energy Melting point Hardness Conductivity Comments
Ionic
bonding
Large
(150-370kcal/mol)
Very high Hard and
brittle
Poor
-required
moving ion
Nondirectional
(ceramic)
Covalent
bonding
Variable
(75-300 kcal/mol)
Large -Diamond
Small – Bismuth
Variable
Highest –
diamond
(>3550)
Mercury (-39)
Very hard
(diamond)
Poor Directional
(Semiconductors,
ceramic, polymer
chains)
Metallic
bonding
Variable
(25-200 kcal/mol)
Large- Tungsten
Small- Mercury
Low to high Soft to hard Excellent Nondirectional
(metal)
Secondary
bonding
Smallest Low to
moderate
Fairly soft Poor Directional
inter-chain
(polymer)
inter-molecular

Ceramics
(Ionic & covalent bonding):
Metals
(Metallic bonding):
Polymers
(Covalent & Secondary):
secondary bonding
Large bond energy
large Tm
large E
small 
Variable bond energy
moderate Tm
moderate E
moderate 
Directional Properties
Secondary bonding dominates
small T
small E
large 
SUMMARY : PRIMARY BONDING

Exercise : Final Exam [March 2002]
1a] Briefly describe differences between metallic bond and covalent bond.
Support your answer with an example and simple sketch.
(7 Marks)

1.3 CRYSTAL STRUCTURE
Crystal
structure
Crystalline
Material
Single Crystal polycrystal
Noncrsytalline
material
(Amorphous)
* comprised of many single
crystal or grain

• atoms pack in periodic, 3D arrays
• typical of:
Crystalline materials...
-metals
-many ceramics
-some polymers
• atoms have no periodic packing
• occurs for:
Noncrystalline materials...
-complex structures
-rapid cooling
Si Oxygen
crystalline SiO2
noncrystalline SiO2
"Amorphous" = Noncrystalline

•No recognizable long-
range order
•Completely ordered
•In segments
•Entire solid is made up
of atoms in an orderly
array
Amorphous
Polycrystalline
Crystal
•Atoms are disordered
•No lattice
•All atoms arranged on
a common lattice
•Different lattice
orientation for each
grain
Structure of SOLID

• Some engineering applications require single crystals:
--turbine blades
The single crystal turbine blades
are able to operate at a higher
working temperature than
crystalline turbine blade and thus
are able to increase the thermal
efficiency of the gas turbine cycle.

• Most engineering materials are polycrystals.
grain

1a] With the aid of sketches, explain the following terms :
i. Crystalline materials
ii. Amorphous materials
iii. Single crystalline
iv. Polycrystalline
[8 marks]
QUESTION : FINAL EXAM [OCT 2012]

59
Lattice (lines network in 3D) + Motif (atoms are arranged in a repeated pattern)
= CRYSTAL STRUCTURE
Most metals exhibit a crystal structure which show a unique arrangement of atoms
in a crystal.
A lattice and motif help to illustrate the crystal structure.
CRYSTAL STRUCTURE
lattice motif crystal structure
=+

Lattice - The three
dimensional array
formed by the unit cells
of a crystal is called
lattice.
Unit Cell - When a solid
has a crystalline
structure, the atoms are
arranged in repeating
structures called unit
cells. The unit cell is the
smallest unit
that demonstrate the full
symmetry of a crystal.
A crystal is a three-
dimensional repeating
array.
+
=

61
Fig. : The crystal structure (a) Part of the space lattice for natrium chloride (b)Unit cell for natrium
chloride crystal
Unit cell - a tiny box that
describe the crystal structure.
•Crystal structure may be present with any of the
four types of atomic bonding.
•The atoms in a crystal structure are arranged
along crystallographic planes which are designated
by the Miller indices numbering system.
•The crystallographic planes and Miller indices are
identified by X-ray diffraction.
Fig. : The wavelength of the X-ray is
similar to the atomic spacing in crystals.

62
BRAVAIS LATTICE - describe the geometric arrangement of the lattice points and
the translational symmetry of the crystal.
CRYSTAL SYSTEM AND CRYSTALLOGRAPHY
cubic, hexagonal,
tetragonal,
rhombodhedral,
orthorhombic, monoclinic,
triclinic.
•7 crystal systems :
•By adding additional
lattice point to 7 basic
crystal systems –
form 14 Bravais
lattice.

Crystal Structure of Metals
• Simple Cubic (SC) ‐ Manganese
• Body‐centered cubic (BCC) ‐ alpha iron, chromium, molybdenum, tantalum,
tungsten, and vanadium.
• Face‐centered cubic (FCC) ‐ gamma iron, aluminum, copper, nickel, lead, silver,
gold and platinum.
Common crystal structures for metals:
FCCSC BCC

64
SIMPLE CUBIC (SC)
• The atoms lie on a grid: layers of rows and
columns.
• Sit at the corners of stacked cubic
No. of atom at corner
= 8 x 1/8 = 1 atom
Total No. of atom in
one unit cell
= 1 atom
Example : Manganese

Body‐centered Cubic Crystal
Structure
The body-centered cubic (bcc) crystal structure:
(a) hard-ball model; (b) unit cell; and (c) single crystal with many unit cells

66
BODY CENTERED CUBIC STRUCTURE (BCC)
• Cubic unit cell with 8 atoms located at the corner & single atom at cube
center
Example : Chromium, Tungsten,Molybdenum,Tantalum, Vanadium
No. of atom at corner = 8 x 1/8 = 1 atom
No. of atom at center = 1 atom
Total No. of atom in one unit cell = 2 atoms

Face‐centered Cubic Crystal
Structure
The face-centered cubic (fcc) crystal structure:
(a) hard-ball model; (b) unit cell; and (c) single crystal with many unit cells

68
FACE CENTERED CUBIC STRUCTURE (FCC)
Atoms are located at each of the corners and the centers of all the
cube faces. Each corner atom is shared among 8 unit cells,face
centered atom belong to 2.
Example : Cu,Al,Ag,Au, Ni, PtNo. of atom at corner
= 8 x 1/8 = 1 atom
No. of atom at face
= 6 x 12 = 3 atoms
Total No. of atom in
one unit cell
= 4 atoms

69
1.4 EFFICIENCY OF ATOMIC
PACKING,DENSITY COMPUTATION
AND MILLER INDEX

70
APF = no. of atom, n x volume of atoms in the unit cell, (Vs)
volume of the unit cell, (Vc)
ATOMIC PACKING FACTOR
•Atomic packing factor (APF) is defined as the efficiency of atomic arrangement
in a unit cell.
•It is used to determine the most dense arrangement of atoms. It is because how
the atoms are arranged determines the properties of the particular crystal.
•In APF, atoms are assumed closely packed and are treated as hard spheres.
•It is represented mathematically by :

71
close-packed directions
a
R=0.5a
contains 8 x 1/8 =
1 atom/unit cell
EXAMPLE
Calculate the APF for Simple Cubic (SC)?

72
EXERCISE
a) BCC b) FCC
Calculate the APF for BCC and FCC ?

73
a (lattice constant) and
R (atom radius)
Atoms/unit
cell
Packing
Density
(APF)
Examples
Simple
cubic a = 2R
1 52% CsCl
BCC
a = 4R/√3
2 68% Many metals:
α-Fe, Cr, Mo, W
FCC
a = 4R/√2
4 74% Many metals : Ag,
Au, Cu, Pt
Table : APF for simple cubic, BCC, FCC and HCP

74
1a] Give the definition of a unit cell. Briefly describe lattice constant in the unit cell.
[ 4 marks]
1b] Give the definition of APF for a unit cell and calculate the APF for FCC.
[4 marks]
QUESTION : FINAL EXAM [Oct 2010]

75
DENSITY COMPUTATIONS
• A knowledge of the crystal structure of a metallic
solid permits computation of its density through the
relationship :
Where
ρ = n A
Vc NA
n = number of atoms associated with each unit cell
A = atomic weight
Vc = volume of the unit cell
NA = Avogadro’s number (6.023 x 1023 atoms/mol)

76
Calculate the density for nickel (simple cubic structure).
Note that the unit cell edge length (a) for nickel is 0.3524 nm.
EXAMPLE

77
Copper has an atomic radius of 0.128 nm, FCC crystal structure and an atomic
weight of 63.5 g/mol. Compute its density and compare the answer with its
measured density.
EXERCISE
Element Symbol Atomic
weight
(amu)
Density of
solid, 20oC
(g/cm3)
Crystal
Structure,
20oC
Atomic
radius
(nm)
Copper Cu 63.55 8.94 FCC 0.128

78
1b] Platinum has a FCC structure, a lattice parameter of 0.393 nm and an atomic weight
of 195.09 g/mol. Determine :
i. Atomic radius [in cm]
ii. Density of platinum
[ 6marks]
QUESTION : TEST 1 [August 2012]

79
Miller indices is used to label the planes and directions of atoms in a crystal.
Why Miller indices is important?
To determine the shapes of single crystals, the interpretation of X-ray
diffraction patterns and the movement of a dislocation , which may determine
the mechanical properties of the material.
MILLER INDICES
Miller indices
• (h k l) : a specific crystal plane or face
• {h k l} : a family of equivalent planes
• [h k l] : a specific crystal direction
• <h k l> : a family of equivalent directions
Figure : Planes of the form {110} in cubic systems

80
POINT COORDINATES
- The position of any point located within a unit cell may be
specified in terms of its coordinates (x,y,z)
z
y
x
Example : BCC structure
Point
Number x axis y-axis z-axis
Point
Coordinated
1
2
3
4
5
6
7
8
9

81
MILLER INDICES OF A DIRECTION
How to determine crystal direction indices?
i) Determine the length of the vector
projection on each of the three axes,
based on .
ii) These three numbers are expressed as the
smallest integers and negative quantities
are indicated with an overbar.
iii) Label the direction [hkl]. Figure : Examples of direction
Axis X Y Z
Head (H) x2 y2 z2
Tail (T) x1 y1 z1
Head (H) –Tail (T) x2-x1 y2-y1 z2-z1
Reduction (if necessary)
Enclosed [h k l]
* No reciprocal involved.

82
EXAMPLE : CRYSTAL DIRECTION INDICES
0,0,0
1,1,01
1
1,0,0
0,1,0

83
EXERCISE : CRYSTAL DIRECTION INDICES
0½
1
1
0
1
1
1
1

84
EXERCISE : CRYSTAL DIRECTION INDICES
0
½
0
¾
½
½
½

Determine the direction indices of the cubic
direction between the position coordinates
TAIL (3/4, 0, 1/4) and HEAD (1/4, 1/2, 1/2)?

Draw the following Miller Indices
direction.
a) [ 1 0 0 ]
b) [ 1 1 1 ]
c) [ 1 1 0 ]
d) [ 1 1 0 ]

87
i) Determine the points at which a given crystal plane
intersects the three axes, say at (a,0,0),(0,b,0), and (0,0,c). If
the plane is parallel an axis, it is given an intersection ∞.
ii) Take the reciprocals of the three integers found in step (i).
iii) Label the plane (hkl). These three numbers are expressed
as the smallest integers and negative quantities are indicated
with an overbar,e.g : a.
MILLER INDICES OF A PLANE
How to determine crystal plane indices?
Figure : Planes with different Miller
indices in cubic crystals
Axis X Y Z
Interceptions
Reciprocals
Reduction (if necessary)
Enclosed (h k l )

+x
+y
+z
_
z
_
y
_
x
(1 , 0 , 0)
(0 , 1 , 0)
(0 , 0 , 1)
_
(0 , 0 , 1)
_
(0 , 1 , 0)
_
(1 , 0 , 0)

89
EXERCISE. : CRYSTAL PLANE INDICES

0
½

Determine the Miller Indices plane for the
following figure below?

Draw the following Miller Indices
plane.
a) ( 1 0 0 )
b) ( 0 0 1 )
c) ( 1 0 1 )
d) ( 1 1 0 )

94
NOTE (for plane and direction):
• PLANE
Make sure you enclosed your final answer in brackets (…) with no
separating commas → (hkl)
• DIRECTION
Make sure you enclosed your final answer in brackets (…) with no
separating commas → [hkl]
• FOR BOTH PLANE AND DIRECTION
Negative number should be written as follows :
-1 (WRONG)
1 (CORRECT)
Final answer for labeling the plane and direction should not have fraction
number do a reduction.

95
1.5 RELATIONSHIP BETWEEN
ATOMIC STRUCTURE, CRYSTAL
STRUCTURES AND PROPERTIES OF
MATERIALS

96
PHYSICAL PROPERTIES OF METALS
•Solid at room temperature (mercury is an exception)
•Opaque
•Conducts heat and electricity
•Reflects light when polished
•Expands when heated, contracts when cooled
•It usually has a crystalline structure
Physical properties are the characteristic responses of materials to
forms of energy such as heat, light, electricity and magnetism.
The physical properties of metals can be easily explained as follows :

Mechanical Properties
 Terminology for Mechanical Properties
 The Tensile Test: Stress‐Strain Diagram
 Properties Obtained from a Tensile Test
 Hardness of Materials

98
MECHANICAL PROPERTIES OF METALS
Mechanical properties are the characteristic dimensional changes in response to
applied external or internal mechanical forces such as shear strength, toughness,
stiffness etc.
The mechanical properties of metals can be easily explained as follows :

99
Tensile Test
specimen
machine

Terminology
 Load ‐ The force applied to a material during testing.
 Strain gage or Extensometer ‐ A device used for
measuring change in length (strain).
 Engineering stress ‐ The applied load, or force,
divided by the original cross‐sectional area of the
material.
 Engineering strain ‐ The amount that a material
deforms per unit length in a tensile test.

Stress-Strain Diagram
Strain ( ) (L/Lo)
4
1
2
3
5
Elastic
Region
Plastic
Region
Strain
Hardening Fracture
ultimate
tensile
strength
Elastic region
slope =Young’s (elastic) modulus
yield strength
Plastic region
ultimate tensile strength
strain hardening
fracture
necking
yield
strength
UTS
y
εEσ 
ε
σ
E 

12
y
εε
σ
E



Stress-Strain Diagram (cont)
• Elastic Region (Point 1 –2)
- The material will return to its original shape
after the material is unloaded( like a rubber band).
- The stress is linearly proportional to the strain in
this region.
εEσ 
: Stress(psi)
E : Elastic modulus (Young’s Modulus) (psi)
: Strain (in/in)
σ
ε
- Point 2 : Yield Strength : a point where permanent
deformation occurs. ( If it is passed, the material will
no longer return to its original length.)
ε
σ
E or

• Strain Hardening
- If the material is loaded again from Point 4, the
curve will follow back to Point 3 with the same
Elastic Modulus (slope).
- The material now has a higher yield strength of
Point 4.
- Raising the yield strength by permanently straining
the material is called Strain Hardening.

• Tensile Strength (Point 3)
- The largest value of stress on the diagram is called
Tensile Strength(TS) or Ultimate Tensile Strength
(UTS)
- It is the maximum stress which the material can
support without breaking.
• Fracture (Point 5)
- If the material is stretched beyond Point 3, the stress
decreases as necking and non-uniform deformation
occur.
- Fracture will finally occur at Point 5.

Figure : Stress strain diagram
Typical regions that can
be observed in a stress-
strain curve are:
• Elastic region
• Yielding
• Strain Hardening
• Necking and Failure
• This diagram is used to determine how material will react under a certain load.

108
Important Mechanical Properties
from a Tensile Test
• Young's Modulus: This is the slope of the linear portion
of the stress‐strain curve, it is usually specific to each
material; a constant, known value.
• Yield Strength: This is the value of stress at the yield
point, calculated by plotting young's modulus at a
specified percent of offset (usually offset = 0.2%).
• Ultimate Tensile Strength: This is the highest value of
stress on the stress‐strain curve.
• Percent Elongation: This is the change in gauge length
divided by the original gauge length.

The stress-strain curve for an aluminum alloy.

1100.2
8
0.6
1
Magnesium,
Aluminum
Platinum
Silver, Gold
Tantalum
Zinc, Ti
Steel, Ni
Molybdenum
Graphite
Si crystal
Glass-soda
Concrete
Si nitride
Al oxide
PC
Wood( grain)
AFRE( fibers)*
CFRE*
GFRE*
Glass fibers only
Carbon fibers only
Aramid fibers only
Epoxy only
0.4
0.8
2
4
6
10
20
40
60
80
100
200
600
800
1000
1200
400
Tin
Cu alloys
Tungsten
<100>
<111>
Si carbide
Diamond
PTFE
HDPE
LDPE
PP
Polyester
PS
PET
CFRE( fibers)*
GFRE( fibers)*
GFRE(|| fibers)*
AFRE(|| fibers)*
CFRE(|| fibers)*
Metals
Alloys
Graphite
Ceramics
Semicond
Polymers
Composites
/fibers
E(GPa)
Eceramics
> Emetals
>> Epolymers
109 Pa Composite data based on
reinforced epoxy with 60 vol%
of aligned carbon (CFRE),
aramid (AFRE), or glass (GFRE)
fibers.
Young’s Moduli: Comparison

111
T
E
N
S
I
L
E
P
R
O
P
E
R
T
I
E
S

112
Graphite/
Ceramics/
Semicond
Metals/
Alloys
Composites/
fibers
Polymers
Yieldstrength,y(MPa)
PVC
Hardtomeasure,
sinceintension,fractureusuallyoccursbeforeyield.
Nylon 6,6
LDPE
70
20
40
60
50
100
10
30
200
300
400
500
600
700
1000
2000
Tin (pure)
Al (6061)a
Al (6061)ag
Cu (71500)hr
Ta (pure)
Ti (pure)a
Steel (1020)hr
Steel (1020)cd
Steel (4140)a
Steel (4140)qt
Ti (5Al-2.5Sn)a
W (pure)
Mo (pure)
Cu (71500)cw
Hardtomeasure,
inceramicmatrixandepoxymatrixcomposites,since
intension,fractureusuallyoccursbeforeyield.
HDPE
PP
humid
dry
PC
PET
¨
Room T values
a = annealed
hr = hot rolled
ag = aged
cd = cold drawn
cw = cold worked
qt = quenched & tempered
Yield Strength: Comparison

113
tensile stress, 
engineering strain, 
y
p = 0.002
Yield Strength, y
tensile stress, 
engineering strain, 
Elastic
initially
Elastic+Plastic
at larger stress
permanent (plastic)
after load is removed
p
plastic strain

114
F

bonds
stretch
return to
initial
1. Initial 2. Small load 3. Unload
F

Linear-
elastic
Non-Linear-
elastic
Elastic Deformation
• Atomic bonds are stretched but not
broken.
• Once the forces are no longer
applied, the object returns to its
original shape.
• Elastic means reversible.

115
Typical stress-strain
behavior for a metal
showing elastic and
plastic deformations,
the proportional limit P
and the yield strength
σy, as determined
using the 0.002 strain
offset method (where there
is noticeable plastic deformation).
P is the gradual elastic
to plastic transition.

116
1. Initial 2. Small load 3. Unload
.
F

linear
elastic
linear
elastic
plastic
planes
still
sheared
F
elastic + plastic
bonds
stretch
& planes
shear
plastic
Plastic Deformation (Metals)
• Atomic bonds are broken and new
bonds are created.
• Plastic means permanent.

117
Permanent Deformation
• Permanent deformation for metals is
accomplished by means of a process called
slip, which involves the motion of
dislocations.
• Most structures are designed to ensure that
only elastic deformation results when stress
is applied.
• A structure that has plastically deformed, or
experienced a permanent change in shape,
may not be capable of functioning as
intended.

118
• After yielding, the stress necessary to
continue plastic deformation in metals
increases to a maximum point (M) and
then decreases to the eventual fracture
point (F).
• All deformation up to the maximum
stress is uniform throughout the tensile
sample.
• However, at max stress, a small
constriction or neck begins to form.
• Subsequent deformation will be confined
to this neck area.
• Fracture strength corresponds to the
stress at fracture.
Region between M and F:
• Metals: occurs when noticeable necking starts.
• Ceramics: occurs when crack propagation starts.
• Polymers: occurs when polymer backbones are aligned and about to break.
Tensile Strength, TS

119
In an undeformed
thermoplastic polymer
tensile sample,
(a) the polymer chains are
randomly oriented.
(b) When a stress is
applied, a neck
develops as chains
become aligned locally.
The neck continues to
grow until the chains
in the entire gage
length have aligned.
(c) The strength of the
polymer is increased

120
Room T values
Si crystal
<100>
Graphite/
Ceramics/
Semicond
Metals/
Alloys
Composites/
fibers
Polymers
Tensilestrength,TS(MPa)
PVC
Nylon 6,6
10
100
200
300
1000
Al (6061) a
Al (6061) ag
Cu (71500) hr
Ta (pure)
Ti (pure)a
Steel (1020)
Steel (4140) a
Steel (4140) qt
Ti (5Al-2.5Sn) a
W (pure)
Cu (71500) cw
LDPE
PP
PC PET
20
30
40
2000
3000
5000
Graphite
Al oxide
Concrete
Diamond
Glass-soda
Si nitride
HDPE
wood ( fiber)
wood(|| fiber)
1
GFRE(|| fiber)
GFRE( fiber)
CFRE(|| fiber)
CFRE( fiber)
AFRE(|| fiber)
AFRE( fiber)
E-glass fib
C fibers
Aramid fib
Based on data in Table B4, Callister 6e.
a     = annealed
hr   = hot rolled
ag = aged
cd = cold drawn
cw = cold worked
qt   = quenched & tempered
AFRE, GFRE, & CFRE =
aramid, glass, & carbon
fiber‐reinforced epoxy
composites, with 60 vol%
fibers.
Tensile Strength: Comparison

121
• Tensile stress, : • Shear stress, :
Area, A
Ft
Ft
 
Ft
Ao
original area
before loading
Area, A
Ft
Ft
Fs
F
F
Fs
 
Fs
Ao
Stress has units: N/m2 or lb/in2
Engineering Stress

122
VMSE
http://www.wiley.com/college/callister/0470125373/vmse/strstr.htm
http://www.wiley.com/college/callister/0470125373/vmse/index.htm

123
Engineering tensile strain, 
Engineering
tensile
stress, 
smaller %EL
(brittle if %EL<5%)
larger %EL
(ductile if
%EL>5%)
• Another ductility measure: 100% x
A
AA
AR
o
fo 

• Ductility may be expressed as either percent elongation (% plastic strain at fracture)
or percent reduction in area.
• %AR > %EL is possible if internal voids form in neck.
Lo Lf
Ao
Af
100% x
l
ll
EL
o
of 

Ductility, %EL
Ductility is a measure of the plastic
deformation that has been sustained at
fracture:
A material that
suffers very
little plastic
deformation is
brittle.

124
Toughness
Lower toughness: ceramics
Higher toughness: metals
Toughness is
the ability to
absorb
energy up to
fracture (energy
per unit volume of
material).
A “tough”
material has
strength and
ductility.
Approximated
by the area
under the
stress-strain
curve.

• Energy to break a unit volume of material
• Approximate by the area under the stress-strain
curve.
21
smaller toughness-
unreinforced
polymers
Engineering tensile strain, 
Engineering
tensile
stress, 
smaller toughness (ceramics)
larger toughness
(metals, PMCs)
Toughness

126
Linear Elastic Properties
Modulus of Elasticity, E:
(Young's modulus)
• Hooke's Law:  = E 
• Poisson's ratio:
metals:  ~ 0.33
ceramics:  ~0.25
polymers:  ~0.40

Linear-
elastic
1
E

Units:
E: [GPa] or [psi]
: dimensionless
F
F
simple
tension
test
xy

127
Engineering Strain
Strain is dimensionless.

128
Axial (z) elongation (positive strain) and lateral (x and y) contractions (negative strains) in
response to an imposed tensile stress.

True Stress and True Strain
 True stress The load divided by the actual cross-sectional
area of the specimen at that load.
 True strain The strain calculated using actual and not
original dimensions, given by εt ln(l/l0).
•The relation between the true stress‐true
strain diagram and engineering stress‐
engineering strain diagram.
•The curves are identical to the yield point.

(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
The stress-strain behavior of brittle materials compared with
that of more ductile materials

131
‐‐brittle response (aligned chain, cross linked & networked case)
‐‐plastic response (semi‐crystalline case)
Stress-Strain Behavior: Elastomers
3 different responses:
A – brittle failure
B – plastic failure
C ‐ highly elastic (elastomer)
initial: amorphous chains are
kinked, heavily cross-linked.
final: chains
are straight,
still
cross-linked
0
20
40
60
0 2 4 6
(MPa)
 8
x
x
x
elastomer
plastic failure
brittle failure
Deformation
is reversible!

132
Stress-Strain Results for Steel
Sample

133
Metals can fail by brittle or ductile fracture.
FRACTURE MECHANISM OF METALS
Ductile fracture is better than brittle fracture because :
Ductile fracture occurs over a period of time, where as brittle fracture is fast
and can occur (with flaws) at lower stress levels than a ductile fracture.
Figure : Stress strain curve for brittle and ductile material

Ductile Vs Brittle Fracture

(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
• Localized deformation of a ductile material during a tensile test produces a
necked region.
• The image shows necked region in a fractured sample
Ductile Fracture

136
1c] Ductility is one of the important mechanical properties.
i] Define the ductility of a metal.
ii] With the aid of schematic diagrams, describe elastic and plastic deformations.
[6 marks]
QUESTION : FINAL EXAM [April 2011]

Ductile fracture Brittle fracture
What are the differences between
ductile fracture & brittle fracture?

Hardness of Materials
 Hardness test - Measures the resistance of a material to
penetration by a sharp object.
 Macrohardness - Overall bulk hardness of materials
measured using loads >2 N.
 Microhardness Hardness of materials typically measured
using loads less than 2 N using such test as Knoop
(HK).
 Nano-hardness - Hardness of materials measured at 1–
10 nm length scale using extremely small (~100 µN)
forces.

139
Hardness
• Hardness is a measure of a material’s resistance
to localized plastic deformation (a small dent or
scratch).
• Quantitative hardness techniques have been
developed where a small indenter is forced into
the surface of a material.
• The depth or size of the indentation is measured,
and corresponds to a hardness number.
• The softer the material, the larger and deeper the
indentation (and lower hardness number).

140
• Resistance to permanently indenting the surface.
• Large hardness means:
‐‐resistance to plastic deformation or cracking in
compression.
‐‐better wear properties.
e.g.,
10mm sphere
apply known force
(1 to 1000g)
measure size
of indent after
removing load
dD
Smaller indents
mean larger
hardness.
increasing hardness
most
plastics
brasses
Al alloys
easy to machine
steels file hard
cutting
tools
nitrided
steels diamond
Hardness

Indentation Geometry for Brinnel
Testing
Figure Indentation geometry in
Brinell hardness testing: (a)
annealed metal; (b) work-
hardened metal; (c) deformation
of mild steel under a spherical
indenter. Note that the depth of
the permanently deformed zone
is about one order of magnitude
larger that the depth of
indentation. For a hardness test
to be valid, this zone should be
developed fully in the material.

Hardness
Scale
Conversions
Figure Chart for converting
various hardness scales. Note
the limited range of most scales.
Because of the many factors
involved, these conversions are
approximate.

146
Conversion of
Hardness Scales
Also see: ASTM E140 - 07
Volume 03.01
Standard Hardness Conversion
Tables for Metals Relationship
Among Brinell Hardness, Vickers
Hardness, Rockwell Hardness,
Superficial Hardness, Knoop
Hardness, and Scleroscope
Hardness

147
Correlation
between
Hardness and
Tensile
Strength
• Both hardness and tensile
strength are indicators of
a metal’s resistance to
plastic deformation.
• For cast iron, steel and
brass, the two are roughly
proportional.
• Tensile strength (psi) =
500*BHR

148
1c] Hardness is one of the important mechanical properties
in engineering. Describe FOUR [4] types of hardness
measurement method in terms of name and types of
indenter.
[ 4 marks]
QUESTION : FINAL EXAM [Oct 2012]

149
•  Stress and strain:  These are size‐independent
measures of load and displacement, respectively.
•  Elastic behavior:  This reversible behavior often
shows a linear relation between stress and strain.
To minimize deformation, select a material with a
large elastic modulus (E or G).
•  Plastic behavior:  This permanent deformation
behavior occurs when the tensile (or compressive)
uniaxial stress reaches y.
•  Toughness:  The energy needed to break a unit
volume of material.
•  Ductility:  The plastic strain at failure.
Summary

CHAPTER 2
METALLIC MATERIALS
(14 hours)
1

CONTENTS
2.1 Solidification of Pure Metal and Alloys
2.2 Phase diagram: Microstructure development,
Microconstituent of phases.
2.3 Fe‐Fe3C system: Microstructure development,
2.4 Ferrous and Non‐Ferrous Metals
2

2.1 Solidification of Pure Metal
and Alloys
3

2.1 Solidification of Pure Metal and Alloys
• Terminology
• Solution
– Metal Solid Solution
– Type of Solid Solution
• Substitutional Solid Solution
• Interstitial Solid Solution
– The Solubility Limit
• Solidification
4
• Cooling Curve
– Cooling Curve of Pure Metal
– Cooling Curve of Alloys
– Development of Phase
Diagram
– Cooling Curve for Binary
Isomorphous

LEARNING OBJECTIVE
Students should be able to :
• Understand the phase transformation during
solidification process.
• Differentiate between cooling curve for pure
metal and alloys.
5

6
Solvent
In an alloy, the element or compound present in greater amount.
Solute
In an alloy, the element or compound present in lesser amount.
Solution
When two components combine to form a single phase.
Solubility
Degree to which the two components mix.
Solubility limit
The max. concentration of solute that may be added without forming
a new phase.
TERMINOLOGY

Components:
The elements or compounds which are mixed initially
(e.g., Al and Cu)
Phases:
The physically and chemically distinct material regions
that result (e.g.,  and ).
Example :
Liquid
L (liquid) + α (alpha-solid)
Aluminum-Copper Alloy
(darker
phase)
 (lighter
phase)
TERMINOLOGY
* Note that solid, gas and liquid is a phase. 7

SOLUTION
• When 2 components combined they can
either remain separate or combine to form
a single phase which is referred to as a
solution.
• i.e.
– Alcohol and water – completely soluble
– Hot choc – powder mix soluble in water but
limited extent
– Oil and vinegar – insoluble liquids can be
temporarily mixed
8

9
• Most metals are combined to form alloy in order to impart
specific characteristic.
• An alloy is a combination of two or more elements (added
impurity atoms), at least one of which is a metal.
• The addition of impurity atoms to a metal will result in the
formation of a solid solution.
• A solid solution is a solid‐state solution of one or more
solutes in a solvent.
• E.g : Steel/Cast Iron (Iron base alloys),
Bronze/Brass (Copper base alloys),
Al alloys, Ni base alloys, Mg base alloys, Ti alloys.
METALLIC SOLID SOLUTION

10
Characteristic of solid solution:
• Form when solute atoms are added to the host material.
• Crystal structure is maintained.
• No new structure formed.
• Compositionally homogeneous.
Solute
Used to denote an
element/compound present in a
minor concentration
Solvent
Element / compound that is
present in the greatest amount
(host atoms)
METALLIC SOLID SOLUTION

11
TYPES OF SOLID SOLUTION
i. Substitutional solid solution
ii. Interstitial solid solution
Known as point defects
(where an atom is missing or
is in an irregular place in the
lattice structure).

Substitutional Solid Solution
Hume -Rothery Rules
Substitutional solid solution with complete solubility exists when :
RULE PROPERTIES CONDITIONS
1 Atomic radius Less than about ± 15% difference in atomic radii
2 Crystal structure Same crystal structure (e.g : BCC, FCC or HCP).
3 Electronegativity Similar electronegativity/ smaller diff.
4 Valence electron Similar valance electron
12
Note:
Not all alloys
systems that fit these rules
will form appreciable solid
solutions
Host atoms are replaced/substitute with solute/ impurity atoms.

13
EXAMPLE 1 : Cu-Ni system
• Both metals are completely soluble in each other
because all the requirement of Hume Rothery Rules
have been satisfactorily fulfilled.
• The solid phase is a substitutional solid solution.
System RULE 1
Atomic radius, R (nm)
RULE 2
Crystal structure
RULE 3
E/negativity
RULE 4
Valences
Cu
Ni
0.128
0.125
FCC
FCC
1.90
1.80
+2
+2

14
EXAMPLE 2: Cu-Ag system
• Both metals are partially soluble in each other because
one of the requirement of Hume Rothery Rules have not
been satisfactorily fulfilled.
• The solid phase is a substitutional solid solution.
System RULE 1
Atomic radius, R (nm)
RULE 2
Crystal structure
RULE 3
E/negativity
RULE 4
Valences
Cu
Ag
0.128
0.144
FCC
FCC
1.90
1.80
+2
+1

15
The atoms of the parent or
solvent metal are bigger
than the atoms of the
alloying or solute metal. In
this case, the smaller atoms
fit into spaces between the
larger atoms.
Interstitial Solid Solution exists when :
• Impurity atoms fill the voids in the solvent atom lattice.
• It interstices among the host atoms.
• Atomic diameter of an interstitial impurity must be smaller
than host atoms.
• Normal max. allowable concentration of interstitial
impurity atom is low (<10%).
Interstitial Solid Solution

• Solubility Limit: Max concentration for which only a solution
occurs.
• Question : What is the solubility limit at 20oC?
Answer :
If Co < 65wt% sugar:
If Co > 65wt% sugar:
• Solubility limit increases with T:
Ex: Phase Diagram:
Water-Sugar System
Pure
Sugar
Temperature(°C)
0 20 40 60 80 100
Co=Composition (wt% sugar)
L
(liquid solution
i.e., syrup)
Solubility
Limit L
(liquid)
+
S
(solid
sugar)
65
20
40
60
80
100
Pure
Water
THE SOLUBILITY LIMIT
16

SOLIDIFICATION OF PURE METAL
& ALLOYS
COOLING CURVE
PHASE DIAGRAM
17

SOLIDIFICATION
• Solidification is the most important phase transformation
because most of metals/alloys undergo this transformation
before becoming useful objects.
• Solidification involve liquid-solid phase transformation,
e.g : casting process.
• The solidification process differs depending on whether
the metal is a pure element or an alloy.
18

19
Liquid
Nucleus
Liquid
Grain
Grain boundaries
(means region between crystals)
Crystals growing
(irregular grain)
(a) (b) (c)
Nucleation
of Crystals
Crystal
Growth
Crystals Grow
Together and Form
Grain Boundaries
Solution
(Liquid State)
SOLIDIFICATION
Solidification of Pure Metal and Alloys
1. The formation of stable nuclei in the melt (nucleation)
2. The growth of nuclei into crystal
3. The formation of a grain structure

& ALLOYS
COOLING CURVE
PHASE DIAGRAM
20

• Used to determine phase transition temperature.
• Temperature and time data of cooling molten metal is
recorded and plotted.
• Produce a graph known as PHASE DIAGRAM which
shows the relationship among temperature,
composition and phases present in alloy
COOLING CURVE
21

22
A pure metal solidifies at a constant temperature
equal to its freezing point, which is the same as its
melting point.
Figure : Cooling curve for a pure metal during casting
Cooling Curve of Pure Metal

23
Most alloys freeze over a temperature range rather than at
a single temperature.
Figure : a) Phase diagram for a copper-nickel alloy system and
b) Associated cooling curve for a 50%Ni-50%Cu composition during casting
Cooling Curve of Alloys

• Series of cooling curves at different metal composition are
first constructed.
• Points of change of slope of cooling curves (thermal arrests)
are noted and phase diagram is constructed.
• More the number of cooling curves, more accurate is the
phase diagram.
Development of Phase Diagram
24

25
• For pure metal, the cooling
curves show horizontal
thermal arrest at their
freezes points, as seen for
pure A and pure B (at AB
and CD).
• Different composition will
give different cooling
curves.
• The slope changes at L1‐L9
are correspond to the
liquidus point.
• The slope changes at S1‐S9
are correspond to the
solidus points.
Freezing
zone
Cooling Curve For
Binary Isomorphous
L1
S1
A B
D
C
L9
S9
1

26
L1
S1
By removing the time axis and
replacing it with composition
get straight lines
Connection of points on a phase
diagram representing the temp. at
which each alloy in the system begins
to solidify --- obtain liquidus line
Join all the points where the liquid has
solidified is complete --- obtain solidus
line
Red regions – material is liquid
Green regions – solid and liquid
phases are in equilibrium.
Blue regions – material is solid
2
3

& ALLOYS
COOLING CURVE
PHASE DIAGRAM
27

2.2 Phase diagram
Microstructure development,
28

2.2 Phase diagram: Microstructure development,
• Phase Diagram
• The Lever Rule
• Binary Phase Diagram
– Binary Isomorphous Phase Diagram
(COMPLETE SOLID SOLUTION)
– Binary Eutectic Phase Diagram
(NO SOLID SOLUTION)
– Binary Eutectic Phase Diagram
(LIMITED SOLID SOLUTION)
29
• Invariant Equilibrium
• Terminology

Learning objective:
Students should be able to:
• Schematically sketch and label the various phase
regions for simple binary phase diagrams.
• Determine the phase(s) present, composition(s)
and relative amount of phase(s).
• Discuss the development of the microstructures,
upon cooling, for several situations.
• Locate the invariant point and write reaction for
all the transformations for either heating or
cooling.
30

31
What is PHASE DIAGRAM?
A graphical representations of what phases are present in a
materials system at various temperature (T), pressure (P) and
composition (C).
Why do I need to know about PHASE DIAGRAM?
1. Because there is a strong correlation between microstructure
and mechanical properties.
2. Besides, development of alloy microstructure is related to the
characteristics of its phase diagram.
Applications of PHASE DIAGRAM?
1. Casting
2. Soldering
PHASE DIAGRAM

32
Types of PHASE DIAGRAM?
1. Unary – Consists of One components in an alloy
2. Binary – Consists of two components in an alloy
3. Ternary‐ Consists of three components in an alloy
Example: Unary Phase Diagram
PHASE DIAGRAM

33
What do I need to know about BINARY PHASE DIAGRAM?
Definition : Consists two components in an alloy.
Types :
1. Complete solid solution (e.g. Cu and Ni are completely soluble)
2. No solid solution (e.g. Pb insoluble in copper)
3. Limited solid solution (e.g. Sn has limited solubility in Pb)
PHASE DIAGRAM

34
There are three(3) types of binary phase diagram :
1) Complete solid
solution
2) No solid solution 2) Limited solid
solution
Alcohol and water Oil and water Pepper powder and water
−Complete solubility in
liquid and solid
- Result in single phase
- Result in multi phase −Often soluble up to limit
- Result in multi phase
Cu and Ni Pb and Copper Zinc and Copper,
Sn and Pb
BINARY PHASE DIAGRAM

35
BINARY
ISOMORPHOUS
PHASE DIAGRAM
(COMPLETE SOLID SOLUTION)

36
Isomorphous
• Complete liquid & solid solubility
• Only one solid phase forms
• Same crystal structure
Example : Cu-Ni system
• 2 phases: L (liquid), α (FCC solid solution)
• 3 different phase fields/regions
1) Liquid phase(L)
 homogeneous liquid solution (Cu + Ni)
2) Two phases
 α (FCC solid solution) + liquid (L)
3) α phase (FCC solid solution)
 substitutional solid solution (consists both Cu-Ni)
Figure : Cu-Ni system
Note that :
• Liquidus is line above which all of alloy is liquid
• Solidus is line below which all of alloy is solid
BINARY ISOMORPHOUS
PHASE DIAGRAM

37
• Rule 1: If we know T and Co, then we know:
--the # and types of phases present.
wt% Ni20 40 60 80 1000
100 0
110 0
120 0
130 0
140 0
150 0
160 0
T(°C)
L (liquid)

(FCC solid
solution)
L + 
liquidus
solidus
A(1100,60)
B(1250,35)
Cu-Ni system
Some common features of
phase diagrams
 “α”,“β” and “γ” and etc. are used
to indicate solid solution
phases.
 “L” represents a liquid.
BINARY ISOMORPHOUS PHASE DIAGRAM:
# and types of phases

--the composition of each phase (weight percent, wt%).
wt% Ni
20
1200
1300
T(°C)
L (liquid)

(solid)L + 
liquidus
solidus
30 40 50
TA
A
D
TD
TB
B
tie line
L + 
433532
CoCL C
Cu-Ni system
Determination of phase compositions
1. Locate the temperature.
2. If one phase present, the composition
= overall composition (Co) of alloy.
3. If two phase present, use tie line.
composition of phases
38

• Sum of weight fractions:
• Conservation of mass (Ni):
• Combine above equations:
WL  W  1
Co  WLCL  WC
 R
R  S
W 
Co  CL
C  CL
 S
R  S
WL
 C  Co
C  CL
• A geometric interpretation:
Co
R S
WWL
CL C
moment equilibrium:
1 W
solving gives Lever Rule
WLR  WS
THE LEVER RULE
Let WL = fraction of liquid and Wα = fraction of solid (unknown)
39

Ask yourself ?
Larger distanceSmaller
distance
Higher mass
smaller mass
40
THE LEVER RULE

--the amount of each phase [e.g: Single phase (1.0 or 100%)].
Cu-Ni system
SR
Note
•Within single phase alloy, the alloy is completely
(100%) that phase.
•If two phase alloy exists, use Lever Rule
41
weight fractions of phases
wt% Ni
20
1200
1300
T(°C)
L (liquid)

(solid)
L + 
liquidus
solidus
30 40 50
TA
A
D
TD
TB
B
tie line
L + 
433532
CoCL C
R S

42
EXAMPLE : Calculate the amounts of α and L at 1250°C in the
Cu-35% Ni alloy?
THE LEVER RULE

43
EXERCISE : Determine the relative amount on each phase in the Cu 40% Ni alloy
shown in Figure below at 1300°C, 1270°C, 1250°C and 1200°C ?
THE LEVER RULE

44
Consider Co = 35wt% Ni
Figure : Cooling of Cu-Ni alloy
Microstructure
A
B
C
D
E
Microstructure
wt% Ni
20
1200
1300
30 40 50
1100
L (liquid)

(solid)
L + 
L + 
T(°C)
A
D
B
35
Co
L: 35wt%Ni
: 46wt%Ni
C
E
L: 35wt%Ni
46
43
32
24
35
36
: 43wt%Ni
L: 32wt%Ni
L: 24wt%Ni
: 36wt%Ni

45
BINARY EUTECTIC
PHASE DIAGRAM
(NO SOLID SOLUTION)

46
•Region above line ced = liquid solution
•Line ce and ed = liquidus
•Line cfegd = solidus
•Region below line feg = mixture of solid A & B
•Point e = eutectic point
(the lowest temp. at which a liquid solution can exist)
BINARY EUTECTIC PHASE DIAGRAM
(NO SOLID SOLUTION)
Eutectic: the composition of
a mixture that has the lowest
melting point where the
phases simultaneously
crystallize from molten solution
at this temperature.
From the Greek 'eutektos',
meaning ‘easily melted’.
No solid solution where the
components are completely
soluble in the liquid state
but complete insoluble in
the solid state.
Example : Pb-Cu system

47
Determination of phase and phase composition:
Same as in binary isomorphous system.
Determination of weight fraction
Weight fraction of liquid,
WL= R/(R+Q)
Weight fraction of solid A,
WA = Q/(R+Q)
(NO SOLID SOLUTION)
HYPOEUTECTIC HYPEREUTECTIC
Three phases in equilibrium at
eutectic point compositions and
temperature
Eutectic reaction
L A+ B

48
EUTECTIC
HYPOEUTECTIC
HYPEREUTECTIC

The eutectic microstructure forms in the alternating layers which
is known as lamellar:
→ atomic diffusion of lead and tin only occur over relatively short distances
in solid state.
Eutectic α Eutectic β
Figure : Lamellar eutectic structure
(NO SOLID SOLUTION)
49

50
Liquid
Hypoeutectic alloy Hypereutectic alloy
When the composition of an
alloy, places it to the left of the
eutectic point
When the composition of an
alloy, places it to the right of
the eutectic point
First solid to form : Primary α
(a.k.a. proeutectic α)
First solid to form : Primary β
(a.k.a. proeutectic β)
β
(NO SOLID SOLUTION)

51
1. Cooling curve at eutectic alloy
Same as single component (pure metal) because solidification
takes place at a single temperature.
A Solid
Liquid
Figure : Cooling curve at eutectic alloy
(NO SOLID SOLUTION)

52
2. Cooling curve at hypo/hypereutectic alloy
Once the liquid reach TE, it will have the eutectic composition and
will freeze at that temperature to form solid eutectic mixture of
two phases.
TL = temperature of liquid
TE = temperature at eutectic point
Figure : Cooling curve at hypo/hyper
eutectic alloy
Liquid + Solid
Liquid
Solid
(No SOLID SOLUTION)
TE
HYPOEUTECTIC HYPOEUTECTIC

53
BINARY EUTECTIC
PHASE DIAGRAM

54
Limited solid solution where the components are completely
soluble in the liquid state but limited solubility in the solid state.
E.g : Sn‐Pb system, Cu‐Ag system
α, β = solid solution
ae, be = liquidus
ac, cd, bd = solidus
cf, dg = solvus
• 3 single phase region = α, β, L
• Solvus cf denotes the solubility
limit of B in A
• Solvus dg shows the solubility
limit of A in B

CE
Eutectic temp.
(TE) : below TE
form 2 different
solid phases.
Eutectic point
a.k.a. triple point.
Eutectic composition (CE)Figure : Copper-silver phase diagram
Solvus
Liquidus
Solidus
TM Ag
TM Cu
55

56
Determination of phase and phase composition:
Same as in binary isomorphous system
Determination of weight fraction
Weight fraction of liquid,
WL= Q/(R+Q)
Weight fraction of β,
Wβ = R/(R+Q)

• 3 single phase regions
(L, )
• Limited solubility:
: mostly Cu
: mostly Ni
• TE: No liquid below TE
• CE: Min. melting T
composition
Ex.: Cu-Ag system
L (liquid)
 L +  L+

Co, wt% Ag
20 40 60 80 1000
200
1200
T(°C)
400
600
800
1000
CE
TE 8.0 71.9 91.2
779°C
Cu-Ag system
Eutectic reaction
L α + β
(Liq.) (s.s) (s.s)
57

EXERCISE:
1) Label each phase region
(i), (ii) and (iii).
2) Determine Tm for pure
Sn and Bi.
3) Determine the eutectic
temperature and
eutectic composition.
Sn-Bi phase diagram
β
α + L
(i)
(ii)
(iii)
58

EXAMPLE: Pb‐Sn EUTECTIC SYSTEM
For a 40wt%Sn-60wt%Pb alloy at
150oC, find...
--the phases present:
--the compositions of
the phases:
--the relative amounts
of each phase:
Pb-Sn system
L + 
L+

200
T(°C)
18.3
Co, wt% Sn
20 40 60 80 1000
Co
300
100

L (liquid)
 183°C
61.9 97.8
150
59

EXERCISE:
a) What is the TE and CE?
b) Consider an alloy contains 20wt%
Cu, what are the compositions of
the  phase in equilibrium at just
above and  below TE?
c) Consider Ag‐Cu alloy contains
40 wt% Ag, at temperature just
below TE, find the relative amounts
of each phase. 60

(LIMITED SOLID SOLUTION) : MICROSTRUCTURES
1. Consider Co < 2wt% Sn
Liquid
Pb-Sn system
Liquid +  grains of solid
polycrystal of  grains
phase solid)
61

: Cowt%Sn
L + 
200
T(°C)
Co, wt% Sn
10
18.3
200
Co
300
100
L

30
L: Cowt%Sn
 + 
400
(sol. limit at TE)
TE
2
(sol. limit at Troom)
L



2. Consider 2wt%Sn < Co < 18.3wt%Sn
 polycrystal + fine  crystals
phase solid)
Pb-Sn system
Liquid
Liquid +  grains of solid
polycrystal of  grains
phase solid)
62

L + 
200
T(°C)
Co, wt% Sn
20 400
300
100
L

60
L: Cowt%Sn
 + 
TE
: 18.3wt%Sn

0
80 100
L + 
CE18.3 97.8
61.9
183°C
: 97.8wt%Sn
3. Consider Co = CE = 61.9 wt% Sn
Eutectic reaction
L α + β
(Liq.) (s.s) (s.s)
Pb-Sn system
EUTECTIC
Cooling Curve
63

L + 
200
T(°C)
Co, wt% Sn
20 400
300
100
L

60
L: Cowt%Sn
 + 
TE

0
80 100
L + 
Co18.3 61.9
L

L
primary 
97.8
S
S
R
R
eutectic 
eutectic 
4. Consider 18.3wt%Sn < Co < 61.9wt%Sn
HYPOEUTECTIC
64

5. Consider 61.9wt%Sn < Co < 97.8wt%Sn
HYPEREUTECTIC
Pb‐Sn system
65

HYPOEUTECTIC & HYPEREUTECTIC
Eutectic
mixture

Proeutectic β
or Primary β
Proeutectic 
or Primary 
Eutectic
mixture

: Co=85wt%Sn
Eutectic reaction
L α + β
(Liq.) (s.s) (s.s)
66

67
Different systems have different types of alloy transformation.
Invariant equilibrium involve :
– 3 phases co‐exist in.
– Exist only at one temperature / fixed temp.
– Composition for 3 phases co‐exist is fixed at the point.
– Zero degree of freedom.
Below are the example of alloy transformation at invariant
equilibrium :
1.Eutectic
2.Eutectoid
3.Peritectic
4.Peritectoid
INVARIANT EQUILIBRIUM
5. Metatectic
6. Monotectic
7. Synthectic

Invariant Point Reaction Example System
Eutectic l α + β Ag‐Cu,
Pb‐Sn
Eutectoid γ α + β Fe‐C,
Al‐C
Peritectic l + α β Cu‐Zn
Peritectoid β + α γ Al‐Ni,
Cu‐Zn
Metatectic α l + β U‐Mn
Monotectic l1 α + l2 Cu‐Pb
Syntectic l1 + l2 α K‐Zn,
Na‐Zn
L
α β
L + α L + β
α + β
γ
α βγ + α γ + β
α + β
α
β
L
β + α
L + α
L + β
α β
γγ + α
α + β
γ + β
α
β L
β + α L + α
L + β
L1
α L2
L1 + α L1 + L2
L2 + α
L1
α
L2L1 + α
L1 + L2
L2 + α
68

69
EXAMPLE: Find the eutectoid and peritectic reactions in the
Cu-Zn system?

70
Exercise: Refer to zinc‐copper phase diagram. Specify temperature‐composition points at
which all eutectics, eutectoids and peritectics phases’ transformation occurs. Also for each,
write the reaction upon cooling.

71
• Liquidus : Line above which all of alloy is liquid.
• Solidus : Line below which all of alloy is solid.
• Solvus : Boundaries between solid phase regions.
• Invariant point : It is a point at which three phases are in
equilibrium.
• Eutectic structure : The resulting microstructure consists of
alternating layers, called lamellae, of α and β that form during
eutectic reaction.
• Proeutectic : Form before (higher temperature) eutectic.
• Terminal solid solutions : Phases containing the pure components
which situated at the end of the phase diagram.
• Hypoeutectic : Having a composition less than eutectic.
• Hypereutectic : Having a composition greater than eutectic.
TERMINOLOGY

2.3 Fe‐Fe3C system
Microstructure development,
72

2.3 Fe‐Fe3C system: Microstructure development,
• Fe‐Fe3C Phase Diagram
− Allotropy Transformation
− Solid Phases
− Phase Transformation Reactions
− Microstructural Changes
Eutectoid
Hypoeutectoid Steel
Hypereutectoid Steel
73

LEARNING OBJECTIVE
74
• Sketch and label in the iron‐iron carbide phase
diagram up to eutectic isotherm.
• Specify whether the alloy is hypoeutectoid or
hypereutectoid.
• Identify the proeutectoid phase.
• Explain the development of the microstructure
at a temperature just below the eutectoid.
• Compute the relative amount of pearlite and
proeutectoid phase.

The effect of
adding C into Fe
will introduce
various types of
steel and cast iron
which are
represented by
the iron‐iron
carbide phase
diagram.
75
Fe-Fe3C PHASE DIAGRAM

ALLOTROPIC TRANSFORMATION
• A material that can exist in more than one lattice structure
(depending on temperature‐heating@cooling) allotropic.
• An allotropic material is able to exist in two or more forms having
various properties without change in chemical composition.
• E.g : Upon heating, pure iron experiences two changes in crystal
structure:
– At room temperature, it exists as ferrite,or α iron (BCC).
– When we heat it to 912°C, it experiences an allotropic
transformation to austenite,or γ iron (FCC).
– At 1394°C, austenite reverts back to a BCC phase called δ ferrite.
77

912°C
1538°C
768°C
ALLOTROPIC TRANSFORMATION
Allotropy of iron(Fe)
Delta iron Austenite Alfa iron
(BCC) (FCC) (BCC)
High Temp
Low Temp
Moderate
Temp
78

Phases present in Fe-Fe3C system :
1) δ Ferrite
2) γ (Austenite)
3) α Ferrite
4) Fe3C (Cementite)
5) α + Fe3C (Pearlite)
79
SOLID PHASES

80
1) δ Ferrite
• This is a solid solution of carbon in iron and has a BCC crystal
structure (same structure as α‐ferrite).
• It is a phase which exists at extreme temperature (>1400°C)
and stable only at high temperature, above 1394 °C.
• It melts at 1538 °C.
• The maximum solubility of C in Fe is 0.09% at 1495°C. This has
no real practical significance in engineering.
SOLID PHASES
Figure : δ Ferrite crystal structure

81
2) Austenite (γ Iron)
• It is also known as (γ) gamma‐iron, which is an interstitial solid
solution of carbon dissolved in iron with a face centered cubic
crystal (FCC) structure.
• Transforms to BCC δ‐ferrite at 1394°C.
• The maximum solubility of carbon in austenite, 2.14%.
• Austenite is normally unstable below eutectoid temperature
unless cooled rapidly.
• It is a non magnetic material.
Figure :Austenite (γ iron) crystal structure
SOLID PHASES

82
3. α Ferrite
• It is also known as alpha(α) iron, which is an interstitial solid
solution of a small amount of carbon dissolved in iron with a Body
Centered Cubic (BCC) crystal structure.
• It is the softest structure on the iron‐iron carbide diagram.
• Stable form of iron at room temperature.
• The maximum solubility of C is 0.022 wt%.
• Ferrite dissolves considerably less carbon than austenite.
• Transforms to FCC γ‐austenite at 912°C.
• α ‐ferrite is magnetic (below 768°C).
Figure : Ferrite (α iron) crystal structure
SOLID PHASES

83
3) Cementite (Fe3C)
• Cementite is also known as iron carbide which has a chemical formula,
Fe3C.
• Fe3C is an intermetallic compound. It is because a fixed amount of C
and a fixed amount of Fe are needed to form cementite (Fe3C).
• It is a hard and brittle material, low tensile strength and high
compressive strength.
• It contains 6.70 wt% C and 93.3 wt% Fe.
• This intermetallic compound is metastable,
it remains as a compound indefinitely at room
temperature, but decomposes (very slowly,
within several years) into α-Fe and C
(graphite) at 650 - 700°C.
SOLID PHASES

84
5) α + Fe3C (Pearlite)
• It is resulted from transformation of austenite of eutectoid
composition on very slow cooling.
• Pearlite is a laminated structure (lamellar structure) formed of
alternate layers of ferrite (white matrix‐ferritic background) and
cementite (thin plate).
• In most steels, the microstructure consists of both α+Fe3C
(pearlite) phases.
• It has intermediate mechanical properties between α and Fe3C.
Cementite
(hard)
Ferrite
(soft)
Figure : Pearlite microstructure
(Light background is the ferrite matrix,
dark lines are the cementite network)
SOLID PHASES

85
PhaseTransformation Reactions
C
3 Invariant points:

The iron-carbon diagram exhibits three phase transformation reactions :
86

%
Carbon
HYPOHYPO HYPER HYPER
EUTECTOID EUTECTIC
Peritectic
87

MICROSTRUCTURAL CHANGES
• Microstructure that exists in those reactions depends on :
− Composition(carbon content)
− Heat treatment
• Three significant regions can be made relative to the steel portion of the
diagram which known as:
1) Eutectoid
− Carbon content 0.76% and temperature 727°C.
− It entirely consists of pearlite (α + Fe3C).
2) Hypoeutectoid
− Carbon content from 0.022 to 0.76%.
− It consist of pearlite and primary (proeutectoid) ferrite.
3) Hypereutectoid
− Carbon content from 0.76 to 2.14%.
− It consist of pearlite and primary (proeutectoid) cementite.
88

89
EUTECTOID STEEL
γ α +    Fe3C
austenite                 pearlite
αFe3C
Pearlite
Figure : Photomicrograph of a
eutectoid steel showing the pearlite
microstructure consisting of
alternating layers of α ferrite
(thick layers, light phase) and
Fe3C (thin layers most of which
appear dark).
Note :
• Many cementite layers are so thin
that adjacent phase boundaries are
indistinguishable (appear dark).
• Alternating layers  of α and Fe3C
form pearlite.

90
The layers of alternating phases in
pearlite are formed for the same
reason as layered structure of eutectic
structures:
Redistribution C atoms
between ferrite (0.022 wt%)
and cementite (6.7wt%) by
atomic diffusion.
The pearlite exist as grains,
often termed as colonies.
EUTECTOID STEEL

91
HYPOEUTECTOID STEEL
Figure : Microstructures for Fe-Fe3C system of
hypoeutectoid composition Co
α’        +       α +    Fe3C
(proeutectoid ferrite)    +      (pearlite)
Note :
Eutectoid α = Ferrite that is present in the
pearlite.
Proeutectoid (meaning pre- or before
eutectoid) = Formed above eutectoid
temperature.
α’               +             γ
(proeutectoid ferrite)   +      (Austenite)
γ
(Austenite)

93
EXERCISE
Consider an Fe – C alloy containing 0.25 wt% C, at a
temperature just below the eutectoid temperature.
Determine
a) the mass fractions of proeutectoid ferrite and pearlite
HYPOEUTECTOID STEEL
b) the mass fractions of total ferrite, eutectoid ferrite
and cementite.

94
HYPEREUTECTOID STEEL
Note :
Eutectoid Fe3C= Cementite that is present in
the pearlite
Figure : Microstructures for Fe-Fe3C system of
hypereutectoid composition
Fe3C’           +        γ
(proeutectoid cementite)  +  (Austenite)
γ
(Austenite)
Fe3C’      +     α +    Fe3C
(proeutectoid cementite)  +         (pearlite)

96
EXERCISE
Consider an Fe – C alloy containing 1.25 wt% C, at a
temperature just below the eutectoid temperature. Determine
a) the mass fractions of proeutectoid cementite and pearlite
HYPEREUTECTOID STEEL
b) the mass fractions of total ferrite, cementite and
eutectoid cemmentite.

97
Figure : Photomicrograph of a 1.4wt% C steel
having a microstructure consisting of a white
proeutectoid cementite network surrounding
the pearlite colonies.
Hypereutectoid steel
+Fe3C (pearlite)
+
proeutectoid cementite(Fe3C)
Hypoutectoid steel
+Fe3C (pearlite)
+
proeutectoid ferrite(α)
Figure : Photomicrograph of a 0.38wt% C
steel having a microstructure consisting of
pearlite and proeutectoid ferrite.
HYPO vs HYPER EUTECTOID STEEL

2.4Ferrous and
Non‐Ferrous Metals
98

2.4 Ferrous and Non‐Ferrous Metals
• Introduction
• Classification of Metal Alloys
• Classification of Ferrous Alloys
– Steel
• Plain Carbon Steel
• Low Carbon Steel
• Medium Carbon Steel
• High Carbon Steel
• Stainless Steel
• Tool Steel
– Cast Iron
• Gray Cast Irons
• Nodular (Ductile) Cast Irons
• White Cast Irons
• Malleable Cast Irons
99
• Non‐Ferrous Alloys
– Aluminium and its alloys
– Copper and its alloys
– Magnesium and its alloys
– Titanium and its alloys
– The Noble Metal
– The Refractory Metals

LEARNING OBJECTIVE
• Differentiate the differences between ferrous
and nonferrous metals.
• Describe the characteristics of white, gray,
ductile and malleable cast irons.
• Understand the properties and applications of
metals and its alloys.
100

INTRODUCTION
1. Ferrous
• Metal alloys that
contain iron as a prime
constituent.
• E.g : steels, cast iron.
• Tend to have a higher
chance of corrosion.
2) Nonferrous
• Metal alloy contain less
@ no iron.
• E.g : Cu, Al, Mg, Ti and
its alloys
• Have a much higher
resistance to corrosion.
Metal alloys can be divided into two categories :
Note :
The word ferrous is derived from the Latin term "Ferrum" which means
"containing iron".
101

INTRODUCTION
Advantages of Ferrous
alloys over Non‐Ferrous
alloys:
Advantages of Non‐Ferrous
alloys over ferrous alloys:
– Generally greater strength.
– Generally greater stiffness.
– Better for welding
– Good resistance to corrosion.
– Casting and cold working
processes and are often
easier.
– High ductility.
– Higher thermal and electrical
conductivities.
– Colors.
102

CLASSIFICATION OF METAL ALLOYS
103
Al Cu Mg Ti
Noble
Metal
Refractory
metal

CLASSIFICATION OF FERROUS ALLOY
 Definition : Those of which iron is the prime constituent.
 Advantages :
1. Iron ores exist in abundant quantities within the earth’s
crust.
2. Produced from economical process : Extraction, refining,
alloying and fabrication techniques are available.
3. Versatile material : Wide range of mechanical and physical
properties.
 Disadvantages :
1. Tends to corrode.
2. High density.
3. Low electrical conductivity.
104

• The ferrous alloys are classified based on the
percentage of carbon present in the ferrous.
(steel <2.14 %C, cast iron 2.14 ‐ 4.3%C)
• Carbon is the most important commercial
steel alloy (↑C, ↑hardness, ↑ strength,
↑bri leness, ↓ weldability)
105

106

Low Alloy High Alloy
low carbon
<0.25wt%C
med carbon
0.25-0.6wt%C
high carbon
0.6-1.4wt%C
Uses auto
struc.
sheet
bridges
towers
press.
vessels
crank
shafts
bolts
hammers
blades
pistons
gears
wear
applic.
wear
applic.
drills
saws
dies
high T
applic.
turbines
furnaces
V. corros.
resistant
Example 1010 4310 1040 4340 1095 4190 304
Additions none
Cr,V
Ni, Mo
none
Cr, Ni
Mo
none
Cr, V,
Mo, W
Cr, Ni, Mo
plain HSLA plain
heat
treatable
plain tool
austentitic
stainless
Name
Hardenability 0 + + ++ ++ +++ 0
TS - 0 + ++ + ++ 0
EL + + 0 - - -- ++
increasing strength, cost, decreasing ductility
Steels
107

Steels
• Are iron carbon alloys that may contain carbon
less than 2.14%.
• Classification by carbon content
– Low, medium and high carbon type
• Subclasses by concentration of other alloying
elements :
– Plain carbon steel
– Alloy steel
• The microstructures of steel are normally ferrite
and relatively soft and weak but good ductility
and toughness.
108

First digit indicates the family to which the steel belongs (a.k.a. the major alloying elements) :
Second digit indicate % of major alloying elements (1 means 1%).
Last two digits(3rd and 4th number) indicate amount of carbon in steel (10 means 0.10% C).
Example
• SAE 5130 means alloy chromium steel, containing 1% of chromium and 0.30% of Carbon.
• AISI 1020 which means 10 indicates plain carbon steel with 0.2% amount of Carbon.
*SAE : Society of Automotive Engineers *AISI : American Iron and Steel Institute 109
Steels

Plain Carbon Steels
• Iron with less than 1% carbon alloy contains a
small amount of manganese, phosphorous, sulfur
and silicon.
• Disadvantages of plain carbon steel:
– Hardenability is low
– Loss of strength and embrittleness
– Subjected to corrosion in most environments
• 3 groups:
– Low carbon steels
– Medium carbon steels
– High carbon steels
110

Low Carbon Steels (< 0.25%C )
Plain carbon steels
• unresponsive to heat treatments
intended to form martensite.
• Microstructures consist of ferrite
and pearlite
• Properties:
– Relatively soft and weak, but
possess high ductility and toughness
– Good formability, Good weldability
– Low cost
– Rated at 55‐60% machinability
• Application: Auto‐body
components, structural shapes,
sheets for pipelines, building,
bridges, tin cans, nail, low
temperature pressure vessel.
High‐strength low alloy (HSLA)
steels
• Low Carbon Steel combine with 10
wt% of alloying elements, such as
Mn, Cr, Cu, V, Ni, Mo
• Properties:
– higher strength than plain low
carbon steels.
– ductile, formable and machinable
– More resistance to corrosion
• Strengthening by heat treatment.
• Application : bridges, towers,
support columns in high rise
building, pressure vessels.
111

Medium Carbon Steel
• Composition: 0.25 ‐ 0.6% C
• Advantages:
– Machinability is 60‐70%.
Both hot and cold rolled
steels machine better
when annealed.
– Good toughness and
ductility
– Fair formability
– Responds to heat
treatment but often used
in natural condition.
• Plain medium carbon steel
−Low hardenability
− Heat treatment:
quenching and tempering
• Heat treatable steel
−Containing Cr, Ni and Mo
−Heat treated alloy stronger
than Low Carbon Steel, lower
ductility and toughness than
Low Carbon Steel
Applications : Couplings, forgings, gears, crankshafts other high‐strength
structural components.
: Steels in the 0.40 to 0.60% C range are also used for rails,
railway wheels and rail axles. 112

High Carbon Steels
• Composition: 0.6% ‐ 1.4% C
• Properties:
– hardest
– strongest
– least ductile of the carbon
steels
• Application:
– Used for withstanding wear.
– A holder for a sharp cutting
edge.
E.g : drills, woodworking tools,
axes, turning and planning tools,
milling cutters, knives.
– Used for spring materials,
high‐strength wires, cutting
tools, and etc.
• Advantages:
– Hardness is high
– Wear resistance is high
– Fair formability
• Disadvantages:
– Low toughness, formability
– Not recommended for
welding
– Usually joined by brazing with
low temperature silver alloy
making it possible to repair or
fabricate tool steel parts without
affecting their heat treated
condition.
113

Stainless Steels
• Primary alloying element is chromium (>11%)
• Others element : Nickel, Manganese, Molybdenum.
• Called stainless because in the presence of oxygen, they develop a
thin, hard, adherent film of chromium oxide (Cr2O3) that protect the
metal from corrosion.
• Highly resistance to corrosion.
• 3 basic types of stainless are
– Martensite
– Ferritic
– Austenitic
• Applications
− Decorative trim, nozzles.
− Springs, pump rings, aircraft fittings.
− Cookware, chemical and food processing equipment.
− Turbine blades, steam boilers, parts in heating furnaces.
− Temporary implant devices such as fractures plates, screw and hip nails.
− The best choice for the walls of a steam boiler because it is corrosion resistant
to the steam and condensate.
114

Tool Steels
• High carbon steel alloys (containing Cr, V, W and Mo)
that have been designed to prevent wear resistance
and toughness combined with high strength.
• Have excess carbides (carbon alloys) which make
them hard and wear resistant.
• Most tool steels are used in a heat treated state
generally hardened and tempered.
• Applications:
– gauges, shear knives, punches, chisels, cams, mould for
die casting.
– Best choice for a drill bit because it is very hard and wear
resistant and thus will retain a sharp cutting edge.
115

Cast Irons
• Carbon contents : Greater than 2.14wt% C.
• Si content : 0.5‐3wt%Si
(used to control kinetics of carbide formation)
• Commercial range : 3.0‐4.5 wt% C + other alloying elements.
• The differences between cast irons and steels :
– Carbon content.
– Silicon content.
– Carbon microstructure (stable form and unstable form).
• Properties :
– Low melting points (1150‐1300°C).
– Some cast iron are brittle.
• Microstructure:
– Most commonly graphite (C) & ferrite.
116

Cast Irons
• Properties of cast iron is controlled by three main factors:
– The chemical composition of the iron
– The rate of cooling of the casting in the mould
– The type of graphite formed
• Advantages:
– Low tooling and production cost
– Ready availability
– Good machinability without burring
– Readily cast into complex shapes
– High inherent damping
– Excellent wear resistance and high hardness
• Types of cast irons :
• Gray Cast Irons
• Nodular (Ductile) Cast Irons
• White Cast Irons
• Malleable Cast Irons
117

Gray Cast Irons
• Composition : Carbon content : 2.5 ‐ 4.0 wt% C and
Silicon content : 1.0 ‐ 3.0wt% Si.
• Microstructure : Graphite flakes surrounded by α‐ferrite or
pearlite matrix.
• The formation of graphite occurs because of the cooling rate is
too slow where austenite in unstable position and brake down
to give graphite microstructure.
• Properties:
– Less hard and brittle (easy to machine)
– Very weak in tension due to the pointed and sharp end of graphite flake
– Good during compression (high compressive strength)
– Low shrinkage in mould due to formation of graphite flakes
– High damping capacity
– Low melting temperature (1140‐1200oC).
• Applications: Base choice for milling machine base because it
effectively absorbs vibration (good vibration damping).
118

THE MICROSTRUCTURE OF
GRAY CAST IRONS
Graphite flakes
* Graphite flakes shows fracture surface (gray appearance).
Figure : Dark graphite flakes in a‐Fe matrix.
119

Ductile (Nodular) Cast Irons
• Composition: Mg or Ce is added to the gray iron composition
before casting occurs (to prevent the formation of
graphite flakes during the slow cooling of the iron)
• Microstructure : Nodular or spherical‐like graphite structure in
pearlite or ferritic matrix.
• Properties :
– Significant increase in material ductility.
– Tensile strength > gray cast iron.
– Others mechanical properties ≈ steel.
• Applications : Valves, pump bodies, gear and other automotive
and machine components.
• A HT can be applied to pearlite nodular iron to give
microstructure of graphite nodules in ferrite (ferrite structure is
more ductile and weldable but less tensile strength)
120

DUCTILE (or NODULAR) CAST IRONS
Figure : Dark graphite nodules in α‐Fe matrix.
Graphite nodules (a.k.a. spherical‐like)
* Note that the carbon is in the shape of small sphere, not flakes.
121

White Cast Iron
• Composition: 2.5 < C < 4.0%C and Si<1%
• Microstructure : Pearlite and cementite
(due to rapid cooling).
• An intermediate metal for the production of malleable cast
iron.
• Properties:
– Relatively very hard, brittle and not weldable compare
to gray cast iron
– When it is annealed, it become malleable cast iron
– Not easily to machine
– Fracture surface: white appearance
122

WHITE CAST IRONS
Figure : Light Fe3C regions surrounded by pearlite.
Pearlite
Fe3C
(Light regions)
123

Malleable Cast Irons
• Is produced by the HT of white cast irons
− Heating temperature: 800oC – 900oC
− Duration : 2 or 3 days (50 hours)
− Heating environment: Neutral atmosphere
• Microstructure : A clumps (rossette) of graphite
(due to decomposition of cemmentite) surrounded by a
ferrite or pearlite matrix
• Properties:
− Similar to nodular cast iron and give higher strength
− More ductile and malleability
• Applications : Pipe fittings, valve parts for railroad, marine
and other heavy duty.
124

MALLEABLE CAST IRONS
Figure : Dark graphite rosettes in α‐Fe matrix.
Graphite rosettes
125

Non‐Ferrous Alloys
• Definition: Used for alloys which do not have iron as
the base element.
• Examples: Al alloys, Cu alloys, Mg alloys, Ti alloys,
Noble metals, Refractory metals, etc.
• Advantages of Ferrous alloys over Non‐Ferrous alloys:
– Generally greater strength
– Generally greater stiffness ( ↑E)
– Better for welding
• The advantages of Non‐Ferrous alloys over ferrous
alloys:
– Good resistance to corrosion
– Much lower density
– Casting is often easies ( ↓ mel ng points)
– Cold working processes are often easier (ductility)
– Higher thermal and electrical conductivities
– colors 126

NonFerrous
Alloys
• Cu Alloys
Brass: Zn is subst. impurity
(costume jewelry, coins,
corrosion resistant)
Bronze: Sn, Al, Si, Ni are
subst. impurity
(bushings, landing
gear)
Cu-Be:
precip. hardened
for strength
• Al Alloys
-lower : 2.7g/cm3
-Cu, Mg, Si, Mn, Zn additions
-solid sol. or precip.
strengthened (struct.
aircraft parts
& packaging)
• Mg Alloys
-very low : 1.7g/cm3
-ignites easily
-aircraft, missles
• Refractory metals
-high melting T
-Nb, Mo, W, Ta• Noble metals
-Ag, Au, Pt
-oxid./corr. resistant
• Ti Alloys
-lower : 4.5g/cm3
vs 7.9 for steel
-reactive at high T
-space applic.
NON‐FERROUS ALLOYS
127

Aluminium and its alloys
• Atomic weight 26.97; Crystal structure: FCC
• Appearance: silvery white metal
• Tm=660oC, boiling point 2270oC
• Relatively low density 2.7 g/cm3
(very light i.e. light weight vehicle, vessels, etc.)
• Tensile strength= 45 MPa, E : 7.5 GPa
• Ductile and malleable
• High resistant to corrosion
(Al naturally produces a fine oxidized surface film which protect it from corrode)
• Stable against normal condition but attacked by both acids & alkalis.
• Nonmagnetic
• High electrical and thermal conductivities (second to copper)
• Non toxic (widely used as packing materials (food))
Characteristics
128

The All Aluminum
Audi A8
(a) The Audi A8 automobile
which has an all‐aluminum
body structure.
(b) The aluminum body
structure, showing various
components made by
extrusion, sheet forming,
and casting processes.

Disadvantages :
• Difficult to weld.
• Prone to severe spring back.
• Abrasive to tooling.
• Expensive than steel.
• Low melting point 660oC.
Applications :
• Used in applications that required
lightness, high corrosion resistance,
electrical and thermal conductivities.
• E.g : cooking utensil, container,
appliances, building materials and
etc.
130

• Al alloys are alloys in which Al is the predominant metal.
• The typical alloying elements are : Cu, Mg, Mn, Si and Zn.
• Al alloys can be divided into 2 groups:
1. Wrought alloys ‐ shaping by working process.
E.g. forging, extrusion, rolling
– Non‐heat treatment wrought alloys
– Heat treatable wrought alloys
2. Cast alloys – shaping by casting process. E.g : die casting.
– Non‐heat treatment cast alloys
– Heat treatable cast alloys
 Non‐heat treatment alloys ‐An alloy which cannot be improved by heat treatment
 Heat treatable alloys ‐ An alloy whose mechanical strength is improved by
precipitation hardening/ martensitic transformation.
131

Non‐heat treatment wrought alloys
• Do not respond to HT but have their properties controlled by the
extent of the working to which they are subjected.
• E.g. Al with Mn (increase the TS of Al)
• Applications: kitchen utensil, tubing, constructional material for
boats and ship
Heat treatable wrought alloys
• The properties changed by HT.
• Common alloy addition: Cu, Zn, Si
• E.g : 4% Cu, 0.8% Mg, 0.5% Si, 0.7% Mn and 94% Al is known as
Duralumin. The heat treatment process used is quenching and then
precipitation hardening at room temperature for about 4 days. This
alloys is widely used in aircraft, bodywork.
• Applications: aircraft, bodywork, container bodies etc.
132

Non‐heat treatment cast alloys
• E.g. Al with 9 –13% Si. The addition of Si will
increase its fluidity for casting purposes.
• Applications: oil sumps, gear boxes, radiators.
Heat treatable cast alloys
• The addition of Cu, Mg to Al alloys enable the
alloys to be heat treated to give a high
strength casting materials.
133

Aluminum Alloy Properties and
Applications

Copper and its alloys
• Atomic weight 63.57
• Appearance: Reddish metal of bright luster
• Highly malleable and ductile
• High electrical and heat conductivitiy
• Excellent corrosion resistance.
• Relatively high strength.
• Tm=1083oC, boiling point 2336oC
• Sg = 8.94
• Used in pure state as sheet, tube, rod and wire
Disadvantages :
1) Difficult to machine.
2) Expensive.
Characteristics
Applications : Jet aircraft landing gear bearing, radiator parts for cars and trucks,
surgical and dental instruments.
135

i. Brasses
• Cu based alloys in which Zn is the
principal added element.
• Harder and stronger that Cu or Zn
• Malleable and ductile
• Develops high tensile with cold
working
• Ease of working
• Colour
• Resistance to atmospheric and
marine corrosion
• Used for electrical fittings,
ammunition cases, screws,
household fittings, and
ornaments.
ii.Bronze
• Contain up to 8% Sn and can be
cold work
• Softer and weaker than steels
• It resists corrosion (especially in
seawater)
• Wrought bronzes are stronger,
better corrosion resistance but
high cost compare brass
• Widely used for spring, bearing,
bushing and similar fittings
136
• Alloyed by other elements: Zn, Al, Sn and Ni
• Examples: Brass, Bronze, Copper‐Nickel, Copper‐Zinc‐Nickel, Aluminium Bronze

iii. Copper – Nickel alloys
• a.k.a – cupronickels
• It is an alloy of Cu that
contains Ni and strengthening
elements, such as Fe and Mn
• Good resistance to seawater,
alkalies, sulphric acids and
alkaline solution.
• Poor resistance to nitric acids,
cyanide solution
• Application: turbine blade,
valve parts, pump rod liners
and impellers
iv. Aluminium Bronze
• Cu rich Al alloys
• Properties
− High strength
− Resistance to corrosion
and wear
− High resistance to
fatigue
− Fine golden colour
− Possibility of heat
treatment in manner
similar to steel
• Application: heat exchanger
tubes
137

Wrought Copper and Brasses:
Properties and Applications

Wrought Bronzes: Properties and
Applications

Magnesium and its alloys
• Atomic weight 24.302, Appearance: Silver‐white
• Density 1.7 g/cm3, Tm = 627oC
• Light, malleable, ductile metallic element
• Low TS, relatively soft, low E.
• At Troom difficult to deform. Most fabrication is by casting or hot working.
• Corrosion resistance in natural atmosphere. On the other hand relatively
unstable especially susceptible to corrosion in marine environments.
– E.g. Mg anode provide effective corrosion protection for water heaters,
underground pipelines, ship hulls and ballast tanks.
• Mg alloys are used in applications where lightness is primary consideration,
e.g. aircraft components, missile application.
• Replaced engineering plastics that have comparable densities since Mg are
stiffer, more recyclable and less costly to produce. Example
– in a variety of handheld devices (chain saws, power tools, hedge clippers),
– in automobiles (steering wheel and column, seat frames, transmission cases) and
– in audio‐video‐computer‐communications equipment (laptop computers, cam
recorders, TV sets, cellular telephones)
140

Wrought Magnesium Alloys:
Properties and Forms

Titanium and its alloys
• Relatively low density (4.5 g/cm3)
• High Tm = 1668oC, E= 107 GPa
• Low strength when pure but alloying gives a considerable
increase in strength, highly ductile and easily forged and
machined
• Expensive metal: excellent corrosion resistance (immune
to air, marine and a variety of industrial environment);
high cost reflecting the difficulties in extraction and
formation of material.
• Limitation: chemical reactivity with other materials at
elevated temperature
• Applications:
– pure Ti – chemical plant components, surgical implants, marine
and aircraft engine parts.
– Ti alloys – steam turbine blades, rocket motor cases
142

Wrought Titanium Alloy Properties and
Applications

The Noble Metals
• Characteristics/Properties:
– Expensive (precious)
– Soft, ductile
– Resistant to corrosion and oxidation
– Good electrical conductivity
• A group of 8 elements:
– The precious metal group: silver (Ag) and gold (Au).
– The six platinum metals: platinum (Pt), palladium (Pd),
iridium (Ir), rhodium (Rh), ruthenium (Ru), and
osmium (Os).
144

The Noble Metals
Element Properties Application
Ag ↑electrical & themal
conductivity
jewelry, dental restoration materials, brazing solder,
coins, silver coatings (reflectors), electrical contact
Au ↑ corrosion resistance,
nontarnishing characteristics,
good electrical conductivity
jewelry, electric wiring, colored‐glass production,
dentistry, electronics, brazing solder, heat shielding
foil in the engine compartment
Pt ↑ corrosion resistance,
↑ Tm,
ductility
thermocouple, thermometer elements, electrical
contact, electrodes, jewelry, catalyst in the
production of sulfuric acid
Pd properties ≈Pt
however lower cost
telephone relay contacts, catalyst to remove oxygen
from heat treating atmosphere
Ir most corrosion resistant,
↑temperature
crucibles, extrusion dies
Rh ↑ reﬂec vity,
↑corrosion resistance
reflector for motion picture projectors and aircraft
searchlight, alloying addition to Pt and Pd
Ru corrosion resistant catalyst for synthesis of hydrocarbon, a hardener
for Pt and Pd
Os ↑ hardness,
↑wear resistance,
good corrosion resistance
fountain‐pen nibs, phonograph needles, electrical
contacts instrument pivots
145

The Refractory Metals
• Five elements widely used: niobium (Nb), molybdenum (Mo),
tantalum (Ta), tungsten (W), and rhenium (Re).
• Properties:
– Tm above 2000 °C
– High hardness at room temperature.
– Chemically inert
– Relatively high density
– Resistance to heat and wear
– Resistant to corrosion (ability to form a protective layer), although
they do readily oxidize at high temperatures.
– Resistance to creep (the tendency of metals to slowly deform under
the influence of stress)
– Resistant to thermal shock (repeated heating and cooling will not
easily cause expansion, stress and cracking).
– Good electrical and heat conducting properties
• Applications include tools to work metals at high temperatures, wire
filaments, casting molds, and chemical reaction vessels in corrosive
environments. 146

The Refractory Metals
Element Properties Application
Nb
↑ Tm, immune to attack by
most acids, a wide variety
of strengths and elasticity
Superconducting alloys for electronic applications;
High strength alloys for aerospace applications;
Moderately strengthened alloys for nuclear applications;
aircraft gas turbines, aerospace rocket engines, insulators
Mo
↑ Tm,
↑ resistance to arc erosion
Missile and rocket engine components, Die‐casting dies,
Alloying additions; Electric furnace heating elements,
boats, heat shields
Ta
corrosion resistant,
high temperature strength,
low vapor pressure
Crucibles for handling molten metal and alloys,
Electrolytic capacitors, Heat exchangers, Cutting tools,
Surgical implants, Aerospace engine components,
Vacuum tube filaments,
W
↑ Tm (3410oC),
↑ density
Lamp filaments, Anodes and targets for x‐ray tubes,
Electrodes for inert gas arc welding, Forming dies,
Catalysts in chemical and petrochemical processes,
Lubricants, Cutting tools for metal machining
Re
↑ Tm,
↑ density,
ductile to brittle transition
temperature
Catalysts for reforming in conjunction with platinum,
nuclear reactors, semiconductors, electronic‐tube
components, thermocouples, gyroscopes, miniature
rockets, electrical contacts, thermionic converters,
aerospace applications. 147

Nonferrous Metals and Alloys

1
Thermal Treatment
of Metallic Material
(8 hours)


3.1 Time-Temperature Transformation Diagram (TTT)
 Products of cooling austenite
 Factors affecting position of the TTT diagram
3.2 Heat Treatment of Ferrous Metals
 Annealing
 Normalizing
 Quenching/Hardening
 Tempering
3.3 Hardenability of Steel
 Hardenability curve
 Martensite microstructure
2
HEAT TREATMENT


After completing this chapter, students should be
able to :
• Explain the principles of heat treatment.
• Explain the differences among the various kinds of
heat treatment processes.
3
OBJECTIVE


What is TTT diagram?
 Time-temperature transformation (TTT) diagram is also known as isothermal
transformation (IT) diagram or Bain S Curve.
 It shows the effect of time and temperature on the microstructure of steel.
 Generated from the % transformation vs log. times measurements.
 Plot as temperature vs. the log. of time for a steel alloy of definite composition.
 (why log. of time so that times of 1 min, 1 day or 1 week can be fitted into a reasonable space).
 Knowledge of the TTT diagram of steels is important in the processing of steels.
Why used TTT diagram?
 Because the iron-iron carbide phase diagram shows no time axis.
 To show the transformation of the microstructure especially for martensite, bainite
structure.
 To determine the microstructure produced in a steel at various rates of cooling.
4
3.1 TIME-TEMPERATURE-
TRANSFORMATION (TTT) DIAGRAM

5
3.1 TIME-TEMPERATURE-TRANSFORMATION
(TTT) DIAGRAM
• Fe-C system, Co = 0.77wt%C
• Transformation at T = 675C.
400
500
600
700
1 10 102 103 104 105
0%pearlite100%
50%
Austenite (stable)
TE (727°C)Austenite
(unstable)
Pearlite
T(°C)
100
50
0
1 102 104
T=675°C
y,
%transformed
time (s)
time (s)
isothermal transformation at 675°C
A plot of temperature vs. the logarithm of time for a steel alloy of definite composition

6
Products Of Cooling Austenite :
PEARLITE MORPHOLOGY
10m
- Smaller T:
colonies are
larger
- Larger T:
colonies are
smaller
• Ttransf just below TE
--Larger T: diffusion is faster
--Pearlite is coarser.
Two cases:
• Ttransf well below TE
--Smaller T: diffusion is slower
--Pearlite is finer.


The time-temperature transformation curves correspond to the start and finish
of transformations which extend into the range of temperatures where
austenite transforms to pearlite.
Refer Figure :
• Line AB indicate the rapid cooling of austenite.
• Horizontal line C-D marks the beginning and end of isothermal
transformations (isothermal means temperature stay constant).
• At point C, the transformation of austenite to pearlite begins.
• At point D, the transformation is complete. 7

8
Products Of Cooling Austenite
• Bainite:
-- lathes (strips) with long
rods of Fe3C
--diffusion controlled.
• Isothermal Transf. Diagram
Fe3C
(cementite)
5 m
(ferrite)
10 103 105
time (s)
10-1
400
600
800
T(°C)
Austenite (stable)
200
P
B
TE
0%
100%
50%
100% bainite
pearlite/bainite boundary
100% pearlite
A
A

9
11
• Martensite:
--(FCC) to Martensite (BCT)
x
x x
x
x
x
potential
C atom sites
Fe atom
sites
(involves single atom jumps)
time (s)10 103 10510-1
400
600
800
T(°C)
Austenite (stable)
200
P
B
TE
0%
100%
50%
A
A
S
M + A
M + A
M + A
0%
50%
90%
Martentite needles
Austenite
60m
•  to M transformation..
-- is rapid!
-- % transf. depends on T only.


 Formed when austenitized iron-carbon alloys are quenched to a
relative low temperature.
 Non-equilibrium single phase
 Rapid quenching will prevent the carbon diffusion.
 Carbon remain as interstitial impurities in martensite
 Instantaneously transformation
 Martensite:
 ƴ (FCC) to Martensite (BCT)
10
MARTENSITE MICROSTRUCTURE


• Two types of martensite microstructure
1. Lath ( massive martensite)
2. Lenticular (needlelike/platelike)
1. Lath martensite
– For alloy < about 0.6 wt% C
– Long and thin plates, form side by side, aligned parallel to one another
– Lath group form block
2. Lenticular martensite
– For alloy > ≈ 0.6 wt% C.
– Needlelike / platelike appearance.
– Under m/scope observation appears as a dark regions
11
MARTENSITE
MICROSTRUCTURE


Unit Cells
The unit cells for (a) austenite, (b) ferrite, and (c) martensite.  The effect of percentage of
carbon (by weight) on the lattice dimensions for martensite is shown in (d).
Note the interstitial position of the carbon atoms. Also note, the increase in dimension c
with increasing carbon content:  this effect causes the unit cell of martensite to be in the
shape of a rectangular prism.


• Fine Pearlite vs Martensite:
• Hardness: fine pearlite << martensite.
13
Pearlite vs martensite


In TTT diagram for iron-carbon alloy, there are 5 regions to observe :
1. Stable austenite
2. Unstable austenite (to the left of the transformation start curve)
3. Pearlite and austenite region (upper side inside nose-shaped curve)
4. Austenite and bainite region (lower side inside nose-shaped curve)
5. Martensite region (below ≈ 200˚C)
 Below 200 down to -20°C = Martensitic start temperature (Ms)
 Below -20°C = Martensitic finish temperature (Mf) 14


15


16
Example:


17


18


19


20


1. Using the time-temperature-transformation
diagram given in of eutectoid composition, sketch
the time-temperature paths to produce the
following microstructures at room
temperature:(October 2010)
a) 100% Bainite,
b) 100% Martensite,
c) 50% Pearlite and 50% Martensite, and
d) 25% Pearlite and 75% Bainite.
21
EXERSICE


22
TIME-TEMPERATURE-TRANSFORMATION DIAGRAM


2. Using the Time-Temperature Transformation diagram
given in for eutectoid steel, draw and label time-
temperature cooling paths that will produce the
following microstructures. In each case assume that the
specimen begins at 850°C.(October 2012)
a) 100% tempered martensite,
b) 75% pearlite, 25% lower bainite,
c) 25% fine pearlite, 37.5% upper bainite and 37.5 austenite,
d) 80% upper bainite, 5.00% lower bainite, 15% martensite,
and
e) 50% pearlite, 12.5% bainite, 37.50% martensite.
23
EXERSICE


24

• reduces brittleness of martensite,
• reduces internal stress caused by quenching.
Adapted from
Fig. 10.24,
Callister 6e.
(Fig. 10.24
copyright by
United States
Steel
Corporation,
1971.)
TEMPERING MARTENSITE
• decreases TS, YS but increases %AR
YS(MPa)
TS(MPa)
800
1000
1200
1400
1600
1800
30
40
50
60
200 400 600
Tempering T (°C)
%AR
TS
YS
%AR
9m
• produces extremely small Fe3C particles surrounded by 

26
60 m

(ferrite)
Fe3C
(cementite)
• Spheroidite:
-- crystals with spherical Fe3C
--diffusion dependent.
--heat bainite or pearlite for long times
--reduces interfacial area (driving force)
10 103 105time (s)10-1
400
600
800
T(°C)
Austenite (stable)
200
P
B
TE
0%
100%
50%
A
A
Spheroidite
100% spheroidite
100% spheroidite


3. Using the Time-Temperature Transformation diagram for iron-carbon alloy
of eutectoid composition, specify the nature of the final microstructure (in term
of micro constituents present and approximate percentage) of a small specimen
that has been subjected to the following heat treatment : (March 2013)
a) Heated up to temperature 780°C and held until the
microstructure completely transformed to austenite. Quenched
rapidly to room temperature.
b) Reheated specimen (a) to temperature 55O°C, held for 20 s, then
quenched to room temperature.
c) Reheated specimen (a) to temperature 700°C and held for 24 hrs.
Then, left to cool to room temperature, naturally.
d) Reheated specimen (a) to temperature 800°C and held for 24 hrs.
Cooled rapidly at temperature 600°C, held for 100 s, then
quenched to room temperature.
e) Reheated specimen (a) to 350°C, held for 2 hrs, then cooled to
room temperature in normal air. 27
EXERSICE


28

 The addition of carbon, nickel, manganese, silicon and copper
move the nose-shaped curve to the right
 Molybdenum, chromium and vanadium move the pearlite C-curve
to the right and also displace it upwards to high temperature.
29
Factors affecting position of the TTT diagram


• Heat treatments are widely used in various manufacturing processes to
enhance the quality of a product.
• The basis for the understanding heat Fe-C phase diagram.
WHAT IS HEAT TREATMENT?
Process involved the heating and cooling of metals in the solid state. Heat
treatment can be a primary process in itself (heat in furnace), or as a secondary
phase of another process (casting, welding, forging).
TYPES OF HEAT TREATMENT?
The most common heat treatment process :
• Annealing
• Normalizing
• Spherodizing
• Quenching/Hardening
• Tempering
30
3.2 HEAT TREATMENT OF FERROUS METALS


3.2 HEAT TREATMENT OF
FERROUS METALS
WHY WE NEED TO DO HEAT
TREATMENT?
Heat treatments are usually
applied to :
• Change the mechanical
properties e.g : increase or
decrease the strength/
hardness/ machinability etc. of
metal.
• Relieve the internal stress
Several problems may occur if heat
treatment process is not carefully
performed.
eg : cracking, distortion.
WHEN WE NEED TO DO HEAT
TREATMENT?
Most parts will require heat
treatment either after or during
the processing for proper in-
service properties. Example :
• Before shaping
– To softening a metal for
forming.
• After forming
– To relieve strain hardening.
• Final finish
– To achieve final strength and
hardness. 31


32
TEMPERATURE REGIME OF STEEL
HEAT TREATMENT


 Annealing
 Process anneal
 Full anneal
 Normalizing
 Spherodizing
 Quenching/Hardening
 Tempering
33
FORMS OF HEAT
TREATMENT


Annealing :
A heat treatment in which a material is exposed to an elevated temperature for
an extended time period and then slowly cooled.
When it should be done :
Annealing is done between process steps to allow further working or for final
stress relief.
Purpose :
1) Relieve stress.
- relieve internal stresses induced by some previous treatment (e.g:
machining).
2) Soften the steel.
- improve machinability and respond better to forming operations.
3) Refinement of grain structures.
Three stages of annealing (applicable for all heat treatment under annealing):
1. Heat to the specified temperature.
2. Hold or “soaking” at that temperature for a specified time.
3. Cool slowly, usually to room temperature.
34
ANNEALING
Time and Temperature
are important at all 3
steps


Purpose :
 Used to treat parts made out of low carbon steel (<0.25% Carbon) which
allow the parts to be soft enough to undergo further cold working without
fracturing.
 Commonly employed for wire & sheets steels because it restores the
ductility to cold-worked materials and permit further cold working to
achieve the required deformation.
Process :
 Raise the steel temperature just below the eutectoid region (line A1 at
727°C), about 500°C to 650°C for several hours until the recrystallization of
ferrite phase occur.
 Then, cooled in still air.
Microstructure desired :
 Fine grained structure
35
(i) PROCESS ANNEAL
Pearlite (α+Fe3C)


Purpose :
 Utilized for low, medium & high carbon steels.
 Full annealing is used to soften pieces which have been hardened by plastic
deformation, and which need to undergo subsequent machining/forming.
Process :
 Heat the steel above the austenite temperature either 15-40˚C above line A3
[hypo] – to form austenite or line A1 [hyper] – to form austenite and
cementite phases.
 Cool very slowly in furnace.
Microstructure produced :
 Coarse pearlite that will give soft and ductility properties. 36
(ii) FULL ANNEAL

• Similar to full annealing but performed at a higher
• temperature and cooling at faster rate (e.g: in air) to form fine pearlite.
• Normalizing is a process that makes the grain size normal.
• This process is usually carried out after forging, extrusion, drawing or heavy
bending operations. It is also used to avoid softening steel too much.
Purpose :
• Refine grains (decrease the average grain size).
• More uniform & desirable size distribution of pearlite
(fine-grain size).
• Increase toughness.
37
NORMALIZING


Process :
• Heat the steel above the austenite temperature (either 55-80˚C above line A3
[hypo] or line ACM [hyper]) .
• After sufficient time has been allowed for the alloy to completely transform
to austenite - austenitizing
• Removed from the furnace and cool it in air (at room temperature).
 Fine pearlite (due to faster cooling rate) will give toughness properties &
acceptable softness to the metal.
Properties :
 Faster cooling provides higher strength and hardness but lower ductility if
compared to full annealing.
38
NORMALIZING
UC‐Upper critical temperature
LC‐Lower critical temperature
RT‐Room temperature

Purpose :
• Used for high carbon steels (Carbon>0.6%) that will be machined or cold
formed.
• Applied when more softness is needed.
Process :
• Heat the part to a temperature just below the eutectoid temperature (line A1
at 727°C) or at about 700°C in the α + Fe3C region for several hours (about
20 hours or more) and followed by slow cooling.
• Cementite transforms into soft globes/spheroids which
dispersed throughout the ferrite matrix.
Properties :
• Result in a more ductile material.
• Improve machining in continuous operations such as lathe and screw
machined. These spheroids act as chip-breakers –easy machining.
39
SPHEROIDIZING


40
Effect of treatment
( pearlite vs spherodite)

41
THERMAL PROCESSING OF METALS
Annealing: Heat to Tanneal, then cool slowly.
Types of
Annealing
• Process Anneal:
Negate effect of
cold working by
(recovery/
recrystallization)
• Stress Relief: Reduce
stress caused by:
-plastic deformation
-nonuniform cooling
-phase transform.
• Normalize (steels):
Deform steel with large
grains, then normalize
to make grains small.
• Full Anneal (steels):
Make soft steels for
good forming by heating
to get , then cool in
furnace to get coarse P.
• Spheroidize (steels):
Make very soft steels for
good machining. Heat just
below TE & hold for
15-25h.

Quenching : It is the act of rapidly cooling the hot steel to harden the steel.
Hardenability :
 The ability of an alloy to be hardened by the formation of martensite as a
result of heat treatment.
 A qualitative measure of the rate at which hardness drops of with distance
Purpose :
 To increase strength and wear properties.
Process :
 Heat the steel above the austenite temperature (either 15-40˚C above line A3
[hypo] or line A1 [hyper]) until the austenite composition is form and
cooled very rapidly in the quench media (a.k.a. cooling medium).
 Martensite (hard but brittle).
42
QUENCHING/HARDENING


To produce microstructure of martensite throughout
the cross section need to consider:
1. Composition alloy,
2. Type & character of quenching medium,
3. Geometry of specimen
43
QUENCHING/HARDENING

1. Composition of alloy
– Higher carbon content gives higher hardenability.
– Alloying element gives higher hardenability compare to
plain carbon steel.
44
QUENCHING/HARDENING

Water
• Advantages
– Most efficient quenching media in commercial use where
maximum hardness is required
• Disadvantage:
– Liable to cause distortion and cracking the sample
– Not suitable for higher carbon steel.
– Form soft spot
– Corrosion
45
2. Type & character of quenching medium
QUENCHING/HARDENING

Oil
 Lower efficiency quenching media than water
 Oil such as mineral & cotton seed are used
 Less cracking and distortion compare to water
 Safety factors is required
Air
 Cooling with air pressure
 Less efficiency quenching media
46
2. Type & character of quenching medium
QUENCHING/HARDENING
Medium
air
oil
water
Severity of Quench
small
moderate
large
Hardness
small
moderate
large

 When surface-to-volume ratio increases:
 cooling rate increases
 hardness increases
diameter size hardness value
47
3. Effect of geometry
QUENCHING/HARDENING
Position
center
surface
Cooling rate
small
large
Hardness
small
large

Tempering :
• It is a process of heating a martensitic steel at a temperature below
the eutectoid temperature to make it softer and more ductile.
• Used to reduce brittleness on martensite (tempered martensite).
• Precipitation of fine carbide particle.
• BCT BCC
Purpose :
• To increase ductility and toughness of martensite.
• To relieve the internal stress.
Procedure :
• Immediately after quenching, sample is heated (normally below A1
line at about 250-650˚C)
• Held at that temperature for about 2 hours.
• Lastly removed from the bath and cooled in air (at room
temperature). 48
TEMPERING


 Tempered martensite which is hard but more malleable and ductile
is produced.
 This microstructure consists of extremely small and uniformly
dispersed cementite particle embedded with a matrix of ferrite.
49
TEMPERING


Steel will oxidize (oxygen in air react with iron to form iron oxide)as it is
reheated and begin to show colors. The higher the temperature, the thicker the
oxide layer and the darker the colors. These temper colors sometimes used as a
guide to temperature.
50
TEMPERING


51
Summary : Processing Options


1. Describe the required heat treatment that Tony Stark should do on his mask
after he has finished cold forging process. (October 2012)
2. Explain the influence of quenching medium and specimen size on the
hardenability of steel. (October 2012)
3. Compare between normalizing process and full annealing process in terms of
microstructure, cooling rate, properties, cooling medium and purpose of those
heat treatments. (April 2011)
4. Describe the following heat treatment process in terms of the purpose,
temperature, cooling medium, microstructure produced and properties for
eutectoid steel. (September 2011)
a) Normalizing
b) Annealing
5. Heat treatment is used to change the microstructure and properties of
materials. (October 2010)
a) Differentiate between coarse pearlite and fine pearlite in terms of the type
of heat treatment and the properties of material that may be obtained from
the microstructures.
b) Briefly describe the hardening in terms of the purpose, process,
microstructure and effect to the properties of materials. 52
EXERCISE


Hardenability:
 A measure of the depth to which the metals of an alloy may be hardened by
the formation of martensite as a result of a given heat treatment.
 Hardening process (e.g: quenching) for steels consist of heating and rapid
cooling form martensite.
 The cooling rate depends on the medium used for the quenching, e.g: water
gives a faster cooling rate than oil and air cooling.
 Generally, the faster steel cools, the harder it will be.
The Jominy Test is used to measure the hardenability of a steel. 53
3.3 HARDENABILITY OF STEELS


Jominy test:
 used to measure the hardenability of steels by heat treatment which shows
the effects of cooling rate on steel hardness.
Jominy Process:
 Heating a standard test piece of the steel to a standard austenite state.
 Fixing it in a vertical position and then quenching it with a jet of water at
one end only, thus producing a range of cooling rates along the steel bar.
 After the quenching, a flat portion is ground along one side of the test
piece, 0.38mm deep, and hardness measurements are made along the
length of the test piece from the quench end.
Jominy distance:
 the distance from the quenched end of a Jominy bar which is related to the
cooling rate.
54
3.3 HARDENABILITY OF STEELS


• Hardenability curve is the graph showing the effect of the cooling
rate on the hardness of as-quenched steel.
• The cooling rate at the quench end is very fast but becomes slower
as the distance from the quench end increases.
• Therefore, the distance from the quench end is an equivalent
measure of the cooling rate, and can be used to give the
hardenability of the steel. 55
HARDENABILITY CURVE


• At quenched end - cools most rapidly, therefore it contains
most martensite.
• Cooling rate decrease with distance from quenched end:
greater C diffusion- more pearlite/bainite, lower hardness.
56
HARDENABILITY CURVE


End-Quench
Hardenability
Test
(a) End‐quench test and
cooling rate.
(b) Hardenability curves for
five different steels, as
obtained from the end‐
quench test.  Small
variations in composition
can change the shape of
these curves.  Each curve
is actually a band, and its
exact determination is
important in the heat
treatment of metals for
better control of
properties.

1. Hardenability can be defined as the ability of an alloy to be hardened by the
formation of martensite. (March 2013)
a) Explain THREE (3) factors that influence the hardenability of steel.
b) Figure Q3 (a) shows a sample that has been preceded to Jominy End-
Quench Test. identify the microstructure at point A, B, C and D.
2. Define hardenability of steel. (April 2009)
3. With the aids of neat sketches, describe how hardenability can be
determined experimentally. (April 2009) 58
EXERSICE

CHAPTER 4
ENGINEERING
MATERIALS
(10 HOURS)

CONTENT:
4.1 CLASSIFICATION OF ENGINEERING MATERIALS
4.2 PLASTICS AND ELASTOMER: MOLECULAR,
STRUCTURES, PROPERTIES AND APPLICATIONS
4.3 CERAMIC: STRUCTURE, PROPERTIES AND
APPLICATIONS
4.4 COMPOSITE MATERIALS: TYPES, PROPERTIES AND
APPLICATIONS.

4.1Classification of
Engineering Materials

CLASSIFICATION OF ENGINEERING MATERIALS

RELATIONSHIP BETWEEN PROPERTIES,
STRUCTURE AND PROCESSING

(METALS)
Metals can be further classified as Ferrous & Non-Ferrous, and
some examples include:
Ferrous Non-Ferrous
Steels Aluminium
Stainless
Steels
Copper
Cast Irons Titanium

(POLYMERS)
Polymers can be further classified as
Thermoplastics Thermosets Elastomers
Acrylics Epoxy resins Rubbers
Nylons Phenolic Silicones
PVC Polyesters Polyurethanes
Polyethylene

(CERAMICS)
Ceramics are compounds of metallic and
non-metallic elements, examples include;
• Oxides (alumina – insulation and
abrasives, zirconia – dies for metal
extrusion and abrasives)
• Carbides (tungsten-carbide tools)
• Nitrides (cubic boron nitride, 2nd in
hardness to diamond)

Materials in our lives – electronic & electrical

Materials in our lives – Civil & Structural

METAL AND NON-METAL USE IN
AUTOMOBILES
Some of the metallic and nonmetallic materials used in a typical automobile

Materials in our lives – Aerospace & Mechanical

4.2 PLASTICS AND ELASTOMERS:
Molecular structures,
properties and applications

POLYMER - INTRODUCTION
 Naturally (those derived from plants, animals) –
wood, rubber, cotton, wool, leather, silk.
 Other natural polymer – proteins, enzymes,
starches, cellulose.
 Development of numerous polymer – synthesis
from small organic molecules.
 Synthetic polymer – plastics, rubbers, fibers
(inexpensive & properties managed to degree that
many are superior to their natural counterparts.

POLYMER MOLECULES
 To understand the chemistry of the polymer, we need
to understand the definition of hydrocarbon,
 Hydrocarbon
 are composed of hydrogen and carbon.
 Has covalent bonds for the intramolecular/interatomic
bonds but for intermolecules exist secondary bond, thus
these hydrocarbons have relatively low melting and
boiling points (p’).
 Saturated and unsaturated
 May have different atomic arrangements, isomerism
CnH2n+2 (Molecular formula)

POLYMER MOLECULES
 Most polymer are organic – review concept relating
to structure of their molecules.
 Each C atom has 4 electron that may participate in
covalent bonding, whereas every H atom has only 1
bonding electron.
 A single covalent bond exits when each of the
bonding atoms contributes 1 electron – saturated
(no new atoms may be joined without removal of
others that are already bonded.
 E.g. paraffin family ethane C2H6, propane C2H8
 Covalent bond –strong
 Van de waals bonds exit between molecules – low
melting & boiling temp. T boil rise with increasing
molecular weight

POLYMER MOLECULES
 Saturated hydrocarbon –all bonds are single
 Molecules that have double or triple covalent
bonds are termed unsaturated.
 Double or triple bonds between 2 C atoms involve
sharing of 2 or 3 pairs of electrons, respectively.
 Within molecule, atoms are bonded together by
covalent interatomic bonds.
C C
H
H H
H
H
H

POLMER MICROSTRUCTURE
Mer
• a structural entities or part
• a single mer is called a monomer
Functionality
 no of bonds that a given monomer can form
 Bifunctional mer – 2 covalent bonds with other monomer
forming 2D chainlike molecular structure
 Trifunctional mer – 3 active bonds, form 3D
molecular network structure
Bifunctional mer
OH
CH2
CH2CH2
trifunctional mer

( )n
• Mer – a repeat unit (repeated along the chain)
• Monomer – small molecule from which polymer is synthesized.
• Polymer – many mer
• The repeat units are enclosed in parentheses (), subscript n indicate
the number of times it repeats.
• R depicits either atoms i.e. H, Cl or an organic group i.e. CH3 (methyl),
C2H5 (ethyl), C6H5 (phenyl)
• vinyl or ethenyl is the functional group −CH=CH2, namely the ethene
molecule (H2C=CH2) minus one hydrogen atom

( )n( )n( )n
( )n
C C C C C C
HHHHHH
HHHHHH
Polyethylene (PE)
mer
ClCl Cl
C C C C C C
HHH
HHHHHH
Polyvinyl chloride (PVC)
mer
Polypropylene (PP)
CH3
C C C C C C
HHH
HHHHHH
CH3 CH3
mer
--CH2-CH2 -- -CH2-CHCl-- --CH2-CHCH3--
Polymer = many mers

CHEMISTRY OF POLYMER
Polymers
• Consist of many mers
• Are gigantic/ macromolecules
• Mostly these molecules are long and flexible chain, the
backbone of the chain is a string of carbon atoms.
• A large molecule (macromolecule) built up by repetitive
bonding (covalent) of smaller molecules (monomers)
• Generally not a well defined structure, or molecular weight.
(A A A A ) n n, degree of polymerization

POLYMER
STRUCTURE
Basic structure of some polymer
molecules:
(a) ethylene molecule;
(b) polyethylene, a linear chain of many
ethylene molecules;
(c) molecular structure of various
polymers.
These molecules are examples of the
basic building blocks for plastics.

Name(s)/Tradename Formula Monomer
Poly(vinylidene chloride)
(Saran A)
–(CH2-CCl2)n–
vinylidene chloride
CH2=CCl2
Polystyrene
(PS)
–[CH2-CH(C6H5)]n–
styrene
CH2=CHC6H5
Polyacrylonitrile
(PAN, Orlon, Acrilan)
–(CH2-CHCN)n–
acrylonitrile
CH2=CHCN
Polytetrafluoroethylene
(PTFE, Teflon)
–(CF2-CF2)n–
tetrafluoroethylene
CF2=CF2
Poly(methyl methacrylate)
(PMMA, Lucite, Plexiglas)
–[CH2-C(CH3)CO2CH3]n–
methyl methacrylate
CH2=C(CH3)CO2CH3
Poly(vinyl acetate)
(PVAc)
–(CH2-CHOCOCH3)n–
vinyl acetate
CH2=CHOCOCH3
cis-Polyisoprene
natural rubber
–[CH2-CH=C(CH3)-CH2]n–
isoprene
CH2=CH-C(CH3)=CH2
Polychloroprene (cis + trans)
(Neoprene)
–[CH2-CH=CCl-CH2]n–
chloroprene
CH2=CH-CCl=CH2

• Physical characteristics of a polymer depends on its
molecular weight, shape, differences in the structure of
the molecular chains.
• Covalent chain configurations and strength:
Direction of increasing strength
POLYMER STRUCTURE
Branched Cross-Linked NetworkLinear
secondary
bonding

MOLECULAR STRUCTURE
i. Linear polymers
 the repeat unit are joined together end to end in
single chain
 May have extensive van der Waals and hydrogen
bonding between the chains
 uninterrupted straight chain, spegetti
 These long chains are flexible
 Extensive van de waals and hydrogen bonding
between the chains
 e.g. Polyethylene,
poly(vinyl chloride),
nylon
Linear
secondary
bonding

MOLECULAR STRUCTURE
ii. Branched polymer
 occasional branches off longer chain
 which the side-branch chains are connected to the
main chain
 Lowering of the polymer density
 The branches may result from side reactions that
occur during the synthesis of polymer
 Branches considered to be part of the main chain
molecule
 e.g. high density polyethylene (HDPE) – primary linear
polymer & low density polyethylene (LDPE) – short
chain branches
B ranched

MOLECULAR STRUCTURE
iii. Crosslinked polymer
 The adjacent linear chains are joined one to
another at various positions by covalent bond
 of crosslinking is achieved either during synthesis
or by a non-reversible chemical reaction
 Many in rubber elastic materials
Cross-Linked

iv. Network polymer
 Having three active or more covalent bond, form
three dimensional network
 highly crosslinked
 Distinctive mechanical and thermal properties
 e.g. epoxies, phenol formaldehyde
Network
MOLECULAR STRUCTURE

CLASSIFICATION OF POLYMER
 Thermoplastics - Linear or branched polymers in
which chains of molecules are not interconnected to
one another.
 Thermosetting polymers - Polymers that are heavily
cross-linked to produce a strong three dimensional
network structure.
 Elastomers - These are polymers (thermoplastics or
lightly cross-linked thermosets) that have an elastic
deformation > 200%.

Thermoplastic Thermoset Elastomer

THERMOPLASTIC POLYMERS
Characteristic:
• Soften when heated
• Harden when cooled
• Reprocessable
• Relatively soft
• High viscosity at processing
temperatures
• Difficult to process
• Examples: polyethylene,
polypropylene, polystyrene
Properties:
• relatively soft
• melt processability
• lower thermal
resistance,
• higher creep,
• Higher moisture
absorption

CHARACTERISTICS AND TYPICAL APPLICATIONS
FOR COMMON THERMOPLASTIC
Polymer Major application characteristic Typical application
Polyethelylene
(HDPE, LDPE)
Chemically resistant and electrically
insulating, tough and relatively low
coefficient of fraction, low strength
and poor resistance to weathering
Flexible bottle, toys,
tumblers, battery part, ice
trays, film wrapping
materials
Polypropylene Resistant to heat distortion, excellent
electrical properties and fatigue
strength, chemically inert, relatively
inexpensive, poor resistance to uv
light
Sterilizable bottles,
packaging film, tv
cabinets, luggage, Tanks,
rope
Polyvinyl
cloride (PVC)
Good low cost, general purpose
materials, ordinarily rigid, but may be
made flexible with plasticizer,
susceptible to heat distortion
Floor coverings, pipe,
electrical wire insulation,
garden hose, valve, fitting
Polystyrene Excellent electrical properties and
optical clarity, good thermal and
dimensional stability, relatively
inexpensive
Packaging, wall tile,
battery cases, toys,
appliance housing

THERMOSETTING POLYMERS
(THERMOSETS)
Characteristic:
• do not melt on heating
• ease of their processing
• low cost
• Lose their stiffness properties at the heat distortion temperature
• Examples: rubbers, epoxies, polyester, phenolics
• NETWORK POLYMERS – have covalent cross links between
adjacent molecular chains.
• They become permanently hard during their formation and do not
soften upon heating.
• Only heating to excessive temp will cause severance of these link
bonds and polymer degration.
Properties:
• harder, stronger, better dimensional stability and more brittle than
thermoplastics.

Polymer Major application characteristic Typical application
Epoxies Excellent combination of
mechanical
properties and corrosion resistance;
dimensionally stable;
Good adhesion;
relatively inexpensive;
good electrical properties
Electrical moldings, sinks,
adhesives, protective
coatings,
used with fiberglass
laminates
Polyesters Excellent electrical properties and
low cost; can be formulated for
room- or high-temperature use;
often fiber reinforced
Helmets, fiberglass boats,
auto body components,
chairs, fans
Phenolics Excellent thermal stability to over
150C (300F); may be
compounded with a large number
of resins, fillers, etc.; inexpensive
Motor housings,
telephones, auto
distributors,
electrical fixtures
Characteristics and typical applications for
common thermoset

ELASTOMER (RUBBER)
Characteristic:
• Soft
• have low elastic modulus values
• show great dimensional change when
stressed but it will return to its original
dimensions immediately after the
deforming stress is removed
• low glass transition temperature.

Two type of rubber:
i. natural rubber
ii. synthetic rubber (SBR, NBR)
Classification of Natural rubber:
• R Class,
 composed of unsaturated chain polymers (these
unsaturated materials can have their properties
by modified by cross linking)
• M Class
 which are saturated chain linear polymers,
• U Class or polyerethanes and the Q Class
 silicone rubbers
ELASTOMER (RUBBER)

Characteristic and typical application for
common Elastomers

LOAD-ELONGATION CURVE AND TENSILE-TEST
SPECIMEN
(a) Load-elongation curve for polycarbonate, a thermoplastic.
(b) High-density polyethylene tensile-test specimen, showing uniform elongation (the long,
narrow region in the specimen).

PLASTIC PRODUCT RECOMMENDATIONS

SUMMARY: POLYMER CHAINS
Schematic illustration of polymer chains.
(a) Linear structure; thermoplastics such as acrylics, nylons, polyethylene, and polyvinyl chloride have
linear structures.
(b) Branched structure, such as polyethylene.
(c) Cross-linked structure; many rubbers and elastomers have this structure. Vulcanization of rubber
produces this structure.
(d) Network structure, which is basically highly cross-linked; examples include thermosetting plastics
such as epoxies and phenolics.

4.3 Ceramic:
Structure,
Properties,
and
Applications

47
 Ceramic (burnt stuff)-desirable properties a high temp heat
treatment process (firing)
 composed of at least two elements or more (e.g.,Al2O3, NaCl,
SiC, SiO2)
 Crystal structure more complex than metals
 Inorganic & non metallic materials
 Most ceramics – metallic & nonmetallic element – ionic or
predominantly ionic but having some covalent character
 Types of ceramic materials:
 Oxide
 Aluminum oxide/Alumina (Al203),
 Zirconium oxide/ Zirconia (ZrO2),
 Non-oxide
 Carbide, Silicate
CERAMICS

Class of ceramics
48
Traditional Ceramics:
primary raw materials is clay
Example: porcelain, bricks, tiles, sewer,
glasses, pipe, whiteware, high
temperature ceramics
Engineering Ceramics :
Contain more of pure compounds of oxides,
carbides, nitrides, etc.
Oxygen sensor
Example: refractory tubing, crucibal, spark
plung insulator, advance ceramic,
electroceramic

CERAMICS
49
Properties :
• Generally hard and brittle
• Generally electrical and thermal insulators
exceptions: graphite, diamond, Aluminium nitride (AlN)
• Can be optically opaque, semi-transparent, or transparent
• High chemical stability and high melting temperature
• Corrosion resistant
• Better compressive strength than tensile (5-10 times)
• Tmelt for glass is moderate, but large for other ceramics.
• Small toughness, ductility; large moduli & creep resist.
Applications:
• High T, wear resistant, novel uses from charge neutrality.

50
TAXONOMY OF CERAMICS
-
-
- -
-
-
-Advanced
-valvesreinforce
Glasses Clay
products
Refractories Abrasives Cements
ceramics
-optical
-composite
-containers/
household
-whiteware
-structural:
bricks
-bricks for
high T
(furnaces)
-sandpaper
-cutting
-polishing
-composites
structural
-engine
-rotors
-bearings
-sensors
Ceramic Materials

GLASSES
- Glasses
- Glass-Ceramic
51

GLASSES
• Non crystalline silicates containing network modifiers; Na2O,
CaO, K2O and Al2O3
• Typical example of glass;
soda-lime silica glass  70% SiO2 + soda (Na2O) and lime (CaO)
• Glass is transparent and easy to be fabricated
52

CHARACTERISTICS OF COMMON COMMERCIAL GLASSES
53

APPLICATIONS : GLASSES
54
96% Silica
Laboratory ware
Borosilicate, 81% Silica, 3.5%
Na2O, 2.5% Al2O3 and 3%
B2O3
Pyrex
Laboratory ware/oven ware
Containers,
windows

GLASS CERAMICS
• Most glass are amorphous (non crystalline)
• But can be transformed to crystalline by heat treatment  fine
grained polycrystalline material – glass-ceramics
• The heat treatment process  devertification process
• During the heat treatment process, a nucleating agent is
required to initiate crystallization or devertification process
• Easy to fabricate; mass production.
• Glass ceramic commercially under trade names of Pyroceram,
corning ware, cercor, vision
• Applications: ovenware, tableware, oven windows, range top –
primary coz of their strength & excellent resistant to thermal
shock
55

APPLICATIONS :GLASS CERAMICS
56
Properties/characteristics
High mechanical strength
Low coefficient of thermal expansion
(to avoid thermal shock)
high temperature capabilities
Good biological compatibility
Some optical transparent;
others are opaque
Electrical insulator
Other Applications
Glassware
Electrical insulator
Substrate for printed circuits board
Architectural cladding
Heat exchangers
Generator

CLAY
PRODUCTS
- Structural clay products
- Whitewares
57

CLAY PRODUCT
- Widely used as ceramic raw materials
- Inexpensive ingredient
- Found naturally in great abundance
• Adding water to clay
-- very amenable to shaping (form a plastic mass)
-- enables extrusion
-- enables slip casting
• The formed piece is dried to remove moisture
• Fired at elevated temp to improve its mechanical strength
• 2 broad classification
1) structural clay product - structural integrity is important
2) whitewares
58

CLAY COMPOSITION
A mixture of components used
(50%) 1. Clay
Clay facilitates the forming operation since, when mixed with water,
the mass may be made to become either hydroplastic or form a
slip. Also, since clays melt over a range of temperatures, the
shape of the piece being fired will be maintained.
(25%) 2. Filler – e.g. quartz (finely ground)
(25%) 3. Fluxing agent (Feldspar)
binds it together
The flux facilitates the formation of a glass having a relatively low
melting temperature
59
aluminosilicates + K+, Na+, Ca+

CLAY PRODUCT:
STRUCTURAL CLAY PRODUCT
used mainly in construction
Properties :
load-bearing strength, resistance to wear, resistance to chemical
attack, attractive appearance, and an ability to take a decorative finish.
Products :
facing buildings, surfacing highways, making containers for corrosive
acids, as aggregate for low-density concrete, as conduits for sewage, as
structural arches supporting bridges, as roofs, and as chimney liner
60Tiles bricks
sewer pipe

61
Bricks
conduits for sewage
roofs
chimney liners
Application : structural clay product

CLAY PRODUCT :WHITEWARES
- ceramic products that are white to off-white in appearance
- become white after high temp firing
- frequently contain a significant vitreous, or glassy,
component.
Properties :
imperviousness to fluids, low conductivity of electricity,
chemical inertness, and an ability to be formed into
complex shapes.
Products :
china dinnerware, lavatory sinks and toilets,
dental implants, and spark-plug insulators, 62

63
lavatory sinks and toilets, plumbing
fixture (sanitary ware)
china dinnerware (Porcelain, pottery, tableware)
dental implants
spark-plug insulators
Application : Whiteware

65
Characteristic of Refractory Ceramics
 Can withstand high temperature without
melting or decomposing
 Can remain inert even at sever conditions
 Can provide thermal insulations
 In a form of bricks (most common)
 Use as furnace linings for metal refining,
glass manufacturing heat treatment and
power generation

REFRACTORIES CERAMICS
Several classification
• fireclay-used in furnace construction, to confine hot atm & to
thermally insulate structural members from excessive temp (alumina
& silica)
• Silica (asid refractories) – high temp load bearing capacity (used in
arched roofs of steel & glass making furnace),
• basic – magnesia(MgO) + Ca, Cr, Fe + silica (used in some steel
making furnace)
• special refractories – e.g. (SiC) cruciable material & electrical
resistance heating elements & internal furnace component
Raw ingredients – (both) large & fine particles
• Upon firing, fine particles –formation of bonding phase –increased
strength of the brick
• Control the porosity – porosition reduction incred strength, load
bearing capacity, resistance to corrosive materials
• However, diminished the thermal insulation characteristic and
resistance to thermal shock

67
APPLICATION: REFRACTORIES CERAMICS
• Need a material to use in high temperature furnaces.
• Fireclay bricks, crucible material, internal furnace
components
Fireclay bricks
crucible
internal furnace components

• Consider Silica (SiO2) - Alumina (Al2O3) system.
• Phase diagram shows:
mullite, alumina, and crystobalite (made up of SiO2)
tetrahedra as candidate refractories.
3
Composition (wt% alumina)
T(°C)
1400
1600
1800
2000
2200
20 40 60 80 1000
alumina
+
mullite
mullite
+ L
mullite
Liquid
(L)
mullite
+ crystobalite
crystobalite
+ L
alumina + L
3Al2O3-2SiO2
APPLICATION: REFRACTORIES

ABRASIVE CERAMICS
• Used to wear, grind and cut away other material
• Hardness and wear resistance important
• High degree of toughness – do not want material
which deform or facture during cutting!
• Diamond is the best but expensive
• Other examples; Tungsten carbide (WC) , Alumina
(Al2O3) and Silica(SiO2), SiC, silica sand
70

71
APPLICATION: ABRASIVE CERAMICS
• Abrasive are used in several forms-Bonded to grinded wheels
- as coating abbrasive – the abrasive particles/powder is
coated on some type of paper or cloth material; sand
paper, wood, metal ceramics & plasric
- loose grains – grinding, lapping & polishing wheels often
employ loose abrasive grain that are delivered in some
type of oil or water based vehicles (diamods, SiC, iron
oxide) grinding wheel, sandpaper
grinding wheel

5
• Tools:
--for grinding glass, tungsten, carbide, ceramics
--for cutting Si wafers
--for oil drilling
blades oil drill bits
APPLICATION: ABRASIVE CERAMICS

74
Cements
• Inorganic cements : cement, plaster of paris and lime
• known as binder,
• When mixed with water, forms a paste which harden as
a results of complex hydration reactions
• substance that sets and hardens independently
• can bind other materials together
• The role of cement is similar to glassy bonding when clay
product & refractory brick are fired.
•The different is cementitious bond develop at room temp.
•Lime involved in hardening reaction

APPLICATION : CEMENTS
• as an ingredient in the production of mortar in masonry, and
concrete
75
Mortar
concrete

• Produced in extremely large quantities.
• Portland cement:
--mix clay and lime bearing materials
--calcinate (heat mixture to 1400°C in rotary kiln)
--primary constituents:
tri-calcium silicate
di-calcium silicate
• Adding water
--produces a paste which hardens
--hardening occurs due to hydration (chemical reactions
with the water).
• Forming: done usually minutes after hydration begins.
16
CEMENTS

CEMENTS
 Hydration reactions begin just as soon as water is
added to the cement
1)Setting i.e. stiffening of once plastic phase (several
hours)
2)Hardening –water actually participates in a chemical
bonding reaction
 Porland cement- its hardness develops by
chemical reaction with water
 Used in mortar & concrete to bind aggregated of
inert particles (sand) into cohesive mass
(composite materials)

ADVANCED CERAMIC
 Ceramics that displays unique
electrical, magnetic and optical
properties
 Utilized in microelectromechanical
system (MEMS), Sensors, fuel
cells, superconductors, actuators,
electronics packaging,
semiconductor devices, solar cells,
fibre optics, laser production, etc
79
MEMS
Ceramic cannula in fibre optics

80
Applications: Advanced Ceramics
 Ceramic Armor
 Al2O3, B4C, SiC & TiB2
 Extremely hard materials
 shatter the incoming bullet
 energy absorbent material underneath

81
Applications: Advanced Ceramics
Electronic Packaging
 Chosen to securely hold microelectronics & provide heat transfer
 Must match the thermal expansion coefficient of the
microelectronic chip & the electronic packaging material. Additional
requirements include:
- good heat transfer coefficient
- poor electrical conductivity
 Materials currently used include:
 Boron nitride (BN)
 Silicon Carbide (SiC)
 Aluminum nitride (AlN)
- thermal conductivity 10x that for Alumina
- good expansion match with Si

CERAMIC TYPES AND CHARACTERISTICS

EXERCISE :
1. Describe the main Difference between traditional ceramics and
engineering ceramics.
2. List two example of applications for traditional ceramics and
engineering ceramics.
3. Some of our modern kitchen cookware is made of ceramic materials.
a) List three (3) important characteristics required of a materials to be used
for this application.
b) Choose the material that most suitable for cookware.
83

4.4 COMPOSITE MATERIALS:
TYPES, PROPERTIES AND APPLICATION

COMPOSITE MATERIAL
• Consists of two or more physically and/or chemically
distinct, suitably arranged or distributed phases with an
interface separating them.
• Composite – multiphase materials (metal alloys, ceramics &
polymers) artificially made
• Has characteristics that are not represent by any of the
components in isolation.
• Material have specific & unusual prop in i.e. aerospace,
underwater, bio-engineering & transportation industries.
• e.g. low density, strong, stiff, abrasion, impact resistance & do
not easily corrode.

composites
Particle-reinforced Fiber-reinforced Structural
Large- particle
Dispersion-
strengthened
(0.01 and 0.1 µm)
Continuous
(aligned)
Discontinuous
(short)
aligned
Randomly
oriented
Laminates Sandwich panel
CLASSIFICATION OF COMPOSITE MATERIALS

PARTICLE REINFORCED COMPOSITES
• The particle diameter is typically a few microns (μ)
• Particle reinforced composites are much easier and less
costly than making fiber reinforced composites.
• Particulate phase is harder & stiffer than the matrix.
• Particulate –same dimension in all direction
1. Large - Particle
Particulate Flake Filler

Based on reinforcement or strengthening mechanism.
Example:
• concrete composed of cement (matrix) and sand & gravel
(particulates). Cerment (ceramic metal composite) matrix
“metal” such as Co, Ni, particles “ceramic” such as WC or
TiC.
1. Large - Particle (cont.)
Concrete is a mixture of cement and
aggregate, giving a robust, strong
material that is very widely used

Example:
• Used as cutting tools for hardened steel – carbide is brittle.
toughness is enhanced by inclusion in the ductile metal
matrix. Withstand high temp generate during cutting.

Example:
• automobile tire which has carbon black particles in a
matrix of polyisobutylene elastomeric polymer. Carbon
black evenly distributed though out the rubber (inexpensive
material) – enhanced TS, toughness, tear & abrasion
resistance

• The particle diameter is small particles between 0.01 and 0.1 μm(10–100nm)
• An example : metal matrix composite with a fine distribution.
Metal & metal alloys + dispersed phases (metallic/ nonmetallic/oxide
materials)
• The strengthening mechanism involve the interactions between the particles
dislocation between the matrix. Particle matrix interaction leads to
strengthening
2. Dispersion Strengthen

FIBER-REINFORCED COMPOSITE
•The fibers can be in the form of long continuous fibers, or
discontinuous fibers, particles, whiskers and even weaved
sheets, wires.
•Fiberglass is likely the best know fiber reinforced composite.
(a)
Continuous and aligned
(b)
Discontinuous and aligned
(c)
discontinuous and
randomly oriented

• Aligned Continuous fibers
• Examples:
fracture
surface
matrix: (Mo) (ductile)
fibers:’ (Ni3Al) (brittle)
2m
--Metal: '(Ni3Al)-(Mo)
by eutectic solidification.
--Glass w/SiC fibers
formed by glass slurry
Eglass = 76GPa; ESiC = 400GPa.
(a)
(b)

• Discontinuous, random 2D fibers
• Example: Carbon-Carbon
--process: fiber/pitch, then
burn out at up to 2500C.
--uses: disk brakes, gas
turbine exhaust flaps, nose
cones.
• Other variations:
--Discontinuous, random 3D
--Discontinuous, 1D
fibers lie
in plane
view onto plane
C fibers:
very stiff
very strong
C matrix:
less stiff
less strong

Factor that influence composites properties
1. Fiber length
- short fiber --- less significant improvement in strength
- more effective if continuous fiber
2. Fiber orientation
- parallel alignment - align direction, reinforcement and strength are max;
perpendicular to alignment, they are minimum
- random alignment -Able to support multiple direction forces
3. Fiber concentration
- Better properties when fiber distribution is uniform

MATRIX
COMPONENTS OF COMPOSITE
REINFORCEMENT
INTERFACE

MATRIX
• A bulk phase, which is continuous
• Surrounds the reinforcements
• Providing uniform load distribution to the
reinforcing constituents
• General polymer and metal – ductility is desirable
• Ceramic matrix to improve fracture toughness
• Examples:
metal-, polymer- and ceramic- matrix

Purpose of Matrix :
• to bind and hold the reinforcements.
• to transfer load to and between reinforcements.
• allows the strength of the reinforcements to be used to
their full potential by providing effective load transfer from
external forces to the reinforcement.
• to protect the reinforcements from environments and
handling.
• provides a solid form to the composite which aids handling
during manufacture and is typically required in a finished
part.
• controls the transverse properties, interlaminar strength
and elevated-temperature strength of the composite.
MATRIX

REINFORCEMENT
• Fiber reinforcement are classified as follows
a) Fibers – normally polymer or ceramics (amorphous or
polycrystalline) i.e glass, carbon, boron, aluminum
oxide, SiC
b) Whiskers – thin single crystals that have very small
diameters) i.e. Graphie, SiC, Al2O3
c) Wires – metal/alloys that have relatively large
diameters. . i.e. steel, Molybdenum, W
• provide superior levels of strength and stiffness to
the composite.
• provide thermal and electrical conductivity,
controlled thermal expansion, and wear resistance
in addition to structural properties.

IMPORTANT CHARACTERISTICS FOR REINFORCEMENT
• Diameter size
 strength decreases with an increases of diameter
• A high aspect ratio (l/d)
 allows a very large fraction of the applied load to be
transferred via the matrix to stiff and strong fiber
• High degree of flexibility
 it is a characteristic of material having a high
modulus and a small diameter.
 Permits a variety of techniques for making
composites

INTERFACES
• can be defined as a bounding surface where a
discontinuity of some kind occurs (between matrix and
reinforcement).
• the interface is an essentially two-dimensional region
through which material parameters such as
 concentration of an element, crystal structure,
atomic registry, elastic modulus, density and
coefficient of thermal expansion, change from one
side to another

CLASSIFICATION OF COMPOSITES (MATRIX)
Ceramic-Matrix
Composite
(CMC)
Metal-Matrix
Composite
(MMC)
Polymer Matrix
Composite
(PMC)
composite

POLYMER-MATRIX COMPOSITES (PMC)
 PMCs consist of a polymer resin as the matrix, with
fibers as the reinforcement medium
 They may be reinforced with glass, carbon and aramid
fibers, etc.
 Polymer Matrix
 The most widely used (least expensive) polymer resins are
polyesters and vinyl ester.
 Epoxies (more expensive)
 PMCs for aerospace applications
 Better mechanical properties and resistance to moisture than
polyesters and vinyl ester.
 Polyimide resins for high temperature applications
 aerospace application – polyetheretherketone,
polyphenylene sulfide, polyetherimide

EXAMPLE OF PMC:
Glass Fiber-Reinforced Polymer (GFRP) Composites
• fiberglass consist of glass fibers (continuous or discontinuous) contained
within polymer matrix (polyester resin)
• widely use due to
a. Easily drawn into high strength fiber
b. Easily to be processes to composite ( less cost)
c. Very high specific strength
d. Most type chemical inertness -- variety of corrosion environment
• Limitation
a. High strength but not very stiff (rigidity) – not suitable for structure application
b. Low service temperature ( below 200oC) – improve by adding high temp purity silica
and high temp polymer (polymide) – ( 300oC)
• Application
 plastic pipeline, tanks and vessel for chemical process industry, storage
containers, automotive & marine bodies
 Transportation industries – decrease vehicle weight & boost fuel efficiency
Polymer-Matrix Composites (PMC)

EXAMPLE OF PMC:
Carbon Fiber-Reinforced Polymer (CFRP) Composites
 Carbon widely use as fiber reinforced due to
a. Highest specific modulus and specific strength
b. High tensile modulus and high strength retain at elevated temp
c. At room temp, carbon fiber are not effected by moisture, most
solvent, acids bases
d. Low fabrication cost and effective
 Limitation
a) fabrication of carbon fiber are complex
b) at high temp, carbon - high tendency to oxidized
 Application
sport and recreational equipment ( fishing rod, golf clubs), filament wound
rocket motor casing, pressure vessel, aircraft structural, Helicopters (wing,
body, stabilizer), crank arms for bicycle

EXAMPLE OF PMC:
Aramid Fiber-Reinforced Polymer Composites
 aramid – chemical name : poly paraphenylene (polymer)
 high strength, high modulus materials (outstanding strength to weight ratio)
 good longitudinal tensile strength, toughness, impact resistant,
resistance to creep and fatigue failure
 even polymer group but resist to combustion and stable to relatively high
temperature ( application range -200oC to 200oC)
 degradation to strong acid and base but inert to solvent and other
chemical
 Trade name-Kevlar, Nomex
 Polymer matrix – epoxies, polyesters
 Higher fatigue resistant than carbon PMC
 Application:
 Racing yachts and private boats, helmets, rocket engine cases, gasket,
clutch lining, ballistic products (bulletproof vest & armor), tires, ropes,
sporting goods

EXAMPLE OF PMC:
Boron/epoxies composite
 Golf clubs, tennis rackets, horizontal stabilizers and tail
section of military aircraft, helicopter rotor blade

CROSS-SECTIONS OF FIBER-REINFORCED
MATERIALS
(a) Cross-section of a tennis racket, showing graphite and aramid (Kevlar) reinforcing fibers.
(b) Cross-section of boron fiber-reinforced composite material

POLYMER-MATRIX COMPOSITES (PMC)
Advantages:
 found widespread applications.
 can be easily fabricated into any large complex
shape, do not involve high pressures and temp.
(less degradation of reinforcement)
 Equipment required may be simpler; (hand lay-
up)

Disadvantages:
• low maximum working temperatures.
• high coefficients of thermal expansion and hence
dimensional instability (except: carbon fibre-
reinforced polymers )
• sensitivity to radiation (except: epoxies) and
moisture.
• The absorption of water from the environment may
have many harmful effects which degrade mechanical
performance, including swelling.
• formation of internal stresses and lowering of the
glass transition temperature.

Properties
• improve strengths and stiffnesses,
• Ease of molding for complex shapes,
• high environmental resistance all coupled with low densities
• Make the resultant composite superior for many
applications
Fibre Reinforced Polymer

Application in general:
• Aramid and carbon fibers have strengths and low
densities and are used in many applications,
particularly aerospace, in spite of their higher
cost.
• In electronic applications, glass fiber from E-glass
type is used as reinforcement in substrate
application because the fibers have very desirable
and stable electrical properties.

BOEING 757-200
Application of advanced composite materials in Boeing 757-200 commercial aircraft.

METAL-MATRIX COMPOSITE (MMC)
Examples of metal-matrix composite parts.

METAL-MATRIX COMPOSITES (MMC)
 The matrix is a ductile metal
 The most common metals employed in MMC are
aluminum, copper, titanium and magnesium.
 Typical fibers used in the composite systems are
carbon and silicon carbide
 Metals are mainly reinforced to increase or decrease
their properties to suit the needs of the design.
 Reinforcement
- may improve sp stiffness, sp strength, abrasion resistance,
creep resistance, thermal conductivity, dimensional stability
- Particulate, fibers (continuous & discontinuous), wiskers
- E.g. Boron aluminum oxide, refractory metals

Advantages :
• higher application temperature ranges,
• higher transverse stiffness and strengths,
• high electric and thermal conductivities and can be fabricated with
conventional metal working equipment
• high toughness values. higher strength-to-density, stiffness-to-density ratios as
well as better fatigue resistances, lower coefficients of thermal expansion (CTE)
and better wear resistances as compared with monolithic metals
Disadvantages:
 most metals are heavy
 Susceptible to interfacial degradation at
the reinforcement and matrix interface
 susceptible to corrosion
 high material and fabrication costs
 exhibit degradation of properties at very
high temperatures

EXAMPLE MMC
• Automotive
• Al alloy MMC; reinforced with aluminum oxide
and carbon fibers
• drive shaft ( higher vibration rotational speed)
• Extruded stabilizer bars
• Forged suspension and transition
components
• Aerospace (MMC is light in weight)
• Al alloys MMC;
• boron fibers – space shuttle orbiter
• Continuous graphite fibers – Hubble Space
telescope

METAL MATRIX AUTOMOTIVE BRAKE CALIPER
Aluminum-matrix composite brake caliper
using nanocrystallyne alumina fiber
reinforcement

METAL-MATRIX COMPOSITE MATERIALS AND
APPLICATIONS

CERAMIC-MATRIX COMPOSITES (CMC)
• Contains a ceramic matrix such as alumina and
calcium alumino silicate reinforced by fibers such as
carbon or silicon carbide.
• The main objective is to increase the toughness,
strength and stiffness of the material.
• High temp & severe stresses applications –
automobile & aircraft gas turbine engines
• Fracture toughness value for ceramic materials are
low.
• Reinforced – particulates, fibers, whiskers of one
ceramics
• Matrix- another ceramics

Advantages:
• a very high application temperature range (>2000ºC).
• provide advanced heat engine applications
• low density and usually have very high elastic modulus
values
• chemical inertness.
Disadvantages:
 brittleness which makes them easily susceptible to flaw
 Only employed high temperature reinforcement
 High temperature for processing (High production costs)
 lack uniformity in properties and have low thermal and mechanical
shock resistances as well as low tensile strengths

EXAMPLE OF CMC
• SiC wiskers reinforced Al2O3
 cutting tool material replacing the metallic carbide
cutting tool
 Cutting tool insert for machining hard metal alloys
 Resistance to thermal shock
 Improve strength, fracture toughness
• C/C composite
 Disc brake, hot pressing mold

CHARACTERISTICS OF COMPOSITE
MATERIALS

• CMCs: Increased toughness • PMCs: Increased E/
• MMCs:
Increased
creep
resistance
20 30 50 100 200
10-10
10-8
10-6
10-4
6061 Al
6061 Al
w/SiC
whiskers (MPa)
ss (s-1)
E(GPa)
G=3E/8
K=E
Density,  [Mg/m3]
.1 .3 1 3 10 30
.01
.1
1
10
102
103
metal/
metal alloys
polymers
PMCs
ceramics
fiber-reinf
un-reinf
particle-reinf
Force
Bend displacement
COMPOSITE BENEFITS

STRUCTURAL COMPOSITES
Laminate composite
1. Laminar Composites
• Composed of 2D sheets or panels.
• Sheets (panels) with different orientation of high strength
directions are stacked and glued together
• Examples : plywood and modern ski, application more in
aircraft
The layers are stacked & cemented
together such that the orientation of high
strength direction varies with each
successive layer

1. Laminar Composites (cont.)
• # of laminated – varies
• joined by plastic adhesive ( for glass, effect from adhesive more
importance)
• improve corrosion resistance with low cost , high strength & light
weight
• improve thermal expansion characteristic
• improve fatigue failure
• Stacked and bonded fiber-reinforced sheets
-- stacking sequence: e.g., 0/90
-- benefit: balanced, in-plane stiffness
A small sample of Aerospace grade
Carbon-fibre/Epoxy laminate
Plywood is used widely in
construction

2. Sandwich Panels
• Consist of two strong and stiff sheet (faces) separated by a layer of less-
dense material (core materials) or structure for instance honeycomb which
provides strength to shear.
• Benefit: These structure combine relatively high strength and stiffness with
low density.
• Application
 roofs, walls, and aircraft structures (wings, fuselage, tailplane skin).

2. Sandwich Panels (cont.)
Diagram of an assembled composite sandwich (A), and its
constituent face sheets or skins (B) and honeycomb core (C)
(alternately: foam core)
• Faces – support all load (relatively stiff & strong materials)
- example : Al alloy, fiber reinforce plastic,
titanium, steel, plywood
- thick enough to withstand tensile & compressive
stresses from loading
• Core –lightweight, low modulus of Elasticity.
-Rigid polymeric foams (phenolics, epoxy,
polyurethanes), synthetic rubber
 Wood (balsa wood)
 Honeycomb ( thin foils that have been formed
into interlocking hexagonal cells)
- Function:
1) provide continuous support for faces
2) have sufficient shear strength to withstand
transverse shear stresses
3) thick enough to provide high shear stiffness(to
resist buckling of the panel) shear rigidity

COMPOSITE SAILBOARD CROSS-SECTION
Cross-section of a composite sailboard, an example of advanced materials construction.

SUMMARY:
METHODS OF REINFORCING PLASTICS
Schematic illustration of methods of reinforcing plastics (matrix) with (a) particles, (b)
short or long fibers or flakes, and (c) continuous fibers. The laminate structures shown in
(d) can be produced from layers of continuous fibers or sandwich structures using a foam
or honeycomb core.

• Composites are classified according to:
-- the matrix material (CMC, MMC, PMC)
-- the reinforcement geometry (particles, fibers, layers).
• Composites enhance matrix properties:
-- MMC: enhance y, TS, creep performance
-- CMC: enhance Kc
-- PMC: enhance E, y, TS, creep performance
• Particulate-reinforced:
-- Elastic modulus can be estimated.
-- Properties are isotropic.
• Fiber-reinforced:
-- Elastic modulus and TS can be estimated along fiber dir.
-- Properties can be isotropic or anisotropic.
• Structural:
-- Based on build-up of sandwiches in layered form.
SUMMARY: COMPOSITE

SUMMARY:

Mec281 lecture

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