2. 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
3. • 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
4. • Course Work : 40%
Test 1 : 15%
Test 2 : 15%
Quiz (x4) : 10% (i-Learn)
• Final Examination : 60%
• Total : 100%
ASSESSMENT
5. 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.
6. 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.
12. 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.
13. 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
15. 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.
+
16. 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
17. 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.
18. 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
19. 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.
20. 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
27. • 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
28. 20
d- orbital
- There is 5 distinct d – orbitals
- Max 10 electrons
- Third energy level
29. 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
30. • 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
31. 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.
32. 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-
37. 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 .
38. 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
40. • 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
41. • 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
44. • 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)
45.
46. Properties
• Gases, liquids, or solids (made of molecules)
• Poor electrical conductors in all phases
• Variable ( hard , strong, melting temperature, boiling point)
2) COVALENT BONDING
47. • 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
49. • 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??
56. 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
57. 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
62. • 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
63. •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
64. • 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.
66. 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]
67. 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
=+
68. 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.
+
=
69. 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.
70. 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.
71. 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
74. 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
76. 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
78. 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 :
81. 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
82. 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]
84. 76
Calculate the density for nickel (simple cubic structure).
Note that the unit cell edge length (a) for nickel is 0.3524 nm.
EXAMPLE
85. 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
86. 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]
87. 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
88. 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
89. 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.
93. 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)?
94. Draw the following Miller Indices
direction.
a) [ 1 0 0 ]
b) [ 1 1 1 ]
c) [ 1 1 0 ]
d) [ 1 1 0 ]
95. 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 )
101. Draw the following Miller Indices
plane.
a) ( 1 0 0 )
b) ( 0 0 1 )
c) ( 1 0 1 )
d) ( 1 1 0 )
102. 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.
104. 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 :
106. 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 :
109. 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.
110. 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
111. 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
112. • 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.
Stress-Strain Diagram (cont)
113. • 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.
Stress-Strain Diagram (cont)
114. 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.
122. 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.
123. 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.
124. 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.
125. 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.
128. 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
129. 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
131. 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.
133. • 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
134. 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
xy
137. 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.
138. (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
141. 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
143. (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
144. 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]
145. Ductile fracture Brittle fracture
What are the differences between
ductile fracture & brittle fracture?
146. 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.
147. 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).
152. 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.
153. 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.
154. 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
156. 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]
161. 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
163. 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
164. 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
167. 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
168. 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).
169. 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.
170. 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
Substitutional Solid Solution
171. 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
Substitutional Solid Solution
172. 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
173. • 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
175. 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
176. 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
178. • 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
179. 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
180. 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
181. • 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
183. 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
191. 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
193. 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
194. 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
195. • Rule 2: If we know T and Co, then we know:
--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.
BINARY ISOMORPHOUS PHASE DIAGRAM:
composition of phases
38
196. • Sum of weight fractions:
• Conservation of mass (Ni):
• Combine above equations:
WL W 1
Co WLCL WC
R
R S
W
Co CL
C CL
S
R S
WL
C Co
C CL
• A geometric interpretation:
Co
R S
WWL
CL C
moment equilibrium:
1 W
solving gives Lever Rule
WLR WS
THE LEVER RULE
Let WL = fraction of liquid and Wα = fraction of solid (unknown)
39
198. • Rule 3: If we know T and Co, then we know:
--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
BINARY ISOMORPHOUS PHASE DIAGRAM:
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
199. 42
EXAMPLE : Calculate the amounts of α and L at 1250°C in the
Cu-35% Ni alloy?
THE LEVER RULE
200. 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
201. 44
Consider Co = 35wt% Ni
Figure : Cooling of Cu-Ni alloy
Microstructure
A
B
C
D
E
BINARY ISOMORPHOUS PHASE DIAGRAM:
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
203. 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
204. 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)
BINARY EUTECTIC PHASE DIAGRAM
(NO SOLID SOLUTION)
HYPOEUTECTIC HYPEREUTECTIC
Three phases in equilibrium at
eutectic point compositions and
temperature
Eutectic reaction
L A+ B
206. 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
BINARY EUTECTIC PHASE DIAGRAM
(NO SOLID SOLUTION)
49
207. 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
BINARY EUTECTIC PHASE DIAGRAM
(NO SOLID SOLUTION)
212. 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
BINARY EUTECTIC PHASE DIAGRAM
(LIMITED SOLID SOLUTION)
TM Ag
TM Cu
55
213. 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)
BINARY EUTECTIC PHASE DIAGRAM
(LIMITED SOLID SOLUTION)
214. • 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
BINARY EUTECTIC PHASE DIAGRAM
(LIMITED SOLID SOLUTION)
Eutectic reaction
L α + β
(Liq.) (s.s) (s.s)
57
215. 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)
BINARY EUTECTIC PHASE DIAGRAM
(LIMITED SOLID SOLUTION)
58
216. 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
BINARY EUTECTIC PHASE DIAGRAM
(LIMITED SOLID SOLUTION)
59
228. 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
234. 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
240. 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
241. 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
245. 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
246. 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.
248. 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)
250. 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.
251. 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)
253. 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.
254. 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
256. 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
258. 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".
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261. 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.
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268. 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.
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269. 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
270. 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.
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271. 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.
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273. 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.
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274. 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
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275. 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).
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276. THE MICROSTRUCTURE OF
GRAY CAST IRONS
Graphite flakes
* Graphite flakes shows fracture surface (gray appearance).
Figure : Dark graphite flakes in a‐Fe matrix.
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278. THE MICROSTRUCTURE OF
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.
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279. 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
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280. THE MICROSTRUCTURE OF
WHITE CAST IRONS
Figure : Light Fe3C regions surrounded by pearlite.
Pearlite
Fe3C
(Light regions)
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281. 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.
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