Chp 1 Material Science 281 Uitm Em110
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Transcript of Chp 1 Material Science 281 Uitm Em110
Rasdi Deraman, FKM UiTM NPP 1
MEC281 MATERIALS SCIENCE
CHAPTER 1
THE STRUCTURE AND PROPERTIESOF MATERIALS
Rasdi bin DeramanFakulti Kejuruteraan Mekanikal
UiTM Pulau Pinang
Rasdi Deraman, FKM UiTM NPP 2
What are Materials?
• Materials may be defined as substance of which something is composed or made.
• We obtain materials from earth crust and atmosphere.
Examples :- Silicon and Iron constitute 27.72 and 5.00
percentage of weight of earths crust respectively. Nitrogen and Oxygen constitute 78.08 and 20.95
percentage of dry air by volume respectively.
Introduction to Materials Science
and Engineering
Rasdi Deraman, FKM UiTM NPP 3
Why the Study of Materials is Important?
• Production and processing of materials constitute a large part of our economy.
• Engineers choose materials to suite design.
• New materials might be needed for some new applications.
Example :- High temperature resistant materials.
Space station and Mars Rovers should sustain conditions in space.
* High speed, low temperature, strong but light.
• Modification of properties might be needed for some applications.
Example :- Heat treatment to modify properties.
Rasdi Deraman, FKM UiTM NPP 4
Materials Science and Engineering
• Materials science deals with basic knowledge about the internal structure, properties and processing of materials.
• Materials engineering deals with the application of knowledge gained by materials science to convert materials to products.
Materials Science Materials Science and Engineering
Materials Engineering
Basic Knowledge of
Materials
Resultant Knowledge
of Structure and Properties
Applied Knowledge
of Materials
Rasdi Deraman, FKM UiTM NPP 5
• Metallic Materials - Composed of one or more metallic elements.
Example: Iron, Copper, Aluminum.- - Metallic element may combine with nonmetallic elements.
Example: Silicon Carbide, Iron Oxide.
• Polymeric (Plastic) Materials - Poor conductors of electricity and hence used as insulators. - Strength and ductility vary greatly. - Low densities and decomposition temperatures.
Examples : Poly vinyl Chloride (PVC), Polyester.
Applications : Appliances, DVDs,
Fabrics etc.
Types of Materials:
Rasdi Deraman, FKM UiTM NPP 6
• Ceramic Materials
- High hardness, strength and wear resistance.
- Very good insulator. Hence used for furnace lining for heat treating and melting metals.
Other applications : Abrasives, construction materials, utensils etc.
Example: Porcelain, Glass, Silicon nitride.
• Composite Materials
- Mixture of two or more materials.
- Consists of a filler material and a binding material.- Materials only bond, will not dissolve in each other.
Examples : Fiber Glass ( Reinforcing material in a polyester or epoxy matrix).
Concrete ( Gravels or steel rods reinforce
in cement and sand). Applications : Aircraft wings and engine, construction.
Rasdi Deraman, FKM UiTM NPP 7
• Electronic Materials
- Not Major by volume but very important.
- Silicon is a common electronic material.
- Its electrical characteristics are changed by adding impurities.
Examples: Silicon chips, transistors
Applications : Computers, Integrated Circuits, Sattelites etc.
Rasdi Deraman, FKM UiTM NPP 8
Atomic Structure And Bonding- Structure of Atoms
ATOMBasic Unit of an Element
Diameter : 10 –10 m.Neutrally Charged
NucleusDiameter : 10 –14 m
Accounts for almost all massPositive Charge
ProtonMass : 1.673 x 10 –24 g
Charge : 1.602 x 10 –19 C
NeutronMass : 1.675 x 10 –24 g
Neutral Charge
Electron CloudMass : 9.109 x 10 –28 gCharge : -1.602 x 10 –9 CAccounts for all volume
Rasdi Deraman, FKM UiTM NPP 9
Protons and neutrons join together to form the nucleus – the central part of the atom
+
+ --
Electrons move around the nucleus
Neutron
Proton
Electron
Nucleon
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?
Rasdi Deraman, FKM UiTM NPP 10
ATOMIC NUMBER and ATOMIC MASS
1)1) ATOMIC NUMBER ATOMIC NUMBER 2) ATOMIC MASS2) ATOMIC MASS
Atom can be described using :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
Rasdi Deraman, FKM UiTM NPP 11
- Atomic Number and Atomic Mass
• Atomic Number = Number of Protons in the nucleus
• Unique to an element Example :- Hydrogen = 1, Uranium = 92
• Relative atomic mass = Mass in grams of 6.203 x 1023
( Avagadro Number) Atoms. Example :- Carbon has 6 Protons and 6 Neutrons. Atomic
Mass = 12.
• One Atomic Mass unit is 1/12th of mass of carbon atom.
• One gram mole = Gram atomic mass of an element. Example :-
One gramMole ofCarbon
12 Grams Of Carbon
6.023 x 1023
Carbon Atoms
Rasdi Deraman, FKM UiTM NPP 12
Periodic Table
Rasdi Deraman, FKM UiTM NPP 13
•Atoms which have the same number of protons but different numbers of neutrons.
•Atoms which have the same atomic number but different mass number.
•Eg : Hydrogen has 3 isotopes.
Natural Natural
IsotopeIsotope
ProtonProton NeutronNeutron Mass Mass numbernumber
Hydrogen 1
(hydrogen)
1 0 1
Hydrogen 2
(deuterium)
1 1 2
Hydrogen 3
(tritium)
1 2 3
H11 H (D)2
1 H (T)31
Same atomic no. @ no. of protons
Different mass number
ISOTOPES
Rasdi Deraman, FKM UiTM NPP 14
ELECTRON CONFIGURATIONS OF THE
ELEMENTSElectron configurationElectron configuration – the ways in which electrons are arranged around the nucleus of atoms.
Electron capacity = 2n2
The Pauli principle can be used to show the maximum number of electron permitted in any sub-shells.
Rasdi Deraman, FKM UiTM NPP 15
Based on the Aufbau principle, which assumes that electrons enter orbitals of lowest energy first.
Sub-shells
No. of electron
s
s 2
p 6
d 10
f 141s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6p6, 7s2, 5f14, 6d10, 7p6
Rasdi Deraman, FKM UiTM NPP 16
Each orbital holds a max. of 2 electrons.
Rasdi Deraman, FKM UiTM NPP 17
The number of available electron states in some of the electrons shells and sub-shells.
Rasdi Deraman, FKM UiTM NPP 18
ATOMIC BONDING
Chemical bonding between atoms occurs since there is a lower potential energy of atoms to achieve more stability arrangements than they exist as an individual atoms.
2) Secondary Atomic Bonding Van der WaalsVan der Waals
1) Primary Inter-atomic Bonding:Metallic, Ionic and CovalentMetallic, Ionic and Covalent
Chemical bonds can be divided into 2 categories :
Rasdi Deraman, FKM UiTM NPP 19
- a) Ionic Bonding• Ionic bonding is due to electrostatic force of attraction
between cations (+ ve charge) and anions (- ve charge).
• Ionic bonds are nondirectional.
• It can form between metallic and nonmetallic elements.
• Electrons are transferred from electropositive to electronegative atoms.
Cation+ve charge
ElectropositiveElement
ElectronegativeAtom
Electron Transfer
Anion-ve charge
IONIC BOND
ElectrostaticAttraction
Rasdi Deraman, FKM UiTM NPP 20
Ionic Bonding - Example
• Ionic bonding in NaCl
SodiumAtomNa=11
ChlorineAtomCl=17
Chlorine IonCl -
IONIC
BOND
Rasdi Deraman, FKM UiTM NPP 21
b) Covalent Bonding
• Large interatomic forces are created by the sharing of electrons to form directional bonds.
• In Covalent bonding, outer s and p electrons are shared between two atoms to obtain noble gas configuration.
• Takes place between elements
with small differences in
electronegativity and close by
in periodic table.
• In Hydrogen, a bond is formed between 2 atoms by sharing their 1s1 electrons
H + H H H
1s1
Electrons
ElectronPair
HydrogenMolecule
Overlapping Electron Clouds
Rasdi Deraman, FKM UiTM NPP 22
Covalent Bonding - Examples
• In case of F2, O2 and N2, covalent bonding is formed by sharing p electrons
• Fluorine gas (Outer orbital – 2s2 2p5) share one p electron to attain noble gas configuration.
• Oxygen (Outer orbital - 2s2 2p4) atoms share two p electrons
• Nitrogen (Outer orbital - 2s2 2p3) atoms share three p electrons
H H
F + F F F F F
O + O O O O = O
N + N N N N N
Bond Energy=160KJ/mol
Bond Energy=54KJ/mol
Bond Energy=28KJ/mol
Rasdi Deraman, FKM UiTM NPP 23
Covalent Bonding in Benzene
• Chemical composition of Benzene is C6H6.
• The Carbon atoms are arranged in hexagonal ring.
• Single and double bonds alternate between the atoms.
• Chemical composition of Benzene is C6H6.
• The Carbon atoms are arranged in hexagonal ring.
• Single and double bonds alternate between the atoms.
CC
C
C
CH
H
H
H
H
Structure of Benzene Simplified Notations
Rasdi Deraman, FKM UiTM NPP 24
Covalent Bonding in Carbon
• A carbon atom can form symmetrically toward the corners of a tetrahedron.
Rasdi Deraman, FKM UiTM NPP 25
c) Metallic Bonding
• Atoms in metals are closely packed in crystal structure.• Loosely bounded valence electrons are attracted towards
nucleus of other atoms.• Electrons spread out among atoms forming electron clouds.• These free electrons are reason for electric conductivity and ductility.• Since outer electrons are shared by many atoms, metallic bonds are Non-directional bonding.
Positive Ion (ion cores)
Valence electron charge cloud
• The electron cloud act as “glue” to hold the ion cores together.
Rasdi Deraman, FKM UiTM NPP 26
• Overall energy of individual atoms are lowered by metallic bonds
• Minimum energy between atoms exist at equilibrium distance a0
• Fewer the number of valence electrons involved, more metallic the bond is.
Example:- Na Bonding energy 108KJ/mol,
Melting temperature 97.7oC
• Higher the number of valence electrons involved, higher is the bonding energy.
Example:- Ca Bonding energy 177KJ/mol,
Melting temperature 851oC
Metallic Bonds (Cont..)
Rasdi Deraman, FKM UiTM NPP 27
2) Secondary Atomic Bonding: Van der Waals
Occur when there is no exchanging @ sharing of electrons, eg : inert gases.The atom behaves like a dipole.
Rasdi Deraman, FKM UiTM NPP 28
Characteristic of BONDING
Rasdi Deraman, FKM UiTM NPP 29
Crystal Structures• Atoms, molecules, or ions are arranged in repetitive
3-D pattern, in long range order (LRO) give rise to crystal structure.
• Properties of solids depends upon crystal structure and bonding force.
• Generally, fluid substances form crystals when they undergo a process of solidification. Under ideal conditions, the result may be a single crystal, where all of the atoms in the solid fit into the same lattice.
• However, many crystals form simultaneously during solidification, leading to a polycrystalline solid.
Rasdi Deraman, FKM UiTM NPP 30
STRUCTURE OF SOLIDS
•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
Turbine blades
Rasdi Deraman, FKM UiTM NPP 31
The Space Lattice and Unit Cells
• An imaginary network of lines, with atoms at intersection of lines, representing the arrangement of atoms is called space lattice.
• Unit cell is that block of
atoms which repeats itself
to form space lattice.
• Materials arranged in short range order are called amorphous materials. Unit Cell
Space Lattice
Rasdi Deraman, FKM UiTM NPP 32
• Cubic Unit Cell a = b = c α = β = γ = 900
• Tetragonal a =b ≠ c α = β = γ = 900
Simple Body Centered
Face centered
SimpleBody Centered
Crystal Systems and Bravais LatticeAccording to Bravais (1811-1863), fourteen standard unit cells can describe all possible lattice networks.
Rasdi Deraman, FKM UiTM NPP 33
Types of Unit Cells (Cont..)
• Orthorhombic a ≠ b ≠ c α = β = γ = 900
• Rhombohedral a =b = c α = β = γ ≠ 900
Simple Base Centered
Face CenteredBody Centered
Simple
Rasdi Deraman, FKM UiTM NPP 34
Types of Unit Cells (Cont..)
• Hexagonal a ≠ b ≠ c α = β = γ = 900
• Monoclinic a ≠ b ≠ c α = β = γ = 900
• Triclinic a ≠ b ≠ c α = β = γ = 900
Simple
Simple
Simple
BaseCentered
Rasdi Deraman, FKM UiTM NPP 35
Simple Cubic (SC) Crystal Structure
• Represented as one atom at each corner of cube.
• Each atom has 8 nearest neighbors.
• Therefore, coordination number is 8.
close-packed directions
a
R=0.5a
contains 8 x 1/8 = 1 atom/unit cell
Rasdi Deraman, FKM UiTM NPP 36
Body Centered Cubic (BCC) Crystal Structure
• Represented as one atom at each corner of cube and one at the center of cube.
• Each atom has 8 nearest neighbors.
• Therefore, coordination number is 8.
• Examples :- Chromium (a=0.289 nm) Iron (a=0.287 nm) Sodium (a=0.429 nm)
Rasdi Deraman, FKM UiTM NPP 37
BCC Crystal Structure (Cont..)
• Each unit cell has eight 1/8
atom at corners and 1
full atom at the center.
• Therefore each unit cell has
• Atoms contact each
other at cube diagonal
(8x1/8 ) + 1 = 2 atoms
3
4RTherefore, lattice constant a BCC =
Rasdi Deraman, FKM UiTM NPP 38
Face Centered Cubic (FCC) Crystal Structure
• FCC structure is represented as one atom each at the corner of cube and at the center of each cube face.
• Coordination number for FCC structure is 12
• Atomic Packing Factor is 0.74
• Examples :- Aluminium (a = 0.405) Gold (a = 0.408)
Rasdi Deraman, FKM UiTM NPP 39
FCC Crystal Structure (Cont..)
• Each unit cell has eight 1/8 atom at corners and six ½ atoms at the center of six faces.• Therefore each unit cell has • Atoms contact each other across cubic face diagonal
(8 x 1/8)+ (6 x ½) = 4 atoms
2
4R• Therefore, lattice constant a =
Rasdi Deraman, FKM UiTM NPP 40
Atomic packing factor (APF) is defined as the efficiency of atomic arrangement in a unit cell.
ATOMIC PACKING FACTOR
APF = Volume of atoms in unit cell*
Volume of unit cell
*assume hard spheres
What is the APF for SC, FCC and BCC?
Rasdi Deraman, FKM UiTM NPP 4114
n AVcNA
# atoms/unit cell Atomic weight (g/mol)
Volume/unit cell
(cm3/unit cell)Avogadro's number (6.023 x 1023 atoms/mol)
DENSITY,
vVolume/Unit cellMass/Unit cellDensity of metal = =
Rasdi Deraman, FKM UiTM NPP 42
Example:
Copper (FCC) has atomic mass of 63.54 g/mol and atomic radius of 0.1278 nm. Determine:
a) the density of copper.
b) the APF of copper.
a = =2
4R
2
1278.04 nm
Volume of unit cell = V= a3 = (0.361nm)3 = 4.7 x 10-29 m3
v
FCC unit cell has 4 atoms.
Mass of unit cell = m =
Rasdi Deraman, FKM UiTM NPP 43
Characteristics of Selected Elements at 20°C
Rasdi Deraman, FKM UiTM NPP 44
Isotropy and Anisotropy
• Anisotropy : when the properties of a material vary with different crystallographic orientations, the material is called to be anisotropic.
• Isotropy: when the properties of a material are the same in all directions, the material is said to be isotropic.
Long Transverse
Rolling directionLongitudinal
Short Transverse
Rasdi Deraman, FKM UiTM NPP 45
Miller Indices
Miller Indices are used to refer to specific lattice planes of atoms in a crystal.
Why Miller indices is important?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 M. I of a DIRECTION
M. I of a PLANE
Rasdi Deraman, FKM UiTM NPP 46
MILLER INDICES OF A DIRECTION
How to determine crystal direction How to determine crystal direction indices?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.
iii) Place a ‘bar’ over the Negative indices and enclose with square parentheses
Rasdi Deraman, FKM UiTM NPP 47
EXAMPLE : CRYSTAL DIRECTION INDICESAxis X Y Z
Head (H)
Tail (T)
Projection (H-T)
Reduction
Enclose [ ]
Axis X Y Z
Head (H)
Tail (T)
Projection (H-T)
Reduction
Enclose [ ]
1/3
Rasdi Deraman, FKM UiTM NPP 48
Miller Indices of a Plane - Procedure
Choose a plane that does not pass through origin
Determine the x,y and z intercepts
of the plane
Find the reciprocals of the intercepts
Clear fractions bymultiplying by an integerto determine smallest set
of whole numbers
Enclose in round parenthesis Eg: crystal plane for x, y and z axes (111).
Place a ‘bar’ over theNegative indices
Fractions?
Rasdi Deraman, FKM UiTM NPP 49
z
x
y
Miller Indices - ExamplesAxis X Y Z
Intercept 1 ∞ 1
Reciprocal 1 0 1
Reduction - - -
Enclose (1 0 1)
Axis X Y Z
Intercept
Reciprocal
Reduction
Enclose ( )
Rasdi Deraman, FKM UiTM NPP 50
Exercise:Determine the Miller Indices plane for the following figure below?
Rasdi Deraman, FKM UiTM NPP 51
Quiz # 1 03/08/2009
a) Plane A intersects with x-axis and y-axis at ¾ and 1/3 respectively. It is also parallel to the z-axis. Determine the Miller Indices and sketch plane A in a cubic unit cell to support the answer.
b) Determine the direction Indices for vector Q in figure below.
Rasdi Deraman, FKM UiTM NPP 52
PHYSICAL PROPERTIES Physical properties are properties that can be recorded without changing the identity of the substance such as solubility in water, volume, length, colour, odour, melting point, mass, etc.
Example:• Metals have relatively high melting points and remain in the liquid state over a wide temp. range. • Metals conduct electricity and heat. • Metals are usually shiny when polished.
Rasdi Deraman, FKM UiTM NPP 53
MECHANICAL PROPERTIES
Generally, the common mechanical properties of metals are as follows:Toughness - the ability to resist / withstand repeated bending. Ductility – a property of material which can be easily
drawn into wiresBrittle – a property of material which easily
breaks when subjected to impacts.Hardness - resistance to scratching or indentation.Elasticity - ability to return to its original shape.Plasticity - does not return to its original shape.
Rasdi Deraman, FKM UiTM NPP 54
STRESS-STRAIN CURVEA stress–strain curve is a graph derived from measuring load (stress – σ) versus extension (strain – ε) for a sample of a material tested using tensile machine.
Typical regions that can be observed in a stress-strain curve are:
Elastic region, Yielding, Strain
Hardening, Necking
&Failure
Rasdi Deraman, FKM UiTM NPP 55
STRESS-STRAIN CURVE (Cont…)
Rasdi Deraman, FKM UiTM NPP 56
HARDNESS• Hardness is defined as a measure of the resistance of
a material to permanent deformation (plastic deformation).
• The hardness of a material is measured by forcing an indenter into its surface. The indenter can be either a ball, pyramid or cone type which is made of a material much harder than the material being tested such as hardened steel, tungsten carbide or diamond.
Rasdi Deraman, FKM UiTM NPP 57
IndenterIndenter LoadLoad ApplicationsApplications
Knoop Knoop (HK)(HK)
A very small rhombic pyramidal diamond. Between 1 and 1000 g •Measure the hardness of small specimen, very hard brittle materials (ceramic), very thin sections and small elongated areas.
Vickers Vickers (HV)(HV)
A very small square pyramidal diamond. Between 1 and 1000 g •Measure the hardness of small specimen, thin materials and small rounded areas. •More sensitive to measurement errors than Knoop test •Less sensitive to surface conditions than Knoop test
Brinell Brinell (HB)(HB)
Hardened steel or tungsten carbide ball.
Hardness-Width of indentation.
Use much higher loads than Rockwell.
•Steel parts.
Rockwell Rockwell (HR)(HR)
Conical diamond or hardened steel balls.
Hardness-Depth of penetration.
An initial minor load (10kg) followed by a larger major load (60, 100 or 150 kg)
•Measuring many materials from soft bearing metals to carbides.
Rasdi Deraman, FKM UiTM NPP 58
End of the Chapter 1
Rasdi Deraman, FKM UiTM NPP 59
Kemudahan Yang Disediakan:
Jaminan Pekerjaan.Biasiswa disediakan.Asrama disediakan.Makanan harian disediakan.Tenaga Pengajar yang Bertauliah.Pelajar akan ditempatkan di Kampus Golden Hope Academy, Carey Island, Banting Selangor D.E.
Cara Permohonan:
Sila dapatkan borang permohonan di Pejabat FKM, P.PTarikh tutup permohonan:
16 Feb 2007
Projek Usahasama antara UiTM-Golden Hope Academy
Diploma Kejuruteraan Mekanikal
“Palm Oil Mill Technology”
Syarat Permohonan:
Terbuka kepada semua pelajar semester 4
EM 110 UiTM CGPA 2.50 dan keatas dengan lulus semua
kursus sehingga semester 4.Pelajar yang berjaya selepas ditemuduga akan diberikan tawaran.
Keterangan Lanjut:
En. Juri SaedonTel: 03 5543 5167Fax: 03 5543 5160
E-mail: [email protected]