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


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


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.


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


• 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:


• 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.


• 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.


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


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?


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


- 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


Periodic Table


•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


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.


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


Each orbital holds a max. of 2 electrons.


The number of available electron states in some of the electrons shells and sub-shells.


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 :


- 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


Ionic Bonding - Example

• Ionic bonding in NaCl

SodiumAtomNa=11

ChlorineAtomCl=17

Chlorine IonCl -

IONIC

BOND


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


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




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


Covalent Bonding in Carbon

• A carbon atom can form symmetrically toward the corners of a tetrahedron.


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.


• 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..)


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.


Characteristic of BONDING


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.


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


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


• 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.


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


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


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


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)


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 =


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)


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 =


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?


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 = =


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 =


Characteristics of Selected Elements at 20°C


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


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


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


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


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?


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 ( )


Exercise:Determine the Miller Indices plane for the following figure below?


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.


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.


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.


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


STRESS-STRAIN CURVE (Cont…)


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.


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.


End of the Chapter 1


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]

Documents

Chp 1 Material Science 281 Uitm Em110