chapter 1 material science

8/12/2019 Chapter 1 Material Science

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SUBCONTENT :

1.1 ATOMIC STRUCTURE.

1.2 INTERATOMIC BONDING AMORPHOUS AND CRYSTALLINE SOLID.

1.3 CRYSTAL STRUCTURES.

1.4 EFFICIENCY OF ATOMIC PACKING, DENSITY COMPUTATION,MILLER INDICES.

1.5 RELATIONSHIP BETWEEN ATOMIC STRUCTURE, CRYSTAL STRUCTURES AND PROPERTIES OF MATERIAL.


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You should be able:

Describe an atomic structureConfigure electron configurationDifferentiate between each atomic bonding

Briefly describe ionic, covalent, metallic, hydrogenand van der waals bondsRelate the atomic bonding with materialproperties

LEARNING OBJECTIVE


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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 powerfulmicroscopes, scientists rely upon on models to help us to understand theatom.

Even with the worlds best

microscopes we cannot clearlysee the structure or behavior

of the atom.


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Even though we do not know what anatom looks like, scientific modelsmust be based on evidence. Many ofthe atom models that you have seenmay look like the one below whichshows the parts and structure of theatom.

Is this really an ATOM ?

This model represents themost modern version of the

atom.

Bo hr TheoryWave Mechanical Ato m ic Mod el


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Protons and neutrons join togetherto form the nucleus the central

part of the atom

+

+ --

Electronsmovearound thenucleus

Neutron

Proton

Electron

Nucleon orNucleus

Fig. : A simplified diagram of atom

Shell @ Orbital @ Energy level

Atoms are made of a nucleus that contains protons , neutrons and electrons that

orbit around the nucleus at different levels, known as shells.

What does an ATOM look like?
http://sites.google.com/site/danamaterials/atom/atom01.jpg?attredirects=0


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These particles have the following properties:

Particle Charge Location Mass (amu) Symbol

Proton Positive (+ve) Nucleus 1.0073

Neutron Neutral Nucleus 1.0087

Electron Ne gative (-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 determineits properties and identity. This is what makes one atom different from another.

+
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Most atoms are electrically neutral, meaning that they have an equal number ofprotons 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


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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 protons11 electrons Na+ 11 protons10 electrons

Cl 17 protons17 electrons Cl- 17 protons18 electrons

Cations are smaller than their parent atom becausethere is less e-e repulsion

Anions are larger than their parent atom because there ismore e-- e repulsion

If an atom gains or loses electrons, the atom is no longer neutral andit become electrically charged . The atom is then called an ION .


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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


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ATOMIC NUMBER tells how many PROTONS (Z) are in its atoms which determine theatoms identity.

The list of elements (ranked according to an increasing no. of protons) can be looked upon the Periodic Table . So, if an atom has 2 protons (atomic no. = 2), it must be helium(He).

ATOMIC MASS tells the sum of the masses of PROTONS (Z) and NEUTRONS (N) within thenucleus E.g :

Lithium:Atomic number = 33 protons, Z

4 neutrons, NAtomic mass, A = 3 + 4 = 7

BUT... although each element has a defined number of protons, the number of neutrons

is not fixed i so topes
http://d/nota%20METALURGI/periodictablehttp://d/nota%20METALURGI/periodictable


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Atoms which have the samenumber of protons but differentnumbers of neutrons.

Atoms which have the same atomic number but different atomicmass .

Eg : Hydrogen has 3 isotopes.

Natural

Isotope

Proton Neutron AtomicMass

Hydrogen 1(hydrogen)

1 0 1

Hydrogen 2(deuterium)

1 1 2

Hydrogen 3(tritium)

1 2 3

H11 H (D)21 H (T)

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Same atomic no. @ no. of protons

Different mass number

ISOTOPES


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Example of isotopes :

Element Name Number ofProton

NucleonNumber

Number ofNeutron

HydrogenHydrogen 1 1 0Deuterium 1 2 1

Tritium 1 3 2

OxygenOxygen-16 8 16 8Oxygen-17 8 17 9Oxygen-18 8 18 10

CarbonCarbon-12 6 12 6Carbon-13 6 13 7

Carbon-14 6 14 8Chlorine

Chlorine-35 17 35 18Chlorine-37 17 37 20

SodiumSodium-23 11 23 12Sodium-24 11 24 13


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Naturally occurring carbon consists of three isotopes,12C, 13C, and 14C. State the number of protons,

neutrons, and electrons in each of these carbon atoms .12C 13C 14C6 6 6

#p _______ _______ _______#n _______ _______ _______

#e _______ _______ _______

EXERCISE


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The electron cloud that surrounded the nucleus is divided into 7 shells (a.k.a energy level)K (1st shell, closest to nucleus) followed by L, M, N, O, P, Q.

Each of the shell, hold a limited no. of electrons.E.g : K (2 electrons), L (8 electrons), M (18 electrons), N (32 electrons).

3 rd shell

4 th shell

2nd shell

1st shell

K (2 electrons)

L (8 electrons)

M (18 electrons)

N (32 electrons)

ELECTRON SHELLS


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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


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d orbital - There is 5 distinct d orbitals

- Max 10 electrons - Third energy level


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Table : The number of available electron states in some of the electronsshells and subshells.

The max. no. of electrons that can occupy a specific shell can be foundusing the following formula:

Electron Capacity = 2 n2


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The following representation is used :

Example: it means that there are two electrons in the s orbital of thefirst energy level. The element is helium.

ELECTRON CONFIGURATIONS Electron configuration the ways in which electrons are arrangedaround the nucleus of atoms. The following representation is used :


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Based on the Aufbau principle , which assumes that electrons

enter orbital of lowest energy first.

The electrons in their orbital are represented asfollows :

1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10

4p6

5s2

4d10

5p6

6s2

4f 14

5d 10 6p 6 7s 2 5f 14 6d 10 7p 6


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By following these rules, we can build up the electron shell structure of all the atoms.

The key to the properties of atoms is the electrons in the outer shell.

Atom Symbol Atomic Number Electron configuration

Hydrogen H 1 1s 1 Helium He 2 1s 2 Lithium Li 3 1s 2 2s 1 Beryllium Be 4 1s 2 2s 2 Chlorine Cl 17 1s 2 2s 2 2p 6 3s 2 3p 5 Argon Ar 18 1s 2 2s 2 2p 6 3s 2 3p 6 Potasium K

19 1s2

2s2

2p6

3s2

3p6

4s1

Calcium Ca 20 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2


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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-

3 rd shell(energy

level)

Magnesium (12 electrons)

e-

e-

Answer : 1s 2 2s 2 2p 6 3s 2

e-

Answer : ls 2 2s 1


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TRANSITION ELEMENT

Cr [Z = 24] 4s 1 3d 5 (correct) halfly filled

Mo [Z = 42] 5s 1 4d 5 (correct) halfly filled

Cu [Z = 29] 4s 1 3d 10 (correct) completely filled Ag [Z = 47] 5s 1 4d 10 (correct) completely filled

Au [Z = 79] 6s 1 5d 10 (correct) completely filled


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ExerciseWrite the electron configuration for below element.

a) K

b) K1+

c) Fe

d) Fe 3+

1s2 2s

22p

63s

23p

64s

23d

10

4p 6 5s 2 4d 10 5p 6 6s 2 4f 14

5d 10 6p 6 7s 2 5f 14 6d 10 7p 6


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Exercise : TEST 1 [July 2011]

1a] With the aid of sketches, describe the Bohr Model of the

sodium [Na] and its ion in terms of valence electron , number ofelectron and shell.

[4 marks]


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1.2 INTERATOMIC BONDING AMORPHOUSAND CRYSTALLINE SOLID

2) Secondary Atomic BondingVan d er Waals

1) Primary Interatomic BondingMetal l ic , ion ic and cov alent

The forces of attraction that hold atoms together are called chemical bonds which canbe divided into 2 categories :

Chemical reactions between elements involve either the releasing/receiving or sharing ofelectrons .


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How is ionic bonding formed??

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1) IONIC BONDING

PRIMARY INTERATOMIC BONDING

Often found in compounds composed of electropositiveelements (metals) and electronegative elements (non metals)

Electron are transferred to form a bond

Large difference in electronegativity required


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Example: NaCl


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Properties :

Solid at room temperature (made of ions)High melting and boiling pointsHard and brittle

Poor conductors of electricity in solid state

Good conductor in solution or when molten

IONIC BONDING


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Predominant bonding in Ceramics

Give up electrons Acquire electrons

He-

Ne-

Ar-

Kr-

Xe-

Rn-

F4.0Cl

3.0

Br2.8

I2.5

At2.2

Li1.0

Na0.9K

0.8

Rb0.8Cs0.7

Fr0.7

H2.1

Be1.5

Mg1.2

Ca1.0Sr

1.0

Ba0.9

Ra0.9

Ti1.5

Cr1.6

Fe1.8

Ni1.8

Zn1.8

As2.0

CsCl

MgOCaF 2

NaCl

O3.5

EXAMPLE : IONIC BONDING


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EXERCISE

Draw the following ionic bonding ?

IONIC BONDING :Group 1 metal + Group 7 non metal, eg : NaClGroup 2 metal + Group 7 non metal, eg : MgF, BeF, MgBr, CaCl or CaI Group 2 metal + Group 6 non metal, eg : CaO , MgO, MgS, or CaS


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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 (H , Cl , F , etc)


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Properties

Gases, liquids, or solids (made of molecules)

Poor electrical conductors in all phases Variable ( hard , strong, melting temperature, boiling point)

2) COVALENT BONDING


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Molecules with nonmetals Molecules with metals and nonmetals Elemental solids

Compound solids (about column IVA )

He-

Ne-

Ar

-Kr

-

Xe-

Rn-

F4.0Cl

3.0Br

2.8I

2.5

At2.2

Li1.0Na

0.9K

0.8

Rb0.8Cs0.7

Fr0.7

H2.1

Be1.5Mg

1.2Ca1.0

Sr1.0Ba0.9

Ra0.9

Ti1.5

Cr1.6

Fe1.8

Ni1.8

Zn1.8

As2.0

SiC

C(diamond)

H2O

C2.5

H2

Cl2

F2

Si

1.8Ga1.6

GaAs

Ge1.8

O2.0

c o l u m n I V A

Sn1.8Pb1.8

EXAMPLE : COVALENT BONDING


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Draw the following covalent bonding?

SINGLE BOND :HydrogenFluorineWater

DOUBLE BOND :Oxygen

TRIPLE BOND :

Nitrogen

EXERCISE


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Occur when some electrons in the valence shell separatefrom their atoms and exist in a cloud surrounding all thepositively 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??


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3) METALLIC BONDING

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Properties:Good electrical conductivity

Good heat conductivity

Ductile

Opaque

3) METALLIC BONDING


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Explain why are metals ductile and can conductelectricity?


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Arise from atomic or molecular dipoles

VAN DER WAALS

SECONDARY INTERATOMIC BONDING


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Three bonding mechanism

Fluctuating Induced Dipole Bonds

Eg: Inert gases, symmetric molecules (H 2, Cl2)

Polar molecule-Induced Dipole Bonds

Asymmetrical molecules such as HCl, HF

Permanent Dipole Bonds

Hydrogen bonding Between molecules

H-F, H-O, H-N


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Molecule is considered the smallest particle of a purechemical substance that still retains its compositionand chemical properties.

Most common molecules are bound together bystrong covalent bonds .

E.g. : F2, O2, H2.

The smallest molecule : Hydrogen molecule .

MOLECULE


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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 ond energy Melting point Hardness Conductivity Comments

Ionicbonding

Large(150-370kcal/mol)

Very high Hard andbrittle

Poor-required

moving ion

Nondirectional(ceramic)

Covalentbonding

Variable(75-300 kcal/mol)

Large -DiamondSmall Bismuth

Variable

Highest diamond(>3550)

Mercury (-39)

Very hard(diamond)

Poor Directional(Semiconductors,ceramic, polymer

chains)

Metallicbonding

Variable(25-200 kcal/mol)

Large- TungstenSmall- Mercury

Low to high Soft to hard Excellent Nondirectional(metal)

Secondarybonding

Smallest Low tomoderate

Fairly soft Poor Directionalinter-chain(polymer)

inter-molecular


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Ceramics (Ionic & covalent bonding):

Metals (Metallic bonding):

Polymers (Covalent & Secondary):

Large bond energylarge T m large Esmall a

Variable bond energymoderate T m moderate Emoderate a

Directional PropertiesSecondary bonding dominatessmall Tsmall Elarge a

SUMMARY : PRIMARY BONDING


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E i Fi l E [M h 2002]


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Exercise : Final Exam [March 2002]

1a] Briefly describe differences between metallic bond and covalent bond.Support your answer with an example and simple sketch.

(7 Marks)


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1.3 CRYSTAL STRUCTURE

Crystalstructure

CrystallineMaterial

Single Crystal polycrystal

Noncrsytallinematerial

(Amorphous)

* comprised of many single crystal or grain


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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

crystalline SiO 2

noncrystalline SiO 2

"Amorphous" = Noncrystalline


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No recognizable long-range order

Completely ordered

In segments

Entire solid is made upof atoms in an orderlyarray

Amorphous

Polycrystalline

Crystal

Atoms are disordered

No lattice All atoms arranged ona common lattice

Different latticeorientation for eachgrain

Structure of SOLID


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Some engineering applications require single crystals:

--turbine blades


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Most engineering materials are polycrystals.

grain


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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]


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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 atomsin a crystal.

A lattice and motif help to illustrate the crystal structure.

CRYSTAL STRUCTURE

lattice motif crystal structure

=+


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Lattice - The threedimensional arrayformed by the unit cells

of a crystal is calledlattice.

Unit Cell - When a solidhas a crystallinestructure, the atoms are

arranged in repeatingstructures called unitcells. The unit cell is thesmallest unitthat demonstrate the fullsymmetry of a crystal.

A crystal is a three-dimensional repeatingarray.

+

=
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Fig. : The crystal structure (a) Part of the space lattice for natrium chloride (b)Unit cell for natriumchloride crystal

Unit cell - a tiny box thatdescribe the crystal structure.

Crystal structure may be present with any of thefour types of atomic bonding.

The atoms in a crystal structure are arranged

along crystallographic planes which are designatedby the Miller indices numbering system.

The crystallographic planes and Miller indices areidentified by X-ray diffraction.

Fig. : The wavelength of the X-ray issimilar to the atomic spacing in crystals.
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B RAVA IS LATTICE - describe the geometric arrangement of the lattice points andthe translational symmetry of the crystal.

CRYSTAL SYSTEM AND CRYSTALLOGRAPHY

cub ic , hexagonal,tetragonal,rhombodhedral,orthorhombic, monoclinic,triclinic.

7 crystal systems :

By adding additionallattice point to 7 basiccrystal systems form 14 Bravais

lattice.


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FCCSC

BCC

CRYSTAL STRUCTURE OF METALS


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Types of crystal structure Examples

Simple cubic Cube where atoms lie on a grid.

No. of atom at corner = 8x1/8 = 1 atomTotal no. of atom in 1unit cell = 1 atom

Manganese

Body centered cubic(BCC)

*Lower ductility but a higher yieldstrength than FCC metals.

Cube with an atom at each corner and one in the center.

No. of atom at corner = 8x1/8 = 1 atomNo. of atom at center = 1 atomTotal no. of atom in 1unit cell = 2 atoms

Chromium,Tungsten(W),Molybdenum,Vanadium(V)

CRYSTAL STRUCTURE OF METALSMost metals (about 90%) crystallize upon solidification into three densely packed crystal

structures as shown below (BCC,FCC,HCP). However, simple cubic crystals are rather rare whichcan be viewed as simple cubic grid.

Types of crystal structure Examples


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Face centered cubic (FCC)

* Atoms for FCC are more densely packed than BCC.

Cube with an atom at each corner, one in the center and one in thecenter of each side of the cube.

No. of atom at corner = 8x1/8 = 1 atomNo. of atom at face = 6x1/2 = 3 atomTotal no. of atom in 1unit cell = 4 atoms

Aluminum,Nickel, Copper,Gold, Lead,Platinum


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1.4 EFFICIENCY OF ATOMIC

PACKING,DENSITY COMPUTATIONAND MILLER INDEX


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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 arrangementin a unit cell.

It is used to determine the most dense arrangement of atoms . It is because howthe 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 :


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a (lattice constant) and

R (atom radius)

Atoms/unit

cell

Coordi-

nationNo.

Packing

Density(APF)

Examples

Simplecubic a = 2R

1 6 52% CsCl

BCC a = 4R/ 3 2 8 68% Many metals:-Fe, Cr, Mo, W

FCCa = 4R/ 2

or

a = 2R2

4 12 74% Many metals : Ag, Au, Cu, Pt

Table : APF for simple cubic, BCC, FCC and HCP


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EXAMPLE

APF for a simple cubic structure = 0.52

Calculate the APF for Simple Cubic (SC)?


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EXERCISE

a) BCC b) FCC

Calculate the APF for BCC and FCC ?


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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]


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DENSITY COMPUTATIONS

A knowledge of the crystal structure of a metallicsolid permits computation of its density through therelationship :

Where

= n AVc N A

n = number of atoms associated with each unit cell

A = atomic weightVc = volume of the unit cellN A = Avogadros number (6.023 x 10 23 atoms/mol)


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

EXAMPLE

The volume (V) of the unit cell is equal to the cell-edge length (a) cubed.

V = a3

= (0.3524 nm)3

= 0.04376 nm3

Since there are 10 9 nm in a meter and 100 cm in a meter, there must be 10 7 nm in a cm.10 9 x 1m = 10 7 nm/cm1 m 100 cm

We can therefore convert the volume of the unit cell to cm 3 as follows.4.376 x 10 -2 nm 3 x [1 cm ] 3 = 4.376 x 10 -23 cm 3

107

nmThe mass of a nickel atom can be calculated from the atomic weight of this metal and Avogadrosnumber.

58.69g Ni x 1 mol = 9.746 x 10 -23 g/atom 1(9.746 x 10 -23 g/unit cell) = 2.23 g/cm 3 1 mol 6.023 x 10 23 atoms 4.376 x 10 -23 cm 3 /unit cell


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Crystalstructure, 20 oC

Density of solid,20 oC (g/cm 3)

Copper has an atomic radius of 0.128 nm, FCC crystal structure and an atomicweight of 63.5 g/mol. Compute its density and compare the answer with itsmeasured density.

Element Symbol AtomicNumbe

r

Atomicweight(amu)

Atomicradius(nm)

Chromium Cr 24 52.00 7.19 BCC 0.125

Cobalt Co 27 58.93 8.9 HCP 0.125

Copper Cu 29 63.55 8.94 FCC 0.128

EXERCISE


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1b] Platinum has a FCC structure, a lattice parameter of 0.393 nm and an atomic weightof 195.09 g/mol. Determine :

i. Atomic radius [in cm]

ii. Density of platinum

[ 6marks]Solution :

= n AVc N A

QUESTION : TEST 1 [August 2012]


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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-raydiffraction patterns and the movement of a dislocation , which may determinethe 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 : a family of equivalent directions

Figure : Planes of the form {110} in cubic systems


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i) Determine the points at which a given crystal planeintersects the three axes, say at (a,0,0),(0,b,0), and (0,0,c). Ifthe 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 expressedas the smallest integers and negative quantities are indicatedwith an overbar,e.g : a.

MILLER INDICES OF A PLANE

How to determine crystal plane indices?

Figure : Planes with different Millerindices in cubic crystals

Axis X Y Z

Interceptions

Reciprocals

Reduction (if necessary)

Enclosed (h k l )


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+x

+y

+z

_z

_y

_x

(1 , 0 , 0)

(0 , 1 , 0)

(0 , 0 , 1)

_

(0 , 0 , 1)

_(0 , 1 , 0)

_(1 , 0 , 0)


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Axis X Y ZInterceptions

Reciprocals


Enclosed ( )

EXERCISE. : CRYSTAL PLANE INDICES

0

1

1

1

10

1

1

Axis X Y Z

Interceptions

Reciprocals


Enclosed ( )


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0

1

1

1

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Two numbers in one axes

0

Axis X Y Z

Interceptions

Reciprocals


Enclosed ( )

Axis X Y Z

Interceptions

Reciprocals


Enclosed ( )



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Plane pass through o r ig in

0

0

Axis X Y Z

Interceptions

Reciprocals


Enclosed ( )

Axis X Y Z

Interceptions

Reciprocals


Enclosed ( )



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Determine the Miller Indices plane for thefollowing figure below?

Answer :

a) (121)

b) (210)

c) (111)

a)

0

1

1

1

Answer :

a) (121)

b) (210)c) (111)

b)

0

1

1

Answer :a) (121)

b) (210)

c) (111) c)

0

1

1

1

c)

Answer :a) (121)

b) (210)

c) (111) c)

0

1

1

1

d)


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Determine the Miller indices of a cubic crystal plane that intersects the position coordinates

A (1, 1/4, 0), B (1, 1, 1/2) ,C (3/4, 1, 1/4) and D (1/2, 1, 0) ?

Axis x y z

InterceptsReciprocals

Reduction

(if necessary)

Enclosed ( )


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Draw the following Miller Indices plane.

a) ( 1 0 0 )

b) ( 0 0 1 )

c) ( 1 0 1 )

d) ( 1 1 0 )


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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 thesmallest integers and negative quantitiesare indicated with an overbar.

iii) Label the direction [hkl]. Figure : Examples of direction

Axis X Y Z

Head (H) x 2 y2 z 2

Tail (T) x 1 y1 z 1

Head (H) Tail (T) x 2-x 1 y2-y1 z 2-z1


Enclosed [h k l]

* No reciproc al involved.


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Axis X Y Z

Head (H)

Tail (T)

Projection (H-T)

Reduction (if necessary) _____________________

Enclosed [ ]

EXERCISE : CRYSTAL DIRECTION INDICES

0

1

1

01

1

1

1Axis X Y Z

Head (H)

Tail (T)

Projection (H-T)


Enclosed [ ]


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Axis X Y Z

Head (H)

Tail (T)

Projection (H-T)


Enclosed [2 4 3 ]

EXERCISE : CRYSTAL DIRECTION INDICES

0

0

Axis X Y Z

Head (H)

Tail (T)

Projection (H-T)


Enclosed [ 4 3 6 ]


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Determine the Miller Indices direction

for the following figure below? a)

Answer :

a) (121)b) (210)

c) (111) c)

0

1

1

1Answer :

a) (121)

b) (210)

c) (111)

b)

0

1

1

1

Answer :

a) (121)

b) (210)

c) (111)c)

0

1

1

1c)


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Determine the direction indices of the cubicdirection between the position coordinates

TAIL (3/4, 0, 1/4) and HEAD (1/4, 1/2, 1/2) ?

Axis x y z

Head

Tail

Projection(Head Tail)

Reduction(if necessary)

Enclosed [ ]


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Draw the following Miller Indicesdirection.

a) [ 1 0 0 ]

b) [ 1 1 1 ]

c) [ 1 1 0 ]

d) [ 1 1 0 ]


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NOTE (for p lane and direct io n):

PLANE

Make sure you enclosed your final answer in brackets () with noseparating commas (hkl)

DIRECTIONMake sure you enclosed your final answer in brackets () with no

separating commas [hkl]

FOR BOTH PL A NE AND DIRECTION

Nega t ive num ber shou ld be wr i t t en as fo l lows :

-1 (WRO NG)

1 (CORRECT)

Fina l answer fo r l abe l ing the p lane and d i rec t ion sho u ld no t have f rac t ionn u m b e r do a reduction.


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1.5 RELATIONSHIP BETWEENATOMIC STRUCTURE, CRYSTAL

STRUCTURES AND PROPERTIES OFMATERIALS


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PHYSICAL PROPERTIES OF METALS

Solid at room temperature (mercury is an exception)

Opaque

Conducts heat and electricity

Reflects light when polishedExpands when heated, contracts when cooled

It usually has a crystalline structure

Physical properties are the characteristic responses of materials toforms of energy such as heat, light, electricity and magnetism.

The physical properties of metals can be easily explained as follows :

MECHANICAL PROPERTIES OF METALS


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MECHANICAL PROPERTIES OF METALSMechanical properties are the characteristic dimensional changes in response toapplied external or internal mechanical forces such as shear strength, toughness,stiffness etc.

The mechanical properties of metals can be easily explained as follows :


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TOUGHNESS - the ability to resist/withstand repeated bending. DUCTILITY - can be easily drawn into wires BRITTLE - easily breaks when subjected to impacts or

elongation HARDNESS - resistance to scratching or indentation ELASTICITY - ability to return to its original shape PLASTICITY - does not return to its original shape

MECHANICAL PROPERTIES OF METALS


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STRESS-STRAIN CURVEA stress strain curve is a graph derived from measuringload (stress ) versus extension (strain ) for a sampleof a material tested using tensile machine.

The stress-strain curve characterizes the behavior of thematerial tested.

Different materials show different shapes of graph.

Figure : Stress strain curve of differentmaterials.

Mild steel(ductile)

Cast iron(brittle)

Because of grain structure atomicbonding(strong- fine microstructures /weak- coarsemicrostructures).

WHY

This diagram is used to determine how material will react under a certain load


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Figure : Stress strain diagram

Typical regions that canbe 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.


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DEFORMATION OF METALS1. Elast ic deform ation

Atomic bonds are stretched but not broken.Once the forces are no longer applied, the objectreturns to its original shape.This means that elastic deformation is revers ib le .

2. Plast ic deform ation Atomic bonds are broken and new bonds are created .Therefore, the change is permanent .

Slip plane


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F

bondsstretch

return toinitial

1. Initial 2. Small load 3. Unload

__________ deformation

__________ deformation

Elastic/plastic?


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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 fastand can occur (with flaws) at lower stress levels than a ductile fracture.

Figure : Stress strain curve for brittle and ductile material


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Du ctile fr acture Br ittle fr acture Plastic deformation Small/ no plastic deformation

High energy absorption before fracture Low energy absorption before fracture Characterized by slow crack propagation Characterized by rapid crack propagation

Detectable failure Unexpected failure

Eg: Metals, polymers Eg: Ceramics, polymers

What are the differences betweenductile fracture & brittle fracture?


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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]


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HARDNESS

Resistance to plastic (permanent) due to impactfrom a sharp object .

It is usually measured by loading an indenter ofspecified geometry onto the material andmeasuring the dimensions of the resultingindentation .

Indenter Load Applications

Knoop A very small rhombic pyramidal diamond. Between 1 and 1000 g Measure the hardness of small


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p(HK)

y py gspecimen, very hard brittle materials(ceramic), very thin sections and smallelongated areas.

Vickers(HV) A very small square pyramidal diamond. Between 1 and 1000 g Measure the hardness of smallspecimen, thin materials and smallrounded areas.More sensitive to measurement errorsthan Knoop testLess sensitive to surface conditionsthan Knoop test

Brinell(HB)

Hardened steel or tungsten carbide ball.

Hardness-Width of indentation.

Use much higher loadsthan Rockwell.

Steel parts.

Rockwell(HR)

Conical diamond or hardened steel balls.

Hardness-Depth of penetration.

An initial minor load (10kg)followed by a larger majorload (60, 100 or 150 kg)

Measuring many materials from softbearing metals to carbides.


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1c] Hardness is one of the important mechanical propertiesin engineering. Describe FOUR [4] types of hardness

measurement method in terms of name and types ofindenter.

[ 4 marks]

QUESTION : FINAL EXAM [Oct 2012]


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THE END

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STRUCTURE ANDPROPERTIES PowerPoint.

chapter 1 material science

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