CHAPTER 1 : STRUCTURE

(10 hours)

SUBCONTENT :

1.1 ATOMIC STRUCTURE.

1.2 INTERATOMIC BONDING AMORPHOUS AND CRYSTALLINE SOLID.

1.3 CRYSTAL STRUCTURES.

1.4 EFFICIENCY OF ATOMIC PACKING, DENSITY COMPUTATION,

MILLER INDICES.

1.5 RELATIONSHIP BETWEEN ATOMIC STRUCTURE, CRYSTAL STRUCTURES

AND PROPERTIES OF MATERIAL.

You should be able:

Describe an atomic structure

Configure electron configuration

Differentiate between each atomic bonding

Briefly describe ionic, covalent, metallic, hydrogen and Van der Waals bonds

Relate the atomic bonding with material properties

LEARNING OBJECTIVE

4

1.1 ATOMIC STRUCTURE

All matter is made up of tiny particles called atoms.

What are ATOMS?

Since the atom is too small to be seen even with the most powerful

microscopes, scientists rely upon on models to help us to understand the

atom.

Even with the world’s best microscopes we cannot clearly see the structure or behavior

of the atom.

5

Even though we do not know what an

atom looks like, scientific models

must be based on evidence. Many of

the atom models that you have seen

may look like the one below which

shows the parts and structure of the

atom.

Is this really an ATOM?

This model represents the most modern version of the

atom.

Bohr Theory

Wave Mechanical Atomic Model

6

Protons and neutrons join together to form the nucleus – the central part of the atom

+

+ --

Electrons move around the nucleus

Neutron

Proton

Electron

Nucleon or Nucleus

Fig. : A simplified diagram of atom

Shell @ Orbital @ Energy level

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

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

What does an ATOM look like?

7

•These particles have the following properties:

Particle Charge Location Mass (amu) Symbol

Proton Positive (+ve) Nucleus 1.0073

Neutron Neutral Nucleus 1.0087

Electron Negative (-ve) Orbital 0.000549

-

To describe the mass of atom, a unit of mass called the atomic mass unit (amu) is used.

•The number of protons, neutrons and electrons in an atom completely determine its properties and identity. This is what makes one atom different from another.

+

8

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

protons and electrons. The positive and negative charges cancel each other out.

Therefore, the atom is said to be electrically neutral.

Why are all ATOMS are ELECTRICALLY NEUTRAL?

+

-

Neutron

Proton

Electron

++

+-

-

+ -

Fig. : Beryllium atom

Proton = 4

Electron = 4

NEUTRAL

CHARGE

9

cation - ion with a positive charge- If a neutral atom loses one or more electrons, it becomes a cation.

anion - ion with a negative charge - If a neutral atom gains one or more electrons, it becomes an anion.

Na11 protons11 electrons Na+ 11 protons

10 electrons

Cl17 protons17 electrons Cl-

17 protons18 electrons

Cations are smaller than their “parent atom” because there is less e-e repulsion

Anions are larger than their “parent atom” because there is more e-- e repulsion

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

10

periodic: a repeating pattern

table: an organized collection of information

Periodic Table (P.T.)

An arrangement of elements in

order of atomic number;

elements with similar

properties are in the same

group.

Basics of the PERIODIC TABLE

11

The periodic table below is a simplified representation which

usually gives the :

1) period: horizontal row on the P.T.

•Designate electron energy levels

2) group or family: vertical column on the P.T.

Two main classifications in P.T.

12

ATOMIC NUMBER and ATOMIC MASS

1) ATOMIC NUMBER 2) ATOMIC MASS

Atom can be described using :

The element helium has the atomic number 2, is represented by

the symbol He, its atomic mass is 4 and its name is helium.

ATOMIC MASS , A = no. of protons (Z) + number of neutrons (N)

SYMBOL

ATOMIC NUMBER, Z = no. of protons

PERIODIC TABLE

14

ATOMIC NUMBER tells how many PROTONS (Z) are in its atoms which determine the atom’s identity.

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

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

Lithium:Atomic number = 33 protons, Z4 neutrons, NAtomic mass, A = 3 + 4 = 7

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

is not fixed isotopes

15

•Atoms which have the same number of protons but different numbers of neutrons.

•Atoms which have the sameatomic number but different atomic mass .

•Eg : Hydrogen has 3 isotopes.

Natural

Isotope

Proton Neutron Atomic

Mass

Hydrogen 1

(hydrogen)

1 0 1

Hydrogen 2

(deuterium)

1 1 2

Hydrogen 3

(tritium)

1 2 3

H11 H (D)2

1 H (T)31

Same atomic no. @ no. of protons

Different mass number

ISOTOPES

Exercise of isotopes :

Element NameNumber of

ProtonNucleon Number

Number of Neutron

Hydrogen

Hydrogen

Deuterium

Tritium

Oxygen

Oxygen-16

Oxygen-17

Oxygen-18

Carbon

Carbon-12

Carbon-13

Carbon-14

ChlorineChlorine-35

Chlorine-37

SodiumSodium-23

Sodium-24

Example of isotopes :

Element NameNumber of

ProtonNucleon Number

Number of Neutron

Hydrogen

Hydrogen 1 1 0

Deuterium 1 2 1

Tritium 1 3 2

Oxygen

Oxygen-16 8 16 8

Oxygen-17 8 17 9

Oxygen-18 8 18 10

Carbon

Carbon-12 6 12 6

Carbon-13 6 13 7

Carbon-14 6 14 8

ChlorineChlorine-35 17 35 18

Chlorine-37 17 37 20

SodiumSodium-23 11 23 12

Sodium-24 11 24 13

18

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

19

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 6 6 6

#n 6 7 8

#e 6 6 6

ANSWER

20

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

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

3rd shell

4th shell

2nd shell

1st shell

K (2 electrons)

L (8 electrons)

M (18 electrons)

N (32 electrons)

ELECTRON SHELLS

• Within each shell, the electrons occupy sub shell (energy sublevels)

– s, p, d, f, g, h, i. Each sub shell holds a different types of orbital.

• Each orbital holds a max. of 2 electrons.

• Each orbital has a characteristic energy state and characteristic shape.

• s - orbital

–Spherical shape

–Located closest to nucleus (first energy level)

–Max 2 electrons

• p - orbital

- There is 3 distinct p - orbitals (px, py, pz)

- Dumbbell shape

- Second energy level

- 6 electrons

ORBITAL

22

d- orbital

- There is 5 distinct d – orbitals

- Max 10 electrons

- Third energy level

Table : The number of available electron states in some of the electrons

shells and subshells.

The max. no. of electrons that can occupy a specific shell can be found

using the following formula:

Electron Capacity = 2n2

• The following representation is used :

• Example: it means that there are two electrons in the ‘s’ orbital of the

first energy level. The element is helium.

ELECTRON CONFIGURATIONSElectron configuration – the ways in which electrons are arranged

around the nucleus of atoms. The following representation is used :

1s2

Energy level @

Principal

quantum no.

Orbital

No. of electrons

in the orbital

Based on the Aufbau principle, which assumes that electrons

enter orbital of lowest energy first.

The electrons in their orbital are represented as follows :

1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p6

The sequence of addition of the

electrons as the atomic number

increases is as follows with the

first being the shell number the

s, p, d or f being the type of

subshell, the last number being

the number of electrons in the

subshell.

26

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-

Answer : 1s2 2s2 2p6 3s2

e-

Answer : ls2 2s1

27

Exercise: Electron Configurations

Atom Symbol Atomic Number Electron configuration

Hydrogen H

Helium He

Lithium Li

Beryllium Be

Chlorine Cl

Argon Ar

Potasium K

Calcium Ca

28

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 1s1

Helium He 2 1s2

Lithium Li 3 1s2 2s1

Beryllium Be 4 1s2 2s2

Chlorine Cl 17 1s2 2s2 2p6 3s2 3p5

Argon Ar 18 1s2 2s2 2p6 3s2 3p6

Potasium K 19 1s2 2s2 2p6 3s2 3p6 4s1

Calcium Ca 20 1s2 2s2 2p6 3s2 3p6 4s2

TRANSITION ELEMENT

Cr [Z = 24] 1s2 2s2 2p6 3s2 3p6 4s1 3d5 (correct) halfly filled

Mo [Z = 42] … 5s1 4d5 (correct) halfly filled

Cu [Z = 29] 1s2 2s2 2p6 3s2 3p6 4s1 3d10 (correct) completely filled

Ag [Z = 47] … 5s1 4d10 (correct) completely filled

Au [Z = 79] …6s1 5d10 (correct) completely filled

ExerciseWrite the electron configuration for below element.

a) K

b) K1+

c) Fe

d) Fe3+

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2

4f14 5d10 6p6 7s2 5f14 6d10 7p6

Answer : EXERCISEWrite the electron configuration for below element.

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2

4f14 5d10 6p6 7s2 5f14 6d10 7p6

a) K 1s2 2s2 2p6 3s2 3p6 4s1

b) K1+ 1s2 2s2 2p6 3s2 3p6

c) Fe 1s2 2s2 2p6 3s2 3p6 4s2 3d6

d) Fe3+ 1s2 2s2 2p6 3s2 3p6 3d5

Answer: TEST 1 [July 2011]

1a] With the aid of sketches, describe the Bohr Model of the sodium [Na] and its ion in terms of valence electron , number of electron and shell.

[4 marks]

Answer: TEST 1 [July 2011]

1a] With the aid of sketches, describe the Bohr Model of the sodium [Na] and its ion in terms of valence electron , number of electron and shell.

[4 marks]

Na Na+

Drawing

Valence electron 1 8

Number of electron 11 10

Number of shell 3 2

1.2 INTERATOMIC BONDING AMORPHOUS AND CRYSTALLINE SOLID

2) Secondary Atomic Bonding

Van der Waals

1) Primary Interatomic Bonding

Metallic, ionic and covalent

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

• Chemical reactions between elements involve either the releasing/receiving or sharing of electrons .

How is ionic bonding formed??

35

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

• Example: NaCl

• Properties :

Solid at room temperature (made of ions)

High melting and boiling points

Hard and brittle

Poor conductors of electricity in solid state

Good conductor in solution or when molten

IONIC BONDING

• Predominant bonding in Ceramics

Give up electrons Acquire electrons

He -

Ne -

Ar -

Kr -

Xe -

Rn -

F 4.0

Cl 3.0

Br 2.8

I 2.5

At 2.2

Li 1.0

Na 0.9

K 0.8

Rb 0.8

Cs 0.7

Fr 0.7

H 2.1

Be 1.5

Mg 1.2

Ca 1.0

Sr 1.0

Ba 0.9

Ra 0.9

Ti 1.5

Cr 1.6

Fe 1.8

Ni 1.8

Zn 1.8

As 2.0

CsCl

MgO

CaF2

NaCl

O 3.5

EXAMPLE : IONIC BONDING

Exercise : Final Exam [April 2008]

1c] With the aid of sketches, describe how Sodium and Chlorine atoms are joined.

[3 marks]

Answer : Final Exam [April 2008]

1c] With the aid of sketches, describe how Sodium and Chlorine atoms are joined.

[3 marks]

Draw = 1 mark

Single electron on the outer shell of sodium atom is

transferred to the outer shell of chlorine [with 7

electrons] to make it 8, thus filling [valence] the outer

shell of chlorine.

[½ mark]

Transfer of electron from sodium making it a positive

ion while chlorine atom change to negative ion after

receiving the electron.

[½ mark]

Different charge in both ion attracted to each other to

form ionic bonds.

[½ mark]

Sodium and chlorine bonds together by ionic bonding.

[½ mark]

EXERCISE

Draw the following ionic bonding?

IONIC BONDING :Group 1 metal + Group 7 non metal, eg : NaClGroup 2 metal + Group 7 non metal, eg : MgF₂, BeF₂, MgBr₂, CaCl₂ or CaI₂Group 2 metal + Group 6 non metal, eg : CaO, MgO, MgS, or CaS

• Electrons are shared to form a bond.

• Most frequently occurs between atoms with similar electronegativities.

• Often found in:

2) COVALENT BONDING

How is covalent bonding formed??

• Molecules with nonmetals

• Molecules with metals and nonmetals

(Aluminum phosphide (AlP)

• Elemental solids (diamond, silicon, germanium)

• Compound solids (about column IVA)

(gallium arsenide - GaAs, indium antimonide - InSb

and silicone carbide - SiC)

• Nonmetallic elemental molecules (H₂, Cl₂, F₂, etc)

Properties

• Gases, liquids, or solids (made of molecules)

• Poor electrical conductors in all phases

• Variable ( hard , strong, melting temperature, boiling point)

2) COVALENT BONDING

• Molecules with nonmetals• Molecules with metals and nonmetals• Elemental solids • Compound solids (about column IVA)

He -

Ne -

Ar -

Kr -

Xe -

Rn -

F 4.0

Cl 3.0

Br 2.8

I 2.5

At 2.2

Li 1.0

Na 0.9

K 0.8

Rb 0.8

Cs 0.7

Fr 0.7

H 2.1

Be 1.5

Mg 1.2

Ca 1.0

Sr 1.0

Ba 0.9

Ra 0.9

Ti 1.5

Cr 1.6

Fe 1.8

Ni 1.8

Zn 1.8

As 2.0

SiC

C(diamond)

H2O

C 2.5

H2

Cl2

F2

Si 1.8

Ga 1.6

GaAs

Ge 1.8

O 2.0

co

lum

n I

VA

Sn 1.8

Pb 1.8

EXAMPLE : COVALENT BONDING

Draw the following covalent bonding?

SINGLE BOND :HydrogenFluorineWater

DOUBLE BOND :Oxygen

TRIPLE BOND :Nitrogen

EXERCISE

• 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??

3) METALLIC BONDING

48

Properties:

Good electrical conductivity

Good heat conductivity

Ductile

Opaque

3) METALLIC BONDING

Explain why are metals ductile and can conduct electricity?

Explain why are metals ductile and can conduct electricity?

Good electrical conductivity- When an electrical potential difference is applied, the electrons move freelybetween atoms, and a current flows.

Ductile- The valence electrons are not closely associated with individual atoms, butinstead move around amongst the atoms within the crystal.

- The individual atoms can "slip" over one another yet remain firmly held togetherby the electrostatic forces exerted by the electrons.

- This is why most metals can be hammered into thin sheets (malleable) or drawninto thin wires (ductile).

• Arise from atomic or molecular dipoles

VAN DER WAALS

SECONDARY INTERATOMIC BONDING

• Three bonding mechanism

– Fluctuating Induced Dipole Bonds

• Eg: Inert gases, symmetric molecules (H2, Cl2)

– Polar molecule-Induced Dipole Bonds

• Asymmetrical molecules such as HCl, HF

– Permanent Dipole Bonds

• Hydrogen bonding

• Between molecules

• H-F, H-O, H-N

• Molecule is considered the smallest particle of a pure chemical substance that still retains its composition and chemical properties.

• Most common molecules are bound together by strong covalent bonds.

• E.g. : F2, O2, H2.

• The smallest molecule : Hydrogen molecule .

MOLECULE

55

Summary of BONDING

* Directional bonding – Strength of bond is not equal in all directions

* Nondirectional bonding – Strength of bond is equal in all directions

Type Bond energy Melting point Hardness Conductivity Comments

Ionic

bonding

Large

(150-370kcal/mol)

Very high Hard and

brittle

Poor

-required

moving ion

Nondirectional

(ceramic)

Covalent

bonding

Variable

(75-300 kcal/mol)

Large -Diamond

Small – Bismuth

Variable

Highest –

diamond

(>3550)

Mercury (-39)

Very hard

(diamond)

Poor Directional

(Semiconductors,

ceramic, polymer

chains)

Metallic

bonding

Variable

(25-200 kcal/mol)

Large- Tungsten

Small- Mercury

Low to high Soft to hard Excellent Nondirectional

(metal)

Secondary

bonding

Smallest Low to

moderate

Fairly soft Poor Directional

inter-chain

(polymer)

inter-molecular

Ceramics

(Ionic & covalent bonding):

Metals

(Metallic bonding):

Polymers

(Covalent & Secondary):

Large bond energylarge Tm

large E

small a

Variable bond energymoderate Tm

moderate E

moderate a

Directional PropertiesSecondary bonding dominates

small Tm

small E

large a

SUMMARY : PRIMARY BONDING

Exercise : Final Exam [March 2002]

1a] Briefly describe differences between metallic bond and covalent bond.

Support your answer with an example and simple sketch.

(7 Marks)

Answer : 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)

Type Bond energy Melting point Hardness Conductivity

Metallic

bonding

[e.g. :

Zinc]

Variable

(25-200 kcal/mol)

Large- Tungsten

Small- Mercury

Low to high Soft to hard Excellent

Type Bond energy Melting point Hardness Conductivity

Covalent

bonding

[e.g. :

Hydrogen]

Variable

(75-300 kcal/mol)

Large -Diamond

Small – Bismuth

Variable

Highest –

diamond

(>3550)

Mercury (-39)

Very hard

(diamond)

Poor

1.3 CRYSTAL STRUCTURE

1.3 CRYSTAL STRUCTURE

Crystal structure

Crystalline Material

Single Crystal polycrystal

Noncrsytallinematerial

(Amorphous)

* comprised of many single

crystal or grain

• atoms pack in periodic, 3D arrays

• typical of:

Crystalline materials...

-metals

-many ceramics

-some polymers

• atoms have no periodic packing

• occurs for:

Noncrystalline materials...

-complex structures

-rapid cooling

crystalline SiO2

noncrystalline SiO2

"Amorphous" = Noncrystalline

•No recognizable long-

range order

•Completely ordered

•In segments

•Entire solid is made up

of atoms in an orderly

array

Amorphous

Polycrystalline

Crystal

•Atoms are disordered

•No lattice

•All atoms arranged on

a common lattice

•Different lattice

orientation for each

grain

Structure of SOLID

• Some engineering applications require single crystals:

--turbine blades

The single crystal turbine blades

are able to operate at a higher

working temperature than

crystalline turbine blade and thus

are able to increase the thermal

efficiency of the gas turbine cycle.

• Most engineering materials are polycrystals.

grain

1a] With the aid of sketches, explain the following terms :

i. Crystalline materials

ii. Amorphous materials

iii. Single crystalline

iv. Polycrystalline

[8 marks]

QUESTION : FINAL EXAM [OCT 2012]

ANSWER : FINAL EXAM [OCT 2012]

Answer Mark[s]

Crystalline materials Atoms, molecules or ions are packed in a regularly ordered @ repeating pattern 1

Draw [1]

Amorphous materials Atoms, molecules or ions are packed in a irregularly ordered @ unrepeating pattern, 1

Draw [1]

Single crystalline Crystalline materials which is composed by one unit crystal or grain extends throughout its entire without interruption 1

Draw [1]

Polycrystalline Crystalline materials that are composed of more than one crystal or grain [ consist of grain boundary] 1

Draw [1]

crystalline SiO2

noncrystalline SiO2

68

Lattice (lines network in 3D) + Motif (atoms are arranged in a repeated pattern)

= CRYSTAL STRUCTURE

Most metals exhibit a crystal structure which show a unique arrangement of atoms

in a crystal.

A lattice and motif help to illustrate the crystal structure.

CRYSTAL STRUCTURE

lattice motif crystal structure

=+

Lattice - The three

dimensional array

formed by the unit cells

of a crystal is called

lattice.

Unit Cell - When a solid

has a crystalline

structure, the atoms are

arranged in repeating

structures called unit

cells. The unit cell is the

smallest unit

that demonstrate the full

symmetry of a crystal.

A crystal is a three-

dimensional repeating

array.

+

=

70

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.

71

BRAVAIS LATTICE - describe the geometric arrangement of the lattice points and

the translational symmetry of the crystal.

CRYSTAL SYSTEM AND CRYSTALLOGRAPHY

cubic, hexagonal,

tetragonal,

rhombodhedral,

orthorhombic, monoclinic,

triclinic.

•7 crystal systems :

•By adding additional

lattice point to 7 basic

crystal systems –

form 14 Bravais

lattice.

Crystal Structure of Metals

• Simple Cubic (SC) - Manganese

• Body-centered cubic (BCC) - alpha iron, chromium, molybdenum, tantalum, tungsten, and vanadium.

• Face-centered cubic (FCC) - gamma iron, aluminum, copper, nickel, lead, silver, gold and platinum.

Common crystal structures for metals:

FCCSC BCC

73

SIMPLE CUBIC (SC)• The atoms lie on a grid: layers of rows and

columns.

• Sit at the corners of stacked cubic

No. of atom at corner

= 8 x 1/8 = 1 atom

Total No. of atom in

one unit cell

= 1 atom

Example : Manganese

Body-centered Cubic Crystal Structure

The body-centered cubic (bcc) crystal structure:

(a) hard-ball model; (b) unit cell; and (c) single crystal with many unit cells

75

BODY CENTERED CUBIC STRUCTURE (BCC)

• Cubic unit cell with 8 atoms located at the corner & single atom at cube

center

Example : Chromium, Tungsten,Molybdenum,Tantalum, Vanadium

No. of atom at corner = 8 x 1/8 = 1 atom

No. of atom at center = 1 atom

Total No. of atom in one unit cell = 2 atoms

Face-centered Cubic Crystal Structure

The face-centered cubic (fcc) crystal structure:

(a) hard-ball model; (b) unit cell; and (c) single crystal with many unit cells

77

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

1.4 EFFICIENCY OF ATOMIC PACKING,DENSITY COMPUTATION

AND MILLER INDEX

79

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 :

80

EXAMPLE

• APF for a simple cubic structure = 0.52

Calculate the APF for Simple Cubic (SC)?

81

EXERCISE

a) BCC b) FCC

Calculate the APF for BCC and FCC ?

82

aR

• APF for a body-centered cubic structure = 0.68

Unit cell contains: 1 + 8 x 1/8 = 2 atoms/unit cell

ATOMIC PACKING FACTOR: BCC

‘a = 4R/3

83

Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell

a

• APF for a face-centered cubic structure = 0.74

ATOMIC PACKING FACTOR: FCC

‘a = 2R2

84

a (lattice constant) and

R (atom radius)

Atoms/unit

cell

Packing

Density

(APF)

Examples

Simple

cubica = 2R

1 52% CsCl

BCCa = 4R/√3

2 68% Many metals:

α-Fe, Cr, Mo, W

FCCa = 4R/√2

4 74% Many metals : Ag,

Au, Cu, Pt

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

85

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]

86

1a] Give the definition of a unit cell. Briefly describe lattice constant in the unit cell.

[ 4 marks]

ANSWER : FINAL EXAM [Oct 2010]

Answer Mark [s]

The unit cell represent a repeating unit of atom position.

It is a small building block or a structure that can describe the crystal structure.

Lattice constants or lattice parameters are the magnitudes and directions of three lattice vectors such as a, b and c.

Angle = α, β, γ

1

1

1

1

Unit cell - a tiny box that

describe the crystal structure.

87

1b] Give the definition of APF for a unit cell and calculate the APF for FCC.

[ 4 marks]

ANSWER : FINAL EXAM [Oct 2010]

Answer Mark [s]

APF can be defined as the volume of atoms in a selected unit cell with respect to the volume of the unit cell

Or;

An efficiency of an atomic arrangement in a unit cell

APF for FCC = 0.74

1

½

2½

APF = no. of atom, n x volume of atoms in the unit cell, (Vs)

volume of the unit cell, (Vc)

88

DENSITY COMPUTATIONS• A knowledge of the crystal structure of a metallic

solid permits computation of its density through the relationship :

Where

ρ = n A

Vc NA

n = number of atoms associated with each unit cell

A = atomic weight

Vc = volume of the unit cell

NA = Avogadro’s number (6.023 x 1023 atoms/mol)

89

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 109 nm in a meter and 100 cm in a meter, there must be 107 nm in a cm.

109 x 1m = 107 nm/cm

1 m 100 cm

We can therefore convert the volume of the unit cell to cm3 as follows.

4.376 x 10-2 nm3 x [1 cm ]3 = 4.376 x 10-23 cm 3

107 nm

The mass of a nickel atom can be calculated from the atomic weight of this metal and Avogadro’s

number.

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/cm3

1 mol 6.023 x 1023 atoms 4.376 x 10-23 cm 3/unit cell

90

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

Density of Copper = 8.89 g/cm3

• crystal structure = FCC: 4 atoms/unit cell

• atomic weight = 63.55 g/mol (1 amu = 1 g/mol)

• atomic radius R = 0.128 nm (1 nm = 10 cm)

91

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]Solution :

QUESTION : TEST 1 [August 2012]

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]Solution :

ρ = n A

Vc NA

ANSWER : TEST 1 [August 2012]

a = 4R/√2

R = 0.139 nm @ 0.139 x 10-7cm @ 1.39 x 10-8cm

ρ = 21.345 g/cm3

93

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

94

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 0 0 0 0 0 0

2 1 0 0 1 0 0

3 1 1 0 1 1 0

4 0 1 0 0 1 0

5 1/2 1/2 1/2 1/2 1/2 1/2

6 0 0 1 0 0 1

7 1 0 1 1 0 1

8 1 1 1 1 1 1

9 0 1 1 0 1 1

95

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.

96

Axis X Y Z

Head (H) 1 1 0

Tail (T) 0 0 0

Projection (H-T) 1 1 0

Enclosed [1 1 0]

EXAMPLE : CRYSTAL DIRECTION INDICES

0,0,0

1,1,01

1

Axis X Y Z

Head (H) 1 0 0

Tail (T) 0 1 0

Projection (H-T) 1 1 0

Enclosed [1 1 0]

1,0,0

0,1,0

97

Axis X Y Z

Head (H) 1 0 1/2

Tail (T) 0 0 0

Projection (H-T) 1 0 1/2

Reduction (x2) 2__________0 _____1_

Enclosed [ 2 0 1 ]

EXERCISE : CRYSTAL DIRECTION INDICES

0½

1

1

01

1

1

1Axis X Y Z

Head (H) 1 0 0

Tail (T) 0 1 0

Projection (H-T) 1 -1 0

Reduction (if necessary) 1__________-1_ _____0_

Enclosed [ 1 1 0 ]

98

Axis X Y Z

Head (H) ½ 1 0

Tail (T) 0 0 ¾

Projection (H-T) ½ 1 -3/4

Reduction (x4) 2_________4_________-3_

Enclosed [2 4 3 ]

EXERCISE : CRYSTAL DIRECTION INDICES

0

½

0

Axis X Y Z

Head (H) ½ 0 0

Tail (T) 0 ½ 1

Projection (H-T) ½ - ½ -1

Reduction (x2) 1_________-1 _ - 2

Enclosed [ 1 1 2 ]

¾

½

½

½

Determine the direction indices of the cubic

direction between the position coordinates

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

Axis x y z

Head 1/4 1/2 1/2

Tail 3/4 0 1/4

Projection(Head – Tail)

-2/4 1/2 1/4

Reduction(x4)

-2 2 1

Enclosed [ 2 2 1 ]

Draw the following Miller Indices

direction.

a) [ 1 0 0 ]

b) [ 1 1 1 ]

c) [ 1 1 0 ]

d) [ 1 1 0 ]

ANSWER : Draw the following Miller Indices

direction.

a) [ 1 0 0 ]

b) [ 1 1 1 ]

c) [ 1 1 0 ]

d) [ 1 1 0 ]

e) [ 1 1 2 ]

½

½

a) b)

c) d) e)

102

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 PLANEHow to determine crystal plane indices?

Figure : Planes with different Miller

indices in cubic crystals

Axis X Y Z

Interceptions

Reciprocals

Reduction (if necessary)

Enclosed (h k l )

+x

+y

+z

_

z

_

y

_

x

(1 , 0 , 0)

(0 , 1 , 0)

(0 , 0 , 1)

_

(0 , 0 , 1)

_

(0 , 1 , 0)

_

(1 , 0 , 0)

104

Axis X Y Z

Interceptions ½ 1 ½

Reciprocals 2 1 2

Reduction 2 1 2

Enclosed ( 2 1 2 )

EXERCISE. : CRYSTAL PLANE INDICES

Axis X Y Z

Interceptions 1 1 1

Reciprocals 1 1 1

Reduction 1 1 1

Enclosed ( 1 1 1 )

Not intercept at y axis, find new axis

Axis X Y Z

Interceptions 1 1 ∞

Reciprocals 1 1 0

Reduction 1 1 0

Enclosed ( 1 1 0 )

Axis X Y Z

Interceptions 1 ∞ 1/2

Reciprocals 1 0 2

Reduction 1 0 2

Enclosed ( 1 0 2 )

EXERCISE. : CRYSTAL PLANE INDICESParallel with z axis,

give ∞

Not intercept at x axis, find new axis

Parallel with y axis, give ∞

z = 5/6 – 1/3

0

Axis X Y Z

Interceptions ∞ -1 1/2

Reciprocals 0 -1 2

Reduction 0 -1 2

Enclosed ( 0 1 2 )

Axis X Y Z

Interceptions ∞ 1 ∞

Reciprocals 0 1 0

Reduction 0 1 0

Enclosed ( 0 1 0 )

EXERCISE. : CRYSTAL PLANE INDICES

½

Plane pass through origin,

find new axis

Parallel with x axis, give ∞

Determine the Miller Indices plane for the

following figure below?

Draw the following Miller Indices

plane.

a) ( 1 0 0 )

b) ( 0 0 1 )

c) ( 1 0 1 )

d) ( 1 1 0 )

ANSWER : Draw the following Miller Indices plane.

a) ( 1 0 0 )

b) ( 0 0 1 )

c) ( 1 0 1 )

d) ( 1 1 0 )

110

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.

111

1.5 RELATIONSHIP BETWEEN ATOMIC STRUCTURE, CRYSTAL

STRUCTURES AND PROPERTIES OF MATERIALS

112

PHYSICAL PROPERTIES OF METALS

•Solid at room temperature (mercury is an exception)

•Opaque

•Conducts heat and electricity

•Reflects light when polished

•Expands when heated, contracts when cooled

•It usually has a crystalline structure

Physical properties are the characteristic responses of materials to

forms of energy such as heat, light, electricity and magnetism.

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

Mechanical Properties

Terminology for Mechanical Properties

The Tensile Test: Stress-Strain Diagram

Properties Obtained from a Tensile Test

Hardness of Materials

114

MECHANICAL PROPERTIES OF METALSMechanical 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 :

115

Tensile Test

specimen

machine

116

Tensile Test

Terminology

Load - The force applied to a material during testing.

Strain gage or Extensometer - A device used for

measuring change in length (strain).

Engineering stress - The applied load, or force,

divided by the original cross-sectional area of the

material.

Engineering strain - The amount that a material

deforms per unit length in a tensile test.

Stress-Strain Diagram

Strain ( ) (DL/Lo)

41

2

3

5

Elastic

Region

Plastic

Region

Strain

Hardening Fracture

ultimatetensile strength

Elastic region

slope =Young’s (elastic) modulus

yield strength

Plastic region

ultimate tensile strength

strain hardening

fracture

necking

yieldstrength

UTS

y

εEσ

ε

σE

12

y

ε ε

σE

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

σE or

• Strain Hardening

- If the material is loaded again from Point 4, the

curve will follow back to Point 3 with the same

Elastic Modulus (slope).

- The material now has a higher yield strength of

Point 4.

- Raising the yield strength by permanently straining

the material is called Strain Hardening.

Stress-Strain Diagram (cont)

• 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)

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.

123

124

Important Mechanical Propertiesfrom a Tensile Test

• Young's Modulus: This is the slope of the linear portion of the stress-strain curve, it is usually specific to each material; a constant, known value.

• Yield Strength: This is the value of stress at the yield point, calculated by plotting young's modulus at a specified percent of offset (usually offset = 0.2%).

• Ultimate Tensile Strength: This is the highest value of stress on the stress-strain curve.

• Percent Elongation: This is the change in gauge length divided by the original gauge length.

The stress-strain curve for an aluminum alloy.

1260.2

8

0.6

1

Magnesium,

Aluminum

Platinum

Silver, Gold

Tantalum

Zinc, Ti

Steel, Ni

Molybdenum

Graphite

Si crystal

Glass-soda

Concrete

Si nitrideAl oxide

PC

Wood( grain)

AFRE( fibers)*

CFRE*

GFRE*

Glass fibers only

Carbon fibers only

Aramid fibers only

Epoxy only

0.4

0.8

2

4

6

10

20

40

6080

100

200

600800

10001200

400

Tin

Cu alloys

Tungsten

<100>

<111>

Si carbide

Diamond

PTFE

HDPE

LDPE

PP

Polyester

PSPET

CFRE( fibers)*

GFRE( fibers)*

GFRE(|| fibers)*

AFRE(|| fibers)*

CFRE(|| fibers)*

Metals

Alloys

Graphite

Ceramics

Semicond

PolymersComposites

/fibers

E(GPa)

109 Pa Composite data based on

reinforced epoxy with 60 vol%

of aligned carbon (CFRE),

aramid (AFRE), or glass (GFRE)

fibers.

Young’s Moduli: Comparison

127

T

E

N

S

I

L

E

P

R

O

P

E

R

T

I

E

S

128

Room T valuesa = annealed

hr = hot rolled

ag = aged

cd = cold drawn

cw = cold worked

qt = quenched & tempered

Yield Strength: Comparison

129

tensile stress,

engineering strain,

y

p = 0.002

Yield Strength, y

130

F

bonds

stretch

return to

initial

1. Initial 2. Small load 3. Unload

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.

131

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.

132

1. Initial 2. Small load 3. Unload

.

F

linear elastic

linear elastic

plastic

Plastic Deformation (Metals)

• Atomic bonds are broken and new

bonds are created.

• Plastic means permanent.

133

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.

134

• After yielding, the stress necessary to continue plastic deformation in metals increases to a maximum point (M) and then decreases to the eventual fracture point (F).• All deformation up to the maximum stress is uniform throughout the tensile sample. • However, at max stress, a small constriction or neck begins to form.• Subsequent deformation will be confined to this neck area.• Fracture strength corresponds to the stress at fracture.

Region between M and F:• Metals: occurs when noticeable necking starts.• Ceramics: occurs when crack propagation starts.• Polymers: occurs when polymer backbones are aligned and about to break.

Tensile Strength, TS

135

In an undeformedthermoplastic polymer tensile sample, (a) the polymer chains are

randomly oriented. (b) When a stress is

applied, a neck develops as chains become aligned locally. The neck continues to grow until the chains in the entire gage length have aligned.

(c) The strength of the polymer is increased

136

Room T values

Based on data in Table B4, Callister 6e.

a = annealedhr = hot rolledag = agedcd = cold drawncw = cold workedqt = quenched & temperedAFRE, GFRE, & CFRE =aramid, glass, & carbonfiber-reinforced epoxycomposites, with 60 vol%fibers.

Tensile Strength: Comparison

137

• Tensile stress, : • Shear stress, t:

Ft

Aooriginal area

before loading

Stress has units: N/m2 or lb/in2

Engineering Stress

138

VMSE

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

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

139

• Another ductility measure: 100% xA

AAAR

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.

100% xl

llEL

o

of

Ductility, %ELDuctility is a measure of the plastic deformation that has been sustained at fracture:

A material that suffers very little plastic deformation is brittle.

140

Toughness

Lower toughness: ceramics

Higher toughness: metals

Toughness is

the ability to

absorb

energy up to

fracture (energy

per unit volume of

material).

A “tough”

material has

strength and

ductility.

Approximated

by the area

under the

stress-strain

curve.

• Energy to break a unit volume of material

• Approximate by the area under the stress-strain

curve.

21

smaller toughness- unreinforced polymers

Engineering tensile strain,

Engineering

tensile

stress,

smaller toughness (ceramics)

larger toughness (metals, PMCs)

Toughness

142

Linear Elastic Properties

Modulus of Elasticity, E:

(Young's modulus)

• Hooke's Law: = E

• Poisson's ratio:metals: n ~ 0.33

ceramics: n ~0.25

polymers: n ~0.40

Units:

E: [GPa] or [psi]

n: dimensionless

n x/y

143

Engineering Strain

Strain is dimensionless.

144

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

True Stress and True Strain

True stress The load divided by the actual cross-sectional

area of the specimen at that load.

True strain The strain calculated using actual and not

original dimensions, given by εt ln(l/l0).

•The relation between the true stress-true strain diagram and engineering stress-engineering strain diagram. •The curves are identical to the yield point.

(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

The stress-strain behavior of brittle materials compared with that of more ductile materials

147

--brittle response (aligned chain, cross linked & networked case)--plastic response (semi-crystalline case)

Stress-Strain Behavior: Elastomers

3 different responses:

A – brittle failure

B – plastic failure

C - highly elastic (elastomer)

148

Stress-Strain Results for Steel

Sample

149

Metals can fail by brittle or ductile fracture.

FRACTURE MECHANISM OF METALS

Ductile fracture is better than brittle fracture because :

Ductile fracture occurs over a period of time, where as brittle fracture is fast

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

Figure : Stress strain curve for brittle and ductile material

Ductile Vs Brittle Fracture

(c)2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

• Localized deformation of a ductile material during a tensile test produces a

necked region.

• The image shows necked region in a fractured sample

Ductile Fracture

152

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]

153

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]

ANSWER : FINAL EXAM [April 2011]

Ductility is an ability of a material to have large plastic deformation before fracture or area under plastic deformation in stress strain diagram

1

Elastic deformation

The deformation is non permanent or reversible.In terms of atomic level, the bonding between atoms are stretched and it will return back to its original shape after force is released.

1

1½

F

bonds

stretch

return to

initial

1. Initial 2. Small load 3. Unload

Plastic deformation

The deformation is permanent.In terms of atomic level, the bonding between atoms will break and the atom bonded with new atom. As a result, permanent deformation will occur.

1

1½

1. Initial 2. Small load 3. Unload

Ductile fracture Brittle fracture

•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 between

ductile fracture & brittle fracture?

Hardness of Materials

Hardness test - Measures the resistance of a material to

penetration by a sharp object.

Macrohardness - Overall bulk hardness of materials

measured using loads >2 N.

Microhardness Hardness of materials typically measured

using loads less than 2 N using such test as Knoop

(HK).

Nano-hardness - Hardness of materials measured at 1–

10 nm length scale using extremely small (~100 µN)

forces.

157

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).

158

• Resistance to permanently indenting the surface.• Large hardness means:

--resistance to plastic deformation or cracking incompression.

--better wear properties.

Hardness

159

Hardness Testers

Hardness Testers

Indentation Geometry for Brinnel Testing

Figure Indentation geometry in

Brinell hardness testing: (a)

annealed metal; (b) work-

hardened metal; (c) deformation

of mild steel under a spherical

indenter. Note that the depth of

the permanently deformed zone

is about one order of magnitude

larger that the depth of

indentation. For a hardness test

to be valid, this zone should be

developed fully in the material.

Hardness Scale

Conversions

Figure Chart for converting

various hardness scales. Note

the limited range of most scales.

Because of the many factors

involved, these conversions are

approximate.

164

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

165

Correlation between Hardness and Tensile Strength

• Both hardness and tensile

strength are indicators of

a metal’s resistance to

plastic deformation.

• For cast iron, steel and

brass, the two are roughly

proportional.

• Tensile strength (psi) =

500*BHR

166

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]

167

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]

ANSWER: FINAL EXAM [Oct 2012]

Rockwell

Indenter type : Diamond [cone] [120° angle and 0.2mm tip radius] or steel sphere

½

½

Brinell

Indenter type : 10mm sphere of steel / tungsten carbide

½

½

Vickers

Indenter type : Diagonal pyramid diamond

½

½

Knoop

Indenter type : Elongated pyramid diamond

½

½

168

• Stress and strain: These are size-independentmeasures of load and displacement, respectively.

• Elastic behavior: This reversible behavior oftenshows a linear relation between stress and strain.To minimize deformation, select a material with alarge elastic modulus (E or G).

• Plastic behavior: This permanent deformationbehavior occurs when the tensile (or compressive)uniaxial stress reaches y.

• Toughness: The energy needed to break a unitvolume of material.

• Ductility: The plastic strain at failure.

Summary