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Transcript of MEC281 Chapter 1.pdf
![Page 1: MEC281 Chapter 1.pdf](https://reader036.fdocuments.net/reader036/viewer/2022082211/55cf8f13550346703b98b494/html5/thumbnails/1.jpg)
CHAPTER 1 : STRUCTURE
(10 hours)
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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 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
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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 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.
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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
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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?
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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 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.
+
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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
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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.
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.
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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
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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.
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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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PERIODIC TABLE
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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
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•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
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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
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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
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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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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
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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).
3rd shell
4th 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 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
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• 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
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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 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.
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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-
3rd shell
(energy
level)
Magnesium (12 electrons)
e-
e-
Answer : 1s2 2s2 2p6 3s2
e-
Answer : ls2 2s1
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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
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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 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
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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
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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
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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
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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]
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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
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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 .
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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
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• Example: NaCl
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• 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
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• 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
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Exercise : Final Exam [April 2008]
1c] With the aid of sketches, describe how Sodium and Chlorine atoms are joined.
[3 marks]
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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]
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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 : MgF₂, BeF₂, MgBr₂, CaCl₂ or CaI₂Group 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 (H₂, Cl₂, F₂, 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 -
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
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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 conduct electricity?
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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).
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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 (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
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• 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
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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 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
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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
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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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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
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1.3 CRYSTAL STRUCTURE
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1.3 CRYSTAL STRUCTURE
Crystal structure
Crystalline Material
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 SiO2
noncrystalline SiO2
"Amorphous" = Noncrystalline
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•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
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• 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.
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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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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
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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 atoms
in a crystal.
A lattice and motif help to illustrate the crystal structure.
CRYSTAL STRUCTURE
lattice motif crystal structure
=+
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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.
+
=
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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.
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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.
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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
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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
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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
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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
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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
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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
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1.4 EFFICIENCY OF ATOMIC PACKING,DENSITY COMPUTATION
AND 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 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 :
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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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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
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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
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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
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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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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.
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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)
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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)
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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 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
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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)
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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]
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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
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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-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
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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
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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 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.
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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
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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 ]
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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
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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 ]
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Draw the following Miller Indices
direction.
a) [ 1 0 0 ]
b) [ 1 1 1 ]
c) [ 1 1 0 ]
d) [ 1 1 0 ]
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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)
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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 )
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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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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
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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
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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 ∞
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Determine the Miller Indices plane for the
following figure below?
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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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ANSWER : 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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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.
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111
1.5 RELATIONSHIP BETWEEN ATOMIC STRUCTURE, CRYSTAL
STRUCTURES AND PROPERTIES OF MATERIALS
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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 :
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Mechanical Properties
Terminology for Mechanical Properties
The Tensile Test: Stress-Strain Diagram
Properties Obtained from a Tensile Test
Hardness of Materials
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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 :
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115
Tensile Test
specimen
machine
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116
Tensile Test
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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.
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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
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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
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• 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)
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• 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)
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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.
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123
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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.
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The stress-strain curve for an aluminum alloy.
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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
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127
T
E
N
S
I
L
E
P
R
O
P
E
R
T
I
E
S
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128
Room T valuesa = annealed
hr = hot rolled
ag = aged
cd = cold drawn
cw = cold worked
qt = quenched & tempered
Yield Strength: Comparison
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129
tensile stress,
engineering strain,
y
p = 0.002
Yield Strength, y
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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.
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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.
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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.
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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.
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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
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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
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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
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137
• Tensile stress, : • Shear stress, t:
Ft
Aooriginal area
before loading
Stress has units: N/m2 or lb/in2
Engineering Stress
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VMSE
http://www.wiley.com/college/callister/0470125373/vmse/strstr.htm
http://www.wiley.com/college/callister/0470125373/vmse/index.htm
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• 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.
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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.
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• 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
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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
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Engineering Strain
Strain is dimensionless.
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144
Axial (z) elongation (positive strain) and lateral (x and y) contractions (negative strains) in response to an imposed tensile stress.
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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.
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(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
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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)
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Stress-Strain Results for Steel
Sample
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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
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Ductile Vs Brittle Fracture
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(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
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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]
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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
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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
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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?
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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.
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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).
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• Resistance to permanently indenting the surface.• Large hardness means:
--resistance to plastic deformation or cracking incompression.
--better wear properties.
Hardness
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Hardness Testers
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Hardness Testers
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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.
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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.
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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
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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
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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]
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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
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• 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