Instructor: Dr. Emmanuel Kwesi Arthur Dept. of Materials … · 2019. 2. 5. · 4 Scope: Provide an...

MSE 282: MECHANICAL ENIGINEERING MATERIALS II

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Instructor:

Dr. Emmanuel Kwesi Arthur

Dept. of Materials Engineering

Office: 325 Petroleum Building

E-mail: [email protected]

Office Hours: Monday, 1-3 pm.

Meeting times: 10:00-10:50 a.m.

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“WITHOUT MATERIALS, THERE IS NO ENGINEERING.”

Why is glass brittle, while copper is ductile? What is meant by a ductile material?

If we take two rods, one of Al and one of steel, why is it easier to bend the Al rod

as compared to the steel rod?

How can I change properties like hardness, without changing the composition (say

of 0.8% C steel)?

Why is wire of copper conducting, while piece of brick or wood non-conducting?

Why is glass transparent, while any typical metal is opaque?

Why does Iron corrode easily, while Aluminium does not (or does not seem to?!)?

Usually, good thermal conductors are also good electrical conductors. Why is this

so?

Why is diamond a good thermal conductor, but not a good electrical conductor?

If I pull a spring and then release the load, it ‘comes back’ to its original shape.

However, a if I bend an aluminium rod, does not come back to its original shape.

How can one understand these observations?

What are the kind of questions that an engineering students would like to ask on materials?

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Scope: Provide an introduction to concepts in Materials Science

and Engineering (e.g., metals, ceramics, polymers, and

composite).

Objective: Develop an awareness of materials and their

properties that, as an engineer, you must rely in the future.

• To introduce basic concepts, nomenclature, and testing of materials.

• To reveal the relationships between

Processing - Structure - Properties - Performance

• To develop ideas behind phase diagram and heat treatment

The full implication of the aspects presented in this introductory chapter will only become clear after the student has

covered major portions of this course. Hence, students are encouraged to return to this chapter many times during

his/her progress through the course.

Course OutlineFundamentals

Introduction

Classification of materials

Metals

Ceramics

Polymers

Atomic Structure/Bonding

Atomic/Electronic structure

Ionic Bonding

Covalent Bond

Metallic Bond

Crystal Structures

Crystalline and non-crystalline

Simple cubic

Body centered cubic

Face centered cubic

Control and Processing

Phase Diagrams

Phase Transformations

Thermal processing

Mechanical Behavior

Elastic deformation

plastic deformation

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Composite

Electronic

materials

TEXT

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Materials Science and Engineering: An Introduction, William D.

Callister, John Wiley & Sons, 2010.

Fundamentals of Materials Science & Engineering: an integrated

approach, Callister and Rethwisch (Special 3rd edition for

MSE281)

Materials Science and Engineering, V. Raghavan,

Fifth Edition, Prentice Hall of India Pvt. Ltd., New Delhi, 2004.

Syllabus:

• Attendance is your job – come to class!

• Or our regularly scheduled time (Thursday, 2:00 – 4:00 pm)

• Assignments• Don’t copy from others; don’t plagiarize – its just the right thing to do!!

• Tutorials – by Adwoa Owusu Bia (TA)

• Grading

• Class Attendance, Pop Quizzes and Assignments – (10% of your grade!)

• Mid Semester Exams – (20%)

• End of Semester Exams (70%)

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MSE 280: INTRODUCTION TO ENGINEERING MATERIALS“BECAUSE WITHOUT MATERIALS, THERE IS NO ENGINEERING.”

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• Engineering Requires Consideration of Materials

The right materials for the job - sometimes need a new one.

• We will learn about the fundamentals of

Processing Structure Properties Performance

• We will learn that sometime only simple considerations of

property requirements chooses materials.

Consider in your engineering discipline what materials that

are used and why. Could they be better?

WHAT ARE MATERIALS? Materials may be defined as substance of which something is composed or made.

We obtain materials from earth crust and atmosphere.

Examples :-

Silicon and Iron constitute 27.72 and 5.00 percentage of weight of earths crust respectively.

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WHY THE STUDY OF MATERIALS IS IMPORTANT?

Production and processing of materials constitute a large part of developed country’s economy.

Engineers choose materials to suite design.

New materials might be needed for some new applications. Example :- High temperature resistant materials.

Space station and Mars Rovers should sustain conditions in space.

* High speed, low temperature, strong but light.

Modification of properties might be needed for some applications.

Example :- Heat treatment to modify properties.

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MATERIALS SCIENCE AND ENGINEERINGMaterials science deals with basic knowledge about the internal structure, properties and processing of materials.

Materials engineering deals with the application of knowledgegained by materials science to convert materials to products.

Resultant

Knowledge

of Structure and

Properties

Applied

Knowledge

of Materials

Materials ScienceMaterials Science and

Engineering Materials Engineering

Basic

Knowledge

of

Materials

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UNIVERSE

PARTICLES

ENERGYSPACE

FIELDS

STRONG

WEAK

ELECTROMAGNETIC

GRAVITY

METAL

SEMI-METALSEMI-CONDUCTOR

INSULATOR

nD + t

HYPERBOLIC

EUCLIDEAN

SPHERICAL

GAS

BAND STRUCTURE

AMORPHOUS

ATOMICNON-ATOMIC

STATE / VISCOSITY

SOLID LIQUIDLIQUID CRYSTALS

QUASICRYSTALS CRYSTALSRATIONAL

APPROXIMANTS

STRUCTURE

NANO-QUASICRYSTALS NANOCRYSTALS

SIZE

Materials Science

Entropic force

Note: Some fields are hyperlinks

Solid Electrolytes

12

Based on state (phase) a given material can be Gas, Liquid or Solid (based on the thermodynamic variables: P, T,…).

Intermediate states are also possible.

Based on structure (arrangement of atoms/molecules/ions) materials can be

Crystalline, Quasicrystalline or Amorphous.

Intermediate states (say between crystalline and amorphous; i.e. partly

crystalline) are also possible. Polymers are often only partly crystalline.

Liquid Crystals (‘in some sense’) are between Liquids and Crystals.

Similarly Solid Electrolytes (also known as* fast ion conductors and superionic conductors) are also

between crystals and liquids. These materials have a sublattice which is ‘molten’

and the ions in this sublattice are highly mobile (these materials are similar to liquid electrolytes in this sense).

Based on Band Structure we can classify materials into Metals, Semi-metals,

Semiconductors and Insulators.

Based on the size of the entity in question we can have Nanocrystals,

Nanoquasicrystals etc. There are other classifications we will encounter during the course.

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Physical Classification of Materials by State

15

Physical Classification of Materials by

Morphological Structure

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Physical Classification of

Materials by Atomic Structure

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Interrelationships Between Classes of Materials

TYPES OF MATERIALS

Metallic Materials

Composed of one or more metallic elements.

Example:- Iron, Copper, Aluminum.

Metallic element may combine with nonmetallic elements.

Example:- Silicon Carbide, Iron Oxide.

Inorganic and have crystalline structure.

Good thermal and electric conductors.

Metals and Alloys

Ferrous

Eg: Steel,

Cast Iron

Nonferrous

Eg:Copper

Aluminum

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Metals

Properties

• good conductors of

electricity and heat

• lustrous appearance

• susceptible to

corrosion

• strong, but

deformable

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Atoms arranged in a regular repeating

structure

Periodic Table of Elements

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From http://64.224.111.143/handbook/periodic/

Metal Applications

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Applications

• Electrical wiring

• Structures: buildings, bridges, etc.

• Automobiles: body, chassis, springs, engine block, etc.

• Airplanes: engine components, fuselage, landing gear assembly,

etc.

• Trains: rails, engine components, body, wheels

• Machine tools: drill bits, hammers, screwdrivers, saw blades, etc.

• Shape memory materials: eye glasses

• Magnets

• Catalysts

Examples

• Pure metal elements (Cu, Fe, Zn, Ag, etc.)

• Alloys (Cu-Sn=bronze, Cu-Zn=brass, Fe-C=steel, Pb-Sn=solder, NiTinol)

• Intermetallic compounds (e.g. Ni3Al)

TYPES OF MATERIALSPolymeric (Plastic) Materials

Organic giant molecules and mostly noncrystalline.

Some are mixtures of crystalline and noncrystalline regions.

Poor conductors of electricity and hence used as insulators.

Strength and ductility vary greatly.

Low densities and decomposition temperatures.

Examples :- Poly vinyl Chloride (PVC), Polyester.

Applications :- Appliances, DVDs, Fabrics etc.

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Polymers

Covalent Bond

electrons are shared between adjacent

atoms, each contributing at least one

electron

shared electrons belong to both atoms

directional bond

shared

electron

from carbon

shared

electron

from

hydrogenCH

H

H

H

methane (CH4)

Polymers Properties

• very large molecules

• low density, light weight materials

• maybe extremely flexible

• low melting point

• chemical stability: inert to corrosive

environments 23

Composed primarily of C and H (hydrocarbons)

with select additives to elucidate specific properties

Polymers are typically disordered (amorphous) strands

of hydrocarbon molecules.

Some are crystals, many are not.

CLASSES AND PROPERTIES: POLYMERS

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Two main types of polymers are thermosets and thermoplastics.

• Thermosets are cross-linked polymers that form 3-D networks, hence are strong and

rigid.

• Thermoplastics are long-chain polymers that slide easily past one another when heated,

hence, they tend to be easy to form, bend, and break.

Polymer Applications

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Applications and Examples

• Adhesives and glues

• Containers

• Moldable products (computer casings, telephone handsets, disposable

razors)

• Clothing and upholstery material (vinyls, polyesters, nylon)

• Water-resistant coatings (latex)

• Biodegradable products (corn-starch packing “peanuts”)

• Biomaterials (organic/inorganic intefaces)

• Liquid crystals

• Low-friction materials (teflon)

• Synthetic oils and greases

• Gaskets and O-rings (rubber)

• Soaps and surfactants

TYPES OF MATERIALS

Ceramics and Glasses

Metallic and nonmetallic elements are chemically bondedtogether.

Inorganic but can be either crystalline, noncrystalline or mixture of both.

Except for glasses, atoms are regularly arranged

Typically covalent. In some cases highly direction covalent

bonding. Ionic in case of SiO2 glasses and slags

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Ceramics and Glasses

+ +

+ +

+

++

+

Coulombic bonding forceProperties

• Thermally and electrically

insulating

• Stronger than metals

• Lower density than most metals

• Low resistance to fracture: low

toughness or brittle

• Low ductility or malleability: low

plasticity

• Resistant to high temperatures and

harsh environments

• Wear resistant (hard), but brittle

• Chemical stability: corrosion

resistant

• Good optical properties

• Adhesives

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Ceramic Applications

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Applications

• Electrical insulators

• Abrasives

• Thermal insulation and coatings

• Windows, television screens, optical fibers (glass)

• Corrosion resistant applications

• Electrical devices: capacitors, varistors, transducers, etc.

• Highways and roads (concrete)

• Biocompatible coatings (fusion to bone)

• Self-lubricating bearings

• Magnetic materials (audio/video tapes, hard disks, etc.)

• Optical wave guides

• Night-vision

Examples

• Simple oxides (SiO2, Al2O3, Fe2O3, MgO)

• Mixed-metal oxides (SrTiO3, MgAl2O4, YBa2Cu3O7-x, having vacancy

defects.)

• Nitrides (Si3N4, AlN, GaN, BN, and TiN, which are used for hard coatings.)

TYPES OF MATERIALSComposite Materials

Mixture of two or more materials.

Consists of a filler material and a binding material.

Materials only bond, will not dissolve in each other.

types :

Particulate reinforced – discontinuous type with low aspect ratio

Whisker/rod reinforced - discontinuous type with high aspect ratio

Fiber reinforced - continuous type with high aspect ratio (naturally)

Laminated composites - layered structures (surf boards, skate boards)

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• Properties: Depends on composites

– High melting points with improved high temperature strength: ceramic-

ceramic

– High strength and ductile with improved wear resistance: metal-ceramic

– High strength and ductile: polymer-polymer

Classes and Properties: Composites

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Distinguishing features

• Composed of two or more different materials (e.g., metal/ceramic, polymer/polymer,

etc.)

• Properties depend on amount and distribution of each type of material.

• Collective properties more desirable than possible with any individual material.


• Sports equipment (golf club shafts, tennis rackets, bicycle frames)

• Aerospace materials

• Thermal insulation

• Concrete

• "Smart" materials (sensing and responding)

• Brake materials

Examples

• Fiberglass (glass fibers in a polymer)

• Space shuttle heat shields (interwoven ceramic fibers)

• Paints (ceramic particles in latex)

• Tank armor (ceramic particles in metal)

TYPES OF MATERIALS

Electronic Materials

Not Major by volume but very important.

Silicon is a common electronic material.

Its electrical characteristics are changed by adding impurities.

Examples:- Silicon chips, transistors

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ELECTRONIC MATERIALS

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Classes and Properties: Semiconductors

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• Computer CPUs

• Electrical components (transistors, diodes, etc.)

• Solid-state lasers

• Light-emitting diodes (LEDs)

• Flat panel displays

• Solar cells

• Integrated Circuits

• Satellites

• Examples: Si, Ge, GaAs, and InSb

COMPETITION AMONG MATERIALSMaterials compete with each other to exist in new market

Over a period of time usage of different materials changes depending on cost and performance.

New, cheaper or better materials replace the old materials when there is a breakthrough in technology

Example:-

0

200

400

600

800

1000

1200

1400

1600

lb/C

ar

1985 1992 1997

Model Year

Aluminum

Iron

Plastic

Steel

Predictions and use of

materials in US automobiles.

After J.G. Simon, Adv. Mat. & Proc., 133:63(1988) and new data1-10

FUTURE TRENDS

Metallic Materials

Production follows US economy closely.

Alloys may be improved by better chemistry and process control.

New aerospace alloys being constantly researched.

o Aim: To improve temperature and corrosion resistance.

o Example: Nickel based high temperature super alloys.

New processing techniques are investigated.

o Aim: To improve product life and fatigue properties.

o Example: Isothermal forging, Powder metallurgy.

Metals for biomedical applications

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FUTURE TRENDS

Polymeric (Plastic Materials)

Fastest growing basic material (9% per year).

After 1995 growth rate decreased due to saturation.

Different polymeric materials can be blend together to produce new plastic alloys.

Search for new plastic continues.

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FUTURE TRENDS

Ceramic Materials

New family of engineering ceramics are produced last decade

New materials and applications are constantly found.

Now used in Auto and Biomedical applications.

Processing of ceramics is expensive.

Easily damaged as they are highly brittle.

Better processing techniques and high-impact ceramics are to be found.

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FUTURE TRENDS

Composite Materials

Fiber reinforced plastics are primary products.

On an average 3% annual growth from 1981 to 1987.

Annual growth rate of 5% is predicted for new composites such as Fiberglass-Epoxy and Graphite-Epoxy combinations.

Commercial aircrafts are expected to use more and more composite materials.

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FUTURE TRENDS

Electronic Materials

Use of electronic materials such as silicon increased rapidly from 1970.

Electronic materials are expected to play vital role in “Factories of Future”.

Use of computers and robots will increase resulting in extensive growth in use of electronic materials.

Aluminum for interconnections in integrated circuits might be replaced by copper resulting in better conductivity.

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FUTURE TRENDS

Smart Materials : Change their properties by sensing

external stimulus.

Shape memory alloys: Strained material reverts back to its original shape above a critical temperature.

Used in heart valves and to expand arteries.

Piezoelectric materials: Produce electric field when exposed to force and vice versa.

Used in actuators and vibration reducers.

MEMS AND NANOMATERIALS

MEMS: Microelectromechanical systems.

Miniature devices

Micro-pumps, sensors

Nanomaterials: Characteristic length < 100 nm

Examples: ceramics powder and grain size < 100 nm

Nanomaterials are harder and stronger than bulk materials.

Have biocompatible characteristics ( as in Zirconia)

Transistors and diodes are developed on a nanowire.

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THE WIDE WORLD OF MS&E!

ALL of Engineering is impacted by Materials!

CLASSIFICATION OF MATERIALS

Bar-chart of room temperature density values for metals, ceramics, polymer and composite materials


Bar-chart of room temperature stiffness(elastic modulus) values for metals, ceramics, polymer and composite materials


Bar-chart of room temperature resistance to fracture for metals, ceramics, polymer and composite materials


Bar-chart of room temperature tensile strength for metals, ceramics, polymer and composite materials


Bar-chart of room temperature electrical conductivity ranges for metals, ceramics, polymer and composite materials

Common type of materials

Metals Ceramics Polymers

Hybrids (Composites)

The type of atomic entities (ion, molecule etc.) differ from one class to another,

which in turn gives each class a broad ‘flavour’ of properties.

● Like metals are usually ductile and ceramics are usually hard & brittle

● Polymers have a poor tolerance to heat, while ceramics can withstand high

temperatures

●Metals are opaque (in bulk), while silicate glasses are transparent/translucent

●Metals are usually good conductors of heat and electricity, while ceramics are

poor in this aspect.

● If you heat semi-conductors their electrical conductivity will increase, while

for metals it will decrease

● Ceramics are more resistant to harsh environments as compared to Metals

A Common Description

& Glasses

Diamond is poor electrical

conductor but a good thermal

conductor!! (phonons are

responsible for this)

Bonding and structure are key

factors in determining the

properties of materials

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bonding_and_properties.ppt

What determines the properties of materials?Funda Check

There are microstructure ‘sensitive’ properties (often called structure sensitive properties)

and microstructure insensitive properties (note the word is sensitive and not dependent).

Microstructure ‘sensitive’ properties → Yield stress, hardness, Magnetic coercivity…

Microstructure insensitive properties → Density, Elastic modulus…

Hence, one has to keep in focus:

Atomic structure

Bonding

Microstructure

to understand the properties.

From an alternate perspective:

Electronic interactions are responsible for most the material properties.

From an understanding perspective this can be broken down into Bonding and Structure.

Electronic Interactions

Bonding Structure

In materials

Hydrogen bond

Van der WaalsEtc.

Weak

InteractionsStrong

Interactions

COVALENT

IONIC

METALLIC

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Atomic Structure: Recap

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What do you know about: atoms

the periodic table

Atomic Structure

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NEUTRON –neutral, same

mass as proton (“1”)

PROTON –positive, same

mass as neutron (“1”)

ELECTRON –negative, mass nearly nothing

Goals Describe the atomic structure Describe how atomic or electronic structure affect key

material properties Describe how the atomic structure describes

macroscale properties (mechanical, structural, thermal, electrical, optical, ...)

Atomic Structure and Periodic Table

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The Atom

Particle Mass (g) Relative Mass Relative Charge

Proton 1.6727 x10 -24 1 1

Neutron 1.6750 x10 -24 1 0

Electron 9.110 x 10 -28 0 -1 53

Structure of Atom

• All matter are made up of tiny particles called atoms• The structure of atom given by various scientists is called

their atomic model• Most important models proposed by different scientists

are:(i) Dalton’s atomic model• (ii)Thomson’s atomic model• (iii) Rutherford’s atomic model• (iv) Bohr’s atomic model• (v) Modern atomic model

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Dalton’s Atomic Theory (1808)

All matter is composed of atoms Atoms of a given element are identical in size, mass, and other properties;, but are different from atoms of other elements Atoms can’t be subdivided, created, or destroyed. Envisioned atoms as “tiny, solid balls”

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Thomson’s Atomic Model

In 1910, Sir J. J. Thomson discovered that atom is spherical in shape Every atom has negatively charge particles called electrons Atom has positive charge equal and opposite to that of electrons Electrons being embedded in the positive charge distributed uniformly over the entire body of the atom

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Rutherford’s Atomic Model

Rutherford observed that atom has all the positive charge concentrated in a very small central region called nucleus

The nucleus is surrounded by negatively charged particles calledelectrons revolving in orbits

He proved that the force of attraction between electron andnucleus was counter balanced by the centrifugal force acting on the revolving atom

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Modern Atomic Theory: Models

In 1913, Neils Bohr, a student of Rutherford's, developed a new model of the atom

He proposed that electrons are arranged in concentriccircular orbits around the nucleus

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Electrons occupy only certain orbits around the nucleus. Those orbits are stable and are called "stationary" orbits.

Each orbit has an energy associated with it. The orbit nearest the nucleus has an energy of E1, the next orbit E2, etc.

Energy is absorbed when an electron jumps from a lower orbit to a higher one and energy is emitted when an electron falls from a higher orbit to a lower orbit.

The energy and frequency of light emitted or absorbed can be calculated by using the difference between the two orbital energies.

• The model can be summarized by the following four principles:

Bohr’s Atomic Model (1913)

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The behavior of electrons can be correctly described only by the theory of quantum mechanics

For this class, we will simply borrow some of the major results of the theory so that we can adequately understand the behavior of electrons

Notably, the Bohr model of the atom is incorrect as it does not describe the dual particle (e.g., photo-electric effect) and wave (e.g., electron diffraction) features of electron scattering

Quantum Theory

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Major Results from Quantum Mechanics

Electrons sometimes behave like particles but sometimes they behave like waves They don't exist uniquely at any given location, but are

spread out in space They can interfere with each other, and even form

standing wavesWhen orbiting an atomic nucleus, electrons are best

described by energy waves with a specific amplitude and wavelength, and these specific values may be found by solving Schrodinger's equation

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Modern Theory: Quantum Mechanical Model

The quantum mechanical model is based on quantum theory, which says matter also has properties associated with waves

According to quantum theory, it’s impossible to know the exact position and momentum of an electron at the same time. This is known as the Uncertainty Principle

The quantum mechanical model of the atom uses complex shapes of orbitals (sometimes called electron clouds), volumes of space in which there is likely to be an electron.

So, this model is based on probability rather than certainty.

Description of this model:

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Until 1932, the atom was believed to becomposed of a positively charged nucleussurrounded by negatively charged electrons.

In 1932, James Chadwick bombardedberyllium atoms with alpha particles.

An unknown radiation was produced. Chadwickinterpreted this radiation as being composedof particles with a neutral electrical chargeand the approximate mass of a proton.

This particle became known as the neutron. With the discovery of the neutron, an

adequate model of the atom became availableto chemists.

Chadwick’s Atomic Model

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Electron Quantum Numbers

Principal, n: Allowed values of n are 1, 2, 3…• Roughly associated with electron's energy

– higher values of n mean larger (less favorable) energies.• The value of n also tells how "far" from the nucleus the

electron is.• n is also called the electron shell (n=1 is the first shell, etc.)

Angular momentum, l: Allowed values of l are 0 to (n-1)• related to angular momentum of the electron in its orbit• l is also called the electron orbital within a given shell• Each value of l is assigned a letter of the alphabet: 0=s

(sharp), 1=p (principal), 2=d (diffuse) and 3=f (fundamental)

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Magnetic, m: Allowed values of m are -l to +l (2l + 1 distinct values)

Spin, ms: Allowed values of ms are -1/2 (spin-up) and +1/2 (spin-down)

Pauli Exclusion Principle: No two electrons in the same system can share the same four (n, l, m, ms) quantum numbers!

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Quantum Numbers

Number and Energy of Available Electron Sites in Shells & Sub-Shells

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Number of electrons per subshell = 2(2l+1)

Number of electrons per shell = 2n2

Number of states (orbitals) = 2l+1

l = 0 = sl = 1 = pl = 2 = dl = 3 = f

Number and Energy of Available Electron Sites in Shells & Sub-Shells

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1s2s2p3s3p4s3d4p5s4d5p4f5d5f

HHelium

Lowest Energy levels filled first.

Pauli rule says only two spins per state

Hund’s rule says fill energy equivalent level such that paired electrons are minimized

AUFBAU (Filling)

B

C

Orbital Filling

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1s2s2p3s3p4s3d4p5s4d5p4f5d5f

Electronic Configurations

ex: Fe - atomic # =

69

26

Adapted from Fig. 2.4,

Callister & Rethwisch 8e.

1s

2s2p

K-shell n = 1

L-shell n = 2

3s3p M-shell n = 3

3d

4s

4p4d

Energy

N-shell n = 4

1s2 2s2 2p6 3s2 3p6 3d6 4s2

Electronic Configuration for Some Common Elements

70

Answer

Name Protons (p or +) Neutrons (n) Electrons (e-)

Charge +1 No charge -1

Location in nucleus in nucleus in shells around nucleus

RelativeMass

≈ 1 amu ≈ 1 amu ≈ 0

“Job” Determines identity of element

Supplies proper mass to hold

nucleus together

Determines bonding/ how it

reacts

Number Atomic # Atomic mass –atomic # =

# of neutrons

Same as # of protons

Periodic Table

73

Horizontal rows - elements in increasing atomic number Vertical columns - similar chemical properties Group IA = Alkali metals Group IIA = Alkaline earth metals Group IIIB – IIB = Transition metals (partially filled d

states) Group IIIA, IVA and VA are in between metals & non-

metals LHS = electropositive RHS = electronegative

Periodic Table: Orbital Filling

• 4s is slightly lower in energy and fills first, but is removed first as it is less tightly bound (remove electron in higher quantum levels first).• Electrons in higher states, which resides farther from nuclei, come off first.

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Quiz 2

(1) Calculate the total number of orbitals in a shell with n=6.

(2) Without referring to the periodic table, identify which of the following elements is a metal, alkali metal, alkaline metal, inert gas, halogen or transition metal. Explain the reasons for your choices.

(a) 1s2 2s22p6 3s23p63d3 4s2

(b) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

(c) [Kr] 4d10 5s2 5p5

(d) [Ar] 4s2

(e) [Ar] 4s23d7

(f) 1s22s22p63s23p64s1

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(1) Briefly describe the 4 types of quantum numbers and what they represent.

(2) State the Pauli’s exclusion principle and what it means for the periodic table of elements.

(3) Briefly explain why magnesium is a metal and fluorine is a non-metal.

(4) How is the periodic table arranged?

(5) Work out, and name, all the possible orbitals with principal quantum number n=5. How many

orbitals have n=5?

(6) Name the electron configurations of the Sc and Cr atoms. Which is more stable and why?

Assignment 1

CHEMICAL BONDING

Atoms Molecules Solids

Bonds form so as to produce filled outer shells

Some atoms are a few electrons short

Electronegative: readily pick up a few electrons from other atoms,

become negatively charged

Some atoms have a few electrons too many

Electropositive: readily give up a few electrons to other atoms,

become positively charged

Noble gases: filled outer shell, limited chemistry

BONDING BETWEEN ATOMS

Forces between atoms are like little springs:

Determines macroscopic properties Melting Temperature

Thermal Expansion Coefficient

Elastic (Young’s) Modulus

N.B. These are fundamental properties which are not altered by processing

INTERATOMIC BONDING AND MELTING POINT

Types of bonds:

Ionic bonding:Forms between a metal and non-metalHorizontal extremes of the periodic tableExamples are NaCl, CsCl, MgO, CaF2

Schematic representation of ionic bonding in sodium chloride (NaCl).

1). Ionic bond – electron from Na is transferred to Cl, this causes

a charge imbalance in each atom. The Na becomes (Na+) and the

Cl becomes (Cl-), charged particles or ions.

COVALENT BONDBOND FORMED BY THE SHARING OF ELECTRONS

Covalent BondingFormation of covalent bond Cooperative sharing of valence electrons Core electrons are strongly bound to the nucleus Valence electrons can participate in bonding Described by orbital overlap

Example: Methane (CH4) molecule

Bonding is directional

1) Non-metallic elemental moleculesH2, Cl2, F2

2) Molecules of dissimilar elementsH2O, CH4, HF

3) Elemental solidsC (diamond), Si, Ge

4) Compound solidsSiC, GaAs, BN, CdTe

2. Covalent bonds- Two atoms share one or more pairs of outer-shell electrons.

Oxygen Atom Oxygen Atom

Oxygen Molecule (O2)

http://www.usfca.edu/fac-staff/Courses/BIOL104_USF/104_Fall03_ppt/Text%20Chapter%2002/OxgnMol.swf

METALLIC BOND

Formed between atoms of metallic elements

Electron cloud around atoms

Valence electrons free to drift through the entire material forming a “sea of electrons” surrounding net positive ionic cores

non-directional bond

Good conductors at all states, lustrous, very high melting points

+ + +

+

+

+ + +

+

++++

+ ++

“sea of electrons”

ionic

cores

Examples; Na, Fe, Al, Au, Co

WHY DO METALS HAVE HIGH MELTING POINTS?

Metals often have high melting points and boiling points. Gold, for example, has a melting point of 1064 oC and a boiling point of 2807 oC.

The properties of metals are related to their structure.

This property is due to the strong attraction between the positively-charged metal ions and the sea of electrons.

HOW DO METALS CONDUCT HEAT AND ELECTRICITY?

Delocalized electrons in metallic bonding allow metals to conduct heat and electricity.

This makes heat transfer in metals very efficient.

Delocalized electrons also conduct electricity through metals in a similar way.

For example, when a metal is heated, the delocalized electrons gain kinetic energy.

These electrons then move faster and so transfer the gained energy throughout the metal.

heat

Metals are usually strong, not brittle. When a metal is hit, the layers of metal ions are able to slide over each other, and so the structure does not shatter.

WHY ARE METALS STRONG?

The metallic bonds do not break because the delocalized electrons are free to move throughout the structure.

metal after it is hit

forceforce

This also explains why metals are malleable (easy to shape) and ductile (can be drawn into wires).

metal before it is hit

METALLIC BONDING

Quiz

FORCES AND ENERGIES

90

(a) Force vs r and (b) Potential energy vs r.

From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap (© McGraw-Hill, 2002)

EN = EA + ER

mnN

r

B

r

AE

11

mn r

mB

r

nAF

A,B,m,n → constants

m > n

dr

dEF

91

Steep slope at equilibrium position x0

(or r0) gives a high modulus

Force-distance curve for two materials, showing the relationship between atomic bonding and the modulus of elasticity

Interatomic Bonding and Elastic modulus

2

2

dr

Ud

dr

dFY

92

Bond length, r

Bond energy, Eo

Melting Temperature, Tm

Tm is larger if Eo is larger.

r or

Energy

r

larger Tm

smaller Tm

Eo =

“bond energy”

Energy

r or

unstretched length

Interatomic Bonding and Melting Point

93

Coefficient of thermal expansion, α

• a ~ symmetric at ro

α is larger if Eo is smaller.

- DL

length, Lo

unheated, T 1

heated, T 2

r or

smaller a

larger a

Energy

unstretched length

E

oE

o

Interatomic Bonding and Coeff. Thermal Exp.

The stronger the interatomic bonding (deeper the potential

energy curve), the smaller is the thermal expansion.

94

Physical origin of thermal expansion

Rising temperature results in the increase of the average amplitude of atomic vibrations. For an anharmonicpotential, this corresponds to the increase in the average value of interatomic separation, i.e. thermal expansion.

Thermal expansion is related to the asymmetric (anharmonic) shape of interatomic potential. If the interatomic potential is symmetric (harmonic), the average value of interatomic separation does not change, i.e. no thermal expansion.

95

Ceramics

(Ionic & covalent bonding): Large bond energylarge Tm

large Esmall a

Metals(Metallic bonding):

Variable bond energymoderate Tm

moderate Emoderate a

SUMMARY: PRIMARY BONDS

Polymers(Covalent & Secondary):

Directional PropertiesSecondary bonding dominates

small Tm

small Elarge a

Two important contributing factors to the properties of materials is the nature of bonding

and the atomic structure.

Both of these are a result of electron interactions and resulting distribution in the material.

Effect of Bonding on properties: a broad flavour

BondBond

Energy eV

Melting

point

Hardness

(Ductility)

Electrical

ConductivityExamples

Covalent5

(8×1019 J)High Hard (poor) Usually Low

Diamond, Graphite,

Ge, Si

Ionic 1-3 High Hard (poor) Low NaCl, ZnS, CsCl

Metallic 0.5 Varies Varies High Fe, Cu, Ag

Van der

Waals0.001-0.1 Low Soft (poor) Low Ne, Ar, Kr

Hydrogen Low Soft (poor) Usually Low Ice

* For comparison thermal energy at RT (300K) is 0.03 eV97

The Materials Tetrahedron

A materials scientist has to consider four ‘intertwined’ concepts, which are schematically shown

as the ‘Materials Tetrahedron’.

When a certain performance is expected from a component (and hence the material constituting

the same), the ‘expectation’ is put forth as a set of properties.

The material is synthesized and further made into a component by a set of processing methods

(casting, forming, welding, powder metallurgy etc.).

The structure (at various lengthscales*) is determined by this processing.

The structure in turn determines the properties, which will dictate the performance of the

component.

Hence each of these aspects is dependent on the others.

The Materials Tetrahedron

* this aspect will be considered in detail later

The broad goal of Materials Science &

Engineering is to understand and ‘engineer’ this

tetrahedron

99

Materials Science-Investigating the relationship

between structure and properties of materials.

Materials Engineering - Designing the structure

to achieve specific properties of materials.

Chapter_1b_Lengthscales.ppt

Properties influenced by

Atomic structure

Electromagnetic structure

(Bonding characteristics)

Properties of a material are determined by two important characteristics*:

Atomic structure

(The way atoms, ions, molecules arranged in the material).

Electromagnetic structure – the bonding character

(The way the electrons**/charge are distributed and spin associated with electrons).

(Bonding in some sense is the simplified description of valence electron density distributions).

Essentially, the electromagnetic structure and processing determine the atomic structure.

* Both these aspects are essentially governed by (properties of) electrons and how they talk to each other!

** Including sharing of electrons.100

ISSUES TO ADDRESS...

How do atoms assemble into solid structures?

How does the density of a material depend onits structure?

How does the mechanical properties of a material depend onits structure?

1

Structures of Materials

How do the structures of metals differ from those of ceramic materials?

102

Physical Classification of Materials by Atomic Structure

Metallic crystal structures tend to be densely packed.

Reasons for dense packing:

Typically, only one element is present, so all atomicradii are the same.

Metallic bonding is not directional. Nearest neighbor distances tend to be small in

order to lower bond energy. Electron cloud shields cores from each other

Have the simplest crystal structures (74 elements).

We will examine three such structures (those of engineering importance) called: FCC, BCC and HCP – with a nod to Simple Cubic

Crystal Structures

2

Non dense, random packing

Dense, regular-packed structures tend to have lower energy.

Energy and Packing

• Dense, ordered packing

105

• atoms pack in periodic, 3D arraysCrystalline materials...

-metals

-many ceramics

-some polymers

• atoms have no periodic packing

Noncrystalline materials...

-complex structures

-rapid cooling

crystalline SiO2

noncrystalline SiO2"Amorphous" = NoncrystallineAdapted from Fig. 3.23(b),


Adapted from Fig. 3.23(a),


Materials and Packing

Si Oxygen

• typical of:

• occurs for:

106

Metallic Crystal Structures • How can we stack metal atoms to minimize empty

space?

2-dimensions

vs.

Now stack these 2-D layers to make 3-D structures

The Crystal Lattice

A point lattice is made up of regular, repeating points in space. An atom or group of atoms are tied to each lattice point

Point lattice + atom/atom group (Motif) = Crystal

Metallic Crystal Structure

108

Unit CellsUnit cell is the smallest unit of a crystal, which, if

repeated, could generate the whole crystal.

Usually the smallest unit cell which exhibits the greatest symmetry is chosen. If repeated (translated) in 3 dimensions, the entire crystal is recreated.

A crystal’s unit cell dimensions are defined by six numbers, the lengths of the 3 axes, a, b, and c, and the three interaxial angles, a, and .

Crystal Lattice

• One can deduce the pattern in a crystalline solid by thinking of the substance as a lattice or repeating shapes formed by the atoms in the crystal

110

Crystal lattice is an imaginative grid system in three dimensions in which every point (or node) has an environment that is identical to that of any other point or node.

System Interaxial Axes

Angles

Triclinic a ≠ ≠ ≠ 90° a ≠ b ≠ c

Monoclinic a = = 90° ≠ a ≠ b ≠ c

Orthorhombic a = = = 90° a ≠ b ≠ c

Tetragonal a = = = 90° a = b ≠ c

Cubic a = = = 90° a = b = c

Hexagonal a = = 90°, = 120° a = b ≠ c

Trigonal a = = 90°, = 120° a = b ≠ c

Crystal System

# of Atoms/Unit Cell

For atoms in a cubic unit cell:

• Atoms in corners are ⅛ within the cell



• Atoms on faces are ½ within the cell

Locations in Lattices: Point Coordinates

z

x

ya b

c

000

111

Position Coordinate

O 0,0,0 (Origin)

A 1,0,0

B 0,1,0

C 0,0,1

D 1,1,1

E 1,1,0

F 0,1,1

G 1,0,1

H ½,1,½

4

Rare due to poor packing (only Po has this structure) The atoms occupy the corners of a cube. Close-packed directions are cube edges. The coordination number is 6, and the packing efficiency is only 52.4%

• Coordination # = 6(# nearest neighbors)

(Courtesy P.M. Anderson)

Simple Cubic Structure (SC)

5

• APF for a simple cubic structure = 0.52


Callister 6e.

Atomic Packing Factor: SC

Body-Centered Cubic (BCC)

• Body-centered cubic unit cells have an atom in the center of the cube as well as one in each corner

• Close packed directions are cube diagonals

• The packing efficiency is 68%, and the coordination number = 8

--Note: All atoms are identical; the center atom is shadeddifferently only for ease of viewing.

aR

9

APF for a body-centered cubic structure = 0.68

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

Adapted from

Fig. 3.2,

Callister 6e.

Atomic Packing Factor: BCC

Face-Centered Cubic (FCC)

A face-centered cubic unit cell contains a total of 4 atoms: 1 from the corners, and 3 from the faces

Close packed directions are face diagonals.

# of Atoms/FCC Unit Cell

• Coordination # = 12

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

a

7

APF for a body-centered cubic structure = 0.74

Adapted from

Fig. 3.1(a),

Callister 6e.

Atomic Packing Factor: FCC

122

coordination number = 12 total 6 atoms per unit cell APF = 0.74 the closest packing possible of

spherical atoms c/a ratio for ideal HCP structure

is 1.633

Hexagonal Close-Packed Structure (HCP)

A sites

B sites

A sites

Periodic Table: Close-Packed Structures

123

Theoretical Density, r

where n = number of atoms/unit cellA = atomic weight VC = Volume of unit cell = a3 for cubicNA = Avogadro’s number

= 6.023 x 1023 atoms/mol

Density = r =

VCNA

n Ar =

CellUnitofVolumeTotal

CellUnitinAtomsofMass

Theoretical Density, r

• Ex: Cr (BCC)

A = 52.00 g/mol

R = 0.125 nm

n = 2 a = 4R/3 = 0.2887 nm

r = a3

52.002

atoms

unit cellmol

g

unit cell

volume atoms

mol

6.023 x 1023

rtheoretical

ractual

= 7.18 g/cm3

= 7.19 g/cm3

Try

126

Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its theoretical density with the experimental value. You may assume that the experimental density value of iron is 7.87 g/cm3.

Exercise

Cu has FCC structure, atomic radius of 0.1278 nm, atomic mass of 63.54 g/mol

calculate the density of Cu in Mg/m3 .

127

Ex

128

AAs NR

nA

NV

nA3)2(

= = r

Density for HCP α-Co with lattice spacing of a = 2.51 ˚A and c = 4.07 ˚A

For HCP, there are the equivalentof six spheres per unit cell, and thus

The unit cell volume can now be calculated as:

)( atoms/mol 10 6.022cell)/(unit cm 10 1.26(2)

g/mol) cell)(70.4atom/unit (6 =

2338-

)(

r

Crystallographic Directions

1. Vector is repositioned (if necessary) to pass through the Unit Cell origin.2. Read off line projections (to principal axes of U.C.) in terms of unit cell dimensions a, b, and c3. Adjust to smallest integer values

4. Enclose in square brackets, no commas [uvw]

ex: 1, 0, ½ => 2, 0, 1 => [ 201 ]

-1, 1, 1

families of directions <uvw>

Algorithm

where ‘overbar’ represents a negative index[ 111 ]=>

z

x

y

What is this Direction ?????

Projections:Projections in terms of a,b and c:Reduction:

Enclosure [brackets]

x y z

a/2 b 0c

1/2 1 0

1 2 0

[120]

132

HCP Crystallographic Directions

• Hexagonal Crystals• 4 parameter Miller-Bravais lattice coordinates

are related to the direction indices (i.e., u'v'w') as follows.

'ww

t

v

u

)vu( +-

)'u'v2(3

1-

)'v'u2(3

1-

]uvtw[]'w'v'u[

Fig. 3.8(a), Callister & Rethwisch 8e.

-a

3a

1

a2

z

Miller-Bravais indices

Determine indices for the directions shown in the following hexagonal unit cells:

projections on the a1, a2, and z axes are 1, 1/2, and 1/2, which when multiplied by the factor 2 become the smallest set of integers: 2, 1, and 1

This means thatu’ = 2, v’ = 1, w’ = 1

the u, v, t, and w indices become

u =1

3(2u' v' )

1

3(2)(2) (1)

3

3 1

0 )2()1)(2(3

1)2(

3

1= u’v’v

t (u v) 1 0 1

w = w’ = 1

No reduction is necessary inasmuch as all of these

indices are integers; therefore, this direction in the

four-index scheme is

[10 1 1]

To determine the Miller-Bravais indices of the crystallographic direction indicated on the a1-a2-z plane of an h.c.p. unit cell, the vector must be projected onto each of the three axes to find the corresponding component, such that:

To determine the Miller-Bravais indices of the crystallographic direction indicated on the a1-a2-z plane of an h.c.p. unit cell, the vector must be projected onto each of the three axes to find the corresponding component, such that:

Miller-Bravais indices: Example

Determine indices for the directions shown in the following hexagonal unit cells:

projections on the a1, a2, and z axes are a, a, and c/2 or, in terms of a and c the projections are -1, -1, and 1/2, which when multiplied by the factor 2 become the smallest set of integers: -2, -2, and 1.

This means thatu’ = -2, v’ = -2, w’ = 1

the u, v, t, and w indices become

w = w’ = 1

Now, in order to get the lowest set of integers, it is

necessary to multiply all indices by the factor 3, with

the result that this direction is a

u 1

3(2u' v)

1

3(2)(2) (2)

2

3

v 1

3(2v' u' )

1

3(2)(2) (2)

2

3

t (u v) 2

3

2

3

4

3

[2 2 43]

ex: linear density of Al in [110]

direction

a = 0.405 nm

Linear Density

• Linear Density of Atoms LD =

a

[110]

Unit length of direction vector

Number of atoms

# atoms

length

13.5 nma2

2LD

# atoms CENTERED on the direction of interest!Length is of the direction of interest within the Unit Cell

Miller Indices (hkl)

Crystallographic Planes

The (210) surface

Intercepts : ½ a , a , ¥ Fractional intercepts : ½ , 1 , ¥ Miller Indices : (210)

The (111) surface

Intercepts : a , a , aFractional intercepts : 1 , 1 , 1 Miller Indices : (111)

Miller Indices (hkl)

138

Procedure:

1.Find the intersections, r and s, of the plane with any

two of the basal plane axes.

2.Find the intersection t of the plane with the c axis.

3.Evaluate the reciprocals 1/r, 1/s, and 1/t.

4.Convert the reciprocals to smallest set of integers

that are in the same ratio.

5.Use the relation i = -(h + k), where h is associated with

a1, k is associated with a2, and i is associated with a3.

6.Enclose all four indices in parentheses: (h k i l)

Miller Indices in Four-Index Notation

• For hcp crystal structure• Planes (h k i l), directions [h k i l]• Sum of the first three indices h + k + i = 0

139

Example: What is the designation, using four Miller indices, of the lattice plane shaded pink in the figure?

•The plane intercepts the a1, a3, and c axes at r = 1, u = 1, and t = ∞, respectively. •The reciprocals are

•1/r = 1 •1/u = 1 •1/t = 0

•These are already in integer form. •We can write down the Miller indices as (h k i l) = (1 k 1 0), where the k index is undetermined so far. •Use i = (h + k). That is, k = 2. •The designation of this plane is (h k i l) =

Miller-Bravais Indices: Example

)1021(

140

Planar Atomic Density

Exercise

calculate linear atomic density ρl in [110] direction in Cu crystal lattice in atoms/mm (Cu is FCC and lattice constant a = 0.361 nm)

calculate planar atomic density ρp on (110) plane of the α-Fe in BCC lattice in atoms/mm2 . (lattice constant a = 0.287 nm)

141

142

(1) Describe the difference in atomic structure between crystalline and non-crystalline materials.(2) Draw unit cells and derive the relationships between unit cell edge length and atomic radius for face-centered cubic, body-centered cubic, and hexagonal close-packed crystal structures (3) Compute the densities for metals having face-centered cubic and body-centered cubic crystal structures given their unit cell dimensions.(4) Given three direction index integers, sketch the direction corresponding to these indices within a unit cell: [111], [211] and [321].(5) Specify the Miller indices for a plane that has been drawn within a unit cell.

Assignment 3

143

Basic Material Properties

General

Weight: Density r, kg/m3

Expense: Cost/kg Cm, $/kg

Mechanical

Stiffness: Young’s modulus E, GPa

Strength: Elastic limit y , MPa

Fracture strength: Tensile strength ts , MPa

Brittleness: Fracture toughness KIc , MPa·m1/2

Fatigue

Thermal

Expansion: Expansion coeff. a, 1/K

Conduction: Thermal conductivity , W/m·K

Specific Heat (Capacity), cp or cv, J/kg·K

Electrical

Conductor? Insulator? Conductivity σ, S/m

Dielectric Capacity, F/m

Young’s modulus, E

Elastic limit, y

Strain

Str

ess

Ductile materials

Brittle materials

Young’s

modulus, E

Tensile (fracture)

strength, ts

Strain

Str

ess

Thermal expansion

o

Expansion

coefficient, a

Temperature, T

Ther

mal

str

ain

xT1 To

Q joules/secArea A

Thermal conduction

Mechanical properties

Thermal

conductivity,

(T1 -T0)/x

Hea

t fl

ux

, Q

/A

Basic Materials Properties

144

Dielectric properties-it indicates the energy storing capacity of a material (by means of polarization

A. Polarizability - ability to form instantaneous dipoles

B. Capacitance - ability to store an electric charge

C. Ferroelectric properties - exhibit spontaneous electric polarization that can be reversed in direction by the application of an appropriate electric field.

D. Piezoelectric properties - ability to generate an electric charge in response to applied mechanical stress

E. Pyroelectric properties – ability to generate electricity when heated

Magnetic propertiesA. Paramagnetic propertiesB. Diamagnetic propertiesC. Ferromagnetic properties

Basic Materials Properties

145

Optical properties – response to lighto index of refraction – bending of lighto transparent – light passes througho translucent – some light passes through but no distinct

imageo opaque – no light passes through

Corrosion properties

Deteriorative properties

Biological properties A. Toxicity B. bio-compatibility

ELECTRICAL: RESISTIVITY OF COPPER

146

Increase resistivity of Cu• by adding impurities • by mechanical deformation

Fig. 19.8 Callister

Resist

ivity

10

-8Ohms-

m

T (0C)

scattering of e- by microstructure

scattering of e- impurities scattering of e- by phonons

• low from ceramic oxide (structure and conduction properties)• changes due to alloying in metals (even though same structure)

Silica (SiO2) fibres

in space shuttle tiles

Fig. 23.18 Callister

THERMAL: CONDUCTION OF BRASS

147

Brass, Cu-Zn

Fig. 20.4 Callister

Wt % ZnC

on

du

ctiv

ity

(W

/m-K

)

e.g., Light transmission of Alumina (Al2O3).single crystal, polycrystals (low and high porosity)

Fig. 1.2 Callister

Which one is single crystal?Why?

These reflect the effects of processing.

OPTICAL: TRANSMISSION OF LIGHT

148


single-crystal

(transparent)

polycrystalline,

fully dense

(translucent)

polycrystalline,

5% porosity

(opaque)

Aluminum Oxide (Al2O3)

149


Performance Goal: increased strength from a metallic material

In actuality, crystals are NOT perfect. There are defects!In metals, strength is determined by how easily defects can move!

quenching

slow cooling

OFF

150

e.g., Stress, corrosive environments, embrittlement, incorrect structures from improper alloying or heat treatments, …

http://www.uh.edu/liberty/photos/liberty_summary.html

USS Esso Manhattan 3/29/43Fractured at entrance to NY harborbcc Fe Fig. 6.14 Callister

Strain

Str

ess

(M

Pa

)

- 200 C

- 100 C

+ 25 C

DETERIORATION AND FAILURE

151

Mechanical Behaviour

A material changes in shape when forces are applied.This change in dimensions is called deformation.

Strain: is the amount of deformation per unitlength.

Stress: is the force per unit area.

Energy is absorbed during deformation of a materialbecause work must be done by applying force along thedeformation distance.

152

Types of Stresses/Forces

• Tension– pulling

– examples: tug-of-war, slingshot

• Compression– pushing together or squeezing

– examples: bed springs, can crusher, bench vise

Mechanical Properties- Deformation

• elasticity

– ability to return to original shape after being deformed by stress

– rubber ball or piece of elastic

• plasticity

– retains new shape after being deformed by stress

– wet clay ball or piece of saran wrap

156

Elastic means reversible!

Elastic Deformation

2. Small load

F

d

bonds

stretch

1. Initial 3. Unload

return to

initial

F

d

Linear-

elastic

Non-Linear-

elastic

157

Plastic means permanent!

Plastic Deformation (Metals)

F

d

linear

elasticlinear

elastic

dplastic

1. Initial 2. Small load 3. Unload

planes

still

sheared

F

delastic + plastic

bonds

stretch

& planes

shear

dplastic

158

(at lower temperatures, i.e. T < Tmelt/3)

Plastic (Permanent) Deformation

• Simple tension test:

engineering stress,

engineering strain,

Elastic+Plastic

at larger stress

p

plastic strain

Elastic

initially



permanent (plastic)

after load is removed

160

Stress-Strain Testing

• Typical tensile test

machine

Adapted from Fig. 6.3, Callister & Rethwisch 8e. (Fig. 6.3 is taken from H.W. Hayden,

W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III,

Mechanical Behavior, p. 2, John Wiley and Sons, New York, 1965.)

specimenextensometer

• Typical tensile

specimen

Adapted from

Fig. 6.2,

Callister &

Rethwisch 8e.

gauge

length

161

Linear Elastic Properties

• Modulus of Elasticity, E:(also known as Young's modulus)

• Hooke's Law:

= E

Linear-

elastic

E

F

Fsimple tension test


Initially, strain is proportional to stress and shows reversiblebehaviour.

i.e. When stress is removed the strain disappears.

The modulus of elasticity (Young’s Modulus) is the ratio between thestress and the reversible strain .

E = / The units of young’s modulus is psi or pascals.

The value of elastic modulus is a measure of the interatomic bonding forces and relates to the rigidity of engineering designs.

at higher stresses, permanent displacement can occur among the atoms within a material. In this case much of the strain is not reversible when the applied stresses are removed. In contrast with previous (elastic) strain this is called plastic strain.

162

163

Mechanical Properties

Slope of stress-strain plot (which is proportional to the elastic modulus) depends on bond strength of metal



164

Poisson's ratio, n

• Poisson's ratio, n:

Units:

E: [GPa] or [psi]

n: dimensionless

L

-n

n L

metals: n ~ 0.33

ceramics: n ~ 0.25

polymers: n ~ 0.40

• Tensile strain: • Lateral strain:

d

Lo

dL L

w o

165

Engineering stress-strain curve

166

Engineering stress-strain curve

Longitudinal tensile stress

linear strain

167

Factors affecting shape andmagnitude of stress-strain curve

168

Modulus of elasticity

169

Metals

Alloys

Graphite

Ceramics

Semicond

PolymersComposites

/fibers

E(GPa)

Based on data in Table B.2,


Composite data based on

reinforced epoxy with 60 vol%

of aligned

carbon (CFRE),

aramid (AFRE), or

glass (GFRE)

fibers.

Young’s Moduli: Comparison

109 Pa

0.2

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

10 0

200

600800

10 001200

400

Tin

Cu alloys

Tungsten

<100>

<111>

Si carbide

Diamond

PTF E

HDP E

LDPE

PP

Polyester

PSPET

CFRE( fibers) *

GFRE( fibers)*

GFRE(|| fibers)*

AFRE(|| fibers)*

CFRE(|| fibers)*

170

Yielding

171

• Stress at which noticeable plastic deformation has

occurred.when p = 0.002

Yield Strength, y

y = yield strength

Note: for 2 inch sample

= 0.002 = Dz/z

Dz = 0.004 in



tensile stress,

engineering strain,

y

p = 0.002

172

Yield strength of materials

173


174


175

Room temperature

values



a = annealed

hr = hot rolled

ag = aged

cd = cold drawn

cw = cold worked

qt = quenched & tempered

Yield Strength: Comparison

Graphite/

Ceramics/

Semicond

Metals/

Alloys

Composites/

fibersPolymers

Yie

ld s

tren

gth

,

y(M

Pa)

PVC

Har

d t

o m

easu

re,

since

in t

ensi

on

, fr

actu

re u

sual

ly o

ccurs

bef

ore

yie

ld.

Nylon 6,6

LDPE

70

20

40

6050

100

10

30

200

300

400

500600700

1000

2000

Tin (pure)

Al (6061) a

Al (6061) ag

Cu (71500) hrTa (pure)Ti (pure) aSteel (1020) hr

Steel (1020) cdSteel (4140) a

Steel (4140) qt

Ti (5Al-2.5Sn) aW (pure)

Mo (pure)Cu (71500) cw

Har

d t

o m

easu

re,

in c

eram

ic m

atri

x a

nd e

pox

y m

atri

x c

om

posi

tes,

sin

ce

in t

ensi

on

, fr

actu

re u

sual

ly o

ccurs

bef

ore

yie

ld.

HDPEPP

humid

dry

PC

PET

¨

176

Tensile strength

177

Tensile Strength: Comparison

Si crystal<100>

Graphite/

Ceramics/

Semicond

Metals/

Alloys

Composites/

fibersPolymers

Ten

sile

stre

ng

th, T

S(M

Pa)

PVC

Nylon 6,6

10

100

200

300

1000

Al (6061) a

Al (6061) ag

Cu (71500) hr

Ta (pure)Ti (pure) a

Steel (1020)

Steel (4140) a

Steel (4140) qt

Ti (5Al-2.5Sn) aW (pure)

Cu (71500) cw

LDPE

PP

PC PET

20

3040

2000

3000

5000

Graphite

Al oxide

Concrete

Diamond

Glass-soda

Si nitride

HDPE

wood ( fiber)

wood(|| fiber)

1

GFRE (|| fiber)

GFRE ( fiber)

CFRE (|| fiber)

CFRE ( fiber)

AFRE (|| fiber)

AFRE( fiber)

E-glass fib

C fibersAramid fib



a = annealed

hr = hot rolled

ag = aged

cd = cold drawn

cw = cold worked

qt = quenched & tempered

AFRE, GFRE, & CFRE =

aramid, glass, & carbon

fiber-reinforced epoxy

composites, with 60 vol%

fibers.

Room temperature

values

178

Ductility

– ductility – can be drawn into wire (stretched), bent, or extruded

– malleability – can be flattened

179

Measures of ductility

181

Solution

182

Resilience

183

Toughness

184

True-stress-true-strain curve

185

True Stress & Strain

Note: S.A. changes when sample

stretched

• True stress

• True strain

iT AF

oiT ln

1ln

1

T

T



True Stress-strain curve

The diagram below compares the nominal and true stress-strain

curves

186

a) non-ductile material with no plastic deformation

e.g. Cast Iron


b) Ductile material with yield point.

e.g. Low carbon steel

BS: Breaking Stength, YP: yield point and TS : Tensile strength

187


Ductile material without marked yield point

e.g. aluminium True stress-strain curve

188

193

A tensile test was performed on a specimen with a cylindrical cross-section with a gage diameter of 4 mm and a gage length of 45 mm. At fracture, the gage length was 55 mm, and the reduction in area was 90%. The load at fracture was 1200 N.

(i) Compute the engineering strain at failure

(ii) Compute the engineering fracture stress

(iii)Compute the true fracture stress at the neck

(iv) Compute the true fracture strain at the neck

Example

194

Tensile Strength, TS

• Metals: occurs when noticeable necking starts.

• Polymers: occurs when polymer backbone chains are

aligned and about to break.



y

strain

Typical response of a metal

F = fracture or

ultimate

strength

Neck – acts

as stress

concentrator

engin

eeri

ng

TSst

ress

engineering strain

• Maximum stress on engineering stress-strain curve.

196

• Plastic tensile strain at failure:

Ductility

• Another ductility measure: 100xA

AARA%

o

fo-

=

x 100L

LLEL%

o

of

Lf

AoAf

Lo



Engineering tensile strain,

Engineering

tensile

stress,

smaller %EL

larger %EL

197

• Energy to break a unit volume of material

• Approximate by the area under the stress-strain curve.

Toughness

Brittle fracture: elastic energy

Ductile fracture: elastic + plastic energy



very small toughness

(unreinforced polymers)

Engineering tensile strain,

Engineering

tensile

stress,

small toughness (ceramics)

large toughness (metals)


Strength (and Hardness).

The ability of a material to resist plastic deformation is called the

yield strength (ys).

This is calculated by dividing the force required to initiate the yield

(first plastic deformation) by the cross-sectional area.

In mild steel the yield strength is marked by a definite yield point is

not well defined, the yield strength is defined as that stress

required to produce 0.2% plastic offset.

198


Hardness is the resistance of a material to penetration

of its surface. The hardness and strength are related so

that for a wide range of metallic materials strength

increases with hardness.

199

200

Hardness

• Resistance to permanently indenting the surface.

• Large hardness means:-- resistance to plastic deformation or cracking in

compression.

-- better wear properties.

e.g.,

10 mm sphere

apply known force measure size

of indent after

removing load

dDSmaller indents

mean larger

hardness.

increasing hardness

most plastics

brasses Al alloys

easy to machine steels file hard

cutting tools

nitrided steels diamond


Toughness: is the measure of the energy to break the materials. H

is force x distance and is expressed in Joules.

A ductile material with the same strength as a non-ductile (brittle)

material will require more energy for breaking and will be tougher.

There are several ways of measuring this quantity.

Charpy and Izod are two examples.

The shape of lost piece, method of applying the energy and geometry

of stress concentrations are important.

201

202

• Design uncertainties mean we do not push the limit.

• Factor of safety, N

N

y

working

Often N is

between

1.2 and 4

• Example: Calculate a diameter, d, to ensure that yield does

not occur in the 1045 carbon steel rod below. Use a

factor of safety of 5.

Design or Safety Factors

220,000N

d2 / 4 5

N

y

working

1045 plain

carbon steel: y = 310 MPa

TS = 565 MPa

F = 220,000N

d

Lo

d = 0.067 m = 6.7 cm

203

Microstructure + Phase Transformations in Multicomponent Systems

Chapter Outline: Phase Diagrams

Definitions and basic concepts

Phases and microstructure

Binary isomorphous systems (complete solid solubility)

Binary eutectic systems (limited solid solubility)

Binary systems with intermediate phases/compounds

The iron-carbon system (steel and cast iron)

Not tested: The Gibbs Phase Rule

SEE YOU

IN THE

EXAM

Instructor: Dr. Emmanuel Kwesi Arthur Dept. of Materials … · 2019. 2. 5. · 4 Scope: Provide an...

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Transcript of Instructor: Dr. Emmanuel Kwesi Arthur Dept. of Materials … · 2019. 2. 5. · 4 Scope: Provide an...