nmc exam brush-up · nmc exam brush-up november 2009. ... microdefects (point defects, line defects...

NMC EXAM BRUSH-UPNOVEMBER 2009

OVERVIEW

1. Crystal Structures

2. Solidification and Crystal Defects

in Solids

3. Heat Treatments

4. Electrical Properties of Materials

5. Magnetic Properties of Materials

CRYSTAL STRUCTURES

Chapter 3

CRYSTAL STRUCTURES

What do I need to know?

Main Metallic Crystal Structures

FCC, BCC, HCP, BCT

Polymorphism

Unit Lattices and Bravais Lattices

Density Tool Box

Close-packed Crystal Structures

MAIN METALLIC CRYSTAL STRUCTURES

BCC (α-Fe, Na, Li and K)

Coordination Number = 8

Effective number of atoms = 2

Lattice Parameter a = 4R/√3

Fraction of Unit Cell that is occupied by

atoms by Volume

APF = Volume of Atoms/Volume of Unit Cell

APF = 0.68


FCC (γ-Fe, Au, Ag, Pt)



Lattice Parameter a = 4R/√2


atoms by Volume


APF = 0.74


HCP (C, Cd, Co, Zn)



Lattice Parameter a = 2R & c = 1.633a


atoms by Volume


APF = 0.74


Examples1. Calculate the radius of an iridium atom. Ir has

an FCC crystal structure and a density of

22.4g/cm3 and an atomic weight of 192.2g/mol

(R = 0.136nm)

POLYMORPHISM

Crystal structure transformation in

materials due to temperature or

pressure change Fe

@ Room Temperature - BCC

Above 727°C – FCC

Above 1394°C – BCC

C

@ Room Temperature – HCP

@ Very high pressures and temperatures – Diamond Cubic

UNIT AND BRAVAIS LATTICES

14 Bravais lattices (RELAX you don’t have to

know them all!)

These include the geometrical shape of lattice and atom

placement

Cubic, Tetragonal, Hexagonal, Orthorhombic,

Monoclinic, Rhombohedral and Triclinic

4 Types of unit lattices

Only concerned with placement of atoms in lattice

Simple

Body-centred

Face centered

End-centred

DENSITY TOOL BOX

Volumetric Density of Materials

ρv= (NR)(MR)/ [(Vcell)(NA)] (g/cm3)

Planar Atomic Density

ρp= (NR(intersected atoms))/Aplane (atoms/mm2)

Linear Atomic Density

ρl= (NR(atom diameters on line))/Lline (atoms/mm)

DENSITY TOOL BOX

Examples1. Consider the FCC crystal structure of Al.

Determine the planar atomic density of the (111)

plane. (0.91)

2. Cobalt has an HCP crystal structure with an

atomic radius of 0.1253nm and a c/a ratio of

1.623.

a) Compute the volume of the unit cell for Co

V = 0.0664nm3

b) Explain in your own words why the c/a

ratio in question a is not equal to the

theoretical value of 1.633.

CLOSE-PACKED CRYSTAL STRUCTURES

FCC and HCP are both close-packed (APF

0.74)

Closely packed plane has the highest

planar density

Packing sequence differs

ABC ABC ABC – FCC and AB AB AB – HCP

FCC has more closely packed planes than

HCP and BCC does not have a close

packed plane

SUMMARY: TIPS AND FURTHER EXAMPLES

Exam questions will most likely be more focused

on calculations than the theory of this chapter

Familiarise yourself with the sketches of FCC,

BCC, HCP and BCT

If you can sketch it, CN, Atoms per unit cell

and APF can be UNDERSTOOD

Principle for Ionic crystals are similar just note

that the cation and anion valences HAVE TO

BALANCE

Prove to yourself that the c/a ratio = 1.633 for an

ideal HCP crystal

ALWAYS DRAW A PICTURE!!!

IMPERFECTIONS IN SOLIDS

Chapter 4

CRYSTAL STRUCTURES

What do I need to know? Process of Solidification

Polycrystalline Metals (Sketch ingot

solidification mechanism)

Single Crystals (Chozkralski process)

Defects in Solids

Influencing factors

Types of Defects

Calculating Grain Size

PROCESS OF SOLIDIFICATION

Heat is extracted from

mold walls (high cooling

rate, small grains)

Towards centre of ingot

molten metal starts to

cool (columnar zone)

Centre of ingot, last

metal solidifies (large equiaxed

grains)

DEFECTS IN SOLIDS

Influencing Factors

Mechanical Properties

Ductility

Electrical Properties

Conductivity

Heat conductivity ability

Diffusion of atoms

Corrosion resistance

Types of Defects

Microdefects (point defects, line defects and surface defects)

Macrodefects (Cracks, pores, inclusions and blow holes)

DEFECTS IN SOLIDS

Microdefects – Point Defects

Vacancies

Nc = Ne(-Qv/kT)

Self-interstitial Defects

Impurities

Most materials are used in alloy form

Simplest alloy is that of solid solution

Substitutional (Alloying atoms replaces that of paremt

atoms)

Interstitial (Alloying atoms positions between parent

atoms)

Excess alloying elements above saturation limit –

two-phase solid

Solid solution also depends on Hume-Rothery criteria

DEFECTS IN SOLIDS

Microdefects – Point Defects

Hume-Rothery Criteria

Rparent and Ralloy difference < 15%

Parent and alloy crystal structure must be similar

Electron negativity of 2 elements must be about equal

2 Elements must have the same number of valence

electrons

Schottky Defects (Ceramics)

Missing cation AND anion

Frenkel Defects (Ceramics)

Cation vacancy

DEFECTS IN SOLIDS

Formation of Point Defects

Vacancies

During solidification

Rapid cooling

Cold work

Radioactive bombardment

Self-interstitial Atoms

Radioactive bombardment

Impurities

Solid solutions

Diffusion of rogue species

DEFECTS IN SOLIDS

Microdefects – Line defects

Two primary types

Screw defects (Forms through shear)

Edge dislocation

⥜ - Positive dislocation

⥝ - Negative dislocation

Formation of Edge dislocations (Usually forms through tension)

Solidification

Cold work (Enhances slip of dislocations on close-packed planes)

Vacancy condensation

DEFECTS IN SOLIDS

Microdefects – Surface defects

Grain boundaries

Due to neighboring grains with different geometrical

orientation

Grain boundary is area of high energy capacity

Always present in polycrystalline materials/alloys

Twinning

Plane that has a mirror image

Forms through cold work (mechanical twins) or during

annealing (annealed twins)

DEFECTS IN SOLIDS

Macrodefects

Cracks

Due to rapid cooling during solidification

Due to mechanical deformation

Pores or blow holes

Due to decrease in gaseous solubility in the molten metal,

gas escapes through partially solidified surface

Inclusions

Rogue particles that intrude material during

manufacturing

CALCULATING GRAIN SIZE

N = 2n-1

N = Average number of grains per square inch

(@100x)

n = Grain size number

Examples

1. For an ASTM grain size of 6, how many grains

would there be per square inch at

a) 100X? (32)

b) Without any magnification? (320 000)

2. Determine the ASTM grain size number if 25 grains

per square inch are measured at a magnification of

75. (4.8)

ANOTHER EXAMPLE

3. Calculate the fraction of atom sites that are vacant

for Pb at its melting temperature of 327°C. Assume

an energy for vacancy formation of 0.55eV/atoms.

(2.41x10-5)

SUMMARY: TIPS

This chapter contains mainly theory but the

concepts are of utter importance

Expect a few graphs in the exam on this chapter

Number of calculations in this chapter will

probably be limited

0

2

4

6

8

10

12

Grain Size

Strength Creep Resistance

Resis

tiv

ity

Ndefects, CW, %Alloying

Elements

HEAT TREATMENT

Chapter 9

HEAT TREATMENT

What do I need to know? Fe-C phase system

Interpretation of binary phase diagram

Phases present at specific temperature and composition

Lever-rule for calculating percentage of different phases at

temperatures and compositions

Phase transformations (peritectic, eutectic, eutectoid and

peritectoid reactions)

Equilibrium phases and reactions

Non-equilibruim phases

Heat treatments and microstructures

HEAT TREATMENT

Fe-C phase system Only to be used for

equilibrium cooling

conditions

Phase diagram shows

all of the reactions,

compositions and

temperatures

Phases with

equilibrium cooling

Ferrite, cementite

and pearlite

HEAT TREATMENT

Fe-C phase system Examples

1. By using the Fe-C phase diagram, answer the

following questions applicable to a 0.5%C

hypoeutectoid steel that is cooled slowly from

950°C to just below 727°C.

a) Calculate the amount of proeutectoid ferrite in the

steel (38.71%)

b) Calculate the amounts of eutectoid ferrite and

eutectoid cementite in the steel (54.17% and 7.1%)

HEAT TREATMENT

Fe-C phase system Examples

2. Determine the chemical compositions of steels

containing the following microstructural

components after cooling

a) 92% Ferrite and 8% Cementite (0.559%C)

b) 48.2% Proeutectoid ferrite (0.426% C)

c) 4.7% Proeutectoid cementite (1.0773% C)

d) 10.45% Eutectoid cementite (Hypereutectoid

composition) (1.388% C)

HEAT TREATMENT

Non-equilibrium Phases Transformations

Increase in cooling rate – non-equilibrium phases Bainite – T from 250 – 550°C

Fine dissemination of cementite in ferrite matrix

Good toughness, strength and hardness properties

Martensite – Rapid cooling (quenching in water or brine)

C atoms don’t have time to diffuse out of FCC structure, are trapped in BCT cell

Due to high amount of distortion associated with phase transformation, hardness and strength of martensite is very high

Temper treatment is often need to restore ductility of martensite

Tempering occurs below 650°C and allows C to precipitate out – also known as spherodising – internal stresses are relieved and ductility is improved

HEAT TREATMENT

Types of Treatments

Annealing

Steel is austenitised, cooled at equilbrium condictions

Large grains and coarse pearlite

Good ductility

Normarlising

Air-quench

Finer grain size and pearlite

Harder component than annealed sample

Hardening

Rapid queching in brine, oil, water or even liquid nitrogen

Martensite forms

Excessively high hardness

HEAT TREATMENT

Types of Treatments

Stress RelieftreatmentCold worked, quenched, welded or machined

components experience stress fluctuations due to internal stresses

Heat component below eutectoid temperature to relieve internal stress

SpherodisingProcess at which componenet is heated to allow the

rediffusion of C atoms out of the grains to form spheres)

Good machinability and good ductility

Spheres have the lowest surface to volume energy therefore precipitates grow in geometry to mimic this shape

RECAP: TIPS

You will most likely HAVE to use the lever rule

You may expect some application type questions

If you have to design a heat treatment remember FIT FOR PURPOSE

Cementite is highly brittle therefore any application that requires good toughness, the amount in the matrix must be reduced – propose spherodising treatment

With hypoeutectoid steels, pearlite can be a problem for applications that require high strength – CW can resolve this to a degree

If a rapid quench (water, brine, oil or liquid nitrogen is proposed, you will propably end up with martensite – tempering is essential)

Bainite can be produced by quenching in a molten Pb or salt bath and will give excellent mechanical properties but time constraints have to be taken into account

ELECTRICAL PROPERTIES OF

MATERIALS

Chapter 14

ELECTRICAL PROPERTIES OF MATERIALS

What do I need to know? Relationship between resistivity and conductivity

3 Groups of electrical conductivity

Factors that influence resistivity and conductivity

Energy gap model for metals and isolators

Intrinsic semi-conductors

Extrinsic semi-conductors

Dielectric character


Resistivity and Conductivity

Inversely proportional to each other

Resistance of material is dependent on the type of

material, length and cross-sectional area of

component

Ohm’s law can be used to determine Resistance and

the micro-law can be used to determine conductivity

or resistivity

3 Types of Conductors

Conductor (e.g. Metals with high conductivity)

Semi-conductors (e.g. Si with moderate conductivity)

Isolators (e.g. Ceramics with poor conductivity)


Factors that influence resisitivity

Temperature

Linear relationship between resistivity and temperature

Purity of metal

Alloying elements increase resistivity as electrons have less

mobility in the crystal structure

Crystal Defects

An increase in the crystal defects will facilitate an increase

in the resistivity as they will form barriers against the

movement of electrons

Resistivity can be reduced by heat treatments (HX)


Energy gap model

Metals

Small amount of energy needed to fill energy gap with

metals

Therefore most metals have good conductivity

Isolators

Energy gap is separated from a filled band and an empty

band

Electrons need a lot more energy to cross energy gap

therefore conductivity is lower


Intrinsic Semi-Conductors (A-B-C)

Pure, semi-conductors (Si and Ge)

Negative and positive electrons contribute to the

conductivity of semi-conductors

With an increase in temperature , the CONDUCTIVITY of

the material increases for semi-conductors since certain

valence electrons are excited and their mobility increases

Extrinsic Semi-Conductors

Differentiate between p- and n-type

Positive (Group 3 and 4 elements) and negative semi-

conductors (Group 4 and 5 elements)

By doping, impurities decrease the energy gap and through

that, conductivity increases


Extrinsic Semi-Conductors

n-Type

Group 5 substitutes one of Group 4 atoms

Majority of conductors are electrons – minority are vacancies

p-Type

Group 3 replaces one of Group 4 atoms

Majority of conductors are vacancies – minority are electrons

Dielectric Character

Ceramics, ionics and some polymers – mostly isolators but

in some cases also semi-conductors

Capacitor chambers

Pizo-electric ceramics

Ceramics that can convert electrical pulses to mechanical

vibrations or vice versa

RECAP

This chapter consists of 90% theory

The few electrical formulas – Ohm’s law etc have been

covered extensively at high school level but if you have any

questions please don’t hesitate to ask

Do some exercises on extrinsic semi-conductors just to

familiarise yourself with the equations

It’s literally plug-and-play equations with very little

complicated calculations

Once again it’s important to UNDERSTAND the factors

that will influence conductivity and resistivity

You can expect maybe two graphs on this chapter, some

monkey puzzle questions and maybe 1 calculation

(Probably from the extrinsic semi-conductors section)

MAGNETIC PROPERTIES OF

MATERIALS

Chapter 20

MAGNETIC PROPERTIES OF MATERIALS

What do I need to know? Basic Principles (Theory)

Magnetic field strength and magnetic density

Relative permeabilities

Types of magnetism

Diamagnetic, paramagnetic, ferromagnetic, antiferromagne

tic and ferrimagnetic

Influence of Temperature on Ferromagnetics

Hysteresis

Magnetisation and demagnetisation and hysteris loops

Differentitate between hard and soft magnetics


Types of Magnetism Types of magnetism

Diamagnetic - μr < 1

Paramagnetic – particles move toward external magnetic

field but loses their magnetism when field is removed

Ferromagnetic – Magnetisation can be permanent due to

the half-filled orbital of elements. It is essential that

electrons in the 3d orbital are unpaired

Antiferromagnetic – Elements have a magnetic moment but

the a/d ratio is does not range between 1.4 and 2.7 – no

magnetism

Ferrimagnetism – Traces magnetic moments – usually ionic

bonds- spine of electrons are anti-parrallel but not magnetic


Influence of Temperature on

Ferromagnetics At the Curie temperature, the 3d-electrons’

orientations changes and the parallel spin of the

electrons decrease

At this temperature the ferromagnetic nature of the

material is destroyed


Hysteresis Domains on atomic level can be altered via a solenoid

– causes parallel movement of 3d-electrons

Magnetisation occurs with ferromagnetic and

ferrimagnetic materials due

Domains (which have the correct orientation) start to grow

at the expense of incorrect orientated domains

Incorrect orientated domains can be rotated if the applied

field strength is strong enough

Demagnetisation will occur if the material is heated

above its Curie temperature, by applying an opposite

directed field strength or increasing the dislocation

density of the material


Hysteresis

So-called hysteresis loop shows the

life-cycle of a ferromagnetic

material

The larger the area of the

curve, the easier magnetisation is

possible


Hard VS Soft magnetics

HARD SOFT

High Hc and Br values Easy to magnetise and

demagnetise

Large negative magnetic

field need to demagnetise

Needs high Bs value and

high pearmeability

Small magnetic field to

magnetise

Induced current due to the

magnetic field

RECAP

The theory of this chapter is the

most important since there are very

few types of calculations that can be

asked of you

Types of magnets is quite important

also the mechanism that allows

ferromagnetism

B-H curve is very easy to

understand – just follow your notes

nmc exam brush-up · nmc exam brush-up november 2009. ... microdefects (point defects, line defects...

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