Nmc Exam Brush-up

8/8/2019 Nmc Exam Brush-up

1/52

NMC EXAM BRUSH-UPNOVEMBER 2009


2/52

Overview

1.Crystal Structures

2.Solidification and Crystal Defectsin Solids

3.Heat Treatments

4.Electrical Properties of Materials

5.Magnetic Properties of Materials


3/52

CRYSTAL STRUCTURESChapter 3


4/52

Crystal Structures

What do I need to know?Main Metallic Crystal

StructuresFCC, BCC, HCP, BCT

Polymorphism

Unit Lattices and BravaisLattices

Density Tool Box

Close-packed CrystalStructures


5/52

Main Metallic Crystal Structures

BCC (-Fe , Na, Li and K)Coordination Number = 8

Effec tive numbe r o f ato ms = 2

Lattic e Parame ter a = 4R/3

Frac tio n o f Unit Ce ll that is o c c upie dby atoms by Volume

APF = Volume of Atoms/Volume of UnitCell

APF = 0.68


6/52


FCC (-Fe, Au, Ag, Pt)Coordination Number = 12


Lattic e Parame ter a = 4R/2

Frac tio n o f Unit Ce ll that is o c c upie dby atoms by Volume

APF = Volume of Atoms/Volume of UnitCell

APF = 0.74


7/52


HCP ( , , ,C Cd Co Zn)Coordination Number = 12


Lattic e Parame ter a = 2R & c =1.633a

Frac tio n o f Unit Ce ll that is o c c upie d

by atoms by VolumeAPF = Volume of Atoms/Volume of Unit

Cell

APF = 0.74


8/52


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

an FCC crystal structure and a density of22.4g/cm3 and an atomic weight of192.2g/mol (R = 0.136nm)


9/52

Polymorphism

Crystal structure transformation inmaterials due to temperature orpressure change Fe

@ Room Temperature - BCC

Above 727C FCC

Above 1394C BCC

C @ Room Temperature HCP @ Very high pressures and temperatures

Diamond Cubic


10/52

Unit and Bravais Lattices

14 Bravais lattices (RELAX you dont have toknow 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


11/52

Density Tool Box

Volumetric Density of Materials

v= (NR)(MR)/ [(Vcell )(NA)] ( /cm3) Planar Atomic Density

p= (N ( )R intersected atoms )/A plane( /toms mm2)Linear Atomic Density

l= (N ( )R atom diameters on line )/L line( / )toms mm


12/52

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 withan atomic radius of 0.1253nm and ac/a ratio of 1.623.

a)Compute the volume of the unit cell

for Co V = 0.0664nm3b)Explain in your own words why the

c/a ratio in question a is not equalto the theoretical value of 1.633.

3.


13/52

Close-Packed Crystal Structures

FCC and HCP are both close-packed(APF 0.74)

Closely packed plane has the highest

planar densityPacking sequence differs

ABC ABC ABC FCC and AB AB AB HCP

FCC has more closely packed planes

than HCP and BCC does not have aclose packed plane


14/52

Summary: Tips and Further Examples

Exam questions will most likely be morefocused on calculations than the theory ofthis 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 TOBALANCE

Prove to yourself that the c/a ratio = 1.633 foran ideal HCP crystal

ALWAYS DRAW A PICTURE!!!


15/52

IMPERFECTIONS IN SOLIDSChapter 4


16/52

Crystal Structures

What do I need to know?Process of Solidification

Polycrystalline Metals (Sketch ingot

solidification mechanism)Single Crystals (Chozkralski process)

Defects in SolidsInfluencing factors

Types of Defects

Calculating Grain Size


17/52

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)


18/52

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


19/52

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 thatof paremt atoms)

Interstitial (Alloying atoms positions betweenparent atoms)

Excess alloying elements above saturation limit two-phase solid

Solid solution also depends on Hume-Rothery

criteria


20/52

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 ofvalence electrons

Schottky Defects (Ceramics) Missing cation AND anion

Frenkel Defects (Ceramics) Cation vacancy


21/52

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


22/52

Defects in Solids

Microdefects Line defectsTwo 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


23/52

Defects in Solids

Microdefects Surface defects Grain boundaries

Due to neighboring grains with differentgeometrical 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) orduring annealing (annealed twins)


24/52

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 partiallysolidified surface

Inclusions

Rogue particles that intrude material duringmanufacturing


25/52

Calculating Grain Size

N = 2n-1

N = Average number of grains per square inch(@100x)

n = Grain size number

Examples1.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 25grains per square inch are measured at amagnification of 75. (4.8)


26/52

Another Example

3.Calculate the fraction of atom sites that arevacant for Pbat its melting temperature of327C. Assume an energy for vacancyformation of 0.55eV/atoms. (2.41x10-5)


27/52

Summary: Tips

This chapter contains mainly theory but theconcepts are of utter importance

Expect a few graphs in the exam on thischapter

Number of calculations in this chapter willprobably be limited


28/52

HEAT TREATMENTChapter 9


29/52

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


30/52

Heat Treatment

Fe-C phase system

Only to be used forequilibrium coolingconditions

Phase diagram show

all of the reactions, compositions and

temperatures

Phases with

equilibrium cooling Ferrite, cementite

and pearlite


31/52

Heat Treatment

Fe-C phase system Examples

1.By using the Fe-C phase diagram, answer thefollowing questions applicable to a 0.5%Chypoeutectoid steel that is cooled slowly

from 950C to just below 727C.a)Calculate the amount of proeutectoid ferrite in

the steel (38.71%)

b)Calculate the amounts of eutectoid ferrite andeutectoid cementite in the steel (54.17%and 7.1%)

c)


32/52

Heat Treatment

Fe-C phase system Examples

2.Determine the chemical compositions of steelscontaining the following microstructuralcomponents 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)e)

f)


33/52

Heat Treatment

Non-equilibrium Phases Transformations Increase in cooling rate non-equilibrium

phases Bainite T from 250 550C

Fine dissemination of cementite in ferrite

matrix Good toughness, strength and hardness

properties Martensite Rapid cooling (quenching in water or

brine) C atoms dont have time to diffuse out of FCC

structure, are trapped in BCT cell Due to high amount of distortion associated

with phase transformation, hardness andstrength of martensite is very high

Temper treatment is often need to restoreductility of martensite

Tempering occurs below 650C and allows Cto precipitate out also known as


34/52

Heat Treatment

Types of Treatments Annealing

Steel is austenitised, cooled at equilbriumcondictions

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 liquidnitrogen

Martensite forms

Excessively high hardness


35/52

Heat Treatment

Types of TreatmentsStress Relieftreatment

Cold worked, quenched, welded or machinedcomponents experience stress fluctuationsdue to internal stresses

Heat component below eutectoidtemperature to relieve internal stress

Spherodising Process 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 ingeometry to mimic this shape

a)


36/52

Recap: Tips

You will most likely HAVE to use the lever ruleYou may expect some application type questions

If you have to design a heat treatment remember FITFOR PURPOSE

Cementite is highly brittle therefore any application

that requires good toughness, the amount in thematrix must be reduced propose spherodisingtreatment

With hypoeutectoid steels, pearlite can be a problemfor applications that require high strength CWcan resolve this to a degree

If a rapid quench (water, brine, oil or liquidnitrogen is proposed, you will propably endup with martensite tempering isessential)

Bainite can be produced by quenching in a molten Pbor salt bath and will give excellent mechanical

properties but time constraints have to be takeninto account


37/52

ELECTRICAL PROPERTIES OF

MATERIALSChapter 14


38/52

Electrical Properties of Materials

What do I need to know? Relationship between resistivity and

conductivity

3 Groups of electrical conductivity

Factors that influence resistivity andconductivity

Energy gap model for metals and isolators

Intrinsic semi-conductors

Extrinsic semi-conductors Dielectric character


39/52


Resistivity and Conductivity Inversely proportional to each other

Resistance of material is dependent on the typeof material, length and cross-sectional area of

component Ohms law can be used to determine Resistance

and the micro-law can be used to determineconductivity 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)


40/52


Factors that influence resisitivityTemperature

Linear relationship between resistivity andtemperature

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)


41/52


Energy gap model Metals

Small amount of energy needed to fill energy gapwith 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 energygap therefore conductivity is lower


42/52


Intrinsic Semi-Conductors (A-B-C) Pure, semi-conductors (Si and Ge)

Negative and positive electrons contribute to theconductivity of semi-conductors

With an increase in temperature , the

CONDUCTIVITY of the material increases forsemi-conductors since certain valence electronsare excited and their mobility increases

Extrinsic Semi-Conductors Differentiate between p- and n-type

Positive (Group 3 and 4 elements) and negativesemi-conductors (Group 4 and 5 elements)

By doping, impurities decrease the energy gapand through that, conductivity increases


43/52


Extrinsic Semi-Conductors n-Type

Group 5 substitutes one of Group 4 atoms

Majority of conductors are electrons minority arevacancies

p-Type Group 3 replaces one of Group 4 atoms

Majority of conductors are vacancies minority areelectrons

Dielectric Character

Ceramics, ionics and some polymers mostlyisolators but in some cases also semi-conductors

Capacitor chambers

Pizo-electric ceramics Ceramics that can convert electrical pulses to

mechanical vibrations or vice versa


44/52


45/52

MAGNETIC PROPERTIES OF

MATERIALSChapter 20


46/52

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,

antiferromagnetic and ferrimagnetic

Influence of Temperature on Ferromagnetics

Hysteresis Magnetisation and demagnetisation and hysteris

loops

Differentitate between hard and soft magnetics


47/52


Types of MagnetismTypes of magnetism

Diamagnetic - r < 1

Paramagnetic particles move toward external

magnetic field but loses their magnetism whenfield is removed

Ferromagnetic Magnetisationcan be permanentdue to the half-filled orbital of elements. It isessential that electrons in the 3d orbital areunpaired

Antiferromagnetic Elements have a magneticmoment but the a/d ratio is does not rangebetween 1.4 and 2.7 no magnetism

Ferrimagnetism Traces magnetic moments usually ionic bonds- spine of electrons are anti-

parrallel but not magnetic


48/52


Influence of Temperature onFerromagnetics At the Curie temperature, the 3d-electrons

orientations changes and the parallel spin ofthe electrons decrease

At this temperature the ferromagnetic nature ofthe material is destroyed


49/52


Hysteresis Domains on atomic level can be altered via a

solenoid causes parallel movement of 3d-electrons

Magnetisation occurs with ferromagnetic andferrimagnetic materials due

Domains (which have the correct orientation) startto grow at the expense of incorrect orientateddomains

Incorrect orientated domains can be rotated if theapplied field strength is strong enough

Demagnetisation will occur if the material isheated above its Curie temperature, byapplying an opposite directed field strength

or increasing the dislocation density of thematerial


50/52


HysteresisSo-called hysteresis loop

shows the life-cycle of a

ferromagnetic materialThe larger the area of the

curve, the easier

magnetisation is possible


51/52


Hard VS Soft magnetics

HARD SOFT

High Hc and Br values Easy to magnetise anddemagnetise

Large negative magnetic fieldneed to demagnetise

Needs high Bs value and highpearmeability

Small magnetic field tomagnetise

Induced current due to themagnetic field


52/52

Recap

The theory of this chapter is themost important since there arevery few types of calculations

that can be asked of youTypes of magnets is quite

important also the mechanism

that allows ferromagnetismB-H curve is very easy to

understand just follow yournotes

Documents

Nmc Exam Brush-up