NMC EXAM BRUSH-UP NOVEMBER 2009

Transcription

1 NMC EXAM BRUSH-UP NOVEMBER 2009

2 OVERVIEW 1. Crystal Structures 2. Solidification and Crystal Defects in Solids 3. Heat Treatments 4. Electrical Properties of Materials 5. Magnetic Properties of Materials

3 CRYSTAL STRUCTURES Chapter 3

4 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

5 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

6 MAIN METALLIC CRYSTAL STRUCTURES FCC (γ-fe, Au, Ag, Pt) Coordination Number = 12 Effective number of atoms = 4 Lattice Parameter a = 4R/ 2 Fraction of Unit Cell that is occupied by atoms by Volume APF = Volume of Atoms/Volume of Unit Cell APF = 0.74

7 MAIN METALLIC CRYSTAL STRUCTURES HCP (C, Cd, Co, Zn) Coordination Number = 12 Effective number of atoms = 6 Lattice Parameter a = 2R & c = 1.633a Fraction of Unit Cell that is occupied by atoms by Volume APF = Volume of Atoms/Volume of Unit Cell APF = 0.74

8 MAIN METALLIC CRYSTAL STRUCTURES Examples 1. Calculate the radius of an iridium atom. Ir has an FCC crystal structure and a density of 22.4g/cm 3 and an atomic weight of 192.2g/mol (R = 0.136nm)

9 POLYMORPHISM Crystal structure transformation in materials due to temperature or pressure change Fe Room Temperature - BCC Above 727 C FCC Above 1394 C Room Temperature Very high pressures and temperatures Diamond Cubic

10 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

11 DENSITY TOOL BOX Volumetric Density of Materials ρ v = (N R )(M R )/ [(V cell )(N A )] (g/cm 3 ) Planar Atomic Density ρ p = (N R(intersected atoms) )/A plane (atoms/mm 2 ) Linear Atomic Density ρ l = (N R(atom diameters on line) )/L line (atoms/mm)

12 DENSITY TOOL BOX Examples 1. 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 nm and a c/a ratio of a) Compute the volume of the unit cell for Co V = nm 3 b) Explain in your own words why the c/a ratio in question a is not equal to the theoretical value of

13 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

14 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 = for an ideal HCP crystal ALWAYS DRAW A PICTURE!!!

15 IMPERFECTIONS IN SOLIDS Chapter 4

16 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

17 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 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 DEFECTS IN SOLIDS Microdefects Point Defects Vacancies N c = 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

20 DEFECTS IN SOLIDS Microdefects Point Defects Hume-Rothery Criteria R parent and R alloy 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

21 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 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

23 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)

24 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

25 CALCULATING GRAIN SIZE N = 2 n-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? ( ) 2. Determine the ASTM grain size number if 25 grains per square inch are measured at a magnification of 75. (4.8)

26 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 )

27 Resistivity SUMMARY: TIPS 12 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 Strength Grain Size Creep Resistance Ndefects, CW, %Alloying Elements

28 HEAT TREATMENT Chapter 9

29 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 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

31 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%)

32 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)

33 HEAT TREATMENT Non-equilibrium Phases Transformations Increase in cooling rate non-equilibrium phases Bainite T from 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

34 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

35 HEAT TREATMENT Types of Treatments Stress Relieftreatment Cold worked, quenched, welded or machined components experience stress fluctuations due to internal stresses Heat component below eutectoid temperature 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 in geometry to mimic this shape

36 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

37 ELECTRICAL PROPERTIES OF MATERIALS Chapter 14

38 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

39 ELECTRICAL PROPERTIES OF MATERIALS 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)

40 ELECTRICAL PROPERTIES OF MATERIALS 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)

41 ELECTRICAL PROPERTIES OF MATERIALS 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

42 ELECTRICAL PROPERTIES OF MATERIALS 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 semiconductors (Group 4 and 5 elements) By doping, impurities decrease the energy gap and through that, conductivity increases

43 ELECTRICAL PROPERTIES OF MATERIALS 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

44 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)

45 MAGNETIC PROPERTIES OF MATERIALS Chapter 20

46 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

47 MAGNETIC PROPERTIES OF MATERIALS Types of Magnetism Types of magnetism Diamagnetic - μ r < 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

48 MAGNETIC PROPERTIES OF MATERIALS 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

49 MAGNETIC PROPERTIES OF MATERIALS 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

50 MAGNETIC PROPERTIES OF MATERIALS 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

51 MAGNETIC PROPERTIES OF MATERIALS Hard VS Soft magnetics HARD High Hc and Br values Large negative magnetic field need to demagnetise Small magnetic field to magnetise SOFT Easy to magnetise and demagnetise Needs high Bs value and high pearmeability Induced current due to the magnetic field

52 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