# Materials Exam 2

 elastic means _____ plastic means _____
 elastic means REVERSIBLE plastic means PERMANENT
 common states of stress: – simple tension – torsion – simple compression – bi-axial tension – hydrostatic compression match: pressurized tank, balanced rocks, cable, fish under water, drive shaft
 common states of stress: – simple tension – CABLE– torsion – DRIVE SHAFT– simple compression – BALANCED ROCKS – bi-axial tension – PRESSURIZED TANK – hydrostatic compression – FISH UNDER WATER
 units of stress
 units of stress:N/m^2lb/in^2
 units of strain
 units of strain:NONE
 what is hooke’s law?
 ? = ? ?   ? = stress  ? = modulus of elasticity (young’s modulus)  ? = strain
 Slope of stress strain plot (which is proportional to the elastic modulus) depends on what property?
 Slope of stress strain plot (which is proportional to the elastic modulus) depends on what property? BOND STRENGTH Steeper the slope, stronger the bond
 What is yield strength?
 yield strength: ?y, stress at which noticeable plastic deformation has occurred
 what is tensile strength?
 tensile strength: TS, MAXIMUM STRESS ON ENGINEERING STRESS-STRAIN CURVE metals – occurs when noticeable necking starts polymers – occurs when polymer backbone chains are aligned and about to break
 what is ductility? how can we measure it?
 ductility is PLASTIC TENSILE STRAIN at FAILURE. measure it by %EL or %RA
 what is toughness?
 toughness is the ENERGY to BREAK a unit volume of material. approximate by the area under the stress-strain curve
 What is hardness?
 hardness is RESISTANCE TO PERMANENTLY INDENTING THE SURFACE. large hardness means: – resistance to plastic deformation or cracking in compression. – better wear properties
 2 ways to measure hardness
 hardness measurement:ROCKWELLHB = BRINELL HARDNESS
 Why do “true” stress and strain exist?
 Cross sectional area changes when sample is stretched
 Stress measures _____Strain measures _____
 Stress measures LOADStrain measures DISPLACEMENT(both size independent measures)
 What is hardening?
 Hardening is an INCREASE in YIELD STRENGTH due to PLASTIC DEFORMATION.
 Elastic modulus is a _____ property.
 Elastic modulus is MATERIAL property • Critical properties depend largely on sample flaws (defects,etc.). Large sample to sample variability.
 Describe dislocation motion in metals.
 • Metals (Cu, Al): Dislocation motion EASIEST – non-directional bonding – close-packed directions for slip
 Describe dislocation motion in covalent ceramics.
 • COVALENT Ceramics (Si, diamond) Motion DIFFICULT – directional (angular) bonding
 Describe dislocation motion in ionic ceramics.
 • Ionic Ceramics (NaCl): Motion DIFFICULT – need to avoid nearest neighbors of like sign (- and +)
 Metals – plastic deformation occurs by ____ – an edgedislocation (extra half-plane of atoms) slides over adjacentplane half-planes of atoms.
 Metals – plastic deformation occurs by SLIP – an edgedislocation (extra half-plane of atoms) slides over adjacentplane half-planes of atoms.
 If dislocations ___ ____, plastic deformation doesn’t occur!
 If dislocations CAN’T MOVE, plastic deformation doesn’t occur!
 A dislocation moves along a slip plane in a slip direction______ to the dislocation line• The slip direction is the same as the _____ vector direction
 A dislocation moves along a slip plane in a slip directionPERPENDICULAR to the dislocation line (This is on his notes but would anyone like to explain that to me)• The slip direction is the same as the BURGERS vector direction
 Slip System – Slip plane – plane on which ______ slippage occurs• Highest _____ densities (and large interplanar spacings) – Slip directions – directions of _____ movement• Highest _____ densities
 Slip System – Slip plane – plane on which EASIEST slippage occurs• Highest PLANAR densities (and large interplanar spacings) – Slip directions – directions of DISLOCATION movement• Highest LINEAR densities
 What is meant by resolved shear stress? ?R
 What is meant by resolved shear stress? ?R results from applied tensile stresses — factors in the angles between the applied stress and the slip direction & slip plane normal Condition for dislocation motion: ?R > ?CRSS (Critcally resolved shear stress) ?R = ?*cos?*cos? ?R maximum at ? = ? = 45 degrees __________________________________________ ? = applied tensile stress = F/A ? = angle between F and slip direction (Fs) ? = angle between F and slip plane normal (As)
 For deformation to occur the _____ stress mustbe greater than or equal to the ______ stress
 For deformation to occur the APPLIED stress mustbe greater than or equal to the YIELD stress
 SLIP MOTION in POLYCRYSTALS • Polycrystals stronger than single crystals – ____ _____ are barriers to dislocation motion. • Slip planes & directions (?, ?) change from one grain to another. • ____ will vary from one grain to another. • The grain with the ______ ?R yields first. • Other (less favorably oriented) grains yield later.
 SLIP MOTION in POLYCRYSTALS • Polycrystals stronger than single crystals – GRAIN BOUNDARIES are barriers to dislocation motion. • Slip planes & directions (?, ?) change from one grain to another. • ?R will vary from one grain to another. • The grain with the LARGEST ?R yields first. • Other (less favorably oriented) grains yield later.
 Four Strategies for Strengthening:
 Four Strategies for Strengthening:1: Reduce Grain Size2: Form Solid Solutions3: Precipitation Strengthening4: Cold Work (Strain Hardening)
 Why would we want to Reduce Grain Size
 1. Reduce Grain Size – A strategy for STRENGTHENING – Grain boundaries are barriers to slip.– Barrier “strength“ increases with increasing angle of misorientation.– Smaller grain size: more barriers to slip.
 Why would we want to Form Solid Solutions
 2: Form Solid Solutions– strategy for STRENGTHENING – IMPURITY ATOMS DISTORT the lattice & generate lattice STRAINS.– These strains can act as barriers to dislocation motion.
 – Small impurities tend to concentrate at dislocations in regions of _____ strains – Large impurities tend to concentrate at dislocations in regions of _____ strains. – both ____ mobility of dislocations and ____ strength
 – Small impurities tend to concentrate at dislocations in regions of COMPRESSIVE strains – Large impurities tend to concentrate at dislocations in regions of TENSILE strains. – both REDUCE mobility of dislocations and INCREASE strength
 Alloying increases _____ strength and _____ strength
 Alloying increases YIELD strength and TENSILE strength
 Why is Precipitation Strengthening useful
 Precipitation Strengthening: Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum). basically means it just cuts thru the precipitate Either shears them or bows out around them
 What are the 4 common forming operations that reduce the cross sectional area during Cold Work (Strain Hardening)
 • Cold worDeformation at room temperature (for most metals).• Common forming operations reduce the cross-sectional area:– forging – rolling – drawing – extrusion %CW = Ao-Ad/Ao x 100
 How does the dislocation structure change during cold working?
 Dislocations entangle with one another during cold work.• Dislocation motion becomes more difficult.
 What happens to – yield strength – tensile strength – ductility as cold work is INCREASED
 as COLD WORK is INCREASED: – YIELD STRENGTH (?y) INCREASES.– TENSILE STRENGTH (TS) INCREASES.– DUCTILITY (%EL or %AR) DECREASES. can check values by looking at graph
 Effect of heat treating after cold work?
 Heat treating after cold work… • 1 hour treatment at Tanneal… decreases TS and increases %EL.• Effects of cold work are nullified!
 What are the 3 annealing stages?
 3 Annealing stages: 1. Recovery2. Recrystallization 3. Grain growth
 The 1st step of annealing, RECOVERY, does what?
 RECOVERY REDUCES the number of DISLOCATIONS
 The 2nd step of annealing, RECRYSTALLIZATION, does what?
 RECRYSTALLIZATION forms new grains that: – have LOW dislocation densities – are SMALL in size – consume and replace parent cold-worked grains
 The 3rd step of annealing, GRAIN GROWTH, does what?
 GRAIN GROWTH: at longer times, average grain size increases – small grains shrink and ultimately disappear – large grains continue to grow
 TR = RECRYSTALLIZATION TEMPERATURE = temperature at which recrystallization just reaches completion in ___.   0.3Tm < TR < 0.6Tm   For a specific metal/alloy, TR depends on: • %CW — TR DECREASES with _____ %CW • Purity of metal — TR DECREASES with _____ purity
 TR = RECRYSTALLIZATION TEMPERATURE = temperature at which recrystallization just reaches completion in 1 HOUR.   0.3Tm < TR < 0.6Tm   For a specific metal/alloy, TR depends on: • %CW — TR DECREASES with INCREASING %CW • Purity of metal — TR DECREASES with INCREASING purity
 STRENGTHENING IN METALS SUMMARY: • Dislocations are observed primarily in ____ and _____.• Strength is _____ by making dislocation motion difficult.• Strength of metals may be increased by: — ______ grain size — solid _____ strengthening — _______ hardening — ____ working• A cold-worked metal that is ____ treated may experience recovery, recrystallization, and grain growth – its properties will be altered.
 STRENGTHENING IN METALS SUMMARY: • Dislocations are observed primarily in METALS and ALLOYS.• Strength is INCREASED by making dislocation motion difficult.• Strength of metals may be increased by: — DECREASING grain size — solid SOLUTION strengthening — PRECIPITATE hardening — COLD working• A cold-worked metal that is HEAT treated may experience recovery, recrystallization, and grain growth – its properties will be altered.
 Ductile fracture– Accompanied by significant _____ deformation Brittle fracture– Little or no _____ deformation– Catastrophic
 Ductile fracture– Accompanied by significant PLASTIC deformation Brittle fracture– Little or no PLASTIC deformation– Catastrophic
 Ductile vs Brittle Failure (picture it)
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 Ductile fracture is usually ____ desirable than brittle fracture
 Ductile fracture is usually MORE desirable than brittle fracture
 • Ductile failure: — [one/many] piece(s) — [small/large] deformation(s) • Brittle failure: — [one/many] piece(s) — [small/large] deformation(s)
 • Ductile failure: — ONE piece — LARGE deformation • Brittle failure: — MANY pieces — SMALL deformations
 Brittle fracture surfaces: 1. ______ (between grains) 2. ______ (through grains)
 Brittle fracture surfaces: 1. INTERGRANULAR (between grains) 2. TRANSGRANULAR (through grains)
 Moderately ductile failure stages
 [image]
 TS engineering materials [>>/<<] TS perfect materials
 TS engineering materials << TS perfect materials
 DaVinci (500 yrs ago!) observed… — the longer the wire, the smaller the load for failure.• Reasons: — —
 DaVinci (500 yrs ago!) observed… — the longer the wire, the smaller the load for failure.• Reasons: — FLAWS CAUSE PREMATURE FAILURE — larger samples contain longer flaws!
 _____ are stress concentrators!
 FLAWS are stress concentrators!
 Cracks having [sharp/blunt] tips propagate easier than cracks having [sharp/blunt] tips [Brittle/ductile] materials have sharp crack tips[Brittle/ductile] materials have blunt crack tips
 Cracks having SHARP tips propagate EASIER than cracks having BLUNT tips BRITTLE materials have sharp crack tipsDUCTILE materials have blunt crack tips
 Energy balance on the crack:– ______ _____ energy-• Energy stored in material as it is elastically deformed.• This energy is released when the crack propagates.• Creation of new surfaces requires energy
 Energy balance on the crack:– ELASTIC STRAIN energy-• Energy stored in material as it is elastically deformed.• This energy is released when the crack propagates.• Creation of new surfaces requires energy
 what will cause a crack to propagate?
 Crack propagates if crack-tip stress (?m) EXCEEDS a critical stress (?c) the critical stress is that equation with a = half length of internal crack & specific surface energy & modulus of elasticity
 How do we find out how much energy goes into failing a material? (Hint: A mechanical way of measuring toughness)
 Impact Testing 1 example is charpy testing (actually kinda neat, maybe watch the online lecture, idk if its important tho) – severe testing case – makes material more brittle– decreases toughness
 Toughness can change depending on ductile-to-brittle transition temperature (DBTT)
 Toughness can change depending on ductile-to-brittle transition temperature there is a graph on the powerpoint Higher temp = more ductile Design strategy: stay above the DBTT (so things dont get brittle and break)
 Fatigue = failure under applied ____ stress. Stress varies with ____. • Key points: Fatigue… –can cause part failure, even though ?max < ?y. –responsible for ~ 90% of mechanical engineering failures.
 Fatigue = failure under applied CYCLIC stress. Stress varies with TIME. • Key points: Fatigue… –can cause part failure, even though ?max < ?y. –responsible for ~ 90% of mechanical engineering failures.
 Types of fatigue behavior: S = STRESS AMPLITUDE – Fatigue limit, Sfat: –no fatigue if S [] Sfat – For some materials, there is no fatigue limit! (Al)
 Types of fatigue behavior: S = STRESS AMPLITUDE – Fatigue limit, Sfat: –no fatigue if S < Sfat – For some materials, there is no fatigue limit! (Al)
 Cracks grow incrementally: — crack grows faster as • change in stress [increases/decreases] • _____ gets longer • loading freq. [decreases/increases].
 Cracks grow incrementally: — crack grows faster as • change in stress INCREASES • CRACK gets longer • loading freq. INCREASES.
 Improving fatigue life: 1. 2.
 Improving fatigue life: 1. Impose compressive surface stresses (to SUPPRESS SURFACE CRACKS FROM GROWING) 2. Remove stress concentrators (AVOID SHARP EDGES)
 Primary Creep: slope (creep rate) _____ with time. Secondary Creep: steady-statei.e., constant slope (??/?t). Tertiary Creep: slope (creep rate)_____ with time, i.e. _____ of rate.
 Primary Creep: slope (creep rate) DECREASES with time. Secondary Creep: steady-statei.e., constant slope (??/?t). Tertiary Creep: slope (creep rate)INCREASES with time, i.e. ACCELERATION of rate.
 Creep is temperature dependent Occurs at elevated temperature, T > ___ Tm (in K)
 Creep is temperature dependent Occurs at elevated temperature, T > 0.4 Tm (in K)
 Secondary Creep: Strain rate is _____ at a given T, ? strain rate increases with _____ T, ?
 Secondary Creep: Strain rate is CONSTANT at a given T, ? strain rate increases with INCREASING T, ?
 SUMMARY OF MATERIAL FAILURES: • Sharp corners produce ____ stress concentrations and premature failure. • Engineering materials [as strong / not as strong] as predicted by theory • Flaws act as stress concentrators that cause failure at stresses lower than _____ values. • Failure type depends on T and s : -For simple fracture (noncyclic s and T < 0.4Tm), failure stress decreases with: – ____ maximum flaw size, – ____ T, – ____ rate of loading. – For _____ (cyclic s): – cycles to fail decreases as Ds increases. – For _____ (T > 0.4Tm): – time to rupture decreases as s or T increases.
 SUMMARY OF MATERIAL FAILURES: • Sharp corners produce LARGE stress concentrations and premature failure. • Engineering materials NOT AS STRONG as predicted by theory • Flaws act as stress concentrators that cause failure at stresses lower than THEORETICAL values. • Failure type depends on T and s :-For simple fracture (noncyclic s and T < 0.4Tm), failure stress decreases with: – INCREASED maximum flaw size, – DECREASED T, – INCREASED rate of loading.(logical^) – For FATIGUE (cyclic s): – cycles to fail decreases as Ds increases. – For CREEP (T > 0.4Tm): – time to rupture decreases as s or T increases.
 Ceramics: compounds of ____ and _____ elements
 Ceramics: compounds of METALLIC and NON-METALLIC elements e.g. Al2O3, ZrO2, MgO, TiO2, SiC, WC, B4C, ZnS, TiB2 and carbon…
 _______ ceramics:– china, porcelain, bricks, tiles, glasses– many based on clays ______ ceramics – oxides, carbides, nitrides– used in electronics, computers, aerospace…
 TRADITIONAL ceramics:– china, porcelain, bricks, tiles, glasses– many based on clays TECHNICAL ceramics – oxides, carbides, nitrides– used in electronics, computers, aerospace…
 Ceramic Structure/Composition:• At least __ elements – sometimes more (BaTiO3, Ti3, SiC2)? [more/less] complex structures than metals• Determined by bonding:– ranges from ~purely ____ to ~purely ____– many exhibit combination of both
 Ceramic Structure/Composition:• At least 2 elements – sometimes more (BaTiO3, Ti3, SiC2)? MORE complex structures than metals• Determined by bonding:– ranges from ~purely IONIC to ~purely COVALENT– many exhibit combination of both
 Ceramic bonding: — Mostly _____ with varying degrees of ____. — % ionic character increases with difference in_____ of atoms.
 Ceramic bonding: — Mostly IONIC with varying degrees of COVALENCY. — % ionic character increases with difference inELECTRONEGATIVITY of atoms.
 (Ceramics) If Predominantly Ionic: • Consider as electrically charged ____, not ____• Positively charged ____ ions (cations), e.g. Ca2+– Given up valence e to _____ anions– Generally _____ than anions • Negatively charged ______ ions (anions),e.g. F– Accepted valence e from _____ cations– Generally _____ than cations
 (Ceramics) If Predominantly Ionic: • Consider as electrically charged IONS, not ATOMS • Positively charged METALLIC ions (cations), e.g. Ca2+– Given up valence e to NON-METALLIC anions– Generally SMALLER than anions • Negatively charged NON-METALLIC ions (anions),e.g. F– Accepted valence e from METALLIC cations– Generally LARGER than cations
 Key factors influencing ceramic structure: 1. 2.
 Key factors influencing ceramic structure: 1. Magnitude of electrical charges – crystal must be ELECTRICALLY NEUTRAL – required charge balance dictates ratio of # cations to anions 2. Relative sizes (IONIC RADII) of cations to anions – each cation prefers max # nearest neighbor anions and vise versa, this determines how ions pack in crystal
 Non-metallic anions (-) much ____ than metallic cations (+)• Pack like atoms in metals ______ metallic cations may thus fit into the spaces (i.e. the_____ sites)
 Non-metallic anions (-) much LARGER than metallic cations (+)• Pack like atoms in metals SMALLER metallic cations may thus fit into the spaces (i.e. the INTERSTITIAL sites)
 What are the two possible types of ceramic interstitial sites? Give some info on both.
 Tetrahedral Sites: 4 anion spheres as nearest neighbors • 3 in 1 plane + 1 in adjacent plane surround site• CN = 4• rcation/ranion -> .225 – .414 Octahedral Sites: 6 anion spheres as nearest neighbors • 3 each, in 2 planes• CN = 6• rcation/ranion -> .414 – .732
 What determines which site the cations will occupy?
 What determines which site the cations will occupy? 1. Size of sites:– does the cation fit in the site? 2. Stoichiometry:– if all of one type of site is full, the remainder have to go into other types of sites 3. Bond Hybridization:– % covalency
 What determines which site the cations will occupy? 1. Size: Must form stable structures:• ____ the # of oppositely charged ion neighbors.• CN related to cation-anion radius ratio, CN ____ with increasing rcation/ranion
 What determines which site the cations will occupy? 1. Size: Must form stable structures:• MAXIMIZE the # of oppositely charged ion neighbors.• CN related to cation-anion radius ratio, CN INCREASES with increasing rcation/ranion
 Crystal structure stoichiometry: If, for a specific ceramic, each unit cell has 6 cations and these prefer the (octahedral) OH sites, then:– __ will go into (octahedral) OH– __ will go into (tetrahedral) TD
 Crystal structure stoichiometry: If, for a specific ceramic, each unit cell has 6 cations and these prefer the (octahedral) OH sites, then:– 4 will go into (octahedral) OH– 2 will go into (tetrahedral) TD
 List the AX-Type crystal structures AX-Type means EQUAL # of cations and anions
 List the AX-Type crystal structures: 1. NaCl (ROCK SALT) 2. CsCl (Cesium Chloride)
 Describe NaCl (ROCK SALT) structure: – can be considered as 2 interpenetrating ___ lattices: one of cations and one of anions – CN = __– cations prefer ____ sites
 1. NaCl (ROCK SALT) – can be considered as 2 interpenetrating FCC lattices: one of cations and one of anions – CN = 6– cations prefer OCTAHEDRAL sites
 Describe CsCl (Cesium Chloride) structure: – Each ion has __ neighbors – ___ sites preferred – like a ___ but not ___ since there are two different kind of ions.
 Describe CsCl (Cesium Chloride) structure – Each ion has 8 neighbors – CUBIC sites preferred – like a BCC but not BCC since there are two different kind of ions.
 Describe Zinc Blende (ZnS) structure: – small Zn cations sit between __ S anions, CN = __– Happens when .225– Zn is in the ____ interstitial site – __ of the possible ______ sites are occupied – The bonding is usually highly ______
 Describe Zinc Blende (ZnS) structure: – small Zn cations sit between 4 S anions, CN = 4– Happens when .225– Zn is in the TETRAHEDRAL interstitial site – 4 of the possible TETRAHEDRAL sites are occupied – The bonding is usually highly COVALENT (ZnS, ZnTe, SiC).
 AmXp – Type Crystal Structures Charges on cations and anions are not equal (m and/or p ? 1) • Calcium Fluorite (CaF2) • rc/ra ~ 0.8, CN = __ • Cations in ____ sites • Similar to ____ but only half of cubic sites are occupied • Unit cells consist of 8 “sub-cubes”– e.g. UO2, ThO2, ZrO2, CeO2 • ______ structure – positions of cations and anions reversed– e.g. Li2O, Na2O
 AmXp – Type Crystal Structures Charges on cations and anions are not equal (m and/or p ? 1) • Calcium Fluorite (CaF2) • rc/ra ~ 0.8, CN = 8 • Cations in CUBIC sites • Similar to CsCl but only half of cubic sites are occupied • Unit cells consist of 8 “sub-cubes”– e.g. UO2, ThO2, ZrO2, CeO2 • ANTIFLUORITE structure – positions of cations and anions reversed– e.g. Li2O, Na2O
 AmBnXp Crystal Structures • >1 type of ___• ____ structure • Ex: complex oxide BaTiO3• ____ crystal structure• Ba2+ cations at corners• Ti4+ cation in center• O2- anions at each facecenter
 AmBnXp Crystal Structures • >1 type of CATION• PEROVSKITE structure • Ex: complex oxide BaTiO3• CUBIC crystal structure• Ba2+ cations at corners• Ti4+ cation in center• O2- anions at each facecenter
 Diamond (Similar to ___ ___) – A _____ form of carbon – ____ bonding of carbon• hardest material known• very ___ thermal conductivity – Large single crystals – gem stones– Small crystals – used to grind/cut other materials– Diamond thin films• hard surface coatings – used for cutting tools, medical devices, etc.
 Diamond (Similar to ZINC BLENDE) – A POLYMORPHIC form of carbon – TETRAHEDRAL bonding of carbon• hardest material known• very HIGH thermal conductivity – Large single crystals – gem stones– Small crystals – used to grind/cut other materials– Diamond thin films• hard surface coatings – used for cutting tools, medical devices, etc.
 ______: – A POLYMORPHIC form of carbon – Layered structure – parallel hexagonal arrays of carbonatoms — Weak Van der Waal’s forces between layers– Planes slide easily over one another — good lubricant
 GRAPHITE: – A POLYMORPHIC form of carbon – Layered structure – parallel hexagonal arrays of carbonatoms — Weak Van der Waal’s forces between layers– Planes slide easily over one another — good lubricant
 2 more polymorphic forms of carbon: • _____ – spherical cluster of 60 carbon atoms, C60– Like a soccer ball• _____ – sheet of graphite rolled into a tube– Ends capped with fullerene hemispheres
 2 more polymorphic forms of carbon: • FULLERENES – spherical cluster of 60 carbon atoms, C60– Like a soccer ball• CARBON NANOTUBES – sheet of graphite rolled into a tube– Ends capped with fullerene hemispheres
 Point defects in ceramics: • _____ exist in ceramics for both cations and anions• ______ exist for cations, but are not normally observed for anions because anions are large relative to the _____ sites
 Point defects in ceramics: • VACANCIES exist in ceramics for both cations and anions• INTERSTITIALS exist for cations, but are not normally observed for anions because anions are large relative to the INTERSTITIAL sites (not saying its impossible, just far less likely)
 • ____ Defect — a paired set of cation and anion vacancies.• ____ Defect — a cation vacancy-cation interstitial pair.
 • SHOTTKY Defect — a paired set of cation and anion vacancies.• FRENKEL Defect — a cation vacancy-cation interstitial pair.
 _________ must be maintained when impurities are present
 ELECTRONEUTRALITY (charge balance) must be maintained when impurities are present When there is a SUBSTITUTIONAL cation or anion, the substitution must have equal charge to what is being taken away (so if 2 Na+ is replaced with 1 Ca2+, there will be a substitutional cation and a cation vacancy)
 Diffusion of ____ is more complicated than in ____ due to the motion of two ionic species with opposite charges. • Diffusion usually occurs by a ___ mechanism.• Local charge neutrality must be maintained, so diffusion of an ion must be accompanied by the diffusion of an ion(s) with equal and opposite charge.• Therefore diffusion is rate limited by the ____ moving ion.
 Diffusion of CERAMICS is more complicated than in METALS due to the motion of two ionic species with opposite charges. • Diffusion usually occurs by a VACANCY mechanism.• Local charge neutrality must be maintained, so diffusion of an ion must be accompanied by the diffusion of an ion(s) with equal and opposite charge.• Therefore diffusion is rate limited by the SLOWER moving ion.
 Ceramic materials are ____ brittle than metals: Why is thisso? • Consider mechanism of deformation: – In crystalline, by ____ motion– In highly ionic solids, ___ motion is difficult • few slip systems • resistance to motion of ions of like charge (e.g., anions) past one another
 Ceramic materials are MORE brittle than metals: Why is thisso? • Consider mechanism of deformation: – In crystalline, by DISLOCATION motion– In highly ionic solids, DISLOCATION motion is difficult • few slip systems • resistance to motion of ions of like charge (e.g., anions) past one another
 In general, why do ceramics fail?
 In general, ceramics fail due to BRITTLE FRACTURE before any plastic deformation occurs. • Cracks propagate either transgranularly or intergranularly. • Initiation sites are from small flaws that are present in the material which serve as stress risers. • Because there is no mechanisms for plastic deformation, crack blunting cannot occur, leading to cracks propagating un-impeded above a certain stress. Sometimes cracks can propagate slowly at stresses lower thatdefined by the equation above (Kic = Y(delta)(sqrt(pi*a)) in moist environments leading to stress corrosion process at the crack tip.
 Chalk cut from bulk material shows how different pieces could have different flaws.• This leads to a statistical variation in the tensile strength.• This distribution of cracks also creates a ____ dependence on strength since more cracks would be in a larger sample.• A brittle rod will be stronger in ___ than in ____.
 Chalk cut from bulk material shows how different pieces could have different flaws.• This leads to a statistical variation in the tensile strength.• This distribution of cracks also creates a VOLUME dependence on strength since more cracks would be in a larger sample.• A brittle rod will be stronger in BENDING than in TENSION. (tension is applied to the entire sample and cracks propagate in tension… with bending there is tension at some points, compression in others, so the tension is being applied to a smaller amount of the sample, making it less likely to have these flaws.
 ____ _____ Ps(Vo) is the fraction of identical samples, with volume Vo which survive loading to a tensile stress, (Delta) who can you thank for this
 SURVIVAL PROBABILITY Ps(Vo) is the fraction of identical samples, with volume Vo which survive loading to a tensile stress, (Delta) as stress increases, Survival probability decreases THANKS WEIBULL
 Modulus of rupture, DELTAr is larger than tensile strength for two reasons:
 Modulus of rupture, DELTAr is larger than tensile strength for two reasons:• In bending half the beam is subjected to compression leadingto flaws closing up.• The peak tensile stress is below the center load with stresslower in other areas leading to a small volume beingsubjected to high tensile stress.
 Give 3 examples of applications for ceramics
 Give 3 examples of applications for ceramics 1. DIE BLANKS 2. CUTTING TOOLS3. SENSORS
 What are refractories?
 Refractories are materials to be used at HIGH TEMPERATURES.
 Materials for Automobile engines: Advantages:– Operate at high temperatures – high efficiencies– Low frictional losses– Operate without a cooling system– Lower weights than current engines Disadvantages:– Ceramic materials are brittle– Difficult to remove internal voids (that weaken structures)– Ceramic parts are difficult to form and machine • Potential candidate materials: Si3N4, SiC, & ZrO2• Possible engine parts: engine block & piston coatings
 Probably wont ask this directly as a question?, but it has important observations about applying ceramics: Materials for Automobile engines: Advantages:– Operate at high temperatures – high efficiencies– Low frictional losses– Operate without a cooling system– Lower weights than current engines Disadvantages:– Ceramic materials are brittle– Difficult to remove internal voids (that weaken structures)– Ceramic parts are difficult to form and machine • Potential candidate materials: Si3N4, SiC, & ZrO2• Possible engine parts: engine block & piston coatings
 SUMMARY OF CERAMICS: • Interatomic bonding in ceramics is ____ and/or ____.• Ceramic crystal structures are based on: — maintaining ____ ____ — cation-anion ____ ___ .• Imperfections — Atomic point: vacancy, interstitial (cation), Frenkel, Schottky — Impurities: substitutional, interstitial — Maintenance of charge neutrality• Room-temperature mechanical behavior – flexural tests — ____-____; measurement of elastic modulus — brittle fracture; measurement of ____ ____
 SUMMARY OF CERAMICS: • Interatomic bonding in ceramics is IONIC and/or COVALENT.• Ceramic crystal structures are based on: — maintaining CHARGE NEUTRALITY — cation-anion RADII RATIOS .• Imperfections — Atomic point: vacancy, interstitial (cation), Frenkel, Schottky — Impurities: substitutional, interstitial — Maintenance of charge neutrality• Room-temperature mechanical behavior – flexural tests — LINEAR-ELASTIC; measurement of elastic modulus — brittle fracture; measurement of FLEXURAL MODULUS.
 what is a polymer?
 poly = many mer = repeat unit
 • Originally natural polymers were used– Wood – Rubber– Cotton – Wool– Leather – Silk • Oldest known uses– Rubber balls used by Incas– Woven cloth
 idk how to make this a question but its probably good to kno • Originally natural polymers were used– Wood – Rubber– Cotton – Wool– Leather – Silk • Oldest known uses– Rubber balls used by Incas– Woven cloth
 Polymers since WWII • Most plastics, rubbers, fibers today are _____ polymers…– Some of the first? ___ & ___• Since WW II, synthetic polymers have revolutionized thematerials field:– ____ to produce– ____ can be controlled as desired– Often superior to natural materials– Supplanted metals & wood in some applications: • lower cost & superior properties• _____ concerns…recycling, resource usage &conservation…sustainability
 Polymers since WWII • Most plastics, rubbers, fibers today are SYNTHETIC polymers…– Some of the first? BAKELITE & NYLON• Since WW II, synthetic polymers have revolutionized thematerials field:– CHEAP to produce– PROPERTIES can be controlled as desired– Often superior to natural materials– Supplanted metals & wood in some applications: • lower cost & superior properties• ENVIRONMENTAL concerns…recycling, resource usage &conservation…sustainability
 Most polymers are ______ – i.e., made up of H and C
 Most polymers are HYDROCARBONS – i.e., made up of H and C
 Polymer composition: • Strong covalent _____-molecular bonds– 4 valence e- per C atom– 1 valence e- per H atom– Recall CH4…4 x single covalent bonds… • Weaker hydrogen and van der Waals ____molecularbonds:– Thus polymers have ____ melting/boiling points
 Polymer composition: • STRONG covalent INTRA-molecular bonds– 4 valence e- per C atom– 1 valence e- per H atom– Recall CH4…4 x single covalent bonds… • WEAKER hydrogen and van der Waals INTERmolecularbonds:– Thus polymers have LOW melting/boiling points
 Saturated hydrocarbons– Each carbon singly bonded to ___ other atoms.– All covalent bonds are ____ bonds.– Example:• Ethane, C2H6– No new atoms can be joined without removal of others already bonded.
 Saturated hydrocarbons– Each carbon singly bonded to FOUR other atoms.– All covalent bonds are SINGLE bonds.– Example:• Ethane, C2H6– No new atoms can be joined without removal of others already bonded.
 Unsaturated Hydrocarbons• Share _ or _ pairs of electrons.• Double & triple bonds somewhat _____ – can form new bonds• Each carbon not bonded to 4 other atoms– Other atoms or groups can bond to the originalmolecule
 Unsaturated Hydrocarbons• Share 2 or 3 pairs of electrons.• Double & triple bonds somewhat UNSTABLE – can form new bonds• Each carbon not bonded to 4 other atoms– Other atoms or groups can bond to the originalmolecule Double bond found in ethylene or ethene – C2H4Triple bond found in acetylene or ethyne – C2H2
 Polymer molecules are very large – _____– ____ bonds within each molecule– Backbone = string of __ atoms singly bonded to each other– Remaining 2 valence electrons are used for side-bonds to atoms and radicals– ex. Ethylene (C2H4) is a gas at room temp but can be a polymer building block
 Polymer molecules are very large – MACROMOLECULES– COVALENT bonds within each molecule– Backbone = string of C, CARBON atoms singly bonded to each other– Remaining 2 valence electrons are used for side-bonds to atoms and radicals– ex. Ethylene (C2H4) is a gas at room temp but can be a polymer building block
 Chemistries other than pure HC are possible:– Polymerize (CF2=CF2) –> PTFE (Teflon®)– Fluorocarbons– PVC which is similar to ethylene where 1 in 4 H replacedwith Cl General forumla: -CH2-CHR- R = ____ or _____
 Chemistries other than pure HC are possible:– Polymerize (CF2=CF2) –> PTFE (Teflon®)– Fluorocarbons– PVC which is similar to ethylene where 1 in 4 H replacedwith Cl General forumla: -CH2-CHR- R = ATOMS or GROUPS
 Chain Repeat Units– If all the same: _____– If different: _____
 Chain Repeat Units– If all the same: HOMOPOLYMERS– If different: COPOLYMERS
 _______ polymerization – Monomer unit daisy-chained 1 at a time to form linearmacromolecule– 3 stages: initiation, propagation & termination – Chain forms via sequential addition of monomer units– Active site (unpaired e-) transfers to each successive endmonomer • Termination:– Active ends of 2 propagating chains may link: ____– Dead chains are possible, e.g. via H atom: _____
 FREE RADICAL polymerization – Monomer unit daisy-chained 1 at a time to form linearmacromolecule– 3 stages: initiation, propagation & termination – Chain forms via sequential addition of monomer units– Active site (unpaired e-) transfers to each successive endmonomer • Termination:– Active ends of 2 propagating chains may link: COMBINATION– Dead chains are possible, e.g. via H atom:DISPROPORTIATION
 ________ polymerization – Intermolecular reactions involving >1 monomer species giving off a low MW condensate (e.g. H2O)– E.g. Nylon 6,6
 CONDENSATION polymerization – Intermolecular reactions involving >1 monomer species giving off a low MW condensate (e.g. H2O)– E.g. Nylon 6,6
 ______ ______ Polymer (3-D Crosslinking)– Phenol-Formaldehyde (Phenolic)
 THERMOSET NETWORKED Polymer (3-D Crosslinking)– Phenol-Formaldehyde (Phenolic)
 Describe Isomerism
 ISOMERISM: – Two compounds with same chemical formula can have quite different structures ex. C8H18 can be normal octane or 2,4-dimethylhexane
 Thermoplastics vs. Thermosets
 THERMOPLASTICS• Linear polymers with 2 links/mer• They can be softened (and melted) repeatedly by raising the temperature.• Weak secondary bonds between chains.• Strong bonds within chains. THERMOSETS• Crosslinked with 3 links/mer• Rigid 3-D molecules• After a thermoset is formed, it CANNOT be reshaped or remelted.
 Not all chains in a polymer are of the same ____ — i.e., there is a distribution of molecular weights
 Not all chains in a polymer are of the same LENGTH — i.e., there is a distribution of molecular weights
 • MW can be defined in several ways:• Number averaged,
 DP = average number of repeat units per chain
 Molecular Shape (or _______):– Chain bending and twisting are possible by ____ of carbon atoms around their chain single bonds– note: not necessary to ____ chain bonds to alter molecular shape
 Molecular Shape (or CONFORMATION):– Chain bending and twisting are possible by ROTATION of carbon atoms around their chain single bonds– note: not necessary to BREAK chain bonds to alter molecular shape
 Chain ________ = r r << total chain length Large # of chains:• Each may bend, coil or kink• Extensive chain entanglement• Governs some mechanical properties,e.g. in elastomers
 Chain END-TO-END DISTANCE = r r << total chain length Large # of chains:• Each may bend, coil or kink• Extensive chain entanglement• Governs some mechanical properties,e.g. in elastomers
 Physical characteristics of polymers depend on:1. 2.3.
 Physical characteristics of polymers depend on:1. MOLECULAR WEIGHT2. SHAPE3. STRUCTURE
 List the 4 covalent chain configurations in order of increased strength
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 • Linear polymer: ? Repeat units joined end-to-end? Long flexible chains (“spaghetti”)? Extensive _____ and ____ bonds betweenchains? PE, PVC, PS, MPPA, Nylon, fluorocarbons
 • Linear polymer: ? Repeat units joined end-to-end? Long flexible chains (“spaghetti”)? Extensive VAN DER WAALS and HYDROGEN bonds between chains? PE, PVC, PS, MPPA, Nylon, fluorocarbons
 Branched polymer: • Side-branch chains connected to main chain• Chain packing efficiency ____• ____ density• LDPE
 Branched: • Side-branch chains connected to main chain• Chain packing efficiency REDUCED• LOWER density• LDPE
 • Crosslinked polymer: ? Adjacent chains joined by ____ bonds? Occurs during ____? Non-reversible? Rubbers (via vulcanization using S atoms)
 • Crosslinked polymer: ? Adjacent chains joined by COVALENT bonds? Occurs during SYNTHESIS? Non-reversible? Rubbers (via vulcanization using S atoms)
 Network polymer: • Monomers forming __ or more covalent bonds• Form 3-D networks• Epoxies, polyurethanes, phenol-formaldehyde
 Network polymer: • Monomers forming 3 or more covalent bonds• Form 3-D networks• Epoxies, polyurethanes, phenol-formaldehyde
 ______ are mirror images – can’t superimpose without breaking a bond
 STEREOISOMERS are mirror images – can’t superimpose without breaking a bond
 Configurations – to change must ____ ____
 Configurations – to change must BRAEAK BONDS
 ______ – stereoregularity or spatial arrangement of R units along chain_____ – all R groups on same side of chain_____ – R groups alternate sides_____ – R groups randomly positioned
 TACTICITY – stereoregularity or spatial arrangement of R units along chainISOTACTIC – all R groups on same side of chainSYNDIOTACTIC – R groups alternate sides ATACTIC – R groups randomly positioned
 Geometrical isomerism within repeat units with ____bonds between chain C atoms: – cis: H atom and CH3 group on the same side of the chain– trans: H atom and CH3 group on opposite sides of the chain Isomers have the same or different properties? Can you convert from cis to trans?
 Geometrical isomerism within repeat units with DOUBLEbonds between chain C atoms: – cis: H atom and CH3 group on the same side of the chain– trans: H atom and CH3 group on opposite sides of the chain Isomers have the same or different properties? DIFFERENTCan you convert from cis to trans? NO
 What are 4 possible arrangements for copolymers (two or more monomers polymerized together)
 Crystallinity in Polymers: • Involves _____ as opposed to atoms in metals & ceramics• More complex atomic arrangements• Crystallinity due to packing of polymer chains to generate anordered atomic array.• Complex ____ ____• Chain twisting, coiling & kinking all promote disorder: ? Disorder = amorphous regions ? Most polymers thus only partially crystalline ? Combination of ____ + ____ regions • Range from 100% amorphous to ~95% crystalline • Density = fn.(% crystallinity)
 Crystallinity in Polymers: • Involves MOLECULES as opposed to atoms in metals & ceramics• More complex atomic arrangements• Crystallinity due to packing of polymer chains to generate anordered atomic array.• Complex UNIT CELLS• Chain twisting, coiling & kinking all promote disorder: ? Disorder = amorphous regions ? Most polymers thus only partially crystalline ? Combination of CRYSTALLINE + AMORPHOUS regions • Range from 100% amorphous to ~95% crystalline • Density = fn.(% crystallinity)
 How does an increase in % crystallinity of a polymer affect: TS and E How does heat treating affect % crystallinity?
 How does an increase in % crystallinity of a polymer affect: TS (INCREASE) and E (INCREASE) How does heat treating affect % crystallinity? It causes crystalline regions to grow and % crystallinity to INCREASE
 Single polymer crystals – only for ____ and carefully ____ growth rates
 Single polymer crystals – only for SLOW and carefully CONTROLLED growth rates
 Some semicrystalline polymers form spherulite structures• Alternating chain-folded crystallites and amorphous regions• Spherulite structure for relatively ___ growth rates– has a ______ site at the center
 Some semicrystalline polymers form SPHERULITE structures• Alternating chain-folded crystallites and amorphous regions• Spherulite structure for relatively RAPID growth rates– has a NUCLEATION site at the center
 Strengths of polymers ~__% of those for metals
 Strengths of polymers ~10% of those for metals
 Picture the deformation of BRITTLE crosslinked and network polymers
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 Picture the deformation of SEMICRYSTALLINE (PLASTIC) polymers
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 Picture the deformation of ELASTOMERS
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 What does the stress strain curve look like for BRITTLE polymers, PLASTIC polymers, and ELASTOMERS
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 Influence of T on Thermoplastics: Decreasing T… — ____ E— ____ TS — ____ %EL
 Influence of T on Thermoplastics: Decreasing T… — INCREASES E— INCREASES TS — DECREASES %EL
 Influence of strain rate on Thermoplastics: Increases strain rate… — ____ E— ____ TS — ____ %EL
 Influence of strain rate on Thermoplastics: Increases strain rate… — INCREASES E— INCREASES TS — DECREASES %EL
 What is Tm and Tg?
 Tm = MELTING TEMP Tg = GLASS TRANSITION TEMP
 What factors affect Tm and Tg?
 What factors affect Tm and Tg? Both Tm and Tg increase with increasing chain stiffness • Chain stiffness increased by presence of1. Bulky sidegroups2. Polar groups or sidegroups3. Chain double bonds and aromatic chain groups • Regularity of repeat unit arrangements – affects Tm only
 How can we increase chain stiffness, in turn increasing Tm and Tg?
 Chain stiffness increased by presence of 1. Bulky sidegroups2. Polar groups or sidegroups3. Chain double bonds and aromatic chain groups
 Regularity of repeat unit arrangements – affects [Tm/Tg] only
 Regularity of repeat unit arrangements – affects Tm only
 Craze formation prior to cracking: – during crazing, plastic deformation of spherulites – and formation of ____ and ____ bridges
 Craze formation prior to cracking: – during crazing, plastic deformation of spherulites – and formation of MICROVOIDS and FIBRILLAR bridges
 Processing of Polymers: _______– Can be reversibly cooled & reheated, i.e. recycled– Heat until soft, shape as desired, then cool– Examples: polyethylene, polypropylene, polystyrene. ________– When heated forms a molecular network (chemical reaction)– Degrades (doesn’t melt) when heated– Prepolymer molded into desired shape, then chemical reaction occurs– Examples: urethane, epoxy
 Processing of Polymers: THERMOPLASTIC– Can be reversibly cooled & reheated, i.e. recycled– Heat until soft, shape as desired, then cool– Examples: polyethylene, polypropylene, polystyrene. THERMOSET– When heated forms a molecular network (chemical reaction)– Degrades (doesn’t melt) when heated– Prepolymer molded into desired shape, then chemical reaction occurs– Examples: urethane, epoxy
 List 4 methods of processing polymers and which type (thermoplastic or thermoset) it applies to
 List 4 methods of processing polymers and which type (thermoplastic or thermoset) it applies to: 1. COMPRESSION MOLDINGthermoplastics and thermosets 2. INJECTION MOLDING thermoplastics and some thermosets 3. EXTRUSION thermoplastics BLOWN-FILM EXTRUSION for plastic bags*** we watched a video on this, might actually be important
 polymer fibers are formed by _____. – extrude polymer through a spinneret (a die containing many small orifices)– the spun fibers are drawn under tension– leads to highly aligned chains– fibrillar structure primary use in textiles
 polymer fibers are formed by SPINNING. – extrude polymer through a spinneret (a die containing many small orifices)– the spun fibers are drawn under tension– leads to highly aligned chains– fibrillar structure primary use in textiles
 polymer fiber characteristics
 polymer fiber characteristics 1. high tensile strengths 2. high degrees of crystallinity 3. structures containing polar groups
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