MECH270 Handout Weeks 5-7 (Material Science & Engineering) PDF

Recap Dislocations are the primary mode of plastic deformation in metals and alloys. Strength is therefore increased by making dislocation motion difficult. Particular ways to impede dislocation motion and increase strength are: -- solid solution strengthening -- particle strengthen...

Recap Dislocations are the primary mode of plastic deformation in metals and alloys. Strength is therefore increased by making dislocation motion difficult. Particular ways to impede dislocation motion and increase strength are: -- solid solution strengthening -- particle strengthening -- cold-work -- decrease grain size Dislocation behavior seems to change with temperature…  MECH 270 Material Science and Engineering Diffusion in Solids Chapter 5 ISSUES TO ADDRESS... How does diffusion occur? Why is it an important part of processing? How can the rate of diffusion be predicted for some simple cases? How does diffusion depend on structure and temperature?  Diffusion Diffusion - Mass transport by atomic motion Mechanisms Gases & Liquids – random (Brownian) motion Solids – vacancy diffusion or interstitial diffusion  Diffusion Interdiffusion: In an alloy, atoms tend to migrate from regions of high conc. to regions of low conc. Initially After some time Adapted from Figs. 5.1 and 5.2, Callister 7e.  Point defects in crystals substitutional interstitial vacancy self-interstitial solute atoms atom (SIA)  Diffusion Mechanisms Interstitial diffusion – smaller atoms can diffuse between atoms. Adapted from Fig. 5.3 (b), Callister 7e. More rapid than vacancy diffusion Interstitial diffusion  Interstitial diffusion flux to left/right jump frequency  J i  1 i n1 Number of atoms 6  concentration gradient J i  1 i n2 6     net flux 1 2 Ci Ji  Ji  Ji   i 6 x Ci J i   Di x jump distance Fick’s first law – steady state diffusion diffusion coefficient – units of m2s-1  Diffusion Self-diffusion: In an elemental solid, atoms also migrate. Label some atoms After some time C C A D A D B B Vacancy and self-diffusion For an atom to self diffuse it needs a vacancy next to it.... Vacancy Diffusion: atoms exchange with vacancies applies to both self diffusion and diffusion of substitutional impurity atoms rate depends on: -- number (actually density) of vacancies -- activation energy to exchange -- frequency of jumping increasing elapsed time  Substitutional diffusion in dilute binary alloys (solute concentrations < 1%) the rate of diffusion of the solute is proportional to the vacancy concentration but the equilibrium vacancy concentration can be increased if the solutes are attracted to the vacancies attraction of the solute and vacancy increases probability that the solute has a vacancy in an adjacent site the energy barrier for the jump of the solute atom to the vacant site may be different for a solute than for the host atom Interstitial diffusion more rapid than vacancy dependent diffusion  Diffusion distance - diffusion of a single atom or vacancy occurs by random walk - the direction of each jump is uncorrelated with the previous one - statistically, after n jumps, it moves a distance n1/2 jumps from its initial position, i.e., after a time t the distance from the initial position is: r   t r  k Dt ~ 6 Dt Eqn 5.6b in Callister  Example: Chemical Protective Clothing (CPC) Methylene chloride is a common ingredient of paint removers. Besides being an irritant, it also may be absorbed through skin. When using this paint remover, protective gloves should be worn. If butyl rubber gloves (0.04 cm thick) are used, what is the breakthrough time (tb), i.e., how long could the gloves be used before methylene chloride reaches the hand? Data – diffusion coefficient of methylene chloride in butyl rubber at room temperature: D = 110x10-8 cm2/s  CPC Example (cont.) Solution – assuming linear conc. gradient glove Time t, to travel distance r C1 r2 paint skin t remover 6D C2 x1 x2 r  x 2  x1  0.04 cm D = 110 x 10-8 cm2/s ( 0. 04 cm) 2 t -8 2  240 s  4 min ( 6 )(110 x 10 cm /s) Time required for breakthrough ca. 4 min  Processing Using Diffusion Case Hardening: -- Diffuse carbon atoms into the host iron atoms at the surface. -- Example of interstitial diffusion is a case hardened gear. Result: The presence of C atoms makes iron (steel) harder.  Consequences of Unexpected Diffusion! Oxygen diffusion Along grain boundaries Alloy 600 in high T water Steam Generator in Nuclear Plant (280°C-320°C) Environment and Temperature were the key! Diffusion is Thermally Activated Energy needed energy barrier Called the activation energy (Q) Energy Q Vacancy Atom Distance  Diffusion and Temperature Diffusion coefficient increases with increasing T.  Qd  D  Do exp   RT  D = diffusion coefficient [m2/s] Do = pre-exponential [m2/s] Qd = activation energy [J/mol or eV/atom] R = gas constant [8.314 J/mol-K] T = absolute temperature [K]  Diffusion and Temperature D has exponential dependence on T, so we plot on log scale 1500 1000 600 T(C) 300 10-8 D (m2/s) Dinterstitial >> Dsubstitutional C in -Fe Al in Al 10-14 C in -Fe Fe in -Fe Fe in -Fe 10-20 1000 K/T  Q  0.5 1.0 1.5 D  D0 exp   RT  Q ln D  ln D0  RT  Diffusion How do we measure the amount or rate of diffusion? moles (or mass) diffusing mol kg J  Flux   or surface area time  2 cm s m 2s Measured empirically – Make thin film (membrane) of known surface area – Impose concentration gradient – Measure how fast atoms or molecules diffuse through the membrane M= M l dM mass J slope J   diffused At A dt time  Steady-State Diffusion Rate of diffusion independent of time dC Flux proportional to concentration gradient = dx C1 C1 Fick’s first law of diffusion dC C2 C2 J  D dx x1 x2 x D  diffusion coefficient dC  C C 2  C1 (at that temperature) if linear   dx x x 2  x1  Non-steady State Diffusion Most practical situations are not steady state …. the concentration of diffusing species is then a function of both time and position C = C(x,t) In this case Fick’s Second Law is used C C 2 Fick’s Second Law D 2 t x  Non-steady State Diffusion Copper diffuses into a bar of nickel. Surface conc., C s of Cu atoms bar pre-existing conc., Co of copper atoms Cs Adapted from Fig. 5.5, Callister 7e. B.C. at t = 0, C = Co for 0  x   at t > 0, C = CS for x = 0 (const. surf. conc.) C = Co for x =   Solution: C x , t   C o  x   1  erf   Cs  Co  2 Dt  C(x,t) = Conc. at point x at CS time t erf (z) = error function 2 z C(x,t)  y 2  e dy  0 Co erf(z) values are given in Table 5.1 of Callister  Non-steady State Diffusion Sample Problem: An FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere that gives a surface carbon concentration constant at 1.0 wt%. If after 49.5 h the concentration of carbon is 0.35 wt% at a position 4.0 mm below the surface, determine the temperature at which the treatment was carried out. C ( x, t )  C o  x   1  erf   Cs  Co  2 Dt   C ( x ,t )  C o  x  Solution (cont.):  1  erf   Cs  Co  2 Dt  – t = 49.5 h x = 4 x 10-3 m – Cx = 0.35 wt% Cs = 1.0 wt% – Co = 0.20 wt% C ( x, t )  Co 0. 35  0. 20  x    1  erf    1  erf ( z ) Cs  Co 1. 0  0. 20  2 Dt   erf(z) = 0.8125 We must now determine the value of z for which the error function is 0.8125. If using a table, an interpolation is necessary as follows z erf(z) z  0. 90 0. 8125  0. 7970  0. 95  0. 90 0. 8209  0. 7970 0.90 0.7970 z 0.8125 z  0.93 0.95 0.8209 Now solve for D x x2 z D  2 Dt 4 z 2t  x2  3 2 ( 4 x 10 m) 1h D     2. 6 x 10 11 m 2 /s  4 z 2t  ( 4 )( 0. 93 ) 2 ( 49. 5 h) 3600 s   Now to solve for the temperature at which D has this value…  Q  D  D0 exp   RT  Q ln D  ln D0  RT Q T Rln D0  ln D   Qd T  R ( ln D o  ln D ) from Table 5.2, for diffusion of C in FCC Fe Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol 148 ,000 J/mol  T  ( 8. 314 J/mol - K)(ln 2. 3 x10  5 m 2 /s  ln 2. 6 x10 11 m 2 /s) T = 1300 K = 1027ºC  Summary In solids diffusion occurs by the random motion of point defects For interstitial atoms (self or solute) this depends on the energy for migration from one interstitial site to another between lattice atoms For host atoms and substitutional solute atoms it depends on the energy for migration of vacancies and the vacancy concentration (i.e., the vacancy formation energy). By random walk, the mean distance that an atom diffused is proportional to the square root of the product of the diffusion coefficient and time The energies can be determined from plots of the logarithm of diffusion rate vs 1/T (called an Arrhenius plot). Steady state diffusion obeys Fick’s first law. Transient diffusion obeys Fick’s second law.  MECH 270 Material Science and Engineering Recovery, Recrystallization and Grain Growth Chapter 7  Recap Plastic deformation occurs by dislocation motion. Dislocations accumulate during strain causing “work hardening” Each dislocation has an associated energy – due to elastic strain field (stretching and compression of inter-atomic bonds). Hence some of the energy of deformation (force x distance) is stored in the form of dislocation density. Lattice Strains Around Dislocations  Recap In solids diffusion occurs by random motion of point defects For interstitial atoms (self or solute) depends on energy for migration from one interstitial site to another For host and substitutional solute atoms depends on energy for migration of vacancies and the vacancy concentration. By random walk, mean distance that an atom diffuses is proportional to the square root of the product of the diffusion coefficient and time  Free energy, driving force, and system mobility Three important concepts: free energy: (thermodynamic quantity). The internal energy of a system. In addition to the randomness (or the degree to which its atoms and/or molecules are “spread out” i.e. entropy) – nature always tries to minimize the free energy of a given system (through chemical reactions, phase transformations, etc.). BUT - rate at which system moves towards a lower state of free energy depends on two factors, (i) driving force and (ii) system mobility: rate of process = driving force x system mobility driving force: a measure of the overall drop in free energy caused by the reaction or transformation diffusion coefficient, D, is a measure of system mobility  Cold Work Dislocation structure in Ti after cold working. Dislocations entangle with one another during cold work. Dislocation motion becomes more difficult. Adapted from Fig. 5.11, Callister & Rethwisch 4e. (Fig. 5.11 is courtesy of M.R. Plichta, Michigan Technological University.)  Cold-worked Material stored energy associated with each dislocation  vacancy and self-diffusion characterized by diffusion coefficients vacancy diffusion self-diffusion  H mv  (H v  H m ) Dv  D0v  exp Dsd  D0 sd  exp kT kT Dislocations can ‘climb’ by vacancy absorption, generation Climb of edge dislocation Out of plane motion – edge dislocations ‘climb’ Vacancies formation and migration requires activation – therefore more active at high temperatures (or high stress)  Effect of Heating After %CW 1 hour treatment at Tanneal... decreases TS and increases %EL. Effects of cold work are reversed! annealing temperature (ºC) 100 200 300 400 500 600 700 tensile strength (MPa) 600 60 tensile strength ductility (%EL) 50 500 3 Annealing 40 stages to discuss... 400 30 ductility 20 300 Cold-worked Recovery some recovery Low-angle boundary Planar array of dislocations This is a lower energy configuration than a random array because the compressive and tensile strain fields above and below each dislocation are partly cancelled by the stress field of the next dislocation in the array  Energy of grain boundary Low angle: described by dislocations High angle: effectively random (usually) High angle Low angle large areas of poor fit large areas of good fit separated by misfit dislocations open structure with lots little free-volume of free volume interatomic bonds either interatomic bonds are broken or highly only slightly distorted distorted at the boundary  Recovery some recovery highly recovered  Recovery Annihilation reduces dislocation density. Scenario 1 extra half-plane of atoms Dislocations Results from annihilate diffusion atoms and form diffuse to regions a perfect of tension atomic extra half-plane plane. of atoms Scenario 2 3. “Climbed” disl. can now R move on new slip plane 2. grey atoms leave by 4. opposite dislocations vacancy diffusion meet and annihilate allowing disl. to “climb” 1. dislocation blocked; Obstacle dislocation can’t move to the right  Effect of Heating After %CW 1 hour treatment at Tanneal... decreases TS and increases %EL. Effects of cold work are reversed! annealing temperature (ºC) 100 200 300 400 500 600 700 tensile strength (MPa) 600 60 tensile strength ductility (%EL) 50 500 3 Annealing 40 stages to 400 30 discuss... ductility 20 300  Recrystallization New grains are formed that: -- have a small dislocation density -- are small -- consume cold-worked grains. 0.6 mm 0.6 mm 33% cold New crystals worked nucleate after brass 3 sec. at 580C.  Growth of a new crystal atoms are more likely to jump from high free energy position in the strained crystal than in the other direction new, perfect crystal strained crystals  Further Recrystallization All cold-worked grains are consumed. 0.6 mm 0.6 mm After 4 After 8 seconds seconds f23_07_pg198  º TR = recrystallization temperature TR º  Grain Growth At longer times, larger grains consume smaller ones. Why? Grain boundary area (and therefore energy) is reduced. 0.6 mm 0.6 mm Adapted from Fig. 7.21 (d),(e), Callister 7e. (Fig. 7.21 (d),(e) are courtesy of J.E. Burke, General Electric Company.) After 8 s, After 15 min, 580ºC 580ºC coefficient dependent Empirical Relation: on material and T. exponent typ. ~ 2 grain diam. elapsed time at time t. d n  d on  Kt f25_07_pg201 Example question: A piece of brass is cut into two pieces, each are held at 600°C for different times. Once the samples have been cooled to room temperature, their yield stresses and grain sizes are determined, as shown in the table below. Time (min) Grain diameter (mm) Yield stress (MPa) 12 0.040 250 60 0.064 208 Calculate the grain diameter you would predict for a sample held at 600°C for 150 minutes. Calculate the yield stress that the material in (a) would have, assuming that only the grain size is changing. Example question: A piece of brass is cut into two pieces, each are held at 600°C for different times. Once the samples have been cooled to room temperature, their yield stresses and grain sizes are determined, as shown in the table below. Time (min) Grain diameter (mm) Yield stress (MPa) 12 0.040 250 60 0.064 208 Calculate the grain diameter you would predict for a sample held at 600°C for 150 minutes. Calculate the yield stress that the material in (a) would have, assuming that only the grain size is changing.  Recap Work done: e.g. 300MPa,  = 0.1 Gives 3 x 107 Jm-3… approx. 95% heat, 5% as stored defects Effect of temperature Annealing… changes in dislocation structure due to cold work by subsequent heating. recovery: arrangement of dislocations into sub-grains (polygonisation) recrystallisation: reduction in number of “excess” dislocations, introduced during cold working, with an associated marked decrease in the stored energy. Grain growth is a possible third stage. Reduction in the grain boundary energy - the process occurs by atomic diffusion. A note on recovery  Summary Thermodynamically driven changes require – A driving force, i.e., stored energy – System mobility, the ability for atoms to move Cold work provides the stored energy Self- or vacancy diffusion provides the mobility Leading to – Recovery – Recrystallization with increasing temperature Grain boundary area also provides a driving force – Leads to grain growth at high temperatures  MECH 270 Material Science and Engineering Creep Chapter 8 Observation of ‘creep’ High temperature deformation ‘high’ depends on material T > 0.3 to 0.4Tm for metals T > 0.4 to 0.5Tm for ceramics Observation: time dependent response (at constant stress) Damage initiates Damage accumulates Steady state (dynamic dislocation structures) Observation: influence of stress on secondary creep rate (const T)   B n Change in value of n with stress: Transition is slightly T & grain size dependent Observation: influence of T on secondary creep rate (const )  Q    C exp   RT  Combining T and stress dependence  Q    A exp n   RT  Very material dependent; also somewhat stress / temperature dependent… The creep strain rate equation  Summary of Creep Mechanisms Each region has a different, A, n, and Q Medium T High T Low stress Diffusion creep – Diffusion creep – grain boundary diffusion bulk diffusion Medium stress Dislocation creep – Dislocation creep – core diffusion bulk diffusion High stress Power law breakdown – dislocation glide significant  Mechanism I Dislocation creep (or power law creep): T regime where dislocation climb is very easy Occurs by movement of dislocations – but different from plasticity (slip) Climb occurs by diffusion of vacancies Since climb dominated, don’t see slip plane dependence, even though creep is still dislocation controlled 0.3 – 0.5Tm ‘core’ diffusion is dominant diffusion along ‘easy paths’ such as the cores of dislocations (dilated crystal structure allows easy travel) 0.5Tm + ‘bulk’ diffusion is dominant diffusion within the crystal bulk is sufficient to be dominant – provides many more potential diffusion paths than dislocation cores  Mechanism II Diffusion creep (or linear-viscous): Occurs at low stresses – too low to move dislocations Atoms diffuse around grain, providing stress relief mechanism Dislocations not involved Low T/Tm : atoms diffuse along grain boundaries (Coble creep). Strongly affected by grain size High T/Tm : atoms diffuse through the bulk (Nabarro-Herring creep). Affected by grain size  Mechanism III Power law breakdown: Occurs at high stresses Strain rate varies roughly exponentially with stress Transition from climb controlled to glide controlled flow (i.e. much closer to ‘normal’ plasticity) Hot working often occurs in this zone Summary of Creep Mechanisms Medium T High T Low stress Diffusion creep – Diffusion creep – grain boundary diffusion bulk diffusion Medium stress Dislocation creep – Dislocation creep – core diffusion bulk diffusion High stress Power law breakdown – dislocation glide significant  Idealised deformation mechanism map Depends on rate End result: In each of these zones, you will have a different set of parameters in the creep strain rate equation  Example An alloy tie-bar in a chemical plant is designed to withstand a stress of 25MPa at T=620C. Creep tests carried out on specimen of the alloy show a strain rate of 3.1x10-12/s under these conditions, with a stress exponent of 5, and activation energy of 160kJ/mol. In service it was found that 30% of the time the stress and T were actually 30MPa and 650C respectively. What is the average creep rate?  Q    A n exp   RT    160000    A exp  5   8.314T    160000  3.1 10 12  A255 exp   8.314620  273  A  7.26 10 10    160000    7.26 10 10305 exp   8.314650  273   1.554 10 11   0.3 1.554 10 11  0.7  3.1 10 12  6.8  10 12  Creep strength vs rupture strength Creep strength: The stress to cause a given strain, at a given temperature and time: e.g. the creep strength of 316 stainless steel at 800C based on 1% extension in 1000 hours is 210MPa. (principally secondary creep) Rupture strength: Stress required to cause failure, at given temperature in a given time. Note: for creep and rupture data, typically extrapolate data from limited tests. i.e. fit equation to known data, and assume equation valid for other regimes  MECH 270 Material Science and Engineering Phase Diagrams Chapter 9  Phase Diagrams Definitions A simple binary phase diagram Tie lines The lever rule  Phase Diagrams - used to define equilibrium “phases” within an alloy “system” a given set of conditions, eg temperature, pressure, composition – material scientists - typically refer to phase diagrams plotted in terms of temperature and composition –we will limit ourselves to binary (i.e. two-component) phase diagrams of metal alloys, such as steel (Fe-C) and brass (Cu-Zn) Phase B Phase A Different composition Different crystal structure from Phase B X-Ni-Cu Nickel atom Copper atom system  Phase Diagrams To understand phase diagrams, we first need several terms which are commonly used system: a series of possible metal mixtures (i.e., alloys) consisting of the same components, but without regard to alloy composition much the (eg. The Fe-C system) solvent+solute pretty have solid solution: a mixture of two↑ same atomic radius. (or more) types of atoms in which solute atoms occupy either substitutional or interstitial positions within the solvent lattice; the crystal structure of the solvent is maintained solubility limit: maximum concentration of solute atoms that can dissolve in a solvent to form a single phase. If this limit is exceeded, another solid solution or compound of different composition forms phase: a homogeneous portion of a system with uniform physical, chemical and structural characteristics, e.g., every pure metal is a phase, so too is any particular solid solution  Phase Diagrams solubility limit: maximum concentration of solute atoms that can dissolve in a solvent to form a single phase. If this limit is exceeded, another solid solution or compound of different composition forms single-phase a liquid > - limit solubility Iwater Sugar syrup liquid plus solid Temperature - - two phase - > sugar Composition - % phase: a homogeneous portion of a system that has uniform physical, chemical and structural characteristics, i.e. every pure metal is a phase, and so too is any particular solid solution polishing Microstructure > - determines Chemical + physical properties the structure of a prepared surface of material as revealed by a microscope above 25× magnification* - The microstructure of a material can strongly influence properties such as strength, toughness, ductility, hardness, corrosion resistance, high/low temperature behavior, wear resistance, etc > d-phase - - pro- pearlite Fe-C - Alloy & > - FezC cemen tide P z ?  Components and Phases Components: The elements or compounds which are present in the mixture (e.g., Al and Cu) Phases: > - homogeneous part of the system The physically and chemically distinct material regions that result (e.g.,  and ). > Al , Cu - Aluminum-  (lighter S / Copper Al phase) have different Alloy Cu chemical (darker composition, physical/mechanical phase) properties but , composed of the same materials  Effect of T & Composition (Co) Changing T can change # of phases: path A to B. Changing Co can change # of phases: path B to D. / B (100°C,70) D (100°C,90) phase I Add more 2 phase - 1 phase > regime 2 phases - sugar 100 > - > regime Sugar + ↑ 80 L water Temperature (°C) t *T (liquid) water- 60 + Sugar L S sugar (liquid solution (solid 40 i.e., syrup) system sugar) 20 A (20°C,70) - - 2 phase 2 phases regime 0 0 20 40 60 70 80 100 Co =Composition (wt% sugar)  G = H - TS Phase Diagrams free energy: thermodynamic quantity representing the internal energy of a system, in addition to the randomness or disorder of its atoms and/or molecules (a.k.a. entropy) equilibrium phase: a phase which minimizes a system’s free energy for a specified combination of temperature, pressure, and composition Note: A phase diagram contains only equilibrium phases metastable (or non-equilibrium) phase: a phase that can be produced by a very rapid change in system conditions (e.g., temperature or pressure).  Phase Diagrams Simplest phase diagrams are those of binary isomorphous systems, meaning systems in which the two components have complete liquid and solid solubility (eg. Cu-Ni) T(°C) 1600 2 phases: Cu-Ni system (two L (liquid) ↳ binary 1500 L (liquid) components)  (FCC solid solution) 1400 ↓ 3 phase fields: L igio solid 1300 g L +  region between > - where liquid  t 1200  solid is in equilibrium 1100 (FCC solid solution) 1000 100% Cu 0 20 40 60 80 100 wt% Ni  Phase Equilibria Simple solution system (e.g., Ni-Cu solution) Crystal electroneg r (nm) Structure Ni FCC 3 FCC = 1.9 : 0.1246 3 1.8 30.1278 : Cu Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume – Rothery rules) suggesting high mutual solubility. Ni and Cu are totally miscible in all proportions. ↳  Phase Diagrams Using a phase diagram, at any given combination of temperature and composition, we can determine: i) phases present: the “temp-composition” point on the phase diagram indicates the equilibrium phase(s) within the material ii) phase composition: if the temp-composition point is in a single phase region, the corresponding composition is simply read from the x-axis. T(°C) 1600 2 phases: L (liquid) 1500 L (liquid) phase  (FCC solid solution) single 3 phase fields: -- A 1400 · L 1300 L+   1200  singlephase 1100 1 ↑ (FCC solid 1050 - Bo solution) said 1000 0 20 40 60 80 100 wt% Ni 100% Cu X  Phase Diagrams BUT, if the temperature-composition point is in a two-phase region, a tie line is used to determine iii) phase compositions and, iv) phase amounts (or percentages) of the equilibrium phases that are present composition is important two-phase region I T(°C) much Ni in liquid ? 2 phases: - now 1600 L (liquid) 1500 L (liquid)  how much Ni (FCC solid solution) > - in solid ? 1400 3 phase fields: L 1300 L+  1250 ----- & Ci  1200  1100 Zigt hase ime "(FCC solid I solution) fraction of liquid of solid in wil and fraction ↳ weight percent 1000 0 20 40 60 80 100 wt% Ni 100% Cu  and > To find composition - Tie Lines fraction of each phase If we know T and Co, then we know: ↳ and lever rule --the composition of both phases. Cu-Ni system T(°C) 1300 L (liquid) tie line horizontal line at given temp.  1200 (solid) composition of liquid ↓ intersection of tie line 20 30 40 50 with liquidous line Co ↓ wt% Ni composition of solid Cliquid Csolid > intersection - of tie line and ?  Phase Diagrams: composition of phases If we know T and Co, then we know the composition of each phase. Cu-Ni T(°C) system Examples: TA A Co = 35 wt% Ni tie line At T A = 1320°C: 1300 L (liquid) Only Liquid (L) B CL = Co ( = 35 wt% Ni) TB  At T D = 1190°C: 1200 D (solid) Only Solid ( ) TD C = Co ( = 35 wt% Ni) 20 3032 35 4043 50 At T B = 1250°C: CLCo C wt% Ni Both  and L CL = C liquidus ( = 32 wt% Ni here) C = C solidus ( = 43 wt% Ni here)  Tie Lines If we know T and Co, then we know: --the amount of each phase (given in wt%). Cu-Ni Examples: T(°C) system C o = 35 wt% Ni TA A tie line At T : Only Liquid (L) 1300 L (liquid) A WL = 100 wt%, W = 0 B At T : Only Solid (  ) TB R S D **  WL = 0, W = 100 wt% 1200 D (solid) At T : Both  and L TD B opposite side WL  QS  43  35  73 wt % 20 3 03235 4043 50 R + S> 43  32 C LC o C  wt% Ni I - total length of the line R W  R +S = 27 wt% - ↳ versa Xvice * Find fraction of solid-look at opposite side  100 % > - read composition off X-axis Ex: Cooling in a Cu-Ni Binary 2-phase-> lever rule ! use 100 % L Phase diagram: T(°C) L (liquid) - L: 35wt%Ni Cu-Ni system. Cu-Ni System is: 130 0 system A --binary L: 35 wt% Ni : 46 wt% Ni B i.e., 2 components: 35 > 46 - cooling down- 32 C increases solid Cu and Ni. 43 within the --isomorphous D liquid 24 36 L: 32 wt% Ni i.e., complete : 43 wt% Ni solubility of one 120 0 E L: 24 wt% Ni component in another;  phase : 36 wt% Ni  field extends from (solid) > - 100% % 0 to 100 wt% Ni. Consider 110 0 Co = 35 wt%Ni. 20 30 35 40 50 Co wt% Ni  Co = 63 wt Ni fraction of L & C (solid) composition of Liquid and solid W Ni composition of solid : -68 - 32 wt % Cu composition of Liquid-57 wt. Ni - 43 wt/ (u - fraction of solid : # === 55w did Cu - S C Cs R fraction of liquid : is ==63 = 45 w of Light thecis = 100  𝜶 𝟎 𝑳 𝜶 𝑳 𝟎 𝑳 𝜶 𝜶 𝑳 C C S R L: 57 wt% Ni : 68 wt% Ni S microstructure Note: 1-WL = Wα 57 63 68  Summary A phase is a homogeneous portion of a system that has uniform physical, chemical and structural characteristics Microstructure is the distribution of phases observed in a microscope A phase diagram shows regions (fields) in which one or more phases is present as a function of composition (horizontal scale) and temperature (vertical scale). In a two phase field the compositions of the phases are determined by constructing a tie line at the temperature of interest – the proportions of the two phase are then determined using the lever rule  MECH 270 Material Science and Engineering Phase Diagrams - Continued Chapter 9  Phase Diagrams - continued So we have looked at a simple binary equilibrium phase diagram (an isomorphous system) Three phase fields Liquidus and solidus lines Tie lines The lever rule  Now Revisit Tie lines The lever rule Microstructure From Solidification – Equilibrium – Non-equlibrium – Micro-segregation Eutectic Phase Diagram  Tie Lines If we know T and Co, then we know: --the composition of each phase. Cu-Ni system T(°C) phases) 1300 L (liquid) tie line phase fields T1 · E  L +x 1200 (solid) 20 30 40 50 Co Cliquid Csolid wt% Ni 31 wt% Ni 42 wt% Ni  Tie Lines If we know T and Co, then we know: --the amount of each phase (given in wt%). Cu-Ni Examples: T(°C) system C o = 35 wt% Ni TA A tie line At T : Only Liquid (L) 1300 L (liquid) A W L = 100 wt%, W = 0 B At TD : Only Solid (  ) TB R S  W L = 0, W = 100 wt% 1200 D (solid) At TB : Both  and L TD look at S WL  solid > 20 3 03235 4043 50 side - R +S C LC o C  wt% Ni look at R liquid W  = 27 wt% side R +S  Binary-Eutectic Systems has a special composition 2 components with a min. melting T. Cu-Ag system phase T(°C) 2 liguidous regiono - in Cu phase region 1200 Key terms: L (liquid) ↓ -> L "R" 1000  L+  I Phases L B : < , , O I L +  O T 800 Liquidus line only liquid : (1) above E 8.0 a 91.2 Entectic point + S temp Solidus line only solid (0) : below 600 , Y phase Soild() two Solvus line phase: solid to one   400 Eutectic Temperature · Eutectic point 200 ↳ entectic isotherm 0 20 40 60 CE 80 100 Co , wt% Ag Q. Can you think of a time you would want to add an alloying element but keep only one phase?  Binary-Eutectic Systems has a special composition 2 components with a min. melting T. Cu-Ag system T(°C) Ex.: Cu-Ag system 1200 3 single phase regions L (liquid)  ) (L, solid phases 1000 Limited solubility:  L+  L +  779°C  : mostly Cu TE 800 8.0 71.9 91.2 Crichin)  : mostly Ag 600 ↑ min CE will TE : No liquid below TE   give min TE CE : Min. melting TE 400 composition 200 0 20 40 60 CE 80 100 Eutectic transition Co , wt% Ag * (CE) + (CE) - L(CE) *  tie the same ~> Apply rule Pb-Sn Eutectic System line to and this lever system For a 40 wt% Sn - 60 wt% Pb alloy at 150°C, find... --the phases present: + Pb-Sn --compositions of phases: T(°C) system CO = 40 wt% Sn C = 11 wt% Sn 300 L (liquid) C = 99 wt% Sn --the relative amount  L+  200 183°C L +  of each phase: 18.3 61.9 97.8 S C - CO 150 W = R+S =  C  - C R S 100 + 99 - 40 59 = = = 67 wt% 99 - 11 88 CO - C  0 11 20 40 60 80 99100 W = R = C Co C R+S C - C C, wt% Sn 40 - 11 29 = = = 33 wt% 99 - 11 88  Pb-Sn Eutectic System For a 40 wt% Sn-60 wt% Pb alloy at 220°C, find... --the phases present: +L Pb-Sn --compositions of phases: T(°C) system CO = 40 wt% Sn C = 17 wt% Sn 300 L (liquid) CL = 46 wt% Sn L+  --the relative amount 220  200 R S L +  of each phase: 183°C CL - CO 46 - 40 W = = 100 CL - C 46 - 17 + 6 = = 21 wt% 29 100 0 17 20 40 46 60 80 C Co CL CO - C  23 C, wt% Sn WL = = = 79 wt% CL - C 29 Example 1 Pb-Sn Fill in the table T(°C) system S C - CO W= = R+S C - C 300 L (liquid) = 99 - 75 99 - 9 = 24 90 = 27 wt% L+  R CO - C 200  L+  W = = R+S C - C 183°C R S = 75 - 9 = 66 = 73 wt% x 99 - 9 90 100 + Phases α β Chemical 9 wt% Sn 99 wt% Sn 0 Cα=9 - 20 60 C0 = 75 80 - Cβ=99 100 Comp. Phase C, wt% Sn Fraction 27 wt% 73 wt% Example 2 Pb-Sn Fill in the table T(°C) system S C - CO W= = R+S C - C 300 L (liquid) = 98 - 24 98 - 12 = 74 86 = 86 wt% L+  R CO - C 200  L+  W = = R+S C - C R S x 183°C 24 - 12 12 = = = 14 wt% 98 - 12 86 100 + Phases α β Chemical 12 wt% Sn 98 wt% Sn 0 Cα=12 20 C0 = 24 60 80 Cβ=98 100 Comp. Phase C, wt% Sn Fraction 86 wt% 14 wt%  cool down - we see changes in the microstructure As we Cooling in a Cu-Ni Binary T(°C) L (liquid) L: 35wt%Ni Cu-Ni 130 0 system A L: 34 wt% Ni. P.F 91 wt% : 45 wt% Ni. P.F 9 wt% B 34 45 32 C 43 25 D L: 32 wt% Ni. P.F 73 wt% 37  : 43 wt% Ni. P.F 27 wt% 120 0 E L: 24 wt% Ni. P.F 17 wt%  : 36 wt% Ni. P.F 83 wt%  (solid) 110 0 20 30 35 40 50 Adapted from Fig. 9.4, Co wt% Ni Callister 7e. P.F = phase fraction  Cooling rates rabe ~ Idiffusion If we know T and Co, then we know: is time --the composition of each phase. dependent Cu-Ni system T(°C) 1300 L (liquid) L: 35 wt% Ni : 46 wt% Ni !  i -y i It : < 35 wt% Ni I & ! · 1200 (solid) 20 30 40 C 50 ↳ Cored Structure (2 wt% Ni > - Due to rapid and not cooling for enough time diffusion or to homogenize  Cored vs Equilibrium Phases C changes as we solidify. Cu-Ni case: First  to solidify has C = 46 wt% Ni. Last  to solidify has C = 35 wt% Ni. Fast rate of cooling: Slow rate of cooling: Cored structure Equilibrium structure higher concentration to lower a > - of Uniform C  : First  to solidify: Ni 46 wt% Ni 35 wt% Ni Last  to solidify: < 35 wt% Ni  Microstructures in Eutectic Systems: I Co < 2 wt% Sn T(°C) L: Co wt% Sn 400 Result: L  --at extreme ends 300 L --polycrystal of  grains i.e., only one solid phase. L+  200  : Co wt% Sn (Pb-Sn TE System) 100 + 0 10 20 30 Co Co , wt% Sn 2 (room T solubility limit)  Microstructures in Eutectic Systems: II L: Co wt% Sn 2 wt% Sn < Co < 18.3 wt% Sn T(°C) 400 Result: L  Initially liquid +  L 300   then  alone L+ finally two phases : Co wt% Sn   polycrystal 200 TE  -phase inclusions   The process of forming these 100 inclusions is called “solid-state” +  Pb-Sn precipitation. system 0 10 20 30 2 Co Co , wt% Sn (sol. limit at T room ) 18.3 (sol. limit at TE)  Microstructures in Eutectic Systems: III At Entectic Temp Co = CE ~ Result: Eutectic microstructure (lamellar structure) --alternating layers (lamellae) of  and  crystals. Micrograph of Pb-Sn T(°C) eutectic L: Co wt% Sn microstructure 300 L Pb-Sn system L+  200  183°C L  TE 100 160 m  : 97.8 wt% Sn : 18.3 wt%Sn 0 20 40 60 80 100 18.3 CE 97.8 61.9 C, wt% Sn  -diffusion , must find a short path why ? ~> Lamellar Eutectic Structure Pb-Sn T(°C) system 300 L (liquid) lead rejects fin -> L+  rejects ↓ 200  L +  183°C Pb rich Sn rich rejects more adjacent-results in 100 + alternating layers 0 20 60 80 100 C, wt% Sn  Microstructures in Eutectic Systems: IV 18.3 wt% Sn < Co < 61.9 wt% Sn Result:  crystals and a eutectic microstructure Just above TE : T(°C) L: Co wt% Sn  L L C  = 18.3 wt% Sn 300  CL = 61.9 wt% Sn L Pb-Sn S L+  W = = 50 wt% system R+S 200  R S L+   WL = (1- W  ) = 50 wt% TE S R Just below TE : 100 + C  = 18.3 wt% Sn primary  C  = 97.8 wt% Sn eutectic  eutectic  W = S = 73 wt% 0 20 40 60 80 100 R+S 18.3 61.9 97.8 W  = 27 wt% Co, wt% Sn  Hypoeutectic & Hypereutectic 300 L T(°C)  L+  200 L+  (Pb-Sn TE + System) 100 0 20 40 60 80 100 Co, wt% Sn eutectic hypoeutectic: If Co = 50 wt% Sn 61.9 hypereutectic: (illustration only) eutectic: Co = 61.9 wt% Sn             175 m 160 m pro-eutectic eutectic micro-constituent or prior  Example 3 Pb-Sn At each x (BLUE): What is the phase fraction? T(°C) system CO - CL 80 - 70 300 x Wβ = = L (liquid) Cβ - CL - 97 - 70 L+  = 10 27 = 37 wt% 200  x L+  R S 183°C Cβ - CO WL = 100 CL - C + 17 = = 63 wt% C0= 80 27 CL= 70 Cβ= 97 0 20 60 80 100 C, wt% Sn The β formed before the eutectic line amount of pro-eutectic β P.F = phase fraction Example 3 Pb-Sn At each x (GREEN): What is the phase fraction? T(°C) system S C - CO W= = 300 R+S C - C x L (liquid) = 98 - 80 = 18 = 20 wt% L+  98 - 8 90 200  x L+  183°C R x S R CO - C W = = 100 R+S C - C + 80 - 8 72 = = = 80 wt% 98 - 8 90 C0= 80 Cβ= 98 Cα = 8 0 20 60 80 100 C, wt% Sn Includes β formed AFTER the eutectic line TOTAL amount of β P.F = phase fraction Example 3 Pb-Sn At each x: What is the phase fraction? T(°C) system How will the microstructure look? 300 x L (liquid) x L: P.F 100 wt% L+  200  x L+  R S 183°C L: P.F 63 wt% x x β: P.F 27 wt% R S 100 + C0= 80 Cβ= 98 Cα = 8 CL= 70 Cβ= 97 α: P.F 20 wt% 0 20 60 80 100 x β: P.F 80 wt% C, wt% Sn hypereutectic P.F = phase fraction  Fe-C Phase Diagram An example of a slightly more complex phase diagram.  Summary If we cool isomorphous system from the liquid single-phase field to the solid single-phase field, the result depends upon the cooling rate Cool slowly enough that equilibrium is maintained, and single phase formed has uniform composition If not, their will be “segregation”, i.e., a non-uniform composition. A “binary eutectic” system is more complex than an isomorphous system, having six phase fields, three single-phase and three two-phase fields If we cool from the liquid phase of a eutectic system into one of the single- phase fields (i.e., through a liquidus and then a solidus), the result is similar to an isomorphous system If we cool from a single phase-field (solid) into the two-phase field (solid) inclusions of the second phase form by solid-state precipitation If we cool through the eutectic composition we get a lamellar microstructure of two phases with different compositions / crystal structures. If we cool through a liquid/solid two phase field directly into the low temperature two phase field we get a triplex microstructure structure consisting of islands of a“pro-eutectic”within the lamellar structure of the eutectic.  MECH 270 Material Science and Engineering Phase Diagrams - Continued Chapter 9  The story so far Isomorphous systems (3 phase fields) – L, +L,  – Tie lines – Lever rule – Microstructures from cooling (non-uniform composition = segregation) Binary eutectic system (6 phase fields) – L, +L, +L, , , + – Solid-state precipitation, layered structures, mixed structures  Microstructures in Eutectic Systems: III Co = CE Result: Eutectic microstructure (lamellar structure) --alternating layers (lamellae) of  and  crystals. Micrograph of Pb-Sn T(°C) eutectic L: Co wt% Sn microstructure 300 L Pb-Sn system L+  200  183°C L  TE 100 160 m  : 97.8 wt% Sn : 18.3 wt%Sn 0 20 40 60 80 100 18.3 CE 97.8 61.9 C, wt% Sn Lamellar Eutectic Structure  Microstructures in Eutectic Systems: IV 18.3 wt% Sn < Co < 61.9 wt% Sn Result:  crystals and a eutectic microstructure Just above TE : T(°C) L: Co wt% Sn  L L C  = 18.3 wt% Sn 300  CL = 61.9 wt% Sn L Pb-Sn S L+  W = = 50 wt% system R+S 200  R S L+   WL = (1- W  ) = 50 wt% TE S R Just below TE : 100 + C  = 18.3 wt% Sn primary  C  = 97.8 wt% Sn eutectic  eutectic  W = S = 73 wt% 0 20 40 60 80 100 R+S 18.3 61.9 97.8 W  = 27 wt% Co, wt% Sn  Hypoeutectic & Hypereutectic 300 L T(°C)  L+  200 L+  (Pb-Sn TE + System) 100 0 20 40 60 80 100 Co, wt% Sn eutectic hypoeutectic: Co = 50 wt% Sn 61.9 hypereutectic: (illustration only) eutectic: Co = 61.9 wt% Sn             175 m 160 m pro-eutectic eutectic micro-constituent or prior   Now More complicated binary systems Iron-carbon system (steels)  Intermetallic Compounds Mg2Pb Note: intermetallic compound forms a very narrow area – almost a line, because stoichiometry (i.e. composition) is almost exact.  Eutectoid & Peritectic Eutectic - liquid transforms to two solid phases L cool  +  (For Pb-Sn, 183ºC, 61.9 wt% Sn) heat Eutectoid – one solid phase transforms to two other solid phases intermetallic compound S2 S1+S3 - cementite  cool  + Fe3C (For Fe-C, 727ºC, 0.76 wt% C) heat Peritectic - liquid and one solid phase transform to a second solid phase S1 + L S2 cool  +L heat  (For Fe-C, 1493ºC, 0.16 wt% C)  Eutectoid & Peritectic Peritectic transition +L  Cu-Zn Phase diagram Eutectoid transition  +  Even for very complicated phase-diagrams Tie lines give the composit