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ISSUES TO ADDRESS... Transforming one phase into another takes time. How does the rate of transformation depend on time and T ? How can we slow down the.

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Presentation on theme: "ISSUES TO ADDRESS... Transforming one phase into another takes time. How does the rate of transformation depend on time and T ? How can we slow down the."— Presentation transcript:

1 ISSUES TO ADDRESS... Transforming one phase into another takes time. How does the rate of transformation depend on time and T ? How can we slow down the transformation so that we can engineer non-equilibrium structures? Are there other means to improve mechanical behavior? Fe (Austenite) Eutectoid transformation C FCC Fe 3 C (cementite) (ferrite) + (BCC) Chapter 10: Phase Transformations – Considering Kinetic and Heat Treatment

2 Nucleation –nuclei (like biological seeds) act as template to grow crystals –for nuclei to form, the rate of addition of atoms to any nucleus must be faster than rate of loss –once nucleated, the seed must grow until they reach the predicted equilibrium Phase Transformations

3 Driving force to nucleate increases as we increase T –supercooling (eutectic, eutectoid) –superheating (peritectic) With a Small amount of supercooling few nuclei - large crystals With a Large amount of supercooling rapid nucleation - many nuclei, small crystals

4 Solidification: Nucleation Processes Homogeneous nucleation –nuclei form in the bulk of liquid metal (as native chemistry) –requires sufficient supercooling (typically °C max) Heterogeneous nucleation –much easier since stable nucleus is already present (they are non-native chemically) Could be wall of mold or impurities in the liquid phase –allows solidification with only minimal supercooling (0.1-10ºC)

5 r* = critical nucleus: nuclei r* grow (to reduce energy) Homogeneous Nucleation & Energy Effects G T = Total Free Energy = G S + G V Surface Free Energy- destabilizes the nuclei (it takes energy to make an interface) = surface tension Volume (Bulk) Free Energy – stabilizes the nuclei (releases energy)

6 Solidification r* decreases as T increases For typical T r* is around 100Å (10 nm) Note: H S is a strong function of T is a weak function of T H S = latent heat of solidification T m = melting temperature = surface free energy T = T m - T = supercooling r* = critical radius

7 Rate of Phase Transformations Kinetics - measures approach to equilibrium vs. time Hold temperature constant & measure conversion vs. time –How is the amount of conversion measured? X-ray diffraction – have to do many samples electrical conductivity – follows a single sample sound waves (insitu ultrasonic) – follows a single sample

8 Thus, the rate of nucleation is a product of two curves that represent two opposing factors (instability and diffusivity).

9 Rate of Phase Transformation –Modeled by the Avrami Rate Equation: t 0.5 All out of material – done! Fixed Temp. maximum rate reached – now amount unconverted decreases so rate slows rate increases as surface area increases & nuclei grow Log Time

10 Avrami Equation Avrami rate equation y = 1- exp (-kt n ) k & n are fit for any specific sample By convention we define: r = 1 / t 0.5 as the rate of transformation – it is simply the inverse of the time to complete half of the transformation The initial slow rate can be attributed to the time required for a significant number of nuclei of the new phase to form and begin growing. During the intermediate period the transformation is rapid as the nuclei grow into particles and consume the old phase while nuclei continue to form in the remaining parent phase.parent phase Once the transformation begins to near completion there is little untransformed material for nuclei to form in and the production of new particles begins to slow. Further, the particles already existing begin to touch one another, forming a boundary where growth stops.

11 Rate of Phase Transformations In general, rate increases as T r = 1/t 0.5 = A e -Q/RT –R = gas constant –T = temperature (K) (higher causes higher rate too) –A = preexponential rate factor –Q = activation energy r is often small so equilibrium is not possible! Arrhenius expression adapted from B.F. Decker and D. Harker, "Recrystallization in Rolled Copper", Trans AIME, 188, 1950, p C119 C113 C102 C88 C43 C

12 Transformations & Undercooling M. Eng. Can make it occur at:...727ºC (cool it slowly)...below 727ºC (supercool or Undercool it!) Eutectoid transf. (Fe-C System): + Fe 3 C 0.76 wt% C wt% C 6.7 wt% C Fe 3 C (cementite) L (austenite) +L+L +Fe 3 C +Fe 3 C L+Fe 3 C (Fe) C o, wt%C 1148°C T(°C) ferrite 727°C Eutectoid: Equil. Cooling: T transf. = 727 º C T Undercooling by T transf. < 727 C adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990

13 Eutectoid Transformation Rate Coarse pearlite formed at higher T - softer Fine pearlite formed at low T - harder Diffusive flow of C needed Growth of pearlite from austenite: Adapted from Fig. 9.15, Callister 7e. pearlite growth direction Austenite ( ) grain boundary cementite (Fe 3 C) Ferrite ( ) Recrystallization rate increases with T. 52°C ( ˚C ) 0 50 y (% pearlite) 127°C ( ˚C ) 77°C 100

14 Reaction rate is a result of nucleation and growth of crystals. Examples from previous slide: Nucleation and Growth T: just belowTETE Nucleation rate low Growth rate high pearlite colony T: moderately belowTETE Nucleation rate med Growth rate med. Nucleation rate high T: way belowTETE Growth rate low

15 The ideas of reality and the ideal meet in the Material Engineerings Transformation Curves

16 adapted from H. Boyer (Ed.) Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1977, p Isothermal Transformation (TTT) Diagrams Fe-C system, C o = 0.76 wt% C Transformation at T = 675°C T = 675°C y,y, % transformed time (s) %pearlite 100% 50% Austenite (stable) T E (727 C) Austenite (unstable) Pearlite T(°C) time (s) isothermal transformation at 675°C

17 Eutectoid composition, C o = 0.76 wt% C Begin at T > 727°C Rapidly cool to 625°C and hold isothermally. adapted from H. Boyer (Ed.) Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1997, p. 28. Effect of Cooling History in Fe-C System %pearlite 100% 50% Austenite (stable) T E (727 C) Austenite (unstable) Pearlite T(°C) time (s)

18 Transformations with Proeutectoid Materials Hypereutectoid composition – proeutectoid cementite C O = 1.13 wt% C T E (727°C) T(°C) time (s) A A A + C P A + P Adapted from Fig , Callister 7e. Adapted from Fig. 9.24, Callister 7e. Fe 3 C (cementite) L (austenite) +L+L +Fe 3 C +Fe 3 C L+Fe 3 C (Fe) C o, wt%C T(°C) 727°C T

19 Non-Equilibrium Transformation Products in Fe-C Bainite: -- lathes (strips) with long rods of Fe 3 C --diffusion controlled. Isothermal Transf. Diagram adapted from H. Boyer (Ed.) Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1997, p. 28. from Metals Handbook, 8th ed., Vol. 8, Metallography, Structures, and Phase Diagrams, American Society for Metals, Materials Park, OH, 1973.) Fe 3 C (cementite) 5 m (ferrite) time (s) T(°C) Austenite (stable) 200 P B TETE 0% 100% 50% pearlite/bainite boundary A A 100% bainite 100% pearlite

20 From: George Krauss, Steels: Processing, Structure, and Performance, ASM International, TTT Curves showing the Bainite Transformation (a) Plain Carbon Steels; (b) Alloy Steel w/ distinct Bainite Nose

21 Martensite: -- (FCC) to Martensite (BCT) courtesy United States Steel Corporation. Isothermal Transf. Diagram to M transformation.. -- is rapid! -- % transf. depends on T only. Martensite: Fe-C System Martensite needles Austenite 60 m time (s) T(°C) Austenite (stable) 200 P B TETE 0% 100% 50% A A M + A 0% 50% 90% x x x x x x potential C atom sites Fe atom sites (involves single atom jumps)

22 Transformation to Martensite Martensite formation requires that the steel be subject to a minimum – Critical – Cooling Rate (this value is TTT or CCT chart dependent for alloy of interest) For most alloys it indicates a quench into a RT oil or water bath

23 (FCC) (BCC) + Fe 3 C Martensite Formation slow cooling tempering quench M (BCT) M = martensite is body centered tetragonal (BCT) Diffusionless transformation BCT if C > 0.15 wt% BCT few slip planes hard, brittle

24 Martensite Transformation Crystallography: FCC Austenite to BCT Martensite From: George Krauss, Steels: Processing, Structure, and Performance, ASM International, 2006.

25 Austenite to Martensite: Size Issues and Material Response From: George Krauss, Steels: Processing, Structure, and Performance, ASM International, 2006.

26 Spheroidite: -- grains with spherical Fe 3 C --diffusion dependent. --heat bainite or pearlite for long times (below the A C1 critical temperature) --driven by a reduction in interfacial area of Carbide Spheroidite: Fe-C System (Fig copyright United States Steel Corporation, 1971.) 60 m (ferrite) (cementite) Fe 3 C

27 Phase Transformations of Alloys Effect of adding other elements Change transition temp. Cr, Ni, Mo, Si, Mn retard + Fe 3 C transformation delaying the time to entering the diffusion controlled transformation reactions – thus promoting Hardenability or Martensite development

28 Continuous Cooling Transformations (CCT) Isothermal Transformations are Costly requiring careful gymnastics with heated (and cooling) products CC Transformations change the observed behavior concerning transformation –With Plain Carbon Steels when cooled continuously we find that the Bainite Transformation is suppress(see Figure 10.26)

29 Cooling Curve Plot: temp vs. time

30 CCT for Eutectoid Steel Figure: 10-26

31 CCT for Eutectoid Steel

32 Alloy Steel CCT Curve – again note distinct Bainite Nose Adapted from Fig , Callister 7e.

33 Dynamic Phase Transformations On the isothermal transformation diagram for 0.45 wt% C Fe-C alloy, sketch and label the time-temperature paths to produce the following microstructures: a)42% proeutectoid ferrite and 58% coarse pearlite b)50% fine pearlite and 50% bainite c)100% martensite d)50% martensite and 50% austenite

34 A + B A + P A + A B P A 50% time (s) M (start) M (50%) M (90%) Example Problem for C o = 0.45 wt% a)42% proeutectoid ferrite and 58% coarse pearlite first make ferrite then pearlite coarse pearlite higher T Adapted from Fig , Callister 5e. T (°C)

35 b)50% fine pearlite and 50% bainite first make pearlite then bainite fine pearlite lower T T (°C) A + B A + P A + A B P A 50% time (s) M (start) M (50%) M (90%) Example Problem for C o = 0.45 wt% Adapted from Fig , Callister 5e. NOTE: This 2 nd step is sometimes referred to as an Austempering step, quenching into a heated salt bath held at the temperature of need

36 Example Problem for C o = 0.45 wt% c)100 % martensite: 380 C/s {( )/.7s} Adapted from Fig , Callister 5e. d) A + B A + P A + A B P A 50% time (s) M (start) M (50%) M (90%) c) T (°C) d)50%martensite/50%(retained) austenite

37 CT for CrMo Med-carbon steel Hardness of cooled samples at various cooling rates in bubbles -- Dashed lines are IT solid lines are CT regions

38 Tempering Martensite reduces brittleness of martensite, reduces internal stress caused by quenching. copyright by United States Steel Corporation, decreases TS, YS but increases %RA produces extremely small Fe 3 C particles surrounded by from Fig. furnished courtesy of Republic Steel Corporation.) 9 m YS(MPa) TS(MPa) Tempering T (°C) %RA TS YS %RA

39 The microstructure of tempered martensite, although an equilibrium mixture of α-Fe and Fe 3 C, differs from those for pearlite and bainite. This micrograph produced in a scanning electron microscope (SEM) shows carbide clusters in relief above an etched ferrite. (From ASM Handbook, Vol. 9: Metallography and Microstructures, ASM International, Materials Park, OH, 2004.)

40 Temper Martensite Embrittlement – an issue in Certain Steels Suspected to be due to the deposition of very fine carbides during 2 nd and 3 rd phase tempering along original austenite G. B. from the transformation of retained austenite, From: George Krauss, Steels: Processing, Structure, and Performance, ASM International, 2006.

41 Increasing Strength and Hardness of Alloy involves some Heat Treatment The effect of quenching in steels is determined by the Jominey End Quench Test Precipitation Hardness is a method used for may alloy systems (mostly Non-Ferrous ones) Grain Size control is also an important consideration – which can be controlled by annealing processes Recovery after cold work (cold work can also increase strength of alloys) Recrystalization Grain Growth

42 Schematic illustration of the Jominy end-quench test for hardenability. (After W. T. Lankford et al., Eds., The Making, Shaping, and Treating of Steel, 10th ed., United States Steel, Pittsburgh, PA, Copyright 1985 by United States Steel Corporation.)

43 Hardenability--Steels Ability to form martensite Jominy end quench test to measure hardenability. Hardness versus distance from the quenched end. (adapted from A.G. Guy, Essentials of Materials Science, McGraw-Hill Book Company, New York, 1978.) 24°C water specimen (heated to phase field) flat ground Rockwell C hardness tests Hardness, HRC Distance from quenched end

44 Figure The cooling rate for the Jominy bar (see Figure 10.21) varies along its length. This curve applies to virtually all carbon and low-alloy steels. (After L. H. Van Vlack, Elements of Materials Science and Engineering, 4th ed., Addison-Wesley Publishing Co., Inc., Reading, MA, 1980.)

45 Figure Variation in hardness along a typical Jominy bar. (From W. T. Lankford et al., Eds., The Making, Shaping, and Treating of Steel, 10th ed., United States Steel, Pittsburgh, PA, Copyright 1985 by United States Steel Corporation.)

46 The cooling rate varies with position. (adapted from H. Boyer (Ed.) Atlas of Isothermal Transformation and Cooling Transformation Diagrams, American Society for Metals, 1977, p. 376.) Why Hardness Changes W/Position distance from quenched end (in) Hardness, HRC A M A P T(°C) M(start) Time (s) 0 0% 100% M(finish) Martensite Martensite + Pearlite Fine Pearlite Pearlite

47 Hardenability vs Alloy Composition Jominy end quench results, C = 0.4 wt% C "Alloy Steels" (4140, 4340, 5140, 8640) --contain Ni, Cr, Mo (0.2 to 2wt%) --these elements shift the "nose". --martensite is easier to form. (adapted from figure furnished courtesy Republic Steel Corporation.) Cooling rate (°C/s) Hardness, HRC Distance from quenched end (mm) %M%M 4340 T(°C) Time (s) M(start) M(90%) shift from A to B due to alloying B A TETE

48 Figure Hardenability curves for various steels with the same carbon content (0.40 wt %) and various alloy contents. The codes designating the alloy compositions are defined in Table (From W. T. Lankford et al., Eds., The Making, Shaping, and Treating of Steel, 10th ed., United States Steel, Pittsburgh, PA, Copyright 1985 by United States Steel Corporation.)

49 Effect of quenching medium: Medium air oil water Severity of Quench low moderate high Hardness low moderate high Effect of geometry: When surface-to-volume ratio increases: --cooling rate increases --hardness increases Position center surface Cooling rate low high Hardness low high Quenching Medium & Geometry

50 wt% Cu L +L +L (Al) T(°C) composition range needed for precipitation hardening CuAl 2 A Adapted from Fig , Callister 7e. (Fig adapted from J.L. Murray, International Metals Review 30, p.5, 1985.) Precipitation Hardening Particles impede dislocations. Ex: Al-Cu system Procedure: --Pt B: quench to room temp. --Pt C: reheat to nucleate small crystals within crystals. Other precipitation systems: Cu-Be Cu-Sn Mg-Al Temp. Time --Pt A: solution heat treat (get solid solution) Pt A (soln heat treat) B Pt B C Pt C (precipitate Consider: 17-4 PH St. Steel and Ni-Superalloys too!

51 Figure Coarse precipitates form at grain boundaries in an Al– Cu (4.5 wt %) alloy when slowly cooled from the single-phase (κ) region of the phase diagram to the two-phase (θ + κ) region. These isolated precipitates do little to affect alloy hardness.

52 Figure By quenching and then reheating an Al–Cu (4.5 wt %) alloy, a fine dispersion of precipitates forms within the κ grains. These precipitates are effective in hindering dislocation motion and, consequently, increasing alloy hardness (and strength). This process is known as precipitation hardening, or age hardening.

53 Figure (a) By extending the reheat step, precipitates coalesce and become less effective in hardening the alloy. The result is referred to as overaging. (b) The variation in hardness with the length of the reheat step (aging time).

54 Figure (a) Schematic illustration of the crystalline geometry of a Guinier–Preston (G.P.) zone. This structure is most effective for precipitation hardening and is the structure developed at the hardness maximum shown in Figure 10.27b. Note the coherent interfaces lengthwise along the precipitate. The precipitate is approximately 15 nm × 150 nm. (From H. W. Hayden, W. G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. 3: Mechanical Behavior, John Wiley & Sons, Inc., NY, 1965.) (b) Transmission electron micrograph of G.P. zones at 720,000×. (From ASM Handbook, Vol. 9: Metallography and Microstructures, ASM International, Materials Park, OH, 2004.)

55 Figure Annealing can involve the complete recrystallization and subsequent grain growth of a cold-worked microstructure. (a) A cold-worked brass (deformed through rollers such that the cross-sectional area of the part was reduced by one-third). (b) After 3 s at 580°C, new grains appear. (c) After 4 s at 580°C, many more new grains are present. (d) After 8 s at 580°C, complete recrystallization has occurred. (e) After 1 h at 580°C, substantial grain growth has occurred. The driving force for this growth is the reduction of high-energy grain boundaries. The predominant reduction in hardness for this overall process had occurred by step (d) All micrographs have a magnification of 75×. (Courtesy of J. E. Burke, General Electric Company, Schenectady, NY.)

56 Figure The sharp drop in hardness identifies the recrystallization temperature as ~290°C for the alloy C26000, cartridge brass. (From Metals Handbook, 9th ed., Vol. 4, American Society for Metals, Metals Park, OH, 1981.)

57 Recrystallization Temperature, T R T R = recrystallization temperature = point of highest rate of property change 1.T R T m (K) 2.Due to diffusion annealing time T R = f(t) shorter annealing time => higher T R 3.Higher %CW => lower T R – strain hardening 4.Pure metals lower T R due to easier dislocation movements

58 Figure Recrystallization temperature versus melting points for various metals. This plot is a graphic demonstration of the rule of thumb that atomic mobility is sufficient to affect mechanical properties above approximately 1/3 to 1/2 T m on an absolute temperature scale. (From L. H. Van Vlack, Elements of Materials Science and Engineering, 3rd ed., Addison-Wesley Publishing Co., Inc., Reading, MA, 1975.)

59 Figure For this cold-worked brass alloy, the recrystallization temperature drops slightly with increasing degrees of cold work. (From L. H. Van Vlack, Elements of Materials Science and Engineering, 4th ed., Addison- Wesley Publishing Co., Inc., Reading, MA, 1980.)

60 At longer times, larger grains consume smaller ones. Why? Grain boundary area (and therefore energy) is reduced. After 8 s, 580ºC After 15 min, 580ºC 0.6 mm Adapted from Fig (d),(e), Callister 7e. (Fig (d),(e) are courtesy of J.E. Burke, General Electric Company.) Grain Growth Empirical Relation: coefficient dependent on material & Temp. grain dia. At time t. elapsed time Exponent is typ. 2 This is: Ostwald Ripening

61 Figure Schematic illustration of the effect of annealing temperature on the strength and ductility of a brass alloy shows that most of the softening of the alloy occurs during the recrystallization stage. (After G. Sachs and K. R. Van Horn, Practical Metallurgy: Applied Physical Metallurgy and the Industrial Processing of Ferrous and Nonferrous Metals and Alloys, American Society for Metals, Cleveland, OH, 1940.)

62 Figure Examples of cold-working operations: (a) cold-rolling a bar or sheet and (b) cold-drawing a wire. Note in these schematic illustrations that the reduction in area caused by the cold-working operation is associated with a preferred orientation of the grain structure.

63 We can then find that the cold working of an alloy is an effect tool for improving performance – if done properly! – so as not to cause the material to exceed its % EL which was a fracture deformation limit as we saw earlier –If we impose appropriate intermediate recrystalization (and maybe even grain growth steps) –Finish with a cold working step to achieve the desired hardness and finished size

64 Coldwork Hardening Example A cylindrical rod of brass originally 0.40 in (10.2 mm) in diameter is to be cold worked by drawing. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 55,000 psi (380 MPa) and a ductility of at least 15 %EL are desired. Further more, the final diameter must be 0.30 in (7.6 mm). Explain how this may be accomplished.

65 Coldwork Calculations Solution If we directly draw to the final diameter what happens? D o = 0.40 in Brass Cold Work D f = 0.30 in

66 Coldwork Calc Solution: Cont. For %CW = 43.8% – y = 420 MPa –TS = 540 MPa > 380 MPa 6 –%EL = 6< 15 This doesnt satisfy criteria…… what can we do?

67 Coldwork Calc Solution: Cont. Adapted from Fig. 7.19, Callister 7e For %EL > 15 For TS > 380 MPa > 12 %CW < 27 %CW our working range is limited to %CW = 12 – 27%

68 This process Needs an Intermediate Recrystallization i.e.: Cold draw-anneal-cold draw again For objective we need a cold work of %CW –Well use %CW = 20 Diameter after first cold draw (before 2 nd cold draw) –must be calculated as follows: Intermediate diameter =

69 Summary: 1.Initial Cold work D 01 = 0.40 in D f1 = in 2.Anneal above T R D s2 = D f1 1.Secondary Cold work D s2 = in D f 2 =0.30 in Therefore, we have met all requirements Coldwork Calculations Solution


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