HW-CH11 - 454 ° Chapter H l Phase Transformations are...

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Unformatted text preview: 454 ° Chapter H l Phase Transformations are either bulky or polar. Molecular weight and degree of branching also affect Tm and TS. wPORTANT TERMS AND ENEEPTS Alloy steel Artificial aging diagram Athermal transformation Kinetics Bainite Martensite Coarse pearlite Continuous cooling (polymers) transformation diagram Natural aging Fine pearlite Nucleation Free energy Overaging Glass transition temperature Growth (phase particle) REFERENCES Atkins. M., Atlas of Continuous Cooling Transfor- mation Diagrams for Engineering Steels, British Steel Corporation, Sheffield, England, 1980. Atlas of Isothermal Transformation and Cooling Transformation Diagrams. ASM International, Materials Park. OH, 1977. Billmeyer, F. W.. Jr., Textbook of Polymer Science, 3rd edition, Wiley-Interscience, New York, 1984. Chapter 10. Brooks, C. R., Principles of the Heat Treatment of Plain Carbon and Low Alloy Steels, ASM In- ternational, Materials Park, OH, 1996. Porter, D. A. and K. E. Easterling, Phase Trans- formations in Metals and Alloys, Chapman and Hall, New York, 1992. QUESTIONS AND PROBLEMS Isothermal transformation Melting temperature Phase transformation Plain carbon steel Precipitation hardening Precipitation heat treatment Solution heat treatment Spheroidite Supercooling Superbeating Tempered martensite Thermally activated transformation Ttansformation rate Shewmon, P. 6., Transformations in Metals, McGraw-Hill, New York. 1969. Reprinted by Williams Book Company, Tulsa, OK. Vander Voort, G. (Editor). Atlas of Time— Temperature Diagram for Irons and Steels. ASM International, Materials Park, OH, 1991. Vander Voort, G. (Editor), Atlas of Iime~ Temperature Diagrams for Nonferrous Al- loys, ASM International. Materials Park, OH. 1991. Young, R. J. and P. Lovell, Introduction to Poly- mers. 2nd edition, CRC Press, Boca Raton. FL, 1991. Additional problems and questions for this chapter may be found on both Student and lnstroctor Companion Sites_at_w_ivfley.com/college/callu_ter,m The Kinetics of Phase Transformations 11.1 (a) Rewrite the expression for the total free energy change for nucleation (Equation 11.1) for the case of a cubic nucleus of edge length a (instead of a sphere of radius r). Now differentiate this expres- sion with respect to a (per Equation 11.2) and solve for both the critical cube edge length, a", and also AG‘. (b) Is at? greater for a cube or a sphere? Why? 11.2 (a) For the solidification of nickel, calculate the critical radius r‘ and the activation free energy AG" if nucleation is homo- geneous. Values for the latent heat of fu- sion and surface free energy are —2.53 x 109 Jlm3 and 0.255 mm, respectively. Use the supercooling value found in Ta- ble 11.1. (b) Now calculate the number of atoms found in a nucleus of critical size. Assume a lattice parameter of 0.360 nm for solid nickel at its melting temperature. 11.3 (:1) Assume for the solidification of nickel (Problem 11.2) that nucleation is homo» geneous, and the number of stable nuclei is 106 nuclei per cubic meter. Calculate the critical radius and the number of sta- ble nuclei that exist at the following de- grees of supercooling: 200 K and 300 K. (b) What is significant about the magnitudes of these critical radii and the numbers of stable nuclei? 11.4 Compute the rate of some reaction that obeys Avrami kinetics, assuming that the constants n and k have values of 2.0 and 5 x 10". re- spectively, for time expressed in seconds. 11.5 The kinetics of the austenite«to-pearlite transformation obey (the Avrami relation- ship. Using the fraction transformed—time data given here, determine the total time re» quired for 95 % of the austenite to transform to peariite: Imimlirrsfsmd--. ._ Time} 0.2 230 0.6 _ __ 42;; 11.6 (a) From the curves shown in Figure 11.11 and using Equation 11.18, determine the rate of recrystallization for pure copper at the several temperatures (IJ) Make a plot of ln(rate) versus the recip- rocal of temperature (in K" ), and deter- mine the activation energy for this recrys- tallization process (See Section 6.5.) (c) By extrapolation, estimate the length of time required for 50% recrystallization at room temperature. 20°C (293 K). Metastable Versus Equilibrium States 11.7 (a) Briefly describe the phenomena of super- heating and supercooling. (13) Why do these phenomena occur? isothermal Transformation Diagrams 11.8 Suppose that a steel of eutectoid composi- tion is cooled to 675°C (1250“F) from 760°C (1400°F) in less than 0.5 s and held at this temperature. (a) How long will it take for the austenite- to-pearlite reaction to go to 50% com- pletion? To 100% completion? Questions and Problems I 455 (b) Estimate the hardness of the alloy that has completely transformed to peariite. 11.9 What is the driving force for the formation of spheroidite? 11.10 Using the isothermal transformation dia- gram for an iron—carbon alloy of eutectoid composition (Figure 11.23), specify the na— ture of the final microstructure (in terms of microconstituents present and approximate percentages of each) of a small specimen that has been subjected to the following time— temperature treatments In each case assume that the specimen begins at 760°C (1400°F) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Rapidly cool to 400°C (750°F), hold for 500 s, then quench to room temperature. (13) Reheat the specimen in part (a) to 100°C (1290°F) for 20 h. (c) Cool rapidly to 665°C (1230°F), hold for 103 s, then quench to room temperature. (d) Rapidly cool to 350°C (660°F), hold for 150 s, then quench to room tempera» ture. 11.11 Using the isothermal transformation dia- gram for a 1.13 wt% C steel alloy (Fig- ure 11.49). determine the final microstruc- ture (in terms of just the microconstituents present) of a small specimen that has been subjected to the following time—temperature treatments. In each case assume that the spec- imen begins at 920°C (1690°F) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Rapidly cool to 775°C (1430°F), hold for 500 s. then quench to room temperature. (1)) Rapidly cool to 100°C (1290°F). hold at this temperature for 10'i s. then quench to room temperature. (c) Rapidly cool to 350°C (660°F), hold for 300 s, then quench to room temperature. (d) Rapidly cool to 600°C (1110°F), hold at this temperature for 7 s, rapidly cool to 450°C (840°F). hold at this temperature for4 s, then quench to room temperature. 11.12 For parts (13), (c), and (d) of Problem 11.11, determine the approximate percentages of the microconstituents that form. 456 U Chapter" 1' Phase Transformations Figure 11.49 Isothermal 1500' transformation diagram for a 1.13 wt% C iron—carbon alloy: A. auslenite; B, bainite; C, 1400 . . proeutectord cementite; M, - martensite; P. pearlite. [Adapted 1200 from H. Boyer (Editor). Atlas of Isothermal Transformation and Cooiing Masfonnarion 1000 Diagram, American Society for Temperature 1°C] Metals, 1977, p. 33.] Temperature (’F) 600 Time is} 11.13 Make a copy of the isothermal transforma- tion diagram for a 1.13 wt% C iron—carbon alloy (Figure 11.49), and then onthis diagram sketch and label time—temperature paths to produce the following microstructures: (a) 6.2% proeutectoid cementite and 93.8% coarse pearlite (b) 50% fine pearlitc and 50% bainite (c) 100% martensite (d) 100% tempered martensite Continuous Cooling Tramfonnation Diagrams 11.14 Figure 11.50 shows the continuous cooling transformation diagram for a 0.35 wt% C iron—carbon alloy. Make a copy of this figure and then sketch and label continuous cool- ing curves to yield the following microstruc- tures: in) Fine pearlite and proeutectoid ferrite (b) Martensite and proeutectoid ferrite (c) Martensite. fine pearlite. and proeutec- toid ferrite 11.15 Cite two important differences betWeen con- tinuous cooling transformation diagrams for plain carbon and alloy steels. 200 11.16 Name the microstructural products of 4340 alloy steel specimens that are first com- pletely transformed to austenite, then cooled to room temperature at the following rates: (a) 0.005°C.r's. (1:) 05°05, (11) 0.05°C!s, (d) 5°Cfs. 11.1'1r Briefly describe the simplest continuous cool- ing heat treatment procedure that would be used in converting a 4340 steel from one mi- crostructure to another. (a) (Martensite + ferrite + bainite) to (martensite + ferrite + pcarlite + bai- nite) (b) (Martensite + ferrite + bainite) to spheroiditc (c) (Martensite + bainite + ferrite) to tem- pered martensite Mechanical Behavior of iron-Carbon Ailey: Tempered Martensite 11.18 Briefly explain why fine pearlite is harder and stronger than coarse pearlite, which in turn is harder and stronger than spheroidite. Temperature (“Cl Time (5) 11.19 Rank the following iron-carbon alloys and associated microstructures from the hardest to the softest: (a) 0.25 wt% C with coarse pearlite, (b) 0.80 wt% C with spheroidite, (c) 0.25 wt% C with spheroidite, and (d) 0.80 wt% C with fine pearlite. Justify this rank- ing. 11.20 Briefly describe the simplest heat treatment procedure that would be used in converting a 0.76 wt% C steel from one microstructure to the other, as follows: (a) Spheroidite to martensite (h) Pearlite to bainite (e) Pearlite to spheroidite (d) Baim'te to spheroidite 11.21 Estimate the Rockwell hardnesses for spec- imens of an iron—carbon alloy of eutectoid composition that have been subjected to the heat treatments described in parts (b) and (c) of Problem 11.10. 11.22 Estimate the Brinell hardnesses for speci- mens of a 1.13 wt% C iron—carbon alloy that have been subjected to the heat treatments described in parts (b) and (d) of Problem 11.11. Questions and Problems - 457 Figure 11.50 Continuous cooling transformation diagram for “1.35 wt% C ironvcarbon alloy. 11.23 For a eutectoid steel, describe isothermal heat treatments that would be required to yield specimens having the following Brinell hardnesses: (a) 180 HB, 03) 220 HB, (e) 500 HB. Crystallization (Polymer!) 11.24 Determine values for the constants n and k (Equation 11.17) for the crystallization of polypropylene (Figure 11.46) at 150°C. Melting and Glass Transition Temperatures 11.25 Of those polymers listed in Table 11.3, which polymer(s) would be best suited for use as ice cube trays? Why? Factors That influence Melting and Glass Transition Temperature: 11.26 For each of the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one poly- mer has a higher melting temperature than the other: (2) if it is possible. note which has 458 0 Chapter 1'! 1' Phase Transformations the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (a) Branched polyethylene having a number- average molecular weight of 850,000 g/mol; linear polyethylene having a number-average molecular weight of 850.000 g/mol (b) Polytetrafiuoroethylene having a den- sity of 2.14 g/cm3 and a weight-average molecular weight of 600,000 gfmol; PTF E having a density of 2.20 g/ctn3 and DESIGN PROBLEMS Continuous Cooling Transformation Diagram: Mechanical Behavior of lroanarbon Alloys p.131 Is it possible to produce an ironwcarbon alloy of eutectoid composition that has a minimum hardness of 200 BB and a mini- mum ductility of 25 % RA? If so, describe the continuous cooling heat treatment to which the alloy would be subjected to achieve these properties. If it is not possible, explain why. 11.02 It is desired to produce an iron—carbon al- loy that has a minimum hardness of 200 HB and a minimum ductility of 35% RA. Is such Composition total; Be} 1000 g 800 E. E 600 I— 400 (Cu) Composition lwt‘it. Bel Figure l LS! beryllium phase diagram. [Adapted from anry Alloy Phase Diagrams. 2nd edition. Vol. 2. T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International. Materials Park. OH] a weight-average molecular weight of 600,000 g/mol (c) Linear and syndiotactic poly(vinyl chlo- ride) having a number-average molecular weight of 500,000 glrnol; linear polyethy- lene having a number-average molecular weight of 225,000 glmol (d) Linear and syndiotactic polypropy- lene having a weight-average molecu- lar weight of 500,000 glmol; linear and atactic polypropylene having a weight- average molecular weight of 750,000 gfmol an alloy possible? If so, what will be its com- position and microstructure (coarse and fine pearlites and spheroidite are alternatives)? If this is not possible, explain why. Tempered Martensite 1133 An alloy steel (4340) is to be used in an appli- cation requiring a minimum tensile strength of 1515 MPa (220,000 psi) and a minimum ductility of 40% RA. Oil quenching followed by tempering is to be used. Briefly describe the tempering heat treatment. The copper-rich side of the copper- Design Problems 0 459 Heat Treatments {Preapitafion Hardening) an alloy having a composition of your 11.D4 Copper-rich copper—beryllium alloys are .choosmg, yet lying mtlun the range precipitation hardenablc. After consulting given for part (50' . f th . :11: ff 1.31:3}: fougwfilgfe diagram (Figure Mechanism of Hardening (a) Specify the range of compositions over ILDS Is it possible to produce a precipitation hard- which these alloys may be precipitation ened 2014 aluminum alloy having a mini- hardened- mum yield strength of 380 MP3 (55,000 psi) (13) Briefly describe the heat~treatment pro- and a ductility of at least 15% EL? If so. cedures (in terms of temperatures) that specify the precipitation heat treatment. If would be used to precipitation harden it is not possible then explain why. ...
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