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10_8

Materials Science and Engineering: An Introduction

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234 10.17 This problem asks us to determine the approximate percentages of the microconstituents that form for five of the heat treatments described in Problem 10.16. (a) 100% martensite (c) 100% bainite (d) 100% spheroidite (f) 70% bainite and 30% martensite (h) After holding for 10 s at 625 ° C, the specimen has completely transformed to proeutectoid ferrite and fine pearlite; no further reaction will occur at 400 ° C. Therefore, we can calculate the mass fractions using the appropriate lever rule expressions, Equations (9.20) and (9.21), as
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Unformatted text preview: follows: W α ' = 0.76 - C o ' 0.74 = 0.76 - 0.45 0.74 = 0.42 or 42% W p = C o ' - 0.022 0.74 = 0.45 - 0.022 0.74 = 0.58 or 58% 10.18 Below is shown an isothermal transformation diagram for a 0.45 wt% C iron-carbon alloy, with time-temperature paths that will produce (a) 42% proeutectoid ferrite and 58% coarse pearlite; (b) 50% fine pearlite and 50% bainite; (c) 100% martensite; and (d) 50% martensite and 50% austenite....
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