Materials Science and Engineering: An Introduction

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241 (b) This portion of the problem asks for a hardness of 40 HRC the microstructure could consist of either (1) about 75% fine pearlite and 25% martensite (Figure 10.24), or (2) tempered martensite (Figure 10.27). For case (1), after austenitizing, rapidly cool to about 580 ° C (Figure 10.14), hold at this temperature for about 4 s (to obtain 75% fine pearlite), and then rapidly quench to room temperature. For case (2), after austenitizing, rapidly cool to room temperature in order to achieve 100% martensite. Then temper this martensite for about 2000 s at 535 ° C (Figure 10.27). (c) From Figure 10.22(a), in order for a 0.76 wt% C alloy to have a Rockwell hardness of 27 HRC, the microstructure must be fine pearlite. Thus, utilizing the isothermal transformation diagram for this alloy, Figure 10.14, we must rapidly cool to a temperature at which fine pearlite forms (i.e., at about 580 ° C), allow the specimen to isothermally and completely transform to fine
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Unformatted text preview: pearlite. At this temperature an isothermal heat treatment for at least 7 s is required. 10.38 The (a) and (b) portions of the problem ask that we make schematic plots on the same graph for the tensile strength versus composition for copper-silver alloys at both room temperature and 600 ° C; such a graph is shown below. 100 8 0 6 0 4 0 2 0 Composition (wt% Ag) Room temperature 600 ° C Tensile strength (c) Upon consultation of the Cu-Ag phase diagram (Figure 9.6) we note that silver is virtually insoluble in copper (i.e., there is no α phase region at the left extremity of the phase diagram); the same may be said the solubility of copper in silver and for the β phase. Thus, only the α and β phase will exist for all compositions at room temperature; in other words, there will be no...
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This document was uploaded on 05/04/2010.

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