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Chap 11 Solns-6E


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CHAPTER 11 APPLICATIONS AND PROCESSING OF METAL ALLOYS PROBLEM SOLUTIONS 11.1 This question asks that we list four classifications of steels, and, for each, to describe properties and cite typical applications. Low-Carbon Steels Properties: nonresponsive to heat treatments; relatively soft and weak; machinable and weldable. Typical applications: automobile bodies, structural shapes, pipelines, buildings, bridges, and tin cans. Medium-Carbon Steels Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts. High-Carbon Steels Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades. High-Alloy Steels (Stainless and Tool) Properties: hard and wear resistant; resistant to corrosion in a large variety of environments. Typical applications: cutting tools, drills, cutlery, food processing, and surgical tools. 11.2 (a) Ferrous alloys are used extensively because: 1) Iron ores exist in abundant quantities. 2) Economical extraction, refining, and fabrication techniques are available. 3) The alloys may be tailored to have a wide range of properties. (b) Disadvantages of ferrous alloys are: 1) They are susceptible to corrosion. 2) They have relatively high densities. 3) They have relatively low electrical conductivities. 11.3 Ferritic and austenitic stainless steels are not heat treatable since "heat treatable" is taken to mean that martensite may be made to form with relative ease upon quenching austenite from an elevated temperature. 42
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For ferritic stainless steels, austenite does not form upon heating, and, therefore, the austenite-to-martensite transformation is not possible. For austenitic stainless steels, the austenite phase field extends to such low temperatures that the martensitic transformation does not occur. 11.4 The alloying elements in tool steels (e.g., Cr, V, W, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds. 11.5 We are asked to compute the volume percent graphite in a 3.5 wt% C cast iron. It first becomes necessary to compute mass fractions using the lever rule. From the iron-carbon phase diagram (Figure 11.2), the tie-line in the α and graphite phase field extends from essentially 0 wt% C to 100 wt% C. Thus, for a 3.5 wt% C cast iron W α = Χ Γρ - Χ ο Χ Γρ - Χ α = 100 - 3.5 100 - 0 = 0.965 W Gr = C o - Χ α Χ Γρ - Χ α = 3.5 - 0 100 - 0 = 0.035 Conversion from weight fraction to volume fraction of graphite is possible using Equation (9.6a) as V Gr = W Gr ρ Γρ α ρ α + Γρ ρ Γρ = 0.035 2.3 g/cm 3 0.965 7.9 g/cm 3 + 0.035 2.3 γ / χμ 3 = 0.111 or 11.1 vol% 11.6 Gray iron is weak and brittle in tension because the tips of the graphite flakes act as points of stress concentration. 11.7 This question asks us to compare various aspects of gray and malleable cast irons.
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