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11_9

# Materials Science and Engineering: An Introduction

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252 The center of the steel cylinder will cool the slowest and therefore will be the softest. In moderately agitated water the equivalent distance from the quenched end for a 75 mm diameter bar for the center position is about 17 mm (11/16 in.) [Figure 11.8(a)]. The hardnesses at this position for the alloys cited are given below. Center Alloy Hardness (HRC) 8660 59 8640 42 8630 30 8620 22 Therefore, only 8660 and 8640 alloys will have a minimum of 40 HRC at the center, and therefore, throughout the entire cylinder. (b) This part of the problem asks us to do the same thing for moderately agitated oil. In moderately agitated oil the equivalent distance from the quenched end for a 75 mm diameter bar at the center position is about 25.5 mm (1-1/32 in.) [Figure 11.8(b)]. The hardnesses at this position for the alloys cited are given below. Center Alloy Hardness (HRC)
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Unformatted text preview: 8660 53 8640 37 8630 26 8620 <20 Therefore, only the 8660 alloy will have a minimum of 40 HRC at the center, and therefore, throughout the entire cylinder. 11.D3 A thirty eight millimeter diameter cylindrical steel specimen is to be heat treated such that the microstructure throughout will be at least 80% martensite. We are to decide which of several alloys will satisfy this criterion if the quenching medium is moderately agitated (a) oil, and (b) water. (a) Since the cooling rate is lowest at the center, we want a minimum of 80% martensite at the center position. From Figure 11.8(b), the cooling rate is equal to an equivalent distance from the quenched end of 12 mm (1/2 in.). According to Figure 11.5, the hardness corresponding to 80% martensite for these alloys is 50 HRC. Thus, all we need do is to...
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