HW-09-17-08s-Ch23

# HW-09-17-08s-Ch23 - IE 215 Solutions for Problems due(Ch 23...

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IE 215 Solutions for Problems due Sep 17, 2008 (Ch 23) 23.3 A series of turning tests were conducted using a cemented carbide tool, and flank wear data were collected. The feed was 0.010 in/rev and the depth was 0.125 in. At a speed of 350 ft/min, flank wear = 0.005 in at 1 min, 0.008 in at 5 min, 0.012 in at 11 min, 0.0.015 in at 15 min, 0.021 in at 20 min, and 0.040 in at 25 min. At a speed of 450 ft/min, flank wear = 0.007 in at 1 min, 0.017 in at 5 min, 0.027 in at 9 min, 0.033 in at 11 min, and 0.040 in at 13 min. The last value in each case is when final tool failure occurred. (a) On a single piece of linear graph paper, plot flank wear as a function of time. Using 0.020 in of flank wear as the criterion of tool failure, determine the tool lives for the two cutting speeds. (b) On a piece of natural log-log paper, plot your results determined in the previous part. From the plot, determine the values of n and C in the Taylor Tool Life Equation. (c) As a comparison, calculate the values of n and C in the Taylor equation solving simultaneous equations. Are the resulting n and C values the same? Solution: (a) Flank Wear vs Time 0.000 0.005 0.010 0.015 0.020 0.025 0.030 0.035 0.040 0.045 0 5 10 15 20 25 30 Time (min) Flank Wear (in) Flank v = 350 ft/min Flank v = 450 ft/min Using the graph, at 350 ft/min the tool last about 6.2 min ; at 450 ft/min, it lasts 19.0 min .

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