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HW2_solution2

# HW2_solution2 - Manufacturing Methods Homework 2 Solution...

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Manufacturing Methods Homework 2: Solution Fall 2009 Review Questions: a) Refer to Page 677, 678 of the text book b) A good discussion on adhesion, abrasion, fatigue, and mechanical overloading. Problem 16C-12 a) First, the hardness of the material must be found. The metallurgical condition is not defined; assume either annealed or, for greater strength, cold-worked. It is similar to 302SS; from Table 8-2, TS = 600 MPa, thus HB = 3(600)/9.8 = 183. From Fig. 16-45, taking HB = 183, and taking into account a 25% reduction for austenitic stainless steel, 10 % reduction in speed and feed, since Hole depth/D = 3.33 and a 25% increase for using coated carbides. v s = 1.8 m/s; from Figure 16-45 v = (0.75) (0.9) (v s ) (1.25) = 1.52 m/s. b) Drill (r/min) = v / π *D = (1.52 * 1000) / (6* π ) = 80.57 r/s =4834.33 r/min c) From sec. 16.7-1, feed = 0.01D (0.9) = 0.01* 6 * 0.9 = 0.054 mm/r d) Feed rate = (feed) (r/s) = (0.054) (80.57) = 4.35 mm/s e) The drill has two cutting edges, therefore, h = feed / 2 = 0.054 / 2 = 0.027 mm Note that in Eq. 16-15, the reference thickness is 1 mm, thus, h a = (0.027) -0.4 = 4.24 From Table 16-1, E 1 = 2.3 W.s/ mm 3 E = h a E 1 = (4.24) (2.3) = 9.75 W.s/ mm 3 From Eq. 16-16a, W = (EV t ) / η Vt = ( π D 2 / 4) (feed rate) = (6 2 * π /4) (4.35) = 156.6 mm 3 /s Taking η = 0.7, Power = (9.75) (156.6) / (0.7) = 2181 W

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• Fall '06
• El-Gizawy
• Trigraph, Stainless steel, W.s/ mm

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