MCE311_HW05_sol_fall2011

# MCE311_HW05_sol_fall2011 - 20.16 A slab milling operation...

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20.16 A slab milling operation is performed on the top surface of a steel rectangular workpiece 30 cm long by 6.25 cm wide. The helical milling cutter, which has a 7.5 cm diameter and ten teeth, is set up to overhang the width of the part on both sides. Cutting speed is 37.5 m/min, feed is 0.015 cm/tooth, and depth of cut = 0.75 cm. Determine (a) the actual machining time to make one pass across the surface and (b) the maximum metal removal rate during the cut. (c) If an additional approach distance of 1.25 in is provided at the beginning of the pass (before cutting begins), and an overtravel distance is provided at the end of the pass equal to the cutter radius plus 1.25 cm, what is the duration of the feed motion. Solution : (a) N = v / π D = 37.5(100)/7.5 π = 159.15 rev/min f r = Nn t f = 159.15(10)(0.015) = 23.87 cm/min A = ( d ( D - d )) 0.5 = (0.75(7.5-0.75)) 0.5 = 2.25 cm T m = ( L + A )/ f r = (30 + 2.25)/23.87 = 1.35 min (b) R MR = wdf r = 6.25(0.75)(23.87) = 111.89 cm 3 /min (c) The cutter travels 1.25 cm before making contact with the work. It moves 2.25 cm before reaching full depth of cut. It then feeds the length of the work (30 cm). The overtravel consists of the cutter radius (3.75 cm) plus an additional 1.25 cm. Thus, T f = (1.25 + 2.25 + 30 + 3.75 + 1.25)/23.87 = 1.6 min 20.20 The top surface of a rectangular workpart is machined using a peripheral milling operation. The workpart is 735 mm long by 50 mm wide by 95 mm thick. The milling cutter, which is 60 mm in diameter and has five teeth, overhangs the width of the part equally on both sides. Cutting speed = 80 m/min, chip load = 0.30 mm/tooth, and depth of cut = 7.5 mm. (a) Determine the time required to make one pass across the surface, given that the setup and machine settings provide an approach distance of 5 mm before actual cutting begins and an overtravel distance of 25 mm after actual cutting has finished. (b) What is the maximum material removal rate during the cut? Solution : (a) N = v / π D = 80,000 mm/60 π = 424.4 rev/min f r = Nn t f = 424.4(5)(0.3) = 636.6 mm/min A = ( d ( D - d )) 0.5 = (7.5(60-7.5)) 0.5 = 19.84 mm T m = (735 + 5 + 19.84 + 25)/636.6 = 1.233 min (b) R MR = wdf r = 60(7.5)(636.6) = 286,470 mm 3 /min 21.21 The outside diameter of a cylinder made of a steel alloy is to be turned. The starting diameter is 300 mm and the length is 625 mm. The feed is 0.35 mm/rev and the depth of cut is 2.5 mm. The cut will be made with a cemented carbide cutting tool whose Taylor tool life parameters are: n = 0.24 and C = 450. Units for the Taylor equation are min for tool life and m/min for cutting speed. Compute the cutting speed

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## This note was uploaded on 02/12/2012 for the course MCE 311 taught by Professor Diab during the Fall '11 term at American Dubai.

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MCE311_HW05_sol_fall2011 - 20.16 A slab milling operation...

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