HW_2_IE215_f10_sol

HW_2_IE215_f10_sol - 22.1 A cylindrical workpart 200 mm in...

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22.1 A cylindrical workpart 200 mm in diameter and 700 mm long is to be turned in an engine lathe. Cutting speed = 2.30 m/s, feed = 0.32 mm/rev, and depth of cut = 1.80 mm. Determine (a) cutting time, and (b) metal removal rate. Solution : (a) N = v /( π D ) = (2.30 m/s)/0.200 π = 3.66 rev/s f r = Nf = 6.366(.3) = 1.17 mm/s T m = L / f r = 700/1.17 = 598 s = 9.96 min Alternative calculation using Eq. (22.5), T m = 200(700) π /(2,300 x 0.32) = 597.6 sec = 9.96 min (b) R MR = vfd = (2.30 m/s)(10 3 )(0.32 mm)(1.80 mm) = 1320 mm 3 /s 22.3 A facing operation is performed on an engine lathe. The diameter of the cylindrical part is 6 in and the length is 15 in. The spindle rotates at a speed of 180 rev/min. Depth of cut = 0.110 in, and feed = 0.008 in/rev. Assume the cutting tool moves from the outer diameter of the workpiece to exactly the center at a constant velocity. Determine (a) the velocity of the tool as it moves from the outer diameter towards the center and (b) the cutting time. Solution: (a) f r = fN = (0.008 in/rev)(180 rev/min) = 1.44 in/min (b) L = distance from outside to center of part = D /2; T m = L / f r = D /(2 f r ) = 6/(2 x 1.44) = 2.083 min 22.6 A cylindrical work bar with 4.5 in diameter and 52 in length is chucked in an engine lathe and supported at the opposite end using a live center. A 46.0 in portion of the length is to be turned to a diameter of 4.25 in one pass at a speed of 450 ft/min. The metal removal rate should be 6.75 in 3 /min. Determine (a) the required depth of cut, (b) the required feed, and (c) the cutting time. Solution

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This note was uploaded on 01/24/2011 for the course IE 215 taught by Professor Groover during the Fall '08 term at Lehigh University .

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HW_2_IE215_f10_sol - 22.1 A cylindrical workpart 200 mm in...

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