Homework_4_-_solutions

Homework_4_-_solutions - Homework #4: Due on November 3rd...

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Homework #4: Due on November 3rd at midnight through the course web site. The solution will be posted the day after so that you can check your answers before the exam. Chapter 23 problems : 23.5, 23.6, 23.8, 23.12, 23.13, 23.17, 23.22, 23.24 Chapter 24 problems: 24.1, 24.3, 24.5, 24.7, 24.15, 24.17, 24.21 Chapter 23 problems 23.5 Tool life tests on a lathe have resulted in the following data: (1) at a cutting speed of 375 ft/ min, the tool life was 5.5 min; (2) at a cutting speed of 275 ft/min, the tool life was 53 min. (a) Determine the parameters n and C in the Taylor tool life equation. (b) Based on the n and C values, what is the likely tool material used in this operation? (c) Using your equation, compute the tool life that corresponds to a cutting speed of 300 ft/min. (d) Compute the cutting speed that corresponds to a tool life T = 10 min. Solution : (a) VT n = C ; Two equations: (1) 375(5.5) n = C and (2) 275(53) n = C 375(5.5) n = 275(53) n 375/275 = (53/5.5) n 1.364 = (9.636) n ln 1.364 = n ln 9.636 0.3102 = 2.2655 n n = 0.137 C = 375(5.5) 0.137 = 375(1.2629) C = 474 Check: C = 275(53) 0.137 = 275(1.7221) = 474 (b) Comparing these values of n and C with those in Table 23.2, the likely tool material is high speed steel. (c) At v = 300 ft/min, T = ( C / v ) 1/ n = (474/300) 1/0.137 = (1.579) 7.305 = 28.1 min (d) For T = 10 min, v = C / T n = 474/10 0.137 = 474/1.371 = 346 ft/min 23.6 Tool life tests in turning yield the following data: (1) when cutting speed is 100 m/min, tool life is 10 min; (2) when cutting speed is 75 m/min, tool life is 30 min. (a) Determine the n and C values in the Taylor tool life equation. Based on your equation, compute (b) the tool life for a speed of 110 m/min, and (c) the speed corresponding to a tool life of 15 min. Solution : (a) Two equations: (1) 100(10) n = C and (2) 75(30) n = C . 100(10) n = 75(30) n ln 100 + n ln 10 = ln 75 + n ln 30 4.6052 + 2.3026 n = 4.3174 + 3.4012 n 4.6052 - 4.3174 = (3.4012 – 2.3026) n 0.2877 = 1.0986 n n = 0.2619 C = 100(10) 0.2619 = 100(1.8275) C = 182.75 (b) 110 T 0.2619 = 182.75 T 0.2619 = 182.75/110 = 1.661 T = 1.661 1/0.2619 = 1.661 3.819 = 6.95 min (c) v (15) 0.2619 = 182.75 v = 182.75/(15) 0.2619 = 182.75/2.0322 = 89.9 m/min 23.8 A 15.0 in by 2.0 in workpart is machined in a face milling operation using a 2.5 in diameter fly cutter with a single carbide insert. The machine is set for a feed of 0.010 in/tooth and
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a depth of 0.20 in. If a cutting speed of 400 ft/min is used, the tool lasts for 3 pieces. If a cutting speed of 200 ft/min it used, the tool lasts for 12 parts. Determine the Taylor tool life equation. Solution: N 1 = v D = 400(12)/2.5π = 611 rev/min f r = Nfn t = 611(0.010)(1) = 6.11 in/min A = D /2 = 2.50 / 2 = 1.25 T m = ( L +2 A )/ f r = (15 + 2(1.25))/6.11 = 2.863 min T 1 = 3 T m = 3(2.863) = 8.589 min when v 1 = 400 ft/min N 2 = 200(12)/2.5π = 306 rev/min f r = Nfn t = 306(0.010)(1) = 3.06 in/min T m = (15 + 2(1.25))/3.06 = 5.727 min T 2 = 12
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This note was uploaded on 12/07/2009 for the course E ISE 515 taught by Professor Olaharryson during the Fall '09 term at N.C. State.

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Homework_4_-_solutions - Homework #4: Due on November 3rd...

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