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### HW_1_IE215_Sol_f10

Course: IE IE 215, Fall 2010
School: Lehigh
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Word Count: 798

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In 21.1 an orthogonal cutting operation, the tool has a rake angle = 15. The chip thickness before the cut = 0.30 mm and the cut yields a deformed chip thickness = 0.65 mm. Calculate (a) the shear plane angle and (b) the shear strain for the operation. Solution: (a) r = to/tc = 0.30/0.65 = 0.4615 = tan-1(0.4615 cos 15/(1 - 0.4615 sin 15)) = tan-1(0.5062) = 26.85 (b) Shear strain = cot 26.85 + tan (26.85 - 15) =...

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In 21.1 an orthogonal cutting operation, the tool has a rake angle = 15. The chip thickness before the cut = 0.30 mm and the cut yields a deformed chip thickness = 0.65 mm. Calculate (a) the shear plane angle and (b) the shear strain for the operation. Solution: (a) r = to/tc = 0.30/0.65 = 0.4615 = tan-1(0.4615 cos 15/(1 - 0.4615 sin 15)) = tan-1(0.5062) = 26.85 (b) Shear strain = cot 26.85 + tan (26.85 - 15) = 1.975 + 0.210 = 2.185 21.4 In a turning operation, spindle speed is set to provide a cutting speed of 1.8 m/s. The feed and depth of cut of cut are 0.30 mm and 2.6 mm, respectively. The tool rake angle is 8. After the cut, the deformed chip thickness is measured to be 0.49 mm. Determine (a) shear plane angle, (b) shear strain, and (c) material removal rate. Use the orthogonal cutting model as an approximation of the turning process. Solution: (a) r = to/tc = 0.30/0.49 = 0.612 = tan-1(0.612 cos 8/(1 0.612 sin 8)) = tan-1(0.6628) = 33.6 (b) = cot 33.6 + tan (33.6 - 8) = 1.509 + 0.478 = 1.987 (c) RMR = (1.8 m/s x 103 mm/m)(0.3)(2.6) = 1404 mm3/s 21.10 The shear strength of a certain work material = 50,000 lb/in2. An orthogonal cutting operation is performed using a tool with a rake angle = 20 at the following cutting conditions: cutting speed = 100 ft/min, chip thickness before the cut = 0.015 in, and width of cut = 0.150 in. The resulting chip thickness ratio = 0.50. Determine (a) the shear plane angle, (b) shear force, (c) cutting force and thrust force, and (d) friction force. Solution: (a) = tan-1(0.5 cos 20/(1 - 0.5 sin 20)) = tan-1(0.5668) = 29.5 (b) As = (0.015)(0.15)/sin 29.5 = 0.00456 in2. Fs = AsS = 0.00456(50,000) = 228 lb (c) = 2(45) + - 2() = 90 + 20 - 2(29.5) = 50.9 Fc = 228 cos (50.9 - 20)/cos (29.5 + 50.9 -20) = 397 lb Ft = 228 sin (50.9 - 20)/cos (29.5 + 50.9 -20) = 238 lb (d) F = 397 sin 20 + 238 cos 20 = 359 lb The following problem should have been assigned instead of 21.18. 21.19 In a turning operation on stainless steel with hardness = 200 HB, the cutting speed = 200 m/min, feed = 0.25 mm/rev, and depth of cut = 7.5 mm. How much power will the lathe draw in performing this operation if its mechanical efficiency = 90%. Use Table 21.2 to obtain the appropriate specific energy value. Solution: From Table 21.2, U = 2.8 N-m/mm3 = 2.8 J/mm3 RMR = vfd = (200 m/min)(103 mm/m)(0.25 mm)(7.5 mm) = 375,000 mm3/min = 6250 mm3/s Pc = (6250 mm3/s)(2.8 J/mm3) = 17,500 J/s = 17,500 W = 17.5 kW Accounting for mechanical efficiency, Pg = 17.5/0.90 = 19.44 kW 21.30 A turning operation is performed on an engine lathe using a tool with zero rake angle in the direction of chip flow. The work material is an alloy steel with hardness = 325 Brinell hardness. The feed is 0.015 in/rev, depth of cut is 0.125 in and cutting speed is 300 ft/min. After the cut, the chip thickness is ratio measured to be 0.45. (a) Using the appropriate value of specific energy from Table 21.2, compute the horsepower at the drive motor, if the lathe has an efficiency = 85%. (b) Based on horsepower, compute your best estimate of the cutting force for this turning operation. Use the orthogonal cutting model as an approximation of the turning process. Solution: (a) From Table 21.2, U = Pu = 520,000 in-lb/in3 for alloy steel of the specified hardness. Since feed is greater than 0.010 in/rev in the table, a correction factor must be applied from Figure 21.14. For f = 0.015 in/rev = to, correction factor = 0.95. Thus, U = 520,000(0.95) = 494,000 in-lb/in3 = 41,167 ft-lb/in3. RMR = 300 x 12(.015)(0.125) = 6.75 in3/min Pc = U RMR = 41,167(6.75) = 277,875 ft-lb/min HPc = 277,875/33,000 = 8.42 hp HPg = 8.42/0.85 = 9.9 hp (b) HPc = vFc/33,000. Rearranging, Fc = 33,000 (HPc/v) = 33,000(8.42/300) = 926 lb. Check: Use unit horsepower from Table 21.2 rather than specific energy. HPu = 1.3 hp/(in3/min). Applying the correction factor correction factor = 0.95, HPu = 1.235 hp/(in3/min). RMR = 300 x 12(.015)(0.125) = 6.75 in3/min, same as before HPc = 1.235(6.75) = 8.34 hp HPg = 8.34/0.85 = 9.8 hp (b) Fc = 33,000 (8.3/300) = 913 lb. 21.33 Orthogonal cutting is performed on a metal whose mass specific heat = 1.0 J/g-C, density = 2.9 g/cm3, and thermal diffusivity = 0.8 cm2/s. The cutting speed is 4.5 m/s, uncut chip thickness is 0.25 mm, and width of cut is 2.2 mm. The cutting force is measured at 1170 N. Using Cook's equation, determine the cutting temperature if the ambient temperature = 22C. Solution: C = (2.9 g/cm3)(1.0 J/g-C) = 2.90 J/cm3-C = (2.90x10-3) J/mm3-C K = 0.8 cm2/s = 80 mm2/s U = Fcv/RMR = 1170 N x 4.5 m/s/(4500 mm/s x 0.25 mm x 2.2 mm) = 2.127 N-m/mm3 T = 0.4U/(C) x (vto/K)0.333 T = 22 + (0.4 x 2.127 N-m/mm3/(2.90x10-3) J/mm3-C) [4500 mm/s x 0.25 mm/80 mm2/s]0.333 T = 22 + (0.2934 x 103 C)(14.06).333 = 22 + 293.4(2.41) = 22 + 707 = 729C 21.36 It is desired to estimate the cutting temperature for a certain alloy steel whose hardness = 240 Brinell. Use the appropriate value of specific energy from Table 21.2 and compute the cutting temperature by means of the Cook equation for a turning operation in which the cutting speed is 500 ft/min, feed is 0.005 in/rev, and depth of cut is 0.070 in. The work material has a volumetric specific heat of 210 in lb/in3-F and a thermal diffusivity of 0.16 in2/sec. Assume ambient temperature = 88F. Solution: From Table 21.2, U for alloy steel (310 BHN) = 320,000 in-lb/in3. Since f = 0.005 in/rev, correction factor = 1.25. Therefore U = 320,000(1.25) = 400,000 in-lb/in3. v = 500 ft/min x 12 in/ft/60 sec/min = 100 in/sec. T = Ta + (0.4U/C)(vto/K)0.333 = 88 + (0.4 x 400,000/210)(100 x 0.005/0.16)0.333 = 88 + (762)(3.125)0.333 = 88 + 1113 = 1201F
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