Solution for Homework #4 Chapter 21: Multiple Choice Questions (21.3, 21.4, 21.6, 21.8) Problems (21.10, 21.21, 21.20, 21.35) MCQ: 21.3 (d) 21.4 (c) 21.6 (c) 21.8 (a) Problems: 21.10 Solution: (a) φ = tan-1(.5 cos 20/(1 - .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) β = 2(45) + 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. 21.20 Solution: From Table 21.3, HPu = 0.25 hp/(in3/min) for aluminum. Since feed is greater than 0.010 in/rev in the table, a correction factor must be applied from Figure 21.14. For f = 0.020 in/rev = to, correction factor = 0.9. HPc = HPu x MRR, HPg = HP/E MRR = vfd = 900 x 12(.020)(0.250) = 54 in3/min HPc = 0.9(0.25)(54) = 12.2 hp HPg = 12.2/0.87 = 14.0 hp 21.21 Solution: From Table 21.3, U = 2.8 N-m/mm3 = 2.8 J/mm3

MRR = vfd = (200 m/min)(103 mm/m)(6 mm)f = 1200(103)f mm3/min = 20(103)f mm3/s Available power Pc = Pg E = 25(103)(0.90) = 22.5 (103) = 22,500W Required power Pc = (2.8 N-m/mm3)( 20 x 103) f = 56,000 f Setting available power = required power, 22,500 = 56,000 f f = 22,500/56,000 = 0.402 mm (this should be interpreted as mm/rev for a turning operation) However, for this feed, correction factor in Figure 21.14 = 0.9; thus U = 2.8(0.90) = 2.52 N-m/mm3 and an iterative calculation procedure is required to match the unit power value with the feed, taking the correction factor into account. Required Pc = (2.52)(20 x 103) f = 50,400 f Again setting available power = required power, 22,500 = 50,400 f f = 22,500/50,400 = 0.446 mm/rev One more iteration using the correction factor yields a value around f = 0.45 mm/rev. 21.35 Solution: MRR = vtow, v = MRR/tow = 1.8/(0.01 x 0.100) = 1800 in/min = 30 in/sec U = Fcv/vtow = 300(30)/(30 x 0.010 x 0.100) = 300,000 in-lb/in3.

T = 70 + (0.4U/ρC)(vto/K)0.333 = 70 + (0.4 x 300,000/124)(30 x 0.010/0.18)0.333

= 70 + (968)(1.667)0.333 = 70 + 1147 = 1217° F Chapter 22: Multiple Choice Questions (22.3, 22.7, 22.10) Problems (22.2, 22.9, 22.14) MCQ: 22.3 (a), (c), and (f) 22.7 (c) 22.10 (b) Problems: 22.2 Solution: Starting with Eq. (22.4): Tm = L/fr

Substitute Eq. (22.3) (fr = Nf) into the denominator to obtain Tm = L/Nf Then substituting for N from Eq. (22.1) we get Tm = πDoL/vf (this equation is later used in Chapter 24). Rearranging to determine cutting speed: v = πDoL/fTm

Tm = π(0.4)(0.15)/(0.30)(10-3)(5.0) = 0.1257(103) m/min = 125.7 m/min 22.9 Solution: (a) A = 0.5(25.4) tan(90 – 118/2) = 12.7 tan 31 = 7.65 mm N = 25(103)/25.4π = 313.3 rev/min fr = 313.3(0.25) = 78.3 mm/min Tm = (50 + 7.63)/78.3 = 0.736 min (b) MRR = 0.25π(25.4)2(78.3) = 39,675.2 mm3/min 22.14 Solution: N = 100 x 12/2.5π = 152.8 rev/min. fr = 152.8(8)(0.009) = 11.0 in/min. A = (d(D-d)).5 = (.25(2.5-.25)).5 = 0.75 in Tm = (10.0 + 0.75)/11 = 0.98 min. MRR = 3.0(.25)(11.0) = 8.25 in3/min. Chapter 23: Multiple Choice Questions (23.1, 23.4, 23.10) Problems (23.1, 23.14, 23.16, 23.18) MCQ: 23.1 (a) 23.4 (a) and (c) 23.10 (c) and (d) Problems: 23.1 Solution: At v = 125 m/min, T = 20.4 min using criterion FW = 0.75 mm. At v = 165 m/min, T = 10.0 min using criterion FW = 0.75 mm. (a) and (b) Student exercises. Values of C and n may vary in part (b) due to variations in the plots. The values should be approximately the same as those obtained in part (c) below. (c) Two equations: (1) 125(20.4)n = C, and (2) 165(10.0)n = C (1) and (2) 125(20.4)n = 165(10.0)n

ln 125 + n ln 20.4 = ln 165 + n ln 10.0 4.8283 + 3.0155 n = 5.1059 + 2.3026 n 0.7129 n = 0.2776 n = 0.3894 (1) C = 125(20.4)0.3894 = 404.46 (2) C = 165(10.0)0.3894 = 404.46 C = 404.46 23.14 Solution: In this problem we want Tm = T, where Tm = machining time per piece and T = tool life. Both of these times must be expressed in terms of cutting speed. Tm = πDL/fv and T = (C/v)1/n

Tm = π(500)(1000)(10-6)/0.4(10-3)v = 3926.99/v = 3926.99 (v)-1

T = (400/v)1/.23 = (400/v)4.3478 = 4004.3478(v)-4.3478 = 2057.33(108) (v)-4.3478

Setting Tm = T: 3926.99 v-1 = 2057.33(108) (v)-4.3478

v3.3478 = 0.52389(108) v = {0.52389(108)}1/3.3478 = {0.52389(108)}0.2987 = 202.18 m/min Check: Tm = 3926.99 (202.18)-1 = 19.423 min T = (400/202.18)1/.23 = (400/202.18)4.3478 = 19.423 min 23.16 Solution: (a) Finish turning of unhardened steel. Specify a steel-cutting grade suitable for finishing. This is a grade with TiC and low cobalt. Choose grade 2. (b) Rough milling of aluminum. Specify a non-steel roughing grade. This is a grade with no TiC and high cobalt. Choose grade 4. (c) Finish turning of brass. Specify a non-steel finishing grade. This is a grade with no TiC and low cobalt. Choose grade 1. (d) Machining cast iron. Cast iron is included with the non-steel grades. Specify grade 1 for finishing and grade 4 for roughing. 23.18 Solution: Dry: 90(T)0.12 = 130 T = (130/90)1/.12 = (1.444)8.3333 = 21.42 min. With coolant: 90(T)0.12 = 130(1 + 10%) = 130(1.1) = 143 T = (143/90)1/.12 = (1.5889)8.3333 = 47.40 min. Increase = (47.40 - 21.42)/21.42 = 1.21 = 121% Chapter 24: Multiple Choice Questions (24.1, 24.4, 24.7, 24.9) Problems (24.2, 24.8, 24.18, 24.19) MCQ: 24.1 (b), (d), (e), and (f) 24.4 (a) 24.7 (a) 24.9 (b) Problems: 24.2 Solution: (a) Base material: v30 = 450/30.27 = 179.6 m/min

New material: v30 = 420/30.22 = 198.7 m/min MR = 198.7/179.6 = 1.107 = 110.7% (b) (a) Base material: T150 = (450/150)1/.27 = (3.0)3.704= 58.5 min New material: v10 = (420/150)1/.22 = (2.8)4.545 = 107.8 min MR = 107.8/58.5 = 1.84 = 184% 24.8Solution: For cast iron at 150 m/min, extrapolating Figure 24.2 ratio rai = 1.2 in Eq. (24.3), so Ra = 1.2 Ri = 1.2f2/32NR Rearranging, f2 = Ri(32NR)/1.2 = 1.6(10-6)(32)(0.75)(10-3)/1.2 = 31.96(10-9) = 3.196(10-8) m2 f = 3.196(10-8) m2)0.5 = 1.79(10-4) m = 0.179 mm (here, mm is interpreted mm/rev) 24.18 Solution: (a) vmax = 1200/[(1/.27 - 1)(1.5)].27 = 1200/[2.704 x 1.5].27 = 822 ft/min. (b) Tmax = (1200/822)1/.27 = (1.460)3.704 = 4.06 min. (c) Tm = πDL/fv = π(3)(18)/(.013 x 822 x 12) = 1.323 min. np = 4.055/1.323 = 3.066 pc/tool Use np = 3 pc/tool life Tc = Th + Tm + Tt/np = 3.0 + 1.323 + 1.5/3 = 4.823 min/pc. Co = $33/hr = $0.55/min Cc = 0.55(5.823) + 2.0/3 = $3.32/pc 24.19 Solution: (a) Co = $33/hr = $0.55/min vmin = 1200[.55/((1/.27 - 1)(.55 x 1.5 + 2.00))].27 = 1200[.55/(2.704 x 2.825)].27 = 590 ft/min. (b) Tmin = (1200/590)1/.27 = (2.034)3.704 = 13.89 min. (c) Tm = πDL/fv = π(3)(18)/(.013 x 590 x 12) = 1.843 min. np = 13.89/1.843 = 7.54 pc/tool Use np = 7 pc/tool life Tc = Th + Tm + Tt/np = 3.0 + 1.843 + 1.5/7 = 5.06 min/pc. Cc = 0.55(5.06) + 2.0/7 = $3.07/pc Chapter 25: Multiple Choice Questions (25.3, 25.5, 25.9, 25.12) Problems (25.5, 25.10) MCQ: 25.3 (a) 25.5 (a) and (d) 25.9 (a), (c), and (d) 25.12 (b) Problems: 25.5 Solution: (a) lc = (Dd)0.5 = (8 x 0.002)0.5 = (0.016)0.5 = 0.1265 in. (b) MRR = vwwd = (20 x 12)(0.15)(0.002) = 0.072 in3/min (c) nc = vwC = (5000 x 12)(0.15)(300) = 2,700,000 chips/min. (d) Avg volume/chip = (0.072 in3/min)/(2,700,000 chips/min) = 0.000000026 in3

= 26 x 10-9 in3. (e) U = Fcv/MRR = 10(5000 x 12)/0.072 = 8,333,333 in-lb/in3 = 21 hp/(in3/min). 25.10 Solution: (a) Conventional surface grinding: Time of engagement/pass = 200 x 10-3 m/(12 m/min) = 0.01667 min = 1 s

Forward and backward stroke = 2(1 s)/50% = 4 s Number of passes to remove 25 mm = 25/0.025 = 1000 passes Time to complete 1000 passes = 1000(4) = 4000 s = 66.67 min. (b) Creep feed grinding: Total length of feed = 200 mm + approach = 200 + (d(D-d))0.5

Given D = 250 mm and d = 25 mm, Total feed length = 200 + (25(250-25))0.5 = 275 mm fr = (12 x 103 mm/min)/1000 = 12 mm/min Time to feed = 275/12 = 22.917 min. Note: Creep feed grinding requires about 1/3 the time of conventional surface grinding for the situation defined here.