Báo cáo bài tập lớn kỹ thuật chế tạo gia công cơ khí chương 21theory of metal machining

  TRƯỜNG ĐẠI HỌC BÁCH KHOA THÀNH PHỐ HỒ CHÍ MINH BÁO CÁO BÀI TẬP LỚN KỸ THUẬT CHẾ TẠO GIA CƠNG CƠ KHÍ Họ tên MSSV Nguyễn Văn Hoài Phương 1612708 Lê Xuân Thái 1613150 Lê Mạnh Thạc 1613236 Lê Ngọc Thạch 1613238 Nguyễn Dần Nhật Viên 1614086 Nguyễn Quang Vinh 1614125    Chương 21: THEORY OF METAL MACHINING 21.1.In 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 φ = tan-1 (0.4615 cos 15/(1 - 0.4615 sin 15)) = tan-1 (0.5062) = 26.85° (b) Shear strain γ = cot 26.85 + tan (26.85 - 15) = 1.975 + 0.210 = 2.185 21.2. In Problem 21.1, suppose the rake angle were changed to Assuming that the friction angle remains the same, determine (a) the shear plane angle, (b) the chip thickness, and (c) the shear strain for the operation Solution: From Problem 21.1, α = 15° and φ = 26.85° Using the Merchant Equation, Eq (21.16): φ = 45 + α/2 - β/2; rearranging, β = 2(45) + α - 2φ β = 2(45) + α - 2(φ) = 90 + 15 – 2(26.85) = 51.3° Now, with α = and β remaining the same at 51.3°, φ = 45 + 0/2 – 51.3/2 = 19.35°  (b) Chip thickness at α = 0: tc = to/tan φ = 0.30/tan 19.35 = 0.854 mm (c) Shear strain γ = cot 19.35 + tan (19.35 - 0) = 2.848 + 0.351 = 3.199 21.3 In an orthogonal cutting operation, the 0.250 in wide tool has a rake angle of 5° The lathe is set so the chip thickness before the cut is 0.010 in After the cut, the deformed chip thickness is measured to be 0.027 in Calculate (a) the shear plane angle and (b) the shear strain for the operation  Solution: (a) r = t o /t c = 0.010/0.027 = 0.3701 φ = tan -1 (0.3701 cos 5/(1 0.3701 sin 5)) = tan -1 (0.3813) = 20.9° (b)Shear strain γ = cot 20.9 + tan (20.9 – 5) = 2.623 + 0.284 = 2.907 21.4  In a turning operation, spindle speed is set to provide a cutting speed of 1.8 m/s The feed and depthof cut of cut are 0.30 mm and 2.6 mm, respectively The tool     rake angle is After the cut, thedeformed chip thickness is measured to be 0.49 mm Determine (a) shear plane angle, (b) shearstrain, and (c) material removal rate Use the orthogonal cutting model as an approximation of theturning process Solution : (a) r =t o/t c  = 0.30/0.49 = 0.612 φ = tan(0.612 cos 8/(1 – 0.612 sin 8)) = tan-1(0.6628) = 33.6  (b) γ = 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 21.5. The cutting force and thrust force in an orthogonal cutting operation are 1470 N and 1589 N, respectively The rake angle = 5°, the width of the cut = 5.0 mm, the chip thickness before the cut = 0.6, and the chip thickness ratio = 0.38 Determine (a) the shear strength of the work material and (b) the coefficient of friction in the operation Solution: (a) φ = tan-1(0.38 cos 5/(1 - 0.38 sin 5)) = tan-1(0.3916) = 21.38°  Fs = 1470 cos 21.38 – 1589 sin 21.38 = 789.3 N As = (0.6)(5.0)/sin 21.38 = 3.0/.3646 = 8.23 mm2 S = 789.3/8.23 = 95.9 N/mm2 = 95.9 MPa (b) φ = 45 + α/2 - β/2; rearranging, β = 2(45) + α - 2φ β = 2(45) + α - 2(φ) = 90 +  – 2(21.38) = 52.24° μ = tan 52.24 = 1.291 21.6 The cutting force and thrust force have been measured in an orthogonal cutting operation to be 300 lb and 291 lb, respectively The rake angle = 10 °, width of cut = 0.200 in, chip thickness before the cut = 0.015, and chip thickness ratio = 0.4 Determine (a) the shear strength of the work material and (b) the coefficient of friction in the operation Solution: φ= tan-1(0.4 cos 10/(1 - 0.4 sin 10)) = tan-1(0.4233) = 22.94 °  Fs= 300 cos 22.94 - 291sin 22.94 = 162.9 lb  As= (0.015)(0.2)/sin 22.94 = 0.0077 in2 S= 162.9/0.0077 = 21,167 lb/in2 β= 2(45) + α - 2(φ) = 90 + 10 - 2(22.94) = 54.1 ° μ= tan 54.1 = 1.38 21.7:An orthogonal cutting operation is performed using a rake angle of 15 °, chip thickness before the cut = 0.012 in and width of cut = 0.100 in The chip thickness ratio is measured after the cut to be 0.55 Determine (a) the chip thickness after the cut, (b) shear angle, (c) friction angle, (d) coefficient of friction, and (e) shear strain Solution: (a) r= to/tc, tc= to/r= 0.012/0.55 = 0.022 in (b) φ= tan-1(0.55 cos 15/(1 - 0.55 sin 15)) = tan-1(0.6194) = 31.8° (c) β= 2(45) + α - 2(φ) = 90 + 15 - 2(31.8) = 41.5° (d) μ= tan 41.5 = 0.88(e) γ= cot 31.8 + tan(31.8 - 15) = 1.615 + 0.301 = 1.92   21.8  The orthogonal cutting operation described in previous Problem 21.7 involves a work materialwhose shear strength is 40,000 lb/in Based on your answers to the previous problem, compute (a)the shear force, (b) cutting force, (c) thrust force, and (d) friction force Solution : (a) A s = (0.012)(0.10)/sin 31.8 = 0.00228 in2  F s = A sS = 0.0028(40,000) = 91.2 lb  (b) F c= 91.2 cos (41.5 - 15)/cos (31.8 + 41.5 -15) = 155 lb (c)  F t = 91.2 sin (41.5 - 15)/cos (31.8 + 41.5 -15) = 77.2 lb (d)  F = 155 sin 15 - 77.2 cos 15 = 115 lb  21.9. In an orthogonal cutting operation, the rake angle = -5°, chip thickness before the cut = 0.2 mm and width of cut = 4.0 mm The chip ratio = 0.4 Determine (a) the chip thickness after the cut, (b) shear angle, (c) friction angle, (d) coefficient of friction, and (e) shear strain  Solution: (a) r = to/tc, tc = to/r = 0.2/.4 = 0.5 mm  (b) φ = tan-1(0.4 cos(–5)/(1 - 0.4 sin(–5))) = tan-1(0.3851) = 21.1° (c) β = 2(45) + α - 2(φ) = 90 + (-5) - 2(21.8) = 42.9°  (d) μ = tan 42.9 = 0.93 (e) γ = cot 31.8 + tan(31.8 - 15) = 2.597 + 0.489 = 3.09 21.10 The shear strength of a certain work material = 50,000 lb/in 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 Solution: (a) φ = 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 in Fs = As S = 0.00456(50,000) = 228 lb 21.11 Solve previous Problem 21.10 except that the rake angle has been changed to -5° and the resulting chip thickness ratio = 0.35 Solution: (a) φ = tan -1 (0.35 cos(–5)/(1 - 0.35 sin(-5))) = tan -1 (0.3384) = 18.7° (a) As = (0.015)(0.15)/sin 18.7 = 0.00702 in Fs = As S = 0.00702(50,000) = 351 lb   21.12. A carbon steel bar 7.64 inch in diameter has a tensile strength of 65,000 lb/in2 and a shear strength of 45,000 lb/in2 The diameter is reduced using a turning operation at a cutting speed of 400 ft/min The feed is 0.011 in/rev and the depth of cut is 0.120 in The rake angle on the tool in the direction of chip flow is 13° The cutting conditions result in a chip ratio of 0.52 Using the orthogonal model as an approximation of turning, determine (a) the shear plane angle, (b) shear force, (c) cutting force and feed force, and (d) coefficient of friction between the tool and chip Solution: (a) φ = tan-1(0.52 cos 13/(1 - 0.52 sin 13)) = tan-1(0.5738) = 29.8°  (b) As = tow/sin φ = (0.011)(0.12)/sin 29.8 = 0.00265 in2  Fs = AsS = 0.00587(40,000) = 119.4 lb  (c) β = 2(45) + α - 2(φ) = 90 + 10 - 2(29.8) = 43.3° Fc = Fscos (β – α)/cos (φ + β – α) Fc = 264.1 cos (43.3 - 13)/cos (29.8 + 43.3 - 13) = 207 lb  Ft =  Fssin (β – α)/cos (φ + β – α) Ft = 264.1 sin (43.3 - 13)/cos (29.8 + 43.3 - 13) = 121 lb  (d) μ = tan β = tan 43.3 = 0.942 21.14  A turning operation is made with a rake angle of 10, a feed of 0.010 in/rev and a depth of cut =0.100 in The shear strength of the work material is known to be 50,000 lb/in, and the chipthickness ratio is measured after the cut to be 0.40 Determine the cutting force and the feed force.Use the orthogonal cutting model as an approximation of the turning process Solution : φ = tan-1(0.4 cos 10/(1 - 0.4 sin 10)) = tan-1(0.4233) = 22.9   As = (0.010)(0.10)/sin 22.9 = 0.00257 in2  F s = A sS = 0.00256(50,000) = 128 lb   β = 2(45) +α- 2(φ ) = 90 + 10 - 2(22.9) = 54.1 °   F c= 128 cos (54.1 - 10)/cos (22.9 + 54.1 - 10) = 236 lb  F t = 128 sin (54.1 - 10)/cos (22.9 + 54.1 - 10) = 229 lb 21.15Show how Eq (21.3) is derived from the definition of chip ratio, Eq (21.2), and Figure 21.5(b) Solution : Begin with the definition of the chip ratio, Eq (21.2): r =t o/t c = sinφ /cos (φ -α )Rearranging,r cos (φ -α ) = sinφ Using the trigonometric identity cos( φ -α ) = cosφ cosα + sinφ sinα r (cosφ cosα + sinφ sinα ) = sinφ    Dividing both sides by sin φ , we obtain r cosα /tanφ +r sinα = r cosα /tanφ = -r sinα Rearranging, tan φ =r cosα/(1 -r sinα ) 21.16: Show how Eq (21.4) is derived from Figure 21.6 Solution : In the figure, γ = AC / BD = ( AD + DC )/ BD = AD/ BD + DC / BD  AD/ BD = cotφ and DC / BD= tan ( φ -α )Thus, γ = cotφ + tan (φ -α ) 21.18:  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 21.3 to obtain theappropriate specific energy value Solution : From Table 21.3, U = 2.8 N-m/mm3 = 2.8 J/mm3  RMR =vfd = (200 m/min)(10 mm/m)(0.25 mm)(7.5 mm) = 375,000 mm3/min = 6250mm3/s  P c = (6250 mm3/s)(2.8 J/mm3) = 17,500 J/s = 17,500 W = 17.5 kWAccounting for mechanical efficiency,  P g = 17.5/0.90 = 19.44 kW 21.19 :In previous Problem 21.18, compute the lathe power requirements if feed = 0.50 mm/rev Solution : This is the same basic problem as the previous, except that a correction must be made forthe “size effect.” Using Figure 21.14, for   f = 0.50 mm, correction factor = 0.85.From Table 21.3, U = 2.8 J/mm3 With the correction factor, U = 2.8(0.85) = 2.38 J/mm3    =vfd = (200 m/min)(103 mm/m)(0.50 mm)(7.5 mm) = 750,000 mm3/min =  R MR 12,500mm3/s   = (12,500 mm3/s)(2.38 J/mm3) = 29,750 J/s = 29,750 W = 29.75  P c kWAccounting for mechanical efficiency,  = 29.75/0.90 = 33.06 kW  P  g   21.20: In a turning operation on aluminum, cutting conditions are as follows: cutting speed = 900 ft/min, feed = 0.020 in/rev, and depth of cut = 0.250 in What   horsepower is required of the drive motor, if the lathe has a mechanical efficiency = 87%? Use Table 21.2 to obtain the appropriate unit horsepower value Solution: From Table 21.3, HPu = 0.25 hp/(in /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 = t o , correction factor = 0.9 HPc = HPu x R MR, HP g = HPc /E R MR = vfd = 900 x 12(.020)(0.250) = 54 in /min HPc = 0.9(0.25)(54) = 12.2 hp HP g = 12.2/0.87 = 14.0 hp 21.22 :A turning operation is to be performed on a 20 hp lathe that has an 87% efficiency rating Theroughing cut is made on alloy steel whose hardness is in the range 325 to 335 HB The cuttingspeed is 375 ft/min The feed is 0.030 in/rev, and the depth of cut is 0.150 in Based on these values,can the job be performed on the 20 hp lathe? Use Table 21.3 to obtain the appropriate unithorsepower value Solution :  From Table 21.3,  HP u = 1.3 hp/(in3/min)Since the uncut chip thickness (0.030 in) is different from the tabular value of 0.010, a correctionfactor must be applied From Figure 21.14, the correction factor is 0.7 Therefore, the corrected    = 0.7*1.3 = 0.91 hp/(in3/min)  HP u   =vfd = 375 ft/min(12 in/ft)(0.03 in)(0.150 in) = 20.25 in3/min  R MR = (20.25 in3/min)(0.91 hp/(in3/min)) = 18.43 hp  HP c  required.At efficiency  E = 87%, available horsepower = 0.87(20) = 17.4 hp  Since required horsepower exceeds available horsepower, the job cannot be accomplished on the 20hp lathe, at least not at the specified cutting speed of 375 ft/min 21.23:Suppose the cutting speed in Problems 21.7 and 21.8 is cutting speed = 200 ft/min From youranswers to those problems, find (a) the horsepower consumed in the operation, (b) metal removalrate in in3/min, (c) unit horsepower (hp-min/(in3), and (d) the specific energy (in-lb/in3) Solution : (a) From Problem 21.8,   = 155 lb  F c    = 155(200)/33,000 = 0.94 hp  HP c  (b)  =vfd = (200 x 12)(0.012)(0.100) = 2.88 in3/min  R MR = 0.94/2.88 = 0.326 (c)  hp/(in3/min)  HP u (d)U = 155(200)/2.88 = 10,764 ft-lb/in3 = 129,167 in-lb/in3 21.24 :For Problem 21.12, the lathe has a mechanical efficiency = 0.83 Determine (a) the horsepowerconsumed by the turning operation; (b) horsepower that must be generated by the lathe; (c) unithorsepower and specific energy for the work material in this operation Solution : (a) From Problem 21.12,  = 207 lb  F c  = F cv/33,000 = 207(400)/33,000 = 2.76 hp    HP c   = HP c/E = 2.76/0.83 = 3.33 hp  (b)   HP  g    = 12vfd = (400 x 12)(0.0.01101)(0.120) = 6.34 in3/min (c)   R MR  = HP c/ R MR = 1.63/3.6 = 0.453 hp/(in3 /min)    HP u U =   F c  v/    = 207(400 x 12)/6.34 = 157,000 in-lb/in  R MR 21.26 :Solve Problem 21.25 except that the feed = 0.0075 in/rev and the work material is stainless steel(Brinell Hardness = 240 HB) Solution : (a) From Table 21.3,    = 1.0 hp/(in3/min) for stainless steel Since feed is lower than0.010 in/rev in  HP u the table, a correction factor must be applied from Figure 21.14 For     f = 0.0075in/rev =   t o , correction factor = 1.1   = HP u x R MR  HP c   R MR = 400 x 12(0.0075)(0.12) = 4.32 in3/min  HP c = 1.1(1.0)(4.32) =4.75 hp (b)  HP g = 5.01/0.83 = 5.73 hp 21.27 In a turning operation is carried out on aluminum (100 BHN), the cutting conditions are as follows:cutting speed = 5.6 m/s, feed = 0.25 mm/rev, and depth of  cut = 2.0 mm The lathe has a mechanicalefficiency = 0.85 Based on the specific energy values in Table 21.3, determine (a) the cutting power and (b) gross power in the turning operation, in Watts Solution : (a) From Table 21.3, U = 0.7 N-m/mm3  for aluminum  R MR =vfd = 5.6(103)(.25)(2.0) = 2.8(103) mm3/s  P c =U  R MR = 0.7(2.8)(103) = 1.96(103) n-m/s = 1960 W  (b) Gross power   P g = 1960/0.85 = 2306 W 21.28:  Solve Problem 21.27 but with the following changes: cutting speed = 1.3 m/s, feed = 0.75 mm/rev,and depth = 4.0 mm Note that although the power used in this operation is virtually the same as inthe previous problem, the metal removal rate is about 40% greater Solution : (a) From Table 21.3, U = 0.7 N-m/mm3  for aluminum Since feed is greater than 0.25mm/rev in the table, a correction factor must be applied from Figure 21.14 For   f = 0.75 mm/rev =t o , correction factor = 0.80  RMR=vfd = 1.3(103)(.75)(4.0) = 3.9(103) mm3/s  P c =U RMR= 0.8(0.7)(3.9)(103) = 1.97(103) n-m/s = 1970 W  (b) Gross power     P g = 1970/0.8 =  2460 W 21.29:  A turning operation is performed on an engine lathe using a tool with zero rake angle in thedirection of chip flow The work material is an alloy steel with hardness = 325 Brinell hardness.The feed is 015 in/rev, the depth of cut is 0.125 in and the cutting speed is 300 ft/min After the cut,the chip thickness ratio is measured to be 0.45 (a) Using the appropriate value of specific energyfrom Table 21.3, 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 21.3, U = P u = 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 Figure21.14 For   f = 0.015 in/rev =t o , correction factor = 0.95 Thus, U = 520,000(0.95) = 494,000 in-lb/in3 = 41,167 ft-lb/in3  R MR= 300 x 12(.015)(0.125) = 6.75 in3/min  P c =U R MR = 41,167(6.75) = 277,875 ft-lb/min  HP c = 277,875/33,000 = 8.42 hp  HP g = 8.42/0.85 = 9.9 hp 21.30 :A lathe performs a turning operation on a workpiece of 6.0 in diameter The shear strength of thework is 40,000 lb/in2 and the tensile strength is 60,000 lb/in2 The rake angle of the tool is ° Themachine is set so that cutting speed is 700 ft/min, feed = 0.015 in/rev, and depth = 0.090 in Thechip thickness after the cut is 0.025 in Determine (a) the horsepower required in the operation, (b)unit horsepower for this material under these conditions, and (c) unit horsepower as it would belisted in Table 21.3 for a t oof 0.010 in Use the orthogonal cutting model as an approximation of theturning process Solution :  (a) Must find  F c and v to determine HP r = 0.015/0.025 = 0.6 φ = tan-1(0.6 cos 6/(1 - 0.6 sin 6)) = tan-1 (0.6366) = 32.5 °  β = 2(45) +α - 2(φ) = 90 + -2(32.5) = 31.0 °  As=t ow/sinφ = (0.015)(0.09)/sin 32.5 = 0.00251 in2  F s =SAs = 40,000(0.00251) = 101 lb  F c = F s cos (β  – α)/cos (φ + β – α)   24.12 A face milling operation is to be performed on a cast iron part to finish the surface to 36 m-in The cutter uses four inserts and its diameter is 3.0 in The cutter rotates at 475 rev/min To obtain the best possible finish, a type of carbide insert with 4/ 64 in nose radius is to be used Determine the required feed rate (in/min) that will achieve the 36 m-in finish V=πDN=π(3/12)(500)=393 ft/min 23.2, so Ra=1.25 Ri For cast iron at 393 ft/min, the ratio rai=1.25 in Figue Ri= Ra/1.25=32/1.25=25.6 uin Ri=f 2/32NR Rearranging, f 2=32Ra(NR)= 51.2 x 10 -6 in2 f = (51.2 x 10-6)=0.00716 in/tooth f r=Nntf=500(6)(0.00716)=21.5 in/min 24.13 A face milling operation is not yielding the required surface finish on the work The cutter is a four-tooth insert type face milling cutter The machine shop foreman thinks the problem is that the work material is too ductile for the job, but this property tests well within the ductility range for the material specified by the designer Without knowing any more about the job, what changes in (a) cutting conditions and (b) tooling would you suggest to improve the surface finish? (a) Changes in cutting conditions: (1) decrease chip load (2) increase cutting speed v, (3) use cutting fluid (b) Changes in tooling: (1) increase nose radius NR, (2) increase rake angle, and (3) increase relief  angle Items (2) and (3) will have a marginal effect 24.14 A turning operation is to be performed on C1010 steel, which is a ductile grade It is desired to achieve a surface finish of 64 m-in, while at the same time maximizing the metal removal rate It has been decided that the speed should be in the range 200 ft/min to 400 ft/min, and that the depth of cut will be 0.080 in The tool nose radius = 3/64 in Determine the speed and feed combination that meets these criteria   Increasing feed will increase both RMR and Ra Increasing speed will increase RMR and reduce Ra therefore, it stands to reason that we should openate at the highest possible v Try v=400 ft/min, from fig 25.45,r ai=1.15  Ra=1.15Ri R i=R a/1.15=55.6 uin Ri=f 2/32NR f 2=Ra(32NR)=83.4 x 10 -6 in2 f=0.0091 in/rev RMR=3.5 in3/min Compare at v=300 ft/min from fig 25.46,r ai=1.26 Ra=1.26Ri R i=R a/1.26=50.8 uin Ri=f 2/32NR f 2=Ra(32NR)=76.2 x 10 -6 in2 f=0.0087 in/rev RMR=2.51 in3/min Optimum cutting conditions are v=400 ft/min and f=0.0091 in/rev, which max R MR=3.5 in3/min 24.15 A high-speed steel tool is used to turn a steel workpart that is 300 mm long and 80 mm in diameter The parameters in the Taylor equation are: n = 0.13 and C = 75 (m/min) for a feed of 0.4 mm/rev The operator and machine tool rate = $30/hr, and the tooling cost per cutting edge = $4 It takes 2.0 to load and unload the workpart and 3.50 to change tools Determine (a) cutting speed for maximum production rate, (b) tool life in of cutting, and (c) cycle time and cost per unit of product (a) C0=$36/hr=$0.6/min Vmax=75/((1/0.13-1)(4)) 0.13=48.9 m/min   (b) Tmax=(75/48.9) 1/0.13=26.85 (c) Tm= πDL/fv= π(75)(350)/(0.4x48.9x10 3)=4.216 Np=26.85/4.216=6.37 pc/tool life Tc=Th+Tm+Tt/np=7.88 min/pc Cc=0.6(7.88)+4.25/6=$5.44/pc   25: GRINDING AND OTHER ABRASIVE PROCESSES 25.1 In a surface grinding operation wheel diameter = 150 mm and infeed = 0.07 mm Wheel speed = 1450 m/min, workspeed = 0.25 m/s, and crossfeed = mm The number of active grits per area of wheel surface= 0.75 grits/mm2 Determine (a) average length per chip, (b) metal removal rate, and (c) number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work Solution: (a) lc = ( Dd ) 0,5 = (150 * 0, 07 ) 0,5 = 3, 24mm  (b)  R MR = vw wd  = ( 0, 25 m / s ) (103 mm / m ) ( mm)( 0, 07 mm) - 87, mm3 / s = 5250 mm3 / (c) nc = vwC = (1450m / ) (103 mm / m ) ( 5mm ) ( 0, 75 grits / mm2 ) = 5, 437, 500chips / 25.2 The following conditions and settings are used in a certain surface grinding operation: wheel diameter = 6.0 in, infeed = 0.003 in, wheel speed = 4750 ft/min, workspeed = 50 ft/min, and crossfeed = 0.20 in The number of active grits per square inch of wheel surface = 500 Determine (a) average length per chip, (b) metal removal rate, and (c) number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work Solution: (a) lc = (b)  R MR ( Dd ) = 0,5 vw wd (C) nc = vwC = 0,5 ( 6* 0, 003 ) = ( 50 *12 )( 0, )( 0, 003 ) = 0, 36in / = ( 0, 018 ) 0,5 = = 0,1342in ( 4750 *12 )( 0, )( 500 ) = 5, 700, 000chips / 25.3 An internal cylindrical grinding operation is used to finish an internal bore from an initial diameter of 250 mm to a final diameter of 252.5 mm The bore is 125 mm long A grinding wheel with an initial diameter of 150 mm and a width of 20 mm is used After the operation, the diameter of the grinding wheel has been reduced to 149.75 mm Determine the grinding ratio in this operation Solution:     ( 25.3 ) , R MR =  vw wd (in / ) nc = vwC ( chips / )  R MR / nc = vw wd / vwC = v wd / vC  GR = (volume of work material removed)/(volume of wheel removed)Volume of work material removed = (p  / )(125 ) ( 252, 52 - 2502 ) = 123, 332mm3 Volume of wheel removed = ( p  / )( 20  ) (1502 - 149, 752 ) = 1177mm3 GR  = 123,332 /1177 = 104,8 25.4 In a surface grinding operation performed on hardened plain carbon steel, the grinding wheel has a diameter = 200 mm and width = 25 mm The wheel rotates at 2400 rev/min, with a depth of cut (infeed) = 0.05 mm/pass and a crossfeed = 3.50 mm The reciprocating speed of the work is m/min, and the operation is performed dry Determine (a) lengthof contact between the wheel and the work and (b) volume rate of metal removed (c) If there are 64 active grits/cm2 of wheel surface, estimate the number of chips formed per unit time (d) What is the average volume per chip? (e) If the tangential cutting force on the work = 25 N, compute the specific energy in this operation? Solution: (a) lc = ( Dd ) 0,5 = ( 200*0,05) = 3,16mm (b)  R MR = vw wd = ( 6m / ) (103 mm / m) ( 3, mm)( 0, 05 mm) = 1050 mm3 / (c) nc = vwC   v = Np D = ( 2400rev / )( 200p  mm / rev ) = 1,507,964 mm / (1, 507,964mm / )( 3,5mm ) ( 64 grits / cm2 )(10 cm2 / mm2 ) = 3, 377,840 grits / ( = chips / ) nc = U = - 30 (1508 ) / 1050 = 43,1  N - m / mm 25.5 An 8-in diameter grinding wheel, 1.0 in wide, is used in a surface grinding job performed on a flat piece of heat-treated 4340 steel The wheel rotates to achieve a surface speed of 5000 ft/min, with a depth of cut (infeed) = 0.002 in per pass and a crossfeed = 0.15 in The reciprocating speed of the work is 20 ft/min, and the operation is performed dry (a) What is the length of contact between the wheel and the work? (b) What is the volume rate of metal removed? (c) If there are 300 active grits/in2 of wheel surface, estimate the number of chips formed per unit time (d) What is the average volume per chip? (e)   If the tangential cutting force on the workpiece = 7.3 lb, what is the specific energy calculated for this  job? Solution:(a) lc = ( Dd ) 0,5 = (8* 0, 002 ) 0,5 = ( 0, 016 ) 0,5 = 0,1265in (b)  R MR = vw wd nc vwd =   in / ( 20 *12 )( 0,15 )( 0, 002 ) = 0, 072 (C) = = ( 5000 *12 )( 0,15 )(3 00 ) = 2, 700, 000chips / (d) Avg volume/chip = ( 0, 072in / ) /  ( 2, 700, 000chips / ) = 0, 000000026in3 = 26 *10-9 in3 (e) U =F cv/R MR = 10 ( 5000 *12 ) / 0, 072 = 8, 333, 333 in - lb / in3 = 21hp / ( in3 / ) 25.6 A surface grinding operation is being performed on a 6150 steel workpart (annealed, approximately 200 BHN) The designation on the grinding wheel is C-24-D-5-V The wheel diameter = 7.0 in and its width=1.00in.Rotational speed=3000 rev/min.The depth (infeed)=0.002in per pass, and the crossfeed= 0.5 in Workspeed = 20 ft/min This operation has been a source of trouble right from the beginning The surface finish is not as good as the 16m-in specified on the part print, and there are signs of metallurgical 626 Chapter 25/Grinding and Other Abrasive Processes E1C25 11/09/2009 16:46:54 Page 627 damage on the surface In addition, thewheel seems to become clogged almost as soon as the operation begins In short, nearly everything that can go wrong with the job has gone wrong (a) Determine the rate of metal removal when the wheel is engaged in the work (b) If the number of active grits per square inch = 200, determine the average chip length and the number of chips formed per time (c) What changes would you recommend in the grinding wheel to help solve the problems encountered? Explain why you made each recommendation Solution: (a)  R MR = vw wd = ( 20 *12 )( 0, )( 0, 002 ) = 0,  24in / (b) = ( Dd ) 0,5 0,5 ( * 0, 002 ) = 0,1183in v = DN = ( /12 )( 3000 ) = 5498 ft / = 65,973in / nc = vwC = 65,973 ( 0,5 )( 200 ) = 6,597,300 grits / lc p = p  (c) Changes in wheel to help solve problems cited: (1) use Al2O3 oxide abrasive rather than siliconcarbide; (2) use smaller grain size than 24; (3) use shellac bond rather than vitrified bond; and (4)use more open structure than number to reduce wheel clogging 25.7 The grinding wheel in a centerless grinding operation has a diameter = 200 mm, and the regulating wheel diameter = 125 mm The grinding wheel rotates at 3000 rev/min and the regulating wheel rotates   at 200 rev/min The inclination angle of the regulating wheel = 2.5 Determine the throughfeed rate of cylindrical workparts that are 25.0 mm in diameter and 175 mm long Solution: From Eq ( 25,11) ,  f r = p  Dr N r  sin I   f r  = p (125 )( 200 ) sin 2, 5o - 25p  ( 0, 04362 ) = 3426 mm / Parts through-feed rate = ( 3462mm / ) / (175mm / pc ) = 19, 58 pc / 25.8 A centerless grinding operation uses a regulating wheel that is 150 mm in diameter and rotates at 500 rev/min At what inclination angle should the regulating wheel be set, if it is desired to feed a workpiece with length = 3.5 m and diameter = 18 mm through the operation in exactly 30 sec? Solution: From Eq ( 25,11) ,  f r = p  Dr N r  sin I   f r  = 3, 5mper 45 sec = 0, 077778m / s = 4,1667 m /  I   = 1,135o 25.9 In a certain centerless grinding operation, the grinding wheel diameter = 8.5 in, and the regulating wheel diameter = in The grinding wheel rotates at 3500 rev/min and the regulating wheel rotates at 150 rev/min The inclination angle of the regulating wheel = Determine the throughfeed rate of cylindrical workparts that have the following dimensions: diameter = 1.25 in and length = in Solution: From Eq ( 25,11) ,  fr = p Dr N r  sin I = p  (5 )(150 ) sin 3o = 123, 33in / ( 8in / part ) / (123, 33in / ) = 0, 0649 min/ part = 3,9 sec/ part   25.10 It is desired to compare the cycle times required to grind a particular workpiece using traditional surface grinding and using creep feed grinding Theworkpiece is 200 mm long, 30 mm wide, and 75 mm thick To make a fair comparison, the grinding wheel in both cases is 250 mm in diameter, 35 mm in width, and rotates at 1500 rev/min It is desired to remove 25 mm of material from the surface When traditional grinding is used, the infeed is set at 0.025 mm, and the wheel traverses twice (forward and back) across the work surface during each pass before resetting the infeed There is no crossfeed since the wheel width is greater than the work width Each pass is made at a workspeed of 12 m/min, but the wheel overshoots the part on both sides.With acceleration and deceleration, the wheel is engaged in the work for 50% of the time on each pass When creep feed grinding is used, the depth is increased by 1000 and the forward feed is decreased by 1000 How long will it take to complete the grinding operation (a) with traditional grinding and (b) with creep feed grinding?   Solution : (a) Conventional surface grinding:Time of engagement/pass =   25.11 In a certain grinding operation, the grade of the grinding wheel should be ‘‘M’’ (medium), but the only available wheel is grade ‘‘T’’ (hard) It is desired to make the wheel appear softer by making changes in cutting conditions What changes would you recommend? Solution: A hard wheel means that the grains are not readily pulled from the wheel bond Thewheel can be made to appear softer by increasing the force on the individual grits as given by Eq.(25.8) According to this equation, the force on the abrasive grains will be increased by increasingwork speedvw, decreasing wheel speedv, and increasing infeed d 25.12 An aluminum alloy is to be ground in an external cylindrical grinding operation to obtain a good surface finish Specify the appropriate grinding wheel parameters and the grinding conditions for this  job Solution: Grinding wheel specification:Abrasive type: silicon carbide Grain size: small - high grit size numberBond material: shellac bondWheel structure: denseWheel grade: medium to hardWheel specification: C-150-E-5-B  Grinding conditions:Wheel speed: high speed, around 1800 m/min (6000 ft/min)Work speed: low, around 10 m/min (30 ft/min)Infeed (depth of cut): low, around 0.012 mm (0.0005 in)Crossfeed: low, around 1/6 of wheel width 25.13 A high-speed steel broach (hardened) is to be resharpened to achieve a good finish Specify the appropriate parameters of the grinding wheel for this job Solution : Grinding wheel specification:Abrasive type: cubic boron nitrideGrain size: small - high grit size numberBond material: vitrified bondWheel grade: soft to mediumWheel specification:  XX-B150P-XY-V-XZ-1/8, where XX, XY, and XZ are manufacturer’ssymbols 25.14 Based on equations in the text, derive an equation to compute the average volume per chip formed in the grinding process Solution: From Eq ( 25.3) , R MR =  vw wd ( in3 / ) nc = vwC  R MR / nc ( chips / ) = vw wd / vwC = vw d / vC    Chương 26: NONTRADITIONAL MACHINING AND THERMAL CUTTING PROCESSES Application Problems 26.1/ For the following application, identify one or more nontraditional machining processes that might be used, and present arguments to support your selection Assume that either the part geometry or the work material (or both) preclude the use of conventional machining The application is a matrix of 0.1 mm (0.004 in) diameter holes in a plate of 3.2 mm (0.125 in) thick hardened tool steel The matrix is rectangular, 75 by 125 mm (3.0 by 5.0 in) with the separation between holes in each direction =1.6 mm (0.0625 in) Giải: EBM LBM tạo lỗ có kích thước với tỷ lệ độ sâu đường kính lớn tới 0.125/0.004 = 31.25 26.2/ For the following application, identify one or more nontraditional machining processes that might be used, and present arguments to support your selection Assume that either the part geometry or the work material (or both) preclude the use of conventional machining The application is an engraved aluminum printing plate to be used in an offset printing press to make 275 × 350 mm (11 × 14 in) Giải: Có thể ứng dụng quy trình khắc quang hóa 26.3/ For the following application, identify one or more nontraditional machining processes that might be used, and present arguments to support your selection Assume that either the part geometry or the work material (or both) preclude the use of conventional machining The application is a through-hole in the shape of the letter L in a 12.5 mm (0.5 in) thick plate of glass The size of the ‘‘L’’ is 25 × 15 mm (1.0 × 0.6 in) and the width of the hole is mm (1/8 in) Giải: Quy trình thực USM hoạt động kính vật liệu giòn phi kim loại khác 26.4/ For the following application, identify one or more nontraditional machining processes that might be used, and present arguments to support your selection Assume that either the part geometry or the work material (or both) preclude the use of conventional machining The application is a blind-hole in the shape of the letter G in a 50 mm (2.0 in) cube of steel The overall size of the ‘‘G’’ is 25 × 19 mm (1.0 × 0.75 in), the depth of the hole is 3.8 mm (0.15 in), and its width is mm (1/8 in) Giải: Có thể sử dụng quy trình EDM ECM   26.5/ Much of the work at the Cut-Anything Company involves cutting and forming of flat sheets of fiberglass for the pleasure boat industry Manual methods based on portable saws are currently used to perform the cutting operation, but production is slow and scrap rates are high The foreman says the company should invest in a plasma arc cutting machine, but the plant manager thinks it would be too expensive What you think? Justify your answer by indicating the characteristics of the process that make PAC attractive or unattractive in this application Giải: Trong cắt hồ quang plasma, phần làm việc phải vật liệu dẫn điện Sợi thủy tinh khơng dẫn điện PAC khơng lả lựa chọn thích hợp 26.6/ A furniture company that makes upholstered chairs and sofas must cut large quantities of fabrics Many of these fabrics are strong and wear-resistant, which properties make them difficult to cut What nontraditional process(es) would you recommend to the company for this application? Justify your answer by indicating the characteristics of the process that make it attractive Giải: Máy bay phản lực nước trình lí tưởng cho ứng dụng WJC cắt qua loại vải cắt nhanh chóng sẽ, trình dễ dàng tự động hóa Electrochemical Machining 26.7/ The frontal working area of the electrode in an ECM operation is 2000 mm 2  The applied current = 1800 amps and the voltage = 12 volts The material being cut is nickel (valence = 2), whose specific removal rate is given in Table 26.1 (a) If the process is 90% efficient, determine the rate of metal removal in mm3 /min (b) If the resistivity of the electrolyte =140 ohm-mm, determine the working gap Giải: a) Tra bảng ta C = 3.42 × 10-2 mm3/A-s  RMR = f rA = (CI/A).A = CI = (3.42 × 10 -2).(1800) = 61.56 mm 3/s Hiệu 90% => RMR = 55.404 mm3/s b) r = 140 ohm-mm I = E.A/r.g => g = E.A/I.r = (12*2000)/(1800*140) = 0.095 mm 26.8/ In an electrochemical machining operation, the frontal working area of the electrode is 2.5in 2  The applied current = 1500 amps, and the voltage = 12 volts The material being cut is pure aluminum, whose specific removal rate is given in Table 26.1 (a) If the ECM process is 90% efficient, determine the   rate of metal removal in in3 /hr (b) If the resistivity of the electrolyte = 6.2 ohm-in, determine the working gap Giải: a) Tra bảng ta C = 0.000126 in3/A-mm RMR = f rA = (CI/A).A = CI = 0.000126*1500 = 0.189 in 3/min Hiệu 90% => RMR = 0.189*0.9 = 0.1701 in 3/min = 10.2 in 3/hr b) I = E.A/r.g => g = E.A/I.r = (12*2.5)/(1500*6.2) = 0.0032 in 26.9/ A square hole is to be cut using ECM through a plate of pure copper (valence = 1) that is 20 mm thick The hole is 25 mm on each side, but the electrode used to cut the hole is slightly less that 25 mm on its sides to allow for overcu, and its shape includes a hole in its center to permit the flow of electrolyte and reduce the area of the cut This tool design results in a frontal area of 200 mm 2  The applied current = 1000 amps Using an efficiency of 95%, determine how long it will take to cut the hole Giải: a) C = 3.7*10-2 mm3/A-s f r = CI/A = (3.7*10-2)(1000)/200 = 0.185 mm/s hiệu 95% => f r = 0.17575 mm/s thời gian máy = 20/0.17575 = 113.8s = 1.89 b) Diện tích lỗ = 252 = 625 mm2 f r = CI/A = (3.7*10-2)(1000)/625 = 0.0592 mm/s hiệu 95% => f r =0.05624 thời gian máy = 20/0.05624 = 355.6 s = 5.92 26.10/ A 3.5 in diameter through hole is to be cut in a block of pure iron (Valence = 2) by electrochemical machining The block is 2.0 in thick To speed the cutting process, the electrode tool will have a center hole of 3.0 in which will produce a center core that can be removed after the tool breaks through The outside diameter of the electrode is undersized to allow for overcut The overcut is expected to be 0.005 in on a side If the efficiency of the ECM operation is 90%, what current will be required to complete the cutting operation in 20 minutes? Giải: Diện tích A = 0.25π(3.52 – 32) = 2.553 in2 C = 0.000135 in 3/A-min f r = CI/A =0.000135*I/2.553 = 0.000053*I hiểu 90% => f r = 0.0000477*I = 2/20 => I = 2096.4 A Electric Discharge Machining   26.11/ An electric discharge machining operation is being performed on two work materials: tungsten and tin Determine the amount of metal removed in the operation after hour at a discharge current of 20 amps for each of these metals Use metric units and express the answers in mm 3 /hr From Table 4.1, the melting temperatures of tungsten and tin are 3410 0C and 232 0C, respectively Giải: Đối với volfram nhiệt độ nóng chảy tra từ bảng 3410 0C Sử dụng công thức: RMR = KI/Tm1.23 = 664*20/(3410 1.23) = 0.5997 mm 3/s = 2159 mm3/hr Với thiếc có nhiệt độ nóng chảy 232 0C RMR = KI/Tm1.23 = 664*20/(232 1.23) = 16.35 mm 3/s = 58879.8 mm 3/hr 26.12/ An electric discharge machining operation is being performed on two work materials: tungsten and zinc Determine the amount of metal removed in the operation after hour at a discharge amperage = 20 amps for each of these metals Use U.S Customary units and express the answer in in 3 /hr From Table 4.1, the melting temperatures of tungsten and zinc are 6170 0F and 420 0F, respectively Giải: Với volfram, nhiệt độ nóng chảy 6170 0F RMR = KI/Tm1.23 = 5.08*20/(6170 1.23) = 0.0022 in 3/s = 0.1325 in 3/hr Với kẽm, nhiệt độ nóng chảy 420 0F RMR = KI/Tm1.23 = 5.08*20/(420 1.23) = 0.06 in 3/s = 3.61 in 3/hr 26.13/ Suppose the hole in Problem 26.10 were to be cut using EDM rather than ECM Using a discharge current = 20 amps (which would be typical for EDM), how long would it take to cut the hole? From Table 4.1, the melting temperature of iron is 2802 0F Giải: RMR = KI/Tm1.23 = 5.08*I/(T 1.23) = 5.08*20/(2802 1.23) = 0.006 in 3/min Diện tích A = 2.553 in f r = RMR/A = 0.006/2.553 = 0.00235 in/min thời gan cần, Tm = 2/0.00235 = 851 = 14.18 hr 26.14/ A metal removal rate of 0.01 in 3 /min is achieved in a certain EDM operation on a pure iron work part What metal removal rate would be achieved on nickel in this EDM operation if the same discharge current were used? The melting temperatures of iron and nickel are 2802 0F and 2651 0F, respectively   Giải: Với sắt : RMR = KI/Tm1.23 = 5.08*I/(2802 1.23) = 0.000292*I = 0.01 => I = 34.25 A Với nicken : RMR = KI/Tm1.23 = 5.08*I/(2651 1.23) = 0.00031*I = 0.01 => I = 32 A 26.15/ In a wire EDM operation performed on 7-mmthick C1080 steel using a tungsten wire electrode whose diameter = 0.125 mm, past experience suggests that the overcut will be 0.02 mm, so that the kerf  width will be 0.165 mm Using a discharge current =10 amps, what is the allowable feed rate that can be used in the operation? Estimate the melting temperature of 0.80% carbon steel from the phase diagram in Figure 6.4 Giải: Từ bảng 6.4 ta có Tm = 1500 0C cho thép 1080 RMR = KI/Tm1.23 = 644*10/(1500 1.23) = 0.798 mm 3/s = 47.8 mm3/min Diện tích kerf: A = 0.165*7 = 1.155 mm2 f r = 47.8/1.155 = 41.4 mm/min 26.16/ A wire EDM operation is to be performed on a slab of 3/4-in-thick aluminum using a brass wire electrode whose diameter = 0.005 in It is anticipated that the overcut will be 0.001 in, so that the kerf width will be 0.007 in Using a discharge current =7 amps, what is the expected allowable feed rate that can be used in the operation? The melting temperature of aluminum is 1220 0F Giải: RMR = KI/Tm1.23 = 5.08*7/(1220 1.23) = 0.0057 in 3/min Diện tích kerf: A = (¾)*0.007 = 0.00525 in f r = 0.0057/0.00525 = 1.086 in/min 26.17/ A wire EDM operation is used to cut out punch -and-die components from 25-mm-thick tool steel plates However, in preliminary cuts, the surface finish on the cut edge is poor What changes in discharge current and frequency of discharges should be made to improve the finish? Giải: Kết thúc bề mặt EDM cải thiện dịng xả tăng tần suất xả Chemical Machining 26.18/ Chemical milling is used in an aircraft plant to create pockets in wing sections made of an aluminum alloy The starting thickness of one workpart of interest is 20 mm A series of rectangular-   shaped pockets 12 mm deep are to be etched with dimensions 200 mm by 400 mm The corners of each rectangle are radiused to 15 mm The part is an aluminum alloy and the etchant is NaOH The penetration rate for this combination is 0.024 mm/min and the etch factor is 1.75 Determine (a) metal removal rate in mm3 / min, (b) time required to etch to the specified depth, and (c) required dimensions of the opening in the cut and peel maskant to achieve the desired pocket size on the part Giải: a) Diện tích A = 200*400 – ( 30*30 – π(12) 2) = 78647.6 mm RMR = 0.024*78647.6 = 1887.5 mm 3/min b) Thời gian: Tm = 15/0.024 = 625 = 10.42 hr c) Với Fe = 1.75 => u = d/F e = 15/1.75 = 8.57 mm Maskant opening length = L - 2u = 400 – 2*8.57 = 382.86 mm Maskant opening width = W – 2u = 200 – 2*8.57 = 182.86 mm Radius on corners = R – u = 15 – 8.57 = 6.43 mm 26.19/ In a chemical milling operation on a flat mild steel plate, it is desired to cut an ellipse-shaped pocket to a depth of 0.4 in The semiaxes of the ellipse are a = 9.0 in and b = 6.0 in A solution of hydrochloric and nitric acids will be used as the etchant Determine (a) metal removal rate in in 3 /hr, (b) time required to etch to depth, and (c) required dimensions of the opening in the cut and peel maskant required to achieve the desired pocket size on the part Giải: a) Tra bảng ta có Fe = f = 0.001 in/min Diện tích A = πab = π*9*8 = 226.2 in RMR = 0.001*226.2 = 0.2262 in 3/min = 13.572 in3/hr b) Thời gian Tm = 0.4/0.001 = 400 = 6.67 hr c) Với Fe = => u = d/Fe = 0.4/2 = 0.2 in Maskant opening a’ = a - u = – 0.2 = 8.8 in Maskant opening b’ = b – u = 6- 0.2 = 5.8 in 26.21/ In a chemical blanking operation, stock thickness of the aluminum sheet is 0.015 in The pattern to be cut out of the sheet is a hole pattern, consisting of a matrix of 0.100-in diameter holes If photochemical machining is used to cut these holes, and contact printing is used to make the resist (maskant) pattern, determine the diameter of the holes that should be used in the pattern Giải: Tra bảng ta Fe = 1.75 Độ sâu lỗ = 0.015/2 = 0.0075 in u = 0.0075/1.75 = 0.0043 in   đường kính lỗ mẫu = 0.1 – 2*0.0043 = 0.0914 in