H.- J. Steinmetz, Jan. 004 M A N U F A C T U R I N G P R O C E S S E S ME 38.3 A S S I G N M E N T 0 0 4 A machine shop has 15 workers and one shop foreman. The shop has to manufacture a quantity of 800 ball joints, shown in the attached drawing, in the most economical and efficient way on a 0HP CNClathe. All drilling operations have to be done on a drill press. The shop foreman's wage is $4.00/hour. The overhead cost for administration and engineering for this small machine shop is 45%. The cost for the inch (50.8 mm) diameter 4140 bar stock material, with a specific cutting force of 00 N/mm, is $.40 per foot including all taxes. Since the spindle bore of the lathe is only 35mm the bar stock does not feed through the spindle. Therefore each workpiece has to be cut on a circular saw. The hourly rate for a circular saw is $30/hour and the saw blade width =.5mm. The flank wear factor for all tools is 0.3 mm. Leave an equal amount (ap= 0.5 mm) of material on the radius for the finishing tools. The energy cost for the lathe = $0.18/kW. The delay time, tv, is 10% of tg for all time calculations. CNC - Lathe : The Program number is O004, n max. = 3500 RPM, E = 0.80. The replacement cost for this machine = $85,000.00. Cost for space Cs=$1,500.00/year. The maintenance cost Cm=$,000.00/year. Tool and insurance cost Co = $8,500.00/year. The Lathe is used in eight-hour shifts a day, 5 days a week for 5 weeks each year, for a utilization time of 10 years with an 11% interest rate. The $.0/hour wages paid to the CNC machining technician includes all benefits. Xset Zset Vc n Rs D.O.C. f Tool Tool# ID# (mm) (mm) (m/min) (rpm) (mm) (mm) (mm/rev) External roughing 01 5 17.0 50.8 150 na 1. 4.0 0.45 External roughing 0 50 130.0 50.8 150 na 0.8.0 0.5 External finishing 03 55 178.7 51.1 00 na 0.4 Part off tool 04 130 17.3 58.6 80 nmax= 1000 0.4 na 0.08 Drill press: The drill press has an automatic feed rate which can be set to the required feed. The drive motor has 1KW. The set up time for the drill jig and the tools tr=6min. The hourly rate for the drill press is $4.00/hour. The safety distance for positioning all tools is mm and the positioning time ttb= 0.min for each drilling and countersinking operation. The following rpm are available: 10, 40, 480, 960, 1440 and 1880. E=0.7. Diameter L ν Vc n Operation f Tool Tool# (mm) (mm) (mm) (m/min) (rpm) drill (mm/rev) Drill 01 5.0 7.5 4 1 hole 0.5 Drill 01 5.0 10 4 4 holes 0.5 Countersink 0 3.0 18 18 1 hole 0.5 1
THE LAB REPORT HAS TO INCLUDE THE FOLLOWING IN SEQUENCE: H.- J. Steinmetz, Jan. 004 CNC Lathe ( 60/100 marks) 1. Create a CNC Set-up-sheet (Template CNC set-upsheet-0.dwg) for the CNC-machine which can be found at the following site: http://www.engr.usask.ca/classes/me/38/ (Labs, Manufacturing).. Calculate the workpiece datum and the reference points for the CNC-machine. The safety distance for the tool change position u=5.0 mm. 3. Calculate the N.O.C. (number of cuts) for T01, leaving the required amount of material for the finishing tool and calculate the maximum power used for this external roughing tool. 4. Calculate all start and endpoints for each pass for T01 with tool nose radius compensation. Show the geometry for all points. 5. Calculate the N.O.C. (number of cuts) for T0, leaving the required amount of material for the finishing tool and calculate the maximum power used for this external roughing tool. 6. Calculate all start and endpoints for each pass for T0 with tool nose radius compensation. Show the geometry for all points. 7. Calculate the tool nose radius compensation calculation for the external finishing tool T03. Show the geometry for all points. 8. Calculate the feed rate for the external finishing tool T03 based on the required surface finish. Use 0% less federate than calculated in the program for the finishing cuts. 9. Write a Program for the CNC-Lathe. Use CNC-SPACE to format the text from the following site: http://www.engr.usask.ca/classes/me/38/ (Labs, Manufacturing). 10. Calculate the expected tool life for T01. From an exponent table for the insert grade TN5 the following parameters are given: C=198, F=-0.1, E=-0.34, G=-0.14 11. Calculate the machining rate C and the hourly rate for the CNC lathe. Drill press ( 15/100 marks) 1. Calculate if the rpm s used for the drilling and countersinking operation. 13. Calculate if the drill press has sufficient power for the drilling operation. 14. Calculate the cutting time for each hole to drill. 15. Calculate the cutting time for the countersink tool. Design ( 0/100 marks) 16. Design a jig to drill 4 holes, diam. 5mm, L=5mm and 1 hole, diam. 5mm, L=7.5mm. Provide a dimensioned engineers drawing to manufacture this jig. A template drawing can be found at the following site: http://www.engr.usask.ca/classes/me/38/ (Labs, Manufacturing). 17. Describe how the operator of the drill press has to use the jig designed by you in a step by step procedure. Process time and cost ( 5/100 marks) 18. Find the process time for the circular band saw. 19. Calculate the process time for the CNC-Lathe. Use CNC-PLOT from the following site: http://www.engr.usask.ca/classes/me/38/ (Labs, Manufacturing). 0. Calculate the process time for all drilling and countersinking operations. 1. Tabulate all process times and the total cost of this manufacturing process.
3
4 um h7 o 4
. REFERENCE POINT CALCULATION for CNC - PROGRAM No. O004 DATE 10-Jan- 04 What is the ZW to use for these calculation? ZW = 56.000 mm How long is the workpiece? LW = 85.000 mm What is the safety distance? u = 5.000 mm X home 5.77 mm TOOL DIMENSIONS Z home 573.69 mm X-set Z-zet L max REFERENCE TOOL 17.000 50.800 5 T01 17.000 50.800 58.600 50 T0 130.000 50.800 55 T03 178.700 51.100 130 T04 17.000 58.600 CALCULATION of WORKPIECE DATUM TUM (W) X = Z = 395.77 mm 466.89 mm CALCULATION of REFERENCE POINT (RP) for TOOL CHANGE REFERENCE TOOL RP X = 300.000 mm RP Z = 117.800 mm CALCULATION of REFERENCE POINT (RP) for TOOL CHANGE T01 RP X = 300.000 mm RP Z = 117.800 mm T0 RP X = 97.000 mm RP Z = 117.800 mm T03 RP X = 48.300 mm RP Z = 117.500 mm T04 RP X = 300.000 mm RP Z = 110.000 mm 5
3. T01 external roughing tool Power calculation Barstock diameter = 50.8mm Point 1 X = 4.8mm ap = 4.0 mm Point X = 49.0mm ap = 1.4 mm Point 3 X = 34.8mm ap = 4.0 mm Point 4 X = 6.8mm ap = 4.0 mm Point 5 X = 18.8mm ap = 4.0 mm Point 6 X = 10.8mm ap = 4.0 mm Point 7 X =.8mm ap = 4.0 mm Point 8 X = -.4mm ap =.8 mm N.O.C. = 7 cuts ap * f * kc * Vc P= VB 3 60 *10 * E ap depth of cut mm D drill diameter mm f feed rate mm / rev kc specific cutting force for a material N / mm Vc surface speed m / min VB flank wear factor mm E efficiency P power kw 1. The specific cutting force kc is the force necessary to remove a chip with a cross section of 1. Important factors affecting kc are: Strength and alloy components of the workpiece material Cutting geometry, particularly rake angle In addition, the specific cutting force kc is a function of the chip thickness h. The constant, determined for a chip thickness of h = 1 mm and a chip width of b = 1 mm, is the basic value of the specific cutting force denoted k c1. 1 mm. T01 Power requirement for turning ap = 4 mm ap f kc Vc VB E Input 4 0.45 00 150 0.3 0.8 Answers Power kw Power HP Equation 4 0.45 00 150 1.1 0.8 60 1000 13.860 18.586 6
4. T01 external roughing tool nose radius compensation Barstock diameter 50.8mm R+ap+Rs Z 1 X 3 4 5 6 7 8 L = 61.0 mm A X A + = ( R + ap + Rs) ( Rs) ap = 0.5mm Rs = 1.mm Leading edge Z = L + A Rs R = 4.0mm Tool nose radius compensation calculation for T01, 4mm Radius 1 4 4.800 7.036 4 0.5 1. 61 1.36 1.4 49.000 59.800 4 0.5 1. 61 4 49.000-1.00 4 0.5 1. 0 3 4 34.800 77.535 4 0.5 1. 61 17.735 4 4 6.800 80.950 4 0.5 1. 61 1.150 5 4 18.800 83.1 4 0.5 1. 61 3.41 6 4 10.800 84.638 4 0.5 1. 61 4.838 7 4.800 85.368 4 0.5 1. 61 5.568 8 4 -.400 85.500 4 0.5 1. 61 5.700 7
5. T0 external roughing tool Power calculation Point 1 X = 49.0mm ap =.0 mm Point X = 45.0mm ap =.0 mm Point 3 X = 41.0mm ap =.0 mm Point 4 X = 37.0mm ap =.0 mm Point 5 Point 6 X = 33.0mm ap =.0 mm Point 7 X = 9.0mm ap =.0 mm Point 8 X = 5.0mm ap =.0 mm Point 9 X = 1.0mm ap =.0 mm N.O.C. = 8 cuts ap * f * kc * Vc P= VB 3 60 *10 * E ap depth of cut mm D drill diameter mm f feed rate mm / rev kc specific cutting force for a material N / mm Vc surface speed m / min VB flank wear factor mm E efficiency P power kw 3. The specific cutting force kc is the force necessary to remove a chip with a cross section of 1 4. Important factors affecting kc are: Strength and alloy components of the workpiece material Cutting geometry, particularly rake angle In addition, the specific cutting force kc is a function of the chip thickness h. The constant, determined for a chip thickness of h = 1 mm and a chip width of b = 1 mm, is the basic value of the specific cutting force denoted k c1. 1 T0 Power requirement for turning ap = ( 49-45 ) / =.0 mm ap f kc Vc VB E Input 0.5 00 150 0.3 0.8 Answers Power kw Power HP Equation 0.5 00 150 1.1 0.8 60 1000 3.850 5.163 mm. 8
6. T0 external roughing tool nose radius compensation A Diameter 49.0mm j 11 k 10 R-ap-Rs X 8 9 4 6 7 3 5 Z b R+ap+Rs 1 R L = 61.0 mm X A + = ( R + ap + Rs) ( Rs) ap = 0.5mm Rs = 0.8mm Leading edge Z = L A Rs R = 4.0mm α = 90 Intersection between the slope and the arc α 90 a = ( Rsin ) = ( 4sin ) = 16.97056mm a = 33.9411 h = R R a = 4 4 16.971 = 7.099 mm b = ( Rs + ap)cos 45 = ( 0.8 + 0.5) cos 45 = 0.919388 mm Point 5 X = R - h = 33.9411 mm Z = L a b Rs = 61 16.97056 0.919-0.8 = 4.861 mm Calculations on the slope X = ( R h) X X b = tanα c = ap sinα g = Rs tan α 9
Z6 X 33.9411 g c b a R F L=61.0mm F = L R 33.9411 = 44.09 mm Tool nose radius compensation calculation for T0, 4mm Radius Z6 X 33.9411 g c b x Point 6 X = R - h - X = 48 14.0598-0.940 = 33.0 mm X = ( 4 7.099) 33 = 0.4701mm 0.4701 b = tan 45 = 0.4701mm c = 0.5 sin 45 = 0.707mm 45 g = 0.8 tan = 0.33137mm Point 6 Z = F b c g Rs = 44.09-0.4701-0.707 0.33137 0.8 = 41.7053 mm 10
Point DOC X Z R ap Rs L A R+ap+Rs R+ap+Rs X/ 1 4 49.000 60.00 4 0.5 0.8 61 4 45.000 50.341 4 0.5 0.8 61 9.859 5.3 5.3 640.090.500 3.300 3.300 54.89 3 4 41.000 46.547 4 0.5 0.8 61 13.653 5.3 5.3 640.090 0.500 1.300 1.300 453.69 4 4 37.000 43.84 4 0.5 0.8 61 16.358 5.3 5.3 640.090 18.500 19.300 19.300 37.49 5 1.41 34.1796 4.310 4 0.5 0.8 61 17.890 5.3 5.3 640.090 17.090 17.890 17.890 30.04 Point DOC X Z R ap Rs L h F a b DX tan 45 c sin 45 g tan.5 5 1.41 34.1796 4.310 4 0.5 0.8 61 7.099 16.9706 0.919 6 4 33.000 41.704 4 0.5 0.8 61 7.099 44.09 0.5898 0.4701 0.4701 1.000 0.7071 0.7071 0.3314 0.414 7 4 9.000 39.704 4 0.5 0.8 61 7.099 44.09.5898.4701.4701 1.000 0.7071 0.7071 0.3314 0.414 8 4 5.000 37.704 4 0.5 0.8 61 7.099 44.09 4.5898 4.4701 4.4701 1.000 0.7071 0.7071 0.3314 0.414 9 4 1.000 35.704 4 0.5 0.8 61 7.099 44.09 6.5898 6.4701 6.4701 1.000 0.7071 0.7071 0.3314 0.414 Tool nose radius compensation calculation for T0, 10mm Radius 11 k R-ap-Rs j 10 9 Xc = 18 mm Zc = 0 mm j = Xc - R ap Rs = 18-10 0.5 0.8 = 6.7 mm k = ( R ap Rs) j ( 10 0.5 0.8) 6.7 = = 5.54977 mm Point 10 X = 1.0 mm Z = Zc + k Rs = 0 + 5.54977 0.8 = 4.74977 mm Point 11 X = 1.0 mm Z = Zc - k Rs = 0 + 5.54977 0.8 = 13.650 mm 11
7. T03 external finishing tool nose radius compensation e 6 p 5 R-Rsj k q r 4 c 3 d R+Rs b R 1 L = 61.0 mm Rs = 0.4mm R = 4.0mm α = 90 Intersection between the slope and the arc α 90 a = ( Rsin ) = 4sin ) ( = 16.97056 mm h = R R a = 4 4 16.971 = 7.099 mm Calculations of start and end-point on the slope Point e = ( R + Rs) sinα e = ( 4 + 0.4) sin 45 = ( R Rs) cosα b = ( 4 + 0.4) cos45 b + X = b Rs = 33.7068 mm Z = L e Rs = 61 17.534 0.4 = 43.3466 mm Point 3 c = ( Rs) c = ( 0.4) tan α tan 45 =0.165685 mm = 17.534 mm = 17.534 mm d = R h X 3 d 03 = 4 7.099 = 6.9701 mm X = 0.0 mm Z = L a d c Rs = 61 16.97056 6.9701 0.165685-0.4 = 36.4936 mm Point 4 j = Xc r Rs = 18 10 0.4 = 7.6 mm p = r ( 18 r) p = 10 (18 10) = mm q = pr p q = * *10 = 6 mm 1
k = ( r Rs) j k = ( 10 0.4) 7.6 X = 0.0 mm Z = 0 + k Rs = 0 + 5.86515 0.4 = 5.465 mm = 5.86515 mm Point 5 X = 0.0 mm Z = 0 - k Rs = 0 + 5.86515 0.4 = 13.73485 mm 8. Calculate the feed rate for the external finishing tool T03 From drawing: General surface finish = 10 µ m, 4 µ m Theorethical feedrate f th = RthRs 15 R th = 10 µ m R th = 4 µ m R Rs f = th 10*0.4 mm th f th = = 0.17888 minus 0% f = 0.14 15 15 rev 4*0.4 mm f th = = 0.11313 minus 0% f = 0.09 15 rev 13
9. CNC Program % O004 N001 G00 G8 U0.0 W0.0 N00 G99 G1 M70 N003 G50 X395.77 Z466.89 S3500 N004 G00 T0100 N005 G00 X300.000 Z117.8 N006 G50 X300.000 Z117.8 S3500 N007 G96 S150 M04 N008 M08 N009 X54.800 Z85.5 T0101 N010 G01 X-3.0 F0.45 N011 G00 X1.0 Z87.5 N01 X4.8 N013 G04 X1.5 N014 G01 Z7.036 N015 G03 X49.0 Z59.8 R5.7 N016 G01 Z-1. N017 X50.8 N018 G00 Z87.5 N019 X34.8 N00 G01 Z77.535 N01 G03 X4.8 Z7.036 R5.7 N0 G00 Z87.5 N03 X6.8 N04 G01 Z80.95 N05 G03 X34.8 Z77.535 R5.7 N06 G00 Z87.5 N07 X18.8 N08 G01 Z83.1 N09 G03 X6.8 Z80.95 R5.7 N030 G00 Z87.5 N031 X10.8 N03 G01 Z84.638 N033 G03 X18.8 Z83.1 R5.7 N034 G00 Z87.5 N035 X.8 N036 G01 Z85.365 N037 G03 X10.8 Z84.635 R5.7 N038 G00 Z87.5 N039 X-.4 N040 G01 Z85.5 N041 G03 X.8 Z85.368 R5.7 N04 G00 X00.0 M09 N043 X300.0 Z117.8 T0100 M05 N044 G00 U0.0 W0.0 T00 N045 G50 X97.0 Z117.8 S3500 N046 G96 S150 M04 N047 G00 X53.0 Z60. M08 N048 G01 X49.0 F0.5 N049 G03 X45.0 Z50.341 R5.3 N050 G01 Z-0.8 N051 X49.0 N05 G00 Z50.341 N053 G01 X45.0 N054 G03 X41.0 Z46.547 R5.3 N055 G01 Z-0.8 N056 X45.0 N057 G00 Z46.547 N058 G01 X41.0 N059 G03 X37.0 Z43.84 R5.3 N060 G01 Z-0.8 N061 X41.0 14
N06 G00 Z43.84 N063 G01 X37.0 N064 G03 X34.179 Z4.31 R5.3 N065 G01 X33.0 Z41.7 N066 Z-0.8 N067 X37.0 N068 G00 Z41.7 N069 G01 X33.0 N070 X9.0 Z39.7 N071 Z-0.8 N07 X33.0 N073 G00 Z39.7 N074 G01 X9.0 N075 X5.0 Z37.7 N076 Z-0.8 N077 X9.0 N078 G00 Z37.7 N079 G01 X5.0 N080 X1.0 Z35.7 N081 Z-0.8 N08 X5.0 N083 G00 Z4.749 N084 G01 X1.0 N085 G0 X1.0 Z13.65 R8.7 N086 G00 X00.0 M09 N087 X97.0 Z117.8 T000 M05 N088 G00 U0.0 W0.0 T0303 N089 G50 X48.3 Z117.5 S3500 N090 G96 S00 M04 N091 G00 X0.0 Z85.0 M08 N09 G01 X-1.5 F0.14 N093 X-0.8 N094 G03 X33.707 Z43.347 R4.4 N095 G01 X0.0 Z36.494 N096 Z5.465 F0.09 N097 G0 X0.0 Z13.735 R9.6 F0.14 N098 G01 Z-0.4 F0.09 N099 G00 X00.0 M09 N100 X48.3 Z117.8 T0300 M05 N101 G00 U0.0 W0.0 T0404 N10 G50 X300.0 Z110.0350 S1000 N103 G96 S80 M04 N104 G00 X00.0 Z0.0 M08 N105 X54.8 N106 G01 X17.0 F0.08 N107 G00 X4.0 N108 Z3.0 N109 G01 X18.0 Z0.0 N110 X5.0 N111 G00 X00.0 M09 N11 X300.0 Z110.0 T0400 M05 N113 G8 U0.0 W0.0 N114 M30 % 15
10. Tool life calculation T01 Tool life calculation ap = 4 mm Answers ap f Vc Exponent Table Tool life (mm) (mm/rev) (m/min) C F E G min Input 4 0.45 150 198-0.1-0.34-0.14 15.395 ap^f= 4.0-0.1 0.8467 f^e= 0.5-0.34 1.3119 1/G= 1-0.14-7.149 11. Machining rate and hourly rate for the CNC Late 11a. Machining rate C calculation Cw + Ci + Cs + Ce + Cm + Co C = Tu ($/hour) Example: Mean interest 11% Wages $.0 hour Replacement cost $ 85,000.00 Tu Utilization time 10 years, eight hour shifts a day, 5 days a week for 5 weeks each year (4160 hours/year) Cs = Cost for space $ 1500.00 year Ce = Cost for energy, efficiency = 0.8 0 HP = (0.7457*0) = 14.914 KW, $0.18KW hour. = $0.18 4160hours 14.914KW * = $ 13,959.50 year KWhour year * 0.8 Cm = Cost for maintenance and repair $,000.00 per year Co = Other cost (Insurance and Tool cost) $ 8,500.00 year Cw= Calculate write off for 10 years ( $/year ) 16
Interest 11% Utilization time 10 years Depreciation per year = 85,000.00 = $ 8,500.00 10 Depreciation per year Cw= Re placement cos t Utilization time = 85,000 10 = 8,500 Remaining value RV = Sv - Cw Mean value Mv = Sv - [ Cw/ ] = 85,000 - (8,500/) = 80,750.00 Interest = [ Mv * Interestrate ] / 100 Year Sv Mv Rv Interest Remaining f(mean Start Value Mean Value Value Value) 1 85,000.00 80,750.00 76,500.00 8,88.50 76,500.00 7,50.00 68,000.00 7,947.50 3 68,000.00 63,750.00 59,500.00 7,01.50 4 59,500.00 55,50.00 51,000.00 6,077.50 5 51,000.00 46,750.00 4,500.00 5,14.50 6 4,500.00 38,50.00 34,000.00 4,07.50 7 34,000.00 9,750.00 5,500.00 3,7.50 8 5,500.00 1,50.00 17,000.00,337.50 9 17,000.00 1,750.00 8,500.00 1,40.50 10 8,500.00 4,50.00 0.00 467.50 Interest 46,750.00 46,750.00 Ci = Interest per year = = $4,675. 00 10 Cw + Ci + Cs + Ce + Cm + Co C = Tu ($/hour) 8,500.00 + 4,675.00 + 1,500.00 + 13,959.50 +,000.00 + 8,500.00 C = = $ 9.41 hour 4160hrs 17
11b. Fixed hourly rate for machining Wages Machining rate $.0 hour C = $ 9.41 hour $ 31.61 hour $ 31.61 hour Calculate the average hourly shop rate for machining : 1 Foreman for 15 workers = 1 x $ 4.00 hour = $ 4.00 4 / 15 workers= $ 1.60 hour $ 1.60 hour --------------- $ 33.1 hour Overhead cost for administration and engineering department could be 40 to 160% 45 % $ 14.94 hour --------------- $ 48.15 hour Hourly rate $ 49.00 1. Calculate rpm s for drilling and countersinking Vc *1000 n Drill = = d *π 4*1000 n Drill = = 158 rpm use 1440 rpm 5*π Vc *1000 n Counters. = = d *π 18*1000 n Counters. = = 716 rpm use 480 rpm 8*π 13. Power calculation for the drillpress D * f * kc * Vc Solid drilling P= VB 1 *10 4 * E n * d * π Vc Drill = 1000 1440 *5* π = 1000 Vc =.6 m/min Drill 18
Power requirement for drilling Answers D f kc Vc n Pi VB E Input 5 0.5 00.6 1439 3.14 0.3 0.7 Power Power Equation kw HP (1) 5 0.5 00.6 1.1 0.7 1 10000 0.89 1.111 () 5 0.5 00 1439 3.14 1.1 0.7 1 1E+07 0.89 1.111 Feed force for drilling D f kc Edges N lbf 5 0.5 00 1375 309.1 Torque for drilling D f kc Nm lbf-ft 5 00 0.5 8 10 17 16.8 Chip removal rate for drilling D Vc Pi 3 cm / min 3 in / min 5.6 3.14 4000 0.4 0.07 P Drill 5* 0.5* 00*.6 = 1.1 4 1 *10 * 0.7 = 0.8 KW = 1.11 HP 14. Calculate the cutting time for each hole a.) 1 hole diameter 5 mm, L = 7.5+ = 9.5 mm, n=1440 rpm, f=0.5mm/rev b). 4 holes diameter 5 mm, L= 10+ = 1.5 mm ttu = L n * f ttu a = 9.5 1440 *0.5 = 0.06 min 1.5 ttu b = = 0.035 min 1440*0.5 Given: ttb = 0. min/hole 19
15. Calculate the cutting time for the countersink tool Given: chamfer 1.5x45degr. L = 1.5+ = 3.5 mm, n=480 rpm, f=0.5mm/rev ttu = L n * f ttu = 3.5 480 *0.5 = 0.09 min Given: ttb = 0. min/hole 16. Design a jig for drilling 17. Describe the drilling operation 18. Find the process time for the circular band saw From table 7 in Lab manual: tr = 5 min te = 1.8 min 19. Process time for the CAN Lathe from Plot Program The cutting time for each tool is minutes. Tool #1 0.34 Tool # 1.06 Tool #3 0.36 Tool #4 0.48 Tool #5 0.00 Tool #6 0.00 Tool #7 0.00 Tool #8 0.00 The total cutting time is... The rapid traverse time is... The total tool change time is.. The total dwell time is... The overall cutting time Chucking time.3 min 0.39 min 0.06 min 0.03 min. ttu =.71 min ttb = 0.30 min tg = 3.07 min tg *10 tv = tv = 0.307 min 100 te = tg + te te = 3.337 min tr= 3.0 + 5.0 + (3*tg) + (4*4) = 33.1 tr = 34 min 0
0. Process time for drilling and countersinking Drilling horizontal hole ttu = 0.06 min ttb = 0. min Drilling 4 vertical holes (4* 0.035) ttu = 0.14 min (4*0.) ttb = 0.8 min Countersink ttu = 0.09 min ttb = 0. min tg = 1.395 min tg *10 tv = tv = 0.1395 min 100 te = tg + tv te = 1.5345 min 1. Tabulate time and cost tr = 6.0 min Process Time: T = tr + ( te * m ) m = 800 Saw tr = 5min te = 1.8 min T = 5 + ( 1.8 * 800 ) = 1461.00 min CNC tr = 34 min te = 3.337 min T = 34 + (3.337 * 800 ) = 703.60 min Drillpress tr = 6 min te = 1.5576 min T = 6 + ( 1.5345 * 800 ) = 133.60 min Total time = 5398.0 min Process Cost: C = T * hourly rate / 60 Saw C = 1461 * 30 / 60 = $ 730.50 CNC C = 703.60 * 49 / 60 = $ 07.94 Drillpress C = 133.60 * 4 / 60 = $ 493.44 Matirial cost C = ( 91+ 0) mm *$.40 foot 800 = $ 655.98 1* 5.4 Total Cost $ 9,957.86 1