Sugar cane machine
Sugar cane machine MA3001
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Date Created: 09/30/15
TECHNOLOGICAL UNIVERSITY SCHOOL OF MECHANICAL AND AEROSPACE ENGINEERING MA3001 MACHINE ELEMENT DESIGN SUGAR CANE JUICE EXTRACTION MACHINE NG ZHONG JIN U1120878D ANG WEI XUN UXXXXXXXX REVIN UXXXXXXXX ROYDEN UXXXXXXXX Z oz quotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquotquot quot NOLLDEI IEIS AEIN SNouvansavs NsIsaG Gz iHVd SI SNWSVIG iNawow SNIGNHQ 292 U SNWSVIG A909 9399 I92 91 HVHS SNOIWWDWD NSISHG 92 MW SI Awwwns ian DMZ H NOIiWisnm 539909 was 692 H 539909 was ands 892 H SW39 WNOIHGGV 92 SI quotquotquotquot quot EIED VCLS NOIiD GEIH GEIEIdS CINZ was ands NaAma a0 asIOHs 93992 ZI quotquotquotquot quot EIED VCLS NOIiD GEIH GEIEIdS CINZ was ands HEIAIHC a0 asIOHs S39EIZ ZI quotquotquotquot quot EIED VCLS NOIiD GEIH GEIEIdS lsI was ands NaAma a0 asIOHs v39az ZI quotquotquotquot quotEEVLS NOIiD GEIH GEIEIdS m was ands HEIAIHC a0 asIOHs E39EIZ H HHMOd NSISHG 292 H swasmds do adungz H GANG was SNOIWWDWD NSISHG 92 MW 6 Awwwns ian OFVZ 6 savm HVHS 6W 8 SL199 do HagwnN 8W 8 HHMOd Elva GHDHWODQ HHMOd Elva 1 8 HODW NOIDHWOD HHMOd 9 L HLSNE mild Bag 9 HDNVLSIG 99mm 9 L saa wvm HAVHHSWVZ 9 MW Gaadskw 9 NOIDHS 55099 ag 2 9 HHMOd NSISHG FVZ 9 aWC i199 SNOIWWDWD NSISHG VZ MW 17 SNOILWHGISNOD Nmsaagad a 17 SNOIWDHDHCIS NSISHG H 17 NOIDnGOMNI 1 MW SIN3INOD IO 318VJ PART 3 ASSEMBLY DRAWINGS AND PART LIST 21 31 ASSEMBLY DRAWING 21 32 PART LIST 21 APPENDIX 22 PART 1 INTRODUCTION 1 1 DESIGN SPECIFICATIONS These parameters are predetermined and will govern our design choices 0 Power 037 kWfrom assigned motor Input Speed ninitial 1395 rpm Based on catalogue APPENDIX A 0 Final Output Speed nfinal 35 rpm 0 Maximum Dimensions 750 x 750 x 1300 mm The power transmission drive will use a combination of o Aspur or heica gear drive system and o A belt or chain drive system 12 PREDESIGN CONSIDERATIONS We have made some preliminary choices to simplify the design process It is impossible to achieve the desired speed reduction in a single stage 0 Thus we will consider a multi stage speed reduction system We believe that a 2stage speed reduction system will be sufficient A belt drive system is chosen over a chain drive system because 0 It is more suited for high speed and low torque operations 0 It is more versatile in connecting shafts over a large centre distance 0 It is cheaper and produces less noise and vibration Specifically we will use V belts in the first stage of speed reduction 0 Due to space constraints V belts are more suitable Flatround belts are useful if the centre distance between pueys is very long 0 Although less efficient with efficiency of around 70 to 90 V belts are capable of transmitting more power 0 V belts wedge tightly into grooves This increases friction and allows higher torques to be transmitted Some assumptions made are 0 The machine will be used 10 16 hours a day 0 100 power transmission from input to output Tables A1 to A 8 are obtained from Roulunds Roflex V belts Catalogue o In the 2nd stage of speed reduction spur gears will be used because 0 They are cheaper to manufacture o No additional axial thrust load which simplifies the design process 0 Spur gears are selected from NOZAG Gear Catalog PART 2A DESIGN CALCULATIONS BELT DRIVE We will first proceed with design calculations for the V belt drive to obtain an intermediate output speed This intermediate output speed should be achieved by a setup that uses standard industry components and satisfies the space constraints imposed by the design requirements Corresponding loads on shafts will also be calculated 2A1 DESIGN POWER 0 From Table A2 Service Factor 14 Because of these reasons 0 Predicted daily service hours of 10 to 16 hours 0 The driving unit is an AC Motor transmitting normal torque o The driven unit is Crushers 0 Design Power Service Factor Drive Power 0 Drive Power 037kW predetermined in design specifications 0 Therefore Design Power 14037 kW 0518 kW 2A2 VBELT CROSS SECTION 0 From Table A3 SPZ VBelt is chosen because 0 Faster Shaft 1395 rpm Design Power 0518 kW 2A3SPEED RATIO 0 As explained in Section 12 at least a 2stage speed reduction is needed 0 First assume an Intermediate Output Speed m 600 rpm 0 Hence Speed Ratio Assumed 1395 600 2325 0 Feasibility of this assumption will be checked against predetermined design requirements If it does not satisfy iteration will be needed 2A4SHEAVE DIAM ETERS 0 Assume Belt Speed vb 10 ms 0 D1 2vboo D1 210139521160 013691 m o From Table A4 Select D1 140 mm Standard Sheave Diameter Check belt speed vb 0140139527t602 10226 ms 5 ltVblt 33 ms ThereforeD1 140mm is feasible 0 Using Speed Ratio Assumed 2325 D2 can be calculated 0 D2 1402325 3255 mm o From Table A4 Select D2 315 mm Standard Sheave Diameter 0 Calculate Actual Speed Ratio 0 F12 01D1D2 n 1395140315 620 rpm 0 Actual Speed Ratio n1n2 1395620 225 2A5 CENTRE DISTANCE amp BELT PITCH LENGTH 0 Consider Tentative Centre Distance TCD 075m o 075 m is a good starting point as it leaves sufficient room for the rollers and the motor to fit within the height constraint of 13 m 0 Calculate Tentative Belt Length Dz D12 o TBL 2 x TCD 15712 D1 4XTCD Using D1 014 m D2 0315 m from Section 2A4 2 TBL 2 x 075 1 57o315 014 w 4x075 o From Table A1 Belt Length 2240 mm Standard Pitch Length 2225 m 0 Calculate Actual Centre Distance C o Usmg the equation C 2 B B 3126D2 D1 WlthB4L628D2D1 B 42240 6280315 014 61026 m 61026 610262 320 315 0142 16 Center Distance 07578 m 2A6 POWER CORRECTION FACTOR 0 Calculate Angle of Contacton smaller sheave o 91 1800 zen16322 0315 014 207578 Since91gt 120 no slippage will occur 01 180 25in1 16674 0 Determine Correction Factor for Arc of Contact Ce 0 Using91 166740 from Table A 5 C9 097 0 Determine Belt Length Correction Factor CL 0 Using L 2240 mm interpolate from Table A6 CL 106 2A7 RATED POWER amp CORRECTED RATED POWER 0 D1 140 mm m 1395 rpm Speed Ratio 225 Interpolate from Table A 7a Rated Power kW per Belt 42704 kW 0 Corrected Rated Power CeCLRated Power CRP 09710642704 43908 kW 2A8 NUMBER OF BELTS o of Belts DesignPowerCRP Design Power 0518 kW from Section 2A1 Thus of Belts 051843908 0118 We see that only 1 belt is required in this belt drive system 2A9 SHAFT LOADS 0 Calculate o the angle between drive centre line and straight section of belt 0 G 1800912 a 180166742 663 0 Since a is small ie lt 10 FX F1F2cos o 0993F1F2 z F1F2 Fy F1F2sin o 0115F1F2 z 0 0 We can assume parallel forces ie FR F1F2 0 Determine forces F1 and F2 T 0 F1 2 FC 2 where y eelfe FC 2 mvb2 o From Table A8 fe 040as Dmgt 80 mm Therefore y eC16674 180040 3203 0 From Table A1 m 0070 kgm Therefore Fc 0070102262 732 N 0 T1 Pwl 03713952n60 2532 Nm LE 0 Fl FC2y1D1 F 7322 3203 2532 1 39 3203 1 014 F1 5991 N 0 F2 F12T1D1 F2 5991 22532o14 M 0 Determine resultant force FR 0 Since parallel forces are assumed FR F1F2 E3 5991 2374 8365 N 2A10 PART SUMMARY From this section of calculations we have obtained specifications for the belt drive and the forces F1 and F2 This is summarized in the on the next page F2 2374 N F2 2374 N SH EAVE 2 DIAMETER 315MM Fl 5991 N SPZ VBELT 2240MM I CENTER DISTANCE 7578MM l Fl 5991 N SHEAVE 1 DIAMETER 140MM Sheaves D1 140 mm D2 315 mm Center Distance 07578 m Speed Ratio 225 Belts Belt SPZ VBelt No of Belts 1 Belt Length 2240 mm Belt Speed 10226 ms Forces F1 5991 N F2 2374 N FR 8365 N 10 PART 23 DESIGN CALCULATIONS GEAR DRIVE We will now proceed to the second stage of speed reduction using a gear drive in which the intermediate output speedn2 620 rpm is to be further reduced to the required final output speed of 35 rpm A preliminary arrangement is shown below GEARS 3 amp 4 IST SPEED REDUCTION STAGE GEARS 5 amp 6 2ND SPEED REDUCTION STAGE 231 TYPE OF SPUR GEARS o Spur gears made of heat treatable steel C45 heat treated are chosen 0 Higher strength as a result of heat treatment allows smaller gears to be used when transmitting the same torque 233 CHOICE OF DRIVER SPUR GEAR 15139 SPEED REDUCTION STAGE 0 Calculate torque experienced by gear G3 0 G3 rotates at the same speed as n2 n2 n3 620 rpm 0 T3 Pw3 where 03 6202Tl60 64926 rads T3 37064926 5699 Nm 0 Referring to Load Diagram Z 056 0 Consider no of teeth N3 16 Minimum module 125 We choose gear of slightly larger module m 2 o From NOZAG Spur Gear Table Z 108 0 Choose Gear Sg 2016 It has the following specifications D3 32 mm 234 CHOICE OF DRIVEN SPUR GEAR 15139 SPEED REDUCTION STAGE 0 Overall speed reduction required is 62035 177 times 0 Speed ratio has to be approximately 5 0 Number of teeth for driven gear N4 z 5N3 80 We choose Gear Sg 2070 with N4 70 closest to 80 It has the following specifications D4 140 mm 0 004 003N3N4 04 649261670 14840 rads I14 1484060239I39I39 14171 rpm 235 CHOICE OF DRIVER SPUR GEAR 2 SPEED REDUCTION STAGE 0 Calculate torque experienced by gear G5 0 G5 rotates at the same speed as n4 ns n4 14171 rpm 0 T5 Poos where 005 04 14840 rads T5 37014840 24932 Nm 0 Referring to Load Diagram Z 056 12 0 Consider no of teeth N5 16 Minimum module 2 We choose gear of slightly larger module m 25 0 From NOZAG Spur Gear Table Z 109 0 Choose Gear Sg 2516 It has the following specifications D5 40 mm 236 CHOICE OF DRIVEN SPUR GEAR 2 SPEED REDUCTION STAGE 0 Speed reduction required is 1417135 405 times 0 Number of teeth for driven gear N6 405N5 648 We choose Gear Sg with N6 65 closest to 648 It has the following specifications D6 1625 mm o 006 005N5N6 05 148401665 3653 rads I16 365360211 3488 rpm 3 35 rpm satisfies requirement 237 ADDITIONAL GEARS 0 Additional gears are needed to ensure optimum separation between the two 100mm diameter roers We believe that a 40mm separation is optimal Therefore Gear Sg 2060 of d 120 mm is chosen as the additional gears o Considerations 0 Gear 7 will be on the same shaft as Gear 6 0 Gear 7 will mesh with Gear 8 both of which will have the same number of teeth so that speed is kept at 35 rpm 0 Calculate torque experienced by gears G7amp G8 0 where 006 07 008 3653 rads I 3703653 10129 Nm 2B8 SPUR GEAR FORCES 13 o For Gear 3 G3 0 Tangential force Wt 2TD From Section 233 T3 5699 Nm amp D3 32 mm Wt3 256990032 35617 N 0 Radial force Wr Wt tan CD where D 20 Wr3 35617 tan 20 12963 N o For Gear 4 G4 0 Wt4 35617 Nin opposite direction to Wt3 0 Wm 12963 N o For Gear 5 G5 0 Tangential force Wt 2TD From Section 2B5 T5 24932 Nm amp D5 40 mm Wt5 2249320040 124661 N 0 Radial force Wr Wt tan CD where D 20 Wr5 124661 tan 20 45372 N o For Gear 6 G5 0 W55 124661 N 0 Wm 45372 N o For Gear 7 G7 0 Tangential force Wt 2TD From Section 2B7 T7 10129 Nm amp D7 120 mm Wt7 2101290120 168817 N 0 Radial force Wr Wt tan CD where D 20 Wr7 168817 tan 20 61444 N o For Gear 8 G8 0 Wts 168817 N 0 Wm 61444 N 239 GEAR FORCES ILLUSTRATION 0 To facilitate shaft design calculations in the next section the forces on gear pairs G5 amp G6 and G7 amp G8 are illustrated in the diagrams below 14 2310 PART SUMMARY From this section of calculations we have obtained specifications for the 4 spur gears that are used in the 2 stage reduction Tangential and radial forces experienced by the spur gears were also calculated All data is summarized in a table and illustrated in the diagram below Gears used in M3 M4 2 mm 1st reduction D3 N3 32 mm 16 teeth D4 N4 140 mm 70 teeth n3 620 rpm n4 14171 rpm Gears used in M5 M6 25 mm 2nd reduction D5 N5 40 mm 16 teeth D6 N6 1625 mm 65 teeth n5 14171 rpm n6 3488 rpm Forces Wt3 35617 N Wr3 12963 N Wt4 35617 N Wr4 12963 N Wt5 124661 N Wr5 45372 N Wt6 124661 N Wr6 45372 N Wt7 168817 N Wr7 61444 N Wt8 168817 N Wr8 61444 N PART 2C DESIGN CALCULATIONS SHAFT We have already identified spur gear forces and shaft forces from the belt drive Section 2A10 We will now proceed with the shaft design Based on our design there are in total 4 shafts connecting all the machine elements This is illustrated in the diagram below Sample calculations will be carried out for the critical shaft Shaft 3 as it involves the most number of forces Shaft 3 is 290 mm in length A closer view of Shaft 3 together with machine elements mounted is shown below FRAME 16 2C1 FREE BODY DIAGRAMS IEBBEQN 453 T2 N 595iilrlN madam In the x direction 3951 N 10 me G l 40459 453mm EBBBBGN In the y direction The locations of points A to G are as follow Distance A B C D E F G from the 5 25 45 15917 245 265 285 left mm D is a critical point found from the bending moment diagrams in the next section Comments 0 Through multiple online references it is reasonable to assume the roller in operation causes a distributed load of 10 Nm in the y direction 0 Reaction forces at B amp F are solved by taking moment equilibrium about B and then considering force equilibrium ZF O 17 2C2 BENDING MOMENT DIAGRAMS In the xdirectjinn Hmm Ben cling Moment 2493192 In the ydirectjun j Hmrn Bending 39Mumem Ham 1939 n v e ganja IF 18 0 Using Equation 12 24 the minimum required shaft diameters can be calculated The final chosen shaft diameters are based off the nominal shaft diameters available Distance from Moment in x Moment in y Minimum Chosen left end mm direction Nm direction Nm MR Nm T Nm D mm D mm A 5 0 0 0 10129 1680 22 B 25 9074 24932 2653 10129 2013 25 C 45 6241 2098 658 10129 1735 30 D 15917 9936 63075 6385 10129 2794 30 E 245 22097 2624 3430 10129 2313 30 F 265 24932 9074 2653 10129 2013 25 G 285 0 0 0 10129 1680 22 o The following information was used in the calculation Parameters Design Choice Comment Shaft Material AISI 1137 CD N 3 Recommended design factor be at A 2 Profile Keyseat be at B 15 Well rounded Shoulder Fillets be at C 2 Profile Keyseat be at D 2 Profile Keyseat be at E 2 Profile Keyseat be at F 15 Well rounded Shoulder Fillets be at G 2 Profile Keyseat Reliability Factor CR 081 99 desired reliability Table 5 1 Size Factor Cs 086 Shaft diameter is x 28mm Figure 5 9 Endurance Strength Sn 263106 Machined surface finish Figure 5 8 Sn 1832106 Sn SnCsCR Yield Strength Sy 565 106 Based on shaft material Appendix 3 Tensile Strength Su 676 106 Based on shaft material Appendix 3 19 Based on our shaft design above we will do a sample calculation for shaft 3 which requires a keyseat profile for gear 6 and 7 o AISI 1020 Cold drawn steel is used as the material for the key 0 Sy 352Mpa lower than the yield strength for the materials used in the shaft and gear hub 0 ISO standard is used in the design calculations 4TN 4x 10129x3 LAG S DWSy 22x6x352 002615m 2615mm The key should extend over all or a substantial apart of the length of the hub Therefore we choose Key for Gear 635mm 30mm Key for Gear 731mm 30mm PART 2E DESIGN CALCULATIONS BEARING SELECTION From the shaft design procedure shown in Part 2C there are 2 bearings used located at point B and point F respectively Since only spur gears and a roller are mounted on the shaft only radial loads are exerted on the shaft Thus single row radial deepgroove ball bearings will be used 2E1 RADIAL FORCES ON BEARINGS 0 As calculated in Part 2C 0 FBX 138829 N FAY 40458 N o FFX 59541 N FBY 238820 N o Resultant Radial Load FB 1388292 404582 144604 N FF J595412 2388202 246130 N 2E2 BASIC DYNAMIC LOAD RATING 20 0 Design Life Ld 0 Assume operating hours of 10 hours per day 0 Assume that the machine is under warranty for 5 years 0 Total Operating Hours 103655 18250 hours 0 L10 rev L10 hoursn rpm60 minh L10 rev 18250 348860 382107 revolutions 0 Design Load Pd 0 At B Pd FB 144604 N since there is no axial force 0 At F Pd FF 246130 N since there is no axial force c Pad 56M c 14460438213 486996 N at B C 24613038213 828915 N at F 2E3 BEARING SELECTION 0 At points B and F shaft diameter is chosen as 25mm 0 From Table 14 ABearing 6005 is selected since C gtCcalculated PART 3 ASSEMBLY DRAWINGS AND PART LIST 31 ASSEMBLY DRAWING 32 PART LIST 21 APPENDIX A1 MOTOR SELECTION 22