ELATECH® iSync™ high performance timing belts
Transcript
ELATECH® iSync™ high performance timing belts
SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti en .qxd 14/04/2010 11:02 Pagina 143 ELATECH® iSync™ high performance timing belts SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti en .qxd 14/04/2010 11:02 Pagina 144 ELATECH® iSync™ In the spirit of continuos innovation, in order to answer to the increased need of industry in power transmission, ELATECH® has developed the iSync™ range of belts. iSync™ belts are made with special polyurethane compound and high resistance steel tension cords which are processed with a unique and highly sophisticated technology to get a superior polyurethane belt. iSync™ belts offer optimal performances on all type of industrial applications. iSync™ belts are able to transmit up to 30% more than conventional T, AT type of belts in the same space or same power with a more compact drive. 160 Power increase 120 iSync T - AT Power (%) 140 100 80 60 40 20 0 iSync T - AT Polyurethane belt A - AT Belt Type Features Available profile range • High power transmission capabilities • Maintenance free • Superior length stability • Clean power transmission with no dust dispersion • No contamination of object in contact • Very high chemical resistance and particularly to oils, greases and gasoline • Superior abrasion resistance • High quality, thermo-set polyurethane designed specifically for timing belt applications • Available with either steel or Kevlar® reinforcement • Application temperature -30°C - +100 °C ELATECH® iSync™ belts are available in a standard range in the following profile range: Typical application fields ELATECH® iSync™ belts are suitable for power transmission drives where high precision is needed, cleanliness is critical and in difficult environment (presence of chemicals). • Plotters • Office automation • Medical technology • Packaging machines • Swimming pool cleaning robots • Banking machines • Coin dispenser • Vending machines • Optical instruments • Cameras • Machine tools • Robot arms • Home appliances • Vacuum systems • Food processing machines • Textile machines • Gardening equipment and machines Applications with special backing and cleats are specifically designed for special heavy duty conveying drives. 144 T2,5, T5, T10, AT5, AT10 As special the following profile can be manufactured on request MXL, L, H, HTD5M, DD double sided executions. Tension cords ELATECH® iSync™ timing belts are manufactured with high tensile strength steel cords as standard. All technical data shown in the catalogue are valid for standard cords. Belt with special cords have different mechanical and chemical properties. Special type of tension member such as stainless steel, HFE high flexibility or aramid fiber (Kevlar®) are available on request for special applications. Aramid (Kevlar®) tension cords are used where non magnetic drives are requested. Stainless steel used where high corrosion resistance is required. Fiberglass and polyester used where high flexibility and water restistance are required. SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti en .qxd 14/04/2010 11:02 Pagina 145 Standard belt sizes 120 145 160 177,5 180 200 210 230 245 265 277,5 285 290 305 317,5 330 342,5 380 420 480 500 540 600 650 780 915 950 1.20 1 40° 5 Number of teeth z length [mm] 33 37 40 43 44 45 49 50 51 52 54 165 185 200 215 220 225 245 250 255 260 270 length [mm] Number of teeth z length [mm] 170 172 180 188 198 200 215 220 223 228 240 243 263 270 276 288 850 860 900 940 990 1000 1075 1100 1115 1140 1200 1215 1315 1350 1380 1440 55 56 59 60 61 64 65 66 68 70 71 72 73 75 78 80 82 84 85 86 88 89 90 91 92 95 96 100 102 105 109 110 112 115 118 120 122 124 125 126 128 130 132 135 138 140 144 145 150 156 160 163 168 275 280 295 300 305 320 325 330 340 350 355 360 365 375 390 400 410 420 425 430 440 445 450 455 460 475 480 500 510 525 545 550 560 575 590 600 610 620 625 630 640 650 660 675 690 700 720 725 750 780 800 815 840 T10 40° 10 Number of teeth z length [mm] 26 32 35 37 40 41 44 45 50 53 55 56 60 61 63 65 66 260 320 350 370 400 410 440 450 500 530 550 560 600 610 630 650 660 Number of teeth z length [mm] 69 70 72 75 78 80 81 84 85 88 89 90 91 92 95 96 97 98 100 101 105 108 110 111 114 115 120 121 124 125 130 132 135 139 140 142 144 145 146 150 156 160 161 170 175 178 180 188 196 225 690 700 720 750 780 800 810 840 850 880 890 900 910 920 950 960 970 980 1000 1010 1050 1080 1100 1110 1140 1150 1200 1210 1240 1250 1300 1320 1350 1390 1400 1420 1440 1450 1460 1500 1560 1600 1610 1700 1750 1780 1800 1880 1960 2250 ELATECH® iSync™ 48 58 64 71 72 80 84 92 98 106 111 114 116 122 127 132 137 152 168 192 200 216 240 260 312 366 380 1.20 Number of teeth z 2 length [mm] 40° 10 5 2.50 Number of teeth z T5 1 1 5 1.20 0.70 2.5 T10 40° 40° 0.90 40° T5 2 T5 2.50 T2,5 Order example ELATECH iSync™ Timing Belt U ® 420 T5 / 16 145 SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti en .qxd AT5 AT10 50° Number of teeth z length [mm] 45 51 56 60 68 75 78 84 90 91 100 109 120 122 132 142 144 150 156 165 172 195 210 225 300 225 255 280 300 340 375 390 420 450 455 500 545 600 610 660 710 720 750 780 825 860 975 1050 1125 1500 2.50 10 1.20 5 2 1.50 50° Number of teeth z length [mm] 50 53 56 60 61 66 70 73 78 80 84 89 92 96 98 100 101 105 108 110 115 120 121 125 128 130 132 135 136 140 142 148 150 160 170 172 180 186 194 500 530 560 600 610 660 700 730 780 800 840 890 920 960 980 1000 1010 1050 1080 1100 1150 1200 1210 1250 1280 1300 1320 1350 1360 1400 1420 1480 1500 1600 1700 1720 1800 1860 1940 Order example ELATECH® iSync™ Timing Belt U 146 450 AT5 / 16 14/04/2010 11:02 Pagina 146 DT5 DT10 2.50 1.20 Number of teeth z length [mm] 60 70 80 82 90 92 96 100 103 110 118 120 124 130 140 150 160 163 170 172 180 188 206 220 228 278 300 350 400 410 450 460 480 500 515 550 590 600 620 650 700 750 800 815 850 860 900 940 1030 1100 1140 1390 Profile 2.50 10 1.20 5 11:02 Pagina 147 L XL 40° 40° 14/04/2010 Number of teeth z length [mm] 53 60 63 66 70 72 75 80 84 90 98 100 110 120 121 124 125 130 132 135 140 142 150 160 161 170 180 188 530 600 630 660 700 720 750 800 840 900 980 1000 1100 1200 1210 1240 1250 1300 1320 1350 1400 1420 1500 1600 1610 1700 1800 1880 Belt width DT5 [mm] 6 - 8 - 10 - 12 - 16 - 20 - 25 - 32 DT10 [mm] 10 - 12 - 16 - 20 - 25 - 32 - 50 Number of teeth z length [mm] length [Inches] 30 35 38 40 42 45 47 50 51 52 53 55 57 58 60 62 63 64 65 68 70 75 76 77 80 83 85 90 93 95 100 105 106 110 115 120 125 127 130 135 145 150 160 165 172 180 188 192 195 207 230 240 256 282 315 335 152,4 177,8 193 203,2 213,4 228,6 238,8 254 259,1 264,2 269,2 279,4 289,6 294,6 304,8 315 320 325,1 330,2 345,4 355,6 381 386,1 391,2 406,4 421,6 431,8 457,2 472,4 482,6 508,6 533,4 538,5 558,8 584,2 609,6 635 645,2 660,4 685,8 736,6 762 812,8 838,2 873,8 914,4 955 975,4 990,6 1051,6 1168,4 1219,2 1300,5 1432,6 1600,2 1701,8 6 7 7,6 8 8,4 9 9,4 10 10,2 10,4 10,6 11 11,4 11,6 12 12,4 12,6 12,8 13 13,6 14 15 15,2 15,4 16 16,6 17 18 18,6 19 20 21 21,2 22 23 24 25 25,4 26 27 29 30 32 33 34,4 36 37,6 38,4 39 41,4 46 48 51,2 56,4 63 67 Number of teeth z length [mm] length [Inches] 33 40 44 46 50 56 60 64 68 72 76 80 86 92 98 100 104 112 114 120 128 136 144 160 314,3 381 419,1 438,2 476,3 533,4 571,5 609,6 647,7 685,8 723,9 762 819,2 876,3 933,5 952,5 990 1066,8 1084,6 1143 1219,2 1295,4 1371,6 1524,1 12,4 15 16,5 17,3 18,8 21 22,5 24 25,5 27 28,5 30 32,3 34,5 36,8 37,5 39 42 42,7 45 48 51 54 60 Profile Belt width [mm] 6,4 - 7,9 9,5 - 12,7 [inch] 0,25 - 0,31 0,37 - 0,50 [mm] 12,7 - 19,0 25,4 - 38,1 [inch] 0,50 - 0,75 1,00 - 1,50 XL L 147 ELATECH® iSync™ SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti en .qxd SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti en .qxd 14/04/2010 11:02 Pagina 148 ELATECH® iSync™ high performance endless timing belt technical data T2,5 iSync™ 0.90 40° Belt characteristics • Truly endless polyurethane timing belt with steel tension cords according to DIN 7721 T1 • Metric pitch 2,5 mm • Ideal for drives where high belt flexibility is requested • Allows to use small diameter pulleys • Transmissible power up to 5 kW • Rpm up to 10.000 [1/min] 0.70 2.5 • Width tolerance: • Thickness tolerance: Belt width [mm] 4 6 8 10 12 16 25 32 Weight [g/m] 6 9 12 15 18 24 37 48 ±0,3 [mm] ±0,2 [mm] Other widths are available on request Tooth shear strength rpm [min-1] Mspez Pspez [Ncm/cm] [W/cm] rpm [min-1] Mspez Pspez [Ncm/cm] [W/cm] rpm [min-1] Mspez Pspez [Ncm/cm] [W/cm] 0 0,47 0,000 1200 0,29 0,361 3400 0,23 0,810 20 0,45 0,010 1300 0,28 0,385 3600 0,22 0,845 40 0,44 0,018 1400 0,28 0,408 3800 0,22 0,880 60 0,43 0,027 1440 0,28 0,417 4000 0,22 0,914 80 0,42 0,035 1500 0,27 0,431 4500 0,21 0,996 100 0,41 0,043 1600 0,27 0,454 5000 0,21 1,074 200 0,38 0,080 1700 0,27 0,476 5500 0,20 1,150 300 0,36 0,114 1800 0,26 0,498 6000 0,19 1,223 400 0,35 0,145 1900 0,26 0,519 6500 0,19 1,293 500 0,34 0,175 2000 0,26 0,541 7000 0,19 1,360 600 0,33 0,204 2200 0,25 0,582 7500 0,18 1,426 700 0,32 0,232 2400 0,25 0,622 8000 0,18 1,489 800 0,31 0,259 2600 0,24 0,662 8500 0,17 1,551 900 0,30 0,286 2800 0,24 0,700 9000 0,17 1,611 1000 0,30 0,311 3000 0,24 0,715 9500 0,17 1,668 1100 0,29 0,336 3200 0,23 0,738 10000 0,16 1,725 Flexibility Minimum number of teeth and minimum diameter Drive without reverse bending • Driver pulley zmin = 10 • Idler (flat) running on belt teeth dmin = 15 mm Drive with reverse bending and double sided belt • Driver pulley zmin = 18 • Idler (flat) running on belt back dmin = 15 mm 148 The total power “P” and the total torque “M” transmitted by the belt, are calculated with the following formulas: P [kW] = Pspez • ze • zk • b / 1000 M [Nm] = Mspez • ze • zk • b / 100 ( ) t ⋅ z g − zk Zk ⋅ arccos ⋅ 180 2 ⋅ π ⋅ A Ze = P M Pspez Mspez ze = power in kW = torque in Nm = specific power = specific torque = number of teeth in mesh of the small pulley = 12 = number of teeth of the small pulley = belt width in cm = centre distance [mm] = pitch zemax zk b A t SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti en .qxd 11:02 Pagina 149 iSync™ T5 1 40° Belt characteristic • Truly endless polyurethane timing belt with steel tension cords according to DIN 7721 T1 • Metric pitch 5 mm • Ideal for drives where high belt flexibility is requested • Allows to use small diameter pulleys • Rpm up to 10.000 [1/min] 1.20 T5 14/04/2010 5 • Width tolerance: • Thickness tolerance: Belt width [mm] 10 12 16 25 32 50 75 100 Weight [g/m] 24 28 38 60 77 120 180 240 Mspez Pspez [Ncm/cm] [W/cm] rpm [min-1] ±0,5 [mm] ±0,15 [mm] Other widths are available on request Tooth shear strength Mspez Pspez [Ncm/cm] [W/cm] rpm [min-1] Mspez Pspez [Ncm/cm] [W/cm] 0 2,523 0,000 1200 1,607 2,019 3400 1,248 4,444 20 2,458 0,051 1300 1,580 2,151 3600 1,229 4,632 40 2,403 0,101 1400 1,555 2,279 3800 1,209 4,812 60 2,354 0,148 1440 1,545 2,330 4000 1,191 4,988 80 2,312 0,194 1500 1,532 2,406 4500 1,149 5,414 100 2,276 0,238 1600 1,510 2,529 5000 1,111 5,818 200 2,135 0,447 1700 1,489 2,651 5500 1,078 6,206 300 2,032 0,638 1800 1,470 2,770 6000 1,046 6,571 400 1,951 0,817 1900 1,451 2,888 6500 1,017 6,924 500 1,884 0,987 2000 1,433 3,001 7000 0,991 7,262 600 1,829 1,149 2200 1,400 3,226 7500 0,966 7,588 700 1,781 1,306 2400 1,371 3,445 8000 0,943 7,897 800 1,738 1,456 2600 1,342 3,654 8500 0,920 8,191 900 1,701 1,603 2800 1,317 3,860 9000 0,900 8,480 1000 1,667 1,745 3000 1,306 3,940 9500 0,880 8,758 1100 1,635 1,884 3200 1,292 4,059 10000 0,862 9,027 The total power “P” and the total torque “M” transmitted by the belt, are calculated with the following formulas: P [kW] = Pspez • ze • zk • b / 1000 M [Nm] = Mspez • ze • zk • b / 100 ( ) t ⋅ z g − zk Zk ⋅ arccos ⋅ 180 2 ⋅ π ⋅ A Ze = P M Pspez Mspez ze = power in kW = torque in Nm = specific power = specific torque = number of teeth in mesh of the small pulley = 12 = number of teeth of the small pulley = belt width in cm = centre distance [mm] = pitch zemax zk b A t Flexibility Minimum number of teeth and minimum diameter Drive without reverse bending • Driver pulley zmin = 10 • Idler (flat) running on belt teeth dmin = 30 mm Drive with reverse bending and double sided belt • Driver pulley zmin = 15 • Idler (flat) running on belt back dmin = 30 mm 149 ELATECH® iSync™ rpm [min-1] SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti en .qxd 11:02 Pagina 150 iSync™ T10 40° Belt characteristics • Truly endless polyurethane timing belt with steel tension cords according to DIN 7721 T1 • Metric pitch 10 mm • Ideal for drives where high belt flexibility is requested • Allows to use small diameter pulleys • Rpm up to 10.000 [1/min] 2 T10 14/04/2010 2.50 10 • Width tolerance: • Thickness tolerance: Belt width [mm] 10 16 25 32 50 75 100 150 Weight [g/m] 50 77 120 155 240 365 480 725 ±0,5 [mm] ±0,2 [mm] Other widths are available on request Tooth shear strength rpm [min-1] Mspez Pspez [Ncm/cm] [W/cm] rpm [min-1] Mspez Pspez [Ncm/cm] [W/cm] rpm [min-1] Mspez Pspez [Ncm/cm] [W/cm] The total power “P” and the total torque “M” transmitted by the belt, are calculated with the following formulas: 0 8,244 0,000 1200 4,808 6,042 3400 3,460 12,318 20 8,009 0,168 1300 4,708 6,409 3600 3,385 12,761 40 7,805 0,327 1400 4,614 6,764 3800 3,312 13,179 60 7,627 0,479 1440 4,577 6,902 4000 3,245 13,592 M [Nm] = Mspez • ze • zk • b / 100 80 7,472 0,626 1500 4,526 7,109 4500 3,088 14,549 Ze = 100 7,339 0,768 1600 4,444 7,445 5000 2,946 15,424 200 6,804 1,425 1700 4,366 7,771 5500 2,817 16,224 300 6,411 2,014 1800 4,292 8,090 6000 2,701 16,969 400 6,105 2,557 1900 4,222 8,401 6500 2,593 17,646 500 5,857 3,066 2000 4,157 8,706 7000 2,492 18,269 P M Pspez Mspez ze 600 5,648 3,549 2200 4,033 9,291 7500 2,398 18,836 700 5,467 4,007 2400 3,920 9,851 8000 2,311 19,359 800 5,306 4,445 2600 3,815 10,386 8500 2,228 19,832 900 5,163 4,866 2800 3,718 10,901 9000 2,150 20,264 1000 5,034 5,271 3000 3,680 11,097 9500 2,077 20,661 = power in kW = torque in Nm = specific power = specific torque = number of teeth in mesh of the small pulley = 12 = number of teeth of the small pulley = belt width in cm = centre distance [mm] = pitch 1100 4,916 5,663 3200 3,626 11,389 10000 2,007 21,015 Flexibility Minimum number of teeth and minimum diameter Drive without reverse bending • Driver pulley zmin = 12 • Idler (flat) running on belt teeth dmin = 60 mm Drive with reverse bending and double sided belt • Driver pulley zmin = 20 • Idler (flat) running on belt back dmin = 60 mm 150 P [kW] = Pspez • ze • zk • b / 1000 zemax zk b A t ( ) t ⋅ z g − zk Zk ⋅ arccos ⋅ 180 2 ⋅ π ⋅ A SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti en .qxd 11:02 iSync™ AT5 50° 1.20 5 Pagina 151 Belt characteristics • Truly endless polyurethane timing belt with steel tension cords. Metric pitch 5 mm • Tooth profile and dimension are optimised to guarantee uniform load distribution and mini mum deformation under load • High resistance and low stretch steel cords to guarantee high stability and low elongation • Reduced polygonal effect with reduced drive vibration and noise • Rpm up to 10.000 [1/min] 1.50 AT5 14/04/2010 • Width tolerance: • Thickness tolerance: Belt width [mm] 6 10 16 25 32 50 75 100 Weight [g/m] 21 34 54 86 110 175 260 350 Mspez Pspez [Ncm/cm] [W/cm] rpm [min-1] ±0,5 [mm] ±0,15 [mm] Other widths are available on request Tooth shear strength Mspez Pspez [Ncm/cm] [W/cm] rpm [min-1] Mspez Pspez [Ncm/cm] [W/cm] The total power “P” and the total torque “M” transmitted by the belt, are calculated with the following formulas: 0 3,813 0,000 1200 2,668 3,352 3400 1,993 7,096 20 3,758 0,079 1300 2,620 3,566 3600 1,954 7,368 40 3,708 0,155 1400 2,574 3,773 3800 1,917 7,627 60 3,663 0,230 1440 2,557 3,855 4000 1,881 7,879 M [Nm] = Mspez • ze • zk • b / 100 80 3,623 0,304 1500 2,531 3,975 4500 1,799 8,479 Ze = 100 3,586 0,376 1600 2,491 4,173 5000 1,725 9,032 200 3,448 0,722 1700 2,452 4,365 5500 1,658 9,549 300 3,343 1,050 1800 2,416 4,554 6000 1,596 10,029 400 3,235 1,355 1900 2,381 4,737 6500 1,539 10,473 500 3,137 1,642 2000 2,348 4,918 7000 1,485 10,887 P M Pspez Mspez ze 600 3,050 1,916 2200 2,285 5,265 7500 1,436 11,278 700 2,972 2,178 2400 2,229 5,601 8000 1,389 11,635 800 2,900 2,430 2600 2,175 5,923 8500 1,346 11,980 900 2,834 2,671 2800 2,125 6,231 9000 1,304 12,289 1000 2,775 2,905 3000 2,106 6,352 9500 1,264 12,576 = power in kW = torque in Nm = specific power = specific torque = number of teeth in mesh of the small pulley = 12 = number of teeth of the small pulley = belt width in cm = centre distance [mm] = pitch 1100 2,719 3,132 3200 2,079 6,531 10000 1,228 12,854 P [kW] = Pspez • ze • zk • b / 1000 zemax zk b A t ( ) t ⋅ z g − zk Zk ⋅ arccos ⋅ 180 2 ⋅ π ⋅ A Flexibility Minimum number of teeth and minimum diameter Drive without reverse bending • Driver pulley zmin = 15 • Idler (flat) running on belt teeth dmin = 30 mm Drive with reverse bending and double sided belt • Driver pulley zmin = 25 • Idler (flat) running on belt back dmin = 60 mm 151 ELATECH® iSync™ rpm [min-1] SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti en .qxd AT10 14/04/2010 11:02 iSync™ AT10 T10 Belt characteristics • Truly endless polyurethane timing belt with steel tension cords. Metric pitch 10 mm • Tooth profile and dimension are optimised to guarantee uniform load distribution and mini mum deformation under load • High resistance and low stretch steel cords to guarantee high stability and low elongation • Reduced polygonal effect with reduced drive vibration and noise • Rpm up to 10.000 [1/min] 2 50° 2.50 10 Pagina 152 • Width tolerance: • Thickness tolerance: Belt width [mm] 16 25 32 50 75 100 150 Weight [g/m] 101 158 200 316 475 630 950 ±0,5 [mm] ±0,2 [mm] Other widths are available on request Tooth shear strength rpm [min-1] Mspez Pspez [Ncm/cm] [W/cm] rpm [min-1] Mspez Pspez [Ncm/cm] [W/cm] rpm [min-1] Mspez Pspez [Ncm/cm] [W/cm] The total power “P” and the total torque “M” transmitted by the belt, are calculated with the following formulas: 0 15,903 0,000 1200 10,174 12,785 3400 7,019 24,989 20 15,670 0,328 1300 9,945 13,538 3600 6,838 25,778 40 15,452 0,647 1400 9,731 14,266 3800 6,664 26,516 60 15,246 0,958 1440 9,649 14,550 4000 6,500 27,225 M [Nm] = Mspez • ze • zk • b / 100 80 15,053 1,261 1500 9,529 14,968 4500 6,120 28,837 Ze = 100 14,870 1,557 1600 9,340 15,649 5000 5,777 30,248 200 14,103 2,954 1700 9,160 16,305 5500 5,464 31,470 300 13,483 4,236 1800 8,990 16,944 6000 5,179 32,536 400 12,927 5,414 1900 8,828 17,563 6500 4,916 33,460 500 12,439 6,513 2000 8,672 18,162 7000 4,670 34,232 P M Pspez Mspez ze 600 12,008 7,545 2200 8,380 19,305 7500 4,441 34,878 700 11,626 8,522 2400 8,113 20,390 8000 4,227 35,409 800 11,282 9,451 2600 7,866 21,414 8500 4,023 35,808 900 10,969 10,337 2800 7,632 22,378 9000 3,832 36,113 1000 10,683 11,186 3000 7,544 22,751 9500 3,651 36,322 = power in kW = torque in Nm = specific power = specific torque = number of teeth in mesh of the small pulley = 12 = number of teeth of the small pulley = belt width in cm = centre distance [mm] = pitch 1100 10,418 12,000 3200 7,416 23,296 10000 3,479 36,429 Flexibility Minimum number of teeth and minimum diameter Drive without reverse bending • Driver pulley zmin = 15 • Idler (flat) running on belt teeth dmin = 50 mm Drive with reverse bending and double sided belt • Driver pulley zmin = 25 • Idler (flat) running on belt back dmin = 120 mm 152 P [kW] = Pspez • ze • zk • b / 1000 zemax zk b A t ( ) t ⋅ z g − zk Zk ⋅ arccos ⋅ 180 2 ⋅ π ⋅ A = 0,1 [Kg] ⋅ 1 + 2 = ⋅ 1 + = 0,33 kg 9,81 2 153 2 100 2 da d 2 0,61 SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti en .qxd 14/04/2010 11:02 Pagina mFORMULE 28 2 mc + mR + msred = 50 +Peso 0,48 += 00,,1 33[Kg = 50 Kg S UOVE ]N,81 ⋅ 1 + 2PAG = 0 , 33 kg =. 41 ⋅ 1 + 2 9,81 2 DELLA 2 t 100 2 CALCOLO TOTALE 1 z g − zk ⋅ t dMASSA 2000 ⋅ M a L R ≈ ⋅ z g + z k + 2A + ⋅ Ft = 2 π 4A m = mc + mR + msred d=p 50 + 0,48 + 0,33 = 50,81 Kg OLO DELLA FORZA PERIFERICA FU L 2+=01600 2,= mR = 3,2 ⋅ 0,15 = 0,48 Kg R = L m = mc + mR + mLsred =1 +50 ,48 +⋅ 0 333200 = 50,mm 81 Kg 1116,2 ⋅ 101,86 m ⋅ a + Fr = 50,25 ⋅ 20 + 100 = 1128,4 [N] M= = 56,85 Nm β t β CALCOLO DELLA FORZA PERIFERICA FU 2000 LR = 2A ⋅ sin ⋅ + ⋅ z g + zk + 1 − ⋅ z g − z k LR = 2 ⋅ A + π ⋅ d w = 2 ⋅ A + z ⋅ t 2 180 CALCOLO DELLA FORZA2 PERIFERICA FU 2 2 22 1116,2 ⋅ 101,86 d − d ⋅ π ⋅ ρ ⋅ B 100 − 28 ⋅ π ⋅ 2 , 8 ⋅ 30 a Ft ⋅ dp 1105 ⋅ 101,86 Ft = m ⋅ a + Fr = 50,25 ⋅ 20 + 100 = 1128,4 [N] M= = 56,85 Nm = = 0 , 61 Kg 6 6 = = 56,28 [Nm] 2000 4 ⋅ 10 4 ⋅ 10 1116,2 ⋅ 101,86 2000 2000 Ft = m ⋅ a + Fr = 50,25 ⋅ 20 + 100 = 1128,4 [N] M= = 56,85 Nm 2000 t ⋅ z − zk t ⋅ z g − zk F ⋅ d Z k β 1105 ⋅ 101,86 g ze = ⋅ zk β = 2 ⋅ arccos ⋅ M = t p180 = ⋅ arccos ⋅ 2=⋅ π56⋅ A ,28 [Nm] 2 2 2⋅ π ⋅ A 360 8 mS d 0,61 28 2000 2000 Ft ⋅ dp 1105Peso ⋅ 101,=860,1 [Kg] 1 ⋅ + = ⋅ 1 + = 0 , 33 kg 2 2 [ ] M= = = 56 , 28 Nm 9,81 2 da 2 100 2000 2000 F ⋅d 9100⋅ P ⋅ 103 π ⋅n M= t p ω = J = 98,2 ⋅ 10−15 ⋅ B ⋅ ρ ⋅ de4 − d4 PAG 48 1000 ⋅ 20 ⋅ 1,4 J ⋅ ∆n n ⋅ dp 2000 30 = b = = 4 , 86 cm = 48,6 mm M a PAG 48 9,55 ⋅ 12Kg ⋅ 8,572 56,81 m = mc + mR + msred = 50 + 0⋅ ,t48 F ⋅d a + 0,33 = 50 19100⋅ P ⋅ 103 π ⋅n Ft = M= t p ω = J = 98,2 ⋅ 10−15 ⋅ B ⋅ ρ ⋅ de4 − d4 3 n d ⋅ 2000 30 Ft ⋅ dp p 19100⋅ P ⋅ 10 π ⋅n 49 −15 4 4 Ft = M= J = 98,2 ⋅ 10 ⋅ B ⋅ ρ ⋅ de − d ω= t n ⋅ dp CALCOLO2000 DELLA FORZA30 PERIFERICA FU 2 α ⋅ 350 t 1 z g − zk ⋅ t 2000 ⋅ M 100 PAG. 49 LR ,= b= = 5,53 cm = 55 mm ⋅ z g + z k + 2A + ⋅ Ft = 1116 ,2 ⋅ 101 862 ⋅ 1800 + 56 ⋅ 8 = 4048 mm ⋅ 12 ⋅=91128 ,422 ,4 [N] dp ⋅ 2056 2 π 4A + 100 M= = 56,85 Nm Ft = m ⋅ a + Fr = 50,25 PAG. 49 2000 2 t 1 z g − zk ⋅ t 2000 ⋅ M L R ≈ ⋅ z g + z k + 2A + ⋅ Ft = 2 F d ⋅ dp 2 π 4 A 1105 101 , 86 ⋅ 4048 − ⋅ z z t g t p 2 ] 1 2000 ⋅ M k β t d2 ,28 [Nm t ⋅ βz g + zk M + 2=A + L=⋅ = 2 ⋅ A + π d⋅ d1 ==56 Ft = − 2A ⋅ sin ⋅ + ⋅ z g + zk L+R ≈1 2 2 ⋅ A + z ⋅ t ⋅ z g − zk2000 2000 8 R w β 3 4A dp α π 2 2 180 β t β da dw PAG 48 da LR = 2A ⋅ sin ⋅ + ⋅ z g + zk + 1 − ⋅ z g − z k LR = 2 ⋅ A + π ⋅ d w = 2 ⋅ A + z ⋅ t 2 Zk 2 2 180 F = F = 2337 , 5 [ N ] Z B, L R Zg t ⋅ z g − zk β t zZg k+ zk + 13 − tβ⋅v z g⋅3−z gzt k− zk L = 2 ⋅ A + π ⋅ d = 2 ⋅ A + zd⋅ wt β w P ⋅ 10 ⋅ zk β = 2 ⋅ arccosL⋅ R = 2A ⋅ sin ⋅2 + 2 ⋅19100 ⋅⋅arccos ⋅ 180 Ft ⋅ dp π ⋅ nR −15 4 4 M = ⋅ A ω = J = 98,2 ⋅ 10 ⋅ B ⋅ ρ ⋅ de − d 360 2 ⋅ π ⋅ A Ft = 180 2 ⋅ π n ⋅ dp 2000 30 t ⋅ z g − zk t ⋅ z g − zk Zk β ze = ⋅ zk β = 2 ⋅ arccos ⋅ ⋅ arccos ⋅ A 360 180 2 ⋅ π ⋅ A t ⋅ z − z t ⋅ z − z Zk 2 ⋅ π ⋅ A g k g k ⋅ 1β,4 ⋅ z 1000z⋅ 20 J ⋅ ∆n . 49 PAG = ⋅ arccos ⋅ β = 2 ⋅ arccos ⋅ e k b= 360 = 4,86 cm = 48,6 mm 180 2⋅ π ⋅ A 9,55 ⋅ ta 56 ⋅ 12 ⋅ 8,572 2 ⋅ π ⋅ A Ft ⋅ dp ⋅ n 2 2000 ⋅ M 9550 ⋅ P Ft = M = 1000 ⋅ 20 ⋅ 1,4 P = 1 z g − z3 k ⋅ t t 2000 ⋅ M Definitions LR ≈ ⋅ z g + zk + 2A + 19100 ⋅ ⋅ 10 F dtp= Ma = J ⋅ ∆n n b= = 4,86 cm = 48,6 mm dp 9,55 ⋅ ta 2 π 4A 56 ⋅ 12 ⋅ 8,572 1000 ⋅ 20 ⋅ 1,4 J ⋅ ∆n b= = 4,86 cm = 48,6 mm M = 100 ⋅ 350 b a (cm) Belt width FU (N) Peripheral force ⋅ ta 55 ⋅ 12 ⋅ 8 LR = 256 ⋅ 1800 +,572 56 ⋅ 8 = 4048 mm = 5,53 cm = 559,mm (mm) Belt lenght LR 6 ⋅ 12 ⋅ 9,422 M (Nm) Torque β t β = 2A ⋅ sinof ⋅ teeth + ⋅ zofg +the + 1− LRNumber zk belt 2 ⋅ A2 + π ⋅100 d(kW) 2 ⋅ A + zPower ⋅t zR ⋅ z g − z k LR = P w = 2 ⋅ 350 2 ⋅ 180 sa v ⋅ 100 5,53 cm = 55 mm v 2 2 S = at ⋅ tba =⋅ 1000(s) LR = 2 ⋅ 1800 + 56 ⋅ 8 = 4048 mm ta = = =⋅ 9,422 =Acceleration time a B (mm) Pulley width ab 56 12 ⋅ 048 100 ⋅ 350 a a ⋅ 1000 2 2⋅a 2 L = 2 ⋅ 1800 + 56 ⋅ 8 = 4048 mm b = = 5 , 53 cm = 55 mm (s) Deceleration time tav A (mm) Center distance R 8 56 ⋅ 12 ⋅ 9,422 3 (mm) Effective center distance t ⋅ z − z Aeff v (m/s) Peripheral speed t ⋅ z − z Z β k g k k 4048g = ⋅ zk β = 2 ⋅ arccos ⋅ ⋅ arccos d (mm) z e Pulley ze ⋅ 2 ⋅ π ⋅-A 2 N. of teeth in mesh 360 bore diameter 180 2 ⋅ π ⋅ A 8 3 2 1116,2 da 4048(mm) 2 Pulley outside diameter Number of teeth of the small pulley zk Ft = 2337,5 [ N] b = = 1 , 78 cm ≈ 18 mm 3 8 (mm) 3 Small pulley outside 52,21 ⋅ 12 diameter d ( ( ) ) ( ( ( ) ) ) ( ( ( ) ) Drive calculation ) ( ( ( ) ) ( ( () ) ( ) ( ) ) ( ( ( ( ) ) ) ( ) ( ( ( ) ( ) ( ( ) ) ) ) ( ( ) ( ) ) Number of teeth of the large pulley zg ak 2 ⋅ 20 ⋅ 1,4 1000 J ⋅ ∆pulley n (mm) M Large outside diameter dag i ratio ( n1 : n2 ) N] b= = 4,86 cm = 48,6 mm Fv = Ft = 2337,5 [ Drive a = 3 9,55 ⋅pitch ta 56 ⋅ 12 ⋅ 8,5723 Pulley diameter dw 2 (mm) 3 ⋅ M weight 19100 ⋅ P ⋅ 10 F ⋅ d ⋅ρn Specific (kg/dm ) 2000 Fv = Ft = 2337,5 [ N] F = Fu = P= u w 3 Small pulley pitch circle dwk 3 (mm) J (kgm2u) Moment of inertia n ⋅ ddiameter dw 19100 ⋅ 10 w t (mm) Pitch (mm) Large pulley pitch circle diameter dwg 2000 ⋅ M 9550 ⋅ P -1) 100 ⋅ 350 M = Rpm (N) FF t = Wsta LR = 2 ⋅ 1800n+ 56 ⋅ 8 = (min 4048 mm b =Static Shafts load = 5,53n cm = 55 mm dp J ⋅ ∆n ntreibend n -1) 56 ⋅ 12 ⋅ 9 , 422 (min Rpm of driver pulley (N) Pretension force FTV 1 Mabper = belt side i= Ft ⋅ d-1p ⋅ n 2000 ⋅ M 9550 ⋅ P ngetrieben 9,55 ⋅ t ab ω P= (s ) 3 Angular Fspeed M= (N) Allowable tensile load FTzul t = d n 19100 ⋅ 10 Ft ⋅ dp ⋅ n p 9550 ⋅ P β (°) Wrap angle 4048 F =22000 ⋅ M M= P= Ft ⋅ dp ⋅ n 9100 ⋅ 10 3 t dp 19100 ⋅ 10 3 a ⋅ t 2a ⋅ 1000 8 v 2 ⋅ 1003 S = = a Calculation formula 2 2⋅a J = 98,2 ⋅ 10 −15 ⋅ B ⋅ ρ ⋅ d4a − d4 2 Fv = Ft = 2337,5 [ N]Peripheral force Power 3 2 a t 2a ⋅ 1000 2 ⋅ sa v ⋅ c 0 = v ⋅ 100 t a =M ⋅ n= S ab = P ⋅ 1000 3 100 19100 ⋅ P ⋅ 10 a ⋅ 1000 2⋅a 16,2 = a Fuz=k ⋅ z e2⋅ Pspez n = 1,78 cm ≈ 18Pmm 9550 n ⋅ dw 21⋅ 12 v = a ) 2 ⋅ sa a ⋅ 1000 ( ) n 19100 ⋅ v dw 2 ⋅ sa v ta = = Torque a a ⋅ 1000 F ⋅d M= U W 2000 n= Sa = a ⋅ t 2a ⋅ 1000 v 2 ⋅ 100 = 2 2⋅a 1116,2 = 1,78 cm ≈ 18 mm 52 ,21⋅ 12 9550 ⋅P M= n b= 100 ⋅ P ⋅ 10 3 n ⋅ dw 1116 F ⋅ dp ⋅ n2000 ⋅ M 2000 ⋅ M2000 ⋅ M F ⋅ d,2 ⋅ n b Fu mm = P== u w =31,78P =cm ≈t 18 Fu = F = 52 ,21⋅ 12 19100 ⋅ 10 d wt dp 19100 ⋅ 10 3 dw 19100 ⋅ P ⋅ 10 3 F ⋅d ⋅n P= u w 3 FFUu ⋅=C1 n ⋅ dw 19100 ⋅ 10 b= 2000 ⋅ M 19100 ⋅ P ⋅ 10 F ⋅ d ⋅ n J ⋅ ∆n Fu F = = 2000 ⋅ 350 =F4908 Fu i== ntreibend P = u z =w 1503 ⋅ π = Uspez ⋅ z e ,83 N U n ⋅speed dw dw142,6 19100 ⋅ 10 Acceleration torque Angular pheripheral speed ngetrieben ,55 ⋅ t ab 8 a ⋅ t 2a ⋅ 1000 v 2 ⋅ 100 2 ⋅ sa v = π⋅n M t⋅ an = a F=U ⋅ va ⋅ 1000 J ⋅ ∆n n F ⋅d P ⋅ 9550 d ⋅ n S a =n ⋅ z ⋅ t 2 2 ⋅ a Mab = ω= i = treibend = P= = M= U W = v= W 30 n 9 , 55 ⋅ t 2000 19100 60000 n getrieben ab J ⋅ ∆n 9550 1000 ntreibend n1 n Mab = i= i = driver =1 n 9 , 55 t ⋅ 19100 ⋅ v n n getrieben ab 2 driven n= 2 ⋅ 10 −15 ⋅ B ⋅ ρ ⋅ d4a − d4 dw 1116,2 Moment of inertia b = 2000 ⋅ M= 1,78 cmrpm ≈P18 mm 2000 ⋅ M P ⋅ 1000 ⋅ 9550 19100 ⋅ v Ft =52,21⋅ 12 FU = M= = n= J = 98,2 ⋅ 10 −15 ⋅ B ⋅ ρ ⋅ d4a − d4 000 ⋅ c 0 dp d W100 v n dw 19100 ⋅ v F = m ⋅ a + m g ⋅ sin α + m ⋅ g ⋅ µ 19100 ⋅ v u J = 98,2 ⋅ 10 −15 ⋅ B ⋅ ρ ⋅ d4a − d4 n =n = b z e ⋅ Pspez d d F ⋅C w w 2000P⋅ ⋅M 19100 ⋅ P ⋅ 10 3 F b⋅ d= ⋅ nU 1 1000 ⋅ c 0 Fu = b = Fu = P = u wFUspez 100 3 ⋅ ze n ⋅ d d 19100 ⋅ 10 z w w k ⋅ z e ⋅ Pspez P ⋅ 1000 ⋅ c 0 b= 100 000 ⋅ Mπ ⋅ n spez 2 ⋅ s ab v 19100z⋅kv⋅ z e ⋅ P 60000 ⋅v t ab = = n= = dωw= 30 a b ⋅ 1000 dW z⋅t J ⋅ ∆n a b ntreibend veloce giri / min Mab = i = albero RT = 2000 ⋅ M n 9,55 ⋅ t ab giri / mingetrieben albero lento Fu = dw 2000 ⋅ M ⋅π 2000 ⋅ 350 3 ( Fu = 2000 ⋅ M dw ) ( ( ) ) 153 ELATECH® iSync™ ( ) ) v s CALCOLO DELLA FORZA PERIFERICA FU 2000 2000 = =Pagina s ab + s c +154 s av t t cz= − z c s c = v ⋅ t c ⋅ 1000 t av = SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti FU = m ⋅ ab + m gn+ m ⋅ g ⋅ µ zk s ges19100 2 ⋅ s ab v 60000 ⋅ v g v ⋅ 1000 t ⋅ πz⋅gn−11:03 Z⋅kv β en .qxd 14/04/2010 1000 ⋅ 20 ⋅ 1,F4U ⋅ dw J ⋅⋅ ∆ k a av ⋅ v 2 120 = M 2 t = = n= = z e =cm =⋅ z48 ⋅ arccos ⋅ 1116ab β = 2 ⋅ arccos ⋅ω = bd = = 4,86 = , 2 ⋅ 101 , 86 k ,6 mm a mS d FU ==Fm M m ⋅ abmH++9m ⋅ g + Um ⋅ g ⋅ µ 56 ⋅ 12 ⋅ 8,572 z ⋅tM2=⋅ π ⋅ A a b 30 F2000 [ ⋅220 100= 1128d,4W180 N] =a56 85 Nm 360 Ft = m ⋅ a + Fr = 50,25 ⋅ π+⋅ A ,F55 ab + F R ⋅t b ⋅,1000 U ⋅ dw ⋅ 1 + 2 m Sred = Ured = a ⋅ 1 + 2 120 = M PAG 48 2000 2 d a F = F + F + F 2 du Ft ⋅ dp ⋅ n 2000 2000 ⋅ M 9550 ⋅ P U ab H R F = M= 2 P= t 3 3 n d 19100 ⋅ 10 m F d ⋅ d m 19100 F ⋅ d 1105 ⋅ 101 π ⋅ np ,86 ⋅ P 10m = m + m S mUred = U ⋅ 1 Ft =⋅ 20 ⋅ 1,4 = 56,28 Mc=] t R p+ mω ,2Sred ⋅ 10−=15 ⋅ B ⋅ ⋅ρ⋅1 +de4 − 2d4 = + mUredJ = 98m [Nm J ⋅ ∆nM = t p = 1000 Sred 2 2 = b = = = 4 , 86 cm 48 , 6 mm M n d ⋅ 2000 30 d a a ⋅ t 2 ⋅ 1000 2000 2000 a ⋅ tp2 ⋅ 1000 a LR 19100⋅ P100 ⋅ 10⋅3350 = 5,53 cm =F55 v 2 ⋅ 1000 v 2 ⋅ 10 9,R55 ⋅ 12 ⋅ 8,=572 TV ⋅ mm TV ⋅ LR b ab v av L =l⋅2=ta⋅F 1800 + 56 ⋅56 8= 4048 mm ∆ l = ∆ Fu = b = t = t + t + t s = s = = ges ab c av ab av 2F⋅ C⋅ LR n ⋅56 FCspez ⋅ LR F ⋅d 19100 ⋅dPp⋅⋅12 103⋅ 9,422 2 2 ⋅ ab 2 2⋅a ∆l = TV spez ∆l =PAGTV48 120 M= U w Fu = 2000 2 ⋅ C C n ⋅ dp PAG. 49 spez spez a ⋅ t 2 ⋅ 1000 v 2 ⋅ 100 2⋅s v 2 ⋅ sav v =t ⋅ dp FU =a m = ⋅ µ2 a =4 F ⋅ d 4048 mca⋅ g 19100⋅ P ⋅ 103 st a = = F π ⋅⋅nab +t m=⋅ g +sS 2 −15 100 ⋅ 350 w t av = = s + s + s =−vdM ⋅ t c=⋅21000 a a ⋅ 1000 2⋅ sdc e4mm ⋅Ua2000 4 ⋅ C M = mm = 98 , 2 ⋅ 10 ⋅ B ⋅ ρ ab c ω =L av = 2 ⋅ J c LR 120 1800 + 56 ⋅ 8 = 4048 b = spez Ft = n= = 5,19100 53 cm⋅ 5=ges 55 − ⋅ z z t t 1 ⋅ M R g k av a v ⋅ 1000 ,56 FU =2F C= 8⋅ Cspez Cmin = 56 ⋅ 12 ⋅ 9,422 n ⋅ dp L ≈ = ⋅937 3LR = L1 + L 2 30 z2000 Aab ++ FH + ⋅ FR v ⋅ 1000 Ft 2000 = R g + zk + 4 ⋅L CRspez 101,86 L1L⋅RL 2 19100 ⋅5 2 dp π 4A n = = 937 , 56 C = ⋅ C L = L + L C = R 1 2 min F ⋅L F ⋅ L spez 101,86 LR ∆l = TV R ∆l = L1TV⋅ L 2 R 2 ⋅ Cspez Cspez2 4048 PAG. 49 2 Fv = Ft = 2337,5 [ N] mS d2 m d2 1116,2 3 = ⋅ + mUred = U ⋅ 1 + 2 1 m = m + m + m + m 3 mβ b = = 1 , 78 cm ≈ 18 mm β t Sred c R Sred Ured 2 ⋅ L 8 3 F F ⋅ L 19100 ⋅ P ⋅ 10 2 TV R TV R 2 2 LR =52 2,A21⋅ sin ⋅ 12⋅ + ⋅ z + z + 1 − F ⋅ z g − z∆k l= d aLR = 2 ⋅ A + π ⋅ d w = ∆l2=⋅ A + z ⋅ t du t 1 2 zFgu 2−= zk g⋅ tn ⋅ dk 2000⋅ M2 ⋅ C ∆S = U 180 Cspez L R ≈ ⋅ z g + z k + 2A + ⋅ Ft = spez 4 ⋅ CspezFCU p 19100 ⋅ 5 dp 2 π 4A n= = 937,56 + L2 Cmin = ∆ S = 2 C Calculate teeth3in mesh 101,for 86 drive calculation are: F = F = 2337,5 [ N] LR necessary 2000 ⋅ M F ⋅d ⋅n The data v t z 2 19100 n ⋅dP ⋅ 10 1128,4 P t=⋅ z u − zw 3 t 4⋅ ⋅zCg − zk Lb = 2a + π ⋅ d3p = 2 ⋅ a + z ⋅ t b= = 1,80 cm = 18 mm i =Fu = =β 2n=⋅ dw 2 Z k Fu = d g k10 19100 ⋅ L 19100⋅ 5 F = m ⋅ a + m ⋅ g + m ⋅ g ⋅ µ spez w 52 ,21⋅,12 R n21 ⋅ zkdb ww21 β = z ez = arccos 1128 4 w U 1U 2 n= C2=⋅ β ⋅ ⋅C spez2 ⋅π ⋅ ALM = =LF +⋅Ld2w180 ⋅ arccos Cmin ⋅ = β t 120 R 1 = 2 a + π ⋅ d = 2 ⋅ a + z ⋅ t = 1t ,⋅80 i = = = 360 2t ⋅Lπ ⋅ A • bPower to be=F transmitted PL2000 [kW] dpcm ⋅ n = 18 mm b p ⋅ M 9550 ⋅ P 101,86 L ⋅ L = ⋅ ⋅ + ⋅ + + − L 2 A sin z z 1 z z ⋅ − L = 2 ⋅ A + π ⋅ d = 2 ⋅ A + z ⋅ 2000 F1U = F +F R 1 2 g 52,21P⋅ 12 ngab dk Hw1+FR R k R w Ft = M = 2 z2 = 1 -1 180 • Driver rpm⋅ k19100 ⋅ 10 3 100 ⋅ M ⋅ c 0n1 dp [min ] Ft ⋅with kn β [°] = wrap angle P ⋅ 1000 Ftzul ≥ Ft + k ⋅ Ft max b = = b= J ⋅ ∆n n • bMotor [Nm] FF ⋅ kzM = J ⋅ ∆n i =⋅ 1,4treibend ⋅ zstarting ⋅ c 0Mab z100 Pz1⋅ 1000 k torque e ⋅ P⋅sp k⋅ z e ⋅ M spez 1000 ⋅ 20 Ftzul ≥ Ft + k ⋅ Ft max b = tspt Me ab= 9 b = b = ngetrieben ,55 ⋅ t ab • Required center distance A [mm] = b ,6 mm Ftsp ⋅ zea 9,55 ⋅ t t ⋅ zF3 − z56⋅ 12 ⋅ 8,572Z = 4,86 cm =t ⋅ 48 z1 ⋅ ze ⋅ Psp z k⋅ z e ⋅ M spez z g − zk Ft ⋅ dp ⋅ n β gU k 2000 ⋅LP ⋅ k 9550 a⋅ M F ⋅ F ⋅L 19100 ⋅ P ⋅ 10 z n d •FMaximum driver pulley diameter d [mm] z = ⋅ z ⋅ arccos β = 2 ⋅ arccos ⋅ ∆ = TV R w1 FFt == M∆=l = P= e 3 k S ∆l= TV R Lb = 2a + π ⋅ dAp ==22⋅ ⋅FaV +⋅ cos z⋅tβ m i = 2 = 2 = w2 u 360 180 2 ⋅nC dp n ⋅ d 2 ⋅Cπ ⋅ A 19100 ⋅ 10 2 ⋅ π ⋅ A C 2 2 z n d spez spez p 1 w1 FA = 2 ⋅ FV ⋅ cosvβ a ⋅ t ⋅ 1000 v ⋅ 100 2 ⋅ s1a t a =⋅ cos=⋅ β Sa = ⋅ v a = FWsta = 2factors ⋅F ⋅ d n 19100 Safety Tv a p 19100 ⋅ v a ⋅ 1000 2 z 1128,4 −15 4 4 n= v2=⋅ a = J = 98 ,2 ⋅ 10 Lb = 2a + π ⋅ dp = 2 ⋅ a + z ⋅ t b =⋅ B ⋅ ρ ⋅ da − d= 1,80 cm = 18n mm i= 00 ⋅ M ⋅ c 0 100 ⋅ 350 F≥Wsta ⋅iβ=Ft1⋅)k ⋅ nn Determine d for dw ⋅ v working load.19100 19100 Tv ⋅(cos p⋅ ∆ belt width ⋅ ⋅ 1000 20 1 , 4 J d FtzulBelt Ft selection +=k2⋅ ⋅FFt max b 52 , 21 ⋅ 12 z L = 2 ⋅ 1800 + 56 ⋅ 8 = 4048 mm b = = 5 , 53 cm = 55 mm is=made according ton a= constant For w R 4 ⋅ Cspez b =⋅ 12L⋅ 9R,422 = 4,86 cm = 48,6 mm = Mva = 19100⋅ 5 ⋅ ze z e ⋅ M spez 56 FWsta up = 2torque ⋅ FTv ( forand iF=tsp1 19100 n= = 937,56 C =56 ⋅ 12 ⋅⋅8C,572 L R = L1 + L 2 Cmin = w 9 ,55 start in) case of peak loadsdand vibrations must be⋅ ta spez 2 2 101,86 L1 ⋅ L L a 2 ⋅ sa v 100 ⋅ M ⋅ c 0 R F ⋅k considered a safety factor c1. ⋅ 1000 ⋅ kv ⋅ 100 ⋅ c2⋅ t a ⋅P1000 P ⋅ 1000 ta = = 1200 ⋅ 3200 Ftzul ≥ Ft + k ⋅ Ft max b = t b= b = S a = b0= 2 100= 2 ⋅ a 1116 , 2 a a ⋅ 1000 ∆l = = 2,02 mm cm ≅ 0≈,63 F ⋅ z ⋅ M z ⋅ z ⋅ P b= = 1,78 18 mm z ⋅ ⋅ z z P 4048 tsp ⋅ ze e spez e sp k k e spez 2 1 2 ⋅ 952000 1200 ⋅ 3200 52,21=⋅ 12 ∆l = 2,02 mm load ≅ 0,63 Transmission with steady c = 1,0 8 1 3 100 ⋅ 350 dp ⋅ n 2 ⋅ 952000 19100 ⋅ v b= = 5,F53 cm F=A 55 n= v= = 2 mm ⋅ FV ⋅ cos β LR = 2 ⋅ 1800 + 56 ⋅ 8 = 4048 mm U 56 ⋅ 12 ⋅ 9 , 422 ∆ = 19100 dw S PAG.peak NUOVE FORMULE Verifiy allowable tensile load Transmission with C 2000 ⋅ M 19100 ⋅ P41 ⋅ 10 3 or fluctuatingFloads: ⋅ d ⋅ n 2 2000 ⋅ M CALCOLO MASSA TOTALE FuDELLA = P = u w b 3= 1116,2 = 1F,78 u = cm ≈ 18 FWsta ⋅ cos ⋅ β dp ⋅ n UOVE FORMULE PAG. 41 N 19100 ⋅ v Fvmm Ft = 2337 ,5tensile [= N]2 ⋅ FTvload F=u =allowable n ⋅ dw dw The 19100 ⋅ 10 52,21⋅ 12 of the belt must be n =highter than the v= 3 CALCOLO DELLA MASSA TOTALE d z n d 1128 , 4 w 4048 Light c1 = 1,4 for ) F = 2 ⋅ F i = 1 19100 d w ⋅a + z⋅t Wsta Lb = 2a + π ⋅ dp = 2 b2 = corrected = 1 ,80 cm Tv= (18force. mm i = 2 = 2 = w2 total peripheral 0,63 52 , 21 ⋅ 12 z n1 d w1 LR = L1 + L 2 = 1600 ⋅ 2 = 3200 mm 8 = 3 , 2 ⋅ 0 , 15 = 0 , 48 Kg Medium cm = 1,7 3 1 1 R 3 2000 ⋅ M 19100 ⋅ P ⋅ 10 F ⋅ d ⋅ n J ⋅ ∆n⋅ 2 = 3200 mm n LR = L1 + L 2 = 1600 m = 3 , 2 ⋅ 0 , 15 = 0 , 48 Kg Heavy c = 2,0 u w R F = ⋅ 350 c Fu = = 150 FPTzul > c⋅ π0 · F3U 1200 with Mab = i = 1 treibend 2000 ⋅= 4908 Ft ⋅ k P ⋅ 1000 ⋅ k∆l = 0 ≅ 0,,83 F⋅U3200 = = u=100 = n ⋅ dw ⋅ 10 ngetrieben 9,55 ⋅ t ab 2,02⋅dMwmm 63 NF ≥ F + k ⋅ F bz = = 19100 b 2 tzul t t max b = ,6 8 2 ⋅ 952000 z142 2 2 2 2 Fv = Ft = 2337,5 [ Nz]1 ⋅ ze ⋅ Psp Ftsp ⋅ ze ⋅ z e ⋅ M spez − d ⋅ π ⋅ ρ ⋅ B 100 − 28 ⋅ π ⋅ 2 , 8 ⋅ 30 k For dspeed up driver factor c must be considered: a 2 3 = = 0,61 Kg 6 Ft ⋅ dp ⋅ n 2000 ⋅ M 9550 ⋅ P 4 ⋅210⋅ 6π ⋅ 2,8 ⋅ 30 d2a − 4d⋅210⋅ π ⋅ ρ ⋅ B 100 2 − 28 Ft = M= P= 3 = = 0 , 61 Kg Calculate shaft load J ⋅ ∆n n 6 6 F = 2 ⋅ F ⋅ cos β d n 19100 ⋅ 10 treibend V NUOVE FORMULE nA1i = n PAG. 41 p i = from4 ⋅0,66 c42⋅ 10 = 1,1 10 to 1 Mab = i = driver = 1n 19100 ⋅⋅vt ab −15 4 4 9 , 55 m mR = 3,2 ⋅ 0,15 = 0,48 Kg CALCOLO DELLA MASSA TOTALE getrieben n= = 98,2to⋅ 10 n2 ndriven i = fromJ0,40 0,66⋅ B ⋅ρ ⋅ cd2a2 −=d1,2 mS d 0,61 28 2 dw Peso FWsta = 2 ⋅ FTv ⋅ cos ⋅ β dp ⋅ n 19100 ⋅ v = 0,1 [Kg] ⋅ 1 + c22 = =1,3 ⋅ 1 + = 0,33 kg n= v= i < 0,40 22 m2S d a 0,261 100 9,81 28 Peso for FWsta = 2 ⋅ FTv ( i = 1) 19100 d = 0,1 [Kg] ⋅ 1 + ⋅ 1 + LR = L1 + L 2 = 1600 ⋅ 2 = 3200 mmw = m = 3,2 ⋅ 0,15 = 0,48 Kg = 0,33 kg Ft ⋅ dp ⋅ n P ⋅ 1000 ⋅ c 02 d 2a 2000 ⋅ M 95502⋅ PR 9,81 2 100 2 8 2 ⋅ π ⋅ 2,8 ⋅ 30 2 = b 100 F = M = P = a ⋅ t ⋅ 1000 2t⋅ s19100 v ⋅ 100 v The resulting total safety factor is: = 0,61 Kg v m ⋅ g ⋅ µS = a 3 ⋅ ab + n m=⋅ ga ⋅ sin =n zk ⋅ z e ⋅ Pspez d α⋅ + J = 98,2 ⋅ 10 −15 ⋅ B19100 ⋅ ρ ⋅ d4a⋅ 10 − d4tFau == m = ⋅ 10 6 a a ⋅ 3200 a ⋅ 1000 dwp 2 2⋅a m = mc + mR + msred = 50 + 0,48 + 0,33 = 50,81 Kg 1200 ∆l = =22− ,02d2 mm ≅⋅ B0,63100 2 − 28 2 ⋅ π ⋅ 2,8 ⋅ 30 d ⋅ π ⋅ ρ cm c1c ·+ cm2R + msred = 50 + 0,48 + 0,33 = 50,81 Kg 2 ⋅ 952000 a 0 =m = = 0,61 Kg 2 4 ⋅ 10 6 4 ⋅ 10 6 0,61 P ⋅ 1000 ⋅ c 0 28 2 CALCOLO DELLA FORZA PERIFERICA FU = b 100 = ⋅ 1 + = 0 , 33 kg 2 2000 ⋅M zk ⋅ z e ⋅ Pspez Determine installation 2tension 2 belt 2 100 2 Select type a Fu = ofFORZA 1116FORMULE ,2 PAG. 41a ⋅ t ⋅ 1000 CALCOLO DELLA PERIFERICA FU sNaUOVE v ,86 2b⋅ = 1116t ,2=⋅ 101 = 1,78 ≈ 18a mm = v ⋅ 100 2 dw = = 56ACALCOLO Scm DELLA MASSA a a =TOTALE For drive use ,the graph. For initial drive is correctly tensioned when slack ] Ft = the m ⋅ ainitial + Fr = 50 ,25 ⋅selection, 20 + 100 = 1128 4 [Nselection M = , 85 Nm ⋅ athe side tend belt0,61 28 2 is 52,21⋅ 12 Peso a a ⋅ 1000 2000 [Kg2] m S It⋅ 2is 1 +also = 0,veloce 1 conditions. ⋅ 1 + to use = 0,33 kg 1116 ,2 ⋅ 101 ,86 = giriin/ min albero 2 2 pulley choice, it is recommended to use the driver pulley with sioned all working important [ ] F = m ⋅ a + F = 50 , 25 ⋅ 20 + 100 = 1128 , 4 N M = = 56 , 85 Nm RT = 9,81 2 da 2 100 the r 48 + 0,33 = 50,81 Kg t 2000 giri / min albero lento maximum diameter allowable in the application. 2000 ⋅ M minimum necessary tension to minimize shaft loads. Belt tenF ⋅ d 150 LR = L1 + L 2 = 1600 ⋅ 2 = 3200 mm m = 3,2 ⋅ 0,15 = 0,48 Kg 1105 ⋅ π ⋅ 101,86 2000 ⋅ 350 ⋅M 19100 ⋅ P ⋅ 10 3 Fbelt ⋅ n R LR and its 2000 Fu,83 = N M = t zp== = 56F,U28= [Nm] = 4908 = sion is dependent alsoP on u ⋅ d w length Fu = number of F = = F ⋅ d 1105 dw 2000 u 142,6 ⋅ 101,86 8 2000 19100 10 3+of0,teeth, teeth ZnR⋅ .dAccording toR +belt number following msred = ⋅50 48 + 0,33 = 50,81dKg M= t p = = 56,28 [Nm] FU 1116,2 w m = mc + m wtension Calculate ratio b= = 1,78 cm ≈ 18 mm 2000 drive 2000 is suggested: 52,21⋅ 12 d2a − d2 ⋅ π ⋅ ρ ⋅ B 100 2 − 28 2 ⋅ π ⋅ 2,8 ⋅ 30 48 ,2 ⋅ 101,86 PAG 1116 1128,4 [N] M= = 56,85 Nm n = = 0,61 Kg n1 6 6 150 ⋅ π 2000 ⋅ 350 CALCOLO PERIFERICA FU ⋅n 4 n⋅ 10 PAG 48 2000 i = driver =1 J ⋅ 4∆drive FU = 2 shafts =10 4908 ,83 NDELLA FORZA z= = treibend M = i = Ft ⋅ dp n2 ⋅ P ⋅ 103 ndriven 19100 π ⋅n 142,6 8 ab 4 2000 ⋅ M Fu ⋅ d w ⋅ n Ft = M= J = 98,2 ⋅ 10−15 ⋅ B ⋅ ρ ⋅ d19100 − d4 ⋅ P ⋅ 10 3 9,55 ⋅ t ab ngetrieben ω= 1116,2 ⋅ 101,86 Fu = e4 P =F = m F n ⋅⋅dPp ⋅ 103 2000 19100 π30 ⋅n u = = 1128,4 [N] ⋅ a⋅+10 Fr3 = 50F ,25 ⋅= 201/3 +F100 M= = 56,85 t ⋅ dp −15 4 t19100 n ⋅ d d < 75 F Z 8 [Nm] F = M = J = 98 , 2 ⋅ 10 ⋅ B ⋅ ρ ⋅ d − d ω = 2 2 TV U w w R t e 2000 mS Calculate lenght2000 d 0,61 28 Peso n ⋅ dbelt 30 p [ ] 1 = 0 , 1 Kg ⋅ + = ⋅ 1 + = 0 , 33 kg 2 < ZR < 150 = 1/2 n1 n 759,81 2 F 2 FU 100 2 dTV i = driver =1 a Belt lenght for drive with ratio i # 1 ,86 F PAG. 49 Fu = m ⋅ ab + m ⋅ g ⋅ sin α + m ⋅ g ⋅ µ n2 ndriven ZR > 150 M = Ft ⋅ dp = 1105F⋅ 101 TV = 2/3 U = 5619100 ,28 [Nm ⋅v] J ⋅ ∆n treibend n= J = 98,2 ⋅ 10 −15 ⋅i B ⋅nρ2000 ⋅ d4a − d4 2000 PAG. 49 M = = ab 2 d w n 9 , 55 ⋅ t getrieben t 1 z g − zk ⋅ t 2000 ⋅ M ab More m = mthan msred = 50 + 0,48 + 0,33 = 50,81 Kg drive c + mR2+shafts π ⋅n Ft = J = 98,2 ⋅ 10−L15R ⋅ ≈ B2 ⋅tρ⋅ ⋅ zdge4+−zdk4 + 2A + 41A ⋅ z −πz ⋅ t 2 ω= 48 PAG d 2000 ⋅ M g k p 30 L R ≈ ⋅ z g + z k + 2A + ⋅ Fu = m ⋅Fat b= + m ⋅ g ⋅ sin α + m ⋅ gP⋅ µ⋅ 1000 ⋅ c 0 =TV > FU bF 100 dp 2 4A π CALCOLO FORZA PERIFERICA zk ⋅ z e ⋅ PDELLA 19100 ⋅ P ⋅ 103 FU Ft ⋅ dp π ⋅n and more precisely: spez F M= = J = 98,2 ⋅ 10−15 ⋅ B ⋅ ρ ⋅ de4 − d4 ω= t −15 4 4 n ⋅n dp= 19100 ⋅ v 2000 30 giri / min albero veloce J = 98 , 2 ⋅ 10 ⋅ B ⋅ ρ ⋅ d − d β β t 1116,2 ⋅ 101,86 a =⋅ + ⋅ z g + zk + 1 − ⋅ sin LR = 2ART ⋅mt ⋅ a + Fr = 50,25 ⋅ 20 + 100d=w1128,4 [N] ⋅ z g − zk LR = 2 ⋅ A + π ⋅ d w = 2 ⋅ AIn F+t z= M = tension, it is= 56,85 Nm order to ensure the correct drive installation giri /2tmin albero lento180 2 β β 2000 2 LR = 2A ⋅ sin ⋅ + ⋅ z g + zk + 1 − + z⋅t ⋅ z g − zk LR = 2 ⋅ A + π ⋅ d w = 2 ⋅ Arecommended to use the special belt tension meter available 2000 ⋅ M2 2 g − zk ⋅ t 180 2000 ⋅ MPAG. 49 Ft = P ⋅ 1000 ⋅ c 0 from . ,86 F ⋅ELATECH d 1105 ⋅ 101 Fu =100 dp π b= M = t d wp = = 56,28 [Nm] ⋅ ⋅ z z P giri / min albero veloce kt ⋅ z e −spez z t ⋅ z − z 2 Z β 2000 2000 g k g k RT =k ⋅ arccos ⋅ Belt with ratio t 1 z g − zk ⋅ t z e = lenght ⋅ zk for βdrive 2000 ⋅ M = 2 ⋅ arccos ⋅ i = 1 albero lento giri / min L ≈ ⋅ z + z + 2 A + ⋅ Ft = t ⋅ z − z t ⋅2z⋅ gπ −⋅ Azk 360 180 2 ⋅ π ⋅ A Zk β R g k g k dp 2 π 4 A z = ⋅ z ⋅ arccos ⋅ β = 2 ⋅ arccos ⋅ e k β 48⋅ π 180 150 2000 ⋅ 350 1− 2 ⋅ A + π ⋅ dw = 2 ⋅ A + z ⋅ t 2 ⋅ π ⋅ A 2 ⋅ π ⋅ A z PAG ⋅ z g − zk LR =360 FU = = 4908,83 N = = 180 2000 ⋅ M 142,6 8 Ft ⋅ dp 19100⋅ P ⋅ 103 π ⋅n Fu = 1000 ⋅ 20 ⋅ 1,4 J ⋅ ∆n Ft = M = β t ω = J= 98,2β⋅ 10 −15 ⋅ B ⋅ ρ ⋅ de4 − d4 b= = 4,86 cm = 48,6 mm Ma = dw n ⋅ dp LR = 2A ⋅ sin2000 ⋅ + ⋅ z g30 + zk + 1 − 9J ,55 ⋅ 12⋅ 20 ⋅ 8,572 56 ⋅ 1,4 1000 ⋅ ∆⋅nta ⋅ z g − z k LR = 2 ⋅ A + π ⋅ d w = 2 ⋅ A + z b= = 4,86 cm = 48,6 mm M 2 2 180 154= z g − zk Z k a 9,55 ⋅tta⋅ z g − zk 56 ⋅ 12 ⋅ 8,572 n n 1 driver ⋅ arccos ⋅ i= =1 ⋅ π ⋅ 2 A 180 2⋅ π ⋅ A n2 ndriven 150 ⋅ π 2000 ⋅ 350 PAG. 49 FU = = 4908,83 N = z= t ⋅ z g − zk 142,6 β 8 t ⋅ z g − zk Zk 100 ⋅ 350 LR = 2 ⋅ 1800 + 56 ⋅ 8 = 4048 mm b= = 5,53 cm = 55 mm ze = ⋅ zk β = 2 ⋅ arccos2⋅ ⋅ arccos ⋅ − ⋅ z z t 56 ⋅ 12 ⋅ 9 , 422 360 180 2 ⋅ π ⋅ A 100 ⋅ 350 t 1 g k F = 2000 ⋅ M 2 ⋅ π ⋅ A ⋅ 1,4 L = 2 ⋅ 1800 + 56 ⋅ 8 = 4048 mm b = = 5 , 53 cm = 55 mm L ≈ ⋅ z + z + 2 A + ⋅ R R g k t = 4,86 cm = 48,6 mm 56 ⋅ 12 ⋅ 9,422 dp 4A⋅ g ⋅µ π F = m2 ⋅ a + m ⋅ g ⋅ sin α + m ( ( ) ) ( ( ( ( ) ( ) ( ) ( ( ( ( ( ) ) ) ) ( ) ) ) ( ) ( ( ( ( ) ) ) ) ) ) ( ) ) ( ) ( ( ( ( ( ) (( ( )) ) ) ) ) ) ( ( ) ( ( ( ) ) ) ( ( ( ( ) ) ) ( ) ) ( ( ( ) ) ) ( ( ) ) ) ( ( ) ) ( () ) ) ( ( ) ( ) ) ( ) 2000 ⋅ M 19100 ⋅ P ⋅ 10 3 F ⋅d ⋅n SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti Fu =en .qxd Fu = P= u w 3 n ⋅ dw dw 19100 ⋅ 10 Mab = J ⋅ ∆n 9,55 ⋅ t ab i= ( 15/04/2010 07:57 Pagina 155 ntreibend ngetrieben ) J = 98,2 ⋅ 10 −15 ⋅ B ⋅ ρ ⋅ d4a − d4 n= 19100 ⋅ v dw P ⋅ 1000 ⋅ c 0 Calculation example b= 100 zk ⋅ z e ⋅ Pspez Power to be transmitted Driver rpm n1 Driven rpm n2 2000 ⋅ M Motor Fu = start up torque Mab d Requiredw center distance A Max allowable driver pulley diameter dw Safety 150 ⋅ πfactor c1 2000 ⋅ 350 z= = 8 FU = 142,6 z= i= d1 d w d da Zk ZB, LR a dw Zg A 130 ⋅ π z= = 130 10 ⋅ π zL L == =2 2 = ⋅ dπ+ 40 130 ⋅ 400 ⋅ A + = 2⋅ 10 ⋅ A =+ z1200 ⋅ t mm R R 10 w= z = πbelt Calculate width 10 = 4908,83 N ndriver ndriven ⋅ 15 ,4 2 ⋅ A + z ⋅ t LbR == 21000 ⋅A + π ⋅ d⋅w1= cmmm = 29,2 mm LR = 2 ⋅ 400 + 40 ⋅ 10= =2,92 1200 40 ⋅ 12 ⋅ 14,968 LR = 2 ⋅ A + π ⋅ d w = 2 ⋅ A + z ⋅ t LR = 2 ⋅ A + π ⋅ d w = 2 ⋅ A + z ⋅ t A belt width of 32 mm is selected. ⋅ 15+⋅ 140 ,4 ⋅ 10 = 1200 mm LR = 1000 2 ⋅ 400 bThe = belt width is =verified 2,92 cmaccording = 29,2 mm to the peak torque (starting 100 ⋅+200 LbR == 400 40 ⋅ 10== 41200 mm 402⋅⋅12 ⋅ 14,968 ,37 Nm cm =as43,7 mm torque) for2 n = 0 +with 200 start up torque L = ⋅ 400 40 ⋅ 10 = 1200 mm 40R ⋅ 12 ⋅ 9,529 1000 ⋅ 15 ⋅ 1,4 b= = 2,92 cm = 29,2 mm 1000 15 ⋅ 1,4 40 ⋅ 12 ⋅⋅14,968 100 ⋅ 200⋅ 15 = b= 2,92 cm = 29,2 mm 1000 ⋅1 b = 40 =,44,37 cm =cm 43,7 mm mm ⋅ 14,968 b ⋅=⋅12 = 2,92 = 29,2 M ab 2000 ⋅ 40 12 ⋅ 9,529 40 ⋅ 12 ⋅ 14,968 FU = = 3141 N dw 100 ⋅ 200 The mm b = next belt width= 50 4,37 cmis= chosen. 43,7 mm 200 40 ⋅100 12 ⋅ ⋅9,529 b= = 4 , 37 cm = 43,7 mm M 2000 ⋅ ab⋅ 200 100 1200 FUZ= 40 = teeth 3141= N4,37 cm = 43,7 mm ⋅ 12 ⋅ 9,529 b= = 120 R = dw⋅installation Determine tension according to belt number of 40 12 9,529 ⋅ 10 Fu = mselection ⋅ ab + m ⋅ ggraph ⋅ sin α +and m ⋅ gthe ⋅ µ corrected power of 21 kW, a AT10 From pitch is chosen. Calculate pulley diameter From the maximum allowable pulleys diameter, the drive ratio giri / min albero veloce and RT =the type of belt selected, the number of teeth of the driver giri / min albero lento and driven pulley is calculated. teeth 130 ⋅ π = 40,84 - select z = 40 with dw = 127,32 mm 10 2000 ⋅ M ab FU =1 1200 = 3141 N 2000 ab dw=⋅ M120 teeth FZUR=2= = 3141 N M 2000 ⋅ ab FU10 = dw = 3141 N dw 1200 1 ZR = = 120 teeth 1200 Z2R = 10 1200 = 120 teeth Z10 = 120 teeth R = 10 tension per belt side FTV is therefore: The 1 installation 21 1 • FU = 1570 N with zR = 120 FTV2 = 2 The maximum allowable diameter is chosen to minimize belt width. LR = 2 ⋅ A + π ⋅ d w = 2 ⋅ A + z ⋅ t 130 ⋅ π z= = 1 =130 40⋅ π = z z = 10 10 zL2 ==40 2 ⋅ 400 + 40 ⋅ 10 = 1200 mm R Calculate belt length LR = 2 ⋅ A + π ⋅ d w = 2 ⋅ A + z ⋅ t 1000 ⋅ 15 ⋅ 1,4 A + z cm ⋅ t = 29,2 mm bL=R = 2 ⋅ A + π ⋅ dw = 2=⋅ 2,92 40 ⋅ 12 ⋅ 14,968 Verify flexibility The minimum pulley diameters are respected. LR = 2 ⋅ 400 + 40 ⋅ 10 = 1200 mm LR = 2 ⋅ 400 + 40 ⋅ 10 = 1200 mm 100 ⋅ 200 b= = 4,37 cm = 43,7 mm 40 ⋅ 12 ⋅teeth 9,529 Calculate 1000 ⋅ 15 ⋅ 1,4 in mesh b= = 2,92 cm = 29,2 mm 1000 ⋅ 1,4ratio 1, the pulleys have 20 teeth in mesh. Being the⋅ 15 drive b = 40 ⋅ 12 ⋅ 14,968 = 2,92 cm = 29,2 mm ze =40 20⋅ 12 ⋅ 14,968 2000 ⋅ M ab FU = = 3141 N 100 d ⋅ 200 w b= = 4,37 cm = 43,7 mm 100 ⋅ 200 Selection graph b = 40 ⋅ 12 ⋅ 9,529 = 4,37 cm = 43,7 mm1000 40 ⋅ 12 ⋅ 9,529 1200 ZR = = 120 teeth 10 200 2000 ⋅ M ab 100 FU = = 3141 N 1 2000 dw⋅ M ab = 3141 N FU = dw 2 Selected belt ELATECH iSync U1200 AT10 / 50 AT 1 0 T10 AT 5 T5 T 2 ,5 P . k [Kw] 1200 ZR = = 1200 120 teeth ZR = 10 = 120 teeth 10 1 21 2 d2 α 130 ⋅ π z= = LR =102 ⋅ A + π ⋅ dw = 2 ⋅ A + z ⋅ t Select belt type and pitch z= 130 ⋅ π = 10 β Calculate drive ratio n1 =1 n2 t α 15 [kW] 1500 [1/min] 1500 [1/min] 200 [Nm] 400 [mm] 130 [mm] 1,4 ELATECH® iSync™ - 10 1 00 100 1000 5000 10000 n [min-1] 155 SIT_ELA_pag143_156_manicotti_sit_new:Ela_manicotti en .qxd 14/04/2010 11:03 Pagina 156 Belt installation Drive installation When installing belt on pulleys, it must be checked before tensioning the drive, that belt teeth and pulley grooves correctly match. Belt drive tension Correct belt drive tension and alignement are very important to optimize belt life and minimize noise level. In fact improper tension in the belt drive, affect belt fit in the pulley grooves while correct tension minimizes belt pulley interference reducing the noise in the drive. Drive Alignment Pulley misalignment will result in an unequal tension, edge wear and reduction of belt life. Also, misaligned drives are much noisier than correctly aligned drives due to the amount of interference that is created between the belt teeth and the pulley grooves. Proper pulley alignement should be checked with a straight edge or by using a laser alignment tool. Belt width [mm] Allowable pulley misalignment [°] 10 16 32 over 0,28 0,16 0,1 Idlers Idlers are often a mean to apply tension to the drive when the centre distance is fixed but also to increase the number of teeth in mesh of the small pulley. A toothed idler on the inside of the belt on the slack side is recommended with respect to a back side idler. Drives with inside flat idlers are not recommended as noise and abnormal belt wear may occur. • Idler location is on the slack side span of the belt drive • Diameter for inside toothed idler must be ≥ of the diameter of the small pulley in the drive • Idler must be mounted on a rigid support • Idlers both flat and toothed, should be uncrowned with a minimum arc of contact. • Idler should be positioned respecting: 2 • (dwk + dwg)< A • Idlers width should be ≥ of pulley width B Backside idlers, although increase the teeth in mesh on both pulleys in the drive, force counterflexure of the belt thus contributing to premature failure. When such an idler is necessary, it should be at least 1,25 times the diameter of the small pulley in the drive and it must be located as close as possible to the small pulley in the drive in order to maximise the number of teeth in mesh of the small pulley. Belt handling and storage Proper storage is important in order avoid damaging the belts which may cause premature belt failure. Do not store belts on the floor unless in a protective container to avoid damages which may be accidentally caused by people or machine traffic. Belts should be stored in order to prevent direct sunlight and in a dry and cool environment without presence of chemicals in the atmosphere. Avoid belt storage near windows (to avoid sunlight and moisture), near electric motors or devices which generate ozone, near direct airflow of heating/cooling systems. Do not crimp belts while handling or when stored to avoid damage to tensile cords. Belts must not be hang on small pins to avoid bending to a small diameter. Handle belts with care while moving and installing. On installation, never force the belt over the pulley flange. Special belts Special belts with cleats, backing and with special moulded shape are designed and manufactured to maximize application performance. 156