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Pagina 143

ELATECH® iSync™ high performance timing belts

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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.

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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

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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™

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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

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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]

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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 

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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

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Drive calculation

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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

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153

ELATECH® iSync™

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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 − 2d4 = + 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−= zk 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⋅ 2is 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 FU 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 

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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=

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Pagina 155

ntreibend ngetrieben

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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

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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.

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