5_pag_143_157_manicotti:Ela_manicotti en .qxd
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Pagina 143
ELATECH® iSync™
high performance timing belts
5_pag_143_157_manicotti:Ela_manicotti en .qxd
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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.
5_pag_143_157_manicotti:Ela_manicotti en .qxd
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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
length
[mm]
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
T5
1.20
1
40°
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
1
1.20
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
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
Number of
teeth
z
40°
10
2
length
[mm]
5
2.50
Number of
teeth
z
5
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
5_pag_143_157_manicotti:Ela_manicotti en .qxd
AT5
AT10
50°
2
1.50
1.20
10
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
50°
5
13/04/2010
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
146
U 450 AT5 / 16
16:19
Pagina 146
5_pag_143_157_manicotti:Ela_manicotti en .qxd
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Pagina 147
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
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
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 = 15 mm
Drive with reverse bending and double sided belt
• Driver pulley zmin = 18
• Idler (flat) running on belt back dmin = 15 mm
147
ELATECH® iSync™
rpm
[min-1]
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16:19
Pagina 148
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
13/04/2010
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
rpm
[min-1]
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
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
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
5_pag_143_157_manicotti:Ela_manicotti en .qxd
16:19
Pagina 149
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
13/04/2010
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
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
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 = 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
149
ELATECH® iSync™
rpm
[min-1]
5_pag_143_157_manicotti:Ela_manicotti en .qxd
16:19
Pagina 150
iSync™
AT5
50°
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
13/04/2010
1.20
5
• 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
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
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
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
150
P [kW] = Pspez • ze • zk • b / 1000
zemax
zk
b
A
t
(
)
 t ⋅ z g − zk 
Zk
⋅ arccos ⋅ 

180
 2 ⋅ π ⋅ A 
5_pag_143_157_manicotti:Ela_manicotti en .qxd
AT10
13/04/2010
16:19
Pagina 151
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
• 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
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
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 = 50 mm
Drive with reverse bending and double sided belt
• Driver pulley zmin = 25
• Idler (flat) running on belt back dmin = 120 mm
151
ELATECH® iSync™
rpm
[min-1]
= 0,1 [Kg]
⋅ 1 +
⋅ 1 +
=
 = 0,33 kg
2 2
9,81
2  d 2a 
2  100 2 
5_pag_143_157_manicotti:Ela_manicotti
152
m
  d 2d2  0.,61
0,61en .qxd
 13/04/2010 16:19 Pagina
mKg
2828
mc + mR + msred
= 50Peso
+Peso
0,48 += 00,,1
33[Kg
= 50
S FORMULE
UOVE
]N,81
0,33
]
= 0,1 [Kg
⋅S1⋅ +1 + 2 2PAG
= = 41 ⋅ 1⋅ +1 + 2 2= 0=,33
kgkg
2
9,81
2  t 100
100 
CALCOLO
9,81
1  z g − zk ⋅ t 
2 2  DELLA
 d adMASSA
2000 ⋅ M
a  2 TOTALE
L R ≈ ⋅ z g + z k + 2A +
⋅
Ft =

d=p 50 + 0,48 + 0,33 = 50,81 Kg
2
π
4A 
m = m + m + m
(
(
)
)
OLO DELLA FORZA PERIFERICA FU
c
R
sred
L
⋅0
2,=
mR = 3,2 ⋅ 0,15 = 0,48 Kg
R =
1+
= cm+c m
+ Rm+R m
+ sred
mLsred
50
01600
,48
33
= 50,mm
81 Kg
mm
=m
= L=50
+20+=1116
,48
+ ,0+
=3200
2,33
⋅ 101
,50
86,81 Kg
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
2
180


CALCOLO
DELLA
FORZA
PERIFERICA
F
U
2
2
2
2
CALCOLO DELLA FORZA PERIFERICA FU
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
⋅ 101
1116
,2 ,⋅2101
,86,86
2000
2000
=m
= 50
⋅ 20
+ 100
= 1128
= 56
Nm
[N[]N]
Ft F
=t m
⋅ a⋅ +a F+r F
=r 50
,25,25
⋅ 20
+ 100
= 1128
,4 ,4
MM
==
= 56
,85,85Nm
2000
 t ⋅ z − zk 
2000
 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 F
⋅ dt ⋅pdp 1105
1105
⋅ 101
101
,86,=860,1 [Kg]
⋅ Peso
1
⋅
+
=
⋅
1
+
=
0
,
33
kg




2
2
[
]
M
=
=
=
56
,
28
Nm
[
]
M
=
=
=
56
,
28
Nm
9,81
2  da 
2  100 
2000
2000
F ⋅ d 2000
9100⋅ P ⋅ 103
π ⋅ n 2000
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
4848
PAG
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 3
n
d
⋅
2000
30
p
Ft F
⋅ dt ⋅pdp
19100
⋅ 10
49
19100
⋅ P⋅⋅P10
π ⋅πn⋅ n
−
15
4
4
−15
= 98
⋅ 10
ω=
Ft F
MM
==
=t =
J =J98
,2 ,⋅210
⋅ B⋅⋅Bρ⋅⋅ρ d⋅ e4d−e d−4d
ω=
t
2000
n ⋅nd⋅pdp 2 CALCOLO
2000
3030
DELLA FORZA
PERIFERICA FU
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
. 49
PAG
. 49
PAG
2000
2
t
1  z g − zk ⋅ t 
2000 ⋅ M
L R ≈ ⋅ z g + z k + 2A +
⋅
Ft =

2 2
F
d
⋅
dp
2
π
4
A
1105
101
,
86
⋅
4048
−
⋅
z
z
t
p  
z g −g zk k⋅ t   = 56,28 [Nm


2 ] 2000
2000
⋅ M⋅ M
β t 
d2
M =+ t+1 1
 t t z⋅ βz+g z+ z+k 2
⋅L=⋅ = 2 ⋅2000
Ft F
=t =


d
− g  ⋅k z g +−A2zA
2A ⋅ sin ⋅ + ⋅  z g + zLkRL+≈R2≈1⋅2
A
+
π
⋅
d
=
2
⋅
A
+
z
⋅
t
1
8
k2000
R
w
β
3
4A4A  π π  
dpdp α

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


w
β t t  Z
  3 β tβ⋅v z g − zt k  
Zg
t ⋅=A2
z⋅A
−⋅ sin
z⋅kβ⋅+
β
gsin
+ ⋅  ⋅z19100
zg z
k+k z+
LRL⋅ =
1−1 −⋅  ⋅ Fz⋅3t g⋅zd−gpz−k zk  πLR⋅Ln=R 2=⋅2A⋅ A
+ π+ ⋅πd⋅wd=w−152=⋅2A⋅ A
+ z+⋅ztd4⋅ t 4
R 2
k +

g +
P
10
⋅
⋅
⋅ zk β = 2 ⋅ arccos
⋅
arccos






180
22 2 2 
180
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
t ⋅ z g −g zk k


 t ⋅ z g −g zk k 
Z kZ k
β⋅ 1β,4
 2 ⋅ π ⋅ A 
⋅=20
1000
J ⋅ ∆n
.
49
PAG
z
=
⋅
z
⋅
arccos
⋅
β
=
2
⋅
arccos
⋅
z
⋅
z
⋅
arccos
⋅
β
=
2
⋅
arccos
⋅




k

b=
e e
 
360 =k 4,86 cm = 48,6mm
180
360
180
2 ⋅2π⋅⋅πA⋅ A 
 2 ⋅2π⋅⋅πA⋅ A  
9,55 ⋅ ta
56 ⋅ 12 ⋅ 8
,572
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

⋅ 20
1000
⋅ 20
⋅ 1,⋅41,4
1000
J ⋅J∆⋅n∆n
4,86cmcm= =
,mm
6 mm
b =b =
= 4=,86
4848
,6 2
M M= =
100 ⋅ 350
b a a 9,55
Belt width
FU
(N)
Peripheral force
9,mm
55
⋅ 12
8
⋅(cm)
ta⋅ ta
⋅⋅12
⋅ 8⋅,572
LR =56256
1800
+,572
56 ⋅ 8 = 4048 mm 3
= 5,53 cm = 55
(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
100
⋅ 350
2
L
=
2
⋅
1800
+
56
⋅
8
=
4048
mm
b
=
=
5
,
53
cm
=
55
mm
LR =R 2 ⋅ 1800 + 56 ⋅ 8 = 4048 mm
b=
= 5Center
,53 cm distance
= 55 mm
(s)
Deceleration time
tav
A
(mm)
8
9,422
⋅ 12
3
5656
⋅ 12
⋅ 9⋅,422
(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
4048(mm)2 2 Pulley outside diameter
1116,2
da 4048
Number of teeth of the small pulley
zk
Ft = 2337,5 [ N]
b
=
=
1
,
78
cm
≈
18
mm
3
52,21
⋅ 12
diameter
d 8 8 (mm)3 3 Small pulley outside
(
(
(
)
(
(
)
)
(
)
)
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 2 (mm)
3
⋅ M weight
19100 ⋅ P ⋅ 10
F ⋅ d ⋅ρn
Specific
(kg/dm ) 2000
= 2337
Fv F=v = Ft F
=t 2337
,5 ,[5N[]N]
F =
Fu =
P= u w 3
Small pulley pitch
circle
dwk 3 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
p
Ft F⋅ d
2000
9550
⋅ M⋅ M
9550
⋅ P⋅ P β
t ⋅pd⋅pn⋅ n
(°)
Wrap angle
4048 F F= 2000
=2
MM
==
PP
==
Ft ⋅ dp ⋅ n
9100 ⋅ 10 3
t
t
3 3
19100
⋅ 10
dpdp
nn
19100
⋅ 10
a ⋅ t 2a ⋅ 1000 8 v 2 ⋅ 1003
S
=
=
19100 ⋅ v
a
Calculation
formula
2
2⋅a
n=
J = 98,2 ⋅ 10 −15 ⋅ B ⋅ ρ ⋅ d4a − d4
dw
2
2 ⋅ sa
v
Fv = Ft = 2337,5 [ N]Peripheral force
ta = =
Power
Torque
3
2 2
2 2
a
a
⋅ 1000
⋅ 1000 v ⋅ 100
a ⋅at t⋅a1000
2 ⋅2s⋅as a
vv
FU ⋅⋅vc=0 =v ⋅3100
FU=⋅ n
⋅d
=W
=⋅ nP a⋅ 1000
P ⋅ 9550
dW ⋅ n
n⋅z⋅ t
t a t=a =M
S aS=aM
b
100
F
d
⋅
19100
⋅
P
⋅
10
= 2⋅ P
P=
=
M== a a a ⋅a1000
=⋅ 1000
16,2
2 ⋅2a⋅ av =
e
spez
Fuz=k ⋅2z1000
M= U W
9550
2000
19100
60000
n
= 1,78 cm ≈ 18Pmm
9550
n ⋅ dw
2000
21⋅ 12
v
=
a
)
))
2 ⋅ sa
a ⋅ 1000
(
)
Sa =
a ⋅ t 2a ⋅ 1000 v 2 ⋅ 100
=
2
2⋅a
1116,2
= 1,78 cm ≈ 18 mm
52
,21⋅ 12
9550
⋅P
d
M=
n
b=
100 ⋅ P ⋅ 10 3
n ⋅ dw
1116
Ft ⋅ dp ⋅ n2000 ⋅ M 2000
1116
F ,⋅2d,w2 ⋅ n⋅ M
⋅ M2000 ⋅ M P ⋅ 9550 n
P
⋅≈1000
1,78Pcm
cm
≈ 18
mm FFu== 2000
i=
F
b =b
= 1=,78
18
mm
P==F =u 2000
Ft⋅ M
=
=
u= 3
M=
t
⋅ 12
n
d
19100
⋅d10 3 = 19100
5252
,U21,21
⋅ 12
d
dp
⋅ 10
w
dpw
v
n
W
19100 ⋅ P ⋅ 10 3
F ⋅d ⋅n
P= u w 3
FFUu ⋅=C1
n
⋅
d
19100
⋅ 10
b
=
w
2000
⋅
M
19100
⋅
P
⋅
10
F
⋅
d
⋅
n
20002000
⋅ M ⋅ 350 F
19100 ⋅ P ⋅ 10
F ⋅d ⋅n
J ⋅ ∆n
⋅π
= =
Fu F=u F
Fu F=u i== ntreibend
P =P = u u wz =w 150
Uspez ⋅ z e
3 3
=
4908
,
83
N
=
U
n
⋅
d
d
19100
⋅
10
19100
⋅
v
Acceleration
torque
n
⋅
d
d
19100
⋅
10
Angular
speed
pheripheral
speed4w w142
4 ,6
w
w
ngetrieben
,55 ⋅ t ab
8 ,2 ⋅ 10 −15 ⋅ B
J = 98
⋅ ρ⋅ ⋅t 2 d⋅ 1000
a −d
a
2 ⋅ sa
v 2 ⋅ 100 n =
v
a
d
t
=
=
S
=
=
2
⋅
s
v
19100
⋅
v
60000
⋅
v
π
⋅
n
a n⋅z⋅ t
M ⋅ an a FU ⋅ va ⋅ 1000
J ⋅ w∆n
n
FU ⋅ d W
⋅
P ⋅ 9550
d
n
ab
W
2
2 ⋅ a Mab =
ω=
=
i = treibend
=
P =n =
M=
=
v = t ab = = =
a b 60000
a b ⋅ 1000
⋅t
n
9
,
55
⋅
t
2000
19100
n
getrieben
ab
nztreibend
J30
⋅J∆⋅n∆n 9550d W 1000 ntreibend
n1 P ⋅ 1000 ⋅ c 0
n
Mab
Mab
==
i =i =
i = driver
b ==1
100
n
9
,
55
t
⋅
19100
⋅
v
n
9
,
55
t
⋅
n
n
getrieben
getrieben
ab ab
2
driven
zk ⋅ z e ⋅ Pspez
n=
2 ⋅ 10 −15 ⋅ B ⋅ ρ ⋅ d4a − d4
dw
1116,2
Moment of
inertia
2 ⋅M
b = 2000
= 1,78 cmrpm
≈P18
mm
2000 ⋅ M
P ⋅ 1000 2
⋅ 9550
19100 ⋅ v
⋅ 1000
a ⋅ t 2 ⋅ 1000
v 2 ⋅ 1000
−15
4
4
Ft =52=,21v⋅ 12
FU =
Mt=
= s ab = ab ⋅ t ab ⋅ 1000
n=
J = 98
s av = v av
= ,2 ⋅ 10 ⋅ B ⋅ ρ ⋅ da − d
ges = t ab + t c + t av
000 ⋅ c 0
d
d W100
v
n
dw
2
2
2
⋅
a
p2 ⋅ a b
19100
⋅
v
v
F
=
m
⋅
a
+
m
g
⋅
sin
α
+
m
⋅
g
⋅
µ
19100
⋅
v
−15−15
u
b
⋅M
= 98
⋅ 10
⋅ 4ad−4a d−4d4
nF=n == 2000
J =J98
,2 ,⋅210
⋅ B⋅⋅Bρ⋅⋅ρ d
z e ⋅ Pspez
u
d
d
⋅C
w
d wwF ⋅ d ⋅ F
2000
19100 ⋅ P ⋅ 10s3
b = nU 1
1000 ⋅ c 0
⋅ s P⋅ ⋅M
v Fu = 2
Ps= = vu ⋅ t w⋅F1000
c
b =a v
100
3 ⋅z
t ave =
=
s ges = s ab + s c +Fus=av
t
=
Uspez
c 19100
c ⋅ 10
cd
n
⋅
d
z
152
w v ⋅ 1000
w k ⋅ z e ⋅ Pspez
av
a v ⋅ 1000
⋅ 1000
⋅ c⋅0c 0
P ⋅P1000
=
b
100
b=
100
P
⋅π
2000 ⋅ 350
000 ⋅ Mπ ⋅ n
zk z⋅⋅kzv⋅ez⋅ eP⋅spez
spez
⋅ s150
v
19100
60000
⋅v
ab
FU =
= 4908,83 N
z2 =
=
t ab =
=
ω
=
n
=
=
142,6
8
d w 30
a b ⋅ 1000
dW
z⋅t
J ⋅ ∆n a b
n
treibend
2
giri / min albero
veloce
Mab =

m 
d22000
RT== m S ⋅ i1=+n d 
⋅M
55Ured
⋅ t ab m Sred
mUred = U ⋅ 1 +
m = mc + mR + mSred +9,m
getrieben
2
2

giri
/
min
albero
lento
F
=
2  udu  d
2  da 
w
2000
⋅
M
2000
⋅
M
n
n
1
driver
⋅π
2000 ⋅ 350
3 3
(
(
)
)
(
(( ))
)
Fu =
2000 ⋅ M
dw
)
v
s
CALCOLO DELLA FORZA PERIFERICA FU
2000
=
s2000
= s ab + s c + s av  t t cz= − z c
s c = v ⋅ t c ⋅ 1000 t av =
ges19100
5_pag_143_157_manicotti:Ela_manicotti
13/04/2010
16:19 Pagina
FU = m ⋅ ab + m
gn+ m ⋅ g ⋅ µ
2 ⋅ s ab
v
60000 ⋅ v g v ⋅ 1000
⋅ dw
 t ⋅ πz⋅gn− z153
Z⋅kv
β
⋅ 1,F4U.qxd
1000 ⋅ 20en
J ⋅⋅ ∆
k
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 
 153
z g − zk 
Z k a 9,55 ⋅tta⋅ z g − zk  56 ⋅ 12 ⋅ 8,572
n
n
1
driver
⋅ arccos ⋅ 

i=
=1

180
2⋅ π ⋅ A 
n2
ndriven
150 ⋅ π
 2 ⋅ π ⋅ A 
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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5_pag_143_157_manicotti:Ela_manicotti en .qxd
Fu =
19100 ⋅ P ⋅ 10
n ⋅ dw
Mab =
3
P=
J ⋅ ∆n
9,55 ⋅ t ab
Fu ⋅ d w ⋅ n
19100 ⋅ 10 3
i=
(
Fu =
)
n=
Calculation example
FU =
2000 ⋅ M
dw
15 [kW]
1500 [1/min]
1500 [1/min]
200 [Nm]
400 [mm]
130 [mm]
1,4
n
i = driver
ndriven
Fu = m ⋅ ab + m ⋅ g ⋅ sin α + m ⋅ g ⋅ µ
From selection graph and the corrected power of 21 kW, a AT10
pitch is chosen.
Calculate
pulley
diameter
giri / min
albero
veloce
giri / min albero lento
From the maximum allowable pulleys diameter, the drive ratio
and the type of belt selected, the number of teeth of the driver
and driven pulley is calculated.
z=
β
d1
130 ⋅ π
z = dw
=
10
130 ⋅ π
= 40,84 - select z = 40 with dw = 127,32 mm
10
LR = maximum
2 ⋅ A + π ⋅ dw =
2 ⋅ A + z ⋅ t diameter is chosen to minimize belt
The
allowable
width.
zL1 == 40
2 ⋅ 400 + 40 ⋅ 10 = 1200 mm
R
40⋅ π
z2 =130
z=
=
130 ⋅ π
z = 10
=
1000
⋅
15 ⋅ 1,4
b = 10
= 2,92 cm = 29,2 mm
40 ⋅ 12 ⋅ 14,968
Calculate belt length
Zk
da
ZB, LR
dw
Zg
A
LRLR= =2 2
⋅ A⋅ 400
+ π ⋅ d+w40
= 2⋅ 10
⋅ A =+ z1200
⋅ t mm
Calculate belt width
130 ⋅ π
z = 130
1000
⋅ π =⋅ 15 ⋅ 1,4
cmmm
= 29,2 mm
zLbR===10
2 ⋅ 400= + 40 ⋅ 10= =2,92
1200
40 ⋅ 12 ⋅ 14,968
10
130 ⋅ π
z=
=
A belt width
of 32 mm is selected.
1000 ⋅10
15 ⋅ 1,4
bLThe
=
2,92 cmaccording
= 29,2 mm
belt+width
is =verified
to the peak torque (starting
π ⋅⋅ d
R = 2 ⋅ A100
w = 2⋅ A + z⋅ t
200
LbR =402⋅⋅12
Afor
+⋅ 14,968
πn⋅ d=w 0= with
2⋅ A
⋅ t Nm
= +4200
,z37
cm =as43,7
mm
torque)
start
up torque
40 ⋅ 12 ⋅ 9,529
LR = 2 ⋅ A + π ⋅ d w = 2 ⋅ A + z ⋅ t
LR = 2 ⋅ 100
400⋅ 200
+ 40 ⋅ 10 = 1200 mm
1200
bLR= = 2 ⋅ 400 + 40 ⋅ 10
= =4,37
cmmm
= 43,7 mm
2000
⋅ M ab
40 ⋅ 12 ⋅ 9,529
FU =
= 3141 N
LR =⋅ 15
2d
⋅ ⋅w400
1000
1,4 + 40 ⋅ 10 = 1200 mm
b = 1000 ⋅ 15 ⋅ 1,4 = 2,92 cm = 29,2 mm
b = 40 ⋅ 12 ⋅ 14,968 = 2,92 cm = 29,2 mm
⋅2000
12 ⋅belt
14,968
The40next
50 mm is chosen.
⋅ Mwidth
ab
FUZ= = 1200
3141
N
1000
⋅ 120
15=⋅ 1teeth
,4
=
Rb =
dw
= 2,92 cm = 29,2 mm
10
40 ⋅ 12 ⋅ 14,968
100 ⋅ 200
b=
= 4,37 cm = 43,7 mm
200
401200
12 ⋅ ⋅9,529
⋅100
b= 1
= 4,37 cm = 43,7 mm
ZR 2=40 ⋅ 12 ⋅ 9,529
= 120 teeth
Determine
installation
tension according to belt number of
10 100 ⋅ 200
teethb = 40 ⋅ 12 ⋅ 9,529 = 4,37 cm = 43,7 mm
1 2000 ⋅ M ab
FU = 2000 ⋅ M ab = 3141 N
2
dw
FU =
= 3141 N
dw
2000 ⋅ M ab
FU =
= 3141 N
d
1200
ZR = 1200 = 120w teeth
ZR = 10 = 120 teeth
10
1 Z = 1200 = 120 teeth
R
The
10 tension per belt side FTV is therefore:
21 installation
2
1
• FU = 1570 N with zR = 120
FTV =
2
Verify flexibility
2000
ab
1000
⋅ 15⋅ ⋅M
1,4
3141cm
N = 29,2 mm
bF=U =
== 2,92
d
1000
⋅⋅ 14,968
15
Calculate
teeth
w ⋅ 1,4 in mesh
40
⋅
12
b=
= 2,92 cm = 29,2 mm
40 ⋅ 12 ⋅ 14,968
Selected belt
Being the drive ratio 1, the pulleys have 20 teeth in mesh.
1200
zeZ=R 20
= 120 teeth
= 100 ⋅ 200
10
b=
= 4,37 cm = 43,7 mm
100 ⋅ 200
b = 40 ⋅ 12 ⋅ 9,529 = 4,37 cm = 43,7 mm
1 40 ⋅ 12 ⋅ 9,529
2
2000 ⋅ M ab
= 3141 N
2000
dw⋅ M ab = 3141 N
dw
1200
ZR =
=
1200 120 teeth
ZR = 10 = 120 teeth
10
1
154
21
2
da
130 ⋅ π
z=
=
LR =102 ⋅ A + π ⋅ dw = 2 ⋅ A + z ⋅ t
LR = 2 ⋅ A + π ⋅ d w = 2 ⋅ A + z ⋅ t
LR = 2 ⋅ A + π ⋅ d w = 2 ⋅ A + z ⋅ t
100 ⋅ 200
b=
= 4,37 cm = 43,7 mm
40 ⋅ 12 ⋅ 9,529
LR = 2 ⋅ 400 + 40 ⋅ 10 = 1200 mm
LR = 2 ⋅ 400 + 40 ⋅ 10 = 1200 mm
FU =
FU =
d2
α
2000 ⋅ 350
= 4908,83 N
142,6
Select belt type and pitch
RT =
t
α
Calculate drive ratio
n1
=1
n2
Pagina 154
19100 ⋅ v
dw
⋅ c 0 transmitted
P ⋅ 1000
-bPower
to be
=
100
⋅ Pspezn1
zk ⋅ z erpm
- Driver
- Driven rpm n2
- Motor start up torque Mab
- Required
distance A
2000 ⋅center
M
Fu = allowable driver pulley diameter d
- Max
w
dw
- Safety factor c1
150 ⋅ π
=
8
08:01
ntreibend
ngetrieben
J = 98,2 ⋅ 10 −15 ⋅ B ⋅ ρ ⋅ d4a − d4
z=
15/04/2010
The minimum pulley diameters are respected.
ELATECH  iSync 
U1200 AT10 / 50
5_pag_143_157_manicotti:Ela_manicotti en .qxd
13/04/2010
16:19
Pagina 155
Selection graph
1000
AT 1 0
AT 5
T5
200
T 2 ,5
100
10
1
00
100
1000
5000
10000
n [min-1]
ELATECH® iSync™
P . k [Kw]
T10
155
5_pag_143_157_manicotti:Ela_manicotti en .qxd
13/04/2010
16:20
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