1. BELT DRIVES

2. Functions of belts
 Transmit motion between shafts that
are located at a considerable distance
from each other
 They are not used for exact fixed
speed ratio
slipping
 They are very flexible
– distance or
– the angle between the two shafts.

3. Types of Belts

4. Belt Pulleys
•Flat belts:
Crowned pulleys
• Round and V- belts
grooved pulleys or
sheaves
• Timing belts
toothed wheels or
sprockets.

5. Layout of Flat belt drive :
Non-reversing Open Belt
Reversing Crossed Belt
Reversing Open Belt
Quarter Twist Belt drive

6. V-Belts – Standard Sizes
1
1
1 2
1
7 4
1 21 8
32
2
29
3
5 13 17 32
4
16 32 32
A B C D E

7. Selection of V-Belts
 Power to be transmitted
 Speed of the small or large pulley
 Speed ratio
 Field of application.
 Approximate distance between the
centres of the two pulleys

8. Steps of V-belts Selection:
 Determine the service factor Based on
application,
 Obtainthe design power from the
equation:
Design power = transmitted power x service factor
 Select a suitable belt size from fig.(5. )

9. Steps of V-belts Selection:
 Find the diameter of the small pulley
(d) from table (5. ).
 Find the diameter of the large pulley
(D) from the equation:
D= d x speed ratio
 Find the length of the belt using the
equation: ( D d )2
L 2C 1.57( D d)
4C

10. Steps of V-belts Selection:
 Obtain the standard length of the belt
from table (5. )
 Calculate the exact centre distance
from the equation
b b2 32 ( D d )
C
16
 Find the angle of lap (arc of contact),
from the equation
 Angle of lap = 180 ( D d )60
C

11. Steps of V-belts Selection:
 Find the capacity of one belt from the
equation:
YxS
Capacity of one belt =
0.91
 XS ZS 3
de
 The values of X,Y and Z can obtained
from table (5. )
 The equivalent small pulley diameter de
can be obtained from the equation:
 de = diameter of small pulley x coefficient of small
pulley

12. Steps of V-belts Selection:
 Find the power transmitted by one
belt from the equation:
Power of one belt = belt capacity x length
coefficient x coefficient of arc of contact
 The required number of belts can be
obtained from the equation:
No. of belts = Design power/ power of one belt

13. Example:
An engine lathe is driven by a squirrel cage electric
motor through a V-belt. The electric motor runs at
1500 rpm with a maximum power of 3 hp. If the
input speed to the engine lathe is 500 rpm and the
centre distance between the motor pulley and the
lathe pulley is 30 in. Select a suitable size, length
and number of belts if the lathe is expected to be
working for two shifts, 16 hours/day.

14. Solution:
Pitch diameter AC Motor: High torque , High-slip
AC Motor: torque, Standard Groove Dimensions
Squirrel
From table (5.1), service factor for machine tools with Squirrel cage
Size of belt Minimum Cage Groove
,Synchronous, Split
Repulsion-Induction Single-phase
Range W D X S
,Series Wound, Slipping, DC Motor E
electric motor is = 1.2
recommended
Application Phase angle Motor :Shunt
,DC
: series wound Compound Wound
2.6 to 5.4wound, 340Engines 0.494:Multi-
Engine :Single- cylinder Internal 3/8
A 3 Internal, Combustion 0.490 0.125 5/8
Design power = transmitted power x service factor
Over 5.4 cylinder 380 0.504 Combustion, Line shafts :Clutches
Hour in daily service 4.6to 7.0 340 0.637
B 5.4 3--5 8-10 16-24 0.580 3-5 0.1758-10 ¾ 16-24 ½
= 3 x 1.2
Agitators for liquids, Blowers and
Over 7.0 380 0.650
exhausts , Centrifugal pumps and7.0 to 7.99 0
= 3.6 hp 1.0 34 1.1 0.879
1.2 1.1 1.2 1.3
compressor , Fan up9.0 10 hp and 8.0 to 12.0
C to 36 0 0.887 0.780 0.200 1 1 11/16
machine tool, Light-duty conveyors 12.0
Over 38 0 0.895
From figure (5.3) size A is selected
Belt conveyors for sand, grain, etc.
Dough mixers and Fan over 10 hp --12.99
12
Then the recommended diameter of small pulley from table (5.2)
Generators and line-shafts, Laundry and
13.0 -- 17.0
340 1.259
0
printing machinery 13.0
D , Punches, presses 1.1 36 1.2 1.271
1.3 1.0501.2 0.300 1.3 1 7/16 1.4 7/8
d = 3 in.
,shears , Positive displacement rotary
Over 17.0
380 1.283
pumps, Revolving and vibrating screens
speed ratio = 1500/500
Brick and textile machinery 18.0 to 24.0 360 1.527
BucketE elevators and exiters ,Piston
21.0 0 1.300 0.400 1 3/4 1 1/8
Over 24.0 38 1.542
=3
pumps and compressors ,Hammer-mills
and paper-mill beaters , Conveyers and
1.2 1.3 1.4 1.4 1.5 1.6
Diameter of large pulley (D) = d x speed ratio
pulverizers, Positive displacement
blowers, Sawmill and wood-working
machinery =3x3
Crushers ,mills and hoists
1.3 1.4 1.5 1.5 1.6 1.8
= 9 in.
Rubber calendars , extruders and mills

15. The length of the belt can be obtained from the equation:
( D d )2
L 2C 1.57( D d )
4C
( 9 3 )2
L 2 x30 1.57( 9 3 )
4 x30
L = 79.14 in.

16. A B C A B C D E
Standard Standard
Standard Pitch Lengths ,
Designation Designation Standard Pitch Lengths , Inches
Inches
From table (5.3) the standard length of the … …= 79.3 in
26 27.3 …. ……. belt …
31 32.3 …. …. 107.9 … …
with a designation number A78.
33 34.3 …. …. … …
35 36.3 ….. …
The exact centre distance from the …
38
42
39.3
43.3
equation: … …
…..
… …
46 47.3
b… b2 32(D … 2
d) …
48 49.3 C … … …
51 52.3 16 … … …
53 54.3 … … …
56.3 … …
61.3 b 4 L 6.28( D … d)
63.3 … …
65.3 … …
67.3 b 4x 79.3 6.28(9 3)
… …
69.3 …
b = 241.84 72.3
76.3
… …
…
…
…
2 …
79.3
…
…
241.84
…
241.84 …
… …
32(9 3) 2
81.3
….
C … … …
…. … 16 … … …
86.3 … … …
C = 30.08 in.
91.3 … … …
97.3 … … … … … … …

17. The angle of lap (arc of contact), from the equation:
Angle of lap =
( D d )60
180
C
Angle of lap = ( 9 3 )60
180
30.08
Angle of lap = 168.03o

18. Speed Ratio Small Speed Small Speed Small
Range Diameter Regular Quality Belts
Ratio Diameter Ratio Diameter
The capacity of one belt from the equation:
Factor Range Factor
Belt Cross Section Range Factor
1.341 - 1.429
Capacity of one belt =
1.000 - 1.019
FACTORS
1.020 - 1.032
1.00
1.01
1.110 - 1.142
BXS
1.05
A 1.143 - 1.178 0.91 YxS ZS 3D - 1.562 E
C
1.06
1.430
1.10
1.11
de 1.563 - 1.814
1.033 - 1.055 Values of X , 1.179 - 1.222 be Used in H.P. Formula
1.02 Y and Z to 1.07 1.12
1.815 - 2.948
1.056 - 1.081 1.03 1.223 - 1.274 1.08 1.13
2.949 - and
From table(5.4)
1.082 - 1.109 X 1.04 1.945 1.275 - 3.434
1.430 6.372 13.616 19.914 1.14
1.09
over
X =2.684 3.801 9.830 26.899 93.899 177.74
Y
Z 0.0136 0.0234 0.0416 0.0848 0.1222
Y =5.326
Z =0.0136 Premium Quality Belts
S = (3.142 x3 x 1500)/12000 = 1.18
Belt Cross Section
de FACTORS =A
= 3 x 1.14 3.42 B C D E
Values one ,belt = to684Used in91 5.326 x1.18 0.0136 x1.1833
Capacity of of X Y and Z 2. be x1.180. H.P. Formula
X 2.684 4.737 8.792 3.42
18.788 24.478
Y 5.326 13.962 38.819 137.70 263.04
Capacity
Z of one belt = 1.26 hp
0.0136 0.0234 0.0416 0.0848 0.1222

19. Find theStandard Belt Cross Section by one Contact fromCthedrive E
Arc of Contacttransmitted Standard of belt Belt Cross Section
power A Type ofCdrive Length
Length B
Arc
A B
Type of equation:
D
on V Designation on
V to VFactor to Flat V to V Factor to Flat
V
Power of one belt =belt …
Designation
Small sheaves
26
Correction
0.81 …
capacity x sheaves Correction … …
Small length coefficient x
… …
coefficient of arc of … …Factor
31
33 0.86 …
contact
0.84 Correction
…
Correction Factor …
…
… …
… …
Coefficient of arc of contact (from130
35
180
38
0.87
0.88
1.00 …
0.75 table (5.6)= 0.974.
0.86 0.86 …
Length coefficient0.95 … 0.77 (5.7)= 1.03.
170
42
160
46
0.90
0.92
0.98
(from table
… 0.8
120
110
…
…
0.82
0.78
… 0.82 …
0.78 …
… … 0.74 …
Power of one belt = 1.26 0.82
150
48
51
140
0.93 0.92
0.94 0.89
x 0.974 x901.03… 0.74
0.84
100
… 0.69 0.96 …
… … …
53 0.93
= 1.264 hp
… …
…
The required number of…belts can be obtained from the …
… …
equation: … …
No. of belts = Design power/ power…of one belt
… … …
… …
No. of belts = 3.6/ 1.264 … … …
…
= 2.85…… …
…
…
… …
Take 3 belts … … …
…
…
…
…
…
… … …
… … … … … … …

20. Thank You
Dr. Salah Gasim Ahmed
MET 103 20