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1
Belt Drives
Section VIII
2
Belts? & Types of Belts
Angle of Wrap
V-Belt
Belt Design
Talking Points
3
Belts are used to transmit power from one shaft to another where it is not necessary to maintain an exact speed ratio between the two shafts.
Two famous types of belts are: Flat Belts. V-Belts.
Belts? & Types of Belts
Multiple V-belt drive Flat belt drive
4
Power losses due to slip and creep amount is ranging from 3.0 to 5.0 percent for most belt drives.
In this course, it will be assumed that the shafts are parallel.
Belts? & Types of Belts
5
Angle of Wrap
The angles of wrap for an open belt may be determined by:
The angles of wrap for a crossed belt drive may be determined by:
c
rRsin21802180
c
rRsin21802180
c
rRsin
12
11
c
rRsin21802180
c
rRsin
121
V-Belt
6
Multiple V-Belts
V-Belt Configurations
V-Belt
7
Standard cross sections of V-belts. All belts have rubberinpregnated fabric jacket with interior tension cords above a rubber cushion.
a) Standard sizes A, B, C, D, and E
b) High-capacity sizes 3V, 5V, and 8V
8
Belt Design
It involves either the proper belt selection to transmit a required power or the determination of the power that may be transmitted by a given flat belt or by one V-belt. In the first case, the belt dimension is unknown, while in the second case, the belt dimension is known. The belt thickness is assumed for both cases.
The power transmitted by a belt drive is a function of the belt tension and belt speed.
hp ,
550
VTTPower or W V,TTPower 21
21
Where:
m/s speed,belt V
N side, loosein ion belt tens T
N side, in tight ion belt tens T
2
1
The following formula is for determining the stress, 2, for the flat belts applies when the thickness of the belt is given but the width is unknown.
fe
22
21
Vm
Vm
rad. pulley,on belt of wrapof angle
pulley &belt between friction oft coefficien
kg/m section,-crossin m 1.0belt of m 1.0 of mass m
Pa belt, theof sideslack in the stress Pa; stress, allowable maximum 32
21
f
Where:
9
Belt Design – Con.
Where:
The required cross-sectional area of the flat belt for the case of the width unknown may be determined by:
21
21
σσ
TT Area Required
beltflat afor 180 Belt -V thefor angle groove
kg/m density,belt m; thickness,belt t m; width,belt b
kg/m belt, of m 1.0 of masstb m3
2sin2
2
21
VmT
VmT f
e
The required flat belt width b is therefore: Thickness
Area b
The value of (T1 – T2) may be determined from the power requirement,
WV,TTPower 21
The maximum tension in the tight side of the belt depends on the allowable stress of the belt material. The allowable tensile stress for leather belting is usually 2.0 to 3.45 MPa, and the allowable stress for rubber belting will run from 1.0 to 1.7 MPa, depending on the quality of the material. Leather belting can be obtained in various single ply thicknesses. Double and triple ply belts are also available.
The following formula is for determining the value of 2, for both flat & V-belts applies when the width & thickness of the belt are known.
The quantity mV2 is due to centrifugal force, which tends to cause the belt to leave the pulley and reduce the power that may be transmitted.
10
An open belt drive delivers 15 kW when the motor
pulley, which is 300 mm in diameter, turns at 1750
rev/min. The belt is 10 mm thick and 150 mm wide
and has a density of 970 kg/m3. The driven pulley,
which is 1200 mm diameter, has an angle of contact
of 200°.What is the maximum stress in the belt
assuming a coefficient of friction 0.3 for both
pulleys?
EXAMPLE-1
11
EXAMPLE-1
12
d1
d2
Motor
Pulley-1: P = 15 kW
n = 1750 rpm
d1 = 300 mm
Pulley-2: d2 = 1200 mm
2 = 200o
Coefficient of friction, f = 0.3
Leather flat belt:
b
t = 200 kg/m3
b = 1500 mm
t = 10 mm
12
EXAMPLE-1
2sin
f
22
21 e
VmT
VmT
Here, we have the 2nd case, where the belt dimensions are known. So, we use the following
formula:
m = b t = 0.15 x 0.01 x 970 = 1.455 kg/m.
V = dn/60 = ( x 0.300 x 1750)/60 = 27.489 m/s. for flat-belt = 180.
102
180200 i.e. ;20021802
16010218021801
where
The pulley which governs the design is the one with the smaller
. Thus, the smaller pulley governs the design in this
problem since the smaller gives smaller
2sinf
e
2sinf
e
(1)
13
EXAMPLE-1
311.2464.1099T
464.1099T
489.72455.1T
489.72455.1T
2
1
2180sin
1801603.0
22
21
eSubstituting all in (1) gives:
(2)
Power: V
PowerTT V TTPower 2121 N 673.545
27.489
1015 3
673.545T T 12 (3)
Substituting Eq. (3) into (2) gives:
N 364.2061T
311.2137.1645T
464.1099T
311.2464.1099673.545T
464.1099T
1
1
1
1
1
Thus, the maximum stress in the belt is
10150
364.2061
A
T11 1.37 N/mm2 (MPa)
14
EXAMPLE-2
A V-belt drive transmits 11 KW at 900 rev/min of the
smaller sheave. The sheaves pitch diameters are 173 mm
and 346 mm. The center distance is 760 mm. If the
maximum permissible working force per belt is 560 N,
determine the number of belts required if the coefficient
of friction is 0.15 and the groove angle of the sheaves is
34°. The belt mass is 0.194 Kg/m.
15
EXAMPLE-2
12
d1
d2
Motor
V-Belt:
Mass, m = 0.194 kg/m
Groove angle, = 34
Max tension/Belt, T1= 560 NPulley-1 (driver):
Power, P = 11 kW
Speed, n = 900 rpm
Pitch diameter, d1= 173 mmPulley-2 (driven):
Pitch diameter, d2= 346 mmCenter of distance, C = 760 mm
Coefficient of friction, f = 0.16
16
EXAMPLE-2
Here, we can use the 2nd case. So, we use the following formula:
where, T1 = 560 N
m = 0.194 kg/m.
V = ( xd x N)/60 = ( x0.173 x 900)/60 = 8.152 m/s.
for V-belt = 34.
2sin
f
22
21 e
VmT
VmT
(1)
Angles of wrap of smaller and larger pulleys are respectively:
071.19376021732346sin2180
crRsin21802180
929.16676021732346sin2180
crRsin21802180
1
12
1
11
17
EXAMPLE-2
The pulley which governs the design is the one with the smaller
. Thus, the smaller pulley governs the design in this
problem since the smaller gives smaller
2sinf
e
2sinf
e
Substituting all in (1) gives:
458.4892.12T
892.12605
152.8194.0T
152.8194.0605
2
234sin
180929.16615.0
22
2
e
N 617.135T
473.57T458.4108.547
2
2
kW 46.3 W459.573
152.8617.351605 VTTBelt Power/ 21
Thus, the required numbers of belts: Belts 18.346.3
11
Belt Power/
Power TotalBn
Use 4 belts.
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