1. Section VIII
Belt Drives

2. Talking Points Belts? & Types of Belts Angle of Wrap V-Belt
Belt Design

3. Belts? & Types of Belts
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.
Multiple V-belt drive
Flat belt drive

4. Belts? & Types of Belts
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.

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:

6. V-Belt Configurations
Multiple V-Belts
V-Belt Configurations

7. V-Belt
a) Standard sizes A, B, C, D, and E
b) High-capacity sizes 3V, 5V, and 8V
Standard cross sections of V-belts. All belts have rubberinpregnated fabric jacket with interior tension cords above a rubber cushion.

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.
Where:
The following formula is for determining the stress, s2, for the flat belts applies when the thickness of the belt is given but the width is unknown.
Where:

9. Belt Design – Con.
The required cross-sectional area of the flat belt for the case of the width unknown may be determined by:
The required flat belt width b is therefore:
The value of (T1 – T2) may be determined from the power requirement,
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 T2, 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.
Where:

10. EXAMPLE-1
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?

11. Coefficient of friction, f = 0.3
EXAMPLE-1 a1 a2 b d1 d2
Motor
P = 15 kW
n = 1750 rpm
d1 = 300 mm
Pulley-1:
d2 = 1200 mm
a 2 = 200o
Pulley-2:
Coefficient of friction, f = 0.3 b
r = 200 kg/m3 t
Leather flat belt:
b = 1500 mm
t = 10 mm

12. EXAMPLE-1
formula:
Here, we have the 2nd case, where the belt dimensions are known. So, we use the following
(1)
m = b t r = 0.15 x 0.01 x 970 = kg/m.
V = dn/60 = ( x x 1750)/60 = m/s.
q for flat-belt = 180.
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 a gives smaller

13. Substituting all in (1) gives:
EXAMPLE-1
Substituting all in (1) gives:
(2)
Power:
(3)
Substituting Eq. (3) into (2) gives:
Thus, the maximum stress in the belt is
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 Kg/m.

15. EXAMPLE-2 V-Belt: Mass, m = 0.194 kg/m Groove angle, q = 34 d2
Max tension/Belt, T1= 560 N a1 a2 b d1 d2
Motor
Pulley-1 (driver):
Power, P = 11 kW
Speed, n = 900 rpm
Pitch diameter, d1= 173 mm
Pulley-2 (driven):
Pitch diameter, d2= 346 mm
Center of distance, C = 760 mm
Coefficient of friction, f = 0.16

16. Here, we can use the 2nd case. So, we use the following formula:
EXAMPLE-2
Here, we can use the 2nd case. So, we use the following formula:
(1)
where, T1 = 560 N
m = kg/m.
V = (p x d x N)/60 = (p x x 900)/60 = m/s.
q for V-belt = 34.
Angles of wrap of smaller and larger pulleys are respectively:

17. Substituting all in (1) gives: Thus, the required numbers of belts:
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 a gives smaller
Substituting all in (1) gives:
Thus, the required numbers of belts:
Use 4 belts.