# Transmission of Power through Belts

Let T1 = Tension in the tight side of the belt

T2 = Tension in the slack side of the belt

V = Velocity of the belt in meter/s.

The effective tension or force acting at the circumference of the driven pulley is the difference between the two tensions (i.e., T1 – T2).

Problem 3.30. A belt is running over a pulley of diameter 120 cm at 200 r.p.m. The angle of contact is 1650 and co-efficient of friction between the belt and pulley is 0.3. If the maximum tension in the belt is 3000 N. find the power transmitted by the belt.

Sol. Given:

Dia. of pulley, d = 120 cm = 1.2 m

Speed of pulley, N = 200 r.p,m.

Example 3.31. An open-belt drive connects two pulleys 120 cm and 50 cm diameters, on parallel shafts 4 m apart. The maximum tension in the belt is 1855.3 N. The co-efficient of friction is 0.3. The driver pulley of diameter 120 cm runs at 200 r.p.m. Calculate:

(i)                           the power transmitted, and

(ii)                        torque on each of the two shafts.

Sol. Given:

Dia. of larger pulley, d1 = 120 cm = 1.20 m

3.7.1 Centrifugal Tension. The tension caused in the running belt by the centrifugal force is known as centrifugal tension. Whenever a particle of mass m is rotated in a circular path of radius r at a uniform velocity v, a centrifugal force is acting radially outward and its magnitude is equal to  where m is the mass of the particle.

The centrifugal tension in the belt can be calculated by considering the forces acting on an elemental length of the belt (i.e., length MN) subtending an angle δθ at the centre as shown in Fig.3.40.

Let v = Velocity of belt in metre/s.

r = Radius of pulley over which belt runs

M = Mass of elemental length of belt

m = Mass of belt per metre length

Tc = Centrifugal tension acting at M and N tangentially

R = Centrifugal force acting radially outward

The centrifugal force R acting radially outwards is balanced by the components of Tc acting radially inwards. Now elemental length of belt MN = r . δθ

Note.

(i)         From the above equation, it is clear that the centrifugal tension is independent of T1 and T2. It depends upon the velocity of the belt. For lower belt speed (i.e., belt speed less than 10 m/s) the centrifugal tension is very small and may be neglected.

(ii)When centrifugal tension is to be taken into consideration then total tensions on right side and slack side of the belt is given as:

For tight side = T1+ Tc                                                                  … (3.19)

For slack side = T2+ Tc                                                                  … (3.20)

(iii)    Maximum tension ™ in the belt is equal to maximum safe stress in the belt multiplied by cross-sectional are of the belt.

Tm = f × (b × t)                                                                                 … (3.21)

where f = Maximum safe stress in the belt.

b = Width of belt, and

t = Thickness of belt.

Then     Tm = T1 + Tc … if centrifugal tension is to be considered

= T1 … if centrifugal tension is to be neglected

3.7.2 Maximum Power Transmitted by a Belt

Let T1 = Tension on right side,

T2 = Tension on slack side,

v = Linear velocity of belt.

Then the power transmitted is given by equation (3.15) as

P = (T1 × T2) × v                                                                    … (i)

But from equation (3.9), we know that

Problem 3.33. A belt of density 1 gm/cm3 has a maximum permissible stress of 250 N/cm2. Determine the maximum power that can be transmitted by a belt of 20 cm × 1.2 cm if the ratio of the tight side to slack side tension is 2.

Sol. Given:

Note. In this question, speed of belt is not given. Hence the maximum power transmission, speed of the belt is given by equation (3.22). But in question (3.22), the speed of the belt is given and hence there is no need of calculating velocity of belt using equation (3.22). In that question, maximum power transmitted at the given speed is to be calculated.

Problem 3.34. A leather belt is required to transmit 9 kW, from a pully 120 cm in diameter running at 200 r.p.m. The angle embraced is 1650 and the co-efficient of friction between leather belt and pulley is 0.3. If the safe working stress for the leather belt is 140 N/cm2, the density of leather is 1 gm/cm3 and the thickness of the belt is 10 mm. determine the width of the belt taking the centrifugal tension into account.

(AMIE Summer, 1976)

Sol. Given:

Problem 3.35. Determine the maximum power that can be transmitted using a belt of 100 mm × 10 mm with an angle of lap of 1600. The density of belt is 10-3 and co-efficient of friction may be taken as 0.25. The tension the belt should not exceed 1.5 N/mm2.

(AMIE Summer, 1984)

Sol. Given:

Width of belt, b = 100 mm

Thickness of belt, t = 10 mm

3.7.3 Initial Tension in the belt. This tension in the belt which is passing over the two pulleys (i.e., driver and follower) when the pulleys are stationery is known as initial tension in the belt.

When power is transmitted from one shaft to another shaft with the help of the belt, passing over the two pulleys which are keyed to the driver and driven shafts, there should be firm grip between the pulleys and belt. When the pulleys are stationery, this firm grip is increased, by tightening the two ends of the belt. Hence the belt is subjected to some tension. This tension is known as initial tension in the belt.

Problem 3.36. An open belt running over two pulleys 24 cm and 60 cm diameters connects two parallel shafts 3 m apart and transmits 3.75 kW from the smaller pulley that rotates at 300 r.p.m. co-efficient of friction between the belt and the pulleys is 0.3 and the safe working tension in 100 N/cm width. Determine:

(i)                           minimum width of the belt,

(ii)                        initial belt tension, and

(iii)                      length of the belt required.

(AMIE Winter, 1978)

Sol. Given

Problem 3.37. An open belt running over two pulleys 1.5 m and 1.0 m diameters connects two parallel shafts 4.80 m apart. The initial tension in the belt when stationery is 3000 N. If the smaller pulley is rotating at 600 r.p.m. and co-efficient of friction between the belt and pulley is 0.3, determine the power transmitted taking centrifugal tension into account. The mass of belt is given as 0.6703 kg/m length.

Problem 3.38. An open belt connects two flat pulleys, the smaller pulley being of 400 mm in diameter. The angle of lap on the smaller pulley is 1600 and co-efficient of friction between belt and pulley is 0.25.

Which of the following alternatives would be more effective in increasing the power that could be transmitted:

(i) increasing the initial tension by 10%.

(ii) increasing the co-efficient of friction by 10% by the application of suitable dressing to the belt?

(AMIE Winter, 1981)

Sol. Given: