Initial tension in belt

INITIAL TENSION IN BELT

• The tension of the belt when a belt is fitted to a pair of pulleys when the system is stationary, is termed as initial tension of the belt.

• When the driving pulley starts rotating, the pulley pulls the belt from one side and delivers the belt on the other side. Thus tension in one side increases from T0 to T1 and tension in other side decreases from T0 to T2.

Let,

T0 = Initial tension in the belt,

T1 = Tension in the tight side of the belt,

T2 = Tension in the slack side of the belt, and

α = Coefficient of elongation/contraction in length per unit force.

Then, Increase in tension on tight side = α (T1 – T0)

and Decrease in tension on slack side = α (T0 - T2)

Assuming belt material is perfectly elastic and total belt length remains same, we can write

Considering centrifugal tension, we get

Example 8.11 

An open belt running over two pulleys of 1.5 m and 1.0 m diameters connects two paralleĮ shafts 4.80 m apart. The initial tension in the belt when stationary is 3000 N. The smaller pulley is rotating at 600 rpm. The mass of belt is 0.6703 kg/m length. The coefficient of friction between the belt and pulleys is 0.3. Find: (i) the exact length of the belt required, and (ii) the power transmitted taking centrifugal tension into account.

[A.U., Apr/May 2011] 

Given data: Open belt drive; d1 = 1.5 m; d2 = 1 m; x = 4.8 m; T0 = 3000 N; N2 = 600 rpm; m = 0.6703 kg/m; μ = 0.3.

Solution: 

(i) Exact length of the belt required:

We know that exact length of belt for an open belt drive,

(ii) Power transmitted, taking centrifugal tension into account:

We know that TC = m v2 = 0.6703 (31.42)2 = 661.73 N

For an open belt drive, angle of contact,

Considering centrifugal tension, initial tension is given by

On solving equations (i) and (ii), we get

T2 = 1341.14 N and T1 = 3335.4 N

⸫ Power transmitted, P = (T1 - T2) v

Example 8.12 

An open belt running over two pulleys 240 mm and 600 mm diameter connects two parallel shafts 3 metres apart and transmits 4 kW from the smaller pulley that rotates at 300 rpm. μ = 0.3 and the safe working tension is 10 N/mm width. Determine: (i) minimum width of the belt, (ii) initial belt tension, and (iii) length of the belt required. 

[A.U., May/June 2007; Nov/Dec 2012]

Given data: 

d2 = 240 mm 0.24 m; d1 = 600 mm = 0.6 m; x = 3 m; P = 4 kW = 4000 W; N2 = 300 rpm; μ = 0.3; T1 = 10 N/mm width.

Solution:

(i) Minimum width of belt:

We know that for an open belt drive,

On solving equatfons (i) and (ii), we get

T1 = 1779 N and T2 = 718 N

Since the safe working tension is 10 N per mm width, therefore minimum width of the belt,

(ii) Initial belt tension:

We know that initial belt tension,

(iii) Length of the belt required:

We know that the length of the belt required,