Engineering Mechanics

Chapter 33 : Transmission of Power by Belts and Ropes „„„„„ 691

Let T 1 = Tension in the tight side of the belt, and

T 2 = Tension in the slack side of the belt.

We know that velocity of the belt,

v=^2

1400

20·94 m/s

60 60

πdN π× ×

==

and initial tension in the belt when stationary (T 0 )

3=^12

2

TT+

or T 1 + T 2 = 6 kN ...(i)

Now for an open belt drive,

sin α=^12

• 0·75 – 0·5
  0·052
  4·8

rr

l

==

or α=3°

∴ Angle of lap for the smaller pulley,

θ= 180° – 2α = 180° – (2 × 3) = 174° =

174

3·04 rad

180

π× =

We also know that

1

2

2·3 log

T

T

⎛⎞

⎜⎟

⎝⎠

=μθ =0·3×3·04=0·912

∴

1

2

log

T

T

⎛⎞

⎜⎟

⎝⎠

=

0·912

0·3965

2·3

=

or

1

2

T

T = 2·49 or T^1 = 2·49 T^2 ...(Taking antilog of 0·3965)

Substituting the value of T 1 in equation (i),

2·49 T 2 + T 2 =6

∴ T 2 =

6

1·72 kN

3·49

=

and T 1 = 2·49 T 2 = 2·49 × 1·72 = 4·28 kN

∴ Power transmitted by the belt,

P=(T 1 – T 2 ) v = (4·28 – 1·72) 20·94 kN-m/s = 53·6 kW Ans.

EXERCISE 33.2

1. Two pulleys of diameters 500 mm and 300 mm connected by a belt have tensions of 3 kN and
   2.5 kN in the two sides of the belt connecting them. If the power transmitted is 25 kW, find the
   speed of the belt. [Ans. 50 m/s]


2. A pulley is driven by a flat belt running at a speed of 600 m/min. The coefficient of friction
   between the pulley and the belt is 0·3 and the angle of lap is 160°. If the maximum tension in the
   belt is 700 N, find the power transmitted by the belt. [Ans. 3·97 kW]


3. An open belt drive connects two pulleys of 1·2 m and 0·5 diameter on parallel shafts 3·6 m
   apart. The belt has a mass of 0·9 kg/m length and the maximum tension in it is not to exceed 2
   kN. The larger pulley runs at 200 rev/min. Calculate the torque on each of the two shafts and
   the power transmitted μ = 0·3. [Ans. 654·4 N-m ; 272·7 N-m; 13·74 kW]