Engineering Mechanics

Chapter 33 : Transmission of Power by Belts and Ropes „„„„„ 687

Solution. Given: Power to be transmitted (P) = 35 kW; Effective diameter of pulley

(d) = 1·5 m; Speed of pulley (N) = 300 r.p.m; Angle of contact

11

() 2 2·88

24

θ= π× =

rad ; Coefficient of friction (μ) = 0·3; Thickness of belt (t) = 9·5 mm = 0·0095 m; Mass density of the
belt material (ρ) = 1·1 Mg/m^3 = 1100 kg/m^3 and permissible stress (σ) = 2·5 N/mm^2 = 2.5 × 10^6 N/m^2

Let b= Width of the belt,

T 1 = Tension on the tight side of the belt, and

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

We know that velocity of the belt,

v=

1.5 300

23·56 m/s

60 60

ππ××dN

==

and power transmitted (P), 35 = (T 1 – T 2 )v = (T 1 – T 2 ) 23·56

∴ (T 1 – T 2 )=

35

1·486 kN = 1486 N

23·56

= ...(i)

We also know that

1

2

2·3 log

T

T

⎛⎞

⎜⎟⎜⎟

⎝⎠

=μ θ = 0·3 × 2·88 = 0·864

1

2

log

T

T

⎛⎞

⎜⎟⎜⎟

⎝⎠

=

0·864

0·3757

2·3

=

∴

1

2

T

T = 2·375 ...(Taking antilog of 0·3757)

or T 1 = 2·375 T 2

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

2·375 T 2 – T 2 = 1486

∴ T 2 =

1486

1081 N

1·3 7 5

=

and T 1 = 2·375 × 1081 = 2567 N

We know that maximum tension in the belt,

T=σbt = (2·5 × 10^6 ) × b × 0.0095 = 23750 b N

and mass of the belt per metre length,

m= Area × Length × Density

=(b × 0·0095) × 1 × 1100 = 10.45 b kg

∴ Centrifugal tension, TC=mv^2 = 10.45 b × (23·56)^2 = 5800 b N

and tension on the tight side of the belt (T 1 )

2567 =T – TC = 23750 b – 5800 b = 17950 b

∴ b=

2567

17950

= 0.143 m = 143 mm say 150 mm Ans.