Chapter 33 : Transmission of Power by Belts and Ropes „„„„„ 689

Example 33.12. A belt 100 mm × 10 mm thick is transmitting power at 1200 m/min. The net
driving tension is 1·8 times the tension on the slack side. If the safe stress on the belt section is 1·8 N/
mm^2 , calculate the power that can be transmitted at this speed. Assume mass density of the leather as
1 t/m^3.

Also calculate the absolute maximum power that can be transmitted by this belt and the
speed at which this can be transmitted.

Solution. Given: Width of belt (b) = 100 mm ; = 0.1 m ; Thickness of belt (t) = 10 mm;

= 0.01 m ; Velocity of belt (v) = 1200 m/min = 20 m/s; Net driving tension (T 1 – T 2 ) = 1·8 T 2 (where

T 2 is tension in slack side); Safe stress (σ) = 1·8 N/mm^2 = 1.8 × 10^6 N/m^2 and mass density of

leather (ρ) = 1 t/m^3 = 1000 kg/m^3

Power transmitted by the belt

We know that maximum tension in the belt,

T=σ bt = (1·8 × 10^6 ) × 0.1 × 0.01 = 1800 N

and mass of the belt per metre length,

m= Area × Length × Density

= (0·1 × 0·01) × 1 × 1000 = 1 kg

∴ Centrifugal tension, TC=mv^2 = 1 × 20^2 = 400 N

and tension in the tight side, T 1 =T – TC = 1800 – 400 = 1400 N

Now 1400 – T 2 = 1·8 T 2 ...(Given T 1 – T 2 = 1·8 T 2 )

T 2 =

(^11400) 500 N
2·8 2·8
T

and power transmitted by the belt,
P=(T 1 – T 2 ) v = (1400 – 500) 20 N-m/s
= 18 000 W = 18 kW Ans.
Speed at which absolute maximum power can be transmitted
We know that speed of the belt, at which maximum power can be transmitted,
v=
1800
24·5 m/s
331
T
m

×
Ans.
Absolute maximum power that can be transmitted by the belt
We know that for maximum power, the centrifugal tension,
TC=
1800
600 N
33
T

∴ Tension on the tight side, of the belt,
T 1 =T – TC = 1800 – 600 = 1200 N
and tension on the slack side of the belt,
T 2 =^1
1200
428·6 N
2·8 2·8
T

∴ Power transmitted by the belt,
P=(T 1 – T 2 ) v = (1200 – 428·6) 24·5 N-m/s
= 18 900 W = 18·9 kW Ans.