(^686) „„„„„ A Textbook of Engineering Mechanics
Solution. Given: Thickness of belt (t) = 8 mm = 0·008 m; Width of belt (b) = 150
mm = 0·15 m; Diameter of pulley (d) = 1·2 m; Speed of the pulley (N) = 180 r.p.m; Angle of lap
(θ) = 190° =^190 3·316 rad
180
π
×= ; Mass density of belt material (ρ) = 1000 kg/m^3 ; Permissible
stress in the belt (σ) = 1·5 N/mm^2 = 1.5 × 10^6 N/m^2 and coefficient of friction (μ) = 0·3
(i) Power transmitted when the centrifugal tension is considered
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 speed of the belt,
v=
1.2 180
11·31 m/s
60 60
ππ××dN

∴ Maximum tension in the belt,
T=σbt = (1·5 × 10^6 ) × 0.15 × 0.008 = 1800 N
and mass of the belt per metre length,
m= Area × Length × Density
= (0·008 × 0·15) × 1 × 1000 = 1·2 kg
∴ Centrifugal tension, TC=mv^2 = 1·2 × (11·31)^2 = 153·5 N
and tension in the tight side T 1 =T – TC = 1800 – 153·5 = 1646·5 N
We also know that
1
2
2·3 log
T
T
⎛⎞
⎜⎟⎜⎟
⎝⎠
=μθ = 0·3 × 3·316 = 0·9948
1
2
0·9948
log 0·4325
2·3
T
T
⎛⎞
⎜⎟⎜⎟==
⎝⎠
∴
2
1646·5
2·707
T
= ...(Taking antilog of 0·4325)
or T 2 =
1646·5
608·2 N
2·707

and power transmitted, P=(T 1 – T 2 )v = (1646·5 – 608·2) 11·31 N-m/s
= 11 740 W = 11·74 kW Ans.
(ii) Power transmitted when the centrifugal tension is neglected
We know that tension in the tight side (without centrifugal tension),
T 1 = 1800 N
∴ T 2 =
1800
665 N
2·707

and power transmitted, P=(T 1 – T 2 )v = (1800 – 665) 11·31 N-m/s
= 12 840 W = 12·84 kW Ans.
Example 33.11. A flat belt is required to transmit 35 kW from a pulley of 1·5 m effective
diameter running at 300 r.p.m. The angle of contact is spread over
11
24
of the circumference and
coefficient of friction between the belt and pulley surface is 0·3. Taking centrifugal tension into
account, determine the width of the belt. Take belt thickness as 9·5 mm, density as 1·1 Mg/m^3 and
permissible stress as 2·5 N/mm^2.