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A leather is required to transmit 7.5 kw from a pulley 1.2 m in diameter, running at 250 rpm.

The angle embraced is 165 deg and the coefficient of friction between the belt and the pulley is 0.3. If the safe working stress for the leather belt is 1.5 MPa, density of leather 1Mg/m^3 and thickness of belt 10 mm, determine the width of the belt taking centrifugal tension into account.


Subject: Kinematics of Machinery

Topic: Belts, Chains and Brakes

Difficulty: Medium

1 Answer
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P= 7.5 kw =7500 w; d =1.2 m; N =250rpm. ;θ =165 =165 ×(π/180) =2.88 rad; μ =0.3 ; σ =1.5 MPa = 1.5×106 N/m2 ; ρ = 1 Mg//m3 =1 ×10/6 g//m3 =1000 kg//m3 ;t = 10 mm =0.01 m.

​Let b =width of the belt in meters, T1 =Tension in the tight side of the belt in N, and T2 =Tension in the slack side of the belt in N. We know that velocity of the belt, ν=πdN60=π×1.2×25060 = 15.71 m/s. and power transmitted (P), 7500 =(T_{1} –T_{2})ν = (T_{1} –T_{2}) 15.71 T_{1} –T_{2} = 7500/15.71 = 477.4 N. …(i) We know that 2.3log(\frac {T_{1}}{T_{2}}) =\mu\theta =0.3×2.88 =0.864 log(\frac {T_{1}}{T_{2}})=0.864/2.3 or \frac {T_{1}}{T_{2}} =2.375 ….(ii) …(Taking antilog of 0.3756) From equation (i) and (ii) T_{1} =824.6 N T_{2} =347.2 N. We know that mass of the belt per meter length, m = area × length × density =b.t.l.p = b×0.01×1×1000 =10b kg Centrifugal tension, T_{C} = mν^{2} =10b(15.71)^{2} =2468b N And maximum tension in the belt, T =\sigmab.t =1.5×10^{6} × b ×0.01 =15000b N We know that T =T_{1} +T_{C} or 15000b =824.6 +2468b

15000b -2468b =824.6 or 12532b =824.6

b =824.6/12532 =0.0658 m = 65.8 mm.

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