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

P= 7.5 kw =7500 w; d =1.2 m; N =250rpm. ;$\theta$ =165$^{\circ}$ =165 ×($\pi$/180) =2.88 rad; $\mu$ =0.3 ; $\sigma$ =1.5 M$P_{a}$ = 1.5×10$^{6}$ N/$m^{2}$ ; $\rho$ = 1 Mg//$m^{3}$ =1 ×10/$^{6}$ g//$m^{3}$ =1000 kg//$m^{3}$ ;t = 10 mm =0.01 m.

​Let $b$ =width of the belt in meters, $T_{1} $ =Tension in the tight side of the belt in N, and $T_{2} $ =Tension in the slack side of the belt in N. We know that velocity of the belt, $ν =\frac{\pi dN}{60}=\frac{\pi\times 1.2 \times 250}{60}$ = 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 =$\sigma$b.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.