Calculate tension in limbs, Mechanical Engineering

Calculate tension in limbs:

Q : A horizontal drum of belt drive carries the belt over semicircle around it. It is rotated counter clockwise to transmit a torque of 300N-m. If coefficient of friction between the belt and rope is 0.3, calculate tension in limbs 1 and 2 of belt shown in figure, and reaction on bearing. The drum is having mass of 20Kg and the belt is assumed to be mass less.


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Sol: Given: Torque(t) = 300N-m

Coff. of friction(µ) = 0.3

Diameter of Drum (D) = 1m, R = 0.5m

Mass of drum(m) = 20Kg.

Since angle of contact = p rad Torque = (T1  - T2).R

300 = (T1  - T2) X 0.5

T1- T2  = 600N                                                                                                                              ...(i)

And,     T1/T2  = eµθ

T1/T2  = e(0.3)p

T= 2.566T2                                                                                                                               ...(ii)

Solve (i) and (ii)

We get,

T1  = 983.14N                                                                .......ANS

T2  = 383.14N   .......ANS

Now reaction on bearing is opposite to mass of body, and it is equal to

R = T1  + T2  + mg

R = 983.14 + 383.14 + 20 X 9.81

R = 1562.484N                                                            .......ANS

Posted Date: 10/19/2012 6:36:04 AM | Location : United States

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