A differential band brake, as shown in fig. has an angle of contact of 2250. The band has a compressed woven lining and bears against a cast iron drum of 350 mm diameter. The brake is to sustain a torque 350 N-m and the coefficient of friction between the band and drum is 0.30. Find: - a) The necessary force (F) for the clockwise and anticlockwise rotation of the drum; and - (b) the value the value of ‘OA’ for the brake to be self locking, when the drum rotates clockwise. -

Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II

Marks: 12 Marks

Year: Dec 2015

$T_B = 350 N-m \\ D = 0.35m \\ r = D/2 = [0.35/2] \\ µ = 0.30$

To find:

1. Force P

2. Distance OA for self locking

Calculating the tension in the band

$θ= 22.5˚ \rightarrow 3.926 rad \\ \dfrac{T_1}{T_2}= e^{uθ} \\ \dfrac{T_1}{T_2}=e^{0.30 \times 3.9269} \\ \dfrac{T_1}{T_2} = 3.9527 \\ T_1 = 3.9527 \times T_2 \\ T_1 – 3.9527T2 = 0$

Also braking torque

$T_B = (T1 – T2) \times r \\ 350 = (T1 – T2) \times 0.35/2 \\ 700/0.35 = (T1 – T2) \\ T1 – T2 = 2000$

$3.9527 \times T2 – T2 = 2000 \\ 2.9527T2 = 2000 \\ T2 = 2000/2.9527 \\ T2 = 677.346 N$

$T1 = 3.9527 \times 677.346 \\ T1 = 2677.34 N$

Calculation of applied Force for clock wise rotation

Case 1:

For Clockwise Rotation

Considering the equilibrium position of the level as shown in free body diagram:

$∑Mo = 0 = -T2 \times 150 + T1 \times 35 + P \times 500 \\ 0 = 677.34 \times 150 + 2677.34 \times 35 + P \times 500 \\ = - 101601 + 93707 + P \times 500 \\ P = 15.788 N$

Case 2:

For rotating anticlockwise direction

Considering the equilibrium of the system and taking momentum at fulcrum

0 = -T2 × OA + T1 × 150 + P × 550

Case 2: anticlockwise rotation

$= - T2 \times OA + T1 \times OB \\ = -T2 \times OA + 2677.34 \times 35$

Since, for locking condition

$P = 0 \\ 0 = -677.346 \times OA + 2677.34 \times 35 \\ \therefore, OA = 2677.35 \times 35/677.346 \\ OA = 138.344 mm$