A band and block brake, having 14 clocks each of which subtends an angle 150 at the Center, is applied to a drum of 1m effective diameter. The drum and the flywheel mounted on the same shaft has a mass of 2000 kg and a combined radius of gyration of 500 mm. The two ends of the band are attached to the pins on opposite sides of the brake lever at distances of 30 mm and 120 mm from the fulcrum. If a force 200 N is applied at a distances of 750 mm from the fulcrum, find: A) Maximum braking torque. B) Angular retardation of of the drum, and C) time taken by the system to come to rest from the rated of 360 rpm.

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

Marks: 2016

Year: May 2016

Given:

• Type of brake – Band and Block type
• No. of blocks = 14
• Angle of subtend 15˚=θ=15˚
• D= 1 m
• The drum and the Flywheel –M = 200kg
• Radius of gyration 500mm
• N = 360˚

To find:

1. Max Breaking Torque
2. Angular retardation
3. Time required to come rest at 360 rpm

$n = 14 \\ θ=15˚ \\ D=1m \\ re = 0.5 \\ M = 2000kg$

$\therefore \text{mass moment of inertia} = I = m/C^2 \\ = 2000 ×0.5^2 \\ = 500 kg-m^2$

Distance OA = 30 mm = 0.03 m

Distance OB = 120 mm = 0.12 m

Distance OC = 750 mm = 0.75 m

$µ = 0.25 \rightarrow \text{assumed not given} \\ P = 200 N \\ N = 360 rpm \\ \dfrac{To}{Tn}= \dfrac{[1+ µ \tan⁡θ]^n}{[1- µ \tan⁡θ]} \\ Since,2θ=15˚ \\ θ = 7.5˚ \\ \dfrac{To}{Tn}= \dfrac{[1+ 0.25 \tan⁡(7.5)]{14}}{[1- 0.25 \tan⁡(7.5)]} \\ \dfrac{To}{Tn}=2051 \\ To = 2.51Tn \\ To – 2.51Tn = 0$

For equilibrium of the system, and taking moment about 0:

$∑ Mo= 0 = Tn × 120 – To × 30 – P × 750 \\ 30To – Tn × 120 = -200 × 750$

$\therefore, To = 8422 \\ Tn = 3355 \\ Tb = (To – Tn) re \\ = (8422 – 3355) 0.5 \\ Tb = 2533 N-m$

Calculating the Angular Retardation

$Tb = Iα \\ 2533 = 500 × α \\ a = 5.067 rad/s^2$

Calculating the time taken for system to come at rest at 360 rpm

$w = \dfrac{2πN}{60} = \dfrac{2π×360}{60} = 37.699 rad/sec$

Also, the flywheel come to rest therefore, wf = 0

We know that wf = wi – αt

= 37.699 – 5 × t

t = 37.699/5

t = 7.44 sec