A single block brake having asbestos lining without brass wires subjected to a torque capacity of 250 N-m is shown in fig. The brake drum rotates at 100 rpm.

Calculate

(i) actuating force and the hinge-pin reaction for clockwise rotation of the drum;

(ii) rate of heat generated during the braking action

(iii) dimensions of the block

(iv) check whether the brake is self-energizing

Given Data:

Brake Drum radius $R=200m$

Torque $Mt=250Nm$

Speed-100rpm

Assumptions - short shoe brake $(2 \theta=45$, projected area $A=b(2 R \sin \theta),$ $\theta=45 / 2, \mu=0.35$ and $p=1 M P a$ for low intensity brake having asbestos lining without brass wires [PSG 7.97]

Solution:

1) Pin reactions for clockwise rotation of drum

Braking torque, $Mt =\mathrm{FD} / 2,$ but $\mathrm{F}=\mu \mathrm{N}, \therefore 250=0.35 \mathrm{N} \times 0.2, \mathrm{N}=3.571 \mathrm{kN}$

Taking moment at pin,

$\mu \mathrm{N}(50)+\mathrm{P}(500)-\mathrm{N}(200)=0$

$0.35(3.5)(50)+\mathrm{P}(500)-3.57(200)=0$

$\mathrm{P}=1.31 \mathrm{kN}$

The reactions are

Horizontal reaction at pin, $\mathrm{H}=\mu \mathrm{N}=0.35(3.57)=1.250 \mathrm{kN}$ ,

Vertical reaction at pin, $\mathrm{V}=\mathrm{N}-\mathrm{P}=3.57-1.31=2.26 \mathrm{kN}$

Resultant reaction at pin, $\mathrm{R}=2.6 \mathrm{kN}$

2) Heat generation rate (for maximum velocity)

Rate of heat generated = frictional force x velocity $= 1.25 \times (\omega \mathrm{R})=1.25(2 \pi \times 100 / 60) \times 0.2=2.61 \mathrm{kW}$

3) Block dimensions

Let b = width of block, l = length of block, using

$1=2 \mathrm{R} \sin \theta=2 \times 200 \times \sin 22.5=153 \mathrm{mm}$

Pressure = normal reaction$/ b\times 2 \mathrm{R} \sin \theta$

$1 M P a=3570 \mathrm{N} / \mathrm{b} \times 153$

Width of shoe-b $=23 \mathrm{mm}$

4) As $200 \mathrm{mm} \gt 0.35 \times 50\left(\mu \mathrm{e}^{\prime}\right),$ the brake is self energizing