A rope drum of an elevator having 650 mm diameter is fitted with a brake drum of 1 m diameter. The brake drum is provided with four cast iron brake shoes each subtending an angle of $45°$. The mass of the elevator when loaded is 2000 kg and moves with a speed of 2.5 m/s. The brake has a sufficient capacity to stop the elevator in 2.75 meters.

Find

(i) Width of the shoe

(ii) Heat generated in stopping the elevator.

Given Data:

Rope drum diameter $D_r =650mm$

Brake Drum diameter $D=1000mm$

Brake Drum radius $R=500mm$

Mass $m=2000kg$

velocity $v=2.5mps$

Height $h=2.75m$

CI short shoe brake

Solution:

Let b = Width of the shoe in mm

Find out the acceleration of the rope (a).

Using equations of motion under gravity

$\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{ah}$ or $(2.5)^{2}-0=2 \mathrm{a} \times 2.75=5.5 \mathrm{a}$

$\therefore \mathrm{a}=(2.5)^{2} / 5.5=1.136 \mathrm{m} / \mathrm{s}^{2}$

and accelerating force = Mass $\times$ Acceleration $=\mathrm{m} \times \mathrm{a}=2000 \times 1.136=2272 \mathrm{N}$

$\therefore$ Total load acting on the rope while moving,

$\mathrm{W}=$ Load on the elevator in newtons +Accelerating force

$=2000 \times 9.81+2272=21892 \mathrm{N}$

Torque acting on the shaft,

$\mathrm{M}_{\mathrm{t}}=\mathrm{W} \times \mathrm{D}_{\mathrm{r}} / 2=21892 \times 0.325=7115 \mathrm{N}-\mathrm{m}$

Tangential force acting on the drum, $\mathrm{F}=\mathrm{T} / \mathrm{R}=7115 / 0.5=14230 \mathrm{N}$

The brake drum is provided with four cast iron shoes, therefore tangential force acting on each shoe,

$\mathrm{F}=14230 / 4=3557.5 \mathrm{N}$

Normal reaction on each shoe, $\mathrm{N}=\mathrm{F} / \mu=3557.5 / 0.35=10164.28 \mathrm{N}$

Assuming hydraulically compressed asbestos with brass wires, $\mu=0.35$ and bearing pressure on the shoe $p =0.2 \mathrm{MPa}$ for lowering brake [PSG 7.97]

The projected bearing area of each shoe,

$A=b(2 r \sin \theta)=b\left(2 \times 500 \sin \frac{45^{\circ}}{2}\right)=382.7 b m m^{2}$

bearing pressure on the shoe, $\mathrm{p}=\mathrm{N} / \mathrm{A}$

Therefore, $0.2=10164.28 / 382.7 b, \mathrm{b}=133.3 \mathrm{mm}$

Heat generated in stopping the elevator,

= Total energy absorbed by the brake

= Kinetic energy + Potential energy $=0.5 \mathrm{mv}^{2}+\mathrm{m.g.h}$

$=0.5 \times 2 \times 2000(2.5)^{2}+2000 \times 9.81 \times 2.75=60205 \mathrm{N}-\mathrm{m}$

$=60.205 \mathrm{kN}-\mathrm{m}=60.205 \mathrm{kJ}$