A brake is a device by means of which artificial frictional resistance is applied to a moving machine member, in order to retard or stop the motion of a machine. The primary purpose of the brake is to slow down or completely stop the motion of a moving system, such as a rotating drum, machine or vehicle.

Brake may be classified as

i) Mechanical brakes

ii) Hydraulic brakes

iii) Pneumatic brakes

iv) Electrical brakes.

Brake capacity depends upon the following factors:

(i) The unit pressure between braking surfaces

(ii) The contacting area of braking surface

(iii) The radius of the brake drum

(iv) The coefficient of friction

(v) The ability of the brake to dissipate heat that is equivalent to the energy being absorbed.

Design of block or shoe brake

A simple block or shoe brake is shown in fig. It consists of a block or shoe which is pressed against the rim of a revolving brake wheel. The friction between the block and the wheel causes the tangential braking force to act on the wheel. The block is pressed against the wheel by a force applied to one end of the lever which is pivoted at the fulcrum. The block is rigidly fixed to the lever. The angle of contact between the block and the brake drum is usually small. When it is less than $45^°$, the intensity of pressure between the block and brake drum is uniform. It is called as short shoe brake. If angle is greater than $45^°$, it is called as long shoe brake or pivoted shoe brake.

General case

Considering the forces acting on the brake drum, Then

Let, $\mu$ -coefficient of friction

a,b -brake lever dimensions,

$\mathrm{F}$ - friction force,

$\mathrm{M}_{\mathrm{t}}$ - braking torque,

$\mathrm{P}$ - force applied on the lever,

$\mathrm{D}$ - diameter of the brake drum,

$\mathrm{N}$ - normal reaction between drum and shoe,

Then, Braking torque acting on drum is $M_{t}=\mu N R=\mu N D / 2$

Taking moment of forces acting on the lever about the hinge point.

$$ P \times a-N \times b+F \times e^{\prime}=0 $$

Where $e^{\prime}$ is the distance above fulcrum of lever

But $$\mathrm{F}=\mu N \ Then, P \times a-N \times b+\mu N \times e^{\prime}=0$$

$$P \times a-N\left(b-\mu e^{\prime}\right)=0$$

$$\therefore P=\frac{N\left(b-\mu e^{\prime}\right)}{a}$$

Thus, if $b\gt \mu e^{\prime},$ the friction force helps to reduce the magnitude of the actuating force P and a brake is called a partially self-energizing brake. It is desirable condition.

If $\mathrm{b}=\mu e^{\prime}, \mathrm{P}=0$ and the frictional force is great enough to apply the brake with no external force, then the brake is said to be self-locking brake. It is not a desirable condition.

Case I

When the line of action of tangential braking force passes through the fulcrum of the lever,

Then,

$\mathrm{F}=\mu(\mathrm{a} / \mathrm{b}) \mathrm{P}$

$\mathrm{Mt}=\mathrm{M}_{\mathrm{t}}=\mathrm{FD} / 2=\mu \mathrm{a} \mathrm{DP} / 2 \mathrm{b}$

Case II

When the line of action of tangential braking force passes at a distance $e'$ below fulcrum of lever,

$\mathrm{F}=\mathrm{P} \mathrm{a} / (\mathrm{b} / \mu-\mathrm{e}^{\prime} ) \quad$ For anticlockwise rotation of drum

$\mathrm{F}=\mathrm{P} \mathrm{a} /\left(\mathrm{b} / \mu+\mathrm{e}^{\prime}\right) \quad$ For clockwise drum rotation

$\mathrm{Mt}=\mathrm{FD} / 2$

Case III

When the line of action of tangential braking force passes at a distance $e'$ above fulcrum of lever,

$\mathrm{F}=\mathrm{P} \mathrm{a} / (\mathrm{b} / \mu+\mathrm{e}^{\prime} )\quad$ For anticlockwise rotation of drum

$\mathrm{F}=\mathrm{P} \mathrm{a} /\left(\mathrm{b} / \mu-\mathrm{e}^{\prime}\right) \quad$ For clockwise drum rotation

$\mathrm{Mt}=\mathrm{FD} / 2$

Case IV

For a normal pressure distribution as shown in fig.

$p=p_{max} \cos \alpha$

$\mathrm{L}=\frac{4 r \sin (\phi / 2)}{(\phi+\sin \phi)}$

$\mathrm{N}=\frac{p_{max } W r}{2}(\phi+\sin \phi)$

$M_{t}=\mu p_{\max } W r^{2} \int_{-\emptyset / 2}^{\emptyset / 2} \cos \alpha \cdot d \alpha=\mu N L$

Where,

$\mathrm{p}$ - Normal pressure on the shoe at any angle $\alpha, \mathrm{kgf} / \mathrm{cm}^{2}$

$\mathrm{P}_{max}$ - Maximum normal pressure on the shoe, $\mathrm{kgf} / \mathrm{cm}^{2}$

$\mathrm{W}$ - Face width of the brake drum, $\mathrm{cm}$

$\mathrm{L}$ - Distance from the drum centre to the line of action of force, cm

$\mathrm{r}$ - radius of the brake, $\mathrm{cm}$

$\mathrm{M}_{t}$ - braking torque, $\mathrm{kgf} \ \mathrm{cm}$