Braking Torque Required to stop the vehicle

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. In the process of performing this function, the brake absorbs either the kinetic energy of the moving member or potential energy given up by objects being lowered by hoists, elevators, etc. In the previous articles, we discussed how we can calculate the energy absorbed by a brake and Heat Dissipation during Braking. Let us discuss how to calculate the Braking Torque Required to stop the vehicle.

As we mentioned above that the brake is used to retard or stop the motion of a machine. This action is called Braking. In the process of braking, the brake absorbs either the kinetic energy of the moving member or potential energy given up by objects being lowered by hoists, elevators etc. The energy absorbed by brakes is dissipated in the form of heat. This heat is dissipated in the surrounding air (or water which is circulated through the passages in the brake drum) so that excessive heating of the brake lining does not take place.

The design or capacity of a brake depends upon the following factors :

1. The unit pressure between the braking surfaces
2. The coefficient of friction between the braking surfaces
3. The peripheral velocity of the brake drum
4. The projected area of the friction surfaces
5. The ability of the brake to dissipate heat is equivalent to the energy being absorbed.

The major functional difference between a clutch and a brake is that a clutch is used to keep the driving and driven member moving together, whereas brakes are used to stop a moving member or to control its speed.

Example Problem on Braking Torque Calculation

Example Problem Statement: A vehicle of mass 1200 kg is moving down the hill at a slope of 1: 5 at 72 km/h. It is to be stopped in a distance of 50 m. If the diameter of the tyre is 600 mm, determine the average braking torque to be applied to stop the vehicle, neglecting all the frictional energy except for the brake. If the friction energy is momentarily stored in a 20 kg cast iron brake drum, What is the average temperature rise of the drum? The specific heat for cast iron may be taken as 520 J / kg°C.

Determine, also, the minimum coefficient of friction between the tyres and the road in order that the wheels do not skid, assuming that the weight is equally distributed among all four wheels.

Answer:

Given data
Mass of the Vehicle m = 1200 kg
Slope = 1: 5
velocity of vehicle v = 72 km/h = 20 m/s
Distance h = 50 m
Diameter of the tyre d = 600 mm or r = 300 mm = 0.3 m
Brake drum mass m_b = 20 kg
Specific heat c = 520 J / kg°C

Average braking torque to be applied to stop the vehicle

(Read this article to understand the formulas used here)

We know that the kinetic energy of the vehicle,

E_k = (1/2) mv² = E₁ = (1/2) × 1200 × (20)² = 240000 N-m

and potential energy of the vehicle,

E₃ = m × g × h × Slope = 1200 × 9.81 × 50 × (1÷5) = 117 720 N-m

∴ The total energy of the vehicle or the energy to be absorbed by the brake,

E = E_k + E_P = 240 000 + 117 720 = 357720 N-m

Since the vehicle is to be stopped at a distance of 50 m, therefore tangential braking force is required,

F_t = 357720 ÷ 50 = 7154.4N

We know that average braking torque to be applied to stop the vehicle,

T_B = F_t  × r = 7154.4 × 0.3 = 2146.32 N-m

The average braking torque to be applied to stop the vehicle is 2146.32 N-m.

Average temperature rise of the drum

(Read this article to understand the formulas used here)

Let
∆ t = Average temperature rise of the drum in °C.

We know that the heat absorbed by the brake drum

H_g = Energy absorbed by the brake drum = 357720 N-mm = 357 720 J

We also know that the heat absorbed by the brake drum (H_g),

357720 = m_b × c × ∆t
357720 = 20 × 520 × ∆t
357720 = 10400 ∆t
∆t = 357720 ÷ 10400
∆t = 34.4°C

The average temperature rise of the drum is 34.4°C.

Minimum coefficient of friction between the tyre and road

Let
μ = Minimum coefficient of friction between the tyre and road, and
R_N = Normal force between the contact surface. This is equal to the weight of the vehicle = m.g

R_N = 1200 × 9.81 = 11772N

We know that tangential braking force (F_t),

7154.4 = μ. R_N 
7154.4 = μ × 11772
μ = 7154.4 ÷ 11772
μ = 0.6

The minimum coefficient of friction between the tire and the road is 0.6.