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Chapter 7 · 4 hours

Braking Systems

Practice questions

Practice questions and answers

4 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 4+4 marks

(a) Explain the working of a hydraulic braking system with the help of a layout. (b) Describe a pneumatic (air) brake system and state why it is used in heavy vehicles.

Answer

(a) Hydraulic brake system

It works on Pascal's law: pressure applied to a confined fluid is transmitted equally in all directions.

 Pedal -> [Master cylinder] ==brake fluid in pipes==> 
                       |-> wheel cylinder (drum) x2 (rear)
                       |-> caliper piston (disc) x2 (front)
  1. When the driver presses the pedal, the push rod moves the piston in the master cylinder, which builds up the fluid pressure (about 40 to 100 bar at hard braking).
  2. The pressure goes through steel pipes and flexible hoses to the wheel cylinders (drum brakes) or caliper pistons (disc brakes).
  3. The pistons push brake shoes against the drum or pads against the disc; friction slows the wheel.
  4. When the pedal is released, return springs push the pistons back and fluid returns to the master cylinder.

A vacuum booster multiplies the pedal force. A tandem master cylinder gives two separate circuits (front/rear or diagonal) so that braking still works if one fails. Brake fluid is DOT 3/4 glycol-based, high boiling point.

Force multiplication: Fout=Fin×AwheelAmasterF_{out} = F_{in} \times \dfrac{A_{wheel}}{A_{master}}.

(b) Pneumatic (air) brake

 Compressor -> reservoir (8 bar) -> brake valve (pedal)
        -> air lines -> brake chambers -> push rod/slack adjuster
        -> S-cam -> brake shoes on drum
  • An engine-driven compressor fills air tanks to about 8 bar.
  • The pedal opens a brake valve that lets air into the brake chambers; the diaphragm pushes a rod that turns the cam (or wedge) and applies the shoes.
  • When released, air is exhausted and springs return the shoes.
  • In trailers a relay valve applies the brakes quickly. Spring brakes apply automatically if air pressure is lost, a safety feature.

Reasons for use on heavy vehicles: unlimited supply of air, a high force through large chambers is possible, no fluid leakage hazard, easy coupling of trailers, and the driver effort is light. The drawbacks are bulk, cost, and a slight delay in application.

  • Practice · 6 marks

Describe the construction of a disc brake and a drum brake (leading-trailing shoe type) and differentiate between them.

Answer

Disc brake

   caliper (fixed to knuckle)
   |  [pad] |disc| [pad]  |
   |<- piston pushes pads against the rotating disc ->|
  • A cast-iron disc (rotor) rotates with the wheel hub (vented in front brakes).
  • A caliper, mounted on the steering knuckle, holds two friction pads and one or two hydraulic pistons. Fluid pressure pushes the pads against both faces of the disc.
  • Floating calipers have a piston on one side only; fixed calipers have pistons on both sides.

Drum brake

   brake drum (rotates with wheel)
     shoe A (leading)   shoe B (trailing)
        \  wheel cylinder  /
   anchor pin  +  return springs
  • A cast-iron drum rotates with the wheel. Two curved shoes with friction linings are on a fixed backing plate.
  • The wheel cylinder pushes the shoes outward against the inside of the drum. In a leading-trailing design, the leading shoe is self-energised by drum rotation.
  • Return springs retract the shoes; an adjuster compensates for wear; a handbrake link is easy to add.

Comparison

PointDisc brakeDrum brake
Heat dissipationGood (open to air)Poor (enclosed)
Fade resistanceHighLow (drum expands)
Wet performanceGood (water thrown off)Poor
Self-energisingNone; needs higher force or boosterPresent; less force needed
Wear check / serviceEasyHarder
Weight and costLight for performance, costlierHeavy, cheaper
Parking brakeDifficultEasy
UseFront wheels of nearly all cars, all wheels in sports carsRear wheels of small cars, trucks
  • Practice · 6 marks

Explain the working of an anti-lock braking system (ABS) with its main components. What is electronic brake force distribution (EBD)?

Answer

Anti-lock braking system (ABS)

ABS prevents the wheels from locking during hard braking. A locked wheel loses steering control and has less friction than a wheel that is just rolling (about 10 to 20 % slip gives the maximum grip). ABS keeps the slip near this optimum.

 Wheel speed sensors (4) --> [ECU] --> [Hydraulic unit]
                                  |    valves + pump
                         Master cylinder -> brake lines

Main components

  1. Wheel speed sensors (magnetic or Hall) at each wheel with toothed ring.
  2. ECU which compares wheel speeds and calculates deceleration and slip.
  3. Hydraulic modulator with solenoid inlet and outlet valves for each wheel and a return pump with accumulator.
  4. Warning lamp on the dashboard.

Working (three-phase cycle, 10 to 15 times per second)

  • Pressure build-up: normal braking; inlet valve open.
  • Hold: if the ECU senses a wheel decelerating too fast (about to lock), it closes the inlet valve so that pressure is held.
  • Release: the outlet valve opens, pressure falls, and the wheel speeds up again.
  • Then pressure is raised again. Pedal pulsation is felt.

Advantages: shorter stopping distance on wet roads, steering control during braking.

Electronic brake force distribution (EBD)

EBD is an extension of ABS software. During braking, weight shifts to the front axle, so the rear wheels can lock early. EBD uses the same sensors and valves to reduce the pressure on the rear brakes (and balance left/right) according to load, speed and road, so that all four wheels use their available grip. It replaces the mechanical proportioning valve, and permits shorter stopping distance, particularly when carrying load.

  • Practice · 8 marks

A car of mass 1200 kg travels at 72 km/h on a level road with a tyre-road friction coefficient of 0.7. Assuming that all four wheels are on the verge of locking, calculate (i) the maximum deceleration, (ii) the braking force, (iii) the braking distance and time, (iv) the total stopping distance if the driver's reaction time is 1 s, and (v) the energy to be absorbed by the brakes, and the temperature rise of four discs of 6 kg each (specific heat 460 J/kg K) if all this energy is absorbed in them.

Answer

Given: m=1200m = 1200 kg, u=72u = 72 km/h =20= 20 m/s, μ=0.7\mu = 0.7, g=9.81g = 9.81 m/s², tr=1t_r = 1 s.

(i) Maximum deceleration

a=μg=0.7×9.81=6.87 m/s2a = \mu g = 0.7 \times 9.81 = 6.87\ \text{m/s}^2

(ii) Braking force

Fb=ma=1200×6.867=8240 N(=μmg)F_b = m a = 1200 \times 6.867 = 8240\ \text{N} \quad (= \mu m g)

(iii) Braking distance and time

sb=u22a=2022×6.867=29.1 mtb=ua=206.867=2.91 s\begin{aligned} s_b &= \frac{u^2}{2a} = \frac{20^2}{2 \times 6.867} = 29.1\ \text{m} \\ t_b &= \frac{u}{a} = \frac{20}{6.867} = 2.91\ \text{s} \end{aligned}

(iv) Total stopping distance with reaction time

Distance during reaction =u tr=20×1=20= u\, t_r = 20 \times 1 = 20 m.

stotal=20+29.1=49.1 ms_{total} = 20 + 29.1 = 49.1\ \text{m}

(v) Energy and disc temperature rise

E=12mu2=12(1200)(20)2=240 000 J=240 kJE = \tfrac{1}{2} m u^2 = \tfrac{1}{2}(1200)(20)^2 = 240\,000\ \text{J} = 240\ \text{kJ} ΔT=Emdc=240 000(4×6)×460=21.7 K\Delta T = \frac{E}{m_d c} = \frac{240\,000}{(4 \times 6) \times 460} = 21.7\ \text{K}

In practice discs heat much more in repeated stops; the front brakes absorb 60 to 70 % of the energy.

Answer: a=6.87a = 6.87 m/s²; Fb=8.24F_b = 8.24 kN; sb=29.1s_b = 29.1 m in 2.912.91 s; total stopping distance =49.1= 49.1 m; E=240E = 240 kJ and ΔT≈21.7\Delta T \approx 21.7 K.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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