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Chapter 6 · 4 hours

Suspension and Steering

Practice questions

Practice questions and answers

3 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 · 8 marks

State the purposes of a vehicle suspension system. Describe the types of springs used in suspension. Explain the working of a telescopic hydraulic shock absorber and the function of a stabilizer (anti-roll) bar.

Answer

Purposes of suspension

  • Isolate the body from road shocks for ride comfort.
  • Keep the tyres in contact with the road for good traction and braking.
  • Support the vehicle weight and carry loads.
  • Maintain correct wheel alignment and control body roll, pitch and squat.
  • Transmit driving, braking and cornering forces to the frame.

Types of springs

SpringDescriptionUse
Leaf springSeveral steel leaves of unequal length (semi-elliptic); damping by inter-leaf frictionTrucks, buses, rear axles
Coil springHelically wound steel bar; light, compact; needs links to locate axleMost cars (MacPherson, wishbone)
Torsion barSteel bar twisted when the arm movesSome light trucks and cars
Air springRubber bellows with compressed air; stiffness and height adjustableBuses, luxury cars
Rubber / hydro-pneumaticRubber block, fluid + gasSmall cars, special cars

Telescopic shock absorber (damper)

A spring stores energy and would keep oscillating, so a damper is fitted in parallel to convert oscillation energy into heat.

   top mount (body)
        |
   [ piston rod ]
   |  +--------+  | <- oil
   |  | piston |  |    valves in piston
   |  +--------+  |
   |   oil + gas  |
   bottom mount (axle)

The cylinder is filled with oil. During bump the piston moves down and oil is pushed through small orifices and valves into the upper chamber. During rebound it flows back through other valves, giving more resistance (higher in rebound than bump, typically 2:1). The restriction of oil flow creates the damping force. A gas-charged (monotube) design avoids foaming.

Stabilizer (anti-roll) bar

A U-shaped steel torsion bar connected between the left and right suspension arms. In a corner, the outer wheel moves up and the inner wheel moves down; the bar twists and resists this. It reduces body roll, improves handling and keeps both tyres on the road.

  • Practice · 4+4 marks

Explain the layout of a rack and pinion steering gear with a sketch. How does a hydraulic power steering system differ from an electric power steering (EPS) system?

Answer

Rack and pinion steering

 Steering wheel
      |
  Steering column (universal joints)
      |
  [Pinion]---meshes--- [Rack]
                      |      |
                tie-rod    tie-rod
                  |            |
              steering arm   steering arm
                  |            |
              left wheel    right wheel
  • Turning the wheel rotates a small pinion gear that meshes with a toothed rack; the rotary motion becomes linear motion of the rack.
  • The rack moves the tie rods (with ball joints) which turn the stub axle arms of the wheels.
  • Steering ratio is typically 14:1 to 20:1 (steering wheel turns vs wheel angle).
  • Advantages: simple, compact, direct feel, low cost, few joints. Used in nearly all cars and many SUVs.
  • Limitation: passes road shocks back to the driver (kickback), unsuitable for heavy vehicles (a recirculating-ball box is used there).

Hydraulic and electric power steering

Power steering reduces the driver's effort at low speed and for parking.

PointHydraulic power steering (HPS)Electric power steering (EPS)
Source of powerEngine-driven vane pump, fluidElectric motor on column or rack
FluidHydraulic oil, reservoir, hosesNone
OperationRotary valve directs fluid to either side of the rack pistonTorque sensor signal to ECU, ECU drives motor
Engine loadPump runs continuously, uses fuelDraws current only when steering
EfficiencyLowerAbout 3 % fuel saving
TuningFixed assistanceVariable with speed; lane assist, park assist possible
MaintenanceLeaks, fluid changeAlmost none
UseHeavy vehiclesModern cars, EVs
  • Practice · 5 marks

A car has a wheelbase of 2.7 m and a distance of 1.4 m between the king-pin (pivot) centres. State the Ackermann condition for correct steering. When the inner front wheel is turned through 20 degrees, find the angle of the outer front wheel for true rolling, and the turning radii of the inner and outer front wheels (measured at the wheel centre line to the instantaneous centre).

Answer

Condition for correct steering

For pure rolling (no tyre scrub) during a turn, the axes of all four wheels must meet at one point on the extended line of the rear axle (the instantaneous centre). For this the inner wheel turns more than the outer wheel and, with inner angle θ\theta and outer angle ϕ\phi,

cot⁡ϕ−cot⁡θ=cL\cot\phi - \cot\theta = \frac{c}{L}

where cc is the distance between the pivots and LL the wheelbase. The Ackermann linkage (trapezium) approximates this.

 Instantaneous centre O ---------- rear axle line
        |   \               |
        |    \  R_in        |  L
        |      \ theta      |
     inner wheel         outer wheel (phi)
        |<-------- c ------->|

Numerical

Given: L=2.7L = 2.7 m, c=1.4c = 1.4 m, θ=20∘\theta = 20^\circ.

cot⁡20∘=2.7475cL=1.42.7=0.5185cot⁡ϕ=2.7475+0.5185=3.2660ϕ=tan⁡−1(13.266)=17.02∘\begin{aligned} \cot 20^\circ &= 2.7475 \\ \frac{c}{L} &= \frac{1.4}{2.7} = 0.5185 \\ \cot\phi &= 2.7475 + 0.5185 = 3.2660 \\ \phi &= \tan^{-1}\left(\frac{1}{3.266}\right) = 17.02^\circ \end{aligned}

Turning radii (wheel centre-line to instantaneous centre, R=L/sin⁡R = L/\sin of wheel angle):

Rin=Lsin⁡θ=2.7sin⁡20∘=7.89 m,Rout=Lsin⁡ϕ=2.7sin⁡17.02∘=9.22 mR_{in} = \frac{L}{\sin\theta} = \frac{2.7}{\sin 20^\circ} = 7.89\ \text{m}, \qquad R_{out} = \frac{L}{\sin\phi} = \frac{2.7}{\sin 17.02^\circ} = 9.22\ \text{m}

Check: the difference Rout−RinR_{out} - R_{in} projected along the axle is 9.22cos⁡17.02∘−7.89cos⁡20∘=8.82−7.42=1.409.22\cos 17.02^\circ - 7.89\cos 20^\circ = 8.82 - 7.42 = 1.40 m =c= c. Correct.

Answer: outer wheel angle =17.02∘= 17.02^\circ; Rin=7.89R_{in} = 7.89 m; Rout=9.22R_{out} = 9.22 m.

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

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