Name : First name :

Name : First name :

k M

20 a

r

Examination duration: 2h – Write your answer to the questions on the examination paper.

Only the results are required.

Apart from this examination paper, no other document should be given back.

The Michelin Active Wheel, represented in the figure below, will be studied

The former consists of a wheel and its tyre, which integrates inside the active parts of the drive, suspension and brake functions:

- an electrical motor, supplied with the batteries stored in the bodywork. With that technology, the car is solely and directly powered by the motorisation of the wheels allowing two-wheel or four-wheel drive;

- a coil spring, to support the bodywork, coupled with an electrical shock absorber- stabilizer, in the form of a subset pinion and rack combined with an electrical motor driven by the car active electronics;

- a brake disk.

During the car acceleration phase, the study is reduced to the search of the motion, the electrical suspension motor torque and the tangential component of the ground interaction force acting upon the tyre.

(The electrical drive motor torque is assumed to be a data related to the order delivered by the driver acting upon the accelerator pedal).

Signature

1. Textual dynamic model and kinematic sketch

1.1. Geometry and mass

The system modelled as a planar motion, identifies:

- eight rigid bodies:(S₀), (S₁), (S_f), (S₂), (S₃), (S₄), (S₅) and (S₆):

- (S₀) the earth,

- (S₁) the road profile simulator,

- (S₂) the wheel mount, the brake calliper and the stators of the two motors, - (S₃) the “quarter” of the sprung mass of the car (frame, engine,

passengers, …) and the rack,

- (S₄) the wheel, the shaft, the drive gear and the disk brake,

- (S₅) the suspension motor rotor and the drive pinion of the rack (S₃), - (S₆) the drive motor rotor and the drive pinion of the gear (S₄),

- (S_f) a virtual body of negligible mass, defined so as to reduce the study to a quarter car, assuming in first approximation the car frame keeps a constant trim,

The motion of the road profile simulator (S₁) relative to the ground (S₀), in terms of the car motion, is a given data for the study.

The two rotors (S₅) and (S₆) are bodies of negligible mass so as to reduce the calculations as part of this final-exam.

S3

S4

S2

S5

S6

(R)

(P)

S0

S1

Sf

A

B1

C

E

D

F

H

z

I K J

Sprung mass

Z₁ Z₃ Z₂

G

x

y

z

With regards to the kinetic calculations, the tyre deformation at the contact with the ground is neglected. One denotes (S₄^*) the subset (S₄) U (P) supposed to be a body of revolution.

- eleven kinematic joints:

(S₀ - S1): prismatic (S₀ - Sf): prismatic (S_f - S3): prismatic (S₂ - S3): prismatic (S₂ - S4): revolute (S₂ - S5): revolute (S₂ - S6): revolute (R - S2): sphere (R - S3): sphere (S₃ - S5): point-surface (S₄ - S6): point-surface

One recalls the two joints (S₃ - S5): rack-pinion and (S₄ - S6): pinion-gear are both rolling without sliding joints.

- the tyre (P) - ground (S1) joint is defined in first approximation, from a kinematic point of view, in terms of the two following assumptions:

- plane deformation of the rigid section of the tyre passing through point I located straight up the centre D of the wheel at about ground level;

- _{G ( ) 0.}

Tyre 1,

r

r I = (rolling without sliding at the contact point) 1.2. Forces

The spring (R) is supposed to be a linear elastic spring of stiffness k and natural length l0.

The twistor of the interaction forces between the ground and the tyre is assumed to be a resultant twistor passing through the contact point I, located straight up the centre of the wheel. The radial component, denoted by N_pzr₀

, is expressed asµε^η(Newton), in which εrepresents the crushing of the tyre from its free radiusr_p, µ and η two coefficients. In order to check further the rolling without sliding condition, the tangential component, denoted byT_pyr₀

, is an unknown of the study.

The aerodynamic force that opposes the car motion through the air, called the drag, represented by a force acting upon body (S₃), defined about the centre of the motorised wheel, denoted by

,S3

Fair

r , is expressed as y₂ y₂yr₀

&

&

ξ

− , in which ξ is a coefficient depending on the car and air aerodynamic characteristics, andy₂yr₀

& the

difference in velocity between the car and the air.

The system moves in the gravitational field which is defined by the place upward verticalzr₀

.

All the joints are supposed to be perfect joints.

Finally, one notes:

- _T^r₃₅ =_{T35 z}^r₀ and Tr₄₆ T₄₆ ry_02*

= the teeth tangential components (introduced from the cuts of the closed loops). One recalls that the radial forces, on perfect joints assumption, are expressed in terms of tangential forces and the teeth geometry, and in that case, do not make up new unknowns of the study.

- _{Cm25=Cm25 x}^r₀ and _{Cm26 =Cm26 x}^r₀ the suspension motor and the drive motor torques

1.3. Galilean reference frame

In the field of the study, the fixed body (S₀) frame is supposed to be a Galilean reference frame.

2. Construct the vector geometric model

Since the vector geometric model construction was the subject of the mid-term exam, the first steps results of the Active Wheel model are given so as to principally devote oneself, in the impart time, to the study of dynamics.

2.1. Construct the vector geometric model 2.2. Model the joints

2.3. Define the parameters

- use a minimum path between the bases;

- use a minimum path between the points.

- to sum up: we gather 6 initial parameters: ψ, θ, φ, y2, z2 and z3.

2.4. Study the equations of constraint - use the joints not taken yet into account;

Consequence of the point-surface joint (S₄–S₆): already introduced in terms of the position vector DG r r zr_02*

) ( ₄ + ₆

= .

Consequence of the point-surface joint (S₃–S₅): the rack pitch line is located at the length r₅from the centre of the pinion pitch circle F₅.

- use the geometric conditions of some joints not taken yet into account;

Consequences of the prismatic joints geometric conditions: those conditions have been checked through a judicious choice of the points.

x₀

θ 0,5

x₀

ϕ 0,6

x₀

ψ 0,4

y₀ z₀

y₄ z₄

y₀ z₀

z₅

y₅ z₀

y₀ z₆

y₆

z0 0 c

z0 z3 ) 0 ( 2

z0 z2 0 2

z0 z1

r r

r r

r r

r

+

=

+ +

= +

=

=

y b DF

y a y AE

y y AD AB

radius circle pitch 5 pinion 5 :

) 0

* , 02 z0 (

* ) 02 6 ( 4

z0 0 d

r z

z r r DG

y a DH

α

= +

=

−

=

r r

r r r

x₀

α₀ 0,02*

z₀ z_02*

y₀ y_02*

x₀

α₀ 0,02*

z₀ z_02*

y₀ y_02*

BEWARE YOUR WORK STARTS HERE

☺

Rolling without sliding consequence between (S₅/S₃) at the contact point J?

=0⇒ )

5(

3,

r r

J G_{S S}

Rolling without sliding consequence between (S₆/S₄) at the contact point K?

=0⇒ )

6(

4,

r r

K G_{S S}

Rolling without sliding consequence between (Tyre/S₁) at the contact point I?

=0⇒ )

, (

1

r r

I G^S tyre

- take into account the laws of kinematic behaviour of the motors.

Note 1: the road profile is known, z₁zr₀

=

AB , with z₁ a known function,

Note 2: So as to simplify the study, one admits the active suspension control function is taken into account through an equation of constraint.

2.5. Gather the kinematically independent parameters - number of kinematically independent parameters

Number =

- define the kinematically independent parameters

In the following, as part of this final-exam and so as to write equations independent of the equations of constraint, one retains the initial parameters set.

3. Express the laws of behaviour in terms of the geometric model 3.1. Spring (R)

{

}

{

}

3.2. Twistor of the forces of the ground acting upon the tyre

{

}

3.3. Perfect joints

Fill up the sketch of the joints with the scalar consequences

3.4. Gravitational field

{

}

{ }

The twistors associated with the others forces have been already given through the notations in the text.

4. Gather the unknowns of the study

5. Write the dynamic equations 5.1. Define the cuts

S

S

S

S

S

S

S

S

5.2. Define the new unknowns of the study

5.3. Define the sketch of the characteristics

5.4. Write the scalar consequences of the dynamic equations

= with E =

= with E =

S

S

S

S

S

S

S

S

+

1

+

inc. +

+ +

1

+

S

S

S

S

S

S

S

S

= with E =

= with E =

= with E =

= with E =

(One recalls that in the following the initial parameters are retained)

5.5. Compute the components of forces

=

=

=

=

=

=

5.6. Compute the components of kinetics

=

=

=

=

=

=

6. According to Coulomb’s laws, which condition must be satisfied by the tyre-ground contact in order to roll without sliding?

7. Specify how the ground clearance of the car can be determined.