lab 5 modeling of a car suspension system_v3

ME 413: System Dynamics & ControlME 413: System Dynamics & ControlME 413: System Dynamics & ControlME 413: System Dynamics & Control

Mechanical Mechanical Mechanical Mechanical Systems (Systems (Systems (Systems (2222)))) Modeling of a Car Suspension SystemModeling of a Car Suspension SystemModeling of a Car Suspension SystemModeling of a Car Suspension System

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ME 413: System Dynamics and Control Lab Manual

Mechanical Systems (2): Modeling of a Car Suspension System 1

MECHANICAL SYSTEMS (2)

MODELING OF A CAR SUSPENSION

SYSTEM

OBJECTIVES

1. To suggest a simple model of a car suspension system of a pickup truck and

2. to simulate its response for a given profile of the road.

PROCEDURE

The motion of mechanical elements can be described in various dimensions as

translational, rotational, or combination of both. The equations governing the

motion of mechanical systems are often formulated from Newton’s law of motion.

The following steps are necessary in the modeling and formulation of linear

mechanical systems

1. Construct a model for the system containing interconnecting elements.

2. Draw the free-body diagram.

3. Write equations of motion of all forces acting on the free body diagram.

Translational Motion

Suppose that forces are acting on a body of mass m . If ∑∑∑∑F is the sum of all

forces acting on a mass m through the center of mass in a given direction, then

=∑F ma (1)

where a is the resulting absolute acceleration in that direction. The line of action of

the force acting on a body must pass through the center of mass of the body.

Otherwise, rotational motion will also be involved.

Rotational Motion

For a rigid body in pure rotation about a fixed axis, Newton’s second law states that

α=∑T J (2)



where ∑T is the sum of all torques acting about a given axis, J is the moment of

inertia of a body about that axis, and α is the angular acceleration of the body.

TABLE 1

Summary of Elements Involved in Linear Mechanical Systems

Translation Rotation

Spring k

FF

1x

2x

kxxxkF =−= )(21

k

T T

2θ1θ

θθθ kkT =−= )(21

Inertia

Damper b

FF

1x� 2

x�

xbxxbF �� =−= )(21

b

T T

2θ�1θ�

θθθ �� bbT =−= )(21

Element

m1

F

x

2F

3F

4F

∑ = amFT

θJ

∑ = αJT



Part 1: Background

Purpose of the Suspension System

The purpose of the suspension system is

•••• to provide a smooth ride in the car and

•••• to help maintain control of the vehicle over rough terrain or in case of sudden

stops.

The two basic types of suspension systems used in most cars today are:

•••• Strut suspension and

•••• Conventional suspension.

Suspension systems control the movement of the car and keep the tires in contact

with the road, providing a better and safer ride, [1].

Basic Functions of Suspension System

The components of the suspension system perform six basic functions [2]:

1. Maintain correct vehicle ride height

2. Reduce the effect of shock forces

3. Maintain correct wheel alignment

4. Support vehicle weight

5. Keep the tires in contact with the road

6. Control the vehicle’s direction of travel

Shock Absorber

Shock absorbers are attached to the car's frame near each wheel on most cars with

conventional suspension systems. Shock absorbers are either housed inside coil

springs, or mounted beside them.

When a car hits a bump, the spring contracts, the shock absorber works with a

piston and thick fluid inside it to keep the spring from rebounding too quickly,

making your car ride smoother, improving control and reducing tire wear.

Coil Spring

Coil Springs are a component in both Strut and conventional suspension systems.

When a car goes over a bump or has to stop quickly, the coil spring contracts then

rebounds. It is the job of the Strut or shock absorber to keep the coil spring from

rebounding too quickly.



Strut Assembly (Strut and Cartridge)

Most front wheel drive cars have some kind of strut suspension. The strut is a

special type of shock absorber that helps to minimize the motion of the suspension.

The coil spring surrounds the strut assembly. Struts and shock absorbers work in

very similar ways. However, struts function as a structural part of the suspension

system, and therefore also affect things like wheel alignment, vehicle control and

suspension wear.

Strut Suspension System

Most front wheel drive cars have some sort of strut suspension System. In the strut

suspension system, a coil spring with a bearing plate at each end supports the car.

The spring assembly rests on a seat. The strut suspension system helps to cushion

the movement of the coil spring when a car hits a bump or a pothole and works to

hold the tires to the road and provide better vehicle control. See Figure 1.

Figure 1 Strut suspension system.

Conventional (Shock Absorber) Suspension System

In conventional (Shock Absorber) suspension systems, two control arms are

attached to the frame of the car and a coil spring and shock absorber are attached

to the frame and one of the control arms. The spring and shock absorber work

together to smooth the ride of the car over the roughness of the road. This



suspension system works to hold the tires to the road and provide better vehicle

control. See Figure 2.

Figure 2 Shock absorber system.



Part 2: Assignment

The suspension system for one wheel of an old-fashioned pickup truck can be

illustrated as shown in Figure 3 a. The mass of the vehicle is M and the mass of

the wheel is m . The suspension spring has a constant 1k and the tire has a spring

constant 2k The damping constant of the shock absorber is b .

1x

2x

y

Figure 3 Car suspension model

1. Based on the description above suggest a suitable model for this system.

2. Draw the free body diagram of the suggested model.

3. Write the equations of motion of the system.

4. For zero initial conditions, take Laplace transform of both sides of the

equations obtained in (3). Obtain the transfer functions

11

( ) _1( )

( )= =X s num

G sY s den

and 22

( ) _ 2( )

( )= =X s num

G sY s den

in function of the system parameters 1

, , ,M m b k and 2k . In the above

1( )X s and 2 ( )X s represent Laplace transform of 1( )x t and 2 ( )x t ,

respectively and ( )Y s represents Laplace transform of ( )y t . The

quantities _1num , _ 2num , and den are numerator and denominator

polynomials in s .



5. Write 1( )G s and 2 ( )G s in the form

11 4 3 2

( )( )

( )=

+=

+ + + +

… …

… … … … …

X s sG s

Y s s s s s

2

22 4 3 2

( )( )

( )=

+ +=

+ + + +

… … …

… … … … …

X s s sG s

Y s s s s s

6. Use MATLAB to plot the responses 1( )x t and 2 ( )x t for each case of the

road profiles shown in Figure 4 above for the given values of system

parameters:

a) 1 2, , , ,500Kg 20 Kg 250 N/m 1 KN/m and 1.0 KN-s/m. = = = = =M m k k b

b) 1 2, , , ,500 Kg 20 Kg 500 N/m 1 KN/m and 500 N-s/m. = = = = =M m k k b

c) 1 2, , , ,200Kg 20Kg 100N/m 500N/m and 750N-s/m. = = = = =M m k k b

d) Suggest any other values of 1

, , ,M m c k and 2k .

( )y t

1

1

t

( )y t

0 1.

( )y t

( )y t

0 5.

0 25.

t

t t

Figure 4. Profile of the road

7. Referring to the previous question (6), which choice of system

parameters 1

, , ,M m b k and 2k is the best one and why? Justify your

answer.



8. Use the Final Value Theorem (FVT) to obtain the steady state values 1ssx

and 2ssx of the system that appears to you to be the best one.



Part 3: MATLAB Program

1. In MATLAB command windw click: File � New � M-File. The Matlab

editor window opens. In this window write the following MATLAB program.

Matlab Program

% --------------------- System Parameters ---------------------

%

M=500;m=20;k1=250;;k2=1000;b=1000; %M=500;m=20;k1=500;;k2=1000;b=500; M=200;m=20;k1=100;;k2=500;b=750; %t=[0:0.001:12]; num1=[……. …….]; num2=[……. ……. …….]; den=[……. ……. ……. ……. …….]; sys1=tf(num1,den) sys2=tf(num2,den) % %---------------------- Case a: Step Input ----------------- % subplot(2,2,1);step(sys1,'r:',sys2); grid; title('case a: Step Input') legend('y1','y2') % %---------------------- Case b: Impulse Input ----------------- % subplot(2,2,2);impulse(sys1,'r:',sys2); grid; title('case b: Impulse Input') legend('y1','y2') % %---------------------- Case C : Arbitrary Input ----------------- % t=[0:0.001:12]; slope=1; for k=1:length(t) if t(k)<=1 r(k)=slope*t(k); elseif t(k)>=1 r(k)=1; end end y1=lsim(sys1,r,t); y2=lsim(sys2,r,t); subplot(2,2,3);plot(t,y1,'r:',t,y2,t,r,'g:'); legend('y1','y2');grid; xlabel('time');ylabel('y(t)'); title('Case C: Arbitrary Input') % %---------------------- Case D : Arbitrary Input ----------------- % for k=1:length(t) if t(k)<=3 r(k)=0.25; else r(k)=0;



end end y1=lsim(sys1,r,t); y2=lsim(sys2,r,t); subplot(2,2,4);plot(t,y1,'r:',t,y2,t,r,'g:'); legend('y1','y2');grid; xlabel('time');ylabel('y(t)'); title('Case D: Arbitrary Input')

2. Save your file as: File Save as A window opens as shown below. In

the space provided for file name type: xxx.m where xxx is the file name and .m is the extension for Matlab files.

3. To run your program you need to press F5 or copy the program and paste it

into MATLAB prompt and press ENTER

References

[1] http://www.meineke.com/stuff_about_cars/how_work_suspension.asp#text

[2] http://www.monroe.com/tech_support/tec_suspsystemfund.asp

[3] K. Ogata, System Dynamics, Fourth Edition, Pearson Prentice Hall, 2004.

[4] http://users.pandora.be/educypedia/education/careducation.htm [5] http://www.innerauto.com/Automotive_Systems/Steering_And_Suspension_System/

[6] http://www.sciences.univ-

nantes.fr/physique/perso/gtulloue/Meca/Oscillateurs/suspension.html

[6] http://www.careersnet.org/automotive/PHS/ppt's/Steering/suspension/

lab 5 modeling of a car suspension system_v3

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