vehicle dynamics ii v.c

8/2/2019 Vehicle Dynamics II v.C

1/49

AE 467/567 Race Car Vehicle Dynamics

Spring 2010

Vehicle Dynamics Part II


2/49

Vehicle Dynamics II

Roadmap Develop/understand a Spring-Mass-Damper (SMD) model Develop/understand a dual SMD modelApply to:

1/4 car model one suspension assembly (of 4)

Vehicle dynamics corner entry/exit transientsAugment with nonlinear effects

Various

chapters in the book, starting with Ch.6

Basic theorySimple math

modelsHigh fidelitysimulations


3/49

Single SMD Model

m=mass (kg or slugs)F=force (N or lbs)x=displacement (m or ft)k=spring constant (N/m or lbs/ft)c=damping constant (N/m/s or lbs/ft/s)

Draw free-body diagram:

mF(t)

kx

xc)(

)(

tFkxxcxm

xmxckxtF

=++=

Linear ODE, solution techniques well-known


4/49

SMD solution

( )( )

km

c

m

k

xxx

onmanipulatiSome

iSinCosee

decayorGrowthe

m

mkcc

kcm

exkcm

exxexxformSolution

tFkxxcxm

iaginary

al

t

t

t

4

;

02

:

2

4

0

0

:

)(

2

0

2

00

Im

Re

2

2

0

2

0

0

==

=++

+=

=

=++

=++

==

=++


5/49

SMD solution

km

c

m

k

m

k

m

c

m

c

mkcIf

m

mkcc

imaginaryreal

4;

4;

2

4

2

4

2

0

2

2

2

2

==

=

=

> than

then the solution tends to:

1

1

m

k

2

2

m

k

2

2

1

1

1

2 ;mkf

mkf


8/49

What Are We Solving?

An automobile suspension resembles

a dual spring-mass-damper system ateach corner of the vehicle Spring/mass/damper #1 is the tire

itself, plus the effective mass of thecomponents that move along with thewheel (the unsprung mass)

Spring/mass/damper #2 is the actualspring and shock, plus the 25% ofcars body mass (the sprung weight)


9/49

Simulation Approach (SMD)

Convert the differential equation into a difference equation:

Basically a simple numerical integration technique

All differential equations (& difference equations) require

Initial Conditions for solution

( )( )

repeatandii

txxx

txxx

kxxctFm

x

smallttttt

tFkxxcxm

iii

iii

iiii

i

1

1

;

)(

1

1

0

+

++

=

=++

+

+

21

y

x

xdx

dyyy +

12

x


10/49

Simulation Approach (SMD)% Quick m-file for Spring-Mass-Damper problem

%

m=input('Mass (kg)? ');

k=input('Spring constant (N/m)? ');

%

zeta=[0.1 0.3 0.5 1 2];

c=sqrt(4*k*m)*zeta;

%

x=zeros(1000,5); xdot=zeros(1000);

xddot=zeros(1000);

%

for j=1:5

x(1)=0; xdot(1)=0; time(1)=0;dt=0.01;

%

for i=1:1000

if i


11/49

Simulation Approach (Dual SMD)

% m-file for Dual Spring-Mass-Damper problem

clear

highlo=input('damper factor? ');m1=22.7;k1=350000;c1=1;m2=227;

k2=81000;c2=7710;

c2=c2*highlo;

%

time(1)=0;dt=0.001;

%

for i=1:3000

if i1100

x0(i)=1;xdot0(i)=0;

else

x0(i)=(i-1000)/100;xdot0(i)=10;

end

xddot1(i)=(k1*(x0(i)-x1(i))-k2*(x1(i)-x2(i))+c1*

(xdot0(i)-xdot1(i))-c2*(xdot1(i)-xdot2(i)))/m1;

xddot2(i)=(k2*(x1(i)-x2(i))+c2*(xdot1(i)-xdot2(i)))/m2;

xdot1(i+1)=xdot1(i)+xddot1(i)*dt;xdot2(i+1)=xdot2(i)+xddot2(i)*dt;

x1(i+1)=x1(i)+xdot1(i)*dt;

x2(i+1)=x2(i)+xdot2(i)*dt;

time(i+1)=i*dt;

end

x0(3001)=1;

plot(time,x0,time,x1,time,x2);title('Step Response of Dual Spring-Mass-Damper')

grid;xlabel('Time (s)');ylabel('Position')

legend('X0','Mass 1','Mass2')


12/49

1/4 Car Model

Spring/mass/damper #1 is the tire

itself, plus the effective mass of thecomponents that move along with thewheel (the unsprung mass)

Spring/mass/damper #2 is the actualspring and shock, plus the 25% ofcars body mass (the sprung weight)

Base displacement (X0) correspondsto road surface bumps

It is found that for typical values, thetire damping is not important

The damping of the overall vehiclemotion AND of the tire/wheelcombination is taken care ofby the single shock absorber


13/49

1/4 Car Model - Example

Sprung mass = 2000 lbs, so 500 lbs per wheel (15.53 slugs, 227 kg)

Suspension system: Spring rate = 462 lb/in (81,000 N/m) Damping coefficient = 44 lb/in/sec (7710 N/m/s)

Unsprung mass (wheel+tire) = 50 lbs (1.55 slugs, 22.7 kg) Tire characteristics:

Spring rate = 2000 lb/in (350,000 N/m) Damping coefficient = not specified in the text; so guess that the

damping ratio for the wheel+tire is relatively low, say 0.1, so C 3.2 lb/in/sec (564 N/m/s)

9.0

1.0

3/9.18

76.19/124

2

1

20

10

=

=

==

==

Hzsrad

Hzsrad

Mode 1 is wheel hopMode 2 is vehicle heave


14/49

1/4 Car Model - Example

All parameters as specified

Tire damping set to zero

The wheel+tire dynamics are

well controlled by the vehicleshock, so c1 may be neglected!


15/49

1/4 Car Model Classical form


16/49

Springs in Series

1

1k

Fx =

2

2k

Fx =

21

21

3

321

213

kk

kkk

k

F

k

F

k

F

xxx

+=

=+

=+=

FF

F

(p.239)

(Result for dampers is equivalent)

F

x

k


17/49

Nonlinear Springs

F

x

k

Since the spring rate for springs in series is always lower than the

highest rated spring, a stiffening effect for a compression spring canbe created by (for example) a coil spring where coils begin to close upAlternatively, bump stops can be added to shocks or elsewhere


18/49

Transient (and Frequency) Responses

Performance measures include: delay time (td), rise time (tr), peak time (tp),maximum overshoot (Mp), settling time (ts), [response time (1/e final value)]


19/49

Transient Responses

Start with SMD system

Several other graphsin our textbook


20/49

Transient Responses

Slight modification for ramp input, rather than strict step input:


21/49

Frequency Responses

Non-typical approach shown here, to avoid Laplace Transforms, etc.:

( )

( ) ( )22221

2

2

1;

:

)()()()()(

)()()(

)(

)(

)(

)()(

)(

ckmA

B

F

x

km

cTan

onmanipulatiSome

bSinaCosbCosaSinbaSinNow

tASintCoscBtBSinmk

tSinBx

tCosBx

tBSinx

tASintFSuppose

tFcxxkxm

+=

=

+=+

=+++

+=+=

+=

=

=++

Gain

Phase


22/49

Frequency Responses

SMD system Notice resonance at a

particular frequency Resonance peak is a strong

function of damping ratio The slope of the gain curve

changes around theresonance frequency

The phase (between inputand output) also changesdramatically at the same

frequency These observations generalize

to multi-SMD (and other)systems


23/49

Frequency Responses


24/49

Where Are We Heading ?

Full vehicle model (1/4 car model at each corner )

More elaborate than dual SMD system

Reducible to SMDs or dual SMDs (with limited D-O-Fs)

Not well covered in book, one of the RCVD software packages?

Dynamics of bicycle model SMD or dual SMDs

Chapter 6, then 21, 22, 23

Ride quality

SMD or dual SMDs

Chapter 16 and various

Others ?


25/49

Full Vehicle Model

Start with rigid vehicle Study load transfer due to

acceleration/braking and cornering) RCVD 4-wheel model

Add suspension dynamics Add vehicle compliances (!) Add driver inputs VERY complex problem


26/49

Equations of Motion (recap)

VVrvR

Va

rINIT

maYmaF

y

z

y

+=+=

==

==

2

:

:Lateral force, yawing moment

(V constant)

Lateral accn can arise from pathcurvature and true rectilinear accn

Tire slips

Strictly, slip is +ve to right(Table 5.1)

Often drawn in ve sense(Figure 5.17, etc.!)

Mostly we use F = Ci.e. reversed


27/49

Equations of Motion II (recap)

bYaYN

YYY

CYCY

V

ar

V

br

RF

RF

FFFRRR

FR

=

+=

==

+==

;

;Evaluate forces at front and rear tires

(careful with signs!)

Forces on vehicle become

(Eqns 5.5, 5.6)


28/49

Equations of Motion III (recap)

( )

?

1

.

BuAXX

mVY

I

NR

V

mV

Y

mVY

I

N

I

N

r

YrYYrmV

NrNNrI

etcNN

z

r

zz

r

r

rZ

+=

+

=

++=+

++=

Derivative notation

Eqns 5.7, 5.8

Can be written as

The B matrix is appliedforce and moment ?

Solution can now follow different paths:1. Analytic textbook, pages 150 and beyond2. Numerical work from basic eqns or matrix form?3. Simulation numerical integration of eqns


29/49

Derivatives (recap)

Yaw moment induced by steering (large, +ve?)

Yaw moment due to yaw rate (damping) (large, -ve?)

Yaw moment due to sideslip (small?)

YY

r

Y

Y

YY

NN

r

NN

NN

r

r

Side force induced by steering (fairly large, +ve?)

Side force due to yaw rate (small?)

Side force due to sideslip (damping) (large, -ve?)

ControlDampingCoupling


30/49

Derivatives (recap)

aCaV

arC

N

V

arCYbYaYN

NN

FF

FFFFRF

=

+

=

+===

;;

Example 1

Example 2

(eqns 5.9)

( )V

bCaC

V

brC

V

arC

rr

Y

V

brCY

V

arCYYYY

r

YY

RFRF

RRRR

FFFFRF

r

=

+

+

=

==

+==+=

;

;;

(eqns 5.9)


31/49

Steady-State Responses I (recap)

Set higher derivatives to zero!

yaRrNF ,,,,

( ) ( )( ) ( ){ } ( )

( ) ( )

( ) aCrV

bC

V

aCbCaC

bV

brCaV

brC

bYaYN

mC

mCC

VmaC

VmbCVr

mV

brCV

brCVr

m

YY

m

YaVr

R

Va

FRFRF

RF

BF

FFRFR

FR

rFyy

+=

+=

==

+=+

++

=

+====

0

0

0

;

2

Gets quite messy!


32/49

Steady-State Responses II (recap)

( )

YrYYmVrrmV

NrNNrI

r

rZ

++==+

++==

0

Try derivative form instead :

Solve as simultaneous equations, see textbook- or perhaps ?

Once derivatives are known, solution is easy

BuAXBuAXX

mVY

IN

RV

mV

Y

mVY

I

N

IN

rz

r

zz

r

=+==

+

==

10

1

0

Careful here withradians and degrees!


33/49

Steering Responses (recap)

Path Curvature

Angular Rate

After some tricky algebra we find:

K=Stability Factor (eqn. 5.42)

NSforl

Vr

NSforl

R

=

=

1

1

+=

+=

2

2

1

1

1

11

1

KVl

Vr

KVl

R


34/49

Application Examples


35/49

Equations of Motion (new)

Gather equations of motion for bicycle model:

Steady-state responses (eqns 5.13, 5.14, 5.17 in book)

( )

YrYYYrmV

NrNNNrI

r

rZ

++==+

++==

(Vehicle slip) (Angular rate) (Steering input)

( )( )( )rr

rr

rr

NYmVNYN

mVYNNYNYmVNYNV

YNNYVr

R

=

==

1

(Turn response)

(Sideslip response)


36/49

Equations of Motion (new) - II

Derivatives are known as functions of vehicle characteristics:

Simple Example NS car with a=b, same tires front and rear:

( )

F

RFr

RF

CY

bCaCV

Y

CCY

=

=

+=

1 ( )

F

RFr

RF

aCN

CbCaV

N

bCaCN

=

+=

=

221

F

r

CYY

CY

==

=

0

2

( )aCN

CaV

N

N

r

=

=

=

221

0


37/49

Corner Entry Example #1

NS car with a=b=5 ftC

F

=CR

=-630 lb/deg; V=88 ft/s (or 120, 50)m=60 slugs Iz=600 slug-ft2

Y= -1260 lb/deg = -321,330 N/rad N=0Yr = 0 Nr = 357.95

= 27,827 Nm/rad/s

Y = 630 lb/deg = 160,665 N/rad N = 3150 ft-lb/deg= 244,845 Nm/rad

r (SS) = 8.8 deg/s/deg R = 573 ft with 1 degree steer (SS) = -0.144 deg/deg

Since some derivatives are zero, the order of the governing equationsis lower, so we do not expect oscillatory responses


38/49


88 ft/s, 1 degree

Vehicle slip is closeto zero

Front tire loadfluctuatesconsiderably!

Rear tire load buildsup smoothly


39/49

Corner Entry Example #1b

120 ft/s, 1 degree

Yaw and slipdegrees of freedomexhibit differenttime constants

Front tire loadfluctuatesconsiderably!

Vehicle slip is

negative (nose in)


40/49

Corner Entry Example #1c

50 ft/s, 1 degree

Vehicle slip ispositive (nose out)

Time constantsappear to be speeddependent?


41/49


Change vehicle to OS configuration:a=6 ft, b=4 ftCF=CR=-630 lb/deg; V=88 ft/sm=60 slugs Iz=600 slug-ft2

Y= -1260 lb/deg = -321,330 N/rad N=97,941 Nm/radYr = -14.318 lb-sec/deg Nr =-372.27 ft-lb/deg/s

= -3651.5 N/rad/s = 28,927 Nm/rad/sY = 630 lb/deg = 160,665 N/rad N = 3780 ft-lb/deg

= 292,830 Nm/rad

r (SS) = 11.85 deg/s/deg R = 425 ft

(SS) = -0.501 deg/deg

The order of the governing equations suggests a more complexresponse than the NS case


42/49

Corner Entry Example #2b

88 ft/s, 1 degree

Vehicle slip isnegative (nose in)

Responses confirma more complexbehavior than theNS case

Still not oscillatory


43/49

Wheel Load Distributions (some thoughts on )

The classic 4-wheeler is statically over-determinedThis implies that the load distribution due to applied forces or moments

can be controlled by varying the compliance at each corner

Forces/moments reacted by wheel loads:

Vertical (weight)

Pitch moment (about y)Roll moment (about x)

The diagonal load is undefined

RF

LR

RR

LF

y

x z

( )

( )RRRFLRLFt

N

RRLRRFLFl

M

RRLRRFLFWZ

+=

+=

+++==

2

2

M-Z

N


44/49

Wheel Load Distributions in Corners

Use the RCVD software to save some time/effortStart from nominal NASCAR vehicle, aero turned off75% roll stiffness assigned to the front88 ft/s (60 mph)Realistic tire behavior!


45/49


The weight transfer from RL is quite dramatic75% of total transfer occurs at the front (anti-roll bar presumed in play)Rather nonlinear with realistic tire characteristics


46/49


Front anti-roll bar is increasing lateral weight transfer at frontHelps keep the rear wheels (driven?) in contact with ground?

What happens if we eliminate the anti-roll bar?

Guess that weight transfer from RL will now be evenly shared FB?


47/49


48/49

Vehicle Pitch and Heave

The bicycle model can be extended to address vehicle heave and pitchresponse to bumps:

m

L

2 modes are expected, pureheave and pure pitch for anideal NS configuration,otherwise a pitchy heave

and a heavy pitch

HEAVEss

PITCH

y

sHEAVE

fLmL

K

mL

LKf

mLI

m

Kf

624

121

24

12

1

4

2

2

Pitch and heave frequencieswill tend to be same OOMbut not identical


49/49

An Unusual Case of Manipulation of the Pitch Response

The original Mini employed Hydrolastic suspensionAffected stiffnesses for heave and pitch motions in different waysAlso limited pitch motions, analogous to an anti-roll bar for roll

vehicle dynamics ii v.c

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