Mechanical Systems I Mechanical Systems I

Mechanical Systems I Mechanical Systems I

System Analysis Spring 2011

1 Seoul National University

Mechanical and Aerospace Engineering

Vehicle Suspension p

Vehicle Suspension p

System Analysis Spring 2011

3 Seoul National University

Mechanical and Aerospace Engineering

Powertrain – Powertrain Model

• Engine Model

• Torque Converter

• Transmission

• Axle Shaft

• Differential Gear

Powertrain and Brake System

System Analysis Spring 2011

5 Seoul National University

Mechanical and Aerospace Engineering

5 / 46

Vehicle Suspension p

Y a w

P itc h R o ll

X

Z

Y

4 - Q u a rte r C a r M o d e l

R o ll b a r

Y a w

P it h

Z P itc h

R o ll X

Z

Y

4 - Q u a rte r C a r M o d e l

R o ll b a r

Ride Quality

Suspension

spring&damper Suspension

System Analysis Spring 2011

7 Seoul National University

Mechanical and Aerospace Engineering

Newton’s Laws Newton’s Laws

1)First law : conservation of momentum no external force

 no momentum change

linear momentum : mv angular momentum : 2) S d l

J 

F ma m dv 2) Second law :  F ma m

dt T J  J d 

 



T J J

 dt

 



Three Basic Elements in Modeling Mechanical Systems Three Basic Elements in Modeling Mechanical Systems

i Inertial elements _{) (} kinetic energy _):

masses M : moments of inertial : J ii Spring elements _{) (} Potential energy ₎

X

) (₁ ₀

   

k k

T

X



 Torsional spring

T

 xb

F

iii Damper elements _{) (} Energy dissipation ₎



b



x b



x

 x T 

b_

T

System Analysis Spring 2011

9 Seoul National University

Mechanical and Aerospace Engineering

Spring Elements Spring Elements

4

64 3

k Gd nR

d wire diameter





4

T 64 k Ed

 nD nnumber of coils

k EA

 L

4 4

( )

G D d L

3 3

k 4Ewh

 L

4 4

( )

T 32

G D d

k L

 



L

3

4 3

k Ewh

 L ⁴

T 32 k GD

L



3 3

16Ewh k  L

T 32 L

2Ewh3

k 

Equivalent masses of common elements

4Ewh3

0 38

e

3

k 4Ewh

 L m_e

 0.38

md

m

I

I

  I I

/ 3

3

4 3

k Ewh

 L

GD4

e

0.23

m



md ^{e d}

3 3

16Ewh k L

4

T 32 k GD

L



e

0.5

m



md

System Analysis Spring 2011

11 Seoul National University

Mechanical and Aerospace Engineering

Damping Elements Damping Elements

Piston damper p

Pneumatic door closer

Oleo strut

Rotary damper

Damper with spring loaded valves

System Analysis Spring 2011

13 Seoul National University

Mechanical and Aerospace Engineering

Examples of Modeling Mechanical Systems Examples of Modeling Mechanical Systems

bw b

ex ₁₎ t  0,  (0)  

J

d

J    b 

equation of motion _:

0 0

J b

dt

d d b

J b

dt dt J



   



    

, ( )

b

0

let t ce e e

J b

  

 



     ( ) t

( ) , 0, (0)

J

J

t ce t C





 

 

   

1

0e 0.368 0

 ^  

( )

bt J

t

e

 

 

T Time constant time to reach _{= : 63.2%}

Examples of Modeling Mechanical Systems Examples of Modeling Mechanical Systems

Spring

Spring--Mass Mass

ex ₂₎

x(t) k

0 ) 0 ( )

0 ( ,

0  ₀ 

 x x x^ t

m

) ( )

(

, _0,

( ) cos k cos

x t  x t  x  t

2 2

0 0

md x F kx dt

mx kx x k x

  

    



 

0 0

( ) cos cos

x t x t x t

m 

 

2 [sec]

/ T

k m

 



Period

( ) cos sin

(0)

m

k k

x t A t B t

m m

x A x

 

 

1 1

[ ] 2

f k Hz

T  m

 



Frequency

Natural frequency

(0)

(0) sin cos 0

x A xo

k k k k k

x A t B t B

m m m m m

 

    



2 [ / sec]

n

f k rad

    m



Natural frequency

System Analysis Spring 2011

15 Seoul National University

Mechanical and Aerospace Engineering

Examples of Modeling Mechanical Systems Examples of Modeling Mechanical Systems

Spring

Spring--mass mass--damper damper

0 0

(0) (0)

x x

x x





 

2 2

m d x kx bx dt    

b k 0

x x x

m m

  

 

k b b Exponential decay freq

2

0, ( )

x ^   x ^ x ^ x t ^{ } c e

    

^ 2 ^

x ^ 0

,

, 2 ,

n n

2

k b b

let m   m   mk   Damping ratio

p y q Natural frequency

2

0, ( )

x ^   x ^ x x t c e    ^ 2

^

x 0

 

¹

n n n n n

      

        

Examples of Modeling Mechanical Systems Examples of Modeling Mechanical Systems

b k 0 x  x  x 

 

0

(0)

(0) 0

x x

x x



 

 

m m

,

, 2 ,

n n

2

k b b

let m   m   mk  

( )

2

0

x   x   x 

 

m m 2 mk

2 2

( s X s ( )    s x x  )  2  ( sX s ( )  x )   X s ( )  0 Laplace Transform

0 0 0

2 2

0 0

2 2

( ( ) ) 2 ( ( ) ) ( ) 0

( 2 ) ( ) 2 0

( 2 ) ( ) 2

n n

n n n

s X s s x x sX s x X s

s s X s s x x

s s X s s x x

 

  

  

      

     

    

0 0

2 2

( 2 ) ( ) 2

( ) 2

2

n n n

n

n n

s s X s s x x

s x x

X s s s

  



 

    

  

 

System Analysis Spring 2011

17 Seoul National University

Mechanical and Aerospace Engineering

n n



Examples of Modeling Mechanical Systems Examples of Modeling Mechanical Systems

0 0 0 0

2 2 2 2 2 2

2 2

( ) s x

x s x

x

X s 

  









2 2

2 2

( ) 2 2 2

2 : charateristic polynomial

2 0 h i i i

n n n n n n

n n

s s s s s s

s s

     

 



     

 

 

2 2

2 2 2

2 0 : characteistic equation

1 : charateristic roots

n n

n n n n n

s s

s

 

     

  

       

2 2

1) 1 underdamped

1 1

n n n n

s j



     



       

0 0

2 2 2 2 2 2

( )

( ) ( ) (1 ) ( ) (1 )

n n n n

n n

n n n n

j

s x x

X s s s

   

 

     

 

 

     

2 2

0 2 0

( )

cos 1

sin

x t e

x   t  x   t





 

 

   

  

  

 

Examples of Modeling Mechanical Systems Examples of Modeling Mechanical Systems

2

2) 1 overdamped 1 s



  



   







1

( 2 )

( )

1 1

n n

n

s

s x

X s

s s

  



     

   

 

         



 



1 2

1 1

( )

n n n n

t t

s s

x t k e

k e

     

     

  

 

3)   1 critically damped

 

( 2 )

( )

n

n

s

s x

X s







 

 

   

   

1 2

( )

n

t t

s

x t k e

k te



 



 

System Analysis Spring 2011

19 Seoul National University

Mechanical and Aerospace Engineering

Examples of Modeling Mechanical Systems Examples of Modeling Mechanical Systems

Underdamped

1)

  1

2 2

0 0

( ) cos 1

sin

1

nt

n n

x t e

x   t  x   t





 

 

        

: :





 



1

d n

     natural frequency

damping ratio

Overdamped ( )

2)

  1



 



1 2

( )

x t  k e

 k e

( ) x t

2

1

n n

       

1 2

( )

 1

 1

1

     

3)

( )

x t  k e

 k te

 1

t

Dry Friction (no lubricant) Dry Friction (no lubricant)

 Fs ₌

Static Friction Force

Fk ₌

Kinetic Friction Force

Fk ₌

Kinetic Friction Force

Fs = sN s _:

Static Friction Coefficient

F k k N k

Kinetic Friction Coefficient

Fk = kN k _:

Kinetic Friction Coefficient

0

x   x   0

if sgn( ) if

s

s s

F F F

F F F F F



  

 

   

sgn( ) F

 F

x 

System Analysis Spring 2011

21 Seoul National University

Mechanical and Aerospace Engineering

 



Friction (with lubricant) Friction (with lubricant)



sgn( )

if and F ( )

bx G N x x

F F x F G N





    

      

  

if  and F ( )

( )sgn( ) i.e., if and F ( )

s

s s

F F x F G N

F G N F otherwise x F G N







    

      

 

b: viscous friction coefficient G: load-dependent factor N: normal force

Fs: the maximum static friction Fs: the maximum static friction

: a small bound for zero velocity detection

