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Suspension system

Q

S0

1

2

3

spring+damper

Q

S0

1

2

3

0:2 =++ 1232 FQFlink

points of

application only

?

Q

S0

1

2

3

0:3 =+ 0323 FFlink

Q

S0

1

2

3

0=++ 1232 FQF

QF12

F32

Q

S0

1

2

3

0:1 =++ SFF 0121link

F21

F01

S

RRTR example (1)

3

01

2

RRTR example (2)

3

01

2

M3

M1=?

RRTR example (3)

3

01

2

M3

M1=?

0.3 =+ 0323 FF

RRTR example (3)

3

01

2

M3

M1=?

0.2 =+ 3212 FF

RRTR example (3)

3

01

2

M3

M1=?

0.2 =+ 3212 FF

0.3 =+ 0323 FF

RRTR example (4)

3

01

2

M3

M1=?

...

0

0.3

23

3323

3

=

=−

=

F

MhF

M C

B

A

C

h3

F23

F03

RRTR example (5)

3

01

2

M3

M1=?

...

0

0.3

23

3323

3

=

=−

=

F

MhF

M C

B

A

C

h3

F23

F03

F32

F30

RRTR example (6)

3

01

2

M3

M1=?

0.1 =+ 0121 FF

B

A

C

h3

F23

F03

F32

F30

F12

F21

RRTR example (7)

3

01

2

M3

M1

0

0.1

1121

1

=−

=MhF

M A

B

A

C

h3

F23

F03

F32

F30

F12

F21

h1

F01

F10

dynamics, kinetostatics 16

IN KINETOSTATICS

THE MOTION OF MASS BODIES (LINKS)

IS REPRESENTED BY INERTIA FORCES

Inertia forces

Newton, Euler, D’Alembert

εM Sb I−=00 =+=− bSI MMεM

εM SI=

aF mb −=00 =+=− bm FFaF

aF m=

dynamics, kinetostatics 18

Mass moment of inertia

+=m

Z dmyxI )( 22

)( 22

SSSZ yxmII ++=

dynamics, kinetostatics 19

aF mb −=

εM Sb I−=

ma

I

F

M

F

Mh S

b

b

M

b ===

FORCE (Fb) and MOMENT (Mb) OF INERTIA

Given:

m, IS, a, ε

dynamics, kinetostatics 20

FORCE (Fb) and MOMENT (Mb) OF INERTIA

resultant inertia

force

dynamics, kinetostatics 21

Center of percussion

ASAS ==

+=

t

S

2n

S

t

S

n

S

; aa

aaaS

S

A

B

aS

SS

dynamics, kinetostatics 22

Center of percussion

S

A

B

aS

SSW

Fb

h

SI

m

εM

aF

b

Sb

−=

−=

S

S

b

b

ma

I

F

Mh

==

B

S

A aS

SSW

Fb

h

cos(?)==S

t

S

a

a

SW

hCenter of percussion

=

=

=

==

t

S

S

S

t

SS

t

S

S

S

S

t

S

S

a

a

ma

AS

aI

a

a

ma

I

a

ahSW

mAS

I

a

a

ASma

aI S

t

S

S

S

t

SS ==

B

S

A aS

SSW

Fb

h

AS

i

mAS

mi

mAS

ISW SSS

22

===

Center of percussion

iS – radius of inertia

dynamics, kinetostatics 25

-100000

-80000

-60000

-40000

-20000

0

20000

40000

60000

80000

100000

0 0,004 0,008 0,012

czas [s]

Angular accel

of BC

20000

25000

30000

35000

40000

45000

50000

55000

60000

Accel of point

S

m, J

A

C

B

S

BC = 0,2 m

1 = 500 s-1 (1 = const)

(n1 = 5000 rev/min)

dynamics, kinetostatics 26

m, J

A

C

B

S

aSmax = 55000 ms-2

Bcmax = 90 000 s-2

m=0,2 kg, J=0,01 kgm2

Fbmax = - ma = 11000 N !!!

Mbmax = - J = 900 Nm !!!

dynamics, kinetostatics 27

NmaFb 250=−=25=STATIC

b

F

F

S – center of gravity

e

S

Fb

22 /250

/1500

001,0

smea

s

me

S ==

=

=

kgm 1=

Fb

Example of analysis (kinetostatics)

m, J

M1=?

3

0

2

1 1

Data:

m, J – mass, mass moment of inertia

1 – angular velocity of link 1

M1 = ? and joint forces

2

Mb

a

Fb

maFb −=

JM b −=

ma

J

F

Mh

b

b ==

bM Couple of forces Fb

bFbF h

2

Mb

a

Fb

2

Fb

a

2

Fb

Fb

a

Fb

h

M1

3

0

2

1

Fb

F03

c

F23

c

0

A

0=+ 0323 FF

Link 3:

M1

3

0

2

1

Fb

F03

c

F23

c

0

A

0=+ 0323 FF

Link 3:

0=++ 12b32 FFF

Link 2:

M1

3

0

2

1

Fb

F30

c

F32

c

F12

c

0

A

F32

c

F12

c

0=++ 12b32 FFF

M1

3

0

2

1

Fb

F30

c

F32

c

F12

c

F10

c

0

A

h1

c

F32

c

F12

c

=−→==+ 000 12111 hFMM A

0121 FF

m, J

M1=?

3

0

2

1

Fc

c

1

Data:

FC – input force (driver)

m, J – mass, mass mom. of inertia

1 – angular vel of crank 1

M1 = ? and joint forces = ?

Example of analysis (kinetostatics)

M1 =?

3

0

2

1

Fc

c

Fb

F03

c

0

0=++ 12b32 FFF

0=++ 2303C FFF

???

Slider 3:

Coupler 2:

Statically determined group (of links)

kinematic system in equilibrium

every link in equilibrium

any group of links in equlibrium

forces may be fully determined using the six (three) equations of 3D (2D) statics without

using deflection and stiffness criteria.

j

i

Fijx

Fijy

i

j

Fijx

Fijy

i

j

Fijx

Fijy

1 unknown 2 unknowns 2 unknowns

Number of

equations

= Number of unknowns

3k = 2p1 + p2

In kinetostatics unknowns are joint forces

( ) 21 1213 ppnDOF −−−=

0123 21 =−− ppk

Mobility

Condition of static

determination

k 1 2

...

4

p1 1 3

...

6

p2 1 0

...

0

(k-p1–p2) (1-1-1) (2-3-0) (460)

Examples of static. determined groups

group (1-1-1) group (2-3-0)

II class - R or T II class - K or J or Z

I kl - R lub T

1-1-1 stat. determined group (one link)

3

1

2

F12 F2

F32

F32 + F12 + F2 = 0

2-link groups (k-p1-p2 = 230)

RRR

RTR RTT

RRT

TRT