kN 1000newton ora 3600sec

v 40kmora

viteza de deplasare

d 2m d 2 m Lungimea de unda a denivelarii cosinusoidale

h0 2cm Inaltimea denivelarii cosinusoidale

mtot 553.1kgmb 0.93 mtot mb 514.383 kg Masa suspendata

mw mtot mb mw 38.717 kg Masa nesuspendata

kt 210kNm

Rigiditatea pneului

ks 29kNm

Rigiditatea arcului

ct 0.18kN secm

Coeficientul de amortizareal pneului

ct 0.1kN secm

cs 2.5kN secm

Coeficientul de amortizare al amortizorului

bksmb

b 7.509radsec

Pulsatia proprie neamortizata inferioara (acaroseriei)

fbb

2 rad fb 1.2 Hz Frecventa proprie neamortizata inferioara(acaroseriei)

wktmw

w 73.648radsec

Pulsatia proprie neamortizata inferioara (a rotii)

fww

2 rad fw 11.72 Hz Frecventa proprie neamortizata superioara (a rotii)

k1

1kt

1ks

k 25.481 kN

m Rigiditatea echivalenta a suspensiei

c1

1ct

1cs

c 0.096 kN m

s

1

Coeficientul de amortizare echivalent alsuspensiei

Ecuatiile de miscare i 1

mbd2

dt2zb cs

ddt

zwddt

zb ks zw zb =

mwd2

dt2zw ct

ddt

h ddt

zw kt h zw cs ddt zw ddt zb ks zw zb =

mbd2

dt2zb cs

ddt

zb ks zb csddt

zw ks zw=

mwd2

dt2zw ct cs ddt zw kt ks zw ct ddt h kt h cs ddt zb ks zb=

2 mb zb i cs zb ks zb i cs zw ks zw=

2 mw zw i ct cs zw kt ks zw i ct h kt h i cs zb ks zb=

2 mb i cs ks zb i cs ks zw=

2 mw i ct cs kt ks zw i ct kt h i cs ks zb=

zw2 mb i cs ksi cs ks

zb=

2 mw i ct cs kt ks

2 mb i cs ksi cs ks

i cs ks

zb i ct kt h=

zbi cs ks

2 mb i cs ks

zw=

2 mw i ct cs kt ks i cs ks

2

2 mb i cs ks

zw i ct kt h=

Rrbzbh

= Transmisibilitatea (raspunsul in frecventa) drum - masa suspendata

Rrb ( )i ct kt

2 mw i ct cs kt ks

2 mb i cs ksi cs ks

i cs ks

rb ( ) arg Rrb ( ) defazajul Rrw

zwh

= Transmisibilitatea (raspunsul in frecventa) drum masa nesuspendata

Rrw ( )i ct kt

2 mw i ct cs kt ks i cs ks

2

2 mb i cs ks

rw ( ) arg Rrw ( ) defazajul

Rwbzbzw

= Transmisibilitatea (raspunsul in frecventa) masa nesuspendata - masasuspendata

Rwb ( )i cs ks

2 mb i cs ks

wb ( ) arg Rwb ( ) defazajul In lipsa amortizarii

Md2

dt2 Q K Q 0=

Obtinerea pulsatiei proprii

1kmb

i i

1 7.038radsec

12 rad 1.12 Hz

Timpul in care e parcursa lungimea unei denivelari d 2 m v 40 kmora

t0 dv

= 2

=

2 rad vd

34.907 radsec

f 2 rad f 5.556 Hz

h ha ei t= Miscare armonicaddt

h i ha ei t= i h= Viteza miscarii armonice

d2

dt2h i ( )2 ha ei t= 2 h= Acceleratia miscarii armonice

Echilibrul de forte

F k c i( ) h z( )=

mtotd2

dt2 z c d

dth d

dtz k h z( )=

F K i C 2 M Q=Q K 2 M i C 1 F=mtot d2

dt2 z c d

dt z k z c d

dt h k h=

2 mb i c k z i c k( ) h=

Functia de transmisibilitate ( raspunsul in frecventa ) este

Trbzh

= i c kk 2 mtot i c

=AF

Fmtot g

=

Trb ( )i c k

k 2 mtot i c transmisibilitatea

c 96.154 kg s 1

Drb ( ) arg Trb ( )

Trb 1( ) 12.533 valoarea maxima a transmisibilitatii

f 0.01 Hz 0.11 Hz 50 Hz

0 2 4 6 8 10 12 14 16 18 200

0.222

0.445

0.667

0.889

1.111

1.334

1.556

1.778

2.001

2.223

Rrb-2glRrw-2glRwb-2gl

Rrb 2 f( )Rrw 2 f( )Rwb 2 f( )

f

0 1.4 2.8 4.2 5.6 7 8.4 9.8 11.2 12.6 140

0.3

0.6

0.9

1.2

1.5

1.8

2.1

2.4

2.7

3

Rrb-1glRrb-2glRrw-2gl

Trb 2 f( )Rrb 2 f( )Rrw 2 f( )

f

0 5 10 151

0.5

0

0.5

Rrb-2glRrw-2glRwb-2gl

rb 2 f( )

rw 2 f( )

wb 2 f( )

f

0.01 0.1 1 10 1000.01

0.1

1

10

100

transmisibilitateacabinascaun

Rrb 2 f( )rb 2 f( ) 0.1

Trb 2 f( )

f