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[ ] [ ] [ ][ ] [ ][ ] [ ][ ] [ ] [ ][ ]

=

QJQASQ

QSAmMq

11 TTT

T1

O

, (10)

the kinetic energy of body 1 becomes [1, 2, 7]

{ } [ ]{ }1T1 121T qMq q&= , (11)the Lagrange equations read [4, 5, 6, 8] as

{ }1

11

TT

d

dqF

qq=

&t

, (12)

in whichT

1 61q

T

q

TT

=

&L

&&q, (13)

T

1 61q

T

q

TT

=

L

q, (14)

{ } Tqqq 6211 FFF L=qF , (15)

where within the generalized forces1

qF are both the

given forces and the bond forces.On the other hand, we can write [7]

[ ]{ }11

1

TqM

qq &

&=

, (16)

{ }

[ ]{ } { }

[ ]{ }

=

11

T

11

T

1

1

1

1

1 2

1

T

qM

qqM

q

q

qq&&L&&

OX

(17)

and if we call

{ } [ ]{ }

+=

11

T~

11 qqMF qq &

& , (18)

it results the Lagrange equations [7]

[ ]{ } { } { }111

~1 qqq FFqM +=&& (19)

III. Elastic forces

If we call the undistorted AB spring length, 0l (fig.

1), then the force of this spring is

( )AB

lABk ABF 0= (20)

with its components

( )AB

XXlABk AB

= 0XF ,

( )AB

YYlABk AB

= 0YF ,

( )AB

ZZlABk AB

= 0ZF ,

(21)

where

( ) ( ) ( )222 ABABAB ZZYYXXAB ++= . (22)

Given that the force F derives from the potential

( )202

lABk

V = , (23)

and denoting

{ } [ ]TAAAA zyx=r , [ ]

=

0

0

0

AA

AA

AA

A

xy

xz

yz

r ,(24)

{ } [ ]TBBBB zyx=r , [ ]

=

0

0

0

BB

BB

BB

A

xy

xz

yz

r ,(25)

it results the generalized forces column matrix [7]:

{ } ( ) [ ]

[ ] [ ][ ]

=

AB

AB

AB

A ZZ

YY

XX

AB

lABkT

1

T0

ArQ

IF , (26)

where

[ ]{ }AO

O

O

A

A

A

Z

Y

X

Z

Y

X

rA1

1

1

1

+

=

, (27)

[ ]{ }BO

O

O

B

B

B

Z

Y

X

Z

Y

X

rA2

2

2

2

+

=

. (28)

In the case of n springs, the equation (26) is rewritten as

{ } ( )

[ ][ ] [ ][ ] .T1T

1

0

==

ii

ii

ii

iAB

AB

AB

A

n

i ii

iiii

ZZ

YY

XX

BA

lBAk

ArQ

I

F

. (29)

IV. Small oscillations around the equilibrium position

In fig. 2 the rigid is represented in equilibrium position

with continuous line. It is connected by several springs

iiBA (in fig. 2 is represented only one, AB ) with the

base which is also represented in continuous line.

O Y

X

Z

C

Amg

uC

u

k

BO0

Base

O'Y'

X'

Z'

A'

B'O'0

A

B

Fig. 2. The small oscillations.

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At one moment the rigid body and the base reach to

the positions represented with discontinuous line, points

A and B corresponding now to points A and B .We will denote:

k, the stiffness of the spring AB ;

OXYZ, the rigid body reference system at the

considered equilibrium and the general reference system;

, the point O displacement;

X , Y , Z , the projections onto the OX , OY ,OZaxes;

, the rigid body angular displacement (considered

to be small);

X , Y , Z , the projections onto the OX , OY ,OZaxes;

B , the base point B displacement;

AB , the relative displacement defined by

BAAB = ; (30) a , b , c , the AB direction unit vector uprojections

onto the OX , OY , OZaxes;

L , the length of the AB segment;

L~

, the undistorted length of the spring AB ;

s , the equilibrium spring elongation

LLs~

= ; (31) Ar , the OA vector;

AX , AY , AZ , the OA vectors projections onto the

OX , OY , OZaxes;

ABF , the elastic force in the spring AB ;

ABM , the momentum defined byABAB FOAM = ; (32)

XAB

F ,YAB

F ,ZAB

F ,XAB

M ,YAB

M ,ZAB

M , the

ABF , ABM , vectors projections onto the OX , OY ,

OZaxes;

{ }U , { }ABF , { } , { }u , { }Ar matrices given by:

{ } [ ]Tfedcba=U , (33)

{ } [ ]TZYXZYX ABABABABABABAB

MMMFFF=F ,(34)

{ } [ ]TZYXZYX = , (35)

{ }

=0

0

0

ab

ac

bc

u , { }

=0

0

0

AA

AA

AA

A

XY

XZ

YZ

r ; (36)

C, the weight centre of the rigid body;

m , the mass of the rigid body;

{ }Cu , the unit vector of the weight force atequilibrium;

CX , CY , CZ , the OC vector projections onto the

OX , OY , OZaxes;

Ca , Cb , Cc , the Cu vector projections onto the

OX , OY , OZaxes;

Cd , Ce , Cf , the CuOC vector projections ontothe OX , OY , OZaxes;

{ }CU , the column matrix{ } [ ]TCCCCCCC fedcba=U ; (37)

[ ]Cu , [ ]Cr , (36) relation type matrices; [ ]0 , [ ]I , the null matrix, respectively the 3rd unity

matrix;

)1(BK , [ ])2( BK , [ ])3(ABK , [ ]ABK , [ ]CK , [ ]K , the

rigidity matrices given by [3]

[ ] { }{ }T)1( 1 UUK

=L

skAB ,

[ ] [ ] [ ]

[ ] [ ]

=

2

T)2(

AA

AAB

L

ks

rr

rIK ,

[ ] [ ] [ ][ ] [ ][ ] [ ][ ]

+=

AA

AB

ks

ruur0

00K

2

)3(,

(38)

[ ] [ ] [ ] [ ])3()2()1( ABABABAB KKK ++=K , (39)

[ ] [ ] [ ][ ] [ ][ ] [ ][ ]

+=

CCCC

C

mg

ruur0

00K

2, (40)

[ ] [ ] [ ]=

+=n

i

BAC ii1

KKK ; (41)

XJ , YJ , ZJ , XYJ , XZJ , YZJ , the inertial

moments of the rigid body; [ ]OJ , [ ]M , the matrix of the inertial moments,

respectively the inertial matrix

[ ]

=

ZZYZX

YZYYX

XZXYX

O

JJJ

JJJ

JJJ

J ,

[ ] [ ] [ ][ ] [ ]

=

OC

C

rm

rmm

J

IM

T

;

(42)

, the displacement of the 0O base point;

X , Y , Z , the projections of the vector onto

the OX , OY , OZaxes; , the angular displacement (considered small) of

the base;

X , Y , Z , the vector projections onto theOX , OY , OZaxes;

BX , BY , BZ , the BO0 vector projections onto the

OX , OY , OZaxes;

d~

, e~ , f~

, the uBO 0 vector projections onto theOX , OY , OZaxes;

{ }U , { } , { }Br , the matrices given by [3]

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{ } [ ]T~~~ fedcba=U , (43){ } [ ]TZYXZYX = , (44)

{ }

=0

0

0

BB

BB

BB

B

XY

XZ

YZ

r ; (45)

[ ])1(~BK , [ ])2(

~ABK , [ ]ABK

~, [ ]K~ , the matrices given by

[ ] { }{ }T)1( ~1~ UUK

=L

skAB

, (46)

[ ] [ ] [ ][ ] [ ][ ]

=

BAA

BAB

L

ks

rrr

rIK

)2(~, (47)

[ ] [ ] [ ])2()1( ~~~ ABABAB KKK += , (48)

[ ] [ ]

==

n

i

BA ii1

~KK ; (49)

V. Column matrix of excitation force

The components of the { }ABF column matrix aregiven by the partial derivatives of the BV spring potential

energy

( )2~''2

LBAk

VAB = . (50)

The deformed spring length is successively written:

( ) ( )[ ]

( )2

22

''

AB

BA

L

BA

u

ABB'A'

=

==(51)

and if we proceed on series development and we keep the

maximum second degree nonlinear terms, the following

formula is achieved

( ) ( )

( ) ,

122

2

22

+

+=

AB

ABABAB

L

s

L

sss

kV

uu

(52)

where the displacement A reads

( )OAOA ++=2

1A . (53)

Further on, taking into account that the AB vectorspartial derivatives with respect to the arguments X , Y ,

Z , X , Y , Z , are equal to the A vectors partialderivatives with respect to the same parameters and if we

call { }eABF the column matrix of the excitation bycomponents

( )

+

i

AB

i

AB

qL

s

qL

ks uu

1 , (54)

where iq , 6,1=i , represent the displacements X , Y ,

Z , X , Y , Z , we get

{ } { } [ ]{ } { }eABAB ks FKUF += . (55)The column matrix of the excitation force { }

eAB

F ,

based on the expressions like

i

=

X

A ,

( ) ( )[ ]OAiOAiOAi

++=

2

1

X

A ,

(56)

by keeping the linear terms and taking into account the

(43)(47) notations, it results from (54) and we can write

{ } [ ]{ }KF ABeAB~

= . (57)

If the base has no rotational motion ( )0= the (57)relation becomes

{ } [ ]{ }KF ABeAB = , (58)and if the equilibrium elongations are insignificant( )0s , then

[ ] [ ] { }{ }T~ UUKK kABAB == . (59)

VI. Vibration differential equation

The potential energy of the gravity force is

CCG mgV u = , (60)where [3],

( )CCC rr ++=2

1, (61)

and from here, using the partial derivatives with respectto X , Y , Z , X , Y , Z , and taking into accountthe previous relations, it results the gravity force column

matrix

{ } { } [ ]{ }KUG CCmg = . (62)At equilibrium, the displacements { } , { } , being

null, the following equation is obtained

{ } { } { }0UU =+=

n

i

Cii mgsk

1

. (63)

The movement equation is obtained from the dynamic

equilibrium equality

{ } { } { } { }0FG =++=

i

n

j

BA jjF

1

, (64)

where, for the small movements case

{ } [ ]{ }MF &&=i , (65)and taking into account the relation (63), it results the

expression[6,7,8]

[ ]{ } [ ]{ } [ ]{ }KKM ~=+&& , (66)

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VII. Application

For the plate of mass m (Fig. 3) is requested the

excitation force knowing that the 0O base point has the

displacement t= cos0 , parallel with the OX axis,

and the base rotation is t= cos0 , about the OZaxis. Let us consider that the AB , CD springs are

identical, by the stiffnessl

mgk= , and at the

equilibrium position the 0EO spring (by stiffness k~

) has

zero elongation.

AC

E

D B

F

X

Y

O

mg

k k

k~

3l

l

l

2l

3l

l

l

2l

2l 2l 3l 3l

O0

Fig. 3. Application.

From equilibrium condition it results that the AB ,

CD spring elongations are equal and considering that

their common value is s , it results

22

l

k

mgs == (67)

and knowing that ,22lL = it follows .4

1=

L

s

For the AB spring the following relations were

reached

{ } [ ]T2000112

2l=U ,

{ } [ ]T2000112

2~l=U , [ ]

=

03

300

00

ll

l

l

Ar ,

[ ]

=

057

500

700

ll

l

l

Br ,

(68)

[ ][ ]

=2

22

22

2200

01521

057

l

ll

ll

BA rr , (69)

[ ]

=

2

)1(

400022

000000

000000

000000

200011

200011

8

3~

lll

l

l

kABK ,

[ ]

=

2

22

22

)2(

220003

01521300

05700

057100

500010

700001

4

~

lll

lll

lll

ll

l

l

kABK ,

(70)

and, in the same way, for the CD spring the successive

expressions were obtained

{ } [ ]T2000112

2l=U ,

{ } [ ]T2000112

2~l=U , [ ]

=

03

300

00

ll

l

l

Cr ,

[ ]

=

057

500

700

ll

l

l

Dr ,

(71)

[ ][ ]

=

2

22

22

2200

01521

057

l

ll

ll

DC rr , (72)

[ ]

=

2

)1(

400022

000000

000000

000000

200011

200011

8

3~

lll

l

l

kCDK ,

[ ]

=

2

22

22

)2(

220003

01521300

05700

057100

500010

700001

4

~

lll

lll

lll

ll

l

l

k

CDK .

(73)

The EFspring elongation being null, it results

{ } { } [ ]T000010~ == UU (74)

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[ ]

=

000000

000000

000000

000000

000010

000000

~~kEFK , (75)

and the [ ]K~ matrix is[ ] [ ] [ ] [ ] [ ] [ ]EFCDCDABAB KKKKKK

~~~~~~ )2()1()2()1( ++++= , (76)

wherefrom

[ ]

+

=

2

2

2

3200004

03000000014200

0014200

0000

~4

50

2000005

4

~

ll

lll

lk

k

l

kK . (77)

On the other hand, we can write

{ } [ ] t= cos0000 T00 , (78)resulting the excitation force

{ } [ ]{ } t

ll

l

ke

+

== cos

324

0

0

0

0

205

4

~

02

0

00

KF (79)

VIII. Conclusions

In our work we presented, in a specific multibody type

form, the dynamics of a rigid body linked with linear

elastic springs by another one, with imposed motion. Weobtained the equations of motion and we studied the case

of small oscillations around the equilibrium position. For

the theory described we completely solved a practical

application. The presented method can be used in most

general situations such as vehicle suspensions, seismicexcitations, etc.

Acknowledgement

The second authors contribution to this paper is based on

the European Program Dezvoltarea colilor doctorale

prin acordarea de burse tinerilor doctoranzi cu frecvenPOSDRU/88/1.5/S/52826.

References

[1] Shabana, A., A., Dynamics of Multibody Systems, 3 rd edition,

Cambridge University Press, New York, 2005.

[2] Amironache, F., Fundamentals of multibody Dynamics. Theory and

Applications, Birkhuser, Boston, Basel, Berlin, 2006,

[3] Pandrea, N., Elements of the Solids Mechanics in Plckerian

Coordinates, The Publishing House of the Romanian Academy,

Bucharest, 2000.

[4] Lurie, A., I., Analytical Mechanics, Springer-Verlag, Berlin,Heidelberg, New York, 2002.

[5] Stnescu, N.-D., Munteanu, L., Chiroiu, V., Pandrea, N.,

Dynamical Systems. Theory and Applications, Vol. 1, ThePublishing House of the Romanian Academy, Bucharest, 2007.

[6] Stnescu, N.-D., Munteanu, L., Chiroiu, V., Pandrea, N.,

Dynamical Systems. Theory and Applications, Vol. 2, ThePublishing House of the Romanian Academy, Bucharest, 2010 (in

press).

[7] Pandrea, N., Stnescu, N.-D., Ogaru, S., Multibody Dynamics, The

Publishing House of the Romanian Academy, Bucharest, 2011 (inpress).

[8] Stnescu, N.-D., Pandrea, N., Theoretical Mechanics, The

Publishing House of the Romanian Academy, Bucharest, 2011 (in

press).