0699_0714

EIGENVALUES AND EIGENMODES OF AN INCLINED HOMOGENEOUS TRUSS IN A ROTATIONAL FIELD

S. VLASE

TRANSILVANIA University of Braşov, RO-500036, B-dul Eroilor, 29, Romania E-mail: [email protected]

Received January 24, 2014

The technical applications of the last decades are associated to mechanical systems which work with powerful forces at high velocities. The rigid elements hypothesis, which is used in most engineering applications, does not work in this situation, where the elasticity of the bodies leads to sensibly different behaviors, where big deformations, intense vibrations and mechanical instability phenomena can arise. In this paper we aim to use finite elements in order to obtain the model of a homogenous sloping truss in a rotational field.

Key words: eigenvalues, finite element method, linear elastic elements, Lagrange’s equations, one-dimensional finite element, rotational field.

1. INTRODUCTION

Any mechanical system is composed of solid, more or less elastic, elements. The rigid elements hypothesis, which is usually used when a mechanical system is studied, represents an initial approximation in the dynamic analysis and it could or could not be satisfactory, depending of the elements elasticity, the existing velocities and forces, determined by the operating mode of the studied mechanical system. If these velocities and forces have significantly big values, the elasticity becomes significant also and it can qualitatively change the dynamic response of the system. The vibrations phenomena, the big deformations which modify the geometry and the loss of stability will represent, in fact, forms of manifestation of the system's elasticity and can determine the damage or the malfunction of the system.

Theoretically speaking, the study of such a system can be developed using continuous mathematical systems, using the fundamental theorems of the continuum mechanics. The main disadvantage of this approach is represented by the resulting partial derivatives differential equations, which are difficult to integrate [11, 12]. A much more convenient way to approach this problem comes from using numerical methods to discretize the elastic domain. The finite element method represents, under this approach, the most convenient study method. The advantages of this approach result from [8, 10, 21].

Rom. Journ. Phys., Vol. 59, Nos. 7–8, P. 699–714, Bucharest, 2014

S. Vlase 700

The papers which developed this approach used one-dimensional finite elements in a plane motion [1, 2, 5, 7, 13], two-dimensional finete element [24] or in a spatial motion [22, 23]. The papers [9, 18] synthesize the results in this domain.

The aim of this paper is to apply the finite elements method in order to study the behavior of a inclined truss in a rotational field. The shear effect on bending is also considered.

2. MOTION’S EQUATIONS

2.1. DESCRIPTION OF THE MECHANICAL MODEL

In the following we will determine the motion equations which describe the behavior, in time, of a truss finite element in a rotating motion around an axis, along with the joist which it discretizes (Fig. 1).

Fig. 1 – Inclined truss element in a rotational motion.

In order to obtain the equations of motion, we will use the Lagrange equations. This means that we need to determine the kinetic energy, the internal energy which arises due to the elastic behavior of the joist and the mechanical work of the distributed and concentrated forces in order to be able to determine the Lagrangian of the studied mechanical system. The independent coordinates of the nodes which we will write the motion equations for depend both on the type of the chosen finite element and on the shape

Eigenvalues and eigenmodes of an inclined homogeneous truss 701

functions. If we chose a finite truss element with two nodes, the independent coordinates will be the vectors of the nodal displacements which will include the displacements under the three main directions and their derivatives, which have the signification of some rotations or of some curvatures. The number of chosen independent coordinates will depend on the chosen bar hypothesis and on the types of the chosen finite elements. We will obtain the general motion equations for a finite element in a centrifugal field, by customizing the results obtained in [21].

2.2. FORM OF THE LAGRANGIAN

We will use the hypothesis that the elastic deformations are small enough not to influence the general motion of the system. We will suppose that we know the angular velocity and the angular acceleration for the rotation. Let's consider a truss finite element with i and j being the end nodes. The iδ and jδ vectors represent the independant displacements of the truss's ends (Fig.2). The finite element will be referenced to a local coordinates system Oxyz which is in a known rotating motion. We can determine the velocity and the acceleration for the origin of the mobile coordinates system in rapport with the fixed coordinates system OXYZ.

Fig. 2 – One-dimensional finite element with a three-dimensional motion.

S. Vlase 702

A random point M has its displacement vector f (u,v,w) which can be written, in the terms of nodal displacements, as:

==

=

2

1

δδ

NδNf e

wvu

(1)

and: ( )

*1 1 2 2 eN N N αα = α + α = δ

The nodal displacements vector eδ :

=2

1

δδ

δe ; (2)

The N matrix contains shape functions. The lines of the matrix N corresponding to the displacements u, v and w are named ( )uN , ( )vN and ( )wN :

( )

( )

( )

=

w

v

u

NNN

N (3)

We can then write:

( )

( )

( )

( )

( )

( )

=

==

=

2

1

3

2

1

2

1

δδ

NNN

δδ

NNN

δNf

w

v

u

e

wvu

. (4)

The equations for the rotations of the ends of the joist β and γ are written in correlation with the displacements of the truss under Oy and, respectively, Oz directions [22]:

ddwx

β = − and dd

vx

γ = . (5)

If we consider the shape functions, the rotations of the ends of the joist will have the form:

( ) ;dxd

e'

)w(e)w( δNδN −=−=β ( ) ;dxd

e'

)v(e)v( δNδN ==γ (6)

In the end, we can write:


;*

*)(

*)(

*)(

')(

')(

*)(

e

v

w δNδδ

NNN

δδ

NN

N=

=

−=

2

1

2

1

γ

β

αα

γβα

where: ( ) ( )

( )

( )

=

−=*

*

*

'

'

*

*

γ

β

αα

NNN

NN

NN

v

w . We noted: ( )* '

wN Nβ = − and ( )* '

vN Nγ = .

In what follows index G is applied to a vector or a matrix with the values expressed in the global reference system and index L is applied to a vector or a matrix with the values expressed in the local reference system.

The position vector of point M becomes in the deformed position M'’ [21]:

++=

+=+=

wv

ux

wvu

LoLMLMLM ,,,,' rrδrr , (7)

or, with respect to the global reference system:

LeGoGoGMGM

x

wvux

wvu

r ,,,,,'

00

00 δNRRrRRrRr +

+=

+

+=

+= (8)

The matrix R expresses the transition of the vectors from the mobile reference system Oxyz to the fixed reference system O'XYZ . The kinetic energy expression for the whole finite element finite is:

( )∫ +=L

GTGGM

TGMc dxAE

0,',' ''

21 ωIωrrρ (9)

where:

−−=

zzyz

yzyy

xx

IIII

I

00

00I . (10)

yyI and zzI represent moments of inertia of the bar cross section about co-ordinate axis Oy and Oz respectively of a reference system with its origin in the mass center of the element Adxdm ρ= (ρ - density); yzI is the centrifugal moment

S. Vlase 704

of inertia and xxI is the inertia moment about the co-ordinate axis Ox. Since we have chosen y and z as principal directions of inertia 0yzI = , we have:

=

z

y

x

II

I

000000

I , (10’)

For simplicity, we noted: xx xI I= , yy yI I= , zz zI I= . In the written relations we can notice the occurrence of derivatives of the

matrix for positional relation to the global reference system R . These derivatives represent angular velocities and accelerations. For a better understanding we shall express these derivatives according to [21]:

1kG ω=ω , 1kG εε = , (11)

=

ω00

Gω ;

=

ε00

Gε . (12)

The angular velocity and acceleration vectors shall have in the local reference system the components:

sin ; cos ;x zω = ω α ω = ω α sin ; cos .x zε = ω α ε = ω α (13)

The angular velocity of the element dm is:

e

z

x

L δNω *

cos0

sin0' +

=

+

=

α

αω

γβα

ω

ω (14)

where: ω is the components of the angular velocity vector along the Oz axis and.

eδN *=

γβα

; (15)

The matrix operator angular velocity will be, in this case:

ω~ L

0 0 0 00 0

0 0 0 0

z

z x

x

cc s

s

−ω − = ω −ω = ω − ω

(16)


in the local coordinates system and:

ω~ G

0 0 0 1 00 0 1 0 0

0 0 0 0 0 0

z

z

−ω − = ω = ω

(17)

ω~ G = R’ RT ; ω~ L = RT R’ . (18)

We shall also have the angular acceleration tensor ε~ G, defined as:

ε~ G = R’’ RT + R’ R’T ; ε~ L = R’T R’ + RT R’’ . (19)

In order to obtain the Lagrangian of the system, we need to determine the deformation energy. The energy due to the bending of the element can be written using the expression:

2 22 2

2 22 2

0 0

1 d d 1d ' d2 d d 2

L L

pi y z y zw vE EI EI x EI EI x

x x

= + = β + γ ∫ ∫ (20)

If we consider (4), explicitly written:

( ) eww δN= ; ( ) evv δN= .

then:

( ) e

L''

)v(T''

)v(z''

)w(T''

)w(ypi dxEIEIE δNNNN

+= ∫

021

(20’)

or:

eiT

epiE δkδ= (21)

where:

ik ( )∫ +=L

vT

vzwT

wy dxEIEI0

'')(

'')(

'')(

'')( NNNN . (22)

The energy due to axial deformation is:

( ) e

L

uT

uT

e

L

pa δEAdxNNδdxdxduEAE ∫∫ =

=

0

2

0 21

21 '' (23)

S. Vlase 706

or:

eaT

epa δkδE = (24)

where:

( )∫=L

uT

ua EAdxNNk0

'' . (25)

The deformation energy due to torsion is:

( ) ( ) e

L

xTT

e

L

xpt δdxGINNδdxdxdGIE

=

= ∫∫

0

2

0 21

21 '' **

ααα

(26)

or:

etT

ept δkδE = (27)

where:

( ) ( )∫=L

xT

t dxGIk0

** '' αα NN . (28)

The deformation energy due to the bending, traction-compression and torsion is:

eeT

eetaiT

ept δkδδkkkδE =++= )( (29)

where:

taie kkkk ++= . (30)

The axial force P which appears in the transversal section of the truss will give, in hypothesis, that, in a first approximation, the axial deformation can be neglected, and the deformation energy is:

2 2

0

1 d d d2 d d

L

a totv wE P xx x

= + ∫ (31)

where Ptot is the axial force which resides in the bar cross section at the distance x from the end of the bar. The force components acting at the right bar end considered in the local coordinate system are represented by Px=0, Py=0, Pz=0. Beside these components, the value of P and the components of the inertia forces


acting upon the portion of the bar between x and L are being determined. The influence of the inertia forces in the transversal deformation is neglected. In order to calculate them we determine the acceleration of the current point of the bar in the rigid motion (fig.3):

−−

+=

ϕϕ

ϕϕ

cossinx

sinxsinxL,L 00

0 2

2ωεaa o (32)

Fig. 3 – Determination of the axial inertia force.

The inertial force in the x section, which stretches the truss, is given by the formulae:

( ) ( )( )

( )

{ }

2

2,

2 2 2

22 2

2 2

0 sinF a d a d sin d 0 d

0 sin cos

sin0sin 0

2 20 sin cos0

L L L L

i G o G

x x x x

o

o

x

y

z

sm A s s A s A s

s

L xXY A L x L x A A

L x

ϕ = − = − ρ − ε ϕ ρ + ω ρ = − ϕ ϕ

− ϕ ε ω = − ρ − − − ϕ ρ + ρ

− − ϕ ϕ

µ µ = µ µ

∫ ∫ ∫ ∫

22

2 2

0 sinsin 0

2 20 0 sin cos

o

o

XAL Y AL AL

ϕ ε ω = −ρ − ϕ ρ + ρ

− ϕ ϕ

(33)

{ }0

x o

y o

z

XA Y λ λ = λ = ρ

λ

; (34)

S. Vlase 708

{ }2

20 sin

sin 02 2

0 sin cos

x

y

z

A A ν ϕ

ε ω ν = ν = ϕ ρ − ρ ν − ϕ ϕ

. (35)

We will obtain then the internal energy due to the axial force which arises in the bar due to the rotation:

( )( ) e

LTT

xxxT

ea δdxNNNNxxδE

+++= ∫

0

2

21 *

)(*

)(*

)(*

)( ββγγνλµ (36)

or:

e

Ge

TeaE δkδ

21

= (37)

where

Gek = ( )( ) =

+++∫

LTT

xxx dxNNNNxx0

2 *)(

*)(

*)(

*)( ββγγνλµ

+

++

+= ∫∫

L'w

T'w

'v

T'vx

L'w

T'w

'v

T'vx xdxNNNNdxNNNN

00

λµ

∫

++

L'w

T'w

'v

T'vx dxxNNNN

0

2ν 210 ,,, Ge

Ge

Ge kkk ++= (38)

where the notations for 210 ,,, ,, Ge

Ge

Ge kkk are obvious.

The total internal energy is:

( ) ( ) eGee

Tee

Getai

TepE δkkδδkkkkδ +=+++= . (39)

The external work of distributed loads is:

( )

[ ] dxNN

mmmppp

dxmmmwpvpupW

L

ezyxzyx

L

zyxzyx

∫

∫

=

=+++++=

0*

0

δ

γβα (40)


and that of concentrated loads q TeL in the nodes is:

eTLe

cW δq= (41)

Using all of these, we can now build the Lagrangian for the studied finite element:

cc p aL E E E W W= − − + + . (42)

2.3. SINGLE ELEMENT’S MOTION EQUATIONS

The motion equations, in terms of independent coordinates, are obtained by applying the Lagrange equations [3, 6, 15, 16, 17]:

d 0d ee

L Lt ∂ ∂ − =

∂δ∂δ . (43)

If we note:

dxATL

t NNm ∫= 0ρ (inertia of translation); ( ) ( )

0

LT

ij i j Adx= ρ∫m N N ; i,j=1,2,3;

( ) ( ) ( ) ( ) ( ) ( )∫∫ ++==L T

zTT

yT

xL T

r dxIIIdx0

*******0

* )( γγββαα ρρρρ NNNNNNNINm ;

(influence of inertia effects at cross-section rotation)

rte mmm += ;

( ) ∫=L

LT

e AdxNωNc0

ρω ~ ; ( ) ∫=L

LT

e AdxNεNk0

ρε ~ ;

( ) ∫∫ =−=L

LLT

LTT

e AdxAdx00

2 ~~ ρρω NωωNNRRNk ;

0

Li Toe A dx= ρ∫m N ; dx

LiEe ∫= 0

*Nm ; ( )0

LT

ix i x Adx= ρ∫m N ; i,j=1,2,3.

[ ] dxNN

mmmpppL

zyxzyxe

= ∫ *0q ; (44)

the motion equations, taken from the local coordinates system, can be written in a simple form:

S. Vlase 710

( ) ( ) ( )( )( ) ( ) Go

TioeL

iEe

ie

ieee

eGeeeeeeee

rRmεImqqqq

δkkkkδcδm

−−−−+=

=+++++2*

22

ωε

ωεω (45)

3. GLOBAL COORDINATE SYSTEM

The considered truss is composed of finite elements linked between them using nodes. The forces which arise in these nodes will be removed during the assembly process.

The unknown parameters which appear in an elastic system are of two kinds: nodal displacements and liaison forces. The solving methods for the differential and algebraic systems of equations (DAE) focus on trying to remove the algebraic unknowns [14, 19, 20] (which usually are liaison forces); in this situation, the only unknowns left are the nodal independent displacements. The mechanical links between different finite elements are usually defined by linear relationships between the nodal displacements, as:

GeGe qA∆ = (46)

or, for the whole mechanical system (where G, which represents the global coordinates system, will be discarded in the following equations):

=

=

n∆

∆∆

∆ 2

1

qA

A

AA

=

n0

0

2

1

(47)

In rel.(47) ∆ represents the vector of nodal displacements, e∆ represents the vector of nodal displacements of an element e with the components expressed in the fixed coordinates system G and q is the vector of independent coordinates:

[ ]Tsqqq 21=q .

If we differentiate (47) two times, we obtain:

qA=∆ (48)

The equation which determines the link between the components of the nodal displacements vector in the fixed coordinate system G and the mobile coordinate system is:

eee δR=∆ . (49)


If we consider (45),(47), after simple mathematical operations, we will obtain the motion equations for the whole system:

( ) ( )( )

inertialiaisonliaisonextext QQQQQKKKCM

++++=

=∆+++∆+∆**

22 ωε (50)

If we consider (47), (48) and (50), we can write the equations in relation to the independent coordinates q :

( ) ( )( )inertialiaisonextext QQQQ

qAKKKqACqAM+++=

=++++*

22 ωε (51)

If we pre-multiply the relation (51) by AT [4, 19, 20] then we obtain:

( ) ( )( )

inertiaTextTextT

inertiaTliaisonTliaisonTextTextT

TTT

QAQAQAQAQAQAQAQA

qAKKKAqCAAqMAA

++=

=++++=

=++++

*

**

2ωε (52)

In the previous equation (52), we considered that the liaison forces will disappear as one of their orthogonality effect with the displacements [4], [20]. Finally, we will get:

Qqkqcqm =++ (53)

The system of differential equations which we will obtain is non-linear, the matrix coefficients depending, at every moment, on the geometrical configuration of the studied mechanical system.

The matrix coefficients for the differential equations which we obtained have the following properties: the inertial matrix m is symmetric, defined as positive, the c matrix is skew-symmetric and the k matrix represents the sum of two symmetric matrix, defined as non-negative, with a skew-symmetric matrix. The resulting system is strongly non-linear, the matrix coefficients depending on the time parameter. The easiest way to solve the system is by linearization, where we consider some extremely small time intervals so that the general motion can be considered as being the sum of some motions which take place in a very small time interval. On these very small time intervals obtained, we can apply modal analysis, so we can determine the eigenvectors and the eigenvalues.

4. EIGENVALUES AND EIGENVECTORS

The problem of determining the eigenvectors and the eigenvalues for a mechanical system is one of a particular practical importance. If, in (53), we neglect the damping Coriolis matrix and the external forces, the eigenvalues problem reduces to:

S. Vlase 712

02 =− )det( mk ω (54)

If we consider the Coriolis damping effect, then the equations system defined in (53) can be brought to a first order differential equations system. Using the substitution:

21 XqXq == ; (55)

and

=21

XX

X ; (56)

we obtain the system of differential equations:

−−

=

−− 210

21

11 XX

cmkmE

XX p ; (57)

which has the corresponding eigenvalues problem:

.det

−

−− −− p

p Ecmkm

E211

0λ (58)

The system of differential equations which we obtained (53) has the following important property:

In the Rayleigh Quotient, the Eigenvalues Do Not Depend Directly on the Damping Matrix

Let's consider a very short time interval when the matrix coefficients of the system can be considered constant. Let's consider a solution for the system (53), in form of:

=q A cos( )tω + ϕ (59)

By successive differentiations we will obtain:

=q – A ω sin( )tω + ϕ ; =q – A 2ω cos( )tω + ϕ (60)

and if we write these in (53) we will obtain:

02 =+++−+− )cos()sin()cos( ϕωϕωωϕωω tAktActAm (61)

(when we determine the eigenvalues, we use Q =0). If we pre-multiply the relationships with AT, we obtain:

02 =+++−+− )cos()sin()cos( ϕωϕωωϕωω tAkAtAcAtAmA TTT (62)


and because c is skew-symmetric we obtain:

0=AcAT (63)

from which we obtain the pulsation (the Rayleigh quotient) [16]:

AmAAkA

T

T

=2ω (64)

Therefore the c matrix, having the significance of a damping matrix, will not bring any real dampening in the system. The skew symmetry signify that this term does not lead to any kind of exterior energy dissipation. The role of this matrix is to modify the conservative energy inside the system.

5. CONCLUSIONS

In case of the motion of an inclined elastic truss in a centrifugal field, the particular nature of the angular velocity and angular acceleration vectors determines a simplification of the equations of motion and, consequently, the possibility to easily determine the eigenvalues and the eigenvectors for such a problem. The velocities and accelerations field becomes a particular field, identical to that corresponding to the rotation of the fixed-axis rigid. This is a very common kind of problem in the technical field, for example at the helicopter propeller's level. For this reason, the effective definition of the motion equations for these particular situations has great practical value. Furthermore, the specific situations which arise in practical engineering applications lead to these equations to have remarkable properties and particularities, which help ease the calculus effort in order to determine the eigenvalues.

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0699_0714