Spinning finite elements

Journal of Sound and Vibrarion

(1988) 125(3), 523-537

SPINNING A. Y. T.

FINITE LEUNG

AND

ELEMENTS T. C.

Department of Civil and Structural Engineering,

FUNG

University of Hong Kong

(Received 7 May 1987, and in revised form 8 March 1988)

By means of Lagrangian equations, the finite element method is extended to spinning structures. Skeletal structures consisting of straight beam members are taken as examples. Explicit element matrices are given. Numerical solution procedures based on the Newtonian algorithm are suggested. The orientation of a beam member is arbitrary with reference to a rotating frame, so that general spinning space structures can be considered. Numerical examples show the well known fact that the axial force in a beam induced by the centrifugal force has destabilizing effects while gyroscopic motion has stabilizing effects. The combined effect may be either stabilizing or destabilizing depending on the boundary conditions.

1. INTRODUCTION structures has been receiving extensive attention recently analysis of spinning in space technology, and in the jet engine and automobile industries. The gyroscopic or Coriolis effects due to the rotating frame of reference make the natural vibration of a spinning structure deviate from that of the classical theory for non-spinning structures. Lumped mass models for the spinning system can be found in standard .texts [ 1,2]. The finite element method has been proved to be a powerful tool in structural dynamics. Bauer [3] studied the vibration of a rotating beam orientated in the axis of rotation. Nelson [4] established a finite element model for a rotating shaft with a rigid disk. Christensen and Lee [5] suggested a general finite element formulation for spinning structures. Wittrick and Williams [6] considered a beam member generally orientated in a rotating frame using a continuous mass method. However, the resulting partial differential equations with variable coefficients are difficult to solve. In this paper, a skeletal structure in a rotating frame of reference is discretized by means of the finite element method. Equations of motion are established by means of the Lagrangian equations. Explicit element matrices are derived and the explicit gyroscopic element matrices for a general orientation are given for the first time. The solution procedure to determine the steady state displacement is discussed. The oscillation about the steady state displacement is of interest; it is a quadratic eigenvalue problem. The methods of solution for this quadratic eigenvalue problem are well established. Gupta [7] solved the problem by a Sturm sequence method for the resulting Hermitian matrix associated with the state space vector. Wittrick and Williams [6] extended the method to the continuous mass model and corrected the derivation errors in Gupta’s paper. Simpson [ 8,9] suggested an iteration technique and Bauchau [lo] used the Lanczos algorithm to solve the extended Hermitian matrix. Meirovitch [2] transformed the extended matrices to symmetric forms to avoid complex arithmetic. In this paper, a generalized inverse iteration method is recommended to solve the quadratic eigenvalue problem. The order is kept to a minimum, the sparsity of the original matrices is taken advantage of and thecomputational efficiency is greatly improved. The method is particularly useful The dynamic

523 0022-460X/88/180523+ 15 $03.00/O

Q 1988 Academic Press Limited

524

A. Y. T. LEUNG

AND

T. C. FUNG

in parametric studies; for example, in studying the influence of the rotational velocity on the natural frequencies and mode shapes. 2. EQUATIONS

OF

MOTION

Consider a beam element inspace, as shown in Figure 1, with three sets of mutually orthogonal axes describing its motions: (1) xyz along the local principal axes of the beam;

Figure 1. General co-ordinate systems.

(2) Ej% moving with the beam so that its jj axis lies along the spinning axis; (3) an inertial frame XYZ such that the Y axis is coincident with the jj axis. Since the undeformed beam is at rest relative to the frame TjE, any point on the beam is given by {rO]= {rg]+S({rh]--{rg]),

O<[Gl,

(la)

or = {rp] + s{s],

O~sSl,

(lb)

where {r,,}, {r,,}, {rg} are the position vectors of the point and the ends of the beam with reference to frame 1?j?5respectively, {s} is the unit vector along the beam axis with {s]={rg~]/lrg~I and {rgh]={rhI-{rg]. The beam motion can be described fully with reference to the inertial frame XYZ in space. With the frame %jE assumed to be rotating about the 1 axis with constant speed R with respect to the inertial frame, a spinning matrix [a] is defined as

(2) The kinematic state for any point on the beam can be evaluated with respect to the inertial frame XYZ in terms of the co-ordinate %j5 as follows: absolute position vector {r} = [x, f, z],

G-4

absolute velocity vector {v}= {i} + [f&](r),

(3b)

vector {a} = {i}+

(3c)

absolute acceleration

2[ti]{i}

+

[~][fb]{r}.

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ELEMENTS

Let {u} and {ii} be the displacement vectors in the principal beam axes with reference to the local co-ordinates xyz and the moving co-ordinates XjZ respectively, such that {u}= [u

2, WIT,

Qll 1

@I =

[RITW,

(435)

where u, V, w are the displacements along the xyz axes respectively, [R] is the transforma--tion matrix between the global co-ordinates xyz and local co-ordinates xyz such that 1x1 = UUW, (3 = [RI%),

[RI=

(~21

[ a31

ff12

ffl3

(~22

(~23

ff32

a33

,

ix>

=

and

Y9 A’,

[x,

{%}= [n, 7, Z]‘.

Note that the unit vector {s} along the beam is [c.Y,~(Y,~(~~~1~. --- . The position vector of a point on the deformed beam with reference to frame xyz IS

id = id + ifi) = id + [RIT{uI,

(6)

{i} = [RIT{i},

(7)

and since {i,,} = (0).

3. LAGRANGIAN

EQUATIONS

The Lagrangian equations are

where T and U are the kinetic and strain energies, respectively, and {F} is the generalized force vector corresponding to the generalized co-ordinate {u}. For free vibration {F} = (0). The kinetic energy T and strain energy U are given respectively by T=4

~A{v}~{v} ds,

(9)

(10) where E, p, A, Iy, Zz, F(s) and ds are the elastic modulus, mass density, cross-sectional area, the two second moments of area with respect to the principal axes y and z, the axial force, and the elementary length along the beam element. The integrations are performed over the entire structure, which implies the summation of the integration on each beam member of the structure. When using the finite element method, the displacement vector {II,} of a beam element is interpolated by its nodal co-ordinate vector {qe}, so that

N,

fuel = [NlheI

and

where the shape function matrix

WI=

{u,] =

Nz

[

[Nl{ileI,

1

N3

,

(11)

526

A. Y. T. LEUNG

AND

T. C. FUNG

and the nodal co-ordinate vector {qe} = [{ql}T, {q2}T, {q3}T]T, in which {ql}’ = [u, , UJ are the axial nodal co-ordinates, {qZ}T = [u, , &, , u2, O,,] are the bending nodal co-ordinates in the xy plane, {q3}T = [ wl, fly, , w2, 19,,] are the bending nodal co-ordinates in the xz plane, and [N,] is 1 by 2 and [NJ and [NJ are 1 by 4 shape function submatrices. 4. KINETIC

ENERGY

AND STRAIN

ENERGY

From equations (3b), (6) and (7), the absolute velocity vector of any point on the beam and is {v}=[R]~{~~}+[~I]({~,}+[R]~{u}) {v]‘{vI =

~~lTIRIIRIT@l + ~~~‘~~1~~1’~~1~~1’~~~ + M’[~l’[~lkJ -C2{ti~T[R1[~lhJ+ 2WT[RI[JZI[RlT~ul+2{ro}‘[~lT[al[RlT{u}. (12)

Note that 0 0

[0

0

[n]‘[sz]=[a’]=n’

and

[R1[RIT= [II

1 0

By using equation (12), the contribution kinetic energy is found to be T, = kJTW,lbie~

1

0 0 . 1

of a beam element, with length 1, to the total

+h-lTIKn,lh-~ + Toe+ bielTVe)

+ bLITIGJhe~ + Wn,~Tb-I,

(13)

where

&,I =

[Mel = 1’pNm1 ds,

[Gel = j-’PAM ds, 0

0

If,) = I’ wWITVU~l~~o~ ds,

Wd = j’ pAINIT[fn,,fn,,fn,lTds, 0

0

To. = 1

I0

’~A{ro~T[~21{ro~ ds,

1 1 [ a,,N%, a&N, a&N, 1 [ 0

N:N

3

N;Nz

[ml =

= f2

kl

N:N, [kn] = R2

a&N2

~&NJ

symm.

a 11= 011 2 +a:,, a 22=&+&3,

The

=

f12(xow1

a21a31

+

ZO~l,),

_&a,

=

,

,

&3=~,1~31+~13~33, +

(y23a33,

a2(xO~21

a 33 b3

b2=a11a33--a13a31,

b,=ffll~23-~l3~21r fn,

=

W:N b&N 0

a,,N:N,

al*=(Y,1~21+~13a23,

a23

WTNz 0 skew symm.

+

zOa23),

=

hz;

a312

+43,

a21a33

-

=

=

fi2(Xo~3,

ff23ff31, +

zoa33).

total kinetic energy is obtained by the usual assembly technique, T = C Te= f~i~TIMl{il~+~~q}TIKnI~q~ + e

To+

biIT{fl +

{4~TlWq~+ {hdTbd,

where {q} is the global nodal displacement vector with reference to frame 375.

(14)

SPINNING

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ELEMENTS

The strain energy can be considered similarly. For a particular element, as defined in equation (lo), one has u = [N,]{q,}, u = [NJ{q 2} , and w = [N3]{q3}. The contribution of a beam element to the total strain energy is u, = ~~dNLel+

(15)

[&-lh~,

where

We,1= EAKa EI,Kb E12K.], [ U&l = ’[N:ITNI [&I

=

I0

[Kb] =

ds,

I0

’

I0

EKgel=[ ’ Kgb Kgc],

’[N,“lTIN;] ds,

CK,l= I’ N’lTN1 ds, 0

~Wl?%ITNI ds,

Kc1 = 1’ FWNITWSI ds, 0

and a prime denotes differentiation with respect to s. The total strain energy is obtained by the usual assembly technique, LJ = C u, = i{dTWel

e

5. SOLUTION

OF STEADY

(161

+ [K,lHq).

STATE EQUATIONS

The nodal displacement vector {q} can be divided into two parts: the steady state displacement {qS} and the oscillation about steady state {q”}, so that {q} = {qS}+ {q”}. When the steady state deformation is considered, {q”} = {&,} = {&} = {a} = {O}, and {q} = {qS}. From equations (14) and (16), the kinetic and strain energy are respectively

T = f{qJT&l{qJ+

To+UWT{qJ,

u=r{qS}T([K,l+CKsl){qs}.

(17,181

The Lagrangian equations (8) reduce to -aTlG,I

+ a ~l&-bI = W,

([K,l+[K,l-[K,l){q,}={F,}.

or

(19320)

For a statically determinate structure, the internal forces are predetermined. [K,] is therefore independent of the steady state deformation. Hence, by assuming that the change of geometry is small, the centripetal force per unit length acting on the beam is given by -pA[~][~]{ro}. The component normal to the beam is resisted by the bending action while the component along the beam induces the axial force in the beam element. The centripetal force per unit length along the beam is T = -pA{s}‘[fb][fl]{r,}

= a-t bs

(21)

where a =pAf22(a,,~o+~,3ZO)r b=pAR’(af,+&, and [~o,~o, TolT={rg}. If the tensile force at one end of the beam element is F, , the tensile force at any section is given by, as shown in Figure 2, T(S) ds = F, -as --

F(s) = F1 -

k

5

4

Figure 2. Axial force in a section.

b

2

s2

(22)

528

A. Y. T. LEUNG

AND

T. C. FUNG

For statically determinate spinning structures, the tensile force F, can be obtained easily from equilibrium considerations. For statically indeterminate structures, however, the internal force at any section depends on the steady state deformation. Hence [K,] is not known in advance. Iterative schemes may be used to determine {qs}. [K,] and F, for any section are then determined and F(s) can be obtained from equation (22).

6. OSCILLATION ABOUT STEADY STATE Consider small oscillation about steady state; then (4) = {a”}, and the kinetic and strain energy are respectively

T=t{ilJ’DWiJ+th

+dWnl~~s+qu~+

To+{iJT{f~

+ {4JT[Glh + so1+ {WTh +&Iv u

(23)

=t{cls+q”}T([K,l+[Kgl){qs+sJ.

(24)

The Lagrangian equations give, upon using equation (20),

Note that [M] and [Ke] are the well known mass and elastic stiffness matrices for a non-rotating structure. [G] is the skew-symmetric gyroscopic matrix, [Kc] is the stiffness matrix arising from the spinning action, and [K,] is the stiffness matrix arising from the internal steady state axial force. 7. BEAM ELEMENT MATRICES So far, the finite element formulation is quite general. If the shape functions used satisfy exactly the governing partial differential equation, the finite element model is exact. Since such shape functions are not generally available, the following interpolating shape functions are adopted, with .$= s/l;

(264

[NJ = Cl- 5,51, [N2]=[1-3[2+2[3, [N,]=[1-3[2+2[3,

5(52-25+1)1, -5(5’-25+1)5

3t2-2t3,

(e3-.$2)1],

(26b)

3e2-2e3,

-(t3--e*)1].

(26~)

By using equations (13) and (15), the matrices [MJ, [K&j, [Gel, [K,,], and [Kg,] can be evaluated explicitly. Since the torsion effect is known to be uncoupled [6], this can be combined to give the complete three-dimensional beam model. With the nodal parameters rearranged so that they are in the same order as shown in Figure 3, the element matrices are given explicitly in the Appendix.

Figure

3. General

nodal

co-ordinates.

SPINNING

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529

8. INVERSE ITERATION

To solve equation (25), assume {q(t))= ($1 eiw'and substitute into equation (25), to obtain [D(o,

0)1{41= ([&I + E&l -&I

+ 2iw[Gl- w*[MI){4}= WI,

(27)

where i = a, and u and (4) are the natural frequency and mode shape. In equation (27), 0 is a parameter since the matrices [K,], [Kn] and [C] are spinning velocity dependent. The influence of R on w is best studied by the following inverse iteration method. When 0 = 0, the problem (27) reduces to [D(w,

0)1{4)= ([Kel - ~*W1){4L

(28)

whose solutions can be obtained by any standard method. When R f 0, the eigenvalue problem (27) is equivalent to the solution of the set of non-linear equations {f(o, 0)) = [D(w,

~)I{41 = @I,

(29)

with a suitable normalization condition for the eigenvector (4). One can solve equation (29) by means of a Newtonian procedure with an initial approximation o. and {40}, which may be a solution of the standard eigenvalue problem (28) when 0 = 0 or a previous solution with a slightly different 0. Let w = w,+ dw and (4) = {4,+d4}, where dw and {d+} are the correction increments to be determined. Substituting into equation (29) and retaining only first order terms, one has [D(wo,

fl)l{4o+d41= -(doNB’(wo, fl)l{40)

(30)

where D’= dD/dw. Equation (30) constitutes the inverse iteration procedure which is well documented for symmetric matrices [ 111. Equation (30) can be solved by real arithmetic only if one writes the equation as [A+iB]{JI+iO}={a+ib} where has

(31)

[Al, IX W, i@, {a), and @I are real. Comparing real and imaginary parts, one (32)

for the solution of {#} and (0). However, the symmetry and sparsity of [A] and [B] are destroyed. As complex arithmetic is not complicated in modem programming languages, equation (30) is preferred to equation (32). A number of examples will be presented in the next section. The relation between the natural frequencies and the spinning velocity can be studied by varying 0 as a parameter.

9. NUMERICAL 9.1.

STRAIGHT

EXAMPLES

BEAM

Consider a straight beam with section properties I, = I, and with one end clamped at the origin of frame Zj?Z The configuration can be described fully by the unit vector along the x axis. As the first example, consider the natural vibration of a shaft (s = [0 1 OIT) with axial deformation neglected as shown in Figure 4. The analytic relation between the natural frequency w and the spinning frequency R is [3] w = wof 0, where w. is the natural frequency at rest. It can be seen that there are no axial forces in the beam element

A. Y. T. LEUNG

530

AND

T. C.

FUNG

and hence [K,]= [O]. Results obtained by using a 4-element model can be compared with the analytic results. For the other end being clamped, the percentage errors are shown in Table 1. For the other end being free, the response curves are shown in Figure 5. Note that the solutions obtained preserve the linear relationship between the natural frequency and the spinning frequency. As the spinning frequency increases, the initially coinciding natural frequencies respond differently. One of them continues to decrease and, at a certain spinning frequency, the lowest natural frequency drops below zero resulting in an instability by divergence. TABLE Percentage

error

between

analytic

solution

and

1

a 4-element

for clamped-clamped

solution

beam Spin rate

0.0 O-5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0 12.5 13.0 13.5 14.0 14.5 15.0 15.5 16.0 16.5 17.0 17.5 18-O 18.5 19.0 19.5 20.0

01

%

“2

%

“3

%

w4

%

22.403 21.903 21.403 20.903 20.403 19.903 19.403 18.903 18.403 17.903 17.403 16.903 16.403 15.903 15403 14.903 14.403 13.903 13403 12.903 12.403 11.903 11.403 10.903 10.403 9.903 9.403 8.903 8.403 7.903 7.403 6.903 6,403 5.903 5.403 4.903 4.403 3.903 3.403 2.903 2.403

0.13 0.14 0.14 0.14 o-15 0.15 0.15 0.16 0.16 0.17 0.17 0.18 0.18 0.19 0.20 0.20 0.21 o-22 o-22 0.23 0.24 0.25 0.26 0.28 0.29 0.30 0.32 0.34 O-36 0.38 0.41 0.44 0.47 o-51 0.56 0.62 0.69 0.77 0.89 1.04 1.26

22.403 22.903 23.403 23.903 24.403 24.903 25.403 25.903 26.403 26.903 27.403 27.903 28.403 28.902 29.404 29.903 30.403 30.903 31.403 31.903 32.403 32.902 33.404 33.903 34.403 34.903 35.403 35.903 36.403 36.902 37.404 37.903 38.403 38.903 39.403 39.903 40.403 40.902 41.404 41.903 42.403

0.13 o-13 0.13 0.13 0.12 0.12 0.12 o-12 0.11 0.11 0.11 0.11 0.11 0.10 o-11 o-10 0.10 0.10 0.10 0.09 0.09 0.09 0.09 0.09 0.09 0.09 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.07 0.07 0.07 0.07 0.07

62.243 61.743 61.243 60.743 60.244 59.744 59.244 58.744 58.244 57.744 57.245 56.742 56.242 55.742 55.243 54.743 54.244 53.744 53.244 52.743 52.243 51.743 51.243 50.743 50.244 49.744 49.244 48.743 48.243 47.743 47.243 46.743 46.244 45.744 45.244 44.743 44.243 43.743 43.243 42.743 42.243

0.92 0.93 0.93 0.94 0.95 0.96 0.97 0.98 0.98 0.99 1.00 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10 1.11 1.12 1.13 1.14 1.16 1.17 1.18 1.19 1.20 1.21 1.23 1.24 1.26 1.27 1.28 1.30 1.31 1.33 1.34 1.36

62.243 62.743 63.243 63.742 64.242 64.742 65.242 65.744 66.244 66.744 67.244 67.744 68.244 68.743 69.243 69.743 70.243 70.743 71.243 71.742 72.242 72.742 73.242 73.745 74.244 74.744 75.244 75.744 76.244 76.744 77.243 77.743 78.243 78,743 79.243 79.743 80.242 80.742 81.242 81.745 82.244

0.92 0.91 0.90 0.90 O-89 0.88 0.88 O-87 0.86 0.86 0.85 0.85 0.84 0.83 0.83 0.82 0.81 0.81 0.80 0.80 0.79 0.78 0.78 0.78 0.77 0.77 0.76 0.76 0.75 0.75 0.74 0.73 0.73 0.73 O-72 0.72 0.71 0.71 0.70 0.70 0.70

FINITE

SPINNING

Figure 4. Vertical

531

ELEMENTS

cantilever.

16

lT?gIl 0

2

Spinningrate dpAP/

Figure

5. Response

Figure

curve for cantilever

6. Horizontal

4

3

El

shaft.

cantilever.

50Gi

0

0

““l”““‘I”‘1” 2 4

6

8

IO

12

14

16

I8

Spinningrate fh$ii+

Figure

7. Response

curve for cantilever

lying in plane

perpendicular

to spinning

axis.

532

A. Y. T. LEUNG

AND

T. C. FUNG

When the orientation of the beam element is changed so that the spinning axis is perpendicular to the beam axis, as shown in Figure 6, the direction unit vector s is [ 1 0 01’ and the other end is free. Since the structure is statically determinate, the axial force at the support is given by F, = pAL!‘12/2. Again a 4-element model was used, and the response curves are shown in Figure 7. The lowest natural frequency increases with the spinning frequency. Now [C] = [0] and [K,] f [O], and the tensile axial force acts in a stabilizing fashion.

Figure 8. Inclined cantilever.

Spinning rate dpAP/EI

Figure 9. Response curve for cantilever with inclination angle 45”.

Spinning rate R J&/

Figure 10. Response curve for both-ends-clamped

El

beam lying in plane perpendicular to spinning axis.

SPINNING

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ELEMENTS

When the orientation of the beam is changed, different response curves are obtained. For example, when s is [l/a l/a OIT and the other end is free, as shown in Figure 8, the response curves are as shown in Figure 9. Both [Kn] and [K,] are non-vanishing; the destabilizing effect seems to be dominating. Consider as a fourth example, the case when s is [ 1 0 01’ and the other end is clamped. The structure is statically indeterminate, and an iterative procedure is necessary to determine the steady state response. For simplicity, no axial deformation has been assumed, and hence the axial force at the origin is F1 = pA12R2/6. Since the axial force become compressive as the section moves away from the origin, the effect of [K,] tends to destabilize the structure. The response curves are as shown in Figure 10. 9.2.

L-BEAM

For a more complex structure, consider a general L-shape cantilever, as shown in Figure 11, which consists of two arms of length L, and L2 subtending an angle 8. If the end A is clamped and is attached to the origin, the structure is statically determinate and the internal force can be found without evaluating the steady state deflection.

Figure

11. L-beam.

If the supporting arm AB has a unit direction vector {R,l, R12, R,,}, and the hanging arm BC has a unit direction vector {Q1i, QIZ, Q13}, the tensile force at the end of the hanging arm is F, = pAR’( Qllx,,,+ Qi32,,,)L2+ pAR2( Q:, + Q:,)L:/2, and the tensile force at the end of the supporting arm is Fl = pAR2(R,,x,,+ R,3zo)L, +pAO’(Rf, + &)L:/2+

PA~~(R,~~,~+R~~z,~)L~+

pAfi2(R11QII+

R13Q13)L%

where {x0, yo, zol is

the position vector of support A, and {xlo, y ,o, zlo} is the position vector of junction B.

01 0 ’

’ ’

4

’ ’

6

’ ’

‘2

Spinning

Figure

12. Response

curve for L-beam

’ ’

16

rate

’ ’

20

L ’

24

’

20

1

bad/s)

with CI= 0”, p = O”, y = O”, 0 = 90”.

534

A. Y. T. LEUNG AND T. C. FUNG TABLE 2

Sectional and physical properties for L-beam Circular section Elastic modulus Shear modulus Arm AB Arm BC Density

with r = 12.5 mm E = 2OOE3 N/mm2 G = 80E3 N/mm’ L,=lOOOmm L, = 500 mm = 7.85 kg/m/cm2

Figure 13. General orientation of L-beam.

100 - (a) 60-

60 -

60 -

12 Spanning

16 rate

20

24

26

kadh)

Figure 14. Response curve for Abeam with Q = 0”, p = O”, 0 = 30” and (a) y = O”,and (b) y = 45”.

SPINNING

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535

With the physical properties as shown in Table 2, different orientations of the spinning L-beam were studied. When the support arm is horizontal and the hanging arm is vertical, the response curves are as shown in Figure 12, as produced by a 3-element model. For a more general orientation, as shown in Figure 13, the response curves can be found without difficulty. Some of them are shown in Figure 14.

10. CONCLUSIONS

As obtained on the basis of the Lagrangian equations, the spinning beam finite element matrices are presented explicitly. The natural vibration of spinning skeletal structures under the influence of the spinning can be studied in detail. A Newtonian iteration procedure is recommended to study the influence of the spinning frequency on the original non-spinning system. The present method can be extended to other finite elements without particular difficulties.

REFERENCES 1. H. MCCALLION 2. L. MEIROVITCH

1973 Vibration of Linear Mechanical Systems. London: 1980 Computational Methods in Structural Dynamics. The

Longman. Netherlands: Sitjhoff

and Noordhoff. 3. H. F. BAUER 1980 Journal ofSound and Vibrarion 72, 177-189. Vibration of a rotating uniform beam, Part I: orientation in the axis of rotation. 4. H. D. NELSON 1985 American Society of Mechanical Engineers, Journal of Vibration, Acoustics, Stress and Reliability in Design, 107, 460-461. Rotor dynamics equations in complex form. 5. E. R. CHRISTENSEN and S. W. LEE 1986 Computers and Structures 23, 819-829. Nonlinear finite element modeling of the dynamics of unrestrained flexible structures. 6. W. H. WITTRICK and F. W. WILLIAMS 1982 Journal ofSound and Vibration 82, 1-15. On the free vibration analysis of spinning structures by using discrete or distributed mass models. 7. K. K. GUPrA 1973 International Journal of Numerical Methods in Engineering 5, 395-418. Free vibration analysis of spinning structural systems. of natural frequencies 8. A. SIMPSON 1973 Aeronautical Quarterly (May), 139-146. Calculation and modes of steadily rotating systems, a teaching note. of Kron’s 9. A. SIMPSON 1973 Journal of Sound and Vibration 26, 129-139. A generalization eigenvalue procedure. 10. D. A. BAUCHAU 1986 Znternational Journal of Numerical Methods in Engineering 23,1705- 1713. A solution of the eigenproblem for undamped gyroscopic systems with the Lanczos algorithm. 11. J. H. WILKINSON 1965 The Algebraic Eigenoalue Problem. Oxford University Press.

APPENDIX: I

2

3

ELEMENT 4

5

BEAM 6

140

7

MATRICES x

70 156

221 156

54

-221 1407 412 41’

[M] = ,,A//420

III 140 156

Y=

‘,,/A

536

A. Y. T. LEUNG 0

AND

T. C. FUNG 0

1476,

1476,

0

-21/b,

0

156b,

0

-22/b,

21/b, 0

-63b,

0

0

0

-22/b,

-636,

0

0

412b, 0

[C]=(pAl/420)0

0 -546,

0

0

0

63b,

0

14/b, -14/b, 0

13/b, 0 1476, 0

Skew Symmetrical

63b,

0

14/b,

546,

0

13/b,

0

0

0 0

0

0

0

0

0

-14/b, 0 n/b, 0 -3i2b,

13lb,

0

312b,

0

1476,

0

21/b,

-21/b,

1566,

0

22/b,

0

0

0

0

0

0 221b, 0

0

4f2b, 0

L

2

3

4

5

36 36

6

8

3I

-36

9

II

I

12

31

-31

-36

412

31

0 [K,]=F,/301

10

;

I

-31 0 -12

-31

412

36

Symmetrical

36

.

31 0 412 41'

0

-0

36

0

-36

61

36 -61

-36

21'

61

0 0

0

-12 -12

-61

212

-a/60

0 0

36 0

36 0

612 612

-1 .

2

3

4

5

6

8

9

IO

II

12

0 -61

-72

72

61

-72

72

0 -312

1SI

-31'

-b1/420

61

72 Symmetrical

-61

72 0

1812

SPINNING

FINITE

ELEMENTS

1

EA I

htl, I-bEI.

-IZEI,

I’ -GJ

-6EI,

4EI, I

F EA

Symmetrical

bEI,

114fr

I2

I’ GJ

4EI, I

r

I

2

3

4

5

6

‘%,

147a,,

1470,,

0

-21/a,,

2110,,

7%

6301,

6301,

156a,,

1560,,

0

-22h,,

22la,,

63%

54%

540,,

156a,,

0

-22b,,

ZZIO,,

63%

54%

54%

0

0

0 -4&z,,

[Kn] =(pA//420)R2

4Pa,,

Symmetrical

7 I

0 -14b,,

0

0

-1310,,

-1310,,

14/a,,

13la*,

13’%

‘40%,

1470,,

147a,,

1560,,

‘56% ‘5%