General Solution of the Forced In-plane Vibration Problem for a Spinning Disk

General Solution of the Forced In-Plane Vibration Problem for a Spinning Disk Natalie Baddour & Jean Zu Department of Mechanical and Industrial Engineering University of Toronto Toronto, Ontario Abstract The in-plane problem for a spinning disk is considered. Orthogonality properties of the free-motion mode shapes are derived by recasting the problem in state-space. These orthogonality properties permit a simple formulation to the general solution of the forced in-plane vibration problem.

1

Introduction

Vibrations of a spinning disk is an old problem that has attracted the attention of investigators to various aspects of the problem. Spinning disks have numerous applications such as turbines, floppy disks, gyroscopes and circular saws. Both in-plane and transverse vibrations of spinning disks have been investigated in the literature. For linear vibrations these are uncoupled and can be considered separately. In the following, attention will be confined to the in-plane vibrations. In [1], the axisymmetric in-plane vibrations of a solid spinning disk are considered and two types of instability are discussed. Burdess, Wren and Fawcett [2] do not limit themselves to the axisymmetric problem and solve the general problem of in-plane vibrations. The non-symmetric motion has features that are not revealed by the analysis of the symmetric motion. Similarly, Chen and Jhu [3] consider the in-plane vibrations of a spinning annular disk.

As previously mentioned, Burdess, Wren and Fawcett [2] solve the general non-symmetric in-plane vibration problem. The free motion mode shapes that are derived in their paper can be used to construct a general solution to the forced vibration problem. Following the example of Wickert and Mote [4], recasting the problem in state space allows for orthogonality properties of the aforementioned mode shapes to be easily derived. Once complete orthogonality properties have been derived, construction of solutions becomes a simple matter. This paper investigates the orthogonality properties of the free motion mode shapes and demonstrates how these properties can be put to use in constructing general solutions to the forced vibration problem.

2

Statement of the Problem

Assume that the motion of a particle in the disk only occurs in the plane of the disk and is given by u = (ur , uθ ), where ur and uθ are the radial and tangential displacements respectively. For small displacements, linear stress-strain and linear strain-displacement relationships can be assumed. Furthermore, for a thin disk plane stress conditions can be assumed. The preceding assumptions lead to the following equations of motion : 1 F Lu = u ¨ −Ω2 u+2ΩDu− ˙ ρ

1

(1)

where Ω is the spin rate and L is the matrix operator

E L11 L12 L= L21 L22 ρ(1 − ν 2 ) 2 1 ∂ 1 (1 − ν) ∂ 2 ∂ − 2+ L11 = 2 + ∂r r ∂r r 2r2 ∂θ2 (1 + ν)in ∂ (3 − ν) ∂ L12 = − 2r ∂r 2r2 ∂θ (1 + ν)in ∂ (3 − ν) ∂ L21 = + 2r ∂r 2r2 ∂θ 2 1 ∂2 1 ∂ (1 − ν) ∂ 1 − + .

L22 = − 2 ∂r2 r ∂r r2 r2 ∂θ2 √ ∂ = in, where i = −1. In the above, E is Young’s modulus, The operator Ln is derived from L by setting ∂θ Poisson’s ratio and ρ is the density of the disk . Furthermore, D, u and F are given by

D= u= F=

0 1

ur uθ

Fr Fθ

−1 0

,

(2) (3) (4) (5) (6) ν is

(7) (8) (9)

where (Fr , Fθ ) are the radial and circumferential body forces applied at a point in the disk The boundary conditions for a disk with a free boundary are given by σrr = σrθ = 0 at the boundary of the disk, r = a. Using linear stress-strain and strain-displacement relationships, these equations can be written in terms of ur and uθ as

∂uθ ur ∂ur ur + =0 (10) + ∂r r ∂θ uθ ∂uθ 1 ∂ur − + =0 (11) r ∂θ r ∂r Note that for the remainder of what follows, the following inner product will be used : Z a T (u⋆1 ) u2 rdr, hu1 , u2 i =

(12)

0

where the ∗ denotes the complex conjugate.

3

Free Motion Mode Shapes

The free motion mode shapes are required for the construction of the general solution to the forced vibration problem. When F = 0, then general solution to equation (1) can be found through use of Helmholtz’s theorem.

This development is outlined in [2] and will be included here for completeness. To solve equation (1), u may be expressed in terms of the functions φ and ψ such that

∂φ 1 ∂ψ + ∂r r ∂θ

1 ∂φ ∂ψ uθ = . − r ∂θ ∂r ur =

2

(13) (14)

If these equations ae substituted into equation (1), some algebra yields the following two equations for ψ and φ : λ21 ∇2 φ − (φ¨ + 2Ωψ˙ − Ω2 φ) = 0 λ2 ∇2 ψ − (ψ¨ − 2Ωφ˙ − Ω2 ψ) = 0, 2

(15) (16)

where E (1 − ν 2 )ρ E λ22 = .

2(1 + ν)ρ

λ21 =

(17) (18)

The general solution of the above equations for a solid plate can be written in terms of Bessel functions of the first and second kind as φ = Rnφ (r)ei(nθ−pt) = An {Jn (β1 r) + s1 Bn Jn (β2 r)} ei(nθ−pt)

(19)

and ψ = −iRnψ (r)ei(nθ−pt)

= −iAn {s2 Jn (β1 r) + Bn Jn (β2 r)} ei(nθ−pt) .

(20)

In the above, n in an integer and is the circumferential node number, p is an unknown frequency and βj (j = 1, 2) are the positive roots of

and

λ21 λ22 β 4 − (p2 + Ω2 )(λ21 + λ22 )β 2 + (p2 − Ω2 )2 = 0 2pΩ p2 + Ω2 − λ21 β22 2pΩ s2 = − 2 p + Ω2 − λ22 β12

(21)

(22)

s1 = −

(23)

The factors An and Bn are constants to be found from the boundary conditions. Note that equations (19) and (20) are solutions only if p2 6= Ω2 . When p2 = Ω2 , then β1 = 0 and the Bessel functions Jn (β1 r) are undefined.

The displacements ur and uθ are thus given by

u=

ur uθ

= An

=

"

Un iVn

φ dRn n ψ dr ψ + r Rn dRn i dr + nr Rnφ

#

ei(nθ−pt)

ei(nθ−pt) .

(24)

To derive particular values for p and u the boundary conditions must be satisfied. Substituting equation (24) into the boundary conditions (10) and (11) yields equations for An and Bn given by

1 An Q(p, Ω) = 0, (25) Bn where the elements of the matrix Q are given by

Q11 = n(n − 1)(1 − ν)(1 + s2 ) − β12 Jn (β1 ) + β1 (1 − ν)(1 − ns2 )Jn+1 (β1 )

Q12 = n(n − 1)(1 − ν)(1 + s1 ) − s1 β22 Jn (β2 ) − β2 (1 − ν)(n − s1 )Jn+1 (β2 )

Q21 = 2n(1 − n)(1 + s2 ) + s2 β12 Jn (β1 ) + 2β1 (n − s2 )Jn+1 (β1 )

Q22 = 2n(1 − n)(1 + s1 ) + β22 Jn (β2 ) − 2β2 (1 − ns1 )Jn+1 (β2 ).

3

(26)

To obtain nontrivial values of An and Bn , the frequency p must be chosen so that det Q(p, Ω) = 0.

(27)

For each n, the frequency equation (27) has an infinite number of solutions. Corresponding to each root pnk , the factor Bnk can be found and it defines a mode shape (Unk , iVnk ). These mode shapes are important in that they will be used to construct solutions to the general forced vibration problem.

Note that the solutions of the free motion problem are of the form u = [ u iv ]T where both u and v are real functions. Thus, for the remainder of this paper, attention will be confined to functions of this form.

4

State-Space Formulation

Let us now write the equations of motion in state-space. Let   ur

 uθ  u  w= =  u˙ r  . u˙ u˙ θ Then equation (1) can be rewritten as

u˙ 0 w ˙ = = u ¨ Ln + Ω2 I

I −2ΩD

This can further be rewritten as

0 u˙ Ln + Ω2 I 0 = Ln + Ω2 I u ¨ 0 I

(28)

u u˙

Ln + Ω2 I −2ΩD

+

u u˙

0 1 ρF

+

(29)

.

0 1 ρF

.

(30)

This becomes An w ˙ − Bn w = q

(31)

where An =

Ln + Ω2 I 0

0 Ln + Ω2 I

0 q= .

1 ρF

Bn =

Let p =

T

Ln + Ω2 I −2ΩD

(32)

(33) (34)

T

be a vector in state-space. Again, note the special form of the T T constituent vectors p1 and p2 . Similarly, let w = w1 w2 = w11 iw12 w21 iw22 . Here the pij and wij are all real functions. It then follows that Z ah i T T (p⋆1 ) (Ln w1 + Ω2 w1 ) + (p⋆2 ) w2 rdr hp, An wi = Z0 a h i T T T (w1⋆ ) Ln p1 + Ω2 (w1⋆ ) p1 + (w2⋆ ) p2 rdr = p1

p2

=

0 I

p11

ip12

p21

ip22

0

= hw, An pi ,

(35)

where the self-adjointness of Ln and the special form of p1 , p2 , w1 and w2 have been employed. The self-adjointness of the operator Ln is demonstrated in the appendix. In particular, we used the fact that

4

T

⋆

(p⋆1 ) w1 =

p11

ip12

=

w11

−iw12

w11 = p11 w11 + p12 w12 iw12

p11 T = (w1⋆ ) p1 . ip12

(36)

For the above to be true, recall that we require the pj and wj to satisfy the boundary conditions and equation (90) so that the Ln is a self-adjoint operator. Using the same argument,

Z a

⋆ T Ln w2 + Ω2 w2 p1 p2 rdr hp, Bn wi = Ln w1 + Ω2 w1 − 2ΩDw2 Z0 a ⋆ T

(p1 ) Ln w2 + Ω2 (p⋆1 )T w2 + (p⋆2 )T Ln w1 + Ω2 (p⋆2 )T w1 − (p⋆2 )T 2ΩDw2 rdr = Z0 a ⋆ T

(w2 ) Ln p1 + Ω2 (w2⋆ )T p1 + (w1⋆ )T Ln p2 + Ω2 (w1⋆ )T p2 − (w2⋆ )T 2ΩDp2 rdr =

Z0 a

⋆ T Ln p2 + Ω2 p2 w1 w2 rdr = Ln p1 + Ω2 p1 − 2ΩDp2 0 = hw, Bn pi .

(37)

Hence An and Bn are self-adjoint operators on the state-space of functions that satisfy the boundary conditions, T T equation (90) and are of the form p = p1 p2 = p11 ip12 p21 ip22 , pij ∈ R. Now suppose that

Φ1 (r) w= e−ipt (38) Φ2 (r) in state-space so that w ˙ = −ipw. For the free vibration problem then An w−B ˙ n w = 0 becomes −ipAn w−Bn w = 0.

This can be expanded as

ip Ln Φ1 + Ω2 Φ1 + Ln Φ2 + Ω2 Φ2 = 0 (39) ipΦ2 + Ln Φ1 + Ω2 Φ1 − 2ΩDΦ2 = 0.

(40)

If we put Φ2 = −ipΦ1 then equation (39) is satisfied and equation (40) reduces to Ln Φ1 + Ω2 Φ1 + p2 Φ1 + 2ipΩDΦ1 = 0,

(41)

which is exactly the same as equation (1) with F = 0, u = Φ1 (r)e−ipt and the operator L replaced with Ln . This indicates that the function Φ1 is exactly the eigenfunction of the in-plane problem written in normal space (that

T be the state-space eigenfunction corresponding to the kth is, not in state-space). Let Φnk = φnk −ipnk φnk eigenvalue of the nth operator. That is, ipnk An Φnk + Bn Φnk = 0.

(42) Consider the following : hΦnk , Bn Φnj i = hΦnk , −ipnj An Φnj i = −ipnj hΦnk , An Φnj i = −ipnj hΦnj , An Φnk i ,

(43)

where we have used the self-adjointness of An . Similarly, hΦnk , Bn Φnj i = hΦnj , Bn Φnk i = hΦnj , −ipnk An Φnk i = −ipnk hΦnj , An Φnk i

(44)

Subtracting equation (43) from equation (44) yields 0 = −i(pnk − pnj ) hΦnj , An Φnk i .

(45)

Thus for j 6= k and distinct pnj , pnk we obtain hΦnj , An Φnk i = 0, j 6= k.

5

(46)

Furthermore, for nonzero eigenvalues, we also obtain hΦnj , Bn Φnk i = 0, j 6= k.

Recall that Φnk =

φnk

−ipnk φnk

T

. Hence Φ⋆nk =

φ⋆nk

+ipnk φ⋆nk

(47) T

. Therefore, for j 6= k

0 = hΦnj , An Φnk i

Z a ⋆ Ln + Ω2 I 0 φnk φnj ipnj φ⋆nj = rdr −ipnk φnk 0 I 0 Z a ⋆

φnj Ln φnk + Ω2 φ⋆nj φnk + pnj pnk φ⋆nj φnk rdr. =

(48)

0

This implies for j 6= k

Ajk + Ω2 αjk − pnj pnk αjk = 0,

(49)

where Ajk =

Z

a

0

αjk =

Z

φ⋆nj Ln φnk rdr

(50)

φ⋆nj φnk rdr.

(51)

a

0

Similarly, 0 = hΦnj , Bn Φnk i can be expanded to read

where

(pnj − pnk ) Ajk + Ω2 αjk + pnj pnk 2Ωβjk = 0, −iβjk =

Z

0

(52)

a

φ⋆nj Dφnk rdr.

(53)

From the self-adjointness of Ln , it follows that Ajk = Akj . Furthermore, if the φnj are in the space of functions of T , then by symmetry the form Unj iVnj αjk = αkj βjk = βkj .

(54) (55)

If j and k are interchanged in equation (52) and the symmetry relationships (54) and (55) are also employed, then we can also write

(pnk − pnj ) Ajk + Ω2 αjk + pnk pnj 2Ωβjk = 0.

(56) Adding equation (52) and equation (56) gives

4Ωpnj pnk βjk = 0,

j 6= k.

Hence for j 6= k and nonzero eigenvalues, it follows that βjk = 0. Equation (56) then implies that

(pnj − pnk ) Ajk + Ω2 αjk = 0.

(57)

(58)

Hence for distinct eigenvalues, it follows that

Ajk + Ω2 αjk = 0,

j 6= k.

(59)

This last relationship in conjunction with equation (49) implies that for nonzero eigenvalues and j 6= k that αjk = 0.

In summary, for nonzero, distinct eigenvalues and j 6= k, the following relations are true : Ajk = 0, αjk = 0, βjk = 0.

6

5

Eigenfunction Expansions

Given the above orthogonality relationships, arbitrary functions can be expanded in terms of the eigenfunctions.

Consider expanding an arbitrary function G(r) as G(r) =

f (r) ig(r)

=

∞ X

cnk φnk =

∞ X

cnk

k=1

k=1

Unk iVnk

.

(60)

To find the coefficients cnk , premultiply both sides of the above equation by φ⋆nj r and integrate from r = 0 to r = a.

Z

a

0

φ⋆nj G(r)

rdr = =

∞ X

k=1 ∞ X

cnk

Z

a

0

φ⋆nj φnk rdr

cnk αjk = cnj αjj.

(61) (62)

k=1

The last line follows since αjk = 0 for j 6= k. Hence, the coefficient cnj is given by cnj =

6

hφnj , Gi . αjj

(63)

Forced Inplane Problem

We now consider the problem of forced in-plane vibrations. The forcing function F can be written as a fourier series in θ.

F=

∞ X

Fn (r, t)einθ .

(64)

n=−∞

The response of the disk to this forcing function can be found by finding the response of the disk to a typical harmonic Fn (r, t)einθ and then using the principle of superposition. In other words, u=

∞ X

(65)

un

n=−∞

where un is the reponse of the disk to Fn (r, θ)einθ .

Now consider the solution of this ’simplified’ forced inplane problem for un using the free motion mode shapes and using the fourier transform. Recall that the fourier transform of an arbitrary function q(t) is given by Z +∞ q(t)eipt dt.

(66) ̥(q(t); p) = qˆ(p) = −∞

The inverse transform is given by −1

̥

1 (ˆ q (p); t) = q(t) = 2π

Z

+∞

qˆ(p)e−ipt dp.

(67)

−∞

It is also useful to note that

du ̥ ; p = −ip̥(u(t); p) dt

2 d u ; p = −p2 ̥(u(t); p). ̥ dt2

(68) (69)

The PDE in question is given by 1 Ln un = u ¨ n −Ω2 un +2ΩDu˙ n − Fn . ρ 7

(70)

ˆ n = ̥(Fn (r, t); p) = F ˆ n (r, p) Let u ˆ n = ̥(un (r, t); p) = u ˆ n (r, p) be the fourier transform of un (r, t). Similarly, let F be the fourier transform of Fn (r, t). Taking the fourier transform of equation (70) yields 1ˆ Ln u ˆ n = −p2 u ˆ n −Ω2 u ˆ n −2ΩipDˆ un − F n. ρ

(71)

Now suppose that an expansion for u ˆ can be found in terms of the eigenfunctions. That is, suppose that ! ∞ X u ˆ n (r, p) = c˜nk (p)φnk (r) einθ .

(72)

k=1

Note that to find the solution to the equation (70), it suffices to find the coefficients in the expansion. In the above equation, the c˜nk (p) are the coefficients of the expansion of u ˆ n , the fourier transform of un , not those of un itself. Clearly the c˜nk (p) are the fourier transforms of the cnk (t), which are the coefficients in the expansion of un itself. The use of the fourier transform only affects the time domain. The space domain is unaffected by the ˆn fourier transform and thus the eigenfunctions can still be used for the space domain. Since the forcing function F is known, its expansion in terms of the eigenfunctions can be explicitly found: ˆ n (r, p) = F

∞ X

fˆnk (p)φnk (r),

(73)

k=1

where fnj =

E D ˆn φnj , F αjj

(74)

.

Substituting equation (72) and equation (73) into equation (71) gives ! ∞ ∞ ∞ ∞ X X X 1Xˆ c˜nk φnk − c˜nk φnk −2ΩipD c˜nk φnk = −(p2 +Ω2 ) Ln fnk φnk . ρ k=1

k=1

k=1

(75)

k=1

Premultiplying both sides of the preceding equation by φ⋆nj r and integrating from r = 0 to r = a yields ∞ X

k=1

c˜nk Ajk = −(p2 + Ω2 )

∞ X

k=1

c˜nk αjk − 2pΩ

∞ X

k=1

∞

c˜nk βjk −

1Xˆ fnk αjk. ρ

(76)

k=1

However, recall that αjk = βjk = αjk = 0 for j 6= k. Thus the preceding equation can be rewritten as c˜nj =

h

ρ (p +

Ωβjj 2 αjj )

−fˆnj

+ Ω2 +

Ajj αjj

−

2 Ω2 βjj α2jj

i.

(77)

Observe that if fˆnj = 0 then c˜nj = 0. That is, the forcing only excites the same modes that are present in the forcing function. If the c˜nj are known, then u ˆ n is known. To find un itself, the inverse transform of equation (77) is required. To find the inverse transform of equation (77), two important facts are required.

Convolution Theorem Z +∞ f (u)g(t − u) du (78) ̥−1 (f˜ g˜; t) = f ⋆ g = −∞

Inverse Fourier Transform

̥−1 Let

1 (p + b)2 + a2

ǫj =

=

eibt e−|at| .

2a

Ωβjj αjj

γj2 = Ω2 +

(79)

(80) 2 Ω2 βjj Ajj − .

2 αjj αjj

8

(81)

Hence, taking γj as the positive square root of γj2 , then cnj (t) = ̥−1 (˜ cnj (p); t) = −

1 ρ

Z

+∞

−∞

fnj (u) iǫj (t−u) −|γj (t−u)| e e du 2γj

Z t 1 fnj (u)eiǫj (t−u) e−γj (t−u) du =− 2ργj −∞ Z ∞ 1 fnj (u)eiǫj (t−u) e−γj (u−t) du − 2ργj t Z eiǫj t e−γj t t =− fnj (u)e−iǫj u eγj u du 2ργj −∞ Z eiǫj t eγj t t − fnj (u)e−iǫj u e−γj u du.

2ργj −∞

(82)

Once the coefficients cnj are known, then un , and thus the solution to the forced vibration problem is known.

7

Conclusions

The free motion mode shapes for the in-plane vibrations have been previously derived. Here, detailed orthogonality properties are derived by recasting the original problem in state-space. Using these orthogonality properties, a general analytic solution to the in-plane forced vibration problem can be constructed. The details of this construction were given.

References [1] P.G. Bhuta and J.P Jones. Symmetric planar vibrations of a rotating disk. The Journal of the Acoustical Society of America, 35(7):982–989, 1963.

[2] J.S. Burdess, T. Wren, and J.N. Fawcett. Plane stress vibrations in rotating disks. Proceedings of the Institution of Mechanical Engineers, 201(C1):37–44, 1987.

[3] J.S. Chen and J.L. Jhu. On the in-plane vibration and stability of a spinning annular disk. Journal of Sound and Vibrations, 195(4):585–593, 1996.

[4] J.A.. Wickert and C.D. Jr. Mote. Classical vibration analysis of axially moving continua. The Journal of Applied Mechanics, 57:738–744, 1990.

[5] R.Grammel and C.B. Biezano. Engineering Dynamics. D.Von Nostrand Company, Inc., Princeton, New Jersey, 1954.

[6] K. Yamada. Vibration of a turbine disk in its plane. Proceedings of the Second Japanese National Congress on Applied Mechanics, pages 343–347, 1952.

[7] R. Doby. On the elastic stability of coriolis-coupled oscillations of a rotating disk. Journal of the Franklin Institute, 288(3):203–212, 1969.

[8] L.C. Andrews. Elementary Partial Differential Equations. Academic Press, New York, 1986.

8

Appendix : Self-Adjointness of the Operator Ln

Lemma 1 Ln is a self adjoint operator in the space of functions that satisfy the boundary conditions (10) and (11)

T √ and that are of the form u iv where u and v are real functions. Here i = −1.

9

ivj ]T where uj and vj are real will be considered. Consider

Z a u1 rdr [ u2 −iv2 ]Ln hu2 , Ln u1 i = iv1 0

Z a ∂u1 (1 + ν) (1 − ν) ∂ 2 u1 ∂ 2 v1 u2 + + nv2 + ru2 rv = 2 2 ∂r ∂r2 2 ∂r2 0

1 ∂v1 u1 (1 − ν) 2 (3 − ν) + ((1 − ν)v2 − (1 + ν)nu2 ) + nv1 − − n u1 u2 2 ∂r 2r r 2r

(3 − ν) (1 − ν)v1 1 + nu1 − − n2 v1 v2 dr 2r 2r r

∂u1 (1 + ν) ∂u2 (1 − ν) ∂v2 (1 + ν) u1 + ru2 v1 = nv2 − r − nu2 + r 2 ∂r ∂r 2 2 ∂r

a Z a (1 − ν) ∂u2 (1 + ν) ∂v1

∂ 2 u2 ∂v2 + u + − rv2 + ru nu1 1 1

2 2 ∂r 0 ∂r ∂r 2 ∂r 0

Proof. Only functions of the form uj = [ uj

(1 − ν) ∂ 2 v2 (1 + ν) ∂u2 (3 − ν) (1 − ν) ∂v2 v1 + rv1 2 + nv1 + n(v1 u2 + u1 v2 ) 2 ∂r 2 ∂r 2 ∂r 2r 1

(1 − ν) 2 − n u 1 u 2 + v 1 v2 − u1 u2 + n2 v1 v2 dr 2r r

(83)

(84)

+

(85)

It may also be verified

that the integral (i.e. non boundary) portion of equation (85) is equivalent to hu1 , Ln u2 i = u2 rdr. Hence it follows that [ u1 −iv1 ]Ln 0 iv2

Ra

hu1 , Ln u2 i = hu2 , Ln u1 i

(86)

provided that the boundary term in equation (85) disappears. Thus the operator is self-adjoint provided that

u1 (1 + ν)n

u1 u2

(1 + ν)n

u1 u2

u2

− − a ∂u1 ∂u2

0=

v1 v2

v 1 v2

2 2 ∂r ∂r r=a r=0 r=a

a(1 − ν)

v1 v2

.

(87) −

∂v1 ∂v2

2 ∂r ∂r r=a

Now suppose that each of u1 = [ u1 iv1 ]T and u2 = [ u2 (10) and (11). That is, for j = 1, 2 the following are true

iv2 ]T satisfy the boundary conditions, equations

∂uj ν + (uj − nvj )

=0 ∂r r

r=a vj ∂vj

nuj − + = 0. r r ∂r r=a

(88) (89)

It then follows that equation (87) reduces to

(1 + ν)n

u1 −

v1 2

u2

= 0. v2 r=0

(90)

Hence, provided that u1 and u2 satisfy the boundary conditions and equation (90) is satisfied, the operator Ln is self-adjoint.

Remark 2 The solutions of the free in-plane vibration problem are of the form u = [ u iv ]T where both u and v are real functions. It is for this reason that our attention is confined to functions of this form.

10