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Applied Mathematical Modelling 31 (2007) 54–77 www.elsevier.com/locate/apm

Nonlinearly coupled in-plane and transverse vibrations of a spinning disk Natalie Baddour *, Jean W. Zu Department of Mechanical and Industrial Engineering, University of Toronto, 5 King’s College Road, Toronto, Ont., Canada M5S 3G8 Received 1 January 2003; received in revised form 1 March 2005; accepted 2 August 2005 Available online 21 September 2005

Abstract Previous nonlinear spinning disk models neglected the in-plane inertia of the disk since this permits the use of a stress function. This paper aims to consider the effect of including the in-plane inertia of the disk on the resulting nonlinear dynamics and to construct approximate solutions that capture the new dynamics. The inclusion of the in-plane inertia results in a nonlinear coupling between the in-plane and transverse vibrations of the spinning disk. The full nonlinear partial differential equations are simplified to a simpler nonlinear two degrees of freedom model via the method of Galerkin. A canonical perturbation approach is used to derive an approximate solution to this simpler nonlinear problem. Numerical simulations are used to evaluate the effectiveness of the approximate solution. Through the use of these analytical and numerical tools, it becomes apparent that the inclusion of in-plane inertia gives rise to new phenomena such as internal resonance and the possibility of instability in the system that are not predicted if the in-plane inertia is ignored. It is also demonstrated that the canonical perturbation approach can be used to produce an effective approximate solution. Ó 2005 Elsevier Inc. All rights reserved. Keyword: Nonlinear spinning disk coupled vibrations

1. Introduction Spinning disks can be found in many engineering applications. Common industrial applications include circular sawblades, turbine rotors, brake systems, fans, flywheels, gears, grinding wheels, precision gyroscopes and computer storage devices. Spinning disks may experience severe vibrations which could lead to fatigue failure of the system. Thus, the dynamics of spinning disks has attracted much research interest over the years. Many authors have investigated the vibrations of spinning disks using linear theory. The original papers are by Lamb and Southwell [1] and by Southwell [2] where the disk was modelled as a spinning membrane with added bending stiffness. Another popular approach in the literature is to model the spinning disks as a pure membrane with no bending stiffness [3–6]. The incorporation of both the bending stiffness of the disk and the effect of rotation leads to a fourth order PDE that is difficult to solve. As a result, various researchers have *

Corresponding author. Tel./fax: +1 709 753 3628. E-mail address: baddour@mie.utoronto.ca (N. Baddour).

0307-904X/$ - see front matter Ó 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.apm.2005.08.004


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55

applied different solution techniques to the linear analysis of the free transverse vibrations of spinning disks [7–10]. In addition to considering the transverse vibrations of a spinning disk, the problem of free planar vibrations has also been investigated using linear theory [11–15]. Although less extensive, nonlinear models have also been employed since it is known from the theory of stationary disks that the linear theory breaks down when the transverse displacement is on the order of the thickness of the disk [16–19]. However, even in the nonlinear case the in-plane inertia of the disk has always been neglected and the authors could not find any papers that considered its inclusion in the nonlinear dynamics of the rotating disk. This paper thus aims to investigate the effect of the inclusion of the in-plane inertia of the disk on the resulting dynamics. In this paper, an approximate solution to the nonlinear coupled vibrations of a spinning disk is considered and analyzed. The main focus is to investigate the effect of including the in-plane inertia. This is done by representing the solution as one mode for each of the transverse and in-plane vibrations and assuming a form for the space-dependent part of the solution. The time-dependent parts of the in-plane and transverse problems are then solved for. Through this method, three nonlinear partial differential equations become condensed into two nonlinear ordinary differential equations, for the time dependence of the transverse and in-plane vibrations. If the in-plane inertia of the plate is neglected, the same approach would yield a one degree of freedom model. The two degree of freedom model can be used to determine the effect of the inclusion of in-plane inertia on the dynamics of the problem. The thrust of this paper is to derive the simplified two degree of freedom model, to consider analytical and numerical solutions and to compare these solutions to the one degree of freedom model that results if in-plane inertia is neglected. 2. Equations of motion Since all previous models in the literature ignore the in-plane inertia of the disk in order that a stress function may be used, it is necessary to rederive the equations of motion so that the in-plane inertia of the disk may be kept. This results in three equations of motion for the dynamics instead of the usual two—a considerable complication. This derivation was performed in detail in [20]. The resulting equations and accompanying boundary conditions are given below as   qð1  m2 Þ o2 u oX ov 2  X  v ðr þ uÞ  2X E ot2 ot ot " #  2  2 ð1  mÞ o2 u ð1 þ mÞ o2 v o2 u 1 ð1  mÞ ow ð3  mÞ ov ð1 þ mÞ ow u ou þ  ¼ þ þ   þ 2r2 oh2 2r oroh or2 r 2 or 2r oh 2r2 oh r or ð1  mÞ ow o2 w ow o2 w ð1 þ mÞ ow o2 w ; þ þ 2r2 or oh2 or or2 2r2 oh oroh   qð1  m2 Þ o2 v oX ou 2  X þ ðr þ uÞ v þ 2X E ot2 ot ot   2 2 ð1 þ mÞ ow o w ð1  mÞ o v 1 ð1  mÞ ow ow ð1  mÞ ov ð3  mÞ ou ð1  mÞ v ¼ þ þ þ  þ 2r or oroh 2 or2 r 2r oh or 2 or 2r oh 2 r þ

ð1 þ mÞ o2 u 1 o2 v 1 ow o2 w ð1  mÞ ow o2 w þ 2 2þ 3 þ ; 2r oroh r oh r oh oh2 2r oh or2   qð1  m2 Þ o2 w h2 X2 2 h2 o2 2 r w þ rw E ot2 3 3 ot2  h2 1 ow 3 ow o2 w o2 v r ow ow ð1 þ mÞ ou ð1 þ mÞ 2 o2 u þ r þ r ¼  r4 w þ 4 þr 2 2 r oh 2 oh oh 2 oh or 2 oh 2 oroh 3 oh  2 2 2 2 ð1  mÞ r ow o w o w ow ð1  mÞ 3 o v ð1  mÞ 2 ov þ rv þ þ r 2 r þ r2 2 2 oroh or 2 or 2 or 2 oh or þ

ð1Þ

ð2Þ


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  2 1 ow 2 o2 w ow ð1 þ mÞ 3 o2 v ð1  mÞ 2 o2 u r3 ow 3 ou r þ ð1 þ mÞr þ r r þ þ r4 or oroh oh or 2 oroh 2 oh2 2 or     2 r2 ow o2 w 3r4 ow o2 w ð1  mÞ 2 ov 1 o2 w ov 4o u 2 ou r þr uþ þ þr þ  þ 4 2 mr or2 2 oh r oh or oh 2 or oh2 2 or or2       o2 w ou m ov ð1  mÞ o2 w ou ov þ uþ vþr þ 2 þ . or or r oh r2 oroh oh or þ

ð3Þ

Note that the full nonlinear representation of the spinning disk problem requires the solution of three nonlinearly coupled partial differential equation, Eqs. (1)–(3). By way of contrast, the nonlinear model used by other researchers consists of two coupled partial differential equations [16]. The difference arises as a result of the use of Lagrangian coordinates as well as the inclusion of in-plane inertia and Coriolis terms. 2.1. Nonlinear boundary conditions The boundary conditions at a free boundary obtained from applying HamiltonÕs principle and integration by parts are given below. Recall that here r must be evaluated on the free boundary. Thus, for a solid disk, r below is the outer radius of the disk.  2  2 u ou m ov 1 ow m ow m þ þ þ þ 2 ¼ 0; ð4Þ r or r oh 2 or 2r oh   1 ou ov 1 ow ow v þ þ ¼ 0; ð5Þ r oh or r oh or o2 w m o2 w m ow ¼ 0; ð6Þ þ þ or2 r2 oh2 r or     o 2 ð1  mÞ o2 ow w ð1  m2 Þq 2 ow o3 w r wþ  X  þ ¼ 0. ð7Þ or r2 oh2 or r E or orot2 For an annulus, the outer radius is free and the boundary conditions would be as given above. The inner ¼ 0. boundary is clamped and the boundary conditions there would be w ¼ ow or 3. Simplification of solution to a 2 DOF system via Galerkin’s method 3.1. Strategy of solution To consider the effect of the inclusion of in-plane inertia, we choose to proceed with GalerkinÕs procedure to produce a simplified problem. The result will be a two degree of freedom nonlinear model—one for each of the time dependences of the in-plane and transverse vibrations. To this purpose, let us choose the displacements as uðr; h; tÞ ¼ ueq ðrÞ þ U ðrÞ cosðmhÞcðtÞ; vðr; h; tÞ ¼ V ðrÞ sinðmhÞcðtÞ; wðr; h; tÞ ¼ W ðrÞ cosðnhÞsðtÞ;

ð8Þ ð9Þ ð10Þ

where ueq, U, V and W are known and will be discussed momentarily. The only unknowns here are the time dependences, c(t) and s(t). The ueq portion is the in-plane equilibrium displacement due to the spin of the disk. The implication here is that there is a symmetrical time-independent in-plane displacement that occurs in the plane of the disk as a result of its rotation (that is, due to the centrifugal force). The form for u and v follows from the solution of the in-plane linear problem [21,22,13] giving   U mk iðmhp tÞ mk umk ¼ Amk ; ð11Þ e iV mk where Umk and Vmk are the in-plane mode shapes of the linear spinning disk and pmk is the in-plane frequency of vibration. Taking the real part of Eq. (11) leads to the forms for u and v indicated in Eqs. (8) and (9). Now,


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57

U and V will be taken to be Umk and Vmk, respectively, where Umk and Vmk are the (m, k)th in-plane mode shapes found by solving the free linear in-plane vibration problem. Other choices for U and V can also be made and we are not limited to choosing them the be the in-plane mode shapes. However, the advantages of taking U and V to be the in-plane mode shapes are that (i) there are proven orthogonality properties for the in-plane mode shapes [21] and more importantly (ii) the in-plane mode shapes satisfy the boundary conditions rrr = rrh = 0 at the boundary of the disk. Similarly, W will be taken to be Wnk, where Wnk is the (n, k)th mode shape of the linear strain model of the spinning disk, obtained by dropping ALL nonlinear terms in Eq. (3), see [23]. That is,   r r W nk ðrÞ ¼ J n P nk þ Bn I n P nk ; ð12Þ a b a a nk where P nk a and P b are the (n, k)th solutions of the frequency equation for the linear strain model of the spinnk ning disk [20]. Recall that the appropriate relationship between P nk a and P b must be specified. Other choices for the space-dependent portion of the solution can easily be made and substituted into the above expressions. Once a choice for the space-dependent part of the solution has been made, it still remains to use this choice to derive the equations of motion for the temporal part of the solution. One option is to substitute the assumed forms of the solution into the equations of motion and then apply the method of Galerkin. A completely equivalent calculation is to substitute the assumed form of the solution into the expressions for kinetic and potential energies and using LagrangeÕs equations. The advantage of actually calculating the potential and kinetic energies of the system is that conserved quantities are more readily apparent. Such conservation laws can be used to simplify the equations of motion and could aid in the finding of an analytic solution. Furthermore, the machinery of HamiltonÕs equations and canonical perturbation theory is also available in this formulation. For these reasons, the simplified 2 DOF ordinary differential equations of motion are derived in this alternative manner. 3.2. Galerkin method Using the assumption of plane stress, along with linear stress–strain relationships and nonlinear von Karman strains, the potential and kinetic energies of the system can then be found in terms of the two degrees of freedom, c and s to given by PE ¼ Q1 c2 þ Q2 s4 þ Q3 s2 þ Q4 þ Q10 cs2 ; 2

2

2

ð13Þ

2

KE ¼ Q5 c þ Q6 s þ Q7 s_ þ Q8 c_ þ Q9 .

ð14Þ

The coefficients in Eqs. (13) and (14) can be expressed as 2  2 Z a "  2  hEp dU U V ð1  mÞ oV V U Q1 ¼  m  þ m þ þ ð1  m2 Þ 0 dr r r 2 or r r  # dU U V m þ 2m r dr; dr r r  2  2  4 ) Z a (  4 hEp dW dW 4 W 2 W 3n þ 2n þ3 r dr; Q2 ¼ 16ð1  m2 Þ 0 r r dr dr 3

Q3 ¼ Qh3 þ Qh3 ;

ð15Þ ð16Þ ð17Þ

where 3 Qh3

h3 Ep ¼ 3ð1  m2 Þ

Z 0

 þ2ð1  mÞ

a

(

d2 W dr2

2

n dW W n 2 r dr r

d2 W þ 2m 2 dr )

   2 1 dW 1 dW 2W 2W n 2 þ n 2 r dr r r dr r

2

r dr;

ð18Þ


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N. Baddour, J.W. Zu / Applied Mathematical Modelling 31 (2007) 54–77

Qh3

2    ) Z a ( hEp dW dueq ueq n2 W 2 ueq dueq þm þ 2 þm r dr ¼ ð1  m2 Þ 0 dr dr r r r dr ) 2 Z a(  2 2 eq n W eq dW ¼ ph rrr þ rhh 2 r dr. dr r 0

ð19Þ

eq Here req rr ; rhh are the equilibrium stresses in the rotating disk resulting from the equilibrium displacement ueq.  2 ) Z a( 2hEp ueq dueq ueq 2 dueq þ Q4 ¼ 2m þ r dr; ð20Þ ð1  m2 Þ 0 r dr r dr Z a  2 U þ V 2 r dr; Q5 ¼ qhpX2 ð21Þ 0 " #  2 Z qh3 pX2 a dW W2 þ n2 2 r dr; Q6 ¼ ð22Þ dr 3 r 0 " #  2 Z Z a 2 qh3 p a dW 2W Q7 ¼ þ n 2 r dr þ qhp W 2 r dr; ð23Þ 3 dr r 0 0 Z a  2 ð24Þ Q8 ¼ qhp U þ V 2 r dr; 0 Z a  2 Q9 ¼ 2qhpX2 ueq þ r r dr; ð25Þ 0

Q10 ¼ 0

for m 6¼ 2n;     Z a(  hEp U 2W dW W dW dW 2W n 2   2n Q10 ¼ þm : 2ð1  m2 Þ 0 r r dr r dr dr r      V W W dW dW W dW n 2n2  2 þ þ nm r r r dr dr r d ) " 2  2 # dU dW dV W dW 2 W þ  mn r dr otherwise. þ nð1  mÞ d d r dr r dr

ð26Þ ð27Þ

ð28Þ

The above calculations indicate that Q10 = 0 for m 5 2n. This point is worth some consideration since it considerably affects the resulting dynamics if Q10 is indeed zero. In this case, Q10 = 0 follows as a result of our choosing trigonometric functions for the h component of the trial solution to the problem and the fact that the sine and cosine functions possess orthogonality properties. A slightly different choice for the h component of the solution would not necessarily lead to Q10 = 0. Note that for a stationary disk, where X = 0, it would follow that Q4 = 0 unless the disk is pre-stressed and that Q5 = Q6 = Q9 = 0. All other expressions are the same for a stationary disk, although the actual mode shapes used in these expressions will be different. Once expressions for the kinetic and potential energies are found, the Lagrangian can be constructed and the equations of motion for the simplified system can be found. If new variables are introduced such that cI c ¼ pffiffiffiffiffiffiffiffi ; 2Q8

ð29Þ

sI s ¼ pffiffiffiffiffiffiffiffi . 2Q7

ð30Þ

Then the Lagrangian from equation can be written as  I 2  I 2  2 s_ c_ x2  2 k 3  I 4 x2s  I 2 þ  ðQ4  Q9 Þ  c cI  s s L¼   k 1 c I sI ; 2 2 2 4 2

ð31Þ


N. Baddour, J.W. Zu / Applied Mathematical Modelling 31 (2007) 54–77

59

where Q1  Q5 ; Q8 Q  Q6 x2s ¼ 3 ; Q7 Q10 pffiffiffiffiffiffiffiffi ; k1 ¼ 2Q7 2Q8 Q k 3 ¼ 22 . Q7

x2c ¼

ð32Þ ð33Þ ð34Þ ð35Þ

Thus the equations of motion for the system become €cI þ x2c cI þ k 1 ðsI Þ2 ¼ 0; I

€s þ

x2s sI

I I

ð36Þ I 3

þ 2k 1 c s þ k 3 ðs Þ ¼ 0.

ð37Þ

In future discussions, cq and sq will be replaced with c and s for ease of notation. It is important to note that if Q10 = 0 then k1 = 0 and the two equations of motion (36) and (37) are uncoupled. That is, for Q10 = 0, the system behaves like two independent oscillators, where the in-plane oscillator is linear and the transverse one is a nonlinear duffing oscillator. What are the physical implications of this? Recall that to arrive at Eqs. (36) and (37), two particular modes were selected at the outset. If k1 = 0, the implication is that the vibration of these particular modes are not coupled. Since there is an infinite number of in-plane and transverse modes of vibration, this would not necessarily imply that none of the in-plane and transverse modes of vibrations are coupled. However, it does not follow that all the transverse modes are coupled to all the in-plane modes and this is the physical implication of k1 = 0. 3.3. Simplifying the coefficients After some calculation and simplification, it can be rigorously shown that xc and xs are exactly what we would expect them to be—namely the frequencies of vibration of the linear in-plane and transverse vibrations. 3.3.1. Calculating xc It can be shown that Q1 reduces to the sum of a boundary term and an integral term Q1 ¼ Qboundary þ Qintegral ; 1 1

ð38Þ

¼ hprU rrr ja0 þ hprV rrh ja0 ¼ 0 by virtue of the boundary conditions. If U = Umk, V = Vmk is where Qboundary 1 substituted into Qintegral , denoting the (m, k)th modeshape, then it can be shown that Qintegral becomes 1 1 Qintegral ¼ hqpAkk ; 1

ð39Þ

where Akk ¼ ðp2mk  X2 Þakk  pmk for a spinning disk and Akk ¼ p2mk akk for a stationary disk. Here pmk is the linear in-plane frequency of vibration and Z a  ajk ¼ U mj U mk þ V mj V mk r dr. ð40Þ 0

Thus for a spinning disk, x2c can be written as Q1  Q5 Q1 ¼  X2 Q8 Q8 

 p2k  X2 akk  pmk hpq  X2 ¼ hpqakk   p 1  pmk ; ¼ mk  p2mk ¼ pmk akk akk

x2c ¼

ð41Þ ð42Þ ð43Þ


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N. Baddour, J.W. Zu / Applied Mathematical Modelling 31 (2007) 54–77

which is precisely the linear in-plane frequency of vibration as shown in [21]. For a stationary disk, it can be shown that x2c ¼ p2mk . In other words, xc is precisely the frequency of vibration of the in-plane mode that was chosen. This is a naturally intuitive result, which the calculations justify. 3.3.2. Calculating xs Let us now consider how to calculate xs. Recall that xs is defined by x2s ¼

Q3  Q6 Q7

ð44Þ

3

¼

Qh3  Q6 Qh3 þ . Q7 Q7

ð45Þ

are proportional to h3 and That is, x2s can be considered to have two components; a ÔplateÕ part where all terms 3 Qh Q

a ÔmembraneÕ part which is proportional to h. Clearly the plate part is given by 3 Q 6 and the membrane part is 7 Qh given by Q3 . If we take W to be the eigenfunctions of the linear strain model of the spinning disk [23], then 7  r  r W ¼ W nk ¼ Ank J n P a þ C nk I n P b . ð46Þ a a Let knk denote the (n, k)th frequency of vibration of the linear strain model of the spinning disk. Then following some algebra it can be shown that the plate portion of x2s is given by 3

Qh3  Q6 ¼ k2nk . Q7

ð47Þ

That is, the plate portion of x2s is the square of the frequency of the linear strain model of the spinning plate. Considering the fact that the Wnk were chosen to be the mode shapes of this model, this is an intuitive result. Now consider the membrane portion x2s . o R a n eq dW 2 eq n2 W 2 h ph r þ r r dr hh r2 rr dr 0 Q3 i . ð48Þ ¼ 3 R a h  2 R 2 a dW Q7 qh p þ n2 W r dr þ qhp W 2 r dr 3

0

dr

r2

0

Now, for W = Wnk, where the Wnk are the eigenfunctions of the linear strain model of the spinning plate, the above expression does not permit as nice an interpretation as the plate component of x2s did. However, for the sake of illustration, suppose that in Eq. (48) we take W ¼ W mnk , the eigenfunctions of the linear spinning membrane. First, the term proportional toRh3 in the denominator of Eq. (48) would not be present for a membrane. a Thus, the denominator becomes qhp 0 W 2 r dr. Furthermore, if the eigenfunctions of the spinning membrane are normalized, then Eq. (48) can be shown to reduce to  m 2  2 Qh3 qph knk ¼ kmnk ; ¼ Q7 qph

ð49Þ

that is, the square of the natural frequency of the (n, k)th mode of the spinning membrane. Again, this is a natural result. However, in this development, it is the eigenfunctions of the linear strain model of the spinning disk that are being used and thus the membrane component of x2s will not give exactly ðkmnk Þ2 but rather some approximation to it. As another note, suppose that we had taken W ¼ W NLSM , the eigenfunctions of the nonlinear strain model, nk obtained by dropping only the terms nonlinear in w and its derivatives in Eq. (3). Note that these eigenfunctions are not actually available to us, but it is instructive to see what would happen is we could use them in this calculation. A quick review of the above developments would indicate that the result to be expected is x2s ¼ ðkNLSM Þ2 . nk


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61

4. Canonical perturbation approach of the simplified 2 DOF system The coupled 2 DOF equations of motion derived in the above sections are given by €c þ x2c c þ k 1 s2 ¼ 0; €s þ

x2s s

ð50Þ 3

þ 2k 1 cs þ k 3 s ¼ 0.

ð51Þ

Clearly, xc and xs are the natural frequencies of the linear in-plane and transverse modes that were chosen. The expressions for k1 and k3 are complicated, but it suffices to note that they will depend on the amplitudes of the chosen modes. Eqs. (50) and (51) can also be arrived at from the point of view of HamiltonÕs equations. That is, Eqs. (50) and (51) are equivalent to oH ; opc oH s_ ¼ ; ops oH ; p_ c ¼  oc oH ; p_ s ¼  os c_ ¼

ð52Þ

where pc and ps are the conjugate momenta to c and s, and the Hamiltonian is given by p2c p2s x2c 2 x2s 2 k3 þ þ c þ s þ k 1 cs2 þ s4 . ð53Þ 2 2 2 2 4 Eq. (52) represent a general Hamiltonian system. Since the equations of motion of the 2 DOF model can be derived from a Hamiltonian, they are a Hamiltonian system. The general objective here is to transform the given Hamiltonian system into a simpler Hamiltonian system and to use the result to construct an approximate solution. H¼

4.1. Transformation of the Hamiltonian Generally, a transformation of variables from generalized coordinates q, p to qq, pq can be of the form qI ¼ qI ðq; p; tÞ; pI ¼ pI ðq; p; tÞ

ð54Þ

with corresponding transformed differential equations q_ I ¼ q_ I ðq; p; tÞ; p_ I ¼ p_ I ðq; p; tÞ.

ð55Þ

Suppose that Eq. (55) has the same symmetry as the original equations. That is, suppose that there exists a function Hq(qq, pq, t) such that oH I ; opI oH I p_ I ¼  I . oq

q_ I ¼

ð56Þ

Then the transformation (54) is known as a canonical transformation [24], transforming Hamiltonian differential equations into another system of differential equations of the same type. One possible approach to solving the equations of motion is to find a canonical transformation such that the transformed Hamiltonian Hq is a constant function, independent of time and any of the generalized


62

N. Baddour, J.W. Zu / Applied Mathematical Modelling 31 (2007) 54–77

coordinates or momenta. Then Eq. (56) reduce to the simple form of q_ I ¼ p_ I ¼ 0 and can be easily integrated. In reality, finding such a transformation can be just as difficult as solving the original equations of motion. However, it is possible to use a canonical transformation that effectively transforms a simpler Hamiltonian to aid in the construction of an approximate solution for a problem with a more complicated Hamiltonian. Even though such a transformation will not solve the more complicated problem, it will help in the construction of an approximate solution. This is the approach that will be undertaken here. Consider the Hamiltonian of the 2 DOF model: H¼

p2c p2s x2c 2 x2s 2 k3 þ þ c þ s þ k 1 cs2 þ s4 . 2 2 2 2 4

ð57Þ

This Hamiltonian can be expressed as H ¼ H ð0Þ þ H ð1Þ ;

ð58Þ

where p2c p2s x2c 2 x2s 2 þ þ c þ s; 2 2 2 2 k 3 ¼ k 1 cs2 þ s4 . 4

H ð0Þ ¼

ð59Þ

H ð1Þ

ð60Þ

The motivation for doing this is that H(0) can be recognized as the Hamiltonian of two uncoupled linear harmonic oscillators. H(1) brings in coupling by way of the third order k1 term. The fourth order term in H(1) does not couple the two oscillators. Rather, it has the effect of turning the linear s oscillator into a nonlinear duffing oscillator. The k1 term is the only term that couples the two oscillators. Thus the first step in our approximate solution will be to find a canonical transformation that transforms H(0) into a constant. This can easily be done by solving the Hamilton–Jacobi equation associated with H(0) [24]. Now this canonical transformation is equally valid no matter which Hamiltonian is employed. It just has the particular advantage of simplifying this particular Hamiltonian H(0), but it can be used with the entire original Hamiltonian H. After thus transforming the Hamiltonian, approximate solutions for the ensuing Hamiltonian equations will be sought. Solving the Hamilton–Jacobi equation associated with H(0) yields pffiffiffiffiffiffiffi 2a1 c¼ sin xc ðt  b1 Þ; ð61Þ xc pffiffiffiffiffiffiffi 2a2 sin xs ðt  b2 Þ. ð62Þ s¼ xs Furthermore, the generalized momenta can also be transformed qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi oS ¼ 2a1  x2c c2 ¼ 2a1 cos xc ðt  b1 Þ; pc ¼ oc qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi oS ¼ 2a2  x2s s2 ¼ 2a2 cos xs ðt  b2 Þ. ps ¼ os

ð63Þ ð64Þ

Eqs. (61)–(64) are essentially a canonical transformation from the old variables c, s, pc, ps to the new variables a1, a2, b1, b2, where these are constants of integration. It can be verified that if (61)–(64) are substituted into H(0), the result is the constant a1 + a2. This is not coincidence: the transformation was specifically constructed to do so. In fact, Eqs. (61)–(64) are precisely the solution to the uncoupled two oscillator problem. The transformed Hamiltonian for the uncoupled two oscillator problem is given by H ð0Þ þ oS , which is precisely zero (by ot construction). Now, while this canonical transformation transforms H(0) into zero, it does not do so for the entire Hamiltonian H = H(0) + H(1). The transformed Hamiltonian becomes H I ¼ H ð0Þ þ oS þ H ð1Þ ¼ H ð1Þ ða1 ; ot (1) a2 ; b1 ; b2 ; tÞ. Using these results, the Ôperturbing HamiltonianÕ H can be expressed as


N. Baddour, J.W. Zu / Applied Mathematical Modelling 31 (2007) 54–77

63

pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi a2  2 8k sinðx t  x b Þ þ 4k 2a x 2a1 x2s sinððxc þ 2xs Þt  xc b1  2xs b2 Þ 1 1 c c 1 1 s 8xc x4s pffiffiffiffiffiffiffi þ 4k 1 2a1 x2s sinððxc  2xs Þt  xc b1 þ 2xs b2 Þ  3k 3 a2 xc  k 3 a2 xc cosð4xs t  4xs b2 Þ  þ 4k 3 a2 xc cosð2xs t  2xs b2 Þ .

H ð1Þ ¼ 

Note from this definition of H(1) that there is a difference in its form depending on whether or not xc = 2xs. Although it may not be obvious yet, if xc = 2xs, then the system will display internal resonance and H(1) will take on a different form. From the above, using the new variables a1,a2, b1 and b2, HamiltonÕs equations can be written pffiffiffiffiffiffiffi 2a1 oH ð1Þ 1 a_ 1 ¼ ¼ a2 k 1 2 ð2 cos ðxc t þ xc b1 Þ xs 2 ob1 þ cos ðxc t þ xc b1  2xs t þ 2xs b2 Þ þ cos ðxc t  xc b1  2xs t þ 2xs b2 ÞÞ; a_ 2 ¼

ð65Þ

pffiffiffiffiffiffiffi oH ð1Þ 1 a2 ¼ ð2k 2a1 x2s cos ðxc t þ xc b1  2xs t þ 2xs b2 Þ 1 2 x1 x32 ob2 pffiffiffiffiffiffiffi  2k 1 2a1 x2s cos ðxc t  xc b1  2xs t þ 2xs b2 Þ  k 3 a2 xc sin ð4xs t þ 4xs b2 Þ

þ 2k 3 a2 xc sin ð2xs t þ 2xs b2 ÞÞ; pffiffiffi oH ð1Þ 1 2 b_ 1 ¼  ¼ a2 k 1 2 pffiffiffiffiffi ð2 sin ðxc t þ xc b1 Þ 4 xs xc a1 oa1  sin ðxc t þ xc b1  2xs t þ 2xs b2 Þ þ sin ðxc t  xc b1  2xs t þ 2xs b2 ÞÞ;

ð66Þ

ð67Þ

pffiffiffiffiffiffiffi oH ð1Þ 1 b_ 2 ¼  ¼ ð2k 1 2a1 x2s sin ðxc t þ xc b1  2xs t þ 2xs b2 Þ 4 4xc xs oa2 pffiffiffiffiffiffiffi pffiffiffipffiffiffiffiffi 2 þ 4k 1 2 a1 xs sin ðxc t þ xc b1 Þ þ 2k 1 2a1 x2s sin ðxc t  xc b1  2xs t þ 2xs b2 Þ  3k 3 a2 xc  k 3 a2 xc cos ð4xs t þ 4xs b2 Þ þ 4k 3 a2 xc cos ð2xs t þ 2xs b2 ÞÞ.

ð68Þ

Eqs. (65)–(68) are exact—no approximations have been made at this point. We have used a valid transformation to change the variables and obtained the equations of motion for the transformed variables. If these equations could be solved, they would provide the solution to the original problem although practically this is not easier than the original problem. The utility of the transformation is that Eqs. (65)–(68) offer a good starting point for approximate solutions and make certain features of the solutions easy to spot. To construct an approximate solution, a1, a2, b1, b2 on the right hand side of Eqs. (65)–(68) are replaced by their initial constant values. Eqs. (65)–(68) are then integrated to yield the first order canonical perturbation solution. For a second order canonical perturbation solution, a1, a2, b1, b2 on the right hand side of Eqs. (65)– (68) are replaced with the first order perturbation solution found in the previous step. This procedure can be continued indefinitely, although in practice this gets unwieldy after a few iterations. 4.2. Canonical perturbation solution If the initial conditions for (50) and (51) are given by sð0Þ ¼ s0 ; dsð0Þ ¼ 0; dt cð0Þ ¼ c0 ; dcð0Þ ¼ 0; dt

ð69Þ ð70Þ ð71Þ ð72Þ


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then the corresponding initial values for a1, a2, b1, b2 can be worked out from the transformation equations (61)–(64) to be 1 a01 ¼ x2c c20 ; 2 p 0 b1 ¼  ; 2xc 1 a02 ¼ x2s s20 ; 2 p b02 ¼  . 2xs

ð73Þ ð74Þ ð75Þ ð76Þ

Once the a1, a2, b1, b2 are replaced in the right hand side of Eqs. (65)–(68) with their initial values a01 ; a02 ; b01 ; b02 , then Eqs. (65)–(68) can be easily integrated to give the first order canonical perturbation solutions for a1, a2, b1, b2: k 1 xc c0 s20 cosððxc þ 2xs ÞtÞ k 1 xc c0 s20 cosððxc  2xs ÞtÞ k 1 c0 s20   cosðxc tÞ 4ðxc þ 2xs Þ 4ðxc  2xs Þ 2 k 1 c0 s20 ðx2c  2x2s Þ x2c c20 ; þ þ x2c  4x2s 2

a11 ¼ 

k 3 s40 k 3 s40 k 1 xs c0 s20 cosððxc þ 2xs ÞtÞ cosð4xs tÞ  cosð2xs tÞ  2ðxc þ 2xs Þ 32 8 2 2 2 k 1 xs c0 s0 cosððxc  2xs ÞtÞ 2xs k 1 c0 s0 5k 3 s40 x2s s20 þ  2 þ ; þ 2ðxc  2xs Þ xc  4x2s 32 2 k 1 s2 sinðxc tÞ k 1 s20 sinððxc þ 2xs ÞtÞ k 1 s20 sinððxc  2xs ÞtÞ p b11 ¼  0    ; 2c0 x3c 4c0 x2c ðxc þ 2xs Þ 4c0 x2c ðxc  2xs Þ 2xc k 1 c0 sinðxc tÞ k 1 c0 sinððxc þ 2xs ÞtÞ k 1 c0 sinððxc  2xs ÞtÞ   b12 ¼  xc x2s 2x2s ðxc þ 2xs Þ 2x2s ðxc  2xs Þ 2 2 2 k 3 s0 sinð2xs tÞ k 3 s0 sinð4xs tÞ 3k 3 s0 t p     . 4x3s 32x3s 8x2s 2xs

ð77Þ

a12 ¼ 

ð78Þ ð79Þ

ð80Þ

Clearly, a quick glance at some of the values in the denominators reveals that the above expressions for a11 ; a12 ; b11 ; b12 are not valid if the resonance condition of xc = 2xs holds. In that case, the limit of the above expressions as xc ! 2xs is taken. This yields the first order canonical perturbation solution for the internal resonance case: a11;IR ¼ k 1 c0 s20 ð1  cos4 ðxs tÞÞ þ 2x2s c20 ;

ð81Þ k 3 s40



x2s s20

1  cos4 ðxs tÞ þ ; a12;IR ¼ k 1 c0 s20 ðcos2 ðxs tÞ  cos4 ðxs tÞÞ þ 4 2   s20 4c0 px2s 1 3 b1;IR ¼  2k 1 sinðxs tÞcos ðxs tÞ þ k 1 sinðxs tÞ cosðxs tÞ þ þ k 1 xs t ; 16c0 x3s s20 ð4k 1 c0 þ k 3 s20 Þ ð4k 1 c0 þ 3k 3 s20 Þ b12;IR ¼  sinðxs tÞcos3 ðxs tÞ  sinðxs tÞ cosðxs tÞ 3 4xs 8x3s 4pxs þ 4k 1 c0 t þ 3k 3 s20 t  . 8x2s

ð82Þ ð83Þ

ð84Þ

This procedure thus gives a first order solution for both the internal resonance case and the non-internal resonance case. To find the second order solution, a1, a2, b1, b2 on the right hand side of Eqs. (65)–(68) are replaced with their first order values a11 ; a12 ; b11 ; b12 . Once again, the right hand sides of Eqs. (65)–(68) can be integrated to yield the second order perturbation solution a21 ; a22 ; b21 ; b22 , although this can be difficult to achieve analytically. In general, the canonical perturbation approach easily provides a first order approximate solution while higher order approximations are possible but more difficult with this approach.


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65

Recall that a1,a2, b1,b2 do not provide the solution to the original problem directly. These are obtained from the transformation equations, namely pffiffiffiffiffiffiffi 2a1 sin xc ðt  b1 Þ; ð85Þ c¼ xc pffiffiffiffiffiffiffi 2a2 sin xs ðt  b2 Þ. ð86Þ s¼ xs Recall that a01 ; a02 ; b01 ; b02 are constants, thus the resulting solution for c and s for the two uncoupled oscillators are simply periodic trigonometric functions, namely the linear solution for simple harmonic oscillators that are so familiar. For the coupled problem, a11 ; a12 ; b11 ; b12 are actually periodic functions themselves. Thus the resulting solutions for c and s are now periodic trigonometric functions with periodically changing amplitudes. 5. Exact solution of the 1 DOF model Recall that the nonlinear model of a spinning disk used in the literature neglects the presence of in-plane inertia. The model derived here, Eqs. (50) and (51), can be reduced to the model used in the literature by ignoring the in-plane inertia in Eq. (50). Eq. (50) would then become x2c c þ k 1 s2 ¼ 0.

ð87Þ

This equation can be solved for c and the result substituted into Eq. (51) to yield the one degree of freedom model (1 DOF):   2k 2 €s þ x2s s þ k 3  21 s3 ¼ 0. ð88Þ xc Note that this is the equation of motion of a nonlinear one degree of freedom system and will be referred to as the one degree of freedom (1 DOF) model, describing the dynamics of the system that ignores the effect of in-plane inertia. The Hamiltonian associated with Eq. (88) can be written as   p2s x2s 2 k3 k 21 4 H 1 DOF ¼ þ s þ  ð89Þ s. 2 2 4 2x2c Recall that ps ¼ ds and that the Hamiltonian must be conserved since it does not explicitly depend on time, dt that is H1 DOF is a constant. Using those two facts, Eq. (88) can be integrated to give s as a function of time. If the initial conditions are given by sð0Þ ¼ s0 ; dsð0Þ ¼ 0; dt

ð90Þ ð91Þ

then the solution is given by sðtÞ ¼ s0 cnðxð1Þ t; k ð1Þ Þ;

ð92Þ

where sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   2k 2 x ¼ x2s þ s20 k 3  21 ; xc sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   s0 1 2k 21 ð1Þ k3  2 . k ¼ ð1Þ 2 x xc ð1Þ

ð93Þ ð94Þ


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In Eq. (92), cn is the Jacobian elliptic function that is a generalization of the cosine function. The function cn(x(1)t, k(1)) is a periodic function with period T ð1Þ ¼

4K ð1Þ ; xð1Þ

ð95Þ

where K(1)(k(1)) = F(k(1), p/2) is the complete elliptic integral of the first kind. Note that the above solution is only valid for real k(1). That is, for k 3 x2c  2k 21 > 0. The solution to the 1 DOF model is thus a periodic function which reduces to the cosine function when k(1) = 0. Furthermore, the frequency of this function is given by x(1) which depends on xs, as we would expect, as well as s0, k1, k3 and xc. Here, k1 and k3 may depend on the amplitudes of vibration and thus we have the familiar result of the frequency of vibration being dependent on the amplitude. 6. Numerical simulations In this section, numerical simulations of the 1 DOF and 2 DOF models are considered, along with the approximate solution. All numerical results were obtained by using MATLAB. For numerical solutions to the nonlinearly coupled equations, fourth and fifth order Runge–Kutta formulas were employed via builtin MATLAB functions. 6.1. Resonance cases For the first case, the following values for the constants are taken: k1 = 2, k3 = 2, xc = 4, xs = 2, c0 = 0.2, s0 = 0.2. This is clearly a case where xc = 2xs, thus internal resonance is expected. Consider first a comparison of the numerical solution of the 1 DOF and 2 DOF models. This is shown for s and c in Figs. 1 and 2, respectively. Since the approximation c ¼  xk12 is used in order to obtain the 1 DOF c model, it is to be expected that there is a large discrepancy between c(t) as obtained from the solutions of the 1 DOF model and the 2 DOF model. This is in fact shown explicitly in Fig. 2. What is interesting and not predicted by the 1 DOF model is the internal resonance. From Fig. 1 it can be seen that while the amplitude of s(t) remains constant for the 1 DOF model, the 2 DOF model displays a periodically varying amplitude, in keeping with the phenomena of internal resonance. It can be seen from Figs. 1 and 2 that energy is

Tau for 1 and 2 DOF models 0.5 0.4 0.3 0.2

Tau

0.1 0

–0.1 –0.2 –0.3

2 DOF 1 DOF

–0.4 –0.5

0

5

10

15

20

25

30

35

40

45

Time Fig. 1. The numerical solution of s for the 1 DOF and 2 DOF models.

50


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67

c for 1 and 2 DOF models 0.2 0.15 0.1 0.05

c(t)

0

–0.05 –0.1 –0.15 –0.2 –0.25

2 DOF 1 DOF 0

5

10

15

20

25

30

35

40

45

50

Time Fig. 2. The numerical solution of c for the 1 DOF and 2 DOF models.

transferred back and forth from the s oscillator to the c oscillator. Indeed, each achieves its maximum amplitude variation when the other achieves its minimum amplitude variation. Now that we have a better understanding of how the exact solution of the 2 DOF model behaves, let us consider how the approximate solution fares in capturing the periodic variation of amplitude. The exact solution is compared to the first and second order canonical perturbation solutions in Figs. 3 and 4. Clearly, the first order canonical perturbation solution does not have the correct amplitude variation. However, it can be seen from the figures that the second order canonical perturbation solution is a much better approximation to the actual solution since it begins to capture the variation of amplitude of the solution. As a second example, consider the case k1 = 4, k3 = 6, xc = 2, xs = 1, c0 = 0.1, s0 = 0.1. This is also a case where xc = 2xs, thus internal resonance is expected.

Tau for Exact, First and Second Order Canonical Perturbation solutions 0.5 Exact First Order Canonical Second Order Canonical

0.4 0.3 0.2

Tau

0.1 0 –0.1 –0.2 –0.3 –0.4 –0.5

0

5

10

15

20

25

30

35

40

45

50

Time Fig. 3. Comparison of s with numerical solution, first and second order canonical perturbation solution.


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c for Exact, First and Second Order Canonical Perturbation solutions 0.25 0.2 0.15 0.1

c

0.05 0 –0.05 –0.1 –0.15 –0.2 –0.25

0

5

10

15

20

25

30

35

40

45

50

Time Fig. 4. Comparison of c with numerical solution, first and second order canonical perturbation solution.

The graphs of s and c for the exact solutions of the 1 DOF and 2 DOF model are shown in Figs. 5 and 6 respectively. Also, the graphs of s and c for the exact solution of the 2 DOF model as well as the first and second order canonical perturbation solutions are shown in Figs. 7 and 8, respectively. Many of the comments made regarding the previous case can also be made here. Note how the 1 DOF model does not predict the variation of the amplitude. In addition, the first order canonical perturbation solution does not correctly reflect this same change in the amplitude of the solution. Note that once again the second order canonical perturbation solution fares better at representing the solution of the 2 DOF model than the first order canonical perturbation solution.

Tau for 1 and 2 DOF models 0.25 2 DOF 1 DOF

0.2 0.15 0.1

Tau

0.05 0

–0.05 –0.1 –0.15 –0.2 –0.25

0

5

10

15

20

25

30

35

40

45

50

Time Fig. 5. The numerical solution of s for the 1 DOF and 2 DOF models.


N. Baddour, J.W. Zu / Applied Mathematical Modelling 31 (2007) 54–77

69

c for 1 and 2 DOF models 0.15 2DOF 1DOF 0.1

c(t)

0.05

0

–0.05

–0.1

–0.15

0

5

10

15

20

25

30

35

40

45

50

Time Fig. 6. The numerical solution of c for the 1 DOF and 2 DOF models.

Tau for Exact, First and Second Order Canonical Perturbation solutions 0.25 0.2 0.15 0.1

Tau

0.05 0

–0.05 –0.1 –0.15 Exact First Order Canonical Second Order Canonical

–0.2 –0.25

0

5

10

15

20

25

30

35

40

45

50

Time Fig. 7. Comparison of s with numerical solution, first and second order canonical perturbation solution.

6.2. Non-resonance cases As another example, consider the case k1 = 2, k3 = 2, xc = 2, xs = 4, c0 = 0.1, s0 = 0.1. This is not a case where xc = 2xs, thus no resonance is expected. The graphs of s and c for the exact solutions of the 1 DOF and 2 DOF model are shown in Figs. 9 and 10, respectively. Also, the graphs of s and c for the exact solution of the 2 DOF model, the first and second order canonical perturbation solutions are shown in Figs. 11 and 12, respectively. As expected, the solutions for c for the exact 1 DOF and 2 DOF models differ significantly, but this is expected as a result of the assumption required to arrive at the 1 DOF model. Other than that discrepancy, all other solutions match perfectly.


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c for Exact, First and Second Order Canonical Perturbation solutions 0.15

0.1

c

0.05

0

–0.05

–0.1

–0.15

0

5

10

15

20

25

30

35

40

45

50

Time Fig. 8. Comparison of c with numerical solution, first and second order canonical perturbation solution.

Tau for 1 and 2 DOF models 0.15 2 DOF 1 DOF 0.1

Tau

0.05

0

–0.05

–0.1

–0.15

0

2

4

6

8

10

12

14

16

18

20

Time Fig. 9. The numerical solution of s for the 1 DOF and 2 DOF models.

Consider a similar example with k1 = 2, k3 = 2, xc = 2, xs = 4, c0 = 0.4, s0 = 0.4. This is the same as the previous case, with a slight change in the initial conditions. The graphs of s and c for the exact solutions of the 1 DOF and 2 DOF model are shown in Figs. 13 and 14, respectively. The graphs of s and c for the exact solution of the 2 DOF model, the first order canonical perturbation solution and the second order perturbation solution are shown in Figs. 15 and 16, respectively. Note that the second order canonical perturbation solutions and the first order canonical perturbation solution produce results that are in good agreement with the exact solution. Once again, there is good agreement between the approximate solutions and the exact solution. Note also how the frequency of the solutions depends on the initial conditions. The only change in this case from the previous case is in the initial conditions.


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c for 1 and 2 DOF models 0.15 2 DOF 1 DOF 0.1

c(t)

0.05

0

–0.05

–0.1

–0.15

0

2

4

6

8

10

12

14

16

18

20

Time Fig. 10. The numerical solution of c for the 1 DOF and 2 DOF models.

Tau for Exact, First and Second Order Canonical Perturbation solutions 0.15 Exact First Order Canonical Second Order Canonical 0.1

Tau

0.05

0

–0.05

–0.1

–0.15

0

2

4

6

8

10

12

14

16

18

20

Time Fig. 11. Comparison of s with numerical solution, first and second order canonical perturbation solution.

Consider a similar example with k1 = 2, k3 = 4, xc = 1, xs = 3, c0 = 1, s0 = 1. There is no internal resonance predicted in this case. However, a quick calculation will reveal that potential energy function has a maximum and that the initial energy in the system is greater than this maximum. Thus, we should expect instability in the system. The graphs of s and c for the exact solutions of the 1 DOF and 2 DOF model are shown in Figs. 17 and 18, respectively. Also, the graphs of s and c for the exact solution of the 2 DOF model, the first and second order canonical perturbation solutions are shown in Figs. 19 and 20, respectively. A quick glance at the figures reveals that the 2 DOF model does indeed display the expected instability. Indeed, the solutions for s and c both blow up in finite time for the 2 DOF model. However, it should be noted


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c for Exact, First and Second Order Canonical Perturbation solutions 0.15 Exact First Order Canonical Second Order Canonical 0.1

c

0.05

0

–0.05

–0.1

–0.15

0

2

4

6

8

10

12

14

16

18

20

Time Fig. 12. Comparison of c with numerical solution, first and second order canonical perturbation solution.

Tau for 1 and 2 DOF models 0.5 0.4 0.3 0.2

Tau

0.1 0 –0.1 –0.2 –0.3 –0.4 –0.5

2 DOF 1 DOF 0

2

4

6

8

10

12

14

16

18

20

Time Fig. 13. The numerical solution of s for the 1 DOF and 2 DOF models.

that neither the 1 DOF model, nor any of the approximate solutions display the same behaviour. The 1 DOF model solutions and the approximate solutions remain bounded. On the other hand, if the same case is considered with only a change in initial conditions to c0 = 0.1, s0 = 0.1, the only thing this changes is the initial energy in the system so that the initial energy is now less then the maximum value of the potential energy. With the initial energy being less than the maximum value of the potential energy, there is no instability in the system. Only periodic solutions are observed with the 1 DOF and 2 DOF models producing the same solution for s. The graphs of the solutions resemble those of the previous case and are thus not included. As for the previous case, the approximate solutions and the exact solutions agree quite well.


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c for 1 and 2 DOF models 0.4 0.3 0.2 0.1

c(t)

0 –0.1 –0.2 –0.3 –0.4 –0.5

0

2

4

6

8

10

12

14

16

18

20

Time Fig. 14. The numerical solution of c for the 1 DOF and 2 DOF models.

Tau for Exact, First and Second Order Canonical Perturbation solutions 0.5 0.4 0.3 0.2

Tau

0.1 0 0.1 0.2 0.3 0.4 0.5

0

2

4

6

8

10

12

14

16

18

20

Time Fig. 15. Comparison of s with numerical solution, first and second order canonical perturbation solution.

7. Summary of comparison between 1 DOF and 2 DOF models By employing analytical and numerical tools, a better picture of the similarities and differences between the 1 DOF and 2 DOF models begins to emerge. First, the 1 DOF model predicts a periodic solution for the time dependence of the transverse vibrations, s. Although the frequency of vibration depends on various parameters of the solution, the amplitude of the solution is constant. For this model, the time dependence of the inplane vibrations, c is linearly related to s since the in-plane inertia is dropped. Although this will not accurately represent the dynamics of the in-plane vibrations, it simplifies the overall problem enough to permit a closedform solution to the transverse problem. If k1 = 0, for the 1 DOF model we get the seemingly nonsensical result that c = 0. This is actually a reasonable statement which follows from the underlying assumption.


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c for Exact, First and Second Order Canonical Perturbation 0.4 0.3 0.2 0.1

c

0 –0.1 –0.2 –0.3 –0.4 –0.5 0

2

4

6

8

10

12

14

16

18

20

Time Fig. 16. Comparison of c with numerical solution, first and second order canonical perturbation solution.

Tau for 1 and 2 DOF models 250 2 DOF 1 DOF 200

Tau

150

100

50

0

–50

0

5

10

15

20

25

30

Time Fig. 17. The numerical solution of s for the 1 DOF and 2 DOF models.

Namely, by ignoring the in-plane inertia, we are essentially stating that the in-plane problem is a static problem. Further assuming that k1 = 0 states that there is no coupling between the in-plane and transverse vibrations. Hence, no coupling to a static problem leads to the prediction that the static problem remains unchanged. In other words, c = 0. It becomes quickly apparent that solutions of the 2 DOF model that includes the effect of in-plane inertia will exhibit a broader range of possible behaviour than for the 1 DOF model. This model is best viewed as a nonlinear duffing oscillator for the transverse vibrations and a linear harmonic oscillator for the in-plane vibrations. These are nonlinearly coupled. For k1 = 0, there is no coupling between the two oscillators. Thus, the two oscillators can oscillate independently. The first difference between the 1 DOF and 2 DOF oscillator


N. Baddour, J.W. Zu / Applied Mathematical Modelling 31 (2007) 54–77

c for 1 and 2 DOF models

4

x 10

1

75

0

–1

c(t)

–2

–3 2 DOF 1 DOF

–4

–5

–6

–7

0

5

10

15

20

25

30

Time Fig. 18. The numerical solution of c for the 1 DOF and 2 DOF models.

Tau for Exact, First and Second Order Canonical Perturbation solutions 250 Exact First Order Canonical Second Order Canonical

200

Tau

150

100

50

0

–50

0

5

10

15

20

25

30

Time Fig. 19. Comparison of s with numerical solution, first and second order canonical perturbation solution.

becomes apparent: depending on the modes in question, it is possible that these modes are coupled. It is also possible that the vibrations of the chosen modes are not coupled. For the case where the modes are not coupled, the results for the vibrations of the transverse oscillator are very similar to the results of the 1 DOF transverse oscillator. In both cases, they are independent duffing oscillators. For the case of coupled oscillators, the possibility of internal resonance between the oscillators arises. This represents a marked difference from the 1 DOF model where there is no possibility of internal resonance since there is only one degree of freedom. In the case of internal resonance, there is a sort of Ôenergy sharingÕ between the two oscillators. The amplitudes of vibrations of both are not constant but rather vary periodically. This is reminiscent of beats between two linear harmonic oscillators. For the linear harmonic oscillator, the


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c for Exact, First and Second Order Canonical Perturbation 4

1

x 10

0

–1

c

–2

–3

–4

–5

Exact First Order Canonical Second Order Canonical

–6

–7

0

5

10

15

20

25

30

Time Fig. 20. Comparison of c with numerical solution, first and second order canonical perturbation solution.

beat frequency is easily found, whereas for the nonlinearly coupled oscillators it is not. For the non-resonant case numerical simulations showed that the predictions of the 1 DOF and 2 DOF models may vary in their predictions of the amplitude of the solution. For the 2 DOF model, energy analysis and numerical simulations predict the possibility of instability in the system. It can also be seen that sometimes the 1 DOF model predicts a periodic solution while the 2 DOF model predicts instability, a marked difference. 8. Conclusion In conclusion, by including the effect of in-plane inertia the dynamics of the system includes phenomena not previously predicted. New phenomena such as internal resonance or non-uniformity of the amplitude of the solution arise. Depending on initial conditions, system instability is a possibility of the 2 DOF model that is not predicted by the 1 DOF model. Such phenomena are not predicted by the simpler model where in-plane inertia is neglected. Further, the efficacity of a canonical perturbation approach in deriving approximate solutions to the nonlinear problems was investigated. It was generally observed that a second order canonical perturbation solution can effectively capture the nonlinear system dynamics in the cases where the system is expected to remain stable. References [1] H. Lamb, R.V. Southwell, The vibrations of a spinning disk, Proc. Roy. Soc. London, Ser. A 99 (1921) 272–280. [2] R.V. Southwell, Free transverse vibrations of a uniform circular disk clamped at its centre, Proc. Roy. Soc. London, Ser. A 101 (1922) 133–153. [3] J.G. Simmonds, The transverse vibrations of a flat spinning membrane, J. Aeronaut. Sci. 29 (1962) 16–18. [4] W. Eversman, Transverse vibrations of a clamped spinning membrane, AIAA J. 6 (1968) 1395–1397. [5] J.G. Simmonds, Axisymmetric, transverse vibrations of a spinning membrane clamped at its centre, AIAA J. 1 (5) (1963) 1224–1225. [6] M.W. Johnson, On the dynamics of shallow elastic membranes, in: Proceedings of the Symposium of Thin Elastic Shells, pp. 281–300, 1960. [7] S. Barasch, Y. Chen, On the vibration of a rotating disk, J. Appl. Mech. 39 (1972) 1143–1144. [8] W. Eversman, R.O. Dodson, Free vibration of a centrally clamped spinning disk, AIAA J. 7 (10) (1969) 2010–2012. [9] K.A. Cole, R.C. Benson, A fast eigenfunction approach for computing spinning disk deflections, J. Appl. Mech. 55 (1988) 453–457.


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Nonlinearly Coupled Transverse and In-plane Vibrations of a Spinning Disk  

Mathematical analysis of vibrations.