Time-Averaged Canonical Perturbation of Nonlinear Vibrations of a Spinning Disk by Natalie Baddour

Nonlinear Analysis: Real World Applications 7 (2006) 319 – 340 www.elsevier.com/locate/na

Time-averaged canonical perturbation of nonlinear 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, Ontario, Canada, M5S 3G8 Received 17 January 2003; accepted 24 March 2005

Abstract This paper considers the dynamics of a spinning disk with the inclusion of in-plane inertia. The inclusion of the in-plane inertia couples the in-plane and transverse vibrations of the spinning disk. By considering one mode of each vibration, the problem can be reduced to a non-linear 2 DOF model. An analytical solution of this 2 DOF model is then developed using a time-averaged canonical perturbation approach and is compared with that of a numerical procedure. Through the use of these analytical and numerical tools, it becomes apparent that the inclusion of the in-plane inertia leads to the possibility of internal resonance between the oscillators, depending on the relationship between the natural frequencies of the two oscillators. 䉷 2005 Elsevier Ltd. All rights reserved. Keywords: 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. ∗ Corresponding author. Tel./fax: +1 416 978 1287.

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Many authors have investigated the vibrations of spinning disks using linear theory. The original papers are by Lamb and Southwell [19] and by Southwell [26] 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 [24,13,25,16]. 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 applied different solution techniques to the linear analysis of the free transverse vibrations of spinning disks [7,14,11,20]. In addition to considering the transverse vibrations of a spinning disk, the problem of free planar vibrations has also been investigated using linear theory [27,18,8,12,10]. Although Linear models have been extensively used in the literature for many years, 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 [21,23,1,2]. However, in the nonlinear case the in-plane inertia of the disk has always been neglected [21,22,15] and the authors could not find any papers that considered its inclusion in the nonlinear dynamics of the rotating disk. With the inclusion of the in-plane inertia, the simplest possible representation of the nonlinear dynamics result in a nonlinear 2 DOF system. In this paper, the process by which this simplified 2 DOF system is obtained is outlined. The main thrust of the paper is to develop an analytical solution to this simplified 2 DOF system and to verify whether this analytical solution succeeds in capturing the various dynamics of the system. 2. Equations of motion Since all previous models [21,22,15] 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 [5]. To consider the effect of the inclusion of in-plane inertia, Galerkin’s procedure is used in this paper to produce a simplified problem. The result will be a two degree of freedom model—one for each of the time dependences of the in-plane and transverse vibrations. When the in-plane inertia is included, the time dependence of the in-plane vibrations is an extra independent quantity that must be taken into account. To this purpose, let us choose the displacements as u(r, , t) = ueq (r) + U (r) cos(m )c(t),

(1)

v(r, , t) = −V (r) sin(m )c(t),

(2)

w(r, , t) = W (r) cos(n ) (t),

(3)

where ueq , U, V and W are assumed to be known and will be discussed momentarily. The only unknowns here are the time dependences, c(t) and (t). The ueq portion is the in-plane equilibrium displacement due to the spin of the disk. That is, ueq is a symmetrical equilibrium in-plane displacement that occurs in the plane of the disk as a result of its rotation

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(that is, due to the centrifugal force). The form for U and V are chosen from the solution of the in-plane linear problem [4,9,8]. Similarly, W is taken to be one of the mode shapes of the linear strain model of the spinning disk, derived by using purely linear strains [6]. 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. Since the spatial form of the solution is known, the integration can be performed explicitly over space. The kinetic and potential energies are thus reduced to functions of c(t) and (t) given in Eqs. (1)–(3), that is we have a 2 DOF system. Lagrange’s equations are then applied to yield the two equations of motion that are sought. The advantage of actually calculating the potential and kinetic energies of the system, as opposed to directly using Galerkin’s method 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 Hamilton’s equations and canonical perturbation theory is also available in this formulation. For these reasons, the simplified 2 DOF ordinary differential equations of motion are calculated by first constructing the kinetic and potential energies, integrating over space and then using Lagrange’s equations. The potential and kinetic energies can be found from 1 h 2 a PE = ( rr rr + + 2 r r )r dr d dz, (4) 2 −h 0 0 2 a (hKE1 + h3 KE3 )r dr d , (5) KE = 0

where the integration over z has already been incorporated into the expression for the kinetic energy, and KE1 and KE3 are given by [5] ju jv jv 2 2 2 2 KE1 = (v + u + r + 2ru) + 2 u −v +r jt jt jt 2 2 2 ju jv jw , (6) + + + jt jt jt 2 2 2 2 jw 2 j2 w 2 2 jw 2 j w KE3 = 2 + 2 +r +r 3r j jr 3r j jt jrjt 2 2 j w jw j2 w jw + − . (7) jrjt j 3r j jt jr Linear stress–strain relations along with nonlinear von Karman strains are used in the expression for potential energy to complete the derivation. Once the integrations over z, and r are performed, the resulting expressions for the kinetic and potential energies become simple functions of c and and resemble the expressions of a two degree of freedom system. Indeed, by taking only one modeshape for each of the in-plane and out of plane vibrations, we have restricted the system to having only two degrees of freedom.

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After having obtained the expressions for kinetic and potential energy, it is a simple matter to form the Lagrangian and to derive the equations of motion of the system using Lagrange’s equations. The coupled equations of motion can be shown to be c¨ + 2c c + k1 2 = 0,

(8)

¨ + 2 + 2k1 c + k3 3 = 0.

(9)

Clearly, c and are the natural frequencies of the linear in-plane and transverse modes that were chosen. The expressions for k1 and k3 are complicated [3], but it sufﬁces to note that they will depend on the amplitudes of the chosen modes. For certain choices of modes, it is possible that k1 = 0, which indicates that the in-plane and transverse modes in question are uncoupled and will vibrate independently. However, for other choices of modes, k1 will not be zero and the in-plane and transverse vibrations will be coupled. Eqs. (8) and (9) will be referred to as the two degrees of freedom (2 DOF) model since the equations of motion are essentially those of a nonlinear two degree of freedom system. Thus, current work considers Eqs. (8) and (9) to be the fundamental equations describing the dynamics of the system. 3. Time-averaged canonical perturbation approach Eqs. (8) and (9) can also be arrived at from the point of view of Hamilton’s equations. That is, Eqs. (8) and (9) are equivalent to jH , jpc jH ˙ = , jp jH p˙ c = − , jc jH p˙ = − , j c˙ =

(10)

where pc and p are the conjugate momenta to c and , and the Hamiltonian is given by H=

pc2 p2 2 2 k3 4 + + c c2 + 2 + k1 c 2 + . 2 2 2 2 4

(11)

Eqs. (10) 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. Generally, a transformation of variables from generalized coordinates q, p to a new set of coordinates q , p can be of the form: q = q (q, p, t), p = p (q, p, t),

(12)

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323

with the corresponding transformed differential equations q˙ = q˙ (q, p, t), p˙ = p˙ (q, p, t).

(13)

Suppose that Eqs. (13) have the same symmetry as the original equations. That is, suppose that there exists a function H (q , p , t) such that jH jp jH p˙ = − . jq q˙ =

(14)

Then the transformation (12) is known as a canonical transformation [17]. That is, canonical transformations transform 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 H is a constant function, that is a function that does not depend on time or any of the generalized coordinates or momenta. Then Eqs. (14) reduce to the simple form of q˙ = p˙ = 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 simple Hamiltonian to aid in the construction of an approximate solution for a problem with a more complicated Hamiltonian. That is, take a transformation that essentially solves a simple related problem and apply it to the more complicated problem. 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 2DOF model: H=

pc2 2 k3 4 p2 2 + + c c2 + 2 + k1 c 2 + . 2 2 2 2 4

(15)

This Hamiltonian can be expressed as H = H (0) + H (1) ,

(16)

where H (0) =

p2 2 pc2 2 + + c c2 + 2 , 2 2 2 2

H (1) = k1 c 2 +

k3 4 . 4

(17) (18)

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 oscillator into a nonlinear Dufﬁng oscillator. The k1 term is the only term that couples the two oscillators.

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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) [17]. Now this canonical transformation is equally valid no matter which Hamiltonian is employed. It just has the 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. As previously mentioned, the first step in the procedure is to seek the complete solution S(c, , pc , p , t) of the Hamilton–Jacobi equation associated with H (0) (c, , pc , p ). Namely, solve jS jS jS H (0) c, , , ,t + = 0. (19) jc j jt That is, solve 1 jS 2 1 jS 2 2c 2 2 2 jS c + + = 0. + + 2 2 jt 2 j 2 jc

(20)

To achieve this, write S = −ht + W1 (c, 1 ) + W2 ( , 2 )

(21)

where h = 1 + 2 is a constant. If Eq. (21) is substituted into Eq. (20), then Eq. (20) can be satisﬁed if 1 jW1 2 2c 2 c = 1 , + (22) 2 jc 2 1 jW2 2 2 2 = 2 . + 2 j 2 The preceding equations can be integrated directly to yield 2 1 − 2c c2 dc + 2 2 − 2 2 d . S = − 1 t − 2 t + The required transformation can now be found from jS dc = −t +

1 = j 1 2 − 2c c2 1 1 c c = −t + , arcsin √ c 2 1 or equivalently √ 2 1 c= sin c (t − 1 ). c

(23)

(24)

(25) (26)

(27)

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Similarly, we obtain √ 2 2 = sin (t − 2 ).

325

(28)

Furthermore, the generalized momenta can also be transformed jS = 2 1 − 2c c2 = 2 1 cos c (t − 1 ), pc = jc jS = 2 2 − 2 2 = 2 2 cos (t − 2 ). p = j

(29) (30)

Note that Eqs. (27)–(30) are essentially a canonical transformation from the old variables c, , pc , p to the new variables 1 , 2 , 1 , 2 . It can be veriﬁed that if (27)–(30) are substituted into H (0) , the result is the constant 1 + 2 . This is no coincidence: the transformation was speciﬁcally constructed to do so. In fact, Eqs. (27)–(30) are precisely the solution to the uncoupled two oscillator problem where 1 , 2 , 1 , 2 are to be taken as constants of integration. The transformed Hamiltonian for the uncoupled two oscillator problem is given by H (0) + jS/jt, which is precisely zero (by construction). With the transformed Hamiltonian being zero, Hamilton’s equations assume a particularly simple, easily-integrable form. Namely, they give the result that 1 , 2 , 1 , 2 must be constants to satisfy the new Hamilton’s equations. While the canonical transformation given by Eqs. (27)–(30) transforms H (0) into zero, it does not do so for the entire Hamiltonian H = H (0) + H (1) . The transformed total Hamiltonian becomes H = H (0) + jS/jt + H (1) = H (1) ( 1 , 2 , 1 , 2 , t). The ‘perturbing Hamiltonian’ H (1) can be expressed as H (1) ( 1 , 2 , 1 , 2 , t) = k1 c( 1 , 1 , t) 2 ( 2 , 2 , t) +

k3 4 ( 2 , 2 , t) 4

√ √ k1 2 1 sin( c (t − 1 )) 2 sin( (t − 2 ))2 =2 c 2 +

k3 22 sin( c (t − 2 ))4 , 4

2 (−8k1 2 1 2 sin( c t − c 1 ) 4 8 c + 4k1 2 1 2 sin(( c + 2 )t − c 1 − 2 2 ) + 4k1 2 1 2 sin(( c − 2 )t − c 1 + 2 2 ) − 3k3 2 c

H (1) = −

− k3 2 c cos(4 t − 4 2 ) + 4k3 2 c cos(2 t − 2 2 )). Note from the above deﬁnition of H (1) that there is a difference in its form depending on whether or not c = 2 . Although it may not be obvious yet, if c = 2 , the system will display internal resonance and H (1) will take on a different form.

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From the above, using the new variables 1 , 2 , 1 and 2 , Hamilton’s equations (10) become √ 2 1 jH (1) 1 ˙ 1 = = 2 k1 (−2 cos(− c t + c 1 ) 2 2 j 1 + cos(− c t + c 1 − 2 t + 2 2 ) + cos( c t − c 1 − 2 t + 2 2 )), (31) jH (1) 1 2 (2k 2 1 2 cos(− c t + c 1 − 2 t + 2 2 ) = 1 j 2 2 1 32 − 2k1 2 1 2 cos( c t − c 1 − 2 t + 2 2 ) − k3 2 c sin(−4 t + 4 2 ) + 2k3 2 c sin(−2 t + 2 2 )), √ (1) 2 ˙ 1 = − jH = 1 2 k1 √ (2 sin(− c t + c 1 ) 2 j 1 4 c 1 − sin(− c t + c 1 − 2 t + 2 2 ) + sin( c t − c 1 − 2 t + 2 2 )), ˙ 2 =

(32)

(33)

jH (1) 1

˙ 2 = − (−2k 2 1 2 sin(− c t + c 1 − 2 t + 2 2 ) = 1 4 c 4 j 2 √ √ + 4k1 2 1 2 sin(− c t + c 1 ) + 2k1 2 1 2 sin( c t − c 1 − 2 t + 2 2 ) − 3k3 2 c − k3 2 c cos(−4 t + 4 2 ) + 4k3 2 c cos(−2 t + 2 2 )). (34) If the initial conditions for (8) and (9) are given by (0) = 0 ,

(35)

d (0) = 0, dt

(36)

c(0) = c0 ,

(37)

dc(0) = 0, dt

(38)

the corresponding initial values for 1 , 2 , 1 , 2 can be worked out from the transformation equations (27 )–(30) to be 1 2 2 c , 2 c 0

01 = − , 2 c 01 =

(39) (40)

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1 2 2 , 2 0 .

02 = − 2 02 =

327

(41) (42)

Now, Eqs. (31)–(34) 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. In practice though, these equations are no easier to solve than the original equations of motion. However, Eqs. (31)–(34) offer a good starting point for approximate solutions and make certain features of the solutions easy to spot. To construct an approximate solution, approximations will be made fairly early in the game, namely at H (1) itself. Here the perturbing Hamiltonian can be further separated and replaced with its temporal mean. Hamilton’s equations for ˙ 1 , ˙ 2 , ˙ 1 , ˙ 2 are then found using this time-averaged Hamiltonian instead of the original Hamiltonian. The beneﬁt of this approach is that for our problem the ensuing Hamilton’s equations can be solved exactly. It should be noted that this approach is referred to in the literature as a canonical perturbation approach, although it is not the only approach that is referred to as such. Consider again the expression for H (1) from which Eqs. (31)–(34) are derived. The temporal mean of H (1) can easily be calculated. Note that the temporal mean of H (1) will have two different values depending on whether or not c = 2 . If c = 2 then sin(( c − 2 )t + ( c 1 − 2 2 )) is not a function of time, and will not average out to zero. Thus, for the no-resonance case where c = 2 we obtain (1) Hav =

3k3 22 . 8 4

(43)

Similarly, for the internal resonance case where c = 2 , we obtain √ 3k3 22 k1 2 2 1 (1) Hav,IR = − sin(2 ( 2 − 1 )). 8 4 2 c 2 (1)

(44)

(1)

To construct an approximate solution to our problem, Hav and Hav,IR in Eqs. (43) and (44) are used to derive the Hamilton equations of motion instead of the exact value of H (1) . There are thus two possible sets of equations of motion depending on whether or not the condition for internal resonance is met. However, the advantage of this formulation is that in both cases the Hamilton equations of motion can be solved exactly. 4. Solution of the time-averaged approximation 4.1. Case 1: No internal resonance For the case when there is no internal resonance, that is where c = 2 , then the Hamilton equations of motion are given by (1)

˙ 1 =

jHav = 0, j 1

(45)

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˙ 2 =

jHav = 0, j 2

(46)

(1)

jHav = 0,

˙ 1 = − j 1

(47)

(1)

jHav 3k3 2 =− .

˙ 2 = − j 2 8 4

(48)

Integrating these four equations yields 1 = 01 ,

(49)

2 = 02 ,

(50)

1 = 01 ,

(51)

2 = −

3k3 02 t + 02 . 4 4

(52)

Thus the solution to the original problem becomes 2 01 c= sin( c (t − 01 )) c

2 02 3k3 02 0 = 1+ t − 2 . sin 4 4

(53)

(54)

Note that for this case, since 01 , 02 , 01 , 02 are all constants, the solution for c and have constant amplitudes. Furthermore, note that the frequency of depends on k3 so that we have the familiar result of frequency depending on amplitude. Substituting from Eqs. (39)–(42) for 01 , 02 , 01 , 02 yields c = c0 cos( c t)

= 0 cos t 1 +

3k3 20 8 2

(55)

(56)

Note that the solution for c in this case is exactly what it would be if the oscillators were uncoupled—it is not affected by the coupling. 4.2. Case 2: Internal resonance For the case when there is internal resonance, that is where c = 2 , then the Hamilton equations of motion are given by √ (1) jHav,IR k1 2 2 1 ˙ 1 = = cos(2 ( 1 − 2 )), (57) j 1 2 2

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˙ 2 =

jHav,IR j 2

˙ 1 = −

˙ 2 = −

329

√ k1 2 2 1 cos(2 ( 1 − 2 )), =− 2 2

(58)

√ k1 2 2 sin(2 ( 1 − 2 )), =− √ 8 1 3

(59)

(1)

jHav,IR j 1 (1)

jHav,IR j 2

=−

√ 3k3 2 k1 2 1 − sin(2 ( 1 − 2 )). 8 4 4 3

(60)

(1)

Since Hav,IR does not contain the time t explicitly, it must be a conserved quantity, that is, (1)

a constant. Let us denote this constant as g1 , so that Hav,IR = g1 . Adding Eqs. (57) and (58) yields ˙ 1 + ˙ 2 = 0, or 1 + 2 = g2, where g2 is a constant of integration. Both constants g1 and g2 can be found from the initial values of 1 , 2 , 1 and 2 . This yields g1 =

3k3 k1 + , 32 4

(61)

g2 =

5 2 . 2

(62)

Solving Eq. (44) for the sine term, using the fact that 1 =g2 − 2 , and using the trigonometric identity sin2 (x) = 1 − cos2 (x), it can be shown that 2 must satisfy 9k32 4 k12 3 (320k12 + 192k3 k1 + 72k32 ) 2 − + 2 2 16 6 2 2 2 256 2 (9k32 + 48k3 k1 + 64k12 ) 2 − . 256

˙ 22 = −

(63)

Note that Eq. (63) is an equation purely in terms of 2 . That is, this ODE can be integrated for 2 . Integrating this ODE will be discussed subsequently. Once 2 is known, the other variables can be found by integration from 1 = g2 − 2 ,

(64)

−12k3 22 + 3k3 4 + 8k1 4

˙ 1 = , 32(2 2 − 5 2 ) 4

(65)

12k3 22 + 3k3 4 + 8k1 4

˙ 2 = − . 32 4

(66)

Thus, the crux of the problem is to solve Eq. (63) for 2 . Recall that 2 is the amplitude . Before even solving for 2 , it is worthwhile to remark that in the non-resonance case 2 was a constant and in this (resonance) case, it will clearly not be a constant. This is the most important observation that can be made here since even though we can solve for 2 exactly, the calculations are not very intuitive.

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It can be veriﬁed that (2a2 − 2 ) is a factor of the quartic in Eq. (63). Thus, to solve Eq. (63), consider a change of variables by letting 1 2 1 2 = + , (67) 2 y or equivalently 2 2 . 2 2 − 2

(68)

Then Eq. (63) can be written as −

dy = ± (y − y1 )(y − y2 )(y − y3 ), 14k12 + 12k1 k3 dt 4

(69)

where y1, y2 and y3 satisfy y3 > y2 > y1 and are the roots of the cubic y3 +

(8k12 + 12k1 k3 − 9k32 ) 14k12 + 12k1 k3

y2 −

(8k12 + 18k32 ) 14k12 + 12k1 k3

y−

9k32 14k12 + 12k1 k3

(70)

Note that this assumes that the roots to the above cubic are real. Should two of the roots be complex, a different procedure must be considered and will be discussed after this one. The solution of Eq. (69) is given by y = (y2 − y1 )sn2 (M(t − t0 ), k) + y1 , where

(y3 − y1 )(14k12 + 12k1 k3 ) 8

(y2 − y1 ) , (y3 − y1 )

(71)

(72)

(73)

and sn denotes the Jacobian elliptic function with parameter k. Note that sn is a generalization of the trigonometric sine function. In fact, sn reduces to the sine function for k = 0. The jacobian elliptic function can be deﬁned from x dx u= (74) = sn−1 (x), 2 2 2 0 (1 − x )(1 − k x ) where sn−1 (x) denotes the inverse Jacobian elliptic function. That is, x = sn(u, k). By differentiating, it can be shown that du 1 = . 2 dx (1 − x )(1 − k 2 x 2 )

(75)

Note the similarity between Eqs. (75) and (69). √ In fact, Eq. (75) and Eq. (69) are related by a simple change of variable of the form x = A + By.

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331

It still remains to ďŹ nd t0 . This can be found from the initial conditions. Recall that 2 (t = 0) = 2 /2 if 0 = 1. This implies that y â&#x2020;&#x2019; â&#x2C6;&#x17E; at t = 0. Thus sn(â&#x2C6;&#x2019;Mt 0 , k) â&#x2020;&#x2019; â&#x2C6;&#x17E;, implies that â&#x2C6;&#x2019;Mt 0 = iK where

K = 0

(1 â&#x2C6;&#x2019; x 2 )(1 â&#x2C6;&#x2019; (k )2 x 2 ) d , = F k , 2 1 â&#x2C6;&#x2019; (k )2 sin2

(76) (77)

and (k )2 + k 2 = 1. Note that F denotes the complete elliptic integral of the ďŹ rst kind with parameter k . It follows that sn(M(t â&#x2C6;&#x2019; t0 ), k) = sn(Mt + iK , k) 1 1 = k sn(Mt, k) 1 = ns(Mt, k). k Thus ďŹ nally we have that 1 1 + , 2 = 2 2 y

(78) (79) (80)

(81)

y = (y3 â&#x2C6;&#x2019; y1 )ns 2 (Mt, k) + y1 .

(82)

For the more general initial condition where 0 = 1, it follows that 2 (t = 0) = 2 20 /2. Thus y(t = 0) = 2/ 20 â&#x2C6;&#x2019; 1. This yields

sn(â&#x2C6;&#x2019;Mt 0 , k) =

2 â&#x2C6;&#x2019; y1 ( 20 â&#x2C6;&#x2019; 1) (y2 â&#x2C6;&#x2019; y1 )( 20 â&#x2C6;&#x2019; 1)

(83)

Hence the solution for y becomes y = (y2 â&#x2C6;&#x2019; y1 )sn2 (M(t â&#x2C6;&#x2019; t0 ), k) + y1

2 â&#x2C6;&#x2019; y1 ( 20 â&#x2C6;&#x2019; 1) 1 F k, arcsin t0 = . M (y2 â&#x2C6;&#x2019; y1 )( 20 â&#x2C6;&#x2019; 1)

(84) (85)

The above solution is only valid if y1, y2 , and y3 are real. For imaginary roots the solution proceeds as follows. The equation for 2 is written as â&#x2C6;&#x2019;16 d 2 2 2 ( 2 â&#x2C6;&#x2019; x1 )[( 2 â&#x2C6;&#x2019; x4 )2 + x52 ], = â&#x2C6;&#x2019; (86) 2 dt 2 9k32

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where x1, x4 and x5 are real and ( 2 − x1 )[( 2 − x4 )2 + x52 ] = 32 + − Let Z1 = ( 2 −

2 2 )( 2

(16k12 + 9k32 ) 18k32

2 22

(3k3 + 8k1 )2 4 (3k3 + 8k1 )2 6 − . 2 36k32 72k32

(87)

− x1 ) and Z2 = ( 2 − x4 )2 + x52 . Now consider Z1 − Z2 :

2 2 x1 Z1 − Z2 = (1 − ) 22 + 2 x4 − − x1 2 + − (x42 + x52 ). 2 2

(88)

Clearly, Z1 − Z2 is a quadratic in 2 . This quadratic is a perfect square if we choose

such that the discriminant of the quadratic is zero. Thus, let 1 and 2 denote the roots of the discriminant: 2 2 2 2 2 2 2 2 −4x5 + 4x4 − (89) + x1 x4 + 4x5 + 2 x1 + − x1 = 0. 2 2 Eq. (89) is itself a quadratic in and as such has two possible solutions. Further, it is not a quadratic if x5 = 0, as this case would reduce to the previously discussed case of real yi. . Solving for i yields 1

1,2 = 2 − 2 x4 − 2x1 x4 + 2x42 + 2x52 + 2 x1 4x5 (90) ± (x52 + (x1 − x4 )2 )(4x52 + (2x4 − 2 )2 ) . The expression under the square root sign is always positive so 1 and 2 are both real quantities. Take 2 to be the quantity associated with the positive sign in Eq. (90) while 1 is associated with the negative sign. Since the coefﬁcient of 2 in Eq. (89) is negative while the constant term is positive, it is easily seen that 1 < 0. Solving the expression for 2 in Eq. (90) for 2 , it can easily be shown that for 2 to be a real quantity that 2 1. It can be shown that 2 = 1 results in Z1 and Z2 being proportional, clearly an impossibility since one has real roots while the other has imaginary roots. Thus, it follows that 2 > 1. It follows that Z1 − 1 Z2 =

P12 , 4(1 − 1 )

(91)

Z1 − 2 Z2 =

P22 , 4(1 − 1 )

(92)

where P1 = 2(1 − 1 ) 2 −

2 − x1 + 2 1 x4 , 2

(93)

N. Baddour, J.W. Zu / Nonlinear Analysis: Real World Applications 7 (2006) 319 – 340

2 − x1 + 2 2 x4 . 2 Solving for Z1 and Z2 gives

1 P22

2 P12 1 Z1 = − , 4( 2 − 1 ) (1 − 1 ) (1 − 2 ) P12 P22 1 Z2 = − . 4( 2 − 1 ) (1 − 1 ) (1 − 2 ) P2 = 2(1 − 2 ) 2 −

333

(94)

(95)

(96)

Now consider the change of variables y = P1 /P2 . Then after some algebra, Eq. (86) can be written as 2 dy (97) = N 2 2 (y 2 + g 2 )(h2 − y 2 ), dt where N2 =

9k32 ( 2 + 2x1 − 4x4 )2 256(1 − 1 )2

g2 =

1 − 1 ,

2 − 1

h2 =

1 (1 − 1 ) .

2 (1 − 2 )

(98) (99) (100)

In its present form, Eq. (97) can be solved to give ⎤ ⎡

1 (1 − 1 ) ⎣ ( 2 − 1 )(1 − 1 )

1 ⎦ y= . cn N (t − t0 ),

2 (1 − 2 ) ( 2 − 1)

1 − 2

(101)

Since 2 − x1 + 2 1 x4 P1 2 y= = , P2 2 2(1 − 2 ) 2 − − x1 + 2 2 x4 2 this can be solved for 2 to give 2 + x1 (1 − y) 2x4 ( 2 y − 1 ) + 2 2 = . 2( 2 y − 1 ) + 2(1 − y) 2(1 − 1 ) 2 −

(102)

(103)

It remains to ﬁnd t0 . Using the initial condition 2 (t = 0) = 2 0 /2, deﬁne y0 as the value of y at t = 0 such that y0 = y(0) =

2 20 (2 1 − 1) + 2x1 − 4 1 x4 2 20 (2 2 − 1) + 2x1 − 4 2 x4

(104)

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It then follows that â&#x17D;Ą â&#x17D;¤

y ( 2 â&#x2C6;&#x2019; 1)

1 0 1 â&#x17D;Ś. F â&#x17D;Łarccos , t0 = â&#x2C6;&#x2019; N ( 2 â&#x2C6;&#x2019; 1 )(1 â&#x2C6;&#x2019; 1 ) h

1 â&#x2C6;&#x2019; 2

(105)

5. Numerical simulations In this section, numerical simulations of the 2 DOF models are considered to show the effectiveness of the approximate analytical solution. All numerical results were obtained by using MATLAB. Numerical solutions to the nonlinearly coupled equations are included for comparison and were obtained using built-in fourth and ďŹ fth order Rungeâ&#x20AC;&#x201C;Kutta procedures in MATLAB. 5.1. Resonance case For the ďŹ rst case, the following values for the constants are taken: k1 = 2, k3 = 2, c = 4, = 2, c0 = 0.2, 0 = 0.2. This is clearly a case where c = 2 , thus internal resonance is expected. Let us now consider how the approximate solution fares in capturing the expected periodic variation of amplitude. In Figs. 1 and 2, the numerical solution for and c are compared to the approximate solution generated by the time-averaged canonical perturbation theory. It can be seen that the solution produced by the time-averaged perturbation solution does track the actual amplitude variation of the solution. In fact, in this case the time-averaged perturbation solution closely matches the numerical solution. As a second example, consider the case k1 = 4, k3 = 6, c = 2, = 1, c0 = 0.1, 0 = 0.1. This is also a case where c = 2 , thus internal resonance is expected. The graphs of and c for the numerical solution of the 2 DOF model and the time-averaged Hamiltonian perturbation solution are shown in Figs. 3 and 4, respectively. Again, for this case, the time-averaged Hamiltonian canonical perturbation solution is a good approximate solution to the 2 DOF model. 5.2. Non-resonance case As another example, consider the case k1 = 2, k3 = 2, c = 2, = 4, c0 = 0.1, 0 = 0.1. This is not a case where c = 2 , thus no resonance is expected. The graphs of and c for the numerical solution of the 2 DOF model and the time-averaged Hamiltonian perturbation solution are shown in Figs. 5 and 6, respectively. It can be seen from the graphs that there is good agreement between the numerical and time-averaged canonical perturbation solutions. Consider a similar example with k1 = 2, k3 = 2, c = 2, = 4, c0 = 0.4, 0 = 0.4. This is the same as the previous case, with a slight change in the initial conditions. The graphs of and c for the numerical solution of the 2 DOF model and the time-averaged Hamiltonian perturbation solution are shown in Figs. 7 and 8, respectively. Once again, there is good agreement between the approximate solution and the exact solution. Note also how the frequency of the solutions depends on the initial conditions.

N. Baddour, J.W. Zu / Nonlinear Analysis: Real World Applications 7 (2006) 319 â&#x20AC;&#x201C; 340

335

tau for numerical and Time-Averaged Hamiltonian Solutions 0.5 Numerical Time-Av Hamiltonian

0.4 0.3 0.2

tau

0.1 0 -0.1 -0.2 -0.3 -0.4 -0.5

Time Fig. 1. Comparison of with numerical solution and time-averaged canonical perturbation solution. c for numerical and Time-Averaged Hamiltonian Solutions 0.2 0.15 0.1 0.05

0 -0.05 -0.1 -0.15 -0.2 -0.25

Numerical Time-Av Hamiltonian

Time Fig. 2. Comparison of c with numerical solution and time-averaged canonical perturbation solution.

336

N. Baddour, J.W. Zu / Nonlinear Analysis: Real World Applications 7 (2006) 319 â&#x20AC;&#x201C; 340 tau for numerical and Time-Averaged Hamiltonian Solutions 0.25

Numerical Time-Av Hamiltonian

0.2 0.15 0.1

tau

0.05 0 -0.05 -0.1 - 0.15 -0.2 -0.25 0

Time Fig. 3. Comparison of with numerical solution and time-averaged canonical perturbation solution. c for numerical and Time-Averaged Hamiltonian Solutions 0.15 Numerical Time-Av Hamiltonian

0.1

0.05

-0.05

-0.1

-0.15

Time Fig. 4. Comparison of c with numerical solution and time-averaged canonical perturbation solution.

N. Baddour, J.W. Zu / Nonlinear Analysis: Real World Applications 7 (2006) 319 â&#x20AC;&#x201C; 340

337

tau for numerical and Time-Averaged Hamiltonian Solutions 0.15

0.1

tau

0.05

-0.05

-0.1 Numerical Time-Av Hamiltonian

-0.15

Time Fig. 5. Comparison of with numerical solution and time-averaged canonical perturbation solution. c for numerical and Time-Averaged Hamiltonian Solutions 0.15 Numerical Time-Av Hamiltonian

0.1

0.05

-0.05

-0.1

- 0.15

Time Fig. 6. Comparison of c with numerical solution and time-averaged canonical perturbation solution.

338

N. Baddour, J.W. Zu / Nonlinear Analysis: Real World Applications 7 (2006) 319 â&#x20AC;&#x201C; 340 tau for numerical and Time-Averaged Hamiltonian Solutions 0.5 0.4 0.3 0.2

tau

0.1 0 -0.1 -0.2 -0.3 -0.4 -0.5

Time Fig. 7. Comparison of with numerical solution and time-averaged canonical perturbation solution. c for numerical and Time-Averaged Hamiltonian Solutions 0.4 0.3 0.2 0.1

0 -0.1 -0.2 -0.3 -0.4 -0.5

Time Fig. 8. Comparison of c with numerical solution and time-averaged canonical perturbation solution.

N. Baddour, J.W. Zu / Nonlinear Analysis: Real World Applications 7 (2006) 319 – 340

339

The only change in this case from the previous case is in the initial conditions. However, note that in this case although the time-averaged canonical perturbation solution correctly captures the frequency of the solution, it slightly underpredicts the minimum of the solution for c. 5.3. Conclusions In conclusion, for the case of coupled oscillators that we examined, the possibility of internal resonance between the oscillators arises, depending on the relationship between the natural frequencies of the two oscillators. In the case of internal resonance, there is a sort of ‘energy sharing’ 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 beat frequency is easily found, whereas for the nonlinearly coupled oscillators it is not. For the non-resonant case numerical simulations showed that the amplitude of the system would not show the same periodic variation. We can see from the simulations of the numerical and time-averaged canonical analytical solution that the analytical solution does a good job of matching the numerical solution for both possible cases of internal resonance or no internal resonance.

6. Summary In this paper, an analytical solution to the nonlinear spinning disk problem is given. The method of time-averaged canonical perturbation solution is developed for the 2 DOF nonlinear spinning disk model. Numerical simulations of this solution along with simulating the 2 DOF system directly show that the analytical solution agrees quite well with the predictions of the numerical solution for both resonance and non-resonance cases. Thus, the time-averaged canonical perturbation solution does a good job of capturing the various dynamics of the 2 DOF model.

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