Hamilton’s Principle for the Derivation ofEquations of Motion

Natalie Baddournbaddour@uottawa.ca

May 30, 2007

Abstract

Hamilton’s principle is one of the great achievements of analyticalmechanics. It offers a methodical manner of deriving equations motionfor many systems, with the additional benefit that appropriate andcorrect boundary conditions are automatically produced as part of thederivation. It allows insight into the manner that the system is modeled,as any modelling assumptions are clear and the effects of changing basicsystem properties become apparent and are accounted for in a consistentmanner. Simplifications may also be made and Hamilton’s principle canbe used as the basis for an approximate solution. Classical mechanicsdictates that Hamilton’s principle can only be used for systems that arealways composed of the same particles. This has been more recentlyextended to include systems whose constitutent particles change withtime, including open systems of changing mass. In this chapter, wereview the principle and its extended version and show through appli-cation to examples how it can lead to insightful observations about thesystem being modelled.

1 Introduction

One of the great accomplishments of analytical mechanics, Hamilton’s varia-tional principle has found use in many disciplines, including optics and quan-tum mechanics. The development of the equations of mechanics via a varia-tional principle allows the use of powerful approximation techniques for the

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solution of problems that may not be otherwise solvable. For example, theRayleigh-Ritz method has found much use in the solution of mechanics prob-lems. For the sake of completeness and to establish the notation, the principleis first derived in its classical form. Subsequent to this, extensions and appli-cations of the classical principle are presented.

2 Hamilton’s Principle - Classical Theory

Hamilton’s principle and the extended Hamilton’s principle permit the deriva-tion of the equations of motion from a definite integral involving kinetic energyand the virtual work performed by the applied forces. Both the virtual workand the kinetic energy are scalar functions. From d’Alembert’s principle for asystem of n particles,

n∑i=1

(mi

d2ri

dt2+

∂V

∂ri

− Fi

)· δri = 0 (1)

where V = V (r1, r2, ..., rn) is the potential energy of the particles, Fi denotesthe non-conservative applied forces acting on the ith particle, ri is the positionvector of the particle of the mass mi and δri is a virtual displacement. Thenotation δ implies a variation in the system - an imaginary change of config-uration that complies with the system constraints. The variation is only onr and its derivatives; the space and time parameters, usually denoted withxi and t, are not affected by the variation, either directly or in the ranges ofintegration. Where the system is prescribed, the variation must be zero. If theconfiguration of the system is prescribed then the variation must necessarilybe be zero because otherwise any change in configuration would result in aconfiguration that is not possible.

It should be noted that the variation of potential energy of the system isgiven by

δV =n∑

i=1

(∂V

∂ri

)· δri (2)

and the variation of the work done by non-conservative forces is

δW =n∑

i=1

Fi · δri. (3)

2

Therefore, by the product rule

d

dt

[n∑

i=1

(mi

dri

dt

)· δri

]=

n∑i=1

(mi

d2ri

dt2

)· δri +

n∑i=1

(mi

dri

dt

)· δdri

dt

=n∑

i=1

(mi

d2ri

dt2

)· δri + δ

n∑i=1

mi

2

dri

dt· dri

dt

=n∑

i=1

(mi

d2ri

dt2

)· δri + δT, (4)

where T is the kinetic energy of the system. On substitution of equations (2),(3) and (4), along with the definition of the Lagrangian as L = T − V, intothe original equation of motion (1), then d’Alembert’s principle becomes

δL + δW =d

dt

[n∑

i=1

(mi

dri

dt

)· δri

]. (5)

The preceding equation is for a discrete system and allows for straightforwardmodification for a continuous system as

δL + δW =d

dt

[∫V

(ρu) · δr dV

](6)

where ρ is the mass density of the system, u =drdt

is the velocity field of thesystem, L is the Lagrangian of the continuous system and δW is the virtualwork performed on the system by the non-conservative forces undergoing avirtual displacment. The subscript V indicates the fixed volume containingthe material in the system, over which the integration is performed.

The next step to obtaining Hamilton’s principle is to integrate either of theprevious two equations with respect to time, from time t1 to time t2, giving∫ t2

t1

δL dt +

∫ t2

t1

δW dt =

[∫v

(ρu) · δr δv

]t2

t1

. (7)

The motion of the system is defined by the position vector of each particlegiven as a function of time, t. For a system of N particles, at any time t, eachpoint has its own 3 dimensional position vector so that the state of the entiresystem is a point in 3N dimensional space known as configuration space. As

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time unfolds, the motion of the entire system of particles traves a curve in theconfiguration space called the true path. A different path, known as the variedpath, results from imagining the system as moving through configuration spaceby a slightly different path defined by the virtual displacement δri.

Of all the possible paths through configuration space, we consider onlythose that coincide with the true path at times t1 and t2. thus, the configura-tion of the system is given at times t1 and t2, and it follows that δr =0 at thosetwo times. Under those conditions, the last term in equation (7) becomes zero,giving the extended Hamilton’s principle as:∫ t2

t1

(δL + δW ) dt = 0. (8)

The extended Hamilton’s principle is very general and can be used to derive theequations of motion for many mechanical systems, as well as the correspondingcorrect boundary conditions by performing the required variations as given byequation (8). The principle is valid for rigid bodies, particles or deformablebodies. However, the virtual displacements must be reversible, implying thatthe constraint forces must do not work. Importantly, the principle cannot beused for systems with friction forces.

In the special case where there are no nonconservative forces, so that δW =0, then equation (8) reduces to Hamilton’s principle for conservative systems∫ t2

t1

δL dt = 0. (9)

Any system where the constraints can be expressed as relations betweenthe coordinates alone is known as holonomic. For holonomic systems, thevaried path is a possible path. However, if the constraint equations cannotbe written so that they involve the coordinates alone, then the varied pathis in general not a possible path. The difference between these two cases isimportant since the integration and variation operations are interchangeablefor holonomic systems. Thus for holonomic systems, Hamilton’s principle forconservative systems can be further simplified using the interchangeability ofintegration and variation as

δ

∫ t2

t1

L dt = 0. (10)

Equation (10) is the most familiar version of Hamilton’s principle. It canbe interpreted as stating that the actual path taken by the system through

4

configuration space is such that the value of the integral∫ t2

t1Ldt is stationary

with respect to all possible variations of the path between the two instants t1and t2, provided that the variation of the path at those two instants is zero.

Remark that L = T−V is the Lagrangian in which T is the kinetic energy ofthe particles in the system at any instant and V is the corresponding potentialenergy. δW is the virtual work performed by generalized forces undergoinggeneralized virtual displacements. If the system is a non-conducting linearelastic solid, then the potential energy V is most usually the strain energy. Thevirtual work is then the contribution from the non-conservative body forces andsurface stresses undergoing a virtual displacement. If thermodynamic effectsare included then the potential energy includes the internal energy and thevirtual work will include contributions due to to virtual temperature changesat the boundary [1].

3 Characteristic Features of Some Conserva-

tive Systems

Conservative systems are described by differential equations whose operatorsare self-adjoint with respect to the boundary conditions. The self-adjointnessimplies the conservation of energy, a fact which shall be demonstrated below.This is useful in that the energy functional then assumes an extremum for thesolution of the system, thus providing the basis for a variational solution.

For a large class of elastic systems, the application of Newton’s law yieldsan equation of motion of the form

w(x, t) + Gw(x, t) + qSw(x, t) = 0, (11)

with corresponding boundary conditions

[Uw(x, t)]B = 0. (12)

For the given elastic system, w(x, t) is the deflection of the system with thespatial coordinate given by x and time given by t. G is a self-adjoint lineardifferential operator that acts with respect to x and represents the elasticforces in the system. The variable q is a load parameter while S is anotherself-adjoint linear differential operator with respect to x. In equation (12), U isalso a linear differential operator with respect to x, evaluated at the boundariesof the system, as denoted by the subsrcipt B.

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The self-adjointness of the operators G and S is defined such that thefollowing hold: ∫

V

(Gu)v dV =

∫V

(Gv)u dV∫V

(Su)v dV =

∫V

(Sv)u dV. (13)

Here, u and v must be admissible functions, namely functions satisfying theboundary conditions in equation (12) and V is the volume of the system.

The energy of the system is given within a constant factor by the functional

E =

∫V

[w2 + (Gw + qSw)w

]dV. (14)

The derivative of the energy E with respect to time is given by

E =

∫V

[2ww + (Gw + qSw)w + (Gw + qSw)w] dV. (15)

Given the self-adjointness of the operators G and S, the preceding equationbecomes

E = 2

∫V

[w + Gw + qSw] w dV. (16)

Clearly, from equation (11), it follows that E=0 and the energy of the systemmust be a constant. Thus, the use of the self-adjoint nature of the operatorsG and S led to the conservation of energy.

Now consider the functional given by

H =

∫ t2

t1

∫V

[w2 − (Gw + qSw)w

]dV dt. (17)

The variation of this functional is given by

δH =

∫ t2

t1

∫V

[2wδw − (Gδw + qSδw)w − (Gw + qSw)δw] dV dt. (18)

Integration by parts gives

δH =

∫V

2wδw|t2t1 −∫ t2

t1

∫V

[2wδw + (Gδw + qSδw)w + (Gw + qSw)δw] dV dt.

(19)

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Imposing the usual constraint of no variation at the temporal endpoints impliesthat δw(t1) = δw(t2) = 0. Furthermore, we once again make use of the self-adjointness of the operators G and S to obtain

δH = −∫ t2

t1

∫V

2 [wδw + (Gw + qSw)δw] dV dt. (20)

Once again, making use of Newton’s equation for the system, equation (11)yields

δH = 0. (21)

It is clear that the self-adjointness of the operators G and S was necessary inorder to arrive at the statement of the variational principle.

4 Principle of Work and Energy

If the virtual displacements are allowed to coincide with the actual displace-ments, then the principle of work and energy can be derived from the equationof virtual work. If the system is in motion at t1 and t2, then its state is notgiven at those times and the virtual displacements cannot be zero at thoseinstants. Since we have allowed the virtual and actual displacements to co-incide, it then follows that δr = rdt, and more generally that the variation ofa quantity can be replaced with differential of that quantity so that δL = dL,for example.

The derivation follows the same path as that for Hamilton’s principle, upuntil the integration from time t1 to t2, which is restated here:∫ t2

t1

δLdt +

∫ t2

t1

δWdt =

[n∑

i=1

(mi

dri

dt

)· δri

]t2

t1

(22)

The variation δri is no longer zero at the endtimes as for the derivation ofHamilton’s principle. However, we now make use of the fact that we havechosen the virtual and actual displacements to be the same in order to simpifyeach term in the preceding equation. Let the virtual displacement dri takeplace during an infinitesimal time dt = ε. For the first term, we recall thatthe variation becomes a differential. Thus∫ t2

t1

δLdt =

∫ t2

t1

dL dt =

∫ t2

t1

εdL

dtdt = εL

∣∣∣∣t2t1

. (23)

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Similarly ∫ t2

t1

δWdt =

∫ t2

t1

dW dt =

∫ t2

t1

εdW

dtdt = εW

∣∣∣∣t2t1

. (24)

The third term can be written as[n∑

i=1

(mi

dri

dt

)· δri

]t2

t1

=

[n∑

i=1

(mir) · rε

]t2

t1

= ε2T

∣∣∣∣∣t2

t1

(25)

where T is the kinetic energy of the system. Given these three simplifications,the original equation becomes after dividing through by ε

L|t2t1 + W |t2t1 = 2T |t2t1 . (26)

Recalling that L = T −V , this simply states that ∆(T +V ) = ∆W . Dividingboth sides by ∆t and taking the limit as ∆t → 0, it then follows that

d(T + V )

dt=

dW

dt. (27)

Clearly if there is no non-conservative work done on the system then W = 0and this becomes a statement of the conservation of energy of the system.In general though, this states that the total change in the system energy isequal to the rate at which non-conservative work is done on the system. Thesedevelopments can be extended to open systems.

5 Extension of Hamilton’s Principle to Open

Systems

The previous development for Hamilton’s principle was extended to an opensystem by McIver [1]. The primary contribution of his work was to incorporatethe concept of an integral control volume as heavily used in fluid mechanics.In fluid mechanics, there are two ways of looking at the world, either throughLagrangian or through Eulerian coordinates. The Lagrangian description is away of looking at motion where the observer follows the motion of individualfluid particles as they move through space and time. This corresponds to theway the world is described in analytical mechanics as the description of themotion of each particle in the system is given or described. In contrast to

8

this, the Eulerian description of motion does not focus on specific particlesbut rather focusses on a specific area of space through which the particlesmove. The system is described by giving vectors at specific locations in space.Clearly, different particles will pass through the given area of interest.

5.1 Reynold’s Transport Theorem

It is instructive to review Reynold’s Transport theorem. First, a system isdefined as a collection of particles that are of interest. The system boundariesare such that the same particles are always contained within the system. Themass of the system is thus constant. The concept of a control volume isdefined as a clearly delineated, although imaginary, space through which theparticles may move. The external boundary of the control volume is calledthe control surface. The boundaries of the control volume are given at alltimes. A control volume may be chosen to be some physically meaningfulboundary so that although the control volume is always given, it is allowedto move. Particles may move in and out of the control volume and thus thecontrol volume may not be of constant mass. If it is of constant mass, then itneed not always consist of the same set of particles.

Briefly put, Reynolds transport theorem states that the rate of change ofan extensive property, N within the system is equal to the rate of change ofN within the control volume and the net rate of change of N through thecontrol surface - the net flux through the control surface. Mathematically,this becomes

d

dt

∫system

(ρN)dV =

∫control vol

∂

∂t(ρN)dV +

∫control surf

(ρN)u · nds. (28)

Here, N is the property of interest per unit mass, ρ is the density, dV andds are the differential volumes and surface area elements, and u = u(r, t) isthe system velocity at any point on the control surface. Thus ρ(u · n) is themass flow rate across a differential element of the control surface. Reynold’stransport theorem essentially says that the net rate of change of any propertyof interest within the system is equal to the change within the control volumeplus the net flux of N across the boundaries of the control volume. The unitnormal n is defined as positive when it points out from the control surface.Hence for flow out of the control volume, u · n is positive.

In the above, u is the velocity of the particles that are entering or leavingthe control volume, in order to give the next flux across the control surface.

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If any part of the control surface is itself moving, then this needs to be sub-traced so that the interpretation of the net flux across the control surface ismaintained. Thus, if the control surface is moving, then u− vcontrol replacesu in Reynold’s transport theorem.

5.2 Hamilton’s Principle for a System of Variable Mass

McIver presented an extension to Hamilton’s principle where the system bound-aries may not necessarily be well defined. In the classical version of Hamilton’sprinciple, the system contains the same material elements at all times. Withthe use of Reynold’s transport theorem, McIver generalized the analysis bymaking use of control volumes where material is allowed to cross the controlvolume boundary. In particular, with the use of Reynold’s transport theorem,we can write

δLsystem + δW =d

dt

[∫(ρu) · δrdV

]=

∫control vol

∂

∂t(ρu) · δrdV +

∫control surf

(ρu) · δr(u− vcontrol) · n ds(29)

where the Lagrangian Lsystem is the Lagrangian of the open control volumeand thus its mass is not necessarily constant. This is the statement of theprinciple of virtual work, generalized to the case of open control volumes wherethe enclosed mass may change as a function of time.

As for the classical Hamilton’s principle, it is assumed that the systemconfiguration is given at times t1 and t2 so that the variation of the system atthose times is zero. Integrating with respect to time from t1 to t2 gives∫ t2

t1

δLsystemdt +

∫ t2

t1

δWdt−∫ t2

t1

dt

∫control surf

(ρu) · δr(u− vcontrol) ·n ds = 0.

(30)This is the statement of Hamilton’s principle for a system of changing mass.Here δW is the virtual work performed by non-conservative forces and δLsystem

is the Lagrangian of the system contained within the open control volume. Thelast integral may be considered to be the virtual momentum transport acrossthe open control surface. If the virtual non-conservative work arises fromsurface stresses over the open and closed boundaries of the control surface, it

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then follows that

δW = δWclosedCS + δWopen CS

=

∫closedCS

(σ · n) · δr ds +

∫open CS

(σ · n) · δr ds (31)

where σ is the stress tensor. The extended Hamilton’s principle then becomes∫ t2

t1

δLsystemdt +

∫ t2

t1

dt

∫closedCS

(σ · n) · δr ds +

+

∫ t2

t1

dt

∫open CS

[(σ · n) · δr − (ρu) · δr(u− vcontrol) · n ]ds = 0. (32)

Here, L = T − V with

Tsystem =

∫control vol

1

2ρu · u dV

Vsystem =

∫control vol

ρ e dV (33)

where e is the potential energy per unit mass. For systems including structuresand fluids, the Lagrangian must include both the structure and the fluid. Theopen control surface represents the open portion of the control surface throughwhich fluid is permitted to flow. The closed section of the control surface isone through which there is no flow, such as at a solid boundary.

5.3 Principle of Work and Energy for a System of Vari-able Mass

As for the derivation of the classical principle of work and energy given above,replacing the virtual displacements with the actual displacements allows theenergy equation to be derived from the virtual work equation. The derivationproceeds along similar lines, by starting with the principle of virtual work,where now Reynold’s transport theorem is also used:

δLsystem + δW =d

dt

[∫(ρu) · δrdV

]=

∫control vol

∂

∂t(ρu) · δrdV +

∫control surf

(ρu) · δr(u− vcontrol) · n ds .(34)

11

As before, the virtual displacement is replaced with the actual displacementso that δr = dr = rdt = udt, which is akin to replacing the variation operatorwith the differential operator.

dLsystem+dW−∫

control vol

∂

∂t(ρu)·u dt dV−

∫control surf

(ρu)·udt(u−vcontrol)·n ds = 0.

(35)Dividing through by dt gives

d

dtLsystem +

dW

dt−

∫control vol

∂

∂t(ρu2)dV −

∫control surf

(ρu2)(u−vcontrol) ·n ds = 0.

(36)At the same time, Reynold’s transport theorem can also be applied to find thetime derivative of the Lagrangian L = T − V :

d

dtLsystem =

d

dt

∫system

(1

2ρu2 − ρe

)dV

=

∫control vol

∂

∂t

(1

2ρu2 − ρe

)dV +

∫control surf

(1

2ρu2 − ρe

)(u− vcontrol) · n ds(37)

Substituting into the previous equation yields

−∫

control vol

∂

∂t

(1

2ρu2 + ρe

)dV +

dW

dt−

∫control surf

(1

2ρu2 + ρe

)(u−vcontrol)·n ds = 0.

(38)The first term can be clearly identifed as the change in total energy within thecontrol volume, so that the preceding equation can be re-written as:

∂

∂t(T + V )control vol =

dW

dt−

∫control surf

(1

2ρu2 + ρe

)(u− vcontrol) ·n ds . (39)

This equation states that the change in energy within the control volume isequal to the rate at which non-conservative work is done plus the gain or lossof energy by virtue of the flow through the control surface and/or the movingcontrol surface engulfing additional particles. This is the principle of workand energy for a system of variable mass.

In the following, some examples of the use of Hamilton’s principle to derivethe equations of motion of a system are now presented.

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6 Example: Flow in a Viscoelastic Curved Pipe

An example of Hamilton’s principle for systems with variable mass is now pre-sented. This was originally derived in [2]. Fluid flowing through a viscoelasticcircular pipe is considered. Let ur(θ, t) and uθ(θ, t) represent the displacementvariables along the radial and tangential directions, respectively. The radiusof the circular pipe is given by r while θ is the angular coordinate, α is theangular size of the section of pipe that is being considered, s is arclength alongthe centreline of the pipe, A is the cross-sectional area of the pipe (fluid), I isthe moment of inertia of a cross-section of pipe, U is the constant-magnitudeflow velocity of the fluid relative to the pipe wall, ms and mf are the massesper unit length of the pipe and fluid respectively, and finally m = ms + mf isthe total mass per unit length of the pipe-fluid system.

Curved pipe flow is clearly a non-conservative system, therefore Hamil-ton’s principle extended for systems of changing mass is required. Hamilton’sprinciple now becomes

δ

∫Ldt−

∫ t2

t1

mfU

(∂R

∂t+ Uτ

)· δRdt = 0, (40)

where L = Ts+Tf−Vs−Vf is the Lagrangian with s and f subscripts denotingstructure and fluid, respectively. The position vector of the deformed pipecentreline is given by R and τ is a unit vector tangential to the free end ofthe deformed pipe centreline.

The strain energy stored in the pipe due to bending is given by

Vs =

∫ α

0

1

2

(E + η

∂

∂t

)Jz

r3

(∂2ur

∂θ2 + ur

) (∂2ur

∂θ2 +∂uθ

∂θ

)dθ, (41)

where Jz is the polar moment of inertia of the pipe when the curved radius ofthe pipe is sufficiently greater than the cross-sectional radius of the pipe, E isYoung’s Modulus and η is the viscosity of the visco-elastic pipe, assuming aKelvin-Voigt model. The kinetic energy of the pipe is given by

Ts =

∫ α

0

1

2ms

[(∂ur

∂t

)2

+

(∂uθ

∂t

)2]

r dθ (42)

If the fluid is assuming to be incompressible and there is no gravitationalpotential energy, then the fluid potential energy is given by

Vf = 0. (43)

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Let ua denote the inertial velocity of the fluid so that the fluid kinetic energyis given by

Tf =

∫ α

0

1

2mfua · ua rd θ =

1

2md

∫ α

0

[U2 +

(∂ur

∂t

)2

+

(∂uθ

∂t

)2

+ 2U∂uθ

∂t

+ 2U

r

(∂ur

∂θ+ uθ

)∂ur

∂t+

U2

r2

(∂ur

∂θ+ uθ

)2]

rd θ. (44)

Assuming that the centreline of the pipe is inextensible, then the radial dis-placement can be related to the tangential displacement via

ur =∂uθ

∂θ. (45)

Let dimensionless variables be introduced so that

w =uθ

r, β = ρA

m, τ =

t

r2

(EI

m

)1/2

u = Ur(

ρAEI

)1/2, H =

1

r2

(EI

m

)1/2η

E. (46)

Putting all these together and performing the variation as indicated, theHamilton’s Principle extended to the pipe system with changing mass givesthe equation of motion as

H∂7w

∂θ6∂τ+

∂6w

∂θ6+ 2H

∂5w

∂θ4∂τ+ 2

∂4w

∂θ4+ u2∂4w

∂θ4+ 2

√βu

∂4w

∂θ3∂τ

+∂4w

∂θ2∂τ 2+ H

∂3w

∂θ2∂τ+ (1 + 2u2)

∂2w

∂θ2+ 2

√βu

∂2w

∂θ∂τ+ u2w − ∂2w

∂τ 2= 0.(47)

7 Example: Equations of motion of a Spinning

Disk

In this section, use of Hamilton’s principle is demonstrated in deriving theequations of motion of a spinning disk. It will be shown that use of Hamil-ton’s principle provides additional insight into the mechanics of the problem,in particular, generating a new term in the equations of motion that would

14

otherwise be omitted if a less structured approach were taken to their deriva-tion. Hamilton’s principle yields the equations of motion for the transverse aswell as in-plane vibrations and ensures correct boundary conditions in orderto obtain a self-adjoint linear operator.

In order to use Hamilton’s principle, expressions for kinetic energy, elasticstrain energy and work done must be formulated. For continuous systems thisis done by considering a small element of volume and then integrating overthe entire volume of the solid in question. In considering large displacements,the shape of the entire volume changes as a function of time, with the solidloooking different at different points in time. Care must therefore be takenwhen integrating over the entire volume. The question arises as to whetherthe integration should be performed over the current volume or over the ini-tial volume. Since the current volume is usually an unknown, this is bestaddressed by referring all quantities to the initial volume and then performingthe integration over the initial volume of the solid. In other words, Lagrangianand not Eulerian coordinates must be used. Usually, it is desirable to havethe equations of motion formulated in terms of displacements. If this is thecase, then the kinetic and strain energies of the system must be formulated interms of displacements.

7.1 Strain Energy of a Spinning Disk

Assuming that a strain energy exists is akin to making a fundamental assump-tion about the material and how it behaves. If the form for the strain energyof any material is known, then the stress-strain relationship for the materialmay be deduced. Conversely, if the stress-strain relationship of the materialis known then so is the form of the strain energy.

There is a nonzero contribution to the strain energy stored in a deformablebody only if the body bends, stretches or otherwise deforms. If it rotates ortranslates like a rigid body without deforming then such displacements willnot contribute to the strain energy of the body, although they will contributeto the kinetic energy.

In measuring the displacements of particles in the body for the purposes ofcalculating strains, it is imperative to measure them with respect to the rigidbody motion. This is accomplished by fixing a coordinate frame to the body.The translation and rotation of this body-fixed frame describe the rigid bodymotions of the body. Displacements with respect to this frame then contributeto the strain energy stored in the body. Since the displacements are measured

15

with respect to the undeformed (not unrotated) body, the discussion for thederivation of the strain energy proceeds in the same manner as for non-rotatingbodies.

To derived an expression for the strain energy in terms of displacements,expressions for the following are required:

• An expression for the strain energy in terms of the stress and strain inthe body.

• An expression giving stress in terms of strain (stress-strain expression).

• An expression giving the strain in terms of the displacements (strain-displacement expression).

The above would yield energy expressions in terms of displacements of anypoint in the body. In thin plate theory, the displacements of an arbitrary pointare further related to the displacements of the middle surface of the plate viaKirchhoff’s hypothesis. The problem thus reduces to one of solving for thedeflections of one particular surface only. Since the ensuing deflections do notdepend on any vertical component, this serves to turn a three-dimensionalcontinuum mechanics problem into one of two dimentions. Hence, four rela-tionships are required to express the strain energy in terms of the displacementsof the middle surface as measured in the body-fixed frame.

The strain energy per unit volume is denoted by W0 and is given by

W0 =1

2σijεij, (48)

where σij and εij are the stress and strain tensors respectively. To find thetotal strain energy, this expression must be integrated over the entire volumeof the body. Small strains are assumed so that stress-strain relationship willbe taken to be Hookean (linear). Note that this implies that the strains aresmall, but does not imply that the displacements are small. For a flat plate, aplane stress condition may be assumed. This implies that σzz = σrz = σθz = 0.

Thus for a plane stress condition, the strain energy per unit volume can beexplicitly written as

W0 =1

2σijεij

=2G(λ + G)

λ + 2G(εrr + εθθ)

2 + 2G(ε2rθ − εrrεθθ

). (49)

16

The strain-displacement relation is required in order to express the strainenergy in terms of displacements. It is at this point in the derivation thatnonlinearities may be introduced into the modelling. Since all quantities areto be referred to the undeformed body, the lagrangian form of the strain tensorthat is required.

The Von Karman plate theory can be shown [3] to lead to the followingnonlinear strain-displacement expressions

εrr =∂ur

∂r+

1

2

(∂uz

∂r

)2

, (50)

εθθ =ur

r+

1

r

∂uθ

∂θ+

1

2r2

(∂uz

∂θ

)2

, (51)

εrθ =1

2r

[∂ur

∂θ− uθ + r

∂uθ

∂r+

∂uz

∂r

∂uz

∂θ

], (52)

where ur, uθ and uz are the displacements of the disk in the r, θ and zdirections, respectively. For the linear Kirchoff theory, the nonlinear termsinvolving uz are dropped from the above expressions, leading to linear strain-displacement relationships.

The last required expressions are those relating the displacements of anarbitrary point in the plate to those of the middle surface of the plate. In thinplate theory, it is usually hypothesized that the linear filaments of the plateinitially perpendicular to the middle surface remain straight and perpedicularand do not contract or extend. Transverse shear effects are thus neglected.This assumption leads to a relationship between the displacements of an ar-bitrary point ur, uθ and uz and the displacements of the middle surface u, vand w. They are given by

uz = w(r, θ),

ur = u(r, θ)− z∂w

∂r,

uθ = v(r, θ)− z

r

∂w

∂θ. (53)

All the required expressions have now been assembled. and the strainenergy of the entire plate can be obtained by integrating over the entire volumeof the plate. The strain energy will be a function of u, v and w and of thevertical coordinate z. Furthermore, u, v and w are themselves functions of

17

in-plane coordinates (r, θ) and of time, t. Thus the strain energy is an explicitfunction of z. This dependence can be eliminated by explicitly carrying outthe integration over the thickness of the plate from z = −h to z = h, where his the distance between the middle surface of the plate and the plate boundingsurface.

This procedure finally yields the strain energy of the plate as an explicitfunction of u(r, θ, t), v(r, θ, t) and w(r, θ, t) only. The strain energy of the plateis thus given by

Wo =

∫ 2π

0

∫ R2

R1

Wp r dr dθ

=

∫ 2π

0

∫ R2

R1

hW1 + h3 W3 r dr dθ, (54)

where

W1 =G

(λ + 2G)

[(λ + 2G)

(∂v

∂r

)2

+ (λ + G)

(∂w

∂r

)4

+ 4(λ + G)u2

r2

+4(λ + G)

r2

(∂v

∂θ

)2

+(λ + 2G)

r2

(∂u

∂θ

)2

− 2(λ + 2G)v

r2

(∂u

∂θ

)+

(λ + G)

r4

(∂w

∂θ

)4

+2(λ + G)

r2

(∂w

∂θ

)2 (∂w

∂r

)2

+ 4(λ + G)∂u

∂r

(∂w

∂r

)2

− 2(λ + 2G)v

r

(∂v

∂r

)+

2(λ + 2G)

r

(∂u

∂θ

) (∂v

∂r

)+

4(λ + G)

r3

∂v

∂θ

(∂w

∂θ

)2

+ 4(λ + G)u

r3

(∂w

∂θ

)2

+2(λ + 2G)

r2

(∂u

∂θ

) (∂w

∂θ

) (∂w

∂r

)+

2(λ + 2G)

r

(∂v

∂r

) (∂w

∂θ

) (∂w

∂r

)+ 8(λ + G)

u

r2

(∂v

∂θ

)+

2λ

r

∂v

∂θ

(∂w

∂r

)2

+ 4λu

r

∂u

∂r+ 2λ

u

r

(∂w

∂r

)2

+4λ

r

(∂u

∂r

) (∂v

∂θ

)− 2(λ + 2G)

v

r2

(∂w

∂r

) (∂w

∂θ

)+ 4(λ + G)

(∂u

∂r

)2

+ (λ + 2G)v

r

2]

(55)

18

W3 =8G(λ + G)

3(λ + 2G)r3

(∂w

∂r

) (∂2w

∂θ2

)+

4G

3r4

(∂w

∂θ

)2

+4G(λ + G)

3(λ + 2G)r2

(∂w

∂r

)2

+4Gλ

3(λ + 2G)r2

(∂2w

∂θ2+ r

∂w

∂r

)∂2w

∂r2+

4G(λ + G)

3(λ + 2G)r4

(∂2w

∂θ2

)2

+4G

3r2

(∂2w

∂r∂θ

)2

− 8G

3r3

(∂w

∂θ

) (∂2w

∂r∂θ

)+

4G(λ + G)

3(λ + 2G)

(∂2w

∂r2

)2

(56)

Here G is the shear modulus and λ is a constant. They are related toYoung’s modulus, E, and Poisson’s ratio ν of the material by

λ =Eν

(1 + ν)(1− 2ν)(57)

G =E

2(1 + ν). (58)

7.2 Kinetic Energy of a Spinning Disk

While the strain energy of a stationary and a rotating disk are the same, itis in formulating the kinetic energy expression that the difference between arotating and stationary disk becomes apparent.

Let us set up two coordinate systems, S and B. Suppose that S is anintertial frame of reference and that B is rigid-body-fixed to the disk, so thatB rotates with the disk at a spin rate of Ω with respect to S. Thus, an observerin S would see the full rigid body rotation and elastic deflections of the disk,while an obvserver in B would only see the elastic deformations of the disk.In using Hamilton’s principle, the inertial kinetic energy of the disk must befound, so it is the kinetic energy as measured in S that is required. Notehowever, that the strain energy is a function of displacement with respect tothe undeflected body, not the undisplaced body so that the strain energy isbest expressed from the point of view of an observer in B. The inertial kineticenergy as measured by an observer in S (the required quantity) needs to beexpressed in terms of measurements made by an observer in B (the availablequantity). The total kinetic energy is then the sum of the kinetic energyof deflection as seen by an observer in B and the kinetic energy due to therotation of the disk.

Let ro denote the undeformed location of a particle in the disk and letu denote the corresponding displacement vector. Hence, the location of a

19

particle originally at ro is given by r = r0 +u at any given time. These vectorsare chosen to be expressed in terms of unit vectors belonging to the B frame.Then the velocity of any particular particle is given by dr

dt, where it must be

remembered that since the unit vectors are fixed in the rotating frame, theirtime derivative must be found as well. In fact, it is the time derivative of therotating unit vectors that provides the portion of the overall velocity of theparticle that is due to the rotation. The rest of the velocity of the particleis due to the elastic deflection only. The time derivative of the rotating unitvectors can be found be taking the cross product of the angular velocity vectorwith the unit vector in question.

Let er be unit vector in the r direction such that er = cos(θ)iB + sin(θ)jB.Note that iB and jB are unit vectors in the x and y directions in the body-fixedframe, B. Similarly, let eθ = − sin(θ)iB + cos(θ)jB be a unit vector in the θdirection, pointing in the direction of increasing θ. Furthermore, let ez be aunit vector pointing in the z direction such that er, eθ, ez form a right-handedcoordinate system.

The angular velocity vector of the body-fixed frame is given by ω = Ωez.Thus the inertial time-derivatives of the body-fixed unit vectors are given by

der

dt= ω × er = Ωeθ

deθ

dt= ω × eθ = −Ωer

dez

dt= ω × ez = 0. (59)

Points within the body are represented by the polar coordinates (r, θ, z). Theoriginal position of a particle is given by ro = rer+zez. The deformed positionof the same particle is given by

r = ro + u = (r + ur)er + uθeθ + (z + uz)ez, (60)

where ur, uθ and uz are the displacements in the er, eθ, ez directions of aparticle. Each of these displacements will be a function of time and the originalposition of the particle in question. The velocity of this particle is given by

v =dr

dt= (ur − Ωuθ)er + [uθ + Ω(r + ur)] eθ + uzez. (61)

Since the unit vectors are orthonormal, the squared speed is given by

v · v = (ur − Ωuθ)2 + [uθ + Ω(r + ur)]

2 + u2z. (62)

20

Once the velocity as measured by an inertial observer of any particle hasbeen found, the kinetic energy of a small element of volume can be expressedas 1

2ρ dV v · v, where ρ is the density of the material and dV is an element

of volume. Thus the total kinetic energy of the body can be found by inte-grating over the entire undeformed volume. Note that since the velocity hasbeen expressed as a function of the undeformed location of the particle, theintegration is to be performed over the undeformed volume of the body, notthe unknown deformed volume.

The kinetic energy expression is now expressed as a function of ur, uθ

and uz, the displacements of an arbitrary point on the disk in the r, θ andz directions respectively. As for the strain energy, equation (53) relating thedisplacements of an arbitrary point on the disk to the displacement of themiddle surface can be used. The explicit dependence of the kinetic energy onz can be eliminated by integrating over the thickness of the disk, from z = −hto z = +h.

As before, this procedure finally yields the kinetic energy of the plate as anexplicit function of u(r, θ, t), v(r, θ, t) and w(r, θ, t) only. The kinetic energyof the plate is thus given by

KE =

∫ 2π

0

∫ R2

R1

KEp r dr dθ

=

∫ 2π

0

∫ R2

R1

hKE1 + h3 KE3 r dr dθ, (63)

where

KE1 = ρΩ2(v2 + u2 + r2 + 2ru

)+ 2ρΩ

[u∂v

∂t− v

∂u

∂t+ r

∂v

∂t

]+ ρ

[(∂u

∂t

)2

+

(∂v

∂t

)2

+

(∂w

∂t

)2]

, (64)

KE3 =ρΩ2

3r2

[(∂w

∂θ

)2

+ r2

(∂w

∂r

)2]

+ρ

3r2

[(∂2w

∂θ∂t

)2

+ r2

(∂2w

∂r∂t

)2]

+2ρΩ

3r

[∂2w

∂θ∂t

(∂w

∂r

)− ∂2w

∂r∂t

(∂w

∂θ

)](65)

Here ρ is the density of the disk, Ω is its angular velocity, h is its half-thickness,and the displacements of the middle surface are given by u, v and w. Notethat the inner and outer radii of the disk are given by R1 and R2 respectively.

21

7.3 Equations of Motion

The equations of motion and corresponding boundary conditions are derivedby applying Hamilton’s Principle.

7.4 Nonlinear Equations of Motion

The full nonlinear equations of motion are given below.

ρ(1− ν2)

E

[∂2u

∂t2− v

∂Ω

∂t− Ω2(r + u)− 2Ω

∂v

∂t

]=

(1− ν)

2r2

∂2u

∂θ2+

(1 + ν)

2r

∂2v

∂r∂θ+

∂2u

∂r2

+1

r

[(1− ν)

2

∂w

∂r

2

− (3− ν)

2r

∂v

∂θ− (1 + ν)

2r2

∂w

∂θ

2

− u

r+

∂u

∂r

]+

(1− ν)

2r2

∂w

∂r

∂2w

∂θ2

+∂w

∂r

∂2w

∂r2+

(1 + ν)

2r2

∂w

∂θ

∂2w

∂r∂θ, (66)

ρ(1− ν2)

E

[∂2v

∂t2+ (r + u)

∂Ω

∂t− Ω2v + 2Ω

∂u

∂t

]=

(1 + ν)

2r

∂w

∂r

∂2w

∂r∂θ+

(1− ν)

2

∂2v

∂r2

+1

r

[(1− ν)

2r

∂w

∂θ

∂w

∂r+

(1− ν)

2

∂v

∂r+

(3− ν)

2r

∂u

∂θ− (1− ν)

2

v

r

]+

(1 + ν)

2r

∂2u

∂r∂θ

+1

r2

∂2v

∂θ2+

1

r3

∂w

∂θ

∂2w

∂θ2+

(1− ν)

2r

∂w

∂θ

∂2w

∂r2, (67)

ρ(1− ν2)

E

[∂2w

∂t2+

h2Ω2

3∇2w − h2

3

∂2

∂t2∇2w

]= −h2

3∇4w +

1

r4

∂w

∂θ

[3

2

∂w

∂θ

∂2w

∂θ2+ r

∂2v

∂θ2

−r

2

∂w

∂θ

∂w

∂r+

(1 + ν)

2r∂u

∂θ+

(1 + ν)

2r2 ∂2u

∂r∂θ+

(1− ν)

2rv +

r2

2

∂w

∂θ

∂2w

∂r2+ r2 ∂2w

∂r∂θ

∂w

∂r

+(1− ν)

2r3∂2v

∂r2− (1− ν)

2r2∂v

∂r

]+

1

r4

∂w

∂r

[r2 ∂2w

∂r∂θ

∂w

∂θ+ (1 + ν)r3∂u

∂r+

(1 + ν)

2r3 ∂2v

∂r∂θ

+(1− ν)

2r2∂2u

∂θ2+

r3

2

∂w

∂r

2

+r2

2

∂w

∂r

∂2w

∂θ2+ r4∂2u

∂r2+

3r4

2

∂w

∂r

∂2w

∂r2− (1− ν)

2r2∂v

∂θ

]+

1

r4

∂2w

∂θ2

[νr2∂u

∂r+ r

(u +

∂v

∂θ

)]+

∂2w

∂r2

[∂u

∂r+

ν

r

(u +

∂v

∂θ

)]+

(1− ν)

r2

∂2w

∂r∂θ

[∂u

∂θ− v + r

∂v

∂r

]. (68)

22

The full nonlinear representation of the spinning disk problem requires thesolution of three nonlinear, coupled partial differential equation, namely equa-tions (66), (67) and (68). This is the result of the use of Lagrangian coordinatesas well as the inclusion of in-plane inertia, coriolis and rotary inertia terms,which are often neglected.

7.5 Equations of Motion Corresponding to Linear Strain

The equations of motion arising from the use of linear (Kirchhoff) strain-displacement expressions are given as :

ρ(1− ν2)

E

[∂2u

∂t2− v

∂Ω

∂t− Ω2(r + u)− 2Ω

∂v

∂t

]=

(1− ν)

2r2

∂2u

∂θ2+

(1 + ν)

2r

∂2v

∂r∂θ

+∂2u

∂r2+

1

r

[∂u

∂r− (3− ν)

2r

∂v

∂θ− u

r

], (69)

ρ(1− ν2)

E

[∂2v

∂t2+ (r + u)

∂Ω

∂t− Ω2v + 2Ω

∂u

∂t

]=

(1 + ν)

2r

∂2u

∂r∂θ+

(1− ν)

2

∂2v

∂r2

+1

r

[(1− ν)

2

∂v

∂r+

(3− ν)

2r

∂u

∂θ− (1− ν)

2

v

r

]+

1

r2

∂2v

∂θ2, (70)

ρ(1− ν2)

E

[∂2w

∂t2+

h2Ω2

3∇2w − h2

3

∂2

∂t2∇2w

]= −h2

3∇4w. (71)

It may be noted that equations (69) and (70) for the in-plane vibrations arethe same equations derived by other authors [4, 5, 6], whereas equation (71)for the linear transverse vibrations is not. The reason for this discrepancy liesin the different assumptions built into the different models. It turns out thatthe equation (71) for the transverse vibrations does not accurately capture thespinning disk dynamics as the stresses induced in the disk due to its rotationare not captured with the use of linear strain-displacement relations. Interest-ingly, nonlinear strains must be used in order to address this shortcoming.

7.6 Linear Equations of Motion - Nonlinear Strain

Now consider the equations obtained by neglecting all nonlinear terms in thenonlinear equations of motion with the exception of terms containing u or ∂u

∂r.

23

When the disk is rotated and allowed to come to equilibrium, there is an equi-librium deflection in the radial direction. It is possible that this equilibriumdisplacement is not small enough to justify neglecting products of these termswith derivatives of w. The resulting equations for the in-plane vibrations areidentical with equations (69) and (70). However, the equation for the trans-verse vibrations is different as that obtained with linear strains and is nowgiven by

ρ(1− ν2)

E

[∂2w

∂t2+

h2Ω2

3∇2w − h2

3

∂2

∂t2∇2w

]= −h2

3∇4w

+∂w

∂r

[(1 + ν)

r

∂u

∂r+

∂2u

∂r2

]+

1

r2

∂2w

∂θ2

[ν∂u

∂r+

u

r

]+

∂2w

∂r2

[∂u

∂r+ ν

u

r

]. (72)

However, note that corresponding to an in-plane purely radial displacementu(r), use of linear stress-strain and linear strain-displacements relationshipslead to the following stress-displacement relationships :

σrr =E

(1− ν2)

[du

dr+ ν

u

r

](73)

σθθ =E

(1− ν2)

[νdu

dr+

u

r

]. (74)

Using these relationships, equations (72) can be rewritten as

ρ

[∂2w

∂t2+

h2Ω2

3∇2w − h2

3

∂2

∂t2∇2w

]= − Eh2

3(1− ν2)∇4w+

1

r

∂

∂r

(rσrr

∂w

∂r

)+

σθθ

r2

∂2w

∂θ2.

(75)With the exception of the ∇2w and ∇2w, this is the same equation as obtainedby Lamb and Southwell [7] for the transverse vibrations of a spinning disk. Thepresence of the ∇2w term is not unexpected; it is simply the term due to therotary inertia of the disk. The physicaly meaning of the ∇2 term will beexplained subsequently.

7.7 Boundary Conditions

After using the 2D analogue of integration by parts to isolate the variation ofu, v and w, the remaining boundary term can be found. From this bound-ary term, suitable boundary conditions to the problem may be written downdirectly. Since u, v and w are the generalized coordinates for the problem,

24

the entire boundary term is required to vanish . Since u, v and w are inde-pendent, the only way the entire boundary term will vanish is if the followingthree conditions hold on the boundary

1. δu = 0 or the coefficient of δu vanishes

2. δv = 0 or the coefficient of δv vanishes

3. δw = 0 or the coefficient of δw vanishes

4. δ ∂w∂r

= 0 or the coefficient of δ ∂w∂r

vanishes

The boundary term obtained from applying Hamilton’s Principle and inte-gration by parts is given below. Note that r here must be evaluated on theboundary. Thus for a solid disk, r below is the radius of the disk. For an an-nulus, a set of boundary conditions is required at each of the inner and outerradii, so r will assume two possible values.

− 3E

∫ 2π

0

r3δu

[2ν

u

r+ 2

∂u

∂r+

2ν

r

∂v

∂θ+

(∂w

∂r

)2

+ν

r2

(∂w

∂θ

)2]

dθ

− 3E(1− ν)

∫ 2π

0

r3δv

[1

r

(∂u

∂θ− v

)+

∂v

∂r+

1

r

∂w

∂θ

∂w

∂r

]dθ

− 2Eh2

∫ 2π

0

δ

(∂w

∂r

)r3

[∂2w

∂r2+

ν

r2

∂2w

∂θ2+

ν

r

∂w

∂r

]dθ

+ 2(1− ν2)ρh2

∫ 2π

0

r3δw

[Ω2∂w

∂r− ∂3w

∂r∂t2

]dθ

+ 2Eh2

∫ 2π

0

r3δw

[∂

∂r∇2w +

(1− ν)

r2

∂2

∂θ2

(∂w

∂r− w

r

)]dθ

− 3E

∫ 2π

0

r3δw

[1− ν)

r

∂w

∂θ

∂v

∂r+

1

r2

∂w

∂r

(∂w

∂θ

)2

+ 2∂u

∂r

∂w

∂r

+

(∂w

∂r

)3

+(1− ν)

r2

∂w

∂θ

(∂u

∂θ+ v

)+

2ν

r

∂w

∂r

(u +

∂v

∂θ

)]dθ (76)

There are a few points that are worth mentioning. First, the above bound-ary term was obtained from the variation of the Lagrangian obtained with thenonlinear (Von Karman) strain-displacement relations. Note that the corre-sponding boundary conditions are also nonlinear and are coupled. Had the

25

linear (Kirchoff) strain-displacement relations been used, the correspondingboundary conditions would also have been linear. They can be obtained fromthe above expression by neglecting all nonlinear terms.

It was previously noted that formulating the problem in this manner auto-matically accounts for the effect of rotary inertia in the equations of motion.The corresponding term in the boundary condition is ∂3w

∂r∂t2. That is, the vari-

ation of some particular part of the kinetic energy expression gives rise to the∇2w term in the equation of motion and to the ∂3w

∂r∂t2term in the boundary

condition. Hence, if the effect of rotary inertia is ignored (or included) inthe equation of motion, then the corresponding term must also be ignored (orincluded) in the boundary condition.

Note also that the variation of the (∇w×∇w) ·k term in the kinetic energygives rise to boundary terms only. In other words, dropping this term from thekinetic energy expression does not change the resulting equation of motion. Itdoes, however, change the boundary conditions. Since the equation of motionremains unchanged with the omission of this term, it is questionable whetherthe corresponding boundary term should be included as well. It turns outthat including the questionable term leads to physically meaningless results insolving the linear vibration problem. This boundary term has thus not beenincluded in the above expression.

From equation (76), the boundary conditions for the special cases of linearin-plane and transverse vibrations can be derived.

Linear Transverse VibrationsFor the linear transverse vibration problem, the boundary conditions for a freeedge become

∂2w

∂r2+

ν

r

[∂w

∂r+

1

r

∂2w

∂θ2

]= 0, (77)

∂

∂r∇2w +

(1− ν)

r2

∂2

∂θ2

(∂w

∂r− w

r

)=

(1− ν2)ρ

E

[∂3w

∂r∂t2− Ω2∂w

∂r

]. (78)

Note that ∫ h

−h

σrrz dz =2Eh3

3(1− ν2)

[∂2w

∂r2+

ν

r

(∂w

∂r+

1

r

∂2w

∂θ2

)]. (79)

Hence, the first boundary condition (77) says that the moment at the freeedge is zero and this is in agreement with boundary conditions given in theliterature.

26

The standard second assumption for a free edge is that the Kirchoff shear,Vr or ’edge reaction’ [8] be set to zero. For stationary plates (neglecting rotaryinertia) the Kirchhoff shear is given by

Vr = −D

[∂∇2w

∂r+

(1− ν)

r2

∂2

∂θ2

(∂w

∂r− w

r

)]= 0, (80)

where D = 2h3

3(1−ν2)is the bending stiffness of the disk. Recall that h denotes the

half-thickness of the disk. This last boundary condition is bascially summingthe transverse forces at the edge of the disk and setting the result equal tozero.

This same expression is also usually applied to the rotating disk. Even whenthe term corresponding to the rotary inertia of the disk is dropped from theequation of motion and boundary condition, the derived boundary condition,equation (78) and the standard boundary condition, equation (80), do notagree. The derived boundary condition (78) differs by the inclusion of a termproportional to ρΩ2 ∂w

∂r. This term is essentially the product of the centrifugal

force and the slope at the edge of the disk. Due to the presence of the Ω2, thisterm would vanish for a stationary disk. It must be observed that the variationof some particular portion in the kinetic energy gives rise to the Ω2∇2w termin the equations of motion and to this Ω2 in the boundary condition.

The presence of the centrifugal term in the balance of forces at the edge(and interior) of the disk should not come as a surprise. This term was obtainednaturally as part of the variation of the kinetic energy of the disk and not in anad-hoc manner. The physical significance of these new terms will be addressedin the discussion.

Linear In-plane VibrationsThe boundary terms for linear in-plane vibrations are given by[

∂u

∂r+

ν

r

(u +

∂v

∂θ

)]δu = 0, (81)[

1

r

∂u

∂θ− v

r+

∂v

∂r

]δv = 0, . (82)

Equations (81) and (82) must hold on the boundary of the disk. For a soliddisk, equations (81) and (82) must be true on the outer radius. For an annulus,equations (81) and (82) must hold at each of the inner and outer radius.

27

Note that ∫ h

−h

σrr dz =2Eh

ν(1 + ν)

[∂u

∂r+

ν

r

(u +

∂v

∂θ

)], (83)∫ h

−h

σrθ dz = h

[1

r

∂u

∂θ− v

r+

∂v

∂r

]. (84)

Thus equation (81) implies that on the boundary either the displacement of themiddle surface in the radial direction must be specified or the integral of thestress in the radial direction over the side of the disk must vanish. Similarly,equation (82) reads that on the boundary either v must be specified, or theintegral of the shear stress over the side of the disk must vanish.

7.8 Discussion

7.8.1 Three Nonlinear Equations vs Two

The nonlinear equations of motion for a spinning disk are given by equations(66),(67) and (68). Note that there are three nonlinear, coupled equations,implying that they must be solved simultaneously. The spin rate Ω is usuallytaken to be constant so that ∂Ω

∂t= 0.

If all in-plane time-derivatives are neglected, then it is possible to use astress function to reduce the three new equations (66),(67) and (68) to twowhere the generalized coordinates are the the transverse displacement and thenewly-introduced stress function. This implies the omission of the in-planeinertias, ∂2u

∂t2and ∂2v

∂t2, as well as the coriolis terms, ∂u

∂tand ∂v

∂t.

We remark that the centrifugal force is not really an external force atall, but rather a consequence of the fact that the reference frame is rotatingand thus non-inertial. The coriolis force is due to the same effect. In-planeinertia is typically ignored for stationary (non-rotating) plates, and this hasbeen shown to be a good approximation for stationary plates, [9]. However,the same calculation fails for the rotating plate because of the presence of thecentrifugal and coriolis forces. The validity of the approximation of rotatingdisk models that reduce the number of nonlinear equations to be solved fromthree to two still needs to be rigorously examined.

7.8.2 Lagrangian vs Eulerian Variables

In the derived nonlinear equations (66), (67), (68) and their linear counterparts(69), (70) and (71), the centrifugal force appears as a term proportional to

28

Ω2(r + u). This is reasonable given that (r + u) is the current radial positionof a given particle. A particle originally at radius r has radius (r + u) afterdeformation takes place. This is consistent with the Lagrangian descriptionof the system that has been employed. In this description the boundaries ofthe disk or annulus are at the original (known) radii. On the other hand,if an Eulerian description of the system is used, the centrifugal force will beproportional to Ω2r. Now the location of the boundaries is an unknown ; if thedisk stretches, the new location of the boundaries is part of the unknowns of theproblem. Most authors use the Eulerian description Ω2r with the boundaries(incorrectly) located at their Lagrangian (original) location, although this ismost likely a negligeable difference for small in-plane displacements. However,while it may seem that for small displacements there should not be muchdifference between r and (r + u), Bhuta and Jones, [4] showed that the actualsolutions obtained for the linear in-plane problem can be quite different.

7.8.3 Decoupled In-plane and Transverse Plate Problems

Consider the linear equations of motion, (69), (70) and (71). Although theequations for the in-plane displacements are still coupled, they are not coupledto the third equation for the transverse displacement. As for the stationaryplate, the in-plane and out-of-plane free vibrations are independent of eachother. It turns out, however, that this is only valid for very small rates of ro-tation, namely rates of rotation that result in very small in-plane (membrane)stresses.

If the spinning disk is modelled as a spinning (linear) membrane, then thein-plane and transverse vibration problems are not independent. To see this,consider the three linear equations of motion (69), (70), (71) and ignore anyterm proportional to h2. That is, since the membrane is assumed thin, thehalf-thickness h2 is considered to be small in comparison to h. While the twoequations for the in-plane vibrations remain unchanged, the equation for thetransverse vibrations becomes ∂2w

∂t2= 0, i.e. no vibrations in the the transverse

direction. In essence, this means that for a membrane the linear transversevibrations are a second-order effect; to first order, small transverse vibrationsdo not affect the in-plane stresses.

This might lead one to wonder why the transverse vibrations of a plateare affected to first order by the rotation while the transverse vibrations ofa membrane are not. The reason for this lies in the description of the in-plane displacements for the two different models. In both cases, a plane stress

29

situation is assumed while the effect of rotation results in primarily in-planeforces. Thus, it is the in-plane displacements of particles that are primarilyaffected by the rotation of the disk. For a plate, the in-plane displacement ofan arbitrary particle is related to the transverse displacement of the middlesurface. Recall that Kirchhoff’s hypothesis relates the displacements ur, uθ

and uz to the displacements of the middle surface u, v and w as follows

uz = w(r, θ),

ur = u(r, θ)− z∂w

∂r,

uθ = v(r, θ)− z

r

∂w

∂θ. (85)

Thus, if ur or uθ changes, so must w. For a membrane, the displacementsof the middle surface are the displacements of the surface itself and are thusindependent of each other. Therefore u and v may change without affectingw and vice-versa.

It should also be noted that the rotary inertia has automatically been takeninto account in both the linear and nonlinear formulations of the problem. Theterm representing the effect of the rotary inertia of the disk is proportional to∇2w. To ignore the effect of rotary inertia, it suffices to drop these terms fromthe equation for transverse vibrations and from the corresponding boundarycondition.

7.8.4 Presence of New Terms

As previously observed, the equation of motion for the linear transverse vibra-tions featured a term proportional to Ω2∇2w while the corresponding bound-ary conditions also contain a term proportional to Ω2. The physical origin ofthis term will now be examined by examining its origin in Hamilton’s principle.

Suppose that the in-plane displacements and rotary inertia are ignored.Then the kinetic energy of the rotating plate becomes

KE =

∫ 2π

0

∫ R2

R1

ρh

(∂w

∂t

)2

+ρh3Ω2

3∇w · ∇w r dr dθ. (86)

The variation of the ∇w · ∇w gives rise to the new terms in the equation ofmotion and in the boundary condition. Where does this term come from?

First note that due to the presence of the h3, this term will not arise in themembrane problem. Furthermore, it turns out that the term in question is a

30

consequence of the use of Lagrangian coordinates. To see this, consider thevelocity of any element of the spinning plate. The contribution to the velocityof the element due to the rotation of the plate is ω×r, where r = ro +u. Nowconsider ω × u, where we consider the contribution to u from the transversedisplacement only. In other words, take

u =

[−z

∂w

∂r,−z

r

∂w

∂θ, w

]. (87)

Since ω = Ωk, it follows that

ω × u = −z ω ×∇w (88)

and(ω × u) · (ω × u) = z2Ω2∇w · ∇w, (89)

which explains the presence of the term in question in the kinetic energy. Itarises as a consequence of the contribution to velocity due to the rotation ofthe disk and the use of Lagrangian coordinates.

But it is known that the ω × r is an in-plane term eventually giving riseto the centrifugal force. Why does it crop up in the equation of transversevibrations? The answer lies in closer examination of equation (88). This isindeed an in-plane term. However, it is linear in z and thus gives rise to abending moment. In other words,

∫ h

−hω × u z dz gives a non-zero contribu-

tion. If the equations of motion were to be derived in the Newtonian way (forexample, see [8]) the equations summing the moments are used to simplifythe equations summing the in-plane and transverse forces. In this way, thebending moment due to the ω×u term would eventually make its way to theequation expressing the balance of forces in the transverse direction.

In short, the presence of the new Ω2 terms in the equation of transversevibrations and its corresponding boundary condition reflects the contributionof the bending moment due to the ω × u term. It is only relevant for plates(as opposed to membranes). It is also a consequence of the use of Lagrangiancoordinates.

8 Example: In-plane Vibrations of a Spinning

Disk

Consider a typical point P in a disk which is rotating about its polar axis withconstant angular velocity Ω. The position of P is defined by polar coordinates

31

(r, θ) which are measured with respect to axes fixed in the disk. Assumethat the motion of a particle in the disk only occurs in the plane of the diskand is given by u = (ur, uθ)

T , where ur and uθ are the radial and tangentialdisplacements respectively. For small displacements, linear stress-strain andlinear strain-displacement relationships can be assumed. Furthermore, for athin disk plane stress conditions can be assumed. The preceding assumptionswere shown in the preceding section to lead to the following equations of motionfor the in-plane vibrations of a spinning disk, rewritten here using operatornotation:

Lu = u−Ω2u+2ΩDu−1

ρF (90)

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

L =E

ρ(1− ν2)

(L11 L12

L21 L22

)(91)

L11 =∂2

∂r2+

1

r

∂

∂r− 1

r2+

(1− ν)

2r2

∂2

∂θ2(92)

L12 =(1 + ν)in

2r

∂

∂r− (3− ν)

2r2

∂

∂θ(93)

L21 =(1 + ν)in

2r

∂

∂r+

(3− ν)

2r2

∂

∂θ(94)

L22 =(1− ν)

2

(∂2

∂r2+

1

r

∂

∂r− 1

r2

)− 1

r2

∂2

∂θ2. (95)

The operator Ln is derived from L by setting ∂∂θ

= in, where i =√−1. In

the above, E is Young’s modulus, ν is Poisson’s ratio and ρ is the density ofthe disk. Furthermore, D,u and F are given by

D =

(0 −11 0

)(96)

u =

(ur

uθ

)(97)

F =

(Fr

Fθ

), (98)

where (Fr, Fθ)T 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

32

and strain-displacement relationships, these equations can be written in termsof ur and uθ as

∂ur

∂r+

ur

r

(ur +

∂uθ

∂θ

)= 0 (99)

1

r

∂ur

∂θ− uθ

r+

∂uθ

∂r= 0 (100)

Note that for the remainder of what follows, the following inner product willbe used :

〈u1,u2〉 =

∫ a

0

(u?1)

T u2 rdr, (101)

where the ? denotes the complex conjugate.

8.1 Self-Adjointness of the Operator Ln

The in-plane vibrations of a spinning disk are a conservative system and theelastic operator can be shown to be self-adjoint, even though it does notfall into the same class of operators as discussed by Leipholz [10]. The self-adjointness is an important property as it leads to the orthogonality of eigen-functions, in turn implying that the eigenfunctions can be used as a basis tofor a general solution to the problem. The importance of Hamilton’s principleis in arriving at the correct combination of differential operator plus boundaryconditions in order to ensure this self-adjointness.

Lemma 8.1. Ln is a self adjoint operator in the space of functions that satisfy

the boundary conditions (77) and (78) and that are of the form[

u iv]T

where u and v are real functions. Here i =√−1.

Proof. Only functions of the form uj = [ uj ivj ]T where uj and vj are real

33

will be considered. Consider

〈u2, Lnu1〉 =

∫ a

0

[ u2 −iv2 ]Ln

[u1

iv1

]rdr (102)

=

∫ a

0

(u2 +

(1 + ν)

2nv2

)∂u1

∂r+ ru2

∂2u1

∂r2+

(1− ν)

2rv2

∂2v1

∂r2

+1

2((1− ν)v2 − (1 + ν)nu2)

∂v1

∂r+

((3− ν)

2rnv1 −

u1

r− (1− ν)

2rn2u1

)u2

+

((3− ν)

2rnu1 −

(1− ν)v1

2r− 1

rn2v1

)v2 dr (103)

=

[((1 + ν)

2nv2 − r

∂u2

∂r

)u1 + ru2

∂u1

∂r−

((1 + ν)

2nu2 +

(1− ν)

2r∂v2

∂r

)v1

+(1− ν)

2rv2

∂v1

∂r

]∣∣∣∣a0

+

∫ a

0

u1∂u2

∂r+ ru1

∂2u2

∂r2− (1 + ν)

2nu1

∂v2

∂r

+(1− ν)

2v1

∂v2

∂r+

(1− ν)

2rv1

∂2v2

∂r2+

(1 + ν)

2nv1

∂u2

∂r+

(3− ν)

2rn(v1u2 + u1v2)

− (1− ν)

2r

(n2u1u2 + v1v2

)− 1

r

(u1u2 + n2v1v2

)dr (104)

It may also be verified that the integral (i.e. non boundary) portion of

equation (104) is equivalent to 〈u1, Lnu2〉 =∫ a

0[ u1 −iv1 ]Ln

[u2

iv2

]rdr.

Hence it follows that〈u1, Lnu2〉 = 〈u2, Lnu1〉 (105)

provided that the boundary term in equation (104) disappears. Thus theoperator is self-adjoint provided that

0 =(1 + ν)n

2

∣∣∣∣ u1 u2

v1 v2

∣∣∣∣r=a

− (1 + ν)n

2

∣∣∣∣ u1 u2

v1 v2

∣∣∣∣r=0

− a

∣∣∣∣ u1 u2∂u1

∂r∂u2

∂r

∣∣∣∣r=a

− a(1− ν)

2

∣∣∣∣ v1 v2∂v1

∂r∂v2

∂r

∣∣∣∣r=a

. (106)

Now suppose that each of u1 = [ u1 iv1 ]T and u2 = [ u2 iv2 ]T satisfythe boundary conditions, equations (77) and (78). That is, for j = 1, 2 the

34

following are true

∂uj

∂r+

ν

r(uj − nvj)

∣∣∣∣r=a

= 0 (107)

nuj

r− vj

r+

∂vj

∂r

∣∣∣∣r=a

= 0. (108)

It then follows that equation (106) reduces to

−(1 + ν)n

2

∣∣∣∣ u1 u2

v1 v2

∣∣∣∣r=0

= 0. (109)

Hence, provided that u1and u2 satisfy the boundary conditions and equation(109) is satisfied, the operator Ln is self-adjoint.

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 ourattention is confined to functions of this form.

9 Conclusion

In summary, Hamilton’s principle has been derived, in its classical form, ex-tended form and also in an extension that permits application to systems ofvariable mass. It is a very powerful principle in that it will yield the cor-rect equations of motion, along with the corresponding boundary boundaryconditions. Examples of such uses have been demonstrated and in particular,for rotating systems, the proper use of inertial velocities and accelerations isensured.

35

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[2] A. Wang, Z. Zhang, and F. Zhao. Stability analysis of viscoelastic curvedpipes conveying fluid. Applied Mathematics and Mechanics, 26(6):807–813, 2005.

[3] Y.C. Fung. Foundations of Solid Mechanics. Prentice-Hall, Inc., Engle-wood Cliffs, New Jersey, 1965.

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

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

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

[7] H Lamb and R.V. Southwell. The vibrations of a spinning disk. Proceed-ings of the Royal Society of London, Series A, 99:272–280, 1921.

[8] A. Leissa. Vibrations of Plates. NASA SP-160, Washington, D.C., 1969.

[9] H.N. Chu and G. Herrmann. Influence of large amplitudes on free flexuralvibrations of rectangular elastic plates. Journal of Applied Mechanics,23:532–540, 1956.

[10] H.H.E. Leipholz. On an extension of hamilton’s variational principle tononconservative systems which are conservative in a higher sense. Inge-nieur Archiv, 47:257–266, 1978.

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