Hamilton’s Principle for the Derivation of Equations of Motion Natalie Baddour nbaddour@uottawa.ca May 30, 2007 Abstract Hamilton’s principle is one of the great achievements of analytical mechanics. It offers a methodical manner of deriving equations motion for many systems, with the additional benefit that appropriate and correct boundary conditions are automatically produced as part of the derivation. It allows insight into the manner that the system is modeled, as any modelling assumptions are clear and the effects of changing basic system properties become apparent and are accounted for in a consistent manner. Simplifications may also be made and Hamilton’s principle can be used as the basis for an approximate solution. Classical mechanics dictates that Hamilton’s principle can only be used for systems that are always composed of the same particles. This has been more recently extended to include systems whose constitutent particles change with time, including open systems of changing mass. In this chapter, we review the principle and its extended version and show through application to examples how it can lead to insightful observations about the system being modelled.
1
Introduction
One of the great accomplishments of analytical mechanics, Hamilton’s variational principle has found use in many disciplines, including optics and quantum mechanics. The development of the equations of mechanics via a variational principle allows the use of powerful approximation techniques for the 1
solution of problems that may not be otherwise solvable. For example, the Rayleigh-Ritz method has found much use in the solution of mechanics problems. For the sake of completeness and to establish the notation, the principle is first derived in its classical form. Subsequent to this, extensions and applications of the classical principle are presented.
2
Hamilton’s Principle - Classical Theory
Hamilton’s principle and the extended Hamilton’s principle permit the derivation of the equations of motion from a definite integral involving kinetic energy and the virtual work performed by the applied forces. Both the virtual work and the kinetic energy are scalar functions. From d’Alembert’s principle for a system of n particles, n X i=1
d2 ri ∂V − Fi · δri = 0 mi 2 + dt ∂ri
(1)
where V = V (r1 , r2 , ..., rn ) is the potential energy of the particles, Fi denotes the non-conservative applied forces acting on the ith particle, ri is the position vector of the particle of the mass mi and δri is a virtual displacement. The notation δ implies a variation in the system - an imaginary change of configuration that complies with the system constraints. The variation is only on r and its derivatives; the space and time parameters, usually denoted with xi and t, are not affected by the variation, either directly or in the ranges of integration. Where the system is prescribed, the variation must be zero. If the configuration of the system is prescribed then the variation must necessarily be be zero because otherwise any change in configuration would result in a configuration that is not possible. It should be noted that the variation of potential energy of the system is given by n X ∂V · δri (2) δV = ∂r i i=1 and the variation of the work done by non-conservative forces is δW =
n X i=1
2
Fi · δri .
(3)
Therefore, by the product rule " n # n n X X d X dri d2 ri dri dri mi · δri = mi 2 · δri + mi ·δ dt i=1 dt dt dt dt i=1 i=1 n n X X mi dri dri d2 ri · = mi 2 · δri + δ dt 2 dt dt i=1 i=1 n X d2 ri = mi 2 · δri + δT, dt i=1
(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, into the original equation of motion (1), then d’Alembert’s principle becomes " n # d X dri δL + δW = mi · δri . (5) dt i=1 dt The preceding equation is for a discrete system and allows for straightforward modification for a continuous system as Z d (ρu) · δr dV (6) δL + δW = dt V where ρ is the mass density of the system, u = dr is the velocity field of the dt system, L is the Lagrangian of the continuous system and δW is the virtual work performed on the system by the non-conservative forces undergoing a virtual displacment. The subscript V indicates the fixed volume containing the material in the system, over which the integration is performed. The next step to obtaining Hamilton’s principle is to integrate either of the previous two equations with respect to time, from time t1 to time t2 , giving Z t2 Z t2 Z t2 δL dt + δW dt = (ρu) · δr δv . (7) t1
t1
v
t1
The motion of the system is defined by the position vector of each particle given as a function of time, t. For a system of N particles, at any time t, each point has its own 3 dimensional position vector so that the state of the entire system is a point in 3N dimensional space known as configuration space. As 3
time unfolds, the motion of the entire system of particles traves a curve in the configuration space called the true path. A different path, known as the varied path, results from imagining the system as moving through configuration space by a slightly different path defined by the virtual displacement δri . Of all the possible paths through configuration space, we consider only those that coincide with the true path at times t1 and t2 . thus, the configuration of the system is given at times t1 and t2 , and it follows that δr =0 at those two times. Under those conditions, the last term in equation (7) becomes zero, giving the extended Hamilton’s principle as: Z t2 (δL + δW ) dt = 0. (8) t1
The extended Hamilton’s principle is very general and can be used to derive the equations of motion for many mechanical systems, as well as the corresponding correct boundary conditions by performing the required variations as given by equation (8). The principle is valid for rigid bodies, particles or deformable bodies. However, the virtual displacements must be reversible, implying that the constraint forces must do not work. Importantly, the principle cannot be used 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 Z t2 δL dt = 0. (9) t1
Any system where the constraints can be expressed as relations between the coordinates alone is known as holonomic. For holonomic systems, the varied path is a possible path. However, if the constraint equations cannot be written so that they involve the coordinates alone, then the varied path is in general not a possible path. The difference between these two cases is important since the integration and variation operations are interchangeable for holonomic systems. Thus for holonomic systems, Hamilton’s principle for conservative systems can be further simplified using the interchangeability of integration and variation as Z t2 δ L dt = 0. (10) t1
Equation (10) is the most familiar version of Hamilton’s principle. It can be interpreted as stating that the actual path taken by the system through 4
Rt configuration space is such that the value of the integral t12 Ldt is stationary with respect to all possible variations of the path between the two instants t1 and 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 of the particles in the system at any instant and V is the corresponding potential energy. δW is the virtual work performed by generalized forces undergoing generalized virtual displacements. If the system is a non-conducting linear elastic solid, then the potential energy V is most usually the strain energy. The virtual work is then the contribution from the non-conservative body forces and surface stresses undergoing a virtual displacement. If thermodynamic effects are included then the potential energy includes the internal energy and the virtual work will include contributions due to to virtual temperature changes at the boundary [1].
3
Characteristic Features of Some Conservative Systems
Conservative systems are described by differential equations whose operators are self-adjoint with respect to the boundary conditions. The self-adjointness implies the conservation of energy, a fact which shall be demonstrated below. This is useful in that the energy functional then assumes an extremum for the solution of the system, thus providing the basis for a variational solution. For a large class of elastic systems, the application of Newton’s law yields an equation of motion of the form w(x, ¨ t) + Gw(x, t) + qSw(x, t) = 0,
(11)
with corresponding boundary conditions [U w(x, t)]B = 0.
(12)
For the given elastic system, w(x, t) is the deflection of the system with the spatial coordinate given by x and time given by t. G is a self-adjoint linear differential operator that acts with respect to x and represents the elastic forces in the system. The variable q is a load parameter while S is another self-adjoint linear differential operator with respect to x. In equation (12), U is also a linear differential operator with respect to x, evaluated at the boundaries of the system, as denoted by the subsrcipt B. 5
The self-adjointness of the operators G and S is defined such that the following hold: Z Z (Gu)v dV = (Gv)u dV V V Z Z (Su)v dV = (Sv)u dV. (13) V
V
Here, u and v must be admissible functions, namely functions satisfying the boundary 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 Z 2 E= w˙ + (Gw + qSw)w dV. (14) V
The derivative of the energy E with respect to time is given by Z ˙ [2w˙ w¨ + (Gw˙ + qS w)w ˙ + (Gw + qSw)w] ˙ dV. E=
(15)
V
Given the self-adjointness of the operators G and S, the preceding equation becomes Z ˙ [w ¨ + Gw + qSw] w˙ dV. (16) E=2 V
˙ Clearly, from equation (11), it follows that E=0 and the energy of the system must be a constant. Thus, the use of the self-adjoint nature of the operators G and S led to the conservation of energy. Now consider the functional given by Z t2 Z 2 H= w˙ − (Gw + qSw)w dV dt. (17) t1
V
The variation of this functional is given by Z t2 Z δH = [2wδ ˙ w˙ − (Gδw + qSδw)w − (Gw + qSw)δw] dV dt. t1
(18)
V
Integration by parts gives Z Z t2 Z t2 δH = 2wδw| ˙ [2wδw ¨ + (Gδw + qSδw)w + (Gw + qSw)δw] dV dt. t1 − V
t1
V
(19) 6
Imposing the usual constraint of no variation at the temporal endpoints implies that δw(t1 ) = δw(t2 ) = 0. Furthermore, we once again make use of the selfadjointness of the operators G and S to obtain Z t2 Z 2 [wδw ¨ + (Gw + qSw)δw] dV dt. (20) δH = − t1
V
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 in order 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 displacements, then the principle of work and energy can be derived from the equation of virtual work. If the system is in motion at t1 and t2 , then its state is not given at those times and the virtual displacements cannot be zero at those instants. Since we have allowed the virtual and actual displacements to co˙ incide, it then follows that δr = rdt, and more generally that the variation of a 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, up until the integration from time t1 to t2 , which is restated here: #t2 " n Z t2 Z t2 X dr i t · δr i (22) δLdt + δW dt = mi d t1 t1 i=1 t1
The variation δr i is no longer zero at the endtimes as for the derivation of Hamilton’s principle. However, we now make use of the fact that we have chosen the virtual and actual displacements to be the same in order to simpify each term in the preceding equation. Let the virtual displacement dr i take place during an infinitesimal time dt = ε. For the first term, we recall that the variation becomes a differential. Thus
t2 Z t2 Z t2 Z t2
dL δLdt = dL dt = ε tdt = εL
. (23) d t1 t1 t1 t1 7
Similarly Z
t2
Z
t2
t2
dW dt =
δW dt = t1
Z
t1
t1
t2
dW ε tdt = εW
. d t1
(24)
The third term can be written as
t2 " n #t2 " n #t2
X X dr i
˙ · rε ˙ mi t · δr i = (mi r) = ε2T
d i=1 i=1 t1
t1
(25)
t1
where T is the kinetic energy of the system. Given these three simplifications, the original equation becomes after dividing through by ε L|tt21 + W |tt21 = 2T |tt21 .
(26)
Recalling that L = T − V , this simply states that ∆(T + V ) = ∆W . Dividing both sides by ∆t and taking the limit as ∆t → 0, it then follows that dW d(T + V ) = . dt dt
(27)
Clearly if there is no non-conservative work done on the system then W = 0 and 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 is equal to the rate at which non-conservative work is done on the system. These developments 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 open system by McIver [1]. The primary contribution of his work was to incorporate the 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 through Lagrangian or through Eulerian coordinates. The Lagrangian description is a way of looking at motion where the observer follows the motion of individual fluid particles as they move through space and time. This corresponds to the way the world is described in analytical mechanics as the description of the motion 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 particles but rather focusses on a specific area of space through which the particles move. 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 is defined as a collection of particles that are of interest. The system boundaries are such that the same particles are always contained within the system. The mass of the system is thus constant. The concept of a control volume is defined as a clearly delineated, although imaginary, space through which the particles may move. The external boundary of the control volume is called the control surface. The boundaries of the control volume are given at all times. A control volume may be chosen to be some physically meaningful boundary so that although the control volume is always given, it is allowed to move. Particles may move in and out of the control volume and thus the control volume may not be of constant mass. If it is of constant mass, then it need not always consist of the same set of particles. Briefly put, Reynolds transport theorem states that the rate of change of an extensive property, N within the system is equal to the rate of change of N within the control volume and the net rate of change of N through the control surface - the net flux through the control surface. Mathematically, this becomes Z Z Z ∂ d (ρN )dV = (ρN )dV + (ρN )u · nds. (28) dt system control vol ∂t control surf Here, N is the property of interest per unit mass, ρ is the density, dV and ds are the differential volumes and surface area elements, and u = u(r, t) is the system velocity at any point on the control surface. Thus ρ(u · n) is the mass flow rate across a differential element of the control surface. Reynold’s transport theorem essentially says that the net rate of change of any property of interest within the system is equal to the change within the control volume plus the net flux of N across the boundaries of the control volume. The unit normal 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 leaving the control volume, in order to give the next flux across the control surface. 9
If any part of the control surface is itself moving, then this needs to be subtraced so that the interpretation of the net flux across the control surface is maintained. Thus, if the control surface is moving, then u − v control replaces u 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 boundaries may not necessarily be well defined. In the classical version of Hamilton’s principle, the system contains the same material elements at all times. With the use of Reynold’s transport theorem, McIver generalized the analysis by making use of control volumes where material is allowed to cross the control volume boundary. In particular, with the use of Reynold’s transport theorem, we can write Z d (ρu) · δrdV δLsystem + δW = dt Z Z ∂ (ρu) · δr(u − v control ) · n (29) ds = (ρu) · δrdV + control vol ∂t control surf where the Lagrangian Lsystem is the Lagrangian of the open control volume and thus its mass is not necessarily constant. This is the statement of the principle of virtual work, generalized to the case of open control volumes where the enclosed mass may change as a function of time. As for the classical Hamilton’s principle, it is assumed that the system configuration is given at times t1 and t2 so that the variation of the system at those times is zero. Integrating with respect to time from t1 to t2 gives Z t2 Z t2 Z t2 Z δLsystem dt + δW dt − dt (ρu) · δr(u − v control ) · n ds = 0. t1
t1
t1
control surf
(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. The last integral may be considered to be the virtual momentum transport across the open control surface. If the virtual non-conservative work arises from surface stresses over the open and closed boundaries of the control surface, it
10
then follows that δW = δWclosed CS + δWopen CS Z Z = (σ · n) · δr ds + closed CS
(σ · n) · δr ds
(31)
open CS
where σ is the stress tensor. The extended Hamilton’s principle then becomes Z t2 Z Z t2 dt (σ · n) · δr ds + δLsystem dt + t1 closed CS t1 Z t2 Z dt [(σ · n) · δr − (ρu) · δr(u − v control ) · n ]ds = 0. (32) + t1
open CS
Here, L = T − V with Z Tsystem = Zcontrol vol
1 ρu · u dV 2 ρ e dV
Vsystem =
(33)
control vol
where e is the potential energy per unit mass. For systems including structures and fluids, the Lagrangian must include both the structure and the fluid. The open control surface represents the open portion of the control surface through which fluid is permitted to flow. The closed section of the control surface is one through which there is no flow, such as at a solid boundary.
5.3
Principle of Work and Energy for a System of Variable Mass
As for the derivation of the classical principle of work and energy given above, replacing the virtual displacements with the actual displacements allows the energy equation to be derived from the virtual work equation. The derivation proceeds along similar lines, by starting with the principle of virtual work, where now Reynold’s transport theorem is also used: Z d δLsystem + δW = (ρu) · δrdV dt Z Z ∂ = (ρu) · δrdV + (ρu) · δr(u − v control ) · n(34) ds . control vol ∂t control surf 11
As before, the virtual displacement is replaced with the actual displacement ˙ = udt, which is akin to replacing the variation operator so that δr = dr = rdt with the differential operator. Z Z ∂ (ρu)·udt(u−v control )·n ds = 0. (ρu)·u dt dV − dLsystem +dW − control surf control vol ∂t (35) Dividing through by dt gives Z Z d dW ∂ 2 Lsystem + − (ρu )dV − (ρu2 )(u − v control ) · n ds = 0. dt dt control vol ∂t control surf (36) At the same time, Reynold’s transport theorem can also be applied to find the time derivative of the Lagrangian L = T − V : Z d 1 2 d Lsystem = ρu − ρe dV dt dt system 2 Z Z ∂ 1 2 1 2 = ρu − ρe dV + ρu − ρe (u − v control ) · n (37) ds 2 2 control vol ∂t control surf Substituting into the previous equation yields Z Z ∂ 1 2 dW 1 2 − ρu + ρe dV + − ρu + ρe (u−v control )·n ds = 0. 2 dt 2 control vol ∂t control surf (38) The first term can be clearly identifed as the change in total energy within the control volume, so that the preceding equation can be re-written as: Z ∂ dW 1 2 (T + V )control vol = − ρu + ρe (u − v control ) · n ds . (39) ∂t dt 2 control surf This equation states that the change in energy within the control volume is equal to the rate at which non-conservative work is done plus the gain or loss of energy by virtue of the flow through the control surface and/or the moving control surface engulfing additional particles. This is the principle of work and energy for a system of variable mass. In the following, some examples of the use of Hamilton’s principle to derive the equations of motion of a system are now presented.
12
6
Example: Flow in a Viscoelastic Curved Pipe
An example of Hamilton’s principle for systems with variable mass is now presented. This was originally derived in [2]. Fluid flowing through a viscoelastic circular pipe is considered. Let ur (θ, t) and uθ (θ, t) represent the displacement variables along the radial and tangential directions, respectively. The radius of the circular pipe is given by r while θ is the angular coordinate, α is the angular size of the section of pipe that is being considered, s is arclength along the centreline of the pipe, A is the cross-sectional area of the pipe (fluid), I is the moment of inertia of a cross-section of pipe, U is the constant-magnitude flow velocity of the fluid relative to the pipe wall, ms and mf are the masses per unit length of the pipe and fluid respectively, and finally m = ms + mf is the total mass per unit length of the pipe-fluid system. Curved pipe flow is clearly a non-conservative system, therefore Hamilton’s principle extended for systems of changing mass is required. Hamilton’s principle now becomes Z Z t2 ∂R + U τ · δRdt = 0, (40) δ Ldt − mf U ∂t t1 where L = Ts +Tf −Vs −Vf is the Lagrangian with s and f subscripts denoting structure and fluid, respectively. The position vector of the deformed pipe centreline is given by R and τ is a unit vector tangential to the free end of the deformed pipe centreline. The strain energy stored in the pipe due to bending is given by 2 Z α 1 ∂ Jz ∂ 2 u r ∂ ur ∂uθ + ur + Vs = E+η dθ, (41) ∂t r3 ∂θ2 ∂θ ∂θ2 0 2 where Jz is the polar moment of inertia of the pipe when the curved radius of the pipe is sufficiently greater than the cross-sectional radius of the pipe, E is Young’s Modulus and η is the viscosity of the visco-elastic pipe, assuming a Kelvin-Voigt model. The kinetic energy of the pipe is given by " 2 2 # Z α 1 ∂ur ∂uθ r dθ (42) Ts = ms + ∂t ∂t 0 2 If the fluid is assuming to be incompressible and there is no gravitational potential energy, then the fluid potential energy is given by Vf = 0. 13
(43)
Let ua denote the inertial velocity of the fluid so that the fluid kinetic energy is given by 2 2 Z α Z α" 1 1 ∂ur ∂uθ ∂uθ 2 Tf = mf ua · ua rd θ = md U + + + 2U 2 ∂t ∂t ∂t 0 2 0 # 2 U ∂ur ∂ur U 2 ∂ur + 2 + uθ + 2 + uθ rd θ. (44) r ∂θ ∂t r ∂θ Assuming that the centreline of the pipe is inextensible, then the radial displacement can be related to the tangential displacement via ur =
∂uθ . ∂θ
(45)
Let dimensionless variables be introduced so that uθ w= , r
1/2 EI β= m 1/2 1/2 η 1 EI u = Ur ρA . , H= 2 EI r m E t τ= 2 r
ρA , m
(46)
Putting all these together and performing the variation as indicated, the Hamilton’s Principle extended to the pipe system with changing mass gives the equation of motion as 4 p ∂7w ∂6w ∂5w ∂4w ∂4w 2∂ w + + 2H + 2 + u + 2 βu ∂θ6 ∂τ ∂θ6 ∂θ4 ∂τ ∂θ4 ∂θ4 ∂θ3 ∂τ p ∂3w ∂2w ∂2w ∂4w ∂2w + 2 2 + H 2 + (1 + 2u2 ) 2 + 2 βu + u2 w − = (47) 0. ∂θ ∂τ ∂θ ∂τ ∂θ ∂θ∂τ ∂τ 2
H
7
Example: Equations of motion of a Spinning Disk
In this section, use of Hamilton’s principle is demonstrated in deriving the equations of motion of a spinning disk. It will be shown that use of Hamilton’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 derivation. Hamilton’s principle yields the equations of motion for the transverse as well as in-plane vibrations and ensures correct boundary conditions in order to obtain a self-adjoint linear operator. In order to use Hamilton’s principle, expressions for kinetic energy, elastic strain energy and work done must be formulated. For continuous systems this is done by considering a small element of volume and then integrating over the 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 solid loooking different at different points in time. Care must therefore be taken when integrating over the entire volume. The question arises as to whether the integration should be performed over the current volume or over the initial volume. Since the current volume is usually an unknown, this is best addressed by referring all quantities to the initial volume and then performing the integration over the initial volume of the solid. In other words, Lagrangian and not Eulerian coordinates must be used. Usually, it is desirable to have the equations of motion formulated in terms of displacements. If this is the case, then the kinetic and strain energies of the system must be formulated in terms of displacements.
7.1
Strain Energy of a Spinning Disk
Assuming that a strain energy exists is akin to making a fundamental assumption about the material and how it behaves. If the form for the strain energy of any material is known, then the stress-strain relationship for the material may be deduced. Conversely, if the stress-strain relationship of the material is known then so is the form of the strain energy. There is a nonzero contribution to the strain energy stored in a deformable body only if the body bends, stretches or otherwise deforms. If it rotates or translates like a rigid body without deforming then such displacements will not contribute to the strain energy of the body, although they will contribute to the kinetic energy. In measuring the displacements of particles in the body for the purposes of calculating strains, it is imperative to measure them with respect to the rigid body motion. This is accomplished by fixing a coordinate frame to the body. The translation and rotation of this body-fixed frame describe the rigid body motions of the body. Displacements with respect to this frame then contribute to the strain energy stored in the body. Since the displacements are measured 15
with respect to the undeformed (not unrotated) body, the discussion for the derivation of the strain energy proceeds in the same manner as for non-rotating bodies. 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 in the body. • An expression giving stress in terms of strain (stress-strain expression). • An expression giving the strain in terms of the displacements (straindisplacement expression). The above would yield energy expressions in terms of displacements of any point in the body. In thin plate theory, the displacements of an arbitrary point are further related to the displacements of the middle surface of the plate via Kirchhoff’s hypothesis. The problem thus reduces to one of solving for the deflections of one particular surface only. Since the ensuing deflections do not depend on any vertical component, this serves to turn a three-dimensional continuum mechanics problem into one of two dimentions. Hence, four relationships are required to express the strain energy in terms of the displacements of the middle surface as measured in the body-fixed frame. The strain energy per unit volume is denoted by W0 and is given by 1 W0 = Ďƒij ij , 2
(48)
where Ďƒij and ij are the stress and strain tensors respectively. To find the total strain energy, this expression must be integrated over the entire volume of the body. Small strains are assumed so that stress-strain relationship will be taken to be Hookean (linear). Note that this implies that the strains are small, but does not imply that the displacements are small. For a flat plate, a plane stress condition may be assumed. This implies that Ďƒzz = Ďƒrz = ĎƒÎ¸z = 0. Thus for a plane stress condition, the strain energy per unit volume can be explicitly written as 1 Ďƒij ij 2 2G(Îť + G) = ( rr + θθ )2 + 2G 2rθ − rr θθ . Îť + 2G
W0 =
16
(49)
The strain-displacement relation is required in order to express the strain energy in terms of displacements. It is at this point in the derivation that nonlinearities may be introduced into the modelling. Since all quantities are to be referred to the undeformed body, the lagrangian form of the strain tensor that is required. The Von Karman plate theory can be shown [3] to lead to the following nonlinear strain-displacement expressions 2 ∂ur 1 ∂uz + , (50) rr = ∂r 2 ∂r 2 1 ur 1 ∂uθ ∂uz , (51) + + 2 θθ = r r ∂θ 2r ∂θ 1 ∂ur ∂uθ ∂uz ∂uz rθ = − uθ + r + , (52) 2r ∂θ ∂r ∂r ∂θ where ur , uθ and uz are the displacements of the disk in the r, θ and z directions, respectively. For the linear Kirchoff theory, the nonlinear terms involving uz are dropped from the above expressions, leading to linear straindisplacement relationships. The last required expressions are those relating the displacements of an arbitrary point in the plate to those of the middle surface of the plate. In thin plate theory, it is usually hypothesized that the linear filaments of the plate initially perpendicular to the middle surface remain straight and perpedicular and do not contract or extend. Transverse shear effects are thus neglected. This assumption leads to a relationship between the displacements of an arbitrary point ur , uθ and uz and the displacements of the middle surface u, v and w. They are given by uz = w(r, θ), ∂w , ∂r z ∂w = v(r, θ) − . r ∂θ
ur = u(r, θ) − z uθ
(53)
All the required expressions have now been assembled. and the strain energy of the entire plate can be obtained by integrating over the entire volume of the plate. The strain energy will be a function of u, v and w and of the vertical 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 explicit function of z. This dependence can be eliminated by explicitly carrying out the integration over the thickness of the plate from z = −h to z = h, where h is the distance between the middle surface of the plate and the plate bounding surface. This procedure finally yields the strain energy of the plate as an explicit function of u(r, θ, t), v(r, θ, t) and w(r, θ, t) only. The strain energy of the plate is thus given by Z
2π
Z
R2
Wp r dr dθ
Wo = 0
Z
R1 2π Z R2
= 0
h W1 + h3 W3 r dr dθ,
(54)
R1
where W1
4 2 ∂w u2 ∂v G + (λ + G) + 4(λ + G) 2 (λ + 2G) = (λ + 2G) ∂r ∂r r 2 2 4(λ + G) ∂v (λ + 2G) ∂u v ∂u + + − 2(λ + 2G) r2 ∂θ r2 ∂θ r2 ∂θ 4 2 2 2 (λ + G) ∂w 2(λ + G) ∂w ∂w ∂u ∂w + + + 4(λ + G) r4 ∂θ r2 ∂θ ∂r ∂r ∂r 2 ∂v v ∂v 2(λ + 2G) ∂u 4(λ + G) ∂v ∂w − 2(λ + 2G) + + r ∂r r ∂θ ∂r r3 ∂θ ∂θ 2 2(λ + 2G) ∂u u ∂w ∂w ∂w + + 4(λ + G) 3 r ∂θ r2 ∂θ ∂θ ∂r 2 2(λ + 2G) ∂v ∂w ∂w u ∂v 2λ ∂v ∂w + + 8(λ + G) 2 + r ∂r ∂θ ∂r r ∂θ r ∂θ ∂r 2 u ∂u u ∂w 4λ ∂u ∂v v ∂w ∂w + 4λ + 2λ + − 2(λ + 2G) 2 r ∂r r ∂r r ∂r ∂θ r ∂r ∂θ 2 ∂u v2 + 4(λ + G) + (λ + 2G) (55) ∂r r
18
W3
2 4G(λ + G) ∂w + 3(λ + 2G)r2 ∂r 2 2 2 ∂ w 4Gλ ∂w ∂ w 4G(λ + G) ∂ 2 w + +r + 3(λ + 2G)r2 ∂θ2 ∂r ∂r2 3(λ + 2G)r4 ∂θ2 2 2 2 2 4G ∂ w ∂ w 4G(λ + G) ∂ 2 w 8G ∂w + + − 3 (56) 3r2 ∂r∂θ 3r ∂θ ∂r∂θ 3(λ + 2G) ∂r2 8G(λ + G) = 3(λ + 2G)r3
∂w ∂r
∂2w ∂θ2
4G + 4 3r
∂w ∂θ
2
Here G is the shear modulus and λ is a constant. They are related to Young’s modulus, E, and Poisson’s ratio ν of the material by Eν (1 + ν)(1 − 2ν) E G = . 2(1 + ν) λ =
7.2
(57) (58)
Kinetic Energy of a Spinning Disk
While the strain energy of a stationary and a rotating disk are the same, it is in formulating the kinetic energy expression that the difference between a rotating and stationary disk becomes apparent. Let us set up two coordinate systems, S and B. Suppose that S is an intertial frame of reference and that B is rigid-body-fixed to the disk, so that B rotates with the disk at a spin rate of Ω with respect to S. Thus, an observer in 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 be found, so it is the kinetic energy as measured in S that is required. Note however, that the strain energy is a function of displacement with respect to the undeflected body, not the undisplaced body so that the strain energy is best expressed from the point of view of an observer in B. The inertial kinetic energy as measured by an observer in S (the required quantity) needs to be expressed in terms of measurements made by an observer in B (the available quantity). The total kinetic energy is then the sum of the kinetic energy of deflection as seen by an observer in B and the kinetic energy due to the rotation of the disk. Let ro denote the undeformed location of a particle in the disk and let u 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 vectors are chosen to be expressed in terms of unit vectors belonging to the B frame. , where it must be Then the velocity of any particular particle is given by dr dt remembered that since the unit vectors are fixed in the rotating frame, their time derivative must be found as well. In fact, it is the time derivative of the rotating unit vectors that provides the portion of the overall velocity of the particle that is due to the rotation. The rest of the velocity of the particle is due to the elastic deflection only. The time derivative of the rotating unit vectors can be found be taking the cross product of the angular velocity vector with 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-fixed frame, B. Similarly, let eθ = − sin(θ)iB + cos(θ)jB be a unit vector in the θ direction, pointing in the direction of increasing θ. Furthermore, let ez be a unit vector pointing in the z direction such that er , eθ , ez form a right-handed coordinate 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 = ω × er = Ωeθ dt deθ = ω × eθ = −Ωer dt dez = ω × ez = 0. (59) dt Points within the body are represented by the polar coordinates (r, θ, z). The original position of a particle is given by ro = rer +zez . The deformed position of the same particle is given by r = ro + u = (r + ur )er + uθ eθ + (z + uz )ez ,
(60)
where ur , uθ and uz are the displacements in the er , eθ , ez directions of a particle. Each of these displacements will be a function of time and the original position of the particle in question. The velocity of this particle is given by dr = (u˙ r − Ωuθ )er + [u˙ θ + Ω(r + ur )] eθ + u˙ z ez . dt Since the unit vectors are orthonormal, the squared speed is given by v=
v · v = (u˙ r − Ωuθ )2 + [u˙ θ + Ω(r + ur )]2 + u˙ 2z . 20
(61)
(62)
Once the velocity as measured by an inertial observer of any particle has been found, the kinetic energy of a small element of volume can be expressed as 12 ρ 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 integrating over the entire undeformed volume. Note that since the velocity has been expressed as a function of the undeformed location of the particle, the integration is to be performed over the undeformed volume of the body, not the 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 and z directions respectively. As for the strain energy, equation (53) relating the displacements of an arbitrary point on the disk to the displacement of the middle surface can be used. The explicit dependence of the kinetic energy on z can be eliminated by integrating over the thickness of the disk, from z = −h to z = +h. As before, this procedure finally yields the kinetic energy of the plate as an explicit function of u(r, θ, t), v(r, θ, t) and w(r, θ, t) only. The kinetic energy of the plate is thus given by Z 2π Z R2 KE = KEp r dr dθ 0
Z
2π
R1 R2
Z
h KE1 + h3 KE3 r dr dθ,
= 0
(63)
R1
where KE1
KE3
∂u ∂v ∂v −v +r = ρΩ v + u + r + 2ru + 2ρΩ u ∂t ∂t ∂t " # 2 2 2 ∂u ∂v ∂w + ρ + + , ∂t ∂t ∂t
ρΩ2 = 3r2
2
"
∂w ∂θ
2
2
2
2
∂w ∂r
2 #
ρ + 2 +r 3r 2 2ρΩ ∂ w ∂w ∂ 2 w ∂w + − 3r ∂θ∂t ∂r ∂r∂t ∂θ 2
"
∂2w ∂θ∂t
2 +r
2
∂2w ∂r∂t
(64) 2 #
(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. Note that 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 derived by applying Hamilton’s Principle.
7.4
Nonlinear Equations of Motion
The full nonlinear equations of motion are given below. ∂v (1 − ν) ∂ 2 u (1 + ν) ∂ 2 v ∂2u ρ(1 − ν 2 ) ∂ 2 u ∂Ω 2 − Ω (r + u) − 2Ω = + −v + E ∂t2 ∂t ∂t 2r2 ∂θ2 2r ∂r∂θ ∂r2 (1 − ν) ∂w ∂ 2 w 1 (1 − ν) ∂w 2 (3 − ν) ∂v (1 + ν) ∂w 2 u ∂u − + − − + + r 2 ∂r 2r ∂θ 2r2 ∂θ r ∂r 2r2 ∂r ∂θ2 ∂w ∂ 2 w (1 + ν) ∂w ∂ 2 w + , (66) + ∂r ∂r2 2r2 ∂θ ∂r∂θ ρ(1 − ν 2 ) ∂ 2 v ∂Ω ∂u (1 − ν) ∂ 2 v (1 + ν) ∂w ∂ 2 w 2 + (r + u) − Ω v + 2Ω + = E ∂t2 ∂t ∂t 2r ∂r ∂r∂θ 2 ∂r2 1 (1 − ν) ∂w ∂w (1 − ν) ∂v (3 − ν) ∂u (1 − ν) v (1 + ν) ∂ 2 u + + − + + r 2r ∂θ ∂r 2 ∂r 2r ∂θ 2 r 2r ∂r∂θ 2 2 2 1 ∂w ∂ w (1 − ν) ∂w ∂ w 1∂ v + , (67) + 2 2+ 3 r ∂θ r ∂θ ∂θ2 2r ∂θ ∂r2 ρ(1 − ν 2 ) ∂ 2 w h2 Ω2 2 h2 ∂ 2 2 h2 4 1 ∂w 3 ∂w ∂ 2 w ∂2v + ∇ w − ∇ w = − ∇ w + + r E ∂t2 3 3 ∂t2 3 r4 ∂θ 2 ∂θ ∂θ2 ∂θ2 2 (1 − ν) r2 ∂w ∂ 2 w r ∂w ∂w (1 + ν) ∂u (1 + ν) 2 ∂ 2 u 2 ∂ w ∂w + r + r + rv + + r − 2 ∂θ ∂r 2 ∂θ 2 ∂r∂θ 2 2 ∂θ ∂r2 ∂r∂θ ∂r 2 2 (1 − ν) 3 ∂ v (1 − ν) 2 ∂v 1 ∂w 2 ∂ w ∂w (1 + ν) 3 ∂ 2 v 3 ∂u + − r r + r + (1 + ν)r + r 2 ∂r2 2 ∂r r4 ∂r ∂r∂θ ∂θ ∂r 2 ∂r∂θ 2 2 3 2 2 2 4 2 (1 − ν) 2 ∂ u r ∂w r ∂w ∂ w ∂ u 3r ∂w ∂ w (1 − ν) 2 ∂v + r + + + r4 2 + − r 2 2 2 ∂θ 2 ∂r 2 ∂r ∂θ ∂r 2 ∂r ∂r2 2 ∂θ 2 2 ∂u 1∂ w ∂v ∂ w ∂u ν ∂v + 4 2 νr2 +r u+ + 2 + u+ r ∂θ ∂r ∂θ ∂r ∂r r ∂θ 2 (1 − ν) ∂ w ∂u ∂v + − v + r . (68) r2 ∂r∂θ ∂θ ∂r 22
The full nonlinear representation of the spinning disk problem requires the solution of three nonlinear, coupled partial differential equation, namely equations (66), (67) and (68). This is the result of the use of Lagrangian coordinates as 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) straindisplacement expressions are given as : ∂Ω ∂v (1 − ν) ∂ 2 u (1 + ν) ∂ 2 v ρ(1 − ν 2 ) ∂ 2 u 2 − v − Ω (r + u) − 2Ω = + E ∂t2 ∂t ∂t 2r2 ∂θ2 2r ∂r∂θ ∂ 2 u 1 ∂u (3 − ν) ∂v u + 2 + − − , (69) ∂r r ∂r 2r ∂θ r ρ(1 − ν 2 ) ∂ 2 v ∂Ω ∂u (1 − ν) ∂ 2 v (1 + ν) ∂ 2 u 2 + (r + u) − Ω v + 2Ω + = E ∂t2 ∂t ∂t 2r ∂r∂θ 2 ∂r2 1 (1 − ν) ∂v (3 − ν) ∂u (1 − ν) v 1 ∂2v + + − + 2 2, (70) r 2 ∂r 2r ∂θ 2 r r ∂θ h2 ∂ 2 2 h2 4 ρ(1 − ν 2 ) ∂ 2 w h2 Ω2 2 + ∇ w − ∇ w = − ∇ w. (71) E ∂t2 3 3 ∂t2 3 It may be noted that equations (69) and (70) for the in-plane vibrations are the same equations derived by other authors [4, 5, 6], whereas equation (71) for the linear transverse vibrations is not. The reason for this discrepancy lies in the different assumptions built into the different models. It turns out that the equation (71) for the transverse vibrations does not accurately capture the spinning disk dynamics as the stresses induced in the disk due to its rotation are not captured with the use of linear strain-displacement relations. Interestingly, 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 the nonlinear 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 equilibrium deflection in the radial direction. It is possible that this equilibrium displacement is not small enough to justify neglecting products of these terms with derivatives of w. The resulting equations for the in-plane vibrations are identical with equations (69) and (70). However, the equation for the transverse vibrations is different as that obtained with linear strains and is now given by h2 ∂ 2 2 h2 4 ρ(1 − ν 2 ) ∂ 2 w h2 Ω2 2 ∇ w− + ∇ w =− ∇ w E ∂t2 3 3 ∂t2 3 2 2 2 1 ∂ w ∂u u ∂ w ∂u ∂w (1 + ν) ∂u ∂ u u + + 2 + 2 2 ν + +ν + 2 . (72) ∂r r ∂r ∂r r ∂θ ∂r r ∂r ∂r r However, note that corresponding to an in-plane purely radial displacement u(r), use of linear stress-strain and linear strain-displacements relationships lead to the following stress-displacement relationships : du u E +ν (73) σrr = (1 − ν 2 ) dr r E du u σθθ = ν + . (74) (1 − ν 2 ) dr r Using these relationships, equations (72) can be rewritten as 2 ∂ w h2 Ω2 2 h2 ∂ 2 2 Eh2 1 ∂ ∂w σθθ ∂ 2 w 4 ρ + ∇ w − ∇ w = − ∇ w+ . rσ + rr ∂t2 3 3 ∂t2 3(1 − ν 2 ) r ∂r ∂r r2 ∂θ2 (75) With the exception of the ∇2 w and ∇2 w, ¨ this is the same equation as obtained by Lamb and Southwell [7] for the transverse vibrations of a spinning disk. The presence of the ∇2 w¨ term is not unexpected; it is simply the term due to the rotary inertia of the disk. The physicaly meaning of the ∇2 term will be explained subsequently.
7.7
Boundary Conditions
After using the 2D analogue of integration by parts to isolate the variation of u, v and w, the remaining boundary term can be found. From this boundary term, suitable boundary conditions to the problem may be written down directly. 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 independent, the only way the entire boundary term will vanish is if the following three 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 = 0 or the coefficient of δ ∂w vanishes 4. δ ∂w ∂r ∂r The boundary term obtained from applying Hamilton’s Principle and integration by parts is given below. Note that r here must be evaluated on the boundary. Thus for a solid disk, r below is the radius of the disk. For an annulus, a set of boundary conditions is required at each of the inner and outer radii, so r will assume two possible values. " 2 2 # Z 2π u ∂u 2ν ∂v ∂w ν ∂w − 3E r3 δu 2ν + 2 + + + 2 dθ r ∂r r ∂θ ∂r r ∂θ 0 Z 2π 1 ∂u ∂v 1 ∂w ∂w 3 r δv − 3E(1 − ν) −v + + dθ r ∂θ ∂r r ∂θ ∂r 0 Z 2π 2 ν ∂ 2 w ν ∂w ∂w 3 ∂ w 2 + 2 2 + − 2Eh δ r dθ ∂r ∂r2 r ∂θ r ∂r 0 Z 2π ∂3w 3 2 ∂w 2 2 r δw Ω + 2(1 − ν )ρh − dθ ∂r ∂r∂t2 0 Z 2π (1 − ν) ∂ 2 ∂w w ∂ 2 2 3 ∇ w+ − + 2Eh r δw dθ ∂r r2 ∂θ2 ∂r r 0 " 2 Z 2π 1 − ν) ∂w ∂v 1 ∂w ∂w ∂u ∂w 3 − 3E r δw + 2 +2 r ∂θ ∂r r ∂r ∂θ ∂r ∂r 0 3 ∂w (1 − ν) ∂w ∂u 2ν ∂w ∂v + + +v + u+ dθ (76) ∂r r2 ∂θ ∂θ r ∂r ∂θ There are a few points that are worth mentioning. First, the above boundary term was obtained from the variation of the Lagrangian obtained with the nonlinear (Von Karman) strain-displacement relations. Note that the corresponding boundary conditions are also nonlinear and are coupled. Had the 25
linear (Kirchoff) strain-displacement relations been used, the corresponding boundary conditions would also have been linear. They can be obtained from the above expression by neglecting all nonlinear terms. It was previously noted that formulating the problem in this manner automatically accounts for the effect of rotary inertia in the equations of motion. ∂3w The corresponding term in the boundary condition is ∂r∂t 2 . That is, the variation of some particular part of the kinetic energy expression gives rise to the ∂3w ∇2 w¨ term in the equation of motion and to the ∂r∂t 2 term in the boundary condition. Hence, if the effect of rotary inertia is ignored (or included) in the equation of motion, then the corresponding term must also be ignored (or included) in the boundary condition. Note also that the variation of the (∇w ×∇w)·k term in the kinetic energy gives rise to boundary terms only. In other words, dropping this term from the kinetic energy expression does not change the resulting equation of motion. It does, however, change the boundary conditions. Since the equation of motion remains unchanged with the omission of this term, it is questionable whether the corresponding boundary term should be included as well. It turns out that including the questionable term leads to physically meaningless results in solving the linear vibration problem. This boundary term has thus not been included in the above expression. From equation (76), the boundary conditions for the special cases of linear in-plane and transverse vibrations can be derived.
Linear Transverse Vibrations For the linear transverse vibration problem, the boundary conditions for a free edge become ∂ 2 w ν ∂w 1 ∂ 2 w + + = 0, (77) ∂r2 r ∂r r ∂θ2 (1 − ν) ∂ 2 ∂w w (1 − ν 2 )ρ ∂ 3 w ∂ 2 2 ∂w = . (78) ∇ w+ − −Ω ∂r r2 ∂θ2 ∂r r E ∂r∂t2 ∂r Note that 2 Z h 2Eh3 ∂ w ν ∂w 1 ∂ 2 w σrr z dz = . (79) + + 3(1 − ν 2 ) ∂r2 r ∂r r ∂θ2 −h Hence, the first boundary condition (77) says that the moment at the free edge is zero and this is in agreement with boundary conditions given in the literature. 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 rotary inertia) the Kirchhoff shear is given by ∂∇2 w (1 − ν) ∂ 2 ∂w w + − = 0, (80) Vr = −D ∂r r2 ∂θ2 ∂r r 3
2h where D = 3(1−ν 2 ) is the bending stiffness of the disk. Recall that h denotes the half-thickness of the disk. This last boundary condition is bascially summing the transverse forces at the edge of the disk and setting the result equal to zero. This same expression is also usually applied to the rotating disk. Even when the term corresponding to the rotary inertia of the disk is dropped from the equation of motion and boundary condition, the derived boundary condition, equation (78) and the standard boundary condition, equation (80), do not agree. The derived boundary condition (78) differs by the inclusion of a term . This term is essentially the product of the centrifugal proportional to ρΩ2 ∂w ∂r force and the slope at the edge of the disk. Due to the presence of the Ω2 , this term would vanish for a stationary disk. It must be observed that the variation of some particular portion in the kinetic energy gives rise to the Ω2 ∇2 w term in 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 obtained naturally as part of the variation of the kinetic energy of the disk and not in an ad-hoc manner. The physical significance of these new terms will be addressed in the discussion.
Linear In-plane Vibrations The boundary terms for linear in-plane vibrations are given by ∂v ∂u ν u+ δu = 0, + ∂r r ∂θ 1 ∂u v ∂v − + δv = 0, . r ∂θ r ∂r
(81) (82)
Equations (81) and (82) must hold on the boundary of the disk. For a solid disk, 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
∂u ν ∂v 2Eh + u+ , σrr dz = ν(1 + ν) ∂r r ∂θ −h Z h 1 ∂u v ∂v σrθ dz = h − + . r ∂θ r ∂r −h Z
(83) (84)
Thus equation (81) implies that on the boundary either the displacement of the middle surface in the radial direction must be specified or the integral of the stress 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 the integral of the shear stress over the side of the disk must vanish.
7.8 7.8.1
Discussion 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 usually = 0. taken to be constant so that ∂Ω ∂t If all in-plane time-derivatives are neglected, then it is possible to use a stress function to reduce the three new equations (66),(67) and (68) to two where the generalized coordinates are the the transverse displacement and the newly-introduced stress function. This implies the omission of the in-plane 2 2 and ∂v . inertias, ∂∂t2u and ∂∂t2v , as well as the coriolis terms, ∂u ∂t ∂t We remark that the centrifugal force is not really an external force at all, but rather a consequence of the fact that the reference frame is rotating and thus non-inertial. The coriolis force is due to the same effect. In-plane inertia is typically ignored for stationary (non-rotating) plates, and this has been shown to be a good approximation for stationary plates, [9]. However, the same calculation fails for the rotating plate because of the presence of the centrifugal and coriolis forces. The validity of the approximation of rotating disk models that reduce the number of nonlinear equations to be solved from three 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 position of a given particle. A particle originally at radius r has radius (r + u) after deformation takes place. This is consistent with the Lagrangian description of the system that has been employed. In this description the boundaries of the 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 be proportional to Ω2 r. Now the location of the boundaries is an unknown ; if the disk stretches, the new location of the boundaries is part of the unknowns of the problem. Most authors use the Eulerian description Ω2 r with the boundaries (incorrectly) located at their Lagrangian (original) location, although this is most likely a negligeable difference for small in-plane displacements. However, while it may seem that for small displacements there should not be much difference between r and (r + u), Bhuta and Jones, [4] showed that the actual solutions 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 the equations for the in-plane displacements are still coupled, they are not coupled to the third equation for the transverse displacement. As for the stationary plate, the in-plane and out-of-plane free vibrations are independent of each other. It turns out, however, that this is only valid for very small rates of rotation, 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 the in-plane and transverse vibration problems are not independent. To see this, consider the three linear equations of motion (69), (70), (71) and ignore any term proportional to h2 . That is, since the membrane is assumed thin, the half-thickness h2 is considered to be small in comparison to h. While the two equations for the in-plane vibrations remain unchanged, the equation for the 2 transverse vibrations becomes ∂∂tw2 = 0, i.e. no vibrations in the the transverse direction. In essence, this means that for a membrane the linear transverse vibrations are a second-order effect; to first order, small transverse vibrations do not affect the in-plane stresses. This might lead one to wonder why the transverse vibrations of a plate are affected to first order by the rotation while the transverse vibrations of a membrane are not. The reason for this lies in the description of the inplane 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-plane forces. Thus, it is the in-plane displacements of particles that are primarily affected by the rotation of the disk. For a plate, the in-plane displacement of an arbitrary particle is related to the transverse displacement of the middle surface. 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), ∂w , ∂r z ∂w = v(r, θ) − . r ∂θ
ur = u(r, θ) − z uθ
(85)
Thus, if ur or uθ changes, so must w. For a membrane, the displacements of the middle surface are the displacements of the surface itself and are thus independent of each other. Therefore u and v may change without affecting w and vice-versa. It should also be noted that the rotary inertia has automatically been taken into account in both the linear and nonlinear formulations of the problem. The term representing the effect of the rotary inertia of the disk is proportional to ∇2 w. ¨ To ignore the effect of rotary inertia, it suffices to drop these terms from the equation for transverse vibrations and from the corresponding boundary condition. 7.8.4
Presence of New Terms
As previously observed, the equation of motion for the linear transverse vibrations featured a term proportional to Ω2 ∇2 w while the corresponding boundary conditions also contain a term proportional to Ω2 . The physical origin of this 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 Z 2π Z R2 2 ∂w ρh3 Ω2 + ∇w · ∇w r dr dθ. (86) KE = ρh ∂t 3 0 R1 The variation of the ∇w · ∇w gives rise to the new terms in the equation of motion 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 the membrane problem. Furthermore, it turns out that the term in question is a 30
consequence of the use of Lagrangian coordinates. To see this, consider the velocity of any element of the spinning plate. The contribution to the velocity of the element due to the rotation of the plate is ω × r, where r = ro + u. Now consider ω × u, where we consider the contribution to u from the transverse displacement only. In other words, take ∂w z ∂w ,− ,w . (87) u = −z ∂r r ∂θ Since ω = Ωk, it follows that ω × u = −z ω × ∇w
(88)
(ω × u) · (ω × u) = z 2 Ω2 ∇w · ∇w,
(89)
and which explains the presence of the term in question in the kinetic energy. It arises as a consequence of the contribution to velocity due to the rotation of the disk and the use of Lagrangian coordinates. But it is known that the ω × r is an in-plane term eventually giving rise to the centrifugal force. Why does it crop up in the equation of transverse vibrations? The answer lies in closer examination of equation (88). This is indeed an in-plane term. However,R it is linear in z and thus gives rise to a h bending moment. In other words, −h ω × u z dz gives a non-zero contribution. If the equations of motion were to be derived in the Newtonian way (for example, see [8]) the equations summing the moments are used to simplify the equations summing the in-plane and transverse forces. In this way, the bending moment due to the ω × u term would eventually make its way to the equation expressing the balance of forces in the transverse direction. In short, the presence of the new Ω2 terms in the equation of transverse vibrations and its corresponding boundary condition reflects the contribution of 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 Lagrangian coordinates.
8
Example: In-plane Vibrations of a Spinning Disk
Consider a typical point P in a disk which is rotating about its polar axis with constant 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. Assume that the motion of a particle in the disk only occurs in the plane of the disk and is given by u = (ur , uθ )T , where ur and uθ are the radial and tangential displacements respectively. For small displacements, linear stress-strain and linear strain-displacement relationships can be assumed. Furthermore, for a thin disk plane stress conditions can be assumed. The preceding assumptions were shown in the preceding section to lead to the following equations of motion for the in-plane vibrations of a spinning disk, rewritten here using operator notation: 1 Lu = u ¨ −Ω2 u+2ΩDu− ˙ F (90) ρ where Ω is the spin rate and L is the matrix operator E L11 L12 L= (91) ρ(1 − ν 2 ) L21 L22 1 ∂ 1 (1 − ν) ∂ 2 ∂2 − 2+ (92) L11 = 2 + ∂r r ∂r r 2r2 ∂θ2 (1 + ν)in ∂ (3 − ν) ∂ L12 = − (93) 2r ∂r 2r2 ∂θ (1 + ν)in ∂ (3 − ν) ∂ L21 = + (94) 2r ∂r 2r2 ∂θ 1 ∂ 1 (1 − ν) ∂ 2 1 ∂2 + − . (95) L22 = − 2 ∂r2 r ∂r r2 r2 ∂θ2 √ ∂ 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 of the disk. Furthermore, D, u and F are given by 0 −1 D= (96) 1 0 ur u= (97) uθ Fr F= , (98) Fθ 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 terms of ur and uθ as ∂ur ur ∂uθ + ur + =0 (99) ∂r r ∂θ 1 ∂ur uθ ∂uθ − + =0 (100) r ∂θ r ∂r Note that for the remainder of what follows, the following inner product will be used : Z a hu1 , u2 i = (u?1 )T u2 rdr, (101) 0
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 the elastic operator can be shown to be self-adjoint, even though it does not fall into the same class of operators as discussed by Leipholz [10]. The selfadjointness is an important property as it leads to the orthogonality of eigenfunctions, in turn implying that the eigenfunctions can be used as a basis to for a general solution to the problem. The importance of Hamilton’s principle is in arriving at the correct combination of differential operator plus boundary conditions in order to ensure this self-adjointness. Lemma 8.1. Ln is a self adjoint operator in the space of functions that satisfy T the boundary conditions (77) and (78) and that are of the form u iv √ 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 Z a u1 [ u2 −iv2 ]Ln rdr (102) hu2 , Ln u1 i = iv1 0 Z a (1 + ν) ∂u1 ∂ 2 u1 (1 − ν) ∂ 2 v1 = u2 + nv2 + ru2 2 + rv2 2 2 ∂r ∂r 2 ∂r 0 ∂v1 (3 − ν) u1 (1 − ν) 2 1 + nv1 − − n u1 u2 + ((1 − ν)v2 − (1 + ν)nu2 ) 2 ∂r 2r r 2r (3 − ν) (1 − ν)v1 1 2 + nu1 − − n v1 v2 dr (103) 2r 2r r ∂u2 (1 + ν) (1 − ν) ∂v2 ∂u1 (1 + ν) nv2 − r − nu2 + r = u1 + ru2 v1 2 ∂r ∂r 2 2 ∂r a Z a ∂u2 (1 − ν) ∂v1
∂ 2 u2 (1 + ν) ∂v2 u1 + + rv2 + ru1 2 − nu1
2 ∂r 0 ∂r ∂r 2 ∂r 0 2 (1 − ν) ∂v2 (1 − ν) ∂ v2 (1 + ν) ∂u2 (3 − ν) + v1 + rv1 2 + nv1 + n(v1 u2 + u1 v2 ) 2 ∂r 2 ∂r 2 ∂r 2r 1 (1 − ν) 2 − n u1 u 2 + v 1 v 2 − u1 u2 + n2 v1 v2 dr (104) 2r r It may also be verified that the integral (i.e. non boundary) of portion Ra u2 rdr. equation (104) is equivalent to hu1 , Ln u2 i = 0 [ u1 −iv1 ]Ln iv2 Hence it follows that hu1 , Ln u2 i = hu2 , Ln u1 i (105) provided that the boundary term in equation (104) disappears. Thus the operator is self-adjoint provided that
u1 u2
(1 + ν)n
u1 u2
(1 + ν)n
u1 u2
0= − − a ∂u1 ∂u2
v1 v2
v1 v2
2 2 ∂r ∂r r=a r=0 r=a
a(1 − ν)
v1 v2
− (106)
∂v1 ∂v2 . 2 ∂r ∂r r=a Now suppose that each of u1 = [ u1 iv1 ]T and u2 = [ u2 iv2 ]T satisfy the boundary conditions, equations (77) and (78). That is, for j = 1, 2 the
34
following are true
∂uj ν + (uj − nvj )
=0 ∂r r r=a
nuj vj ∂vj
− + = 0. r r ∂r r=a It then follows that equation (106) reduces to
(1 + ν)n
u1 u2
− = 0.
v1 v2
2 r=0
(107) (108)
(109)
Hence, provided that u1 and 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 our attention is confined to functions of this form.
9
Conclusion
In summary, Hamilton’s principle has been derived, in its classical form, extended form and also in an extension that permits application to systems of variable mass. It is a very powerful principle in that it will yield the correct equations of motion, along with the corresponding boundary boundary conditions. Examples of such uses have been demonstrated and in particular, for rotating systems, the proper use of inertial velocities and accelerations is ensured.
35
References [1] D.B. McIver. Hamilton’s principle for systems of changing mass. Journal of Engineering Mathematics, 7(3):249–261, 1972. [2] A. Wang, Z. Zhang, and F. Zhao. Stability analysis of viscoelastic curved pipes conveying fluid. Applied Mathematics and Mechanics, 26(6):807– 813, 2005. [3] Y.C. Fung. Foundations of Solid Mechanics. Prentice-Hall, Inc., Englewood Cliffs, New Jersey, 1965. [4] P.G. Bhuta and J.P Jones. Symmetric planar vibrations of a rotating disk. The Journal of the Acoustical Society of America, 35(7):982–989, 1963. [5] J.S. Chen and J.L. Jhu. On the in-plane vibration and stability of a spinning annular disk. Journal of Sound and Vibrations, 195(4):585–593, 1996. [6] J.S. Burdess, T. Wren, and J.N. Fawcett. Plane stress vibrations in rotating disks. Proceedings of the Institution of Mechanical Engineers, 201(C1):37–44, 1987. [7] H Lamb and R.V. Southwell. The vibrations of a spinning disk. Proceedings 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 flexural vibrations 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 to nonconservative systems which are conservative in a higher sense. Ingenieur Archiv, 47:257–266, 1978.
36