1.6 Hamilton's principle 13 d:(t2)=0 Perturbed path True path x(6)=0 Fig.1.5.True and perturbed paths. because it applies to continuous systems and more.as we will see shortly Some authors argue that being a fundamental law of physics it cannot be derived,just accepted.Thus we could have proceeded the opposite way: state Hamilton's principle,and show that it implies Newton's laws.It is a matter of taste,but also of history:150 years separate Newton's Principia om Hamilton's princip e(183 From now on,We will conside stre tha tinction b notivated here by the subs sequentextension to echanicals ms 4 Fig.1.6.Plane pendulum. Consider the plane pendulum of Fig.1.6;taking the altitude of the pivot O as reference,we find the Lagrangian L=T-V=im(10)2+mgl cos 6L=ml2066-mgl sin 060
1.6 Hamilton’s principle 13 t1 t2 îxi(t2) = 0 îxi(t1) = 0 îxi xi(t) True path Perturbed path Fig. 1.5. True and perturbed paths. because it applies to continuous systems and more, as we will see shortly. Some authors argue that, being a fundamental law of physics, it cannot be derived, just accepted. Thus we could have proceeded the opposite way: state Hamilton’s principle, and show that it implies Newton’s laws. It is a matter of taste, but also of history: 150 years separate Newton’s Principia (1687) from Hamilton’s principle (1835). From now on, we will consider Hamilton’s principle as the fundamental law of dynamics. We stress that when dealing with purely mechanical systems, the distinction between the kinetic energy T and the kinetic coenergy T∗ is not necessary; it is motivated here by the subsequent extension to electromechanical systems. ò l g m o Fig. 1.6. Plane pendulum. Consider the plane pendulum of Fig.1.6; taking the altitude of the pivot O as reference, we find the Lagrangian L = T∗ − V = 1 2 m(l ˙ θ) 2 + mgl cos θ δL = ml2 ˙ θδ ˙ θ − mglsin θδθ
14 1 Lagrangian dynamics of mechanical systems Hamilton's principle states that the variational indicator(V.I) V."m-mglsin for all virtual variations 60 such that 60(t1)=60(t2)=0.As always in variational calculus,0 can be eliminated from the variational indicator by integrating by part over t.Using0 ()-6s0 one gets VL=mlPa50账-mPi+mgls0创i0i=0 之egn0uat60=o 】term must vanish for arbit m2+mgl sin=0 which is the differential equation for the oscillation of the pendulum. The Lagrange's equations give a much quicker way of solving this type of problem,as we will see shortly;however,before this,we illustrate the power of Hamilton's principle by deriving the partial differential equation of the lateral vibration of the Euler-Bernoulli beam. 6u(r,t) (x,t) Fig.1.7.Transverse vibration of a beam 1.6.1 Lateral vibration of a beam Consider the transverse vibration of the beam of Fig.1.7,subjected to a transverse distributed load p(). ssumed that the prin cipal axes I are such the vibra e plane
14 1 Lagrangian dynamics of mechanical systems Hamilton’s principle states that the variational indicator (V.I). V.I. = Z t2 t1 [ml2 ˙ θδ ˙ θ − mglsin θδθ]dt = 0 for all virtual variations δθ such that δθ(t1) = δθ(t2) = 0. As always in variational calculus, δ ˙ θ can be eliminated from the variational indicator by integrating by part over t. Using ˙ θδ ˙ θ = d dt(˙ θδθ) − ¨θδθ one gets V.I. = [ml2 ˙ θδθ] t2 t1 − Z t2 t1 (ml2 ¨θ + mglsin θ)δθdt = 0 for all virtual variations δθ such that δθ(t1) = δθ(t2) = 0. Since the integral appearing in the second term must vanish for arbitrary δθ, we conclude that ml2 ¨θ + mglsin θ = 0 which is the differential equation for the oscillation of the pendulum. The Lagrange’s equations give a much quicker way of solving this type of problem, as we will see shortly; however, before this, we illustrate the power of Hamilton’s principle by deriving the partial differential equation of the lateral vibration of the Euler-Bernoulli beam. îv(x; t) v(x; t) 0 L x Fig. 1.7. Transverse vibration of a beam. 1.6.1 Lateral vibration of a beam Consider the transverse vibration of the beam of Fig.1.7, subjected to a transverse distributed load p(x, t). It is assumed that the principal axes of the cross section are such that the vibration takes place in the plane; v(x, t) denotes the transverse displacements; the virtual displacements
1.6 Hamilton's principle 15 )satisfy the geometric (kinematic)boundary conditions and a d(,t)=(z,t)=0 (1.29) (the configuration is fixed at the limit times t and t2). The Euler-Bernoulli beam theory neglects the shear deformations and assumes that the cross section remains orthogonal to the neutral axis; this is equivalent to assuming that the uniaxial strain distribution S is aear function of the distance to the neutral axis,S e strain energy,reads vs)=5人E(SuPav=5人EeP:asd (1.30) hee”is the curvature of the中a2d/ cros nertia,the kinetic T=M (1.31)） where i is the tra velocity,o is the densit and A the the distributed load: (1.32 From(1.30)and(1.31), V-Ensv'az mnd sTpeni sods As in the previous section,6i can be eliminated by integrating by part over t,and similarly,"can be eliminated by integrating twice by part over z;one gets VEIv'sv'de=(EIv's-(Elyodde =EIt"6w6-【Ee"yi6+(EIuy"ud血
1.6 Hamilton’s principle 15 δv(x, t) satisfy the geometric (kinematic) boundary conditions and are such that δv(x, t1) = δv(x, t2) = 0 (1.29) (the configuration is fixed at the limit times t1 and t2). The Euler-Bernoulli beam theory neglects the shear deformations and assumes that the cross section remains orthogonal to the neutral axis; this is equivalent to assuming that the uniaxial strain distribution S11 is a linear function of the distance to the neutral axis, S11 = −zv′′, where v ′′ is the curvature of the beam. Accordingly, the potential energy, which in this case is the strain energy, reads V (Sij ) = 1 2 Z V E(S11) 2dV = 1 2 Z L 0 Z S E(v ′′) 2z2dSdx or V = 1 2 Z L 0 EI(v′′) 2dx (1.30) where v′′ is the curvature of the beam, E the Young’s modulus and I the geometric moment of inertia of the cross section (EI is called the bending stiffness). If one includes only the translational inertia, the kinetic coenergy is T∗ = 1 2 Z L 0 ̺A(˙v) 2dx (1.31) where ˙v is the transverse velocity, ̺ is the density and A the cross section area. The virtual work of the non-conservative forces is associated with the distributed load: δWnc = Z L 0 p δv dx (1.32) From (1.30) and (1.31), δV = Z L 0 EIv′′δv′′dx and δT∗ = Z L 0 ̺Av δ˙ v dx ˙ As in the previous section, δv˙ can be eliminated by integrating by part over t, and similarly, δv′′ can be eliminated by integrating twice by part over x; one gets δV = Z L 0 EIv′′δv′′dx = [EIv′′δv′ ] L 0 − Z L 0 (EIv′′) ′ δv′ dx = [EIv′′δv′ ] L 0 − [(EIv′′) ′ δv] L 0 + Z L 0 (EIv′′) ′′δv dx
16 1 Lagrangian dynamics of mechanical system and similarly The expression in the bracket vanishes because of(1.29).Substituting the above expressions in Hamilton's principle,one gets (1.33 This variational indicator must vanish for all arbitrary variations v com- patible with the kinematic constraints and satisfying (1.29).This implies that the dynamic equilibrium is governed by the following partial differ- ential equation (EI"”+eA=p 1.34) Besides,cancelling the terms within brackets,we find that the following conditions must be fulfilled in r=0 and r=L. EIv".6v'=0 (1.35) (EI").6v=0 (1.36) uation e 0 which is that,at both ends tatio 0.which means that the bending nt ic ual to o Similarly the nd one implies that at both cnds.cither which is the case if the displace ment is fixed,or(EI)=0,which means that the shear force is equal o 0.ov =0 and ov =0 are kinematic(geometric)boundary conditions: EI=0and(EI)=0are sometimes called naturul bounda ry condi tions,because they come naturally from the va end allow this implies (EI A pE hat it foll he k displacement-shear force.rotation-bending moment. conjugat The Euler-Bernoulli beam will be reexamined in chanter 5 when we include a piezoelectric layer.More elaborate beam theories are available, which account for shear deformations and the rotary inertia of the cross section:they are e based on different kinematic assumptions on the dis placement field,leading to ne expressions for the st ple f
16 1 Lagrangian dynamics of mechanical systems and similarly Z t2 t1 δT∗dt = Z t2 t1 dt Z L 0 ̺Av δ˙ v dx ˙ = [Z L 0 ̺Av δv dx ˙ ] t2 t1− Z t2 t1 dt Z L 0 ̺Av δv dx ¨ The expression in the bracket vanishes because of (1.29). Substituting the above expressions in Hamilton’s principle, one gets Z t2 t1 dt{−[EIv′′δv′ ] L 0 +[(EIv′′) ′ δv] L 0 + Z L 0 [−(EIv′′) ′′ − ̺Av¨+ p]δv dx} = 0 (1.33) This variational indicator must vanish for all arbitrary variations δv compatible with the kinematic constraints and satisfying (1.29). This implies that the dynamic equilibrium is governed by the following partial differential equation (EIv′′) ′′ + ̺Av¨ = p (1.34) Besides, cancelling the terms within brackets, we find that the following conditions must be fulfilled in x = 0 and x = L, EIv′′.δv′ = 0 (1.35) (EIv′′) ′ .δv = 0 (1.36) The first equation expresses that, at both ends, one must have either δv′ = 0, which is the case if the rotation is fixed, or EIv′′ = 0, which means that the bending moment is equal to 0. Similarly, the second one implies that at both ends, either δv = 0, which is the case if the displacement is fixed, or (EIv′′)′ = 0, which means that the shear force is equal to 0. δv = 0 and δv′ = 0 are kinematic (geometric) boundary conditions; EIv′′ = 0 and (EIv′′)′ = 0 are sometimes called natural boundary conditions, because they come naturally from the variational principle. A free end allows arbitrary δv and δv′ ; this implies EIv′′ = 0 and (EIv′′)′ = 0. A clamped end implies that δv = 0 and δv′ = 0. A pinned end implies that δv = 0, but δv′ is arbitrary; it follows that EIv′′ = 0. Note that the kinematic and the natural boundary conditions are energetically conjugate: displacement-shear force, rotation-bending moment. The Euler-Bernoulli beam will be reexamined in chapter 5 when we include a piezoelectric layer. More elaborate beam theories are available, which account for shear deformations and the rotary inertia of the cross section; they are based on different kinematic assumptions on the displacement field, leading to new expressions for the strain energy V and the kinetic coenergy T∗; the subsequent application of Hamilton’s principle follows closely the discussion above
1.7 Lagrange's equations 1.7 Lagrange's equations Hamilton's principle relies on scalar work and energy quantities:it does not refer to a particular coordinate system.The system configuration can be expressed in generalized coordinates q.If the generalized coordinates are independent,the virtual change of configuration can be represented by independent virtual variations of the generalized coordinates,6q.This ations which are Lagrange's equations.r ws us to transi et of n inder points of the system follow xi=工(q1,.,gnt) (1.37) We also allow an explicit dependency on time t,which is important for analyzing gyroscopic systems such as rotating machinery.The velocity of the material point i is given by -工胎+费 (1.38) where the matrix of partial derivatives Ori/dgi has the meaning of a Jacobian.From Equ.(1.38),the kinetic coenergy can be written in the form T*=∑mi=T位+T+T6 (1.39) 2hn石石saT云io he generalized velocities the coefficients of T depend on the partial de vatives o whic are thems oof the n12 d To pendency on function of )Bein tho to tial.it is in is responsible for the gyroscopic forces.The general form of the kinetic coenergy is T=T(91.qn,1,.nit) (1.40) The potential energyVdoes not depend on the to be of the general form
1.7 Lagrange’s equations 17 1.7 Lagrange’s equations Hamilton’s principle relies on scalar work and energy quantities; it does not refer to a particular coordinate system. The system configuration can be expressed in generalized coordinates qi. If the generalized coordinates are independent, the virtual change of configuration can be represented by independent virtual variations of the generalized coordinates, δqi. This allows us to transform the variational indicator (1.26) into a set of differential equations which are Lagrange’s equations. First, consider the case where the system configuration is described by a finite set of n independent generalized coordinates qi. All the material points of the system follow xi = xi(q1, ., qn;t) (1.37) We also allow an explicit dependency on time t, which is important for analyzing gyroscopic systems such as rotating machinery. The velocity of the material point i is given by x˙i = X j ∂xi ∂qj q˙j + ∂xi ∂t (1.38) where the matrix of partial derivatives ∂xi/∂qj has the meaning of a Jacobian. From Equ.(1.38), the kinetic coenergy can be written in the form T∗ = 1 2 X i mix˙i.x˙i = T∗ 2 + T∗ 1 + T∗ 0 (1.39) where T∗ 2 , T∗ 1 and T∗ 0 are respectively homogeneous functions of order 2,1 and 0 in the generalized velocities ˙qi; the coefficients of T∗ i depend on the partial derivatives ∂xi/∂qj , which are themselves functions of the generalized coordinates qi. Note that without explicit dependency on t, the last term in (1.38) disappears and T∗ = T∗ 2 (homogenous quadratic function of ˙qi). Being independent of ˙qi, the term T∗ 0 appears as a potential; it is in general related to centrifugal forces, while the linear term T∗ 1 is responsible for the gyroscopic forces. The general form of the kinetic coenergy is T∗ = T∗(q1, ., qn, q˙1, ., q˙n;t) (1.40) The potential energy V does not depend on the velocity; it can be assumed to be of the general form