AA242B: MECHANICAL VIBRATIONS

Transcription

1 AA242B: MECHANICAL VIBRATIONS 1 / 57 AA242B: MECHANICAL VIBRATIONS Dynamics of Continuous Systems These slides are based on the recommended textbook: M. Géradin and D. Rixen, Mechanical Vibrations: Theory and Applications to Structural Dynamics, Second Edition, Wiley, John & Sons, Incorporated, ISBN-13: / 57

2 AA242B: MECHANICAL VIBRATIONS 2 / 57 Outline 1 Hamilton s Principle 2 Wave Propagation in a Homogeneous Elastic Medium 3 Free Vibrations of Continuous Systems and Response to External Excitation 2 / 57

3 AA242B: MECHANICAL VIBRATIONS 3 / 57 Hamilton s Principle Definitions Elastic body S = S σ (where t i = σ ij n j = t i ) Su (where u i = ū i ) 3 / 57

4 AA242B: MECHANICAL VIBRATIONS 4 / 57 Hamilton s Principle Green Strains ds0 2 = dx i dx i square of the original length ds 2 = d(x i + u i )d(x i + u i ) square of the deformed length ds 2 ds0 2 = 2ε ij dx i dx j where ε ij = 1 2 ( ui x j + u j x i + u m x i ) u m x j is the Green symmetric strain tensor Note that ε ij 0 rigid body motion 4 / 57

5 AA242B: MECHANICAL VIBRATIONS 5 / 57 Hamilton s Principle Green Strains Proof Einstein s notation: dx i dx i = 3 i=1 dx 2 i du i = u i dx j, i = 1, 2, 3 x j ds0 2 = dx i dx i ds 2 = d(x i + u i )d(x i + u i ) ds 2 ds0 2 = d(x i + u i )d(x i + u i ) dx i dx i = dx i du i + du i dx i + du i du i = du i dx i + du j dx j + du m du m = u i dx j dx i + u j dx i dx j + u m u m dx i dx j ( x j x i x i x j ui = + u j + u ) m u m dx i dx j x j x i x i x j = 2ε ij dx i dx j 5 / 57

6 AA242B: MECHANICAL VIBRATIONS 6 / 57 Hamilton s Principle Green Strains Linear deformation (geometric linearity) the extension strains remain infinitesimal: u i x i 1 the rotations have small amplitudes: u i 1 x j the above assumptions lead to a linear expression of the infinitesimal strain tensor ε ij = 1 ( ui + u ) j 2 x j x i consider a ds parallel to x 1 ds 2 ds0 2 = (ds ds 0)(ds + ds 0) = 2ε 11dx1 2 = 2ε 11ds0 2 ( ) ( ) ( ds ds0 1 = ε 11 = 1 + ds ) ds 0 2 ds 0 for infinitesimal strains, the above result becomes ε 11 = ds ds0 ds 0 (engineering or Cauchy strain) 6 / 57

7 AA242B: MECHANICAL VIBRATIONS 7 / 57 Hamilton s Principle Stress-Strain Relationships Hyperelastic material: the work of the mechanical stresses is stored in the form of an internal energy and thus is recoverable σ ij = f (ε kl ) Strain energy density: to a strain increment dε ij in the stress state σ ij corresponds a strain energy per unit volume dw = σ ij dε ij σ ij = W (ε ij) ε ij 7 / 57

8 AA242B: MECHANICAL VIBRATIONS 8 / 57 Hamilton s Principle Stress-Strain Relationships σ ij is energetically conjugate to the Green strain ε ij. It is called the second Piola-Kirchhoff stress tensor. It does not represent the true (Cauchy) stresses inside a structure with respect to the initial reference frame. Rather, it describes the stress field in a reference frame attached to the body and therefore subjected to its deformation but is related to the elementary area of the undeformed structure. In other words, the second Piola-Kirchhoff stress tensor relates forces in the reference (undeformed) configuration to areas in the reference (undeformed) configuration. 8 / 57

9 AA242B: MECHANICAL VIBRATIONS 9 / 57 Hamilton s Principle Stress-Strain Relationships Complementary energy density W = σ ij ε ij W (Legendre transformation) Linear material linear elastic properties = W (σ ij ) = σij 0 = ε ij = W σ ij ε ij dσ ij σ ij = c ijkl ε kl (21 coefficients) W = 1 2 c ijklε kl ε ij 9 / 57

10 AA242B: MECHANICAL VIBRATIONS 10 / 57 Hamilton s Principle Stress-Strain Relationships Linear material (continue) Hooke s law for an isotropic linear elastic material σ ij = λ ε }{{} kk δ ij + 2G ε ij }{{} volumetric strain shear strain where and λ = Eν (1 + ν)(1 2ν) G = E 2(1 + ν) 10 / 57

11 AA242B: MECHANICAL VIBRATIONS 11 / 57 Hamilton s Principle Displacement Variational Principle The displacement variational principle is Hamilton s principle for a continuous system Recall Hamilton s principle: among all possible solutions satisfying δu(t 1 ) = δu(t 2 ) = 0, the true solution of the dynamic equilibrium problem is the one which is the stationary point of t2 t2 t2 δ L[u]dt = δ (T V)dt = 0 t 1 t 1 t 1 (T V)dt 11 / 57

12 AA242B: MECHANICAL VIBRATIONS 12 / 57 Hamilton s Principle Displacement Variational Principle T (u) = 1 ρ 0 u i u i dv 2 V = V int + V ext where V ext = X i (t)u i dv t i u i ds, where the displacement field S σ u i must satisfy the essential Boundary Conditions (BCs) u i = ū i (t) on S u (recall that for particles, δw = n Q sδq s W = n Q sq s) s=1 s=1 the essential BCs are those which cannot be derived from Hamilton s principle those which can, are called the natural BCs V int = W (ε ij )dv = 1 c ijkl ε kl ε ij dv 2 12 / 57

13 AA242B: MECHANICAL VIBRATIONS 13 / 57 Hamilton s Principle Equations of Motion t2 t 1 { t2 δ (T V)dt = 0 t 1 ( ρ 0 u i δ u i W ε ij δε ij + X i δu i ) } dv + t i δu i ds dt = 0 S σ Approach consider the nonlinear Green strain tensor ε ij = 1 ( ui + u ) j + um u m 2 x j x i x i x j integrate by parts with respect to both time and space recall δu i (t 1) = δu i (t 2) = 0 account for the symmetry of the tensor σ ij account for the essential BCs u i = ū i (t) on S u W pay special attention to the evaluation of the quantity δε ij dv ε ij 13 / 57

14 AA242B: MECHANICAL VIBRATIONS 14 / 57 Hamilton s Principle Equations of Motion W ε ij δε ij dv = 1 2 = 1 2 S 1 2 V = Sσ σ ij ( δ u i x j + δ u j + δ um x i x i ( u m [n j σ ij δu i + δu m x i [ σij x j δu i + σ ij n i ( σ ij + σ im u j x m δu j + x i x i ) δu j ds u m + um δ um x j x i x j ) ( u m + n i σ ij δu j + δu m x j ( u m σ ij x j x i ) δu m + ) dv x j ( u j σ ij + σ im x m )] ds ( u m σ ij x i ) δu j dv ) δu m ] dv t2 = δ (T V)dt = t 1 + = 0 t2 { t 1 Sσ { t2 ( t 1 ( ( u j t j σ ij + σ im x m x i ) ) } n i δu j ds ( ) ) u j σ ij + σ im ρ 0ü j + X j δu j dv x m } dt 14 / 57

15 AA242B: MECHANICAL VIBRATIONS 15 / 57 Hamilton s Principle Equations of Motion Since δu j is arbitrary inside and on S σ, the previous equation implies ( ) u j σ ij + σ im + X j = ρ 0 ü j in x i ( x m ) u j t j = σ ij + σ im n i = t j on S σ (natural BC) x m The above equations are the equations of dynamic equilibrium of a deformable body in terms of the second Piola-Kirchhoff stresses. More specifically, they express the equilibrium of the deformed body and thus take into account the geometric nonlinearity. 15 / 57

16 AA242B: MECHANICAL VIBRATIONS 16 / 57 Hamilton s Principle The Linear Case and 2nd-Order Effects ε ij = ε (1) ij ( 1 ui + u ) j 2 x j x i }{{} :linear (small displacements & rotations) The pure linear case: ε = ε (1) ij in this case, HP leads to + 1 u m 2 x i u m x j }{{} ε (2) ij :quadratic σ ij x i + X j = ρ 0ü j in t j = σ ij n i = t j on S σ (natural BC) these are the linearized equations of motion for an elastic body undergoing infinitesimal displacements and rotations they express equilibrium in the undeformed state V 16 / 57

17 AA242B: MECHANICAL VIBRATIONS 17 / 57 Hamilton s Principle The Linear Case and 2nd-Order Effects Second-order effect ε ij = ε (1) ij + ε (2) ij strain energy density Example W = 1 2 c ijklε kl ε ij = 1 2 c ijklε (1) kl ε (1) ij }{{} elastic forces + c ijkl ε (1) kl ε (2) ij c ijjklε (2) kl ε (2) ij }{{} second-order effect v x = v M l/2 = 2v M l v x = 2v M l 0 < x < l 2 l 2 < x < l 17 / 57

18 AA242B: MECHANICAL VIBRATIONS 18 / 57 Hamilton s Principle The Linear Case and 2nd-Order Effects Example (continue) if the analysis is limited to transverse motion, the axial strain can be expressed as ε xx = ( ) 2 v = 1 x 2 4 v M 2 l 2 the kinetic and potential energies are given by T = 1 2 M v 2 M V int = 1 2 the HP can then be expressed as t2 t2 δ (T V)dt = δ t 1 = = t 1 l 0 = 2 v 2 M l 2 EAε 2 xxdx ( 1 t 1 2 M v 2 M 1 l ) EAε 2 xx 2 dx dt 0 t2 ( l ) M v M δ v M EAε xx δε xx dx dt t 1 0 t2 { l ( ) EA v 2 ( v M v M δ v M 0 2 x x ) δ ( v x ) } dx dt = 0 18 / 57

19 AA242B: MECHANICAL VIBRATIONS 19 / 57 Hamilton s Principle The Linear Case and 2nd-Order Effects Example (continue) approach: integrate by parts the first term and substitute all partial derivatives by their computed values = [M v M δv M ] t 2 t1 = { M v M + 2 t2 t 1 t2 t 1 ( EA 2 l 2 ) } δv M dx dt { l ( ) 3 ( 2 EA 2vM 2 M v M δv M l l { l ( ) 3 ( ) } EA 2vM 2 δv M dx dt = 0 2 l l ) ( 2vM l ) 3 ( ) ( ) } 2 l δv M = 0 l 2 ) 3 = M v M + ( 2vM EA l }{{} = 0 restoring force is due to second-order effect 19 / 57

20 AA242B: MECHANICAL VIBRATIONS 20 / 57 Hamilton s Principle The Linear Case and 2nd-Order Effects Example (continue) let N x = Aσ xx = EAε xx be the axial force computed from the second Piola-Kirchhoff stress tensor and its conjugate Green strain measure. The true force N is such that its virtual work (true/cauchy stress, engineering/cauchy ( ) strain) is equivalent ( ) to that of N x that is, ds dx ds Nδ = N xδε xx Nδ = N xδε xx dx dx = N δεxx ds dx recall that ε xx = 1 ( ) ds 2 dx 2 = 1 2 dx 2 2 ( ) 2 ds 1 ds δεxx = dx 2 dx δ ( ) ds dx = N x δε xx = N = ds N }{{} x dx }{{} true force force from Piola-Kirchhoff stresses (relative to surface of underformed cable) 20 / 57

21 AA242B: MECHANICAL VIBRATIONS 21 / 57 Hamilton s Principle The Linear Case and 2nd-Order Effects cos α = dx ds N = N x cos α = EAε xx cos α Let F denote the elastic restoring force of the massless cable ( vm = F = 2N sin α = 2EAε xx tan α = 2EA2 l ) 2 2vM l ( 2vM F = EA l ) 3 which is the same as the restoring force due to second-order effect determined from the HP 21 / 57

22 AA242B: MECHANICAL VIBRATIONS 22 / 57 Hamilton s Principle The Linear Case and 2nd-Order Effects Effect of initial stress u i = u 0 i + u i u i = 0 + u i ε ij = ε 0 ij + ε ij δu i = δu i δε ij = δε ij assume that large displacements and rotations can happen during prestress, but only small displacements and rotations occur after that 22 / 57

23 AA242B: MECHANICAL VIBRATIONS 23 / 57 Hamilton s Principle The Linear Case and 2nd-Order Effects Effect of initial stress (continue) the kinetic energy is given by T = 1 ρ u i u i dt = 1 ρ u 2 V 2 V i u i dt = T and the potential energy is given by V int + V ext where V int = 1 c ijkl ε kl ε ij dv = 1 c ijkl (ε 0 2 V 2 V kl + ε kl )(ε0 ij + ε ij )dv = 1 c ijkl ε 0 2 V kl ε0 ij dv + c ijkl ε 0 V kl ε ij dv + 1 c ijkl ε 2 V kl ε ij dv = int V + σ 0 ij ε ij dv + V int = int + σ 0 (1) }{{} V ij (ε ij + ε (2) ij )dv + 1 c ijkl ε (1) 2 V kl ε (1) ij dv +(HOT ) }{{} cst V int and V ext = (X 0i + X i )u i dv V ( t 0i + t i )u i dv = S ext + V ext 23 / 57

24 AA242B: MECHANICAL VIBRATIONS 24 / 57 Hamilton s Principle The Linear Case and 2nd-Order Effects Effect of initial stress (continue) Two cases: case of externally prestressed structures in which the initial stresses result from the external dead loads X 0i and t 0i : the equilibrium of the prestress state implies δ ext + σ 0 (1) V ij δε ij dv = 0 (note the participation of only ε (1) ij in this equilibrium as after prestress, only small deformations are considered here) case of internally prestressed structures in which the initial stresses result from self-equilibrated stresses due to internalforces such as residual stresses arising from the forming or assembly process: σ 0 (1) V ij δε ij dv = 0 and ext = 0 the HP can then be expressed as δ t2 t 1 (T V)dt = δ t2 t 1 ( T V int σijε 0 (2) ij V dv V ext ) dt = 0 the geometric prestress potential (second-order effect) is defined as (2) V g = dv = δ u i t2 σijε 0 ij V t 1 (T V int Vg V ext )dt = 0, δu i (t 1) = δu i (t 2) = 0 (1) 24 / 57

25 AA242B: MECHANICAL VIBRATIONS 25 / 57 Hamilton s Principle The Linear Case and 2nd-Order Effects The theory of prestressing forms the basis of structural stability analysis, which: consists in computing the prestressing forces applied to a structural system which render possible the existence of a static equilibrium configuration distinct from the prestressed state u = 0 under the geometrically linear and nonlinear elastic forces only in this case, the HP is reduced to δ u i (V int + V g ) = 0 Equation (1) reveals that prestressing modifies the vibration eigenfrequencies, and that the limiting case of a vanishing eigenfrequency corresponds to the limit of stability (T = 0) 25 / 57

26 AA242B: MECHANICAL VIBRATIONS 26 / 57 Wave Propagation in a Homogeneous Elastic Medium The Navier Equations in Linear Dynamic Analysis Small displacements and rotations imply linear expression of the infinitesimal strain tensor ε ij = 1 ( ui + u ) j 2 x j x i linear form of the equations of dynamic equilibrium σ ij x i + X j = ρ 0ü j in t j = σ ij n i = t j on S σ (natural BC) Hooke s law for a linear elastic isotropic medium σ ij = λε kk δ ij + 2Gε ij ( ) ( uk ui = λ δ ij + G + u ) j x k x j x i 26 / 57

27 AA242B: MECHANICAL VIBRATIONS 27 / 57 Wave Propagation in a Homogeneous Elastic Medium The Navier Equations in Linear Dynamic Analysis Assuming a homogeneous medium (λ and G constant) leads to ( ) u i (λ + G) + G 2 u j +X j = ρ 0 ü j, j = 1, 2, 3, in x j x }{{} i x i x }{{} i u 2 u j = (λ + G) e x j + G 2 u j + X j = ρ 0 ü j 2 is the Laplacian operator ( ) e = u is the divergence of the displacement field Propagation of free waves X j = 0 = (λ + G) e x j + G 2 u j = ρ 0ü j, j = 1, 2, 3, in Solutions: Plane elastic waves, Rayleigh surface waves, and Love surface waves 27 / 57

28 AA242B: MECHANICAL VIBRATIONS 28 / 57 Wave Propagation in a Homogeneous Elastic Medium Plane Elastic Waves Plane waves u i (x j, t) = u i (x 1 ± ct) at a given time t, the displacement is identical at any point of the plane perpendicular to the direction of wave propagation (here, x 1) the displacement field at the location (x 1, x 2, x 3) and time t is translated to the location (x 1 x 1, x 2, x 3) at time t + t u i (x 1, x 2, x 3, t) = u i (x 1 ± ct) = u i ((x 1 c t) ± c(t + t)) = u i (x 1 x 1, x 2, x 3, t t) where x 1 = c t c is the velocity of the wave propagating in the positive x 1 direction when u i = u i (x 1 ct) and in the negative x 1 direction when u i = u i (x 1 + ct) 28 / 57

29 AA242B: MECHANICAL VIBRATIONS 29 / 57 Wave Propagation in a Homogeneous Elastic Medium Plane Elastic Waves Plane elastic waves: Longitudinal waves, and transverse waves Longitudinal waves the displacements are parallel to the direction of propagation general form u 1 = ( ) 2π A sin (x 1 ± ct) l u 2 = 0 u 3 = 0 constants A and l represent the wave amplitude and length characteristic longitudinal wave speed that verifies the Navier equations c = c L = λ + 2G ρ E(1 ν) = (1 + ν)(1 2ν)ρ 29 / 57

30 AA242B: MECHANICAL VIBRATIONS 30 / 57 Wave Propagation in a Homogeneous Elastic Medium Plane Elastic Waves Transverse waves the displacements are orthogonal to the direction of propagation general form when the displacement field is parallel to x 2 u 1 = 0 u 2 = A sin u 3 = 0 ( ) 2π (x 1 ± ct) l constants A and l represent the wave amplitude and length characteristic transverse wave speed that verifies the Navier equations G c = c T = ρ here, (x 1, x 2) is the plane of polarization the ratio of c L and c T depends only on the Poisson coefficient 1 2ν c T = c L 2(1 ν) 30 / 57

31 AA242B: MECHANICAL VIBRATIONS 31 / 57 Wave Propagation in a Homogeneous Elastic Medium Surface Waves Surface waves: Rayleigh surface waves, and Love surface waves Rayleigh surface waves two-dimensional semi-infinite medium x 2 0 no excitation on x 2 = 0 (stress free surface) supposing the displacement field is the real part of u 1 = A 1e bx 2 e ik(x 1 ct) u 2 = A 2e bx 2 e ik(x 1 ct) u 3 = 0 with A 1, A 2 C, wave number k R, k = ω c, l = ct, T = 2π ω Navier equations b > 0 e bx 2 0 as x 2 c 2 T 2 u j + (c 2 L c 2 T ) e x j = ü j 31 / 57

32 AA242B: MECHANICAL VIBRATIONS 32 / 57 Wave Propagation in a Homogeneous Elastic Medium Surface Waves Rayleigh surface waves (continue) substituting the expression of the displacement field gives [ ct 2 b 2 + (c 2 cl)k 2 2] A 1 i(cl 2 ct 2 )bka 2 = 0 [ i(cl 2 ct 2 )bka 1 + ct 2 b 2 + (c 2 ct 2 )k 2] A 2 = 0 (A 1, A 2) (0, 0) implies that the determinant vanishes solving for b yields two roots b = k 1 c2, b = k cl 2 b real implies that c < c T < c L corresponding amplitudes ( ) A A2 = = b ik, A = A 1 ( A2 1 c2 c 2 T A 1 ) = ik b 32 / 57

33 AA242B: MECHANICAL VIBRATIONS 33 / 57 Wave Propagation in a Homogeneous Elastic Medium Surface Waves Rayleigh surface waves (continue) the general solution becomes u 1 = A e b x 2 e ik(x 1 ct) + A e b x 2 e ik(x 1 ct) u 2 = b ik A e b x2 e ik(x1 ct) + ik b A e b u 3 = 0 x2 e ik(x 1 ct) A, A and c are determined by the free surface conditions σ 22 = σ 21 = σ 23 = 0, at x 2 = 0 33 / 57

34 AA242B: MECHANICAL VIBRATIONS 34 / 57 Wave Propagation in a Homogeneous Elastic Medium Surface Waves Rayleigh surface waves (continue) using Hooke s law and the expression of the linear strain, these conditions become u 1 + u2 = 0 at x 2 = 0 x 2 x ( ) 1 u1 λ + u2 + 2G u2 = 0 at x 2 = 0 x 1 x 2 x 2 34 / 57

35 AA242B: MECHANICAL VIBRATIONS 35 / 57 Wave Propagation in a Homogeneous Elastic Medium Surface Waves Rayleigh surface waves (continue) substitute in the previous two equations the expression of the general solution and use the identities G = ρct 2 and λ = ρ(cl 2 2cT 2 ) take into account the expressions for b and b = ) 2b A + (2 c2 k 2 A = 0 ) ct 2 b (2 c2 A + 2b A = 0 b c 2 T (A, A ) (0, 0) implies that c verifies the characteristic equation ) 2 (2 c2 = 4 1 c2 1 c2 ct 2 cl 2 ct 2 after factorization, the Rayleigh equation is obtained c 2 ct 2 [ c 6 c 6 T 8 c4 c 4 T + c 2 ( 24 c 2 T k remains a free parameter 16 ) ( )] 16 1 c2 T = 0 cl 2 cl 2 35 / 57

36 AA242B: MECHANICAL VIBRATIONS 36 / 57 Wave Propagation in a Homogeneous Elastic Medium Surface Waves Rayleigh surface waves (continue) Rayleigh equation c 2 ct 2 [ c 6 c 6 T 8 c4 c 4 T + c 2 ( 24 c 2 T 16 ) ( )] 16 1 c2 T = 0 (2) cl 2 cl 2 c = 0 u 1 = u 2 = 0 (trivial solution) from c T < c L, it follows that the second factor of (2) is negative for c = 0 and positive for c = c T : hence, it has a real root 0 < c < c T which shows that surface waves with a velocity lower than c T may appear in the solution of a Navier problem 36 / 57

37 AA242B: MECHANICAL VIBRATIONS 37 / 57 Wave Propagation in a Homogeneous Elastic Medium Surface Waves Rayleigh surface waves (continue) In the propagation of a Rayleigh wave the motion is backward elliptic in contrast to the direct elliptic motion in the propagation of a surface wave in a fluid 37 / 57

38 AA242B: MECHANICAL VIBRATIONS 38 / 57 Wave Propagation in a Homogeneous Elastic Medium Surface Waves Love waves the displacements are perpendicular to the plane of propagation (here, (x 1, x 2)) homogeneous layer of material M 1 with thickness H 1 superimposed on a semi-infinite space of a different material M u 3 and σ 23 are continuous at the interface x 2 = 0 38 / 57

39 AA242B: MECHANICAL VIBRATIONS 39 / 57 Wave Propagation in a Homogeneous Elastic Medium Surface Waves Love waves (continue) the displacement field u 1 = 0 u 2 = 0 kx 2 1 c2 u 3 = Ae kx2 u 3 = A1e c 2 T e ik(x 1 ct)) in M 1 ( ) 2 c c (1) T + A 1e ( kx 2 1 c c (1) T ) 2 eik(x 1 ct) in M 1 satisfies the Navier equations and the condition u 3 0 when x 2 39 / 57

40 AA242B: MECHANICAL VIBRATIONS 40 / 57 Wave Propagation in a Homogeneous Elastic Medium Surface Waves Love waves (continue) u 3 and σ 23 are continuous at the interface x 2 = 0 σ 23 = 0 at x 2 = H 1 = A = A 1 + A 1 ( ) GA 1 c2 ct 2 = G 1(A 1 A 1 ) 2 c 1 c (1) T 2 2 kh 1 1 c c A (1) kh 1 1 c 1e T = A 1 e c (1) T eliminating A, A 1 and A 1 leads to the equation governing the propagation velocity c of a surface wave perpendicular to the propagation direction G 1 c2 c 2 T G 1 ( c c (1) T ) 2 ( ) 2 c 1 tan kh 1 1 = 0 c (1) T for c (1) T < c T, the above equation has a root c (1) T < c < c T Love waves k remains a free parameter 40 / 57

41 AA242B: MECHANICAL VIBRATIONS 41 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Eigenvalue Problem Harmonic motion of a linear system not subjected to external force displacement u i (x j, t) = u ai (x j ) cos ωt time interval [t 1, t 2] chosen such that δu i (t 1) = δu i (t 2) = 0, here for instance [ [t 1, t 2] = π 2ω, π ] 2ω linearity assumption kinetic and internal energy are quadratic in the displacement = T = T max sin 2 ωt, V = V max cos 2 ωt where T max = 1 2 ω2 ρ 0u ai u ai dv, V max = 1 c ijkl ε kl ε ij dv 2 41 / 57

42 AA242B: MECHANICAL VIBRATIONS 42 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Eigenvalue Problem Hamilton s principle eliminate the time variables by accounting for π 2ω π 2ω cos 2 ωt dt = π 2ω π 2ω sin 2 ωt dt = π 2ω [ ω 2 = δl[u] = δ ρ 0u ai u ai dv 1 ] c ijkl ε kl ε ij dv = Definitions displacement vector u = [ u a1 u a2 u a3 ] T stress vector σ = [ σ 11 σ 22 σ 33 σ 12 σ 23 σ 13 ] T strain vector ε = [ ε 11 ε 22 ε 33 γ 12 γ 23 γ 13 ] T, where γ ij = 2ε ij matrix H of Hooke s law elastic coefficients σ = Hε for example in 2D H = E 1 ν 2 1 ν 0 ν ν / 57

43 AA242B: MECHANICAL VIBRATIONS 43 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Eigenvalue Problem Definitions (continue) spatial differentiation operator D T x x 0 2 x 3 = 0 x 0 2 x 1 x x 0 3 x 2 x 1 associated matrix of the direction cosines of the outward normal at S σ N T n1 0 0 n2 0 n3 = 0 n 2 0 n 1 n n 3 0 n 2 n 1 Linear kinematics Local dynamic equilibrium ε = Du σ = HDu σ { ij = ρ 0ü j in x i = σ ij n i = 0 on S σ D T σ + ω 2 ρ 0u = 0 in N T σ = 0 on S σ (3) Variational form of Hamilton s principle } δ {ω ρ0ut udv 2 (Du)T H (Du) dv = 0 43 / 57

44 AA242B: MECHANICAL VIBRATIONS 44 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Eigenvalue Problem Using the matrix notation, the equations of local dynamic equilibrium (3) can be re-written as { D T HDu + ω 2 ρ 0 u = 0 in N T HDu = 0 on S σ The homogeneous system of equations defining the local dynamic equilibrium, together with its associated variational form, defines an eigenvalue problem of the Sturm-Liouville type { D T HDu (i) + ωi 2 ρ 0 u (i) = 0 in D T HDu (i) = ωi 2 ρ 0 u (i) N T HDu (i) = 0 on S σ i = 1,, where are the eigenvectors 1 u (1), u (2), u (3), 1 In this chapter, the subscript (i) is used to denote the i-th mode instead of the subscript i to avoid confusion with the i-th direction of a vector 44 / 57

45 AA242B: MECHANICAL VIBRATIONS 45 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Orthogonality of Eigensolutions Orthogonality of the eigenvectors equilibrium equations verified by the eigenmodes D T HDu (i) + ω 2 i ρ 0u (i) = 0 multiply by u T (j) and integrate over the reference volume integrate the first term by parts u T (j)d T HDu (i) dv = u T (j)n T ( ) T HDu (i) ds Du(j) H ( ) Du (i) dv S compatibility of the displacement field and surface equilibriun condition for u (i) u T (i) = 0 on S u = N T HDu (i) = 0 on S σ [ ( Du (j) ) T H ( Du (i) ) + ω 2 i ρ 0u T (j)u (i) ] dv = 0 (E ji ) 45 / 57

46 AA242B: MECHANICAL VIBRATIONS 46 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Orthogonality of Eigensolutions Orthogonality of the eigenvectors (continue) similarly for u (j) [ ( ) T Du (i) H ( ) ] Du (j) + ω 2 j ρ 0u T (i)u (j) dv = 0 (E ij ) (E ji ) (E ij ) if ω 2 j ω 2 i = (ωj 2 ωi 2 ) ρ 0u T (j)u (i) dv = 0 ρ 0u T (j)u (i) dv = 0 if ω 2 j = ω 2 i and i j (multiple eigenfrequency), the eigenmodes can also be orthogonalized as ρ 0u T (j)u (i) dv = 0 normalize the eigenvector u (i) as follows ρ 0u T (i)u (i) dv = 1 46 / 57

47 AA242B: MECHANICAL VIBRATIONS 47 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Orthogonality of Eigensolutions Orthogonality of the eigenvectors (continue) recall (E ij ) ( Du(j) ) T H ( Du (i) ) dv = ω 2 i ρ 0u T (j)u (i) dv = ρ 0u T (j)u (i) dv = δ ij ( Du(j) ) T H ( Du (i) ) dv = δij ω 2 i 47 / 57

48 AA242B: MECHANICAL VIBRATIONS 48 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Response to External Excitation: Modal Superposition Response of a system with homogeneous BCs eigenmodes form a complete set of solutions of the problem with homogeneous BCs u(x j, t) = η s(t)u (s) (x j ) s=1 where η s(t) are the normal coordinates associated with each mode u (s) the general solution u satisfies the linear equilibrium equation and the homogeneous BCs D T HDu + X ρ 0ü = 0 in N T HDu = t = 0 on S σ u = ū = 0 on S u 48 / 57

49 AA242B: MECHANICAL VIBRATIONS 49 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Response to External Excitation: Modal Superposition Response of a system with homogeneous BCs (continue) linear equilibrium equation using the eigenmodes η sd T HDu (s) + X ρ 0 η su (s) = 0 in s=1 premultiply by u T (r) integrate over integrate by parts in space use the normalization of the modal masses and the orthogonality of the eigenmodes apply the BCs s=1 = η r + ω 2 r η r = φ r, r = 1,, φ r is the participation factor of the eigenmode u (r) to the external excitation X φ r = u T XdV (r) 49 / 57

50 AA242B: MECHANICAL VIBRATIONS 50 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Response to External Excitation: Modal Superposition Response of a system with homogeneous BCs (continue) η r + ω 2 r η r = φ r can be integrated in time as sin ωr t η r (t) = η r (0) cos ω r t + η r (0) + 1 ω r ω r where t 0 φ r (τ) sin (ω r (t τ)) dτ η r (0) = ρ 0u T (r)u(0)dv, η r (0) = ρ 0u T (r) u(0)dv therefore, the general solution obtained by modal superposition is u(x j, t) = u (s) cos ω st ρ 0u T (s)u(0)dv + + s=1 sin ω st u (s) s=1 s=1 u (s) ω s ω s t 0 ρ 0u T (s) u(0)dv φ s(τ) sin (ω s(t τ)) dτ 50 / 57

51 AA242B: MECHANICAL VIBRATIONS 51 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Response to External Excitation: Modal Superposition Response of a system with non-homogeneous spatial BCs consider the following problem D T HDu + X ρ 0ü = 0 in with the initial conditions u(0) and u(0), and the non-homogeneous BCs N T HDu = t on S σ u = ū on S u the external forces t applied on the surface S σ and the displacement ū specified on S u can be function of time solution approach: exploit linearity to split the problem into a quasi-static problem with non-homogeneous BCs and a dynamic problem with a source term and homogeneous BCs (which we already know how to solve) 51 / 57

52 AA242B: MECHANICAL VIBRATIONS 52 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Response to External Excitation: Modal Superposition Response of a system with non-homogeneous spatial BCs (continue) quasi-static displacement field u qs (x j, t) resulting from the application of the non-homogeneous BCs D T HDu qs = 0 in N T HDu qs = t on S σ u qs = ū on S u = characterizes u qs modal superposition for the rest of the response leads to u(x j, t) = u qs (x j, t) + η s (t)u (s) (x j ) equilibrium equation η s D T HDu (s) + X ρ 0 η s u (s) = ρ 0ü qs in BCs s=1 s=1 s=1 ( ) N T HD η s u (s) = 0 on S σ s=1 η s u (s) = 0 on S u s=1 52 / 57

53 AA242B: MECHANICAL VIBRATIONS 53 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Response to External Excitation: Modal Superposition Response of a system with non-homogeneous spatial BCs (continue) pre-multiply by u T (r) integrate by parts over account for the orthogonality of the eigenmodes account for the BCs satisfied by the eigenmodes = η r + ωr 2 η r = φ r ρ 0 u T (r)üqsdv, r = 1,, the solution is sin ωr t η r (t) = η r (0) cos ω r t + η r (0) ω r + 1 t [ ] φ r (τ) ρ 0u T ω (r)ü qs(τ) sin (ω r (t τ)) dτ r 0 where η r (0) = ρ 0u T (r) (u(0) uqs (0)) dv, ηr (0) = ρ 0u T (r) ( u(0) uqs (0)) dv 53 / 57

54 AA242B: MECHANICAL VIBRATIONS 54 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Response to External Excitation: Modal Superposition Response of a system with non-homogeneous spatial BCs (continue) general solution u(x j, t) = u t (s) T u qs (x j, t) ρ 0u(s)üqs (τ) sin (ωs (t τ)) dvdτ ω s=1 s 0 + u (s) cos ω s t s=1 ρ 0u T (s) (u(0) uqs (0)) dv + sin ω s t u (s) ρ 0u T (s) ( u(0) uqs (0)) dv ω s=1 s + u t (s) φ s (τ) sin (ω s (t τ)) dτ ω s=1 s 0 differs from the homogeneous case by the contribution of the quasi-static displacement field (and its time-derivatives) 54 / 57

55 AA242B: MECHANICAL VIBRATIONS 55 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Response to External Excitation: Modal Superposition Response of a system with non-homogeneous spatial BCs (continue) integrate by parts twice the terms involving ü qs 1 t ü qs (τ) sin(ω s (t τ))dτ = ω s 0 sin ωs t u qs (0) + u qs (t) cos ω s t u qs (0) ω s t ω s u qs (τ) sin(ω s (t τ))dτ 0 = = + + u t (s) T ρ 0u(s)üqs (τ) sin (ωs (t τ)) dvdτ ω s=1 s 0 u (s) cos ω s t ρ 0u T (s) uqs (0)dV s=1 sin ω s t u (s) ρ 0u T (s) uqs (0)dV ω s=1 s ( u t ) (s) ω 2 s ρ 0u T (s)(τ)uqs (τ)dv sin (ω s (t τ)) dτ ω s=1 s 0 u (s) ρ 0u V T (s) uqs dv 0 s=1 55 / 57

56 AA242B: MECHANICAL VIBRATIONS 56 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Response to External Excitation: Modal Superposition Response of a system with non-homogeneous spatial BCs (continue) express u qs in the basis of the eigenmodes u qs = u (s) ρ 0u V T (s)u qsdv 0 s=1 substitute in previous expression of u(x j, t) to keep dependence on u qs only = u(x j, t) = + u (s) ρ 0u s=1 V T (s) 0 ( u t (s) φ s (τ) + ω 2 s ω s=1 s 0 ( u(0) cos ω s t + u(0) sin ωs t ω s ρ 0u T (s)(τ)uqs (τ)dv ) dv ) sin (ω s (t τ)) dτ 56 / 57

57 AA242B: MECHANICAL VIBRATIONS 57 / 57 Free Vibrations of Continuous Systems and Response to External Excitation Response to External Excitation: Modal Superposition Response of a system with non-homogeneous spatial BCs (continue) recall equilibrium equations, multiply them by u T qs and integrate over V0 = u T qs DT HDu (s) dv + ω 2 s ρ 0u T qs u (s)dv = 0 integrate the first term in the above equation by parts recall equations satisfied by the quasi-static displacement field u qs introduce r (s) = N T HDu (s) and eliminate dependence on u qs = u(x j, t) = ) u (s) (cos ω s t ρ 0u T sin ωs t (s) u(0)dv + ρ 0u T (s) s=1 ω s + u t { } (s) φ s (τ) + u T ω ū(τ) T r (s) ds s=1 s 0 Sσ Su sin(ω s (t τ))dτ w/r to the homogeneous BCs case, the modal participation factor is augmented by ψ s = u T (s) tds + ū(τ) T r (s) ds Sσ Su which is the work produced by the boundary tractions with the eigenmode displacement and the work produced by the eigenmode boundary reaction with ū 57 / 57