AA242B: MECHANICAL VIBRATIONS


 Anthony Black
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1 AA242B: MECHANICAL VIBRATIONS 1 / 50 AA242B: MECHANICAL VIBRATIONS Undamped Vibrations of ndof 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, ISBN13: / 50
2 AA242B: MECHANICAL VIBRATIONS 2 / 50 Outline 1 Linear Vibrations 2 Natural Vibration Modes 3 Orthogonality of Natural Vibration Modes 4 Modal Superposition Analysis 5 Spectral Expansions 6 Forced Harmonic Response 7 Response to External Loading 8 Mechanical Systems Excited Through Support Motion 2 / 50
3 AA242B: MECHANICAL VIBRATIONS 3 / 50 Linear Vibrations Equilibrium configuration q s (t) = q s (0) q s (t) = 0 } s = 1,, n (1) Recall the Lagrange equations of motion d ( ) T + T V D + Q s (t) = 0 dt q s q s q s q s where T = T 0 + T 1 + T 2 Recall the generalized gyroscopic forces F s = r=1 2 T 1 q s q r q r + T1 q s = r=1 ( 2 T 1 q s q r 2 T 1 q r q s ) q r, s = 1,, n Definition: the effective potential energy is defined as V = V T 0 The Lagrange equations of motion can be rewritten as ( ) d T2 T 2 = Q s (t) V D + F s ( ) T1 dt q s q s q s q s t q s 3 / 50
4 AA242B: MECHANICAL VIBRATIONS 4 / 50 Linear Vibrations Recall that T 0(q, t) = 1 N 3 ( ) Uik 2 m k (q, t) (transport kinetic energy) 2 t k=1 i=1 N vk ( q) D = C k f k (γ)dγ (dissipation function) k=1 0 From the Lagrange equations of motion ( ) d T2 T 2 = Q s (t) V D + F s ( ) T1 dt q s q s q s q s t q s it follows that an equilibrium configuration exists if and only if 0 = Q s (t) V Q s (t) and V are function of q only and not q s explicitly dependent on time Q s (t) = 0 and V = V (q) At equilibrium Q s (t) = 0 and V q s = (V T 0) q s = 0, s = 1,, n 4 / 50
5 AA242B: MECHANICAL VIBRATIONS 5 / 50 Linear Vibrations FreeVibrations About a Stable Equilibrium Position Consider first a system that does not undergo a transport or overall motion T = T 2 ( q) The equilibrium position is then given by Q s (t) = 0 and V q s = 0, s = 1,, n Assume next that this system is conservative E = T + V = cst Shift the origin of the generalized coordinates so that at equilibrium, q s = 0, s = 1,, n (in which case the q s represent the deviation from equilibrium) Since the potential energy is defined only up to a constant, choose this constant so that V(q s = 0) = 0 Now, suppose that a certain energy E(0) is initially given to the system in equilibrium 5 / 50
6 AA242B: MECHANICAL VIBRATIONS 6 / 50 Linear Vibrations FreeVibrations About a Stable Equilibrium Position Definition: the equilibrium position (q s = 0, s = 1,, n) is said to be stable if E / E(0) < E, T (t) E(0) Consequences T + V = E = cst = E(0) V(t) = E(0) T (t)) 0 at a stable equilibrium position, the potential energy is at a relative minimum if E(0) is small enough, V(t) will be small enough and therefore deviations from the equilibrium position will be small enough 6 / 50
7 AA242B: MECHANICAL VIBRATIONS 7 / 50 Linear Vibrations FreeVibrations About a Stable Equilibrium Position Linearization of T and V around an equilibrium position (q s = 0, V = 0) q s actually, this means obtaining a quadratic form of T and V in q and q so that the corresponding generalized forces are linear since q s(t) represent deviations from equilibrium, V can be expanded as follows V(q) = V(0) + = V(0) s=1 V q s q=0 q s s=1 r=1 s=1 r=1 2 V q s q r q=0 q sq r + O(q 3 ) 2 V q s q r q=0 q sq r + O(q 3 ) since the potential energy is defined only up to a constant, if this constant is chosen so that V(0) = 0, then a secondorder approximation of V(q) is given by V(q) = 1 2 s=1 r=1 2 V q s q r q=0 q sq r, for q 0 7 / 50
8 AA242B: MECHANICAL VIBRATIONS 8 / 50 Linear Vibrations FreeVibrations About a Stable Equilibrium Position Stiffness matrix let K = [k sr ] where k sr = k rs = 2 V q s q r q=0 = V(q) = 1 2 qt Kq > 0, for q 0 K is symmetric positive (semi) definite 8 / 50
9 AA242B: MECHANICAL VIBRATIONS 9 / 50 Linear Vibrations FreeVibrations About a Stable Equilibrium Position Recall that T 2 = 1 2 N 3 s=1 r=1 k=1 i=1 T 2(q, q) = T 2(0, 0) s=1 r= s=1 r=1 = 1 2 s=1 r=1 m k U ik q s U ik q r q s q r (relative kinetic energy) s=1 T 2 q s q=0, q=0 q s + 2 T 2 q s q r q=0, q=0 q s q r + T 2 q s q=0, q=0 q s s=1 s=1 r=1 2 T 2 q s q r q=0, q=0 q s q r + O(q 3, q 3 ) 2 T 2 q s q r q=0, q=0 q s q r + O(q 3, q 3 ) 2 T 2 q s q r q=0, q=0 q s q r 9 / 50
10 AA242B: MECHANICAL VIBRATIONS 10 / 50 Linear Vibrations FreeVibrations About a Stable Equilibrium Position Hence, a secondorder approximation of T 2 ( q) is given by T 2 ( q) = 1 2 qt M q > 0, for q 0 where[ M = m sr = m rs = 2 T 2 U ik q=0 = m k q s q r q s k=1 i=1 is the mass matrix and is symmetric positive definite N 3 q=0 U ik q r q=0 ] 10 / 50
11 AA242B: MECHANICAL VIBRATIONS 11 / 50 Linear Vibrations FreeVibrations About a Stable Equilibrium Position Freevibrations about a stable equilibrium position of a conservative system that does not undergo a transport or overall motion (T 0 = T 1 = 0) d dt ( T2 q s ) T 2 = V q s q s = d (M q) 0 = Kq dt = M q + Kq = 0 11 / 50
12 AA242B: MECHANICAL VIBRATIONS 12 / 50 Linear Vibrations FreeVibrations About an Equilibrium Configuration Consider next the more general case of a system in steady motion (a transported system) whose equilibrium configuration defined by V q s = (V T 0) q s = 0, s = 1,, n corresponds ( ) to the balance of forces ( ) deriving from a potential V T0 and centrifugal forces q s q s It is an equilibrium configuration in the sense that q s which represent here the generalized velocities relative to a steady motion are zero but the system is not idle 12 / 50
13 AA242B: MECHANICAL VIBRATIONS 13 / 50 Linear Vibrations FreeVibrations About an Equilibrium Configuration Linearizations effective potential energy V effective stiffness matrix K where K = mutual kinetic energy T 1 = where F = = V (q) = 1 2 qt K q > 0, for q 0 [ ( ) ] ksr 2 T 0 = k sr q=0 q s q r N 3 s=1 k=1( i=1 T 1 q s (0) + q s s=1 2 T 1 q r q=0 q s s=1 r=1 [ f sr = 2 T 1 q s q r q=0 U ik t m U ik k q s = q s ] r=1 s=1 q s T 1 q s (q) 2 T 1 q sq r q=0 q r + O(q 2 ) q sq r T 1(q, q) = q T Fq ) 13 / 50
14 AA242B: MECHANICAL VIBRATIONS 14 / 50 Linear Vibrations FreeVibrations About an Equilibrium Configuration Equations of freevibration around an equilibrium configuration the equilibrium configuration generated by a steady motion remains stable as long as V = V T 0 0 this corresponds to the fact that K remains positive definite in the neighborhood of such a configuration, the equations of motion (for a conservative system undergoing transport or overall motion) are d dt ( T2 where F s = q s ) T2 q s }{{} 0 r=1 = Q s(t) }{{} 0 V q s D q s }{{} 0 ( ) 2 T 1 2 T 1 q r = (F T F) q q s q r q r q s M q + G q + K q = 0 +F s ( ) T1 t q s }{{} 0 where G = F F T = G T is the gyroscopic coupling matrix 14 / 50
15 AA242B: MECHANICAL VIBRATIONS 15 / 50 Linear Vibrations FreeVibrations About an Equilibrium Configuration Example X = (a + x) cos Ωt Y = (a + x) sin Ωt v 2 = Ẋ 2 + Ẏ 2 = (a + x) 2 Ω 2 + ẋ 2 V = 1 2 kx 2 T 0 = 1 2 Ω2 m(a + x) 2 T 1 = 0 T 2 = 1 2 mẋ 2 V = 1 2 kx Ω2 m(a + x) 2 equilibrium configuration V x = 0 = kx Ω2 m(a + x) = 0 = x eq = Ω2 ma k Ω 2 m the system becomes unstable for Ω 2 = k m k = 2 V = k Ω 2 m system is unstable for Ω 2 k x 2 m 15 / 50
16 AA242B: MECHANICAL VIBRATIONS 16 / 50 Natural Vibration Modes Freevibration equations: M q + Kq = 0 q(t) = q a e iωt (K ω 2 M)q a = 0 det (K ω 2 M) = 0 If the system has n degrees of freedom (dofs), M and K are n n matrices n eigenpairs (ω 2 i, q a i ) Rigid body mode(s): ω 2 j = 0 Kq aj = 0 For a rigid body mode, V(q aj ) = 1 2 qt a j Kq aj = 0 16 / 50
17 AA242B: MECHANICAL VIBRATIONS 17 / 50 Orthogonality of Natural Vibration Modes Distinct Frequencies Consider two distinct eigenpairs (ωi 2, q a i ) and (ωj 2, q a j ) q T a j Kq ai = q T a j ωi 2 Mq ai (2) q T a i Kq aj = q T a i ωj 2 Mq aj (3) Because M and K are symmetric (2) (3) T 0 = (ω 2 i ω 2 j )q aj T Mq ai since ω 2 i ω 2 j q aj T Mq ai = 0 and q aj T Kq ai = 0 17 / 50
18 AA242B: MECHANICAL VIBRATIONS 18 / 50 Orthogonality of Natural Vibration Modes Distinct Frequencies Physical interpretation of the orthogonality conditions q aj T Mq ai = 0 q aj T ( ω 2 i Mq ai ) = (ω 2 i Mq ai ) T q aj = 0 which implies that the virtual work produced by the inertia forces of mode i during a virtual displacement prescribed by mode j is zero q aj T Kq ai = 0 (Kq ai ) T q aj = 0 which implies that the virtual work produced by the elastic forces of mode i during a virtual displacement prescribed by mode j is zero 18 / 50
19 AA242B: MECHANICAL VIBRATIONS 19 / 50 Orthogonality of Natural Vibration Modes Distinct Frequencies Rayleigh quotient Kq ai = ω 2 i Mq ai q ai T Kq ai = ω 2 i q ai T Mq ai ω 2 i = q a i T Kq ai q ai T Mq ai γ i = generalized stiffness coefficient of mode i (measures the contribution of mode i to the elastic deformation energy) µ i = generalized mass coefficient of mode i (measures the contribution of mode i to the kinetic energy) Since the amplitude of q ai is determined up to a factor only γ i and µ i are determined up to a constant factor only Mass normalization q aj T Mq ai = δ ij q aj T Kq ai = ω 2 i δ ij = γ i µ i 19 / 50
20 AA242B: MECHANICAL VIBRATIONS 20 / 50 Orthogonality of Natural Vibration Modes Degeneracy Theorem What happens if a multiple circular frequency is encountered? Theorem: to a multiple root ω 2 p of the system Kx = ω 2 Mx corresponds a number of linearly independent eigenvectors {q ai } equal to the root multiplicity 20 / 50
21 AA242B: MECHANICAL VIBRATIONS 21 / 50 Modal Superposition Analysis ndof system: M R n n, K R n, and q R n Coupled system of ordinary differential equations M q + Kq = 0 q(0) = q 0 q(0) = q 0 21 / 50
22 AA242B: MECHANICAL VIBRATIONS 22 / 50 Modal Superposition Analysis Natural vibration modes (eigenmodes) Kq ai = ωi 2 Mq ai, i = 1,, n Q = [q a1 q a2 { Q q an ] KQ = Ω 2 Q T MQ = I Truncated eigenbasis Ω 2 = ω ω 2 n Q r = [ { ] Q T q a1 q a2 q ar, r << n r KQr = Ω2 r (reduced stiffness matrix) Q T r MQr = Ir (reduced mass matrix) Ω 2 r = ω ω 2 r 22 / 50
23 AA242B: MECHANICAL VIBRATIONS 23 / 50 Modal Superposition Analysis Modal superposition: q = Q r y = r y i q ai where y = [y 1 y 2 y r ] T and y i is called the modal displacement i=1 Substitute in M q + Kq = 0 = MQ r ÿ + KQ r y = 0 = Q T r MQ r ÿ + Q T r KQ r y = 0 = I r ÿ + Ω 2 r y = 0 Uncoupled differential equations (modal equations) ÿ i + ω 2 i y i = 0, i = 1, r 23 / 50
24 AA242B: MECHANICAL VIBRATIONS 24 / 50 Modal Superposition Analysis ÿ i + ω 2 i y i = 0, i = 1, r Case 1: ω 2 i = 0 (rigid body mode) y i = a i t + b i Case 2: ωj 2 0 y j = c j cos ω j t + d j sin ω j t General case: r b rigid body modes q = r b r r b (a i t + b i )q ai + (c j cos ω j t + d j sin ω j t)q aj i=1 j=1 Initial conditions q(0) = q 0 = Qy(0) and q(0) = q 0 = Qẏ(0) 24 / 50
25 AA242B: MECHANICAL VIBRATIONS 25 / 50 Modal Superposition Analysis q 0 = Qy(0) and q 0 = Qẏ(0) = q T a i Mq 0 = q T a i MQy(0) From the orthogonality properties of the natural mode shapes (eigenvectors) it follows that q T a i Mq 0 = q T a i MQy(0) = q T a i Mq ai y i (0) = y i (0) y i (0) = q T a i Mq(0) Case 1: ω 2 i Case 2: ω 2 j Case 1: ω 2 i = 0 (rigid body mode) a i 0 + b i = q T a i Mq(0) b i = q T a i Mq(0) 0 c j 1 + d j 0 = q T a j Mq(0) c j = q T a j Mq(0) = 0 (rigid body mode) a i = q T a i M q(0) Case 2: ω 2 j 0 d j = q T a M q(0) j Thus, the general solution is q(t) = r b i=1 ([ ] ) q T a M q(0) t + q T i a Mq(0) q ai + i r r b j=1 ( q T a Mq(0) cos ω j t + q T j a M q(0) sin ω ) j t q aj j ω j 25 / 50
26 AA242B: MECHANICAL VIBRATIONS 26 / 50 Spectral Expansions = x R n, x = x R n, x = s=1 α s q as q T a Mx = j s=1 (q T as Mx ) q as = s=1 α s q T a Mq as = α j j s=1 q as ( q T as Mx ) = s=1 ( ( ) ) q as q T Mx = (q as as q T as )M x s=1 s=1 ( ) q as q T M = I as This is the same result as Q T MQ = I A given load p can be expanded in terms of the inertia forces generated by the eigenmodes, Mq aj as follows p = β j Mq aj q T a i p = j=1 β j q T a i Mq aj = β i j=1 = β i = q T a i p = modal participation factor p = j=1 ( ) q T a j p Mq aj 26 / 50
27 AA242B: MECHANICAL VIBRATIONS 27 / 50 Spectral Expansions ( ) Recall that q as q T M = I as s=1 Hence, A R n, A = n Aq as q T as M and A = n s=1 s=1 A = M M = n Mq as q T as M = n Mq as (Mq as ) T s=1 s=1 A = K K = n Kq as q T as M = n ω 2 s Mqas qt as M = s=1 s=1 q as q T as MA A = M 1 M 1 = n q as q T as MM 1 M 1 = q as q T as s=1 s=1 (because M is symmetric) n s=1 ω 2 s Mqas (Mqas )T A = K 1 K 1 = n q as q T as MK 1 = n q as (Mq as ) T K 1 K 1 q as q T as = s=1 s=1 ω 2 s=1 s 27 / 50
28 AA242B: MECHANICAL VIBRATIONS 28 / 50 Forced Harmonic Response M q + Kq = s a cos ωt q(0) = q 0 q(0) = q 0 Solution can be decomposed as q = q H (homogeneous) + q P (particular) q p = q a cos ωt q a = (K ω 2 M) 1 s a where (K ω 2 M) 1 is called the admittance or dynamic influence matrix The forced response is the part of the response that is synchronous to the excitation that is, q p 28 / 50
29 AA242B: MECHANICAL VIBRATIONS 29 / 50 Forced Harmonic Response Rigid body modes: { } u rb ai i=1 q a R n r b n r b, q a = α i u ai + β j q aj i=1 j=1 = s a = (K ω 2 M)q a = r b n r b α i (K ω 2 M)u ai + β j (K ω 2 M)q aj i=1 j=1 r b n r b = s a = α i ω 2 Mu ai + β j (ω 2 j ω 2 )Mq aj i=1 j=1 u T Premultiply by u T a s a j a α j = and premultiply by q T j ω 2 a β i = i r b = q a = i=1 u T n r a s a b i ω 2 ua + i j=1 q T a s r a b j (ωj 2 ω 2 ) qa = j i=1 q T a i s a (ω 2 i ω 2 ) u ai u T n r a b i + ω 2 j=1 q aj q T a j (ωj 2 s ω 2 a ) Since q a = (K ω 2 M) 1 s a = (K ω 2 M) 1 = 1 ω 2 r b i=1 n r b q aj q T u ai u T a j a + i ω 2 (4) j=1 j ω 2 29 / 50
30 AA242B: MECHANICAL VIBRATIONS 30 / 50 Forced Harmonic Response Which excitation s am will generate a harmonic response with an amplitude corresponding to q am? q am = (K ω 2 M) 1 s am s am = (K ω 2 M)q am = s am = Kq am ω 2 Mq am = (ωm 2 ω 2 )Mq am At resonance (ω 2 = ωm) 2 s am = 0 no force is needed to maintain once it is reached q am 30 / 50
31 AA242B: MECHANICAL VIBRATIONS 31 / 50 Forced Harmonic Response The inverse of an admittance is an impedance Z(ω 2 ) = (K ω 2 M) 31 / 50
32 AA242B: MECHANICAL VIBRATIONS 32 / 50 Forced Harmonic Response Application: substructuring (or domain decomposition) the dynamical behavior of a substructure is described by its harmonic response when forces are applied onto its interface boundaries a subsystem is typically described by K and M, has n 1 free dofs q 1, and is connected to the rest of the system by n 2 = n n 1 boundary dofs q 2 where the reaction forces are denoted here by g 2 ( ) ( ) ( ) Z11 Z 12 q1 0 = Z 21 Z 22 q 2 g 2 = Z 11q 1 + Z 12q 2 = 0 q 1 = Z 1 11 Z 12q 2 = (Z 22 Z 21Z 1 11 Z 12)q 2 = Z 22q 2 = g 2 Z 22 is the reduced impedance (reduced to the boundary) 32 / 50
33 AA242B: MECHANICAL VIBRATIONS 33 / 50 Forced Harmonic Response Spectral expansion of Z 22 look at Z 1 11 and let ( ω 2 i, qa i subsystem: from (4), it follows that Z 1 11 = apply twice the relation n 1 = Z 1 11 = K ω2 j=1 ) denote the n1 eigenpairs of the associated dynamical n 1 q aj q T a j ω j 2 ω 2 1 ω 2 j ω 2 = 1 ω 2 j j=1 ω 2 + ω j 2( ω2 j ω 2 ) q aj q T n a 1 q aj q T n j ω j 2( ω2 j ω 2 ) = a 1 j K ω2 ω 4 +ω 4 j=1 j j=1 q aj q T a j ω 4 j ( ω2 j ω 2 ) owing to the M 11orthonormality of the modes and the spectral expansion of K 1 11, the above expression can further be written as Z 1 11 = K ω2 K 1 11 M11K ω4 = Z 22 = K 22 K 21K 1 11 K12 ω 2 [M 22 M 21K 1 11 ω 4 n 1 i=1 n 1 j=1 q aj q T a j ω 4 j ( ω2 j ω 2 ) K12 K21K 1 11 M12 + K21K 1 11 M11K 1 11 K12] [(K 21 ω 2 i M21) qa i ][(K21 ω2 i M21) qa i )]T ω 4 i ( ω2 i ω 2 ) 33 / 50
34 AA242B: MECHANICAL VIBRATIONS 34 / 50 Forced Harmonic Response Z22 = K 22 K 21 K 1 11 K 12 ω 2 [M 22 M 21 K 1 11 K 12 K 21 K 1 11 M 12 + K 21 K 1 11 M 11K 1 n 1 ω 4 [(K 21 ω i 2M 21) q ai ][(K 21 ω i 2M 21) q ai )] T ω i 4( ω2 i ω 2 ) i=1 11 K 12] The first term K 22 K 21 K 1 11 K 12 represents the stiffness of the statically condensed system The second term M 22 M 21 K 1 11 K 12 K 21 K 1 11 M 12 + K 21 K 1 11 M 11K 1 11 K 12 represents the mass of the subsystem statically condensed on the boundary The last term represents the contribution of the subsystem eigenmodes since it is generated by q ai q T a i (K 21 ω 2 i M 21) q ai is the dynamic reaction on the boundary 34 / 50
35 AA242B: MECHANICAL VIBRATIONS 35 / 50 Response to External Loading M q + Kq = p(t) q(0) = q 0 q(0) = q 0 General approach consider the simpler case where there is no rigid body mode eigenmodes (q ai, ωi 2 ), ωi 2 0, i = 1,, n modal superposition: q = Qy = n y i q ai i=1 substitute in equations of dynamic equilibrium = MQÿ + KQy = p(t) Q T MQÿ + Q T KQy = Q T p(t) modal equations ÿ i + ω 2 i y i = q T a i p(t), i = 1,, n y i (t) depend on two constants that can be obtained from the initial conditions q(0) = Qy(0), q(0) = Qẏ(0) y i(0), ẏ i (0) orthogonality conditions 35 / 50
36 AA242B: MECHANICAL VIBRATIONS 36 / 50 Response to External Loading Response to an impulsive force springmass system: m, k, ω 2 = k m impulsive force f (t): force whose amplitude could be infinitely large but which acts for a very short duration of time magnitude of impulse: I = τ+ɛ τ f (t)dt impulsive force = I δ(t) where δ is the delta function centered at t = 0 and satisfying ɛ 0 δ(t)dt = 1 36 / 50
37 AA242B: MECHANICAL VIBRATIONS 37 / 50 Response to External Loading Response to an impulsive force (continue) dynamic equilibrium m τ+ɛ τ d u dt dt + k τ+ɛ τ udt = τ+ɛ τ f (t)dt = I assume that at t = τ, the system is at rest (u(τ) = 0 and u(τ) = 0) τ+ɛ τ+ɛ ü = A/ɛ üdt = A and udt = Aɛ 2 /6 hence τ τ τ+ɛ d u m τ dt dt τ+ɛ = I m u = I u = u(τ + ɛ) u(τ) = I m u(τ + ɛ) = I m udt = 0 u = 0 u(τ + ɛ) u(τ) = 0 u(τ + ɛ) = 0 τ the above equations provide initial conditions for the freevibrations that start at the end of the impulsive load = u(t) = I mω sin ωt 37 / 50
38 AA242B: MECHANICAL VIBRATIONS 38 / 50 Response to External Loading Response to an impulsive force (continue) linear system superposition principle du = f (τ)dτ mω sin ω(t τ) u(t) = 1 mω t 0 f (τ) sin ω(t τ)dτ 38 / 50
39 AA242B: MECHANICAL VIBRATIONS 39 / 50 Response to External Loading Timeintegration of the normal equations let p i (t) = q T a i p(t) = ith modal participation factor modal or normal equation: ÿ i + ωi 2 y i = p i (t) y i (t) = yi H (t) + yi P (t), yi H = A i cos ω i t + B i sin ω i t the particular solution yi P (t) depends on the form of p i (t) however, the general form of a particular solution that satisfies the rest initial conditions is given by the Duhamel s integral y P i (t) = 1 t p i (τ) sin ω i (t τ)dτ ω i complete solution y i (t) = A i cos ω i t + B i sin ω i t + 1 t p i (t) sin ω i (t τ)dτ ω i 0 A i = y i (0), B i = ẏi(0) and y i (0) and ẏ i (0) can be determined from ω i the initial conditions q(0) and q(0) and the orthogonality conditions 0 39 / 50
40 AA242B: MECHANICAL VIBRATIONS 40 / 50 Response to External Loading Response truncation and mode displacement method computational efficiency q = Q r y = r y i q ai, r n what is the effect of modal truncation? consider the case where p(t) = g static load distribution i=1 φ(t) amplification factor for a system initially at rest (q(0) = 0 and q(0) = 0) y i (0) = q T a i Mq(0) = 0 and ẏ i (0) = q T a i M q(0) = 0 A i = B i = 0 = y i (t) = 1 ω i t = q(t) = 0 p i (τ) sin ω i (t τ)dτ = qt a i g ω i [ ] r q ai q T 1 a i g spatial factor ω i i=1 t 0 t 0 φ(τ) sin ω i (t τ)dτ φ(τ) sin ω i (t τ)dτ temporal factor 40 / 50
41 AA242B: MECHANICAL VIBRATIONS 41 / 50 Response to External Loading Response truncation and mode displacement method (continue) general solution for restrained structure initially at rest [ ] r = q(t) = q ai q T 1 t a i g φ(τ) sin ω i (t τ)dτ spatial factor ω i 0 i=1 temporal factor θ i (t) truncated response is accurate if neglected terms are small, which is true if: q T a i g is small for i = r + 1,, n g is well approximated in the range of Q r 1 t φ(τ) sin ω j (t τ)dτ is small for j > r, which depends on the ω j 0 frequency content of φ(t) φ(t) = 1 θ i (t) = 1 cos ω i t 0 for large circular frequencies ω i φ(t) = sin ωt θ i (t) = ω i sin ωt ω sin ω i t ω i (ωi 2 ω 2 ) 41 / 50
42 AA242B: MECHANICAL VIBRATIONS 42 / 50 Response to External Loading Mode acceleration method M q + Kq = p(t) Kq = p(t) M q apply truncated modal representation to the acceleration q(t) = Q r y(t) q(t) = Q r ÿ(t) = Kq = p(t) recall that r Mq ai ÿ i = q = K 1 p(t) i=1 and that for a system initially at rest r i=1 q ai ÿ ωi 2 i ÿ i + ω 2 i y i = q T a i p(t) (5) y i (t) = 1 t q T a ω i p(τ) sin ω i (t τ)dτ (6) i 0 (5) and (6) ÿ i (t) = q T a i ( p(t) ωi t 0 p(τ) sin ω i(t τ)dτ ) 42 / 50
43 AA242B: MECHANICAL VIBRATIONS 43 / 50 Response to External Loading Mode acceleration method (continue) r substitute in q(t) = K 1 q ai p(t) ÿ ωi 2 i = q(t) = r i=1 q ai q T a i ω i t 0 i=1 p(τ) sin ω i (t τ)dτ+ recall the spectral expansion K 1 = n = q(t) = r i=1 q ai q T a i ω i t 0 i=1 q ai q T a i ω 2 i ( p(τ) sin ω i (t τ)dτ + K 1 ( i=r+1 r i=1 q ai q T a i ω 2 i q ai q T a i ω 2 i which shows that the mode acceleration method complements the truncated mode displacement solution with the missing terms using the modal expansion of the static response ) ) p(t) p(t) 43 / 50
44 AA242B: MECHANICAL VIBRATIONS 44 / 50 Response to External Loading Direct timeintegration methods for solving M q + Kq = p(t) q(0) = q 0 q(0) = q 0 will be covered towards the end of this course 44 / 50
45 AA242B: MECHANICAL VIBRATIONS 45 / 50 Mechanical Systems Excited Through Support Motion The general case ( ) q1 q = where q q 2 is prescribed 2 ( ) ( ) ( ) ( ) ( M11 M 12 q1 K11 K + 12 q1 0 = M 21 M 22 q 2 K 21 K 22 q 2 r 2 (t) The first equation gives M 11 q 1 + K 11 q 1 = K 12 q 2 M 12 q 2 q 1 (t) Substitute in second equation = r 2 (t) = K 21 q 1 + M 21 q 1 + K 22 q 2 + M 22 q 2 ) 45 / 50
46 AA242B: MECHANICAL VIBRATIONS 46 / 50 Mechanical Systems Excited Through Support Motion Quasistatic response of q K 11 q 1 = K 12 q 2 0 = q qs 1 = K 1 11 K 12q 2 = Sq 2 Decompose q 1 = q qs 1 + z 1 = Sq 2 + z 1 ( ) ( q1 I S = q(t) = = 0 I q 2 ) ( z1 q 2 ) where z 1 represents the sole dynamics part of the response Substitute in the first dynamic equation and exploit above results = M 11 S q 2 + M 11 z 1 + K 11 q qs 1 + K 11z 1 = K 12 q 2 M 12 q 2 { M11 z = 1 + K 11 z 1 = g 1 (t) g 1 (t) = M 11 q qs 1 M 12 q 2 = (M 11 S + M 12 ) q 2 46 / 50
47 AA242B: MECHANICAL VIBRATIONS 47 / 50 Mechanical Systems Excited Through Support Motion Consider next the system fixed to the ground and solve the corresponding EVP K 11 x = ω 2 M 11 x Q = [ q ai ] = z 1 (t) = Qη(t) Solve M 11 z 1 + K 11 z 1 = g 1 (t) for z 1 (t) using the modal superposition technique 47 / 50
48 AA242B: MECHANICAL VIBRATIONS 48 / 50 Mechanical Systems Excited Through Support Motion Case of a global support acceleration q 2 (t) = u 2 φ(t), where u = [u 1 u 2 ] T denotes a rigid body mode decomposition of the solution into a rigid body motion and a relative displacement y [ ] [ ] [ ] q = q rb q1 u1 ÿ1 + ÿ = φ(t) + q 2 u 2 0 u 1 = Su 2 g 1(t) = (M 11Su 2 + M 12u 2)φ(t) = g 1(t) = (M 11u 1 + M 12u 2)φ(t) 48 / 50
49 AA242B: MECHANICAL VIBRATIONS 49 / 50 Mechanical Systems Excited Through Support Motion Method of additional masses (approximate method) suppose the system is subjected not to a specified q 2(t) but to an imposed ( ) 0 f(t) suppose that masses associated with q 2 are increased to M 22 + M 22 then ( ) ( ) M11 M 12 q1 M 21 M 22 + M 22 ) ( ) ( q1 K11 K 12 + q 2 K 21 K 22 by elimination one obtains q 2 = (M 22 + M 22) 1 (f(t) K 22q 2 K 21q 1 M 21 q 1) and q 2 = ( 0 f(t) M 11 q 1 + K 11q 1 = K 12q 2 M 12(M 22 + M 22 ) 1 (f(t) K 22q 2 K 21q 1 M 21 q 1) ) 49 / 50
50 AA242B: MECHANICAL VIBRATIONS 50 / 50 Mechanical Systems Excited Through Support Motion Method of additional masses (continue) therefore {M 11 M 12(M 22 + M 22 ) 1 M 21 } q 1 + = K 12q 2 M 12(M 22 + M 22 ) 1 (f(t) K 22q 2) {K 11 M 12(M 22 + M 22 ) 1 K 21 } q 1 now if (M 22 + M 22) 1 0 and if f(t) = (M 22 + M 22) q 2, one obtains M 11 q 1+K 11q 1 = K 12q 2 M 12 q 2 = M 11 q 1+M 12 q 2+K 11q 1+K 12q 2 = 0 which is the same equation as in the general case of excitation through support motion for M 22 very large, the response of the system with ground motion is the same as the response of the system with added lumped mass M 22 and the forcing function ( 0 (M 22 + M 22) q 2 ) ( 0 M 22 q 2 where q 2 is given apply same solution method as for problems of response to external loading ) 50 / 50
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