AA242B: MECHANICAL VIBRATIONS 1 / 50 AA242B: MECHANICAL VIBRATIONS Undamped Vibrations of n-DOF 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-13:9780471975465 1 / 50 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 AA242B: MECHANICAL VIBRATIONS 3 / 50 Linear Vibrations Equilibrium configuration qs (t) = qs (0) q̇s (t) = 0 s = 1, · · · , n (1) Recall the Lagrange equations of motion d ∂T ∂T ∂V ∂D − + − − + Qs (t) = 0 dt ∂ q̇s ∂qs ∂qs ∂ q̇s where T = T0 + T1 + T2 Recall the generalized gyroscopic forces Fs = − n n X X ∂ 2 T1 ∂T1 q̇r + = ∂ q̇s ∂qr ∂qs r =1 r =1 ∂ 2 T1 ∂ 2 T1 − ∂qs ∂ q̇r ∂qr ∂ q̇s ! q̇r , s = 1, · · · , n Definition: the effective potential energy is defined as V ? = V − T0 The Lagrange equations of motion can be re-written as d ∂T2 ∂T2 ∂V ? ∂D ∂ ∂T1 − = Qs (t) − − + Fs − dt ∂ q̇s ∂qs ∂qs ∂ q̇s ∂t ∂ q̇s 3 / 50 AA242B: MECHANICAL VIBRATIONS 4 / 50 Linear Vibrations Recall that T0 (q, t) = D = 2 N 3 ∂Uik 1 XX (q, t) mk 2 k=1 i=1 ∂t N Z v (q̇) X k Ck fk (γ)dγ k=1 (transport kinetic energy) (dissipation function) 0 From the Lagrange equations of motion d ∂T2 ∂T2 ∂V ? ∂D ∂ ∂T1 − = Qs (t) − − + Fs − dt ∂ q̇s ∂qs ∂qs ∂ q̇s ∂t ∂ q̇s it follows that an equilibrium configuration exists if and only if ∂V ? 0 = Qs (t) − ⇒ Qs (t) and V ? are function of q only and not ∂qs explicitly dependent on time ⇒ Qs (t) = 0 and V ? = V ? (q) At equilibrium Qs (t) = 0 and ∂V ? ∂(V − T0 ) = = 0, s = 1, · · · , n ∂qs ∂qs 4 / 50 AA242B: MECHANICAL VIBRATIONS 5 / 50 Linear Vibrations Free-Vibrations About a Stable Equilibrium Position Consider first a system that does not undergo a transport or overall motion ⇒ T = T2 (q̇) The equilibrium position is then given by Qs (t) = 0 and ∂V = 0, ∂qs 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, qs = 0, s = 1, · · · , n (in which case the qs represent the deviation from equilibrium) Since the potential energy is defined only up to a constant, choose this constant so that V(qs = 0) = 0 Now, suppose that a certain energy E(0) is initially given to the system in equilibrium 5 / 50 AA242B: MECHANICAL VIBRATIONS 6 / 50 Linear Vibrations Free-Vibrations About a Stable Equilibrium Position Definition: the equilibrium position (qs = 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 AA242B: MECHANICAL VIBRATIONS 7 / 50 Linear Vibrations Free-Vibrations About a Stable Equilibrium Position Linearization of T and V around an equilibrium position ∂V (qs = 0, = 0) ∂qs actually, this means obtaining a quadratic form of T and V in q and q̇ so that the corresponding generalized forces are linear since qs (t) represent deviations from equilibrium, V can be expanded as follows n n n X 1 X X ∂2V ∂V |q=0 qs + |q=0 qs qr + O(q3 ) V(q) = V(0) + ∂q 2 ∂q s s ∂qr s=1 s=1 r =1 = V(0) + n n 1 X X ∂2V |q=0 qs qr + O(q3 ) 2 s=1 r =1 ∂qs ∂qr since the potential energy is defined only up to a constant, if this constant is chosen so that V(0) = 0, then a second-order approximation of V(q) is given by V(q) = n n 1 X X ∂2V |q=0 qs qr , 2 s=1 r =1 ∂qs ∂qr for q 6= 0 7 / 50 AA242B: MECHANICAL VIBRATIONS 8 / 50 Linear Vibrations Free-Vibrations About a Stable Equilibrium Position Stiffness matrix let K = [ksr ] where ksr = krs = =⇒ V(q) = ∂2V |q=0 ∂qs ∂qr 1 T q Kq > 0, 2 for q 6= 0 K is symmetric positive (semi-) definite 8 / 50 AA242B: MECHANICAL VIBRATIONS 9 / 50 Linear Vibrations Free-Vibrations About a Stable Equilibrium Position Recall that n n N 3 1 XXXX ∂Uik ∂Uik T2 = mk q̇s q̇r 2 s=1 r =1 ∂qs ∂qr (relative kinetic energy) k=1 i=1 T2 (q, q̇) n n X X ∂T2 ∂T2 |q=0,q̇=0 qs + |q=0,q̇=0 q̇s ∂qs ∂ q̇s s=1 s=1 = T2 (0, 0) + + n n n X n X ∂ 2 T2 1 X X ∂ 2 T2 |q=0,q̇=0 qs qr + |q=0,q̇=0 qs q̇r 2 s=1 r =1 ∂qs ∂qr ∂q s ∂ q̇r s=1 r =1 + n n 1 X X ∂ 2 T2 3 3 |q=0,q̇=0 q̇s q̇r + O(q , q̇ ) 2 s=1 r =1 ∂ q̇s ∂ q̇r = n n 1 X X ∂ 2 T2 3 3 |q=0,q̇=0 q̇s q̇r + O(q , q̇ ) 2 s=1 r =1 ∂ q̇s ∂ q̇r 9 / 50 AA242B: MECHANICAL VIBRATIONS 10 / 50 Linear Vibrations Free-Vibrations About a Stable Equilibrium Position Hence, a second-order approximation of T2 (q̇) is given by T2 (q̇) = where" 1 T q̇ Mq̇ > 0, 2 for q̇ 6= 0 N 3 X X ∂ 2 T2 ∂Uik ∂Uik M = msr = mrs = |q=0 = mk |q=0 |q=0 ∂ q̇s ∂ q̇r ∂qs ∂qr i=1 k=1 is the mass matrix and is symmetric positive definite # 10 / 50 AA242B: MECHANICAL VIBRATIONS 11 / 50 Linear Vibrations Free-Vibrations About a Stable Equilibrium Position Free-vibrations about a stable equilibrium position of a conservative system that does not undergo a transport or overall motion (T0 = T1 = 0) ∂T2 ∂V d ∂T2 − =− dt ∂ q̇s ∂qs ∂qs =⇒ d (Mq̇) − 0 = −Kq dt =⇒ Mq̈ + Kq = 0 11 / 50 AA242B: MECHANICAL VIBRATIONS 12 / 50 Linear Vibrations Free-Vibrations 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 ? ∂(V − T0 ) = = 0, ∂qs ∂qs s = 1, · · · , n corresponds to the balance of forces deriving from a potential ∂V ∂T0 and centrifugal forces ∂qs ∂qs It is an equilibrium configuration in the sense that q̇s — which represent here the generalized velocities relative to a steady motion — are zero but the system is not idle 12 / 50 AA242B: MECHANICAL VIBRATIONS 13 / 50 Linear Vibrations Free-Vibrations About an Equilibrium Configuration Linearizations effective potential energy V ? ⇒ effective stiffness matrix K? 1 T ? q K q > 0, for q 6= 0 2 2 ∂ T0 ? ? where K = ksr = ksr − |q=0 ∂qs ∂qr V ? (q) = mutual kinetic energy T1 = n n X N X 3 X X ∂Uik ∂Uik ∂T1 mk q̇s = q̇s (q) ∂t ∂q ∂ q̇s s s=1 s=1 i=1 k=1 = n X n q̇s s=1 n X n X X ∂ 2 T1 ∂T1 (0) + |q=0 qr + O(q2 ) ∂ q̇s ∂ q̇ s qr r =1 ! ∂ 2 T1 qr |q=0 q̇s qr ⇒ T1 (q, q̇) = q̇T Fq ∂ q̇ s s=1 r =1 ∂ 2 T1 where F = fsr = |q=0 ∂ q̇s ∂qr ≈ 13 / 50 AA242B: MECHANICAL VIBRATIONS 14 / 50 Linear Vibrations Free-Vibrations About an Equilibrium Configuration Equations of free-vibration around an equilibrium configuration the equilibrium configuration generated by a steady motion remains stable as long as V ? = V − T0 ≥ 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 ∂V ? ∂D ∂ ∂T1 d ∂T2 ∂T2 = Qs (t) − − +Fs − − dt ∂ q̇s ∂qs ∂ q̇s ∂t ∂ q̇s | {z } ∂qs |{z} |{z} | {z } 0 0 where Fs = n X r =1 0 ∂ 2 T1 ∂ 2 T1 − ∂qs ∂ q̇r ∂qr ∂ q̇s 0 q̇r = (FT − F)q̇ Mq̈ + Gq̇ + K? q = 0 where G = F − FT = −GT is the gyroscopic coupling matrix 14 / 50 AA242B: MECHANICAL VIBRATIONS 15 / 50 Linear Vibrations Free-Vibrations About an Equilibrium Configuration Example X = (a + x) cos Ωt V = 1 2 kx 2 T0 = Y = (a + x) sin Ωt 1 2 2 Ω m(a + x) 2 T1 = 0 2 2 2 2 2 v = Ẋ + Ẏ = (a + x) Ω + ẋ T2 = 1 2 mẋ 2 V ? = 2 1 2 1 2 2 kx − Ω m(a + x) 2 2 equilibrium configuration ∂V ? Ω2 ma = 0 =⇒ kx − Ω2 m(a + x) = 0 =⇒ xeq = ∂x k − Ω2 m the system becomes unstable for Ω2 = k? = k m ∂2V ? k = k − Ω2 m ⇒ system is unstable for Ω2 ≥ ∂x 2 m 15 / 50 AA242B: MECHANICAL VIBRATIONS 16 / 50 Natural Vibration Modes Free-vibration equations: Mq̈ + Kq = 0 q(t) = qa e iωt ⇒ (K − ω 2 M)qa = 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 (ωi2 , qai ) Rigid body mode(s): ωj2 = 0 ⇒ Kqaj = 0 1 For a rigid body mode, V(qaj ) = qT Kqaj = 0 2 aj 16 / 50 AA242B: MECHANICAL VIBRATIONS 17 / 50 Orthogonality of Natural Vibration Modes Distinct Frequencies Consider two distinct eigenpairs (ωi2 , qai ) and (ωj2 , qaj ) T 2 qT aj Kqai = qaj ωi Mqai (2) qT ai Kqaj (3) = 2 qT ai ωj Mqaj Because M and K are symmetric (2) − (3)T ⇒ 0 = (ωi2 − ωj2 )qaj T Mqai since ωi2 6= ωj2 ⇒ qaj T Mqai = 0 and qaj T Kqai = 0 17 / 50 AA242B: MECHANICAL VIBRATIONS 18 / 50 Orthogonality of Natural Vibration Modes Distinct Frequencies Physical interpretation of the orthogonality conditions qaj T Mqai = 0 ⇒ qaj T ωi2 Mqai = (ωi2 Mqai )T qaj = 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 qaj T Kqai = 0 ⇒ (Kqai )T qaj = 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 AA242B: MECHANICAL VIBRATIONS 19 / 50 Orthogonality of Natural Vibration Modes Distinct Frequencies Rayleigh quotient Kqai = ωi2 Mqai ⇒ qai T Kqai = ωi2 qai T Mqai ⇒ ωi2 = qai T Kqai γi = qai T Mqai µi γi = generalized stiffness coefficient of mode i (measures the contribution of mode i to the elastic deformation energy) µi = generalized mass coefficient of mode i (measures the contribution of mode i to the kinetic energy) Since the amplitude of qai is determined up to a factor only ⇒ γi and µi are determined up to a constant factor only Mass normalization qaj T Mqai T qaj Kqai = δij = ωi2 δij 19 / 50 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 ωp2 of the system Kx = ω 2 Mx corresponds a number of linearly independent eigenvectors {qai } equal to the root multiplicity 20 / 50 AA242B: MECHANICAL VIBRATIONS 21 / 50 Modal Superposition Analysis n-dof system: M ∈ Rn×n , K ∈ Rn , and q ∈ Rn Coupled system of ordinary differential equations Mq̈ + Kq = 0 q(0) = q0 q̇(0) = q̇0 21 / 50 AA242B: MECHANICAL VIBRATIONS 22 / 50 Modal Superposition Analysis Natural vibration modes (eigenmodes) Kqai = ωi2 Mqai , Q = [qa1 i = 1, · · · , n T Q KQ = qa2 · · · qan ] ⇒ QT MQ = 2 ω1 .. Ω2 = . 2 ωn Ω2 I Truncated eigenbasis Q r = qa 1 qa 2 ··· qa r , QT r KQr QT r MQr r << n ⇒ 2 Ωr = ω12 = = Ω2r Ir (reduced stiffness matrix) (reduced mass matrix) .. . ωr2 22 / 50 AA242B: MECHANICAL VIBRATIONS 23 / 50 Modal Superposition Analysis Modal superposition: q = Qr y = r P yi qai where i=1 y = [y1 y2 ··· T yr ] and yi is called the modal displacement Substitute in Mq̈ + Kq = 0 =⇒ MQr ÿ + KQr y = 0 T =⇒ QT r MQr ÿ + Qr KQr y = 0 =⇒ Ir ÿ + Ω2r y = 0 Uncoupled differential equations (modal equations) ÿi + ωi2 yi = 0, i = 1, · · · r 23 / 50 AA242B: MECHANICAL VIBRATIONS 24 / 50 Modal Superposition Analysis ÿi + ωi2 yi = 0, i = 1, · · · r Case 1: ωi2 = 0 (rigid body mode) yi = ai t + bi Case 2: ωj2 6= 0 yj = cj cos ωj t + dj sin ωj t General case: rb rigid body modes q= rb rX −rb X (ai t + bi )qai + (cj cos ωj t + dj sin ωj t)qaj i=1 j=1 Initial conditions q(0) = q0 = Qy(0) and q̇(0) = q̇0 = Qẏ(0) 24 / 50 AA242B: MECHANICAL VIBRATIONS 25 / 50 Modal Superposition Analysis q0 = Qy(0) and T i q̇0 = Qẏ(0) T i =⇒ qa Mq0 = qa MQy(0) From the orthogonality properties of the natural mode shapes (eigenvectors) it follows that T i T i T i T i qa Mq0 = qa MQy(0) = qa Mqai yi (0) = yi (0) ⇒ yi (0) = qa Mq(0) T Case 1: ωi2 = 0 (rigid body mode) ⇒ ai × 0 + bi = qT a Mq(0) ⇒ bi = qa Mq(0) i i T Case 2: ωj2 6= 0 ⇒ cj × 1 + dj × 0 = qT a Mq(0) ⇒ cj = qa Mq(0) j j Case 1: ωi2 = 0 (rigid body mode) ⇒ ai = qT a Mq̇(0) i Case 2: ωj2 6= 0 ⇒ dj = qT a Mq̇(0) j Thus, the general solution is q(t) = rb h r −rb i X X sin ωj t T T T T qa Mq̇(0) t + qa Mq(0) qai + qa Mq(0) cos ωj t + qa Mq̇(0) qa j i i j j ωj i=1 j=1 25 / 50 AA242B: MECHANICAL VIBRATIONS 26 / 50 Spectral Expansions n ∀x ∈ R , x= n X T j αs qas ⇒ qa Mx = s=1 n =⇒ ∀x ∈ R , x= n X T j αs qa Mqas = αj s=1 n n n X X X T T T qas Mx qas = qas qas Mx = qas qas Mx = n X T (qas qas )M s=1 s=1 s=1 ⇒ n X s=1 T qas qas ! x M=I s=1 This is the same result as QT MQ = I A given load p can be expanded in terms of the inertia forces generated by the eigenmodes, Mqaj as follows p= n X j=1 βj Mqaj ⇒ qT ai p = n X βj qT ai Mqaj = βi j=1 =⇒ βi = qT ai p = modal participation factor → p = n X qT p Mqaj aj j=1 26 / 50 AA242B: MECHANICAL VIBRATIONS 27 / 50 Spectral Expansions Recall that n P qa s q T M=I as s=1 Hence, ∀A ∈ Rn , n P A= s=1 A=M⇒M= n P s=1 A=K⇒K= n P s=1 Aqas qT as M and A = Mqas qT as M = Kqas qT as M = A = M−1 ⇒ M−1 = n P s=1 A = K−1 ⇒ K−1 = n P s=1 n P s=1 n P s=1 n P s=1 Mqas (Mqas )T ωs2 Mqas qT as M = −1 ⇒ M qas qT as MM −1 = qas qT as MK −1 = qas qT as MA (because M is symmetric) n P s=1 n X ωs2 Mqas (Mqas )T T qas qas s=1 n P s=1 −1 qas (Mqas )T K−1 ⇒ K = n X qas qT as s=1 ωs2 27 / 50 AA242B: MECHANICAL VIBRATIONS 28 / 50 Forced Harmonic Response Mq̈ + Kq = q(0) = q̇(0) = sa cos ωt q0 q̇0 Solution can be decomposed as q = qH (homogeneous) + qP (particular) qp = qa cos ωt ⇒ qa = (K − ω 2 M)−1 sa 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, qp 28 / 50 AA242B: MECHANICAL VIBRATIONS 29 / 50 Forced Harmonic Response Rigid body modes: ∀qa ∈ Rn , qa = uai rb P i=1 r b i=1 αi uai + n−rb P β j qa j j=1 2 =⇒ sa = (K − ω M)qa = rb X n−rb 2 αi (K − ω M)uai + i=1 =⇒ sa = − rb X j =⇒ qa = − rb X uT a sa i i=1 ω2 ua i + uT a sa j ω2 2 βj (K − ω M)qaj j=1 n−rb 2 αi ω Muai + i=1 Premultiply by uT a ⇒ αj = − X X 2 2 βj (ωj − ω )Mqaj j=1 and premultiply by qT a ⇒ βi = i n−rb X qT a sa j=1 (ωj2 − ω 2 ) j qaj = − rb X ua i uT a i i=1 ω2 qT a sa i (ωi2 + − ω2 ) n−rb X qaj qT a j=1 (ωj2 − ω 2 ) j sa Since qa = (K − ω 2 M)−1 sa 2 =⇒ (K − ω M) −1 =− rb n−rb q qT X aj a 1 X j T ua i ua + 2 − ω2 2 i ω i=1 ω j j=1 (4) 29 / 50 AA242B: MECHANICAL VIBRATIONS 30 / 50 Forced Harmonic Response Which excitation sam will generate a harmonic response with an amplitude corresponding to qam ? qam = (K − ω 2 M)−1 sam ⇒ sam = (K − ω 2 M)qam =⇒ sam = 2 Kqam − ω 2 Mqam = (ωm − ω 2 )Mqam 2 At resonance (ω 2 = ωm ) sam = 0 ⇒ no force is needed to maintain qam once it is reached 30 / 50 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 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 n1 free dofs q1 , and is connected to the rest of the system by n2 = n − n1 boundary dofs q2 where the reaction forces are denoted here by g2 Z11 Z12 q1 0 = Z21 Z22 q2 g2 =⇒ Z11 q1 + Z12 q2 = 0 ⇒ q1 = −Z−1 11 Z12 q2 ? =⇒ (Z22 − Z21 Z−1 11 Z12 )q2 = Z22 q2 = g2 ? Z22 is the “reduced” impedance (reduced to the boundary) 32 / 50 AA242B: MECHANICAL VIBRATIONS 33 / 50 Forced Harmonic Response Spectral expansion of Z? 22 2 look at Z−1 11 and let (ω̃i , q̃ai ) denote the n1 eigenpairs of the associated dynamical n1 q̃aj q̃T X aj −1 subsystem: from (4), it follows that Z11 = ω̃j2 − ω 2 j=1 apply twice the relation ω2 1 1 = 2 + 2 2 ω̃j2 − ω 2 ω̃j ω̃j (ω̃j − ω 2 ) −1 −1 =⇒ Z11 = K11 +ω 2 n1 X q̃aj q̃T a j=1 ω̃j2 (ω̃j2 − ω 2 ) j −1 = K11 +ω 2 n1 q̃ q̃T X aj a j j=1 +ω ω̃j4 4 n1 X q̃aj q̃T a j=1 ω̃j4 (ω̃j2 − ω 2 ) j owing to the M11 -orthonormality of the modes and the spectral expansion of K−1 11 , the above expression can further be written as −1 Z11 ? =⇒ Z22 −1 2 −1 −1 = K11 + ω K11 M11 K11 + ω = K22 − K21 K11 K12 − n1 X q̃aj q̃T a j ω̃j4 (ω̃j2 j=1 − ω2 ) −1 −1 2 −1 −1 −1 ω [M22 − M21 K11 K12 − K21 K11 M12 + K21 K11 M11 K11 K12 ] n1 − 4 ω 4 X [(K21 − i=1 ω̃i2 M21 )q̃ai ][(K21 − ω̃i4 (ω̃i2 − ω 2 ) ω̃i2 M21 )q̃ai )]T 33 / 50 AA242B: MECHANICAL VIBRATIONS 34 / 50 Forced Harmonic Response ? Z22 = K22 − K21 K−1 11 K12 −1 −1 −1 −ω 2 [M22 − M21 K−1 11 K12 − K21 K11 M12 + K21 K11 M11 K11 K12 ] n1 X [(K21 − ω̃i2 M21 )q̃ai ][(K21 − ω̃i2 M21 )q̃ai )]T −ω 4 ω̃i4 (ω̃i2 − ω 2 ) i=1 The first term K22 − K21 K−1 11 K12 represents the stiffness of the statically condensed system The second term −1 −1 −1 M22 − M21 K−1 11 K12 − K21 K11 M12 + K21 K11 M11 K11 K12 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 q̃ai q̃T ai (K21 − ω̃i2 M21 )q̃ai is the dynamic reaction on the boundary 34 / 50 AA242B: MECHANICAL VIBRATIONS 35 / 50 Response to External Loading Mq̈ + Kq = q(0) = q̇(0) = p(t) q0 q̇0 General approach consider the simpler case where there is no rigid body mode ⇒ eigenmodes (qai , ωi2 ), ωi2 6= 0, i = 1, · · · , n n P modal superposition: q = Qy = yi qai i=1 substitute in equations of dynamic equilibrium =⇒ MQÿ + KQy = p(t) ⇒ QT MQÿ + QT KQy = QT p(t) modal equations ÿi + ωi2 yi = qTai p(t), i = 1, · · · , n yi (t) depend on two constants that can be obtained from the initial conditions q(0) = Qy(0), q̇(0) = Qẏ(0) ⇒ yi (0), ẏi (0) orthogonality conditions 35 / 50 AA242B: MECHANICAL VIBRATIONS 36 / 50 Response to External Loading Response to an impulsive force k m impulsive force f (t): force whose amplitude could be infinitely large but which acts for a very short Z duration of time spring-mass system: m, k, ω 2 = τ + magnitude of impulse: I = f (t)dt τ impulsive force = I δ(t) where δ isZthe “delta” function centered at t = 0 and satisfying δ(t)dt = 1 0 36 / 50 AA242B: MECHANICAL VIBRATIONS 37 / 50 Response to External Loading Response to an impulsive force (continue) dynamic equilibrium Z τ + Z τ + Z τ + d u̇ m dt + k udt = f (t)dt = I dt τ τ τ assume thatZat t = τ , the system isZat rest (u(τ ) = 0 and u̇(τ ) = 0) τ + τ + üdt = A and udt = A2 /6 ü = A/ ⇒ τ hence τ + Z m τ d u̇ dt dt τ I I ⇒ u̇(τ + ) = m m = I ⇒ m∆u̇ = I ⇒ ∆u̇ = u̇(τ + ) − u̇(τ ) = = 0 ⇒ ∆u = 0 ⇒ u(τ + ) − u(τ ) = 0 ⇒ u(τ + ) = 0 τ + Z u̇dt τ the above equations provide initial conditions for the free-vibrations that start at the end of the impulsive load =⇒ u(t) = I sin ωt mω 37 / 50 AA242B: MECHANICAL VIBRATIONS 38 / 50 Response to External Loading Response to an impulsive force (continue) linear system ⇒ superposition principle du = f (τ )dτ 1 sin ω(t − τ ) ⇒ u(t) = mω mω Z t f (τ ) sin ω(t − τ )dτ 0 38 / 50 AA242B: MECHANICAL VIBRATIONS 39 / 50 Response to External Loading Time-integration of the normal equations let pi (t) = qTai p(t) = i-th modal participation factor modal or normal equation: ÿi + ωi2 yi = pi (t) yi (t) = yiH (t) + yiP (t), yiH = Ai cos ωi t + Bi sin ωi t the particular solution yiP (t) depends on the form of pi (t) however, the general form of a particular solution that satisfies the rest initial conditions is given by the Duhamel’s integral Z t 1 yiP (t) = pi (τ ) sin ωi (t − τ )dτ ωi 0 complete solution yi (t) = Ai cos ωi t + Bi sin ωi t + 1 ωi Z t pi (t) sin ωi (t − τ )dτ 0 ẏi (0) and yi (0) and ẏi (0) can be determined from ωi the initial conditions q(0) and q̇(0) and the orthogonality conditions Ai = yi (0), Bi = 39 / 50 AA242B: MECHANICAL VIBRATIONS 40 / 50 Response to External Loading Response truncation and mode displacement method computational efficiency ⇒ q = Qr y = r P r n yi qai , i=1 what is the effect of modal truncation? consider the case where p(t) = g × static load φ(t) amplification distribution factor for a system initially at rest (q(0) = 0 and q̇(0) = 0) yi (0) = qTai Mq(0) = 0 and ẏi (0) = qTai Mq̇(0) = 0 ⇒ Ai = Bi = 0 =⇒ yi (t) = 1 ωi t Z =⇒ q(t) = pi (τ ) sin ωi (t−τ )dτ = 0 r X i=1 " # qai qTai g spatial factor qTai g ωi t Z φ(τ ) sin ωi (t − τ )dτ 0 1 ωi t Z 0 φ(τ ) sin ωi (t − τ )dτ temporal factor 40 / 50 AA242B: MECHANICAL VIBRATIONS 41 / 50 Response to External Loading Response truncation and mode displacement method (continue) general solution for restrained structure initially at rest # " Z t r X 1 qai qTai g =⇒ q(t) = φ(τ ) sin ωi (t − τ )dτ ωi 0 i=1 spatial factor temporal factor θi (t) truncated response is accurate if neglected terms are small, which is true if: qT ai g is small for i = r + 1, · · · , n ⇒ g is well approximated in the range Z tof Qr 1 φ(τ ) sin ωj (t − τ )dτ is small for j > r , which depends on the ωj 0 frequency content of φ(t) φ(t) = 1 ⇒ φ(t) = sin ωt ⇒ 1 − cos ωi t → 0 for large circular frequencies ωi ωi sin ωt − ω sin ωi t θi (t) = ωi (ωi2 − ω 2 ) θi (t) = 41 / 50 AA242B: MECHANICAL VIBRATIONS 42 / 50 Response to External Loading Mode acceleration method Mq̈ + Kq = p(t) ⇒ Kq = p(t) − Mq̈ apply truncated modal representation to the acceleration q(t) = Qr y(t) ⇒ q̈(t) = Qr ÿ(t) =⇒ Kq = p(t) − r X Mqai ÿi =⇒ q = K−1 p(t) − i=1 r X qai ÿi ωi2 i=1 recall that ÿi + ωi2 yi = qTai p(t) and that for a system initially at rest Z t 1 yi (t) = qTai p(τ ) sin ωi (t − τ )dτ ωi 0 Rt (5) and (6) ⇒ ÿi (t) = qTai p(t) − ωi 0 p(τ ) sin ωi (t − τ )dτ (5) (6) 42 / 50 AA242B: MECHANICAL VIBRATIONS 43 / 50 Response to External Loading Mode acceleration method (continue) substitute in q(t) = K−1 p(t) − r X qai ÿi ωi2 i=1 ! Z r r X X qai qTai qai qTai t −1 p(t) p(τ ) sin ωi (t−τ )dτ + K − =⇒ q(t) = ωi ωi2 0 i=1 i=1 recall the spectral expansion K−1 = n q qT P ai ai 2 i=1 ωi Z r X qai qTai t p(τ ) sin ωi (t − τ )dτ + =⇒ q(t) = ωi 0 i=1 n X qai qTai ωi2 i=r +1 ! p(t) 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 43 / 50 AA242B: MECHANICAL VIBRATIONS 44 / 50 Response to External Loading Direct time-integration methods for Mq̈ + Kq q(0) q̇(0) solving = p(t) = q0 = q̇0 will be covered towards the end of this course 44 / 50 AA242B: MECHANICAL VIBRATIONS 45 / 50 Mechanical Systems Excited Through Support Motion The general case q= M11 M21 M12 M22 q1 q2 q̈1 q̈2 where q2 is prescribed + K11 K21 K12 K22 q1 q2 = 0 r2 (t) The first equation gives M11 q̈1 + K11 q1 = −K12 q2 − M12 q̈2 ⇒ q1 (t) Substitute in second equation =⇒ r2 (t) = K21 q1 + M21 q̈1 + K22 q2 + M22 q̈2 45 / 50 AA242B: MECHANICAL VIBRATIONS 46 / 50 Mechanical Systems Excited Through Support Motion Quasi-static response of q1 0 + K11 q1 = −K12 q2 − 0 −1 =⇒ qqs 1 = −K11 K12 q2 = Sq2 Decompose q1 = qqs 1 + z1 = Sq2 + z1 q1 I =⇒ q(t) = = q2 0 S I z1 q2 where z1 represents the sole dynamics part of the response Substitute in the first dynamic equation and exploit above results =⇒ M11 Sq̈2 + M11 z̈1 + K11 qqs 1 + K11 z1 = −K12 q2 − M12 q̈2 M11 z̈1 + K11 z1 = g1 (t) =⇒ g1 (t) = −M11 q̈qs 1 − M12 q̈2 = −(M11 S + M12 )q̈2 46 / 50 AA242B: MECHANICAL VIBRATIONS 47 / 50 Mechanical Systems Excited Through Support Motion Consider next the system fixed to the ground and solve the corresponding EVP e = [· · · q̃a · · · ] K11 x = ω 2 M11 x ⇒ Q i e =⇒ z1 (t) = Qη(t) Solve M11 z̈1 + K11 z1 = g1 (t) for z1 (t) using the modal superposition technique 47 / 50 AA242B: MECHANICAL VIBRATIONS 48 / 50 Mechanical Systems Excited Through Support Motion Case of a global support acceleration q̈2 (t) = u2 φ(t), where u = [u1 u2 ]T denotes a rigid body mode decomposition of the solution into a rigid body motion and a relative displacement y u1 ÿ1 q̈1 = φ(t) + q̈ = q̈rb + ÿ ⇒ q̈2 u2 0 u1 = Su2 ⇒ g1 (t) = −(M11 Su2 + M12 u2 )φ(t) =⇒ g1 (t) = −(M11 u1 + M12 u2 )φ(t) 48 / 50 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 q2 (t) but to an imposed 0 f(t) suppose that masses associated with q2 are increased to M22 + M?22 then K11 K12 q1 0 q̈1 M11 M12 + = q̈2 K21 K22 q2 f(t) M21 M22 + M?22 by elimination one obtains q̈2 = (M22 + M?22 )−1 (f(t) − K22 q2 − K21 q1 − M21 q̈1 ) and ? M11 q̈1 + K11 q1 = −K12 q2 − M12 (M22 + M22 ) −1 (f(t) − K22 q2 − K21 q1 − M21 q̈1 ) 49 / 50 AA242B: MECHANICAL VIBRATIONS 50 / 50 Mechanical Systems Excited Through Support Motion Method of additional masses (continue) therefore {M11 ? −1 − M12 (M22 + M22 ) = −K12 q2 − M12 (M22 + M22 ) M21 } q̈1 + ? −1 n o ? −1 K11 − M12 (M22 + M22 ) K21 q1 (f(t) − K22 q2 ) now if (M22 + M?22 )−1 → 0 and if f(t) = (M22 + M?22 )q̈2 , one obtains M11 q̈1 +K11 q1 = −K12 q2 −M12 q̈2 =⇒ M11 q̈1 +M12 q̈2 +K11 q1 +K12 q2 = 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 0 ≈ (M22 + M?22 )q̈2 M?22 q̈2 where q̈2 is given ⇒ apply same solution method as for problems of response to external loading 50 / 50