AA242B: MECHANICAL VIBRATIONS - Stanford University
AA242B: MECHANICAL VIBRATIONS1 / 50AA242B: MECHANICAL VIBRATIONSUndamped Vibrations of n-DOF SystemsThese slides are based on the recommended textbook: M. Géradin and D. Rixen, “MechanicalVibrations: Theory and Applications to Structural Dynamics,” Second Edition, Wiley, John &Sons, Incorporated, ISBN-13:97804719754651 / 50
AA242B: MECHANICAL VIBRATIONS2 / 50Outline1 Linear Vibrations2 Natural Vibration Modes3 Orthogonality of Natural Vibration Modes4 Modal Superposition Analysis5 Spectral Expansions6 Forced Harmonic Response7 Response to External Loading8 Mechanical Systems Excited Through Support Motion2 / 50
AA242B: MECHANICAL VIBRATIONS3 / 50Linear VibrationsEquilibrium configurationqs (t) qs (0),q̇s (t) 0,s 1, · · · , n(1)Recall the Lagrange equations of motion d T T V D Qs (t) 0dt q̇s qs qs q̇swhere T T0 T1 T2Recall the generalized gyroscopic forcesFs nnXX T1 2 T1q̇r q̇s qr qsr 1r 1 2 T1 2 T1 qs q̇r qr q̇s!q̇r ,s 1, · · · , nDefinition: the effective potential energy is defined asV ? V T0 V ? (q, t)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 q̇s3 / 50
AA242B: MECHANICAL VIBRATIONS4 / 50Linear VibrationsRecall thatT0 (q, t) D 2 N3 Uik1 XX(q, t)mk2 k 1 i 1 tN Z v (q̇)XkCk fk (γ)dγk 1(transport kinetic energy)(dissipation function)0From the Lagrange equations of motion T2 V ? D T1d T2 Qs (t) (q, t) Fs dt q̇s qs qs q̇s t q̇sit follows that an equilibrium configuration exists if and only if0 Qs (t) V ?(q, t) ! qsHence, at equilibriumQs (t) 0and V ? (V T0 ) 0, s 1, · · · , n qs qs4 / 50
AA242B: MECHANICAL VIBRATIONS5 / 50Linear VibrationsFree-Vibrations About a Stable Equilibrium PositionConsider first a system that does not undergo a transport or overallmotion T T2 (q̇)The equilibrium position is then given byQs (t) 0and V 0, qss 1, · · · , nAssume next that this system is conservative E T V cstShift the origin of the generalized coordinates so that at equilibrium,qs 0, s 1, · · · , n (in which case the qs represent the deviationfrom equilibrium)Since the potential energy is defined only up to a constant, choosethis constant so that V(qs 0) 0Now, suppose that a certain energy E(0) is initially given to thesystem in equilibrium5 / 50
AA242B: MECHANICAL VIBRATIONS6 / 50Linear VibrationsFree-Vibrations About a Stable Equilibrium PositionDefinition: the equilibrium position (qs 0, s 1, · · · , n) is said tobe stable if E ? / E(0) E ? ,T (t) E(0)ConsequencesT V E cst E(0) V(t) E(0) T (t)) 0at a stable equilibrium position, the potential energy is at a relativeminimumif E(0) is small enough, V(t) will be small enough and thereforedeviations from the equilibrium position will be small enough6 / 50
AA242B: MECHANICAL VIBRATIONS7 / 50Linear VibrationsFree-Vibrations About a Stable Equilibrium Position“Linearization” of T and V around an equilibrium position V(qs 0, 0) qsactually, this means obtaining a quadratic form of T and V in q andq̇, respectively, so that the corresponding generalized forces are linearsince qs (t) represent deviations from equilibrium, V can be expandedas followsnnnX1 X X 2V V q 0 qs q 0 qs qr O(q3 )V(q) V(0) q2 qss qrs 1s 1 r 1 V(0) nn1 X X 2V q 0 qs qr O(q3 )2 s 1 r 1 qs qrsince the potential energy is defined only up to a constant, if thisconstant is chosen so that V(0) 0, then a second-orderapproximation of V(q) is given byV(q) nn1 X X 2V q 0 qs qr ,2 s 1 r 1 qs qrfor q 6 07 / 50
AA242B: MECHANICAL VIBRATIONS8 / 50Linear VibrationsFree-Vibrations About a Stable Equilibrium PositionStiffness matrixlet K [ksr ] where ksr krs V(q) 1 Tq Kq 0,2 2V q 0 qs qrfor q 6 0 K is symmetric positive definitein the absence of sufficient boundary conditions – that is, in thepresence of rigid body modes1 Tq Kq 0,2for q 6 0 Kis symmetric positive semi-definite8 / 50
AA242B: MECHANICAL VIBRATIONS9 / 50Linear VibrationsFree-Vibrations About a Stable Equilibrium PositionRecall thatnnN31 XXXX Uik UikT2 mkq̇s q̇r2 s 1 r 1 qs qr(relative kinetic energy)k 1 i 1T2 (q, q̇)nnXX T2 T2 q 0,q̇ 0 qs q 0,q̇ 0 q̇s qs q̇ss 1s 1 T2 (0, 0) nnn XnX 2 T21 X X 2 T2 q 0,q̇ 0 qs qr q 0,q̇ 0 qs q̇r2 s 1 r 1 qs qr qs q̇rs 1 r 1 nn1 X X 2 T233 q 0,q̇ 0 q̇s q̇r O(q , q̇ )2 s 1 r 1 q̇s q̇r nn1 X X 2 T233 q 0,q̇ 0 q̇s q̇r O(q , q̇ )2 s 1 r 1 q̇s q̇r9 / 50
AA242B: MECHANICAL VIBRATIONS10 / 50Linear VibrationsFree-Vibrations About a Stable Equilibrium PositionHence, a second-order approximation of T2 (q̇) is given byT2 (q̇) where"1 Tq̇ Mq̇ 0,2for q̇ 6 0N3XX 2 T2 Uik UikM msr mrs q 0 mk q 0 q 0 q̇s q̇r qs qri 1k 1is the mass matrix and is symmetric positive definite#10 / 50
AA242B: MECHANICAL VIBRATIONS11 / 50Linear VibrationsFree-Vibrations About a Stable Equilibrium PositionFree-vibrations about a stable equilibrium position of a conservativesystem that does not undergo a transport or overall motion(T0 T1 0) T2 Vd T2 dt q̇s qs qs d(Mq̇) 0 Kqdt Mq̈ Kq 011 / 50
AA242B: MECHANICAL VIBRATIONS12 / 50Linear VibrationsFree-Vibrations About an Equilibrium ConfigurationConsider next the more general case of a system in steady motion (atransported system) whose equilibrium configuration defined by V ? (V T0 ) 0, qs qss 1, · · · , ncorrespondsto the balance of forcesderiving from a potential V T0and centrifugal forces qs qsIt is an equilibrium configuration in the sense that q̇s — whichrepresent here the generalized velocities relative to a steady motion— are zero but the system is not idle12 / 50
AA242B: MECHANICAL VIBRATIONS13 / 50Linear VibrationsFree-Vibrations About an Equilibrium ConfigurationLinearizationseffective potential energy V ? effective stiffness matrix K?1 T ?q K q 0,for q 6 02 2 T0?where K ksr ksr q 0 qs qrV ? (q) mutual kinetic energyT1 nn XN X3XX Uik Uik T1mkq̇s q̇s(q) t q q̇sss 1s 1i 1k 1 nXnq̇ss 1n XnXX 2 T1 T1(0) q 0 qr O(q2 ) q̇s q̇s qrr 1! 2 T1qr q 0 q̇s qr T1 (q, q̇) q̇T Fq q̇ss 1 r 1 2 T1where F fsr q 0 q̇s qr 13 / 50
AA242B: MECHANICAL VIBRATIONS14 / 50Linear VibrationsFree-Vibrations About an Equilibrium ConfigurationEquations of free-vibration around an equilibrium configurationthe equilibrium configuration generated by a steady motion remainsstable as long as V ? V T0 0this corresponds to the fact that K? remains positive definitein the neighborhood of such a configuration, the equations of motion(for a conservative system undergoing transport or overall motion)are V ? D T1d T2 T2 Qs (t) Fs dt q̇s qs q̇s t q̇s {z } qs {z} {z} {z}00where Fs n Xr 10 2 T1 2 T1 qs q̇r qr q̇s 0q̇r (FT F)q̇Mq̈ Gq̇ K? q 0where G F FT GT is the gyroscopic coupling matrix14 / 50
AA242B: MECHANICAL VIBRATIONS15 / 50Linear VibrationsFree-Vibrations About an Equilibrium ConfigurationExampleq xV X (a x) cos Ωt1 2kx2T0 2Y (a x) sin Ωt1 22Ω m(a x)2T1 02222v Ẋ Ẏ (a x) Ω ẋT2 12mẋ2V? 21 21 22kx Ω m(a x)22equilibrium configurationΩ2 ma V ? 0 kx Ω2 m(a x) 0 xeq xk Ω2 mthe system becomes unstable for Ω2 k? km 2V ?k k Ω2 m system is unstable for Ω2 x 2m15 / 50
AA242B: MECHANICAL VIBRATIONS16 / 50Natural Vibration ModesFree-vibration equations: Mq̈ Kq 0q(t) qa e iωt (K ω 2 M)qa 0 det (K ω 2 M) 0If the system has n degrees of freedom (dofs), M and K are n nmatrices n eigenpairs (ωi2 , qai )Rigid body mode(s): ωj2 0 Kqaj 01For a rigid body mode, V(qaj ) qTKqaj 02 aj16 / 50
AA242B: MECHANICAL VIBRATIONS17 / 50Orthogonality of Natural Vibration ModesDistinct FrequenciesConsider two distinct eigenpairs (ωi2 , qai ) and (ωj2 , qaj )T 2qTaj Kqai qaj ωi Mqai(2)qTai Kqaj(3) 2qTai ωj MqajBecause M and K are symmetric(2) (3)T 0 (ωi2 ωj2 )qTaj Mqaisince ωi2 6 ωj2 qTaj Mqai 0and qTaj Kqai 017 / 50
AA242B: MECHANICAL VIBRATIONS18 / 50Orthogonality of Natural Vibration ModesDistinct FrequenciesPhysical interpretation of the orthogonality conditions 2T2TqTaj Mqai 0 qaj ωi Mqai (ωi Mqai ) qaj 0which implies that the virtual work produced by the inertia forces ofmode i during a virtual displacement prescribed by mode j is zeroTqTaj Kqai 0 (Kqai ) qaj 0which implies that the virtual work produced by the elastic forces ofmode i during a virtual displacement prescribed by mode j is zero18 / 50
AA242B: MECHANICAL VIBRATIONS19 / 50Orthogonality of Natural Vibration ModesDistinct FrequenciesRayleigh quotient2 T2Kqai ωi2 Mqai qTai Kqai ωi qai Mqai ωi qTγiai Kqai Tqai Mqaiµiγi generalized stiffness coefficient of mode i (measures thecontribution of mode i to the elastic deformation energy)µi generalized mass coefficient of mode i (measures thecontribution of mode i to the kinetic energy)Since the amplitude of qai is determined up to a factor only γiand µi are determined up to a constant factor onlyMass normalizationqTaj Mqai δijqTaj Kqai ωi2 δij19 / 50
AA242B: MECHANICAL VIBRATIONS20 / 50Orthogonality of Natural Vibration ModesDegeneracy TheoremWhat happens if a multiple circular frequency is encountered?Theorem: to a multiple root ωp2 of the systemKqa ω 2 Mqacorresponds a number of linearly independent eigenvectors {qai }equal to the root multiplicity20 / 50
AA242B: MECHANICAL VIBRATIONS21 / 50Modal Superposition Analysisn-dof system: M Rn n , K Rn , and q RnCoupled system of ordinary differential equations Mq̈ Kq 0q(0) q0 q̇(0) q̇021 / 50
AA242B: MECHANICAL VIBRATIONS22 / 50Modal Superposition AnalysisNatural vibration modes (eigenmodes)Kqai ωi2 Mqai ,Q [qa1i 1, · · · , n TQ KQ qa2 · · · qan ] QT MQ 2 ω1 .Ω2 .2ωnΩ2ITruncated eigenbasis Q r qa 1qa 2··· qa r , r n 2Ωr QTr KQrQTr MQrω12 Ω2rIr(reduced stiffness matrix)(reduced mass matrix) .ωr2 22 / 50
AA242B: MECHANICAL VIBRATIONS23 / 50Modal Superposition AnalysisModal superposition: q Qr y rPyi qai wherei 1y [y1y2···Tyr ] and yi is called the modal displacementSubstitute in Mq̈ Kq 0 MQr ÿ KQr y 0T QTr MQr ÿ Qr KQr y 0 Ir ÿ Ω2r y 0Uncoupled differential equations (modal equations)ÿi ωi2 yi 0,i 1, · · · r23 / 50
AA242B: MECHANICAL VIBRATIONS24 / 50Modal Superposition Analysisÿi ωi2 yi 0,i 1, · · · rCase 1: ωi2 0 (rigid body mode)yi ai t biCase 2: ωj2 6 0yj cj cos ωj t dj sin ωj tGeneral case: rb rigid body modesq rbrX rbX(ai t bi )qai (cj cos ωj t dj sin ωj t)qaji 1j 1Initial conditionsq(0) q0 Qy(0)andq̇(0) q̇0 Qẏ(0)24 / 50
AA242B: MECHANICAL VIBRATIONS25 / 50Modal Superposition Analysisq0 Qr y(0)andTiq̇0 Qr ẏ(0)Ti qa Mq0 qa MQr y(0)From the orthogonality properties of the natural mode shapes (eigenvectors) it follows thatTiTiTiTiqa Mq0 qa MQr y(0) qa Mqai yi (0) yi (0) yi (0) qa Mq(0)TCase 1: ωi2 0 (rigid body mode) ai 0 bi qTa Mq(0) bi qa Mq(0)iiTCase 2: ωj2 6 0 cj 1 dj 0 qTa Mq(0) cj qa Mq(0)jjCase 1: ωi2 0 (rigid body mode) ai qTa Mq̇(0)iCase 2: ωj2 6 0 dj 1 Tq Mq̇(0)ωj ajThus, the general solution isq(t) rb hr rb i XXsin ωj tTTTTqa Mq̇(0) t qa Mq(0) qai qa Mq(0) cos ωj t qa Mq̇(0)qa jiijjωji 1j 125 / 50
AA242B: MECHANICAL VIBRATIONS26 / 50Spectral Expansionsn x R ,x nXTjαs qas qa Mx s 1n x R ,x nXTjαs qa Mqas αjs 1n nn X XXTTTqas Mx qas qas qas Mx qas qas Mx nXT(qas qas )Ms 1s 1s 1 n Xs 1Tqas qas !xM Is 1This is the same result as QT MQ IA given load p can be expanded in terms of the inertia forcesgenerated by the eigenmodes, Mqaj as followsp nXj 1βj Mqaj qTai p nXβj qTai Mqaj βij 1 βi qTai p modal participation factor p n X qTpMqajajj 126 / 50
AA242B: MECHANICAL VIBRATIONS27 / 50Spectral ExpansionsRecall that n Pqa s q TM Iass 1Hence, A Rn ,nPA s 1A M M nPs 1A K K nPs 1Aqas qTas M and A Mqas qTas M Kqas qTas M A M 1 M 1 nPs 1A K 1 K 1 nPs 1nPs 1nPs 1nPs 1Mqas (Mqas )Tωs2 Mqas qTas M 1 Mqas qTas MM 1 qas qTas MK 1 qas qTas MA(because M is symmetric)nPs 1nXωs2 Mqas (Mqas )TTqas qass 1nPs 1 1qas (Mqas )T K 1 K nXqas qTass 1ωs227 / 50
AA242B: MECHANICAL VIBRATIONS28 / 50Forced Harmonic Response Mq̈ Kq q(0) q̇(0) sa cos ωtq0q̇0Solution can be decomposed asq qH (homogeneous) qP (particular)qp qa cos ωt qa (K ω 2 M) 1 sa where (K ω 2 M) 1 iscalled the admittance or dynamic influence matrixThe forced response is the part of the response that is synchronousto the excitation — that is, qp28 / 50
AA242B: MECHANICAL VIBRATIONS29 / 50Forced Harmonic ResponseRigid body modes: qa Rn ,qa uairbPi 1rbi 1αi uai n rbPβ j qa jj 12 sa (K ω M)qa rbXn rb2αi (K ω M)uai i 1 sa rbXj qa rbXuTa saii 1ω2ua i uTa sajω22βj (K ω M)qajj 1n rb2αi ω Muai i 1Premultiply by uTa αj XX22βj (ωj ω )Mqajj 1and premultiply by qTa βi i n rbXqTa saj 1(ωj2 ω 2 )jqaj rbXua i uTaii 1ω2qTa sai(ωi2 ω2 )n rbXqaj qTaj 1(ωj2 ω 2 )j saSince qa (K ω 2 M) 1 sa2 (K ω M) 1 rbn rb q qTXaj a1 XjTua i ua 2 ω22iω i 1ωjj 1(4)29 / 50
AA242B: MECHANICAL VIBRATIONS30 / 50Forced Harmonic ResponseWhich excitation sam will generate a harmonic response with anamplitude corresponding to qam ?qam (K ω 2 M) 1 sam sam (K ω 2 M)qam sam 2Kqam ω 2 Mqam (ωm ω 2 )Mqam2At resonance (ω 2 ωm) sam 0 no force is needed to maintainqam once it is reached30 / 50
AA242B: MECHANICAL VIBRATIONS31 / 50Forced Harmonic ResponseThe inverse of an admittance is an impedanceZ(ω 2 ) (K ω 2 M)31 / 50
AA242B: MECHANICAL VIBRATIONS32 / 50Forced Harmonic ResponseApplication: substructuring (or domain decomposition)the dynamical behavior of a substructure is described by its harmonicresponse when forces are applied onto its interface boundariesa 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 boundarydofs q2 where the reaction forces are denoted here by g2 Z11 Z12q10 Z21 Z22q2g2 Z11 q1 Z12 q2 0 q1 Z 111 Z12 q2? (Z22 Z21 Z 111 Z12 )q2 Z22 q2 g2?Z22is the “reduced” impedance (reduced to the boundary)32 / 50
AA242B: MECHANICAL VIBRATIONS33 / 50Forced Harmonic ResponseSpectral expansion of Z?222look at Z 111 and let (ω̃i , q̃ai ) denote the n1 eigenpairs of the associated dynamicaln1q̃aj q̃TXaj 1subsystem: from (4), it follows that Z11 ω̃j2 ω 2j 1apply twice the relationω211 2 2 2ω̃j2 ω 2ω̃jω̃j (ω̃j ω 2 ) 1 1 Z11 K11 ω2n1Xq̃aj q̃Taj 1ω̃j2 (ω̃j2 ω 2 )j 1 K11 ω2n1 q̃ q̃TXaj ajj 1 ωω̃j44n1Xq̃aj q̃Taj 1ω̃j4 (ω̃j2 ω 2 )jowing to the M11 -orthonormality of the modes and the spectral expansion of K 111 ,the above expression can further be written as 1Z11? Z22 12 1 1 K11 ω K11 M11 K11 ω K22 K21 K11 K12 n1Xq̃aj q̃Tajω̃j4 (ω̃j2j 1 ω2 ) 1 12 1 1 1ω [M22 M21 K11 K12 K21 K11 M12 K21 K11 M11 K11 K12 ]n1 4ω4X [(K21 i 1ω̃i2 M21 )q̃ai ][(K21 ω̃i4 (ω̃i2 ω 2 )ω̃i2 M21 )q̃ai )]T33 / 50
AA242B: MECHANICAL VIBRATIONS34 / 50Forced Harmonic Response?Z22 K22 K21 K 111 K12 1 1 1 ω 2 [M22 M21 K 111 K12 K21 K11 M12 K21 K11 M11 K11 K12 ]n1X[(K21 ω̃i2 M21 )q̃ai ][(K21 ω̃i2 M21 )q̃ai )]T ω 4ω̃i4 (ω̃i2 ω 2 )i 1The first term K22 K21 K 111 K12 represents the stiffness of thestatically condensed systemThe second term 1 1 1M22 M21 K 111 K12 K21 K11 M12 K21 K11 M11 K11 K12 representsthe mass of the subsystem statically condensed on the boundaryThe last term represents the contribution of the subsystemeigenmodes since it is generated by q̃ai q̃Tai(K21 ω̃i2 M21 )q̃ai is the dynamic reaction on the boundary34 / 50
AA242B: MECHANICAL VIBRATIONS35 / 50Response to External Loading Mq̈ Kq q(0) q̇(0) p(t)q0q̇0General approachconsider the simpler case where there is no rigid body mode eigenmodes (qai , ωi2 ), ωi2 6 0, i 1, · · · , nnPmodal superposition: q Qy yi qaii 1substitute 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, · · · , nyi (t) depend on two constants that can be obtained from the initialconditionsq(0) Qy(0), q̇(0) Qẏ(0) yi (0), ẏi (0)orthogonality conditions35 / 50
AA242B: MECHANICAL VIBRATIONS36 / 50Response to External LoadingResponse to an impulsive forcekmimpulsive force f (t): force whose amplitude could be infinitely largebut which acts for a very shortZ duration of timespring-mass system: m, k, ω 2 τ magnitude of impulse: I f (t)dtτimpulsive force I δ(t) where δ isZthe “delta” function centered at t 0 and satisfyingδ(t)dt 1036 / 50
AA242B: MECHANICAL VIBRATIONS37 / 50Response to External LoadingResponse to an impulsive force (continue)dynamic equilibriumZ τ Z τ Z τ d u̇mdt kudt f (t)dt Idtτττassume that at t τ , the system is at rest (u(τ ) 0 and u̇(τ ) 0)focus on theZshort (infinitesimal) intervalof time dτZτ τ ü A/ τ Zτd u̇dtdtudt A 2 /6andττhencemüdt AII u̇(τ ) mm I m u̇ I u̇ u̇(τ ) u̇(τ ) 0 u 0 u(τ ) u(τ ) 0 u(τ ) 0τ Zu̇dtτthe above equations provide initial conditions for the free-vibrationsthat start at the end of the impulsive load u(t) Isin ωtmωfor impulses of finite duration37 / 50
AA242B: MECHANICAL VIBRATIONS38 / 50Response to External LoadingResponse to an impulsive force (continue)for the differential time interval dτ (dτ 0), the response analysisof the previous page becomes exact du f (τ )dτsin ω(t τ )mωlinear system superposition principledu f (τ )dτ1sin ω(t τ ) u(t) mωmωZtf (τ ) sin ω(t τ )dτ038 / 50
AA242B: MECHANICAL VIBRATIONS39 / 50Response to External LoadingTime-integration of the normal equationslet pi (t) qTai p(t) i-th modal participation factormodal or normal equation: ÿi ωi2 yi pi (t)yi (t) yiH (t) yiP (t), yiH Ai cos ωi t Bi sin ωi tthe particular solution yiP (t) depends on the form of pi (t)however, the general form of a particular solution that satisfies therest initial conditions is given by the Duhamel’s integralZ t1yiP (t) pi (τ ) sin ωi (t τ )dτωi 0complete solutionyi (t) Ai cos ωi t Bi sin ωi t 1ωiZtpi (t) sin ωi (t τ )dτ0ẏi (0)and yi (0) and ẏi (0) can be determined fromωithe initial conditions q(0) and q̇(0) and the orthogonality conditionsAi yi (0), Bi 39 / 50
AA242B: MECHANICAL VIBRATIONS40 / 50Response to External LoadingResponse truncation and mode displacement methodcomputational efficiency q Qr y rPr nyi qai ,i 1what is the effect of modal truncation?consider the case where p(t) g static loadφ(t)amplificationdistributionfactorfor 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) 1ωitZ q(t) pi (τ ) sin ωi (t τ )dτ 0rXi 1"#qai qTai gspatial factorqTai gωitZφ(τ ) sin ωi (t τ )dτ0 1 ωi tZ0 φ(τ ) sin ωi (t τ )dτ temporal factor40 / 50
AA242B: MECHANICAL VIBRATIONS41 / 50Response to External LoadingResponse truncation and mode displacement method (continue)general solution for restrained structure initially at rest #"Z trX1 qai qTai g q(t) φ(τ ) sin ωi (t τ )dτ ωi 0i 1 spatial factortemporal factor θi (t)truncated response is accurate if neglected terms are small, which istrue if:qTai g is small for i r 1, · · · , n g is well approximated in therangeZ tof Qr1φ(τ ) sin ωj (t τ )dτ is small for j r , which depends on theωj 0frequency content of φ(t)φ(t) 1 φ(t) sin ωt 1 cos ωi t 0 for large circular frequenciesωiωi sin ωt ω sin ωi tθi (t) ωi (ωi2 ω 2 )θi (t) 41 / 50
AA242B: MECHANICAL VIBRATIONS42 / 50Response to External LoadingMode acceleration methodMq̈ Kq p(t) Kq p(t) Mq̈apply truncated modal representation to the accelerationq(t) Qr y(t) q̈(t) Qr ÿ(t) Kq p(t) rXMqai ÿi q K 1 p(t) i 1rXqaiÿiωi2i 1recall thatÿi ωi2 yi qTai p(t)and that for a system initially at restZ t1yi (t) qTai p(τ ) sin ωi (t τ )dτωi 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 VIBRATIONS43 / 50Response to External LoadingMode acceleration method (continue)substitute in q(t) K 1 p(t) rXqaiÿiωi2i 1!ZrrXXqai qTaiqai qTai t 1 q(t) p(τ ) sin ωi (t τ )dτ K p(t)ωiωi20i 1i 1recall the spectral expansion K 1 n q qTPai ai2i 1 ωiZrXqai qTai tp(τ ) sin ωi (t τ )dτ q(t) ωi0i 1nXqai qTaiωi2i r 1!p(t)which shows that the mode acceleration method complements thetruncated mode displacement solution with the missing terms usingthe modal expansion of the static responsehow to deal with the computational cost issue?43 / 50
AA242B: MECHANICAL VIBRATIONS44 / 50Response to External LoadingDirect time-integration methods for Mq̈ Kqq(0) q̇(0)solving p(t) q0 q̇0will be covered towards the end of this course44 / 50
AA242B: MECHANICAL VIBRATIONS45 / 50Mechanical Systems Excited Through Support MotionThe general case q M11M21M12M22 q1q2 q̈1q̈2 where q2 is prescribed K11K21K12K22 q1q2 0r2 (t) The first equation givesM11 q̈1 K11 q1 K12 q2 M12 q̈2 q1 (t)Substitute in second equation r2 (t) K21 q1 M21 q̈1 K22 q2 M22 q̈245 / 50
AA242B: MECHANICAL VIBRATIONS46 / 50Mechanical Systems Excited Through Support MotionQuasi-static response of q10 K11 q1 K12 q2 0 1 qqs1 K11 K12 q2 Sq2Decompose q1 qqs1 z1 Sq2 z1 q1I q(t) q20SI z1q2 where z1 represents the sole dynamics part of the responseSubstitute in the first dynamic equation and exploit above results M11 Sq̈2 M11 z̈1 K11 qqs1 K11 z1 K12 q2 M12 q̈2 M11 z̈1 K11 z1 g1 (t) g1 (t) M11 q̈qs1 M12 q̈2 (M11 S M12 )q̈246 / 50
AA242B: MECHANICAL VIBRATIONS47 / 50Mechanical Systems Excited Through Support MotionConsider next the system fixed to the ground and solve thecorresponding EVPe [· · · q̃a · · · ]K11 x ω 2 M11 x Qie z1 (t) Qη(t)Solve M11 z̈1 K11 z1 g1 (t) for z1 (t) using the modalsuperposition technique47 / 50
AA242B: MECHANICAL VIBRATIONS48 / 50Mechanical Systems Excited Through Support MotionCase of a global support acceleration q̈2 (t) u2 φ(t), whereu [u1 u2 ]T denotes a rigid body modedecomposition of the solution into a rigid body motion and a relativedisplacement y u1ÿ1q̈1 φ(t) q̈ q̈rb ÿ q̈2u20u1 Su2 g1 (t) (M11 Su2 M12 u2 )φ(t) g1 (t) (M11 u1 M12 u2 )φ(t)48 / 50
AA242B: MECHANICAL VIBRATIONS49 / 50Mechanical Systems Excited Through Support MotionMethod of additional masses (approximate method)suppose the system is subjected not to a specified q2 (t) but to animposed 0f(t)suppose that masses associated with q2 are increased to M22 M?22then K11 K12q10q̈1M11M12 q̈2K21 K22q2f(t)M21 M22 M?22by elimination one obtainsq̈2 (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 VIBRATIONS50 / 50Mechanical Systems Excited Through Support MotionMethod of additional masses (continue)therefore{M11? 1 M12 (M22 M22 ) K12 q2 M12 (M22 M22 )M21 } q̈1 ? 1no? 1K11 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 obtainsM11 q̈1 K11 q1 K12 q2 M12 q̈2 M11 q̈1 M12 q̈2 K11 q1 K12 q2 0which is the same equation as in the general case of excitationthrough support motion for M?22 very large, the response of the system with groundmotion is the same as the response of the system with added lumpedmass M?22 and the forcing function 00 (M22 M?22 )q̈2M?22 q̈2where q̈2 is given apply same solution method as for problems ofresponse to external loading50 / 50
AA242B: MECHANICAL VIBRATIONS 1/50 AA242B: MECHANICAL VIBRATIONS Undamped Vibrations of n-DOF Systems These slides are based on the recommended textbook: M. G eradin 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 1/41 AA242B: MECHANICAL VIBRATIONS Direct Time-Integration Methods These slides are based on the recommended textbook: M. G eradin and D. Rixen, \Mechanical Vibrations: Theory and Applications to Structural Dynamics," Second Edition, Wiley, John & Sons, Incorporated, ISBN-13:9780471975465 1/41
SEISMIC: A Self-Exciting Point Process Model for Predicting Tweet Popularity Qingyuan Zhao Stanford University qyzhao@stanford.edu Murat A. Erdogdu Stanford University erdogdu@stanford.edu Hera Y. He Stanford University yhe1@stanford.edu Anand Rajaraman Stanford University anand@cs.stanford.edu Jure Leskovec Stanford University jure@cs.stanford .
Computer Science Stanford University ymaniyar@stanford.edu Madhu Karra Computer Science Stanford University mkarra@stanford.edu Arvind Subramanian Computer Science Stanford University arvindvs@stanford.edu 1 Problem Description Most existing COVID-19 tests use nasal swabs and a polymerase chain reaction to detect the virus in a sample. We aim to
Domain Adversarial Training for QA Systems Stanford CS224N Default Project Mentor: Gita Krishna Danny Schwartz Brynne Hurst Grace Wang Stanford University Stanford University Stanford University deschwa2@stanford.edu brynnemh@stanford.edu gracenol@stanford.edu Abstract In this project, we exa
Stanford University Stanford, CA 94305 bowang@stanford.edu Min Liu Department of Statistics Stanford University Stanford, CA 94305 liumin@stanford.edu Abstract Sentiment analysis is an important task in natural language understanding and has a wide range of real-world applications. The typical sentiment analysis focus on
Mechanical vibrations. (Allyn and Bacon series in Mechanical engineering and applied mechanics) Includes index. 1. Vibrations. I. Morse, Ivan E., joint author. Hinkle, Theodore, joint author. Title. 1978 620.3 77-20933 ISBN ISBN (International)
Vibration is a continuous cyclic motion of a structure or a component. Generally, engineers try to avoid vibrations, because vibrations have a number of unpleasant effects: . It is thought that the vibrations were a form of self-excited vibration known as flutter, or ‘galloping’’ A similar form of vibration is known to occur .
Sector shutdowns during the coronavirus crisis: which workers are most exposed? Authors: Robert Joyce (IFS) and Xiaowei Xu (IFS) Summary The lockdown in response to the Covid-19 pandemic has effectively shut down a number of sectors. Restaurants, shops and leisure facilities have been ordered to close, air travel has halted, and public transport has been greatly reduced. Our analysis shows .
