同济大学：《建筑结构抗震》课程教学资源（课件讲稿）Chapter 3 Seismic Responses of SDOF and MDOF（1/2）

Earthquake action is a 'time-varying"action. Chapter3 Seismic Responses of SDOF and MDOF 国 20151013 Keywords Key Points SDOF(单自由度)，MDOF(多自由度) How to determine theeq.load(action)? Free Vibration,Forced Vib ton(受迫振幼 nsform a dynamic action to an equivalent h1 Excitation(h},Response(反应） What's kind of equivalent condition? Time History(时程)，Spectrum(反应潘) .Action equivalent?Response equivalent? Base Shear method(底剪力法) .Acceleration/velocity/displacement equivalent? 叠加反应语法) 201510r13 21s10r13 Outline Dynamics 米 3.1 Free Vibration of SDOF Systems 3.1 Free Vibration of SDOF Systems 3.1.Dynamic Model and Equilibrium Equation 3.3 Numerical Analysis of Seismic Response of SDOF 3.4Spectrum of Seismic Response of SDOF 3.1.Undamped Free Vibration 3.5 Response of Non ear SDOF Systems (* 3.2 Bee Vibr of M 3.1.3 Damped Free Vibration ethod of MDOF TerytMO 3.8 Earthquake Action and Res 。见大出木智有 1

1 Chapter 3 Seismic Responses of SDOF and MDOF 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 1  Earthquake action is a “time-varying” action.  The structural responses varied with time history, which are all time-dependant varieties．  Solving these kinds of issues need dynamic analysis method.  A structure can be simplified as a Single-Degree-Of￾Freedom (SDOF) system or Multiply-Degree-Of-Freedom (MDOF) system.  The response may be linear response or nonlinear response which depended on the structural material stiffness (Nonlinear Material ) and structural deformations (Nonlinear Structural Element). 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 2 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 3 Keywords SDOF（单自由度）, MDOF（多自由度） Free Vibration, Forced Vibration（受迫振动） Harmonious Vibration（谐振）, Impulsive Vibration（冲击振动）, Excitation（激励）, Response（反应）, Time History（时程），Spectrum (反应谱) Base Shear method（底部剪力法）, Response Spectrum Method（振型叠加反应谱法） Time History Method（时程分析法） 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 4 Key Points  How to determine the eq. load (action) ?  How to transform a dynamic action to an equivalent static load?  What’s kind of equivalent condition?  Action equivalent？Response equivalent？  Acceleration / velocity / displacement equivalent ? 2015/10/13 熊海贝，同济大学土木工程学院 5 xionghaibei@tongji.edu.cn Outline 3.1 Free Vibration of SDOF Systems 3.2 Forced Vibration of SDOF Systems 3.3 Numerical Analysis of Seismic Response of SDOF 3.4 Spectrum of Seismic Response of SDOF 3.5 Response of Nonlinear SDOF Systems （*） 3.6 Free Vibration of MDOF Systems 3.7 Response Specturm Method of MDOF 3.8 Earthquake Action and Responses of MDOF 3.9 Time History Method of MDOF Dynamics 3.1 Free Vibration of SDOF Systems 3.1.1 Dynamic Model and Equilibrium Equation 3.1.2 Undamped Free Vibration 3.1.3 Damped Free Vibration 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 6

3.Dynamic Model and Equilibrium Equation 3.1.1 Dynamic Model and Equilibrium Equation .How to determine theDegreeof-freedom .How to determine theDegree-of-freedomfor such eteha .How to determine the"Degree-of-freedom"for such 3.1.1 Dynamic Model and Equilibrium Equation kinds of system Newton's 2nd Law 而 e8 my· sheouc of thebody mass 路四培贩回 3.1 Dynamic Model and Equilibrium Equation 3.1.2 Undamped Free Vibration D'Alambert Principle Inertial force Based on D'Alambert's principle: F-ma=0 成+x=0 2015/101

2  How to determine the “Degree-of-freedom” In mechanics, the degree of freedom (DOF) of a mechanical system is the number of independent parameters that define its configuration. It is the number of parameters that determine the state of a physical system. Degree of freedom is a fundamental concept central to the analysis of systems of bodies in mechanical engineering, aeronautical engineering, robotics, and structural engineering. 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 7 3.1.1 Dynamic Model and Equilibrium Equation  How to determine the “Degree-of-freedom” for such kinds of system 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 8 3.1.1 Dynamic Model and Equilibrium Equation  How to determine the “Degree-of-freedom” for such kinds of system Adjustable Slippage Element 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 9 Newton’s 2nd Law “The force acting on a body and causing its movement, is equal to the rate of change of momentum in the body.” • Momentum Q, is equal to the product of the body mass by its velocity. mx dt dx Q  mv  m   3.1.1 Dynamic Model and Equilibrium Equation mx ma dt dx m dt dv m dt d( mv ) dt dQ F         D’Alambert suggested that Newton’s Law can be written in a similar way that principle of equilibrium in statics (F=0): Inertial force D’Alambert Principle F ma  0 3.1.1 Dynamic Model and Equilibrium Equation 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 11 x m k 0 mx kx Based on D’Alambert’s principle: m  x   kx  0 3.1.2 Undamped Free Vibration Ordinate that describes the mass movement stiffness mass This is Dynamic Equilibrium Equation of undamped free vibration of SDOF mass

3.1.2 Undamped Free Vibration 3.1.2 Undamped Free Vibration Dividing by m and calling/m 求+o2x=06 The solution is: t x()=Asin(or)+Bcos(ot) It,the initial conditions is: frequency inadians/secnd (rads) =frequency in cycles/second (Hz) Then: T=2/=f=period in seconds (s) x(t)=Ya sinot)+x cos(ot) 20151013 3.1.3 Damped Free Vibration 3.1.3 Damped Free Vibration But, Damping There are several types of da mping ends to Viscous-Proportional to movement velocity Coulomb Caused by friction Hysteretic -For materials working in the em this loss of energy is knowr 3.1.3 Damped Free Vibration 3.1.3 Damped Free Vibration Viscous =cx c CX Hysteretic Damping Force D'Alambert's Princlple:+Cx+=( N Inertia Force- Restoring Force 3

3 Dividing by m and calling : The solution is: If, the initial conditions is: Then: 0 0 0 (0) (0) t x x v v    2 k m  homogeneous linear differential equation with constant coefficients. 常系数齐次线性微分方程 0 2 x    x  xt  Asint Bcost   sin t x cos t v x t    0 0   3.1.2 Undamped Free Vibration = frequency in radians/second (rad/s) = frequency in cycles/second (Hz) = period in seconds (s) Period T t x 0 x 0 v  f  2 T  2  1 f 3.1.2 Undamped Free Vibration 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 14 But, movement tends to decrease with time. This reduction is associated with loss of energy present in the system. Energy, kinetic or potential, transforms in other forms of energy, such as noise, heat, etc. In dynamic system this loss of energy is known as, Damping. 3.1.3 Damped Free Vibration 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 15 There are several types of damping: Viscous - Proportional to movement velocity Coulomb - Caused by friction Hysteretic - For materials working in the nonlinear range where loading curve is different from the unloading curve. Damping 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 16 3.1.3 Damped Free Vibration F a Viscous F a x F =cx a u F F y Hysteretic F a uN N N Coulomb 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 17 3.1.3 Damped Free Vibration 3.1.3 Damped Free Vibration x m k 0 c mass mx kxcx m  x  cx   kx  0 Damping Force Inertia Force Restoring Force 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 18 D’Alambert’s Principle:

3.1.3 Damped Free Vibration 3.1.3 Damped Free Vibration When the radical is zero,we have Cer m2+c2+k=0 c2-4mt-0 ce-plmt Define: Its roots: 1--ctve-imk 2m We obtain:c=25m Its solution: And the roots become: x(t)=Ae+Be 大01： 3.1.3 Damped Free Vibration 3.1.3 Damped Free Vibration When: When: 5=>1 Over damped The solution is: Ce The solution is: Ce x(1)=Ae-+Bte- )= No mare vibration ,、No mare vibration t t 大生 3.1.3 Damped Free Vibration 3.1.3 Damped Free Vibration (<1 When: =C<1 damped Solving for the initial conditions: The imaginaryso )9 -.-cu o.--5o Using Euler transformation: x()-eeod-giot)Dsim-go 。 2π 2015/101 移际大林Ξ 01610w1

4 Characteristic equation is: Its roots: Its solution: 常系数齐次线性微分方程的特 征方程   t t x t Ae Be 1 2   m c c mk 2 4 2      0 2 m  c  k  3.1.3 Damped Free Vibration 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 19 When the radical is zero, we have Cc： Define: We obtain： And the roots become: 4 0 2 cc  mk  c c c   c  2m     1 2     Cc  2 mk  3.1.3 Damped Free Vibration 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 20 The solution is : x t x0 v0   t t x t Ae Bte     2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 21 3.1.3 Damped Free Vibration   1 cc c When:  No mare vibration critical damped x t x0 v0          t  t   t x t e Ae t Be       1 1 2 2 The solution is : Over damped 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 22 3.1.3 Damped Free Vibration When:  1 c c c  No mare vibration The imaginary solution is: Using Euler transformation:          t i  t i  t x t e Ae Be      2 2 1 1 xt e Ccos t D sin t t      2 2  1  1  2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 23 3.1.3 Damped Free Vibration  1 c c c When:  damped Solving for the initial conditions: x t x0 v0 T a  1                        sin t v x x t e x cos t a a a t       0 0 0    2 a  1      2 1 2 2    a Ta Damped frequency 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 24 3.1.3 Damped Free Vibration

3.1.3 Damped Free Vibration (< Example [Solution] en by weight?) 大 201510y13 Example 3.2 Forced Vibration [Solution] Harmonic loading Two basic→ Impulse loading vibeaion 0=628 .Harmonic vibration -e--6.281-005-627 Natural vibrafion frequenoy: f=a/2x=lH拉 Impulse vi tion Natural vibrafion period: T-l/f=ls 3.2.1 Harmonic Vibration 3.2.1 Harmonic Vibration Equation of motion m+kx=Fosin at Where Response xt)=e(acoso,t+bsin@t)+ccosot+dsino ,the accelerationc be represented as F(() The equtvalent force amplitude as F and the frequency ratio 大杠 5

5 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 25  1 3.1.3 Damped Free Vibration [3.1] A single story concrete factory with total mass concentrated on the roof, is 204 t, and the lateral stiffness of all columns is 8048.6 kN/m. Please determine the building’s natural period. [Solution] single story —— SDOF Concrete factory —— Damping ratio=5% ( if it given by steel frame? , masonry wall?) Roof mass —— M= 204 t，( if it given by weight? ) Example 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 26 39 43 204 0 2 8048 6 . . . m k  ＝ ＝ ＝ Undamped natural vibration circular frequency: ＝6.281 6.28 1 0.05 6.27 ＝  2＝  2＝ f＝ 2＝1Hz T 1 f＝1s Damped natural vibration circular frequency: Natural vibration frequency: Natural vibration period: 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 27 Example [Solution]  Harmonic vibration : a train of sinusoidal waves having a given amplitude．  Impulse vibration : during a short time, the wave amplitude can be simplified as a constant value, or a triangular shape． Two basic excitation Harmonic loading Impulse loading No. 28 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 28 3.2 Forced Vibration  undamped system subjected to simple harmonic loading Where is the amplitude and a is the circular frequency of the harmonic load． For a ground acceleration, the acceleration can be represented as The equivalent force amplitude as and the frequency ratio 0 F ( ) sin F t mg t 0 0    F mg oe 0      a / No. 29 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 29 3.2.1 Harmonic Vibration m  x   kx  F0 sin at  Equation of motion  Response sin 2 0 n n F x 2 x t m       xt e a t b t c t d t d d t n       cos  sin  cos  sin  Homogeneous solution a and b satisfy i.c.’s Decays exponentially in time “Transient” response 瞬态反应 Particular solution c and d satisfy F(t) Does not decay “Steady-state” response 稳态反应 3.2.1 Harmonic Vibration 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 30