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

.Steady-state response)Dsin(a) and D= 1-(alo)T'+(25alo)i 。Response is ha Frequncy®B=aa 3.2.1 Harmonic Vibration 3.2.2 Impulsive Vibration onic excitation The 6th National Structural Competition 3.2.2 Impulsive Vibration 3.2.2 Impulsive Vibration ■运动平衡方程 ■冲击荷收作用下的结构的近似响应 E+F+F+P)=0 ,瞬时冲量S=Pdr 式中： ,初始速度。=S/m 厂惯性力=m 用尼力"G 。结构反应 制始度下的有图尼由动 ):弹性铁复力= =时效力（随着时间而改变的外力】 为总加速度x和称是相对于地面的位移和速度。 x0=e.Psno mo

6 Steady-State Response  Steady-state response where and  Response is harmonic; lags behind load by  F0 /k is static load displacement, so D is dynamic amplification factor    D at  k F x t sin 0            2 1 1 2 tan     a a     2 2 2 n n 1 D 1 2             2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 31 a  a  Dynamic amplification factor, 𝐷 0 1 2 3 0 1 2 3 Frequency ratio, / n D  = 0.2  = 0.1  = 0.5  = 0.3 Resonance 1 D 2  Static 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 32 𝛽 = 𝑎/𝜔 ξ ξ=0.1 ξ=0.3 ξ=0.5 t t R(t) Undamped system R(t) Damped system 0 0 Resonance response 共振 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 33 3.2.1 Harmonic Vibration Harmonic excitation No. 34 The 6th National Structural Competition 3.2.2 Impulsive Vibration (3-1) 式中： =惯性力= =阻尼力= =弹性恢复力= =时效力（随着时间而改变的外力） 为总加速度, 和 是相对于地面的位移和速度。 Fi d s     F F P t 0   No. 35 f d cx s f pt x x i f mxx kx  运动平衡方程 3.2.2 Impulsive Vibration  冲击荷载作用下的结构的近似响应 No. 36   ' ' sin t Pd x t e t m         瞬时冲量 S Pd    初始速度 / 0 v S m   结构反应 —— 初始速度下的有阻尼自由振动 3.2.2 Impulsive Vibration

3.2.2 Impulsive Vibration 3.2.2 Impulsive Vibration A serics impulsive excitation-Base Excitation mi0）+ct)+k)=城() ()+25)+x0)=-.() P()=-m,() 一系列冲古使产生的构 地面运动可表示为付四dr Integrate of a series impulsive excitation 3.2.2 Impulsive Vibration 3.2.2 Impulsive Vibration k0）=eoi-P(sin-rdr ()P(sin(-dr 7e0 mo o ()sinu-dr me) e0=e,mou-t 3.2.3 Arbitrary excitation 3.2.3 Arbitrary excitation Arbitrary excitation (随机漾励) Dynamic Equilibrium Equation w mass P( P(t) m++kx=p(t) c where a dot represents differentiation with respect to time D'Alambert's Equilibrium Equation: mc+kx=P(t) 7

7 冲击荷载下的自由振动 一系列冲击荷载产生的结构振动   .. 0 ( ) t d t g P x m   ＝ 地面运动 可表示为 3.2.2 Impulsive Vibration  A series impulsive excitation —— Base Excitation ( ) ( ) ( ) - ( ) mx t cx t kx t mx t    g No. 38 2 ( ) 2 ( ) ( ) - ( ) g x t x t x t x t      = 2 c m   k = m  P( ) - ( ) g t mx t  Integrate of a series impulsive excitation 3.2.2 Impulsive Vibration At moment, τ, the impulsive loading is P(τ), under this action, the responses of a SDOF is ；     ' ' ' ( )sin ( ) t P dx t e t d m            .. : ( )= - g if P m x      ' ' ' 0 ( )sin ( ) t t P x t e t d m                 ' .. ' ' 0 1 - sin ( ) t t x t e x t d g            Duhamel integral 3.2.2 Impulsive Vibration At moment, τ, the impulsive loading is P(τ), under this action, the responses of a SDOF is ；     ' ' ' ( )sin ( ) t P dx t e t d m            .. : ( )= - g if P m x      ' ' ' 0 ( )sin ( ) t t P x t e t d m                 ' .. ' ' 0 1 - sin ( ) t t x t e x t d g            Duhamel integral 3.2.2 Impulsive Vibration D’Alambert’s Equilibrium Equation: m  x   cx   kx  P(t) Pt() Pt() 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 41 Arbitrary excitation （随机激励） 3.2.3 Arbitrary excitation x k 0 c mass mx kxcx  Dynamic Equilibrium Equation  For a dynamic or time-dependant force p (t)，the equation of static equilibrium becomes one of dynamic equilibrium: where a dot represents differentiation with respect to time.  Equation must be satisfied at each instant of time during the time interval under consideration． mx cx kx p t      No. 42 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 42 3.2.3 Arbitrary excitation

3.2.3 Arbitrary excitation 3.2.3 Arbitrary excitation +,+/=p0) Displacement Response under Arbitrary excitatior Duhamel Intogral(:哈查积分) c 0= mom)r f■elastic restoring force■kx time-dependent appled force rim.ic.h- 3.2.4 Base excitation 3.2.4 Base excitation Base excitation(Earthquakes) Base excitation (Earthquakes) +c欧+=- And dividing by m R+2Em+o2x=-式 Then +c+=-m 是大生林杠情 3.3 Numerical Analysis of Seismic Response 3.3 Numerical Analysis of Seismic Response of SDOF Basic Principles From then 0,e,0)A 0=)+2[)6】 Derivate the moment,k 【04,0e x),) 。年 0小取e小-取- 0=ta小60-w小rl-e- 移具大华持红精

8 Where =inertia force= = damping force = = elastic restoring force = k x = time-dependent applied force is the total acceleration of the mass, and , x are the velocity and displacement of the mass relative to the base. f f f pt i  d  s  cx d f s f pt x x i f mx No. 43 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 43 3.2.3 Arbitrary excitation Displacement Response under Arbitrary excitation Duhamel Integral (杜哈密积分)                  F e sin t d m x t t t        2 0 2 1 1 1 d  t F Ft() p( )  d  p(t) dv( ) t  t'=t-  2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 44 3.2.3 Arbitrary excitation Base excitation (Earthquakes) x y ground mass k c y x ground mass k c g x g m  x   cx   kx  m  x  2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 45 3.2.4 Base excitation And dividing by m g m  x   cx   kx  m  x  Then g  x   x   x   x  2 2                   x t x e sin t d t t g         2 0 2 1 1 1  2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 46 Base excitation (Earthquakes) x ground mass k c g x 3.2.4 Base excitation  Basic Principles  From the moment, (k-1) Derivate the moment, k principles：linear acceleration increase Method: linear acceleration method, Newmark-β method, and Wilson-θ method. -1 -1 -1 ( ), ( ), ( ) kkk x t x t x t ( ), ( ), ( ) kkk x t x t x t x t x t x t ( ), ( ), ( ) -1 -1 -1 ( ), ( ), ( ) kkk x t x t x t( ), ( ), ( ) kkk x t x t x t t k -1 t kt 3.3 Numerical Analysis of Seismic Response 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 47         k-1 k-1 k k-1 k k-1 t - t x t x t + x t - x t t - t   ＝             k-1 k-1 k-1 t t t k k-1 k-1 k-1 t t t k k-1 x t - x t x t dt x t dt + t - t dt t - t    ＝              2 k-1 k-1 k-1 k-1 k k-1 k k-1 1 t - t x t x t + x t t - t + x t - x t 2 t - t   ＝                   2 3 k k-1 k-1 k-1 k-1 k-1 k-1 k-1 k k-1 1 1 x t - x t x t x t + x t t - t + x t t - t + t - t 2 6 t - t ＝ t 区段[ tk-1 , t ]积分 3.3 Numerical Analysis of Seismic Response of SDOF 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 48

3.3 Numerical Analysis of Seismic Response of SDOF 3.3 Numerical Analysis of Seismic Response of SDOF ,When t—tlet x),x=x),=)=)口t=4r Let, 主+片式AN=B, Then x+元+rP=4 0=）+-[6-6门 3 1g-1 Then,we can get, 或=或+w1+伐-或) =8,+线当 (3.20) =w++我 移之 (3.21) x=u+w1+号0r+ 6 裤具，裤大学土*红服学接 闺 德典其，同济大学士木工塑学接 国 201510w13 49 模dun 2015/1013 50 3.3 Numerical Analysis of Seismic Response of SDOF 3.3 Numerical Analysis of Seismic Response of SDOF .satisfy the dynamic equilibrium equation, [Example 3.2]Consider a SDOF system with a mass of 200t. a lateral stiffness k of 7200 kN/m,and a damping ratio =0.05. 线+2wB+2xw二+r4+ 6= The record of ground motion acceleration is shown below. Please calculate the displacement,velocity,acceleration, Let, 51+at+ 64 maximum absolute acceleration,and maximum horizontal seismic action of the mass using linear acceleration method then (Take the time step as 0.1s and ends at 1.2s). =民4+25oB+aoA) Table 3-1 s 1a1020.30.405 060.7080.91.01.112 We can obtain,x()use same way to solve all results cms 108 207 300 200 150-198-120 ✉1 00 20151013 2015/1013 3.3 Numerical Analysis of Seismic Response of SDOF Please [Solution]Natrual frequency Review Dynamics o=√/m=√7200/200=6 Conduct iterative calculation according to Eqn.(3.20) (3.21)and (3.22)(Please refer to the textbook for detail to obtain the displacement,velocity,acceleration in Table 3.2) 5a1+58A1+gA12.1+005x6x01+gx01'=109 6 6 S,-求+，l.-十54.03+0昨54.03cm/6 Fm=S.·m=544.03x102×200=1088kN 牌风，济大学土*工表学精 2015/1013 54 9

9  When let, k t t ＝ = = = = = -         k k k-1 k-1 k k k k k k-1 x x t x x t x x t x x t t t t ， ， ， ， ， k k-1 k-1 k k-1   k k-1 k-1 k 1 x x + x t + x - x t 2 1 1 x x + x t + x t 2 2     ＝ ＝  Then 2 2 k k-1 k-1 k-1 k 1 1 x x + x t + x t + x t 3 6 ＝            k-1 k-1 k k-1 k k-1 t - t x t x t + x t - x t t - t   ＝   3.3 Numerical Analysis of Seismic Response of SDOF 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 49 Let, k-1 k-1 K-1 1 x + x t B 2   ＝ 2 k-1 k-1 k-1 k-1 1 x + x t + x t A 3   ＝ Then, we can get, k k-1 k t x B + x 2  ＝  （3.20） 2 k k-1 k t x A + x 6  ＝  （3.21） 3.3 Numerical Analysis of Seismic Response of SDOF 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 50  satisfy the dynamic equilibrium equation, 2 2 2 k k-1 k k-1 k g k t t x +2x w B +2x w x +w A +w x - x 2 6   ＝ 2 2 t 6 s 1 t    Let, ＝  then  B A k k  s 2 k g k 1 1 1 x x 2     ＝   We can obtain, , use same way to solve all （3.22） results ( ), ( ) k k x t x t 3.3 Numerical Analysis of Seismic Response of SDOF 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 51 [Example 3.2] Consider a SDOF system with a mass of 200t, a lateral stiffness k of 7200 kN/m, and a damping ratio ζ=0.05. The record of ground motion acceleration is shown below. Please calculate the displacement, velocity, acceleration, maximum absolute acceleration, and maximum horizontal seismic action of the mass using linear acceleration method (Take the time step as 0.1s and ends at 1.2s). Table 3-1   2 cm/s g x t 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 108 207 300 200 95 0 -150 -198 -120 -18 0 0 3.3 Numerical Analysis of Seismic Response of SDOF 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 52 [Solution] Natrual frequency   k m＝ 7200 200＝6 0 1 1 09 6 6 1 0 05 6 0 1 6 1 2 2 2 2 s    t   t   .   .   .  .    Conduct iterative calculation according to Eqn. (3.20), (3.21) and (3.22) (Please refer to the textbook for detail to obtain the displacement, velocity, acceleration in Table 3.2). 3.3 Numerical Analysis of Seismic Response of SDOF 2 F S m 544.03 10 200 1088 kN max a  ＝    ＝ ＝ Please  Review Dynamics 2015/10/13 熊海贝，同济大学土木工程学院 xionghaibei@tongji.edu.cn 54