《结构动力学》课程教学课件（讲稿）03 Response to harmonic loading

Wuhan University of Technology3.1 Undamped systemFig.arepresents the steadystate component of responsewhile Figb represents the socalled transient response. It is assumed thatβ = 2/3, that is, the applied loading frequency is twothirds of thefreevibration frequency. The total response R(t), i.e., the sum of bothtypes of response, is shown in Fig.c.Twopointsareofinterest:(1)thetendencyforthetwocomponents to get inphase and then out of phase again, causinga"beating" effect in the total response; and (2) the zero slope of totalresponse at time t=O, showing that the initial velocity of the transientresponse is just sufficient to cancel the initial velocity of the steady-stateresponse;thus,itstatifiesthespecifiedinitialconditioni(0)= 04-11

4-11 Wuhan University of Technology Fig.a represents the steadystate component of response while Fig. b represents the socalled transient response. It is assumed tha t = 2/3, that is, the applied loading frequency is twothirds of the freevibration frequency. The total response R(t), i.e., the sum of both types of response, is shown in Fig.c. Two points are of interest: (1) the tendency for the two components to get in phase and then out of phase again, causing a “beating” effect in the total response; and (2) the zero slope of total response at time t=0, showing that the initial velocity of the transient response is just sufficient to cancel the initial velocity of the steady￾state response; thus, it statifies the specified initial condition . v(0) 0   3.1 Undamped system

Wuhan University of TechnologyX, = A coso,tX = X, + X, = A, cosot+ A2 coso,tX, = A, coso,t00, +0,0X=2Acostcos2Q:(（6)X=X1CA4-72

4-12 Wuhan University of Technology X A t 1 1 1  cos X A t 2 2 2  cos X X X A t A t 1 2 1 1 2 2    cos  cos X A t t 2 cos 2 2 cos2 1 1 2 

Wuhan University of Technology3.2 System with viscous damping*y(t)*p(t)m000k()mi(t)+ci(t)+kv(t)= psin@t=250m(t)+250v(t)+0v(t)=P0sinot4-13

4-13 Wuhan University of Technology 3.2 System with viscous damping 2 c m      2 0 2 sin p vt vt vt t m       y(t) c m p(t) k y(t) c m p(t) k 0 mv t cv t kv t p t   ( ) ( ) ( ) sin    

Wuhan University of Technology3.2 System with viscous dampingThe complementary solutionVo(t) =[Acos Opt + Bsin Opt]exp(-Eot)The particular solutiony,(t)=G, cost +G,sinat1- B2P-2EBPG,人(1-β) +(25B)2(1-β) +(2EB)4-14

4-14 Wuhan University of Technology The complementary solution vt A t B t t 0 ( ) cos sin exp     D D      1 2 ( ) cos sin p vt G tG t     The particular solution     0 1 2 2 2 2 1 2 P G k                    2 0 2 2 2 2 1 1 2 P G k                3.2 System with viscous damping

Wuhan University of Technology3.2 System with viscous dampingPv(t)=(Acoso,t+BsinO,t)exp(-Eot)+1-β2)sinot-2BcosatThefirsttermontherighthand sideofthiseguationrepresentsthetransient response, which damps out in accordance with exp(-5ot)whilethe secondtermrepresentsthe steady-stateharmonicresponse,whichwillcontinueindefinitely.TheconstantsAandBcanbeevaluated for any given initial conditions, v(O) and v(O). However,since the transient response damps out quickly, it is usually of littleinterest.4-15

4-15 Wuhan University of Technology         0 2 2 2 2 1 ( ) ( cos sin ) exp 1 sin 2 cos 1 2 D D P vt A t B t t t t k                             3.2 System with viscous damping The first term on the right hand side of this equation represents the transient response, which damps out in accordance with while the second term represents the steady-state harmonic response, which will continue indefinitely. The constants A and B can be evaluated for any given initial conditions, v(0) and v(0). However, since the transient response damps out quickly, it is usually of little interest. expt