结构力学(第4版)教学课件9-9完整

第9章结构动力分析§3.3受迫振动mnyn(t)一.振型分解法Pn(t)1.位移计算my(t)cy(t)ky(t)P(t)m2m1P2(t)y2(t)y1(t)设方程的解为y(t)nXD(t)P1(t)i1iiM*jDj(t)C*jDj(t)K*jDj(t)Pj*(t)(j1,2,n)PtXPtj*()()Tj---广义荷载Dj(t)2jjDj(t)2jDj(t)Pj*(t)/M*j第9章结构动力分析Dj(t)2jjDj(t)2jDj(t)Pj*(t)/M*jM*jPj*(t)*KjDj(t)jjtDj(0)jjDj(0)Dj(t)e[Dj(0)cosDjtsinDjt]DjjtPj*()jj(t)0M*jDjesinDj(t)dDj(0)XTjmy(0)/M*jDj(0)XTjmy(0)/M*j计算步骤:1.求振型、频率;2.求M*j,Pj*3.确定振型阻尼比;;4.求广义坐标;5.求位移y(t);第9章结构动力分析2.动内力计算SPmycyky若略去阻尼力Sky3.简谐荷载Psintmnyn(t)n广义荷载PtXPtXPtj*()()sinTTjjPj*sintm2m1P*XTPP2sinty2(t)y1(t)jj*2PjDj(t)2jjDj(t)jDj(t)*sintP1sintMj第9章结构动力分析2Pj*Pj*sintM*jDj(t)2jjDj(t)jDj(t)*sintMjK*jDj(t)（1）稳态解Dj(t)Dj,stjsin(tj)jPj*Pj*2222Dj,st**2j1/(1j)4jjKjMjjtanj2jj/(12j)ny(t)XiDi,stisin(ti)in1yj(t)xjiDi,stisin(ti)i1nBisin(ti)Asin(t)i1nnnnA(Bicosi)2(Bisini)2tanBisini/Bicosii1i1i1i1第9章结构动力分析（2）瞬态解D(t)ejjt[ccostcsint]Dsin(t)j1Dj2Djj,stjj4.算例例9-36.求各楼面最大位移。m2EI1m1m210200kg120.05kP0sint2m1k13000kN/mk22000kN/mEI1P00.217kN11.451/sk119.8991/s224.251/s11X1,X221/2解.1）计算广义质量10m10200kg01M1*X1TmX151000kg第9章结构动力分析例9-36.求各楼面最大位移。m2EI1120.0519.8991/s24.251/sX1,X1P00.217kN11.451/sPsintk20m1EI121221/2k1解.1）计算广义质量M1*X1TmX151000kgM2*XT2mX212750kg2）计算广义荷载PXP1*1T0.217kNPXP2*T20.217kN3）计算DstDP1*4.34105mD2.89105m1,stM*212,st1第9章结构动力分析例9-36.求各楼面最大位移。m2EI1120.0519.8991/s24.251/sX1,X1P00.217kN11.451/sPsintk20m1EI121221/2k1解.4）计算1/11.162/20.47211/(112)2412122.7421/(122)2422221.2855）计算1,2tan1211/(112)0.335tan2222/(122)0.06071161.4723.48第9章结构动力分析例9-36.求各楼面最大位移。m2EI1120.0519.8991/s24.251/sX1,X1P00.217kN11.451/sPsintk20m1EI121221/2k1解.6）计算D1,D2D1(t)D1,st1sin(t1)11.89105sin(t161.47)D2(t)D2,st2sin(t2)3.727105sin(t3.48)7）计算位移y(t)X1D1X2D211D1D221/2第9章结构动力分析7）计算位移m2EI1y(t)X1D1X2D2Psintk2m111EI1D1D2k121/2y1(t)11.89105sin(t161.47)3.727105sin(t3.48)A1sin(t1)A1(11.89cos161.473.727cos3.48)2(11.89sin161.473.727sin3.48)210557.03516.0671058.55105mtan116.067/57.0350.531152.1y1(t)8.55105sin(t152.1)y2(t)25.52105sin(t163.04)第9章结构动力分析3）计算D1,D2DtP1*()sin(t)dP1*(1cost)10M1*11M1*1210.0032(1cos1t)D20.000534(1cos2t)4）求位移y111D1D2y221/2y(t)2D1D0.006670.0064cos9.9t0.000267cos24.25t2122第9章结构动力分析二.简谐荷载稳态振动分析的直接解法不计阻尼时的情况mnyn(t)Pnsintmy(t)ky(t)PsintP1P2Pm2m1PnPsinty2(t)y1(t)2P1sint设方程的解为y(t)Asint(k2m)AP第9章结构动力分析以两个自由度体系为例：1P1sint(m1y1)P2sintf11f21m1m2yEIy1m1y112m2y2(m2y2)P1P2f12f22sint1P2P运动方程y1f11(m1y1)f12(m2y2)Δ1Psintyf(my)f(my)Δsint2211122222P设特解为y1A1sint(m1f111/2)A1m2f12A2Δ1P/2yAsint22mfA(mf1/2)AΔ/2121122222P第9章结构动力分析D1D2Dm1f111/2Δ1P/2解方程,得A1A22mfΔ/21212PDD其中Δ/2mfmf1/2mfD11P22122Δ2P/m2f221/D1112122m1f21m2f221/1.在平稳阶段,作简谐振动,振动频率与荷载同。(m1f1121)A1m2f12A22Δ1P2.当0时2mfA(2mf1)AΔ121122222PA1Δ1PA2Δ2P3.当4.当1或2A时10A20A1时，A2n自由度体系有n个共振区设特解为y1A1sint(m1f111/2)A1m2f12A2Δ1P/2yAsint22mfA(mf1/2)AΔ/2121122222P第9章结构动力分析5.求稳态振幅可列幅值方程y1A1sintyAsint222P1P2I1(t)m1y1m1A1