结构动力学(单自由度体系的振动分析)

1、自由振动自由振动-由初位移、初速度引起的由初位移、初速度引起的, ,在振动中无动荷载作用的振动。在振动中无动荷载作用的振动。一一. .运动方程及其解运动方程及其解EIl)(ty)(tym )()(11tymty )()(11tymtyk 0)()(2tyty 111121mmk一一. .运动方程及其解运动方程及其解EIl)(ty)(tym )()(11tymty )()(11tymtyk 0)()(2tyty 111121mmk其通解为其通解为tctctysincos)(21由初始条件由初始条件0)0(yy0)0(yy可得可得01yc /02yctytytysincos)(00sin0Ay c

2、os/0Ay)sin()(tAty22020yyA00tanyy二二. .振动分析振动分析其通解为其通解为tctctysincos)(21由初始条件由初始条件0)0(yy0)0(yy可得可得01yc /02yctytytysincos)(00sin0Ay cos/0Ay)sin()(tAty22020yyA00tanyy单自由度体系不计阻尼时的自由振动是简谐振动单自由度体系不计阻尼时的自由振动是简谐振动. .)2sin()sin()(tAtAty)2()2(sintytA2T自振周期自振周期2T自振周期自振周期21T自振园频率自振园频率( (自振频率自振频率) )与外界无关与外界无关, ,体系

3、本身固有的特性体系本身固有的特性A 振幅振幅初相位角初相位角三三. .自振频率和周期的计算自振频率和周期的计算1.1.计算方法计算方法(1)(1)利用计算公式利用计算公式111121mmk11,WmgWststg2(2)(2)利用机械能守恒利用机械能守恒)(cos21)(21)(2222tmAtymtT)(sin21)(21)(2211211tAktyktUmaxmaxUT20202121kvymEI)(21)(21)(22tkytymtE2002cossin-21)(tvtymtE200sincosk21tvty20202121) t (kvymE=常数常数(3)(3)利用振动规律利用振动规

4、律)sin()(tAty)sin()(2tAty )sin()()(2tmAtymtI 位移与惯性力同频同步位移与惯性力同频同步. .211mAAk111kEIl)(ty2mAA幅值方程幅值方程mk1122.2.算例算例例一例一. .求图示体系的自振频率和周期求图示体系的自振频率和周期. .3117121mlEIm)221213221(111lllllllllEIEImlT127223EIlEIl=111=1ll/2l解解: :EIl3127例二例二. .求图示体系的自振频率和周期求图示体系的自振频率和周期. .3332231mlEIEIlmEIl31132EImlT32=1解解: :23lE

5、IEIllm/2EIEIll例三例三. .质点重质点重W,求体系的频率和周期求体系的频率和周期. .3113lEIkk解解: :EIkl11k111kk33lEIgWm/gWlEIk33例四例四. .求图示体系的自振频率和周期求图示体系的自振频率和周期. .222222max29)2(21)(21)2(21mllmlmlmT解解: :mk95lmEImlllkk)(t1.1.能量守恒能量守恒2222max25)2(21)(21kllklkUmaxmaxUT2.2.列幅值方程列幅值方程ml22ml22ml2lklk2A 0AM0222222222lklllmlmllkllml059222klml

6、mk95c-阻尼系数阻尼系数 ( (damping coefficient ）)()(tyctR011ykycym )(ty)(tym )(11tyk)(tycmc2/022yyy tAety)(022221i)2(1mc 21D)cossin()(21tctcetyDDt小阻尼情况小阻尼情况00)0(,)0(vyyy02001,/)(ycyvcD)sin()(DDttAety20020)(DyvyA)/(tan000yvyDD011ykycym )(ty)(tym )(11tyk)(tycmc2/022yyy tAety)(0222临界阻尼情况临界阻尼情况)2(1mc )()(000ytyv

7、etyt超阻尼情况超阻尼情况)2(1mc )()(000tchytshyvetyccct12c)2(1mc 21D小阻尼情况小阻尼情况00)0(,)0(vyyy02001,/)(ycyvcD)sin()(DDttAety20020)(DyvyA)/(tan000yvyDD)cossin()(21tctcetyDDt1mcr2-临界阻尼系数临界阻尼系数mcccr2-阻尼比阻尼比 大量结构实测结果表明，大量结构实测结果表明，对于钢筋混凝土和砌体结构对于钢筋混凝土和砌体结构 ，钢结构，钢结构 。各种坝体的。各种坝体的，拱坝，拱坝 ，重力坝（大头坝），重力坝（大头坝） ，土，土坝、堆石坝坝、堆石坝 。

8、05. 004. 003. 002. 02 . 003. 005. 003. 01 . 005. 02 . 01 . 0mc2/DDT221D小阻尼情况小阻尼情况)sin()(DDttAety20020)(DyvyA)/(tan000yvyDDttytycdyfEtDD)d()(10DDiiTTttiieAeAeAA)(1)sin()(DDttAetyDiiTAA1ln22D1ln21iiAAniiAAnln21niiBBnln21kN4 .160276.012ln421)/(102 .802.0104 .165311mNk) s (5 .04/2DT) s (4998.012DTT)s/1

9、(57.122T)kg(5190/211km)kN(86.50 mgW)s/mN(36012mc)s/1 (89.1368005190102 . 8252)s/1 (70.11)s (537.0/2T0257.02/mcPROBLEMSPROBLEMS：3.A mass m is at rest,partially supported by a spring and 3.A mass m is at rest,partially supported by a spring and partially by stops.In the position shown,the spring force

10、 is partially by stops.In the position shown,the spring force is mg/2. At time t=0 the stops are rotated,suddenly releasing the mg/2. At time t=0 the stops are rotated,suddenly releasing the mass.Determine the motion of the mass. mass.Determine the motion of the mass. kmPROBLEMSPROBLEMS：5.A mass m5.

11、A mass m1 1 hangs from a spring k and is in static equilibrium. hangs from a spring k and is in static equilibrium. A second mass mA second mass m2 2 drops through a height h and sticks to m drops through a height h and sticks to m1 1 without rebound.Determine the subsequent motion y(t) measured wit

12、hout rebound.Determine the subsequent motion y(t) measured from the static equilibrium position of mfrom the static equilibrium position of m1 1 and k. and k.1mk2mhPROBLEMSPROBLEMS：6.For a system with damping ratio ,determine the number of 6.For a system with damping ratio ,determine the number of free vibration cycles required to reduce the displacement free vibration cycles required to reduce the displacement amplitude to 10% of the initial amplitude,the initial velocity amplitude to 10% of the initial amplitude,the initial velocity is zero.is zero.7.7.若系统作自由衰减振动，经过若系统作自由衰减振动，