l结构动力学-2-单自由度体系自由振动

1、第二章第二章单自由度体系的振动分析单自由度体系的振动分析 第二章第二章单自由度体系的振动分析单自由度体系的振动分析 第二章第二章单自由度体系的振动分析单自由度体系的振动分析 第二章第二章单自由度体系的振动分析单自由度体系的振动分析 第二章第二章单自由度体系的振动分析单自由度体系的振动分析 运动方程运动方程运动方程的通解运动方程的通解)()(11tymty )()(11tymtyk 0)()(2tyty 令令111121mmk第二章第二章单自由度体系的振动分析单自由度体系的振动分析 EIl)(ty)(tym mkm12( )cossiny tctct运动方程的特解运动方程的特解令令AATty自振

2、周期自振周期自振频率自振频率与外界无关与外界无关, ,体系本身固有的特性体系本身固有的特性00(0), (0)yyyv00( )cossinvy tytt00sin ,/cosyAvA( )sin()y tAt22002vAy00tanyv2T12T自振频率和周期的计算自振频率和周期的计算1.1.利用计算公式利用计算公式EIlEIl=111=1ll/2l例例1.1.求图示体系的自振周期求图示体系的自振周期. .解解: :211111kmm2T1111211()23222ll lll l lllEI 3712lEI3111127EImml327212mlTEIEIlEIl=111=1ll/2l一

3、次超静定结构，用立法可一次超静定结构，用立法可以解出简支撑内力为以解出简支撑内力为1.51.5，方向向上，然后及可以画出方向向上，然后及可以画出内力弯矩图内力弯矩图(2)(2)利用机械能守恒利用机械能守恒例例2.2.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. .解解: :lmEImlllkk)(t( )( )T tU t 常数222211( )( )cos ()22T tmytmAt222111111( )( )sin ()22U tk ytk AtmaxmaxTU22max222211(2)()2219(2)22Tmlm lmlml 2222max115()(2)222Uk lk

4、lklmaxmaxTU59km(3)(3)利用振动规律利用振动规律位移与惯性力同频同步位移与惯性力同频同步. .例例3.3.质点重质点重W,求体系的频率求体系的频率. .解解: :EIkl2mAAkAAlEI332mA( )sin()y tAt2( )sin()y tAt 2( )( )sin()I tmy tmAt 233EImAkAAl/mW g33EIklgWlmEImlllkkml22ml22ml2lklk2A例例4.4.求自振频率和周期求自振频率和周期. .解解: : 列幅值方程列幅值方程转动惯量转动惯量刚度系数刚度系数自振频率自振频率0AM2222222220lmlkllmlllm

5、ll kl 222950lmkl 222(2 )29Jmlmlml2225klkllklkl 211kkmJc-阻尼系数阻尼系数 ( (damping coefficient ）)(ty)(tym )(11tyk)(tyc( )( )R tcy t 110mycyk y/2cm220yyy( )ty tAe2220)(ty)(tym )(11tyk)(tyc小阻尼情况小阻尼情况110mycyk y/2cm220yyy( )ty tAe22201 (2)cm21i 21D00(0), (0)yyyv10020()/,Dcvycy( )sin()tDDy tAet22000()DvyAy12( )

6、(sincos)tDDy tectct000tan/()DDyvy小阻尼情况小阻尼情况临界阻尼情况临界阻尼情况超阻尼情况超阻尼情况ty1111 (2)cm21D12( )(sincos)tDDy tectct00(0), (0)yyyv10020()/,Dcvycy( )sin()tDDy tAet22000()DvyAy000tan/()DDyvy1(2)cm000( )()ty tevy ty1(2)cm000( )()tcccvyy teshty cht21c 1mcr2-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

7、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 。05. 004. 003. 002. 02 . 003. 005. 003. 01 . 005. 02 . 01 . 0it1itDTt)(tyiA1iADDT221D小阻尼情况小阻尼情况)sin()(DDttAety20020)(DyvyA)/(tan000yvyDDDDiiTTttiieAeAeAA)(1DiiTAA1ln22D阻尼测量阻尼测量DiiTAA1ln22D1ln21iiAAniiAAnln21kN4 .160276.012l

8、n421)/(102 .802.0104 .165311mNk) s (5 .04/2DT) s (4998.012DTT)s/1 (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/mckN4 .160276.012ln421)/(102 .802.0104 .165311mNkPROBLEMSPROBLEMS：1.A heavy table is supported by flat steel legs.Its natural 1

9、.A heavy table is supported by flat steel legs.Its natural period in lateral vibration is 0.5sec.When a 50kg plate is period in lateral vibration is 0.5sec.When a 50kg plate is clamped to its surface.the natural period in lateral vibration clamped to its surface.the natural period in lateral vibrati

10、on is lengthened to 0.75sec.What are the weight and the effective is lengthened to 0.75sec.What are the weight and the effective lateral stiffness of the table?lateral stiffness of the table?T=0.5secT=0.5secT=0.75secT=0.75secPROBLEMSPROBLEMS：2.An electromagnet weighing 400kg and suspended by a sprin

11、g 2.An electromagnet weighing 400kg and suspended by a spring having a stiffness of 100kg/cm.lifts 200kg of iron scrap. having a stiffness of 100kg/cm.lifts 200kg of iron scrap. Determine the equation describing the motion when the electric Determine the equation describing the motion when the elect

12、ric current is turned off and the scrap is dropped.current is turned off and the scrap is dropped.kkkPROBLEMSPROBLEMS：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 is parti

13、ally 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.A mass m1

14、 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 without rebo

15、und.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