1、11机械振动习题2xokmfmg0mx kx+ =mk=1.一弹簧振子放在一斜面上，如图所示求振动周期。解以质量块的静平衡位置为原点固有频率为：它只决定于系统的物理特性与，与坐标无关。m K32.书 14页，例 2.2.1，机械拾振仪原理图42.14 图示轮子可绕水平轴 o转动，对转轴的转动惯量为 Io，轮缘绕有一软绳，下端挂有质量 P的物体，绳与轮缘之间无滑动。在图示位置，由水平弹簧k维持平衡。半径 R与 a都是已知的，求系统绕微幅振动的周期。5解：（ 求等效质量）系统动能为:注意 : , nullx ayR=2eqJmPR=+（求等效刚度）系统势能为 :22eqakkR=6系统微幅振动的周
2、期为 :22222eqnoeq oakkkaRrad sImIRPPR = =+273：试建立图示系统的微振动方程， m仅竖向运动。kI1, R1m I2, R2kr1解： 取 m向下运动位移 x为广义坐标，坐标原点位于静平衡位置处。系统动能：2121222112eIIRmmrrR=+势能212212ek RkkrR=+x8微振动固有频率微振动方程或者( )d0dTUt+=22 212 122 2 212 12 210kr R k RxxmrRIRIR+ =+22121 1222 22112 120kIIR RmxxrrR rR+ + =9SolutionGiven:Unbalanced me
3、 = 0.23 kg-mRunning speed = 1000 rpm = 104.72 rad/sUnbalanced force =Mass of the fan (M) = 45 kgDamping ratio ( ) = 0.2Transmissibility (TR) = 0.12sinme t A centrifugal fan weighing 45 kg and having a rotating unbalance of 0.23 kg-m, is supported on the floor by a spring and a damper. At a running s
4、peed of 1000 rpm, the transmitted force to the floor is only 10% of the unbalanced force. If the damping ratio ( ) of the system is given as 0.2, determine the natural frequency and spring stiffness of the system.2-2.10Note:The magnitude of the unbalanced force is not required is the calculation. Th
5、e magnitude of the transmitted force to the ground is (0.1me )22 Equation of motion is: Mx + cx + kx = me sin t22 22120.112nrrr+=+Transmissibility is given as: TR where r=22.18 /4.72nkrad sM = =Therefore, natural frequency of the system 2nHence, the spring stiffness of the system k=M = 22,138N/m11例：
6、图示小车中放置一设备，二者用弹簧连接，小车从静止以加速度 a撞向墙壁，撞后无反弹。mkxyz小车质量 M，设备质量 mlM远大于 m忽略 m对 M运动的影响小车初始位置与墙壁距离为 l试确定（ 1）碰撞前设备相对于小车的振动响应。（ 2）碰撞后设备的振幅弹簧刚度 k12解：运动方程 mz kz ma+ =（ 1）碰撞之前对于零初始条件，方程解为12cos sinnnmazctctk = + +解：()cos 1nmaztk= 313解：碰撞后设备做自由振动，初始条件为() ()()111 1cos 1 ; sinnnma max ttxt takk= +（ 2）碰撞之后12tla= ：小车接触
7、墙壁的时刻() () ( )( )()111 1cos sinnnnxtx txt tt tt=+()( )22111cos 1 sinnnmatttk=+()()()2211nxtXxt=+14例：图示小车中放置一设备，二者用弹簧连接，小车做简谐运动 。 mkxyz设备质量 m若设备对速度非常敏感，要求设备速度不超过sinyY t=c阻尼比0.1 =试确定弹簧刚度 k。0.1Y15解：由题意：故可取222 21(2 )0.1(1 ) (2 )VY Y +=+20.1113km20.11km=165.01=lzaxf2sin=l =5 mmk/2cx0k/2xfalxfz6：汽车的拖车在波形道
8、路上行驶，已知拖车的质量满载时为m1=1000 kg, 空载时为 m2=250 kg, 悬挂弹簧的刚度为 k =350 kN/m,阻尼比在满载时为 , 车速为 v =100 km/h,路面呈正弦波形，可表示为 ，求： 拖车在满载和空载时的振幅比。170.12112=mm87.11011=kms 93.02022=kms 解：因此得到空载时的阻尼比为：满载和空载时的频率比：1868.0)2()1()2(12112212111=+=sssaB13.1)2()1()2(12222222222=+=sssaB60.021=BB记：满载时振幅 B1，空载时振幅 B2有：因此满载和空载时的振幅比：4192
9、-4. A single cylinder vertical petrol engine of total mass 400-kg is mounted upon a steel chassis frame, which can be assumed to behave as a spring. The mounted enginecauses a vertical static deflection of 0.25 cm. The reciprocating parts of the enginehave a mass of 5 kg and move through a vertical
10、stroke of 12 cm with simpleharmonic motion. A dashpot is provided, the damping resistance of which is directlyproportional to the velocity and amounts to 20 kN at 1 m/s. If steady state vibrationshas been reached, determine the undamped natural frequency, stiffness of the systemand hence the amplitu
11、de of forced vibrations when the driving shaft of the enginerotates at 540 rev/min. Calculate the maximum dynamic force transmitted to theground.SolutionParameters:()2 2g9.813924 (rad/s)0.0025n= =400 kgM =(a). Undamped Natural frequency Out of balance mass: ;Eccentricity: ;5 kgm = 0.12 / 2 0.06 me =
12、Static deflection: ;0.0025 m = 20000 N s/mC =20Therefore ()2 22.39850.0008216 m3924 3198 50 3198X =+(b). Maximum transmitted forceg62.6418 rad/sn=Stiffness 2400 3924 1569600 N/mnkM=Running speed: ()22218 3198 (rad/s)=Equation of motion:2sinMx Cx kx me += 50 3924 2.3985sinxx x + += Amplitude of force
13、d vibration at 540 rpm:Therefore Max. Transmitted force is: ()( )221589.4817 NkX c X+=212-5. A vibration instrument can be represented by the system shown in the figure. A mass, suspended by a spring from the top of the instrument case, operates a piston in a dashpot. The damping is viscous and is 0
14、.6 times the critical value. When the dashpot isremoved, the natural frequency of vertical oscillation of the system is 10 Hz. With the dashpot in operation, the instrument is attached to a body moving vertically with simple harmonic motion at a frequency of 6 Hz. If the amplitude recorded by the in
15、strument (the amplitude of relative motion between the mass and the case) is 0.10 cm, calculate the maximum acceleration of the body with which the instrument is in contact.Figure 5SolutionThe basis for vibrating instrument is govern by:222212nnnYZ= + 22where:0.001 m , 10 Hz = 20 rad/s , =0.6nZ = =
16、6 Hz =12 rad/s , 0.6n=Substitute, all relevant parameters gives223.803 m/sY =237： P346习题 3.13，阻尼系数 c不为零解： 小车运动方程()()110my c y y k y y+= 振幅为：11 1 1cos sinmy cy ky cy ky cY t kY t += + = + 2 vL =() ()221sinmy cy ky Y c k t += + + 1tanck=21 22 21(2 )(1 ) (2 )YY +=+无阻尼时：222224nP vkgL=211 22 2114kgL YYYk
17、gL P v=2cPk g =248： P346习题 3.13，路面波形为解： 小车运动方程( ) ( )110my c y y k y y+ + = 强迫振动解11 1 11cos sinmy cy ky cy ky cY t kY kY t + +=+= + 2 vL =() ()2211sinmy cy ky kY Y c k t + += + + 1tanck=()211 22 21(2 )sin(1 ) (2 )yYY t += +121sinxYL122tan1=5253.17某洗衣机机器部分重 W=2.2*103N，用四个螺旋弹簧在对称位置支撑，每个弹簧的螺圈平均半径 R=51m
18、m，弹簧丝直径 d=18mm，圈数 n=10，剪切弹性模量 G=8*105N/cm2。同时装有四个阻尼器，总的阻尼可用 =0.1表示。在脱水时转速 N=600r/min，此时衣物偏心重为 10N，偏心距为 40cm，试求：（1）洗衣机机器部分的最大振幅；（2）传递率及隔振效率。26解： 根据螺旋弹簧变形的计算公式,可计算每个弹簧的刚度:()()()44334523882810 / 1.8810 25.19892.12 /Gd GdKnDnRNcm cmcmNm=224 4 9892.12 9.8176.3 1/2200nKsM= =222 2 6003948 1/60 60Ns = = 222
19、394822.4176.3n= =27()()22222100.4 39489.82200 109892.123 3948 4 0.01 9892.123 39489.8 9.80.184meXkM ccm=+= + =洗衣机机器部分的的最大振幅为:28传递率为：()()()()2222 222222 22()()()12 1 4 0.1 22.41 22.4 4 0.1 22.4120.0643XkcTRYkm c+=+ + =+=隔振效率为： ( )1 100% 93.57%TR = =294.1 试证明：有阻尼振系对作用于 t=0时的冲量 的响应，其出现峰值的时刻为2max11arcta
20、nqt=并求峰值 。maxxF30对 h（t） 求导：由题意可知，系统的脉冲响应函数为：sin , 0()ntddFettmxt=()() sin cosndndd dtFmxt ett= +令导数为零，得：2max11arctandt=() 0xt=122arctanmax11,aadFxem =带入 h（ t） ,得：6314.2 用脉冲响应函数法求无阻尼振动系统对图示激扰力的响应。设 时， ， 。()f t0t =00x =00x =解：线性振动系统满足叠加原理：激振力 引起的系统响应等于时间区间上所有脉冲激发起的系统响应总和。()f t激振力 可视作图示三个阶跃函数激励 的叠加。()f
21、 t单自由度无阻尼线性振动系统的脉冲响应函数如下所示：1( ) sin( ) ( 0)nnht t tm=123(), (), ()f tftft3201t2tt()ft1F2F01tt2()f t12()F F +0t1()f t1F()a ()b ()c02tt2F3()f t33对于图(a)阶跃力 ,由 Duhamel积分公式得：11() ( 0)ft F t=0101() ( ) ( )d1sin ( ) d= 1 cos( ) ( 0)ttnnnxt ht ftFmFttk =对于图(b)阶跃力 ,由 Duhamel积分公式得：2121() ( + ) ( )f tFFtt= 112
22、2012 10() ( ) ( )d( ) 0d ( ) ( )d ( )ttttxt ht fht ht F F t t =+ 34同理可得，对于图(c)阶跃力 ，有322() ( )f tFtt=12211() 1 cos( ( ) ( )nFFx ttk+=232() 1 cos( ( ) ( )nFx ttk= 由叠加原理得如下结果：当 时，所求响应为 ，即：10 tt123() () () ()x t xtxtxt= +11212( ) cos( ( ) cos( )cos( ( ) cos( ( ) nnnnFxt t t tkFtt ttk=+364.4 求无阻尼弹簧质量系统对图
23、示斜坡力函数的响应。设t=0时，( )f tat=00, x xxx= =解：（方法 1） 无阻尼弹簧质量振动系统的运动微分方程为( )00, mx kx f tx xxx +=其解为：2nkm 固有频率为：12sin cosnnax AtA ttk=+737由初始条件 t=0, 可得001cos sin sinnnxax xt tt tk =+00, x xxx=0120nnxaAAk= =故：38方法 2： 用 Duhamel积分() ( )()0000001=sin cossin1= sin cos sintnnnntnnnxx ftdxttmxatdxt tm + + +由分部积分，有
24、()011sin sintnnnnaatdt tmk =001cos sin sinnn nxax xt tt tk =+于是有：39方法 3： 无阻尼弹簧质量振动系统的运动微分方程为：( )00, mx kx f tx xxx +=两边取拉普拉斯变换有：()()0022002222 2 22111=nnnnnn namx msxsXsms ksx x akss ss+=+ +2nkm 引入记号做拉普拉斯逆变化得：001cos sin sinnnxax xt tt tk =+403-6. A punch press can be modeled for vibration analysis i
25、n the x-direction as indicated by a two-degree-of-freedom system of Figure 2. The values of the various masses and stiffness coefficients are:a) Ignoring damping, formulate the equations of motion of the system and express them in a matrix form. Hence, find its natural frequencies and mode shapes. N
26、ormalized the mode shapes by setting the displacement for mass to unity.12 1 2400 , 2000 , 300kN , 80kN .m kg m kg k m andk m= = 2m41b) Assume that the whole system is initially at rest. The mass is given an impulse of 1000nullnull Ns, where null is the Dirac delta function. The damping ratios for t
27、he first mr= = kg mEquations of motion:0( 0.2 ) ( 0.25 ) 0( 0.25 )0.25 ( 0.2 )0.2 sinmx k x k xIkx kx M t + + =+ =5000N/mk =56Rearranging gives:222 0.05 0 (1)0.05 (0.25 0.2 ) 0(2)mx kx kIkx k+ =+ =sinsinxA tBt =令：代入振动微分方程 ，可得如下代数方程：( )()2202 0.05 0 (3)0.1025 0.05 (4)mkAkBIkBkAM+ + =+ + =不产生摆动，即 B=0，
28、220mk +=22mk=57试建立转子在平面内微振动的运动方程。9： 转子 -轴系统，转子简化为刚体偏心距 e转子质量为 M， 跨度为 l偏心质量 m M m转子绕 O点平面内转动惯量 I偏心距转子中心的水平距离 b竖向弹簧均为刚度为 k/2ObemM2k2k2k2kl582k2k解：取质心 O点的竖向位移 xo和绕O点的转角 为广义坐标。221122oTMx I=+22112222oollUkx kx=+系统动能( )( )22sin sinWPx Pb me tx meb t =+= +ObxoPPbP势能增量外力做功2k2kxy59代入拉格朗日方程得2222sin1sin2ooMxkx
29、me tIklmeb t +=+=矩阵形式22 2200sin10 02ookMx xmetI kl meb += 无耦合60若偏心质量不能忽略，即 M与 m质量相差不是很大，系统微振动运动方程如何？ObemM2k2k2k2k解： m的坐标为()211cos2oxx by bb = += =系统动能()22 22111222oTMx I mxy=+ +()222111222ooMxI mxb + +222222ooklklUx x=+势能增量2k2kObxoPbP2k2kxy1161代入拉格朗日方程得()()2222sin1sin2oooMmx mb kx me tImb mbx kl meb t +=+=矩阵形式22 2 220sin102ookMm mb x x metmb I mb kl meb+ += + 惯性耦合()222111222ooTMx I mxb + +222222ooklklUx x=+()( )22sin sinWPx Pb me tx meb t =+= +
