机械振动复习思考题1心O 距离为l ϑϑsin 0mgl J -= ϑϑ≈sin 00=+ϑϑmgl J T J mgl n n 02,==ω2204n T mgl J=20ml J J c -=2 半径为r 、质量为的固有频率。

解：ϕϑr r R v c =-=)(ϑϕr r R -= 222121ϕ J mv T c c +=cos 1)((r R h --=(21R mg mgh V ==22ref 2max )(43,)(21m m r R m T r R mg V ϑϑ-=-=)(32refmax r R g T V n -==ω3 举例说明振动现象、振动的危害以及如何有效的利用振动。

答：1）振动现象：心脏的搏动、耳膜和声带的振动等；汽车、火车、飞机及机械设备的振动；家用电器、钟表的振动；地震以及声、电、磁、光的波动等等。

2）振动的危害轻则影响乘坐的舒适性；降低机器及仪表的精度，重则危害人体健康，引起机械设备及土木结构的破坏。

3）振动的利用琴弦振动；振动沉桩、振动拔桩以及振动捣固等；振动检测；振动压路机、振动给料机和振动成型机等。

4何为机械振动及研究目的?答：机械振动：机械或结构在平衡位置附近的往复运动。

研究目的：利用振动为人类造福；减少振动的危害。

5 何为振动系统的自由度？请举例说明。

答：自由度就是确定系统在振动过程中任何瞬时几何位置所需独立坐标的数目。

刚体在空间有6个自由度：三个方向的移动和绕三个方向的转动，如飞机、轮船；质点在空间有3个自由度：三个方向的移动，如高尔夫球；质点在平面有2个自由度：两个方向的移动，加上约束则成为单自由度。

6 如何对机械振动进行分类？答：1）按振动系统的自由度数分类单自由度系统振动——确定系统在振动过程中任何瞬时几何位置只需要一个独立坐标的振动；多自由度系统振动——确定系统在振动过程中任何瞬时几何位置需要多个独立坐标的振动；连续系统振动——确定系统在振动过程中任何瞬时几何位置需要无穷多个独立坐标的振动。

2）按振动系统所受的激励类型分类自由振动——系统受初始干扰或原有的外激励取消后产生的振动；强迫振动——系统在外激励力作用下产生的振动；自激振动——系统在输入和输出之间具有反馈特性并有能源补充而产生的振动。

3）按系统的响应（振动规律）分类简谐振动——能用一项时间的正弦或余弦函数表示系统响应的振动；周期振动——能用时间的周期函数表示系统响应的振动；瞬态振动——只能用时间的非周期衰减函数表示系统响应的振动；随机振动——不能用简单函数或函数的组合表达运动规律，而只能用统计方法表示系统响应的振动。

4）按描述系统的微分方程分类线性振动——能用常系数线性微分方程描述的振动；非线性振动——只能用非线性微分方程描述的振动。

7 简述构成机械振动系统的基本元素答：构成机械振动系统的基本元素有惯性、恢复性和阻尼。

惯性就是能使物体当前运动持续下去的性质。

恢复性就是能使物体位置恢复到平衡状态的性质。

阻尼就是阻碍物体运动的性质。

从能量的角度看，惯性是保持动能的元素，恢复性是贮存势能的元素，阻尼是使能量散逸的元素。

所以，质量、弹簧和阻尼器是构成机械振动系统物理模型的三个基本元件。

9 阐明下列概念。

(a) 振动； (b) 周期振动和周期； (c) 简谐振动；（d ）振幅、频率和相位角。

10 如图3所示，已知m =15t, v 0=20 m/min ，k =5.78MN/m 。

求：钢丝绳的最大拉力。

解：以弹簧在静载作用下变形后的平衡位置为原点建立O x 坐标系 ∑=iixFxmkmg mg x k xm =++st st ,)(δδ＝－mk x xn ==220,0ωω＋ kmg mg x k xm =++st st ,)(δδ＝－mk x xnn==22,0ωω＋0n 0,;0,0,)sin(v xt x t t A x ====+= ϕω t mk km v x v A nsin;0,0===ϕωkN 2.245)(0max max =+=+=mk v mg x k F st δ10 已知⎭⎬⎫⎩⎨⎧=⎭⎬⎫⎩⎨⎧⎥⎦⎤⎢⎣⎡+⎭⎬⎫⎩⎨⎧⎥⎦⎤⎢⎣⎡+⎭⎬⎫⎩⎨⎧⎥⎦⎤⎢⎣⎡00212221121121222112112122211211x x k k k k x xc c c c x xm m m m ，用状态空间法求系统的响应。

解：[][][][][][][][][]⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=-=---01220111I K MC M B A N ，{}⎭⎬⎫⎩⎨⎧=520z {}[]{}{}0=-z N z，{}[]{}z N z = ，[][][][]A P P N =。

特征值问题为[][]0=-I N λ，即00122=----λλ，展开有δs0222=++λλ解之得i 1i 121--=+-=λλ，[]⎥⎦⎤⎢⎣⎡=-+Λt t t)i 1(-)i 1(-e 0e e，[]⎥⎦⎤⎢⎣⎡=+---2i 12i111P ，[]⎥⎦⎤⎢⎣⎡=-+-i -i 2i12i11P [][][][]⎥⎦⎤⎢⎣⎡++=⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡==-++----++--+-++----Λt tt tt t ttN P P )i 1(-2i 1)i 1(-2i 1)i 1(-4i)1()i 1(-4i)1(2i 12i1)i 1(-)i 1(-2i 12i11e eeeXX i -i e 00e 11e e22[]⎥⎦⎤⎢⎣⎡=-+Λt tt)i 1(-)i 1(-e 0e e{}[]{}0ez x x z tN =⎭⎬⎫⎩⎨⎧= [][])sin 7cos 5(e eeee5ee2)(-)i 1(-27i 5)i 1(-27i 5)i 1(-2i 1)i 1(-2i 1)i 1(-4i)1()i 1(-4i)1(22t t t x ttttttt+=+=+++=-++--++----++-10 A foot pedal for a musical instrument is modeled by the sketch in Figure P2.30. With k = 2000 kg, c = 25 kg/s, m = 25 kg and F (t ) = 50 cos 2πt N, compute the steady state response assuming the system starts from rest. Also use the small angle approximation.Solution: Free body diagram of pedal follows:Summing the moments with respect to the point, O:11 A very common example of base motion is the single-degree-of-freedom model of an automobile driving over a rough road. The road is modeled as providing a base motion displacement of y (t ) = (0.01)sin (5.818t ) m. The suspension provides an equivale nt stiffness of k = 4 x 105N/m, a damping coefficient of c = 40 x 103kg/s and a mass of 1007 kg. Determine the amplitude of the absolute displacement of the automobile mass.Solution:From the problem statement we haveωb =5.818, k = 4×105N/m ， c =40×103 kg/s ，Y = 0.01 m, m = 1007 kg12用刚度系数法建立如下三自由度系统的振动微分方程解：1）设x 1=1，x 2=x 3=0，则在m 1上施加的力F 1=1×(k 1+k 2)，即k 11= k 1+k 2 ；在m 2上施加的力F 2=-k 2 × 1 =-k 2 ，即k 21=-k 2 ；在m 3上施加的力为零，即F 3=0或 k 31=0。

2）设x 2=1，x 1=x 3=0，则在m 2上施加的力F’2=1× (k 2+k 3)，即k 22= k 2+k 3 ；在m 3上施加的力F’3=-k 3 即k 32=-k 3 ；由刚度矩阵的对称性得 k 12 =k 21= -k 2。

3）设x 3=1，x 1=x 2=0，则在m 3上施加的力F’’3=1× k 3，即k 33= k 3 ；由刚度矩阵的对称性得 k 13 =k 31 = 0 ， k 23 =k 32= -k 3。

得系统振动微分方程⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--+--++⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡0000000000321333322221321321x x x k k k k k k k k k x x x m m m13 写出图示系统的振动微分方程 解：建立广义坐标如图 方程[]{}[]{}[]{}{}0=++x K x C xM {}[]T4321,,,x x x x x =，{}[]T4321,,,x x x x x =，{}[]T4321,,,x x x x x =。

其中，[]⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=432100000000000m m m m M ，[]⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡+----+=0000000646226262c c c c c c c c c C []⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--+++---+---++=5556543363322626210000k k k k k k k k k k k k k k k k k k K 。

Interpretation of Vibration DataThe key to using vibration signature analysis for predictive maintenance, diagnostic, and other applications is the ability to differentiate between normal and abnormal vibration profiles. Manyvibrations are normal for a piece of rotating or moving machinery. Examples of these are normal rotations of shafts and other rotors, contact with bearings, gear-mesh, etc. However, specific problems with machinery generate abnormal, yet identifiable, vibrations. Examples of these are loose bolts, misaligned shafts, worn bearings, leaks, and incipient metal fatigue.Predictive maintenance utilizing vibration signature analysis is based on the following facts, which form the basis for the methods used to identify and quantify the root causes of failure:1. All common machinery problems and failure modes have distinct vibration frequency components that can be isolated and identified.2. A frequency-domain vibration signature is generally used for the analysis because it is comprised of discrete peaks, each representing a specific vibration source.3. There is a cause, referred to as a forcing function, for every frequency component in a machine-train’s vibration signature.4. When the signature of a machine is compared over time, it will repeat until some event changes the vibration pattern (i.e., the amplitude of each distinct vibration component will remain constant until there is a change in the operating dynamics of the machine-train).While an increase or a decrease in amplitude may indicate degradation of the machine-train, this is not always the case. Variations in load, operating practices, and a variety of other normal changes also generate a change in the amplitude of one or more frequency components within the vibration signature. In addition, it is important to note that a lower amplitude does not necessarily indicate an improvement in the mechanical condition of the machine-train. Therefore, it is important that the source of all amplitude variations be clearly understood.Sources of VibrationLike rotating machinery, the vibration profile generated by reciprocating and/or linear-motion machines is the result of mechanical movement and forces generated by the components that are part of the machine. Vibration profiles generated by most reciprocating and/or linear-motion machines reflect a combination of rotating and/or linear-motion forces.However, the intervals or frequencies generated by these machines are not always associated with one complete revolution of a shaft. In a two-cycle reciprocating engine, the pistons complete one cycle each time the crankshaft completes one 360degree revolution. In a four-cycle engine, the crank must complete two complete revolutions, or 720 degrees, in order to complete a cycle of all pistons.Because of the unique motion of reciprocating and linear-motion machines, the level of unbalanced forces generated by these machines is substantially higher than those generated by rotating machines. As an example, a reciprocating compressor drives each of its pistons from bottom-center to top-center and returns to bottom-center in each complete operation of the cylinder. The mechanical forces generated by the reversal of direction at both top-center and bottom-center result in a sharp increase in the vibration energy of the machine. An instantaneous spike in the vibration profile repeats each time the piston reverses direction.Linear-motion machines generate vibration profiles similar to those of reciprocating machines. The major difference is the impact that occurs at the change of direction with reciprocating machines. Typically, linear-motion-only machines do not reverse direction during each cycle of operation and, as a result, do not generate the spike of energy associated with direction reversal.14 填图，补充完整下列振动分类图。