機械振動復(fù)習(xí)思考題(含答案)

1、機械振動復(fù)習(xí)思考題1 圖示的直升機槳葉經(jīng)實驗測出其質(zhì)量為m，質(zhì)心C距鉸中心O距離為l?，F(xiàn)給予槳葉初始擾動，使其微幅擺動，用秒表測得多次擺動循環(huán)所用的時間，除以循環(huán)次數(shù)獲得近似的固有周期，試求槳葉繞垂直鉸O 的轉(zhuǎn)動慣量。 解：取圖示坐標(biāo)系，將直升機槳葉視為一物理擺，根據(jù)繞固定鉸的動量矩定理得到其擺動微分方程2 半徑為r、質(zhì)量為m的圓柱體在半徑為R的內(nèi)圓柱面上繞最低點作純滾動，試求其微振動的固有頻率。解：3 舉例說明振動現(xiàn)象、振動的危害以及如何有效的利用振動。答：1）振動現(xiàn)象： 心臟的搏動、耳膜和聲帶的振動等；汽車、火車、飛機及機械設(shè)備的振動；家用電器、鐘表的振動；地震以及聲、電、磁、光的波動等等

2、。2）振動的危害 輕則影響乘坐的舒適性；降低機器及儀表的精度，重則危害人體健康，引起機械設(shè)備及土木結(jié)構(gòu)的破壞。3）振動的利用 琴弦振動；振動沉樁、振動拔樁以及振動搗固等；振動檢測；振動壓路機、振動給料機和振動成型機等。4何為機械振動及研究目的?答：機械振動：機械或結(jié)構(gòu)在平衡位置附近的往復(fù)運動。 研究目的：利用振動為人類造福； 減少振動的危害。5 何為振動系統(tǒng)的自由度？請舉例說明。答：自由度就是確定系統(tǒng)在振動過程中任何瞬時幾何位置所需獨立坐標(biāo)的數(shù)目。剛體在空間有6個自由度：三個方向的移動和繞三個方向的轉(zhuǎn)動，如飛機、輪船； 質(zhì)點在空間有3個自由度：三個方向的移動，如高爾夫球；質(zhì)點在平面有2個自由度

3、：兩個方向的移動，加上約束則成為單自由度。6 如何對機械振動進(jìn)行分類？答：1）按振動系統(tǒng)的自由度數(shù)分類單自由度系統(tǒng)振動確定系統(tǒng)在振動過程中任何瞬時幾何位置只需要一個獨立坐標(biāo)的振動；多自由度系統(tǒng)振動確定系統(tǒng)在振動過程中任何瞬時幾何位置需要多個獨立坐標(biāo)的振動；連續(xù)系統(tǒng)振動確定系統(tǒng)在振動過程中任何瞬時幾何位置需要無窮多個獨立坐標(biāo)的振動。2）按振動系統(tǒng)所受的激勵類型分類自由振動系統(tǒng)受初始干擾或原有的外激勵取消后產(chǎn)生的振動；強迫振動系統(tǒng)在外激勵力作用下產(chǎn)生的振動；自激振動系統(tǒng)在輸入和輸出之間具有反饋特性并有能源補充而產(chǎn)生的振動。3）按系統(tǒng)的響應(yīng)（振動規(guī)律）分類簡諧振動能用一項時間的正弦或余弦函數(shù)表示系統(tǒng)

4、響應(yīng)的振動；周期振動能用時間的周期函數(shù)表示系統(tǒng)響應(yīng)的振動；瞬態(tài)振動只能用時間的非周期衰減函數(shù)表示系統(tǒng)響應(yīng)的振動；隨機振動不能用簡單函數(shù)或函數(shù)的組合表達(dá)運動規(guī)律，而只能用統(tǒng)計方法表示系統(tǒng)響應(yīng)的振動。4）按描述系統(tǒng)的微分方程分類線性振動能用常系數(shù)線性微分方程描述的振動；非線性振動只能用非線性微分方程描述的振動。7 簡述構(gòu)成機械振動系統(tǒng)的基本元素答：構(gòu)成機械振動系統(tǒng)的基本元素有慣性、恢復(fù)性和阻尼。慣性就是能使物體當(dāng)前運動持續(xù)下去的性質(zhì)?；謴?fù)性就是能使物體位置恢復(fù)到平衡狀態(tài)的性質(zhì)。阻尼就是阻礙物體運動的性質(zhì)。從能量的角度看，慣性是保持動能的元素，恢復(fù)性是貯存勢能的元素，阻尼是使能量散逸的元素。所以，質(zhì)

5、量、彈簧和阻尼器是構(gòu)成機械振動系統(tǒng)物理模型的三個基本元件。 9 闡明下列概念。 (a) 振動； (b) 周期振動和周期； (c) 簡諧振動；（d）振幅、頻率和相位角。10 如圖3所示，已知m=15t, v0=20 m/min，k=5.78MN/m。求：鋼絲繩的最大拉力。v0解：以彈簧在靜載作用下變形后的平衡位置為原點建立Ox坐標(biāo)系10 已知，用狀態(tài)空間法求系統(tǒng)的響應(yīng)。解：，。特征值問題為，即，展開有解之得，10 A foot pedal for a musical instrument is modeled by the sketch in Figure P2.30. With k = 200

6、0 kg, c = 25 kg/s, m = 25 kg and F(t) = 50 cos 2tN, 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

7、 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 equivalent stiffness of k = 4 x 105 N/m, a damping coefficient of c = 40 x 103 kg/s and a mass of 100

8、7 kg. Determine the amplitude of the absolute displacement of the automobile mass.Solution: From the problem statement we have b=5.818, k = 4×105N/m， c=40×103 kg/s ，Y = 0.01 m, m = 1007 kg12 用剛度系數(shù)法建立如下三自由度系統(tǒng)的振動微分方程解：1）設(shè)x1=1，x2=x3=0，則在m1上施加的力F1=1×(k1+k2)，即k11= k1+k2 ；在m2上施加的力F2=-k2 

9、15; 1 =-k2 ，即k21=-k2 ；在m3上施加的力為零，即F3=0或 k31=0。2）設(shè)x2=1，x1=x3=0，則在m2上施加的力F2=1× (k2+k3)，即k22= k2+k3 ；在m3上施加的力F3=-k3 即k32=-k3 ；由剛度矩陣的對稱性得 k12 =k21= -k2。3）設(shè)x3=1，x1=x2=0，則在m3上施加的力F3=1× k3，即k33= k3 ；由剛度矩陣的對稱性得 k13 =k31 = 0 ， k23 =k32= -k3。得系統(tǒng)振動微分方程 13 寫出圖示系統(tǒng)的振動微分方程解：建立廣義坐標(biāo)如圖方程，。其中，。Interpretation

10、 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. Many vibrations are normal for a piece of rotating or moving machinery. Examples of these are

11、 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 maintenan

12、ce 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

13、.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-trains vibration signature.4. When

14、 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 a

15、mplitude 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

16、 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

17、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

18、, 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 comple

19、te 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