案例专业课课件analysis of mechanisms

1、This chapter mainly introduces the tasks and methods of kinematic analysis. The method of instant centres, graphic and analytical methods are discussed. Chapter 3Kinematic Analysis of MechanismsEmphases： The method of instant centres, graphic methodDifficulties: Graphic method3.1 Tasks and Methods o

2、f Kinematic Analysis3.2 Velocity Analysis by the Method of instant Centers3.3 Kinematic Analysis by Graphic Methods3.4 Kinematic Analysis by Analytical MethodsProblems and ExercisesSummary and kernel3.1 Tasks and Methods of Kinematic AnalysisTask: to find positions, velocities and accelerations or a

3、ngular positions, angular velocities and angular accelerations.Position: to determine whether all links will interfere with each other, to find locus Velocity: to calculate the stored kinetic energy or power. Acceleration：To calculate the dynamic forcesMethods: graphical method (图解法): 直观，精度不高 analyt

4、ical method (解析法) : 精度高，计算烦琐 experimental method (实验法) : 需专门的仪器 Instant(瞬时的) centre method for velocityVector(矢量) equation method3.2 Velocity Analysis by the Method of instant Centres3.2.1 Definition of the Instant Centre instant centerP12 or P21Shown in Fig. are two bodies 1 and 2 having relative p

5、lanar motion. At any instant there exists a pair of coincident points, the absolute velocities of which are the same.At this instant, there is no relative velocity between this pair of coincident points. This pair of coincident points with the same velocities is defined as the instantaneous centre o

6、f relative rotation, denoted as P12 or P21.Suppose that the positions of points A and B, the directions of VA2A1 and VB2B1 are known.VA2A1AP. VB2B1BP. instant centerP12 One of bodies is static absolute instant center Both of them are moving relative instant centerAttention:A pair of coincident point

7、s, the absolute velocities of which are the same, in both magnitude and direction.instant centerRelative velocity is zero.3.2.2 Number of Instant Centers of a Mechanismknumber of links. The frame is included in the number k.Each pair of links i and j has an instant centre and Pij is identical to Pji

8、. Thus the number N of instant centres of a mechanism with k links is(1) Revolute Pair3.2.3 The location of the Instant Center of Two Links Connected by a Kinematic Pair(2) Sliding Pair Instant center: lies at infinity in either direction perpendicular to the guide-way. Attention：The common normal m

9、ay pass through any point !(3) Higher Pair: Pure-rolling PairInstant centerthe point of contact (4)Higher Pair :rolling & sliding PairInstant centerlies somewhere on the common normal through the point of contact3.2.4 Theorem of Three Centres (Aronhold-Kennedy Theorem)Any three links have three inst

10、ant centers.They must lie on a straight line CVc 2Vc 3P12P13AB123233.2.5 Applications of Instant CentersExample1: The revolute four-bar linkage. (1) locate all instant centers for the mechanism(2) find the ratio 2/4Find：P12、P23、P34和P14；Solution：Theorem of Three Centres：(1) locate all instant centers

11、 P12 ， P14P24P23 ， P34P12 ， P23P13P14 ， P34(2) find the ratio 2/4P24 : the absolute velocities of which are the same, in both link2 and link 4.Example2:For the following Slider-Crank Mechanism，Given: lAB, lBC,，1 locate all instant centers for the mechanism the velocity of follower link3 V3 。Solution

12、：(1) Find Instant centerP13(2) Determine V3Example3In the cam mechanism as shown in the following fig. the cam2 rotates anti-clockwise at a constant speed ，Determine the velocity of the follower2 for the position shown. (1) Find Instant center P23P23must lie on the straight line connecting P12 and P

13、13.P23P23 must lie along the common normal n-n through the point of contact C(2) Determine V3Solution：3.2.6 Advantages and Disadvantages of the Method of Instant CentersExcellent tool in simple mechanisms. Difficult to find Instant Center in a complex mechanismNot be used in acceleration analysis.A

14、Base point ，B moving pointFor two points A and B on the same body: vBA=lABw， vBA Direction：AB，along with anBA=lABw2，anBA Direction：BA。atBA=lABa，atBA Direction：AB, along with aIn which：3.3 Kinematic Analysis by Graphic Methods1、The velocity and acceleration of the points on the same bodyGiven：The len

15、gth of all links in mechanism, position,Cal：Sol：1) Draw the kinematic diagram of the mechanismSelect a length suitable scale ul .2) Determine the velocity and angular velocityB Base point，C Moving pointGraphic method：Select velocity scale v , construct velocity diagram p-bc：Direction：Magnitude：(Cloc

16、kwise)(Anticlockwise)(1) Point C：CD？AB lABw1BC？ Bbase point，Emoving point vE = vB + vEB(2) Point EP-bec Polygon of velocities P Speed poleDir.： ？Mag.： ？ABlAB w1EBlEB w2On the velocity diagram p-bc，construct be= vEB/mv，point e is got, so： (1) In the polygon of velocities, bce and BCE are similar. The

17、 letters of bce and BCE are in the same direction，clockwise. The triangle bce is called When the velocities of two points on the same link are known, the velocity of any point on the link can be determined by similarity principle 。The image of velocities can only be used by the points on the same li

18、nk。 (2) P Speed pole. Its velocity is zero.Attentions：Absolute speed vector passes through the speed pole and is from P to the point.The subscript of relative speed vector and its image are the opposite.(Anticlockwise)(clockwise)方向：CD大小：lCDw323) Determine the acceleration and the angular acceleratio

19、n Polygon of accelerations？ EB EB ？ lEBw22 lEB2 Construct be”=anEB/ua, get e”；Construct e”e=atEB/ua，get e： Acceleration pole The acceleration image of points B、C、E on the link The method can only be used by the pints on the same link。2. The velocities and accelerations of coincide points on two link

20、s which constitute a sliding pair.Given：The lengths and positions of all links， Link 1 rotate in a constant angular velocity Cal：Sol：1) Draw the kinematic diagram of the mechanismSelect a length suitable scale ul .2)、Determine velocity vB3 = vB2 + vB3B2Dir.： CBMag.： ？Select suitable velocity scale mv.Construct polygon of velocity.So：(Clockwise) B3 moving point， Slider 2 moving reference systemAB lABw1/Cx？3)Determine accelerationCoriolis acceleration a kB3B2:() BC lBCw32Mag：a kB3B2=2evr=22vB3B2， 2=3；Dir.：Turn vr=vB3B2 90 along