Presentation on theme: "Chapter3 Kinematic Analysis of Mechanisms"— Presentation transcript:
1Chapter3 Kinematic Analysis of Mechanisms §3－1 Tasks and Methods of Kinematic Analysis§3－2 Velocity Analysis by the Method of InstantCenters§3－3 Graphical Velocity and AccelerationAnalysis
2§3－1 Tasks and Methods of Kinematic Analysis 一、Tasks of Kinematic AnalysisThe tasks of kinematic analysis are to find angular positions, angular velocities and angular accelerations of driven links and/or positions, linear velocities and linear accelerations of points on driven links, according to input parameters of driving link (s) and the dimensions of all links.In order to determine whether or not all links will interfere with each other, or to determine the stroke of a driven link, or to find the locus of a point, we must analyze positions of the links and/or the points of interest.
3二、Methods of Kinematic Analysis Methods of Kinematic Analysis：method of instant centres、graphical method、analytical method.Kinematic analysis of mechanisms can be carried out by graphical or analytical or experimental methods. By geometric drawing, the positions of all links in a grade II linkage can be determined easily according to the assembly order of Assur groups, step by step. In this chapter, we will introduce one of the graphical methods, named the instant centers method, for velocity analysis. The analytical method has many advantages over graphical methods.
4§3－2 Velocity Analysis by the Method of Instant Centers 12A2(A1)vA2A1Shown in Fig. are two bodies 1 and 2 having relative planar motion. At any instant there exists a pair of coincident points, e.g.P1 and P2, the absolute velocities of which are the same, in both magnitude and direction.. i. e. v P1 =V P2. At this instant, there is no relative velocity between this pair of coincident points, i.e.B2(B1)vB2B1Bi(Bj)vBiBjP12VP1P2= VP2P1=0. Thus, at this instant, either link will have pure rotation relative to the other link about the point. This pair of coincident points with the same velocities is defined as the instantaneous center of relative rotation, or more briefly the instant center, denoted as P 12 or P 21.二、Number of Instant Centers of a MechanismK＝ N ( N - 1) / 2
5三、 Location of the Instant Center 1 . Two Links Connected by a Kinematic Pair12P12Revolute pair12P12P12∞12P12∞Sliding pair12P12vM1M2M1212Higher pair2 . Theorem of Three CentersTheorem of Three Centers——the three instant centers of any three independentlinks in general plane motion must lie on a common straight line.
6四、Applications of Instant Centers Example 3-1 For the four-bar mechanism shown in figure, the angular velocity 2 of driver 2 is given. For the position shown, locate all instant centers for the mechanism and find the angular velocity 4 of the link 4.K＝N(N-1)/2＝6Solution:Number of Instant CentersP13P12 、P23 、P34、P14P13 、P24 （Theorem of Three Centers）P13P12P23P34P14ω4P24P12P23Instant Center P24
7vP24 Solution: Determine the number and location of instant centers Example In the slide-crank mechanism shown in Fig., the angular velocity of crank 2 is known. The velocity vc of slider 4 is to be found for the position shown.。Solution: Determine the number and location of instant centersP24vP24instant center P24P23P12P34P14∞The direction isto the left.
8vP23 Solution: Determine the number and location of instant centers Example 3-3 In the cam mechanism shown in Fig., supposing that the angular velocity 2 of the cam is known, the velocity v3 of the follower 3 is to be found for the position shown.Solution: Determine the number and location of instant centers123ω2Three instant centers P12、P13、P23KvP23P12P13∞P23Advantages and Disadvantages of theMethod of Instant Centers:1）To offer an excellent tool in the velocity analysis of simple mechanisms.2）In a complex mechanism, some instant centres may be difficult to find. In some cases they will lie off the paper.3）The instant center method can not be used in acceleration analysis.
9§3－3 Graphical Velocity and Acceleration Analysis This chapter presents a few simple principles of rigid body planar kinematics. These principles are useful in visualizing and understanding the motions of interconnected rigid bodies.Case 1: Two points velocity and acceleration in the same bodyCase 2: Coincidence point velocity and acceleration in two bodies一、 Two points velocity and acceleration in the same linkThe lengths of all the links and the angular velocity 1 of driver 1 are known. We wish to solve for the linear velocities of points C and E(vc , vE), accelerations of points C and E(ac, aE)，and the angular velocities and accelerations of links 2 and 3 (2 , 3 , 2 and 3)velocity = translation component + rotation component
10ω3 ω2 vc = vB + vCB 1 .Velocity analysis procedures： 4Bw11DE23Cω2vc = vB vCBdirection CD AB CBmagnitude ？ 1lAB ？ω3procedures：1）Select a velocity scalebc2) Select a point p draw pb // vB3) Draw the directions of vCB and vcpbvB bc vCBpevc = vpc vCB = v bc2 = vCB/ lBC (clockwise)3 = vC/ lCD (counter clockwise)vE = vB vEB = vC vECdirection ? √ BE √ CEmagnitude ？ √ ？ √ ？
11ω2 Velocity polygon properties： A4Bw11DE23Cω21）The vector from zero velocity point p (starting point)to point b represents the absolute velocity VB. （ pbvB ）b2）The line from point b to point c represents the velocity difference (relative velocity) VCB. Notice that VCB is represented by the vector from point b to point c .（ bc vCB ）.pec3) ∵ bce ∽BCE，bce is called the velocity image of BCE. Once velocities of two points on a link are calculated, the velocity of any other point can be obtained by similar triangles.
13α3 Acceleration polygon properties： ω2 α2 ω3 4Bw11DE23Cω2α2α3ω3b’c’’’1） The vector from zero acceleration point p’ (starting point) to point b represents the absolute acceleration（ p ’b’ aB）.2）The acceleration difference (relative acceleration) between two points, say B and C, is represented by the line point b’ to point c’. (b’c’ aCB )3）∵ b’ c’ e’ ∽BCE，b’ c’ e’ is called the acceleration image of BCE. Once accelerations of two points on a link are calculated, the acceleration of any other point can be obtained by similar triangles.
14二、Coincidence point velocity and acceleration in two bodies Given lAC , lBC , 1(counter clockwise)，Find 2, 3, 2 , 3.12ω13ABCinstantaneous absolute motion= frame of reference motion + relative motion1 . Velocity analysisω3vB2 = vB vB2B1magnitude ？ 1lAB ？direction BC AB //ABpb2vB2B1 = vb1b vB3 = vB2 = vpb22 = 3 = vB2 / lBC (counter clockwise)b1