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§3 - 1 Tasks and Methods of Kinematic Analysis §3 - 2 Velocity Analysis by the Method of Instant Centers §3 - 3 Graphical Velocity and Acceleration Analysis.

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Presentation on theme: "§3 - 1 Tasks and Methods of Kinematic Analysis §3 - 2 Velocity Analysis by the Method of Instant Centers §3 - 3 Graphical Velocity and Acceleration Analysis."— Presentation transcript:

1 §3 - 1 Tasks and Methods of Kinematic Analysis §3 - 2 Velocity Analysis by the Method of Instant Centers §3 - 3 Graphical Velocity and Acceleration Analysis Chapter3 Kinematic Analysis of Mechanisms Chapter3 Kinematic Analysis of Mechanisms

2 一、 Tasks of Kinematic Analysis The 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 - 1 Tasks and Methods of Kinematic Analysis

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 1 一、 Instant Center Shown in Fig. are two bodies 1 and 2 having relative planar motion. At any instant there exists a pair of coincident points, e.g.P 1 and P 2, 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. Bi(Bj)Bi(Bj) v BiBj 2 A2(A1)A2(A1) v A2A1 P 12 V P1P2 = V P2P1 =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 Mechanism K = N ( N - 1) / 2 § 3 - 2 Velocity Analysis by the Method of Instant Centers B2(B1)B2(B1) v B2B1

5 1 2 三、 Location of the Instant Center 1. Two Links Connected by a Kinematic Pair Revolute pair 1 2 1 2 P 12 Sliding pair 1 2 P 12 ∞ 1 2 ∞ Higher pair 1 2 P 12 v M1M2 M 2. Theorem of Three Centers Theorem of Three Centers——the three instant centers of any three independent links 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. Solution: Number of Instant Centers K = N(N-1)/2 = 6 P 12 、 P 23 、 P 34 、 P 14 P 13 、 P 24 ( Theorem of Three Centers ) P 12 P 23 P 13 1 2 3 P 13 P 12 P 23 P 34 P 14 Instant Center P 24 ω4ω4 P 24

7 Solution: Determine the number and location of instant centers P 23 P 12 P 34 P 14 ∞ P 24 v P24 instant center P 24 The direction is to the left. Example 3- 2 In the slide-crank mechanism shown in Fig., the angular velocity of crank 2 is known. The velocity v c of slider 4 is to be found for the position shown. 。

8 1 2 3 ω2ω2 P 12 P 13 ∞ K v P 23 Example 3-3 In the cam mechanism shown in Fig., supposing that the angular velocity  2 of the cam is known, the velocity v 3 of the follower 3 is to be found for the position shown. Solution: Determine the number and location of instant centers Three instant centers P 12 、 P 13 、 P 23 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. Advantages and Disadvantages of the Method of Instant Centers: P 23

9 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 body  Case 2: Coincidence point velocity and acceleration in two bodies 一、 Two points velocity and acceleration in the same link The 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(v c, v E ), accelerations of points C and E(a c, a E ) , and the angular velocities and accelerations of links 2 and 3 (  2,  3,  2 and  3 ) § 3 - 3 § 3 - 3 Graphical Velocity and Acceleration Analysis velocity = translation component + rotation component

10 v c = v B + v CB direction  CD  AB  CB magnitude ?  1 l AB ? procedures : 1 ) Select a velocity scale 3) Draw the directions of v CB and v c pb  v B bc  v CB 2) Select a point p draw pb // v B e b p A 4 B  1 D E 2 3 C c v c =  v pc v CB =  v bc  2 = v CB / l BC (clockwise) ω2ω2 ω3ω3  3 = v C / l CD (counter clockwise) v E = v B + v EB = v C + v EC direction ? √  BE √  CE magnitude ? √ ? √ ? 1.Velocity analysis

11 c e b p A 4 B  1 D E 2 3 C ω2ω2 3) ∵  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. Velocity polygon properties : 1 ) The vector from zero velocity point p (starting point)to point b represents the absolute velocity V B. ( pb  v B ) 2 ) The line from point b to point c represents the velocity difference (relative velocity) V CB. Notice that V CB is represented by the vector from point b to point c. ( bc  v CB ).

12 A 4 B  1 D E 2 3 C ω2ω2 ω3ω3 magnitude ω 3 2 l CD ? ω 1 2 l AB ω 2 2 l BC ? direction C  D  CD B  A C  B  CB c’’ p’ b’ c’’’ c’ e’ Select a acceleration scale , draw acceleration vector diagram. 2. Acceleration analysis magnitude  ω 2 2 l BE ?  ω 2 2 l CE ? direction  E  B  BE  E  C  CE α2α2 α3α3

13 A 4 B  1 D E 2 3 C ω2ω2 ω3ω3 c’’ p’ b’ c’’’ c’ e’ Acceleration polygon properties : α2α2 α3α3 1 ) The vector from zero acceleration point p’ (starting point) to point b represents the absolute acceleration ( p ’b’  a B ). 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’  a CB ) 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 l AC, l BC,  1 (counter clockwise) , Find  2,  3,  2,  3. 1 2 ω1ω1 3 A B C instantaneous absolute motion = frame of reference motion + relative motion 1. Velocity analysis v B2 = v B1 + v B2B1 magnitude ?  1 l AB ? direction  BC  AB //AB b2b2 b1b1 p v B2B1 =  v b 1 b 2 v B3 = v B2 =  v pb 2  2 =  3 = v B2 / l BC (counter clockwise) ω3ω3

15 a B3 = a B2=  a p 'b 2 ' 1 2 ω1ω1 3 A B C b2b2 b1b1 p ω3ω3 magnitude ω 2 2 l BC ? ω 1 2 l AB 2v B2B1 ? direction B  C  BC B  A  AB // AB p’ b1’b1’ k’ b2’b2’ b 2 ’’ a k B2B1 2. Acceleration analysis (counter clockwise) α3α3


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