4. Graphical Analysis of acceleration and velocity.pptx

graphical analysis of velocity and acceleration. The notes help to achieve the better analysis of velocity and acceleration. So in order to analyze graphical analysis these notes help you to attain better analysis and help to study it.

1. Kinematic Analysis of four bar linkage Theory of Machines (ME225) Instructor: Bilawal Ramzan

2. Kinematic Analysis of Four bar linkage Summary Chapter 4,6, 7 + Lecture Notes Coordinate Systems and coordinate transformation Translation and Rotation Graphical Position, Velocity and Acceleration Analysis Velocity Analysis using Instant Centers

3. 4.1 Coordinate system Inertial Frame of Reference: No acceleration Global coordinate system: often taken to be attached to the Earth, though it could as well be attached to another ground plane such as the frame of an automobile (GCS – X , Y ). Also called as the absolute coordinate system Local coordinate system are typically attached to a link to some point of interest. This might be a pin joint, a centre of gravity, or a line of centres of a link (LCS – x , y )

4. Position A position of a point in the plane can be defined by the use of “position vector” The choice of reference axis is arbitrary and is selected to suit the observer A two dimensional vector has two attributes, which can be expressed in either Polar or Cartesian coordinates The Polar form provides the magnitude and angle of the vector The Cartesian form provides the X and Y components of the vector

5. Relationship between polar and Cartesian

6. 4.1 Coordinate Transformation It is often necessary to transform the coordinates of a point defined in one coordinate system to coordinate system in another If the position of point A is expressed in the local xy system as R x , R y and it is desired to transform its coordinated to R X , R Y in the global XY system, the equations are: R x R y

7. 4.2 Displacement Displacement of a point is the change in its position and can be defined as the straight line distance between the initial and final position of a point which has moved in the reference frame The position of B with respect to A is equal to the (absolute) position of B minus the (absolute) position of A, where absolute means with respect to global reference frame R BA = R B – R A R BA = R BO – R AO

8. 4.3 Translation All points on the body have the same displacement Note that translation need not be along a straight path If the path happens to be a straight line then its called the rectilinear translation , otherwise it is the curvilinear translation R A’A = R B’B

9. Rotation Different points in the body undergo different displacements and thus there is a displacement difference between any two points chosen R B’B = R BA – R B’A

10. Complex motion Sum of translation and rotation components Total displacement = translation component + rotation component R B’’B = R B’B + R B’’B’ R B’’A = R A’A + R B’’A’

11. Theorems Euler’s theorem (Pure Rotation): The general displacement of a rigid body with one fixed point is a rotation about the some axis. Chasles ’ theorem (Complex Motion): Any displacement of a rigid body is equivalent to the sum of a translation of any one point on that body and a rotation of the body about an axis through that point.

12. Kinematic Analysis variables Synthesis: For the required motion find the a, b , c and d Position Analysis : Given  2 , a, b, c and d, find  3 and  4 Graphical Analytical

13. 4.4 Graphical Position Analysis of Linkages Given the length of the links ( a,b,c,d ), the ground position, and q 2 find q 3 and q 4 b c a d q 2 b c q 3 q 4 A O 2 O 4 B 1 B 2

14. Problem 4-7a (4-6 in 5 th edition)

15. Graphical velocity analysis To solve any mechanism graphically for velocity, two equations are needed

16. Example 6-1 Given  2 ,  3 ,  4 ,  2 , find  3 ,  4 , V A , V B , V C by graphical method

17. Graphical Velocity Analysis (w3 & w4) Given linkage configuration, w 2 , find w 3 and w 4 It is known that V A and direction of V B and V BA are perpendicular to their corresponding links V BA ? V B V BA Direction V BA V B V B Direction V A

18. Graphical Velocity Analysis (w3 & w4) After finding V B and V BA , find w 3 & w 4

19. Graphical Velocity Analysis (VC) After finding w 3 and w 4 , find V C V CA = V C - V A V CA V C V A Double Scale V CA V C

20. 7.2 Graphical acceleration analysis For solving any acceleration analysis problem graphically, three equations are needed

21. Graphical Acceleration Analysis ?  ?          q q p p

22. Acceleration of Coupler Point ? ?     A CA

23. Example 7-1 Given  2 ,  3 ,  4 ,  2 ,  3 ,  4 ,  2 , Find  3 ,  4 , A A , A B , A P by graphical method L 1 = 6” L 2 = 2” L 3 = 5.5” L 4 = 3.5”  2 = 60 o  4 = 86.56 o  2 = 2 rad/sec 2  2 = 1 rad/sec  3 = 1 rad/sec  4 = 1.2 rad/sec

24. Graphical solution of linkage in a pin-jointed linkage with a negative (CW) 2 and a positive (CCW) 2

25. 6.3 Instant Centers A point common to two bodies in plane motion, which has the same instantaneous velocity in each body. Both links sharing the instant center will have identical velocity at that point Instant centers are sometimes called centros or poles It takes two bodies or links to create an instant center (IC), one can easily predict the quantity of instant centers to expect from any collection of links For a combination of n things taken r at a time is Where r = 2 (minimum number of links to make I.C.) n = number of total links in a mechanism

26. Instant Centers Kennedy’s rule: any three bodies (links) will have exactly three instant centers and they will lie on the same straight line The pins are instant centers I 13 is from links 1,2,2,3 and 1,4,4,3 I 24 is from links 1,2,1,4 and 2,3,3,4 I 13 I 24

27. Instant Centres (Linear Graph) A linear graph is useful for keeping track of which IC’s have been found Can be created by drawing a circle on which we mark off as many points as there are links in assembly Lines are then drawn between the dots representing the links pair each time an instant circle is found The resulting linear graph is a set of lines connecting the dots The circle is not included which was used to place the dots A dual way to represent lines with dots or vice versa

28. Rectilinear slider’s instant centre at infinity

29. Instant Centres in the slider-crank linkage

30. 6.4 Velocity Analysis using instant centres Link 1 is at zero velocity in global coordinate system I 1,3 is between link 1 and 3 stating link 3 at zero velocity at that instant So, link 3 is pivoted around I 1,3 to move in pure rotation instantaneously

31. Velocity Analysis using instant centres The magnitude of V A can be computed as  2 and length of link 2 are known Note that point A is also instant centre I 2,3 It has the same velocity as part of the link 2 and as part of link 3 Since link 3 is effectively pivoting about I 1,3 at this instant, the angular velocity  3 can be found by  3 = v A /AI 1,3 Once  3 is known, the magnitude of V B can also be found; v B = (BI 1,3 ) 3 Once V B is known,  4 can also be found as  4 = v B /BO 4 Finally the magnitude of V C (or any other point on the coupler) can be found v C = (CI 1,3 )w 3

32. Velocity Analysis using instant centres Since the location of I 1,3 , which is instantaneous “fixed” pivot for link 3 , all of that link’s absolute velocity vectors for this instant will be perpendicular to their radii from I 1,3 to the point in question V B and V C can be seen to be perpendicular to their radii from I 1,3 V B is also perpendicular to the radius from O 4 because B is also pivoting about that point as part of link 4 Also the magnitude of velocities of V B’ and V C’ can be found by drawing arcs from point B and C to line extended from I 1,3 I 1,2 and then extending their tips to line I 1,3 V A

33. 6.5 Centrodes Path or locus of the instant center is called the centrode. Reading Assignment.

34. Assignment Session # 03 Problem Numbers (5 th Edition) 4.6 (Graphical Method) 4.9 (Graphical Method) 6.1  6.4 6.6 6.8 6.12 6.18 7.1 7.3 7.5 7.13 7.15 7.21 7.24