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- 1. Kinematics of Machinery (ME-224) Module-3 By Sajan Kapil Assistant Professor Indian Institute of Technology Guwahati Guwahati-781039, Assam, India Email: sajan.Kapil@iitg.ac.in Phone: 003612582652
- 2. Elements of kinematic chain, mechanisms, their inversions, mobility (Kutzhbach criteria) and range of movements (Grashof's law); Miscellaneous mechanisms: straight line generating mechanism, intermittent motion mechanism; Displacement, velocity and acceleration analysis of planar mechanisms by graphical, analytical and computer aided methods; Dimensional synthesis for motion; function and path generation; Cam profile synthesis and determination of equivalent mechanisms; Gears (spur, helical, bevel and worm); gear trains: simple, compound and epicyclic gearing. Texts: 1. K. J, Waldron and G. L Kinzel, Kinematics, Dynamics and Design of Machinery, 2nd Ed., Wiley Student Edition, 2004. 2. A. Ghosh and A. K. Mallik, Theory of Mechanisms, and Machines, 3rd Ed., East West Press Pvt Ltd, 2009 References: 1. J. J Uicker (Jr), G. R Pennock and J. E Shigley, Theory of Machines and Mechanisms, 3rd ed., Oxford International Student Edition. 2. S. S. Rattan, Theory of Machines, 3rd Ed., Tata McGraw Hill, 2009. 3. R. L. Norton, Kinematics and Dynamics of Machinery, Tata Mcgraw Hill, 2009. 4. J. S. Rao, R. V. Dukkipati, Mechanism and Machine Theory, 2nd Ed., New Age International, 2008. 5. A. G. Erdman and G. N. Sandor, Mechanism Design, Analysis and Synthesis Volume 1, PHI, Inc., 1997. 6. T. Bevan, Theory of Machines, CBS Publishers and Distributors, 1984 ME 224 Kinematics of Machinery (2-1-0-6)
- 4. Lecture-1
- 5. Introduction Synthesis of Linkages β’ Generating circular motion is an easiest task. β’ Many times you are asked to generate an irregular motion on the output side. β’ After solving the assignment-1, you must be appreciating the importance of 1 DOF in the mechanisms. β’ The two main type of 1 DOF mechanisms are CAM and linkages. β’ In order to generate irregular path, CAM are easy to design but expensive, and subjected to wear. β’ Linkages are more difficult to design but are relatively inexpensive and reliable. β’ The simplest linkage capable of generating the desirable motion is 4-link mechanism. β’ We shall consider the design of 4-link mechanism and the technique can be extended for the design of 6 & 8 link mechanism
- 6. Introduction 4-link Mechanism RRRR RRRP RRPP RPRP 4 Bar Mechanism Slider Crank Mechanism Elliptical-trammel
- 7. Introduction 4 Bar Mechanism β’ The double rocker problem. β’ The rocker-amplitude problem. β’ The motion-generating problem. β’ The path generating problem.
- 8. 4 Bar Mechanism β’ Simplest linkage design problem and very common in industry β’ The problem is to design a 4-bar mechanism that will move its output link through an angle π while the input link moves through an angle π B A C Bβ Cβ π D π Cββ Dββ Double Rocker Problem: Graphical Method
- 9. B A C Bβ Cβ π D π Cββ Dββ 4 Bar Mechanism Double Rocker Problem: Graphical Method
- 10. B A C Bβ Cβ π D π Cββ 4 Bar Mechanism Double Rocker Problem: Graphical Method
- 11. 4 Bar Mechanism Double Rocker Problem: Analytical Method B A C Bβ Cβ π D π π1 π1 β’ The angles are measured +ve in counter clockwise direction. β’ This is consistent with the standard right hand rule. (x, y) coordinates of C π₯π = r1 + r4 Γ cos π1 π¦π = r4 Γ sin π1 (x, y) coordinates of Cβ π₯πβ² = r1 + r4 Γ cos π + π1 π¦πβ² = r4 Γ sin π + π1 (x, y) coordinates of B π₯π΅ = π2 Γ cos π1 π¦π΅ = r2 Γ sin π1 (x, y) coordinates of Bβ π₯π΅β² = π2 Γ cos π + π1 π¦π΅β² = r2 Γ sin π + π1 π1 π2 π3 π4
- 12. 4 Bar Mechanism Double Rocker Problem: Analytical Method B A C Bβ Cβ π D π π1 π1 Note that distance BC=BβCβ π3 = π₯π β π₯π 2 + π¦π β π¦π 2 = π₯πβ² β π₯πβ² + π¦πβ² β π¦πβ² 2 β β 1 π2 cos π1 β π1 β π4 cos π1 2 + π2 sin π1 β π4 sin π1 2 = π2 cos(π + π1) β π1 β π4 cos π + π1 2 + π2 sin(π + π1) β π4 sin π + π1 2 βπ1π2 cos π1 β π2π4 cos π1 cos π1 + π1π4 cos π1 β π2π4 sin π1 sin π1 = βπ1π2 cos π + π1 β π2π4 cos π + π1 cos π + π1 + π1π4 cos π + π1 β π2π4 sin π + π1 sin π + π1 Or π1 π2 π3 π4
- 13. 4 Bar Mechanism Double Rocker Problem: Analytical Method B A C Bβ Cβ π D π π1 π1 The only unknown in that equation is π2. Solving the equation for π2: π2 = π1π4 cos π + π1 β cos π1 βπ4 cos π1 β π1 + π1 cos π + π1 β cos π1 + π4 cos π + π1 β π β π1 Using equation 3. π3 = π2 cos π1 β π1 β π4 cos π1 2 + π2 sin π1 β π4 sin π1 2 π1 π2 π3 π4
- 14. Introduction 4 Bar Mechanism: Double Rocker Problem: Example A*B*=20 mm B*B1=10 mm Design the double rocker mechanism shown in Figure:
- 15. Lecture-2
- 16. 4 Bar Mechanism π΅1 π΄1 πΆ1 πΆ2 πΌ π·1 π The rocker-amplitude problem: Graphical Method π΅2 π π1 π3 π4
- 17. 4 Bar Mechanism π΅1 π΄1 πΆ1 πΆ2 πΌ π·1 π The rocker-amplitude problem: Graphical Method π΅2 π π1 π3 π4 π = π 360Β° β π π = 180Β° + πΌ 180Β° β πΌ If the crank is rotating with a constant velocity then the ratio of time in forward and backward stroke will be: or If in the synthesis of linkage time ratio Q and output link oscillation angle is give. Then the first step is to calculate the angle πΌ. πΌ = π β 1 π + 1 180Β°
- 18. 4 Bar Mechanism π΄1 πΆ1 πΆ2 πΌ π·1 π The rocker-amplitude problem: Graphical Method π π3 π4 π΄1 β² π¦1 β² π₯1 β² π¦1 π₯1 Once πΌ is known, there are several methods to proceed with the design. The simplest way is shown in Figure. β’ Choose location of π·1 β’ Select π β’ Draw two positions of the rocker separated by angle π β’ Draw any line π₯ passing through πΆ1 β’ Draw line y passing through πΆ2 at an angle πΌ β’ The interaction between π₯ and π¦ will be pivot π΄1
- 19. 4 Bar Mechanism π΄1 πΆ1 πΆ2 πΌ π·1 π The rocker-amplitude problem: Graphical Method π π3 π4 π΄1 β² π¦1 β² π₯1 β² π¦1 π₯1 Next compute value the value of π2 and π3. This can be done as: π2 + π3 = π΄1πΆ1 π3 β π2 = π΄1πΆ2 π2 = π΄1πΆ1 β π΄1πΆ2 2 π3 = π΄1πΆ1 + π΄1πΆ2 2 Note that during the design procedure many choices were made such as: starting angle π and slop of line π. There are infinite number of such choices which results in different linkages. But not all are valid.
- 20. 4 Bar Mechanism The rocker-amplitude problem: Graphical Method But, all the solutions are not valid. In particular, the pivots πΆ1 and πΆ2 may not cross the line of center through the fixed pivot π΄1 and π·1. If this happens the desired oscillation of the output link can not be obtained and the linkage must dissembled to reach the two position. π΄1 πΆ1 πΆ2 πΌ π·1 π π π¦ π₯ π΄1 πΆ1 πΆ2 πΌ π·1 π π π¦ π₯
- 21. 4 Bar Mechanism The rocker-amplitude problem: Graphical Method π΄1 πΆ1 πΆ2 πΌ π π π¦ π₯ 2πΌ π΄1moves on a circle as in triangle πΆ1πΆ2π΄1 base and apex angle is fixed
- 22. 4 Bar Mechanism The rocker-amplitude problem: Graphical Method
- 23. 4 Bar Mechanism The rocker-amplitude problem: Graphical Method Transmission Angle β’ When transmission angle: π = Β± π 2 , the force acting on the out link will be maximum. β’ While, π = 0Β°, π = 180Β°, the coupler force will produce zero force. β’ Make such linkage so that the transmission angle remains close to + π 2 β’ Typically, poor transmission angle is correspond to a large value of π 2 β ππππ₯/πππ ππππ₯ β² = cosβ1 π4 2 β π1 + π2 2 + π3 2 2π3π4 ππππ β² = cosβ1 π4 2 β π1 β π2 2 + π3 2 2π3π4 If ππππ₯ β² is negative, then ππππ₯ = π + ππππ₯ β² If ππππ₯ β² is positive, then ππππ₯ = ππππ₯ β² Same with ππππ Different combination of π2 and π3
- 24. 4 Bar Mechanism The rocker-amplitude problem: Graphical Method Unscaling the Solution It was earlier assumed that the value of π4 is known, however it may not be the case always. Although, it is required to know the size of at least one link initially. Through a scaling factor we can determine the size of other links. Assume that the link length are: π 1, π 2, π 3 πππ π 4. π 1 = πΎπ1 π 2 = πΎπ2 π 3 = πΎπ3 π 4 = πΎπ4 Where K is the scaling factor. After the design procedure is completed we know the value of π1, π2, π3 πππ π4. Therefore we need to defined only one of π 1, π 2, π 3 ππ π 4 to find out the value of K. Knowing K will help in calculating other links.
- 25. 4 Bar Mechanism The rocker-amplitude problem: Graphical Method Range for πΌ ππ π·1 π π·1 π 2πΌ = 0 β 2πΌ = 0 β π΅1 π΅2 2πΌ πππ ππ‘ππ£π πΌπππ₯ = π 2 + π 2 Negative πΌπππ₯ = β π 2 + π 2
- 26. 4 Bar Mechanism The rocker-amplitude problem: Graphical Method: Example A crank-rocker mechanism with a time ratio 2 1 3 and a rocker oscillation angle of 72Β° is to be designed. The oscillation is to be symmetric about a vertical line through π΅β. Draw the mechanism in any position. If the length of the base link is 2 in, give the length of the other three links. Also show the transmission angles in the position in which the linkage is drawn.
- 27. Thank You