airwalkbooks.comairwalkbooks.com/images/pdf/pdf_113_1.pdf · 2019-02-23 · Dr. S. Ramachandran, M.E., Ph.D., Professor - Mech Sathyabama Institute of Science and Technology Chennai

Dr. S. Ramachandran, M.E., Ph.D.,

Professor - Mech

Sathyabama Institute of Science and TechnologyChennai - 119

(Near All India Radio)

80, Karneeshwarar Koil Street,

Mylapore, Chennai – 600 004.

Ph.: 2466 1909, 94440 81904Email: [email protected],

[email protected]

www.airwalkbooks.com, www.srbooks.org

As Per R-2017 Syllabus of Anna University

FOR IV SEMESTER B.E. MECHANICAL ENGINEERING STUDENTSFOR IV SEMESTER B.E. MECHANICAL ENGINEERING STUDENTS

R - 2017

KINEMATICS

OF MACHINERY(FAST READING)

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First Edition: 2-2-2019

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KINEMATICS OF MACHINERY

Unit I: Basics of Mechanisms

Classification of mechanisms - Basic kinematic concepts and definitions- Degree of freedom, Mobility - Kutzbach criterion, Grubler’s criterion -Grashof’s Law - Kinematic inversions of four-bar chain and slider crankchains - Limit positions - Mechanical advantage - Transmission Angle -Description of some common mechanisms - Quick return mechanisms,Straight line generators, Universal Joint - rocker mechanisms.

Unit II: Kinematics of Linkage Mechanisms

Displacement, velocity and acceleration analysis of simple mechanisms– Graphical method – Velocity and acceleration polygons – Velocity analysisusing instantaneous centres – kinematic analysis of simple mechanisms –Coincident points – Coriolis component of Acceleration – Introduction tolinkage synthesis problem.

Unit III: Kinematics of Cam Mechanisms

Classification of cams and followers - Terminology and definitions -Displacement diagrams - Uniform velocity, parabolic, simple harmonic andcycloidal factions - Derivatives of follower motions - Layout of plate andcam profiles - Specified contour cams - Circular arc and tangent cams -pressure angle and undercutting - sizing of cams.

Unit IV: Gears and Gear Trains

Law of toothed gearing - Involutes and cycloidal tooth profiles - SpurGear terminology and definitions - Gear tooth action - contact ratio -Interference and undercutting. Helical, Bevel, Worm, Rack and Pinion gears[Basics only]. Gear trains - Speed ratio, train value - Parallel axis gear trains- Epicyclic Gear Trains.

Unit V: Friction in Machine Elements

Surface contacts - Sliding and Rolling friction - Friction drives -Friction in screw threads - Bearings and lubrication - Friction clutches - Beltand rope drives - Friction in brakes - Band and Block brakes.AIRW

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Table of ContentsUnit I: Basics of Mechanisms

1.1 Basic Kinematic Concepts and Definitions . . . . . . . . . . . . . . . . . . . . 1.1

1.2 Kinematic Link, Pair and Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2

1.3 Kinematic Link (or) Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2

1.4 Kinematic Pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3

1.4.1 Types of kinematic pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3

1.4.1.1 Kinematic Pairs-based on nature of relative motion. . . . . 1.4

1. Sliding pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4

2. Turning pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5

3. Cylindrical pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5

4. Rolling pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6

5. Spherical pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6

6. Screw Pair (or) Helical Pair . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7

1.4.1.2 Kinematic Pair-Based on nature of contact . . . . . . . . . . . 1.7

1. Lower pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7

2. Higher pair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7

1.5 Kinematic Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8

1.7 Number of Degrees of Freedom (or) Mobility . . . . . . . . . . . . . . . 1.10

1.8 Kutzbach Criterion and Grubler’s Criterion . . . . . . . . . . . . . . . . . . 1.12

1.8.1 Grubler’s Criterion for Plane Mechanism . . . . . . . . . . . . . 1.14

1.9. Classification of Mechanisms - Three Important Kinematic Chains: (with four lower pairs) – Various mechanisms. . . . . . . . . . . . . . . 1.17

1.10 Four Bar Chain: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.17

1.10.1 Grashof’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.18

1.10.2 Kinematic Inversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.18

1.10.3 Kinematic Inversions of Four Bar Chain . . . . . . . . . . . . . 1.21

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1. Beam Engine: (Crank and Lever Mechanism) . . . . . . . . . . . . 1.21

2. Coupling Rod Of A Locomotive (double Crank Mechanism) 1.22

3. Watt’s Indicator Mechanism (Double Lever Mechanism) . . . 1.22

1.11 Single Slider Crank Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.23

1.11.1 Kinematic Inversions of Single Slider Crank Chain . . . . 1.24

1. Pendulum Pump (or) Bull Engine . . . . . . . . . . . . . . . . . . . . . . 1.24

2. Oscillating Cylinder Engine . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.25

3. Rotary Internal Combustion Engine (or) Gnome Engine . . . 1.26

4. Crank And Slotted Lever Quick Return Motion Mechanism . 1.26

5. Whitworth Quick Return Motion Mechanism . . . . . . . . . . . . . 1.27

1.12 Double Slider Crank Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.28

1.12.1 Kinematic Inversions of Double Slider Crank Chain . . . 1.28

1. Elliptical Trammel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.29

2. Scotch Yoke Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.30

3. Oldham’s Coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.30

1.14 Transmission Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.32

1.15 Mechanical Advantage (MA) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.33

1.15 Description of Common Mechanism . . . . . . . . . . . . . . . . . . . . . . . 1.35

1.15.1 Offset Slider crank mechanism as a quick return Mechanism.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.35

1.17 Straight Line Motion Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . 1.37

1.18 Peaucellier Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.39

1.25 Pantograph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.40

1.26 Steering Mechanisms Condition for Correct Steering . . . . . . . . . 1.43

1.28 Ackermann Steering Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . 1.45

1.29 Hooke’s Joint (or) Universal Joint (Coupling) . . . . . . . . . . . . . . . 1.48

1.30 Hooke’s Joint (or) Universal Coupling - Velocity Ratio . . . . . . 1.49

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1.32 Toggle Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.52

Unit II: Kinematics of Linkage Mechanisms

2.2 Linear Displacement, Velocity & Acceleration Analysis . . . . . . . . 2.1

2.4 Key for Construction of the Velocity Polygons . . . . . . . . . . . . . . . . 2.2

2.5 Solved Problems on Relative Acceleration - Graphical Method . . 2.6

2.6 Analytical Method for Relative Velocity and Acceleration . . . . . 2.41

2.7 Instantaneous Centre Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.45

2.7.1 Kennedy Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.46

2.7.2 Locating Instantaneous Centers . . . . . . . . . . . . . . . . . . . . . . 2.46

2.8 Coriolis Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.57

2.9 Introduction to Linkage Synthesis Problem . . . . . . . . . . . . . . . . . . 2.71

2.10 Tasks of Kinematic Synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.71

2.11 Type Synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.72

2.12 Number Synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.72

2.13 Dimensional Synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.74

2.17 Coupler Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.76

2.19 Graphical Synthesis - Linkage Synthesis Problem . . . . . . . . . . . . 2.77

Unit III: Kinematics of Cam Mechanisms

3.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1

3.2 CLassification of Cam and Followers . . . . . . . . . . . . . . . . . . . . . . . . . 3.1

3.3 Terms Used in Radial Cams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4

3.4 Motion of the Follower . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5

3.5 Displacement Diagram, Velocity and Acceleration Analysis when theFollower Moves with Uniform Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6

3.6 Displacement Diagram, Velocity and Acceleration Analysis when theFollower Moves with Shm (Simple Harmonic Motions). . . . . . . . . . . . . 3.7

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3.7 Displacement Diagram, Velocity and Acceleration Analysis when theFollower Moves with Uniform Acceleration and Retardation (or) ParabolicMotion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10

3.8 Displacement Diagram, Velocity And Acceleration Analysis when theFollower Moves with Cycloidal Motion . . . . . . . . . . . . . . . . . . . . . . . . . 3.13

3.11 Cam Profile With Knife Edge Follower . . . . . . . . . . . . . . . . . . . . . 3.16

3.12 Cam Profile with Roller Follower . . . . . . . . . . . . . . . . . . . . . . . . . . 3.30

3.13 Cam Profile With Flat Faced Follower . . . . . . . . . . . . . . . . . . . . . 3.54

3.14 Cam Profile With Oscillating Follower . . . . . . . . . . . . . . . . . . . . . 3.58

3.15 Cam Profile With Cycloidal Motion . . . . . . . . . . . . . . . . . . . . . . . . 3.60

3.16 Specified Contour Cams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.65

3.17 Tangent Cam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.65

3.18 Circular Arc Cam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.72

3.20 Sizing of Cams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.80

3.21 Pressure Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.80

3.22 Undercutting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.81

Unit IV: Gears and Gear Trains

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1

4.4 Terminology of Spur Gears and Definitions . . . . . . . . . . . . . . . . . . 4.2

4.5 Gear Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4

4.6 Law of Gearing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4

4.7 Involute Spur Gears Involutometry . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8

4.7.1 Velocity of sliding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8

4.7.2 Length of path of contact . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8

4.7.3 Length of arc of contact . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.10

4.7.4 Contact ratio (or) Number of pairs of teeth in contact. . 4.10

4.8 Important Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.11


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4.9 Interference in Involute Gears . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.12

4.10 Undercutting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.14

Circular pitch pc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.16

4.11 Helical Gears . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.52

4.12 Bevel Gears . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.53

4.13 Worm and Worm Wheel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.54

4.14 Rack and Pinion Gears . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.56

4.15 Gear Trains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.59

4.16 Parallel Axes Simple Gear Train . . . . . . . . . . . . . . . . . . . . . . . . . . 4.59

4.16.1 Velocity ratio (or) Speed ratio : . . . . . . . . . . . . . . . . . . . . 4.60

4.17 Parallel Axes Compound Gear Train . . . . . . . . . . . . . . . . . . . . . . . 4.62

4.18 Epicyclic Gear Train (or) Planetary Gear Train . . . . . . . . . . . . . 4.65

4.19 Solution of Planetary Gear Train Problems: . . . . . . . . . . . . . . . . . 4.66

4.16 Differential Gear Using Epicyclic Gear Train . . . . . . . . . . . . . . . 4.94

Unit V: Friction in Machine Elements

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1

5.2 Types of Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1

5.3 Sliding Friction - Limiting Friction . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2

5.4 Coloumb’s Law of Dry Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3

5.5 Angle of Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4

5.6 Rolling Friction (or) Rolling Resistance . . . . . . . . . . . . . . . . . . . . . . 5.4

5.7 Inclined Plane and Angle of Repose . . . . . . . . . . . . . . . . . . . . . . . . . 5.7

5.8 To Solve the Problem Related to Inclined Plane . . . . . . . . . 5.10

5.9 Friction in Screw Jack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.14

5.10 Friction in V Threads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.28

5.11 Bearings and Lubrications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.30

5.12 Flat Pivot Bearing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.34

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5.13 Conical Pivot Bearing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.35

5.14 Flat Collar Bearing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.35

5.15 Friction Clutches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.39

5.16 Three Important Types of Clutches . . . . . . . . . . . . . . . . . . . . . . . . 5.39

5.17 Single Plate Clutch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.40

5.18 Multiple Disc Clutch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.43

5.19 Cone Clutch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.51

5.20 Centrifugal Clutch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.58

5.21 Belt Drives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.65

5.23 Flat Belt: Types of Belt Drives . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.66

5.26 Compound Belt Drive . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.67

5.29 Velocity Ratio of the Belt Drive . . . . . . . . . . . . . . . . . . . . . . . . . . 5.68

5.30 Slip and Creep of Belt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.70

5.31 Centrifugal Tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.76

5.32 Maximum Tension in the Belt (t) . . . . . . . . . . . . . . . . . . . . . . . . . 5.78

5.33 Condition For Maximum Power Transmission . . . . . . . . . . . . . . . 5.79

5.34 Initial Tension in the Belt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.81

5.35 V Belt Drive . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.94

5.36 Ratio of Driving Tension for V Belts . . . . . . . . . . . . . . . . . . . . . 5.95

5.37 Rope Drive . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.104

5.38 FRiction in Brakes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.109

5.39 Types of Brakes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.109

5.40 Simple Band Brake . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.110

5.41 Differential Band Brake . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.114

5.42 Band and Block Brake . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.118


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UNIT I

BASICS OF MECHANISMS

Classification of mechanisms - Basic kinematic concepts and definitions- Degree of freedom, Mobility - Kutzbach criterion, Grubler’s criterion -Grashof’s Law - Kinematic inversions of four-bar chain and slider crankchains - Limit positions - Mechanical advantage - Transmission Angle -Description of some common mechanisms - Quick return mechanisms, Straightline generators, Universal Joint - rocker mechanisms.

1.1 BASIC KINEMATIC CONCEPTS AND DEFINITIONS

Kinematics of Machinery (or) simply kinematics is a branch ofengineering science which deals with the study of relative motion betweenvarious parts of machine without considering forces.

Note:

Dynamics of machines deal with the relative motion between the partsby considering forces.

1.1.1 MachineA machine is an apparatus for applying mechanical power consisting

of a number of interrelated parts, each having a definite function.A machine consists of so many links.

1

2

3

L ink -1

Link - 2 Link - 3

Fig. 1.1 Structure

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A Machine is an assemblage of rigid bodies that transmits and/ortransforms forces, motion and energy in a predetermined manner, to do work.

A structure is an assemblage of number of resistant bodies havingno relative motion between them and structures are meant for taking up loads.

In structure, there will not be any relative motion between the links.Forces are transmitted through links but there are no motion.

Difference between Machine and Structure

Sl.No. Machine Structure

1. There is relative motion betweenthe parts/ members of machine.

There is no relative motionbetween the parts/ members ofstructure.

2. Machine converts available energyinto useful work.

Structure does not convertavailable energy into usefulwork.

3. Members of machine are meantto transmit motion and forces.

Members of structure are meantto-take up loads only.

4. Machine modifies and transmitsthe mechanical work.

Structure modifies and transmitsforces only

5. Examples are Car, Lathe, Scooter,Shaping machines, Press etc.

Examples are Bridges, Frames,trusses, buildings etc.,

1.2 KINEMATIC LINK, PAIR AND CHAIN

Three important terms to be considered are

1. Kinematic link (or) Element

2. Kinematic pair

3. Kinematic chain

1.3 KINEMATIC LINK (OR) ELEMENT

It is a resistant body (or) assembly of resistant bodies of a machineconnecting other parts of the machine with relative motion between them.

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1.4 KINEMATIC PAIRA joint of two links (elements) that permits relative motion is called

a pair. Consider the slider crank mechanism. Refer Fig 1.2.

A is a turning pair joining fixed link 1 (crankshaft bearing) and link2 (crank shaft).

B is a turning pair joining link 2 (crankshaft) and link 3 (connecting rod)

C is a turning pair joining link 3 (connecting rod) and link 4 (piston)

D is a sliding pair joining link 4 (piston) and fixed link 1 (frame andcylinder)

1.4.1 Types of kinematic pairs

Classify kinematic pairs based on degrees of freedom(AU. May/June 2016)

The kinematic pairs are classified based on the 1. Nature of relative motion between the links.

2. Nature of contact between the links.

3. Nature of the mechanical arrangement for complete (or) successfulconstraint between the elements.

Based on the nature of relative motion between the links, thekinematic pairs are classified as follows.

1. Sliding pair

2. Turning pair

L ink3L ink2

L ink4

DL ink1

Fram e &cylinder(fixed)

Connecting rod

Fly wheel

C rank sha ft

P iston

Fig.1.2

A

B

C

www.srbooks.org Basics of Mechanisms 1.3

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3. Cylindrical pair

4. Rolling pair

5. Spherical pair

6. Helical pair (or) Screw pair

Based on nature of contact between the links, the kinematic pairs areclassified as follows.

1. Lower pairs

2. Higher pairs.

Based on the nature of mechanical constraint, the kinematic pairsare classified as follows.

1. Closed pair

2. Unclosed pair.

1.4.1.1 Kinematic Pairs-based on nature of relative motion1. Sliding pair

A rectangular bar A (Link A)sliding in a rectangular hole B (linkB) is considered as sliding pair. Thisbar can move linearly but will notrotate. So it has only one degree offreedom. Since it has sliding motiononly, it is considered as completelyconstrained motion. Fig.1.3 Sliding Pair

A

B

Fig.1.4

Link3 L ink2

L ink4

DLink1

Fram e &cylinde r

C onn ecting rod

Fly w heel

C rank sha ft

P is ton

Sliding Pair

A

BC


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In a slider crank mechanism, the piston and cylinder form a sliding pair,since there is a sliding motion of piston surface relative to cylinder surface. Herethe piston will not rotate in the cylinder since it is connected with connectingrod by gudgeon pin. So it is called successfully constrained motion.

2. Turning pairIt is also called Revolute pair (or) Hinged pair. If a link has only

rotary motion relative to another link, then these links form a turning pair.It allows only rotary motion. So it has single degree of freedom. i.e., this

rotational motion can be expressed in terms of only.

In slider crank mechanism, A,B and C are the turning pairs. Thefollowing turning pair shows a shaft A with two collars and a bearing B inwhich it rotates.

Since it has only rotary motion and no linear motion, it is consideredas completely constrained motion.

3. Cylindrical pairIf the collar is removed in a turning pair, then the shaft has two

motions-translation as well as rotation. This pair is called cylindrical pair. Ithas two degrees of freedom. These motions have no relationship with eachother. Since the shaft can move linearly (or) rotate, this motion is known asincompletely constrained motion.

A

B

H ere A & B fo rm a tu rn ing pair

Turn ing pair

Fig. 1.5 Turn ing pair

N o line arm otion

O nly tu rn ing


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4. Rolling pair

A wheel and surface on which it rolls form a rolling pair at the lineof contact. Consider the belt and pulley. The connection between the beltsurface and pulley surface constitutes a rolling pair.

5. Spherical pair

A ball and a socket joint form a spherical pair. Here A is a ball elementand B is a socket element.

This pair has three degrees of freedom.

Fig. 1.7

Rolling whee l

Floor

Pu lley PulleyBelt

Rolling P air

A

B

Cylindrical pairFig. 1.6

Rotary m otionA lso linea rm otion

x

A

B

z

x

yL ink A

L ink B

Ba ll and Socket jo in tFig. 1.8

Spherical P air Rotary motions abou t and axes

x , y z


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The coordinates , and are needed to describe the relative motionof link A with respect to B.

6. Screw Pair (or) Helical Pair

If a screw is rotating andmoving inside a nut, then it is knownas screw pair. Here both rotationaland linear motions of A relative to Boccur. Even then, it is considered ascompletely constrained motion.Because, a specified amount ofrotation of A relative to B makes aproportional amount of axial motion(linear motion) of A relative to B.

1.4.1.2 Kinematic Pair-Based on nature of contact

1. Lower pair

A pair having surface contact or area contact between the twoelements while in motion, is called a lower pair. The relative motion inlower pair is only turning or sliding. In slider-crank mechanisms (Fig. 1.4)a pairs A, B, C and D are lower pairs. All turning pairs, sliding pairs, screwpairs, spherical apirs, cylindrical pairs and flat pairs are known as lower pairs.

2. Higher pair

If a pair has a point (or) line contact between the two elements whilein motion, it is called a higher pair. Consider Fig 1.10 A. In this slidercrank mechanism, the piston is replaced by a sphere. The cylinder and spherehave line contact. The motion between the sphere and the cylinder is slidingas well as turning. So it is a complicated motion.

The cam and followers (Fig 1.10 B), tooth gears, pawl and ratchet,ball bearings and roller bearings are the examples of higher pair.

A

B

Screw Pair or Helical PairFig. 1.9

x


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1.5 KINEMATIC CHAINIf the last link is joined to first link to transmit definite motion,

then it is known as kinematic chain. It is a combination of kinematicpairs and the relative motion between the links is completely (or)successfully constrained.

If a number of links are connected in space so that relative motion ofone link with respect to another link follows a law, then the chain is calledkinematic chain.

Note: Successfully constrained motion (Fig. 1.11)

Consider a foot step bearing. In this case, the shaft may rotate inbearing or it can move upwards. If the load is applied on the shaft to preventupward movement, then the shaft will only rotate. Now this motion isconsidered as successfully constrained motion.

The relation between number of pairs (p) and number of links l isgiven below.

l 2p 4 ... (a)

A

B

(B)

Sliding &

turningPair Cylinder

Link3Sphere

Crankshaft

Connec ting rod Flyw he el

(A)

Fig. 1.10F

ollo

wer

Cam


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The relation between number of links l and number of joints (j) isalso given below.

j 32

l 2...(b)

The above equations (a) and (b) are used only for lower pair. To makethese formulae to valid for higher pair, we have to consider each higher pairas equivalent to two lower pairs with an additional link.

Determine, whether the following are kinematic chain (or) not?

1. Arrangement of three links

Here no.of pair, p = 3

No.of links l = 3Apply equation,

l 2p 4

3 2 3 4

3 2

Since LHS RHS, it is not a kinematic chain. So no relative motionis possible. It is known as structure. (or locked chain).

2. Arrangement of four bar chain

Here

No.of links l 4; No.of pair p 4

Load

Shaft

Foot step bearing

Fig. 1.11

link 1

link 2

link

3

Fig. 1.12


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Apply the equation (a)

l 2p 4

4 2 4 4

4 4

Since LHS = RHS, it satisfies theequation (a). Also, Apply the equation (b)

j 32

l 2

4 32

4 2

4 4

Since LHS = RHS, it satisfies the equation (b).

Since it satisfies the both equations (a) and (b), it is kinematic chain.

1.7 NUMBER OF DEGREES OF FREEDOM (OR) MOBILITY

Number of degrees of freedom is also known as movability (or) mobility.

If a body is in space, it has 6 degrees of freedom.

1. The body may move in x (or) -x directions.

2. It may move in y (or) -y directions.

3. It may move in z (or) -z directions.

4. Also the body may rotate clockwise (or) anticlockwise about x axis.

5. It may rotate clockwise (or) anticlockwise about y axis.

6. It may rotate clockwise (or) anticlockwise about z axis.

So totally it has 6 degrees of freedom.

If the body is resting on a floor, then it may not move in -y direction.And also, it may not rotate about x axis and z axis. In this case, the bodyhas only 3 degrees of freedom. Similarly, we can arrest the motions (linearmotion and rotary motion) and reduce the number of degrees of freedom.

Fig. 1.13link 1

link

2

link 3

lin k 41.10 Kinematics of Machinery - (FR) - www.airwalkbooks.com

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If four bar chain is considered, it has only one degree of freedom. i.e.,

The link 4 can only rotate. So only is sufficient to define the relativeposition of all other links.

But in case of five bar chain, 1 and 2 are required to define the

relative position of all other links. So the number of degrees of freedom for5 bar chain is 2.

C

BA

D

Lin

k 2

L ink

4

Link 3

L ink 1

(a) Four bar chain

A B

E

D

Link 4

L ink 1

L ink 3

L ink 5 L ink 2

(b) Five bar chain

C

Fig. 1.15

+x- x

+y -z

+z -y

c.c.w

c.c.w

c.c.w

c.w

c.w

c.w

Fig. 1.14

c.w = c lockwise

c.c.w = counter clockw ise


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1.8 KUTZBACH CRITERION AND GRUBLER’S CRITERION

Kutzbach criterion is used to determine the number of degrees of freedom(or) movability (n) of a mechanism. This number of degrees of freedom can bedetermined directly from the number of links, number of joints and type of pair.

Number of degrees of freedom n 3 l 1 2j h – Kutzbach Criterion

where

l Number of links

j Number of binary joints or lower pairs.

h Number of higher pairs

If there is no higher pair,then h = 0

For three bar mechanism,

l 3; j 3; h 0

So, n 3 l 1 2j h

3 3 1 2 3 0

6 6 0

Number of degrees of freedom

for three bar mechanism n 0

For four bar mechanism,

l 4, j 4; h 0

Then

n 3 l 1 2j h

3 4 1 2 4 0

9 8 1

Number of degrees

of freedom for 4 bar mechanism

n 1

A B

C

1

23

Three bar m echanismFig. 1.16

A B

D

3

4

C

2

1

Four bar mechanism

Fig. 1.17


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For five bar mechanism,

l 5; j 5; h 0

The n 3 l 1 2j h

3 5 1 2 5 0

12 10 2

Number of degreesof freedom

n 2

Consider this three link mechanism

Here number of links l = 3 ; binary joints j = 2

higher pair h = 1

We know, Kutzbach cr iter ion,

n 3 l 1 2j h

3 3 1 2 2 1

6 4 1 1

Consider the four link mechanism

Here, no. of links 4

Binary joints 3

higher pair h 1

n 3 l 1 2j h

3 4 1 2 3 1

9 6 1 2

A B1

2

C

D

E

5

4 3

Five bar m echanism

Fig. 1.18

1

2

3

Three link mechanism

M echanism with a higher pair

Fig. 1.19

32

1

wheel

M echanism w ith a higher pair

Four link m echanism

4

Fig. 1.20


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1.8.1 Grubler’s Criterion for Plane Mechanism

The Grubler’s criterion applies to mechanisms with only single degreeof freedom joints, where the overall movability of the mechanism is equal toone.

By substituting n 1 and h 0 in Kutzbach equation,

n 3 l 1 2j h

we get 3l 2j 4 0

This equation is known as the Grubler’s criterion for plane mechanismwith constrained motion.

State the inconsistencies of Grubler’s criterion. (AU. May/June 2016)

Grubler’s criterion gives incorrect results when

The mechanism has a lower pair which could be replaced by ahigher pair, without influencing output motion.

The mechanism has a kinematically redundant pair and

There is a link with redundant degree of freedom.

A plane mechanism with a movability of one and only single degreeof freedom joints cannot have odd number of joints. The example of thismechanism is

Four-bar mechanism

Slider-crank mechanism.

where link l 4 and joint j 4

Explain turning pair, sliding pair, screw pair, cylidrical pair, spherical pairand flat pair with respect to degress of freedom. (FAQ)

Turning pair

A

B

Here A & B form a turning pair

Turn ing pair

Rotation

x

No Linear M otion

Fig. 1.21(a )


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The turning pair designated by R can only rotate about x axis, but

there is no linear motion. Hence DOF 1. (Refer Fig. 1.21(a))

Cylindrical pair

Cylindrical pair is denoted by C. It can rotate about x axis well as

move in x direction. Hence DOF 2. (Refer Fig. 1.21(b))

Prismatic pair

It is denoted by P. It has only sliding motion. Hence DOF 1 onlyslinding motion. (Refer Fig. 1.21(c))

Spherical pair

Spherical pair is denoted by G. The lever can rotate about x, y and z

axes. Hence DOF 3. (Refer Fig. 1.21(d))

A

B

Fig. 1.21(b ) Cylindrical Pair

R otation

L inea r M otion x

S

Fig. 1.21(c) P rism atic (Slider) Pair P D.O.F. = 1


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Screw pair

Screw pair is denored by S. It has rotary motion as well as linearmotion. But both rotary motion and linear motion are interrelated (through

helix angle ), so it has only one degree of freedom. Hence DOF 1. (ReferFig. 1.21(e))

Flat pair

The flat object can move in x andy direction. But it won’t move in zdirection, but it will rotate above z axis.

Hence DOF 3. (Refer Fig. 1.21(f))

A

B

z

x

yL ink A

L ink B

Ba ll and Socket jo in t

Fig. 1.21(d) Spherical Pair

D O F = 3

A

B

Fig. 1.21(e) Screw Pair or Helical Pair

y L inear M otion

Rotation

z

y x

12

Fig. 1.21(f) Flat (P laner) Pair FD.O.F. = 3


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1.9. CLASSIFICATION OF MECHANISMS - THREE IMPORTANTKINEMATIC CHAINS:

(with four lower pairs) – Various mechanisms.

1. Four bar chain (or) quadric cyclic chain (or) mechanism (4 turningpairs)

2. Single slider crank chain (or) mechanism (Three turning pairs andone sliding pair)

3. Double slider crank chain (or) mechanism (Two turning pairs andtwo sliding pairs)

1.10 FOUR BAR CHAIN:Normally, rotating link is known as crank. Oscillating link is known

as Rocker (or) Lever. Connecting link is known as Coupler (or) Connectingrod. And coupler will not be connected to the fixed frame.

The simplest and the basic kinematic chain is a four bar chain. Itconsists of four links

link (1): Fixed link (frame)

link (2): Rocker (lever (or) follower)

link (3): Connecting rod (coupler)

link (4): Crank (driver)

1

2

3

4

A B

C

D

fixed link(fram e)

R ocker(lever)

(fo llow er)

c rank(Drive r)

C onn ecting rod(o r) coupler

Fig. 1.22 Four bar chain


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1.10.1 GRASHOF’S LAW

For a four bar mechanism,

Grashof’s law states that the sum of the shortest and longest linklengths should not be greater than the sum of the remaining two link lengths,if there is to be continuous relative motion between the two links.

In this four bar mechanism,

link 4 is shortest length.

link 3 is longest length

According to Grashof’s law,

Length link 4 link 3 length link 1 link 2

In four bar chain, when the crank (link 4 ) is the driver, the mechanismis transforming rotary motion into oscillating motion.

1.10.2 KINEMATIC INVERSIONS

What do you understand by inversion of a kinematic chain? Describe themechanism obtained by inversion of the four bar chain.

(AU. Apr/May 2015)

If link 1 is fixed and other links are in relative motion, then it isknown as one mechanism.

If link 2 is fixed and other links are in relative motion, then it isknown as another mechanism.

BA

D

Link

2

L ink

4

L ink 3

L in k 1

C

Fig. 1.23


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If different links are fixed to get different mechanisms, then thedifferent mechanisms are known as kinematic inversions.

By fixing, in turn, different links in a kinematic chain, kinematicinversion is obtained.

Explanation

1. When shortest link is fixed, then the adjacent links (connected toshortest link) a and c rotate full revolution continuously. The mechanismthus obtained is known as crank-crank mechanism (or) double-crankmechanism (or) rotary-rotary mechanism. (Refer Fig. 1.24)

s shortest length

l longest length

[Fig. (a) and (b)] A mechanism in which the sum of the lengths ofthe longest and the shortest links is less than the sum of the lengths ofthe other two links, is known as a class-I, four-bar mechanism. Heres l lengths of remaining 2.

C ’’ ’

bB

a

C

c

B ’

C ’A Dd

B ’’ ’B ’’

C ’’

Cra

nk

Cra

nk

C ouple r

Fram e

Shortest L ink

in F ixed

Class I mechanism

(or) Drag link mechanism

Fig. 1.24 Crank - crank m echanism (or) Double crank m echanism


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2. If any one of theadjacent links of shortestlink i.e., a or c is fixed,d can have a fullrevolution (crank) andthe link opposite to itoscillates (rocks). InFig. 1.25(a), a is fixed,d is the crank and b isrocker to oscillatewhereas in Fig. 1.25(b),c is fixed, d is the crankand b (the rocker)oscillates. The mechanism is known as crank-rocker or crank-levermechanism or rotary-oscillating converter. If any one of adjacent links

to the shortest link a or c is fixed, then the mechanism is calledcrank-rocker mechanism (or) crank lever mechanism.

3. If the link opposite to the shortestlink, i.e., link b is fixed and theshortest link d is made a coupler,then the other two links a and cwould oscillate [Fig. 1.25(c)]. Themechanism is known as arocker-rocker or double-rocker ordouble-lever mechanism oroscillating-oscillating converter.

If the sum of the longest near (Fig.(c)) and the shortest links is more than thesum of the lengths of the other two links,then mechanism is known as a class-II,four-bar mechanism. In such a mechanism, fixing of any of the links alwaysresults in a rocker-rocker mechanism.

By using the above concepts, Grashof’s law states that a four-barmechanism has at least one revolving link if the sum of the lengths of the largestand the shortest links is less than the sum of lengths of the other two links.

a c

d

b

RockerRocker

F ixed Fram e

Coup le r

(c)

Class II mechanism

Fig. 1.25 Rocker-rocker mechanism

(or) Double lever m echanism

(or) Double-rocker m echanism

ac

b

dC rank

C oup le rF ixed Fram e

O sc illa ting

Rocker

aC rank

d

Rocker b

c

Oscilla ting

Co

uple

r

Fix

e d

Fra

me

(a) (b)

Fig. 1.25 Crank-rocker mechanism (or)Crank lever m echanism


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1. Further, if the shortest links is fixed, it will become double-crankmechanism where the links adjacent to the fixed link will havecomplete revolutions.

2. If the link adjacent to the shortest links is fixed, the mechanism willbecome crank-rocker mechanism where the shortest link will revolveand the other link adjacent to the fixed link will oscillate.

3. If the link opposite to the shortest link is fixed, it will becomedouble-rocker mechanism where links adjacent to the fixed link willoscillate.

1.10.3 KINEMATIC INVERSIONS OF FOUR BAR CHAIN

Sketch and explain any three kinematic inversion of four-bar chain.(AU. May/June 2014)

Important inversions of four bar chain are

1. Beam engine (crank and lever mechanism)

2. Coupling rod of a locomotive (Double crank mechanism)

3. Watt’s indicator mechanism (Double lever mechanism)

1. BEAM ENGINE: (CRANK AND LEVER MECHANISM)

Pistonrod

O scilla tinglever (link 4)

E

DC

L ink 3cylinde r

A

BFram e(L ink 1)

crank(L ink 2)

(link1 )

(connectingrod)

Fig. 1.26 Beam engine


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It consists of four links. This mechanism converts the rotary motioninto reciprocating motion. The crank (link 2) rotates about A. This motion istransmitted to oscillating lever (link 4) through connecting rod (link 3). Theoscillating lever oscillates about D and it makes the piston to reciprocateinside the cylinder.

2. COUPLING ROD OF A LOCOMOTIVE (DOUBLE CRANK MECHANISM)

This chain consists of four links as shown in Fig. 1.27

This mechanism is used to transmit rotary motion from one wheel tothe other wheel. In train, the power from leading wheel is transmitted totrailing wheel. Here AD and BC are two cranks having equal length. Theyare connected to two different wheels. The link CD (link 3) is the couplingrod coupling both wheels. The fixed link (1) maintains the constant centre tocentre distance between the two wheels.

3. WATT’S INDICATOR MECHANISM (DOUBLE LEVER MECHANISM)

It is used to indicate the intensity of the steam (or) gas pressure insidethe cylinder. It consists of four links.

1. Fixed link (link 1) at A

2. Link AC (link 2)

3. Link CE (link 3)

4. Link BFD (The link BF and FD have no relative motion with eachother. So both are considered as one link).

L ink 4 L ink 2

L ink 1

W heels

L ink 3

A

DC

B

Fig. 1.27 Coupling rod of a locom otive


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The displacement of the link BFD is directly proportional to the gaspressure inside the cylinder. This displacement is shown by the tracing pointE at the end of link CE.

The continuous (solid) lines show the initial position of the mechanismand the dotted lines show the extreme position when the pressure of gas issufficiently more.

1.11 SINGLE SLIDER CRANK CHAIN

This mechanism is used to convert the rotary motion into reciprocatingmotion and vice versa. It is a modification of basic four bar chain. It consistsof one sliding pair and three turning pairs.

Link 1 and 2 - Turning pair

link 2 and 3 - Turning pair


link 4 and 1 - Sliding pair

Link 1 : Frame (Fixed link)Link 2 : CrankLink 3 : Connecting rodLink 4 : Cross head (or) slider (or) piston

Link2

L ink4

F

B

B’

C

DF’

A E

E’

L ink1

L ink3

D’

C ’

Indica torcylinder

Indica torp lunger

Fig. 1.28 W att’s indicator m echanism


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1.11.1 KINEMATIC INVERSIONS OF SINGLE SLIDER CRANK CHAIN

What is kinematic inversion? Discuss any three applications of inversionsof single slider crank mechanism.

(AU. Apr/May 2017, 2015, Nov/Dec 2013, 2010)

Sketch and explain four inversions of single slider crank mechanism.(AU. Nov/Dec 2012, May/June 2009)

By fixing, in turn, different links in a kinematic chain, an inversion isobtained.

Important inversions:

1. Pendulum pump (or) Bull engine

2. Oscillating cylinder engine

3. Rotary internal combustion engine (or) Gnome engine.

4. Crank and slotted lever quick return motion mechanism.

5. Whitworth quick return motion mechanism.

1. PENDULUM PUMP (OR) BULL ENGINEHere, the link 4 (cylinder) is fixed, when the crank (link 2) rotates,

the connecting rod oscillates about a pin pivoted to the fixed link 4 at A andthe piston attached to the piston rod reciprocates. This type of mechanism isused in duplex pump to supply feed water to the boilers.

Piston(Link 4)

Guides

Connecting rod(Link 3)

Crank(Link 2)

Fram e (L ink 1)

Fram e (L ink 1)

Fig. 1.29 Single slider crank chain


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2. OSCILLATING CYLINDER ENGINE

This mechanism is used to convert rotary motion into oscillatorymotion. Here the connecting rod (link 3) is fixed. When the crank rotates,the piston attached to piston rod reciprocates and the cylinder oscillates abouta pin pivoted to the fixed link at A.

A

Piston rod(L ink 1 )

C ylind er(L ink 4 )

C onnecting rod(L ink 3 )

Pendulum pum p

C rank(L ink 2 )

C ylind er(L ink 4 )

Fig. 1.30

cy linder

(L ink 4)

P iston rod(L ink 1)

C o nnecting rod(L ink 3)

Cra nk

(L in k 2 )

A

Fig. 1.31 Oscillating cy linder engine


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3. ROTARY INTERNAL COMBUSTION ENGINE (OR) GNOME ENGINE

It consists of seven cylinders and all revolve about fixed centre D. Inthis mechanism, the crank (link 2) is fixed. When the connecting rod (link4) rotates, the piston (link 3) reciprocates inside the cylinders (link 1).

4. CRANK AND SLOTTED LEVER QUICK RETURN MOTION MECHANISM

Sketch and describe the working of two different types of quick returnmechanism. Give example of their application. (AU. Apr/May 2015, 2014)

This mechanism is used in shaping machines and slotting machines.Here, the link (3) (This link (3) corresponds to connecting rod in

reciprocating engine mechanism) is fixed. The crank CB (link 2) rotates withuniform speed. A slider (link 1) is attached to the crank pin B. It slides alongslotted bar AP and so the slotter bar AP oscillates about the pivoted pointA. The link (or) a connecting rod connects the AP with ram which carriesthe tool and reciprocates.

The cutting stroke occurs when the crank rotates from position CB1

to CB2 (i.e., through an angle ). The return stroke occurs when the crank

rotates from position CB2 to CB1 (i.e., through an angle ).

Connecting rod(L ink 4)

Piston(L ink 3)

Fixed crank(L ink 2)

Cylinder(L ink 1) D

Rotary internal com bustion engine Fig. 1.32


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Note that is greater than . i.e., The angle made by cutting stroke

is greater than the angle made by return stroke. Since the crank rotateswith uniform angular speed, the return stroke occurs quickly. That is why, itis called quick return motion mechanism.

5. WHITWORTH QUICK RETURN MOTION MECHANISM

This mechanism is also mostly used in shaping machines and slottingmachines.

Here the link 2 is fixed. (This link 2 corresponds to crank in areciprocating steam engine). The driving crank CA (link 3) rotates withuniform speed. The slider (link 4) is attached to the crank pin A. It slidesalong the slotted bar PA (link 1). So the slotted bar PA oscillates about apivoted point D. The connecting rod PR connects the slotted bar and ramwhich carries the tool and reciprocates.

Connectingrod

Cutting stroke

Return stroke L ine o f s troke

Too lRam

P 1

R 2

P 2

P

Q

R 1

B

B 2B 1

C

Slider (Link 1 )

Fixed(Link 3)

(90o -

2)

S lotted bar(L ink 4)

AFig. 1.33 Crank and slotted lever quick return motion mechanism

R


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When the driving crank CA moves from CA1 to CA2 (clockwise)

through an angle , the tool is moving forward stroke (cutting stroke).When the driving crank CA moves from CA2 to CA1 through an angle

(clockwise), the tool is moving back (return stroke).

Since is greater than and the driving crank CA rotates with uniformangular velocity, the return stroke is quicker than the cutting stroke. Thus itexecutes the quick return motion.

What is Inversion of mechanism? With neat sketch and explain any twoinversions of double slider crank mechanism. (AU. Nov/Dec 2011)

1.12 DOUBLE SLIDER CRANK CHAIN

It consists of two turning pairs and two sliding pairs

1.12.1 KINEMATIC INVERSIONS OF DOUBLE SLIDER CRANK CHAIN

Three important inversions

1. Elliptical trammel

2. Scotch yoke mechanism

3. Oldham’s coupling

connectingrod

cutting stroke

R eturn s troke

Too lR am

R R 2R 1

L ine o f s troke

A 1A 2 P 1 P 2D

C

D rivingcrank

(L ink 3)

F ixed(L ink 2)

S lotted bar(L ink 1)

S lider(L ink 4)

A

P

W hitworth quick return motion m echanismFig. 1.34


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1. ELLIPTICAL TRAMMEL

Elliptical trammel is an instrument used for drawing ellipses. Refer theFig. 1.35

This inversion is obtained by fixing the link 4 (slotted plate). This slottedplate has two straight grooves at right angle to each other. (Just like + sign).

The slider (link 1) and slotted plate (link 4) form one sliding pair.The slider (link 3) and slotted plate (link 4) form another sliding pair.

The connecting bar (link 2) and slider (link 1) form one turning pair;The connecting bar (link 2) and slider (link 3) form another turning pair.When the sliders slide along their respective grooves, any point on ‘connectingbar’ (say point P) traces out an ellipse.

Here AP is the half of major axis of the ellipse and BP is the half ofminor axis of the ellipse.

A

C onnectingBa r

(L ink 2)

S ide r(L ink 1)

S lottedp la te (Link 4 )P

B

F ig. 1.35 Elliptical tram mels

Side r(L ink 3)

F ixe d


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2. SCOTCH YOKE MECHANISM

Scotch yoke mechanism is used to convert rotary motion intoreciprocating motion. The inversion is obtained by fixing any one of thesliders link 1 or link 3.

In this mechanism, link 1 is fixed. In such arrangement, the wholeframe i.e., link 4 will reciprocate. When the crank (link 2) rotates about Bas centre, the whole frame (link 4) reciprocates. The fixed link 1 guides theframe.

3. OLDHAM’S COUPLING

This coupling is used for connecting two parallel shafts whose axesare at a small distance apart. In this mechanism, when one shaft rotates, theother shaft also rotates at the same speed.

The link 1 (flange) is rigidly fastened to the end of the driving shaftby forging.

The link 3 (another flange) is rigidly fastened to the end of the drivenshaft.

The supporting frame (link 2) is fixed.

The link 1 and link 3 have diametrical slots cut in their inner faces.

1

L ink 3

A

B

F ig. 1.36 Scotch Yoke Mechanism

L ink 1

L ink 4L ink 2


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In between to these two flanges, an intermediate piece (link 4) issliding.

It has two tonges-two diametrical projections on both faces - at rightangles to each other.

The link 4 slides on the slots of the two links 1 and 3.

The rotary motion is given to link 1. So the intermediate piecealso rotates with the same speed. Hence the link 3 also rotates withthe same speed, i.e., links 1, 3 and 4 have the same angular velocityat every instant.

The distance between the axes of the shafts is constant and thereforethe centre of the intermediate piece will follow the path of a circle withradius equal to the distance between the axes of the two shafts.

Therefore, maximum sliding speed of each tongue of the intermediatepiece in the slot will be given by the peripheral velocity of the centre of thedisc along its circular path.

Peripheral velocity of the disc Angular velocity of the shaft

distance between the axes of the shafts

V r

Fig. 1.37 O ldham ’s coupling

Flange(L ink 1)

D rivingsha ft

A

C

In te rm ed ia tep iece

(L ink 4)

D Flange(L ink 3)

B D rivensha ftSuppor ting

fram e(L ink 2) In te rm ed ia te

p iece(L ink)

4

Tongue

Tongue

H orizon ta l

Vertica l

L ink 2

x


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1.14 TRANSMISSION ANGLE

Explain mechanical advantage and transmission angle related to four barmechanism.

Sketch and Define Transmission angle for a four bar mechanism

(AU. Apr/May 2017)

It is the angle between the coupler and the follower links at anyposition other than dead centre positions. If this angle is small, thenmechanical advantage approaches zero. (Refer Fig. 1.38 (a))

Transmission angle less than 20 is not acceptable for a high speedmechanism.

Applying cosine law to triangle ABD [Refer Fig. 1.38 (a)]

AB2 AD2 2AB AD cos BD2

l22 l1

2 2l2 l1 cos BD2...(1)

Similarly, for triangle BCD,

l32 l4

2 2l3 l4 cos BD2...(2)

Equating Eqs. (1) and (2), we get

l22 l1

2 2l2 l1 cos l32 l4

2 2l3 l4 cos

cos l1

2 l22 l3

2 l42 2l2 l1 cos

2l3 l4 ...(3)

To find minimum (or) maximum , differentiate eqn (3) with respect

to and equate to zero.

Then we get l1 l2 sin 0[Here l1 and l2 are not zero]

0 or 180

Therefore, will be minimum, when 0 (or) 180


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1.15 MECHANICAL ADVANTAGE (MA)The main objective of any mechanism is to transmit torque (or) force

effectively.

Mechanical Advantage Output torque by driven linkInput torque by driving link

Mechanical advantage of 4 bar mechanism varies directly with

sin and inversely with sin

Mechanical Advantage sin

Mechanical Advantage 1

sin where Angle between coupler and follower link

Angle between coupler and crank (dr iving link)

These angles and change as the mechanism goes on changingits position.

When is very small, sin also will become very small and MAbecomes infinity.

A 4

A 2

A 1O

A 3

C

B 2

B 4

B 3

B 1

B

Coupler

Cra

nk

Fo llower L ink(o r)

O utput D rivenL ink Inpu t

D riving L ink

Fig. 1.39 Transmission Angle is an Index of Mechanical Advantage

A


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The meaning is that very small input torque is sufficient toovercome large resisting torque of driven link.

This position is obtained when OA and AB are in line. i.e., atposition OA1 and at position OA2.

In the above positions, crank positions are at CB1, (for OA1

position) and CB2 for OA2 position.

The above position of linkage is called toggle position and thismechanism is called toggle mechanism ie, for small amount ofinput torque, large output torque can be developed.

Similarly,

If the transmission angle is too small, then sin will becomezero and mechanical advantage also approaches zero.

Meaning is that, if the is too small, then mechanical advantagewill become small. And hence output torque may not be developedby using given input torque and it may cause the mechanism tolock. i.e.,Total mechanism will not move and it will jam.

Therefore, shouold not be less than 45 50.

In the position OA3, ie when crank lies on OC line, the smallest

occurs.

In the position OA4, the highest occurs.

Hence by making transmission angle as maximum as possible,the maximum MA will be possible.

Therefore, transmission angle is an index of mechanicaladvantage.


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1.15 DESCRIPTION OF COMMON MECHANISM

Derive an expression for the ratio of time taken in forward and returnstroke. (AU. Apr/May 2015)

1.15.1 Offset Slider crank mechanism as a quick return Mechanism.

Offset slider mechanismThis mechanism is used to convert the rotary motion into reciprocating

motion and vice versa. It is a modification of basic four bar chain. It consistsof one sliding pair and three turning pairs.

Link 1 and 2 - Turning pair



link 4 and 1 - Sliding pair

Link 1 : Frame (Fixed link)

Link 2 : Crank

Link 3 : Connecting rod

Link 4 : Cross head (or) slider.

In some specific applications the common line of slider and crank centreis offset by ‘e’ distance as shown in Fig 1.40 which is known as offset slidercrank mechanism. The same mechanism can be used as a quick return mechanismbecause of the slider has different velocities during forward and return stroke.

C onn ecting rod(L in k 3)

Fram e(L in k 1)

C rank(L in k 2)

(L in k 4)P iston

e = offset

G uides

Fig. 1.36 Off-set slider crank chain


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Let

OA r Crank radius

AB l Length of the connecting rod

e off set distance

The farthest position of slider is obtained when the crank OAR and

connecting rod ARBR of slider are at a distance r l from crank centre O.

OBR OAR ARBR

r l ...(i)

similarly the closest position of the slider is obtained, when

OBF BFAF OAF

l r ...(ii)

from BFOM and BROM

cos MOBF OMOBF

e

l r

cos MOBR OMOBR

e

r l

Hence Q cos 1

er l

cos 1

el r

e

A R

B F B R

O

O A= r = C rank R adiusA F

120 m m

�

AB= =C onnec ting rod�

M

Fig. 1.41


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Thus

return stroke angle

QR 180

forward stroke angle

QF 180

The ratio of forward to return stroke timing is given as

Time of forward strokeTime of return stroke

angle of forward strokeangle of return stroke

QF

QR

Application:

1. It’s used in automatic packing machines.

2. In assembly line it is used to push parts to work area.

3. It is also used in punching / riveting press.

1.17 STRAIGHT LINE MOTION MECHANISM

The mechanisms designed to produce a straight line motion are knownas straight line motion mechanisms. The straight line motion mechanism isdivided into three categories:

What are the types of straight line motion mechanism(AU. Nov/Dec 2012)

ii(i) Exact straight line motion mechanism

i(ii) Approximate straight line motion mechanism

(iii) Straight line copying mechanism

Explain exact straight line motion mechanism (FAQ)

Exact Straight Line Motion Mechanism

Exact straight line motion mechanism follows a mathematical relationwhich holds true for all positions of input link such that output link followsstraight line.


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O, B, A are three points of a mechanism which are always constrainedto lie along a common straight line as the link OA turn about centre O andthe position of the point A is such that the product OA. OB is constant. Thenthe path of the point A will be straight line perpendicular to diameter OC ofthe circle on which circumference B moves (Refer Fig. 1.42).

Proof: Draw AD perpendicular to OC produced. Joint CB (r toOB) to form triangle OBC. Then the triangles OBC and ODA are similar

right angle triangle OBC ODA.

Therefore OBOC

ODOA

OD OA OBOC

But OC is constant since it is the diameter of the circle. If the product

OA OB is constant, then the distance OD will be constant and the point A

moves along the straight line AD which is perpendicular to OD.

Several mechanisms have been proposed to connect points O, B and

A in such a way as to satisfy the condition OA OBOC

constant (i.e., OD)

The above condition is satisfied in Peaucellier and Hart mechanismwhich are explained here

B

A

C DO

Fig. 1 .42 Straight Line Motion


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1.18 PEAUCELLIER MECHANISM

What are straight line mechanisms? Sketch the Peaucellier straight-linemotion mechanism and prove that the generating point moves in a straightline. (AU. Nov/Dec 2012)

(Or)

Prove that the peaucellier mechanism generates a straight line mechanism.(FAQ)

(Or)

Sketch and describe the Peaucellier mechanism. Show that it can be usedto trace a straight line. (FAQ))

State application of Peaucellier mechanism. (FAQ)

A Peaucellier mechanism consists of eight links as shown inFig.1.43

OQ QC ; OB OA ;

BP PA AC CB

O

Q

C

A

P

B

R N

Fig. 1.43 Peucellier Mechanism.

E


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OQ is fixed link and QC is rotating link. It can be proved that as thelink QC moves around Q, P moves in a straight line perpendicular to QO.

Hence,

OC OP constant. It can be proved as follows.

From the triangle OEA and PEA

OA2 OE2 EA2...(1)

PA2 EP2 EA2...(2)

Subtracting equation (2) from (1)

OA2 PA2 OE2 EP2

OE EP OE EP

OP OC

Since OA and PA are constant length, therefore, the product of

OP OC remains constant. Hence the Point P traces a straight line path,perpendicular to the diameter OR.

1.25 PANTOGRAPH

What is pantograph? (FAQ)

Explain with a neat sketch, pantograph mechanism. (AU. Nov/Dec 2010)

State its applications. What is pantograph? Show that it can produce pathsexactly similar to the ones traced act by a point on link on an enlarged ora reduced scale. (FAQ)

(Or)

Sketch a pantograph, explain its working and show that it can be used toreproduce to an enlarged scale of a given figure. (FAQ)

(Or)

Show that the pantograph can produce paths exactly similar to the onestraced out by a point on a link (FAQ)

(Or)

Sketch a pantograph and explain how the mechanism would be used toenlarge a drawing. (FAQ)


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(Or)

What is the purpose of pantograph? Explain its working with sketch.Provide the mathematical proof for working of pantograph. (FAQ)

A pantograph is an instrument having four-bar links with lower pairsused to re-produce a path exactly similar to one traced out by a point onthe linkage. The path may be on an enlarged (or) reduced scale as required.

It is made up of bars connected by turning pairs. The bars BA and

BC are extended to O and E respectively, such that OA/OB AD/BE. Thus,for all relative positions of the bars, the triangles OAD and OBE are similarand the points O, D and E are in one straight line. It may be noted that pointE traces out the same path as described by point D.

From similar triangles OAD and OBE, we find that OD/OE AD/BELet point O be fixed and the points D and E move to some new positions

D and E. Then OD/OE OD/OE (Refer Fig. 1.52)

Now we can know that the straight line DD is parallel to the straight

line E E. Hence, if O is fixed to the frame of a machine by means of aturning pair and D is attached to a point in the machine which has rectilinearmotion relative to the frame, then E will also trace out a straight line path.

O

D �

B �

C �

E �

A

D

A �

E

C

B

Fig. 1.52


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Similarly, if E is constrained to move in a straight line, then D will traceout a straight line parallel to the former.

What are the applications of a pantograph? (FAQ)

A pantograph is mostly used for the reproduction of plane areas andfigures such as maps, plans etc., on enlarged or reduced scales.

It is, sometimes, used as an indicator rig in order to reproduce to asmall scale the displacement of the crosshead and therefore of the piston ofa reciprocating steam engine.

It is also used to guide cutting tools. A modified form of pantographis used to collect power at the top of an electric locomotive.

Problem 1.1: Two points A and B, 40 mm apart are to be connected by apantograph. The motion of A to the motion of B is 13 : 7. Find the distanceof B from the fixed point O of the pantograph such that the point A movesatleast 12.7 mm in either direction of line OBA when it is horizontal. Findalso the main dimensions of the pantograph. (FAQ)

Given

AB 40 mm

OAOB

13/7

To find: OA and OB

In a pantograph,

Q

RP

B AO40 m m


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Since PQRO is parallelogram and from similar triangles BPO andBQA

OBBA

BPBQ

POQA

OBAB

OB

OB OA [. . . AB OA OB]

ABOB

OB OA

OB

1 OAOB

1 137

given

ABOB

207

40OB

207

OB 40 7

20

OB 14 mm

And, OA AB OB

40 14

OA 26 mm

The distance of B from fixed point O of the pantograph is

OB 14 mm

1.26 STEERING MECHANISMS CONDITION FOR CORRECT STEERING

Derive the condition for correct steering. (FAQ)

What is the condition for correct steering? Sketch and show the two maintypes of steering gears and discuss their relative advantage. (FAQ)

(Or)


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What is automobile steering gear? What are its types? Which steering gearis preferred and why? (FAQ)

Steering mechanism in automobile is used to control the direction ofa vehicle’s motion. The perfect steering is achieved when all the four wheelsare rolling perfectly under all conditions of running. While taking turns thecondition of perfect rolling is satisfied if the axes of the front wheel whenproduced meet the rear wheel axis at one point. Fig. (1.53). Then this point(I) is the instantaneous centre of the vehicle.

Perfect steering angle - Condition for correct steering

From the Fig. 1.53 it is clear that the inside wheel is required to turnthrough a greater angle than the outer wheel. Larger the steering angle, smallerthe turning circle; and by referring Fig. 1.53 for correct steering,

cot Y C

b

Yb

Cb

cot Cb

cot cot Cb ...(1)

Whe

el B

ase

b

C

W heel Track

Y

Fig. 1.53. Steering Mechanism

�

Stub ax leS tub ax le

Inside w

hee l ax is

Ou tside wheel a

x is


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where Angle of outside wheel axis

Angle of inside wheel axis

Equation 1 represents the basic condition for the steering mechanismfor perfect rolling of all wheels. It is the condition for correct steering. Ifthis condition is satisfied, then there is no lateral slip of wheel, when vehicleis taking turn.

A mechanism which satisfies the equation of steering

cot cot wheel trackwheel base

at all radii of curvature of the path followed by

the vehicle is called steering mechanism.

There are two types of steering mechanism - Davis steering mechanismand Ackermann steering mechanism. The difference between these two typesis that the Davis mechanism has sliding pairs compared to turning pairs usedin Ackermann steering mechanism.

1.28 ACKERMANN STEERING MECHANISM

Explain Ackermann steering gear mechanism with a neat sketch.

(FAQ)

(or)

For an Ackermann steering gear, derive the expression for the angle ofinclination of the track arms to longitudinal axis of the vehicle.

(FAQ)

In Ackermann steering mechanism, the sliding pairs (in Davis) arereplaced by turning pairs to minimise friction and thereby wear and tear alsominimized.

Ackermann steering mechanism is based upon the four bar chainmechanism with two longer links having equal lengths and two shorter linkshaving equal lengths and it involves turning pairs only.

If the vehicle has to travel in a straight path, the longer links should

be parallel to each other and the shorter links are having an angle ‘’ inclinedwith the longitudinal axis. (Fig. 1.55(a))


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So the turning of the vehicle is done by varying the angle of ‘. If

‘’ increases, then the vehicle turns right (Fig. 1.55(b)) and if decreases thenit turns left.

From the geometry of Fig. 156

cot cot AQ CQ

IQ

A CD

K L

FE

(a ) Fo r S traigh t D rive

B

+

C

K �

A

L �

�

R ear Axis

(b ) Fo r R igh t D riveFig. 1.55

whe

el b

ase

w hee l track w id th

Q

Fig. 1.56 Plan View of Ackermann Steering Gear

B

A

B ’

K ’

K

W

LL ’

D

CD ’

Q

�

Track W id th

bW heel Base


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cot cot ACIQ

...(1)

I is the point of intersection of stub-axles B A and CD produced.Clearly, for correct steering, the point of intersection I of front stub axle linesshould lie on rear axle line produced if necessary. In other words, equation(1) should conform with

cot cot track widthwheel base

For private cars with AC:AK 8.5:1 and 18, if the distance ACis 0.4 times the wheel-base, Ackermann steering mechanism would give

correct steering for 24. Thus with above proportions, Ackermann steeringgear provides correct steering in the following three positions only

(i) When moving straight, i.e., when 0.

(ii) When turning to the left at 24.

(iii) When turning to the right at 24.

At all other values of angle , wheels follow path along circular arcswhich do not have a common centre. And the instantaneous centre I doesnot lie on the axis of rear wheels but lies on a line parallel to the rear axletowards the front side. Hence some amount of skidding is bound to occur at

steering angles other than 0 and 24. Hence, angle and the proportionsof the mechanism should be chosen carefully to reduce the tendency to skidto a minimum, at high speeds.

Despite these limitations, Ackermann steering gear is still widely useddue to the use of pin joint instead of sliding pairs.

Also, deviation from condition of correct steering is small inAckermann steering mechanism.

To find the value of

Let and denote angles of steering corresponding to correct steeringposition. Now, for satisfactory operation of Ackermann steering gear,movement in a direction parallel to AC (Fig. 1.56) of ends K and L ofcross-bar KL must always be same. From Fig. 1.56.


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Projection of arc LL on AC projection of arc KK on AC.

Treating arcs LL and KK to be equal to chords, for small angles and , we have

CL [sin sin ] AK [sin sin ]

As AK CL, simplifying further,

sin cos cos sin sin sin sin cos cos sin

sin cos cos 2 cos sin sin

tan sin sin

cos cos 2 (2)

By knowing and steering angles for the two wheels in correct

steering position, We can find

To obtain desired value of , Wb

ratio should be known. It can then

be calculated from, [ W C track width]

cot cot Wb

1.29 HOOKE’S JOINT (OR) UNIVERSAL JOINT (COUPLING)

Explain Hooke’s joint with necessary derivation. (FAQ)

What is hooke’s joint? (FAQ)

A Hooke’s joint is used to connect intersecting driving and driven

shafts, while rotating. The shafts make small angle which depends on thespeed of two shafts as shown in Fig. 1.57 Hooke’s joint consists of twosemi-circular yokes (forks 2 and 4) at the end of each shafts and each yokecontains two bearings for the arms of a cross. This arrangement ensures pinconnection between the cross arm and yokes. The power is transmittedthrough a cross between the two shafts. The inclination of two shafts areconstant but while running, it varies depending on the speed of shafts.

The joint permits some amount of angular misalignment in connectedshafts. This becomes a very desirable feature because misalignment is arising outof elastic frame and also due to spring in suspension system of automobiles.


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Eventhough both shafts should complete one revolution in the sametime-interval, the angular velocity ratio of the two shafts is not constant allthroughout the revolution, and the angular velocity ratio varies as a function

of Angle between driver and driven axis

Angle of rotation of driving shaft

What are the applications of hooke’s joint? (FAQ)

Applications:1. It’s used in transmitting power from gear box to rear axle (or)

differential in automobiles.

2. In milling machine, it is used as a knee joint.

3. In multiple drilling machine, it is used to transmit the power to differentspindles.

1.30 HOOKE’S JOINT (OR) UNIVERSAL COUPLING - VELOCITY RATIO

Speed ratio N1

N2 1

2

1 cos2 sin2

cos ...(6)

Vertica l P lane

D riven Shaft

4

A

3

DB

2

O

Horizon ta l P

lane

D irec tio

n o f View ing

D riving Sha ft

A rm

Sem i C ircula r Yokes (o r) Forks

Fig. 1.57 Hooke’s (Cardan) Join t (or) Universal Coupling

cross

C


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Derive the condition for equal speeds of driving and driven shafts.(AU. Nov/Dec 2012)

Angle of inclination of driven shaft w.r.t the driving shaft

Angle turned by driving shaft

The ratio of the speeds of the driven and driving shafts is

2

1

cos 1 cos2 sin2

1 2 1 cos2 sin2

cos

For equal speeds, 1 2

Therefore, cos 1 cos2 sin2 or cos2 sin2 1 cos

Therefore,

cos2 1 cos

sin2 ...(7)

We know that

sin2 1 cos2 1 1 cos

sin2 1

1 cos 1 cos2

1 1 cos

1 cos 1 cos 1

11 cos

cos

1 cos

sin2 cos

1 cos ...(8)

Dividing equation (8) by equation (7),

sin2 cos2

cos

1 cos

sin2 1 cos

or tan2 cos sin2 1 cos2

cos sin2

sin2 cos

tan cos

This is the condition for the equal speeds of the driving and drivenshafts.


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To find 2 min and 2 max

Maximum angular velocity of driven shaft:

2 max 1

cos (or) N2 max

N1

cos ...(9)

Minimum angular velocity of driven shaft:

2 min 1 cos (or) N2 min N1 cos ...(10)

Note:

The angle and mass moment of inertia should be as low as possibleto minimise the alternate stresses developed in the joint due to the alternateacceleration and retardation.

To keep the angle very small, a double universal joint can be used.

Problem 1.2: The driving shaft of a Hooke’s joint rotates at a uniformspeed of 400 rpm. If the total maximum variation in speed of the driven shaftis 12% of the mean speed, determine greatest permissible angle between theaxes of the shaft. Also find the maximum and minimum speeds of the drivenshaft. (FAQ)

Solution:

N2 max N1

cos

N2 min N1 cos

N2 max N2 min 12% of N1 0.12 N1

Max. variation of speed of driven shaft N2 max N2 min 0.12 N1

N1

cos N1 cos 0.12 N1

N1

1cos

cos 0.12 N1

1 cos2 cos

0.12


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1 cos2 0.12 cos

cos2 0.12 cos 1 0

cos 0.12 0.122 4 1 1

2 1

0.941798

19.64

N2 max N1

cos

400cos 19.64

424.71 rpm

N2 min N1 cos 400 cos 19.64 376.73 rpm

1.32 TOGGLE MECHANISM

Write short notes on Toggle mechanism. (AU. Nov/Dec 2013, 2010)

As discussed in Section 1.15, in Toggle mechanism, for small amountof input torque, large output torque can be developed.

A mechanical linkage is a series of rigid links connected with jointsto form a closed chain, or a series of closed chains. Each link has two ormore joints, and the joints have various degrees of freedom to allow motionbetween the links. A linkage is called a mechanism if two or more links are

1 2O 2

A

P C6 5 B 4

O 1

F 3

1 1Fig. 1.60 Toggle Mechanism

P �

B �


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movable with respect to a fixed link. Mechanical linkages are usually designedto take an input and produce a different output, altering the motion, velocity,acceleration, and applying mechanical advantage.

Many practical multilink mechanisms are made up of basic linkagecombinations such as the slider-crank and the four-bar mechanism. The togglelinkage shown in Fig. 1.60.

Toggle mechanisms are used, where large resistances are to beovercome through short distances. Here, effort applied will be small but actsover large distance. In this mechanism 2 is the input link to which power issupplied and 6 is the output link, which has to overcome external resistance.Links 4 and 5 are of equal length.

This mechanism is used in rock crushers, presses, riveting machinesetc.

An important feature of the toggle mechanism is its ability to producehigh values of force at the slider with relatively low torque input. While thestudy of mechanisms is concerned primarily with motion, forces are of greatimportance to the designer and are intimately related to motion analysis. Ifa rigid mechanism has a single input and a single output with negligiblelosses, the rate of energy input equals the rate of energy output. Force ratiosare the inverse of velocity ratios when inertia effects are negligible.

Taking moment about C by force P in connecting rod and tangential

force F on crank. From CBB

cos PP

P P cos ...(i)

Then, P cos BB F2

B C

BBBC

F/2

P cos

tan F/2P

[. . . P P cos ]PF

1

2 tan or F 2P tan


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Problem 1.3: The distance between the two parallel shafts connected byoldham’s coupling is 30 mm. The speed of the driving shaft is 300 rpm. Findthe maximum speed of sliding of the tongue of the intermediate piece in theslot in the flange.

Solution:

r 0.03 m

Angular speed of the shaft

2 N

60

2 30060

31.42 rad/s

Maximum speed of sliding

V r 31.42 0.03 0.942 m/s

Problem 1.4: Define transmission angle of a four bar linkage. What is theeffect of transmission angle on mechanical advantage?

Transmission angle

The angle between the output link and the coupler (connecting rod) is

known as transmission angle .

Effect: When the transmission angle 90, the mechanical advantage is

maximum. When is maximum, MA is more. When is minimum, MA isless and mechanism will lock and jam.

Mechanical AdvantageIt is defined as the ratio of driven link torque to driver link torque.

In a four bar mechanism as shown in Fig. the link PQ is driving link andRS is driven link.

The mechanical advantage of four bar mechanism varies depending on

the position of the links. So both angles (,) changes continuously. When

the is too small, the mechanical advantage becomes infinite.

TR Driven torque

TQ Driving torque

M.A.Ideal TRTQ

Q

R


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M.AActual TR

TQ considering friction

where Transmission efficiency.

Problem 1.5: Briefly explain various types of constrained motions.

Completely constrained motionsWhen the motion between a pair is limited to a definite direction

irrespective of the direction of the force applied, then the motion is said tobe completely constrained motion. Ex: Motion of a square bar in a squarehole and screw pair, turning pair. Refer Fig. 1.3 and Fig. 1.4 and Fig. 1.5

Incompletely constrained motion

When the motion between a pair can take place in more than onedirection, then the motion is called an incompletely constrained motion. Thechange in the direction of the impressed force may alter the directions ofrelative motion between the pair.

Example: Motion of circular bar in a circular hole. Refer Fig. 1.6

(cylindrical pair)

Successfully constrained motion

When the motion between the elements, forming a pair is such thatthe constrained motion is not completed by itself, but by some other means,then the motion is said to be successfully constrained motion Ex: A shaft ina foot-step bearing and piston and cylinder in slider crank mechanism. ReferFig. 1.4 and Fig. 1.11

P S

Q

R

Q R

D riving link

D rivinglink


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Problem 1.6: Find the degrees of freedom for the mechanisms shown inFig. (AU. Apr/May 2017)

Solution:

(i) n 3 l 1 2j h

l 5; j 5; h 0

n 3 5 1 2 5 0

2

DOF n 2

(ii) n 3 l 1 2j h

l 6; j 7; h 0

n 3 6 1 2 7 0

1

DOF n 1

(iii) n 3 l 1 2j h

l 11; j 11; h 0

n 3 11 1 2 11 0

8

DOG n 8

3 4

1 1

5 2 4

5

3

6

1

2 3

4

5 8

11

10

97

6

(1) (2 ) (3 )

Fig:

2

n Number of degreeof freedomWhere,

l Number of links

j Number of joints (or) lower pairs

h Number of higher pairs


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Problem 1.7: Find the degrees of freedom of the mechanisms shown infig. (AU. May/June 2016)

The Kutzbhach criterion for determining the number of degrees offreedom or movability (n) of a plane mechanism is

n 3 l 1 2j h

where,

n number of degrees of freedom

l number of links

j number of lower pairs or binary joints

h number of higher pairs.

2

3

4

56

78

10

9

A

B ( c)

1

2

3

6

5

9 4

8R

RG

J7R

H

R

E

FR

BR

AR

P

CR

RD

(d)

Fig:11 (b)

A

1

2

3B

CH 5

G6 7

DF

E

R olle r P in

(e)

(a)

9

7 6 5

4

3

21

2 1

3

4

56

78

(b)

8


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(a) l 9 ; j 11 ; h 0

n 3 9 1 2 11 24 22 2 degrees of freedom

(b) Number of links l 8

One ternary joint is equivalent to twobinary joints. ie., D has double joint.

j 9 2 11 (2 is to account for double joint)

n 3 8 1 2 11 21 22 1

Therefore this mechanism is a staticallyindeterminate structure.

(c) l 10

Two ternary joints at A and B. ieA has double joints and B has doublejoints.

j 9 2 2 13[2 (2) is to account for 2 double-joints]

n 3 10 1 213

27 26 1 degree of freedom

(d) l 9

1 ternary joint at J. i.e., Onedouble joint at J, one sliding pairmarked P

j 9 2 11(2 is to account for double joint atJ)

n 3 9 1 2 11 24 22 2 degrees of freedom

(e) Spring H is a device to apply force but it is not considered as link.

Therefore, total number of links l 7

(a)

9

7 65

4

3

21

8

2 1

3

4

56

78

(b)

DDoublejo in t

2

3

4

56

78

10

9

A

B ( c)

1

2

3

6

5

9 4

8R

RG

J7R

H

R

E

FR

BR

AR

P

CR

RD

(d)


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We have six turning pairs atA, B, D, C, F and G, one sliding pair betweenlink 7 and link 1 and one rolling pair (highestpair) at E

j 6 1 7 and h 1

(h 1 because one rolling pair)

n 3 7 1 2 7 1

18 14 1 3 degrees of freedom

Problem 1.8: Find out degrees of freedom of the mechanism shown in Fig(a), (b) and (c)

Solution:

(a) Number of links 8 l 8

Number of binary joints

9 j 9 and h 0

A

1

2

3B

CH 5

G6 7

DF

E

Ro ller P in

(e)

7

65

4 2

8

3

1(a)

7

6

8

9

5

4

3

2

1(b)

2

1

34 5

6

(c)

Am putee Knee-M echan ism

(Artificial L im bs)

7

65

4 2

8

3

1(a)


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Degree of freedom n 3 l 1 2j h

3 8 1 2 9 3

(b) l No. of links = 9

j No. of binary joints = 10

h 0

DOF n 3 l 1 2j h

3 9 1 2 10

4

(c) h 0

Here l 6

j 7

DOF n 3 l 1 2j h

3 6 1 2 7

1

Problem 1.9 Find the maximum and minimum transmission angles for themechanisms shown in fig. of case 1 and case 2. The figures indicate thedimensions in standards units of length. (AU. Nov/Dec 2013)

7

6

8

9

5

4

3

2

1(b)

2

1

34 5

6

(c)

Am putee K nee-M echanism

(Artificial Limbs)

Case 1 Case 2


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Transmission angle

The angle between the output link and the coupler (connecting rod) is

known as transmission angle .

For minimum transmission angle position,

Crank AB1, should lie on AD as shown in Fig. (i) B1, C1, D gives

minimum transmission angle.

Procedure (Refer Fig. (i))

Draw AD 3 cm

Draw AB1 1 cm on right side of A. i.e., lie along AD.

B1 as centre, draw arc of radius 3 cm and D as centre, draw arc

of radius 2 cm. Both arcs intersect at C1.

Now, measure B1 C1 D 41.58 which gives minimum .

For maximum transmission angle position (Fig. (ii))

Crank AB2 should lie as an extension of AD as shown in Fig. (ii).

B2 C2 D gives maximum transmission angle.

Procedure

Draw AD 3 cm

Draw AB2 1 cm on left side of A, ie as an extension of DA.

A DB 1

41.58 O

C 1

M in im um transmiss ion angle( ) = 41.58 O

Coupler

Fo l

low

er l i

nk

13 2

3

M in im um transmiss ion angle( ) Pos ition

A D

=104.72O

C 2

M ax 104 .72Oim um transm ission ang le ( ) =

Coupler F ol lower link

1

3 2

3

M axim um transm ission ang le ( ) P os ition

B2

(i) (i i)

Case 1


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Now B2 as centre, draw an arc of radius 3 cm and D as centre

draw an are of radius 2cm. Both arcs intersect at C2.

Now measure B2 C2 D 104.72 which gives maximum .

Note: The above problem can be solved by analytical method.

To find minimum Refer Fig. (i)In this, mechanism,

Longest link length 3 l

Shortest link length 1 s Length of other links = 3 and 2

As per Grand law,

s l other 2 links lengths.

Since 3 1 3 2, it is coming under class I mechanism. And sincethe link adjacent to the shortest link is fixed, it is called crank-rockermechanism.

B1 D2 B1 C12 C1 D2 2 B1 C1 C1 D cos

22 32 22 2 3 2 cos 4 9 4 12 cos

cos 912

0.75

41.4 Minimum transmission angle

To find maximum : (Refer Fig. (ii))

B2 D2 B2 C22 C2 D2 2 B2 C2 C2 D cos

42 32 22 2 3 2 cos

16 9 4 12 cos

12 cos 3

cos 312

0.25

104.48 Maximum

As per Grass of law,

s l < other 2 links lengths.


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Case 2

In this mechanism,

Longest link = 10

Shortest link = 6

Length of other links = 8 and 7

Since 10 6 8 7, it is comingunder class II mechanism and hence it isdouble-rocker (Rocker-rocker mechanism).

Maximum is obtained when

180

From Fig (iii), To find maximum

a d2 b2 c2 2bc cos

152 62 102 2 6 10 cos

225 36 100 120 cos

cos 0.7416

137.87 137.9

To find minimum , should be 180

Refer Fig. (iv)

d2 a b2 c2 2 a b c cos

82 132 102 2 13 10 cos

64 169 100 260 cos

260 cos 205

cos 205260

0.788

37.96 ~ 38

Case 2


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M ax 137.9Oim um transm ission ang le( ) =

Coupler Follower link

M axim um transm ission angle( ) P osition

=137.9o

d8a 7

b

6

c10

Fig.iii

The m axim um position for maximum can be obta ined by the previous procedure.

Graphical M ethod

M in 38Oim um transm ission angle( ) =

Coupler

Follow

er link

M in im um transmission ang le( ) Position

d8

a

7

b

6

c

10

Fig.iv

=38 o

Q

RP


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To find minimum (Refer Fig. (iv))

Draw d 8 cm

Draw an arc of radius 13 cm, keeping left end of d as centre

P

Draw another arc of radius of 10 cm, keeping right end of d as

centre R.

Both arcs intersect at Q.

Now measure PQR 38 which is minimum transmissionangle.

Problem 1.10: In a crank and slotted lever quick return motion mechanism,the distance between the fixed centre is 240 mm and the length of the drivingcrank is 120 mm. Find the inclination of the slotted bar with the vertical inthe extreme position and the time ratio of cutting stroke to the return stroke.If the length of the slotted bar is 450 mm, find the length of the stroke ifthe line of stroke passes through the extreme positions of the free end of thelever. (AU. Nov/Dec 2011)

Solution:

Given: AC 240 mm; CB1 120 mm; AP1 450 mm

Inclination of the slotted bar with the vertical

Let CAB1 Inclination of the slotted bar with the vertical.


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sin CAB1 sin 90

2

B1 C

AC

120240

0.5

CAB1 90 2

sin 1 0.5 30

Time ratio of cutting stroke to the return stroke

We know that

90 2

30

2

90 30 60

or 2 60 120

Time of cutting strokeTime of return stroke

360

360 120120

2

Length of the stroke

We know that length of the stroke,

R1 R2 P1 P2 2 P1 Q 2 AP1 sin 90

2

2 450 sin 90 60 900 0.5 450 mm

Problem 1.11: In a quick return motion mechanism of crank and slottedlever type, the ratio of the maximum velocities is 2. If the length of strokeis 25 cm find (i) The length of the slotted level, (ii) The ratio of times ofcutting and return strokes (iii) The maximum cutting velocity per second ifthe crank rotates at 300 rpm. (AU. May/June 2009)


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Solution:

sin CAB1 sin 90

2

CB1

AC

12

0.5

C AB1 90 2

sin 1 0.5 30

2

90 30 60

2 60 120

We know that

Length of stroke 2 AP1 sin 90

2

0.25 2 AP1 sin 90

1202

AP1 0.25 m

(i) Length of slotted lever

AP1 0.25 m

(ii) Time of cutting strokeTime of return stroke

360

360 120120

2

(iii) Maximum cutting velocity/s if crank rotates at 300 rpm

Angular velocity, 2N60

2 300

60

31.42 rad/s


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Problem 1.12: Extend Grublers criterion for planar mechanism to obtainthe ‘Degree of freedom’ of a space mechanism as

F 6 L 1 5g 4c 4s.

Where g Total number of sliding pairs

c Total number of cylindrical pairs

s Total number of spherical pairs

L Total number of links, (AU. May/June 2009)

The Grubler’s criterion applies to mechanisms with only single degreeof freedom joints where the overall movability of mechanism is unity.

Applying F 1 and C 0, (higher pair)

F 6 L 1 5g 4c 4s

1 6L 6 5g 4s

6L 5g 4s 7 0

The simplest possible mechanism of this type are a four bar mechanism

and a slider crank mechanism in which l 4 and j 4

64 5g 4s 7 0

5g 4s 17 0

5g 4s 17 0


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UNIT II

KINEMATICS OF LINKAGE MECHANISMS

Displacement, velocity and acceleration analysis of simple mechanisms– Graphical method – Velocity and acceleration polygons – Velocity analysisusing instantaneous centres – kinematic analysis of simple mechanisms –Coincident points – Coriolis component of Acceleration – Introduction tolinkage synthesis problem.

2.2 LINEAR DISPLACEMENT, VELOCITY & ACCELERATION ANALYSIS

Linear Displacement(s) is defined as the distance moved by a body withrespect to a certain fixed point. The displacement may be along a straightline or a curved path. Displacement of a body is a vector quantity.

Linear velocity (v) may be defined as rate of change of linear displacementwith respect to time. It is a vector quantity.

It is given by: Linear velocity (v) = ds /dt

Note: Speed is rate of change of displacement with respect to timeirrespective of direction , hence it is a scalar quantity.

Linear Acceleration (a) may be defined as rate of change of linear Velocitywith respect to time. It is a vector quantity.

It is given by :

Linear Acceleration (a) dv/dt d ds/dt/dt d2s/dt2

Linear Acceleration (a) dv ds/ds dt v dv/ds.

Note: Acceleration can be positive or negative and Negative acceleration iscalled as deceleration.

2.2.1 Angular Displacement, Velocity & Acceleration

Angular Displacement: is defined as the angle described by the particleof the body from one point to another with respect to time. It is a vector quantity.

Angular velocity: may be defined as rate of change of angulardisplacement with respect to time. It is a vector quantity.

It is given by : Angular velocity d /dt

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Angular Acceleration: may be definedas rate of change of angular velocity withrespect to time. It is a vector quantity.

It is given by : Angular Acceleration

d /dt d d /dt/dt d2 /dt2

2.4 KEY FOR CONSTRUCTION OF THE VELOCITY POLYGONSThe velocity of any point in a link relative to any other point in the

same link is always perpendicular to the line joining these two points.

Refer Fig 2.8(a). A and B are the end points of a link AB. Assume

the link AB is rotating about A with angular velocity in clockwise direction;since the distance between points A and B remains constant, therefore therewill not be any relative motion between A and B, along line AB. So therelative motion of B with respect to A must be perpendicular to AB.

VBA - It should be read as “Velocity of B with respect to A”.

This VBA (Velocity of B with respect to A) is represented by a vector

ab

(Refer Fig. 2.8(b)).

CB

A

O

Fig 2.4 Angular

A

BV

=xAB

BA

=

VBA

a

b

A

B

V=

xAC

CA

=V

CA

a

cC

Fig (a) Fig (b)

Fig (d)Fig (c)

V=

xAB

BA

V=

xAC

CA

Fig. 2.8


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VBA distance between two points A and B

VBA AB ... (i)

Similarly, the “Velocity of C with respect to A” VCA Refer

Fig 2.8(c).

VCA AC ... (ii)

This is represented by vector ac

. Refer Fig 2.8(d). Divide (i) by (ii)

VBA

VCA

ab

ac

AB

AC

ABAC

i.e., ab

ac

ABAC

or ac

ab

ACAB

From this, we can note that, the point c on the vector ab divides itin the same ratio as C divides the link AB. (Fig. 2.9)

2.4.1 Velocity polygon

The construction of velocity polygon can be explained by the followingexamples.

1. Four bar mechanism and

2. Slider crank mechanism

a

c

b

Velocity diagramA

B

V=

xAB

BA

V=

xAC

C A

Link

C

AB Fig. 2.9

www.srbooks.org Kinematics of Linkage Mechanisms 2.3

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2.4.2 Graphical method - Solved Problems

Problem 2.1: Draw the velocity polygon for the four bar mechanism shown

in fig. and determine VBA, VCB and VCD. Also find the AB, BC and CD.

In this four bar chain ABCD, AD is fixed. The crank AB rotates at 120 rpm clockwise. (AU. Nov/Dec 2013)

Solution

First of all, draw space diagram to some suitable scale.

Velocity diagram

The linear velocity VBA or VB can be found out by VB AB AB

[Note: Here VBA can be denoted by VB, because A is a fixed point]

Angular velocity of link AB AB

AB 2 120

60 12.57 rad/s

VB AB AB 12.57 0.2 2.51 m/s

i.e., VBA 2.51 m/s

The direction of VBA is perpendicular to the line AB.

We know,

1. The magnitude and direction of VBA (VBA 2.51 m/s and it is

perpendicular to AB)

60o

A D

B

C

20cm

75cm

40cm

75cm

AB


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2. The direction of VCB (i.e., perpendicular to BC)

3. The direction of VCD (i.e., perpendicular to CD)

By knowing these above, we can draw velocity diagram (or velocitypolygon).

1. Since the link AD is fixed, the points A and D are fixed points. Alwaysfixed points should be marked at one place in velocity diagram (alsoin acceleration diagram which is explained later).

V = 1.91m /sCD

V=1m

/sC

B

V=2.51m /s

BA

a ,d

c

b

to AB

r

rto C D

rto BC

Velocity Diagram

60 o

A D

B

C

20cm

75cm

40cm

75cm

Space Diagram


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2. Through a draw line (vector) ab representing velocity VBA, perpendicular

to AB and the length equal to 2.51 m/s to some suitable scale (i.e.,choose 1 m/s = 1 cm. So 2.51 m/s can be drawn as 2.51 cm).

3. Now from point b, draw line (vector) perpendicular to BC to get VCB

(Here length is not known).

Similarly, from point d, draw line (vector) perpendicular to CD to getVCD (Here also, length is not known). But both the above two lines intersect

at c.Now, we can measure.

VCB bc 1 m/s

VCD dc 1.91 m/s

To find angular velocity of BC and CD.

BC VCB

BC

10.75

1.333 rad / s anti clockwise about B

CD VCD

CD

1.910.4

4.775 rad / s (clockwise about D)

2.5 SOLVED PROBLEMS ON RELATIVE ACCELERATION - GRAPHICAL METHOD

VQP Relative velocity of Q with respect to P

Similarly,

aQP Relative acceleration of Q with respect to P.

By refering following problems, we can understand better about relativeacceleration,

Problem 2.2: PQRS is a four bar chain with link PS fixed. The lengths of the

links are PQ 62.5 mm; QR 175 mm; RS 112.5 mm; and PS 200 mm.The crank PQ rotates 10 rad/s clockwise. Draw the velocity and acceleration

diagram when angle QPS 60 and Q and R lie on the same side of PS.Find the angular velocity and angular acceleration of links QR and RS. (FAQ)


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P

Q

R

S

62.5mm

175m m

112.5m m

200m m60o

10rad/s

p ,s r

q

V R S

V QP

V R Q

0.426m /s

0.625m /s

I r

Ir

Ir

0 .333m /s

Velocity Diagram

Space Diagram

a =1.613m /sr 2RS

a =5.3m /st 2RS

a=

6.25

m/s

QP

2

a = 0.634m /sr 2RQ

a= 4 .1m

/st

2

RQ

a RS

aRQ

r

p,s

q

I�

Ir

I�

I

y

I�I

Ir

I

x

Acceleration D iagram


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Solution:

Procedure

1. Draw space diagram first.

2. Then draw velocity diagram as follows.To draw velocity diagram.

3. The points p and s are fixed points. So these are marked in one placein velocity diagram.

To represent -‘Velocity of Q with respect to P’ = “VQP”

VQP QP PQ 10 0.0625 0.625 m/s

4. VQP is perpendicular to the link PQ. So draw vector pq perpendicular

to PQ to some suitable scale to represent VQP. (i.e., draw

VQP 9.375 cm to represent 0.625 m/s i.e., scale 15:1)

5. From point q, draw a line perpendicular to QR to represent VRQ. Here

length is not known.

From point s, draw line perpendicular to SR to represent VRS. Here

also length is not known. Both lines intersect at ‘r.’

6. Thus we have drawn velocity diagram pqrs which represents the velocitydiagram.

Now measure

sr VRS 6.4 cm 6.415

0.426 m/s.

qr VRQ 5 cm 515

0.333 m/s[. . . scale 15:1]

Result

Angular velocity of link QR QR

QR VRQ

RQ

0.3330.175

1.9 rad/s (anti clockwise)

Angular velocity of link RS RS


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RS VRS

RS

0.4260.1125

3.8 rad/s (clockwise)

To draw Relative acceleration diagram (or) simply acceleration diagram

Normally, acceleration can be resolved into two components.

1. Radial component ar (or) Normal component an

2. Tangential component at

aQP can be resolved into aQPr and aQP

t . Always, aQPr is acting along

the link QP and towards the centre.

Always aQPt is acting perpendicular to QP.

Angular velocity of PQ is constant. Hence PQ 0.

Since the angular acceleration of the crank PQ PQ is zero, aQPt 0.

. . . PQ

aQPt

QP

aQPr = radial component of acceleration of Q with respect to P.

aQPr

VQP2

QP

0.6252

0.0625 6.25 m/s2

aRQr = radial component of acceleration of R with respect to Q.

aRQr

VRQ2

RQ

0.3332

0.175 0.634 m/s2

aRSr radial component of acceleration of R with respect to S.

aRSr

VRS2

RS

0.4262

0.1125 1.613 m/s2

The acceleration diagram can be drawn as follows.

7. Since P and S are fixed points, these points are marked at one place inthe acceleration diagram.


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Note: Always, radial component of acceleration ar acts parallel to the link

and tangential component of acceleration at acts perpendicular to the link.

8. Draw the vector pq to represent aQPr 6.25 m/s2 (parallel to link PQ).

9. From point q, draw vector qx parallel to QR to represent

aRQr 0.634 m/s2

ie aQPr qx 0.634 m/s2

10. From point x, draw vector perpendicular to QR to represent aQRt (length

is not known).

11. From point s, draw vector sy parallel to SR to represent

aRSr 1.613 m/s2

i.e., aRSr sy 1.613 m/s2.

12. From point y, draw vector perpendicular to SR to represent aSRt (length

is not known).

13. Both the lines from x and y intersect at r.

14. Now measure, aRQt 4.1 m/s2

aRSt 5.3 m/s2

15. To get total acceleration aRS, join s and r

To get total acceleration aRQ, join q and r.

Result:

Angular acceleration of link QR,

QR aRQ

t

RQ

4.10.175

23.43 rad / s2

(Anti clockwise)

Angular acceleration of link RS

RS aRS

t

RS

5.30.1125

47.1 rad/s2

(Anti clockwise)


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. . . VAB VBA 5.586 m/s

AG 275 mm

AB 500 mm

VAB AB AB

AB VAB

AB

5.5860.5

11.1 rad/sec

AB 500 mm 0.5 m

Result:Velocity of slider OA VOA 6.48 m/s

Velocity at point G VOG 6.75 m/s

Angular velocity of connecting rod AB AB 11.1 rad/sec

Problem 2.9: An engine mechanism is shown in Fig. The CB = 200 mmand the connecting rod BA = 600 mm. In the position shown, the crankshaft

has a speed of 50 rad/s and an angular acceleration of 800 rad/s2. Find: (i)angular velocity of AB and (ii) angular acceleration ofAB. (AU. Apr/May 2017)

Given data:

Angular velocity BC 50 rad/s

Angular acceleration AB 800 rad/s2

To find:

(i) Angular velocity AB

(ii) Angular acceleration AB

A

B

C

200 m m120o

Fig:


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Solution:

Velocity diagram

vBC vB BC CB 50 0.2 10 m/s. . . BC 50 r/s

vB 10 m/s which is used to set the velocity scale

5 cm = 10 m/s

scale; 1 cm = 2 m/s[. . . 10/5]

1. Mark any point c. From c draw a line Cb VBC 10 m/s perpendicular

to CB.

2. From c, draw a horizontal line (parallel to the direction of motion ofpiston) (length not known)

3. From b, draw a line perpendicular to BA, to represent VBA (length not

known)

4. Both the above lines intersect at a

By measurement, vAB 5 m/s

. . . vAB ab

scale 2.5 2 5 m/s

5. To find angular velocity of AB AB

AB vAB

BA

50.6

8.33 rad/sec

AB 8.33 rad/sec

Acceleration diagram

1. Draw bc aBCr 500 m/s2 parallel to BC to some suitable scale.

. . . aBC

r vBC

2

CB 102

0.2 500 m/s2

5 cm 500 m/s2

Scale 1 cm 100 m/s2


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2. From point b, draw b

x aAB2 41.66 m/s2 parallel to BA

. . . aAB

r vAB

2

BA

52

0.6 41.66 m/s2

Space Diagram

Scale 1cm = 100m m

Velocity Diagram

Scale 1cm = 2m /s

Acceleration Diagram

Scale 1cm = 100m /s2

A

B

C120

o600mm

200m

m

(a)

c

b

aV A

V A BV B

(b)

a ’ c ’a A

aA

Bt

aA

B

aB

C

b ’’

b’a BC

t

aBCr

a ABr

(c) a BC = x CB = 800 x 0.2 = 160m /st

B C 2

b b = ’’ ’160100 = 1.6cm


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41.66100

0.41 cm b x

3. From point x, draw line perpendicular to BA (length not known) to

represent aBAt

4. Both lines intersect at a

By measurement, aBAt 520 m/s2

. . . aBA

t Xa

scale 5.2 100 520 m/s

5. To find angular acceleration of AB AB

AB aBA

t

BA

5200.6

866.66 rad/sec2

AB 866.66 rad/s2

Problem 2.10 The diagram shows part of a quick return mechanism. Thepin A slides in the slot when the disc is rotated. Calculated the angular

velocity and acceleration of link BC when 60 and 100 rad/s. (AU. May/June 2016)

C

20 N

70 m m

B

A

O A = 40m mBC = 180 m m

O


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Derive the expression for coriolis component of acceleration with neatsketch and give its direction for various conditions. (AU. May/June 2016)

The acceleration of a moving point relative to a moving coordinatesystem (moving body) may have two components of acceleration. One istangential and the other one is perpendicular to the point. The perpendicularcomponent of acceleration of the point is known as coriolis component ofacceleration.

Consider a slider B on a link OA such that when the link OA is rotating

clockwise with angular velocity , the slider B mass outward with linearvelocity V.

Let the angle turned through by link OA be d in time dt, and let thelink occupy new position ‘OA’. The slider moves to position E.

The slider can be considered to move from B to E as follows:

1. From B to C due to angular velocity of link OA.

2. C to D due to outward velocity V of the slider.

3. D to E due to acceleration perpendicular to the rod due to corioliscomponent of acceleration.

arc DE arc EF arc FD

F

B

V

A’

E

C

d

d

O


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arc EF arc BC

OF d OB d

OF OB d

BF d

CD d

CD v dt

& d dt

arc DE v dt dt

arc DE chord DE as d is very small

DE v w dt2 ... (i)

But, DE 1/2 fcor dt2 ... (ii)

Where fcor (or) Acor is the constant coriolis acceleration of the particle

From Eqn (i) & (ii), v dt2 1/2 fcor dt2

fcor 2 vw

2.8.1 Coriolis component of acceleration

Coriolis acceleration can also be defined as - when a point on onelink is sliding along another rotating link, then the corioliscomponent of acceleration is formed.

Examples of Coriolis component formed (mechanism)

Slotter lever mechanism

Whitsworth quick return mechanism

Swivelling joint mechanism

Various directions of Coriolis acceleration

There are four possible direction of coriolis component ofacceleration, with respect to slider velocity (B) vector and theangular velocity of link OA.


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A

V

B

2 V = Acor 2 V = A cor

V

A

B

O O

2 V = Acor 2 V = A cor

O O

A

V

B

A

VB

Case (i) Case (ii)

Case (iii) Case (iv)

Fig. 2.16


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From the above four possible cases, Fig. 2.16 it is clearly seenthat, the coriolis component of acceleration changes its sign, if

either (or) v is reversed in direction.

But the direction of coriolis component of acceleration will not be

changed in sign if both and v are reversed in direction.

It is concluded that, the direction of f coriolis (or) A coriolis is

obtained by rotating v, at 90 about its origin in the same directionof angular velocity.

Importance of Coriolis component of acceleration:

Whenever the point on one link is sliding along another rotatinglink, then there is a coincident point exists in the mechanism.

The total acceleration of that coincident point about another pointin the link is the vector sum of Radial acceleration, Tangentialacceleration and the Coriolis acceleration.

Coriolis component of acceleration is necessary for dynamicbalancing of the system.

If we neglect the coriolis component of acceleration in dynamicanalysis, then the system will be dynamically unbalance.

Problem 2.19: In a whitworth quick return mechanism, as shown in figure,crank OA rotates at 30 rpm in clockwise direction. The dimensions of various

links are OA 150 mm, OC 100 mm, CD 125 mm and DR 500 mm.Determine the acceleration of the sliding block R and the angular accelerationof the slotted lever BD. (AU. Nov/Dec 2013)

Solution

Given: NAO 30 r.p.m. or AO 2 30/60 3.142 rad/s;

OA 150 mm 0.15 m; OC 100 mm 0.1 m;

CD 125 mm 0.125 m; DR 500 mm 0.5 m

Velocity of A with respect to O or velocity of A,

VAO VA AO OA 3.142 0.15 0.47 m/s

(Perpendicular to OA)


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First of all draw the space diagram, to some suitable scale, as shownin Fig.(a). Now the velocity diagram, as shown in Fig.(b), is drawn asdiscussed below:1. Since O and C are fixed points, therefore these are marked at the same

place in velocity diagram. Now draw vector oa perpendicular to OA,to some suitable scale, to represent the velocity of A with respect toO or simply velocity of A i.e. VAO or VA, such that vector

oa VAO vA 0.47 m/s

2. From point c, draw vector cb perpendicular to BC to represent thevelocity of the coincident point B with respect to C i.e. vBC or vB and

from point a draw vector ab parallel to the path of motion of B (whichis along BC) to represent the velocity of coincident point B with respectto A i.e. vBA. The vectors cb and ab intersect at b.

Note: Since we have to find the coriolis component of acceleration of sliderA with respect to coincident point B, therefore we require the velocity of Awith respect to B i.e. vAB. The vector ba will represent vAB as shown in

Fig.(b).

3. Since D lies on BC produced, therefore divide vector bc at d in thesame ratio as D divides BC in the space diagram. In other words,

bd/bc BD/BC

O

45o

D

C

A

B

R

All D im ensions are in m m


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(b) Acceleration Diagram .

(a) Space Diagram .

d

O ,C

V Rr

VR D

V A O

a

b

V BC

V BA

V BA

(b) Velocity Diagram.

C

AB

V ABac AB

B C

(c) Direction of Coriolis Component.

acA B

arA Ba �

b�

y

x

aA

o ,c� �

a Rr�

d �z

100

O45o

D

C

A

B

500R

All D im ension in m m .

Slider (A)

Coincident pt. (B )

A on OA; B on BD

V A O

VB C


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4. Now from point d, draw vector dr perpendicular to DR to represent thevelocity of R with respect to D i.e. vRD, and from point c draw vector

cr parallel to the path of motion of R (which is horizontal) to representthe velocity of R i.e. vR.

By measurement, we find that velocity of B with respect to C.

vBC vector cb 0.46 m/s

Velocity of A with respect to B,

vAB vector ba 0.15 m/s

and velocity of R with respect to D

vRD vector dr 0.12 m/s

We know that angular velocity of the link BC,

BC vBC

CB

0.460.24

1.92 rad/s Clockwise

Acceleration of the sliding block R

We know that the radical component of the acceleration of A withrespect to O,

aAOr

vAO2

OA

0.472

0.15 1.47 m/s2

Coriolis component of the acceleration of slider A with respect tocoincident point B,

aABc 2BC vAB 2 1.92 0.15 0.576 m/s2

Radial component of the acceleration of B with respect to C,

aBCr

vBC2

CB 0.462

0.24 0.88 m/s2

Radial component of the acceleration of R with respect to D

aRDr

vRD2

DR 0.122

0.5 0.029 m/s2

Now the acceleration diagram, as shown in Fig. is drawn as discussedbelow:


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1. Since O and C are fixed points, therefore these are marked at the same

place in the acceleration diagram. Draw vector o a parallel to OA, tosome suitable scale, to represent the radial component of the acceleration

of A with respect to O i.e. aAOr or aA such that

vector o a aAOr aA 1.47 m/s2

2. The acceleration of the slider A with respect to coincident point B hasthe following two components:

(i) Coriolis component of the acceleration of A with respect to B i.e.

aABc , and

(ii) Radial component of the acceleration of A with respect to B i.e. aABr

These two components are mutually perpendicular. Therefore from

point a draw vector a x perpendicular to BC to represent aABc 0.576 m/s2

in a direction as shown in Fig. and draw vector xb perpendicular to vector

a x (or parallel to BC) to represent aABr whose magnitude is yet unknown.

Note: The direction of aABc is obtained by rotating vAB (represented by vector

ba in velocity diagram) through 90 in the same sense as that of BC which

rotates in clockwise direction.

3. The acceleration of B with respect to C has the following twocomponents:

(i) Radial component of B with respect to C i.e. aBCr , and

(ii) Tangential component of B with respect to C i.e. aBCt

These two components are mutually perpendicular. Therefore, draw

vector c y parallel to BC to represent aBCr 0.88 m/s2 from point y draw

vector yb perpendicular to c y to represent aBCt . The vectors xb and yb

intersect at b . Join b c .


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4. Since the point D lies on BC produced, therefore divide vector b c at

d in the same ratio as D divides BC in the space diagram. In otherwords,

b d /b c BD/BC

5. The acceleration of the sliding block R with respect to D has also thefollowing two components.

(i) Radial component of R with respect to D i.e. aRDt , and

(ii) Tangential component of R with respect to D i.e. aRDr .

These two components are mutually perpendicular. Therefore from

point d , draw vector d z parallel to DR to represent aRDr 0.029 m/s2 and

from z draw zr perpendicular to d z to represent aRDt whose magnitude is

yet unknown.

6. From point c , draw vector c r parallel to the path of motion of R

(which is horizontal). The vector c r intersects the vector zr at r .The vector c r represents the acceleration of the sliding block R.

By measurement, we find that acceleration of the sliding block R,

aR vector c r 0.18 m/s2

Angular acceleration of the slotted lever CABy measurement from acceleration diagram, we find that tangential

component of B with respect to C.

aBCt vector yb 0.14 m/s2

We know that angular acceleration of the slotted lever CA

CA BC aCB

t

BC

0.140.24

0.583 rad/s2 Anticlockwise

Problem 2.20: A mechanism of a crank and slotted lever quick returnmotion is shown in Figure. If the crank rotates counter clockwise at 120 rpm.Determine for the configuration shown, the velocity and acceleration of theram D. Also determine the angular acceleration of the slotted lever. Crank,

AB 150 mm; Slotted arm, OC 700 mm and link CD 200 mm. (AU. Nov/Dec 2011)


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2.11 TYPE SYNTHESISType synthesis is used to select the kind of mechanism with

Gear combination (or)

Belt-pulley combination (or)

Cam mechanism and so on by considering design aspects like

Space consideration

Safety aspects

Economic consideration

Manufacturing process

Selection of materials and so on.

2.12 NUMBER SYNTHESISThe combination of different links to meet certain requirements for

obtaining specified motion is called as mechanism.

If there is no relative motion between links, then it is called structure.

2.12.1 Degrees of freedom DOF (or) simply nThe number of degrees of freedom n is the number of independent

variables needed to define completely the condition of the system.

We have studied about DOF and Kutzbach criterion (or) Grubler’sequation and Grashof’s law in Chapter 1.

Fig. 2.17(a) has zero DOF ie n 0 ie it is locked hence it is a structure.

Fig. 2.17(b) is Four-bar chain having 1 DOF ie n 1 ie 1 variable

is needed to define relative positions of all links.

(a ) Truss , n = 0 (Locked) (b ) Four ba r chain, n = 1(C onstra ined)

(c) F ive ba r cha in , n = 2

Fig. 2.17 Exam ples of K inematic Chains w ith n = 0,1 and 2

12


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Fig 2.17(c) is a 5 bar chain having 2 DOF ie n 2 ie 2 variables

1 and 2 are needed to define relative positions of all links.

Application of Grubler’s criterion to various closed kinematic chainswill reveal DOF

1. 2 links, 2 joints: (2 overlapped links)

l 2 ; j 2

n 3 l 1 2j

n 3 2 1 2 2 1

This is called statically indeterminate structure.

2. 3 links, 3 joints: (Truss) [Fig. 2.17 (a)]

l 3 ; j 3

n 3 l 1 2j

3 3 1 2 3 0

This is called statically determinate structure

3. 4 links, 4 joints: (Movable 4 bar chain) [Fig. 2.17 (b)]

l 4 ; j 4

n 3 l 1 2j

3 4 1 2 4 1

This is constrained motion and one variable (input) is required,provided it should satisfy the Grashof’s law to have continuous relativemotion between the two links.

Note: Grashof’s law states that the sum of the shortest and longest linklength should not be greater than the sum of the remaining two links length,if there is to be continuous relative motion between the two links.

4. Five links, five joints: [Fig. 2.17 (c)]

l 5; j 5

n 3 l 1 2 j

3 5 1 2 5 2


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This is unconstrained motion since it requires 2 input. Also, it shouldsatisfy the Grashof’s law for being movable.

At the outset, a kinematic chain is said to be movable, when its DOF

is one (or) greater than 1 n 1. Otherwise, it is locked (when n 0)

2.13 DIMENSIONAL SYNTHESIS

It is used to determine the dimensions of parts - lengths and angles -necessary to generate mechanism to obtain desired motion.

Generating a cam profile to obtain follower motion is dimensionalsynthesis.

Dimensional synthesis determines

The distance between hinge pins

Length of links

Angle between adjacent links

Cam profile dimension

Radius of roller follower

Gear ratios

Eccentricities and so on.

Dimensional synthesis is used to solve the following problems in ascientific way.

1. Path generation – Guiding along a specified path withprescribed timing.

2. Function generation – Coordinating the positions of inputand output links.

3. Motion generation (Rigidbody Guidance)

– Guiding the rigid body through anumber of prescribed positions.

2.13.3 Motion Generation (or) Rigid body guidance

By using motion generation, the rigid body is guided through apredetermined sequence of motion. For example, in bulldozer, the bucketsshould be moved by bucket loader mechanism, while the top of bucket should


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be guided to follow a path to have a scooping trajectory followed by liftingand dumping trajectory.

The path generation is concerned with the path of a tracer point, whilethe motion generation is concerned with the guidance of the entire motion ofrigid body.

Hence, Chebyshev’s spacing called best spacing of precision points in

the range xs x xf is given by

xj 12

xs xf 12

xf xs cos 2j 1

2n

Problem 2.21: y x1.7 in the range of 1 x 4. Determine the dependentvariable y with Chebyshev’s spacing with 3 number of precision points.

Solution

xs 1 ; xf 4

y x1.7 ; 180

We know xj 12

xs xf 12

xf xs cos 2j 1

2n

First precision point x1 12

1 4 12

4 1 cos 180 2 1 1

2 3

x1 2.5 1.299 1.2

Second precision point x2 12

1 4 12

4 1 cos 180 2 2 1

2 3

x2 2.5

Third precision point x1 12

1 4 12

4 1 cos 180 6 1

6

2.5 1.29

x3 3.79

The precision points x1, x2 and x3 can be found out by Graphical

method and discussed in section 2.15.


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For the function, y x1.7

y1 x11.7 1.21.7 1.36

y2 x21.7 2.51.7 4.75

y3 x31.7 3.791.7 9.63

2.17 COUPLER CURVES

Coupler curve is a curve (or) path traced out by a point located in theplane of the coupler link. Coupler link connects the input and output linksthrough turning pairs. When the input link rotates, any point on couplergenerates a path (or) curve called coupler curve.

++

A

L ink 2

L ink 3

L ink 4

B

C o up le r(o r)

C o up le r link

O 2 O 4

L ink 1Fig. 2.20


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2.19 GRAPHICAL SYNTHESIS - LINKAGE SYNTHESIS PROBLEM

2.19.1 Two position synthesis - Overlay method - Problems

Problem 2.22: Rocker output - Two position with Angular Displacement(Function): Design a four bar Grashof crank-rocker speed motor input to

give 45 of rocker motion with equal time forward and back, from a constant

speed motor input.

1. Draw the output link O4 B1 given length (3

cm) and O4 B2 given length (3 cm) both

extreme positions, in any convenient location,such that the desired angle of motion

4 45 is subtended.

L ink 1

L ink 3

L ink 3

L ink 2

A 1

A 2

B 2

B 1

L ink 4

O 4

O 2

+

(R ocker)

45 o

Step 1:

3cm3cm

45o

B2B 1

O 4

4


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2. Draw the chord B1 B2 and extend it.

3. Select a convenient point O2 on the line B1 B2 extended so that

O4 O2 6 cm (your choice)

4. Bisect line segment B1 B2 to get crank radius and draw a circle of that

crank radius about O2. Mark A1 and A2 on intersection points of circle

and B1 B2 extended. ie B1 B2 A1 A2

O 2

6 cm (YO U R CH OIC E)

3cm3cm

45o

B2B 1

O 4

4

Step 3:

3cm3cm

45 o

B 2B 1

O 4

4

Step 2:


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5. Label the two intersection of the circle and B1 B2 extended, A1 and

A2.

2. Measure the length of the coupler A1 B1 or A2 B2 6 cm.

7. Measure ground length 1 (Link 1), crank length 2, and rocker length 4is already given as 3 cm.

Result:

Link 1 O2 O4 2.59 cm

Link 2 = Crank length = 1.134 cm

Link 3 A1 B1 A2 B2 6 cm

Link 4 = 3 cm 8. Find the Grashof condition. If non-Grashof, redo steps 3 to 8 with O2

further away from O4. Grashof’s law states that the sum of the shortest

and longest link lengths should not be greater than the sum of the remainingtwo link length, if there is continuous relative motion between the twolinks.

Sum of shortest and longest lengths 1.134 2.59 7.724 cm

7.724 < Sum of remaining two link length 6 3 9 cm7.724 < 9 cm Hence satisfied.

A 1A 2

O 2

6 cm

3cm 3cm

45o

B 2B 1

O 4

4

Step 4:CrankRadius


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9. Make model of the linkage and check its function and transmissionangles.

Problem 2.23: Rocker Output - Two Position with Complex Displacement(Motion): Design a four bar linkage to move link CD (3 cm) from C1 D1 to

C2 D2.

1. Draw the link CD 3 cm in its two desired positions, C1 D1 and

C2 D2 in plane as shown.

Link 1

Link 3

Link 3

Link 2

A1

A2

B2

B 1

Link 4

O 4

O 2

+

A 1

A 2

O 2

L ink 2

L ink 1

L ink 4

C 1

L ink 3 B 1 B 2

C 2

D 1

D 2

O 4 R o topo le


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2. Draw construction line from point C1 to C2 and from point D1 to D2.

3. Bisect line C1 C2 and line D1 D2 and extend their perpendicular bisectors

to intersect at O4. Their intersection is known as the rotopole.

C 1

D 1

D 2

C 2

3 cm

3 cm

Step 1

C 1

D 1

D 2

C 23 cm

3 cm

Step 2

O 4R otopole

C 1

D 1

D 2

C 23 cm

3 cm

Step 3


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4. Joint O4 C1 and O4 C2. Joint Select a convenient radius and draw an arc

about the rotopole to intersect line O4 C1 at B1 and line O4 C2 at B2.

5. Draw the chord B1 B2 and extend it.

2. Select a convenient point O2 (Take B1 O2 6 cm) on the line B1 B2

extended.

7. Bisect line segment B1 B2, draw a circle of dia B1 B2 about O2.

8. Mark the two intersection of the circle and B1 B2 extended as A1 and A2.

9. Measure the length of the coupler (Link 3) A1 B1 or A2 B2. [7.45 cm

by measurement]

10. Measure ground length 1 (Link 1), crank length 2 (Link 2), and rockerlength 4 is already given.

A1 B1 (Link 3) 7.45 cm

Link 1 8.19 cm

Link 2 1.42 cm

Link 4 3 cm Given

B 1B 2

O 4R otopole

C 1

D 1

D 2

C 23 cm

3 cm

Step 4


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11. Find the Grashof condition. Sum of shortest and longest link length

1.42 8.19 9.61 cm. 9.61 < sum of remaining link length

[7.45 3] 10.45 cm. Hence Grashof condition is satisfied. Ifnon-Grashof, redo steps 3 to 11 with O2 further away from O4.

12. Make a model of the linkage and articulate it to check its function andits transmission angles.

B 1B 2

O 4R o topo le

C 1

D 1

D 2

C 2

3 cm

3 cm

Step 5

O 2A 1 A 2

L ink 1

C rank rad ius

(Link 2 )

A 1

A 2

O 2

L ink 2

L ink 1

L ink 4

C 1

L ink 3 B 1 B 2

C 2

D 1

D 2

O 4 R o topo le


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Problem 2.24: Coupler output – 3 position with Complex Displacement

(Motion): Design a four bar linkage to move the link CD 3 cm fromposition C1 D1 to C2 D2 and then to position C3 D3. Moving pivots are C

and D. Find the fixed pivot locations.

1. Draw link CD 3 cm in its three positions C1 D1, C2 D2, C3 D3 in the

plane as shown. C1 D1 at 30 ; C1 C2 3.3 cm C1 D2 15 ;

C3 D3 35 as assumed problem.

L ink 2

L ink 3

L ink 4

L ink 1 = O O2 4

O 2

2

2

3

3C 2

C 3

D 3D 2

O 4

44

C 1

D 1

C 1

D 1

D 2

C 2

D 3

C 3

30o

3 .3 cm

15o

1.0

9 cm 0 .32 cm

35o

Step 1


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2. Draw construction lines from point C1 to C2 and from C2 to C3.

3. Bisect line C1 C2 and line C2 C3 and extend their perpendicular bisectors

until they intersect at O2.

4. Similarly, join D1 D2 and D2 D3 by construction lines. Bisect D1 D2 and

D2 D3 and extend their perpendicular bisectors to intersect at O4.

C 1

D 1

D 2

C 2

D 3

C 3

Step 2

O 2

C 1

D 1

D 2

C 2

D 3

C 3

Step 3

+


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5. Connect O2 with C1 to get link 2. Connect O4 with D1 to get link 4.

Three links of different positions are shown in 3 different lines.

O 2

C 1

D 1

D 2

C 2

D 3

C 3

Step 4

+

+O 4

L ink 3

L ink 4

L ink 2

L ink 1 = O O2 4

O 2

C 1

D 1

D 2

C 2

D 3

C 3

Step 5

+

+O 4


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6. Line C1 D1 is link 3. Line O2 O4 is link 1 (Fixed link).

By measurement:O2 C1 Link 2 = 5.22 cm

C1 D1 Link 3 = 3 cm (given)

O4 D1 Link 4 = 3.86 cm

O2 O4 Link 1 = 9.36 cm

7. Check the Grashof condition. Note that any Grashof condition ispotentially acceptable in this case.

8. Construct a model and check its function to be sure it can get frominitial to final position without encountering any limits positions.

Case study - Quick – Return Mechanism

Time ratio TR defines the degree of quick-return of the linkage.

TR

; 360

|180 |

|180 |

Works well for time ratios down to about 1.5

Link 2L ink 3

L ink 4

L ink 1 = O O2 4

O 2

2

2

3

3C 2

C3

D3D2

O 4

44

C 1

D 1


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Problem 2.25: Synthesise a four-bar, mechanism to guide a rod AB throughthree consecutive positions A1 B1, A2 B2 and A3 B3 as shown in Fig.

0 2 4 6 8 10 12

2

4

6

B1

A 1

A 2

B 2

A 3B3

(2 ,6)

(2 ,0)

50 o

(5 ,2)

(6 ,6)

(12 ,6)

Step:1: D raw 3 posit ions o f Gu ide rod(Coupler L ink 2)

X

Y G raph ica l Synthesis fo r M otio n Generat ion(R ig id body G uidance) - 3 P osition Synthesis

0 2 4 6 8 10 12

2

4

6

B 1

A1

A 2

B 2

A3B 3

(2,6)

(2,0)

50o

(5,2)

(6,6)

(12,6)

M id Norm al Of B2 B3

Mid

Nor

ma l

Of A

2 A

3

Mid

Nor

mal

Of A

1 A

2

O B

Mid N

ormal O

f B1 B2

Step 2 : Draw M id - Norm als o f A1A2 and A2A3 to meet at S imila rly draw Mid - Norm als of B1B2 and B2B3 to meet at

O .

O

A

B

O A


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Draw mid-normals of A1 A2 and A2 A3 to meet at OA. Similarly, draw

mid-normals of B1 B2 and B2 B3 to meet at OB.

Step 3: Now OA and OB are the required fixed supports.

Step 4: The four-bar mechanism OA A1 B1 OB represents the required 4-bar

mechanism with coupler link A1 B1 capable of taking positions A2 B2 and A3 B3.

0 2 4 6 8 10 12

2

4

6

B 1

A 1

A 2

B2

A3B3

(2,6)

(2,0)

50o

(5,2)

(6,6)

(12,6 )

M id Normal Of B2 B3

M id Norm al O f B1 B2

Mid

Nor

mal

Of A

2 A

3

Mid

Nor

mal

Of A

1 A

2

O A

O B

Fixed L ink 1

L ink 2

Couple r L ink 3

Lin k 4

Length o f Couple r L ink 3 = 6 cm

Length o f Fixed Link 1 = 17 .9 cm

Length o f Input L ink 2 = 5 .3 cmLength o f Output L ink 4 = 18 cm

Given:

By M easurem ent:

Graphical Synthesis for M otion G enera tion(Rigid body Guidance) - 3 Position Synthesis


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Problem 2.26: Synthesise a mechanism to move AB successively throughpositions 1, 2 and 3 as shown in Fig.

0 2 4 6 8 10 12

2

4

6B 1

A 1

A 2

B 2

A 3

B 3

(2,6)

50o

(8,6)

(12,6)

Step:1: Draw 3 positions of Guide rod(C oupler Link 2) AB

X

Y Graphical Synthesis for Motion Generation(Rigid body Guidance) - 3 Position Synthesis

(12,0)

(9,2)

PO SITION 1

POSIT ION 2

PO

SIT

ION

3

0 2 4 6 8 10 12

2

4

6B 1

A 1

A 2

B 2

A 3

B 3

(2,6)

50o

(8,6)

(12,6 )

X

Y

G raph ical Synthesis fo r M otion G enera tion(R igid body G uidance ) - 3 Position Syn thesis

(12,0 )

(9,2)

OB

Step 2: D raw M id - N orm als of A1A2 and A2A3 to m eet a t S im ila rly d raw M id - N o rm als of B1B2 and B2B3 to m eet a t

O .

O

A

B

OA

Mid

Nor

mal

Of B

2 B3M id N orm al

O f B1 B2

Mid

Nor

ma

l Of

A1

A2

Mid

No r

ma

l Of A

2 A

3


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SolutionAs in the previous example mark point of intersection of mid-normals

of A1 A2 and A2 A3 to locate fixed pivot OA. Similarly, mark point of

intersection of mid-normals of B1 B2 and B2 B3 locates the pivot OB. Then

OA A1 B1 OB is the required 4-bar mechanism which can guide AB through

the three prescribed positions.

0 2 4 6 8 10 12

2

4

6B 1

A 1

A2

B 2

A 3

B3

(2,6)

50o

(8,6)

(12,6 )

X

Y

(12,0 )

(9,2)

O B

O A

Lin k 2

Coup le r L ink 3 Link 4

Length of Coup le r L ink 3 = 6 cm

Length of Fixed Link 1 = 17 .8 cm

Length of L ink 2 = 18.3 cmLength of L ink 4 = 5 .34 cm

Given:

By M easurem ent:

Fix

ed L

ink

1

G raphical Synthesis for M otion Genera tion(Rigid body Guidance) - 3 Position Synthesis


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UNIT III

KINEMATICS OF CAM MECHANISMS

Classification of cams and followers - Terminology and definitions -Displacement diagrams - Uniform velocity, parabolic, simple harmonic andcycloidal factions - Derivatives of follower motions - Layout of plate andcam profiles - Specified contour cams - Circular arc and tangent cams -pressure angle and undercutting - sizing of cams.

3.1 INTRODUCTIONA cam is a mechanical element which drives another element called

the follower, through a reciprocating (or) oscillating motion. The cam andfollower have direct line contact and constitute a higher pair.

3.2 CLASSIFICATION OF CAM AND FOLLOWERSDepict the types of cam. (AU. May/June 2016)

3.2.1 According to Cam Shape

Cams are classified according to their basic shapes. Fig. 3.1 showsdifferent types of cam.

(a )

(b )

To and fro motion

To and fro motion

Types of cam s (a) plate cam (b) wedge cam (or)

Radial cam with Translating follower.

Rolle r

Fig 3.1

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1. Fig 3.1(a) A plate cam, also called a disk cam (or) a radial cam.

2. Fig 3.1(b) A wedge cam

3. Fig 3.1(c) A cylindric cam (or) barrel cam

4. Fig 3.1(d) An end (or) face cam.

Among the four types, the most commonly used type is the plate cam.


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3.2.2 According to Shape of the follower

Cam systems can also be classified according to the basic shape ofthe follower as follows.

3.2.2.1 According to the surface in contact (Shape of the follower)

1. Knife edge follower

2. A flat face follower (or) mushroom follower.

3. A roller follower

4. A spherical-face (or) Curved shoe follower

1. Knife-edge follower: The end of follower has sharp knife edge. Sinceit results in excessive wear, this type of follower is normally not used.

2. Flat-faced (or) mushroom follower: The end of the follower has aflat face. In this system, the side thrust between the follower and guideis greatly reduced. The wear can be reduced by off-setting the axis of thefollower.

3. Roller follower: The end of the follower has a roller. In this system,the rate of wear is greatly reduced. The roller followers are extensively used.

4. Spherical faced followers: The end of the follower has spherical shape.To reduce the high surface stresses, the spherical force followers are used.

3.2.2.2 According to the motion of the follower

The followers according to its motion, are classified as follows.

(b)

y

(a)

y

A recip rocating flat- face fo llower

(c.)

y

An oscilla ting ro ller

follow er

(d)

y

An oscilla ting curved-shoe

fo llower(Spherical follower)

Fig. 3.2

www.srbooks.org Kinematics of Cam Mechanisms 3.3

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1. Reciprocating (or) translating follower: If the follower hasreciprocating motion, when the cam rotates, then the follower is known asreciprocating (or) translating follower. The Fig 3.2(a) and (b) show thereciprocating (or) translating follower.

2. Oscillating (or) rotating follower: The follower oscillates, when thecam rotates with uniform speed. Fig. 3.2 (c).

3.2.2.3 According to the path of motion of the follower

1. Radial follower: If the follower reciprocates along the axis passingthrough the centre of the cam, then it is known as radial follower. Fig 3.2(b)is the radial follower.

2. Off-set follower: If the follower reciprocates along the axis away from(offset from) the axis of the cam centre, then it is known as off-set follower.Fig 3.2(a).

3.3 TERMS USED IN RADIAL CAMS

1. Base circle: The smallest circle that can be drawn to the cam profileis known as base circle. By using minimum radius of the cam, we can drawbase circle.

2. Trace Point: It is a point corresponds to the end point of a knife edgefollower or centre of the roller follower (or) on the surface of a flat-faced follower.

3. Pitch curve: The trace point is used to generate the pitch curve, Forknife edge follower, the pitch curve and cam profile are identical (Both aresame). But for a roller follower, they are separated by a radius of the roller.

4. Prime circle: It is the smallest circle that can be drawn from the centreof the cam and tangent to the pitch curve.

Prime circle radius Base circle radius for knife edge and flat face follower.

Prime circle radius Base circle radius radius of roller for roller follower.

5. Lift (or) Stroke: It is the maximum travel of the follower from itslowest position to the topmost position. The maximum rise is called lift.

6. Pitch point: It is a point on the curve having maximum pressure angle.

7. Pitch circle: It is the circle drawn from the centre of the cam throughthe pitch points.


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8. Pressure angle: It is the angle between the follower motion and anormal to the pitch curve. This is very important angle for designing a camprofile. If the pressure angle is more, then the reciprocating follower will jamin its bearings.

3.4 MOTION OF THE FOLLOWER

The follower excutes any one of the following motion.1. Uniform velocity motion2. Simple harmonic motion3. Uniform acceleration and retardation motion4. Cycloidal motion

Cam pro file

Dwell angle

L

Kn ife edgefo llower

ABCD

E

F

G

Basecirc le

HJ

K

M

N

6�

P5 �

4 �

3 �

2 �

1 �0T

S

123

4

5

6

0

O

O ut Stroke angle

Return S troke ang le

E

F

G

Pitch curve

012

3

4

5

6

0 � 1 �

2 �

3 �

4 �

5 �

6 �

ABCD

HJ

K

L

M

N

P

Roller follower

Base

c ircle

Prim ecirc le

camprofile

O

O ut Stroke angle

Dwell

Re turn S troke ang le

Fig 3.3

Fig 3.4


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During the rotation of the cam through one cycle of input motion, thefollower executes a series of events as shown in displacement diagram.

The abscissa represents one revolution of the cam and is drawn toany convenient scale.

The ordinate represents the follower travel y. When the follower ismoving away from the cam centre, it is known as Outward stroke. Whenthe follower is at rest, it refers to dwell period.

When the follower is moving towards the cam centre, it is known asinward stroke (or) return stroke.

3.5 DISPLACEMENT DIAGRAM, VELOCITY AND ACCELERATIONANALYSIS WHEN THE FOLLOWER MOVES WITH UNIFORM VELOCITY

A

B 1 C 1

B C D

E

Rise Dwell Return Dwell

O ne revolution of camAngu lar d isplacement

Vel

ocity

Acc

eler

atio

nD

ispl

acem

ent

(a) Displacem ent diagram

(b) Velocity diagram

(c) Acceleration diagram

Fig 3.5 when the follower moves withuniform velocity

y


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Since the follower moves with uniform velocity during its outward andreturn stroke, therefore, the slope of the displacement curve is constant. Thefollower remains at rest during dwell periods. The velocity of the followerincreases, gradually to its maximum value at the beginning of each strokeand decreasing gradually to zero at the end of each stroke.

In this case, the acceleration (or) retardation of the follower at thebeginning and at the end of stroke is infinite.

3.6 DISPLACEMENT DIAGRAM, VELOCITY AND ACCELERATIONANALYSIS WHEN THE FOLLOWER MOVES WITH SHM (SIMPLE HARMONIC MOTIONS)

Draw the displacement, velocity and acceleration curves, when the followermoves with simple harmonic motion and derive the expression for maximumvelocity and maximum acceleration. (AU. May/June 2016)

The displacement diagram is drawn as followsAssume the motion of the follower during both outstroke and return

stroke is Simple Harmonic Motion.

Lift of the follower 25 mm

outstroke 120; Dwell 60Return stroke 90; Dwell 90

1. Draw a rectangle of length equal to 90 mm representing 360 (Take

scale 1 mm 4 i.e. 90 mm 360

2. Divide this rectangle in such a way that

Outstroke 120 so 120

4 30 mm

Dwell 60 so 604

15 mm

Return stroke 90 so 904

22.5 mm

Dwell 90 so 904

22.5 mm

Total 360 Total 90 mmas shown in Fig. 3.6(a).

3. Draw a semi circle with its diameter equal to total lift (or) stroke ofthe follower at the left end of the rectangle.


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4. Divide the semicircle into any number of equal parts (Take 6 equal parts)and mark a, b, c, d, e, f and g.

5. Divide the outstroke 120 30 mm into same number of (6) equal

parts. Similarly, divide the return stroke 90 22.5 mm into samenumber of (6) equal parts.

6. Mark the number 0, 1, 2, 3, 4, 5 and 6 in outstroke and

0, 1, 2, 3, 4 5 and 6 in return stroke.

7. From point b, draw horizontal line to cut the line 1 and get B. Sincereturn stroke is also SHM, extend this horizontal line to cut the line

5 at N.

0 1 2 3 4 5 6a

b

c

d

e

BC

D

EF

G HJ

K

L

MN

P

A

fg

120o90o60o 90o

O utstroke D we ll R e turn s troke D we ll

Lift

=2

5m

m

S T

6 �5 �4 �3 �2 �1 �0 �

Fig 3.6


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Problem 3.5: A disc cam used for moving a knife edge follower with SHMduring lift and uniform acceleration and retardation motion during return.Cam rotates at 300 rpm clockwise direction. The line of motion of the followerhas an offset 10 mm to the right angle of cam shaft axis. The minimum radiusof the cam is 30 mm. The lift of the follower is 40 mm. The cam rotation

angles are: lift 60, dwell 90, return 120 and remaining angle for dwell.Draw the cam profile and determine the maximum velocity and accelerationduring the lift and return. (AU. Nov/Dec 2009)

Solution:

The displacement diagram

1. Draw a rectangle of length equal to 90 mm (360).

2. Divide the rectangle in such a way outstroke 15 mm 60), dwell 22.5

mm 90, return 30 mm 120), dwell 22.5 mm 90)

3. Here lift is with simple harmonic motion and return stroke is withuniform acceleration and retardation.

To draw profile of cam

1. Draw the base circle with radius equal to the minimum radius of cam(30 mm) with O as centre.

2. Draw a offset circle of radius 10 mm concentric with the base circle. Nowdraw a vertical axis line tangent to the offset circle. This vertical axis lineis the axis of follower.

3. The axis of follower cuts the base circle at A.

4. Join OA. Mark angle AOS 60 to represent lift, angle SOT 90 to

represent dwell and angle TOP 120 to represent return stroke.

5. Divide the outstroke AOS into six equal parts as in displacement diagramand mark 0, 1, 2, 3, 4, 5 and 6 and similarly divide return stroke.

6. Now draw tangents from all the points.

7. Join all the points ABCDEFGHJKIMNPA by smooth curve. This curveis the profile of cam.

N 300 rpm; 2 300

60 31.42 rad/s


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Maximum velocity VO S

2 o

31.42 0.04

2 60

180

1.8852 m/s

K

M

N

L

H I

B

C

D

E

F

G

Ab

c

d

e

fg

1 2 3 4 5 6

L ift60o

D w ell90 o

R e turn120 o

D w ell90o

O

S H M U A & R

60o

90o

90o

120o

G

F

E

D C BA

6

5

43

2 1

H J

K

L

M

N

P

S

O

T


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Maximum acceleration

ao 2 2 S

2 o2

2 31.422 0.04

2 60

180

177.7 m/s2

VR 2 SR

2 31.42 0.04

120

180

1.2001 m/s

aR 4 2 S

R2

4 31.422 0.04

120

180

2 36.0093 m/s2

3.12 CAM PROFILE WITH ROLLER FOLLOWER

Problem 3.6: Draw the profile of a cam operating a roller reciprocatingfollower with following data.

Minimum radius of the cam 25 mm

Lift 30 mm

Roller diameter 15 mm

The cam lifts the follower for 120 with SHM followed by a dwell period of

30. Then follower lowers down during 150 of the cam rotation with uniformacceleration and retardation followed by a period of dwell.If the cam rotates at a uniform speed of 150 rpm, calculate the maximumvelocity and acceleration during ascent. (AU. Apr/May 2011)

Solution:

Draw displacement diagram as shown in fig.

Here outstroke is with Simple Harmonic motion.

Return stroke is with uniform acceleration and Retardation motion.

Draw profile of the cam as follows

1. Draw a base circle of radius equal to minimum radius of the cam (25mm) with O as centre.


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120 o 150 o

30o

60o

B

C

D

E

FG H

J

K

L

M

NP

A 1 2 3 4 6 0 � 1 � 2 � ��

b

c

d

e

fg

a��

30m

m

outstroke Return stroke

Dw

ell0

01

2

3

4

5

6

0 �1 � 2 �

3 �

4 �

5 �

6 �

ABC

D

E

F

G

H

J

K

L

M

N

PPitch curve

Ro lle r fo llower

120

o

150o

30 o

Base

c irc le R25

Prim ecircle

camprofile

Ro lle r d ia

O

S

T


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2. Draw a prime circle concentric with base circle and radius,

radius OA Minimum radius of cam radius of roller.

25 152

32.5 mm

3. Draw outstroke angle AOS 120, SOT 30, return stroke

angle TOP 150.

4. Divide the angle AOS and TOP into six equal parts as in displacementdiagram.

5. Join the points 1, 2 ... 6 with O and also join the points 0, 1 6 withcentre O and produce these lines beyond the prime circle as shown in fig.

6. Now transfer the lengths 1B, 2C 5N and 6P from displacementdiagram.

7. Join the points A, B N, P, A by smooth curve (using centre line) torepresent pitch curve.

From points A, B, C N, P draw circles with radius equal to the radiusof roller.

Join the bottoms of the roller circles with smooth curve as shown infig. This curve is known as the required profile of the cam.

N 150 rpm ; 2 150

60 15.71 rad/s

Maximum velocity of the follower during ascent (with SHM)

V0 S

20

15.71 0.03

2 120

180

0.353 m/s

Maximum acceleration of the follower during ascent (with SHM)

a0 22S

2 02

2 15.712 0.03

2 120

180

2 8.33 m/s2


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Problem 3.7: Draw the profile of a cam which will lift a roller followerthrough 40 mm. The diameter of the roller is 30 mm and the line of strokeis offset by 20 mm from the axis of the cam. The minimum radius of the camis 50 mm. The outstroke of the follower is to take place with SHM during

90 of can rotation followed by a dwell period of 30. Follower is to returnto its initial position with uniform and equal acceleration and retardation

during 60 of cam rotation and then dwell for the remaining period of

180. (AU. Apr/May 2012)

Solution:

Draw the displacement diagram as shown in fig.

To draw the profile of the cam when the line of stroke is offset by20 mm from the axis of the cam shaft.

1. Draw base circle of radius equal to 50 mm with centre O.

2. Draw a prime circle of radius

OA base circle radius radius of the roller.

OA 50 15 65 mm

The prime circle should be concentric with the base circle.

3. Draw an offset circle of radius equal to 20 mm with centre O.

4. Join OA. From OA, draw angle AOS 90and angle

SOT 30 and angle TOP 60.

5. Divide the angle AOS into six equal parts and mark 0, 1, 2 ... 6 andangle TOP into six equal parts as in displacement diagram.

6. From points 1, 2, 3 6 and 0, 1 6 on the prime circle, drawtangents to the offset circle and produce these tangents beyond the primecircle.

7. Now transfer the lengths 1B, 2C 5N,6P from displacement diagram.

8. Join the points A, B, ,GH, NP by smooth curve (using centre line)which represents the pitch curve.

9. Now draw circles with radius equal to the radius of the roller, keepingA, B, C P as centers.


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40A

O

123

4

5

6

0 �

1 �

2 �

3 �

4 � 5 � 6 �

BCD

E

F

G

H

J

K L M NP

Camprofile

Bas

e c i

rcle

3 0 o

60o

90o

20m m

O ffse t

cir

cle

B

C

D

E

FG H

J

K

k

m

NA

b

c

d

e

fg

a

SHM

Mn

j

Ll

90o1 2 3 4 5 6

o 60o 180o

S T

30o

30

T

S

R65

R50

Prim

e c i

rcle


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10. Join the bottoms of the circles with a smooth curve as shown in fig.This curve is known as profile of the cam.

Problem 3.8: Draw the cam profile for the following data: base circle radius

of cam 50 mm, Lift 40 mm, Angle of ascent with SHM 90, Angle of

descent with uniform acceleration and deceleration 90, Speed of cam 300

rpm, Type of follower , Roller follower (with roller radius 10 mm).

(AU. Nov/Dec 2010)

Given data

Cam radius 50 mm ; Lift 40 mm

Angle 90, 90, 90 and 90

Displacement SHM Ascent ; UA & R Descent

Follower Roller; Roller radius 10 mm

Displacement Diagram

1. Draw a base circle with center O and radius equal to the minimumradius of the cam (i.e., 50 mm)

2. Draw a prime circle with centre O and radius,

Min. radius of cam (50 mm) + Roller radius (10 mm) = 60 mm

3. Draw angle AOB 90 to represent raising or out stroke of the valve,

angle BOC 90 to represent dwell and angle COD 90 to representlowering or return stroke of the valve.

4. Divide the angular displacements of the cam during raising and loweringof the valve (i.e. angle AOB and COD) into the same number of equaleven parts as in displacement diagram.

5. Join the points 1, 2, 3 etc. with the centre O and produce the linesbeyond prime circle as shown in Fig.

6. Set off 1a, 2b, 3c etc. equal to the displacements from displacement diagram.

7. Join the points a, b, c c, b, a. The curve drawn through these pointsis known as pitch curve.

8. From the points a, b, c c, b, a draw circles of radius equal to theradius of the roller.


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90o

90o

90o

Cam rad ius

Prim ecirc le

Rolle rFollower

Aa

bc

d

e

f

1

23

4

5

6B

D

a �

b �

c �

1�

2 �

3 �

4�5 � 6 �

Pitch C urvef�

d �

e�

C

O


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9. Join the bottoms of the circles with a smooth curve as shown in Fig.This is the required profile of the cam.

Problem 3.9: A cam operates on offset roller follower. The least radius ofthe cam is 50 mm, roller diameter is 30 mm and offset is 20 mm, the cam

rotates at 360 rpm. The angle of ascent is 48, angle of dwell is 42 and

angle of descent is 60. The motion is to be SHM during ascent and uniformacceleration and deceleration during decent. Draw the cam profile. (AU. Nov/Dec 2013)

Solution:

Assume Lift = 40 mm

Draw the displacement diagram as shown in fig.

To draw the profile of the cam when the line of stroke is offset by20 mm from the axis of the cam shaft.

1. Draw base circle of radius equal to 50 mm with centre O.

2. Draw a prime circle of radius

OA base circle radius radius of the roller.

OA 50 15 65 mm

The prime circle should be concentric with the base circle.

3. Draw an offset circle of radius equal to 20 mm with centre O.

4. Join OA. From OA, draw angle AOS 90and angle

SOT 30 and angle TOP 60.

5. Divide the angle AOS into six equal parts and mark 0, 1, 2 ... 6 andangle TOP into six equal parts as in displacement diagram.

6. From points 1, 2, 3 6 and 0, 1 6 on the prime circle, draw tangentsto the offset circle and produce these tangents beyond the prime circle.

7. Now transfer the lengths 1B, 2C 5N,6P from displacement diagram.

8. Join the points A, B, ,GH, NP by smooth curve (using centre line)which represents the pitch curve.


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Apr-2001

40A

O

123

4

5

6

0 �

1 �

2 �

3 �

4 � 5 � 6 �

BCD

E

F

G

H

J

K L M NP

Camprofile

Ba s

e c i

rcle

3 0 o

60o

90o

20m m

O ffse t

circ

le

B

C

D

E

FG H

J

K

k

m

NA

b

c

d

e

fg

a

SHM

Mn

j

Ll

90o1 2 3 4 5 6

o 60o 180o

S T

30o

30

T

S

R65

R50

Pr im

e c i

rcle


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Problem 3.24: In a symmetrical tangent cam operating a roller follower,the least radius of the cam 30 mm and the roller radius is 15 mm. The angle

of ascent is 75 and the total lift is 20 mm. The speed of the cam is 600

rpm. Calculate (i) The principal dimensions of the cam. (ii) Theacceleration of the follower at the beginning of lift, where straight flankmerges into the circular nose and at the apex of the nose. (AU. April/May 2017) (AU. Nov/Dec 2012) (AU. Nov/Dec 2011)

Solution:

Given data: r1 30 mm, r2 15 mm, 75,

Total lift 20 mm, N 600 rpm,

2 N60

2 600

60 62.84 rad/sec

r2 D

G

H

J

r3

KA

L

r

r1

EB

O

Fig:3.10 Tangent cam with reciprocating rollerfollower having contact with the nose

1


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Let r OK Distance between cam centre and nose centre

r3 Nose radius

Angle of contact of cam with straight flanks

r r3 r1 Total lift

30 20 50 mm

r 50 r3 OK

OE OP PE

r1 OP r3

OP r1 r3 30 r3

K

75o

r

r3

E

G

OP

B

D

r =30 m m1

R 15 mmJ

HL

r =152

PO K=75 O


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In OKP,

OP OK cos

30 r3 50 r3 cos 75 12.94 0.2588 r3

30 12.94 r3 0.2588 r3

r3 23.019 mm

r OK 50 r3 50 23.019 26.981 mm

ODB,

tan DBOB

KPOB

OK sin

r1 r2

26.981 sin 7530 15

30

Acceleration of follower at the beginning of the lift

amin 2 r1 r2 62.842 30 15

177698.95 mm/s2

amin 177.7 m/s2

Acceleration of follower where straight flank merges into a circularnose.

amax 2 r1 r2 2 cos2

cos3

177.7 2 cos2 30

cos3 30 341.98 m/s2

amax 341.98 m/s2


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Acceleration of the follower at the apex of the circular nose,

a 2 r 1

L2 r

L3

2 r 1

rr2 r3

L r2 r3

62.842 26.98 1

26.9815 23.016

182152.23 mm/s2

a 182.15 m/s2

Problem 3.25: A symmetrical tangent cam with a least radius of 25 mm

operates a roller follower of radius 10 mm. The angle of ascent is 60 andtotal lift is 15 mm. If the speed of the cam is 400 rpm, then calculate:The principal dimensions of cam i.e the distance between the cam centre to nosecentre; nose radius and angle of contact of cam with straight flank. (AU. Apr/May 2008)(AU. Nov/Dec 2009)

Given

r1 25 mm least radius of cam

r2 10 mm Roller radius

60 Angle of ascent

Total lift 15 mm

N 400 rpm

Solution:

2N/60 2 400/60 41.89 rad/s

Let r OK Distance between cam centre and nose centre

r3 Nose radius &

Angle of contact of cam with straight flanks

From the geometry of the figure,


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r r3 r1 Total lift 25 15 40 mm

r 40 r3 OK

and OE OP PE

r1 OP r3

OP r1 r3 25 r3

OP 25 r3 ...(i)

From Right angled triangle OKP,

OP OK cos r cos

OP 40 r3 cos 60 ... (ii)

Now equating (i) & (ii)

25 r3 40 r3 cos 60 40 r3 0.5

25 r3 20 0.5 r3

25 20 r3 0.5 r3

R adius 10

J

HL

D

K

O

60 o

r

P

E

B

25 m m

r 3

r2

r =1

PO K=60 O


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0.5 r3 5

r3 5

0.5 10 mm

Nose radius 10 mm

and r 40 r3 40 10 30 mm

OK r Distance between centres of cam and nose 30 mm Ans

Now, From right angled triangle ODB,

tan DBOB

KPOB

OK sin

r1 r2

30 sin 60

25 10 0.74

tan 1 0.74

36.6

3.18 CIRCULAR ARC CAM

3.18.1 Circular arc cam with flat faced follower:

The Fig. 3.11 shows the circular arc cam with flat-faced reciprocatingfollower. If the flanks of the cam connecting the base circle and nose curvesof convex circle, then the cam is known as circular arc cam.

Let r1 and r2 be the base circle radius and nose circle radius, and R

be the radius of the circular flank. Then O and Q to be the centres of basecircle and nose circle respectively.

We consider the following two cases in circular Arc cam:

1. When the flat face of the follower has contact on the circular flank,and

2. When the flat face of the follower has contact on the nose.


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3.20 SIZING OF CAMS

The determination of cam size is related to the determination of thebase circle of the cam. Nearly in every application the size of the cam beingused is minimized. Large cams are not used mostly due to the followingreasons:

1. More space is required.

2. Unbalanced mass increases.

3. Follower has a longer path to follow for each cycle. Hence the angularvelocity of the follower and the surface velocity increases.

However if we decrease the cam size, the following factors take intoeffect:

1. The force transmission characteristics deteriorate. The steepness of camprofile increases.

2. The curvature of the cam profile decreases (sharp curves).

3. Strength requirements due to the forces and moments are acting on thecam.

In practice, the cam size is determined by considering two factors:Pressure angle and Minimum radius of curvature.

3.21 PRESSURE ANGLE

It is a measure of the steepness of the cam profile. It is the anglebetween the normal to the pitch curve at a point and the direction ofthe motion of the follower.

Pressure angle varies from maximum to minimum at all instantsof the follower motion.

If the pressure angle is higher, then the side thrust is also higherwhich jams the follower in its guide ways.

Pressure angle should be as low as possible based on thelimitations of design.

Pressure angle should be less than 45 for low speed cam with

oscillating followers and should be less than 30 for cams withtranslating followers.


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Pressure angle is reduced by increasing the size of the cam (or)by adjusting the follower offset.

In an offset follower, the pressure angle reduces during rise of thefollower but at the expense of increased pressure during the returnstroke.

3.22 UNDERCUTTING

The cam profile must be continuous curve without any loop. If thecurvature of the pitch curve is too sharp, then the part of the cam shapewould be lost and thereafter the intended cam motion would not be achieved.Such a cam is said to be undercut. Undercutting occurs in the cam becauseof attempting to achieve too great a follower lift with very small cam rotationwith a smaller cam.

4

3 Pitch Curve

y

y0

1

2

O

Prim e Circle

l34

l23

l23

l24

e

l14 at

Fig. 3.16 Determination of Pressure Angle

Cam Pressure Angle

Fo llower Rod

Ro lle r Fo llower

Fixed Link


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Undercutting can be prevented in cam

By decreasing the follower lift.

By increasing cam rotation angle.

By increasing the cam size (i.e., Base circle).

By selecting a large base circle radius or (if strength conditionspermit) reduce the radius of the roller.

Consider the cam profile shown in Fig. 3.17. According to therequirements of cam motion, with the selected roller radius and base circleradius, the positions of the roller are A, B, C, D and it is not possible to drawa smooth curve that is tangent to all the roller circles. With Profile #1, thecam profile is tangent to the circles positions A, C and E and with profile #2 the cam profile is tangent to positions A, B and E of the roller. There isno cam profile that will be tangent to all the positions of the roller. This isknown as “undercutting” and it occurs whenever the radius of curvatureof the cam profile is less than the radius of the roller.

Pro file 1

Pro file 2

AB

C

D

E

Fig: 3.17


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UNIT IV

GEARS AND GEAR TRAINS

Law of toothed gearing - Involutes and cycloidal tooth profiles - SpurGear terminology and definitions - Gear tooth action - contact ratio -Interference and undercutting. Helical, Bevel, Worm, Rack and Pinion gears[Basics only]. Gear trains - Speed ratio, train value - Parallel axis geartrains - Epicyclic Gear Trains.

4.1 INTRODUCTION

When the positive drive (i.e without slipping) is required for someprecision machines, and if the distance between the driver and follower is verysmall, the gears (or) toothed wheels are used.

Driver Fo llower

+

Line o f con tact

Shafts

+

Driver Fo llower

Fig:4.1 A Pair of Gear

Pitch C irc le

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Gears are used for transmitting motion and power from one shaft toanother with constant velocity ratio. When the Driver Gear is rotated by aninput shaft, it will rotate the follower in the opposite direction as shown inFig. 4.1.

4.4 TERMINOLOGY OF SPUR GEARS AND DEFINITIONS

Explain gear nomenclature with neat diagram and define all salient termspertaining to the gear. (Spur gear) (AU. May/June 2016)

The terminology of spur gear tooth is illustrated in Fig. 4.5.

The pitch circle is a theoretical circle on which all calculations areusually based. The pitch circle is an imaginary circle. The pitch circles of apair of mating gears are tangent to each other.

Pitch circle diameter: The diameter of the pitch circle is known as pitchcircle diameter. The size of the gear is usually specified by the pitch circlediameter.

Pinion: Pinion is a smaller of the two mating gears.

Gear (or) wheel: The larger of the two mating gears is called the gear (or) wheel.

Face w

idth

Top la

nd

Face

F lank

Bottom

land

C ircularpitch

Toothth ickness

w id th o fspace

(o r)tooth space

Dedendum

Addendum

Fillet rad ius

c learance c irc le

c learance Dedendum circle

P itch surfacee lement

F ig.4.5. Gear Tooth Terminology


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Pitch point: It is a common point of contact between the two pitch circles.

Pressure angle (or) Angle of obliquity :

It is the angle between the common normal to two gear teeth at thepoint of contact and the common tangent at the pitch point. The standard

pressure angles are 14 12 and 20.

Addendum: It is the radial distance between the top land and pitch circle.

Dedendum: It is the radial distance from the bottom land to the pitch circle.

Whole depth (or) Total depth: It is the sum of addendum and dedendum.

Addendum circle: It is the circle drawn through the top of the teeth andconcentric with the pitch circle.

Dedendum circle: It is the circle drawn through the bottom of the teeth(or) root circle.

Dedendum circle dia or

root circle dia Pitch circle dia cos

Circular pitch pc; It is the distance measured on the pitch circle from a

point on one tooth to the corresponding point on the adjacent tooth.

Circular pitch, pc DT

where, D Dia of pitch circle in mm.

T Number of teeth on wheel

Diametral pitch pd:

It is the ratio of number of teeth to the pitch circle diameter.

Diametral pitch pd TD

pc

Module (m): It is the ratio of the pitch circle diameter to the number ofteeth (or) reciprocal of diametral pitch.

m DT

1

Pd

www.srbooks.org Gears and Gear Trains 4.3

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Clearance circle: It is a circle that is tangent to the addendum circle of themating gear.

Tooth thickness: It is a width of the tooth measured along the pitch circle.

Tooth space: It is a width of space between the two adjacent teeth measuredalong the pitch circle.

Back lash: It is the amount by which the width of a tooth space exceedsthe thickness of the engaging tooth as the pitch circles.

Face of tooth: It is the surface of the gear tooth above the pitch surface.

Flank of tooth: It is the surface of the gear tooth below the pitch surface.

Top land: It is the surface of the top of the tooth.

Face width: It is the width of gear tooth measured parallel to its axis.

Profile: It is a curve formed by the face and flank of the tooth.

Fillet radius: It is the radius that connects the root circle to the profile ofthe tooth.

4.5 GEAR MATERIALS

The gears are manufactured from metallic (or) non-metallic materialsand the materials are selected depending upon the strength and serviceconditions like wear, noise, etc.Metallic gears: Cast iron, Steel and Bronze.

Non-metallic gears: Raw hide, wood, compressed paper, and syntheticresins like nylon, etc.

4.6 LAW OF GEARING

State and prove the law of gearing. Show that the involute curves as theprofiles of making gear satisfy the law of gearing. (AU. Nov/Dec 2012)

Law of gearing states that the pitch point must remain fixed on theline joining centers. In order to have a constant angular velocity ratio, forall position of the wheels, the pitch point P must be a fixed point for thetwo wheels.

(or)The common normal at the point of contact between a pair of teeth

must always pass through the pitch point P.


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T

2

O 2 2

E N

P

Q

M

V

C

D

V1

1

V2

T1

O 1

Fig. 4.6. Law of gearing

Driven

Driver


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Analyse the two profiles which are in contact at Q. Let the profile 1be the driver and 2 be driven. A normal to the profiles at the point of contactQ intersect the line of centers O1 O2 at a point P, where P is a pitch point.

Now, 1

2

r2

r1

This equation is frequently used to define the law of gearing.

Consider the two teeth of two wheels 1 and 2 as shown in Fig. 4.6.The two teeth come in contact at point Q.

Now the TT is the tangent and MN is the normal to the curves at thepoint of contact Q. Join centers O1 and O2. Draw O1M and O2N which is

perpendicular to MN.

When the point Q is considered with respect to wheel 1, its velocity

V1 moves in the direction QC and V1 QC. When the point Q is considered

with respect to wheel 2, its velocity V2 moves in the direction QD and

V2 QD.

For the teeth to remain in contact, the components of velocities

V1 and V2 V1 cos and V2 cos along the common normal MN must be equal.

ie., V1 cos V2 cos

i.e., 1 O1 Q cos 2 O2 Q cos ... (i)

from the Fig. 4.6, cos O1M

O1Q

cos O2N

O2Q

then the equation (i) becomes,

[1 O1Q] O1M

O1Q 2 O2Q

O2N

O2Q

1 O1M 2 O2N


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1

2

O2N

O1M ... (ii)

Consider the similar triangles, O1MP and O2NP

O2N

O1M

O2 P

O1P ... (iii)

combining the equation (ii) and (iii), we get

1

2

O2N

O1M

O2P

O1P ... (iv)

We know, O2P Pitch circle radius r2

O1P pitch circle radius r1

So, 1

2

r2

r1

The teeth of involute form satisfies the law of gearing:

Also, 1

2

r2

r1

T2

T1

Since pitch circle diameter is proportional to the number of teeth onit, it assumes module is same.

N

P

M

O 1

O 2

P

Fig. 4.6(a)


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4.10 UNDERCUTTINGThe tip of rack cutter digs inside flank portion of the gear and material

is chiselled out, such process is known as undercutting.

Under cutting weakens the tooth because of material removed from theinvolute profile, there by reducing the length of path of contact

Problem 4.1: Two gears in mesh have 8 mm module and presure angle of

20. The larger has 57 teeth while the pinion has 23 teeth. If the addendum

on pinion and gear are equal to 1 module, find (i) number of pairs of teethin contact (ii) the angle turned through by the pinion and gear wheel whenone pair of teeth is in contact and (iii) the ratio of sliding to rolling velocity(a). at the beginning of contact (b). at the end of contact (c). at its pitchpoint. (AU. Nov/Dec 2011)

Given data: 20 ; T 57 ; t 23,m 8 mm; Addendum on pinion and

gear are equal to 1 module 8 mm

Solution: To find number of pairs of teeth in contact.

We know, Pitch circle radius of the pinion,

c

r

L

P

K

R

Q

Add circle

Add circle

Base c irc le

Base c irc le

P itch c ircle

P itch c ircle

Fig.


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r m t

2

8 232

92 mm

and pitch circle radius of the gear,

R m T

2

8 572

228 mm

Radius of addendum circle of pinion,

rA r Addendum

92 8 100 mm

Radius of addendum circle of gear,

RA R Addendum

228 8 236 mm

The length of path of approach (i.e., the path of contact when theengagement occurs)

KP RA2 R2 cos2 R sin

KP 2362 2282 cos2 20 228 sin 20

20.98 mm

Length of path of recess (i.e., the path of contact when disengagementoccurs)

PL rA2 r2 cos2 r sin

1002 922 cos2 20 92 sin 20 18.8 mm

Length of path of contact KL

KL KP PL

20.98 18.8 39.78 mm


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and length of arc of contact Length of path of contact

cos

39.78cos 20

42.33 mm

Circular pitch pc

pc m 8 25.13 mm

Number of pairs of teeth in contact ((or) contact ratio)

Length of arc of contact

circular pitch

42.3325.13

1.68 i.e., 2 pairs of teeth

Angle turned through by the pinion and gear wheel when one pair ofteeth is in contact

Angle turned thro’ by the pinion,

Length of arc of contactCircumference of pinion

360 42.33

2 92 360 26.36

Angle turned thro’ by the gear wheel,

Length of arc of contactcircumference of gear

360 42.33

2 228 360 10.64

Ratio of sliding to rolling motion

1 Angular velocity of pinion

2 Angular velocity of gear wheel.

We know,

1

2

Tt or 2 1

tT

2 1 2357

2 0.404 1 rad/s


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Addendum height for smaller fear wheel (pinion)

rA r 173.98 160 13.98 mm

Length of path of contact

KP PL 0.4 MP 0.4 PN

r R sin 0.4

160 240 sin 20 0.4 54.72 mm

Length of arc of contact

Length of path of contact

cos

54.72cos 20

58.23 mm

Problem 4.6: The pressure angle of two gears is 20 and has a module of10 mm. The number of teeth on pinion is 24, on gear is 60. The addendumof pinion and gear is same and equal to one module. Determine (i) thenumber of paris of teeth in contact (ii) the angle of action of pinion andgear and the ratio of sliding to rolling velocity at the beginning of contact. (AU. Nov/Dec 2010)

Given data

20; m 10 mm; t 24 ; T 60; Addendum 1 m

Solution: Refer fig from previous Problem 4.6

Pitch circle radius of pinion,

r mt2

10 24

2 120 mm

Pitch circle radius of Wheel,

R mT2

10 60

2 300 mm

rA r Addendum 120 10 130 mm

RA R Addendum 300 10 310 mm

Length of path of approach

KP RA2 R2cos2 R sin


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3102 3002cos220 300 sin 20

128.95 102.61

26.34 mm

Length of path of recess

PL rA2 r2cos2 r sin

1302 1202cos220 120 sin 20

64.69 41.04

23.64 mm

Length of the path of contact KL KP PL

26.34 23.64

49.98 mm

Length of arc of contact Length of path of contact

cos

49.98

cos 20 53.18 mm

Number of pairs of teeth in contact (Contact ratio)

Contact ratio Length of arc of contact

pc

53.18 10

16.9 ~ 2

Angle turned through by pinion

Length of arc of contact 360

Circumference of pinion

53.18 360

2 120 25.4


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Angle turned through by gear

Length of arc of contact 360

Circumference of gear

53.18 360

2 300 10.15

Ratio of sliding to rolling

1

2

Tt

6024

2.5 2 0.41

VR 1 r 2R 1 120 1201

VS 1 2 KP 1 0.41 26.34 36.8761

Ratio Vs

VR

36.8761

1201 0.3073

Problem 4.7: Two spur gears of 24 teeth and 40 teeth of 8 mm module

and 20 pressure angle are in mesh. Addendum of each gear is 7.5 mm.Determine (a) the angle through which the pinion turns while any pair ofteeth are in contact (b) the velocity of sliding between the teeth when thecontact on the pinion tooth is at a radius of 103 mm. Speed of pinion is300 rpm. (AU. Nov/Dec 2008)

Solution:

Pitch circle radius of pinion

r m t

2

8 242

96 mm

M

C

P

K

W heel

L

p inionO Q

N


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Problem 4.15: A pair of spur gears with involute teeth is to give a gearratio of 3:1. The arc of approach is not be less than the circular pitch and

smaller wheel is the driver. The angle of pressure is 20. (i) What is theleast number of teeth that can be used on each wheel? (ii) What is theaddendum of the wheel in terms of circular pitch? (AU. May/June 2014)

Solution:

Given G T/t Rr

3; 20

1. Least number of teeth on each wheel

Let

t Least number of teeth on smaller wheel

T Least number of teeth on larger wheel

r Pitch circle radius of smaller wheel

Maximum length of patch of approach

Maximum length of path of approach

cos

r sin cos

r tan

and circular pitch, pc m 2r

t

Since the arc of approach is not less than circular pitch,

r tan 2r

t or t

2tan

2

tan 20

17.26 ~ say 18

and T Gt 3 18 54


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2. Addendum of the wheel

mT2

1 tT

tT

2 sin2 1

m 54

2

1 1854

1854

2 sin2 20 1

27 m 0.045 1.2 m

1.2 pc

T1

0.38 Pc

Problem 4.16: A pair of involute spur gears with 16 pressure angle andpitch of module 6 mm is in mesh. The number of teeth on pinion is 16 andits rotational speed is 240 rpm. When the gear ratio is 1.75, find in orderthat the interference is just avoided, (i) addendum on the pinion and gearwheel (ii) the length of path of contact (iii) the maximum velocity of slidingof teeth on either side of the pitch point. (AU. April/May 2013)

Given data:

16 ; m 6 mm ; t 16 ; N1 240 rpm ; G 1.75

Solution:

1 2 240

60 25.13 rad / s

Gear ratio G Tt

1.75

T G t 1.75 16 28

Addendum on pinion and gear wheel

Addendum on pinion


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m t

2 1 G G 2 sin2 1

6 16

2 1 1.75 1.75 2 sin2 16 1 10.76 mm

Addendum on wheel

m T

2

1

1G

1G

2 sin2 1

6 28

2

1

11.75

11.75

2 sin2 16 1

4.56 mm

Length of path of contactPitch circle radius of wheel

R mT2

6 28

2 84 mm

Pitch circle of radius of pinion,

r mt2

6 16

2 48 mm

Addendum circle radius of wheel.

RA R Addendum on wheel

84 10.76 94.76 mm

Addendum circle radius of pinion

rA r Addendum on pinion

48 4.56 52.56 mm

Length of path of approach, = KP

KP RA2 R2 cos2 R sin

94.762 842 cos2 16 84 sin 16

49.59 23.15 26.44 mm


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Length of path of recess = PL

PL rA2 r2 cos2 r sin

52.562 482 cos2 16 48 sin 16

25.17 13.23

11.94 mm

length of path of contact

KL KP PL 26.44 11.94

38.38 mm

Maximum velocity of sliding of teeth on each side of pitch point

2 Angular speed of gear wheel.

2

1

tT

2 1 tT

25.13 1G

25.13 1

1.75

14.36 rad/s

Maximum velocity of sliding of teeth

on left side of pitch point P.. i.e., on KP

1 2 KP 25.13 14.36 26.44

VS 1044.11 mm/s

Maximum velocity of sliding of teethon right side of pitch point P

i.e., on PL

1 2 PL

25.13 14.36 11.94

VS 471.51 mm/s


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4.15 GEAR TRAINSThe combination of two (or) more gears meshing with one another to

transmit power from one shaft to another is called a gear train (or) train oftoothed wheels. A gear train may consist of any (or) all kind of gears likespur, bevel, spiral, etc.

Gear trains are used when,

(i) A large velocity reduction is desired

(ii) The distance between the two shafts is neither too high nor tooshort.

4.15.1 Classification

Gear trains are classified as follows

4.16 PARALLEL AXES SIMPLE GEAR TRAINEach shaft carries only one gear. Such gear train is called as simple

gear train. It is represented by their pitch circles as shown in Fig. 4.18.

Refer Fig. 4.18 (a). The gear 1 drives the gear 2. So the gear 1 iscalled driver and the gear 2 is called driven (or) follower. In this case, themotion of the driven is opposite to the motion of the driver.

G ear trains

Simp le gear train

Compound gear train

Paralle l axes Epicyclic

Compound epicyclic gear tra in

Reverted

gear trainepicyclic

S imp lecom pound gear train

Revertedcom pound gear train


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4.16.1 Velocity ratio (or) Speed ratio :

It is the ratio of speed of the driver to the speed of the driven (or)follower, and it is equal to ratio of no. of teeth on follower to the no.of teethon the driver.

Velocity ratio

orSpeed ratio

Speed of driverSpeed of driven

N1

N2

T2T1

N Speed ; T No. of Teeth

DriverDriven orfo llower1

2 3 4

(c.)

Fig: 4.18 Sim ple gear train

Driver

Driver

D riven orfo llower

Driven orfo llower

12

12

3

(a) (b)


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Speed ratio N1N2

T2T1

The train value is the reciprocal of velocity ratio.

Train value 1

Velocity ratio.

When the distance between the two gears is large, the motion fromone gear to another can be transmitted by providing one (or) more no. of theidle intermediate gears. (Refer Fig 4.18(b) and 4.18(c)).

So idle (or) intermediate gears are used to bridge the gap between thedriver and driven (Follower) Also, idler is used to change the direction ofmotion of the gears.

If the number of idle gears (intermediate gears) are in odd number,then the motion of the follower will rotate in the same sense of the driver.

If the number of intermediate gears are in even number, then themotion of the follower will rotate in opposite direction of the driver.

Refer Fig. 4.18 (b)

1 refers to the driver

2 refers to the intermediate gear

3 refers to the follower

N1

N2

T2

T1 [ It is the speed ratio for gears 1 and 2]

...(i)

N2

N3

T3

T2 [It is the speed ratio for gears 2 and 3]

...(ii)

The speed ratio of the gear train is

i iiN1

N2

N2

N3

T2

T1

T3

T2

N1

N3

T3

T1

From the above, we conclude,


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Speed ratio Speed of driven

Speed of follower

No.of teeth of followerNo.of teeth on driven

From this, there is no effect of intermediate gears (or) idle gears onspeed ratio.

4.17 PARALLEL AXES COMPOUND GEAR TRAIN

Briefly explain the sub classification of compound gear trains with neatsketches. (AU. May/June 2018)

Compound gear trains can be further classified into,

(i) Simple compound Gear train

(ii) Reverted compound Gear train

(i) Simple Compound gear train

When more than one gear is rigidly fixed to a shaft and they rotatetogether with same speed, it is called a compound gear. In Fig. 4.19, thegears 2 and 3 are known as compound gears. Similarly, the gears 4 and 5are also compound gears. The gear 1 is the driver and 6 is follower. All thegears form a compound gear train.

When gear 1 is meshing with gear 2,

Speed ratio N1

N2

T2T1 ...(i)

since 2 and 3 are compound gears,

N2 N3

For gears 3 and 4,

Speed ratio N3

N4

T4

T3 ... (ii)

For gears 5 and 6,

Speed ratio N5

N6

T6

T5 ... (iii)

For getting speed ratio for the compound gear train, we can multiplythe equation (i), (ii) and (iii).


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N1

N2

N3N4

N5

N6

T2

T1

T4

T3

T6

T5

Since N2 N3 Compound gears

and N4 N5 Compound gears,

N1

N6

T2 T4 T6

T1 T3 T5 all gears 2, 4, 6 are followersall gears 1, 3, 5 are drivers

Speed ratio Speed of first driver

Speed of last follower

Product of no.of teeth on followersProduct of no. of teeth on drivers

1

2

2

3

4

5

6

D rive r

D riven

C om p oundgears

A B

C

3

D

5 6

4

F ig. 4.19. Compound Gear Train


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In simple gear trains, the intermediate gears do not have effect on thespeed ratio. In compound gear train, the intermediate compound gears havemuch effect on the speed ratio. So when larger speed reduction is required,compound gear train is used.

(ii) Reverted Compound gear trains

In a Gear train, the first and last gears are coaxial (rotating on thesame sense), such type of gear train is called as Reverted Compound Geartrain as shown in Fig. 4.20.

Gear 1 (input gear) and Gear 4 (output gear) are coaxial Gear 2 andGear 3 are compound gears.

For Gears 1 & 2

Speed ratio N1

N2

T2

T1 ...(i)For Gears 3 & 4

N3

N4

T4

T3 ...(ii)

3

2

4

1

follower

last fo llower

driver

first dr iver

Fig. 4.20. Reverted gear train


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For Reverted compound gear train speed ratio as,

N1

N4

T2 T4

T1 T3

N2 N3 (Compound Gears)

Since the distance between the two shafts are equal,

r1 r2 r3 r4 ...(1)

Assume all the gears have same module, then equation 1 becomes,

T1 T2 T3 T4

Speed ratio Speed of first driver

Speed of last follower

Product of no.of teeth on followersProduct of no.of teeth on drivers.

4.18 EPICYCLIC GEAR TRAIN (OR) PLANETARY GEAR TRAIN

Write short notes on speed ratio of a planetary gear trian(AU. Nov/Dec 2013)

In greek, epi means upon andcyclic means around. When the gearsare arranged in such a manner that oneor more gears rotate upon andaround another gear, then the geartrain is known as epicyclic gear train.This gear train is also called asPlanetary gear train (or) Sun andPlanet gear train.

Refer the Fig. 4.21 when thegear A is fixed, and the arm C isrotated, the gear B rotates about itsown axis O2 and also it revolves upon

and around the gear A. It is similar tothe planets rotate about their own axisand revolve around the sun. Hence thegear B is known as planet gear andthe gear A is known as sun gear. Such a motion is known as epicyclic andthe gear train making this motion is epicyclic gear train.

O 1

O 2

A

B

arm C

Fig. 4.21. Epicyclic gear train


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When the arm is fixed and A is free to rotate, then the gear train willbecome simple gear train. Here the gear A drives B and vice versa.

The epicyclic gear trains are much useful for transmitting high velocityratios.

4.18.1 Applications of Epicyclic gear trains.

1. Wrist watches

2. Back gear of lathe

3. Differential in Automobiles

4. Hoists

5. Pulley blocks

4.18.2 Velocity Ratio for Epicyclic gear train

Velocity ratio for epicyclic gear trains is found by two methods.

(i) Tabulation method ; (ii) Formula (or) Algebraic method.

The tabulation method is much easier and hence it is mostly used forsolving problems in epicyclic gear train. The tabulation method will beexplained clearly while solving the following problems.

4.19 SOLUTION OF PLANETARY GEAR TRAIN PROBLEMS:

Problem 4.17: In an Epicylic gear train, an arm C carries two gears A andB having 36 and 45 teeth respectively. If the arm rotates at 200 rpm in theanti clockwise direction about the centre of the gear A which is fixed,determine the speed of gear B.

Solution:

Given:

TA Number of teeth on Gear A 36

TB Number of teeth on GearB 45

NC Speed of the arm 200 rpm anti clockwiseAnti clockwise in taken as +ve and clockwise is taken as –ve for all

the problems.Prepare the table as below [While preparation of table, just forget

about the given conditions. (given data)]


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Step Operation Nc NA NB

1. Arm C is fixed.

(i.e.NC 0.) Then,

Rotate the gear Athrough +1 revolution,

then NA 1

0 +1 NBNA

TA

TB NB NA

TA

TB

NB 1

TA

TB Note:

When A is rotated in+ve, then B must rotatein -ve for externalgearing.

2. Arm C is fixed. Rotatethe gear A through +xrevolution. [Simplymultiply by x]

0 x x

TATB

3. Add +y revolutions toall elements [simply addy] to the step 2.

0 y y x yy x

TA

TB

O 1

O 2

A

B

arm C


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Explanation: 1. Assume arm is fixed. Once arm is fixed, then it will become simple gear

train. When the gear A makes +1 revolution, the gear B will make TA

TB

ie NA 1 then NB

NA

TA

TB

When A rotates in anti clockwise (+) then NB NA TA

TB

i.e., B rotates in clockwise (–) NB TA

TB (– sign for clockwise)

Enter this in the first row of the table. 2. If the gear A rotates +x revolutions, then the gear B will rotate

x TATB

revolutions.

Enter this in the second row of the table. (or) Simply multiply all themotion elements by x. 3. Add +y revolutions to each element and this should be entered in the

third row.

Now According to table NC y

NA x y

NB y x TA

TB

Now apply the given conditions (given data).

Since A is fixed, NA 0

Since arm C rotates at 200 rpm ; Nc 200

Compare the given condition and tabular column result.

ie NA 0 but NA x y from table (Third row, 4th column)

So NA 0 x y

x y 0 ...(i)


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NC 200 but NC y from table Third row, third column

NC 200 y

y 200 ...(ii)

Solving (i) and (ii) we get,

x y 0

y 200

x 200 ...(iii)

By knowing x and y values, substitute in

NB y x TA

TB

NB 200 200

3645

360 rpm anticlockwise

Problem 4.18: In a reverted epicyclic gear train, the arm A carries twogears B and C and a compound gear D-E. B gears with E and C gears withD. The number of teeth on gears B, C and D are 75, 30 and 90 respectively.Find the speed and direction of gear C when the gear B is fixed and thearm A makes 200 rpm clockwise.

Solution:

TB 75 ; TC 30 ; TD 90 ; NB 0 [. . . fixed] ; NC 200

[. . . clockwise]

To find: TE

From the geometry of the figure, we can find TE, through DB, DC, DD

and DE which are the pitch circle diameters of the gears B, C, D and E

respectively.

From geometry of figure, consider the centre to centre distances.

We know, DB2

DE

2

DC

2

DD

2


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Problem 4.22: In a sun and planet type epicyclic gear train, the pitch circlediameter of the annulus is to be approximately 324 mm and the module 6mm. When the annulus is stationary, the three armed spider makes onerevolution for every five revolutions of the sun wheel S. Find the number ofteeth for all the wheels and exact pitch circle diameter of annulus. If a torqueof 30 N-m is applied to the shaft carrying S, determine the fixing torque ofthe annulus? (AU. April/May 2008)

Solution:

Module is same for all meshing gears,

m 6 mm ; dA ~ 324 mm

We know, dA mTA TA dA

m

TA 3246

54

PP

P

S

A

sun

Annu lus

A rm


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Condition of motion NArm NS NP NA

1. Arm is fixed. Rotate

S through +1 revolution.

0 +1 NP

NS

TS

TP ,

NP TS

TP

NA

NP

TP

TA

NA TS

TP

TP

TA

NA TS

TA

2. Multiply by x 0 x x

TS

TP x

TS

TA

3. Add y y x yy x

TS

TPÙ›é y x

TS

TA

Given data:

Narm 1 revolution

NS 5 revolution ; NA 0 . . . Annulus is stationaryFrom table,

Narm y 1 [given data]

y 1

NS x y 5 [given data]

x 1 5

x 4

The number of teeth 13.5 is not practically possible. It should be 13(or) 14.

Case (i) If TS 13, then, TA 4 TS 4 13 52

We know DA2

DP DS

2 from geometry of Fig.

TA

2 TP

TS

2

NA y x TS

TA

0 1 4 TS

TA

TS 14

TA or TA 4 TS

TS 14

54 13.5


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TP TA _ TS

2

... (i)

TP 52 13

2 19.5 which is not practically possible

So TA can not be 52 and TS can not be 13.

Case (ii)

If TS 14

Then TA 4 TS 4 14 56

TA 56

Substitute in equation (i)

TP TA TS

2

56 142

21 It is practically possible

TP 21

Number of teeth on the gears are,

TA 56 ; TS 14 ; TP 21

When TA 56, the exact pitch circle diameter DA m TA 6 56

DA 336 mm

To find fixing torqueWhen arm rotates 1 rev, sun rotates 5 revolutions.

arm 1 ; S 5

Torque applied to shaft carrying S MS

Ms 30 Nm

Marm arm MS S

Marm MS S

arm 30

51

150

Marm 150 Nm


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9.88 1450

181.25

79.04 Nm

Fixing torque to fix Annulus E

79.04 9.88 69.16 Nm

4.16 DIFFERENTIAL GEAR USING EPICYCLIC GEAR TRAINDuring turning of a vehicle, the outer wheel will have to travel greater

distance as compared to the inner wheel as shown in the Fig. 4.22

Outer wheel distance - O to O

Inner wheel distance - I to I

If the vehicle has a solid rearaxle, then there will be a tendency toskid. Hence in order to avoid thisskidding and maintaining differentspeeds for inner wheels and outerwheels, a device is needed. This deviceis called differential.

Differential in automobile consistsof an epicyclic train of gears designed topermit two or more shafts to rotate atdifferent speeds, by means of whichouter wheel runs faster than inner wheelwhile taking a-turn.

4.16.1 Final Drive using Differential gears

Final drive is the last stage in transferring power from engine towheels, using differential gears. It reduces the speed of the propeller shaft(drive shaft) to that of wheels. It also turns the drive of the propeller shaft

by an angle of 90 to drive the wheels.

The propeller shaft has a small bevel pinion which meshes with crownwheel. (Refer Fig. 4.23 and Fig. 4.24)

O

I

O IFig. 4.22 Distance Travelled by Inner

and Outer Wheels on a Turn.

10 M eters

11.5 Meters

90o


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The crown wheel is connected with cage so that it rotates along withcrown wheel. Small planetary gears freely rotating on the pins of the cage,gets revolving motion around large sun gears. These small planetary gearswill revolve around and rotate the large sun gears connected to the rear axles.

The crown wheel gives rotary motion to rear axles. The size of crownwheel is bigger than that of bevel pinion, therefore, the speed of rear axles(or crown wheel) is lower than the speed of pinion.

Beve l P in ion

P lan etary gears

C row n wheel

Su

n g e

ar

Le ft s ide a xle shaft

D iffe re ntia l sha ft

R igh t s ideaxle sh aft

P inions free to

ro tate onp inion

p in

P lan etary gears

Fig. 4.23. F inal Drive - D ifferential drive

Arm ro ta ting alon g w ith crow n wh eel

Sun

ge a

r

P rope ller sh aft from eng in e


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Be

vel P

inio

n

Cro

wn

Whe

el

Sm

all

Pla

neta

ry G

ear

Rot

atin

g C

age

alo

ng w

ith C

row

n w

heel

Larg

e S

un G

ears

Sm

all

Gea

rP

lane

tary

Out

er (

Rig

ht)

Hal

f axl

e S

haft

Pro

pelle

r S

haft

Inne

r(L

eft)

H

alf a

xle

Sh

aft

Fig

. 4.2

4 3D

vie

w o

f F

inal

Dri

ve -

Dif

fere

nti

al d

rive

- u

sin

g E

pic

yclic

gea

r tr

ain


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Problem 4.30: Fig. shows a differential gear used in a motor car. Thepinion A on the propeller shaft has 12 teeth and gears with the crown gearB which has 60 teeth. The shafts P and Q form the rear axles to which therod wheels are attached. If the propeller shaft rotates at 1000 r.p.m. and theroad wheel attached to axle Q has a speed of 210 r.p.m. while taking a turn,find the speed of road wheel attached to axle P. (AU. May/June 2016)

Given

TA 12

TB 60

NA 1000 rpm

NQ ND 210 rpm

Since the propeller shaft or the pinion A rotates at 1000 rpm. Thereforespeed of crown gear B.

NB NA TA

TB

1000 1260

200 rpm

Step

No

Condition of

motion

Revolutions of elements

Gear B Gear C Gear E Gear D

1.

Gear B fixed - Gear

C rotated through +1

revolution (i.e. 1

revolution clockwise)

0 1 TC

TE

TC

TE TE

TD 1

TC TD

2.

Gear B fixed-Gear C

rotated through + x

revolutions. Multiply

by x

0 x x TC

TE x

3.Add +y revolutions to

all elements y x y y x

TC

TE y

y x


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Since the speed of gear B is 200 rpm, therefore, from the fourth rowof the table.

y 200

Also, the speed of road wheel attached to axle Q or the speed of gearD is 210 rpm. Therefore from the fourth row of the table.

y x 210 (or) x y 210

200 210

410

Speed of road wheel attached to axle P.

Speed of gear C x y

410 200

610 rpm

B

Rear Axle

Arm

A

Propeller Shaft

DRear Axle

EC

P Q

W heel

W heel Sp indle

F

Arm


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UNIT V

FRICTION IN MACHINE ELEMENTS

Surface contacts - Sliding and Rolling friction - Friction drives -Friction in screw threads - Bearings and lubrication - Friction clutches -Belt and rope drives - Friction in brakes - Band and Block brakes.

5.1 INTRODUCTION

Whenever there is a tendancy of sliding motion between two surfacesin contact, a resisting force is developed tangential to the common surfaceof contact. This resisting force is called Frictional Force. The frictional forceexperienced without lubrication is called dry friction.

Friction is the property of two surfaces in contact by virtue of whicheach surface resists the relative motion of the other surface.

The resisting force developed by each surface is tangential to thecommon surface of contact and is opposite to the direction of impendingmotion.

The maximum resisting force that can be developed is called limitingfricitional force.

5.2 TYPES OF FRICTION

List the various types of friction. (AU. Nov/Dec 2013)

Depending on the contacting surface, friction is classified into twotypes.

1. Static friction

2. Dynamic friction

(a) Sliding friction

(b) Rolling friction

(c) Pivot friction

5.2.1 Static friction

The friction experienced when the body is at rest, is known as staticfriction.

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5.2.2 Dynamic friction

When the body is under motion, if the friction is experienced, thensuch friction is known as dynamic friction.

(a) Sliding friction

The friction experienced, when one body slides over another, is knownas sliding friction.

(b) Rolling friction

The friction experienced when roller rolls over the surface, is knownas rolling friction.

(c) Pivot friction

The friction experienced when the pivot is rotating on a bearingsurface, is known as pivot friction.

5.3 SLIDING FRICTION - LIMITING FRICTIONWhen a body is resting

on a floor, its weight W actsdownwards. This force (weightW) is cancelled by the normalreaction RN offered by the

floor.

Now, if an externalforce P is applied pullingrightside, the frictional forceF will come into play to resistthe sliding motion as shown in Fig. 5.1.

i.e., Force P tends to move the body toward the rightside, frictionalforce opposes the motion by giving equal amount of force in oppositedirection.

For example, if we give P 10 N force rightside, F will become

10 N leftside. If we increase P 20 N, then F 20 N. If we increase

P 30 N, then F 30 N. In a similar way, if we go on increasing the P, theF will also increase. But at one stage, F will attain its maximum limit. For

example, if P 100 N, F will be only 99 N. At this moment, the body startsmoving in the rightside direction.

F

W

NR

P

Fig.5.1. Limiting Friction


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So, 99 N is the maximum limit of friction. This is called limitingfriction. At this stage, limiting friction is the maximum frictional forceexerted when the motion is about to begin. Upto limiting friction, it is calledstatic friction.

Once the body starts moving, the friction is called kinetic fricion ordynamic friction. Dynamic friction effect is less than the static friction.

Static friction, Fmax s RNDynamic friction, F k RN

where s coefficient of static friction;

k Coefficient of dynamic friction.

5.4 COLOUMB’S LAW OF DRY FRICTION

These laws are applicable for friction between two dry surfaces. Iflubricating fluid is used then frictional force is reduced and the laws of dryfriction are not applicable.

Following are the laws of dry friction.

1. Frictional force opposes the motion of the body.

2. For two given surfaces, the limiting frictional force depends on thenormal reaction and is independent of the external area of surfaces incontact

3. For two given surfaces, the limiting (maximum) value of Frictional forceis proportional to the normal reaction.

Fmax RN; Fmax RN i.e., Fmax

RN

The ratio “Frictional force to normal reaction” remains constant and this

constant is called coefficient of friction

4. Dynamic frictional force is less than the limiting value of static frictionalforce.

www.srbooks.org Friction in Machine Elements 5.3

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5.5 ANGLE OF FRICTIONIn Fig. 5.2, RN is the normal reaction.

F RN is the maximum limiting frictional

force, and ‘R’ is the resultant reaction.

Resultant reaction is the vector sum ofnormal reaction and limiting frictional force.

The angle that the resultant reactionmakes with the normal reaction is the angleof friction.

In Fig.5.2, is the angle of friction

Also tan RN

RN

Angle of friction, tan 1

Resultant reaction R RN2 2 RN

2

R RN 1 2

5.6 ROLLING FRICTION (OR) ROLLING RESISTANCEWhen a wheel is rolling on the ground it has a point of contact. So

there is no friction. If the above statement is true, then the wheel should rollcontinuously and indefinitely. But in actual practice, the wheel stops aftersometime. This is because of the rolling resistance to the rolling motion.

F= R N

R N

W

R

Fig.5.2. Limiting angle of Friction

BB

Rb

WW

P

RN Line of action of force W

fig . (a) fig . (b)

Fig.5.3. Rolling Friction

r


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Actually, under load W, both wheel and ground deform slightly (referFig. 5.3 (b)), causing friction over a certain area.

In Fig. 5.3 (b), P is applied force

W is weight of wheel

The resultant force of P and W is R. This R will be acting - not exactlyon the line of action of W but slightly in front of line of action of force W.This horizontal distance (b) is called “the forward length of deformation”(or) “the coefficient of rolling resistance”. To balance the moment of Wabout B and to keep the wheel rolling at constant speed, the horizonatal forceP is applied.

Now, MB 0; so, Pr Wb.

5.6.1 Friction acts in the same direction of motion during rolling

A wheel is sunk in the sand, when sufficient friction force is notavailable. If sufficient friction force is given, then the wheel will roll forward.

So when the wheel is given power the torque M rotates the wheel, but thefriction force provides the necessary force to make the wheel move. Thus inthis case, the friction is acting in the same direction of motion as shown inFig. 5.4.

C

M

Sand is thrown back due to the absence o f friction

C M

F

Fig. 5.4


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Problem 5.7: The following data relate to a screw jack:Pitch of the threaded screw = 8 mm; Diameter of the threaded screw = 40mmCoefficient of friction between screw and nut = 0.1; Load = 20 kNAssuming that the load rotates with the screw, determine the(i) Ratio of torques required to raise and lower the load(ii) Efficiency of the machine. (AU. Apr/May - 2017)

Solution: Given:

Pitch, p 8 mm 8 10 3 m;

Dia of screw thread,

d 40 mm 40 10 3 m;

tan p d

8 10 3

40 10 3 0.0637 3.64

Friction angle can be calculated by,

tan 0.1 5.71

Torque required to raise and lower the load:

Torque required to raise the load,

T1 P d2

W tan d2

20 103 tan 3.64 5.71 40 10 3

2

T1 65.861 Nm

Torque required to lower the load,

T2 W tan d2

20 103 tan 5.71 3.64 40 10 3

2

T2 14.457 Nm

tan 0.1;

Load, W 20 kN 20 103 N


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Ratio of torques, T1

T2

65.86114.457

4.556

Efficiency of the machine

screw jack tan

tan

tan 3.64tan 3.64 5.71

0.386 or 38.6%

Problem 5.8: The pitch of a single threaded screw jack is 8mm and itsmean diameter is 70mm. If the coefficient of friction is 0.2, find the forcerequired at the end of a lever 300 mm long (a) to raise a load of 60 kN (b)to lower the same load. Find the torque applied in each case. (AU. May/June 2007)

Solution:

Given: Pitch p 8mm

d 70 mm ; r 702

35 mm

0.2

To find

tan ; so tan 1 ; tan 1 0.2 11.31

11.31

To find

tan pitch

circumference

p d

8

70 0.0364

tan 1 0.0364 2.08

P1 required to raise the load

P1 rl W tan

P1 35

300 60 tan 11.31 2.08 1.667 kN

l length of the lever = 300 mm

W weight of the load = 60 kN

P1 Force required at the end of lever


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To find torque applied during raising of load

T P1 l 1.667 300 500 kN mm

Torque 500 Nm

P2 required to lower the load

P2 rl W tan

35

300 60 tan 11.31 2.08 1.138 kN

To find Torque applied during lowering of load

T P2 l 1.138 300 341.3 kNmm

Torque 341.3 Nm

Problem 5.9: A screw-jack has a square thread of mean diameter 60 mmand pitch 8 mm. The co-efficient of friction at the screw thread is 0.09. Aload of 3 kN is to be lifted through 120 mm. Determine the torque requiredand the work done in lifting the load through 120 mm. Find also the efficiencyof the jack. (AU. Nov/Dec 2012)

Solution: Given: d 60 mm; p 8 mm ; 0.09 tan

Load W 3 kN 3 103 N; tan p d

8

60 0.0424

Force required on the screw to raise the load

P W tan

P W

tan tan 1 tan tan

3

0.0424 0.09

1 0.0424 0.09

3 .13240.996

0.3987

P 0.3987 kN


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Torque required to lift the load

T1 P d2

0.3987 602

11.96 kN mm

T1 11.96 N m

Since the load is lifted through a vertical distance of 120 mm and thedistance moved by the screw in 1 rotation is 8 mm (equal to pitch), thereforenumber of rotation made by the screw, N

N 120

8 15

Workdone in lifting load T 2 N

11.96 2 15 1127.2 Nm

Efficiency of screw jack

tan 0.09; 5.14

tan 0.0424; 2.43

tan

tan

tan 2.43

tan 2.43 5.14

0.319 32%

Problem 5.10: The mean diameter of the screw jack having pitch of 10 mmis 50 mm. A load of 20 kN is lifted through a distance of 170 mm. Find thework done in lifting the load and efficiency of the screw jack when(i) the load rotates with the screw, and(ii) the load rests on the loose head which does not rotate with the screw.The external and internal diameters of the hearing surface of the loose headare 60 mm and 10 mm respectively. The coefficient of friction for the screwas well as the bearing surface may be taken as 0.08. (AU. Nov/Dec - 2011)

Solution

Given: p 10 mm; d 50 mm; W 20 kN 20 103 N; d2 60 mm or

r2 30 mm; d1 10 mm or r1 5 mm; tan 1 0.08 (or) 4.57


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Problem 5.16: A Conical pivot supports a load of 25 kN, the cone angle

being 120, and the intensity of normal pressure does not exceed 0.25 MPa.The external radius is twice the internal radius. Find the outer and innerradii of the bearing surface. If the shaft rotates at 180 rpm and the coefficientof friction is 0.15, find the power lost in friction, assuming uniform pressure. (AU. Nov/Dec - 2010)

Given data: W 25 kN ; 2 120 ; pn 0.25 MPa ; N 180 rpm ;

0.15 ; r1 2r2

Solution:

2 N60

2 180

60 18.85 rad/sec

Let r1 External radii; r2 Internal radii

pn W

[r12 r2

2]

0.25 106 25 103

[2r22 r2

2]

Power absorbed in friction,

T 23

W cosec r13 r2

3

r12 r2

2

23

0.15 25 103 cosec 60 0.3573 0.1783

0.3572 0.1782

1201.5 N.m

P T 1201.5 18.85 22.65 kW

r22 0.03185

r2 0.1785 178.5 mm

r1 2r2 2 178.5 357 mm


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5.15 FRICTION CLUTCHESFriction clutches are used to connect the engine flywheel to the gear

box in automobiles: When the clutch is engaged, the power is transmittedfrom fly wheel to gear box and to the road wheel. When the clutch isdisengaged, the power is not transmitted from engine to fly wheel as shownin Fig. 5.14. Thus we can stop the power to be transmitted to the gear boxand to the road wheel without stopping the engine.

While changing the gears, the clutches are used to disconnect the powerfrom engine. While we are pressing the clutch, the power is disconnected andwill not be transmitted to the gear box. Now, we can change the gears. Then,we can release the clutch so that power is transmitted from engine to gearbox and to the wheels. For the power to be transmitted to the machinespartially (or) fully, the clutches are used.

5.16 THREE IMPORTANT TYPES OF CLUTCHES

1. Disc (or) plate clutches

(a) Single plate clutches

(b) Multiplate clutches

2. Cone clutches

3. Centrifugal clutches.

(a) Before engagement

W

D riving D iscD riving D isc D riven D iscD riven D isc

(b) After engagement

(w h ile the c lu tch peda l is pressed,th e clutch surfaces a re disengaged)

(w h ile the c lu tch peda l is n ot p ressed , the c lutch surfaces a re engaged)

Fric tion su rfaces

N o ro tation

Fig. 5.14


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5.17 SINGLE PLATE CLUTCHIt consists of single clutch plate having both sides coated with friction

materials. So both sides of the clutch plate are friction surfaces. The pressureplate presses the clutch plate along with fly wheel as shown in Fig. 5.15.

During normal running condition, the clutch plate is engaged with thefly wheel. When the clutch pedal is pressed, the pressure plate moves backand the clutch plate becomes free between the fly wheel and the pressureplate, so that power is not transmitted.

T Torque to be transmitted by the clutch.

p Intensity of axial pressure due to which the contact surfaces are held together.

r1 and r2 External and Internal radii of friction faces.

Fly wheel

c lu tch spring

pressurep la teEngine

shaft

c lu tch p late

Friction lin ing

F ig. 5.15. S ingle Plate Clutch

clu tchpedal

c lu tchshaft

Fu lcrumpin

Bearing


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Clutches are designed by considering any one of the followingconditions.

(i) There is uniform pressure all over the friction surfaces. p constant

(ii) There is uniform wear all over the friction surfaces. pr2 constant

5.17.1 Considering Uniform Pressure Theory

Here pressure is distributed all over the friction surfaces uniformly. So

The intensity pressure, p W

r12 r2

2 ... (i)

where W axial thrust (given by spr ing) by which the fr iction sur faces areheld together .

T 23

WR

where 23

R mean radius of friction surface.

23

r13 r2

3

r12 r2

2

P

P

r1

r2

Fric tionsurface

Sing le p late

w

r1

r2

r

d r

F ig.5.16. Forces on a single disc or plate clutch


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5.17.2 Considering uniform wear theory

In this case, the pressure is more at the inner radius and less at theouter radius. So the pressure varies inversely with the distance.

ie., p r constant C

So, p Cr

C W

2 r1 r2

Frictional torque acting on the ring

2Pr2 dr 2 Cr

r2 dr

2 Cr dr [ . . . P Cr

]

Total frictional torque acting on the entire friction surface. T

T W r1 r2

2 WR

where R mean radius offriction surface

r1 r2

2

1. If the no. of friction surfaces are n,

T n WR

where, R 23

r13 r2

3

r12 r2

2 for uniform pressure condition.

R r12 r2

2

2 for uniform wear condition.

2. If both sides of the single plate clutch are effective, then n 2

3. For new clutch - Apply uniform pressure condition

For old clutch - Apply uniform wear condition.


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4. For design purpose, the uniform wear theory should be applied, since ittransmits lesser frictional torque. So for any problem, if nothing is mentioned,then we should use only uniform wear theory.

5.18 MULTIPLE DISC CLUTCHWhen large amount of torque is to be transmitted, the multiple disc

clutch is used. In a multiple disc clutch, the number of friction plates aremore. The multiple disc clutch works in the same way as the single plateclutch.

Here, n1 Number of discs on the driving shaft

n2 Number of discs on the driven shaft.

Number of pair of contact surfaces

n n1 n2 1

Fric tion rings(splined)

Fric tion

spring

driven sha ft

Engine sha ft

D isc (p la te )

Flyw heel

Fig.5.17. Multiplate or Multiple Disc Clutch

Splines


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Problem 5.18: A single plate clutch transmits 25 kW at 900 rpm to the

maximum pressure intensity between the plates is 85 kN/m2. The outerdiameter of the plate is 360 mm. Both the side of the plate are effective andthe coefficient of friction is 0.25 Determine the(i) Inner radius of the plate (ii) Axial force to engage the clutch. (AU. April/May 2017)

Solution: Given data:

Power, P 25 kW

0.25

To find:

(i) Inner radius of the plate.

(ii) Axial force to engage clutch.

Torque T n WR

where R r1 r2

2

Since the intensity of pressure is maximum at inner radius, therefore,

pr2 C C 85 103 r2

Axial thrust transmitted to frictional surface (W)

W 2 C r1 r2

W 2 85 103 r2 r1 r2

So, Torque transmitted, T n WR

Power 2 NT

60 25 103

2 90060

T

T 25 103 60

2 900

T 265.26 Nm

T n WR

N 900 rpm

r1 180 mm 0.18 m

n 2


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265.26 2 0.25 r1 r2

2 [2 85 103 r2] r1 r2

0.25 2 85 103 r2 r12 r2

2

265.26 133450 r2 [0.182 r22]

1.987 10 3 r2 [0.0324] r23

By trial and error method.

r2 0.1315 m (or) 131.5 mm

(ii) The axial force to engage the clutch

W 2C r1 r2 Where C pr2

2 85 103 0.1315 0.18 0.1315

7094.7 0.0485

3404.4 N

W 3.404 kN

Problem 5.19: A rotor is driven by a co-axial motor through a single plateclutch, both sides of the plate being effective. The external and internaldiameters of the plate are respectively 220 mm and 160 mm and the totalspring load pressing the plates together is 570 N. The motor armature andshaft has a mass of 800 kg with an effective radius of gyration of 200 mm.The rotor has a mass of 1300 kg with an effective radius of gyration of 180mm. The coefficient of friction for the clutch is 0.35. The driving motor isbrought up to a speed of 1250 r.p.m. When the current is switched off andthe clutch suddenly engaged. Determine(i) The final speed of motor and rotor(ii) The time to reach this speed, and(iii) The kinetic energy lost during the period of slipping. (AU. Apr/May 2015, Nov/Dec 2011)


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5.30 SLIP AND CREEP OF BELT

Slip is defined as the relative motion between the belt and pulley.

When the friction force between belt and driver pulley is insufficient,the driver will rotate without carrying the belt with it. This is called slip of

the belt between the driver and belt. s1 %

When the friction force between belt and follower pulley is insufficient,the belt moves without rotating the follower pulley. This is called slip of the

belt between the belt and the follower. s2 %

Velocity ratio N2

N1

d1d2

1

s1

100

s2

100

Creep of belt

When the belt passes from slack side to tight side, a part of belt isextended. When the belt passes from tight side to slack side, a part of beltcontracts. This repeated extension and contraction makes the belt move withthe relative motion with respect to pulley. This relative motion is known ascreep.

Velocity ratio, considering creep,

N2

N1

d1

d2

E 2

E 1

where, 1 and 2 stress in the belt on thetight and slack side respectively

E Young’s Modulus for the belt material

Problem 5.27: The speed and dia of driver pulley are 200 rpm and 1000mm respectively. The belt connects the driver with the driven pulley of 600mm dia which is keyed to a line shaft. Another pulley of dia 1200 mm isalso keyed to the same line shaft. Another belt connects the pulley of dia1200 mm with the pulley of 200 mm dia which is keyed to the dynamo shaft.Find the speed of the dynamo shaft, when (i) there is no slip (2). There isa slip of 3% between the driving pulley and belt and slip of 2% between thebelt and follower pulley.


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Solution:

N1 200 rpm ; d1 1000 mm 1 m ; d2 0.6 m ; d3 1.2 m; d4 0.2 m ;

N4 ?

(i) When there is no slip

Speed of last followerSpeed of first driver

Product of dia of drivers

Product of dia of followers

N4

N1

d1 d3

d2 d4

N4

200

1 1.2

0.6 0.2 10

N4 200 10 2000 rpm

(ii) When s1 3% and s2 2%

N4

N1

d1 d3

d2 d4 1

s1

100 1

s2

100

N4

200

1 1.12

0.6 0.2 1 0.03 1 0.02 8.872

N4 8.872 200 1774.5 rpm

First D rive r

1000m m 600mm

1200m m200mm

Dynam o shaft

Last fo llower

1

2

3

4

Engine sha ft L ine shaft


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Problem 5.28: A flat belt rolls over a pulley of a diameter D making contact

angle when the respective tensions are T1 and T2. Derive a relationship

between the tensions, angle of contact and coefficient of friction . Show thatthe diameter of pulley is immaterial. (AU. Oct. 2003, Nov. 2010)

Solution:

Consider a flat belt wrapped around a portion of a pulley.

Angle of wrap (or) Angle of lap (or) Angle of contact.

d

T1T 2

Driven

dR N

dRN

(T+dT)cosd2

(T+dT)sin d2

-Tsin d2 d

2d2

-Tcosd2

y ax is

x ax is

T=Slackside tension (less Tension)

T+dT=Tights ide tension (M ore Tension)

T T+dT

Pu lley Centre

dR =Sm all R forceN N

d =Small angledT=Small tension

d

Fig. (a)


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Assume the pulley is in equilibrium and is about to rotate in clockwise

direction by the difference of tension T1 T2. So T1 T2.

T1 Tight side tension (max) and T2 Slack side tension (min)

Consider an infinetesimal segment of a belt. This segment subtends an

angle d at the pulley center as shown in Fig. (a).

Writing the Equations of Equilibrium

Fx 0; Fy 0

Fx T cos d2

T dT cos d2

dRN 0

So, dT cos d2

dRN

Since d is very small, assume d 0. So cos d2

cos 0 1

dT dRN 1

Fy 0

T sin d2

T dT sin d2

dRN 0

2T sin d2

dT sin d2

dRN 0

Since is very small, we can take sin d2

d2

. So the equation

becomes,

2T d2

dT d2

dRN 0

Since dT d2

is very small and negligible, we write the equation as

T d dRN 0

T d dRN

i.e., dRN T d


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substitute dRN T d in equation (1), we get

dT T d

Rearranging,

dTT

d

Integrating on both sides around the portion of the belt in contact withthe drum,

T2

T1

dTT

0

d

ln T1

T2

Ratio of tension T1T2

e where is in radian

We have got the relation between the tensions and . Here we can

see the ratio of tension depends only on the angle of lap and coefficient

of friction . It does not depend on the diameter of the pulley.

Problem 5.29: Two pulleys, one 450 mm diameter and the other 200 mmdiameter are on parallel shafts 2.1 m apart and are connected by a crossedbelt. The larger pulley rotates at 225 rpm. The maximum permissible tensionin the belt is 1 kN and the co-efficient of friction between the belt and thepulley is 0.25. Find the length of the belt required and the power that canbe transmitted. (AU. Apr 2011)(AU. May/June 2014)

Given data:

d1 450 mm ; d2 200 mm ; x 2.1 m ; N1 225 rpm ; T1 1 kN ; 0.25

Solution:

r1 4502

225 mm 0.225 m ; r2 200

2 100 mm 0.1 m

Angle of lap for cross belt is,


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Where maximum safe stress in N/m2

Since the maximum tension

in the belt T total tension on the

tight side.

When the centrifugal tension is neglected,

Max. tension T T1 (tightside tension)

When the centrifugal tension is considered,

maximum tension T T1 Tc

[tightside tension centrifugal tension]

5.33 CONDITION FOR MAXIMUM POWER TRANSMISSION

We know, power transmitted P T1 T2 V ...(i)

where T1 Tightside tension ; T2 slackside tension

Ratio of tension T1

T2 e or T2

T1

e ...(ii)

Substitute the value T2 T1

e in eqn. (i), we get

P T1

T1

e V

T1 1

1

e V

P T1 V C ...(iii)

t

b

Fig. 5.32


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where C 1

1

e constant

We know T T1 Tc

So T1 T Tc where T maximum tension

substitute T1 T Tc in eqn. (iii)

P T Tc V C

T mV2 V C[ . . . Tc mV2]

P [T V mV3] C ...(iv)

For maximum power, differentiate the power equation (iv) with respectto V and equate to 0.

dPdV

d

dV T V mV3 C 0

T 3mV2 C 0

T 3 mV2 0

T 3 mV2...(v)

V T3m

...(vi)

from eqn (v), T 3mV2 3 Tc

So T 3 Tc or Tc T3

or T1 23

T

Thus we know,

5.33.1 For maximum power transmission,

Maximum tension T 3 Tc and V T3m


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5.34 INITIAL TENSION IN THE BELT

Belt is transmitting power by virtue of friction. To increase the power,the frictional grip between the belt and pulley can be increased. To increasethe frictional grip, the belt is tightened up. (i.e., the centre distance of thetwo pullies is increased). Due to this tightening, the belt is subjected to sometension, known as initial tension,

By derivation, T0 initial tension T1 T2

2 when TC is neglected.

To T1 T2 2 Tc

2 when Tc is considered.

Problem 5.31: A flat belt is required to transmit 15 kW at 300 rpm. The

angle of lap over smaller pulley is 160 and the coefficient of friction is 0.2.

Diameter of smaller pulley is 900 mm. Belt thickness 6 mm. Mass density ofbelt 1 gm/cc. Permissible intensity of stress in the belt material is

2 N/mm2. Find the width of the belt. (AU. May 2005)

Solution: Power P 150 1000 W ; N 300 rpm; Angle of lap

160

180 2.793 rad ; 0.2; d 0.9 m ; t 6 mm 0.006 m ; 1 gm/cc

1 10 3 106 kg/m3

1 103 kg/m3

i.e., 1000 kg/m3

2 N/mm2 2 106 N/m2

Velocity of the belt V dN60

0.9 300

60

14.14 m/s

Power transmitted P 15000 T1 T2 V

15000 T1 T2 14.14

T1 T2 1061.03 N ... (i)

Let, b width of the belt in meters

T1 Tightside tension of the belt in N

T2 Slackside tension of the belt in N


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We know, T1T2

e

T1T2

e[0.2 2.793] 1.748

T1T2

1.748...(ii)

Solving (i) and (ii),

T1 1.748 T2

1.748 T2 T2 1061.03

T2 1061.060.748

1418.5 N

T2 1418.5 N

Then, T1 1.748 1418.5

T1 2479.5 N

Centrifugal tension Tc mV2 6 b 14.142

Tc 1200 b N

Maximum tension in the belt,

T b t 2 106 b 0.006

T 12000 b ...(iii)

But we know, T T1 Tc 2479.5 1200 b ...(iv)

Equate (iii) and (iv) equation,

12000b 2479.5 1200 b

10800 b 2479.5

b 2479.510800

0.23 m

Width of the belt b 0.23 m

Mass of the belt per meter length,

m Area length density

b t l

b 0.006 1 1000

m 6b kg/m


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T2 T1

2.187

5452.187

249.18 N

T2 249.18 N

To find power transmitted by 1 rope

Power transmitted by one rope T1 T2 V

545 249.18 23.56

6969.6 W 6.97 kW

To find no. of ropes required

No.of ropes required Total power transmitted

Power transmitted by one rope

2006.97

28.69

29 ropes required

Problem 5.43: Following data is given for a rope pulley transmitting23.628 kW. Dia of pulley = 40 cm; speed = 110 rpm, angle of groove

45 ; angle of lap 60, co-efficient of friction = 0.28, No. of ropes =

10. Mass in kg/m length of ropes 0.0053 C2 and working tension is

limited 12.2 C2 N where C = girth of rope in cm. Find (i) initial tension,and (ii) diameter of each rope. (AU. May/June 2016)

Solution: Given: P 23.628 kW, d1 40 cm 0.40 m, N1 110 r.p.m

2 45, 22.5, 160 160

180 radians , 0.28 No. of ropes = 10

Mass (m) per metre length 0.0053 C2

Max, tension, Tmax 12.2C2 N

Velocity of rope is given by,

V d1 N1

60 0.4 110

60 2.303 m/s

We know for centrifugal tension,


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TC m V2

0.0053 C2 2.3032 0.0281 C2 N ... (i)

Now using equation for the ratio of tension in rope,

T1

T2 e cosec B

e0.28 160

180 cosec 22.5 e2.046 7.71

T1 7.71 T2 (ii)

Now power transmitted by one rope

Total powerNo. of ropes

23.628

10 2.3628 kW

But power transmitted by one rope is also

T1 T2 V

1000

2.3628 T1 T2 2.303

1000

T1 T2 2.3628 1000

2.303 1026 N

7.71 T2 T2 1026 [. . . From equation ii T1 7.71 T2]

6.71 T2 1026

T2 10266.71

153 N

From equation (ii),

T1 7.71 153 1179 N

Now Tmax T1 TC

12.2 C2 1179 0.0281 C2

[. . . From equation i, TC 0.0281 C2]

12.2 0.0281 C2 1179


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C2 1179

12.2 0.0281 96.86

C 96.86 9.84 cm (iii)

Substituting the value of C in equation (i), we get

TC 0.0281 9.842 2.72 N

(i) Initital tension To.

To T1 T2 2TC

2

To 1179 153 2 2.72

2 668.72 N

(ii) Dia. of each rope (d).

C Girth of one rope d

From equation (iii), C 9.84 cm

9.84 d

d 9.84

3.132 cm.

5.38 FRICTION IN BRAKES

A brake is a device used to retard (or) stop the motion of a body.This brake develops frictional resistance to the moving body.

Generally, the kinetic energy of the moving body is absorbed by the brakesand this energy is dissipated in the form of heat to the surrounding air.

5.39 TYPES OF BRAKES

1. Single block (or) shoe brake

2. Pivoted block (or) shoe brake

3. Double block (or) shoe brake

4. Simple band brake

5. Differential band brake

6. Band and Block brake.

Among the above, the last three types are widely used and these areexplained with exam point of view.


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5.40 SIMPLE BAND BRAKESimple band brake is shown in Fig. 5.36. Here one end of the band

is attached to the fixed pin called fulcrum (O) of the lever and the other endis attached to the pin B on the lever at a distance b from the fulcrum (O).

Upward Effort (or) force is applied at the end of the lever C to tightenup the band on the drum and hence brakes are applied. The braking force isprovided by virtue of friction between the band and drum.

5.40.1 To find the braking torque TB

T1 Tightside tension

T2 Slackside tension

Angle of lap of the band on the drum.

Ratio of tension T1

T2 e

Braking force on the drum T1 T2

Fig. 5.36 Sim ple band brake

(a )C lockwise rota tion of drum

�

P

CB

T1

b

OT2

M

L

O

tBrake d rum

shaft

Band

r

�

P

CBT 1

bO

T2

M

L

O

t Brake d rum

shaft

Band

r

coefficient of friction between the band and drum.

r radius of the drum

t thickness of the band

re effective radius of the drum r t2


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Braking Torque on the drum

TB T1 T2 r

If the thickness of the belt is considered, then TB T1 T2 re

Note: When the drum rotates in the clockwise sense, then the end of bandattached to B will have tight side tension T1 and the other side will have

slackside tension T2 Fig. 5.36 (a)

But when the drum rotates in the anti clockwise direction, then theend attached to fulcrum O will have tightside tension T1 and other side will

have slackside tension T2. Fig. 5.36 (b)

Take moment about O : Refer Fig. 5.36 (a) and (b)

for clockwiserotation of the drum

Pl T1 b

for anticlockwiserotation of the drum

P l T2 b

where l length of the lever OC

b Distance of B from O. distance between line of action of T1 and T2

Problem 5.44: A band brake acts on the 3/4th of circumference of a drumof 450 mm diameter which is keyed to the shaft. The band brake provides abraking torque of 255 N-m. One end of the band is attached to a fulcrumpin of the lever and the other end to a pin 100 mm from the fulcrum. If theoperating force is applied at 500 mm from the fulcrum and the coefficientof friction is 0.25, find the operating force when the drum rotates in the (a)anti clockwise direction and (b) clockwise direction. (AU. Nov 2007)

Solution: Given: 34

th of the circumference

34

360 270 270

180 4.72 rad

d 0.45 m ; r 0.225 m ; TB 225 Nm ; OB b 0.1 m

OC l 0.5 m ; 0.25


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So the operating force required at the end of the lever for theanticlockwise rotation of the drum is 2261N.

5.42 BAND AND BLOCK BRAKE

The band and block brake is shown in Fig. 5.38. Assume ‘n’ number

of blocks are given and each block is subtending an angle of 2 at the centreand if the drum rotates in anti clockwise direction, then the band attached toA will have T1 (tightside tension) and the other end will have T2 (slackside

tension). It is similar to differential band brake.

Here also, the OA OB, for the brakes to be applied when thedownward force P is applied at C.

T1 Tension in the band between the first and second block.

T1 , T3

Tension in the band between the second and

third block, between the third and fourth block etc.

A

BO

P

C

T1

T2

1

n

n-1

2

3

T 1 T �1

R N

Block

R N

Fig. 5.38 Band and block brake


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By derivation, T1

T1

1 tan1 tan

T1

T2

T2

T3

T3

T4

Tn1

T2

1 tan 1 tan

T1

T2 1 tan 1 tan

n

Braking torque T T1 T2 rIf the band thickness is considered, then

T T1 T2 re

Problem 5.46: In a band and block brake, the band is lined with 14 blocks

each of which subtends an angle of 20 at the centre of the drum. One endof the band is attached to the fulcrum of the brake lever and other end toa pin 15 cm from the fulcrum. Find the force required at the end of the lever100 cm long from the fulcrum centre to give a torque of 4 k N-m. Thediameter of brake drum is 1 m and coefficient of friction is 0.25.

Solution:

Number of blocks n 14 ; Angle 2 20 10

l 1 m ; Braking torque TB 4 103 Nm

Dia of brake drum 1m ; 2.5

We know,

T1

T2 1 tan 1 tan

n

T1

T2 1 0.25 tan 10

1 0.25 tan 10

14

1.0440.956

14

3.435

T1 3.435 T2

Taking moment about O,

For anticlockwise rotation of the drum,


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P l T1a T2b [here b 0]

So P l T1 a

P l 0.15 T1

P 0.15 T1

Also, Braking Torque TB T1 T2 r

4 103 3.435 T2 T2 0.5[. . . T1 3.435 T2]

T2 3285.5 N

T1 3.435 T2 3.435 3285.5 11286 N

T1 11,286 N

Operating force required at the end of lever (P)

P 0.15 T1 0.15 11,286 1692.9 N

A

B

O

P

C

T 1

T 2

1

2

3

Band and block brake

14

13

12

2 = 20 o

� = 100 cm

15 cm


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Problem 5.47: A band and block brake having 12 blocks each of which

subtends an angle of 15 at the centre is applied to a rotating drum ofdiameter 600 mm. The blocks are 75 mm thick. The drum and the fly wheelmounted on the same shaft have mass of 1800 kg and have a combined radiusof gyration of 600 mm. The two ends of the band are attached to pins onthe opposite sides of the brake fulcrum at a distances of 40 mm and 150mm from the fulcrum. If a force of 250 N is applied at a distance of 900mm from the fulcrum, find.(i) The maximum braking torque(ii) The angular retardation of the drum and(iii) the time taken by the system to come to rest from the rated speed of300 rpm.

Take 0.3 (AU. Apr 2002)

Solution:

Given : n 12 ; 2 15 7.5; d 0.6 m

r 0.3 m; m 1800 kg; k 0.6 m; 0.3

a 0.04 m ; b 0.15 m

900

250N

A

T 1

T2

O B C

AO =40

O B > O A

1

12

11

0.6m All d im ensions a re in m m

10

9

8

76 5

4

3

2


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To find maximum braking torque

The braking torque will be maximum, when OB OA and the drumrotates in anticlockwise sense as shown in Fig. So the force P must actupwards and the end of the band attached to A have tight side tension T1,

and the end of the band attached to B have slackside tension T2.

Taking moment about O

P l T1 a T2 b

250 0.9 T1 0.04 T2 0.15

225 0.04 T1 0.15 T2 ...(i)

Also we know,

T1T2

1 tan 1 tan

n

1 0.3 tan 7.51 0.3 tan 7.5

12

1.03950.9605

12

2.582

T1 2.582 T2

Substitute T1 value in eqn (i),

225 0.04 2.582 T2 0.15 T2

T2 4814.6 N

T1 2.582 T2 2.582 4814.6 12,431.3

T1 12,431.3 N

Maximum braking torque TB T1 T2 re

12,431.3 4814.6 0.3

0.0752

. .

. Equivalent radius re radius of drum

thickness of band2

TB 2570.64 Nm


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To find the angular retardation

Braking torque TB I

where I mk2

1800 0.62 648 kg.m2

2570.64 648

3.967 rad /s2

To find the time taken by the system to come to rest

The system is to come to rest N2 0, from the rated speed N1 of

360 rpm.

So N1 360 rpm ; N2 0

Initial angular speed

1 2N1

60

2 36060

37.69 rad /s; 2 0

We know,

2 1 t

0 37.69 3.967 t

[ 3.967 rad/s2 because it is retardation ie., negative acceleration]

t 37.693.967

9.503 sec

time t 9.503 sec

Problem 5.48: The brake whose dimensions are shown in figure has aco-efficient of friction of 0.3 and is to have a maximum pressure of 1000kPa against the friction material.(1) Using an actuating force of 1750 N, determine the face width of theshoes (both shoes have same width) and(2) What torque will the brake absorb? (AU. May 2007)


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Solution:

Given: 0.3

Pb 1000 kPa 1 N/mm2 ; F 1750 N

2 100 100

180 1.75 rad

d 200 mm 0.2 m

r 100 mm

4 sin

2 sin 2

4 0.3 sin 501.75 sin 100

0.336

For the left hand side block, taking moments about O,

F 250 200 RN1 250 RN1 r c 0

1750 450 RN1 250 0.336 RN1 100 40 0

787500 250RN1 20.16 RN1 0

229.84RN1 787500

FF

F=1750 N

200

200 m m

100 o

100o

Ft1 RN 1

R n2

Ft2

O 1 O 2

40 40

b

250 m m

b


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RN1 787500229.84

3426.3 N

Taking moments about O2 for the right hand side block.

1750 450 RN2 250 0.336 RN2 100 40 0

787500 250RN2 20.16RN2 0

RN2 787500270.16

2914.94 N

Let b width of brake shoes in mm.

Bearing area of one shoe, Ab b 2r sin

b 2 100 sin 50

Ab 153.21 b

Taking the maximum normal force, RN1.

Bearing pressure RN1

Ab

3426.3153.21b

i.e., 1 3426.3

153.21b

22.36b

b 22.36 mm

Torque, TB RN1 RN2 r

0.336 3426.3 2914.94 100

213065.664 N mm

TB 213.0657 N m

Problem 5.49: A car moving along a level road is having the following

data: Wheel base of car 2.85 m, Height of C.G. from road surface

600 mm; Perpendicular distance of CG from rear axle 1.2 m; speed

60 km/h; 0.1. Find the minimum distance in which the car may bestopped which brakes are applied (i) to the rear wheels; (ii) to all the fourwheels (AU. Apr 2008)


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Given

Speed 60 km/h 16.67 m/s

Wheel base, L 2.85 m

Distance of C.G from ground, h 600 mm 0.6 m

Distance of C.G from rear a b x 1.2 m

0.1

Solution:

Car moves on a level road

1. When brakes are applied to the rear wheels

Retardation, a g L x

L h

0.1 9.81 2.85 122.85 0.1 0.6

a 0.556 m/s2

Minimum distancetravelled

s

V2

2a

16.672

2 0.556 249.9 m

2. When brakes are applied to all four wheels

Retardation of car a g 9.81 0.1

0.981 m/s2

Minimum distancetravelled

s

V2

2a

16.672

2 0.981

s 141.64 m

Problem 5.50: A bicycle and rider and mass 100 kg are travelling at therate of 16 km/hr on a level road. A brake is applied to the rear wheel whichis 0.9 m in diameter and this is the only resistance acting. How many turnswill it make before it comes to rest? The pressure applied on the brake is

100 N and 0.05. (AU. Nov 2009)

Solution:

Given:

V 16 km/hr 4.44 m/s; D 0.9 m; 0.05; m 100 kg


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Pressure applied on the brake RN 100 N ; m 100 kg

Distance travelled by the bicycle before it comes to rest

Let x be the distance travelled by the bicycle.

Tangential Braking force, Ft RN

0.05 100 5 N (i)

Work done Ft x

5x Nm (ii)

Kinetic energy 12

mV2

12

100 4.442

986.7 N m

To bring bicycle to rest, work done must be equal to kinetic energyFrom (i) and (ii)

5x 986.7 x 986.7

5

x 197.3 m

Number of turns made by the bicycle before it comes to rest

N Number of turns

x DN

0.9 N

2.83 N

N 197.32.83

69.78 ~ 70 turns


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Documents

airwalkbooks.comairwalkbooks.com/images/pdf/pdf_113_1.pdf · 2019-02-23 · Dr. S. Ramachandran, M.E., Ph.D., Professor - Mech Sathyabama Institute of Science and Technology Chennai