ME 264 Mechanisms Synthesis & Analysis I
Supplementary Lecture Notes
Analysis Unit
Josh Ampofo
ii
PREFACE
This handout is supplementary notes for ME 264 Mechanisms Synthesis and Analysis I. It is prepared to suit Aerospace and Mechanical Engineering students of Kwame Nkrumah University of Science and Technology, Kumasi. It covers formalized graphical dimensional mechanisms synthesis, kinematic analysis of machines and mechanisms using both graphical and vector analytical methods, and transmission of rotary motion in including belt drive configuration design.
The style of presentation is clear, informal and easy to read. In addition, sample problems and alternative solutions, if any, are given at the end of each subtopic. Students must note that this lecture handout is not a substitute for textbooks and use of either the Engineering or the main libraries, which have abundant books on the course.
Josh Ampofo, PhD
Mechanical Engineering Department
KNUST, Kumasi
January 2012
iii
TABLE OF CONTENTS
PREFACE ....................................................................................................................................... ii
RECOMMENDED TEXTBOOKS FOR FURTHER READING ................................................. iv
Course Outline ............................................................................................................................... vi
CHAPTER 1 Fundamentals OF MECHANISMS ....................................................................... 1
1.1 Introduction ..................................................................................................................... 1
1.2 Terminologies and Definitions ........................................................................................ 1
1.2.1 Rigid Body ................................................................................................................ 1
1.2.2 Kinematic Link or Element ....................................................................................... 1
1.2.3 Types of Links .......................................................................................................... 2
1.2.3.1 Rigid Link .......................................................................................................... 2
1.2.3.2 Flexible Link ...................................................................................................... 2
1.2.3.3 Fluid Link .......................................................................................................... 2
1.2.4 Joint or Kinematic pair .............................................................................................. 2
1.2.5 Kinematic Chain........................................................................................................ 3
1.2.6 Mechanism ................................................................................................................ 3
1.2.7 Machine ..................................................................................................................... 3
1.2.8 Inversion .................................................................................................................... 3
1.2.9 Differences between a Machine and a Structure ....................................................... 4
1.3 Degree of Freedom (Mobility) in Planar Mechanisms ................................................... 4
1.4 Mechanisms and Structures............................................................................................. 6
1.5 Inconsistencies of Gruebler’s Equation .......................................................................... 6
CHAPTER 2 POSITION ANALYSIS ........................................................................................ 9
2.1 Introduction ..................................................................................................................... 9
2.2 Position and Displacement .............................................................................................. 9
2.2.1 Position ...................................................................................................................... 9
2.2.2 Displacement ............................................................................................................. 9
2.3 Rigid Body Motions ...................................................................................................... 10
2.3.1 Translation............................................................................................................... 10
2.3.2 Rotation ................................................................................................................... 11
2.3.3 Complex Motion ..................................................................................................... 11
2.4 Graphical Position Analysis of Linkages ...................................................................... 11
2.5 Analytical Method ......................................................................................................... 12
2.5.1 Position Analysis of Crank-Slider Linkage ............................................................ 12
2.5.2 Vector Loop Position Analysis of Four-Bar Linkage ............................................. 13
2.5.3 Algebraic Position Analysis of Four-Bar Linkage .................................................. 18
CHAPTER 3 VELOCITY ANALYSIS .................................................................................... 25
3.1 Graphical Velocity Analysis: Instant Centre Method ................................................... 25
3.1.1 Instant Centres ......................................................................................................... 25
3.1.1.1 Kennedy’s Theorem ......................................................................................... 26
3.1.1.2 Application of Kennedy’s Theorem ................................................................ 26
3.1.1.3 Successive Tracking of Instant Centres ........................................................... 26
3.1.1.4 Instant Centres of Sliding Joint ........................................................................ 27
3.1.1.5 Velocity Analysis using Instant Centres .......................................................... 29
3.2 Relative Velocity Method ............................................................................................. 31
3.2.1 General Motion without Sliding ............................................................................. 31
3.2.2 General Motion with Sliding on Fixed Link ........................................................... 33
3.2.3 General Motion with Sliding on Rotating Link ...................................................... 35
iv
3.3 Velocity Analysis Using Vector Method ...................................................................... 38
3.3.1 Pure Translational Motion....................................................................................... 39
3.3.2 General Motion without Sliding on Rotating Link ................................................. 39
3.3.3 General Motion with Sliding on Fixed Link ........................................................... 40
3.3.4 General Motion with Sliding on Rotating Link ...................................................... 44
3.4 Mechanical Advantage of Mechanism .......................................................................... 47
CHAPTER 4 Acceleration Analysis .......................................................................................... 51
4.1 General Motion ............................................................................................................. 51
4.1.1 Pure Translational Motion....................................................................................... 51
4.1.2 General Motion without Sliding on Rotating Link ................................................. 51
4.1.3 General Motion with Sliding on Rotating Link ...................................................... 54
CHAPTER 5 Rotary Motion Transmission ............................................................................... 59
5.1 Introduction ................................................................................................................... 59
5.2 Gear Drive ..................................................................................................................... 59
5.2.1 Types of Gear .......................................................................................................... 59
5.2.1.1 Parallel-Axis Gears .......................................................................................... 59
5.2.1.2 Nonparallel-Axis Coplanar Gears .................................................................... 60
5.2.1.3 Nonparallel-Axis, Noncoplanar Gears ............................................................. 60
5.2.2 Gear Trains .............................................................................................................. 61
5.2.2.1 Ordinary Gear Train......................................................................................... 62
5.2.2.2 General Gear Train Equation ........................................................................... 63
5.2.2.3 Planetary (Epicyclic) Gear Trains.................................................................... 66
Tabular Method ................................................................................................................. 75
5.3 Belt Drives .................................................................................................................... 79
5.3.1 Types of Belt ........................................................................................................... 79
5.3.2 Length and Wrap Angle of Belt Drive .................................................................... 79
5.3.3 Types of Belt Drive Configurations ........................................................................ 80
5.3.4 Slip of Belt .............................................................................................................. 81
RECOMMENDED TEXTBOOKS FOR FURTHER READING
[1] Robert L. Norton, Design of Machinery An introduction to Synthesis and Analysis of Mechanisms and Machines, 2nd Edition, McGraw-Hill. 1999
[2] C. E. Wilson and J. P. Sadler, Kinematics and Dynamics of Machinery, Prentice Hall, 2003.
[3] A. G. Erdman and G. N. Sandor , Mechanism Design, Analysis and Synthesis, Vol. I 2nd Edition, Prentice Hall, 1991.
[4] Joseph E. Shigley and John J. Uicker, Jr, Theory of Machines and Mechanisms, McGraw-Hill, 1981.
v
[5] Charles E. Wilson and J. Peter Sadler, Kinematics and Dynamics of Machinery, Harper Collins College Publishers.
[6] J. Hannah & R. C. Stephens, Mechanics of machines -Elementary theory and Examples, Edward Arnold.
[7] R. S. Khurmi & J. K. Gupta, Theory of Machines, Eurasia Publishing House (PVT.) Ltd, 2005.
vi
COURSE OUTLINE
1. Mechanisms-Definitions & terminology
2. Formalized Dimensional Graphical Synthesis of Linkage
(a) Two Position synthesis without quick-return
(b) Addition of Drive dyad and use of Drive cylinders
(c) Three Position synthesis with and without specified pivots
3. Displacement and Velocity Analysis of planer mechanisms (up to six-bar linkage)
(a) Analytical method using vectors
(b) Graphical
(i) Instantaneous centre method
(ii) Relative velocity method
4. Acceleration Analysis (Vector analytical method only)
5. Transmission of Rotational Motion
(a) Gear trains including planetary gear (epicyclic) trains
(b) Belt and rope drives
(c) Chain drives
1
CHAPTER 1 FUNDAMENTALS OF MECHANISMS
1.1 Introduction
Mechanisms are basic units of machines. Sample machines include cranes, wrapping machines, block-making machines, car jacks, bottling machines, agricultural machines, optical drives, sewing machines, timing mechanisms, standing fans, etc.
Machines consist of simple to complicated mechanisms. In practice, selecting a mechanism to perform a function in a machine is very challenging. Mechanism design and analysis require experience and good background in mathematics and dynamics. Kinematics is the study of the relationship between time, position, velocity and the acceleration of interacting rigid bodies.
1.2 Terminologies and Definitions
1.2.1 Rigid Body
A rigid body is an idealised model of an object that does not deform or change in shape. Thus, the distance between any two points on a rigid body remains constant under the action of forces. Machine components that do not fit this assumption such as springs in either tension or compression and, belt or rope under compressive load are not links.
1.2.2 Kinematic Link or Element
It is a part of machine, which moves relative to some other part. A link may consist of several parts, which are rigidly fastened together, and hence there is not relative motion between each pair of parts of the link.
A link or element needs not to be rigid body, but it must be capable of transmitting the required force with negligible deformation. The main characteristics of a link are:
(a) It should have relative motion between itself and other links
(b) It should be capable of transmitting a force, torque and/or motion with little or no deformation.
2
1.2.3 Types of Links
1.2.3.1 Rigid Link
A rigid link is capable of transmitting motion, force and/or torque with no or negligible deformations. It possesses at least two nodes, which are points for attachment to other links. Figure 2-1 shows a rigid connection between two or more elements of different kinematic pairs.
Figure 1-1: Types of Rigid Links
1.2.3.2 Flexible Link
It is capable of transmitting motion, force and/or torque with deformation that does not affect its performance. Examples of flexible links include belts, ropes, chains and wires. However, belts, chains and ropes have one-way rigidity. Therefore, they are considered as links when they are in tension and non-link when they are in compression.
1.2.3.3 Fluid Link
It is formed by a fluid in a receptacle and the motion, force or torque is is transmitted through the fluid pressure. E.g Hydraulic and pneumatic devices
1.2.4 Joint or Kinematic pair
It is connection between two or more links, which permit some or potential motion between the links. Shown in Figure 2-1 are examples of planer joints. Joints may be classified by:
(i) the type of contact between the links, e.g. point, line or surface.
(ii) the number of degree of freedom allowed by the joint.
(iii) the type of physical closure at the joint: The closure be either force or form closed.
(iv) the number of links joined together.
3
Figure 1-2: Various Types of Planer Joints
1.2.5 Kinematic Chain
It is an assembly of links and joints, which are interconnected such that it has internal mobility and able to transform a supplied motion into a controlled output.
1.2.6 Mechanism
It is a kinematic chain with at least one link grounded or fixed to a reference frame. Three general tasks for mechanisms include the generation of path, motion and function.
1.2.7 Machine
It is an assembly of rigid bodies such that it has internal mobility and capability to apply power, transform motion or/and do work. The principal aims of machines are to produce a desired motion, force, torque, work or power.
1.2.8 Inversion
The absolute motion of a linkage depends on which link is fixed. If two equivalent linkages have different fixed links, then each linkage is an inversion of the other.
4
1.2.9 Differences between a Machine and a Structure
Machine Structure
Relative motion between parts No relative motion between parts
Energy is transformed into useful work No energy is transformed into useful work
Members of a machine may transmit both
power (or force, torque) and motion
Members of a structure may transmit only
force and moments
1.3 Degree of Freedom (Mobility) in Planar Mechanisms
The number of independent inputs required in order to predict the output of a mechanism
is called mobility or degree of freedom. The mobility of a mechanism with one ground link is
given by
1 2 m 3 n 1 2 j j Equation 1-1
where n = number of links, j1 = number of 1 dof joints, and j2 = number of 2 dof joints.
Equation 2-1 is Kutzbach criterion or modified Gruebler’s equation.
Mechanisms may be either open or closed, as illustrated in Figure 2-3. Open mechanisms
have open nodes or attachment points while closed mechanisms lack open nodes. An open
mechanism with more than two links has more than 1 dof. Examples of open mechanisms are
backhoe, front-end loader and industrial robots. A dyad is a two-bar chain, which is used with
motor to drive other linkages.
In Figure 2-3 (a), the mechanism chain has four links (3 links + 1 ground link) and three
1 dof joints. Using Equation 2-1, the mobility of the open mechanism is
m 34 1 23 0 m 3
(a) (b)
Figure 1-3: (a) Open and (b) Closed Mechanism Chains
5
Similarly, the closed mechanism chain in Figure 2-3(b) has four links (3 links + 1 ground link)
and four j1 joints. Using Equation 2-3, the mobility of the mechanism chain is
m 34 1 24 0 m 1
Figure 1-4: Reciprocating Mechanism
Figure 2-5 shows mechanism chains with multiple joints. At joint C of both mechanisms,
three links are joined by a single pin. If one of the links is fixed, then there are two independent
joints at the nodes. Hence, the joints at C are counted as two j1 joints. It may be generalised that
the number of j1 (or 1 dof) joints at a node connecting n links is
1 1 j n , n≥ 2 Equation 1-2
(a) (b)
Figure 1-5: Mechanism chain with multiple Joints
6
1.4 Mechanisms and Structures
Degree of freedom (dof) of an assembly predicts whether an assembly is a mechanism or a structure. If dof is positive (i.e. m ≥ 1), then the assembly is a mechanism and there is relative motion between its links. Figure 2-6 (a) shows an example of a four-link mechanism. The assemblies shown in Figure 2-6 (b and c) have zero dof, which implies that they are structures and no motion is possible. In Figure 2-6 (d and e), the assemblies have negative dof. This means that the assemblies are preloaded or redundant and no motion is possible. Preloaded structures refer to structures having stresses at the time of assembly. Redundant structures are statically indeterminate, and have redundant link(s).
Figure 1-6: Mechanism and Structures
1.5 Inconsistencies of Gruebler’s Equation
Sometimes, the Gruebler’s equation gives false mobility. In Figure 2-7(a), the calculated dof using the Gruebler’s equation is zero though the assembly has one dof. However, similar assembly with different dimensions as shown in Figure 2-7(b) has 1 dof, which agrees with the Gluebler’s equation. The difference in the values of dof is due to omission of dimensions of links in the Gruebler’s equation. Inclusion of dimensions would have made the equation lengthy and complicated, thus making it less useful for dof calculation. Figure 2-7(c) is an example of a mechanism with dof of one, though Gruebler’s equation predicts dof of zero.
7
(a) (b)
Figure 1-7: Linkages having Mobility that disagrees with Gruebler Equation
(c)
Figure 1-8: Mechanism having Mobility that disagrees with Gruebler Equation
Problems
2-1. Determine the mobility of the drum foot-pedal mechanism shown in Figure P2-1
Ans: 1
Figure P1-1 Figure P2-2
2-2. Figure P2-2 is a schematic of a front-loader construction machinery. Determine the degree of freedom of the mechanism. Ans: 2
2-3. In Figure P2-3, how many actuator(s) are required to control position (a) A on the boom, and (b) B on the bucket. Ans: (a) 1, (b) 2 1 dof pure rolling no slip n = 3, j1 = 3, j2 = 0 m = 0
8
Figure P2-3 Figure P2-4
2-4. Find the mobility of the mechanism shown in Figure P2-4. Ans: 2
Figure P2-5
2-5. Find the mobility of the mechanism shown in Figure P2-5. Ans: 1
2-6. What is a machine?
2-7. With examples, differentiate between a machine and a structure.
2-8. Sketch and explain the various inversion of a crank slider chain.
9
CHAPTER 2 POSITION ANALYSIS
2.1 Introduction
A tentative mechanism design may be analysed graphically or analytically to determine displacement, velocity and acceleration of moving parts. Note that graphical position and/or displacement analysis is a truly trivial exercise, while the algebraic approach to position analysis is much more complicated. However, graphical velocity and acceleration analyses are quite complex and difficult.
Position analysis is required for velocity, acceleration and forces analysis. Usually, the lengths of members of a linkage and the position(s) of input(s) are given during position analysis. However, in machine design it may be required to find lengths of members of a linkage that give a desired motion. Complete motion analysis include virtually moving the linkage through its intended path. Doing so requires examination of the linkage at various positions. If a mechanism is to be examined for only one or two positions, the graphical method is convenient. However, for multiple positions analysis, analytical method is more convenient and accurate than the graphical method.
From the second law of Newton, the dynamic force is proportional to acceleration. Therefore, kinematics analysis of mechanism is necessary for calculation of forces and stresses in components.
2.2 Position and Displacement
2.2.1 Position
A position of point in a plane can be defined as the position vector of the point from a given reference axes (or a point) as shown in Figure 3-1. The choice of the reference axes or point is arbitrary or/and may selected to suit the observer. A 2D position vector has two attributes, which may be expressed in polar or Cartesian coordinates. In the polar form, the position vector is expressed in terms of its magnitude and angle. The Cartesian form is expressed in the x and y components of the vector.
2.2.2 Displacement
It defined as the change in position vector or a straight-line distance between the initial and final points of a point, which has moved in a reference frame.
10
Figure 2-1: A vector position in a plane
2.3 Rigid Body Motions
Rigid bodies can exit three types of motion: translation, rotation and complex motion.
2.3.1 Translation
In translational motion, a body moves such that all points on the body have the same displacement (or parallel), as shown in Figure 3-2(a). When a body translates along a straight path, the motion is rectilinear translation. When a body translates along a curved path, the motion is curvilinear translation.
Figure 2-2: Rigid Body Motions
Rx Ry RA A θ x y x y A B RBB' x y A B A' B' RAA' x y B A B' A' x y A B A' B' (a)Translational motion (b)Rotation motion (c)Complex motion
11
2.3.2 Rotation
In rotational motion, a body moves such that one point remains at the origin and all other points move with the same angular displacement, as shown in Figure 3-2(b).
2.3.3 Complex Motion
The complex motion is a combination of translational and rotation motions, as shown in Figure 3-2(c)
2.4 Graphical Position Analysis of Linkages
Graphical displacement analysis is trivial exercise, but very tedious as each of the link position must redone. The algebraic or vector approach is less tedious as one expression is required for all positions, but it is complex and difficult.
Displacement analysis of a single dof mechanism such as a four-bar mechanism may be performed graphically by generating a number of positions of the mechanism. A quick method for generating a number of positions of a mechanism is illustrated in Figure 3-1. The method is as follows:
1. Construct the starting or initial position of the mechanism and marked each joint with the subscript.
2. Draw all known paths of joints. In the figure, a circle about A with radius AB1 and arc about D with radius DC1 are the paths of joints C and D, respectively. In addition, line GG0 is the path of the sliding joint F, which is parallel to the sliding surface.
3. Draw the jth position of the input link ABj. The moving joint B of the input link AB is always determined first.
4. With the radius B1C1, draw an arc about Bj to intersect the path of joint C at point Cj.
5. Locate joint Ej by intersecting the arc of radius C1E1 centred at Cj and the arc of radius DE1 centred at D.
6. Locate joint Fj by intersecting the arc of radius E1F1 centered at Ej and the path GG) of the slider at Fj.
12
Figure 2-3: Graphical Displacement Analysis
2.5 Analytical Method
2.5.1 Position Analysis of Crank-Slider Linkage
Consider an offset crank-slider linkage, which has crank of length r and at angle θ to the
positive x-axis, and connecting rod of length l and sliding along the horizontal direction as
shown in Figure 3-2. The crank length r, angular position θ and connecting rod length l are
normally the inputs for determine the position of the slider. The vertical position of point B is
r sin e l sin
which leads to
l
r e
sin
sin 1 Equation 2-1
Figure 2-4: Positions of a Crank-Slider Linkage
Then, the position of the slider is given as
x r cos l cos Equation 2-2
r
l
e
θ
ϕ
x
y
B
C
A
13
Differentiating the above equation once and twice with respect to time respectively give velocity
and acceleration of the slider.
Example 2-1
Find the horizontal position x0 of the slider C and the inclination of the connecting rod
BC to the horizontal axis of the crank-slider mechanism shown in Figure E4-1, which has the
following lengths:
Crank: r1 = 30 mm
Connecting rod BC: r2 = 100 mm
Vertical offset of slider from horizontal axis: yo = 10 mm
Orientation of crank to horizontal axis: θ1 = 45o
Figure E4-1
Solution 2-1
Using Equation 3-1,
100
30sin 45 10
sin
sin
sin 1
2
1 1 1
2 r
r yo
o 6.44 2
Using Equation 3-2,
cos cos 30cos45 100cos(6.44) 0 1 1 2 2 x r r 120.58 mm 0 x
2.5.2 Vector Loop Position Analysis of Four-Bar Linkage
Consider a four-bar linkage shown in Figure 3-3. The vector of diagonal BD is given by the
vector loop equation
r1
r2
θ1
θ2
y
x
x0
C
B
A y0
14
Figure 2-5: Vector Loop representation of a Four-Bar Linkage
0 1 r r r d
(1)
Taking the dot product of each side of the above equation, we have
d d a r .r r r . r r r r 2r r cos 0 1
2
1
2
0 1 0 1 0
From which we have
0 1 1
2
1
2
0
2 r r r 2r r cos 180 d
Equation 2-3
or 0 1 1
2
1
2
0
2 r r r 2r r cos d
which is basically the cosine law. The direction of the diagonal is given in terms of the x and y
components of r0 and r1 as
d
y y
d r
r r0 1 sin
Equation 2-4
d
x x
d r
r r0 1 cos
Equation 2-5
where subscript i stands for link i and
cos ix i i r r
sin iy i i r r
If the solution of Equations 3-4 and 3-5 give values greater than 1, then the links cannot connect
at a given input angle or at all. The orientation of the diagonal vector given by θd may be
determined from Table 2 or from the relation
d
d d
sin
1 cos
2
tan
Equation 2-6
Using Equation 3-6 does not require the determination of the quadrant of the angle.
r1
r2
θ1
θd
Link 1 B
r3
r0
Link 2
Link 3
Fixed Link
rd
θ2
θ3
θt
A
C
D
θb
θa
15
Table 1: Quadrant and Corresponding Value of θd
cos θd sin θd θd
positive positive
d
d
d
cos
sin
tan 1
negative positive
d
d
d
cos
sin
180 tan 1
negative negative
d
d
d
cos
sin
180 tan 1
positive negative
d
d
d
cos
sin
360 tan 1
Considering the loop BC CD DB 0
, then
3 2 r r r d (2)
Taking the dot product of each side of the above equation, we have
3 3 2 2 r .r r r . r r d d
From which we have
2
2
2
2 2
3
2
cos
r r
r r r
d
d
b
Equation 2-7
Then,
d b 2
Equation 2-8
where θb is negative when the vector loop r2r3rd is clockwise, otherwise positive. The x and y
components of r3 are determined from the loop equation for the entire linkage. Substituting
equation (1) into (2), we have
3 0 1 2 r r r r
Equation 2-9
The above equation may be written in terms of the vertical and horizontal components as
3 0 1 1 2 2 r r r cos r cos x x
Equation 2-10
3 0 1 1 2 2 r r r sin r sin y y
Equation 2-11
As a check, the calculated length of link 3, r3, must be equal to the given length.
2
3
2
3 3x y r r r
16
If the fixed link is aligned to the horizontal, then
0 0 y r
and 0 0 r r x
If the fixed link is inclined at angle θ0 to the horizontal axis, then
0 0 0 r r sin y
0 0 0 r r cos x
The orientation of link 3 is determined from horizontal and vertical lengths of link 3 as
3
3
3 sin
r
r y
Equation 2-12
3
3
3 cos
r
r x
Equation 2-13
Then, θ3 is determined using Table 2.
Example 2-2
A four-bar linkage has the following lengths:
Fixed link: r0 = 45 mm
Crank: r1 = 10 mm
Coupler: r2 = 50 mm
Rocker: r3 = 20 mm
Find the positions of links 2 and 3 when the crank is at θ1 = 45o to the x-axis. Assume that the
fixed link is aligned to the x-axis.
Solution 2-2
Using Equation 3-3,
2 cos180 45 10 24510cos180 45 2 2
0 1 1
2
1
2
0
2 r r r r r d
38.58 mm d r
Using Equations 3-4 and 3-5
38.58
0 10sin 45
sin 0 1
d
y y
d r
r r
sin 0.18328 d
38.58
45 10cos 45
cos 0 1
d
x x
d r
r r
cos 0.98312 d
Using Equation 3-6,
0.18328
1 0.98312
sin
1 cos
2
tan
d
d d
o
d 169.44
17
Alternatively, since sin θd is positive and cos θd is negative, the diagonal vector lies in second
quadrant. Hence, using Table 2, θd is given by
0.98312
0.18328
180 tan 1
d
o
d 169.44
Using Equation 3-7,
238.5850
20 38.58 50
2
cos
2 2 2
2
2
2
2 2
3
r r
r r r
d
d
b
o
b 158.45
For assembly mode 1 and using Equation 3-8,
169.4 158.45 2,mod 1 e d b
o
e 10.95 2,mod 1
For assembly mode 2
,
169.4 158.45 2,mod 2 e d b
o
e 327.85 2,mod 2
The horizontal and vertical components of r3 are
cos cos 45 10cos45 50cos10.95 3 0 1 1 2 2 r r r r x x
11.16 mm 3 x r
sin sin 0 10sin 45 50sin10.95 3 0 1 1 2 2 r r r r y y
16.57 mm 3 y r
Then
2 2 2
3
2
3 3 11.16 16.57 x y r r r
19.978 mm 3 r
which is approximately equal to r3
20
16.57
sin
3
3
3
r
r y
sin 0.8285 3
20
11.16
cos
3
3
3
r
r x
cos 0.558 3
Using Equation 3-6,
0.8285
1 0.558
sin
1 cos
2
tan
3
3 3
o 124 3
o o 360 124 3
o 236 3
Alternatively, both sine and cosine are negative, which implies that the angle is in the third
quadrant. Hence, θ3 is given by
0.558
0.8285
180 tan 1
3
o 236 3
For the alternative mode, the horizontal and vertical components of r3 are
cos cos 45 10cos45 50cos327.85 3 0 1 1 2 2 r r r r x x
4.4 mm 3 x r
sin sin 0 10sin45 50sin327.85 3 0 1 1 2 2 r r r r y y
19.54 mm 3 y r
18
Check
2 2 2
3
2
3 3 4.4 19.54 x y r r r
20.0 mm 3 r
which is approximately equal to r3. The orientation of link 3 is determined from
20
19.54
sin 3
sin 0.977 3
20
4.4
cos 3
cos 0.22 3
Then, using Table 2, θ3 is given by
0.22
0.977
180 tan 1
3
o 102.6 3
(a) Assembly Mode 1
(b) Assembly Mode 2
Figure E3-2: Modes of Assembly of Example 4-2
2.5.3 Algebraic Position Analysis of Four-Bar Linkage
Consider a four-bar linkage shown in Figure 3-3. The coordinates of point B are
r1
r2
θ1= 45o
θd=169.4o
B
Link 1
r3
r0
Link 2
Link 3
Fixed Link
rd
θ2= 10.95o
θ3 =236o
A
C
D
r1
r2
θ1= 45o
θd=169.4o
B
Link 1
r3
r0
Link 3
Fixed Link
rd
θ2= 327.85o
θ3 =102.6o
A
C
D
Link 2
19
1 1 B r cos x
1 1 B r sin y
Equation 2-14
The coordinates of point C are found using the equations of circles about B and D,
2 2 2
2 x x y y r C B C B
Equation 2-15
2 2
0
2
3 x y r C r C
Equation 2-16
which provide a pair of simultaneous equations in Cx and Cy.
Figure 2-6: Algebraic Positions of a four-bar Linkage
Subtracting Equation 3-17 from 3-16 gives an expression for
0 0 0
2
0
2
3
2
2
2
1
2
2
2
2
2 B r
B C
S
B r
B C
B r
r r r r
C
x
y y
x
y y
x
x
Equation 2-17
Substituting 3-18 into 3-17 gives a quadratic equation in Cy,
0
2
2 2
3
2
0
2
r
B r
B C
C S
x
y y
y
Equation 2-18
which has two solutions corresponding to the modes of the assembly shown in Figure 3-6.
Equation 3-19 can be solved for the roots of a quadratic equation, given as
P
Q Q PR
Cy 2
4 2
Equation 2-19
where
r1
r2
θ1
θ3
B
Link 1
r3
r0
Link 2 Link 3
Fixed Link or Link 1
θ2
A
C
D
θ2'
θ3'
C'
20
1 2
0
2
B r
B
P
x
y
0
0 2
B r
B r S
Q
x
y
2
3
2
0 R r S r
0
2
0
2
3
2
2
2
1
2 B r
r r r r
S
x
The solution to this set can be real or imaginary. If the solution is imaginary, then the links
cannot connect at a given input angle or at all. The two real values of By can be substituted into
Equation 3-18 to find the corresponding x component (Cx). The link angles for this position can
be determined from
x x
y y
C B
C B 1
2 2 tan
0
1
3 3 tan
C r
C
x
y
Equation 2-20
where β and plus or minus sign are determined from Table 2.
Example 2-3
Re-solve Example 3-2 using the algebraic method.
Solution 2-3
cos 10cos45 1 1 B r x 7.07 mm x B
sin 10sin 45 1 1 B r y
7.07 mm x B
Then
27.07 45
10 50 20 45
2
2 2 2 2
0
2
0
2
3
2
2
2
1
B r
r r r r
S
x
S 53.059
1
7.07 45
7.07
1 2
2
2
0
2
B r
B
P
x
y
P 1.006
7.07 45
2 2 7.07 45 53.059
0
0
B r
B r S
Q
x
y
Q 3.005
2 2 2
3
2
0 R r S r 45 53.059 20
R 335
Using Equation 3-18,
21
21.006
3.005 3.005 4 1.006 335
2
4 2 2
P
Q Q PR
Cy
16.817 1 , y C and 19.804 ,2 y C
Substituting the above into Equation 3-16, we have
27.07 45
2 7.07 16.817
53.059
2
2
0
1 ,
1 ,
B r
B C
C S
x
y y
x
56.19 1 , x C
27.07 45
2 7.07 19.804
53.059
2
2
0
,2
,2
B r
B C
C S
x
y y
x
49.37 1 , x C
Now
16.817 7.07 1 , y y C B 9.746 ve 1 , y y C B
56.19 7.07 1 , x x C B 49.126 ve 1 , x x C B
Since the above values are both positive, θ2 lies in the first quadrant. Hence
49.126
9.746
tan 0 tan 1 1
2 2
x x
y y
C B
C B
o 11.22 2
For mode 2, we have
19.804 7.07 ,2 y y C B 26.875 ve ,2 y y C B
49.37 7.07 ,2 x x C B 42.3 ve ,2 x x C B
which is the 4th quadrant
42.3
26.875
360 tan 1
2
o
o 327.6 2
16.817 ve 1 , y C
56.19 45 1 , x o C r 11.19 ve 1 , x o C r (1st quadrant)
11.19
16.817
tan 0 tan 1
0
1
3 3 C r
C
x
y
o 56.36 3
19.804 - ve ,2 y C
49.126 45 ,2 x o C r 4.126 ve ,2 x o C r (4th quadrant)
4.126
19.804
tan 360 tan 1
0
1
3 3
o
x
y
C r
C
o 281.8 3
22
The modes of this example are shown in Figure E3-3. The results from vector loop and algebraic
methods are same. Note the in vector loop method angle o 102.6 3
is
measured from the
positive horizontal axis at point C to CD. In the algebraic method, the angle is measured from the
positive horizontal axis at D to DC. Hence, the
180 3,algebraic 3,vector loop
Figure E3-3
Problems
4-1. A mechanism, as shown in Figure P3-1, has the following dimensions: AB = 50 mm; BC =
250; CD = 130 mm; CD = 200 mm, AD = 300 mm and CE = 200 mm. Slider C can only
move vertically. If the crank AB rotates from θ = 25o to θ = 90o, determine (a) angular
displacement of link CD, and (b) displacement of slider E. Ans. (a) 32.4o (b) 72.6 mm
Figure P3-1 Figure P3-2
45o 56.4o
B
10 mm
45 mm
50 mm 20 mm 11.22o
A
C
D
327.6o
281.8o
C'
23
4-2. In the swiveling joint mechanism, shown in Figure P3-2, the driving crank is rotating clock clockwise. The lengths of various links are : OA = 50 mm; AB 350 mm; AD = DB; DE=EF = 250 mm; CB = 125; OC = 300 mm and vertical distance between F and C is 250 mm. If OA rotates from θ = 45o to 135o, Determine (a) displacement of block F, (b) vertical displacement of DE, (c) angular displacement of DE, (d) angular displacement of CB. Ans: (a) 144.8 mm, (b) 14.5 mm, (c) 51.7o, (d) 38.5o
4-3. The mechanism shown in Figure P3-3 has the following dimensions: BC = CD= 150 mm; BD = 100 mm; DE = 200 mm; AC = 200 mm. The block E slides horizontal and the vertical distance of the sliding path from point C is 120 mm. Link BCD is a ternary link, which is pivoted at point C. If the slider, which is initially at horizontal distance of 170.7 mm from point C, slides 80 mm to the right, determine (a) angular displacement of the actuator AB, (b) extension or contraction of the actuator and (c) angular displacement of the ternary link BCD. Ans. (a) 5.3o (b) 76.2 mm, (c) 30o
Figure P3-3
4-4. The 50-mm crank AB of the inverted-slider-crank mechanism shown in Figure P3-4 is rotating clockwise. If crank AB rotates through 60o from the vertical position, determine (a) the angular displacement of the inverted-slider, and (b) displacement of the slider B along the sliding path. Use graphical method. Ans. (a) 5.8o (b) 38.8 mm
4-5. Figure 3-5 shows a mechanism used in one of the shaping machines at the Mechanical Engineering Department’s machine shop. The slotted yoke AD is pinned to the ground at A while the end D is used to move the ram R horizontally. The disc BC has radius 30 cm and other dimensions of the mechanism are: AB = 54.6 cm, AD = 132.3 cm and DR =90 cm. Draw the positions of the mechanism when pin C is at C0 and C1. Use drawing to determine displacement of the ram R when pin C moves from C0 to C1. Ans: 79.5 cm
24
Figure 3-4 Figure 3-5
25
CHAPTER 3 VELOCITY ANALYSIS
3.1 Graphical Velocity Analysis: Instant Centre Method
3.1.1 Instant Centres
An instant (or instantaneous) centre (IC) is a point where there is no relative velocity
between two links of a mechanism at that instant. Instant centres are useful for calculation of
linear and angular velocities, and mechanical advantage of mechanisms.
In any mechanism, obvious instant centres are joints. In the four-link mechanism shown
in Figure 4-1, there is no relative motion between link 1 and link 2 at joint A. Similarly, there are
no relative motions between links 2 and 3 at joint B, between links 3 and 4 at joint C, and
between links 4 and 1 at joint D. Therefore, the joints at A, B, C and D are instant centres.
Figure 3-1: Instant Centres Coinciding at the Joints
Normally, an instant centre is labelled with numbers of the pair of links that forms it. For
example, (1, 2) is the instant centre of links 1 and 2. The orders in which the numbers of the links
appear in an instant label are irrelevant. An instant centre may not coincide on any joint. Such an
instant centre needs to be determined using the Kennedy’s theorem. The number of instant
centres of a linkage is determined by pairing all the links. Thus,
2
1
2
n n
N Cn
, n ≥ 2 Equation 3-1
where N = number of instant centres and n = number of links
Example 3-1
Find the number of instant centres of a
(a) four-link mechanism, and (b) six-link mechanism
Solution 3-1
From Equation 4-1
26
(a)
6
2
4 4 1
2
1
n n
N (b)
15
2
6 6 1
2
1
n n
N
3.1.1.1 Kennedy’s Theorem
Kennedy’s theorem deals with three instant centres of three links in a mechanism. It states that
three instant centres of three links moving relative to another must lie on a straight line.
3.1.1.2 Application of Kennedy’s Theorem
In Figure 4-2, the obvious centres (2, 3) and (3, 4) must lie on the same line and are connected to
the links 2, 3 and 4. Since the number 3 is common to centres (2, 3) and (3, 4), the left-out
combination is (2, 4). Hence, the non-obvious centre on the line is (2, 4). Similarly, the nonobvious
instant centre on the line connecting instant centres (1, 2) and (4, 1) is (2, 4). The point
of intersection of lines through (2, 3) and (3, 4), and through (1, 2) and (4, 1) is the instant centre
(2, 4). Similarly, the instant centre (1, 3) is the point of intersection of line through (3, 4) and (4,
1), and through (2, 3) and (1, 2).
Figure 3-2: Non-obvious Instant Centres of a Four-Link Mechanism
3.1.1.3 Successive Tracking of Instant Centres
For complex mechanisms (i.e. more than four links), keeping track of instant centres
already obtained and those that are obtainable is very helpful. First, the tracking chart is drawn.
The chart consists of a circle marked with the numbers of links in the mechanism. Considering
the four-link mechanism shown in Figure 4-2, four marks 1 to 4 are indicated on a tracking chart
27
circle, which is illustrated in Figure 4-3. These four marks on the circle represent the four links, and the continuous lines connecting the marks represent the identified instant centres. For example, the continuous line connecting 1 and 2 represents the instant centre (1, 2). Whenever an instant centre is about to be located, the corresponding marks on the tracking chart are connected with dotted lines. The dotted lines are immediately changed to continuous lines as soon as the instant centre are located. The process is repeated until all the instant centres are located. For a four-link mechanism, tracking chart may not be necessary.
Figure 3-3: Instant Centre tracking Chart
The tracking chart also indicates the centres that may be connected in order to locate an instant centre. An instant centre is only obtainable if the two links connected to the centre form two independent path triangles on the tracking chart. In Figure 4-3, the marks 1, 2 and 3 form a triangle, while 1, 4 and 3 form another triangle. This implies that the centre (1, 3) is at the point of intersection of lines through (1, 2) and (2, 3), and through (1, 4) and (3, 4). Similarly, the instant centre (2, 4) is at the point of intersection of the lines through (1, 4 and (1, 2), and through (3, 4) and (2, 3).
3.1.1.4 Instant Centres of Sliding Joint
The instant centre of slider joint is not necessary at the point where the two links meet. It is off at infinity in the direction perpendicular to the sliding direction. Thus, the centre is at any point on any line perpendicular to the direction of sliding. Figure 4-4 shows a slider mechanism with sliding joint at (4, 1). In the figure, the instant centre (4, 1) lies on the two dotted lines. The instant (1, 3) is located at the point of intersection of the lines through (1, 2) and (2, 3), and through (4, 1) and perpendicular to sliding direction. Similarly, instant centre (2, 4) is identified 2 1 3 4 (a) 2 1 3 4 (b)
28
by construction a lines through (2, 3) and (3, 4), and through (1, 2) and perpendicular to the
sliding direction.
Figure 3-4: Instant Centre of Mechanism with Slider
Example 3-2
Determine the location of all the instant centres for the six-bar link mechanism shown in
Figure E4-2.
Figure E4-2
Solution 3-2
The mechanism has six links and therefore the number of instant centres is, by equation 4-1
15
2
6 6 1
2
1
n n
N
By inspection, the obvious instant centres are indicated in the tracking chart shown in
Figure S4-1(a). In the figure, the continuous lines are the determined instant centres, and dotted
lines are the next obtainable centres. It is clear from the figure that the instant centres that are
obtainable are (1, 3), (2, 4), (1, 5) and (4, 6). In addition, instant centres such as (2, 6), (3, 6) and
(2, 5) are unobtainable. The final tracking chart is illustrated in Figure S4-1(b). The rest of the
29
instant centres, shown in Figure S4-1(c), are obtained through the same process of linking ticks
on the tracking chart.
(a) (b)
Figure S4-1: (a) Initial, and (b) Almost Completed Tracking Chart
Figure S4-1(c): Construction of all instant centres of Example 4-1
3.1.1.5 Velocity Analysis using Instant Centres
Consider a pair of gears shown in Figure 4-5 that produce the same instant angular velocity ratio
as a pair of links. From the figure, the pitch radii of gears 2 and 3 are
AB and
CB , respectively.
At the point of contact of the two gears, the tangential VB is the same for both gears. Thus,
1
2
4
5
6
3
1
2
4
5
6
30
(a) (b)
Figure 3-5: Graphical Displacement Analysis
V AB CB B 2 3
1,2 2,3
1,3 2,3
3
2
AB
CB
1,2 2,3
1,3 2,3
3
2
From the figure, the equation is negative when the instant centre (2, 3) lies between the instant
centres (1,2) and (1,3), and positive when the instant centre (2, 3) lies between the instant centres
(1, 2) and (1, 3). Generalisation of the above equation gives
n n m
m n m
m
n
1, ,
1, ,
, Equation 3-2
where n and m are numbers of interacting links.
Example 3-3
If bar AB in Figure E4-3 rotates at 12 rad/s in the anti-clockwise direction, determine the
instantaneous angular velocity of bars BC and CD.
Figure E4-3
2
)
(1, 2)
3
)
(1, 3)
1
)
(2, 3)
ω2
)
ω3)
A
)
B
) C
)
VB
2
)
(1, 2)
3
)
(1, 3)
1
)
(2, 3)
ω2
)
ω3)
A
)
B
) C
)
VB
(2, 3)
31
Solution 3-3
Using the Kennedy’s theorem, the instant centres relating to links BC and CD are located,
as shown in Figure S4-3. Using Equation 4-2
6
2
1,3 2,3
1,2 2,3
2
3
12 4
6
2
3 rad/s 4 BC rad/s
[Hint: Note (2, 3) lies between (1, 2) and (1, 3)]
Similarly,
9
3
1,4 2,4
1,2 2,4
2
4
12 4
9
3
3 rad/s 4 CD rad/s
(Hint: Note (2, 4) outside of (1, 2) and (1, 4)
Figure S4-3: Instant Centres of Example 4-2
3.2 Relative Velocity Method
3.2.1 General Motion without Sliding
Consider the ends of a rigid link AC in Figure 4-6 (a) moving with absolute velocities va
and vc, respectively. Note that these velocities are relative to the fixed point O. If the link is not
extending, then, difference in velocity is due to only rotation about a point. Therefore, the link
may be regarded as rotating about a point A with linear velocity given by
v AC ac x Equation 3-3
where vab is the velocity of point C relative A and ω is angular velocity of the link AC. The
direction of vac is perpendicular to the line connecting points A and C because it is due to
rotation. The velocity of point B, which is located on the line from A to C, is also perpendicular
the line AC. The velocity of point B relative to point A is shown in Figure 4-6(b) is
32
v AB ab Equation 3-4
Substituting ω in Equation 3-3 into Equation 3-4 yields
AC
AB
v vab ac Equation 3-5
The relative velocity of point B is located on the line ac of the velocity diagram using
proportionality relationship given by Equation 4-5. Note that lower case letters are used for
points on relative velocity diagrams.
Figure 3-6 (a) Link having Rotating Motion, and (b) its Relative Velocity Diagram.
Example 3-4
If the bar AB in Figure E4-4 rotates at 10 rad/s in the anti-clockwise angular direction,
determine the instantaneous angular velocity of bar CD.
Figure E4-4
10 rad/s
3 m 2 m
2 m
A
B
C
D
33
Solution 3-4
The fixed (or zero-velocity) points are A and D, and are indicated with lower cases a and d,
respectively. Using Equation 4-4, the velocity of point B is
v AB 10x2 20 m/s ab .
Since AB is rotating in the anti-clockwise direction, vector ab points to the left and is
perpendicular to AB. The velocity of C relative B is perpendicular to BC and the velocity of C
relative D is perpendicular to CD. Hence, the intersection of lines through b and d, perpendicular
to BC and DC respectively, yields the point c, as shown in Figure S4-4. Note that the points A
and D are on the ground link and have zero velocities. Therefore, the two points have the same
relative velocity diagram. From the relative velocity diagram, velocity of point C relative d is
28.24 m/s. The velocity of point C is
v w r w CD C CD C D CD x x /
10 rad/s
2 2
28.28
2 2
CD
vC
CD
Figure S4-4: Relative Velocity Diagram of Problem 3-5
3.2.2 General Motion with Sliding on Fixed Link
Consider a rigid bar B sliding on a rigid link AC, as shown in Figure 4-7(a). The velocity
of B relative to the link AC is in the same direction as (or parallel to) the sliding direction AC,
which is illustrated in Figure 4-7(b). If B is moving towards the left, then b will be on the left of
a and c. Note that A and C are fixed (zero-velocity) points, and therefore have zero velocities.
b a, d
c
34
Figure 3-7 (a) Rigid Body in Sliding Planar Motion, and (b) Corresponding Relative
Velocity Diagram
Example 3-5
In Figure E4-5, the angle θ = 45o and the link AB has a constant anti-clockwise angular
velocity of 5 rad/s. Determine the sliding velocity of C.
Figure E4-5
Solution 3-5
The velocity of point B is, using Equation 3-4
v AB 5x0.2 1.0 m/s ab .
The line ab is perpendicular to AB with length equivalent to 1.0 m/s. The velocity of C relative
to D is parallel to the sliding direction, which is given by line dc. To locate point c, a line
through b and perpendicular to BC is construction and the intersection of bc and dc is c. Figure
S4-5 shows the constructed relative velocity diagram. From the relative velocity diagram, the
velocity of the slider is dc, which is 0.974 m/s.
Figure S4-5: Relative velocity diagram Example 4-5
(a) (b)
A
B
C
a,c b
A
35
3.2.3 General Motion with Sliding on Rotating Link
In Figure 4-8(a), link OA is sliding and rotating about a fixed point O with angular
velocity ω, as shown. Let A and A be the points on the rotating link and sliding link,
respectively. The two points are coinciding at point A. Using equation 3-4, the velocity of A
relative to point O is given is
v OA A .
The velocity A v is perpendicular to link OA, and the velocity of A relative to A is
parallel to OA. The velocities of these coinciding points are illustrated on the velocity diagram
shown in Figure 4-8 (b).
Figure 3-8 (a) Rigid Body Sliding and Rotating, and (b) Corresponding Relative Velocity
Diagram
Example 3-6
In Figure E4-6, the hydraulic actuator BC of the crane is extending at a constant rate of 0.2 m/s.
When the angle β = 35o, what is the angular velocity of the crane’s boom AD.
Figure E4-6
o
a'
a
O
va
A'
A
ω
(a) (b)
36
Solution 3-6
The equivalent mechanism and velocity diagram are shown in Figure S4-6. In the figure,
point B' is an imaginary point on the link BB'. First, the velocity of B' relative to B is drawn
perpendicular to the line B- B’ and on the left side of point (a, b) on the velocity diagram. The
point b’ on the relative velocity is on the left-hand side of b because as the actuator BC expands,
BB' turns in the anti-clockwise direction. The velocity of C relative to A is drawn perpendicular
to AC and on the left-hand side of C because AC also turns in the anti-clockwise direction.
Along the direction of BC, the velocity of C relative B' is drawn through b to say point c' with
dotted or construction lines. The length of b'c has an equivalent magnitude of 0.2 m/s. A line
through c and parallel to bc' is draw to meet line ac. The point where the line through c' meets
line ac is point c. From the diagram, the velocity of point C is, which is the length of ac, is
vC = 0.3104 m/s.
0.1035 rad/s
3
0.3104
AC
vC
AD
Figure S4-6 (a) Equivalent Mechanism and (b) Relative Velocity
Example 3-7
Link AB of the mechanism shown in Figure E4-7 is rotating at 20 rad/s in the anti-clockwise
direction. Using the relative velocity method, determine the angular velocity of ADE and vertical
sliding velocity of node E.
37
Figure E4-7
Solution 3-7
Using Equation 3-4, the velocity of point B relative point A is computed:,
1000
45
v x AB 20 x B A AB 0.9 B A v m/s
The ground a and d are located and a line of equivalent length equal to 0.9 B A v m/s is
constructed perpendicular to link AB, through point a, d and points in the downward direction.
The constructed relative velocity diagram is shown in Figure S4-7. In the figure, the point c is at
intersection of lines perpendicular to BC and through b and through point a, d and perpendicular
to link DC. Two points are coinciding at point E. Let E and E’ be points on link BCE and slider
E. Relative to point B, points C and E have the same angular velocities. Point e is located the line
through b and c and be is computed using Line bc using Equation 3-6.
150
251.66
0.554
BC
BE
be bc be 0.92946 m/s
Point E’ slides on link BCE and therefore a line is constructed through point e and parallel to link
BCE. Since point E’ slides in the vertical direction through the ground, line is constructed
through point a,d and parallel to the vertical. The point of intersection of the two lines is point
e’. From the relative velocity diagram, the vertical velocity of the slider is ae’, which is 1.31 m/s.
Thus,
1.31 e v m/s
38
Figure S4-7
3.3 Velocity Analysis Using Vector Method
Shown in Figure 4-9 are two points A and B on a body, and point O as the reference
frame. The position of B is
B A B A r r r / Equation 3-6
where B A r / is the position of point B relative to point A, which may written as
r xi yj zk B A /
Differentiation Equation 3-7 with respect to time gives
dt
dk
k z
dt
dz
dt
dj
j y
dt
dy
dt
di
i x
dt
dx
v vB A
dt
dk
z
dt
dj
y
dt
di
k x
dt
dz
j
dt
dy
i
dt
dx
v vB A (a)
Differentiating the unit vectors with respect to time only causes rotation of the vectors. Thus,
w i
dt
di
x w j
dt
dj
x w k
dt
dk
x (b)
39
Figure 3-9: General Motion of a Rigid Body
Substituting the above equation (b) into equation (a), we have
k x x i y x j z x k
dt
dz
j
dt
dy
i
dt
dx
v vB A
k xi yj zk
dt
dz
j
dt
dy
i
dt
dx
v vB A
x
/ B/A v v v w x r B A B A Equation 3-7
where
k
dt
dz
j
dt
dy
i
dt
dx
v B / A
where vA and vB are respectively velocities of point A and B relative to the reference frame, and
B A v / is the sliding velocity of point B relative to point A.
3.3.1 Pure Translational Motion
For pure translational motion of a rigid body, 0 / B A v andw 0 , hence Equation 4-7
becomes: B A v v .
3.3.2 General Motion without Sliding on Rotating Link
For general motion without sliding on a rotating link, the sliding velocity, vB/A , is zero.
Hence, Equation 4-7 becomes
B A B A v v w r / x Equation 3-8
40
Equation 4-8 is applicable to both 2D and 3D motions. For 2D motions in x-y plane, the angular
velocity is always in the z direction and, therefore, unit vector k is attached to the expression
of angular velocity.
3.3.3 General Motion with Sliding on Fixed Link
When a body slides on a fixed link, the velocity of the body is determined by imposing
motion along the direction of sliding on the body. Consider a body sliding on a fixed link that is
inclined at θ to the horizontal in x-y plane, as shown in Figure 4-10. If the magnitude of the
sliding velocity is V, then Equation 4-9 is imposed on the slider. Equation 4-9 is only applicable
to planar (2D) motions.
v V i j slider cos sin . Equation 3-9
Figure 3-10: Rigid Body Sliding on Fixed Link (2D Motion)
Example 3-8
Crank AB of crank-slider mechanism in Figure E4-8 is rotating at 15 rad/s in counter-clockwise
direction. Determine the velocity of slider C and angular velocity of connecting rod BC.
Figure E4-8
Slider Direction of Motion
Fixed Link
θ
x
y
41
Solution 3-8
First, we determine the velocity of point B from that of A since the point A is stationary
and angular velocity of AB is known. Imposing horizontal motion on the point C (or slider), the
velocity of the slider can be computed.
Using Equation 4-8,
B A B A v v w r / x
B A B A v v wk r / x
0.2 0.3 0
0 0 0 15
i j k
v 3i 4.5 j (m/s) B
Also
C B BC C B v v w r / x
0.6 0.2 0
3 4.5 0 0 C BC
i j k
v i i j
v i j C BC BC 0.2 3.0 0.6 4.5 (1)
Using Equation 4-9
v V i j v i j V i C
o o
C c c cos sin cos 0 sin 0 (2)
where C V is the magnitude of sliding velocity of C. Equating the i and j components of equation
(2) yields
0 0.6 4.5
0.2 3.0
BC
C BC v
Solving the above two equations simultaneously gives
4.5 m/s
7.5 /
C
BC
v
rad s
Example 3-9
The crank AB in Figure E4-9 rotates 200 rpm in the clockwise direction. Determine the
angular velocity of the link CD in rpm and the vertical velocity VE of the rack-and-pinion gear if
average diameter of the pinion is 0.6 m.
42
Figure E4-9
Solution 3-9
The Equation 4-8 is applied between points A and B. From the figure
r i j B A 0.3 0.6 /
x rad s AB 20.94 /
60
2
200
The velocity of point B is
v v w k r k i j B A AB B A x 0 29.94 x 0.3 0.6 /
v 12.564i 6.282 j (m/s) B
Also
1.2 0.2 0
x 12.564 6.282 0 0 C B BC C / B BC
i j k
v v w r i j
v i j C BC BC 12.564 0.2 6.2821.2 (1)
Using Equation 4-8 between points C and D gives
C D CD C D v v w r / x
0.3 1.0 0
0 0 0 CD
i j k
v i 0.3 j (m/s) D CD CD (2)
Equating (2) and (1) gives
i j i j CD CD BC BC 0.3 12.564 0.2 6.282 1.2 (3)
Equating i and j components of the above equation yields two equations (4) and (5).
43
CD BC 12.564 0.2 (4)
CD BC 0.3 6.2821.2 (5)
Solving the above equations simultaneously gives
12.12 rad/s CD
2.204 rad/s BC
The vertical velocity of the rack is equal to the velocity of the point on the gear where it makes
contact with the rack.
12.12 x0.3 -3.64 m/s E v
Example 3-10
Slider C of the mechanism shown in Figure E4-10 is moving horizontally at 10 m/s towards the
left-hand side. Determine the angular velocities of crank AB and connecting rod BC.
Figure E4-10
Solution 3-10
Using Equation 4-8 between points A and B, the velocity of point A is
v v w k r i j B A AB B A AB x 0 k x 0.5 1.0 /
v i j B AB AB 0.5
Also
x 0.5 x 1.7i - 2.2j / v v w r i j k C B BC C B AB AB BC
v i j C AB BC AB BC 2.2 0.5 1.7 (2)
The velocity of point C
v i j i C 10 cos0 sin0 10 (3)
Equating i and j components of equations (2) and (3) gives
2.2 10 AB BC (4)
44
0.5 1.7 0 AB BC (5)
Solving the above equations simultaneously gives
6.071 rad/s AB and 1.786 rad/s BC
3.3.4 General Motion with Sliding on Rotating Link
Equation 4-8 is application to both planar (2D) and spatial (3D) motions.
Example 3-11
For the mechanism shown in Figure E4-11, determine the angular velocity of bar BC and the
magnitude of the sliding velocity of the pin at B if crank AB rotates at 4 rad/s in the
anticlockwise direction.
Figure E4-11
Solution 3-11
Using Equation 4-8 between points A and B, we have
B A B A v v w r / x 0 A v 4 k (rad/s)
r 0.5i 1.2 j (m) B A
v k i j B 0 4 0.5 1.2
v i j B 4.8 2 (1)
Using Equation 3-8 between points C and B, we have
B C B C CB B C v v v w r / x
0 C v and r 2.0i 1.2 j (m) B C
where B C v is the sliding velocity and CB is the angular velocity of the link CB
v v k i j B B C CB 0 2.0 1.2
45
v v i j B B C CB CB 1.2 2 (2)
Let V be the magnitude of the sliding velocity of the pin at B. Then,
v V i j B C cos sin (3)
where θ is the angle the sliding direction makes with the horizontal. The angle may be calculated
from Figure S 4-11 as
o 149.04
2
1.2
180 tan 1
Substituting θ = 149.03o into equation (3) gives
v V i j B C cos149.03 sin149.03
v V( 0.8575i 0.5145 j) B C (4)
Figure S4-11
Substituting the above equation into equation (2) yields
v V i j i j B CB CB (0.8575 0.5145 ) 1.2 2
v Vi V j B CB CB 1.2 0.8575 2 0.5145 (5)
Equating equations (1) and (4), we have
i j Vi V j CB CB 4.8 2 1.2 0.8575 2 0.5145
Equating i and j components of the above equation gives
V CB 4.8 1.2 0.8575 (6)
V CB 2 2 0.5145 (7)
Solving the equations (6) and (7) simultaneously gives
1.79 CB rad/s
V 3.09m/s
Example 3-12
Solve Example 4-7 using the vector analytical method.
46
Solution 3-12
Using Equation 4-8 between points A and C, we have
C A AC C A v v w r / x
where 0 A v and r i i i j o o
B A 3 cos35 sin35 2.457 1.721
v k i j C AC 0 x 2.457 1.721 v i j C AC AC 1.7207 2.457 (1)
Using Equation 4-7 between points B and C, we have
C B C B BC C B v v v w r / x (2)
The slider velocity 0.2 C B v m/s needs to be resolved into the i and j components.
From Figure S4-12 and using the cosine rule,
o R 3 2 2 3x2cos35 2 2 2 R 1.78m
Using the sine rule
3
sin35 sin
R
o
o 104.89 o o o o 180 180 104.89 75.11
Using Equation 3-12
v i j o o
C B 0.2 cos75.11 sin75.11 v i j C B 0.0514 0.1933 (3)
Figure S4-12
Also 0 B v (4)
r i j o o
C B 3cos35 2 3sin35 / r i j o
C B 0.4575 1.7207 / (5)
[ Hint: Also r i j o o
C B 1.78cos75.11 1.78sin75.11 / ]
Substituting equations (3) to (5) into equation (2) gives
v i j w i j o
C BC 0 0.0514 0.1933 k x 0.4575 1.7207
v i j C BC BC 1.7207 0.0514 0.4575 0.1933 (6)
Equating equations (1) and (6) yields equation (7).
47
i j i j AC AC BC BC 1.7207 2.457 1.7207 0.0514 0.4575 0.1933 (7)
Equating i and j components of the above equation (7)
1.7207 1.7207 0.0514 AC BC (8)
2.457 0.4575 0.1933 AC BC (9)
Solving the equations (8) and (9) simultaneously gives
0.1333 BC rad/s
0.1034 AC rad/s
3.4 Mechanical Advantage of Mechanism
Assuming that a mechanism is conservative (i.e. there is no energy loss through friction,
heat, noise, etc.), then from Figure 4-13, the power in, in P ,is equal to the power out, out P . Thus
in in in out out out P T T P
From the above equation,
out
in
in
out
T
T
Equation 3-10
By definition, the mechanical advantage (MA) is the ratio of the magnitude of output
force, out F , to the magnitude of input force, in F . Thus,
in
in
out
out
in
out
r
T
r
T
F
F
MA
out
in
in
out
r
r
T
T
MA
Substituting Equation 3-10 into the above equation gives
out
in
out
in
r
r
MA
Equation 3-11
Using Equation 4-3, the angular velocity ratio of the input link to that of the output link is
1,2 2,2
1,4 2,4
4
2
Substituting the above equation into Equation 4-11, the mechanical advantage of the mechanism
shown in Figure 4-13 is
out
in
r
r
MA
1,2 2,2
1,4 2,4
48
Figure 3-11
Problems
5-1. In Figure P4-1, the rotation of the dumping bin BC of the truck about point C is operated by the extension of the hydraulic cylinder AB. At the instant shown bin BC is rotating about point C at constant angular velocity of 2.5 rad/s in the clockwise direction. Determine
(a) the rate of extension of the hydraulic cylinder AB
(b) the angular velocity of the hydraulic cylinder AB
Ans: (a) 7.49 m/s (b) 2.42 rad/s
Figure P4-1
5-2. Find the angular velocity of link AB and sliding velocity of point B relative to link AB of the mechanism shown in Figure P4-2 if slider B is moving towards the right at a constant velocity V= 5 m/s and θ = 60o. Ans: (a) 2.5 rad/s, (b) 4.33 m/s
49
Figure P4-2 Figure P4-3
5-3. Link AB of the mechanism shown in Figure P4-3 is rotating at 20 rad/s in the anti-clockwise direction. Using the vector analytical method, determine the sliding velocity of slider C. Ans: -7.37 i (m/s)
5-4. The collar at A in Figure P4-4 slides along the circular bar, causing the pin A to move at a constant speed 5 m/s in a circular path of radius R. Bar AC slides in the collar at B, and the collar is pinned to ground at B. At the instant shown, determine the velocity of bar AC relative to pin B and angular velocity of bar AC. Ans: (a) 2.75 m/s, (b) 0.93 rad/s
Figure P4-4 Figure P4-4
5-5. A mechanism, as shown in Figure P4-5, has the following dimensions: AB = 50 mm; BC = 250; CD = 130 mm; CD = 200 mm, AD = 300 mm and CE = 200 mm. Slider C can only move vertically. Locate all the instant centres at an instant θ = 25o.
50
Figure P4-6 & P4-7 Figure P4-8
4-6. In the swiveling joint mechanism, shown in Figure P4-6, the driving crank is rotating clock clockwise. The lengths of various links are : OA = 50 mm; AB 350 mm; AD = DB; DE=EF = 250 mm; CB = 125; OC = 300 mm and vertical distance between F and C is 250 mm. If crank OA is rotating at 100 rad/s anticlockwise, draw a relative velocity diagram of the mechanism and use it to determine (a) velocity of block F, (b) sliding velocity of DE, (c) angular velocity of CB
4-7. Locate all the instant centres of the mechanism shown in Figure P4-7.
4-8. Figure 3.4 shows a mechanism used in one of the shaping machines at the Mechanical Engineering Department’s machine shop. The slotted yoke AD is pinned to the ground at A while the end D is used to move the ram R horizontally. The disc BC has radius 30 cm and other dimensions of the mechanism are: AB = 54.6 cm, AD = 132.3 cm and DR =90 cm. If BC is rotating at 50 rad/s anticlockwise, draw a relative velocity diagram of the mechanism at the instant point C is at C1 and use it to determine (a) velocity of ram, (b) ) angular velocity of AD.
51
CHAPTER 4 ACCELERATION ANALYSIS
4.1 General Motion
To obtain an equation for the acceleration of point B of Figure 4-1, Equation 3-8 is
differentiated with respect to time. Using Equation 3-7b, the result of differentiation is
/ B/A B/A B/A 2 x v x r x x r B A B A a a a Equation 4-1
The term B/A 2 x v is called Coriolis acceleration, named after French Engineer G. C. Coriolis,
who was the first to determine it. Equation 4-1 is applicable to both planar (2D) and spatial (3D)
motions. For planar motions, Equation 4-1 may written as
B/A
2
/ B/A B/A 2 x v x r r B A B A a a a Equation 4-2
Figure 4-1 Rigid Body in Planar Motion
4.1.1 Pure Translational Motion
For pure translational motion without sliding, the vector ω, α, B A a / and B A v / are zero. Thus,
B A a a . Equation 4-3
4.1.2 General Motion without Sliding on Rotating Link
For general motion without sliding on a rotating link, the components B A a / and B A v / are zero.
Hence, Equation 4-2 becomes
52
B/A
2
B/A x r r B A a a Equation 4-4
The term B/A x r in Equation 4-4 is the tangential component of the acceleration and
B A r / x x is the normal component. The tangential component is perpendicular to link AB
and the normal component is normal or parallel to AB and points towards the point of rotation.
Note that ω2 in Equation 4-2 is a scalar quantity and therefore, no vector component is attached
to it.
Problem 4-1
In Figure 4-2, link AB has an anticlockwise angular velocity of 12 rad/s and a clockwise
angular acceleration of 250 rad/s2. Determine the angular accelerations of links BC and CD.
Figure 4-2
Solution 4-1
Since the angular acceleration formula has angular velocity component, first the
velocities of the links BC and CD are determined. Then, the determined velocities are used to
determine accelerations of the links. The velocity of point B
B/A x r B A AB v v 0 12 k x 2i 2j B v v i j B 24 24 (m/s)
Let BC and CD be the angular velocities of links BC and CD, respectively. The velocity of
point C in terms of the velocity of point B is
C/B x r C B BC v v
v 24i 24 j k x 3i C BC v i j C BC 24 3 24 (1)
Similarly, the velocity of point C in terms of the velocity of point D is
2 m
2 m 3 m
A
B C
D
12 rad/s
250 rad/s2
1
53
C/D x r C D CD v v v k i C CD CD 0 x 2j 2 (2)
Equating the two expressions for the velocity of point C, i.e. equations (1) and (2), yields
i j i BC CD 24 3 24 2
Equation the i and the j components of the above equation and solving for the angular velocities,
we have
for i component:
2 24 CD , 12 CD rad/s
for j component:
3 24 0 BC , 8 BC rad/s
The acceleration of point B in terms of that of point A is
B A B A a a r /
2
B/A x r
a i j B 0 250 k x 2i 2j 12 2 2 2
a i j B 212 788 (m/s).
Let BC and CD be the angular accelerations of links BC and CD, respectively. The acceleration
of point C in terms of that of point B is
C/B
2
C/B x r r C B BC BC a a
a i j i C BC 212 788 k x 3i 8 3 2
a i j C BC 20 3 788 (3)
Similarly, the acceleration of point C from point D is
C/D
2
D/C x r r C D CD CD a a
a j C CD 0 k x 2j 12 2 2
a j C CD 2 i 288 (4)
Equating the two expressions for the acceleration of point C (i.e. equations (3) and (4)) gives
i j j BC CD 20 3 788 2 i 288
Equating the i and the j components of the above equation and solving for the angular velocities,
we have
i component: 2 20 CD , 10 CD rad/s2
54
j component: 3 788 288 BC , 166.7 BC rad/s2
Problem 4-2
Solve Problem 3-11 if slider C is moving horizontally at 10 m/s and 5 m/s2 towards the left-hand
side.
Solution 4-2
From Solution 4-2,
6.071 rad/s AB and 1.786 rad/s BC
The acceleration of point B in terms of that of point A is
B A AB AB B A a a r /
2
B/A x r
0 k x 0.5i 1.0j 6.071 0.5i 1.0j 2 B AB a
18.428i 0.5 36.857j B AB AB a (1)
Similarly, the acceleration of point C in terms of that of B is
C B BC BC C B a a r /
2
C/B x r
18.428i 0.5 36.857 k x 1.7i 2.2j 1.786 1.7i 2.2j 2 C AB AB BC a j
a j C AB BC AB BC 23.851 2.2 i 0.5 43.8741.7 (2)
Using equation 3-11, a i C 5 m/s (3)
Equating the i and j components of equations (2) and (3) gives
AB BC 28.851 2.2 (4)
AB BC 43.874 0.5 1.7 (5)
Solving equations (4) and (5) simultaneously gives
242.6 AB rad/s and 97.2 BC rad/s
4.1.3 General Motion with Sliding on Rotating Link
For general motion with sliding on a rotating link, equations 4-1 and 4-2 are used.
Problem 4-3
For the mechanism shown in Figure P2.12, determine the angular acceleration of bar AC
and acceleration of pin A relative to the slot in bar AB if bar AB has an anticlockwise velocity of
2 rad/s and a clockwise angular acceleration of 10 rad/s2.
55
Figure 4-3
Solution 4-3
First, velocity analysis is performed and the result is use in the acceleration analysis.
Using Equation 3-8 the velocity of point A is given as
A B A B BA A B v v v r / / x
v v i j A A B 0 2 k x 1.3 1.2 / v v i j A A B 2.4 2.6 / (1)
The velocity of the pin A is parallel to the slot on link BA. Using equation 3-11
v v i j A B A B cos sin / / (2)
where o 42.709
1.3
1.2
tan 1
, which is the angle between the sliding direction and the
horizontal axis. Substituting (2) into (1) gives
v v i v j A A B A B 0.7384 2.4 0.6783 2.6 / / (3)
Let CA and CA be the angular velocity and angular acceleration of link CA. The velocity of
point A in terms of that C is
A C CA A C v v r / x
0.5 1.2 0
0 0 0 A CA
i j k
v
v i j A CA CA 1.2 0.5 (4)
Equating the two velocities of A gives
1.2 0.5 0.7384 2.4 0.6783 2.6 CA CA Arel Arel i j v i v
Equating the i and j components yields
56
1.2 0.7384 2.4 CA Arel v (5)
0.5 0.6783 2.6 CA Arel v (6)
Solving equations (5) and (6), we have
2.995 CA rad/s
1.625 / A B v m/s
The velocity and acceleration of A relative B are resolved into i and j components as
v i j i j A B 3.5258 cos42.709 cos42.709 2.5907 2.3915 / (7)
a i j i j A/ B A/ B A/B A/B cos42.709 cos42.709 0.7348 0.6783
(8)
where A/ B is the magnitude of the acceleration.
Using Equation 4-1, the acceleration of point A given the velocity of B is
A/B
2
/ A/B / 2 x x r A B A B AB BA A B a a a v r (9)
Substitution equations (7) and (8) into (9) gives
10 1.3i 1.2j
1.3 1.2 0
0 0 10
2.5907 2.3915 0
0 0.7348 0.6783 2 0 0 2
2
A/B A/B
i j k
i j k
a i j A
0.7348 137.2171 0.6783 112.1815 A/B A/B a i j A (10)
The acceleration of A given the velocity of C is
A/C
2
/ x r A C CA A C a a r
2 0.5i 1.2j
0.5 1.2 0
0 0 0 2
A CA
i j k
a
a i j A CA CA 1.2 2 0.5 1.2 (11)
Equating the i and j components of the two accelerations of point C yields
0.7348 137.2171 1.2 2 A/B CA a
0.6783 112.1815 0.5 1.2 A/B CA a
and solving simultaneously,
57
171.6582 A/B m/s2
10.9017 A//B a rad/s2
Problems
5-1. In Figure P5-1, the rotation of the dumping bin BC of the truck about point C is operated
by the extension of the hydraulic cylinder AB. At the instant shown bin BC is rotating about
point C at constant angular velocity of 2.5 rad/s in the clockwise direction. Determine
(a) the acceleration of extension of the hydraulic cylinder AB
(b) the acceleration angular velocity of the hydraulic cylinder AB
Figure P5-1
5-2. Find the angular acceleration of link AB and sliding acceleration of point B relative to
link AB of the mechanism shown in Figure P5-2 if slider B is moving towards the right at a
velocity of v= 5 m/s, increasing at a = 0.5 m/s2 at the instant θ = 60o
Figure P5-2 Figure P5-3
58
5-3. Link AB of the mechanism shown in Figure P5-3 is rotating at 20 rad/s and 50 rad/s2 , both in the anti-clockwise direction. Determine the sliding acceleration of slider C.
5-4. A mechanism, as shown in Figure P5-5, has the following dimensions: AB = 50 mm; BC = 250; CD = 130 mm; CD = 200 mm, AD = 300 mm and CE = 200 mm. Slider C can only move vertically. If AB is rotating at 10 rad/s anticlockwise and 50 rad/s2 clockwise, find the sliding acceleration of E at the instant θ = 60o. Consider mode 1 only.
Figure P4-4 Figure P4-6
4-9. Figure 3.4 shows a mechanism used in one of the shaping machines at the Mechanical Engineering Department’s machine shop. The slotted yoke AD is pinned to the ground at A while the end D is used to move the ram R horizontally. The disc BC has radius 30 cm and other dimensions of the mechanism are: AB = 54.6 cm, AD = 132.3 cm and DR =90 cm. If BC is rotating at 50 rad/s anticlockwise and 250 rad/s2 clockwise, determine (a) acceleration of the ram, (b) ) angular acceleration of AD.
59
CHAPTER 5 ROTARY MOTION TRANSMISSION
5.1 Introduction
Rotary motions are general transmitted using gears, belts, ropes or /and chain drives. These machine elements provide a convenient means of transferring power from one shaft to another and are necessary to reduce rotational speeds of motors to values required by mechanical equipment. Belts and ropes use friction as useful agent to transmit power, but are subjected to uncertainties in the values of coefficient of friction.
In general, gears and chains transmit power efficiently by eliminating slipping. Slipping is eliminated by replacing roller surfaces with teeth or sprockets, which provide continuous positive engagement of input and output gears and sprockets. Constant velocity ratio is maintained between the two engaged gears when the common normal passes through the line of line of centres (from Kennedy’s theorem).
5.2 Gear Drive
5.2.1 Types of Gear
Gears may be divided into three main groups. The groups are parallel-axis gears, non-parallel-axis but co-planer gears, and nonparallel-axis and noncoplaner gears.
5.2.1.1 Parallel-Axis Gears
Parallel-axis gears are the simples and most popular type of all gear types. They include spur (shown in Figure 5-1) and helical gears. They can transmit large tongue with high efficiency.
Figure 5-1: Spur Gear Drive
60
5.2.1.2 Nonparallel-Axis Coplanar Gears
These types of gears have nonparallel axis of rotation unlike spur and helical gears, but are coplanar. Bevel (Figure 5-2) and spiral gears are type examples of this group of gears. The common feature of this group of gears is redirection of power from one axis to another, though the two pair of gears (input and output) are not parallel but coplanar (same xy-plane).
Figure 5-2: Bevel Gear Drive
5.2.1.3 Nonparallel-Axis, Noncoplanar Gears
These gears are more complex in both geometry and manufacturing than parallel-axis gears. As a result, these gears are more expensive than parallel-axis and nonparallel coplanar gears. They include worm gear drives (shown in Figure 5-3).
Figure 5-3: Worm Gear Drive y
61
5.2.2 Gear Trains
Shown in Figure 5-4 are two roller discs 1 and 2 having a common contact point at B.
The instant centres are (1, 2), (2, 3) and (1, 3). From the Kennedy’s theorem, the three instant
centres lie on the same line. This implies that the line, which passes through the centres of the
discs, must be the common normal to both discs. For continuity, the velocity of the common
point of contact is
B A C B r r 2 / 3 /
(2,3) (1,2)
(2,3) (1,3)
/
/
3
2
IC IC
IC IC
r
r
B A
C B
Equation 5-1
Figure 5-4: Two Circular Discs in Rolling Contact
Since the instant centres lie on the same line, Equation 5-1 is negative and therefore the two
gears rotate in opposite directions. The equation also shows that the angular velocity ratio
2 3 is inversely proportional to the pitch-radius ratio ( 3 2 r r ), the pitch-diameter ratio ( 3 2 d d
), the pitch-circumference ratio ( 3 2 c c ), or the number of teeth ( 3 2 N N ). Thus,
2
3
2
3
2
3
2
3
3
2 sgn .
N
N
c
c
d
d
r
r
Equation 5-2
where sgn =-1 if the two engaging gears are external gears and, sgn = 1 if one of the engaging
gears is external (ring, internal or annular ) gear.
2
3
1
1
(1,2)
(2,3)
(1,3)
Circle Pitch
A
B
C
62
5.2.2.1 Ordinary Gear Train
Sometimes, a design may call for a high angular velocity ratio that will be impractical to
achieve due to space and material strength limitations. An alternate way of achieving such large
angular velocity ratio is the use of ordinary gear train. Ordinary gear train consists of at least two
pairs of meshing gears, where the output angular velocity of one pair of gears is coupled to the
input of the next pair of gears. Thus, the velocity ratio is changed at several stages.
Example 5-1
Find the velocity ratio 8 2 of the gear trains shown in Figure E6-1.
Figure E6-1
Solution 5-1
From the figure, the first stage engagement is between gears 2 and 3. Gears 3 and 4 are
attached one shaft, and so are gears 6 and 7. Using equation (5-2),
8
7
8 7 N
N
(1)
Since gears 6 and 7 are rigidly attached to the same shaft, 7 6 . Thus,
8
7
8 6 N
N
(2)
Using Equation 5-2,
6
5
6 5 N
N
(3)
Substituting equation (3) into (1) yields
6
5
6
5
8 5 N
N
N
N
(4)
Similarly,
5
4
3
2
5 2 N
N
N
N
(5)
2
3
4
6
7
8
5
63
Replacing 5 in equation (5) with equation (5), and dividing through by 2 gives
6
5
6
5
5
4
3
2
2
8
N
N
N
N
N
N
N
N
(6)
or 4
6
5
6
5
5
4
3
2
2
8 1
N
N
N
N
N
N
N
N
(7)
5.2.2.2 General Gear Train Equation
From equation (6), Equation 6-2 may be generalised as
product of number of teeth on driven gears
product of number of teeth on driver gears
driver
driven
Equation 5-3
n
driver
driven x 1
product of number of teeth on driven gears
product of number of teeth on driver gears
Equation 5-4
where n is number of external meshes. In Equation 6-3, the negative sign that accounts for
direction of rotation is omitted in the equation, but it must be accounted for when using it. If an
external gear meshes with an internal (annular or ring), the negative sign is omitted. Equation 6-
2is applicable to all types of gears, belt drives, chain drives and rollers. In belt and chain drives,
the number of teeth (N) is replaced with the diameter of pulley and number of sprockets,
respectively.
In equation (6), N5’s cancels out but the negative sign remains. This implies that the
presence of gear 5 does not affect the magnitude of the angular velocity ratio, but it cause
direction reversal. Gear 5 is known as idler, and its main purpose is to change the direction of
rotation.
Example 5-2
Determine the angular velocity ratio 4 1 of the gear train shown in Figure E6-1.
64
Figure P 5.1
Solution 5-2
Using Equation 6-2
16
1
60
15
80
20
4
3
2
1
1
4
N
N
N
N
Example 5-3
Figure E6-3 shows a gear train that consists of miter (same-size bevel) gears having 16
teeth each, a 4-tooth right hand worm and 40-tooth worm gear. If the input bevel gear 2 rotates at
150 rpm in the anti-clockwise direction, what is the speed of the worm gear 5.
Figure E6-3
1
2
3
4
N1 = 20
N2 = 80
N3 = 15
N4 = 60
y
z
x
ω2
16 T
16 T
10 T
45 T
2
3 4
5
65
Solution 5-3
9
2
45
10
16
16
5
4
3
2
2
5
N
N
N
N
150 33.33
9
2
9
2
5 2 rpm
Example 5-4
In the train shown in Figure E6-4, the inlet gear 2 rotates at 1200 rpm in the anticlockwise
direction as viewed from the left. The angular velocity of outlet shaft is 300 rpm and
the gear contact between gear 1 and gear 2 reduces the speed ratio by 5:3. Find the number of
teeth of gear 2 and diameter of pulley 3.
Solution 5-4
(a)
5
3
2
1
1
2
N
N
30 50
3
5
3
5
2 1 N N
Figure E6-4
(b)
4
3
2
1
1
4
D
D
N
N
1200
5 60
3
300 3
D
In Out
30 T D = 60 cm
1
2
4
3
66
25 3 D cm
5.2.2.3 Planetary (Epicyclic) Gear Trains
Planetary gear trains are often utilised to make more compact gear reducer than an
ordinary compound gear train. In addition, one major advantage of planetary gear train is that it is
two degrees of freedom mechanism.
Planetary gear train has domestic and industrial applications. It is ultilised in both
domestic and industrial food mixers. Other major uses of planetary gear train are in automatic
transmission, packaging of food products into carton boxes, and automated packaging papers and
toilet rolls.
A typical planetary gear train consists a ring (or annular or external) gear, sun gear,
planet gears and arm, as shown in Figure 6-6. The arm carries the planet gears and rotates with
them. As the arm rotates, the planet gears engage both the ring gear and the sun gear. Therefore,
the angular velocity of a planet gear is the sum of the angular velocity of the arm plus its angular
velocity about the arm.
Figure 5-5: Typical Gear Train
5.2.2.3.1 Formula Method
(a) One-Input Planetary Gear Train
Consider a planetary train shown in Figure 6-7. The gear train has no ring gear; sun gear
2 fixed to the ground and the arm 4 carries planet gear 3 while it rotates with angular velocity
Ring gear
Sun gear
Planet gear
Arm (or Carrier)
67
41 with respect to the ground. Then, the absolute velocity of planet gear 3 may be expressed in
terms of 41 and 34 as
31 41 34 (1)
Dividing equation (1) by 41 gives
41
34
41
31 1
(2)
Since the gear 2 is fixed to the ground, 41 42 . Replacing 41 in equation (2) with 42 gives
42
34
41
31 1
or
24
34
41
31 1
Equation 5-5
Figure 5-6 Simple Planetary Gear Train with Fixed Sun Gear (One Input)
The angular velocity ratio 34 24 is the angular velocity ratio with respect to the arm. The arm
may be considered as fixed and treat the gear train as an ordinary gear train in this kinematic
inversion. Thus,
3
2
3
2
41
31 1 1
N
N
N
N
Equation 5-6
3
2
41
31 1
r
r
Equation 5-7
Example 5-5
If the arm 4 in Figure E6-5 has an anti-clockwise velocity of 4 rad/s, determine the
absolute angular velocity of gear 3.
N2
N3
2
3
4
1
68
Figure E6-5
Solution 5-5
Using Equation 6-6),
4
25
40
1 1 41
3
2
31
N
N
10.4 31 rad/s
Example 5-6
In Figure E6-6, arm AB is carrying a planet gear B and crack C, and has an anticlockwise
velocity of 10 rad/s. Determine the absolute angular velocity of bar DE.
Figure E6-6
Solution 5-6
First, the angular velocity of the planet gear is determined. Then, the velocity of bars CD
and DE in that ordered are calculated using vector method. Using equation 6-6
40 T
25 T
2
3
4
1
10 rad/s
0.8 m
0.6 m 36 T
12 T
A
B
C
D
E
0.3 m 1 m 0.4 m
69
10 40
12
36
1 1
arm
planet
sun
planet N
N
rad/s
The absolute angular velocity of point C is 40 rad/s since the roller BC and planet gear are
rigidly attached to the same shaft. The angular velocity of the planet gear relative to the arm is
planet arm planet arm /
40 10 30 / planet arm rad/s
The velocity of point B is given as
B A AB B A v v r / x
v 0 10k x 0.8j 8i B (m/s)
Similarly, the velocity of point C is
C B BC C B v v r / x
v 8i 30k x 0.3i 8i 9j B (m/s)
Let the angular velocity of bar CD and DE be CD and DE , respectively. The velocity of point D
in terms of that of point C is
D C CD D C v v r / x
v 8i 9 j k x 1.0i D CD
8i 9 j D CD v
Similarly,
D E DE D E v v r / x
0.4 0.8 0
0 0 0
D DE
i j k
v
v i j D DE DE 0.8 0.4
Equating the two velocities of point D gives
0.8 0.4 8i 9 j DE DE CD i j
Equating the i component and j components, and solving the resulting simultaneous equations
yields
10 DE rad/s
70
13 CD rad/s
(b) Two-Input Planetary Gear Train
Consider a planetary train shown in Figure 6-8. The gear train has no ring gear; sun gear
2 and arm 4 are the two inputs. Then, the absolute velocity of planet gear 3 and sun gear 2 are,
using equation 6-4,
34 31 41 (4)
24 21 41 (5)
Dividing equation (4) by equation (5) gives
21 41
31 41
24
34
(6)
Figure 5-7: Planetary Gear Train with Two Inputs
If the arm 4 is fixed, then
3
2
24
34
21 41
31 41
N
N
(7)
Equation (7) may be generalized as
product of number of teeth on driven gears
product of number of teeth on driver gears
F A
L A
FA
LA
Equation 5-8
Where
LA = angular velocity of the last gear relative to that of the arm
FA = angular velocity of the first gear relative to that of the arm
N2
N3
2
3
4
1
71
L = absolute angular velocity of the last gear
F = absolute angular velocity of the first gear
A = absolute angular velocity of the arm
Equation is applicable to all gear types, chain drives, rollers and belts drives.
Consider that a planetary gear train shown in Figure 6-9. From the figure, the diameter of
the ring gear 4 is
r s p D D 2D Equation 5-9
As the diameter of a gear is proportional to the number of teeth, it can be deduced that
r s p N N 2N Equation 5-10
Figure 5-8: Planetary Gear Train sun, planet and ring gears
Example 5-7
Shown in Figure E6-7 is planetary gear train with sun gear 2 and arm 6 as inputs.
Determine the angular velocity of sun gear 5 if sun gear 2 rotates at 50 rad/s in and arm 6 rotates
at 75 rad/s, both in anti-clockwise as viewed from the left.
72
Figure E6-7
Solution 5-7
Considering sun gear 2 as first (driver) and sun gear 5 as last (driven) gear and using
Equation 6-8, we have
5
4
3
2
2 6
5 6
26
56
N
N
N
N
15
30
16
20
50 75
75 5 12.5 5 rad/s
Example 5-8
The gear train shown in Figure E6-8 has two inputs: sun gear 2 rotates at 400 rmp and
ring gear 6 rotates at 600 rmp, both in anti-clockwise as viewed from the right. Find the
resulting angular velocity of arm 7.
ω2
6
4
5
3
20 T
15 T
30 T
ω6
2
3
73
Figure E6-8
Solution 5-8
Considering sun gear 2 as first (driver) and ring gear 6 as last (driven) gear and using
Equation 6-8 we have
4 6
2 5
6
5
4
3
3
2
2 7
6 7
27
67
N N
N N
N
N
N
N
N
N
22120
35 30
400
600
7
7
732 5 rpm
Example 5-9
Find the angular velocity of sun gear 2 in Figure E6-9 if ring gear 6 rotates at 600 rmp in
anti-clockwise as viewed from the right.
4
3
35 T
22 T
120 T
30 T
ω2 = 400 rpm
2
4
7 3
6
5
ω6 = 600 rpm
74
Figure E6-9
Solution 5-9
Ring gear 6 becomes the first (driver) gear and ring gear 8 and sun gear 2 are the last
gears. First, we determine the velocity of arm 7 by starting from gear 6 and ending at gear 8.
Since gear 8 is fixed, its velocity is zero. Using Equation 6-8, we have
8
4
5
6
6 7
8 7
67
87
N
N
N
N
7 8
7 22
30
120
600
0
N
(1)
Now
2 35 218 22 115 8 2 3 4 N N N N
Substituting 115 8 N
into equation (1) and solving for 7 gives
1955 7 rpm
Using equation (3.6) and taking the path from gear 6 to gear 2, we have
5 2
6 4
2
3
3
4
5
6
6 7
2 7
67
27
N N
N N
N
N
N
N
N
N
3
35 T
22 T
120 T
30 T
2
4
7
18 T
3
6
5
ω6 = 600 rpm
8
75
3035
120 22
600 1955
1955 2
4469 2 rpm
Tabular Method
Tabular method is based on kinematic inversion. It involves analyzing two easily
described parts of the motion separately, and then adding the results together. The follow steps
should be used:
(1) Disconnect any gears from ground (if any gear/arm is fixed to the ground), and
fixed all gears rigidly to the rotating arm
(2) Motion with arm: Rotate the arm with rigidly attached gears by a number of
revolutions proportional to the angular velocity of the arm.
(3) Motion relative to arm: Separate the gears from the arm while holding the arm
fixed, and rotate the rest of the gears back so that the total rotation (step 2 + step
3) of one or more of the gears matches their given rotations.
Example 5-10
Find the angular velocity of planet gear 3 in Figure E6-10 if arm 4 rotates at 20 rad/s in
clockwise direction.
Figure E6-10
Solution 5-10
In the “Motion with arm” row, the absolute velocity of the arm is input for all cells. We
select gear 2 as the starting gear and 3 as end gear. Since gear 2 is fixed, its angular velocity
40 T
20 T
2
3
4
1
20 rad/s
76
relative to the arm is negative of the velocity of the arm. Using equation 6-2, the motion of gear
3 relative to the arm is
3
2
34 24 N
N
.
Gear 2 Gear 3 Arm 4
Motion with arm 20 20 20
Motion relative to arm -20 +20(N2/N3) 0
Total Motion 0 20+20(N2/N3) 20
From row 3 and column 3 of the above table,
60
20
40
20 20 20 20
3
2
3
N
N
rad/s
Example 5-11
Solve Example 5-6 using the tabular method.
Solution 5-11
Gear 2 Gear 3 Gear 4 Gear 5 Arm 4
Motion with arm 75 75 75 75 75
Motion relative to arm -25 -25(-N2/N3) 25(N2/N3) 25(N2/N3) (-N4/N5) 0
Total Motion 50 75-25(N2/N3) (N4/N5) 75
From the above table.
5
4
3
2
5 75 25
N
N
N
N
15
30
16
20
75 25 5 12.5 5 rad/s
Example 5-12
Solve Example 6-7 using the tabular method.
Work Direction
77
Solution 5-12
Let the absolute angular velocity of arm be x. We start from gear 2 and end at gear 6, and
the complete table is listed in table below. Starting from gear 6 and ending at gear 2 will not
affect the results.
Table 5.1
Gear 2 Gear 3 Gear 4 Gear 5 Gear 6 Arm 7
Motion with arm x x x x x x
Motion relative to arm 400-x (400-x)(-N2/N3) (400-x)(-N2/N3)
(-N3/N4)
Same as Gear 4 (400-x)(-N2/N3) (-
N3/N4) (N5/N6)
0
Total Motion x+(400-x)(-N2/N3)
(-N3/N4) (N5/N6)
x
From the above table
400 600
6
5
4
3
3
2
N
N
N
N
N
N
x x
600
120
30
22
18
18
35
400
x x
x 732rpm
Example 5-13
Solve Example 6-8 using the tabular method.
Solution 5-13
Here, two tables are created for paths from gear 6 to gear 8 and from gear 6 to gear 2. The
first table will be used to determine the angular velocity of the arm, and the second table for
calculating the angular velocity of the gear 2. Since gear 8 is fixed, its velocity is zero. Let the
Work Direction
78
absolute angular velocity of arm be x. We start from gear 2 and end at gear 6, and the complete
table is listed in table below. Starting from gear 6 and ending at gear 2 will not affect the results.
Table 5.2
Gear 8 Gear 4 Gear 5 Gear 6 Arm 7
Motion with arm x x x x x
Motion relative to arm (600-x)(N6/N5) N4/N8) (600-x)(N6/N5) (600-x)(N6/N5) 600-x 0
Total Motion x+(600-x)(N6/N5) N4/N8) x
From the table,
600 0
8
4
5
6
N
N
N
N
x x x 1955 rmp
Table 5.3
Gear 2 Gear 3 Gear 4 Gear 5 Gear 6 Arm
7
Motion with arm -1955 -1955 -1955 -1955 -1955 -1955
Motion relative to arm 2555(N6/N5) (-N4/N3) (-N3/N2) 2555(N6/N5) (-N4/N3) 2555(N6/N5) 2555(N6/N5) 2555 0
Total Motion -1955+2555(N6/N5) (-N4/N3) (-
N3/N2)
-1955
From the above table,
2
4
5
6
2
3
3
4
5
6
2 1955 2555 1955 2555
N
N
N
N
N
N
N
N
N
N
35
22
30
120
1955 2555 2 4469 2 rmp
Work Direction
Work Direction
79
5.3 Belt Drives
5.3.1 Types of Belt
Table 5.4 Belt Types
Shape Figure Joint Centre Distance
Flat Yes No Upper limit
Round Yes No Upper limit
V None Limited
Timing None Limited
5.3.2 Length and Wrap Angle of Belt Drive
(a) Open Belt
Figure 5-9 Open Belt Geometry
The following geometric relationships are applicable to both belt and chain drives.
C
D d
2
sin 1 Equation 5-11
2 L Equation 5-12
2 S Equation 5-13
Belt Length, L S L C D d D d
2
1
4 2 2 Equation 5-14
(a) Crossed Belt
80
Figure 5-10 Open Belt Geometry
The following geometric relationships are applicable to only belts drives.
C
D d
2
sin 1 Equation 5-15
2 Equation 5-16
Belt Length, L C D d D d
2
1
4 2 2 Equation 5-17
5.3.3 Types of Belt Drive Configurations
The power from one pulley to another may be transmitted by any or combinations of the
following types of configurations.
(a) Non-reversing belt drive (b) Reversing crossed belt drive
(a) (b)
Figure 5-11 (a) Non-reversing belt drive, and (b) Reversing crossed belt drive
(c) Reversing open belt drive (d) Quarter-Twist belt drive
Figure 5-12 (a) Reversing open belt drive, and (b) Quarter-Twist belt drive
3
2
2
3
D
D
3
2
2
3
D
D
81
(e) Drive with idlers
It is normally used to transmit motion to many shafts, arranged in parallel as shown in Figure 6-13. The idlers are used to adjust tension in belt and increase angle of wrap.
Figure 5-13 (a) Reversing open belt drive, and (b) Quarter-Twist belt drive
(f) Compound Belt drive
A compound belt drive as illustrated in Figure 6-14, is used when power is transmitted from one pulley to another through a series of pulleys.
Figure 5-14: Compound Belt Drive
5.3.4 Slip of Belt
Sometimes, the frictional force which drive pulleys of a belt maybe insufficient. This may cause the belt to slip over the pulleys. This is called slip of belt and it is generally expresses as a percentage.
Consider a belt drive with the following:
s1 = percentage slip between the driver pulley and the belt;
s2 = percentage slip between the belt and the driven pulley;
D1 , D2 = diameters of driver and driven pulleys, respectively Driver pulley Driven pulleys Idler pulleys Driver Driver Driven Driven
82
Then, the velocity of the belt is
100
1
1 1 1 1
s
v D D
100
1 1
1 1
s
v D
(1)
and the velocity of a point on the driven pulley is
100 100 100 100
1 2
1 1
1
1 1
2
2 2
s s
D
s
D
s
v v D
100 100
1
100
1
100
1 2
1 1
1
1 1
2
2 2
s s
D
s
D
s
v v D
(2)
Substituting (1) into (2), we have
100
1
100
1
100
1 2
1 1
2
2 2
s s
D
s
D v v
100
1
100
1 1 2
2
1
1
2 s s
D
D
Equation 5-18
If the thickness of the belt (t) is considered, the Equation 6-18 becomes
100
1
100
1 1 2
2
1
1
2 s s
D t
D t
Equation 5-19
Example 5-14
A 3-phase electric motor is used to drive a reciprocating compressor through an intermediary line
shaft, as shown in Figure E6-14. The diameters of the pulleys are as indicated on the figure.
Knowing that the motor is run at 1500 rpm, determine the speed of the compressor shaft if
(a) there is no slip at all pulleys, and the thickness of each belts is 10 mm;
(b) there is 2% slip at each driver pulley, and the thickness of each belt is 10 mm;
(c) there are 2 % slip at each driver pulley, and 3 % slip at each driven pulley, thickness of first
and second belts is 15 mm and 10 mm, respectively.
83
Figure E6-14
Solution 5-14
Let the subscripts 1, 2, 3 and 4 denotes motor, driven line shaft, driver line shaft and compressor
shaft, respective. Using Equation 6-19,
100
1
100
1 1 2
2
1
1
2 s s
D t
D t
f
f
100
1
100
1 1 2
2
1
2 1
s s
D t
D t
f
f
(1)
100
1
100
1 3 3
4
3
3
4 s s
D t
D t
s
s
100
1
100
1 3 4
4
3
4 3
s s
D t
D t
s
s
(2)
Since shafts 2 and 3 are same, 3 2 . Substituting equation (1) into (equation (2), we have
100
1
100
1
100
1
100
1 1 2 3 4
4
3
2
1
4 1
s s s s
D t
D t
D t
D t
s
s
f
f
(3)
(a) tf = ts = 10 mm; s1 = s2 = s3 = s4 = 0;
100
0
1
100
0
1
100
0
1
100
0
1
800 10
400 10
1000 10
700 10
1500 4 533.7 rpm 4
(b) tf = ts = 10 mm; s1 = s3 = 2%; s2 = s4 = 0;
100
0
1
100
2
1
100
0
1
100
2
1
800 10
400 10
1000 10
700 10
1500 4 512.6 rpm 4
(c) tf = ts = 10 mm; ts = 15 mm s1 = s3 = 2%; s2 = s4 = 3%;
100
3
1
100
2
1
100
3
1
100
2
1
800 15
400 15
1000 10
700 10
1500 4 485.2 rpm 4
700 mm 1000 mm
400 mm 800 mm
Line Shaft Compressor Shaft
Motor Shaft
Popo
Friends are lyk egg
